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Expanding Thurston Maps

Mario Bonk

Daniel Meyer

Department of Mathematics, University of California,Los Angeles, Los Angeles, CA 90095, USA

E-mail address : [email protected]

Jacobs University Bremen, Campus Ring 1, 28759 Bre-men, Germany

E-mail address : [email protected]

Key words and phrases. Expanding Thurston map,postcritically-finite map, visual metric, invariant curve, Markov

partition.

M.B. was supported by NSF grants DMS 0244421, DMS 0456940,DMS 0652915, DMS 1058772, and DMS 1058283.

D.M. was supported by an NSF postdoctoral fellowship and theAcademy of Finland, projects SA-134757 and SA-118634.

Contents

List of Figures vii

Chapter 1. Introduction 11.1. A Lattes map as a first example 21.2. Cell decompositions and invariant curves 51.3. Fractal spheres 101.4. Visual metrics and the visual sphere 171.5. Further results 201.6. Outline of the presentation 221.7. List of examples for Thurston maps 271.8. Notation 28

Chapter 2. Thurston maps 312.1. Branched covering maps 312.2. Definition of Thurston maps 322.3. Definition of expansion 342.4. Thurston equivalence 362.5. The orbifold associated with a Thurston map 412.6. Thurston’s characterization of rational maps 48

Chapter 3. Lattes maps 533.1. Crystallographic groups and Lattes maps 583.2. Quotients of torus endomorphisms and parabolicity 653.3. Lattes-type maps 733.4. Examples of Lattes maps 80

Chapter 4. Quasiconformal and rough geometry 914.1. Quasiconformal geometry 914.2. Gromov hyperbolicity 964.3. Gromov hyperbolic groups and Cannon’s conjecture 994.4. Quasispheres 102

Chapter 5. Cell decompositions 1075.1. Cell decompositions in general 1075.2. Cell decompositions of 2-spheres 1155.3. Cell decompositions induced by Thurston maps 124

iii

iv CONTENTS

5.4. Labelings 1335.5. Thurston maps from cell decompositions 1415.6. Flowers 1475.7. Joining opposite sides 151

Chapter 6. Expansion 1576.1. Definition of expansion revisited 1576.2. Further results on expansion 1616.3. Lattes-type maps and expansion 168

Chapter 7. Thurston maps with two or three postcritical points 1757.1. Thurston maps with # post(f) ∈ 2, 3 are equivalent to

rational maps 1767.2. Parabolic Thurston polynomials 179

Chapter 8. Visual Metrics 1878.1. The number m(x, y) 1918.2. Existence and basic properties of visual metrics 1948.3. The canonical orbifold metric as visual metric 200

Chapter 9. Symbolic dynamics 205

Chapter 10. Tile graphs 211

Chapter 11. Isotopies 22111.1. Equivalent expanding Thurston maps are conjugate 22111.2. Isotopies of Jordan curves 22811.3. Isotopies and cell decompositions 233

Chapter 12. Subdivisions 24312.1. Thurston maps with invariant curves 24612.2. Two-tile subdivision rules 25612.3. Examples of two-tile subdivision rules 266

Chapter 13. Combinatorially expanding Thurston maps 277

Chapter 14. Invariant curves 29914.1. Existence and uniqueness of invariant curves 30414.2. Iterative construction of invariant curves 31414.3. Invariant curves are quasicircles 325

Chapter 15. The combinatorial expansion factor 331

Chapter 16. Measure-theoretic dynamics 34516.1. Review of measure-theoretic dynamics 34616.2. The measure of maximal entropy 349

CONTENTS v

Chapter 17. The geometry of the visual sphere 36517.1. Linear local connectedness 36717.2. Doubling and Ahlfors regularity 37017.3. Quasisymmetry and rational Thurston maps 374

Chapter 18. Rational expanding Thurston maps: measure-theoretic properties 387

18.1. Ergodicity of Lebesgue measure for rational maps 39018.2. The Jacobian of a measure 39218.3. The invariant absolutely continuous measure 39418.4. Lattes maps, the measure of maximal entropy, and

Lebesgue measure 40718.5. Chordal versus visual metric: a.e. comparisons 411

Chapter 19. A combinatorial characterization of Lattes maps 41919.1. Visual metrics, 2-regularity, and Lattes maps 420

Chapter 20. Outlook 44320.1. Invariant Peano curves, matings, and fractal tilings

associated to expanding Thurston maps 44320.2. Open problems 448

Appendix A. 457A.1. Metric space terminology 457A.2. Orientations on surfaces 459A.3. Covering maps 462A.4. Branched covering maps 464A.5. Quotients of maps 468A.6. Tori and lattices 473A.7. Orbifolds and geometric structures 479A.8. Orbifolds and coverings 483A.9. Orbifold lifts of Thurston maps 488A.10. The canonical orbifold metric 491A.11. The canonical orbifold measure 497A.12. Planar crystallographic groups 500A.13. Quotients of circles and arcs 508

Bibliography 513

Index 519

List of Figures

1.1 The Lattes map g. 4

1.2 The map h. 11

1.3 Polyhedral surfaces obtained from the replacement rule. 13

2.1 The map g. 40

2.2 An obstructed map. 50

3.1 Invariant tiling of crystallographic group (244). 60



3.4 A Lattes map with orbifold signature (2, 4, 4). 81



3.7 Folding a tetrahedron from a triangle. 85

3.8 Construction of Θ = ℘. 85

3.9 A Euclidean model for a flexible Lattes map. 88

3.10 The map f in Example 3.24. 89

4.1 The generator of the snowsphere S. 103

4.2 The set Z. 105

5.1 The cycle of a vertex v. 117

6.1 A map with a Levy cycle. 167

7.1 Model for a Chebychev polynomial. 184

8.1 Separating points via tiles. 188

11.1 J is not isotopic to S1 rel. 1, i ,−1,−i. 228

11.2 Constructing a path through a, b, p. 234

11.3 Construction of the curve C ′. 239

vii

viii LIST OF FIGURES

12.1 The proof of Lemma 12.7. 251

12.2 Two two-tile subdivision rules. 256

12.3 The two-tile subdivision rule for z2 − 1. 267

12.4 Tiles of order 7 for Example 12.17. 267

12.5 The barycentric subdivision rule. 268

12.6 Tiles of order 4 for the barycentric subdivision rule. 269

12.7 The subdivision rule for Example 12.19. 270

12.8 The 2-by-3 subdivision rule. 271

12.9 The subdivision rule realized by 1− 2/z2. 273

12.10 Adding a flap. 273

12.11 The subdivision rule realized by f6. 274

12.12 Tiles of order 4 of the map f6. 275

13.1 Equivalence classes of vertex-, edge-, and tile-type. 283

13.2 The map f is not combinatorially expanding, but equivalentto the expanding map g. 296

14.1 The invariant curve for Example 14.6. 302

14.2 Invariant curves for the Lattes map g. 306

14.3 No invariant Jordan curve C ⊃ post(f). 309

14.4 Iterative construction of an invariant curve. 315

14.5 Iterative construction by replacing edges. 322

14.6 Example where C is not a Jordan curve. 324

14.7 A non-trivial rectifiable invariant Jordan curve. 324

15.1 Replacing k-tiles by (k − 1)-tiles. 337

16.1 Bijection of tiles. 360

17.1 Construction of γ. 369

19.1 Proof of Lemma 19.10. 429

20.1 Construction of γ for the map g. 444

20.2 Fractal tiling obtained from invariant Peano curve. 448

20.3 Another fractal tiling obtained from an invariant Peanocurve. 449

A.1 Group associated to signature (2, 4, 4). 506

LIST OF FIGURES ix



CHAPTER 1

Introduction

In this work we study the dynamics of Thurston maps under iteration.A Thurston map is a branched covering map of a 2-sphere S2 whereeach critical point has a finite orbit. Mostly, we also assume that theThurston map is expanding in a suitable sense.

We will see that every expanding Thurston map f : S2 → S2 is afactor of a shift operator. This link to symbolic dynamics suggests ourultimate goal of finding a combinatorial description of the dynamicsof f in terms of finite data. We approach this question by studyingcell decompositions of the 2-sphere S2 that are related to the givenmap. Relevant for this in turn are existence and uniqueness resultsfor f -invariant Jordan curves C ⊂ S2 containing the set of postcriticalpoints of f . For every sufficiently high iterate fn of an expandingThurston map such an invariant Jordan curve C always exists. If f is

a rational map on the Riemann sphere C, then C is a quasicircle. Thisstatement remains true in the general case if the underlying 2-sphereS2 is endowed with a suitable geometry induced by a “visual” metric% adapted to the dynamics of f .

A consequence of these results is that every sufficiently high iterateof a Thurston map can be described by a particular type of “cellular”Markov partition of S2 arising from a pair of cell decompositions. The2-dimensional cells or tiles in these cell decompositions are quasidisksin (S2, %) and accordingly have good geometric properties.

The space (S2, %) encodes many features of the dynamics of f . Forexample, f : S2 → S2 is topologically conjugate to a rational map if andonly if (S2, %) is quasisymmetrically equivalent to the Riemann sphere

C. Establishing a framework for proving these and other statementsfor expanding Thurston maps is the main purpose of this work.

To discuss some specific results, we start with some basic definitions

(more details can be found in the body of the text). Let f : C → Cbe a rational map on the Riemann sphere C of degree ≥ 2. As usual

we call a point p ∈ C a critical point of f if near p the map f is not alocal homeomorphism. A postcritical point is any point obtained as animage of a critical point under forward iteration of f . So if we denote

1

2 1. INTRODUCTION

by crit(f) the set of critical points of f and by fn the n-th iterate off , then the set of postcritical points of f is given by

post(f) :=⋃n∈N

fn(c) : c ∈ crit(f).

It is a fundamental fact in complex dynamics that much informationon the dynamics of f can be deduced from the structure of the orbitsof critical points. A very strong assumption in this respect is that eachsuch orbit is finite, i.e., that the set post(f) is a finite set. In this casethe map f is called postcritically-finite.

A characterization of such maps is due to Thurston. The frameworkfor his investigations was the setting of branched covering maps of2-spheres. These are continuous maps f : S2 → S2 on an oriented2-sphere S2 that near each point can be written as z 7→ zd near 0after suitable orientation-preserving coordinate changes in source andtarget. For such maps the sets of critical and postcritical points can

be defined in the same way as for rational maps on C. A Thurstonmap is a non-homeomorphic branched covering map f : S2 → S2 witha finite set of postcritical points. We will usually also assume that theThurston map is expanding. For a rational Thurston map f (i.e., for apostcritically-finite rational map) this is satisfied if and only if f hasno periodic critical points. It is also equivalent to requiring that theJulia set J (f) of f is the whole Riemann sphere (see Proposition 2.3).For general Thurston maps we will define the notion of expansion laterin Section 1.2 after we discussed an example that will motivate thedefinition. In contrast to the rational case, an expanding Thurstonmap can have periodic critical points in general (see Example 12.18).

1.1. A Lattes map as a first example

Expanding Thurston maps are abundant and include specific rationalmaps such as f(z) = 1− 2/z2 or f(z) = 1 + (i − 1)/z4. More examplescan be found in Section 12.3, and a complete list of all examples consid-ered in this book in Section 1.7. We will later give a general method forproducing Thurston maps (see Proposition 12.3 in combination withTheorem 13.2); it follows from one of our main results (Theorem 14.1)that at least some iterate of every expanding Thurston map can beobtained from this construction.

A large class of well-understood Thurston maps are Lattes maps.These are rational maps obtained as quotients of holomorphic torus en-domorphisms (see Chapter 3; note that the terminology is not uniformand some authors use the term Lattes map with a slightly differentmeaning). They were the first examples of rational maps whose Julia

1.1. A LATTES MAP AS A FIRST EXAMPLE 3

set is the whole sphere [Lat]. We will now discuss a particular Lattesmap in detail. This will serve as an introduction to some of the mainthemes of this work.

The square [0, 12]2 ⊂ R2 ∼= C can be conformally mapped to the up-

per half-plane in C such that the vertices 0, 12, 1

2+ i

2, i

2of the square are

mapped to the points 0, 1,∞,−1, respectively. By Schwarz reflection

we can extend this to a map Θ: C→ C. Up to post-composition by aMobius transformation this map is a classical Weierstraß ℘-function;it is doubly periodic with respect to the lattice Γ := Z⊕Zi and inducesa double branched covering map of the torus T := C/Γ to the sphere

C.Consider the map

A : C→ C, u 7→ A(u) := 2u.

One can check that there is a well-defined and unique map g : C → Csuch that the diagram

(1.1) C A//

Θ

C

Θ

Cg// C

commutes. The map g obtained in this way is a Lattes map. It is arational map given by

g(z) = 4z(1− z2)

(1 + z2)2for z ∈ C.

One can show that the Julia set of g is the whole sphere.More relevant for us than this explicit formula for g is that one

can describe g geometrically as follows. There is an essentially unique

orbifold metric on C with four conical singularities whose pull-back by℘ is the Euclidean metric on C (see Section 2.5 for the terminology).Geometrically, the sphere equipped with this metric looks like a pillow(in general, a pillow is a metric space obtained from gluing two identicalEuclidean polygons together along their boundary, equipped with the

induced path-metric). In our case, the upper and lower half-planes in Cequipped with the orbifold metric are isometric to copies of the square[0, 1

2]2. If we glue two copies of this square along their boundaries, then

we obtain the pillow P .We color one of these squares, say the upper half-plane, white, and

the other square, the lower half-plane, black. We divide each of thesetwo squares into four smaller squares of half the side-length, and colorthe eight small squares in a checkerboard fashion black and white. If

4 1. INTRODUCTION

0

1

−1

∞

17→07→1

07→0

7→−1

−17→0

7→1

∞7→0

7→−1

g

7→∞

Figure 1.1. The Lattes map g.

we map one such small white square to the large white square by aEuclidean similarity, then this map extends by reflection to the wholepillow. There are obviously many different ways to color and map thesmall squares. If we do this in an appropriate way as indicated inFigure 1.1, then we obtain the map g.

The vertices where four small squares intersect are the criticalpoints of g. They are mapped by g to the set 1,∞,−1, which inturn is mapped to 0. The point 0 is a fixed point of g. So g isa postcritically-finite map with post(g) = 0, 1,∞,−1, and hence aThurston map. The postcritical points of g are the vertices of the pil-low, which are the conical singularities of our orbifold metric. The

extended real line C := R = R ∪ ∞ (corresponding to the commonboundary of the two large squares forming the pillow) is a Jordan curvethat is invariant under g in the sense that g(C) ⊂ C and contains the set0, 1,∞,−1 of postcritical points of g. The set g−1(C) is an embeddedgraph in the pillow consisting of all sides of the small squares on theleft hand side of Figure 1.1 as edges and the points in g−1(post(g)), i.e.,the corners of these squares, as vertices. This graph g−1(C) determinesthe tiling in this picture.

The set g−2(C) is obtained by pulling g−1(C) back by the map g.Since g restricted to any small square S ′ is a homeomorphism ontoone of the two large squares S forming the pillow, in this process S ′ issubdivided in the same way as S was subdivided by the small squares ofside length 1/4. It follows that g−2(C) subdivides the pillow into 4×8 =32 squares of side length 1/8. Proceeding in this way inductively, we seethat g−n(C) subdivides the pillow into 2n+1 squares of side length 2−n−1

for n ∈ N. The complementary components of g−n(C) are the interiorsof these squares. In particular, the diameters of these components tend

1.2. CELL DECOMPOSITIONS AND INVARIANT CURVES 5

to 0 uniformly as n → ∞. This observation will be the basis of ourdefinition of an expanding Thurston map (see the beginning of the nextsection).

For each n ∈ N0 the set g−n(C) forms an embedded graph in thepillow P with the points in g−n(post(g)) as vertices. Note that thisis also meaningful for n = 0, if we interpret g0 as the identity on P .Then this graph is just the Jordan curve C with the points in post(g)as vertices.

The graph g−n(C) is the one-dimensional skeleton or 1-skeleton ofa cell decomposition Dn = Dn(g, C) of the pillow P generated by g andC (see Chapter 5 for the terminology that we use here and below). The2-dimensional cells or tiles of the cell decomposition Dn are squares ofside-length 1/2n and are given by the closures of the complementarycomponents of g−n(C) in P . Since C is g-invariant and so g−n(C) ⊂g−(n+1)(C), the tiles in Dn are subdivided by the tiles in Dn+1, and sothe cell decomposition Dn+1 is a refinement of Dn.

Note that the tiles in D0 are the two initial squares forming thepillow, and D1 is formed by squares of side-length 1/2 subdividingthese squares. Since we repeat the same subdivision procedure in thepassage from Dn to Dn+1, this whole sequence of cell decompositionDn is generated by the initial pair (D0,D1) (see Proposition 5.10 for ageneral result in this direction). This pair (D0,D1) is a cellular Markovpartition for g (see Definition 5.8). The map g sends each cell in D1 toa cell in D0. The cellular Markov partition (D0,D1) together with thisinformation completely determines the map g (up to conjugation). Inthis sense the dynamics of g is described by finite combinatorial data.

1.2. Cell decompositions and invariant curves

The previous example motivates several concepts for general Thurstonmaps. First, we give a precise definition of expansion. We say thata Thurston map f : S2 → S2 is expanding if there exists a Jordancurve C in S2 containing the set of postcritical points of f such thatthe complementary components of f−n(C) become uniformly small indiameter as n → ∞. Here S2 is equipped with any metric inducingthe topology on S2. It is easy to see that this condition is independentof the choice of this base metric. Later we will also show that it isindependent of the choice of C and will give other characterizations ofexpansion (see Chapter 6). We have seen that the condition is satisfiedfor the example g discussed in the previous section, and we concludethat g is an expanding Thurston map.

6 1. INTRODUCTION

We will show that a combinatorial description as in this example ispossible for some iterate of every expanding Thurston map f : S2 → S2.For this we choose a Jordan curve C ⊂ S2 with post(f) ⊂ C, andconsider the preimages f−n(C). Then for each n ∈ N0 one obtains anassociated cell decomposition Dn = Dn(f, C) of S2. Its vertices arethe points in f−n(post(f)), and its 1-skeleton the set f−n(C). Thecondition post(f) ⊂ C ensures that the closure of each complementarycomponent of f−n(C) is a 2-dimensional cell (a tile of level n or an n-tileof the cell decomposition). If f is expanding, then tiles in Dn shrinkto 0 in diameter uniformly as n→∞. The relevance of this conditionis that then the pointwise behavior of the map is determined by thecombinatorics of the tiles in Dn, n ∈ N0.

In general, these cell decompositions Dn are not compatible fordifferent levels n, but if the curve C is f -invariant (i.e., f(C) ⊂ C), thenDn+1 is a refinement of Dn for each n ∈ N0. Some of our main resultsare about existence and uniqueness of such invariant Jordan curves C.

For rational Thurston maps we have the following statement.

Theorem 1.1. Let f : C → C be a rational Thurston map with

Julia set J (f) = C. Then for each sufficiently large n ∈ N there exists

a quasicircle C ⊂ C with post(f) ⊂ C and fn(C) ⊂ C.

(Note that in this introduction we label the results by numbers asthey appear in later chapters.)

If a curve C is invariant for some iterate fn, then one cannot expectit to be invariant for some other iterate fk unless k is a multiple of n(see Remark 14.15). So in general, the curve C in the previous theorem(also in Theorem 14.1 below) will depend on n. For the definition ofa quasicircle see Section 4.1. Theorem 1.1 is a special case of a moregeneral fact.

Theorem 14.1. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Then for each

sufficiently large n ∈ N there exists a Jordan curve C ⊂ S2 that isinvariant for fn and isotopic to C rel. post(f).

See Section 2.2 for a discussion of isotopies and related terminology.

Since C is isotopic to C rel. post(f), it will also contain the set post(f).

The curve C is actually a quasicircle if S2 is equipped with a suitablemetric (see Theorem 14.3 below).

An obvious question is whether one can always find a Jordan curveC ⊂ S2 with post(f) ⊂ C that is invariant under the map f itself,and hence invariant under all iterates fn. This is not true in general


(Example 14.11), but one can give a necessary and sufficient conditionfor the existence of such an invariant curve (see Theorem 14.4 discussedbelow).

Let f : S2 → S2 be an expanding Thurston map, C ⊂ S2 be aJordan curve with post(f) ⊂ C, and Dn = Dn(f, C) the associated celldecomposition of S2 for n ∈ N0. If the curve C is f -invariant, then Dn+1

is a refinement of Dn for each n ∈ N. Then the pair (D1,D0) gives acellular Markov partition for f that determines the combinatorics ofall cell decompositions Dn. This cellular Markov partition for f is ofa particular type, namely coming from a two-tile subdivision rule (seeSection 12.2).

The main consequence of Theorem 14.1 is that we get such a two-tilesubdivision rule for some iterate F = fn of every expanding Thurstonmap.

Corollary 14.2. Let f : S2 → S2 be an expanding Thurston map.Then for each sufficiently large n there exists a two-tile subdivision rulethat is realized by F = fn.

This essentially gives a description of the map F and its dynamicsin terms of finite combinatorial data. A very general setting that allowsone to address similar questions is the recently developed theory of self-similar group actions (see [Ne], in particular Sect. 6). Our approachis more concrete and adapted to Thurston maps. One of its mainfeatures is that we get good geometric control for the cells in the celldecompositions Dn = Dn(f, C) if C is f -invariant. For example, thenthe tiles in Dn are quasidisks with uniform parameters independent ofthe level n.

If we allow more general cellular Markov partitions, it seems verylikely that not only an iterate of f , but f itself has a cellular Markovpartition.

Conjecture. Every expanding Thurston map admits a cellularMarkov partition.

We believe that even though we were not able to settle the con-jecture, the methods established here will be useful for answering thisquestion.

The proof of Theorem 14.1 is based on a general criterion for theexistence of invariant curves. To state this somewhat technical result,we first have to explain some terminology. If f : S2 → S2 is a Thurstonmap and C ⊂ S2 a Jordan curve with post(f) ⊂ C, then the associatedcell decompositions Dn(f, C) can be used to define a combinatorialquantityDn(f, C). It is given by the minimal number of tiles inDn(f, C)

8 1. INTRODUCTION

that are needed to form a connected set joining opposite sides of C, i.e.,two non-adjacent arcs obtained from subdividing C by the points inpost(f) (see (5.14) in Section 5.6; for # post(f) = 3 one has to adjustthis definition). For example, if g is the Lattes map discussed in the

previous section and C = R, then Dn(g, C) = 2n for n ∈ N0.Suppose C is f -invariant. We call a Thurston map f combinatorially

expanding for C if there exists a number n0 ∈ N such that Dn0(f, C) ≥ 2(Definition 12.4). This means that there exists n0 ∈ N so that notile of level n0 joins opposite sides of C. In this case the numbersDn = Dn(f, C) grow at an exponential rate as n→∞ (Lemma 12.7).

It is easy to see that if f is an expanding Thurston map, thenDn(f, C) → ∞ as n → ∞. In particular, if C is f -invariant, then f iscombinatorially expanding for C. The converse is not true in general.We will investigate this in Chapter 13 where we show that every com-binatorially expanding Thurston map is equivalent to an expandingThurston map (Proposition 13.3 and Theorem 13.2).

The intuitive reason behind this important fact is that if the diam-eters of the tiles in Dn(f, C) fail to shrink to zero as n→∞, then onecan “correct” the map f so that this becomes true without affectingthe combinatorics of the cell decompositions Dn(f, C).

We can now formulate our general existence result for invariantcurves.

Theorem 14.4 (Existence of invariant curves). Let f : S2 → S2 bean expanding Thurston map. Then the following conditions are equiv-alent:

(i) There exists an f -invariant Jordan curve C ⊂ S2 with post(f) ⊂C.

(ii) There exist Jordan curves C, C ′ ⊂ S2 with post(f) ⊂ C, C ′ andC ′ ⊂ f−1(C), and an isotopy H : S2× I → S2 rel. post(f) withH0 = idS2 and H1(C) = C ′ such that the map

f := H1 f is combinatorially expanding for C ′.

Moreover, if (ii) is true, then there exists an f -invariant Jordan curve

C ⊂ S2 with post(f) ⊂ C that is isotopic to C rel. post(f) and isotopicto C ′ rel. f−1(post(f)).

In (ii) the most relevant point is that there exists a Jordan curveC that can be isotoped rel. post(f) into its own preimage under f .

If this is true, then one can always extract an invariant set C, but ingeneral this will not be a Jordan curve due to possible self-intersections


of C. This problem is ruled out by the additional requirement that the

auxiliary map f is combinatorially expanding.

Note that the curve C ′ in (ii) is f -invariant, and so it is meaningful

to require that f is combinatorial expanding for C ′. In general, thiscondition is easy to check (see Remark 14.13 (i), Proposition 14.18,and the examples discussed in Section 14).

One can actually formulate a related criterion for the existence of aninvariant curve in a given isotopy class rel. post(f) or rel. f−1(post(f))

(see Remark 14.13 (iii)). Moreover, if an f -invariant Jordan curve Cexists, then it is the Hausdorff limit of a sequence of Jordan curvesCn that can be obtained from a simple iterative procedure (Proposi-tion 14.19).

We will now give a sketch of the proof of Theorem 14.1 above (aboutthe existence of an invariant curves for suitable iterates) in order toillustrate some main ideas. Let f : S2 → S2 be an expanding Thurstonmap, and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. We considerthe cell decompositions Dn = Dn(f, C) of S2 induced by f and C. If nis large enough, one can find a homeomorphism ψ of S2 that maps Cinto the 1-skeleton f−n(C) of Dn and is isotopic to the identity on S2

by an isotopy that fixes the points in post(f). Essentially, this followsfrom the fact that f is expanding and so the 1-skeleton f−n(C) of Dnforms a fine “grid” in S2 for n large. This grid contains post(f) andallows us to trace the curve C closely (see Lemma 11.17 for a precisestatement).

The map F = ψ fn is a Thurston map with an invariant curve

C := ψ(C) ⊇ post(F ) = post(f). Moreover, if n is large enough, then

F is combinatorially expanding. Actually, we can even assume that Fis expanding, because if necessary, the map can be “corrected” to havethis property (Theorem 13.2). The expanding Thurston maps F = fn

and F are Thurston equivalent (see Definition 2.4) and hence topolog-

ically conjugate (Theorem 11.4). Since the map F has the invariant

Jordan curve C, one obtains an F -invariant curve with the desired prop-

erties as an image of C under a homeomorphism that conjugates F andF .

Our existence results are complemented by the following uniquenessstatement for invariant Jordan curves.

Theorem 14.5. Let f : S2 → S2 be an expanding Thurston mapand C and C ′ be f -invariant Jordan curves in S2 that both contain theset post(f). Then C = C ′ if and only if C and C ′ are isotopic rel.f−1(post(f)).

10 1. INTRODUCTION

As a consequence one can prove that if # post(f) = 3, then there areat most finitely many f -invariant Jordan curves C ⊂ S2 with post(f) ⊂C (Corollary 14.8). This is also true if f is rational and not a (flexible)Lattes map. In general, a Thurston map f can have infinitely manysuch invariant curves (Example 14.9), but there are at most finitelymany in a given isotopy class rel. post(f) (Corollary 14.7).

1.3. Fractal spheres

In order to motivate other important concepts of our investigation, wewill now discuss another Thurston map and an associated fractal 2-sphere. As our main purpose here is to provide the reader with someintuition, we will omit the justification of some details.

The map arises from a geometric construction that is similar to theone used to describe the Lattes map in Section 1.1. Again we start witha pillow obtained by gluing together two squares along their boundaries;see the top right of Figure 1.2. This pillow is a polyhedral surface S0

homeomorphic to the 2-sphere. It carries a natural cell decompositionwith the two squares as tiles, the four sides of the common boundaryas edges, and the four common corners of the squares as vertices. Todistinguish them from other topological cells that we will introducemomentarily, we consider them as cells of level 0 and accordingly callthem 0-tiles, 0-edges, and 0-vertices. As in the example of Section 1.1,we assign colors to the tiles; say the square on top of S0 as shown inFigure 1.2 is white, and the bottom square is black.

To obtain cells on the next level 1, each of the two squares, or moreprecisely 0-tiles, is divided into four squares of half the side-length. Wecall these eight smaller squares tiles of level 1, or simply 1-tiles. Theedges of these squares are 1-edges. We slit the sphere along one such1-edge in the white 0-tile and glue in two small squares at the slit, asindicated on the upper left of Figure 1.2. This gives two additional1-tiles and we obtain a polyhedral surface S1 homeomorphic to the2-sphere. The surface S1 carries a cell decomposition given by thetopological cells of level 1 as described. We color the 1-tiles black andwhite in an alternate way as indicated in Figure 1.2.

To define a Thurston map based on our construction, we choosean identification of the polyhedral surface S1 with S0. To do this, werepresent the six 1-tiles that replaced the white 0-tile topologically assubsets of this tile, and similarly the other four 1-tiles as subsets ofthe black 0-tile. So the 0-tiles are “subdivided” by 1-tiles. This isindicated in the lower left of Figure 1.2. Under this identification the

1.3. FRACTAL SPHERES 11

S0S1

h

Figure 1.2. The map h.

cell decomposition of S1 is a refinement of the cell decomposition ofS0.

Now we can define a Thurston map as follows. We map each white1-tile on the polyhedral surface S1 to the white 0-tile in S0, and eachblack 1-tile in S1 to the black 0-tile in S0 by a similarity map (preserv-ing orientation). This is a well-defined and uniquely determined mapon S1 if we do this so that the 1-vertices marked by black and by whitedots in the upper left of Figure 1.2 are sent to 0-vertices in the upperright of the picture with the same markings.

If we identify S1 with S0 as discussed, we get a map h : S2 →S2 on the 2-sphere S2 := S0. Since h restricted to each 1-tile is ahomeomorphism onto a 0-tile, h is a branched covering map. Thecritical points of h are the 1-vertices where at least four 1-tiles intersect.These critical points are all mapped to vertices of the pillow, i.e., to0-vertices. All 0-vertices in turn are mapped to the 0-vertex markedblack. Thus h is postcritically-finite, i.e., a Thurston map. Note thatthe boundary C of the pillow is an h-invariant Jordan curve (i.e., h(C) ⊂

12 1. INTRODUCTION

C) and that the cell decomposition on S1 ∼= S0 is determined by h−1(C).Namely, h−1(C) is a topological graph that gives the 1-skeleton of thiscell decomposition, and each 1-tile is the closure of a complementarycomponent of h−1(C).

There is no rational map that realizes a similar combinatorial pic-ture as our map h. More precisely, h is not Thurston equivalent toa rational map, because h has a Thurston obstruction. This togetherwith the terminology will be explained in Section 2.6.

We will now describe a fractal sphere that is associated with ourconstruction and gives an alternative way to view our map h. Here wedo not identify the surfaces S0 and S1, and consider the passage fromS0 to S1 as a replacement procedure. The white 0-tile is replaced bythe top part of S1, consisting of six 1-tiles, i.e., squares of side-length1/2; we call this top part of S1 the white generator. Similarly, the four1-tiles subdividing the black 0-tile give the black generator. The poly-hedral surface S1 consisting of ten squares is the first approximationof a fractal space S that we are about to construct by iterating thisprocedure.

Namely, the 1-tiles of the black and white generators are also coloredas indicated in Figure 1.2. So if we replace each black or white 1-tile bya suitably scaled copy of the black or white generators, then we obtaina polyhedral surface S2 glued together from squares of side-length 1/4as 2-tiles. Here we have to be careful about how precisely a tile isreplaced by an appropriate generator, because the generators with theirgiven colorings of tiles are not symmetric with respect to rotations. Tospecify the replacement rule uniquely, we use the additional markingsof some points. Each generator carries two points corresponding tothe points on S0 marked black and white. In the replacement processwe require that these points match the corresponding points with thesame markings on 1-tiles.

If we iterate the replacement procedure, then we obtain polyhedralsurfaces Sn for all levels n ∈ N0 glued together from squares of side-length 1/2n. Each surfaces Sn carries a natural cell decompositionDn given by these squares as tiles. Some iterates of this constructionare shown in Figure 1.3. The pictures essentially indicate the gluingpattern of the squares which give the surfaces. One should view them asabstract polyhedral surfaces, and not confuse them with the underlyingsubsets of R3 in these pictures. Each surface Sn is a topological 2-sphere and carries a piecewise Euclidean length metric %n with conicalsingularities.

One can now extract a self-similar “fractal” space S as a limitSn → S for n → ∞ in several ways. One possibility is to pass to a


S1 S2

S3 S4

Figure 1.3. Polyhedral surfaces obtained from the re-placement rule.

Gromov-Hausdorff limit of the sequence (Sn, %n) of metric spaces. Wewill discuss a different method that is closer in spirit to considerationsin Chapter 13.

Namely, given an n-tile X n ⊂ Sn (which is a square of side-length

2−n), and an (n + 1)-tile X n+1 ⊂ Sn+1, we write X n ⊃∼ X n+1 if X n+1

is contained in the scaled copy of a generator that replaced X n in theconstruction of Sn+1 from Sn. We now consider descending sequencesX 0 ⊃∼ X 1 ⊃∼ X 2 ⊃∼ . . . . On an intuitive level the squares in such asequences should shrink to a point in our desired limit space S rep-resented by the sequence. Here we consider two sequences X n andYn as equivalent and representing the same point if X n ∩Yn 6= ∅ forall n ∈ N0. It is not hard to see that this indeed defines an equivalencerelation for descending sequences. By definition our limit space S isnow the set of all equivalence classes.

For x, y ∈ S we set

(1.2) %(x, y) := lim supn→∞

dist%n(X n,Yn),

14 1. INTRODUCTION

where X n and Yn are sequences representing x and y, respectively.Then % is well-defined and one can show that this is a metric on S.

For x, y ∈ S, x 6= y, we define

(1.3) m(x, y) := infn ∈ N : X n ∩ Yn = ∅,where the infimum is taken over all sequences X n and Yn repre-senting x and y, respectively. Then

(1.4) %(x, y) 2−m(x,y)

for x, y ∈ S, x 6= y, where the implicit multiplicative constant is inde-pendent of the points. So roughly speaking, the distance of two pointsin S is given in terms of the minimal level on which two descendingsequences representing the points start seeing the difference betweenthem.

It is intuitively clear (S, %) is a topological 2-sphere. To outline arigorous proof for this fact, we return to the Thurston map h definedabove. The cell decomposition of S2 given by 1-tiles is determined bythe complementary components of h−1(C), where C is the h-invariantJordan given by the common boundary of the 0-tiles. We can iteratethis procedure and consider the pull-back of C by higher iterates of h.Then h−n(C) is an embedded graph in S2, and we call a closure of acomplementary components of h−n(C) an n-tile for (h, C). These n-tilesform a cell decomposition Dn(h, C) of S2. The map hn sends each n-tile to a 0-tile homeomorphically, and we can assign colors to n-tiles sothat hn sends an n-tile to the 0-tile of the same color. In the passagefrom Dn(h, C) to Dn+1(h, C) each n-tile is subdivided by (n + 1)-tilesin the same way as the 0-tile of the same color is subdivided by 1-tiles.From this it is clear there is a one-to-one correspondence between n-tiles in Sn and n-tiles for (h, C). Moreover, these tiles realize identicalcombinatorics. More precisely, we have

X n+1 ⊃∼ X n ⇔ Xn+1 ⊃ Xn, and

X n ∩ Yn 6= ∅ ⇔ Xn ∩ Y n 6= ∅,

where the n-tiles X n and Yn in Sn and the (n + 1)-tile X n+1 in Sn+1

correspond to the n-tiles Xn and Y n and the (n + 1)-tile Xn+1 for(h, C), respectively.

If our Thurston map h is expanding, then the diameters of n-tiles for(h, C) tend to 0 uniformly as n→∞ (with respect to some fixed basemetric on S2 representing the topology). Then one gets a well-definedmap ϕ : S → S2 by sending point in S represented by a descendingsequence X n to the unique point in the intersection

⋂n∈N0

Xn of the


corresponding n-tiles for (h, C). In general, our map h need not beexpanding, but we may assume this if we choose the identification ofS1 with S0 carefully (this hinges on the fact that h is combinatoriallyexpanding and so the map can be corrected if necessary to make itexpanding; see Proposition 13.3 for details). It is then not hard to seethat ϕ is a homeomorphism, and so S is a 2-sphere.

One can turn this consideration around and start with the expand-ing Thurston map h instead of our geometric construction to definethe fractal space S. In our example, this amounts to finding a naturaldescription of a metric on S2 that corresponds to the metric % on Sas defined in (??). The idea for this is the observation that the ex-pression in (1.3) is meaningful for points in x, y ∈ S2 if we replace X n

and Yn by n-tiles containing x and y, respectively. Moreover, m(x, y)is completely determined by the combinatorics of tiles. By (1.4) thiscombinatorial quantity determines a metric on S2 corresponding to themetric % on S, up to a multiplicative ambiguity that is irrelevant formost considerations.

For general Thurston maps one has to allow a more general form of(1.4). Namely, we consider metrics % satisfying

%(x, y) Λ−m(x,y),

for some Λ > 1. We call such a metric a visual metric for the givenThurston map, and Λ its expansion factor. In our particular example,% (under the identification of S with S2 by the homeomorphism ϕ) isa visual metric for h with expansion factor Λ = 2. In this case, weobtain visual metrics with arbitrary expansion factor 1 < Λ ≤ 2 if weconsider a “snow-flaked” metric %α with suitable α ∈ (0, 1), but thereis no visual metric for h with Λ > 2. Indeed, if % is a visual metric withexpansion factor Λ > 1 and Xn is an n-tile, then diam%(X

n) . Λ−n.Now it is easy to see that one can form a connected chain of n-tiles with2n elements that joins two non-adjacent 0-edges (as Figure 1.3 suggestsone obtains such a chain by running along the bottom 0-edge). Thenby the triangle inequality 2n · Λ−n & 1 for all n ∈ N, and so Λ ≤ 2.

We will give a precise definition of a visual metric for an expand-ing Thurston map in Chapter 8. Visual metrics for a given Thurstonmap are not unique, but two different visual metrics with the sameexpansion factor are bi-Lipschitz equivalent, and they are snowflakeequivalent if they have different expansion factors (see Section 4.1 forthis terminology).

If f : S2 → S2 is an arbitrary Thurston map, then the supremumof all Λ > 1 for which there exist visual metrics with expansion factorΛ agrees with the combinatorial expansion factor of Thurston map

16 1. INTRODUCTION

f , denoted by Λ0(f). The quantity Λ0(f) satisfies 1 < Λ0(f) < ∞, isinvariant under topological conjugacy, and well-behaved under iteration(see Chapter 15, and in particular Theorem ??.). In our example,Λ0(h) = 2.

The combinatorial expansion factor can be computed from dataassociated with the cell decompositions determined by the expandingThurston map f . Namely, let C ⊂ S2 be a Jordan curve with post(f) ⊂C. Then we get an associated cell decompositions Dn(f, C), and we usethis to define the combinatorial quantity Dn(f, C) (as discussed in theprevious section) given by the minimal number of tiles in Dn(f, C) thatare needed to join form a connected set joining opposite sides of C, i.e.,two non-adjacent arcs obtained from subdividing C by the points inpost(f). In our example, Dn(h, C) = 2n.

In general, the number Dn(f, C) depends on C. For an expandingThurston map it grows at an exponential rate as n→∞. This growthrate is independent of C, and is determined only by f . Moreover, ifΛ0(f) is the combinatorial expansion factor of f , then (see Proposi-tion 15.1)

(1.5) Λ0(f) = limn→∞

Dn(f, C)1/n.

If f : S2 → S2 is an expanding Thurston map, and % a visual metricfor f , then we call the metric space (S2, %) the visual sphere of f . Fora given map, this space is unique up to snow-flake equivalence.

In our example, the space S equipped with the visual metric % in(1.2) is (isometric to) the visual sphere of h. Note that the space (S, %)is self-similar in the sense that the part of the surface that is “builton top” of some n-tile X n is similar (i.e., is isometric up to scaling bythe factor 2n) to the part of the surface that is “built on top” of thewhite or black 0-tile. In the same fashion we can find visual metricsfor any Thurston map f such that f scales tiles by a constant factor.This means that the metric behavior of the dynamics on tiles becomesvery simple, although the space on which f acts becomes complicated,i.e., a fractal sphere.

This fractal sphere encodes many properties of the map f ; for ex-ample, the statement that our map h is not Thurston equivalent to arational map amounts to the same as saying that (S, %) is not a quasi-sphere, which means that is not quasisymmetrically equivalent to thestandard 2-sphere (i.e., the unit sphere in R3, or equivalently the Rie-

mann sphere C equipped with the chordal metric). This last statementcan be derived from a general result (Theorem 17.1 ((ii))), but one canalso show this directly (we will outline an argument in Section 4.4).

1.4. VISUAL METRICS AND THE VISUAL SPHERE 17

1.4. Visual metrics and the visual sphere

As we discussed, an expanding Thurston map f : S2 → S2 induces anatural class of metrics on S2 that we call visual metrics (see Section 6).Each visual metric % has an associated expansion factor Λ > 1. Itis characterized by the geometric property that for cells σ, τ in thecell decompositions Dn = Dn(f, C) we have diam%(σ) Λ−n if σ hasdimension ≥ 1 and dist%(σ, τ) & Λ−n if σ ∩ τ = ∅ (Proposition 8.4).Any two visual metrics are snowflake equivalent, and the class of visualmetrics for f and any iterate of f are the same.

The range of possible expansion factors of visual metrics for anexpanding Thurston map f is given by the combinatorial expansionfactor Λ0(f) (as in (1.5)).

Theorem 15.3. Let f : S2 → S2 be an expanding Thurston map,and Λ0(f) ∈ (1,∞) be its combinatorial expansion factor. Then thefollows statements are true:

(i) If Λ is the expansion factor of a visual metric for f , then1 < Λ ≤ Λ0(f).

(ii) Conversely, if 1 < Λ < Λ0(f), then there exists a visual metric% for f with expansion factor Λ. Moreover, the visual metric% can be chosen to have the following additional property:

For every x ∈ S2 there exists a neighborhood Ux of x suchthat

%(f(x), f(y)) = Λ%(x, y) for all y ∈ Ux.

In particular, if f is an expanding Thurston map, then we can al-ways find a visual metric % so that f scales the metric % by a constant

factor at each point. The example of the Lattes map g : C → C dis-

cussed in Section 1.1 above illustrates this statement. If we equip Cwith a suitable visual metric for g (a flat orbifold metric with four coni-cal singularities), then g behaves like a piecewise similarity map, wheredistances are scaled by the factor Λ = 2.

If Λ = Λ0(f), then in general one cannot guarantee the existenceof a visual metric with expansion factor Λ; see Example 15.8 which isgiven by a Lattes-type map with an associated affine map on R2 thathas a shear component.

Our choice of the term “visual metric” is motivated by the closerelation of this concept to the notion of a visual metric on the boundaryof a Gromov hyperbolic space (see Section 4.2 for general background;very similar ideas can be found in [HP09]). Namely, if f : S2 → S2 isan expanding Thurston map and C ⊂ S2 a Jordan curve with post(f) ⊂

18 1. INTRODUCTION

C, then one can define an associated tile graph G(f, C) as follows. Itsvertices are given by the tiles in the cell decompositions Dn(f, C) onall levels n, and one joins two vertices by an edge if the correspondingtiles intersect and have levels differing by at most 1 (see Section 10).The graph G(f, C) depends on the choice of C, but if C ′ ⊂ S2 is anotherJordan curve with post(f) ⊂ C ′, then G(f, C) and G(f, C) are rough-isometric (Theorem 10.4); note that this is much stronger than beingquasi-isometric (see Section 4.2 for the terminology).

The graph G(f, C) is Gromov hyperbolic (Theorem 10.1). Its bound-ary at infinity ∂∞G(f, C) can be identified with S2, and under thisidentification the class of visual metrics in the sense of Gromov hyper-bolic spaces coincides with the class of visual metrics for f in our sense(Theorem 10.2).

If f : S2 → S2 is an expanding Thurston map and % a visual metricfor f , then the metric space (S2, %) is the visual sphere of f . If %′ is an-other visual metric, then (S2, %) and (S2, %′) are snowflake-equivalent;so the visual sphere is unique up to this type of equivalence. As wehave seen, one should think of (S2, %) as a fractal space that in general

is quite different from the standard unit sphere C (equipped with thechordal metric); see Section 4.4 for some concrete examples.

Many dynamical properties of f are encoded in the geometry of thevisual sphere. The following statement is one of the main results ofthis work.

Theorem 17.1. Let f : S2 → S2 be an expanding Thurston mapand % be a visual metric for f . Then (S2, %) is

(i) doubling if and only if f has no periodic critical points,

(ii) is quasisymmetrically equivalent to the standard unit sphere

C if and only if f is topologically conjugate to a rational map,

(iii) is snowflake equivalent to the standard unit sphere C if andonly if f is topologically conjugate to a Lattes map.

See Section 4.1 for the terminology.

If f is a rational Thurston map on C, then one cannot expect the

chordal metric on C to be a visual metric for f . As the proof of The-orem 17.1 (iii) will show, this is the case if and only if f is a Lattesmap.

Part (ii) of the previous theorem provides an intriguing analog ofCannon’s conjecture in geometric group theory (see Section 4.3 for amore detailed discussion). According to this conjecture every Gro-mov hyperbolic group G whose boundary at infinity ∂∞G is a 2-sphereshould arise from some standard situation in hyperbolic geometry. The

1.4. VISUAL METRICS AND THE VISUAL SPHERE 19

conjecture is equivalent to showing that ∂∞G equipped with a visualmetric (in the sense of Gromov hyperbolic spaces) is quasisymmetri-cally equivalent to the standard 2-sphere. One of the reasons whyCannon’s conjecture is still open may be the lack of non-trivial exam-ples that guide the intuition (see the recent paper [BouK] though thatin a sense addresses this issue). All examples come from fundamentalgroups G of compact hyperbolic orbifolds where one already has a nat-

ural identification of ∂∞G with C; according to Cannon’s conjecturethere are no other examples. In contrast, the visual spheres of ex-panding Thurston maps provide a rich supply of metric 2-spheres thatsometimes are and sometimes are not quasisymmetrically equivalent to

C (see Section 4.4).The proof of one of the implications in Theorem 14.4 (ii) (the “if”

part) uses some well-known ingredients. Namely, if (S2, %) is quasisym-

metrically equivalent to the standard sphere C, the one can conjugate

f to a map g on C. Since the map f dilates distances with respectto a suitable visual metric by a fixed factor (see Theorem 15.3 (ii)mentioned above), the map g is uniformly quasiregular. Hence g, andtherefore also f , is conjugate to a rational map by a standard theorem.

The converse direction (the “only if” part) is harder to establish.If f is conjugate to a rational map, then we may assume without loss

of generality that f is an expanding rational Thurston map on C tobegin with. If % is a visual metric for f , then one has to show that

the identity map of (C, %) to (C, σ), where σ is the chordal metric, isa quasisymmetry. This follows from a careful analysis of the geome-try of the tiles in the cell decompositions Dn(f, C) with respect to σ(Proposition 17.7). For example, while it fairly obvious from the defi-nitions that adjacent tiles in Dn(f, C) have comparable diameter withrespect to a visual metric % (with uniform constants independent of thelevel n), the same assertion is also true for the chordal metric σ. Ourproof of this and related statements is based on Koebe’s distortion the-orem and the fact that if f has no periodic critical points, then in thecell decompositions Dn(f, C) we see locally only finitely many differentcombinatorial types.

Thurston studied the question when a given Thurston map is rep-resented by a conformal dynamical system from a viewpoint differentfrom the one suggested by Theorem 17.1 (ii) (see Section 2.6 for a shortoverview). He asked when a Thurston map f : S2 → S2 is in a suit-able sense (Thurston) equivalent (see Definition 2.4) to a rational mapand obtained a necessary and sufficient condition [DH]. For expanding

20 1. INTRODUCTION

Thurston maps his notion of equivalence actually means the same astopological conjugacy of the maps (Theorem 11.4).

The proof of part (ii) of Theorem 17.1 does not use Thurston’stheorem. Indeed, none of our statements relies on this, and so ourmethods possibly provide a different approach for its proof.

It is not clear how useful Theorem 17.1 (ii) is for deciding whetheran explicitly given expanding Thurston map is topologically conjugateto a rational map. It is likely that our techniques can be used toformulate a more efficient criterion, but we will not pursue this furtherhere. We will content ourselves with a simple statement that easilyfollows from our results.

Theorem 7.3. Let f : S2 → S2 be a Thurston map with # post(f) =3. Then f is Thurston equivalent to a rational Thurston map. If themap f is expanding, then f is topologically conjugate to a rationalThurston map if and only if f has no periodic critical points.

1.5. Further results

The visual sphere of an expanding Thurston map carries a naturalmeasure adapted to its dynamics.

Theorem 16.1. Let f : S2 → S2 be an expanding Thurston map.Then there exists a unique measure νf of maximal entropy for f . Themap f is mixing for νf .

This theorem follows from results due to Haıssinsky-Pilgrim [HP09,Thm. 3.4.1]. We will present a different proof and give an explicit de-scription of νf in terms of the cell decompositions Dn(F, C), where Cis an invariant curve as in Theorem 14.1 and F = fn. In particu-lar, νf = νF assigns equal mass to all tiles in the cell decompositionsDn(F, C) of a given “color” (see Proposition 16.10 and Theorem 16.11).

The measure νf can be used to study the topological and measuretheoretic dynamics of f under iteration. For example, we will see thathtop(f) = log(deg f), where htop(f) is the topological entropy and deg fthe topological degree of f (Corollary 16.2).

If µ is a Borel measure on a metric space (X, d), then we call themetric measure space (X, d, µ) Ahlfors Q-regular, Q > 0, if

µ(Bd(x, r)) rQ

for all closed balls Bd(x, r) in X whose radius does not exceed thediameter of the space. If an expanding Thurston map has no peri-odic critical points, then its visual sphere together with its measure ofmaximal entropy has this property.

1.5. FURTHER RESULTS 21

Proposition 17.2. Let f : S2 → S2 be an expanding Thurstonmap without periodic critical points, % be a visual metric for f withexpansion factor Λ > 1, and ν = νf be the measure of maximal entropyof f . Then (S2, %, ν) is Ahlfors Q-regular with

(1.6) Q =log(deg f)

log Λ.

It follows that Q as in (1.6) is the Hausdorff dimension of (S2, %),and that the Q-dimensional Hausdorff measure HQ

% of S2 with respect

to % satisfies 0 < HQ% (S2) <∞.

The properties of visual metrics are essential in proving the follow-ing statement (see Chapter 9 for the relevant definitions).

Theorem 9.1. Let f : S2 → S2 be an expanding Thurston map.Then f is a factor of the left-shift Σ: Jω → Jω on the space Jω of allsequences in a finite set J of cardinality #J = deg f .

An immediate consequence of this theorem and its proof is the factthat the periodic points of an expanding Thurston map f : S2 → S2

are dense in S2 (Corollary 9.2).The invariant curve C in Theorem 14.1 equipped with (the restric-

tion of) a visual metric is a quasicircle. This follows from the followinggeneral fact applied to the map fn.

Theorem 14.3. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. If C is f -invariant,then C equipped with (the restriction of) a visual metric for f is aquasicircle.

Recall from Section 1.3 that if f : S2 → S2 is an expanding Thurs-ton map and C ⊂ S2 a Jordan curve with post(f) ⊂ C, then theassociated cell decompositions Dn(f, C) can be used to define a combi-natorial quantity Dn(f, C). Its is given by the minimal number of tilesin Dn(f, C) that are needed to form a connected set joining oppositesides of C.

As we remarked, for an expanding Thurston map Dn(f, C) growsat an exponential rate as n → ∞. This growth rate is independentof C, is determined only by f , and related to the the combinatorialexpansion factor Λ0(f) given by (1.5). One has Dn(f, C) . (deg f)n/2

which implies Λ0(f) ≤ (deg f)1/2.Lattes maps realize the maximal growth rate in the sense that

Dn(f, C) (deg f)n/2, and these maps are characterized by this prop-erty. More precisely, we have the following statement.

22 1. INTRODUCTION

Theorem 19.2. Let f : S2 → S2 be an expanding Thurston map.Then f is topologically conjugate to a Lattes map if and only if thefollowing conditions are true:

(i) f has no periodic critical points.

(ii) There exists c > 0, and a Jordan curve C ⊂ S2 with post(f) ⊂C such that

Dn(f, C) ≥ c(deg f)n/2

for all n ∈ N0.

This theorem is due to Qian Yin [Yi]. It is remarkable that onecan characterize the conformal dynamical systems given by iteration ofLattes maps in terms of essentially combinatorial data.

An important ingredient in the proof of Theorem 19.2 is the fol-lowing characterization of Lattes maps in terms of their measures ofmaximal entropy.

Theorem 18.3. Let f : C→ C be a rational expanding Thurston.Its measure of maximal entropy is absolutely continuous with respect toLebesgue measure if and only if f is a Lattes map.

This is a special case of a more general theorem due to A. Zdunik[Zd] which gives a similar characterization of Lattes maps among all,not only postcritically-finite, rational maps. We will derive Theo-rem 18.3 from some considerations of Lyapunov exponents in Chap-ter 18.

1.6. Outline of the presentation

Our work is an introduction to the subject. We hope that it will moti-vate more research and will serve as a basis for future studies. There-fore, we kept our presentation elementary, self-contained, and ratherdetailed.

The prerequisites from the reader are modest and include some ba-sic knowledge of complex analysis and topology, in particular planetopology and the topology of surfaces. A background in complex dy-namics is helpful, but not absolutely necessary. In later chapters ourdemands on the reader are more substantial. In particular, in Chap-ter 16 and Chapter 18 we require some concepts and results from topo-logical and measure theoretic dynamics, but we will state and reviewthe necessary facts.

We will now give an outline of how of this book is organized andwill briefly discuss some main concepts and ideas.

1.6. OUTLINE OF THE PRESENTATION 23

In the remaining two sections of this introductory chapter the readercan find a list of all examples of Thurston maps that we consider inthis work (Section 1.7) and a summary of notation (Section 1.8).

In Chapter 2 we then turn to Thurston maps, the main objectof our investigation. In Sections 2.1 and 2.2 we review some relatedbasic concepts that are important throughout. In particular, we give aprecise definition of an expanding Thurston map (Definition 2.2). Forrational Thurston maps one can find a characterization when they areexpanding in Section 2.3. We could have postponed this discussion untillater, when we explore the concept of expansion more systematically(in Chapter 8), but readers familiar with complex dynamics may findit helpful to get some perspective early on.

Every Thurston map has a ramification function giving rise to an as-sociated orbifold. It can be parabolic in some exceptional cases (namelyfor Lattes maps) and is hyperbolic in general (Section 2.5). We will notexplore the geometric significance of this orbifold in depth, but ratheruse it as a device to record some branching data.

As we already mentioned, Thurston gave a characterization of whena Thurston map is (in a suitable sense) equivalent to a rational map.We will review this without proofs in Section 2.6. This material is notreally essential for the rest of the book, but we included this discussionfor general background.

In Chapter 3 we discuss a large class of Thurston maps, namelyLattes maps and a related class that we call Lattes-type maps. Lattes-type maps are quotients of torus endomorphisms, but in contrast to theLattes case we do not require that the endomorphism is holomorphic.The classification of all such maps is surprisingly involved. Since themain purpose of Chapter 3 is to give a supply of examples of expandingThurston maps, our definition of a Lattes-type map will be somewhatad hoc leading quickly to the main facts.

In Chapter 4 we collect facts from quasiconformal geometry (Sec-tion 4.1) and from the theory of Gromov hyperbolic spaces (Section 4.2)that will be relevant later on. We then turn to Cannon’s conjecture ingeometric group theory (Section 4.3). As was mentioned in Section 1.4,it gives an intriguing analog to some of the main themes of our study ofexpanding Thurston maps. We illustrate this with an explicit descrip-tion of some examples of fractal 2-spheres that arise a visual spheresof expanding Thurston maps (Section 4.4). We included Sections 4.3and 4.4 mainly to give some motivation for our investigation.

This starts in earnest in Chapter 5, the technical core of our com-binatorial approach. Here we discuss cell decompositions and their re-lation to Thurston maps. We start with some general facts about cell

24 1. INTRODUCTION

decompositions in Section 5.1. In particular, we introduce the conceptof a cellular Markov partition (Definition 5.8).

In Section 5.2 we specialize to cell decompositions on 2-spheres.Lemma 5.22 shows how to construct branched covering maps and Thurs-ton maps from cell decompositions. One can pull-back a cell decompo-sition D of a 2-sphere by a branched covering map f if the vertex setV of D contains the critical values of f (Lemma 5.15). Since the setpost(f) contains the critical values of all iterates of a Thurston map f ,this applies to all iterates fn if post(f) ⊂ V.

We use this in Section 5.3 to show that if f : S2 → S2 is a Thurstonmap and C ⊂ S2 a Jordan curve with post(f) ⊂ C, then we obtain anatural sequence Dn = Dn(f, C) of cell decompositions of S2. Thesecell decompositions are our most important technical tool for studyingThurston maps. Their properties are summarized in Proposition 5.17.Simple applications are Proposition 7.1, which gives a classification ofall Thurston maps f with # post(f) = 2 up to Thurston equivalence,and Corollary 7.2 showing that for an expanding Thurston map f wehave # post(f) ≥ 3.

In general, the cell decompositions Dn = Dn(f, C) are not compat-ible for different levels n unless the Jordan curve C is f -invariant. Toovercome some of the ensuing problems, we introduce the concept of ann-flower W n(p) of a vertex p in the cell decomposition Dn (Section 5.6).The set W n(p) is formed by the interiors of all cells in Dn that meet p(see Definition 5.25 and Lemma 5.26). An important fact is that whilein general a component of the preimage f−n(K) of a small connectedset K will not be contained in an n-tile (i.e., a 2-dimensional cell inDn), it is always contained in an n-flower (Lemma 5.32).

In Section 5.6 we also define the quantity Dn = Dn(f, C) that mea-sures the combinatorial expansion rate of a Thurston map given bythe minimal number of n-tiles needed to form a connected set joiningopposite sides of C (see (5.14) and Definition 5.30).

Visual metrics for expanding Thurston maps are introduced in Chap-ter 8. Their most important properties are stated in Proposition 8.3and Proposition 8.4. In particular, if f : S2 → S2 is an expandingThurston map, and % a visual metric, then the %-diameters of cells inDn will approach 0 at an exponential rate as n → ∞. This impliesthat lifts of paths under fn shrink to 0 exponentially fast if n → ∞(Lemma 8.7). This fact is of crucial importance. For example, it im-plies that every expanding Thurston map is a factor of a shift operator.This is proved in Chapter 9 (see Theorem 9.1) where we explore thelink of our theory to symbolic dynamics.

1.6. OUTLINE OF THE PRESENTATION 25

In Chapter 10 we define the tile graph G(f, C) associated with anexpanding Thurston map f : S2 → S2 and a Jordan curve C ⊂ S2

with post(f) ⊂ S2. The graph G(f, C) only depends on C up to rough-isometry (Theorem 10.4). We show that it is Gromov hyperbolic (The-orem 10.1) and that its boundary at infinity G∞(f, C) can be identifiedwith S2 (Theorem 10.2).

The exponential shrinking of lifts will be used in Chapter 11 toshow that if two expanding Thurston maps are Thurston equivalent,then they are topologically conjugate (Theorem 11.4). The idea forthe proof is well-known in the study of expanding dynamical systems.We lift an initial isotopy repeatedly to obtain a sequence of isotopiesthat form a “tower”. If the Thurston maps are expanding, then thediameters of the tracks of these isotopies shrink fast enough so that theisotopies converge to a time independent homeomorphism that providesthe desired conjugacy. In this chapter we also prove some results onisotopies of Jordan curves (Section 11.2). The subsequent Section 11.3contains some auxiliary statements on graphs. The main result is theimportant, but rather technical Lemma 11.17 which gives a sufficientcriterion when a Jordan curve can be isotoped into the 1-skeleton of agiven cell decomposition of a 2-sphere.

In the next Chapter 12 we study the cell decomposition Dn =Dn(f, C) under the additional assumption that C is f -invariant. In thiscase Dn+k is a refinement of Dn for all n, k ∈ N0; on a more intuitivelevel, each cell in any of the cell decompositions Dn is “subdivided”by cells on a higher level. Moreover, the pair (Dn+k,Dn) is a cellularMarkov partition for fk (Proposition 12.5). If a Thurston map has aninvariant Jordan curve C with post(f) ⊂ C, then it can be described bya two-tile subdivision rule (Definition 12.1) as discussed in Section 12.2.The main result here is Proposition 12.3 which gives a general methodfor constructing Thurston maps with invariant curves from a two-tilesubdivision rule (see Definition 5.20 and the following discussion). Thistheory is illustrated in Section 12.3 where we consider many examplesof Thurston maps from our combinatorial viewpoint.

As we discussed, a Thurston map f is combinatorially expandingfor an invariant curve C if there exists a number n0 ∈ N such that notile in Dn0(f, C) joins opposite sides of C. If f is expanding, then themap is combinatorially expanding for f , but the converse is not true ingeneral. We investigate this in Chapter 13 where we show that everycombinatorially expanding Thurston map is equivalent to an expandingThurston map (Proposition 13.3 and Theorem 13.2).

The intuitive reason behind this important fact is that if the diam-eters of the tiles in Dn(f, C) fail to shrink to zero as n→∞, then one

26 1. INTRODUCTION

can “correct” the map f so that this becomes true without affectingthe combinatorics of the cell decompositions Dn(f, C).

It is somewhat cumbersome to implement this idea. We do this byintroducing an equivalence relation that forces descending sequences ofn-tiles to shrink to points as n→∞. We then invoke Moore’s theorem(Theorem 13.10) to show that the quotient space of the original 2-

sphere S2 by this relation is also a 2-sphere S2. The original Thurston

map f : S2 → S2 descends to a map f : S2 → S2 and one can show

that f is an expanding Thurston map that is equivalent to the originalmap f .

Existence and uniqueness results for invariant Jordan curves areproved in Chapter 14. This chapter constitutes the center of the presentwork. Here we establish Theorems 14.1, 14.4, and 14.5, and Corol-lary 14.2 about existence and uniqueness of invariant curves mentionedabove in Section 1.2. One can obtain invariant curves from an iterativeprocedure discussed in detail in Section 14.2.

In the next Section 14.3 we prove that our type of invariant Jor-dan curves are quasicircles (Theorem 14.3). We also show that iff : S2 → S2 is an expanding Thurston map, S2 is equipped with avisual metric % for S2, and the cell decompositions Dn(f, C), n ∈ N0,are obtained from an f -invariant Jordan curve C, then the edges inthese cell decompositions are uniform quasiarcs and the boundary oftiles are uniform quasicircles (Proposition 14.25).

We revisit visual metrics in Chapter 15. We introduce the combina-torial expansion factor Λ0(f) associated with an expanding Thurstonmap f and prove Theorem 15.3 (mentioned above in Section 1.4) aboutthe existence of visual metrics with given expansion factors.

Chapter 16 is devoted to the measure theoretic dynamics of anexpanding Thurston map. The main result is Theorem 16.11 aboutexistence and uniqueness of a measure of maximal entropy νf for anexpanding Thurston map f : S2 → S2.

The geometry of the visual sphere of an expanding Thurston mapis explored in the next Chapter 17, another central part of our work.If % is a visual metric of an expanding Thurston map f : S2 → S2,then the visual sphere (S2, %) has some properties that are importantin the analysis of metric spaces. For example, (S2, %) is linearly locallyconnected (Proposition 17.4). Moreover, (S2, %) is a doubling metricspace if and only if f : S2 → S2 has no periodic critical points (Theo-rem 17.1 (i)). Actually, the absence of periodic critical points impliesthe much stronger property that (S2, %) is Ahlfors regular (Proposi-tion 17.2).

1.7. LIST OF EXAMPLES FOR THURSTON MAPS 27

In Section 17.3 we prove that f is conjugate to a rational map ifand only if (S2, %) is quasisymmetrically equivalent to the standard2-sphere (Theorem 17.1 (ii)).

In Chapter 18 we measure theoretic properties of rational expanding

Thurston maps f : C → C. We first construct a measure λ on Cthat is f -invariant and absolutely continuous with respect to Lebesgue

measure on C (Theorem 18.1). This allows us to apply methods fromergodic theory. We use this, for example, to establish an extremalityproperty of Lyapunov exponents for Lattes maps (Theorem 18.5), andobtain a description of the expansion behavior of f at typical points in

C (Theorem 18.7). It is an easy consequence of our analysis that themeasure of maximal entropy of f is absolutely continuous with respectto Lebesgue measure if and only if f is a Lattes map (Theorem 18.3).

This last fact is used in the next Chapter 19 where we establish acharacterization of Lattes maps in terms of their combinatorial expan-sion behavior (see Theorem 19.2 mentioned in Section 1.5).

The last Chapter 20 is devoted to some further developments and fu-ture perspectives. We discuss some recent related work (Section 20.1),and finish with a list of open problems (Section 20.2).

1.7. List of examples for Thurston maps

Throughout the book we consider several examples of Thurston mapsto illustrate various phenomena. We list them here with a short descrip-tion for easy reference. The relevant terms used in these descriptionsare later defined in the body of the text.

A Lattes map was considered in Section 1.1.The map in Example 12.17 is f1(z) = z2 − 1; it realizes a two-tile

subdivision rule that is not combinatorially expanding.

In Example 12.18 there are two maps f2 and f2 that both realizethe barycentric subdivision rule. The map f2 is a rational map, but it

is not expanding (i.e., its Julia set is not the whole Riemann sphere C).

The map f2 however is expanding. It is an example of an expandingThurston map with periodic critical points.

The map f3 in Example 12.19 (realizing a certain two-tile subdi-vision rule) is an obstructed map. This means f3 is not Thurstonequivalent to a rational map.

The map f4 in Example 12.20 is again not Thurston equivalent toa rational map. While somewhat easier than the map f3 in Example12.19, it is less generic, since f4 has a parabolic orbifold, whereas f3 hasa hyperbolic orbifold. The map f4 realizes the 2-by-3 subdivision rule.

28 1. INTRODUCTION

With respect to a suitable visual metric for f4, the sphere S2 consistsof two copies of a Rickman’s rug.

In Example 12.21 a whole class of maps is considered. The first oneis the map f5(z) = 1−2/z2 which realizes a simple two-tile subdivisionrule. By “adding flaps” we obtain the other maps. All these maps arerational; in fact they are given by an explicit formula, which makesthem easy to understand and visualize.

In Example 13.21 we consider a Thurston map f that is not combi-natorially expanding, yet Thurston equivalent to an expanding Thurs-ton map g. This shows that the sufficient condition in Proposition 13.3is not necessary.

In Example 14.6 we illustrate the main ideas of Section 14. In

particular, we show how for a specific map f an f -invariant curve Cwith post(f) ⊂ C is constructed; see Figure 14.1.

In Example 14.9 we show that the Lattes map g from Section 1.1has infinitely many distinct g-invariant curves C with post(g) ⊂ C.

In Example 14.11 we consider an expanding Thurston map f forwhich no f -invariant Jordan curve C with post(f) ⊂ C exists.

In Remark 14.15 we show that the fn-invariant curve C given byTheorem 14.1 will in general depend on n.

In Example 14.16 we use another Lattes map to illustrate an iter-ative construction of invariant curves; see Figure 14.4.

Example 14.22 shows what can happen if one of the necessary con-ditions in the iterative procedure for producing invariant curve is vio-

lated. Namely, the “limiting object” C is not a Jordan curve anymore.The map used to illustrate this phenomenon is again a Lattes map.

In Example 14.23 (again a Lattes map) we obtain a non-trivial (inparticular non-smooth) invariant curve that is rectifiable.

In Example 15.8 we exhibit an expanding Thurston map f for whichno visual metric with expansion factor Λ equal to the combinatorial ex-pansion factor Λ0(f) exists. This shows that part (ii) in Theorem 15.3cannot be improved.

1.8. Notation

We denote by N = 1, 2, . . . the set of natural numbers, and by N0 =0, 1, 2. . . . the set of natural numbers including 0. The symbol istands for the imaginary unit in the complex plane C. We define D :=z ∈ C : |z| < 1 as the open unit disk in C.

Let (X, d) be a metric space, a ∈ X and r > 0. We denote byBd(a, r) = x ∈ X : d(a, x) < r and by Bd(a, r) = x ∈ X : d(a, x) ≤r the open and the closed ball of radius r centered at a, respectively.

1.8. NOTATION 29

If A,B ⊂ X, we let diamd(A) be the diameter, A be the closure of Ain X, and

distd(A,B) = infd(x, y) : x ∈ A, y ∈ Bbe the distance of A and B. If p ∈ X, we let distd(p,A) = distd(p, A).If ε > 0 then

N εd(A) := x ∈ X : distd(x,A) < ε

is the open ε-neighborhood of A. We drop the subscript d in Bd(a, r),etc., if the metric d is clear from the context.

The cardinality of a set M is denoted by #M and the identity mapon M by idM . If f : M → M is a map, then fn for n ∈ N is the n-thiterate of f . We set f 0 := idM .

Two non-negative quantities a and b are said to be comparable ifthere is a constant C ≥ 1 depending on some obvious ambient param-eters such that

1

Ca ≤ b ≤ Ca.

We then write a b. The constant C is referred to by C(). We writea . b or b & a, if there is a constant C > 0 such that a ≤ Cb. We referto the constant as C(.) or C(&).

The floor of a real number x, denoted by bxc, is the biggest integerm ∈ Z with m ≤ x. The ceiling of a real number x, denoted by dxe, issmallest integer m ∈ Z with x ≤ m.

CHAPTER 2

Thurston maps

Here we set the stage for the subsequent developments. We defineThurston maps and what it means for such a map to be expanding.We also explain several closely related notions, namely Thurston equiv-alence and the orbifold associated to a Thurston map. In the last sec-tion of this chapter we briefly discuss Thurston’s characterization ofrational maps among Thurston maps.

2.1. Branched covering maps

Branched covering maps are modeled on (non-constant) holomorphicmaps between Riemann surfaces. As our immediate purpose in thissection is to prepare the definition of a Thurston map, we will discussonly some basic facts on this topic and relegate more technical aspectsto the appendix (see Section A.4 in particular).

As usual we call a two-dimensional manifold (without boundary) asurface. All surfaces that we consider will be orientable, and we willassume that some fixed orientation has been chosen on such a surface(for a review of orientation and related concepts see Section A.2).

In the following, X and Y are two compact and connected orientedsurfaces. If f : X → Y is a continuous and surjective map, then f iscalled a branched covering map if we can write it locally as the map z 7→zd for some d ∈ N in orientation-preserving homeomorphic coordinatesin domain and range. More precisely, we require that for each pointp ∈ X there exists d ∈ N, open neighborhoods U ⊂ X of p and U ′ ⊂ Yof q := f(p), such that f(U) ⊂ U ′, open neighborhoods V and V ′ of0 in C, and orientation-preserving homeomorphisms ϕ : U → V andψ : U ′ → V ′ with ϕ(p) = 0 and ψ(q) = 0 such that

(2.1) (ψ f ϕ−1)(z) = zd

for all z ∈ V . This means that the following diagram commutes:

p ∈ U ⊂ Xf//

ϕ

q ∈ U ′ ⊂ Y

ψ

0 ∈ V ⊂ C z 7→zd// 0 ∈ V ′ ⊂ C.

31

32 2. THURSTON MAPS

For the definition of a branched covering map between non-compactsurfaces see Section A.2.

The integer d ≥ 1 in (2.1) is uniquely determined by f and p, andcalled the local degree of the map f at p, denoted by degf (p). Wesometimes write deg(f, p) = degf (p) if our emphasis is on the map fand not on the point p. A point c ∈ X with degf (c) ≥ 2 is called acritical point of f , and a point that has a critical point as a preimagea critical value. The set of all critical points of f is denoted by crit(f).Obviously, if f is a branched covering map on X, then crit(f) has noaccumulation points in X and hence is a finite set, since X is assumedto be compact. Moreover, f is open (images of open sets are open),and finite-to-one (every point in Y has finitely many preimages underf). Actually, if d = deg(f) is the topological degree of f , then

(2.2)∑

p∈f−1(q)

degf (p) = d

for every q ∈ Y (see [Ha, Sect. 2.2]). In particular, if q is not a criticalvalue of f , then p has precisely d preimages.

We denote by χ(X) the Euler characteristic of a compact surfaceX (see [Ha, Theorem 2.44], for example). If X = S2 is a 2-sphere,then χ(S2) = 2. The only other case relevant for us is if X = T is a 2-dimensional torus in which case χ(T ) = 0. The degrees of a branchedcovering map f : X → Y at critical points are related to the Eulercharacteristics of the surfaces by the Riemann-Hurwitz formula (see,for example, [Hu06, Theorem A3.4]); namely,

(2.3) χ(X) +∑

c∈crit(f)

(degf (c)− 1) = deg(f) · χ(Y ).

Suppose Z is another compact and connected oriented surface, andf : X → Y and g : Y → Z are branched covering maps. Then g f isalso a branched covering map and

deg(g f) = deg(g) · deg(f).

Moreover, we have

(2.4) deg(g f, x) = deg(g, f(x)) · deg(f, x)

for all x ∈ X.

2.2. Definition of Thurston maps

We will be mostly interested in the case of branched covering mapsf : S2 → S2 of a 2-sphere S2 to itself. We then denote by fn for n ∈ N0

2.2. DEFINITION OF THURSTON MAPS 33

the n-th iterate of f (where f 0 := idS2). If f is a branched covering mapon S2, then the same is true for fn and we have deg(fn) = deg(f)n.

A postcritical point of f is a point p ∈ S2 of the form p = fn(c)with n ∈ N and c ∈ crit(f). So the set of postcritical points of f isgiven by

post(f) :=⋃n∈N

fn(c) : c ∈ crit(f).

If the cardinality # post(f) is finite, then f is called postcritically-finite. This is equivalent to the requirement that the orbit fn(c) : n ∈N0 of each critical point c of f is finite.

For n ∈ N we have

(2.5) crit(fn) = crit(f) ∪ f−1(crit(f)) ∪ · · · ∪ f−(n−1)(crit(f)).

This implies that post(fn) = post(f) and fn(crit(fn)) ⊂ post(f). Itfollows that fn : S2 \ f−n(post(f)) → S2 \ post(f) is a covering map(see Section A.3 for a review of this topic) and away from post(f) all“branches of the inverse of fn” are defined; more precisely, if U ⊂S2 \ post(f) is a path-connected and simply connected set, q ∈ U , andp ∈ S2 a point with fn(p) = q, then there exists a unique continuousmap g : U → S2 with g(q) = p and fn g = idU (this easily followsfrom Lemma A.1). We refer to such a right inverse of fn informally asa “branch of f−n”.

We can now record the definition of the main object of investigationin this work.

Definition 2.1 (Thurston maps). A Thurston map is a branchedcovering map f : S2 → S2 of a 2-sphere S2 with deg(f) ≥ 2 and a finiteset of postcritical points.

Away from its finitely many critical points, a Thurston map is anorientation-preserving local homeomorphism on the oriented sphere S2.

If S2 = C is the Riemann sphere, and f : C→ C is a non-constantholomorphic map, then f is a rational function and can be representedas the quotient of two polynomials. If in addition f is a Thurston map(i.e., it is postcritically-finite and satisfies deg(f) ≥ 2), then we call fa rational Thurston map.

There are no Thurston maps with # post(f) ∈ 0, 1 (see Re-mark 5.16), and all Thurston maps with # post(f) = 2 are Thurs-ton equivalent (see Section 2.4) to a map z 7→ zk, k ∈ Z \ −1, 0, 1,on the Riemann sphere (see Proposition 7.1). Postcritically-finite ra-tional maps give a large class of Thurston maps. Examples includeP (z) = z2 − 1 (commonly known as the Basilica map, since its Ju-lia set supposedly resembles St Mark’s basilica reflected in the water),

34 2. THURSTON MAPS

f(z) = 1− 2/z2, or g(z) = i/2(z + 1/z). Many other examples can befound throughout this work (see also [BBLPP]).

Since the orbit of each critical point of a Thurston map f is finite, itis often convenient to represent such orbits by the ramification portrait .This is a directed graph, where the vertex set V is the union of theorbits of all critical points. For p, q ∈ V there is a directed edge fromp to q if and only if f(p) = q. Moreover, if p is a critical point withdegf (p) = d we label this edge by “d : 1”.

For example, the map f(z) = 1−2/z2 has the ramification portrait

02:1//∞ 2:1

// 1 // −1

and for g(z) = i/2(z + 1/z) we obtain

12:1

// i)) 0 //∞

−12:1// −i

55

2.3. Definition of expansion

Let f : S2 → S2 be a Thurston map and C be a Jordan curve in S2 withpost(f) ⊂ C. We fix a base metric d on S2 that induces the standardtopology on S2. For n ∈ N we denote by mesh(f, n, C) the supremumof the diameters of all connected components of the set f−n(S2 \ C).

Definition 2.2 (Expansion). A Thurston map f : S2 → S2 iscalled expanding if there exists a Jordan curve C in S2 with post(f) ⊂ Cand

(2.6) limn→∞

mesh(f, n, C) = 0.

We will study the concept of expansion in more detail in Chapter 6after we have built up some methods for a systematic investigation. Forthe moment we summarize some main facts related to this concept.

It is not hard to see that the set f−n(S2 \ C) has only finitely manycomponents; so the supremum in the definition of mesh(f, n, C) is actu-ally a maximum (see Proposition 5.17). We will see in Lemma 6.1 thatif condition (2.6) is satisfied for one Jordan curve C ⊃ post(f), then itactually holds for every such curve. So expansion is a property of themap f alone. Moreover, it is really a topological property, since it is in-dependent of the choice of the base metric d on S2 (as long as d inducesthe standard topology on S2). Our notion of expansion for a Thurs-ton map is equivalent to a similar concept of expansion introduced byHaıssinky-Pilgrim (see [HP09, Sect. 2.2] and Proposition 6.3).

2.3. DEFINITION OF EXPANSION 35

It is immediate that expansion is preserved under topological con-jugacy, i.e., if f and g are topologically conjugate Thurston maps, thenf is expanding if and only if g is expanding. On the other hand, expan-sion is not preserved under Thurston equivalence (see the next sectionfor the terminology). A related fact is that if two expanding Thurs-ton maps are Thurston equivalent, then they are actually topologicallyconjugate (see Theorem 11.4).

Expansion is compatible with iteration of the map. Namely, iff : S2 → S2 is a Thurston map, and F = fn, n ∈ N, is an iterate, thenf is expanding if and only if F is expanding (Lemma 6.4).

A map f : S2 → S2 is called eventually onto, if for any non-emptyopen set U ⊂ S2 there is an iterate fn such that fn(U) = S2. Everyexpanding Thurston map is eventually onto (Lemma 6.5). The conversedoes not hold: there are Thurston maps that are eventually onto, butnot expanding (see Example 6.14).

If f : S2 → S2 is a branched covering map, then a point p ∈ S2 iscalled periodic if fn(p) = p for some n ∈ N. A periodic critical point isa periodic point c ∈ crit(f). The following statement gives a criterionwhen a rational Thurston map is expanding.

Proposition 2.3. Let R : C → C be a rational Thurston map.Then the following conditions are equivalent:

(i) R is expanding.

(ii) The Julia set of R is equal to C.

(iii) R has no periodic critical points.

An immediate consequence of this proposition is that no postcriti-

cally-finite polynomial P : C→ C (with deg(P ) ≥ 2) can be expanding.

Indeed, in this case ∞ ∈ C is both a critical point and a fixed pointof P . So condition (iii) is violated. For a more general related fact seeLemma 6.7.

In general expanding Thurston maps can have periodic criticalpoints (see Example 12.18). On the other hand there are Thurstonmaps that do not have critical periodic points, yet are not Thurstonequivalent to any expanding map (see Example 6.10).

Our proof of Proposition 2.3 relies on facts (in particular, Lemma 6.5,Lemma 6.6, and Lemma A.25) that we will establish later in Chapter 6.We will also use some basic concepts from complex dynamics, whichcan be found in [CG] and [Mi], for example.

Proof of Proposition 2.3. It suffices to establish the chain ofimplications (i)⇒ (ii)⇒ (iii)⇒ (i).

36 2. THURSTON MAPS

(i) ⇒ (ii) Let R : C → C be an expanding rational Thurston map.

Then its Julia set J is non-empty. Let F = C\J be the Fatou set of R.By Lemma 6.5 the map R is eventually onto. So if we assume F 6= ∅,then, since F is open, this implies that Rn(F) = C for a sufficiently

high iterate Rn. Now F is invariant for R; so this means that F = Cand J = ∅. This is a contradiction.

(ii) ⇒ (iii) If the Julia set of R is equal to C, then its Fatou setis empty. This implies that R cannot have periodic critical points,because a periodic critical point of a rational map is part of a super-attracting cycle and belongs to the Fatou set.

(iii) ⇒ (i) Suppose R has no periodic critical points. Then there

exists a geodesic metric ω on C (the canonical orbifold metric of f ; see

Section A.10) such that R expands the ω-length of each path in C by afixed factor ρ > 1 (Proposition A.25). This implies that R is expanding(Lemma 6.6).

2.4. Thurston equivalence

Suppose f : S2 → S2 and g : S2 → S2 are two Thurston maps. Here S2

is another 2-sphere. Often S2 = S2, but sometimes it is important todistinguish the spheres on which the Thurston maps are defined. Wecall the maps f and g topologically conjugate if there exists a homeo-

morphism h : S2 → S2 such that h f = g h. This defines a notionof equivalence for Thurston maps. Topologically conjugate maps haveessentially the same dynamics under iteration up to “change of coordi-nates”.

It is often convenient to consider a weaker notion of equivalence forThurston maps. To define it, we recall the definition of homotopies andisotopies between spaces. Let I = [0, 1], and X and Y be topologicalspaces. An homotopy between X and Y is a continuous map H : X ×I → Y . We define Ht := H(·, t) : X → Y for t ∈ I. The map Ht

is sometimes called the time-t map of the homotopy. A homotopyH : X × I → Y is called an isotopy if Ht is homeomorphism of X ontoY for each t ∈ I.

If H : X × I → Y is a homotopy and A ⊂ X, then we say H isa homotopy relative to A (abbreviated “H is a homotopy rel. A”) ifHt(a) = H0(a) for all a ∈ A and t ∈ I. So this means that the imageof each point in A remains fixed during the homotopy. Similarly, wespeak of isotopies rel. A.

Two homeomorphisms h0, h1 : X → Y are called isotopic rel. A ifthere exists an isotopy H : X×I → Y rel. A with H0 = h0 and H1 = h1.

2.4. THURSTON EQUIVALENCE 37

Definition 2.4 (Thurston equivalence). Two Thurston maps

f : S2 → S2 and g : S2 → S2 are called (Thurston) equivalent if there

exist homeomorphisms h0, h1 : S2 → S2 that are isotopic rel. post(f)and satisfy h0 f = g h1.

In particular, the following diagram commutes:

(2.7) S2 h1//

f

S2

g

S2 h0// S2.

Since S2 and S2 are assumed to be oriented, sometimes one definesThurston equivalence differently by insisting on the homeomorphismsh0 and h1 in Definition 2.4 to be orientation-preserving. In this case,we say that f and g are orientation-preserving Thurston equivalent.This is a stronger notion of Thurston equivalence. So, for example, ac-

cording to our definition, a rational Thurston map R on C is Thurstonequivalent to the map z 7→ R(z) (we use h0(z) = h1(z) = z in Def-inition 2.4), but these maps are in general not orientation-preservingThurston equivalent.

If two Thurston maps are topologically conjugate, then they areThurston equivalent. In general, they will not be Thurston equivalentin the other sense though. This is the main reason why we use ourmore general concept of Thurston equivalence.

Critical and postcritical points of Thurston equivalent maps f andg are related as follows.

Lemma 2.5. Let f and g be Thurston maps that are equivalent asin Definition 2.4. Then

crit(g) = h1(crit(f)),

post(g) = h0(post(f)) = h1(post(f)).

Proof. The first equality is clear from diagram (2.7). Thus

g(crit(g)) = (g h1)(crit(f)) = (h0 f)(crit(f)).

Since h0| post(f) = h1| post(f) and fn(crit(f)) ⊂ post(f) for all n ∈ N,we inductively derive

gn(crit(g)) = h0(fn(crit(f))) = h1(fn(crit(f)))

38 2. THURSTON MAPS

for all n ∈ N. Hence

post(g) =⋃n∈N

gn(crit(g)) =⋃n∈N

h0(fn(crit(f)))

= h0(post(f)) = h1(post(f))

as desired.

The previous lemma implies that the roles of the maps f and g in

Definition 2.4 are symmetric. Indeed, suppose H : S2 × I → S2 is an

isotopy rel. post(f) with H0 = h0 and H1 = h1. Define K : S2 × I →S2 as K(x, t) = (Ht)

−1(x) for x ∈ S2, t ∈ I. The map (x, t) 7→(H(x, t), t) is a continuous bijection between the compact Hausdorff

spaces S2 × I and S2 × I, and hence a homeomorphism. Its inverseis given by (x, t) 7→ (K(x, t), t). This implies that K is continuous,and so obviously an isotopy rel. h0(post(f)) = h1(post(f)) = post(g).Since k0 := h−1

0 = K0 and k1 := h−11 = K1, the homeomorphisms k0

and k1 are isotopic rel. post(g). We also have

k0 g = h−10 g = f h−1

1 = f k1,

and so the conditions in Definition 2.4 are also satisfied if we inter-change f and g.

It is also clear that if f, g, h are Thurston maps, and f is Thurstonequivalent to g, and g is equivalent to h, then f is equivalent to h. Thus,Thurston equivalence leads to a notion of equivalence for Thurstonmaps.

Example 2.6. Up to Thurston equivalence, one can often constructcombinatorial models of maps that are given in some other specific way,for example by an analytic formula. To illustrate this, we consider the

map f : C→ C given by

(2.8) f(z) = 1 + (ω − 1)/z3,

where ω = e4πi/3. Then crit(f) = 0,∞, and f has the ramificationportrait

(2.9) 03:1//∞ 3:1

// 1 // ω

So post(f) = 1, ω,∞, and f is a Thurston map.We now construct a related Thurston map g : S2 → S2 in a similar

fashion as the map h in Section 1.3. For this let T be a right-angledisosceles Euclidean triangle whose hypotenuse has length 1. The an-gles of T are then π/2, π/4, π/4. We also consider a triangle T ′ thatis similar to T by the scaling factor

√2/2. We glue two copies of T

2.4. THURSTON EQUIVALENCE 39

together along their boundaries to form a pillow S0, which is a topo-logical 2-sphere. These two triangles are called the 0-tiles. As beforewe color one of them white, and the other black.

We divide each of the two 0-tiles by the perpendicular bisector ofthe hypotenuse into two triangles similar to T and isometric to T ′.We slit the pillow open along one such bisector and glue two copiesof T ′ into the slit as indicated on left of Figure 2.1. This results ina polyhedral surface S1 consisting of six triangles, each isometric toT ′. These six small triangles are called the 1-tiles. They are coloredin a checkerboard fashion black and white so that triangles sharing anedge have different colors, as indicated in the picture. Note that thereare two points (labeled “0 7→ ∞” and “∞ 7→ 1”) in which all 1-tilesintersect.

Each small triangle in S1 is now mapped by a similarity to thetriangle in S0 of the same color. This determines a unique map S1 →S0 if the vertices of the 1-tiles are mapped as indicated in the top ofFigure 2.1.

We identify S1 with S0 = S2 such that the four triangles shown onthe top right in Figure 2.1 are identified with the white triangle in S0,and the other two small triangles in S1 are identified with the blacktriangle in S0. This yields a Thurston map g : S2 → S2, indicated atthe bottom of Figure 2.1. Here we have cut the pillow S0 along the twolegs of the two triangles. The two pairs of legs marked with the samesymbol have to be identified, i.e., glued together, to form the pillow.

Note that g depends on how precisely S1 is identified with S0. Adifferent identification yields a another Thurston map g, but it is easyto see that g is always Thurston equivalent to g. Thus the notion ofThurston equivalence allows us some latitude in specifying the preciseidentification of S1 with S0 = S2.

Our main point here is the following statement: The map f as de-fined in (2.8) and the map g constructed above are Thurston equivalent.As our purpose is to give the reader some intuition for the concept ofThurston equivalence, we only outline a proof omitting some details.

Note that f(z) = τ(z3), where τ(ζ) = 1 + (ω − 1)/ζ is a Mobiustransformation that maps the upper half-plane to the half-plane abovethe line through the points ω and 1 (indeed, τ maps 0, 1,∞ to∞, ω, 1,respectively).

Let C ⊂ C be the circle through ω, 1,∞ (i.e., the extended linethrough 1 and ω). Then C contains all postcritical points of f . The

closures of the two components of C\C are called the 0-tiles. The 0-tile

40 2. THURSTON MAPS

S0S1

ω

1

∞

17→ω

∞7→1

ω 7→ω

07→∞

ω 7→ω

g

ω 1

∞

∞

17→ω

∞7→1

∞7→1

ω 7→ω

ω 7→ω07→∞

Y1Y2Y3Y4

Y5 Y6

Y1

Y2

Y4

Figure 2.1. The map g.

containing 0 ∈ C (i.e., the half plane above the line through 1 and ω)is colored white, the other 0-tile is colored black.

Since f(z) = τ(z3), we have f−1(C) =⋃k∈1,...,6Rk, where

Rk = reikπ/3 : 0 ≤ r ≤ ∞for k ∈ Z is the ray from 0 through the sixth root of unity eikπ/3 (note

that R0 = R6). These rays divide C into open sectors that are the

complementary components of C \ f−1(C). For k = 1, . . . , 6 let

Xk := reit : 0 ≤ r ≤ ∞, (k − 1)π/3 ≤ t ≤ kπ/3.be the closure of the sector bounded by Rk−1 and Rk. We call thesesets the 1-tiles. Note that f maps each 1-tile Xk homeomorphically to

a 0-tile X0 ⊂ C. We color Xk white if X0 is white, and black otherwise;so then the 1-tile Xk, k = 1, . . . , 6, is colored black or white dependingon whether k is even or odd.

In order to show that f and g are Thurston equivalent, we need twohomeomorphisms h0 and h1 as in Definition 2.4. The homeomorphism

h0 : C → S2 = S0 is defined as follows. We map the white and black

0-tiles in C homeomorphically to the white and black 0-tile in S0,

2.5. THE ORBIFOLD ASSOCIATED WITH A THURSTON MAP 41

respectively. We can do this so that the points 1, ω,∞ ∈ C are mappedto the points labeled 1, ω,∞ ∈ S2 on the top right of Figure 2.1, and sothat the homeomorphisms on these two 0-tiles match along the commonboundary.

The homeomorphism h1 : C→ S1 is constructed similarly by map-

ping 1-tiles in C homeomorphically to corresponding 1-tiles in S1. Wewill then view h1 as a map into the domain S2 = S0 of g by using thegiven identification of S1 with S0

For the precise definition of h1, we denote the 1-tiles in S1 byY1, . . . , Y6 so that they follow in positive cyclic order around the pointlabeled “0 7→ ∞” on the left in Figure 2.1 and such Y1 is the white1-tile containing the point labeled “1 7→ ω”. With our labeling the

1-tile Xk in in C has the same color as the 1-tile Yk in S1. So for fixedk = 1, . . . , 6, the map f |Xk : Xk → X0 is a homeomorphism of Xk

onto a 0-tile X0 in C, and g|Yk : Yk → Y 0 is a homeomorphism of Ykonto a 0-tile Y 0 in S0, where the tiles Xk, Yk, X

0, Y 0 all have the samecolor. In particular, h0 maps X0 homeomorphically onto Y 0, and wecan define

h1|Xk := (g|Yk)−1 h0 (f |Xk).

This is a homeomorphism from Xk onto Yk. It is then straightforward

to check that theses partial homeomorphisms on the 1-tiles Xk of Cpaste together to a well-defined homeomorphism h1 : C → S1 ∼= S2.Moreover, it follows directly from the definition of h1 that h0f = gh1,and so we get a commutative diagram as in (2.7).

It remains to argue that the homeomorphisms h0 and h1 are isotopicrel. post(f). Note that h0 and h1 are orientation-preserving and agreeon the set post(f) = 1, ω,∞. Moreover, h0 maps C (the extendedline through ω and 1) to the boundary of the pillow, while h1 mapsR0 ∪R4 to the boundary of the pillow. So basically we need to deformC to R0 ∪ R4 while keeping post(f) = ω, 1,∞ = C ∩ (R0 ∪ R4) fixedto obtain the desired isotopy between h0 and h1. It is intuitively clearthat this is possible. Since post(f) = 3 the existence of the desiredisotopy actually follows from a general fact (see Lemma 11.11).

2.5. The orbifold associated with a Thurston map

An orbifold is a generalization of a manifold that is especially usefulfor the study of branched covering maps. Every Thurston map hasan associated orbifold Of . The orbifold data are determined by theramification function αf of f . We will first define αf and discuss themain properties of this function before we turn our attention to Of .

42 2. THURSTON MAPS

We use the notation N := N ∪ ∞, and extend the usual order

relations <,≤, >,≥ on N in the obvious way to N. So a < ∞ for

a ∈ N, a ≤ ∞ for a ∈ N, etc. We also extend multiplication of natural

numbers to N by setting a ·∞ =∞· a =∞ for a ∈ N. For a, b ∈ N we

say that a divides b, written a|b, if there exists k ∈ N such that b = ak.So the relation a|b is an extension of the usual divisor relation in Nwith the additional convention that every value in N∪∞ divides∞.

Suppose A ⊂ N is arbitrary. Then there exists a unique L ∈ N,called the least common multiple of the elements of A and denoted by

lcm(A), with the following properties: a|L for all a ∈ A, and if L′ ∈ Nis such that a|L′ for all a ∈ A, then L|L′. It is easy to see that ifA ⊂ N is a bounded set of natural numbers, then lcm(A) ∈ N is theleast common multiple of the numbers in A in the usual sense, andlcm(A) =∞ otherwise.

If α, β : S2 → N are functions on a 2-sphere S2, then we write α ≤ βand α|β if α(p) ≤ β(p) and α(p)|β(p) for all p ∈ S2, respectively.

Definition 2.7 (Ramification function). Let f : S2 → S2 be a

Thurston map. Then its ramification function is the map αf : S2 → Ndefined as

(2.10) αf (p) = lcmdeg(fn, q) : q ∈ S2, n ∈ N, and fn(q) = pfor p ∈ S2.

The ramification function admits the following characterizations.

Proposition 2.8. Let f : S2 → S2 be a Thurston map and αf : S2 →N be its associated ramification function. Then we have:

(i) deg(f, q) · αf (q)|αf (p) whenever p, q ∈ S2 and f(q) = p,

(ii) if β : S2 → N is any function such that deg(f, q) · β(q)|β(p)whenever p, q ∈ S2 and f(q) = p, then αf |β.

Moreover, αf is the unique function with the properties (i) and (ii).

Before we turn to the proof of this proposition, we point out a factthat follows from repeated application of (i). Namely, suppose thatp, q ∈ S2, n ∈ N, and fn(q) = p. Let qk := fk(q) for k = 0, . . . , n.Then

deg(fn, q) = deg(f, q0) · deg(f, q1) · · · deg(f, qn−1)

as follows from (2.4). Moreover,

αf (qk) · deg(f, qk)|αf (qk+1)

for k = 0, . . . , n − 1. Since q0 = q and qn = p, this implies thatdeg(fn, q) · αf (q)|αf (p).


In the following, we will often use relation (2.4) without mentioningit explicitly.

Proof of Proposition 2.8. Let f : S2 → S2 be a Thurston map,and αf be its ramification function as given in Definition 2.7.

To establish (i) for αf , suppose p, q ∈ S2 and f(q) = p. If αf (p) =∞, then deg(f, q) · αf (q)|αf (p), and there is nothing to prove. So wemay assume that αf (p) ∈ N.

Suppose q′ ∈ S2 is a point with fn(q′) = q for some n ∈ N. Thenfn+1(q′) = p, and

deg(fn+1, q′) = deg(fn, q′) · deg(f, q);

so deg(fn, q′) · deg(f, q) divides αf (p) by definition of αf (p). Thenαf (p)/ deg(f, q) must be a natural number that has deg(fn, q′) as adivisor. This implies that the least common multiple αf (q) of all suchnumbers deg(fn, q′) divides αf (p)/ deg(f, q). We conclude that

deg(f, q) · αf (q)|αf (p),

and (i) follows.

Let β : S2 → N be another function with the property (i). Thenby repeated use of this property we conclude that deg(fn, q)·βf (q)|β(p)whenever p, q ∈ S2, n ∈ N, and fn(q) = p. In particular, deg(fn, q)|β(p).As this is true for all q ∈ S2 with fn(q) = p for some n ∈ N, we con-clude αf (p)|β(p) for all p ∈ S2. This establishes the desired property(ii) of αf .

The uniqueness of a function with properties (i) and (ii) is clear.

To state other important properties of the ramification function weneed a definition. A critical cycle C of a Thurston map f : S2 → S2 isan orbit of a periodic critical point of f ; so then there exists c ∈ crit(f),and n ∈ N with fn(c) = c such that C = fk(c) : k = 0, . . . , n− 1.

Proposition 2.9. Let f : S2 → S2 be a Thurston map and αf : S2 →N be its associated ramification function. Then for p ∈ S2 we have:

(i) αf (p) ≥ 2 if and only if p ∈ post(f),

(ii) αf (p) = ∞ if and only if p is contained in a critical cycle off .

Proof. (i) If p ∈ S2 \ post(f), then deg(fn, q) = 1 whenever q ∈S2, n ∈ N, and fn(q) = p, because none of the points fk(q), k =0, . . . , n− 1, can be a critical point of f . Hence αf (p) = 1 by definitionof αf .

44 2. THURSTON MAPS

If p ∈ post(f), then there exist c ∈ crit(f) and n ∈ N such thatfn(c) = p. Then deg(f, c) ≥ 2 which implies that deg(fn, c) ≥ 2. Bydefinition of αf we have deg(fn, c)|αf (p); so αf (p) ≥ 2.

(ii) If p is contained in a critical cycle, then there exists a periodiccritical point c of f such that fn(c) = p for some n ∈ N0. Thereexists k ∈ N such that fk(c) = c, and so fn+km(c) = p for all m ∈ N.Then the numbers deg(fn+km, c) ≥ deg(f, c)m ≥ 2m divide αf (p) forall m ∈ N. This is only possible if αf (p) =∞.

Conversely, suppose p ∈ S2 and αf (p) = ∞. Then by definitionof αf the set deg(fn, q) : q ∈ S2, n ∈ N, fn(q) = p is unbounded.In particular, there exist q ∈ N and n ∈ N with fn(q) = p such thatdeg(fn, q) > M , where

M :=∏

c∈crit(f)

deg(f, c) ∈ N.

Let qk = fk(q) for k = 0, . . . , n− 1. Then

M < deg(fn, q) =n−1∏k=0

deg(f, qk).

Since deg(f, qk) > 1 only if qk is a critical point and since deg(fn, q) >M , there exists a critical point c of f that appears at least twice inthe list q0, . . . , qn−1. Then c is periodic and p belongs to the orbit of c.Hence p is an element of a critical cycle of f .

An orbifold is a space that is locally represented as a quotient ofa model space by a finite group action (see [Th, Chapter 13]). In ourpresent context we are only interested in the 2-dimensional case wheregroup actions are given by cyclic groups near each point (in general wehave to allow the infinite cyclic group though). Then all the relevant

information is given by a pair (S, α), where S is a surface and α : S → Nis such that the set of points p ∈ S with α(p) 6= 1 is a closed and discreteset. We call such a function α a ramification function on S, and thepair (S, α) an orbifold. Each number α(p) should be thought of asthe order of an associated cyclic group (this is made more precise inSection A.7). The set supp(α) := p ∈ S : α(p) ≥ 2 is the supportof α. It is a closed and discrete subset of S; so if S is compact, thensupp(α) is a finite set.

Our point of view leads to the following definition.

Definition 2.10 (Orbifold of a Thurston map). Let f : S2 → S2

be a Thurston map. The orbifold associated to f is the pair Of :=

(S2, αf ), where αf : S2 → N is the ramification function of f .


If O = (S2, α) is an orbifold, then the Euler characteristic of O isdefined as

(2.11) χ(O) := 2−∑p∈S2

(1− 1

α(p)

).

Here and below we use the convention that a/∞ = 0 for a ∈ N. Thesum in (2.11) is really a finite sum, where only the points in supp(α)give a non-zero contribution, but it is convenient to write it (and similarsums below) in this form, where p ranges over the whole underlyingspace S2.

We call O parabolic if χ(O) = 0, and hyperbolic if χ(O) < 0. Wewant to show that the orbifold Of of a Thurston map f is alwaysparabolic or hyperbolic. We first need a lemma.

Lemma 2.11. Let α, α′ : S2 → N be ramification functions on a 2-sphere S2, and let f : S2 → S2 be a branched covering map satisfyingdegf (p) · α′(p) = α(f(p)) for all p ∈ S2. Then for the orbifolds O′ =(S2, α′) and O = (S2, α) we have χ(O′) = deg(f) · χ(O).

This statement is an orbifold version of a well-known identity χ(X) =deg(f) · χ(Y ) for covering maps f : X → Y between surfaces (or moregeneral spaces) X and Y . For compact surfaces this is a special caseof the Riemann-Hurwitz formula (2.3).

Proof. Let d := deg(f). Then for each point q ∈ S2 we have∑p∈f−1(q)

degf (p) = d.

So the Riemann-Hurwitz formula (2.3) implies that

2− χ(O′) =∑p∈S2

(1− 1

α′(p)

)=∑p∈S2

(1−

degf (p)

α(f(p))

)

=∑p∈S2

(1− degf (p)) +∑p∈S2

(degf (p)−

degf (p)

α(f(p))

)

= 2− 2d+∑q∈S2

∑p∈f−1(q)

(degf (p)−

degf (p)

α(f(p))

)

= 2− 2d+ d∑q∈S2

(1− 1

α(q)

)= 2− 2d+ d(2− χ(O)) = 2− dχ(O).

46 2. THURSTON MAPS

Note that each sum in this computation where the summation rangeis over all points in S2 actually has only finitely many non-zero terms.The claim follows.

Proposition 2.12. Let f : S2 → S2 be a Thurston map, and αf beits associated ramification function. Then

(2.12) χ(Of ) = 2−∑p∈S2

(1− 1

αf (p)

)≤ 0.

Here we have equality if and only if degf (p) · αf (p) = αf (f(p)) for allp ∈ S2.

Proof. For p ∈ S2 we define

α′(p) =

αf (f(p))/ degf (p) if αf (f(p)) <∞,

∞ if αf (f(p)) =∞.

By Proposition 2.8 the function α′ takes values in N and we have α′ ≥αf . Moreover, p ∈ S2 : α′(p) ≥ 2 ⊂ f−1(post(f)) is a finite set,and so α′ is a ramification function on S2. It satisfies degf (p) · α′(p) =αf (f(p)) for all p ∈ S2. Hence by Lemma 2.11 we have χ(O′) =dχ(Of ), where d = deg(f) and O′ = (S2, α′).

On the other hand, the fact that α′ ≥ αf and the definition of theEuler characteristic of an orbifold imply that χ(O′) ≤ χ(Of ). Hence

(d− 1)χ(Of ) = χ(O′)− χ(Of ) ≤ 0.

This implies χ(Of ) ≤ 0. Here we have equality if and only if χ(O′) =χ(Of ) which in view of α′ ≥ αf is in turn equivalent to α′ = αf . Thislast condition is the same as the requirement that degf (p) · αf (p) =αf (f(p)) for all p ∈ S2. The statement follows.

By the previous proposition the orbifold of a Thurston map f : S2 →S2 is parabolic or hyperbolic.

Let O = (S2, α) be an orbifold. If we label the finitely many pointsp1, . . . , pk in supp(α) so that 2 ≤ α(p1) ≤ · · · ≤ α(pk), then the k-tuple

(α(p1), . . . , α(pk))

is called the signature of O. The signature of a Thurston map f : S2 →S2 is the signature of its orbifold Of = (S2, αf ). Note that in thiscase the support of the ramification function αf consists precisely ofthe points in post(f) (see Proposition 2.9), and so the signature of fis determined by αf | post(f).

The Lattes map g from Section 1.1, for example, has signature(2, 2, 2, 2), and accordingly its associated orbifold Og is parabolic.


Proposition 2.13 (Thurston maps with parabolic orbifold).Let f : S2 → S2 be a Thurston map. Then the following conditions

are equivalent:

(i) Of is parabolic,

(ii) the signature of Of is

(∞,∞), (2, 2,∞), (2, 2, 2, 2), (2, 4, 4), (3, 3, 3), or (2, 3, 6),

(iii) the ramification function αf : S2 → N has the following prop-erty: for all p ∈ S2 and q ∈ f−1(p), we have

αf (p) = degf (q)αf (q).

Proof. (i)⇔ (ii) If Of has one of the signatures listed in (ii), thenχ(Of ) = 0, and so Of is parabolic. Conversely, if χ(Of ) = 0 then onefirst notes that f can have at most four postcritical points. Exhaustingall combinatorial possibilities, we are led to the signatures in (ii).

(i)⇔ (iii) This immediately follows from the second part of Propo-sition 2.12.

The signatures of equivalent Thurston maps are the same.

Proposition 2.14. Two Thurston maps that are Thurston equiv-alent have the same signatures.

Proof. Let f : S2 → S2 and g : S2 → S2 be two Thurston maps on

2-spheres S2 and S2, and suppose they are Thurston equivalent. Then

there exist homeomorphisms h0, h1 : S2 → S2 as in Definition 2.4.

Let αf : S2 → N and αg : S2 → N be the ramification functionsof f and g, respectively. The claim will follow if we can show thatαf = αg h0.

To establish this identity, define ν := αg h0. Since αg is supportedon post(g), h0| post(f) = h1| post(f), and h0(post(f)) = h1(post(f)) =post(g) (see Lemma 2.5), we have ν = αg h0 = αg h1.

Now let p ∈ S2 be arbitrary, and define p := h1(p). By what we haveseen, ν(p) = αg(h1(p)) = αg(p ); moreover, the relation h0 f = g h1

implies that deg(f, p) = deg(g, p ) and h0(f(p)) = g(p ). Hence

ν(p) · deg(f, p) = αg(p ) · deg(g, p )

divides

αg(g(p )) = αg(h0(f(p))) = ν(f(p)).

Proposition 2.8 (ii) implies that αf |ν = αg h0. If we reverse theroles of f and g, then a similar argument shows that αg|αf h−1

0 , orequivalently, αg h0|αf . So αf = αg h0 as desired.

48 2. THURSTON MAPS

The ramification function and hence the signature of a Thurstonmap do not change if we pass to any of its iterates.

Proposition 2.15. Let f : S2 → S2 be a Thurston map. Thenαf = αfn for each n ∈ N.

Proof. Fix n ∈ N, and let F = fn. If p ∈ S2, k ∈ N, andq ∈ F−k(p), then p = F k(q) = fnk(q), and so

deg(F k, q) = deg(fnk, q)|αf (p)

by definition of αf (see (2.10)). Since this is true for all k ∈ N andq ∈ F−k(p), this in turn implies αF (p)|αf (p) by definition of αF ; soαF |αf .

On the other hand, suppose p ∈ S2, and let k ∈ N and q ∈ f−k(p)be arbitrary. Then there exist l,m ∈ N such that Fm = fk f l. Wecan find a point q′ ∈ S2 such that f l(q′) = q. Then

deg(Fm, q′) = deg(fk, q) · deg(f l, q′),

and so deg(fk, q)| deg(Fm, q′). Since Fm(q′) = fk(q) = p, we havedeg(Fm, q′)|αF (p) by definition of αF , which implies deg(fk, q)|αF (p).Since k ∈ N and q ∈ f−k(p) were arbitrary, we have αf (p)|αF (p) bydefinition of αf . We conclude that αf |αF ; but we have seen above thatαF |αf , and so αf = αF as desired.

2.6. Thurston’s characterization of rational maps

Thurston’s criterion when a Thurston map is equivalent to a rationalmap is an important theorem in complex dynamics. To formulate thisstatement, we need some definitions.

Definition 2.16 (Invariant multicurves). Let f : S2 → S2 be aThurston map.

(i) A Jordan curve γ ⊂ S2 \ post(f) is called non-peripheral ifeach of the two components of S2 \ γ contains at least twopoints from post(f), and is called peripheral otherwise.

(ii) A multicurve is a non-empty finite set of non-peripheral Jordancurves in S2 \ post(f) that are pairwise disjoint and pairwisenon-homotopic rel. post(f).

(iii) A multicurve Γ is called f -invariant (or simply invariant if f isunderstood) if each non-peripheral component of the preimagef−1(γ) of a curve γ ∈ Γ is homotopic rel. post(f) to a curveγ′ ∈ Γ.

2.6. THURSTON’S CHARACTERIZATION OF RATIONAL MAPS 49

Note that if γ ⊂ S2 is a Jordan curve (i.e., a set homeomorphicto a circle), then by the Schonflies theorem S2 \ γ has precisely twocomponents, each homeomorphic to an open disk.

If # post(f) = 3 every Jordan curve in S2 \ post(f) is peripheral;so there are no multicurves in this case. If # post(f) = 4 then everymulticurve consists of a single Jordan curve.

In (iii) we call two Jordan curves γ, γ′ ⊂ S2 \ post(f) homotopicrel. post(f) if there exists a homotopy H : S2× [0, 1]→ S2 rel. post(f)such that H0 = idS2 and H1(γ) = γ′. Here we also implicitly used thatif γ ⊂ S2 \post(f) is a Jordan curve, then each component σ of f−1(γ)is also a Jordan curve in S2 \post(f). Essentially, this follows from thefact that a suitable branches of f−1 give a local homeomorphisms of γonto σ.

Suppose Γ = γ1, . . . , γn is an invariant multicurve for a givenThurston map f : S2 → S2. Then one can associate an (n× n)-matrixwith f and Γ as follows. Fix i, j ∈ 1, . . . , n, and let σ1, . . . , σk bethe components of f−1(γj) that are homotopic to γi rel. post(f) (herek = k(i, j) ∈ N0). Then each set σl is a Jordan curve in S2 \ post(f),and f |σl is a covering map of σl onto γj. Let

di,j,l := deg(f |σl)

be the (unsigned) topological degree of this map (in this case, this isjust the number of preimages of each point p ∈ γj under the map f |σl).Then the Thurston matrix A = A(f,Γ) = (aij) is the matrix withnon-negative entries

aij =

k(i,j)∑l=1

1

di,j,l

for i, j ∈ 1, . . . , n; if k(i, j) = 0, then the sum is interpreted as theempty sum, in which case aij = 0.

A Thurston obstruction for a Thurston map f is an invariant mul-ticurve Γ such that the spectral radius (which is the largest eigenvalueby the Perron-Frobenius theorem) of the Thurston matrix A(f,Γ) is≥ 1.

With these definitions Thurston’s criterion can be formulated asfollows.

Theorem 2.17 (Thurston’s characterization of rational maps). Letf : S2 → S2 be a Thurston map with a hyperbolic orbifold. Then f isThurston equivalent to a rational map if and only if there exists noThurston obstruction for f .

50 2. THURSTON MAPS

fγ

σ1

σ2

σ3

Figure 2.2. An obstructed map.

The proof can be found in [DH]. We will not use this theorem inany essential way, and included its statement for general backgroundand to put our work into context.

Example 2.18. To illustrate Theorem 2.17 and the concepts weintroduced for its formulation, we consider the Thurston map f = hconstructed in Section 1.3.

Recall that # post(f) = 4, and that the postcritical points of fare given by the corners of the pillow on the left of Figure 2.2. Thesignature of the orbifold Of of f is (2, 6, 6, 6), and so Of is hyperbolic.

Let γ be the Jordan curve indicated to the right in Figure 2.2. Thepreimage f−1(γ) of γ has three components σ1, σ2, σ3 indicated on theleft in Figure 2.2. The Jordan curves σ1 and σ2 are non-peripheral,while σ3 is peripheral. Both curves σ1 and σ2 are homotopic to γ rel.post(f). Thus Γ = γ is an invariant multicurve.

The degree of f |σl : σl → γ is 2 for l = 1, 2; so the Thurston matrixA(f,Γ), which is a 1× 1-matrix, has the single entry 1/2 + 1/2 = 1.

It follows that the spectral radius of Γ = γ is equal to 1. HenceΓ is a Thurston obstruction for f , and f is not Thurston equivalent toa rational map by Thurston’s criterion.

Theorem 2.17 can be interpreted as a condition for the existence ofa Thurston equivalence. It is complemented by the following statementwhich is essentially a uniqueness statement for Thurston equivalences.

Theorem 2.19 (Thurston’s uniqueness theorem). Let f, g : C→ Cbe two rational Thurston maps with hyperbolic orbifolds, and suppose

that there are two orientation-preserving homeomorphisms h0, h1 : C→C that are isotopic rel. post(f) and satisfy h0 f = g h1. Then there

exists a conformal homeomorphism ϕ : C → C that is isotopic to h0

and h1 rel. post(f) and satisfies ϕ f = g ϕ.

2.6. THURSTON’S CHARACTERIZATION OF RATIONAL MAPS 51

A conformal homeomorphism on C is of course a Mobius transfor-mation. So in particular, if two rational Thurston maps with hyper-bolic orbifolds are orientation-preserving Thurston equivalent (see thediscussion after Definition 2.4), then they a conjugate by a Mobiustransformation.

Theorem 2.19 is contained in [DH]; there it is not formulated ex-plicitly, but it can be easily derived from the considerations in thispaper.

CHAPTER 3

Lattes maps

A Lattes map is a rational Thurston map that is expanding and hasa parabolic orbifold. There are other equivalent ways to characterizethese maps. For example, a Lattes map is a quotient of a holomorphicautomorphism of the complex plane by the action of a crystallographicgroup or a quotient of a holomorphic endomorphism on a complextorus. We will explain this more precisely below.

These maps play a special role in the theory. On the one hand, theyare very easy to construct and visualize, and provide a convenient classof examples (one was given in Section 1.1). On the other hand, theyoften show exceptional behavior compared to generic rational Thurs-ton maps (that are expanding). This is already apparent in Thurs-ton’s characterization of rational maps (Theorem 2.17). In general,Lattes maps are distinguished among typical rational Thurston mapsin terms of metric geometry (Theorem 17.1 (iii)), by their measure-theoretic properties (Theorems 18.5, 18.7, 18.8, and Theorem 18.3),or by their “combinatorial expansion rate” (Theorem 19.2). Some ofthese statements are among the main results of this work and so wewill take a closer look at these maps. We will also define the relatedclass of Lattes-type maps. They are Thurston maps with a parabolicorbifold and no periodic critical points, but not necessarily (equivalentto) rational maps.

Some aspects of a thorough treatment of the underlying theory arerather technical. As we do not want to overburden the reader withdetails at this point, we will rely on various results that are more fullydeveloped in the appendix.

To motivate our definition of Lattes maps in terms of three equiv-alent conditions, we will now consider a specific example. For precisedefinitions of the terminology in the ensuing discussion we refer to thebeginning of Section 3.1.

Let f be the map from Section 1.1 (there denoted by g). Thenf is a rational Thurston map that is expanding, or equivalently, hasno periodic critical points. Its orbifold has signature (2, 2, 2, 2), and ishence parabolic.

53

54 3. LATTES MAPS

The map f is a quotient of the automorphism A : C → C, z 7→A(z) = 2z, on C by a holomorphic map Θ: C → C in the sense thatthe following diagram commutes:

(3.1) C A//

Θ

C

Θ

Cf// C.

Here Θ is essentially a Weierstrass ℘-function for the lattice Γ = Z⊕Zi(see Section 3.4 for the definition of ℘ and a related discussion). Notethat Θ(z) = Θ(w) for z, w ∈ C if and only if w = ±z + m + ni withm,n ∈ Z. This last condition can most conveniently be expressed interms of an action of a crystallographic group of orientation-preservingisometries on C.

Indeed, let G be the group of all maps g : C→ C of the form

g(z) = ±z +m+ ni ,

where m,n ∈ Z. Then G is a crystallographic group and the map Θ isinduced by G in the sense that Θ(z) = Θ(w) for z, w ∈ C if and only ifthere exists g ∈ G such that w = g(z). This implies that the quotient

C/G can be identified with C, and the map Θ with the quotient mapC→ C/G.

The property that allows us to pass to a quotient f in (3.1) is thatthe map z 7→ A(z) = 2z is G-equivariant. This means that A mapspoints that are in the same G-orbit to points that are also in the samesame G-orbit, or equivalently that

(3.2) A g A−1 ∈ G for all g ∈ G.

The translations in G form a subgroup Gtr isomorphic (as a group)to Z2 = Z ⊕ Zi . The quotient C/Gtr is naturally a complex torus T,i.e., a Riemann surface whose underlying 2-manifold is a 2-dimensional

torus. The maps A : C → C and Θ: C → C descend to T, and we

obtain holomorphic maps A : T→ T and Θ: T→ C such that f Θ =Θ A. So we have the following commutative diagram:

(3.3) T A//

Θ

T

Θ

Cf// C.

We call a non-constant holomorphic map A : T → T on a complextorus T a holomorphic torus endomorphism. The Riemann-Hurwitz

3. LATTES MAPS 55

formula (2.3) implies that such a map A has no critical points and ishence a covering map.

The relations of our example f to crystallographic groups or tocomplex torus endomorphisms hold for a more general class of rationalmaps, called Lattes maps, as the following statement shows.

Theorem 3.1 (Characterization of Lattes maps). Let f : C → Cbe a map. Then the following conditions are equivalent:

(i) f is a rational Thurston map that has a parabolic orbifold andno periodic critical points.

(ii) There exists a crystallographic group G, a G-equivariant holo-morphic map A : C → C of the form A(z) = αz + β, where

α, β ∈ C, |α| > 1, and a holomorphic map Θ: C→ C inducedby G such that f Θ = Θ A.

(iii) There exists a complex torus T, a holomorphic torus endo-morphism A : T → T with deg(A) > 1, and a non-constant

holomorphic map Θ: T→ C such that f Θ = Θ A.

So in (ii) the map f is given as in (3.1), and in (iii) as in (3.3). Wewill see that we have deg(f) = deg(A) = |α|2 > 1 (see Lemma 3.14).

As we already indicated, the previous theorem motivates the fol-lowing definition.

Definition 3.2 (Lattes maps). A map f : C→ C is called a Lattesmap if it satisfies one of the conditions (and hence every condition) inthe last theorem.

The terminology is not uniform in the literature and some authorsuse the term “Lattes map” with a slightly different meaning (see thediscussion in Section 1.2).

Theorem 3.1 is well-known (see, for example, [Mi]). We will proveit in Sections 3.1 and 3.2. The map A in (ii) is subject to strong furtherrestrictions. See Proposition 3.13 (or [Mi] and [DH84, Appendix]) formore details.

By Lemma 2.3, condition (i) can equivalently be expressed as:

(i’) f is a rational Thurston map that has a parabolic orbifold andis expanding.

This is how we introduced Lattes maps in the beginning of the chapter.

Let f : C→ C be a Lattes map and αf : C→ N be its ramification

function (see Definition 2.7). Then the map Θ: C → C as in (3.1) isa holomorphic branched covering map such that deg(Θ, z) = αf (Θ(z))

56 3. LATTES MAPS

for all z ∈ C. Therefore, Θ is in fact the universal orbifold covering

map of Of = (C, αf ) (see Theorem 3.8 and Section A.8).It is quite natural to consider more general maps f as in (3.1) or as

in (3.3), where the maps involved are branched covering maps, but notnecessarily holomorphic. To state this more precisely, we first recallsome terminology.

As usual, we call a map A : R2 → R2 affine if it has the form

(3.4) A(u) = LA(u) + u0, u ∈ R2,

where LA : R2 → R2 is R-linear and u0 ∈ R2. We call LA the linearpart of A.

Let G be a crystallographic group not isomorphic to Z2. Thenthe quotient R2/G is homeomorphic to a 2-sphere S2 and the quotientmap Θ: R2 → S2 ∼= R2/G is a branched covering map induced by G.Suppose A : R2 → R2 is an affine map that is G-equivariant and whoselinear part LA satisfies det(LA) > 1. In this case there is a Thurstonmap f : S2 → S2 such that the diagram

(3.5) R2 A//

Θ

R2

Θ

S2 f// S2

commutes (see the beginning of Section 3.3). This is the basis of thefollowing definition.

Definition 3.3 (Lattes-type maps). Suppose f : S2 → S2 is mapsuch that there exists a crystallographic groupG, an affine mapA : R2 →R2 with det(LA) > 1 that is G-equivariant, and a branched coveringmap Θ: R2 → S2 induced by G such that f Θ = Θ A. Then f iscalled a Lattes-type map.

So Lattes-type maps are given as in (3.5), where A is affine. It isclear that every Lattes map belongs to this class. For the degree ofsuch a map we have deg(f) = det(LA) (see Lemma 3.14).

Before we discuss some other properties of Lattes-type maps, we willfirst define another generalization of Lattes maps based on (3.3). Wedenote by T 2 a 2-dimensional torus, now considered as a purely topo-logical object with no complex structure. If A : T 2 → T 2 is a branchedcovering map, then again by the Riemann-Hurwitz formula (2.3) themap A cannot have any critical points and must be an orientation-preserving covering map (in the usual topological sense). We call sucha map A a (topological) torus endomorphism. So a continuous map

3. LATTES MAPS 57

A : T 2 → T 2 is a torus endomorphism precisely if it is an orientation-preserving local homeomorphism.

Definition 3.4 (Quotients of torus endomorphisms). Let f : S2 →S2 be a map on a 2-sphere S2 such that there exists a torus endomor-phism A : T 2 → T 2 with deg(A) ≥ 2, and a branched covering mapΘ: T 2 → S2 such that f Θ = Θ A. Then f is called a quotient of atorus endomorphism.

In this case, we have a commutative diagram of the form

(3.6) T 2 A//

Θ

T 2

Θ

S2 f// S2.

Properties of quotients of torus endomorphisms are recorded in Lem-ma 3.10. In particular, every such map is a Thurston map withoutperiodic critical points. Lattes-type maps are in this class.

Proposition 3.5. Every Lattes-type map f : S2 → S2 is a quotientof a torus endomorphism and hence a Thurston map. It has a parabolicorbifold, and no periodic critical points.

This implies that the orbifold of every Lattes-type map has one ofthe signatures (2, 2, 2, 2), (2, 4, 4), (3, 3, 3), or (2, 3, 6). The last threesignatures do not lead to genuinely new maps, as each Lattes-typemap whose orbifold has such a signature is topologically conjugate to aLattes map (Proposition 3.16). The most interesting case is signature(2, 2, 2, 2). More details on these maps can be found in Example 1.2(see also Theorem 1.2). These maps include flexible Lattes maps (seeDefinition 3.23 and the discussion that follows).

The last statement in Proposition 3.5 essentially characterizes Lattes-type maps.

Proposition 3.6. Let f : S2 → S2 be a Thurston map. Then f isThurston equivalent to a Lattes-type map if and only if f has a parabolicorbifold and no periodic critical points.

If f has a parabolic orbifold Of , but also periodic critical points,then the signature of Of is (∞,∞) or (2,∞,∞). It is easy to classifythese maps up to Thurston equivalence as well (see Theorem 1.2).

Each Lattes map is expanding, but this is not always true for aLattes-type map (see Example 6.14). One can state a simple criterionthough when this is the case. Namely, a Lattes-type map is expandingif and only if the two (possibly complex) eigenvalues λ1 and λ2 of

58 3. LATTES MAPS

the linear part LA of the affine map A in (3.5) satisfy |λ1|, |λ2| > 1(Proposition 6.11).

Every Lattes map is a Lattes-type map, and every Lattes-type mapis a quotient of a torus endomorphism. On the other hand, not everyLattes-type map is (conjugate to) a Lattes map. It is a natural ques-tion whether every quotient of a torus endomorphism f is Thurstonequivalent to a Lattes-type map. One can show that this is true if f isexpanding (in this case, f is even conjugate to Lattes-type map), butwe have been unable to answer this question in full generality.

3.1. Crystallographic groups and Lattes maps

In this section we focus on maps as in statement (ii) of Theorem 3.1.We first review some facts related to crystallographic groups. For amore detailed discussion related to group actions and quotient spacessee Section A.12.

We use the notation

Aut(C) = z ∈ C 7→ αz + β : α, β ∈ C, α 6= 0

for the group of all holomorphic automorphisms of C and

Isom(C) = z ∈ C 7→ αz + β : α, β ∈ C, |α| = 1 ⊂ Aut(C)

for the group of all orientation-preserving isometries of C (equippedwith the Euclidean metric).

Let G be a group of homeomorphisms acting on C. If z ∈ C, thenwe denote by Gz := g ∈ G : g(z) = z its stabilizer subgroup andby G(z) = g(z) : g ∈ G its orbit under G or G-orbit. The group Ginduces a natural equivalence relation on C whose equivalence classesare given by the G-orbits. The corresponding quotient space is denotedby C/G.

The group G acts properly discontinuously on C if for each compactset K ⊂ C there are only finitely many maps g ∈ G with g(K)∩K 6= ∅.Then the stabilizer Gz is finite for each z ∈ C. The group G actscocompactly on C if there exists a compact set K ⊂ C such that thesets g(K), g ∈ G, cover C. In this case, C/G is compact.

We call G a (planar) crystallographic group if each element g ∈ Gis an orientation-preserving isometry on C and if the action of G on Cis properly discontinuous and cocompact. Note that this definition ofa crystallographic group is more restrictive than usual, since we insiston the isometries in G to be orientation-preserving.

3.1. CRYSTALLOGRAPHIC GROUPS AND LATTES MAPS 59

We say that two crystallographic groups G and G are conjugate ifthere exists h ∈ Aut(C) such that

G = h G h−1 := h g h−1 : g ∈ G.

The following proposition gives a classification of crystallographic groupsup to conjugacy.

Theorem 3.7 (Classification of crystallographic groups). Let G ⊂Isom(C) be a planar crystallographic group. Then G is conjugate to

one of the following groups G consisting of all g ∈ Isom(C) of the form

(torus) z 7→ g(z) = z +m+ nτ, m, n ∈ Z;

(2222) z 7→ g(z) = ±z +m+ nτ, m, n ∈ Z;

(244) z 7→ g(z) = ikz +m+ ni, m, n ∈ Z, k = 0, 1, 2, 3;

(333) z 7→ g(z) = ω2kz +m+ nω, m, n ∈ Z, k = 0, 1, 2;

(236) z 7→ g(z) = ωkz +m+ nω, m, n ∈ Z, k = 0, . . . , 5.

Here τ ∈ C is a fixed number with Im(τ) > 0 in the first two casesand ω = eiπ/3 in the last two cases.

This classification of (planar) crystallographic groups is classical.Proofs can be found in [Be] and [Ar]; see also [We].

We used Conway’s orbifold notation for planar crystallographicgroups, see [Con] (with one difference: Conway denotes the (torus)case by ()). In Theorem 3.7 the group G determines the type of

the conjugate group G uniquely. Accordingly, we speak of a crystallo-

graphic group G of type (2222), etc., if G belongs to the correspondingclass. This terminology is explained by the fact that the quotient space

C/G is a torus in the first case of the theorem and homeomorphic to Cin the other cases. In addition, then G induces a natural ramification

function α on C = C/G so that the orbifold (C, α) has a signature asindicated by its type (see the discussion below).

If G is a crystallographic group, we denote by Gtr the subgroupconsisting of all translations in G, i.e., Gtr consists of all maps g ∈ G ofthe form z 7→ g(z) = z+γ with γ ∈ C. Theorem 3.7 implies that Gtr isa normal subgroup of finite index in G and that there is a rank-2 latticeΓ ⊂ C such that Gtr = z 7→ z + γ : γ ∈ Γ. We call Γ the underlying

lattice of the crystallographic group G. The underlying lattice Γ of Gas in Theorem 3.7 is equal to Z ⊕ Zτ in cases (torus) and (2222), toZ⊕Zi in case (244), and to Z⊕Zω in cases (333) and (236). Note thata crystallographic group G is of (torus) type if and only if G = Gtr, or,equivalently, if and only if G is isomorphic to Z2.

60 3. LATTES MAPS

Figure 3.1. Invariant tiling of crystallographic group (244).




Crystallographic groups G of type (244), (333), and (236) are rep-resented in Figure 3.1, Figure 3.2, and Figure 3.3, respectively. Foreach type (abc) we see (part of) a tiling of C given by isometric copiesof Euclidean triangles with angles π/a, π/b, π/c. The correspondinggroup G consists of all orientation-preserving isometries of the planeC that keep the tiling invariant. This means that each group elementg ∈ G maps each triangle in the tiling to another one of the same color.Moreover, g maps each point marked by a dot to another such point.If we assume that 0 ∈ C is one of these points, then all points markedby a dot form the underlying lattice Γ of G, i.e., the orbit of 0 ∈ Cby the group of translations Gtr ⊂ G. Actually, the full G-orbit of 0 isequal to its Gtr-orbit Γ.

Suppose G is a crystallographic group and z, w ∈ C are containedin the same G-orbit. Then their stabilizers Gz and Gw have the sameorder #Gz = #Gw, since they are conjugate subgroups of G. Eachnon-trivial stabilizer Gz is cyclic and of order 2, 3, 4, or 6 as can beseen from Theorem 3.7 (actually, an independent proof of this fact isone of the main steps in the proof of Theorem 3.7). The numbers in thelabel for the type of a group indicate the orders of non-trivial stabilizersin distinct orbits. For example, a crystallographic group of type (236)has three distinct orbits consisting of points with non-trivial stabilizersof order 2, 3, or 6.

Crystallographic groups are closely related to parabolic orbifolds

(C, α) satisfying α(p) <∞ for p ∈ C. These orbifolds have one of thesignatures (2, 2, 2, 2), (2, 4, 4), (3, 3, 3), or (2, 3, 6). This correspondsprecisely to the types of crystallographic groups that are not of (torus)type, i.e., not isomorphic to Z2. This relation is based on the orbifolduniformization theorem which is discussed in detail in the appendix(see Section A.8). For the present purpose we will formulate a relevantspecial case explicitly. First, we recall some terminology.

If Θ: C → C is a branched covering map, then a deck transforma-tion of Θ is a homeomorphism g : C→ C such that Θ g = Θ. If Θ isholomorphic, then this is also true for each deck transformation g of Θand so g ∈ Aut(C). The deck transformations of Θ form a group G.

We say that Θ: C → C is a regular branched covering map if itsdeck transformations act transitively on the fibers of Θ; this meansthat if z, w ∈ C and Θ(z) = Θ(w), then there exists g ∈ G such thatw = g(z). Note that Θ is regular if and only if it is induced by itsgroup of deck transformations G.

Theorem 3.8. Let (C, α) be a parabolic orbifold with α(p) < ∞for all p ∈ N. Then there exists a holomorphic branched covering map

62 3. LATTES MAPS

Θ: C→ C such that

(3.7) degΘ(z) = α(Θ(z)) for each z ∈ C.

The branched covering map Θ is regular and its deck transformationgroup G is a crystallographic group.

Moreover, if Θ : C → C is another holomorphic map satisfying

(3.7), then there exists h ∈ Aut(C) such that Θ = Θ h.

By the last part of the statement, the map Θ satisfying (3.7) isessentially unique. We call it the universal orbifold covering map of

the orbifold (C, α) as in the theorem.Since Θ is regular, it is induced by the crystallographic group G

given by its deck transformations. So we can naturally identify C/Gwith C by sending each orbit G(z) of G (considered as a point in C/G)

to Θ(z) ∈ C. Under this identification, the quotient map C → C/Gthen corresponds to Θ. In particular, C/G is a topological 2-sphereand so G is not isomorphic to Z2 (in which case C/G is a torus).

If Θ: C→ S2 is any branched covering map induced by a crystallo-graphic group G (as the universal orbifold covering map just discussed),then

(3.8) degΘ(z) = #Gz

for z ∈ C. Indeed, for each z ∈ C the stabilizer Gz is a cyclic rotationgroup fixing z. So the G-orbit of a point w 6= z sufficiently close to zcontains precisely d = #Gz distinct points near z. Since Θ is inducedby G, this means that near z the map Θ is locally d-to-1 and (3.8)follows.

The relation (3.8) implies that the crystallographic group G arisingin Theorem 3.8 has a type corresponding to the signature of the orbifold

(C, α); so for example, if the signature of (C, α) is (2, 2, 2, 2), then G isof type (2222).

One way to prove the existence of the universal orbifold coveringmap Θ in Theorem 3.8 is to start with a suitable crystallographic groupG and construct Θ from the quotient map C → C/G. Indeed, if G isa not isomorphic to Z2, then the quotient space C/G is a topological2-sphere. One can push down the Euclidean metric d0 on C by thequotient map π : C → C/G to obtain a metric ω on C/G that makesthis space a polyhedral surface, i.e., a locally Euclidean surface withconical singularities. This gives C/G a natural conformal structureand by the uniformization theorem we can identify C/G (non-uniquely)

with the Riemann sphere C by a conformal map. If we denote the coneangle at each point p ∈ C/G of the polyhedral surface by 2π/α(p),


then α(p) ∈ N and (C/G, α) ∼= (C, α) is a parabolic orbifold. We willdiscuss this geometric viewpoint in more detail in Section 3.4.

We are now ready to prove the first implication of Theorem 3.1.

Proof of (ii) ⇒ (iii) in Theorem 3.1. Assume f : C → C is amap as in (ii). Then there exists a crystallographic group G, a G-equivariant map A : C→ C of the form A(z) = αz+β where α, β ∈ C,

|α| > 1, and a holomorphic map Θ: C → C induced by G such that

f Θ = Θ A. Note that then C/G is homeomorphic to C and so Gis not isomorphic to Z2.

Let Gtr ⊂ G be the normal subgroup of translations in G. Considerthe quotient space T := C/Gtr and the quotient map π : C → T =C/Gtr. Then π is a covering map and T a topological torus. Moreover,there is a natural conformal structure on T so that π is holomorphicand T a complex torus (see Section A.6 for more details).

Let g ∈ Gtr with g 6= idC be arbitrary. Since A is G-equivariant, wehave g := A g A−1 ∈ G. So the map g is an orientation-preservingisometry on C. Moreover, this map has no fixed points, because thisis true for its conjugate map g, which is a translation. Hence g ∈ G isa also a translation, i.e., g ∈ Gtr. This shows that A g A−1 ∈ Gtr

whenever g ∈ Gtr.We conclude that A is Gtr-equivariant and so descends to the quo-

tient T = C/Gtr. More explicitly, if we set

A(π(z)) := π(A(z))

for z ∈ C, then A : T → T is a well-defined continuous map satisfyingA π = π A. Since π and π A are holomorphic, the map A is holo-morphic as well (see Lemma A.5). It follows that A is a holomorphictorus endomorphism.

Similarly, the map Θ descends to T. Indeed, suppose z, w ∈ C andπ(z) = π(w). Then there exists g ∈ Gtr such that w = g(z). Since Θis induced by G ⊃ Gtr, we then have Θ(z) = Θ(w). So if we set

Θ(π(z)) := Θ(z)

for z ∈ C, then we get a well defined map Θ: T→ C such that Θπ =Θ. This last relation implies that Θ is non-constant and holomorphic,and a branched covering map (see Lemma A.5).

Note that

f Θ π = f Θ = A Θ = A Θ π.

Since π : C→ T is surjective, it follows that f Θ = A Θ.

64 3. LATTES MAPS

The holomorphic maps considered and their relations can be sum-marized in the following commutative diagram:

(3.9) C A//

π

Θ

Cπ

Θ

T A//

Θ

T

Θ

Cf// C.

This shows that f is a quotient of a holomorphic torus endomorphism.If we have maps as in (3.9), then

(3.10) deg(f) = deg(A) = |α|2

(recall that A(z) = αz + β). We postpone the justification of this toLemma 3.14, where we will establish a more general fact (not relyingon Theorem 3.1, of course). In particular, deg(A) = |α|2 > 1, and sodeg(A) ≥ 2. It follows that f is indeed a map as in (iii).

Remark 3.9. The action of the the crystallographic group G de-scends to the complex torus T = C/Gtr; more precisely, the groupG = G/Gtr acts naturally on T. It easily follows from Theorem 3.7that G is a cyclic group. Accordingly, condition (iii) in Theorem 3.1can be formulated similarly as condition (ii) in terms of a cyclic groupaction on T. This is discussed in [Mi] in more detail. We do not pursuethis point of view here, since the underlying geometry is not as easy tovisualize as for crystallographic groups.

Proof of (i) ⇒ (ii) in Theorem 3.1. Let f : C → C be a mapas in (i), i.e., a rational Thurston map with a parabolic orbifold Of =

(C, αf ) and no periodic critical points. By Proposition 2.9 we then

have αf (p) <∞ for p ∈ C.

Let Θ: C → C be the (holomorphic) universal orbifold coveringmap of the orbifold Of (see Theorem 3.8) and G be the group of decktransformations of Θ. Then G is a crystallographic group and Θ isinduced by G.

We have to find an automorphism A : C → C such that f Θ =Θ A. For this we consider the holomorphic branched covering map

f Θ: C → C. Since Of = (C, αf ) is parabolic, by Proposition 2.13we have

deg(f, p) · αf (p) = αf (f(p))

3.2. QUOTIENTS OF TORUS ENDOMORPHISMS AND PARABOLICITY 65

for all p ∈ C. For each z ∈ C we have deg(Θ, z) = αf (Θ(z)), and so

deg(f Θ, z) = deg(f,Θ(z)) · deg(Θ, z)

= deg(f,Θ(z)) · αf (Θ(z)) = αf(f(Θ(z))

).

This shows that f Θ is another universal orbifold covering map ofOf . The essential uniqueness of the universal orbifold cover (see The-orem 3.8) implies that there is an automorphism A : C→ C satisfyingf Θ = Θ A. In other words, we have a commutative diagram asin (3.1). Since Θ is induced by G, it follows that the map A is G-equivariant.

The map A is necessarily of the form z 7→ A(z) = αz + β, whereα, β ∈ C, α 6= 0. As in (3.10), we have deg(f) = |α|2 ≥ 2, and so|α| > 1. It follows that f is as in (ii).

3.2. Quotients of torus endomorphisms and parabolicity

We now prepare the proof of the implication (iii)⇒ (i) in Theorem 3.1.The main difficulty here is to show that a map as in Theorem 3.1 (iii)has a parabolic orbifold. To address this, we will first establish somegeneral statements for quotients of endomorphisms on a torus T 2 (seeDefinition 3.4).

Lemma 3.10 (Properties of quotients of torus endomorphisms). Letf : S2 → S2 be a quotient of a torus endomorphism, and Θ: T 2 → S2

and A : T 2 → T 2 with deg(A) ≥ 2 be as in Definition 3.4. Then thefollowing statements are true:

(i) The map f is a Thurston map without periodic critical points,and deg(f) = deg(A) ≥ 2.

(ii) The set post(f) is equal to the set of critical values of Θ, i.e.,

post(f) = Θ(crit(Θ)).

(iii) The ramification function of f is given by

αf (p) = lcmdeg(Θ, x) : x ∈ Θ−1

(p)

for p ∈ S2.

Proof. Let f , Θ, and A be as in the statement. Then A andΘA are branched covering maps. Since f A = ΘA, it follows fromLemma A.5 that f is a branched covering map as well.

Note that

deg(f) · deg(Θ) = deg(f Θ) = deg(Θ A) = deg(Θ) · deg(A),

66 3. LATTES MAPS

and sodeg(f) = deg(A) ≥ 2,

as claimed. To show that f is a Thurston map without periodic criticalpoints, we first establish (ii) and (iii).

(ii) Let VΘ = Θ(crit(Θ)) denote the set of critical values of Θ. SinceT 2 is compact and the set of critical points of Θ has no accumulationpoint in T 2, there are only finitely many critical points of Θ. Hencethe set VΘ is also finite. We will first prove that post(f) ⊂ VΘ.

If we denote by Vfn = fn(crit(fn)) the set of all critical values offn, then from (2.5) it follows that

Vfn = fn(crit(fn)) =n⋃k=1

fk(crit(f))

for all n ∈ N. Thus

(3.11) post(f) =⋃n∈N

fn(crit(f)) =⋃n∈N

Vfn .

Now let p ∈ post(f) be arbitrary. Then by (3.11) the point p is acritical value of some iterate of f . So there exist n ∈ N and q ∈ S2

with deg(fn, q) ≥ 2 and fn(q) = p. As a branched covering map, Θ issurjective, and so we can find x ∈ T 2 with Θ(x) = q.

Recall that by the Riemann-Hurwitz formula (2.3) the map A can-not have critical points, and hence is locally injective. In particular,deg(A

n, x) = 1. Since f Θ = Θ A, we have fn Θ = Θ An. It

follows that

deg(Θ, An(x)) = deg(Θ, A

n(x)) · deg(A

n, x)

= deg(Θ An, x) = deg(fn Θ, x)

= deg(fn, q) · deg(Θ, x) ≥ 2.

Thus An(x) is a critical point of Θ. So we have

p = fn(q) = (fn Θ)(x) = (Θ An)(x) ∈ VΘ.

The desired inclusion post(f) ⊂ VΘ follows.To show that actually post(f) = VΘ, we argue by contradiction and

assume that there exists a point p ∈ VΘ \ post(f). Then by (3.11) thepoint p is not a critical value of any iterate fn of f . Since p is a critical

value of Θ, the set Θ−1

(p) contains a critical point c of Θ. This implies

that for each n ∈ N, the set A−n

(c) consists of critical points of Θ.

Indeed, if a ∈ A−n(c), then

fn(Θ(a)) = Θ(An(a)) = Θ(c) = p,


and so deg(fn,Θ(a)) = 1; moreover, An(a) = c, and so

deg(Θ, a) = deg(fn,Θ(a)) · deg(Θ, a)

= deg(fn Θ, a) = deg(Θ An, a)

= deg(Θ, An(a)) · deg(A

n, a) = deg(Θ, c) ≥ 2.

Since A is a covering map with deg(A) ≥ 2, we have

#A−n

(c) = deg(A)n ≥ 2n,

and so there must be at least 2n distinct critical points of Θ. Sincen ∈ N was arbitrary, and the number of critical points of Θ is finite,this is a contradiction showing post(f) = VΘ.

Since post(f) = VΘ is a finite set, we conclude that f is a Thurstonmap.

(iii) We define

ν(p) = lcmdeg(Θ, x) : x ∈ Θ−1

(p)

for p ∈ S2. We claim that the function ν is equal to the ramificationfunction αf of f . To prove this claim, we will show that ν satisfies thecharacterization of αf in Proposition 2.8.

To see this, let p ∈ S2 be arbitrary, and x ∈ Θ−1

(p). If y := A(x),then

Θ(y) = Θ(A(x)) = f(Θ(x)) = f(p).

Hence y ∈ Θ−1

(f(p)), and so

deg(f, p) · deg(Θ, x) = deg(f Θ, x) = deg(Θ A, x)

= deg(Θ, A(x)) · deg(A, x) = deg(Θ, y).

This shows that deg(f, p) · deg(Θ, x) divides ν(f(p)). Since this is true

for all x ∈ Θ−1

(p) we conclude that deg(f, p) · ν(p) divides ν(f(p)). Sothe function ν satisfies the condition in Proposition 2.8 (i).

Now suppose β : S2 → N is another function such that deg(f, p) ·β(p) divides β(f(p)) for each p ∈ S2. Then deg(fn, q) · β(q) dividesβ(fn(q)) for all q ∈ S2 and n ∈ N (see the remarks after Proposi-tion 2.8).

Now let p ∈ S2 and x ∈ Θ−1

(p) be arbitrary. Then #A−n

(x) =deg(A)n ≥ 2n. Since there are only finitely many critical points of Θ,

there exist n ∈ N and y ∈ A−n(x) such that y /∈ crit(Θ). Let q := Θ(y).Then

fn(q) = fn(Θ(y)) = Θ(An(y)) = Θ(x) = p,

68 3. LATTES MAPS

and

deg(Θ, x) = deg(Θ, x) · deg(An, y) = deg(Θ An, y)

= deg(fn Θ, y) = deg(fn, q) · deg(Θ, y)

= deg(fn, q).

Clearly, deg(Θ, x) = deg(fn, q) divides deg(fn, q) · β(q), which in turndivides β(p) = β(fn(q)) by the remark above. Hence deg(Θ, x)|β(p)

for all x ∈ Θ−1

(p). By definition of ν this implies that ν(p)|β(p) forp ∈ S2. This means that ν satisfies condition (ii) in Proposition 2.8.

From the uniqueness property of αf given by Proposition 2.8 weconclude ν = αf as desired.

(i) We have already seen that f is a Thurston map with deg(f) =deg(A). From (iii) it follows that αf (p) < ∞ for all p ∈ S2. Thus fhas no critical periodic points (see Proposition 2.9 (ii)).

Lemma 3.11 (Criterion for orbifold parabolicity). Let f : S2 → S2

be a quotient of a torus endomorphism and Θ: T 2 → S2 be as in Defi-nition 3.4. Then f has a parabolic orbifold if and only if

(3.12) deg(Θ, x) = deg(Θ, y)

for all x, y ∈ T 2 with Θ(x) = Θ(y).

We do not know whether condition (3.12) is always true, or equiva-lently, whether every quotient of a torus endomorphism has a parabolicorbifold. One can show this under that additional assumption that themap is expanding. The proof is rather involved, and so we will notdiscuss it.

Proof. As in Definition 3.4, let A : T 2 → T 2 be a torus endomor-phism with deg(A) ≥ 2 for our given maps f and Θ.

Suppose first that deg(Θ, x) = deg(Θ, y), whenever x, y ∈ T 2 andΘ(x) = Θ(y). This is equivalent to the statement that the local degree

of Θ is constant in each fiber Θ−1

(p), p ∈ S2. Then Lemma 3.10 (iii)

implies that αf (p) = deg(Θ, x) whenever x ∈ Θ−1

(p). If y := A(x),then Θ(y) = f(p), and so

αf (f(p)) = deg(Θ, y) = deg(Θ, y) · deg(A, x)

= deg(Θ A, x) = deg(f Θ, x)

= deg(f, p) · deg(Θ, x) = deg(f, p) · αf (p)

for p ∈ S2. Hence Of = (S2, αf ) is parabolic by Proposition 2.13.


Conversely, suppose that f has a parabolic orbifold. We claimthat the local degree of Θ is constant in each fiber over a point in

S2. For this it suffices to show that if p ∈ S2 and x ∈ Θ−1

(p), thendeg(Θ, x) = αf (p).

Note that the set Θ−1

(post(f)) is finite. So by picking n ∈ N large

enough, we can find a point y ∈ A−n

(x) with q := Θ(y) 6∈ post(f).Then αf (q) = 1 and so deg(Θ, y) = 1. We also have

fn(q) = fn(Θ(y)) = Θ(An(y)) = Θ(x) = p,

and

deg(Θ, x) = deg(Θ, x) · deg(An, y)

= deg(Θ An, y) = deg(fn Θ, y)

= deg(fn, q) · deg(Θ, y) = deg(fn, q).

The parabolicity of Of implies that

αf (p) = αf (fn(q)) = αf (q) · deg(fn, q) = deg(fn, q).

We conclude that

deg(Θ, x) = deg(fn, q) = αf (p)

as desired.

To complete the proof of Theorem 3.1, and to establish the remain-

ing implication (iii)⇒ (i), let f : C→ C be given as in (3.3), where the

corresponding maps A : T → T and Θ: T → C are holomorphic anddefined on a complex torus T. This implies that f is a holomorphic

map as well and hence a rational map on C (see Lemma A.5).The universal cover of T (as a Riemann surface) is C and so there

exists a holomorphic covering map π : C→ T. Actually, we can identifyT with a quotient space C/Γ, where Γ is a suitable rank-2 lattice. Undersuch an identification T ∼= C/Γ, the map π : C→ C/Γ ∼= T is just thequotient map.

The map A can be lifted by π to a holomorphic homeomorphismA : C → C such that A π = π A (see Section A.6 and in particularLemma A.11 (ii)). The map A has to be of the form A(z) = αz+β withα, β ∈ C, α 6= 0. Then we have the following commutative diagram:

70 3. LATTES MAPS

(3.13) C A//

π

Cπ

T A//

Θ

T

Θ

Cf// C.

We are now ready to show the implication (iii)⇒ (i) of Theorem 3.1.As we will see, the proof strongly relies on the fact that the involvedmaps are holomorphic.

Proof of (iii) ⇒ (i) in Theorem 3.1. Suppose f : C → C is amap as in statement (iii) of Theorem 3.1. Then there is a complex torusT, a holomorphic torus endomorphism A : T → T, and a holomorphic

map Θ: T → C such that f Θ = Θ A. As we discussed, f is also

a holomorphic map and hence a rational map on C. Moreover, f is aquotient of a torus endomorphism and so by Lemma 3.10 (i) a Thurstonmap without periodic critical points.

It remains to show that the orbifold of f is parabolic. By Lem-ma 3.10 3.12 it is enough to prove that the local degree of Θ is constant

in each fiber Θ−1

(p), p ∈ C. We argue by contradiction and assume

that there exists p ∈ C, and x, y ∈ Θ−1

(p) with

(3.14) deg(Θ, x) 6= deg(Θ, y).

In particular, one of these degrees must be ≥ 2; so p is a critical valueof Θ and hence belongs to post(f) by Lemma 3.10 (ii).

For all n ∈ N we have An(x), A

n(y) ∈ Θ

−1(fn(p)), and, since A

does not have critical points,

deg(Θ, An(x)) = deg(Θ, A

n(x)) · deg(A

n, x) = deg(Θ An, x)

= deg(fn Θ, x) = deg(fn, p) · deg(Θ, x)

6= deg(fn, p) · deg(Θ, y) = deg(Θ, An(y)).

In other words, the iterates An(x) and A

n(y) lie in the fiber over the

point fn(p) ∈ post(f) and Θ has different local degrees at these points.Since there are only finitely many points in post(f), and each fiber

Θ−1

(p) contains only finitely many points, the points x and y mustbe preperiodic under iteration of A, and p must be preperiodic underiteration of f . This implies that in (3.14) we may in addition assumethat x and y are periodic points for A, and that p is a periodic point


for f . Moreover, by replacing the maps A and f by suitable iterates,we are further reduced to the case where x and y are fixed points of A,and p is a fixed point of f .

If we introduce a suitable holomorphic coordinate w near p suchthat p corresponds to w = 0, then f has a local power series represen-tation of the form

f(w) = λwd + . . . ,

where λ 6= 0, and d ∈ N. Since f has no periodic critical points byLemma 3.10 (i), we actually have d = 1, and so

f(w) = λw + . . .

As we discussed before the proof, the map A lifts to a map A : C→C of the form A(z) = αz + β for z ∈ C, where α, β ∈ C, α 6= 0, suchthat we have a commutative diagram as in (3.13). This implies thatby introducing a suitable holomorphic coordinate u near x such thatu = 0 corresponds to x, the maps A and Θ can be given the formsA(u) = αu and

w = Θ(u) = buk + . . .

near u = 0, where b 6= 0 and k = deg(Θ, x). Similarly, by using asuitable holomorphic coordinate v near y such that v = 0 correspondsto y, we can write A(v) = αv and

w = Θ(v) = cvn + . . .

near v = 0, where c 6= 0 and n = deg(Θ, y). Since f Θ = Θ A nearu = 0, we obtain

f(Θ(u)) = λbuk + · · · = Θ(A(u)) = bαkuk + . . .

In particular, λ = αk. Similarly, by considering the relation f(Θ(v)) =Θ(A(v)) near v = 0, we obtain λ = αn. We conclude that αk = λ = αn.Now 2 ≤ deg(f) = deg(A) = |α|2 by (3.10), and so |α| > 1; butthen αk = αn implies k = n. This contradicts our assumption thatk = deg(Θ, x) 6= deg(Θ, y) = n.

This finishes the proof of Theorem 3.1.

Remark 3.12. If f : C→ C is a Lattes map , then the map A : C→C and the groupG provided by Theorem 3.1 (ii) are not unique. Indeed,

we can conjugate G and A by an arbitrary map h ∈ Aut(C). Then G =

h−1 g h : g ∈ G is also a crystallographic group. If A = h−1 Ahand Θ = Θ h, then we obtain the same map f in Theorem 3.1 (ii), if

72 3. LATTES MAPS

we replace G,Θ, A by G, Θ, A, respectively. The situation is illustratedin the following commutative diagram:

C A//

h

Θ

Ch

Θ

C A//

Θ

C

Θ

Cf// C.

By using a conjugation as in the previous remark, in Theorem 3.1 (ii)

we can always assume that the group G is one of the groups G listed

in Theorem 3.7. The requirement that the map A(z) = αz + β is G-equivariant can then be explicitly analyzed and puts strong restrictionson α and β as the following proposition shows.

Proposition 3.13. Let G = G be a crystallographic group notisomorphic to Z2 as in Theorem 3.7. Let Γ = Z⊕Zτ be the underlyinglattice, where τ ∈ C with Im(τ) > 0 in the case (2222), τ = i in the case(244), and τ = ω = eπi/3 in the cases (333) and (236). Then A : C→ Cgiven by A(z) = αz + β (where α, β ∈ C, α 6= 0) is G-equivariant ifand only if

αΓ ⊂ Γ,

and2β ∈ Γ

(1 + i)β ∈ Γ(1 + eπi/3)β ∈ Γ

(1 + e2πi/3)β ∈ Γ

when G is of type

(2222),(244),(333),(236).

According to this statement, we can always choose α ∈ Z \ 0 andβ = 0 for any lattice Γ. This and Theorem 3.1 (ii) imply that Lattesmaps exist for all signatures (2, 2, 2, 2), (2, 4, 4), (3, 3, 3), and (2, 3, 6).We will discuss more explicit examples for each of these signatures laterin Section 3.4.

Proof. Recall from (3.2) that a map z 7→ A(z) = αz+β as in thestatement is G-equivariant if and only if A g A−1 ∈ G for all g ∈ G.

The maps g ∈ G have the form g(z) = λz+γ, where γ ∈ Γ and λ is aroot of unity depending on the type of G. An elementary computationshows that

(A g A−1)(z) = λz + αγ + (1− λ)β.

3.3. LATTES-TYPE MAPS 73

Using this first for γ = 0 ∈ Γ, we see that (3.2) can only be valid if

(3.15) (1− λ)β ∈ Γ,

and if αγ ∈ Γ for all γ ∈ Γ, i.e., αΓ ⊂ Γ. Conversely, if αΓ ⊂ Γand (3.15) is true for the appropriate roots of unity λ, then A is G-equivariant.

To analyze (3.15) further, we consider several cases depending onthe type of G.

If G is of type (2222), then we have λ = ±1. Thus (3.15) is satisfiedif and only if 2β ∈ Γ.

If G is of type (244), then we have Γ = Z⊕Zi and λ = 1, i ,−1,−i .Thus (3.15) implies that (1 + i)β ∈ Γ. Conversely, suppose this lastcondition is true. Note that −iΓ = Γ and (1−i)Γ ⊂ Γ. So we concludethat −i(1+i)β = (1−i)β ∈ Γ and (1−i)(1+i)β = 2β ∈ Γ. Therefore,(3.15) holds for λ = 1, i ,−1,−i .

If G is of type (333), then we have Γ = Z ⊕ Zω and λ = 1, ω2, ω4

(recall that ω = eπi/3). So (3.15) implies that (1−ω4)β = (1+ω)β ∈ Γ.Conversely, if this last condition is true, then using ωΓ = Γ we obtainω5(1+ω)β = (1−ω2)β ∈ Γ; so (3.15) holds for λ = ωj where j = 0, 2, 4.

Finally, if G is of type (236), then Γ = Z ⊕ Zω and λ = ωj wherej = 0, . . . , 5. Thus (3.15) implies (1−ω5)β = (1+ω2)β ∈ Γ. Conversely,suppose the last condition is true. Since ωΓ = Γ, this implies ω4(1 +ω2)β = (1 − ω)β ∈ Γ. We conclude that ω2(1 − ω2)β = (1 + ω2)β −(1−ω)β ∈ Γ, and so (1−ω2)β ∈ Γ. Thus ω(1−ω2)β = (1−ω4)β ∈ Γ.Finally, we conclude that (1 − ω)β + (1 − ω4)β = 2β ∈ Γ. In sum,(3.15) is satisfied for all λ = ωj where j = 0, . . . , 5.

The claim follows.

3.3. Lattes-type maps

We now consider Lattes-type maps f : S2 → S2 as in Definition 3.3. Iff is such a map, then there exists a crystallographic group G actingon R2 ∼= C, a G-equivariant affine map A : R2 → R2 and a branchedcovering map Θ: R2 → S2 induced by G such that f A = ΘA. SinceΘ is induced by G, the quotient space R2/G is homeomorphic to S2

which implies that G is not isomorphic to Z2.In explicit constructions of Lattes-type maps one usually turns this

around and starts with a crystallographic group G not isomorphic toZ2 and an G-equivariant affine map A : R2 → R2. Then S2 = R2/G isa topological 2-sphere and the quotient map Θ: R2 → R2/G ∼= S2 is abranched covering map induced by G. The G-equivariance of A ensuresthat this maps descends to the quotient R2/G ∼= S2 and so there existsa map f : S2 → S2 such that f A = ΘA. As the considerations below

74 3. LATTES MAPS

will show, the additional condition det(LA) > 1 (as in Definition 3.3)ensures that f is a Thurston map.

Indeed, let Gtr be the subgroup of translations in G. We know thatthen T 2 = R2/Gtr is a (topological) 2-torus. We denote by π : R2 →T 2 = R2/Gtr the quotient map.

The argument in the proof of the implication (ii) ⇒ (iii) in Theo-rem 3.1 (see Section 3.1) shows that A and Θ descend to maps A andΘ on T 2. In this proof the maps were assumed to be holomorphic, butthis played no role in the existence proof for A and Θ. So we obtainmaps A : T 2 → T 2 and Θ: T 2 → S2 such that A π = π A andΘ = Θ π. These last relations imply that A and Θ are branchedcovering maps (see Lemma A.5). As we discussed, this implies that Ais a (topological) torus endomorphism.

Similarly as in (3.9), one can summarize the relations of these mapin the following commutative diagram:

(3.16) R2 A//

π

Θ

R2

π

Θ

T 2 A//

Θ

T 2

Θ

S2 f// S2.

Since Θ and Θ A are branched covering maps, and f Θ = Θ A,it follows from Lemma A.5 that f is also a branched covering map.

It is easy to see that deg(f) = deg(A) (see the beginning of theproof of Lemma 3.10), but the degree of f can also be computed fromA.

Lemma 3.14. Let f : S2 → S2 be a Lattes-type map, A : R2 → R2 bean affine map as in Definition 3.3, LA be the linear part of A, and A bea torus endomorphism as in (3.16). Then deg(f) = deg(A) = det(LA).

In particular, if f : C → C is a Lattes map and A(z) = αz + β isas in Theorem 3.1 (ii), then deg(f) = |α|2.

Proof. Let f : S2 → S2 be a Lattes-type map, and suppose Gis a crystallographic group and A an affine map as in Definition 3.3.Then we have a commutative diagram as in (3.16) and we know thatdeg(f) = deg(A). So we have to prove that deg(A) = det(LA).

If Γ ⊂ R2 ∼= C is the underlying lattice of G, then Gtr consistsof all translations τγ : R2 → R2 with γ ∈ Γ, where τγ(u) = u + γ.Accordingly, we can identify the torus T 2 = R2/Gtr with the quotient


R2/Γ (see Section A.6). Since A descends to T 2, by Lemma A.11 thereexists a linear map L : R2 → R2 with L(Γ) ⊂ Γ such that

A τγ A−1 = τL(γ)

for γ ∈ Γ. This identity implies that L(γ) = LA(γ) for γ ∈ Γ, and soL = LA, since Γ spans R2.

The lattice Γ is isomorphic to the fundamental group of T 2, and wehave essentially shown that LA : Γ→ Γ is the map on the fundamentalgroup induced by A (see the discussion after Lemma A.6). A well-known fact, explicitly formulated in Lemma A.11 (iv), now shows thatdeg(A) = det(LA) as desired.

If f is a Lattes map and A(z) = αz + β as as in Theorem 3.1 (ii),then in complex notation LA(z) = αz for z ∈ C. For the determinantof LA considered as a R-linear map, we have det(LA) = |α|2. So itfollows from the first part of the proof that deg(f) = det(LA) = |α|2as claimed.

Proof of Proposition 3.5. Let f : S2 → S2 be a Lattes-typemap, and G, A, and Θ be as in Definition 3.3. Then we have a diagramas in (3.16). Here det(LA) > 1 by assumption which by Lemma 3.14and its proof translates into deg(f) = deg(A) = det(LA) ≥ 2. We con-clude that f is a quotient of a torus endomorphism (see Definition 3.4).

So we can apply Lemma 3.10 and it follows that f is a Thurstonmap without periodic critical points. It remains to show that f has aparabolic orbifold.

For this we verify the criterion in Lemma 3.10 3.12 with the branchedcovering map Θ: T 2 → S2 as provided by (3.16). So suppose x, y ∈ T 2

and Θ(x) = Θ(y). Since π : R2 → T 2 is surjective, there exist u, v ∈ R2

with π(u) = x and π(v) = y. Then

Θ(u) = (Θ π)(u) = Θ(x) = Θ(y) = (Θ π)(v) = Θ(v).

Since Θ is induced by the crystallographic group G, there exists g ∈ Gwith v = g(u). Now Θ = Θ g and

deg(π, u) = deg(π, v) = deg(g, u) = 1.

We conclude that

deg(Θ, y) = deg(Θ, y) · deg(π, v) = deg(Θ, v)

= deg(Θ, v) · deg(g, u) = deg(Θ, u)

= deg(Θ, x) · deg(π, u) = deg(Θ, x)

as desired.

76 3. LATTES MAPS

Remark 3.15. The considerations in the previous proof actuallyallow us to determine the signature of the orbifold Of = (S2, αf ) asso-ciated with a Lattes-type map f : S2 → S2. Indeed, if p ∈ S2, then theargument given shows that the degree of Θ in each point of the fiber

Θ−1

(p) is the same and is equal to deg(Θ, u) for any point u ∈ Θ−1(p).If we combine this with (3.8) and Lemma 3.10 (iii), we see that

αf (p) = #Gu

for each u ∈ Θ−1(p). As we discussed, the type of a crystallographicgroup G is determined by the orders of the point stabilizers #Gu,u ∈ R2. So the signature of Of corresponds to the type of the crys-tallographic group G. For example, if G is of type (2222), then thesignature of Of is (2, 2, 2, 2).

We will now turn to the characterization of Lattes-type maps up toThurston equivalence.

Proof of Proposition 3.6. Suppose first that f is a Thurstonmap that is Thurston equivalent to a Lattes-type map g. Then fand g have the same orbifold signature (Proposition 1.2). Since g hasa parabolic orbifold and no periodic critical points by Proposition 1.2,the same is true for the map f as follows from Propositions 1.2 and 1.2.

For the converse direction, suppose that f is a Thurston map withparabolic orbifold Of and no periodic critical points. We may assume

that f is defined on C. Let α = αf be the ramification function and

Θ: C ∼= R2 → (C, α) be the universal orbifold covering map as providedby Theorem 1.2. The group G of deck transformations is a crystallo-graphic group whose type corresponds to the signature of Of . Let Γbe the underlying lattice of G and T 2 = R/Gtr = R2/Γ.

We now repeat part of the argument for the proof of the implication(1.2) → (1.2) in Theorem 1.2. The difference is that we do not haveholomorphicity of the maps involved, but this is is mostly inessential.First, the parabolicity of Of in combination with the uniqueness part ofTheorem 1.2 implies that there exists a homeomorphism A : R2 → R2

such that f Θ = Θ A. Again A is G and Gtr-equivarinant, and we

obtain a diagram as in (1.2) (with S2 = C), where Θ: T 2 → C is abranched covering map and A : T 2 → T 2 is a torus endomorphism.

By Lemma 1.2 there exists a linear map L : R2 → R2 with L(Γ) ⊂ Γ(essentially the map induced by A on the fundamental group Γ of T 2)such that

A τγ A−1 = τL(γ)


for γ ∈ Γ, where tauγ denotes the translation u ∈ R2 7→ τγ(u) :=u+ γ.

Since a Lattes-type map has parabolic orbifold, we know by Propo-sition 2.13 that the orbifold of each such map has one of the signatures(2, 2, 2, 2), (2, 4, 4), (3, 3, 3), or (2, 3, 6) (this also follows from the pre-vious remark). The last three cases lead to nothing new and essentiallygive Lattes maps.

Proposition 3.16. Let f : S2 → S2 be a Lattes-type map withorbifold signature (2, 4, 4), (3, 3, 3), or (2, 3, 6). Then f is topologicallyconjugate to a Lattes map.

To prove this statement we need a lemma that gives a criterionwhen an R-linear map L : C → C is C-linear. Here the R-linearity orC-linearity for L means that L(λz) = λL(z) for all z ∈ C and all λ ∈ Ror all λ ∈ C, respectively.

Lemma 3.17. Let L : C → C be an R-linear map with det(L) > 0.Suppose there exist ζ ∈ C \ R and η ∈ C with L(ζz) = ηL(z) for allz ∈ C. Then L is C-linear.

Proof. Since L is R-linear, there exist unique numbers a, b ∈ Csuch that L(z) = az + bz for z ∈ C. Then the determinant of L (asan R-linear map) is given by det(L) = |a|2 − |b|2 > 0. It follows thata 6= 0.

Now for all z ∈ C we have

L(ζz) = ζaz + ζbz = ηL(z) = ηaz + ηbz,

and soζa = ηa and ζb = ηb.

Since a 6= 0, the first equation implies ζ = η. Then the second equationcombined with the fact that ζ 6∈ R gives b = 0. Hence L(z) = az forz ∈ C. This shows that L is C-linear.

Proof of Proposition 3.16. We know that there exists a crys-tallographic group G as well as a G-equivariant affine homeomorphismA : R2 → R2 with det(LA) > 1 such that f arises as in (3.5). By con-jugation with a suitable map in Aut(C), we may assume that G is one

of the groups G listed in Theorem 3.7 (see Remark 3.12). The quotientS2 = R2/G ∼= C/G has a natural complex structure, and so we can

identify it with the Riemann sphere C by the uniformization theorem.We will show that under this identification, f is a Lattes map.

By assumption the signature of the orbifold of f is (2, 4, 4), (3, 3, 3),or (2, 3, 6). By Remark 3.15 this means that G is a crystallographic

78 3. LATTES MAPS

group G of type (244), (333), or (236) in Theorem 3.7. In these cases,G contains a rotation g0 of the form z ∈ C 7→ g0(z) = ζz, whereζ = e2πi/n is a primitive n-th root of unity with n ∈ 3, 4, 6. In

particular, ζ ∈ C \ R. Since A passes to the quotient C ∼= C/G, thismap is G-equivariant and so there exists g1 ∈ G such that

(3.17) A g0 = g1 A.

Let L = LA be the linear part of A. This is an R-linear map satisfyingdet(L) > 0. Since G consists of orientation-preserving isometries, thelinear part of g1 is C-linear, as for every map in G. Comparing linearparts of the maps in (3.17), we see that there exists η ∈ C, |η| = 1,such that

L(ζz) = ηL(z)

for all z ∈ C. This shows that L satisfies the hypotheses of Lemma 3.17and we conclude that L is C-linear. Hence A can be written in the formA(z) = αz + β for z ∈ C, where α, β ∈ C, α 6= 0. In particular, A isholomorphic, and it follows that f is indeed a Lattes map.

By the previous proposition only Lattes-type maps whose orbifoldshave signature (2, 2, 2, 2) give genuinely new maps beyond Lattes maps.We summarize some facts about these maps in the following discussion.For specific maps see Examples 6.14 and 15.8.

Example 3.18 (Lattes-type maps whose orbifolds have signature(2, 2, 2, 2)). We know that each such Lattes-type map arises from acrystallographic group G of type (2222) (see Remark 3.15). In thiscase, G remains a crystallographic group of the same type not only ifwe conjugate G by an isometry on R2, but even if we conjugate it by anaffine homeomorphism h : R2 → R2. Similarly, the class of orientation-preserving affine homeomorphisms A : R2 → R2 is preserved undersuch conjugations. Based on the considerations as in Remark 3.12,we may therefore assume in the construction of Lattes-type maps withorbifold signature signature (2, 2, 2, 2) that the underlying lattice Γ ofG is equal to the integer lattice Γ = Z2 and that the group G consistsof all isometries g : R2 → R2 of the form u ∈ R2 7→ g(u) = ±u + γ,where γ ∈ Γ = Z2.

The quotient space R2/G is a 2-sphere S2. One can identify it witha pillow as follows. Let

R := [0, 1]× [0, 1/2],

S1 := [0, 1/2]× [0, 1/2], and S2 := [1/2, 1]× [0, 1/2].


Then R = S1 ∪ S2 is a fundamental domain for the action of G on R2,i.e., R contains a representative from every orbit, and this representa-tive is unique in R if it lies in the interior of R. So R2/G is obtainedfrom R by identifying certain points on the boundary of R. For thisone folds the rectangle R along the line ` = (x, y) ∈ R2 : x = 1/2 andidentifies corresponding points on the boundaries of the two squares S1

and S2 under this folding operation; so for example the point (0, t) ∈ R2

is identified with (1, t) ∈ R2 for t ∈ [0, 1/2]. The pillow obtained inthis way is our quotient space R2/G.

Let A(u) = LA(u) + u0, u ∈ R2, be an orientation-preserving affinehomeomorphism. Then A induces a map on the quotient R2/G if andonly if A is G-equivariant, or equivalently if A g A−1 ∈ G for eachg ∈ G. It is easy to see that this is the case if and only if 2u0 ∈ Γand LA(Γ) ⊂ Γ. Since Γ = Z2 the latter condition is true precisely ifthe matrix representing LA with respect to the standard basis in R2

has integer coefficients. So we conclude that the homeomorphism Ainduces a map on S2 = R2/G precisely if A has the form

(3.18) A(u) =

(a bc d

)(xy

)+

1

2

(x0

y0

)for u =

(xy

)∈ R2,

where a, b, c, d, x0, y0 ∈ Z and det(LA) = ad − bc ≥ 1. Here the lastinequality follows from our assumption that A is orientation-preserving.

Let f : S2 → S2 be the branched covering map induced by A as in(3.18). If det(LA) = 1, then A−1 also is of the form (3.18), and so fhas a continuous inverse induced by A−1. In this case f : S2 → S2 is ahomeomorphism.

If det(LA) ≥ 2, then deg(f) = det(LA) ≥ 2 by Lemma 3.14. Thenf is a Lattes-type map, and we know that in this case the orbifold off has signature (2, 2, 2, 2) (see Remark 3.15).

An obvious question is when a Lattes-type map f is Thurston equiv-alent to a rational map. This is always true if the signature of Of isequal to (2, 4, 4), (3, 3, 3), or (2, 3, 6), because then f is even conjugateto a Lattes map (Proposition 1.2).

If Of has signature (2, 2, 2, 2), the question is answered by the fol-lowing statement which can be seen as complement to Thurston char-acterization theorem of rational maps with hyperbolic orbifold (Theo-rem 1.2).

Theorem 3.19 (Thurston’s characterization of rational maps withorbifold signature (2, 2, 2, 2)). Let f : S2 → S2 be a Lattes-type mapwith orbifold signature (2, 2, 2, 2) and A : R2 → R2 be an affine map asin Definition 1.2 with linear part LA. Then f is Thurston equivalent

80 3. LATTES MAPS

to a rational map if and only if LA is a multiple of the identity map onR2 or the eigenvalues of LA belong to C \ R.

We will not prove this here, but refer to [DH, Proposition 9.7] foran essentially equivalent statement.

3.4. Examples of Lattes maps

In this section we present some examples of Lattes maps based onTheorem 3.1 (ii).

The map Θ in Theorem 3.1 (ii) is the universal orbifold covering

map of Of = (C, αf ), the orbifold associated to f . We can use Θ to

push the Euclidean metric on C forward to a metric ω on C. Thismetric is called the canonical orbifold metric and is discussed in detailin Section A.10. In the case of Lattes maps, Of is parabolic and thecanonical orbifold metric ω is unique up to a scaling factor. The metric

space (C, ω) is locally isometric to C, except at the postcritical points

of f (i.e., the points with αf (p) ≥ 2) where C is locally isometric to aEuclidean cone of angle 2π/αf (p).

One can reverse this procedure, and construct Θ in a simple geo-metric way. This leads to a very explicit geometric description of Lattesmaps. To explain this, we start with a given crystallographic group Gnot isomorphic to Z2 as in Theorem 3.7. We first consider the types(244), (333), and (236). The type (2222) is different and will be treatedlater.

So let G be the group of type (244) as in Theorem 3.7. We explainthe construction in detail in this case. For the types (333) and (236)the considerations are completely analogous.

We start with the invariant tiling for G shown in Figure 3.1. Recallthat the union of a white and a black triangle with a common edgeforms a fundamental domain for the action of G on C.

Let T be one of the white triangles in this tiling. Then T is aright-angled Euclidean triangle with angles π/2, π/4, π/4. We glue anisometric copy Tw of T , colored white, with another isometric copy Tbof T , colored black, along their boundaries to form a pillow ∆ whichrepresents the quotient space C/G. Then ∆ is a topological 2-sphere.We fix an orientation on ∆, and equip ∆ with the path metric ω inducedby the Euclidean metric on the two copies of T . One can now define

a map Θ: C → ∆ as follows. The map Θ sends each white triangleT ⊂ C from the tiling in Figure 3.1 to (the white triangle) Tw ⊂ ∆and each black triangle T ⊂ ∆ to (the black triangle) Tb ⊂ ∆ by

3.4. EXAMPLES OF LATTES MAPS 81

−1

1

∞

0

17→−1

∞7→1

07→∞

−17→−1

g

Figure 3.4. A Lattes map with orbifold signature (2, 4, 4).

an orientation-preserving isometry. Then Θ : C → ∆ is a well-definedbranched covering map.

The topological 2-sphere ∆ is a polyhedral surface equipped with alocally Euclidean metric with three conical singularities. In particular,∆ carries a natural conformal structure, with respect to which it is

conformally equivalent to C (see Section A.7). Let a, b, c ∈ C be threedistinct points. Then by the uniformization theorem there exists a

conformal map ϕ : ∆ → C that sends the vertices of the pillow to thepoints a, b, c. In fact, one can construct ϕ quite explicitly by firstmapping Tw conformally to the upper half-plane, and Tb to the lowerhalf-plane such that the vertices of the triangles are sent to 0, 1,∞,followed by a suitable Mobius transformation. If we define Θ = ϕ Θ : C→ C, then this map is holomorphic and induced by G.

We may assume that a is the image of the common vertex of Twand Tb with angle π/2. We define the map α : C → N by α(a) = 2,

α(b) = α(c) = 4, and α(p) = 1 for all p ∈ C \ a, b, c. Then Θ is the

universal orbifold covering map of the orbifold O = (C, α). If we equip

C with the canonical orbifold covering metric (scaled appropriately)given by Θ, then the map ϕ is in fact an isometry.

Example 3.20 (A Lattes map with orbifold signature (2, 4, 4)).Based on the previous considerations, it is now easy to give an exampleof a Lattes map with orbifold signature (2, 4, 4). Let ∆ be the pillowas above obtained from gluing together the triangles Tw and Tb. Wecall Tw ⊂ ∆ the white 0-tile, and Tb ⊂ ∆ the black 0-tile.

We divide Tw along the perpendicular bisector of its hypotenuseinto two triangles T1 and T2 that are similar to Tw by the scaling factor√

2. In the same way we divide Tb into two similar triangles. Thesefour small triangles are called 1-tiles; we color them black and whitein a checkerboard fashion so that two 1-tiles with a common edge havedistinct colors as indicated in Figure 3.4.

82 3. LATTES MAPS

−17→∞ 17→10 7→−1

7→1 7→∞

∞7→−1

f

−1 1

∞


A map g : ∆ → ∆ is now given as follows. We map each white1-tile to Tw and each black 1-tile to Tb by a similarity as indicated inFigure 3.4. Then g is a branched covering map. The points labeled0 and ∞ are its critical points and we have the following ramificationportrait:

(3.19) 02:1//∞ 2:1

// 1 // −1.

Thus g is a Thurston map with post(g) = −1, 1,∞ and orbifoldsignature (2, 4, 4). Note that g is given as z 7→ (1 − i)z in suitableflat coordinates (i.e., if we identify Tw and Tb with a triangle T ⊂ Cin the obvious way, so that the fixed point −1 ∈ ∆ is identified with0 ∈ C). In particular, g : ∆ → ∆ is holomorphic with respect to thegiven conformal structure on ∆.

Let ϕ : ∆→ C be the conformal map considered earlier. Then the

map f : C → C given by f := ϕ g ϕ−1 is a rational Thurston map.This map is actually a Lattes map by Theorem 3.1, because its orbifoldsignature is (2, 4, 4). If the conformal conjugacy ϕ is chosen such that

it sends the points labeled −1, 1,∞ ∈ ∆ to −1, 1,∞ ∈ C, respectively,then it is not hard to check that f(z) = 1− 2/z2.

The map f can also be represented as in Theorem 3.1 (ii). Namely,f is the quotient of the map A : z 7→ (1 − i)z by the crystallographic

group G of type (244) in Theorem 3.7.

Example 3.21. In Figure 3.5 we illustrate a Lattes map f withorbifold signature (3, 3, 3). The map given here is obtained as a quotient

of the map A(z) = 2z by the crystallographic group G of type (333)

in Theorem 3.7. The Riemann sphere C equipped with the canonicalorbifold metric is isometric to a pillow obtained by gluing together two


07→0 7→∞ 17→0

∞7→∞

7→1f

0 1

∞


equilateral triangles along their boundaries (equipped with the path-metric). The map f has in fact the form

f(z) = 1− 2(z − 1)(z + 3)3

(z + 1)(z − 3)3.

Example 3.22. A Lattes map f with orbifold signature (2, 3, 6)is illustrated in Figure 3.6. It is obtained as the quotient of the map

A(z) = 12(3 + i

√3)z by a crystallographic group G of type (236) in

Theorem 3.7. The map f is given by

f(z) = 1− (3z + 1)3

(9z − 1)2.

We now turn to crystallographic groups of type (2222) and thecorresponding Lattes maps. For the moment we allow groups G of thistype with an arbitrary underlying (rank-2) lattice Γ ⊂ C. Then Gconsists of all isometries on C of the form z 7→ ±z + γ, where γ ∈ Γ.

Holomorphic maps Θ: C→ C induced by G can be obtained fromthe Weierstraß ℘-function. Recall that the Weierstraß ℘-function for agiven lattice Γ ⊂ C is defined as

(3.20) ℘(z; Γ) =1

z2+

∑γ∈Γ\0

(1

(z − γ)2− 1

γ2

)for z ∈ C.

Then ℘ = ℘(·; Γ) is an even meromorphic function on C with the period

lattice Γ. It descends to a holomorphic map ℘ : T → C on the torusT = C/Γ with deg(℘) = 2. The function ℘ satisfies the differentialequation

(3.21) (℘′)2 = 4(℘− e1)(℘− e2)(℘− e3).

The numbers e1, e3, e3 ∈ C here are three distinct values (dependingon Γ) with

(3.22) e1 + e2 + e3 = 0.

84 3. LATTES MAPS

The critical values of ℘ : C→ C are e1, e2, e3 and∞. Actually, ℘ : C→C is a holomorphic branched covering map satisfying

(3.23) deg(℘, z) =

2 if ℘(z) ∈ e1, e2, e3,∞,1 otherwise.

This shows that ℘ is the universal orbifold covering map of the

orbifold (C, α), where α(p) = 2 if p ∈ e1, e2, e3,∞ and α(p) = 1 for

p ∈ C\e1, e2, e3,∞. If G is the crystallographic group correspondingto Γ as above, then ℘ is induced by G. Indeed, ℘(w) = ℘(z) if andonly if w = ±z + γ for some γ ∈ Γ.

As is well-known and classical, one can reverse this procedure andstart with the differential equation (3.21): if three distinct values e1,e2, e3 ∈ C with (3.22) are given, then there exists a (unique) lattice Γsuch that the corresponding function ℘ = ℘(·; Γ) satisfies (3.21).

This implies that the universal orbifold covering map Θ of an orb-

ifold (C, α) with signature (2, 2, 2, 2) can always be obtained from aWeierstraß ℘-function followed by a suitable Mobius transformation.

Indeed, if p1, . . . , p4 ∈ C are the four distinct points with α(pk) = 2 for

k = 1, . . . , 4, then there exists a Mobius transformation ψ on C suchthat ψ(p1, . . . , p4) = e1, e2, e3,∞, where e1, e2, e3 ∈ C are threedistinct points satisfying (3.22) (first map p4 to ∞ by a Mobius trans-formation and then apply a suitable translation). Then we can finda lattice Γ such that the corresponding function ℘ = ℘(·; Γ) satisfies(3.21) with these values e1, e2, e3. If we define Θ = ψ−1 ℘(·; Γ), then

Θ is the universal orbifold covering map of the given orbifold (C, α).Actually, one can make one more reduction here. Namely, in the last

statement we may assume that the lattice has the form Γ = Z⊕Zτ withτ ∈ C and Im(τ) > 0. This immediately follows from the homogeneityproperty

℘(λz;λΓ) =1

λ2℘(z; Γ), z ∈ C, λ ∈ C \ 0,

of the ℘-function.We will now describe a more geometric construction for the map Θ

and the orbifold C/G similar in spirit to the one given for the crystal-lographic groups of type (244) as discussed above.

For this we fix τ ∈ C with Im(τ) > 0 and consider the crystallo-graphic group G of type (2222) given by all isometries z 7→ ±z + γ,where γ ∈ Γ := Z⊕ Zτ . A fundamental region for G is the Euclideantriangle T ⊂ C with vertices 0, 1, τ . To obtain the quotient space C/Gfrom T , we divide T into four similar triangles by connecting the mid-points of each side and fold T along the edges of these smaller triangles.


0 1

τ

D

∆

Figure 3.7. Folding a tetrahedron from a triangle.

P

τ

10

∆

Figure 3.8. Construction of Θ = ℘.

In this way, we can built a tetrahedron ∆ in Euclidean 3-space as indi-cated in Figure 3.7 (the two halves of each side are identified). Here weallow the degenerate case that ∆ is a pillow. This happens preciselywhen T has a right angle.

We equip ∆ with the path metric ω induced by the Euclidean metricon T , and the orientation induced by the orientation on T ⊂ C. Then∆ is an oriented topological 2-sphere carrying a locally flat metric withfour conical singularities. The cone angles at these singularities are π.If we assign to each of these singularities the value 2 for a ramificationfunction on ∆ and the value 1 to all other points, then we can consider∆ as an orbifold with signature (2, 2, 2, 2).

The parallelogram P ⊂ C spanned by 1 and τ is a fundamentaldomain for the group of translations Gtr ⊂ G given by all maps of theform z 7→ z + γ, where γ ∈ Γ = Z ⊕ Zτ . We can divide P into two

86 3. LATTES MAPS

triangles that are isometric to T , and into 8 smaller triangles similarto T by the scaling factor 2 (see Figure 3.8).

As indicated in Figure 3.8, there is a map Θ : P → ∆ such thateach small triangle in P is mapped isometrically to a face of ∆. This

map extends naturally to a map Θ : C → ∆ that respects the lattice

translations in the sense that Θ g = Θ for all g ∈ Gtr. Note that then

Θ(z) = Θ(−z) for z ∈ C, and it easily follows that Θ is induced by G.So we can identify C/G with the tetrahedron ∆. In fact, C/G equippedwith the canonical orbifold metric (see Section A.10) is isometric to(∆, ω) (up to scaling).

The polyhedron ∆ carries a natural conformal structure (see Sec-tion A.7). By the uniformization theorem there is a conformal map

ϕ : ∆ → C. It is uniquely determined up to post-composition by a

Mobius transformation. Then Θ = ϕ Θ : C → C is a holomorphicmap induced by G. If we choose a suitably normalized map ϕ = ϕ0

here, then Θ = ϕ0 Θ = ℘(·; Γ). Moreover, it follows from our analyticdiscussion above that if we set ϕ = ψ ϕ0 for a suitable Mobius trans-formation ψ and choose τ ∈ C with Im(τ) > 0 appropriately, then a

universal orbifold covering map Θ of any orbifold on C with signature

(2, 2, 2, 2) can be obtained in the form Θ = ϕ Θ.Lattes maps with orbifold signature (2, 2, 2, 2) are obtained from

G-equivariant maps A as in Theorem 3.1 (ii). The conditions on Aspecified in Proposition 3.13 are quite restrictive. For generic τ ∈ Cwith Im(τ) > 0 there are no α ∈ C \ Z with αΓ ⊂ Γ. More precisely,such α ∈ C \Z exists if and only if the lattice allows so-called complexmultiplication and α is an integer in an imaginary quadratic field. Onthe other hand, if α ∈ Z \ 0, then αΓ ⊂ Γ for each lattice Γ. Thisleads to a class of Lattes maps given by the following definition.

Definition 3.23 (Flexible Lattes map). A Lattes map f : C →C is called flexible if its orbifold has signature (2, 2, 2, 2) and can be

represented as in Theorem 3.1 (ii) with a crystallographic group G = Gof type (2222) as in Theorem 3.7 and a map A : C → C of the formA(z) = αz + β with α ∈ Z \ −1, 0, 1 and 2β ∈ Γ, where Γ is theunderlying lattice of G.

Note that we have to rule out α = ±1 due to the requirement thatdeg(f) ≥ 2 for a Thurston map f .

The term “flexible” derives from the fact that by deforming theunderlying lattice Γ (and possibly the parameter β of A) of a flexi-ble Lattes map, one obtains a family of such maps depending on one


complex parameter (for example, the parameter τ in the representa-tion of the lattice Γ = Z ⊕ Zτ). It is not hard to see that this leadsto maps that are topologically conjugate, but in general not conjugate

by a Mobius transformation on C (see the discussion below). FlexibleLattes maps are exceptional in many respects. For example, Thurs-ton’s Uniqueness Theorem (see Theorem 2.19) fails for these maps.They also carry invariant line fields (see [McM94a, Chapter 3.5]).According to a well-known conjecture they are the only rational mapswith this property. Its validity would imply the “density of hyperbolicrational maps”, which is possibly the most famous open problem incomplex dynamics (see [MSS] and [McM94b] for an overview).

Flexible Lattes maps can be considered as generic Lattes maps,because they are the only type of Lattes maps that can be definedfor arbitrary lattices Γ = Z ⊕ Zτ or, equivalently, can be obtained asquotients of maps on arbitrary complex tori T. This is the reason whysome authors use the term “Lattes map” only for maps with orbifoldsignature (2, 2, 2, 2). For example, Milnor uses this more restrictivedefinition in the second edition (and earlier editions), while he uses thedefinition used here in the third edition of [Mi].

To discuss a geometric description of flexible Lattes maps, supposef is such a map as in Definition 3.23 obtained from a map A : C→ Cof the form A(z) = nz+β with n ∈ Z\−1, 0, 1 and 2β ∈ Γ = Z⊕Zτ .We get the same map f if we replace A by g A g′, where g, g′ ∈ G.This allows us to assume that A has the form A(z) = nz + β, where

n ∈ N, n ≥ 2, and β ∈ 0, 1/2, τ/2, (τ + 1)/2. Note that the map Θconsidered above sends the four points 0, 1/2, τ/2, (τ + 1)/2 to the fourvertices of the tetrahedron ∆.

The triangle T from Figure 3.7 can be divided into n2 trianglesthat are similar to T by the scaling factor n. For this we divide eachside of ∆ into n edges of the same length and draw the line segmentsparallel to the sides of T through the endpoints of these edges. Thetetrahedron ∆ has four faces, denoted by T1, T2, T3, T4. Each of themis similar to T , and so we can divide each face of ∆ in the same wayinto n2 triangles similar to T . In the following, we will call them thesmall triangles in ∆.

We now construct a map f : ∆→ ∆ as follows. We consider a smalltriangle S that contains one of the four vertices of ∆ and send S toone of the faces T ′ of ∆ by an orientation-preserving similarity thatscales by the factor n. For given S and T ′ there is only one such mapunless the base triangle T , and hence also S and T ′, are equilateral. Inthis case, there are three such maps, only one of which will lead to a

88 3. LATTES MAPS

f∆ ∆

Figure 3.9. A Euclidean model for a flexible Lattes map.

flexible Lattes map. We will ignore this special case in order to keepour discussion simple.

The similarity between S and T ′ can be uniquely extended to a

continuous map f on ∆ that sends each small triangle to one of thetriangles Tj, j ∈ 1, 2, 3, 4, by an orientation-preserving similarity

with scaling factor n. It is clear that f is a branched covering map. Itscritical points are the vertices of the small triangles with the exception

of the four vertices of ∆. The map f sends each vertex of a smalltriangle to one of the vertices of ∆. It follows that the postcritical

points of f are the four vertices of ∆ and f is postcritically-finite.Since the local degree at each critical point is 2, it is easy to see that

the orbifold signature of f is (2, 2, 2, 2).

If we conjugate f by a uniformizing map ϕ : ∆→ C, then we obtain

a flexible Lattes map f := ϕ f ϕ−1 : C→ C. Conversely, each flexibleLattes map can be obtained in this form. One such map (with n = 3)is illustrated in Figure 3.9.

If we use exactly the same construction as just described with an-other parameter τ ′ ∈ C, Im(τ ′) > 0, then we obtain a different tetra-

hedron ∆′. This leads to a different flexible Lattes map g : C→ C. Itis easy to see that the f and g are always topologically conjugate, butin general not conjugate by a Mobius transformation. So we obtain a1-parameter family of Lattes maps with similar dynamics.

Example 3.24. Not all Lattes maps whose orbifold has signature(2, 2, 2, 2) are flexible Lattes maps. For example, suppose G is thecrystallographic group consisting of the maps g(z) = ±z + γ, whereγ ∈ Γ := Z ⊕ Zi . Then A(z) = (1 − i)z is G-equivariant and sothere is a Lattes map f that arises as the quotient of A by G as in


07→∞

−i 7→0

∞7→∞

i 7→0

17→i

−17→−i

f

0

−i

∞

i

Figure 3.10. The map f in Example 3.24.

Theorem 3.1 (ii). Then the map f has orbifold signature (2, 2, 2, 2).In fact, f is (up to conjugation by a Mobius transformation) given byf(z) = i

2(z + 1/z). Since deg(f) = 2, this map is not a flexible Lattes

map, because the degree of each such map is the square of an integer.A geometric model for f is shown in Figure 3.10. A detailed discussionof this map can be found in [Mi04].

CHAPTER 4

Quasiconformal and rough geometry

The immediate purpose of this chapter is to provide a framework fordiscussing Cannon’s conjecture in geometric group theory and relatedquestions. This will give some motivating background for our study ofexpanding Thurston maps. The chapter will also serve as a referencefor some geometric terminology and facts that we will use throughoutthis work.

4.1. Quasiconformal geometry

In this section we record some material related to quasiconformal geo-metry and the analysis on metric spaces (see [He] for an exposition tothis subject).

Let (X, dX) and (Y, dY ) be metric spaces, and f : X → Y be ahomeomorphism. Then f is called bi-Lipschitz if there exists a constantL ≥ 1 such that

1

LdX(u, v) ≤ dY (f(u), f(v)) ≤ LdX(u, v)

for all u, v ∈ X. If there exists α > 0 and L ≥ 1 such that

(4.1)1

LdX(u, v)α ≤ dY (f(u), f(v)) ≤ LdX(u, v)α

for all u, v ∈ X, then f is called a snowflake homeomorphism. Themap f is called a quasisymmetric homeomorphism or a quasisymmetryif there exists a homeomorphism η : [0,∞)→ [0,∞) such that

(4.2)dY (f(u), f(v))

dY (f(u), f(w))≤ η

(dX(u, v)

dX(u,w)

)for all u, v, w ∈ X with u 6= w. If we want to emphasize η here, thenwe speak of an η-quasisymmetry.

The inverse of a bi-Lipschitz homeomorphism and the compositionof two such maps (when it is defined) is again a bi-Lipschitz homeo-morphism. Similarly, the classes of snowflake homeomorphisms andquasisymmetries are closed under taking inverses or compositions. Thisimplies that all these classes of maps lead to a corresponding notionof equivalence for metric spaces. So X and Y are called bi-Lipschitz,

91

92 4. QUASICONFORMAL AND ROUGH GEOMETRY

snowflake, or quasisymmetrically equivalent if there exists a homeo-morphism between X and Y that is of the corresponding type. Inthe same way, we call two metrics d and d′ on a space X bi-Lipschitz,snowflake, or quasisymmetrically equivalent if the identity map from(X, d) to (X, d′) is a homeomorphism with the corresponding property.Since every bi-Lipschitz homeomorphism is also a snowflake homeo-morphism and every snowflake homeomorphism is a quasisymmetry,this leads to correspondingly weaker notions of equivalence.

A collection of metrics on a given space X that are mutually qua-sisymmetrically equivalent is called a quasisymmetric gauge on X. Onedefines a bi-Lipschitz or snowflake gauge on X similarly (see [He,Ch. 15] for related terminology and further discussion).

As we will later see in Chapter 8, each expanding Thurston mapf : S2 → S2 induces a natural snowflake gauge on its underlying 2-sphere S2. This is related to the fact that if X is a Gromov hyperbolicspace, then there is a natural snowflake gauge on its boundary at in-finity ∂∞X (see Section 4.2 below). In both cases we will later usethe term visual metric for a metric in the natural snowflake gauge.A visual metric % gives the underlying 2-sphere S2 of an expandingThurston map f : S2 → S2 some geometric structure related to the dy-namics of the map. An important question in this context is whetherthis sphere (S2, %) is quasisymmetrically equivalent to the standard 2-sphere, i.e., the Riemann sphere equipped with the chordal metric (seeTheorem 17.1). This is an instance of a more general problem that canbe formulated as follows:

The Quasisymmetric Uniformization Problem. Suppose Xis a metric space homeomorphic to some “standard” metric space Y .When is X quasisymmetrically equivalent to Y ?

This problem is very similar to questions in classical uniformization

theory where one asks whether a given region in C or a given Riemannsurface is conformally equivalent to a “standard” region or Riemannsurface. The quasisymmetric uniformization problem can be seen asa metric space version of this where the class of conformal maps isreplaced by the class of quasisymmetric homeomorphisms.

It depends on the context how the term “standard space Y ” isprecisely interpreted. A satisfactory answer to the quasiymmetric uni-formization problem for given Y is essentially equivalent to character-izing Y up to quasisymmetric equivalence. We will see that this isrelated to Cannon’s conjecture in geometric group theory, for example.One can also pose a similar problem for other classes of maps such asbi-Lipschitz or snowflake maps.

4.1. QUASICONFORMAL GEOMETRY 93

The prime instance for a quasisymmetric uniformization result is atheorem due to Tukia and Vaisala that characterizes quasicircles andquasiarcs. In order to state this result, we have to first discuss someterminology.

The standard unit circle S1 is the unit circle in R2 equipped therestriction of the Euclidean metric. A metric circle, i.e., a metric spacehomeomorphic to S1, is called a quasicircle if it is quasisymmetricallyequivalent to S1. Similarly, a metric arc (a metric space homeomorphicto the unit interval [0, 1]) is called a quasiarc if it is quasisymmetricallyequivalent to [0, 1].

Let (X, d) be a metric space. Then X is called doubling if thereexists a number N ∈ N such that every ball in X of radius R > 0 canbe covered by N balls of radius R/2. We say that X is of boundedturning there is a constant K ≥ 1 such that for all points x, y ∈ X,x 6= y, there exists a continuum γ ⊂ X with x, y ∈ γ such that

(4.3) diamd(γ) ≤ Kd(x, y).

If X is a metric circle, then it is of bounded turning precisely whenthe last inequality is always true for one of the subarcs γ ⊂ X withendpoints x and y. Similarly, if X is a metric arc, then it is of boundedturning precisely when (4.3) is always true for the unique subarc γ ofX with endpoints x and y.

The Tukia-Vaisala-Theorem [TuV, p. 113, Thm. 4.9] can now beformulated as follows.

Theorem 4.1. Let X be a metric circle or a metric arc. Then Xis a quasicircle or a quasiarc, respectively, if and only if X is doublingand of bounded turning.

A metric space X is called linearly locally connected if there exists aconstant λ ≥ 1 satisfying the following conditions: If B(a, r) is an openball in X and x, y ∈ B(a, r), then there exists a continuum in B(a, λr)connecting x and y. Moreover, if x, y ∈ X \ B(a, r), then there existsa continuum in X \B(a, r/λ) connecting x and y.

It is easy to see that a metric circle is of bounded turning if andonly if it is linearly locally connected. This gives an alternative versionof part of the Tukia-Vaisala Theorem: A metric circle is a quasicircleif and only if it is doubling and linearly locally connected.

Every subset of C (equipped with the restriction of the chordal met-

ric) is doubling; this implies that a Jordan curve J ⊂ C is a quasicircleif and only if it is of bounded turning. A similar remark applies to arcs

α ⊂ C.


A quasisymmetric characterization of the standard 1/3-Cantor setcan be found in [DS]. Work by Semmes [Se2, Se3] shows that thequasisymmetric characterization of Rn or the standard sphere Sn forn ≥ 3 is a problem that seems to be beyond reach at the moment.The intermediate case n = 2 is particularly interesting. To formulatea specific result in this direction we need one more definition.

Let (X, d) be a locally compact metric space, and µ a Borel measureon X. Then the metric measure space (X, d, µ) is called Ahlfors Q-regular, where Q > 0, if

(4.4)1

CRQ ≤ µ(B(x,R)) ≤ CRQ

for all closed balls B(x,R) with x ∈ X and 0 < R ≤ diam(X), whereC ≥ 1 is independent of the ball. If (X, d) is understood, we also saythat µ is Ahlfors Q-regular.

If this condition is satisfied, then the Q-dimensional Hausdorff mea-sure HQ actually satisfies an inequality as in (4.4), and µ and HQ aremutually absolutely continuous with respect to each other (for the def-inition of Hausdorff Q-measure HQ see [He], for example). If we want

to emphasize the underlying metric d, we use the notation HQd for the

Hausdorff Q-measure.A locally compact metric space (X, d) is called Ahlfors Q-regular for

Q > 0, if the Q-dimensional Hausdorff measure HQ has this property.For a metric 2-sphere X to be quasisymmetrically equivalent to

the standard 2-sphere (C, σ) it is necessary that X is linearly locallyconnected. This alone is not sufficient, but will be if we add Ahlfors2-regularity as an assumption [BK, Thm. 1.1].

Theorem 4.2. Suppose X is a metric space homeomorphic to C.If X is linearly locally connected and Ahlfors 2-regular, then X is qua-

sisymmetrically equivalent to (C, σ).

A similar result for other simply connected surfaces has been ob-tained by K. Wildrick [Wi].

The assumption of Ahlfors regularity for some exponent Q ≥ 2is quite natural, because it is satisfied in many interesting cases: forexample, for boundaries of Gromov hyperbolic groups (see [Coo]) orfor 2-spheres equipped with visual metrics of an expanding Thurstonmap (under some mild extra conditions; see Proposition 17.2). Thereare metric 2-spheres X though that are linearly locally connected andQ-regular with Q > 2, but are not quasisymmetrically equivalent to

(C, σ) [Va2].

4.1. QUASICONFORMAL GEOMETRY 95

We will now prove a fact that will be useful in Chapter 19, whereTheorem 4.2 is applied to give a characterization of Lattes maps.

Proposition 4.3. Let d be a metric on C that is quasisymmetri-

cally equvialent to the chordal metric σ. Let µ be a Borel measure on Csuch that that (C, d, µ) is Ahlfors 2-regular. Then µ and the Lebesgue

measure λ on C are absolutely continuous with respect to each other.

Proof. The quasisymmetric equivalence of σ and d implies thata metric ball with respect to σ roughly looks like a metric ball withrespect to d (typically with quite different radius); more precisely, there

exists a constant λ ≥ 1 such that for all p ∈ C and all R > 0 thereexists R′ > 0 such that

Bd(p,R′/λ) ⊂ Bσ(p,R) ⊂ Bd(p, λR

′).

One can deduce from this and the fact that µ is Ahlfors 2-regular

on (C, d) that µ is a doubling measure on (C, σ). Actually, µ is a metric

doubling measure on (C, σ): if x, y ∈ C are arbitrary, R := σ(x, y), andwe set

d′(x, y) := µ(Bσ(x,R) ∪Bσ(y,R))1/2,

then d′ is comparable to a metric (in this case to the metric d). See[He] for the terminology employed here.

It is known that a metric doubling measure on (C, σ) is absolutely

continuous with respect to Lebesgue measure λ on C with a Radon-Nikodym derivative that is an A∞-weight (see [Se1]). So there exists an

A∞-weight w on C such that dµ = w dλ, and µ is absolutely continuouswith respect to λ. Since w cannot vanish on a set of positive Lebesguemeasure, this implies that λ is also absolutely continuous with respectto µ.

A homeomorphism f : X → Y between metric spaces (X, dX) and(Y, dY ) is called weakly quasisymmetric if there exists a constant H ≥ 1such that for all u, v, w ∈ X the following implication holds:

dX(u, v) ≤ dX(u,w)⇒ dY (f(u), f(v)) ≤ HdY (f(u), f(w)).

Under mild extra assumptions on the spaces weak quasisymmetry of amap implies its quasisymmetry ([He, Thm. 10.19]).

Proposition 4.4. Let (X, dX) and (Y, dY ) be metric spaces, andf : X → Y be weakly quasisymmetric homeomorphism. If X and Y areconnected and doubling, then f is a quasisymmetry.


This proposition is very useful if one wants to establish that a givenmap is a quasisymmetry.

We will discuss a final variant of quasisymmetric maps that appearsin the context of Gromov hyperbolic groups. If x1, x2, x3, x4 are fourdistinct points in a metric space (X, d), then their cross-ratio is thequantity

[x1, x2, x3, x4] =d(x1, x3)d(x2, x4)

d(x1, x4)d(x2, x3).

Let η : [0,∞) → [0,∞) be a homeomorphism, and f : X → Y ahomeomorphism between metric spaces (X, dX) and (Y, dY ). The mapf is an η-quasi-Mobius homeomomorphism if

[f(x1), f(x2), f(x3), f(x4)] ≤ η([x1, x2, x3, x4]).

for every 4-tuple (x1, x2, x3, x4) of distinct points in X. For boundedmetric spaces the classes of quasisymmetric and quasi-Mobius homeo-morphisms are the same.

4.2. Gromov hyperbolicity

In this section we review some standard material on Gromov hyperbolicspaces. For general background on this topic see [BuS, GH, Gr].

Let (X, d) be a metric space. Then for p, x, y ∈ X the quantity

(x · y)p :=1

2

(d(x, p) + d(y, p)− d(x, y)

)is called the Gromov product of x and y with respect to the basepointp. The space X is called δ-hyperbolic for δ ≥ 0, if the inequality

(4.5) (x · z)p ≥ min(x · y)p, (y · z)p − δ

holds for all x, y, z, p ∈ X. If this condition is true for some basepointp ∈ X (and all x, y, z ∈ X), then it is actually true for all basepoints ifone changes the constant δ to 2δ. We say that X is Gromov hyperbolicif X is δ-hyperbolic for some δ ≥ 0.

The space X is called geodesic if any two point in X can be joinedby a path whose length is equal to the distance of the points. Fora geodesic metric space its Gromov hyperbolicity is equivalent to athinness condition for geodesic triangles [GH].

Roughly speaking, the Gromov hyperbolicity of a space requires itto be “negatively curved” on large scales. Examples for such spaces aresimplicial trees, or Cartan-Hadamard manifolds with a negative uppercurvature bound such as the real hyperbolic spaces Hn, n ≥ 2.

4.2. GROMOV HYPERBOLICITY 97

Suppose (X, dX) and (Y, dY ) are metric spaces. A map f : X → Yis called a quasi-isometry if there exist constants λ ≥ 1 and k ≥ 0 suchthat

(4.6)1

λdX(u, v)− k ≤ dY (f(u), f(v)) ≤ λdX(u, v) + k

for all u, v ∈ X and if

(4.7) infx∈X

dY (f(x), y) ≤ k

for all y ∈ Y . If λ = 1, then we call f a rough-isometry. The spaces Xand Y are called quasi-isometric or rough-isometric if there exists a mapf : X → Y that is a quasi-isometry or a rough-isometry, respectively.In coarse geometry one often considers two metric spaces the same ifthey are quasi-isometric or rough-isometric.

Quasi-isometries form a natural class of maps in theory of Gromovhyperbolic spaces. For example, Gromov hyperbolicity of geodesic met-ric spaces is invariant under quasi-isometries [GH].

A subset A of a metric space (X, d) is called cobounded if thereexists a constant k ≥ 0 such that for every x ∈ X there exists a ∈ Awith d(a, x) ≤ k. Then every point in X lies within uniformly boundeddistance of the set A. With this terminology, the condition (4.7) canbe reformulated by saying that the map f has cobounded image inY . If A is cobounded in X, then X is Gromov hyperbolic if and onlyif A (equipped with the restriction of the ambient metric) is Gromovhyperbolic.

To each Gromov hyperbolic space X one can associate a boundaryat infinity ∂∞X as follows. We fix a basepoint p ∈ X, and considersequences of points xi in X converging to infinity in the sense that

(4.8) limi,j→∞

(xi · xj)p =∞.

We declare two such sequences xi and yi in X as equivalent if

(4.9) limi→∞

(xi · yi)p =∞.

Then ∂∞X is defined as the set of equivalence classes of sequencesconverging to infinity. It is easy to see that the choice of the basepointp does not matter here. Moreover, if A is cobounded in X, then onecan represent each equivalence class by a sequence in A, and so we havea natural identification ∂∞X = ∂∞A.

A metric space X is called proper if every closed ball in X is com-pact. If a Gromov hyperbolic space X is proper and geodesic, thenthere is an equivalent definition of ∂∞X as the set of equivalence classesof geodesic rays emanating from the basepoint p. One declares two such


rays as equivalent if they stay within bounded Hausdorff distance. In-tuitively, a ray represents its “endpoint” on ∂∞X (see [BuS] for thedetails).

The Gromov product on a Gromov hyperbolic spaceX has a naturalextension to the boundary ∂∞X. Namely, if p ∈ X and a, b ∈ ∂∞X,we set

(4.10) (a · b)p := inf

lim infi→∞

(xi · yi)p : xi ∈ a, yi ∈ b.

Note that by definition a point in ∂∞X is an equivalence class (i.e., aset) of sequences in X. So in (4.10) it makes sense to take the infimumover all sequences xi and yi that represent (i.e., are contained in)a and b, respectively. We have (a · b)p ∈ [0,∞] with (a · b)p = ∞ iffa = b.

In (4.10) one can actually use any sequences representing the pointsa and b to determine (a · b)p (up to an irrelevant additive constant).Namely, there exists a constant k ≥ 0 independent of a, b, and p suchthat for all xi ∈ a and yi ∈ b we have

(4.11) lim infi→∞

(xi · yi)p − k ≤ (a · b)p ≤ lim infi→∞

(xi · yi).

The boundary ∂∞X is equipped with a natural class of visual met-rics. By definition a metric % on ∂∞X is called visual if there existp ∈ X, C ≥ 1, and Λ > 1 such that

(4.12)1

CΛ−(a·b)p ≤ %(a, b) ≤ CΛ−(a·b)p

for all a, b ∈ ∂∞X (here we use the convention that Λ−∞ = 0). If Xis δ-hyperbolic, then there exists a visual metric % with parameter ε ifε > 0 is small enough depending on δ.

Later we will also define a notion of a visual metric for an expand-ing Thurston map. So then we have two notions of visual metrics—onefor expanding Thurston maps and one for boundaries of Gromov hy-perbolic spaces. For clarity we will sometimes use the phrases visualmetric in the sense of Thurston maps and visual metric in the sense ofGromov hyperbolic spaces to distinguish between these two notions. Aswe will see in Chapter 10, to each expanding Thurston map f : S2 → S2

one can associate a Gromov hyperbolic space G such that ∂∞G can beidentified with S2 and such that the class for visual metric of f (inthe sense of Thurston maps) is exactly the same as the class for visualmetric on ∂∞G ∼= S2 (in the sense of Gromov hyperbolic spaces). Thisfact was actually the reason for our choice of the term “visual metric”for expanding Thurston maps.

4.3. GROMOV HYPERBOLIC GROUPS AND CANNON’S CONJECTURE 99

We can think of the boundary at infinity of a Gromov hyperbolicspace ∂∞X as a metric space if we equip it with a fixed visual metric.If %1 and %2 are two visual metrics on ∂∞X, then the identity map is asnow-flake equivalence between (∂∞X, %1) and (∂∞X, %2). So the visualmetrics form a snowflake gauge on ∂∞X. In particular, ∂∞X carries awell-defined topology induced by any visual metric, and the ambiguityof the visual metric is irrelevant if one wants to speak of snowflake orquasisymmetric maps on ∂∞X. One should consider the space ∂∞Xequipped with such a visual metric ρ as very “fractal”. For example, ifthe parameter ε in (4.12) is very small, then (∂∞X, %) will not containany non-constant rectifiable curves.

The following fact links the theory of Gromov hyperbolic spaces toquasisymmetric maps (see [BS] for more on this subject).

Proposition 4.5. Let X and Y be proper and geodesic Gromovhyperbolic spaces. Then every quasi-isometry f : X → Y induces anatural quasisymmetric boundary map f : ∂∞X → ∂∞Y .

The boundary map f is defined by assigning to a point a ∈ ∂∞Xrepresented by the sequence xi the point b ∈ ∂∞Y represented bythe sequence f(xi).

This statement lies at the heart of Mostow’s proof for rigidity ofrank-one symmetric spaces [Mo]. The point is that a quasi-isometrymay locally exhibit very irregular behavior, but gives rise to a qua-sisymmetric boundary map that can be analyzed by analytic tools.

4.3. Gromov hyperbolic groups and Cannon’s conjecture

The theory of Gromov hyperbolic spaces can be used to define a class ofdiscrete groups. Here one adopts a geometric point of view by studyingthe Cayley graph of the group. We review some standard definitionsrelated to this, but will not attempt an in-depth treatment of the sub-ject. We will just develop the necessary background to state and discussCannon’s conjecture that served as one of our motivations for studyingexpanding Thurston maps.

Let G be a finitely generated group, and S a finite set of generatorsof G that is symmetric, i.e., if s is in S, then its inverse s−1 is also inS. The Cayley graph G(G,S) of G with respect to S is now definedas follows: the group elements are the vertices of G, and one joins twovertices given by g, h ∈ G by an edge, if there exists s ∈ S such thatg = hs (here we use the common convention that juxtaposition of groupelements means their composition in the group). Since S is symmetric,this “edge relation” for vertices is also symmetric, and so we consider


edges as undirected. If we identify each edge in G(G,S) with a closedinterval of length 1, then G(G,S) becomes a cell complex, where single-ton sets consisting of group elements are the cells of dimension 0 andthe edges are the cells of dimension 1. The graph G(G,S) is connectedand carries a unique path metric so that each edge is isometric to theunit interval [0, 1]. In the following we always consider G(G,S) as ametric space equipped with this path metric.

The group G is called Gromov hyperbolic if the metric space G(G,S)is Gromov hyperbolic for some (finite and symmetric) set S of genera-tors of G. If this is the case, then G(G,S ′) is Gromov hyperbolic for allgenerating sets S ′. This essentially follows from the fact that G(G,S)and G(G,S ′) are quasi-isometric.

Examples of Gromov hyperbolic groups are free groups, fundamen-tal groups of compact negatively curved manifolds, or small cancella-tion groups.

If G is a Gromov hyperbolic group, then one defines its boundaryat infinity as ∂∞G = ∂∞G(G,S). A priori this depends on the choiceof the generating set S, but if S ′ is another generating set, then thereis a natural identification ∂∞G(G,S ′) = ∂∞G(G,S); namely, since Gis cobounded in G(G,S) and G(G,S ′), one can represent points in theboundaries of both spaces by equivalence classes of sequences in Gconverging to infinity where the equivalence relation is independent ofthe generating set. So ∂∞G is well-defined.

One has to be careful though when one considers visual metrics. If% is a visual metric on ∂∞G(G,S), then in general % will not be a visualmetric on ∂∞G(G,S ′); but if %′ is a visual metric on ∂∞G(G,S ′), then %and %′ are quasisymmetrically equivalent, i.e., the identity map between(∂∞G(G,S), %) and (∂∞G(G,S ′), %′) (given by the natural identificationof these spaces as discussed) is a quasisymmetry. This follows from thefact that G(G,S) and G(G,S ′) are quasi-isometric and Proposition 4.5.So ∂∞G carries a natural quasisymmetric gauge.

If we equip ∂∞G with any of these visual metrics %, then we canunambiguously speak of quasisymmetric and quasi-Mobius maps on∂∞G. Another consequence of this is that ∂∞G carries a unique topol-ogy induced by any visual metric % on ∂∞G = G(G,S).

Letting a group element g ∈ G act on the vertices of G(G,S) byleft-translation, we get a natural action G y G(G,S). This action isgeometric, i.e., it is isometric, properly discontinuous, and cocompact.To get a better understanding of the properties of a group, one oftenwants to find a “better” space than G(G,S) on which G admits ageometric action (for a systematic exploration of this point of view see[Kl]).

4.3. GROMOV HYPERBOLIC GROUPS AND CANNON’S CONJECTURE 101

A related question is how the topological structure of the boundary∂∞G of a Gromov hyperbolic group determines the algebraic structureof the group. Since the Cayley graph of G and and of its subgroupsof finite index are quasi-isometric and hence indistinguishable from theperspective of coarse geometry, one is mostly interested in virtual prop-erties of G, i.e., algebraic properties that are true for some subgroupof finite index.

The spaces ∂∞G form a very restricted class. For example, if G isnon-elementary (i.e., if #∂∞G ≥ 2) and has topological dimension 0,then ∂∞G is homeomorphic to a Cantor set. Moreover, in this case Gis virtually isomorphic to a free group (i.e., some finite-index subgroupof G is free).

If ∂∞G is homeomorphic to a circle, then G is virtually Fuchsian,i.e., G is virtually isomorphic to a fundamental group of a compacthyperbolic orbifold, or equivalently, there is a geometric action of G onH2 (see [KB] for an overview on this subject).

If ∂∞G has no local cut points and has topological dimension one,then ∂∞G is a Menger curve or a Sierpinski carpet. Moreover, a conjec-ture due to Kapovich and Kleiner [KK] predicts that in the latter case,G admits a geometric action on a convex subset of H3 with non-emptytotally geodesic boundary.

This conjecture is related to (and implied by) another conjecturedue to Cannon.

Conjecture (Cannon’s conjecture. Version I). Let G be a Gromov

hyperbolic group and suppose ∂∞G is homeomorphic to C. Then thereexists a geometric action of G on hyperbolic 3-space H3.

If this were true, then G would be virtually isomorphic to the fun-damental group of a compact hyperbolic 3-orbifold.

In higher dimensions a corresponding statement is false; there areGromov hyperbolic groups G with ∂∞G homeomorphic to an n-sphere,n ≥ 3, that do not admit geometric actions on Hn+1. One can obtainsuch examples as fundamental groups of Gromov-Thurston manifolds[GT]. These are closed negatively-curved manifolds that do not carrya hyperbolic metric, and exist in dimension n+ 1 ≥ 4.

Cannon’s conjecture can be reformulated in equivalent form as aquasisymmetric uniformization problem.

Conjecture (Cannon’s conjecture. Version II). Let G be a Gro-

mov hyperbolic group and suppose ∂∞G is homeomorphic to C. Then∂∞G equipped with a visual metric is quasisymmetrically equivalent to

C equipped the the chordal metric.


In view of this formulation of the conjecture it is very interesting tostudy the quasisymmetric uniformization problem for metric 2-spheresin general and ask for general conditions under which such a sphere

is quasisymmetrically equivalent to the standard 2-sphere C. Herethe conditions should be similar to those that one can establish forboundaries of Gromov hyperbolic groups.

In all known examples where ∂G is a 2-sphere, G is (essentially) thefundamental group of a hyperbolic orbifold and ∂∞G can naturally beidentified with ∂∞H3 which is the standard 2-sphere. So in these cases,no uniformization problem arises. Cannon’s conjecture predicts thatthere are no other examples. One of the difficulties in making progresson Cannon’s conjecture is this lack of non-trivial examples that mayguide the intuition.

In contrast, the theory of Thurston maps provides a large class ofself-similar fractal 2-spheres that sometimes are, and sometimes arenot, quasisymmetrically equivalent to the standard 2-sphere. By ana-lyzing these examples, one may hope to discover some general featuresthat could also be relevant in the solution of Cannon’s conjecture.

4.4. Quasispheres

A metric space that is quasisymetrically equivalent to the sphere Cequipped with the chordal metric is called a quasisphere. In view ofthe previous discussion of Cannon’s conjecture and the characterizationof rational Thurston maps as given by Theorem 17.1 ((ii)), we now wantto discuss two examples that may guide the reader’s intuition. As thisis our main purpose here, we will skip the justification of most details.

Example 4.6. A snowball is a compact set in R3 that is obtainedin a way analogous to the set in the plane bounded by the classical vonKoch snowflake curve. The boundary of a snowball is a snowsphere S.In many cases this is a quasisphere.

The simplest example is obtained as follows. We start with the unitcube [0, 1]3 ⊂ R3 as the 0-th approximation B0 of the snowball. Theboundary of B0 is a polyhedral surface S0 consisting of six copies of theunit square [0, 1]2 ⊂ R2 as faces. We divide each of these six faces into5× 5 squares of side-length 1/5 (a 1/5-square). We place a cube thathas sidelength 1/5 and sticks out of B0 on the 1/5-square in the middleof each face. This results in a set B1 ⊃ B0. The boundary of B1 is apolyhedral surface S1 consisting of 6× 29 1/5-squares. The procedureis now iterated; namely, each 1/5-square is divided into 5 × 5 squaresof side-length 1/25, on each middle square we put a cube of side-length1/25, and so on. We obtain an increasing sequence B0 ⊂ B1 ⊂ . . . of

4.4. QUASISPHERES 103

Figure 4.1. The generator of the snowsphere S.

compact sets in R3. Their union is the snowball B with the snowsphereS := ∂B as its boundary. It is not hard to see that S is indeed a 2-sphere. For each n ∈ N0 the boundary of Bn is a polyhedral surface Snthat consists of 1/5n-squares. The surfaces Sn give better and betterapproximations of the snowsphere S for larger and larger n.

One can also give a more abstract construction of S by a replace-ment procedure very similar as in Section 1.3. For this we let the gen-erator of the snowball be the polyhedral surface shown in Figure 4.1.The approximation Sn+1 is then obtained from Sn by replacing each5−n-square of Sn by a scaled copy of the generator. If Xn is one of the5−n-squares from which Sn is built, and Xn+1 is a 5−(n+1)-square in thescaled copy of the generator that replaces Xn, we write Xn ⊃∼ Xn+1.

The snowsphere S inherits the Euclidean metric from R3. It is nothard to see that if x, y ∈ S, then there exists a rectifiable path γ ⊂ Sjoining x and y whose length is comparable to |x − y|. If we defined(x, y) for x, y ∈ S as the infimum of the lengths of such paths, thenwe get a length metric d on S that is bi-Lipschitz equivalent to theEuclidean metric (see [Me02]).

One can also obtain the Euclidean metric on S in an intrinsic waysimilar as for the example in Section 1.3. For this we note that for eachpoint x ∈ S there exist sequences X0 ⊃∼ X1 ⊃∼ . . . of 1/5n-squares Xn

such that Xn Hausdorff converges to x. Now for x, y ∈ S, x 6= y,we define (cf. (1.3))

m(x, y) := inf minn ∈ N0 : Xn ∩ Y n = ∅,(4.13)

where the infimum is taken all such sequences Xn for x and Y nfor y. Then Euclidean distance of x and y can essentially be expressedin terms of m(x, y); namely,

(4.14) |x− y| 5−m(x,y),

where C = C() is independent of x and y. So up to a multiplica-tive constant, the Euclidean metric on S can be recovered from thecombinatorics of the sets Xn.

It is a small step from here to the theory of Gromov hyperbolicspaces. We construct a graph G as follows (see Chapter 10 for very


similar considerations). The set of vertices of G is the set of 1/5n-squares Xn for all n ∈ N0. It is convenient to add another vertex X−1.We declare that X−1 ⊃∼ X0 for any 1/50-square X0. Then each vertexof G as represented by Xn has an attached level n ∈ N0 ∪ −1.

The set of (undirected) edges of G is now given as follows. Weconnect two distinct vertices by an edge if they have the same level nand are represented by two 1/5n-squares Xn and Y n with Xn∩Y n 6= ∅.Moreover, we join two vertices represented by Xn and Xn+1 if Xn ⊃∼

Xn+1. There are no other edges in G.If we identify each edge with a copy of the unit interval [0, 1], then G

carries a natural path metric (corresponding to combinatorial distancein G on the set of vertices). It can be shown that with this path metricG is a Gromov hyperbolic metric space.

There is a natural identification of S with the boundary at infinity∂∞G. Namely, if x ∈ S, then we choose a sequence X0 ⊃∼ X1 ⊃∼ . . . of1/5n-squares such that Xn Hausdorff-converges to x. Then Xn,now considered as a sequence of vertices in G, converges at infinity(see Section 4.2). Sending a point x to the equivalence class of Xn(considered as a point on ∂∞G), we get a bijection between S and ∂∞Gthat we use to identify these two sets.

We choose the basepoint p = X−1 in G. Then for the Gromovproduct (x · y)p of two points x, y ∈ S = ∂∞G we have

m(x, y)− c ≤ (x · y)p ≤ m(x, y) + c,

where c ≥ 0 is a constant independent of x and y. From this and (4.14)it follows that the Euclidean metric on S = ∂∞G is a visual metric inthe sense of Gromov hyperbolic spaces; indeed, it satisfies (4.12) withΛ = 5. It easy to see that there are no visual metrics with Λ < 5.

The snowsphere S (equipped with the metric inherited from R3)is a quasisphere. This was shown in [Me02] (see also [Me10] and[Me09a]). For the proof one constructs a rational Thurston map (ina non-obvious way) leading to sets that mirror the combinatorics ofthe sets Xn related to S. One can use this to show directly that Sis a quasisphere, or one invokes the general criterion given by Theo-rem 17.1 ((ii)).

Example 4.7 (A non-quasisphere). Our second example is thesphere S equipped with the metric % as discussed in Section 1.3. As wealready remarked, it follows from the fact that the associated Thurs-ton map h has a Thurston obstruction in combination with Theo-rem 17.1 ((ii)) that S is not a quasiphere. Here we want to outline

4.4. QUASISPHERES 105

π

UZ ⊂ S

[0,1]×y

Figure 4.2. The set Z.

a direct argument for this statement (it emerged in discussions withB. Kleiner). We use the notation from Section 1.3.

We consider the top part of S given by all equivalence classes ofsequences X0 ⊃∼ X1 ⊃∼ . . . where X0 is the top white 0-tile of S0. Fromthis top part we remove all the “flaps” that were successively addedto X0 in the construction of S. What remains is a subset Z ⊂ S thatlooks like the unit square U = [0, 1]2 with countably many slits. Theseslits are all parallel to one of the sides of U , say to [0, 1] × 0. Eachpoint z ∈ Z corresponds to a unique point in U . This gives a surjectivemap π : Z → U . If a point p ∈ U is an interior point of one of the slits,then there are two points in Z (one for each side of the slit) that mapto p. For all other points p ∈ U we have #π−1(p) = 1.

We equip Z with the metric % and U with the Euclidean metric.Then the map π : Z → U is Lipschitz. Actually, π is David-Semmesregular. For π this means that in addition to being Lipschitz, thereexists a number N ∈ N such that the preimage π−1(B(p, r)) of eachball B(p, r) in U can be covered by N balls in Z of the same radius r.

At least on an intuitive level, one can see that π has this last prop-erty as follows. If no slit cuts through B(p, r), then π−1(B(p, r)) iscontained in a ball in Z whose radius is not much larger, and hencecomparable to r. If a slit cuts through B(p, r), then π−1(B(p, r)) issplit into two parts each of which is contained in a ball in Z with ra-dius comparable to r. This implies that in any case, π−1(B(p, r)) canbe covered by a controlled number of balls in Z with radius r.

Since U is Ahlfors 2-regular, and π is David-Semmes regular, Z isAhlfors 2-regular as well ([?]).

Let Γ be the set of all paths in Z that project under π to a linesegment that has the form [0, 1]×y, y ∈ [0, 1], and does not containa slit. Restricted to a path γ ∈ Γ, the map π is bi-Lipschitz with auniform constant independent of γ. This implies that the 2-modulusof Γ in Z (see [He, Sect. 7.3] for the definition of the modulus of a


path family) cannot be much smaller than than the 2-modulus of π(Γ)and is hence positive. Together with the Ahlfors 2-regularity of Z thisimplies that an image of Z under any quasisymmetric homeomorphismhas Hausdorff dimension ≥ 2 ([He, Thm. 15.10]).

One other property of Z will be important. Namely, Z is a poroussubset of S. This means that there exists a constant c ∈ (0, 1) withthe following property: if a ∈ Z and r > 0 with r ≤ diam(Z) arearbitrary, then there exists x ∈ B(a, r) with B(x, cr) ∩ Z = ∅. So theset B(a, r) ∩ Z has the “hole” B(x, cr) of comparable size.

The porosity of Z follows from the fact that each ball B(a, r) inS with r ≤ diam(S) contains a flap of size comparable to r. Since inthe construction of Z we removed all flaps from the top side of S, thismeans that all sufficiently small balls centered in Z contain a hole ofabout the same size.

Now we can see that S is not a quasisphere as follows. We argue bycontradiction and assume that there exists a quasisymmetry ϕ of S onto

C (equipped with the chordal metric). From what we have discussedabove, it then follows that the Hausdorff dimension of ϕ(Z) is ≥ 2.On the other hand, images of porous sets under quasisymmetries are

porous. Hence ϕ(Z) is a porous subset of C ([?]). A porous subsetof an Ahlfors Q-regular space, Q > 0, has Hausdorff dimension < Q.It follows that the Hausdorff dimension of ϕ(Z) is < 2. This is acontradiction, and so S cannot be a quasisphere.

CHAPTER 5

Cell decompositions

In this chapter we discuss some technical, but very crucial aspects of ourwork, namely cell decompositions and their relation to Thurston maps.We start with a general review of cell decompositions for arbitraryspaces (Section 5.1) and then specialize to cell decompositions of 2-spheres (Section 5.2).

If f : S2 → S2 is a Thurston map and C ⊂ S2 is a Jordan curvewith post(f) ⊂ C, then one gets an associated sequence of cell decom-positions Dn = Dn(f, C) of the underlying 2-sphere (see Section 5.3).This was already mentioned in the introduction and is of prime impor-tance for our approach to the investigation of Thurston maps. Propo-sition 5.17 summarizes the most important properties of these cell de-compositions Dn. One can turn this around and use cell decompositionto construct Thurston maps. Sections 5.4 and 5.5 are relevant for this.

Section 5.6 is devoted to flowers—a concept related to the cell de-composition Dn and very useful for dealing with various finer points.In the final Section ?? we introduce a precise definition for a set tojoin opposite sides of a Jordan curve C with post(f) ⊂ C and discussvarious related facts.

At first reading the reader may want to skim through most of thesections to pick up relevant definitions and statements.

5.1. Cell decompositions in general

Here we review some facts about cell decompositions of arbitrary spaces.Most of this material is fairly standard. For the purpose of the presentwork we could have restricted ourselves to cell decompositions of sub-sets of a 2-sphere, but it is more transparent to discuss the topic ingreater generality.

In this section X will always be a locally compact Hausdorff space.A (closed topological) cell c of dimension n = dim(c) ∈ N in X is aset c ⊂ X that is homeomorphic to the closed unit ball Bn in Rn. Wedenote by ∂c the set of points corresponding to ∂Bn under such a home-omorphism between c and Bn. This is independent of the homeomor-phism chosen, and the set ∂c is well-defined. We call ∂c the boundary

107

108 5. CELL DECOMPOSITIONS

and int(c) = c \ ∂c the interior of c. Note that boundary and interiorof c in this sense will in general not agree with the boundary and in-terior of c regarded as a subset of the topological space X . A cell ofdimension 0 in X is a set c ⊂ X consisting of a single point. We set∂c = ∅ and int(c) = c in this case.

Definition 5.1 (Cell decompositions). Suppose that D is a collec-tion of cells in a locally compact Hausdorff space X . We say that D is acell decomposition of X provided the following conditions are satisfied:

(i) the union of all cells in D is equal to X ,

(ii) we have int(σ) ∩ int(τ) = ∅, whenever σ, τ ∈ D, σ 6= τ ,

(iii) if τ ∈ D, then ∂τ is a union of cells in D,

(iv) every point in X has a neighborhood that meets only finitelymany cells in D.

If D is a collection of cells in some ambient space X , then we callD a cell complex if D is a cell decomposition of the underlying set

|D| :=⋃c : c ∈ D.

Suppose D is a cell decomposition of X . By (iv), every compactsubset of X can only meet finitely many cells in D. In particular, if Xis compact, then D consists of only finitely many cells. Moreover, foreach τ ∈ D, the set ∂τ is compact and hence equal to a finite union ofcells in D. It follows from basic dimension theory that if dim(τ) = n,then ∂τ is equal to a union of cells in D that have dimension n− 1.

The union X n of all cells in D of dimension ≤ n is called the n-skeleton of the cell decomposition. It is useful to set X−1 = ∅. Itfollows from the local compactness of X and property (iv) of a celldecomposition that X n is a closed subset of X for each n ∈ N0. By thelast remark in the previous paragraph, we have ∂τ ⊂ X n−1 for eachτ ∈ D with dim(τ) = n.

Lemma 5.2. Let D be a cell decomposition of X . Then for eachn ∈ N0 the n-skeleton X n is equal to the disjoint union of the setsint(c), c ∈ D, dim(c) ≤ n. The sapce X is equal to the disjoint unionof the sets int(c), c ∈ D. Similarly, every cell τ ∈ D is the disjointunion of the sets int(c), where c ∈ D and c ⊂ τ .

So in particular, the interiors of the cells in a cell decompositionpartition the space X . This is of prime importance and will be usedfrequently throughout this work.

5.1. CELL DECOMPOSITIONS IN GENERAL 109

Proof. We show the first statement by induction on n ∈ N0. Sinceint(c) = c for each cell c in D of dimension 0, it is clear that X 0 is thedisjoint union of the interiors of all cells c ∈ D with dim(c) = 0.

Suppose that the first statement is true for X n, and let p ∈ X n+1

be arbitrary. If p ∈ X n, then p is contained in the interior of a cellc ∈ D with dim(c) ≤ n by induction hypothesis. In the other case,p ∈ X n+1 \ X n, and so there exists c ∈ D with dim(c) = n + 1 andp ∈ c. Since ∂c ⊂ X n, it follows that p ∈ c\∂c = int(c). So X n+1 is theunion of the interiors of all cells c in D with dim(c) ≤ n+1. This unionis disjoint, because distinct cells in a cell decomposition have disjointinteriors.

The second statement follows from the first, and the obvious factthat X =

⋃n∈N0X n.

To see the last statement, let Dτ := c ∈ D : c ⊂ τ. Then it isclear that Dτ is a cell decomposition of (the compact Hausdorff space)τ . So the claim follows from the previous statement.

The lemma immediately implies that if τ ∈ D and dim(τ) = n,then each point p ∈ int(τ) is an interior point of τ regarded as a subsetof the topological space X n.

Lemma 5.3. Let D be a cell decomposition of X .

(i) If σ and τ are two distinct cells in D with σ ∩ τ 6= ∅, thenone of the following statements holds: σ ⊂ ∂τ , τ ⊂ ∂σ, orσ ∩ τ = ∂σ ∩ ∂τ and this intersection consists of cells in D ofdimension strictly less than mindim(σ), dim(τ).

(ii) If σ, τ1, . . . , τn are cells in D and int(σ) ∩ (τ1 ∪ · · · ∪ τn) 6= ∅,then σ ⊂ τi for some i ∈ 1, . . . , n.

Proof. (i) We may assume that l = dim(σ) ≤ m = dim(τ), andprove the statement by induction on m. The case m = 0 is vacuous andhence trivial. Assume that the statement is true whenever both cellshave dimension < m. If l = m then by definition of a cell decompositionint(σ) is disjoint from τ ⊂ int(τ)∪Xm−1, and similarly int(τ)∩ σ = ∅.Hence σ ∩ τ = ∂τ ∩ ∂σ. Moreover, both sets ∂σ and ∂τ consist offinitely many cells in D of dimension ≤ m− 1. Applying the inductionhypothesis to pairs of these cells, we see that ∂τ ∩ ∂σ consists of cellsof dimension < m as desired.

If l < m, then σ ⊂ Xm−1 and so σ ∩ int(τ) = ∅. This shows thatσ ∩ τ = σ ∩ ∂τ . Moreover, we have ∂τ = c1 ∪ · · · ∪ cs, where c1, . . . , csare cells of dimension m−1. So we can apply the induction hypothesisto the pairs (σ, ci). If σ = ci or σ ⊂ ∂ci for some i, then σ ⊂ ∂τ ; wecannot have ci ⊂ ∂σ, because ci has dimension m− 1, and ∂σ is a set


of topological dimension < m − 1. So if none of the first possibilitiesoccurs, then σ∩ ci = ∅ or σ∩ ci = ∂σ∩∂ci and this set consists of cellsof dimension < l (by induction hypothesis) contained in ∂ci ⊂ ci ⊂ ∂τfor all i. In this case σ ∩ τ = ∂σ ∩ ∂τ , and this sets consists of cells ofdimension < l as desired. The claim follows.

(ii) There exists i ∈ 1, . . . , n with int(σ) ∩ τi 6= ∅. By the alter-natives in (i) we then must have σ = τi or σ ⊂ ∂τi. Hence σ ⊂ τi.

Lemma 5.4. Let A ⊂ X be a closed set, and U ⊂ X \ A be anonempty open and connected set. If ∂U ⊂ A, then U is a connectedcomponent of X \ A.

Proof. Since U is a nonempty connected set in the complement ofA, this set is contained in a unique connected component V of X \ A.Since ∂U ⊂ A ⊂ X \ V , we have V ∩ U = V ∩ U = U showing that Uis relatively open and closed in V . Since U 6= ∅ and V is connected, itfollows that U = V as desired.

Lemma 5.5. Let D be a cell decomposition of X with n-skeletonX n, n ∈ −1 ∪ N0. Then for each n ∈ N0 the nonempty connectedcomponents of X n \ X n−1 are precisely the sets int(τ), where τ ∈ Dand dim(τ) = n.

Proof. Let τ be a cell in D with dim(τ) = n. Then int(τ) is aconnected set contained in X n\X n−1 that is relatively open with respectto X n. Its relative boundary is a subset of ∂τ and hence contained inthe closed set X n−1. It follows by Lemma 5.4 that int(τ) is equal to acomponent V of X n \ X n−1.

Conversely, suppose that V is a nonempty connected component ofX n \ X n−1. Pick a point p ∈ V . Then p lies in the interior of a uniquecell τ ∈ D with dim(τ) = n. It follows from the first part of the proofthat V = int(τ).

Definition 5.6 (Refinements). Let D′ and D be two cell decom-position of the space X . We say that D′ is a refinement of D if thefollowing two conditions are satisfied:

(i) For every cell σ ∈ D′ there exits a cell τ ∈ D with σ ⊂ τ .

(ii) Every cell τ ∈ D is the union of all cells σ ∈ D′ with σ ⊂ τ .

It is easy to see that if D′ is a refinement of D and τ ∈ D, thenthe cells σ ∈ D′ with σ ⊂ τ form a cell decomposition of τ . Moreover,every cell in D′ arises in this way. So roughly speaking, the refinementD′ of the cell decomposition D is obtained by decomposing each cell inD into smaller cells. We informally refer to this process as subdividingthe cells in D by the smaller cells in D′.


Lemma 5.7. Let D′ and D be two cell decompositions of X , and D′be a refinement of D. Then for every cell σ ∈ D′ there exists a minimalcell τ ∈ D with σ ⊂ τ , i.e., if τ ∈ D is another cell with σ ⊂ τ , thenτ ⊂ τ . Moreover, τ is the unique cell in D with int(σ) ⊂ int(τ).

Proof. First note that if σ ∈ D′, τ1, . . . , τn ∈ D and

int(σ) ∩ (τ1 ∪ · · · ∪ τn) 6= ∅,then σ ⊂ τi for some i ∈ 1, . . . , n. Indeed, by definition of a refine-ment the union of all cells in D′ contained in some τi covers τ1∪· · ·∪τn.Hence this union meets int(σ). It follows from Lemma 5.3 (ii) that σis contained in one of these cells from D′ and hence in one of the cellsτi.

Now if σ ∈ D′ is arbitrary, then σ is contained in some cell of D bydefinition of a refinement, and hence in a cell τ ∈ D of minimal dimen-sion. Then τ is minimal among all cells in D containing σ. Indeed, letτ 6= τ be another cell in D containing σ. We want to show that τ ⊂ τ .

One of the alternatives in Lemma 5.3 (i) occurs. If τ ⊂ ∂τ ⊂ τ weare done. The second alternative, τ ⊂ ∂τ , is impossible, since τ hasminimal dimension among all cells containing σ. The third alternativeleads to leads to σ ⊂ τ ∩ τ = ∂τ ∩ ∂τ , where the latter intersectionconsists of cells in D of dimension < dim(τ). By the first part ofthe proof σ is contained in one of these cells, again contradicting thedefinition of τ . Hence τ is minimal.

We have int(σ) ⊂ int(τ); for otherwise int(σ) meets ∂τ which is aunion of cells in D. Then σ would be contained in one of these cells bythe first part of the proof. This contradicts the minimality of τ .

Finally, it is clear that τ ∈ D is the unique cell with int(σ) ⊂ int(τ),because distinct cells in a cell decomposition have disjoint interiors.

Definition 5.8 (Cellular maps and cellular Markov partitions).Let D′ and D be two cell decompositions of X , and f : X → X be acontinuous map. We say that f is cellular for (D′,D) if the followingcondition is satisfied:

If σ ∈ D′ is arbitrary, then f(σ) is a cell in D and f |σ is ahomeomorphism of σ onto f(σ).

If f is cellular with respect to (D′,D) and D′ is a refinement of D,then the pair (D′,D) is called a cellular Markov partition for f .

Lemma 5.9. Let D′ and D be cell decompositions of X , and f : X →X be a continuous map that is cellular for (D′,D). Suppose that σ′ ∈D′, τ ∈ D, and τ ⊂ f(σ′).

Then there exists τ ′ ∈ D′ with τ ′ ⊂ σ′ and f(τ ′) = τ .


Proof. Note that f |σ′ is a homeomorphism of σ′ onto σ = f(σ′).Pick a point q ∈ int(τ). Then there exists a point p ∈ σ′ with f(p) = q,and a cell τ ′ ∈ D′ with p ∈ int(τ ′). Then int(τ ′) meets σ′ and soτ ′ ⊂ σ′ (Lemma 5.3 (ii)). Moreover, f(τ ′) is a cell in D with q =f(p) ∈ int(f(τ ′)) ∩ int(τ). It follows that f(τ ′) = τ .

The following proposition shows that a cellular Markov partitionfor a map generates a whole sequence of cell decompositions of theunderlying space. This is the key for our combinatorial approach forinvestigating expanding Thurston maps.

Proposition 5.10. Let D′ and D be two cell decompositions of X ,and f : X → X be a continuous map. If (D′,D) is a cellular Markovpartition for f , then there exist unique cell decompositions Dn of X forn ∈ N0 such that

(i) D0 = D, D1 = D′, and Dn+1 is a refinement of Dn for n ∈ N0,

(ii) each pair (Dn+1,Dn), n ∈ N0, is a cellular Markov partitionfor f .

Note that this implies that Dn+k is a refinement of Dn, and fk iscellular with respect to (Dn+k,Dn) for all n, k ∈ N0. So (Dn+k,Dn) isa cellular Markov partition for fk.

Proof. The cell decompositions Dn are constructed inductively.Let D0 = D and D1 = D′.

The idea for constructing the refinement D2 of D1 is very simple:we want to decompose a cell τ ∈ D1 into cells in a similar way, as thecell f(τ) ∈ D0 is decomposed by the cells σ ⊂ f(τ) in D1. Accordingly,we define the set D2 as

D2 = (f |τ)−1(σ) : σ, τ ∈ D1 and σ ⊂ f(τ).Then D2 consists of cells. Indeed, if σ, τ ∈ D1 and σ ⊂ f(τ), then

f |τ is a homeomorphism of τ onto f(τ) ∈ D0. Hence the preimageλ := (f |τ)−1(σ) of the cell σ under this homeomorphism is a cell. Notethat the cell σ = f(λ) is uniquely determined by λ, but τ in general isnot. The map f |λ is a homeomorphism of λ onto σ = f(λ).

We now show that D2 is a cell decomposition of X by verifying theconditions (i)–(iv) of Definition 5.1.

Condition (i): Let x ∈ X be arbitrary. Then there exists τ ∈ D1

with x ∈ τ . The set f(τ) is a cell in D0. Since D1 is a refinement ofD0, there exists a cell σ ∈ D1 with f(x) ∈ σ ⊂ f(τ). Then (f |τ)−1(σ)is a cell in D2 that contains x. It follows that the cells in D2 cover X .

Condition (ii): Let λ1, λ2 ∈ D2 be arbitrary, and assume thatint(λ1) ∩ int(λ2) 6= ∅. We have to show that λ1 = λ2. We have


int(f(λi)) = f(int(λi)) for i = 1, 2. So f(λ1) and f(λ2) are cells inD1 with a common interior point. Hence σ := f(λ1) = f(λ2) ∈ D1.

By definition of D2 there exist cells τi ∈ D1 with σ ⊂ f(τi) andλi = (f |τi)−1(σ) for i = 1, 2. Let σ0 be the minimal cell in D0 thatcontains σ. Then σ ⊂ σ0 ⊂ f(τ1) ∩ f(τ2). By Lemma 5.9 there existcells σi ∈ D1 with σi ⊂ τi and f(σi) = σ0 for i = 1, 2.

By Lemma 5.7 we have int(σ) ⊂ int(σ0). Applying the homeomor-phism (f |τi)−1 to both sets in this inclusion, we obtain int(λi) ⊂ int(σi)for i = 1, 2. It follows that σ1 and σ2 are cells in D1 with a commoninterior point. Hence σ1 = σ2, and so

λ1 = (f |τ1)−1(σ) = (f |σ1)−1(σ) = (f |σ2)−1(σ) = (f |τ2)−1(σ) = λ2

as desired.

Condition (iii): Let λ ∈ D2 and x ∈ ∂λ be arbitrary. Then thereexist cells σ, τ ∈ D1 with σ ⊂ f(τ) and λ = (f |τ)−1(σ). Moreover,σ = f(λ) and so f(x) ∈ ∂σ. By definition of a cell decomposition there

exists a cell σ ∈ D1 with f(x) ⊂ σ ⊂ ∂σ. Then λ = (f |τ)−1(σ) is a cell

in D2 with x ∈ λ ⊂ ∂λ. It follows that ∂λ is a union of cells in D2.

We have verified conditions (i)–(iii) in the definition of a cell de-composition. Before we prove the last condition (iv), we will first showthat D2 has the required properties of a refinement.

Indeed, if λ ∈ D2 and σ, τ ∈ D1 are such that σ ⊂ f(τ) andλ = (f |τ)−1(σ), then

λ ⊂ (f |τ)−1(f(τ)) = τ.

So every cell in D2 is contained in a cell in D1.Moreover, let τ ∈ D1 and x ∈ τ be arbitrary. Then f(τ) is a cell

in D0 containing f(x). Since D1 is a refinement of D0 there existsσ ∈ D1 with f(x) ∈ σ ⊂ f(τ). Then λ = (f |τ)−1(σ) is a cell in D2

with x ∈ λ ⊂ τ . It follows that every cell in D1 is a union of cells inD2.

Moreover, this is a finite union. Indeed, if τ ∈ D1, then f |τ inducesa bijection of the cells in D2 contained in τ and the cells in D1 containedin f(τ) ∈ D0. Since the latter set is finite, the former is finite as well.

Condition (iv): This can now easily be established. If p ∈ X isarbitrary, then there exists a neighborhood U of p that meets onlyfinitely many cells in D1. Every cell in D2 that meets U must becontained in one of these finitely many cells from D1. Since every cellin D1 contains only finitely many cells in D2, it follows that U meetsonly finitely many cells in D2.


We have proved that D2 is cell decomposition of X that is a refine-ment of D1. It immediately follows from the definition of D2 that fis cellular with respect to (D2,D1). Therefore, (D2,D1) is a cellularMarkov partition for f .

To show uniqueness of D2 suppose that D2 is another cell decom-

position of X such that (D2,D1) is a cellular Markov partition for f .

Then D2 ⊂ D2. Indeed, let λ ∈ D2 be arbitrary. Since D2 is arefinement of D1, there exists a cell τ ∈ D1 with λ ⊂ τ . Moreover,σ = f(λ) is a cell in D1 and σ ⊂ f(τ). Since f |τ is a homeomorphismof τ onto f(τ), it follows that

λ = (f |τ)−1(σ) ∈ D2.

If the inclusion D2 ⊂ D2 were strict, then there would be a cell inD2 whose interior would be disjoint from the interior of all the cells in

D2. This is impossible, because D is a cell decomposition of X and so

the interiors of the cells in D form a cover of X . So D2 = D2.We have shown the existence and uniqueness of a cell decomposi-

tion D2 of X with the desired properties. Now D3 is constructed from(D2,D1) in the same way as D2 was constructed from (D1,D0). Con-tinuing in this manner we get the desired existence and uniqueness ofthe cell decompositions Dn for n ∈ N0.

Remark 5.11. The main idea of the previous proof can be sum-marized by saying that if the cell decompositions Dn and Dn−1 havealready been defined for some n ∈ N, then one obtains the elements inDn+1 by subdividing the cells τ ∈ Dn in the same way as the imagesf(τ) in Dn−1 are subdivided by the cells in Dn.

From this description it is clear that the “combinatorics” of the cellsin the sequence Dn, n ∈ N0, that is, their inclusion and intersectionpattern, is determined by the pair (D1,D0) and by the assignment τ ∈D1 7→ f(τ) ∈ D0. Such an assignment of a cell in D0 to each cell in D1

is related to the concept of a “labeling” (see Definition 5.20). So for thecombinatorics of the decompositions Dn the only relevant informationon the map f is its induced “labeling” τ ∈ D1 7→ f(τ) ∈ D0. It is nothard, but somewhat tedious, to formulate a precise statement based ona suitable notion of “combinatorial equivalence” for such sequences ofcell decompositions (see the related Definition 5.13 where we define thenotion of an isomorphism between cell complexes). We will not do this,because it would not add anything of substance, but content ourselveswith the intuitive statement that the “combinatorics” of the sequenceDn, n ∈ N0, is determined by the pair (D1,D0), and the assignmentτ ∈ D1 7→ f(τ) ∈ D0.

5.2. CELL DECOMPOSITIONS OF 2-SPHERES 115

5.2. Cell decompositions of 2-spheres

We now turn to cell decompositions of 2-spheres. We first review somestandard concepts and results from plane topology (see [Mo] for generalbackground and more details).

Let S2 be a 2-sphere. An arc α in S2 is a homeomorphic imageof the unit interval [0, 1]. The points corresponding to 0 and 1 undersuch a homeomorphism are called the endpoints of α. They are theunique points p ∈ α such that α \ p is connected. If p is an interiorpoint of α, i.e., a point in α distinct from the endpoints, then thereexist arbitrarily small open neighborhoods W of p such that W \α hasprecisely two open connected components U and V .

A closed Jordan region X in S2 is a homeomorphic image of theclosed unit disk D. The boundary ∂X of a closed Jordan region X ⊂ S2

is a Jordan curve, i.e., the homeomorphic image of the unit circle ∂D. IfJ ⊂ S2 is a Jordan curve, then by the Schonflies Theorem there exists

a homeomorphism ϕ : S2 → C such that ϕ(J) = ∂D. In particular, theset S2 \ J has two connected component, both homeomorphic to D.Note that arcs and closed Jordan regions are cells of dimension 1 and2, respectively.

LetD be a cell decomposition of S2. Since the topological dimensionof S2 is equal to 2, no cell in D can have dimension > 2. We call the2-dimensional cells in D the tiles, and the 1-dimensional cells in D theedges of D. The vertices of D are the points v ∈ S2 such that v isa cell in D of dimension 0. So there is a somewhat subtle distinctionbetween vertices and cells of dimension 0: a vertex is an element of S2,while a cell of dimension 0 is a subset of S2 with one element.

If c is a cell in D, we denote by ∂c the boundary and by int(c) theinterior of c as introduced in the beginning of Section 5.1. Note that foredges and 0-cells c this is different from the boundary and the interiorof c as a subset of the topological space S2.

We now summarize some facts related to orientation. See Sec-tion A.2 for a more detailed discussion.

We always assume that the sphere S2 is oriented, i.e., one of thetwo generators of the singular homology group H2(S2) ∼= Z (with co-efficients in Z) has been chosen as the fundamental class of S2.

The orientation on S2 induces an orientation on every Jordan regionX ⊂ S2 which in turn induces an orientation on ∂X and on every arcα ⊂ ∂X. On a more intuitive level, an orientation of an arc is justa selection of one of the endpoints as the initial point and the otherendpoint as the terminal point. Let X ⊂ S2 be a Jordan region in theoriented 2-sphere S2 equipped with the induced orientation. If α ⊂ ∂X


is an arc with a given orientation, then we say that X lies to the leftor to the right of α depending on whether the orientation on α inducedby the orientation of X agrees with the given orientation on α or not.Similarly, we say that with a given orientation of ∂X the Jordan regionX lies to the left or right of ∂X.

To describe orientations, one can also use the notion of a flag. Bydefinition a flag in S2 is a triple (c0, c1, c2), where ci is an i-dimensionalcell for i = 0, 1, 2, c0 ⊂ ∂c1, and c1 ⊂ ∂c2. So a flag in S2 is a closedJordan region c2 with an arc c1 contained in its boundary, where thepoint in c0 is an distinguished endpoint of c1. We orient the arc c1 sothat the point in c0 is the initial point in c1. The flag is called positively-or negatively-oriented (for the given orientation on S2) depending onwhether c2 lies to the left or to the right of the oriented arc c1.

A positively-oriented flag determines the orientation on S2 uniquely.

The standard orientation on C is the one for which the standard flag(c0, c1, c2) is positively-oriented, where c0 = 0, c1 = [0, 1] ⊂ R, and

c2 = z ∈ C : 0 ≤ Re(z) ≤ 1, 0 ≤ Im(z) ≤ Re(z).

Since edges and tiles in a cell decomposition D of S2 are arcs andclosed Jordan regions, respectively, it makes sense to speak of orientededges and tiles in D. A flag in D is a flag (c0, c1, c2), where c0, c1, c2

are cells in D. If ci are i-dimensional cells in D for i = 0, 1, 2, then(c0, c1, c2) is a flag in D if and only if c0 ⊂ c1 ⊂ c2.

Cell decompositions of a 2-sphere S2 have additional properties thatwe summarize in the next lemma.

Lemma 5.12. Let D be a cell decomposition of S2. Then it has thefollowing properties:

(i) There are only finitely many cells in D.

(ii) The tiles in D cover S2.

(iii) Let X be a tile in D. Then there exists a number k ∈ N, k ≥ 2,such that X contains precisely k edges e1, . . . , ek and k verticesv1, . . . , vk in D. Moreover, these edges and vertices lie on theboundary ∂X of X, and we have

∂X = e1 ∪ · · · ∪ ek.

The indexing of these vertices and edges can be chosen suchthat vj ∈ ∂ej ∩ ∂ej+1 for j = 1, . . . , k (where ek+1 := e1).

(iv) Every edge e ∈ D is contained in the boundary of precisely twotiles D. If X and Y are these tiles, then int(X)∪int(e)∪int(Y )is a simply connected region.


X1

X2

X3

X4

X5

X6

e1

e2

e3

e4

e5

e6

v

Figure 5.1. The cycle of a vertex v.

(v) Let v be a vertex of D. Then there exists a number d ∈ N,d ≥ 2, such that v is contained in precisely d tiles X1, . . . , Xd,and d edges e1, . . . , ed in D. We have v ∈ ∂Xj and v ∈ ∂ej foreach j = 1, . . . , d. Moreover, the indexing of these tiles andedges can be chosen such that ej ⊂ ∂Xj∩∂Xj+1 for j = 1, . . . , d(where Xd+1 := X1).

(vi) The 1-skeleton of D is connected and equal to the union of alledges in D.

Statement (iii) actually holds for all tiles (i.e., 2-dimensional cells)in each cell decomposition of a locally compact space. If the boundaryof a tile X is subdivided into vertices and edges as in (iii), we say thatX is a (topological) k-gon.

If the edge e and the tiles X and Y are as in (iv), then there existsa unique orientation of e such that X lies to the left and Y to the rightof e.

We say that the cells v, e1, . . . , ed, X1, . . . , Xd as in (v) form thecycle of the vertex v and call d the length of the cycle. We refer toX1, . . . , Xd as the tiles and to e1, . . . , ed as the edges of the cycle.

Proof. (i) By Definition 5.1 (iv) this follows from the compactnessof S2 and the fact that every point in S2 has a neighborhood that meetsonly finitely many cells in D.


(ii) The set consisting of all vertices and the union of all edges hasempty interior (in the topological sense) by (i) and Baire’s theorem.Hence the union of all tiles is a dense set in S2. Since this union is alsoclosed by (i), it is all of S2.

(iii) Let X be a tile in D. Then int(X) does not meet any edgeor vertex, and ∂X is a union of edges and vertices. Since there areonly finitely many vertices, ∂X must contain an edge, and hence atleast two vertices. Suppose v1, . . . , vk, k ≥ 2, are all the vertices on∂X. Since ∂X is a Jordan curve, we can choose the indexing of thesevertices so that ∂X is a union of arcs αj with pairwise disjoint interiorsuch that αj has the endpoints vj and vj+1 for j = 1, . . . , k, wherevk+1 = v1. Then for each j = 1, . . . , k the set int(αj) is connected andlies in the 1-skeleton of the cell decomposition D, it is disjoint fromthe 0-skeleton and has boundary contained in the 0-skeleton. It followsfrom Lemma 5.4 and Lemma 5.5 that there exists an edge ej in D withint(ej) = int(αj). Hence αj = ej, and so αj is an edge in D. It is clearthat ∂X does not contain other edges in D. The statement follows.

(iv) Let e be an edge in D. Pick p ∈ int(e). By (ii) the point p iscontained in some tile X in D. By Lemma 5.3 we have e ⊂ X. Onthe other hand, int(X) is disjoint from each edge and so e ⊂ ∂X. Itfollows from the Schonflies theorem that the set X does not containa neighborhood of p. Hence every neighborhood of p must meet tilesdistinct from X. Since there are only finitely many tiles, it followsthat there exists a tile Y distinct from X with p ∈ Y . By the samereasoning as before, we have e ⊂ ∂Y .

Let q ∈ int(e) be arbitrary. Then there exists a small open andconnected neighborhood W of q such that W \ int(e) consists of twoconnected components U and V . If W is small enough, then U and Vdo not meet ∂X. Since q ∈ int(X), one of the sets, say U , meets int(X),and so U ⊂ int(X). We can also assume that the set W is small enoughso that it does not meet ∂Y either. By the same reasoning, U or Vmust be contained in int(Y ), and, since int(X) ∩ int(Y ) = ∅, we haveV ⊂ int(Y ). Hence M := int(X) ∪ int(e) ∪ int(Y ) is a neighborhoodof each point in int(e) which implies that this set is open. Since thesets int(X), int(e), int(Y ) are connected, and each of them meets theconnected set W , it follows that their union M is also connected. SoM is a region. To see that M is simply connected, first note that theSchonflies theorem implies there exists a homotopy on int(X) ∪ int(e)that deforms this set into int(e) and keeps the points in int(e) fixedduring the homotopy. If we combine this homotopy with a similar


homotopy on int(Y ) ∪ int(e), then we see that M is homotopic toint(e), and hence also to a point. So M is simply connected.

Suppose that Z is another tile in D with e ⊂ ∂Z. Since X ∪ Ycontains an open neighborhood for p, there exists a point x ∈ int(Z)near p with x ∈ X∪Y , say x ∈ X. Since the interior of a tile is disjointfrom all other cells, we conclude X = Z. This shows the uniqueness ofX and Y .

(v) Let v be a vertex of D. If an edge e in D contains v, then v is anendpoint of e and we orient e so that v is the initial point of e. By (ii)there exists a tile X1 in D with v ∈ X1. Then v ∈ ∂X1, and so by (iii)there exist two edges in ∂X1 that contain v. For one of these orientededges, denote it by e1, the tile X1 will lie on the right of e1. Thenv ∈ e1 ⊂ ∂X1 and X1 will lie on the left of the other oriented edge. By(iv) there exists a unique tile X2 6= X1 with e1 ⊂ ∂X2. Then X2 will lieon the left of e1. By (iii) there exists a unique edge e2 ⊂ ∂X2 distinctfrom e1 with v ∈ e2. The tile X2 will lie on the right of e2. We cancontinue in this manner to obtain tiles X1, X2, . . . and edges e1, e2, . . .that contain v and satisfy Xj 6= Xj+1, ej 6= ej+1, and ej ⊂ ∂Xj∩∂Xj+1

for all j ∈ N. Moreover, Xj will lie on the right and Xj+1 on theleft of the oriented edge ej. Since there are only finitely many tiles,there exists a smallest number d ∈ N such that the tiles X1, . . . , Xd

are all distinct and Xd+1 is equal to one of the tiles X1, . . . , Xd. SinceX1 6= X2, we have d ≥ 2.

Moreover, Xd+1 = X1. To see this we argue by contradiction andassume that Xd+1 is equal to one of the tiles X2, . . . , Xd say Xd+1 = Xj.Note that Xd 6= Xd+1, so 2 ≤ j ≤ d − 1. Then e = ed is an edge withv ∈ e that is contained in ∂Xd and in ∂Xd+1 = ∂Xj. Hence e = ej−1

or e = ej. Since Xd+1 = Xj lies on the left of e = ed, we must havee = ej−1. Then e is contained in the boundary of the three distincttiles Xj−1, Xj, Xd which is impossible by (iv). So indeed Xd+1 = X1.

By a similar reasoning we can show that the edges e1, . . . , ed are alldistinct. Indeed, suppose e = ej = ek, where 1 ≤ j < k ≤ d. Thenk > j + 1 and e is contained in the boundary of the three distinct tilesXj, Xj+1, Xk which is again absurd.

To show that there are no other edges and tiles containing v notethat by (iii) the set

U = int(X1) ∪ int(e1) ∪ int(X2) ∪ · · · ∪ int(ed) ∪ int(Xd+1)


is open. Moreover, its boundary ∂U consists of the point v and a closedset

A ⊂d⋃j=1

∂Xj

disjoint from v. Hence v is an isolated boundary point of U whichimplies that W = U ∪ v is an open neighborhood of v.

If c is an arbitrary cell in D with v ∈ c and c 6= v, then v ∈ int(c).This implies that int(c) meets U . Since interiors of distinct cells inD are disjoint, this is only possible if c is equal to one of the edgese1, . . . , ed or one of the tiles X1, . . . , Xd. The statement follows.

(vi) By (v) every vertex is contained in an edge. Hence the 1-skeleton E of D is equal to the union of all edges in D. To show thatE is connected, let x, y ∈ E be arbitrary. Since the tiles in D cover S2,there exist tiles X and Y with x ∈ X and y ∈ Y . The interior of eachtile is disjoint from the 1-skeleton E, and so x ∈ ∂X and y ∈ ∂Y . SinceS2 is connected, there exist tiles X1, . . . , XN in D such that X1 = X,XN = Y , and Xi ∩ Xi+1 6= ∅ for i = 1, . . . , N − 1. The interior of atile meets no other tile. Hence ∂Xi ∩ ∂Xi+1 6= ∅ for i = 1, . . . , N − 1.Since each set ∂Xi is connected, it follows that

K = ∂X1 ∪ · · · ∪ ∂XN

is a connected subset of E containing x and y. Hence E is connected.

Let d ∈ N, d ≥ 2, and the tiles Xj and edges ej for j ∈ N be asdefined in the proof of statement (v) of the previous lemma. Then weshowed that Xd+1 = X1, but it is useful to point out that actuallyXj = Xj+d and ej = ej+d for all j ∈ N.

Indeed we have seen that Xd+1 = X1. Moreover, e1, ed, ed+1 areedges in D that contain v and are contained in the boundary of thetile X1 = Xd+1. Since there are only two such edges, e1 6= ed, anded 6= ed+1, we conclude that ed+1 = e1. Then e1 = ed+1 is an edgecontained in the boundary of the tiles X1, X2, Xd+2. Since there areprecisely two tiles containing an edge in its boundary, X1 6= X2 andX1 = Xd+1 6= Xd+2 it follows that Xd+2 = X2.

If we continue in this manner, shifting all indices by 1 in each step,we see that ed+2 = e2, Xd+3 = X3, etc., as claimed.

Note that if we choose the indexing of the edges ej and Xj as in theproof of statement (v) of the previous lemma, then for each j ∈ N theflag (v, ej, Xj+1) in D is positively-oriented, and flag (v, ej, Xj) isnegatively-oriented. In other words, if ej is oriented so that v is the


initial point of ej, then Xj+1 lies to the left and Xj lies to the right ofej.

We will now discuss how cell decompositions can be used to definehomeomorphisms and isotopies of the underlying spaces. We start witha general definition.

Definition 5.13 (Isomorphisms of cell complexes). Let D and Dbe cell complexes. A bijection φ : D → D is called an isomorphism (ofcell complexes) if the following conditions are satisfied:

(i) dim(φ(τ)) = dim(τ) for all τ ∈ D.

(ii) if σ, τ ∈ D, then σ ⊂ τ if and only if φ(σ) ⊂ φ(τ).

Let h : X → X be a homeomorphism between two locally compact

Hausdorff spaces X and X , and suppose D is a cell decomposition

of X . Then it is easy to see that D := h(c) ⊂ X : c ∈ D is a

cell decomposition of X and φ : D → D given by φ(c) = h(c) for allc ∈ D is an isomorphism. We will see that this procedure can bereversed and one can construct a homeomorphism from a given cellcomplex isomorphism. For simplicity we will restrict ourselves to thecase of 2-spheres. As a preparation for the proof of the correspondingLemma 5.14, we first record some facts about homeomorphisms andisotopies on subsets of 2-spheres.

If α is an arc, then every homeomorphism ϕ : α → α that fixesthe endpoints of α is isotopic to the identity rel. ∂α. Indeed, we mayassume that α is equal to the unit interval I = [0, 1]. Then ϕ(0) = 0,ϕ(1) = 1, and ϕ is strictly increasing on [0, 1]. Define H : I × I → I by

H(s, t) = (1− t)ϕ(s) + ts

for s, t ∈ I. Then for each t ∈ I, the map Ht = H(·, t) is strictlyincreasing on I. It follows that H is an isotopy. We have Ht(0) = 0and Ht(1) = 1 for all t ∈ I, and H0 = ϕ and H1 = idI . Hence ϕ andidI are isotopic rel. ∂I = 0, 1 by the isotopy H.

Let X ⊂ S2 be a closed Jordan region. If h : ∂X × I → ∂X is anisotopy with h(·, 0) = id∂X , then there exists an isotopy H : X×I → Xsuch that H(·, 0) = idX and H(p, t) = h(p, t) for all p ∈ ∂X and t ∈ I.So an isotopy h on the boundary of X with h0 = id∂X can be extendedto an isotopy H on X with H0 = idX . To see this, we may assumethat X = D. Then H is obtained from h by radial extension; moreprecisely, we define

H(reis, t) = rh(eis, t)

for all r ∈ [0, 1] and s ∈ [0, 2π]. Then H is well-defined and it is easyto see that H is an isotopy with the desired properties. By using the


Schonflies theorem and a similar radial extension one can also show thatifX andX ′ are closed Jordan regions in S2, then every homeomorphismϕ : ∂X → ∂X ′ extends to a homeomorphism Φ: X → X ′.

If ϕ : X → X is a homeomorphism with ϕ|∂X = id∂X , then ϕ isisotopic to idX rel. ∂X. Indeed, again we may assume that X = D.Then we obtain the desired isotopy by the “Alexander trick”: for z ∈ Dand t ∈ I we define H(z, t) = tϕ(z/t) if |z| < t, and H(z, t) = z if|z| ≥ t. It is easy to see that H is an isotopy rel. ∂D with H0 = idXand H1 = ϕ.

If ϕ, ϕ : X → X are two homeomorphism with ϕ|∂X = ϕ|∂X, thenwe can apply the previous remark to ψ = ϕ−1 ϕ and conclude that ϕand ϕ are isotopic rel. ∂X.

We are now ready to state and prove a fact that allows us to con-struct homeomorphisms from cell complex isomorphisms.

Lemma 5.14. Let D and D be isomorphic cell decompositions of 2-

spheres S2 and S2, respectively, and let φ : D → D be an isomorphism.Then the following statements are true:

(i) If h : S2 → S2 is a map such that h|τ is a homeomorphism ofτ onto φ(τ) for each τ ∈ D, then h is a homeomorphism of S2

onto S2.

(ii) There exists a homeomorphism h : S2 → S2 such that h(τ) =φ(τ) for all τ ∈ D.

(iii) Let V be the set of vertices of D. If h0, h1 : S2 → S2 are twohomeomorphisms with h0(τ) = φ(τ) = h1(τ) for all τ ∈ D,then h0 and h1 are isotopic rel. V.

If h is as in (ii), then we say that h realizes the cell complex isomor-phism φ. So an isomorphism between cell decompositions of 2-spherescan always be realized by a homeomorphism h and by (iii) this home-omorphism is unique up to isotopy. A similar fact is actually true ingreater generality, but Lemma 5.14 will be enough for our purposes.

Proof. In the following we write τ := φ(τ) for τ ∈ D.

(i) Let h : S2 → S2 be a map such that h|τ is a homeomorphism ofτ onto τ for each τ ∈ D. Then h is continuous, because the restrictionh|τ is continuous for each τ ∈ D and the cells τ ∈ D form a finitecover of S2 by closed sets. The image cells τ = f(τ) form the cell

decomposition D of S2 and hence cover S2. So h is also surjective.

In order to conclude that h : S2 → S2 is a homeomorphism, it suf-fices to show that h is injective. To see this, let x1, x2 ∈ S2 and assumethat y := h(x1) = h(x2). Then there exists unique cells τ1, τ2 ∈ D such


that x1 ∈ int(τ1) and x2 ∈ int(τ2). Since f |τi is a homeomorphismof τi onto τi for i = 1, 2, we have y ∈ int(τ1) ∩ int(τ2). This implies

that τ1 = τ2. Since the map τ ∈ D 7→ τ ∈ D is an isomorphism,it follows that τ1 = τ2. So x1 and x2 are contained in the same cellτ := τ1 = τ2 ∈ D. Since h|τ is a homeomorphism onto τ and henceinjective, we conclude that x1 = x2 as desired.

(ii) By (i) it suffices to find a map h : S2 → S2 such that h|τ is ahomeomorphism of τ onto τ for each τ ∈ D. The existence of h followsfrom the well-known procedure of successive extensions to the skeletaof the cell decomposition D.

Indeed, if v is vertex in D, then there exists a unique vertex v in Dsuch that φ(v) = v. We define h(v) = v. Then h is a bijection of

the set of vertices of D onto the set of vertices in D. To extend h fromthe 0-skeleton of D to the 1-skeleton, let e be an arbitrary edge in Dand u and v be the vertices in D that are the endpoints of e. Then uand v are the endpoints of e. So we can extend h to e by choosing ahomeomorphism of e onto e that agrees with h on the endpoints of e.In this way we can continuously extend f to the 1-skeleton of D1 sothat f |τ is a homeomorphism of τ onto τ , whenever τ is a cell in Dwith dim(τ) ≤ 1. An argument as in the proof of (i) shows that h is a

homeomorphism of the 1-skeleton of D onto the 1-skeleton of D.If X is an arbitrary tile in D, then ∂X is a subset of the 1-skeleton of

D and hence h is already defined on ∂X. Then h|∂X is an injective and

continuous mapping of ∂X into the boundary ∂X of the tile X ∈ D.Since an injective and continuous map of a Jordan curve into anotherJordan curve is necessarily surjective, h|∂X is a homeomorphism of ∂X

onto ∂X. Hence h can be extended to a homeomorphism of X onto X.These extensions on different tiles paste together to a map h : S2 → S2

with the desired property that h|τ is a homeomorphism of τ onto τ for

each τ ∈ D. By (i) the map h is a homeomorphism of S2 onto S2 withh(τ) = τ = φ(τ) for τ ∈ D.

(iii) Suppose h0, h1 : S2 → S2 are as in the statement. Then ϕ =h−1

1 h0 is a homeomorphism on S2 that maps each cell τ ∈ D ontoitself. In particular, ϕ is the identity on the set V of vertices of D.

By successive extensions to the 1- and the 2-skeleton of D we willshow that ϕ is actually isotopic to idS2 rel. V. For this we denote theset of edges of D by E be and by E =

⋃e : e ∈ E the 1-skeleton of

D, and let e ∈ E be an arbitrary edge in D. Since ϕ(e) = e and ϕ is theidentity on V, the map ϕ|e is isotopic to ide rel. ∂e. These isotopieson edges paste together to an isotopy of ϕ|E to idE rel. V. If X is a


tile in D, then this isotopy is defined on ∂X ⊂ E, and we can extendit to an isotopy of a homeomorphism on X that agrees with ϕ|∂X on∂X to idX . These extensions on tiles X paste together to an isotopyΦ: S2 × [0, 1] → S2 rel. V such that ψ|E = ϕ|E, where ψ := Φ(·, 0),and Φ(·, 1) = idS2 .

For each tile X ∈ D the maps ϕ|X and ψ|X are homeomorphisms ofX onto itself that agree on ∂X ⊂ E. As we have seen in the discussionbefore the proof of the lemma, this implies that ϕ|X and ψ|X areisotopic rel. ∂X. Again by pasting these isotopies on tiles together,we can find an isotopy Ψ: S2 × [0, 1] → S2 rel. E with Ψ(·, 0) = ϕand Ψ(·, 1) = ψ. The concatenation of the isotopies Ψ and Φ givesan isotopy rel. V between ϕ = h−1

1 h0 and idS2 . If we postcomposethis isotopy by h1, we get an isotopy between h0 and h1 rel. V asdesired.

5.3. Cell decompositions induced by Thurston maps

In this section we will see how we can use a Thurston map on a 2-sphereS2 to define a sequence of cell decompositions. We start with a lemmathat allows us to pull back cell decompositions by branched coveringmaps.

Lemma 5.15. Let f : S2 → S2 be a branched covering map and D acell decomposition of S2 such that every point in f(crit(f)) is a vertexin D. Then there exists a unique cell decomposition D′ of S2 such thatf is cellular with respect to (D′,D).

Note that in general D′ will not be a refinement of D.

Proof. To show existence, we define D′ to be the set of all cellsc ⊂ S2 such that f(c) is a cell in D and f |c is a homeomorphism of conto f(c). It is clear that D′ does not contain cells of dimension > 2.As usual we call the cells c in D′ edges or tiles depending on whether chas dimension 1 or 2, respectively. The vertices p of D′ are the pointsin S2 such that p is a cell in D′ of dimension 0. It is clear that theset of vertices of D′ is equal to f−1(V), where V is the set of verticesof D.

To show that D′ is a cell decomposition of S2, we first establish twoclaims.

Claim 1. If p ∈ S2 and q = f(p) ∈ int(X) for some tile X ∈ D,then there exists a unique tile X ′ ∈ D′ with p ∈ X ′.

In this case let U = int(X). Then U is an open and simply con-nected set in the complement of V ⊃ f(crit(f)). Hence there existsa unique continuous map g : U → U ′ := g(U) with f g = idU and

5.3. CELL DECOMPOSITIONS INDUCED BY THURSTON MAPS 125

g(q) = p. The map g is a homeomorphism onto its image U ′. HenceU ′ ⊂ S2 is open and simply connected.

We equip S2 with some base metric inducing the standard topology.In the following metric terms will refer to this metric. Recall thatN ε(A) denotes the open ε-neighborhood of a set A ⊂ S2. Then f hasthe following property: for all w ∈ S2 and all ε > 0, there exists δ > 0such that

(5.1) f−1(B(w, δ)) ⊂ N ε(f−1(w)).

Indeed, if for some w ∈ S2 and ε > 0 there is no such δ, then there existsa sequence zi in S2 \ N ε(f−1(w)) such that f(zi) ∈ B(w, 1/i) for alli ∈ N. By passing to a subsequence, we may assume that zi → z ∈ S2.Then f(z) = limi→∞ f(zi) = w, while

dist(z, f−1(w)) = limi→∞

dist(zi, f−1(w)) ≥ ε.

This is a contradiction showing (5.1).We want to prove that g has a continuous extension to U = X.

For this it suffices to show that g(wi) converges whenever wi is asequence in U converging to a point w ∈ ∂U . Since g is a right in-verse of f , it follows that the limit points of g(wn) are contained inf−1(w). Since f is finite-to-one, the point w has finitely many preim-ages z1, . . . , zm under f .

We can choose ε > 0 so small that the sets B(zi, ε), i = 1, . . . ,m,are pairwise disjoint. By (5.1) we can find δ > 0 such that

(5.2) f−1(B(w, δ)) ⊂m⋃i=1

B(zi, ε).

The set U = X is a closed Jordan region, and hence locally connected.So there exists an open connected set V ⊂ U such that V is a neigh-borhood of w in U and V ⊂ B(w, δ). Then g(V ) is connected subsetof f−1(B(w, δ)). Since the union on the right hand side of (5.2) is dis-joint, the set g(V ) must be contained in one of the sets of this union,say g(V ) ⊂ B(zk, ε). Now wi ∈ V for sufficiently large i, and so all

limit points of g(wi) are contained in g(V ) ⊂ B(zk, ε). On the otherhand, the only possible limit points of g(wi) are z1, . . . , zm, and zkis the only one contained in B(zk, ε). This implies g(wi) → zk. So ghas indeed a continuous extension to U . We also denote it by g. It isclear that

(5.3) f g = idU .


This implies that g is a homeomorphism of U = X onto its imageX ′ := g(U) = g(U). Then X ′ is a closed Jordan region, and by (5.3)the map f |X ′ is a homeomorphism of X ′ onto U = X. Hence X ′ is atile in D′ with p ∈ g(U) ⊂ X ′.

So a tile X ′ ∈ D′ containing p exists. We want to show that it isthe only tile in D′ containing p. Indeed, suppose Y ′ ∈ D′ is anothertile with p ∈ Y ′. Then f(Y ′) is a tile in D containing the point q =f(p) ∈ int(X). Hence f(Y ′) = X, and so f |Y ′ is a homeomorphism ofY ′ onto X. Let h = (f |X)−1. Then g and h are both inverse branchesof f defined on the simply connected region U with g(q) = p = h(q).Hence h and g agree on U , and so by continuity also on U . It followsthat X ′ = g(X) = h(X) = Y ′ as desired.

Claim 2. If p ∈ S2 and q = f(p) ∈ int(e) for some edge e ∈ D, thenthere exists a unique edge e′ ∈ D′, and precisely two distinct tiles X ′

and Y ′ in D′ that contain p. Moreover, e′ ⊂ ∂X ′ ∪ ∂Y ′.By Lemma 5.12 (iv) we know that that are precisely two distinct

tiles X, Y ∈ D that contain e in their boundary, and that U = int(X)∪int(e) ∪ int(Y ) is an open and simply connected region in the comple-ment of the set V ⊃ f(crit(f)). Hence there exists a unique continuousmap g : U → S2 with g(q) = p and f g = idU . As before one canshow that the maps g1 := g| int(X) and g2 := g| int(Y ) have continuousextensions to X and Y , respectively. We use the same notation g1 andg2 for these extensions. It is clear that g1|e = g2|e. Moreover, g1 is ahomeomorphism of X onto a closed Jordan region X ′ = g1(X) with in-verse map f |X ′. In particular, X ′ is a tile in D′. Similarly, Y ′ = g2(Y )is a tile in D′. The tiles X ′ and Y ′ are distinct, because f maps themto different tiles in D. Moreover, e′ := g1(e) = g2(e) is an edge in D′with p ∈ e′ ⊂ ∂X ′ ∩ ∂Y ′.

It remains to prove the uniqueness part. If e is another edge in D′with p ∈ e, then f(e) is an edge in D and f |e is a homeomorphismof e onto f(e). Hence q = f(p) ∈ int(e) ∩ f(e) which implies thatf(e) = e. So f |e is actually a homeomorphism of e onto e. Then(f | int(e′))−1 and (f | int(e))−1 are right inverses of f defined on theopen arc int(e) that both map q to p. Hence these right inverses mustagree on int(e). By continuity this implies (f |e′)−1 = (f |e)−1 on e, andso e′ = (f |e′)−1(e) = (f |e)−1(e) = e.

If Z ′ is another tile in D′ with p ∈ Z ′, then f maps ∂Z ′ homeo-morphically to the boundary ∂f(Z ′) of the tile f(Z ′) ∈ D. Moreover,p ∈ ∂Z ′; for otherwise f(p) would lie in the set int(f(Z ′)) which isdisjoint of e. It follows that there is an edge in D′ that contains pand is contained in the boundary of ∂Z ′. Since this edge in D′ is


unique, as we have just seen, we know that e′ ⊂ ∂Z ′. Note thatX ′ ∪ Y ′ ⊃ g(U) ⊃ int(e′), and so X ′ ∪ Y ′ contains an open neigh-borhood for each point in int(e′). Since e′ ⊂ ∂Z ′ there exists a pointx ∈ int(Z ′) near p with x ∈ X ′∪Y ′, say x ∈ X ′. Then f(x) is containedin the interior of the tile f(Z ′) ∈ D. Since x ∈ X ′ ∩ Z ′ and X ′ andZ ′ are both tiles in D′, we conclude X ′ = Z ′ by the first claim. Thiscompletes the proof of Claim 2.

Now that we have established the claims, we can show that D isa cell decomposition of S2 by verifying conditions (i)–(iv) of Defini-tion 5.1.

Condition (i): If p ∈ S2 is arbitrary, then f(p) is a vertex of D orf(p) lies in the interior of an edge or in the interior of a tile in D. Inthe first case p is a vertex of D′, and in the other two cases p lies incells in D′ by Claim 1 and Claim 2. It follows that the cells in D′ coverS2.

Condition (ii): Let σ, τ be cells in D′ with int(σ)∩ int(τ) 6= ∅. Thenf(σ) and f(τ) are cells in D with int(f(σ)) ∩ int(f(τ)) 6= ∅. Henceλ := f(σ) = f(τ). In particular, σ and τ have the same dimension.

If σ and τ are both tiles, then σ = τ by Claim 1, because everypoint in int(σ) ∩ int(τ) 6= ∅ has an image under f in int(λ). Similarly,if σ and τ are edges, then σ = τ by Claim 2.

If σ and τ consist of vertices in D′, then the relation int(σ)∩int(τ) 6=∅ trivially implies σ = τ .

Condition (iii): Let τ ′ ∈ D′ be arbitrary. Then f |τ ′ is a homeomor-phism of τ ′ onto the cell τ = f(τ ′) ∈ D. Note that (f |τ ′)−1(σ) ∈ D′whenever σ ∈ D and σ ⊂ τ . Since ∂τ ′ = (f |τ ′)−1(∂τ) and ∂τ is a unionof cells in D, it follows that ∂τ ′ is a union of cells in D′.

Condition (iv): To establish the final property of a cell decompo-sition for D′, we will show that D′ consists of only finitely many cells.Indeed, let Ni ∈ N be the number of cells of dimension i in D fori = 0, 1, 2. Since the vertices in D′ are the preimages of the vertices ofD, we have at most deg(f)N0 vertices in D′.

Pick one point in the interior of each edge in D. The set M of thesepoints consists of N1 elements. If q ∈ M , then q /∈ V ⊃ f(crit(f)),and so q is not a critical value of f . Hence #f−1(M) = N1 deg(f).It follows from Claim 2 that each element of f−1(M) is contained ina unique edge in D′, and it follows from the definition of D that eachedge in D′ contains a unique point in f−1(M). Hence the number ofedges in D′ is equal to #f−1(M) = N1 deg(f).

Similarly, pick a point in the interior of each tile in D and let M bethe set of these points. Then #f−1(M) = N2 deg(f) and by the same


reasoning as above based on Claim 1, we see that the the number oftiles in D′ is equal to #f−1(M) = N2 deg(f).

We have shown that D′ is a cell decomposition of S2. It followsimmediately from the definition of D′ that f is cellular with respect to(D′,D).

To show uniqueness of D′, suppose that D is another cell decompo-

sition such that f is cellular with respect to (D,D). Then by definition

of D′ every cell in D also lies in D′. So we have D ⊂ D′. If this inclusionwere strict, then there would be a cell τ ∈ D′ whose interior int(τ) 6= ∅is disjoint from the interior of all cells in D. This is impossible, since

these interiors form a cover of S2. Hence D = D′.

Remark 5.16. No Thurston maps f : S2 → S2 with # post(f) ∈0, 1 exist. Indeed, suppose that f is such a map. Then U := S2 \post(f) is simply connected, and so there exists a continuous mapg : U → S2 with f g = idU .

If # post(f) = 0, we have U = S2 and so we conclude that g is ahomeomorphism onto its image. This image must be all of S2. Hence gand f are homeomorphisms, contradicting our assumption deg(f) ≥ 2(see Definition 2.1).

If # post(f) = 1, we have U = S2 \ p for some p ∈ S2. Then byan argument as in the proof of Claim 1 in Lemma 5.15, one can showthat g has a continuous extension to the point p, and hence to S2. Ifwe denote this extension to S2 also by g, then f g = idS2 , and againwe conclude that f is a homeomorphism and obtain a contradiction.

Now let f : S2 → S2 be a Thurston map, and C ⊂ S2 be a Jordancurve such that post(f) ⊂ C. We will discuss how the pair (f, C)induces natural cell decompositions of S2.

By the Schonflies theorem there are two closed Jordan regionsX0

b , X0w ⊂ S2 whose boundary is C. Our notation for these regions

is suggested by the fact that we often think of X0b as being assigned

or carrying the color “black”, represented by the symbol b, and X0w

as being colored “white” represented by w. We will discuss this moreprecisely later in this section (see Lemma 5.19).

The sets X0b and X0

w are topological cells of dimension 2. We callthem tiles of level 0 or 0-tiles. The postcritical points of f are on theboundary of X0

w and X0b . We consider them as vertices of X0

w andX0

b , and the closed arcs of C between vertices as the edges of the 0-tiles. In this way, we think of X0

w and X0b as topological m-gons where

m = # post(f) ≥ 2 (see Remark 5.16). To emphasize that these edges


and vertices belong to 0-tiles, we call them 0-edges and 0-vertices. A 0-cell is a 0-tile, a 0-edge, or a set consisting of a 0-vertex. Obviously, the0-cells form a cell decomposition of S2 that we denote byD0 = D0(f, C).

Since every point in post(f) is a vertex of D0, we can apply Lem-ma 5.15 to obtain a unique cell decomposition D1 = D1(f, C) suchthat f is cellular with respect to (D1,D0). The vertices of D1 are pre-cisely the points whose image is a vertex of D0. In particular, sincef(post(f)) ⊂ post(f), or equivalently post(f) ⊂ f−1(post(f)), it fol-lows that every point in post(f) is a vertex for D1. Hence we can applyLemma 5.15 again and obtain a cell decomposition D2 = D2(f, C) suchthat f is cellular with respect to (D2,D1). Continuing in this manner,we obtain cell decompositions Dn = Dn(f, C) of S2 for n ∈ N0 suchthat f is cellular for (Dn+1,Dn) for all n ∈ N0.

We call the elements in Dn the n-cells for (f, C), or simply n-cells iff and C are understood. We call n the level of an n-cell. When we speakof n-cells, then n always refers to this level and not to the dimensionof the cell. An n-cell of dimension 2 is called an n-tile, and an n-cell ofdimension 1 an n-edge. An n-vertex is a point p ∈ S2 such that p isan n-cell of dimension 0. We denote the set of all n-tiles, n-edges, andn-vertices for (f, C) by Xn(f, C), En(f, C), and Vn(f, C), respectively.If f and C are understood, we simply write Xn for Xn(f, C), etc.

In the following proposition we summarize properties of the celldecompositions Dn.

Proposition 5.17. Let k, n ∈ N0, let f : S2 → S2 be a Thurstonmap, C ⊂ S2 be a Jordan curve with post(f) ⊂ C, Dn = Dn(f, C), andm = # post(f).

(i) The map fk is cellular with respect to (Dn+k,Dn). In particu-lar, if τ is any (n + k)-cell, then fk(τ) is an n-cell, and fk|τis a homeomorphism of τ onto fk(τ).

(ii) Let σ be an n-cell. Then f−k(σ) is equal to the union of all(n+ k)-cells τ with fk(τ) = σ.

(iii) The 0-skeleton of Dn is the set Vn = f−n(post(f)), and wehave Vn ⊂ Vn+k. The 1-skeleton of Dn is equal to f−n(C).

(iv) We have #Vn ≤ m deg(f)n, #En = m deg(f)n, and #Xn =2 deg(f)n for n ∈ N0.

(v) The n-edges are precisely the closures of the connected compo-nents of f−n(C) \ f−n(post(f)). The n-tiles are precisely theclosures of the connected components of S2 \ f−n(C).

(vi) Every n-tile is an m-gon, i.e., the number of n-edges and n-vertices contained in its boundary is equal to m.


In the proof we will use the following fact about open and contin-uous maps g : S2 → S2 (such as iterates of Thurston maps): if A ⊂ S2

is arbitrary, then

(5.4) g−1(A) ⊂ g−1(A).

Indeed, if U is an open neighborhood of a point p ∈ g−1(A), then g(U)is an open neighborhood of g(p) ∈ A. Hence there exists a point p′ ∈ Uwith g(p′) ∈ A, and so U ∩ g−1(A) 6= ∅. The inclusion (5.4) follows.

Proof. (i) This immediately follows from the facts that f is cel-lular with respect to (Dn+1,Dn) for each n, and that compositions ofcellular maps are cellular (if, as in our case, the obvious compatibilityrequirement for the cell decompositions involved is satisfied).

(ii) It follows from (i) and Lemma 5.15 that Dn+k is the unique celldecomposition of S2 such that fk is cellular with respect to (Dn+k,Dn).Moreover, recall from the proof of Lemma 5.15 that a topological cellc ⊂ S2 is an (n + k)-cell if and only if fk(c) is an n-cell and fk|c is ahomeomorphism of c onto fk(c).

This immediately implies the statement if σ = q, where q is ann-vertex.

Suppose σ is equal to an n-edge e. Let M be the union of all(n+ k)-edges e′ with fk(e′) = e. It is clear that M ⊂ f−k(e).

To see the reverse inclusion, first note that because there are onlyfinitely many (n+ k)-edges, the set M is closed.

Let p ∈ f−k(int(e)) be arbitary. Then from Claim 2 in the proofof Lemma 5.15 it follows that there exists an (n + k)-edge e′ withp ∈ e′. Then fk(e′) is an n-edge that contains q = fk(p) ∈ int(e).Hence e = fk(e′), and so f−k(int(e)) ⊂ M . Since fk is an open andcontinuous map and M is closed, it follows from (5.4) that

f−k(e) = f−k(int(e)) ⊂ f−k(int(e)) ⊂M = M.

Hence M = f−k(e) as desired.If σ is equal to an n-tile X, let M be the union of all (n + k)-tiles

X ′ with fk(X ′) = X. Then M ⊂ f−k(X) and M is closed.If p ∈ f−k(int(X)), then by Claim 1 in the proof of Lemma 5.15

there exists an (n+k)-tile with p ∈ X ′. Similarly as above we concludefk(X ′) = X, and so p ∈ M . Hence f−k(int(X)) ⊂ M . Now again by(5.4) we have

f−k(X) = f−k(int(X)) ⊂ f−k(int(X)) ⊂M = M.

We conclude that M = f−k(X) as desired.


(iii) The 0-skeleton of Dn is the set Vn of all vertices of Dn. By (ii)we know that Vn = f−n(V0) = f−n(post(f)). Moreover,

fn+k(Vn) ⊂ fn+k(f−n(post(f))) ⊂ fk(post(f)) ⊂ post(f) = V0,

and so Vn ⊂ Vn+k.The 1-skeleton of Dn is equal to the set consisting of all n-vertices

and the union of all n-edges. As follows from (ii) this set is equal tothe preimage of the 1-skeleton of D0 under the map fn. Since the 1-skeleton of D0 is equal to C, it follows that the 1-skeleton of Dn is equalto f−n(C).

(iv) Note that deg(fn) = deg(f)n, and that #V0 = m, #E0 = m,and #X0 = 2. The statements about Vn,En, and Xn, then follow fromthe corresponding statement established in the last part of the proofof Lemma 5.15.

(v) This immediately follows from (iii) and Lemma 5.5.

(vi) If X is an n-tile, then fn|X is a homeomorphism of X onto the0-tile fn(X). The n-vertices contained in X are precisely the preim-ages of the 0-vertices contained in fn(X); hence X contains exactlym = # post(f) n-vertices, and hence also the same number of n-edges(Lemma 5.12 (iii)). So every n-tile is an m-gon.

Instead of an inequality for #Vn as in (iv) one can easily give aprecise formula for this number; namely, if we set d = deg(f), andm = # post(f), then #Xn = 2dn and En = mdn. Moreover, by Euler’spolyhedral formula we have

#Xn −#En + #Vn = 2,

and so

#Vn = (m− 2)dn + 2.

We record another lemma that relates tiles with the mapping prop-erties of a given Thurston map.

Lemma 5.18. Let k, n ∈ N0, f : S2 → S2 be a Thurston map, andC ⊂ S2 be a Jordan curve with post(f) ⊂ C.

(i) If Z ⊂ S2 is a Jordan region such that fn|Z is a homeo-morphism onto its image and fn(Z) is a k-tile, then Z is an(n+ k)-tile.

(ii) If X ′ is a k-tile and p ∈ S2 is a point with fn(p) ∈ int(X ′),then there exists a unique (n + k)-tile X with p ∈ X andfn(X) = X ′.


Proof. It is understood that tiles are for (f, C). Note that f iscellular for (Dn+k(f, C),Dk(f, C)), and the set f−k(post(f)) ⊃ post(f)of vertices of Dk(f, C) contains the set fn(crit(fn)) ⊂ post(f) (thelast inclusion follows from (2.5)). Hence we are in the situation ofLemma 5.15 with D = Dk(f, C) and D′ = Dn+k(f, C).

Then (i) follows from the uniqueness statement of Lemma 5.15 andthe definition of D′ in the first paragraph of the proof of this lemma.

Moreover, under the assumptions of (ii) it follows from Claim 1 inthe proof of Lemma 5.15 that there exists a unique (n+ k)-tile X withp ∈ X. Then fn(X) is a k-tile containing fn(p) ∈ int(X ′), and sofn(X) = X ′.

It is often useful, in particular in graphical representations, to assignto each tile one of the two colors “black” and “white” represented bythe symbols b and w, respectively. To formulate this, we denote by X∞

the disjoint union of the sets Xn, n ∈ N0 (for given f and C). Moreinformally, X∞ is the set of all tiles. Note that in general, a set canbe a tile for different levels n, so the same tile may be represented bymultiple copies in X∞ distinguished by their levels n.

Lemma 5.19 (Colors of tiles). There exits a map L : X∞ → b, wwith the following properties:

(i) L(X0b ) = b and L(X0

w ) = w.

(ii) If n, k ∈ N0, Xn+k ∈ Xn+k, and Xn = fk(Xn+k) ∈ Xn, thenL(Xn) = L(Xn+k).

(iii) If n ∈ N0, and Xn and Y n are two distinct n-tiles that havean n-edge in common, then L(Xn) 6= L(Y n).

Moreover, L is uniquely determined by properties (i) and (ii).

So with the normalization (i) one can uniquely assign colors “black”or “white” to the tiles so that all iterates of f are color-preserving asin (ii). By (iii) colors of distinct n-tiles are different if they share an n-edge. This implies that the number of white n-tiles is equal to deg(f)n.Indeed, every n-edge is contained in exactly one white n-tile, and byProposition 5.17 (vi) every n-tile contains m = # post(f) n-edges; soby Proposition 5.17 (iv) we conclude that

number of white n-tiles = #En/m = deg(f)n.

In the same way one sees that the number of black n-tiles is equal todeg(f)n.

Our notion of colorings of tiles is related to the more general conceptof a labeling of cells in a cell decomposition (see the next Section 5.4,in particular Lemma 5.21).

5.4. LABELINGS 133

Proof of Lemma 5.19. To define L we assign colors to the two0-tiles X0

b and X0w as in (i). If Zn is an n-tile for some arbitrary level

n ≥ 0, then fn(Zn) is a 0-tile (Proposition 5.17 (i)), and so it alreadyhas a color assigned. We set L(Zn) := L(fn(Zn)).

This defines a map L : X∞ → b, w. By definition, L has prop-erty (i). To show (ii), assume that n, k ∈ N0 and Xn+k ∈ Xn+k.Then by Proposition 5.17 (i), we have Xn := fk(Xn+k) ∈ Xn, andfn+k(Xn+k), fn(Xn) ∈ X0. So by definition of L we have

L(Xn) = L(fn(Xn)) = L(fn(fk(Xn+k)))

= L(fn+k(Xn+k)) = L(Xn+k)

as desired.Let Xn and Y n be as in (iii). Then again by Proposition 5.17 (i),

we have fn(Xn), fn(Y n) ∈ X0. There exists an n-edge e such thate ⊂ ∂Xn ∩ ∂Y n. We orient e so that Xn lies to the left of e. The set ofn-vertices is equal to f−n(post(f)) and disjoint from int(e). It followsthat no point in int(e) is a critical point of fn. In particular, fn is a localhomeomorphism near each point in int(e) which implies that fn(Xn)and fn(Y n) are distinct. We conclude that L(fn(Xn)) 6= L(fn(Y n)),and so by definition of L we have

L(Xn) = L(fn(Xn)) 6= L(fn(Y n)) = L(Y n)

as desired. It follows that L has the properties (i)–(iii).It is clear that L is uniquely determined by (i) and (ii).

If the tiles in a cell decomposition D of a 2-sphere S2 are assignedcolors “black” and “white” so that two distinct tiles sharing an edgehave different colors, then we say the cell decomposition is a checker-board tiling of S2. It is clear that for the existence of such a coloring thelength of the cycle of each vertex in D has to be even. In Lemma 5.21we will see that this necessary condition is also sufficient.

If there exists m ∈ N, m ≥ 2, such that each tile in D is an m-gon, then we say that D is a tiling by m-gons. With this terminolgywe can summarize some of the main results of this section by sayingthat the cell decompositions Dn = Dn(f, C) (with the colorings givenby the previous lemma) are checkerboard tilings by m-gons, wherem = # post(f).

5.4. Labelings

Suppose D0 and D1 are cell decompositions of a 2-sphere S2. In Sec-tion 5.5 we will see that under suitable conditions one can constructa Thurston maps that is cellular for (D0,D1). If one wants to obtain


a unique map up to Thurston equivalence, one needs additional data;namely, for each cell in D0 we have to assign an image in D1. The nec-essary properties of such assignments can be abstracted in the notionof a labeling.

Definition 5.20 (Labelings). Let D1 and D0 be cell complexes.Then a labeling of (D1,D0) is map L : D1 → D0 satisfying the followingconditions:

(i) dim(L(τ)) = dim(τ) for all τ ∈ D1.

(ii) if σ, τ ∈ D1 and σ ⊂ τ , then L(σ) ⊂ L(τ).

(iii) if σ, τ, c ∈ D1, σ, τ ⊂ c, and L(σ) = L(τ), then σ = τ .

So a labeling is a map L : D1 → D0 that preserves inclusions anddimensions of cells, and is “injective on cells” c ∈ D1 in the sense of(iii). In particular, every cell of dimension 0 in D1 is mapped to acell of dimension 0 in D0. If v is a vertex in D1, i.e., if v is a cellof dimension 0 in D1, then we can write L(v) = w, where w is avertex in D0. We define L(v) = w. In the following we always assumethat a labeling L : D1 → D0 has been extended to the set of verticesof D1 in this way; this will allow us to ignore the subtle distinctionbetween vertices and cells of dimension 0, i.e., sets consisting of onevertex.

Let S2 be an oriented 2-sphere, and D be a cell decomposition of S2.Recall (see Section 5.2) that a flag in D is a triple (c0, c1, c2), whereci is a cell in D of dimension i for i = 0, 1, 2 and c0 ⊂ c1 ⊂ c2. IfL : D1 → D0 is a labeling of a pair (D1,D0) of cell decompositions ofS2 and (c0, c1, c2) is a flag in D1, then (L(c0), L(c1), L(c2)) is a flag inD0. This follows from the definition of a labeling. So a labeling maps“flags to flags”. We say that the labeling is orientation-preserving if itmaps positively-oriented flags in D1 to positively-oriented flags in D0.

If f : S2 → S2 is cellular for (D1,D0), then f induces a naturallabeling L : D1 → D0 given by L(τ) = f(τ) for τ ∈ D1. Moreover, ifin addition f |X is orientation-preserving for each tile X in D1 (whichis always true if f is a branched covering map), then this labeling L isorientation-preserving.

If a labeling L : D1 → D0 is given, then we say that a map f : S2 →S2 that is cellular for (D1,D0) is compatible with the labeling L ifL(τ) = f(τ) for each τ ∈ D1, i.e., if the labeling induced by f is equalto the given labeling.

5.4. LABELINGS 135

Let X be a closed Jordan region on the oriented 2-sphere S2 withk ≥ 3 distinct points v0, . . . , vk−1, vk = v0. Here the indices are ele-ments of Zk = 0, 1, . . . , k − 1 = Z/kZ, the cyclic group with k ele-ments. Suppose further that the points v0, . . . , vk−1 are indexed suchthat if we start at v0 and run through ∂X with suitable orientation,then the points v0, . . . , vk−1 are traversed in successive order. If this istrue and if with this orientation of ∂X the region X lies on the left, thenwe call the points v1, . . . , vk in cyclic order on ∂X, and otherwise, if Xlies on the right, in anti-cyclic order on ∂X. If the points v0, . . . , vk−1

are in cyclic or anti-cyclic order on ∂X, then ∂X is decomposed intounique arcs e0, . . . , ek−1; here el for l ∈ Zk is the unique subarc of ∂Xthat has the endpoints vl and vl+1, but does not contain any other ofthe points vi, i ∈ Zk \ l, l + 1. We say that the arcs e0, . . . , ek−1

are in cyclic or anti-cyclic order on ∂X, if this is true for the pointsv0, . . . , vk−1, respectively.

If we have a labeling L : D1 → D0, we should think of each elementτ ∈ D1 as “carrying” the label L(τ) ∈ D0. In applications it is oftenmore intuitive and convenient to allow more general index sets L of thesame cardinality as D0 as labeling sets for the elements in D1. In suchsituations we fix a bijection ψ : D0 → L and call a map L′ : D1 → L alabeling if ψ−1L′ : D1 → D0 is a labeling in the sense of Definition 5.20.

We will discuss this in a case that will be relevant for us later.Namely, suppose that the cell decomposition D0 of the oriented sphereS2 has two tiles X0

b and X0w with common boundary C := ∂X0

b =∂X0

w . We represent X0b by the symbol b for “black” and X0

w by w for“white”. By Lemma 5.12 (iii) the set C is a Jordan curve containingk ≥ 2 vertices and edges, and there are no other edges and vertices inD0. Let us assume that k ≥ 3 and that we have indexed the verticesv0, . . . , vk−1 so that they are cyclically ordered on ∂X0

w . Then they areanti-cyclically ordered on ∂X0

b . As above, we index the edges such thatel is the unique subarc on C = ∂X0

w with endpoints vl and vl+1.There exists a bijection of D0 with the set L consisting of the sym-

bols b and w (for the two tiles in D0) and two copies of Zk, one for theedges and the other one for the vertices in D0. More explicitly, such abijection ψ : D0 → L is given by

(5.5) ψ(X0w ) = w, ψ(X0

b ) = b, ψ(vl) = l and ψ(el) = l for l ∈ Zk.

Now suppose that that in this situation D0 is equal to the celldecomposition D0(f, C) (as defined in Section 5.3) for a Thurston mapf : S2 → S2. In other words, post(f) ⊂ C and the vertex set of D0 isequal to post(f). Let D1 = D1(f, C) be the cell decomposition givenby cells of level 1, and X = X1, E = E1, and V = V1 be the sets of


1-tiles, 1-edges, and 1-vertices, respectively. One can then define threemaps LX : X→ w, b, LE : E→ Zk and LV : V→ Zk as

LX(X) = ψ(f(X)), LE(e) = ψ(f(e)), LV(v) = ψ(f(v))

for X ∈ X, e ∈ E, and v ∈ V. Accordingly, a 1-tile X is called“white” if LX(X) = w, and called “black” if LX(X) = b. Recall fromProposition 5.17 (i) that for each X ∈ X the map f |X is an orientation-preserving homeomorphism onto either X0

w or X0b . This implies that

if we use the map LV : V → Zk to index 1-vertices, then they are incyclic order on the boundary ∂X of a white 1-tile X and in anti-cylicorder on a boundary of a black 1-tile. Similarly, the 1-edges are incyclic order on the boundary of white 1-tiles, and in anti-cyclical orderon the boundary of black 1-tiles. The maps LX, LE and LV can becombined in the obvious way to a map L : D1 → L ∼= D0 (so thatL|X = LX, etc.), and it follows easily from the previous discussionthat L is a labeling.

In the following lemma, we will turn this construction around andask when a labeling with similar properties exists on a given cell decom-position D = D1 of a 2-sphere that is a prori not related to a Thurstonmap. This will later be useful when we want to construct Thurstonmaps.

Lemma 5.21. Let D be a cell decomposition of S2, and denote byV the set of vertices, by E the set of edges, and by X the set of tilesin D. Suppose that the length of the cycle of every vertex in D is evenand that there exists k ≥ 3 such that every tile in X is a k-gon.

Then for each positively-oriented flag (c0, c1, c2) in D there existmaps LV : V → Zk, LE : E → Zk, and LX : X → b, w with thefollowing properties:

(i) LV(p0) = 0, where c0 = p0, LE(c1) = 0, and LX(c2) = w,

(ii) if X, Y ∈ X are two distinct tiles with a common edge on theirboundaries, then LX(X) 6= LX(Y ),

(iii) if X is an arbitrary tile in X, then LV induces a bijection ofthe set of vertices in ∂X with Zk so that the order of thesevertices is cyclic if LX(X) = w and anti-cyclic if LX(X) = b,

(iv) if e ∈ E and l = LE(e), then LV(∂e) = l, l + 1,(v) if X is an arbitrary tile in X, then LE induces a bijection of

the set of edges contained in ∂X so that the order of thesesedges is cyclic if LX(X) = w and anti-cyclic if LX(X) = b,

(vi) if (τ0, τ1, τ2) is a flag in D, then the flag is positively-orientedif and only if there exists l ∈ Zk such that LV(τ0) = l,

5.4. LABELINGS 137

LE(τ1) = l, LX(τ2) = w, or LV(τ0) = l, LE(τ1) = l − 1,LX(τ2) = b.

The maps LV, LE, and LX are uniquely determined by the proper-ties (i)–(iv).

Recall (see the discussion after Lemma 5.12) that the length of thecycle of a vertex v is the number of edges as well as the number of tilesthat contain v. So instead of saying that the length of each cycle ofevery vertex is even, we could have said that every vertex is containedin an even number of tiles (equivalently contained in an even numberof edges).

Condition (ii) says that one of the two tiles containing an edge is“black” and the other is “white”. So if the tiles have been labeled in thisway, then D becomes a checkerboard tiling by k-gons. Condition (ii)is equivalent to the statement that the dual graph of the 1-skeleton ofD is bipartite.

Condition (iii) says that for any white tile X, vertices on ∂X areare labeled cyclically; for any black tile Y , vertices on ∂Y are labeledanti-cyclically. A more precise formulation is as follows: for each i ∈ Zkthere is exactly one vertex v ∈ ∂X with LV(v) = i, and if we write v =vi if LV(v) = i, then the vertices v0, . . . , vk−1 ∈ ∂X are in cyclic orderon ∂X if X is white and in anti-cyclic order if X is black. Condition (v)has to be interpreted in a similar way.

By (iv) the label LE(e) of an edge e ∈ E is determined by the labelsLV(u) and LV(v) of the two endpoints u and v of e (here it is importantthat k ≥ 3).

To prove the previous lemma it is useful to introduce some addi-tional terminology. Let D be a cell decomposition of S2. A chain oftiles X1, . . . , XN in D is called an e-chain if for i = 1, . . . , N − 1 wehave Xi 6= Xi+1 and there exists an edge ei in D with ei ⊂ ∂Xi∩∂Xi+1.The e-chain joins the tiles X and Y if X1 = X and XN = Y . If X isan arbitrary tile if D, then every tile Y in D can be joined to X by ane-chain. This follows from the fact that the union of the tiles Y thatcan be joined to X is equal to S2; indeed, this union is a nonemptyclosed set, and it is also open, as follows from Lemma 5.12 (vi) and(v). Hence the union is all of S2.

The 1-skeleton of D can be considered as a graph with the givenvertices and edges. Then the dual graph has the set of tiles as vertices,and two vertices as represented by tiles are joined by an edge if thetiles both contain an edge e ∈ D in their boundaries. Then an e-chainin D is essentially a path in this dual graph.

Proof of Lemma 5.21. We first establish the following statement.


Claim. Suppose that J ⊂ S2 is a Jordan curve that does notcontain any vertex (in D) and has the property that for every edgee the intersection e ∩ J is either empty, or e meets both componentsof S2 \ J and e ∩ J consists of a single point. Then J meets an evennumber of edges.

To see this, pick one of the complementary components U of S2 \J ,and let v1, . . . , vn be the vertices contained in U , where n ∈ N0 (forn = 0 we consider this as an empty list). For i = 1, . . . , n let di bethe length of the cycle of vi, i.e., the number of edges containing vi.We denote by EJ the set of all edges that meet J and by EU the setof all edges contained in U . From our assumption on the intersectionproperty of J with edges it follows that an edge is contained in U ifand only if its two endpoints are in U , and it meets J if and only if oneendpoint is in U and the other in S2 \ U . Hence

d1 + · · ·+ dn = #EJ + 2#EU ,

because the sum on the left hand side counts every edge in EJ once,and every edge in EU twice. Since all the numbers d1, . . . , dn are evenby our assumptions, it follows that the number #EJ of edges that Jmeets is also even, proving the claim.

To show existence and uniqueness of the map LX we now proceedas follows. For every tile Y there exits an e-chain Y0 = c2, . . . , YN = Yof tiles joining the “base tile” c2 (from the given flag (c0, c1, c2)) to Y .We put LX(Y ) = w or LX(Y ) = b depending on whether N is even orodd. It is clear that if this is well-defined, then it is the unique choicefor LX(Y ). This follows from the normalization (i) and that fact thatby (ii) the labels of tiles have to alternate along an e-chain.

To see that LX is well-defined, it is enough to show that if an e-chain X0, X1, . . . , XN forms a cycle, i.e., if X0 = XN , then N is even.To prove this we may make the additional assumption that N ≥ 3 andthat the chain is simple, i.e., that the tiles X1, . . . XN are all distinct.

We can choose edges ei for i = 1, . . . , N such that ei ⊂ ∂Xi−1∩∂Xi.Then the edges e1, . . . , eN are all distinct. For otherwise, ei = ej forsome 1 ≤ i < j ≤ N . Then ei = ej is contained in the boundary ofthe tiles Xi−1, Xi, Xj−1, Xj which is impossible, because three of thesetiles must be distinct (note that N ≥ 3).

We now construct a Jordan curve J that “follows” our closed e-chain. Formally, for each edge ei pick a point xi ∈ int(ei). Moreover, fori = 1, . . . , N , we can choose an arc αi ⊂ Xi with endpoints xi and xi+1

such that int(αi) ⊂ int(Xi). Here xN+1 := x1. Then J = α1 ∪ · · · ∪ αNis a Jordan curve that has properties as in the claim above. The curve

5.4. LABELINGS 139

J meets the edges e1, . . . , eN and no others. Hence N is even. ThusLX is well-defined, and it has property (ii) and is normalized as in (i).

To show the existence of LV it is useful to quickly recall some basicdefinitions from the homology and cohomology of chain complexes.Denote by Eo the set of oriented edges in D. Let C(X) and C(Eo) bethe free modules over Zk generated by the sets X and Eo, respectively.So C(Eo), for example, is just the set of formal finite sums

∑aiei,

where ai ∈ Zk and ei ∈ Eo. Note that in contrast to other commonlyused definitions of chain complexes we have e + e 6= 0 if e and e areoriented edges with the same underlying set, but opposite orientations.

There is a unique boundary operator b : C(X) → C(Eo) that is amodule homomorphism and satisfies

bX := b(X) =∑e⊂∂X

e

for each tile X, where the sum is extended over all oriented edgese ⊂ ∂X so that X lies on the left of e.

Let e be an oriented edge and X be the unique tile with e ⊂ ∂Xthat is on the left of e. We put α(e) = 1 ∈ Zk or α(e) = −1 ∈ Zkdepending on whether LX(X) = w (X is a white tile) or LX(X) = b

(X is a black tile). If e and e are oriented edges with with the sameunderlying set, but opposite orientations, then α(e) + α(e) = 0 asfollows from property (ii) of LX.

The map α can be uniquely extended to a homomorphism α : C(Eo)→Zk. In the language of cohomology it is a “cochain”. This cochain α isa cocycle, i.e.,

(5.6) α(bX) =∑e⊂∂X

α(e) = ±k = 0 ∈ Zk

for every tile X, considered as one of the generators of C(X). Indeed,by our convention on the orientation of edges e ⊂ ∂X in the abovesum, for each such edge we get the same contribution α(e), and so,since X has k edges, the sum is equal to ±k = 0 ∈ Zk.

Consider an arbitrary closed edge path consisting of the orientededges e1, . . . , en; so the terminal point of ei is the initial point of ei+1

for i = 1, . . . , n, where en+1 := e1. We claim that

(5.7)n∑i=1

α(ei) = 0.

Essentially, this is a consequence of the fact that we have H1(S2,Zk) =0 for the first cohomology group of S2 with coefficients in Zk. Thisimplies that the cocycle α is a coboundary.


We will present a simple direct argument. To show (5.7), it isclearly enough to establish this for simple closed edge paths, i.e., forclosed edge paths where the union of the edges forms a Jordan curveJ ⊂ S2. Let U be the complementary component of S2 \ J so that Ulies on the left if we traverse J according to the orientation given bythe edges ei. If X1, . . . , XM are all the tiles contained in U , then

b(X1 + · · ·+XM) =∑e⊂U

e,

where the sum is extended over oriented edges contained in U . Eachedge on J is equal to one of the edges ei and it appears in the abovesum exactly once and with the same orientation as ei. All other edgesin U appear twice and with opposite orientations. Hence by (5.6),

n∑i=1

α(ei) =∑e⊂U

α(e) =M∑i=1

α(bXi) = 0.

We now define LV : V → Zk as follows. For v ∈ V we can pick anedge path consisting of the oriented edges e1, . . . , en that joins the basepoint p0 to v (this list of edges may be empty if v = p0). The existenceof such an edge path follows from the connectedness of the 1-skeletonof D (see Lemma 5.12 (vi)). Put

(5.8) LV(v) =n∑i=1

α(ei).

This is well-defined, because we have (5.7) for every closed edge path;we also have the normalization LV(p0) = 0.

The definition of LV implies that if e is an oriented edge, and u isthe initial and v the terminal point of e, then

(5.9) LV(v) = LV(u) + α(e).

This means that if we go from the initial point u of e to the terminalpoint v, then the value of LV is increased by 1 or decreased by −1depending on whether the tile on the left of e is white or black. Thedesired property (iii) of LV immediately follows from this.

This shows existence of LV. Conversely, every function LV withproperty (iii) must satisfy (5.9). Together with the normalizationLV(p0) = 0 this implies that LV is given by the formula (5.8), andso we have uniqueness.

To define LE note that if e ∈ E, then by (ii) we can choose a uniqueorientation for e such that the tile on the left is “white”, and the oneon the right is “black”. If u is the initial and v the terminal point of e

5.5. THURSTON MAPS FROM CELL DECOMPOSITIONS 141

according to this orientation, and LV(u) = l ∈ Zk, then LV(v) = l+ 1.Now set LE(e) := l. Then LE has property (iv). Moreover, we alsohave the normalization (i) for LE; indeed, is c1 is oriented so that p0

is the initial point of c1, then c2 lies on the left of c1, because the flag(c0, c1, c2) is positively-oriented. Since LX(c2) = w, the tile c2 is whiteand so LE(e) = LV(p0) = 0. Uniqueness of LE follows from (iii) andthe uniqueness of LV.

We have proved (i)–(iv) and the uniqueness statement. It remainsto establish (v) and (vi).

To show (v) let X ∈ X be arbitrary. Then by (iii) we can assumethat the indexing of the k vertices v0, . . . , vk−1 on ∂X is such thatLV(vi) = i for all i ∈ Zk, and that v0, . . . , vk−1 are met in successiveorder if we traverse ∂X. This implies that for each i ∈ Zk there existsa unique edge ei ⊂ ∂X in D with endpoints vi and vi+1. Hence by (iv)we have LE(ei) = i. Moreover, by (iii) the edges e0, . . . , ek−1 are incyclic or anti-cyclic order on ∂X depending on whether LX(X) = w orLX(X) = b. So (v) holds.

Finally, to see that (vi) is true, let (τ0, τ1, τ2) be a flag in D. Thenτ0 = u for some u ∈ V. The vertex u is the initial point of theoriented edge τ1. Let v ∈ V be the terminal point of τ1, and definel = LV(u).

Depending on whether the flag is positively- or negatively-oriented,the vertex v follows u in cyclic or anti-cyclic order on ∂τ2. So if theflag is positively-oriented, then by property (iii) we have LV(v) = l+ 1if LX(τ2) = w and LV(v) = l − 1 if LX(τ2) = b. Property (iv) impliesthat LE(τ1) = l if LX(τ2) = w and LE(τ1) = l − 1 if LX(τ2) = b.

So if (τ0, τ1, τ2) is positively-oriented, then the cells in this flag carrythe labels l, l, w, or l, l − 1, b, respectively.

Similarly, if (τ0, τ1, τ2) is negatively-oriented, then we get the labelsl, l− 1, w, or l, l, b for the cells in the flag. Statement (vi) follows fromthis.

5.5. Thurston maps from cell decompositions

In Section 5.3 we have seen how to obtain cell decompositions fromThurston maps. In this section we reverse this procedure and askwhen a pair (D1,D0) of cell decompositions gives rise to a Thurstonmap f that is cellular for (D1,D0). In general one cannot expect f tobe determined just by by the cell decompositions alone, but one needsadditional information on how f is supposed to map the cells in D1

to cells in D0. This is given by a (orientation-preserving) labelings as


discussed in the previous section (see Definition 5.20 and the discussionfollowing this definition).

We start with a lemma that allows us to recognize branched coveringmaps.

Lemma 5.22. Let D′ and D be cell decompositions of S2, and f : S2 →S2 be a cellular map for (D′,D) such that f |X is orientation-preservingfor each tile X in D′.

(i) Then f is a branched covering map on S2. Each critical pointof f is a vertex of D′.

(ii) If in addition each vertex in D is also a vertex in D′, thenevery point in post(f) is a vertex of D. In particular, f ispostcritically-finite, and hence a Thurston map if f is not ahomeomorphism.

Proof. (i) We will show that for each point p ∈ S2, there exists k ∈N, an orientation-preserving homeomorphism ϕ of the open unit diskD = z ∈ C : |z| < 1 onto a neighborhood W ′ of p, and an orientation-preserving homeomorphism ψ of a neighborhood W ⊃ f(W ′) of q =f(p) onto D such that ϕ(0) = p, ψ(q) = 0, and

(ψ f ϕ)(z) = zk

for all z ∈ D. The desired relation between the points and maps canbe represented by the commutative diagram

(5.10) p ∈ W ′ f// q ∈ W

ψ

0 ∈ D

ϕ

OO

z 7→zk// 0 ∈ D.

We will use the fact that if f is an orientation-preserving localhomeomorphism near p, then we can take k = 1 and can always findsuitable homeomorphisms ϕ and ψ.

Let p ∈ S2 be arbitrary. Since S2 is the disjoint union of theinteriors of the cells in D′, the point p is contained in the interior ofa tile or an edge in D′, or is a vertex of D′. Accordingly, we considerthree cases.

Case 1. There exists a tile X ′ ∈ D′ with p ∈ int(X ′). ThenW ′ := int(X ′) is an open neighborhood of p, and f |W ′ is an orientation-preserving homeomorphism of W ′ = int(X ′) onto W := int(X), whereX = f(X ′) ∈ D. Hence f is a orientation-preserving local homeomor-phism near p.


Case 2. There exists an edge e′ ∈ D′ with p ∈ int(e′). By Lem-ma 5.12 (iv) there exist distinct tiles X ′, Y ′ ∈ D′ such that e′ ⊂ ∂X ′ ∩∂Y ′. Then W ′ = int(X ′) ∪ int(e′) ∪ int(Y ′) is an open neighborhoodof p. Since f is cellular, X = f(X ′) and Y = f(Y ′) are tiles in D ande = f(e′) is an edge in D. Moreover, e ⊂ ∂X ∩ ∂Y .

We orient e′ so that X ′ lies to the left and Y ′ to the right of e′.Since f is orientation-preserving if restricted to cells in D′, the tile Xlies to the left, and Y to the right of the image e of e′. In particular,X 6= Y , and so the sets int(X), int(e), int(Y ) are pairwise disjoint, andtheir union is open. Since f is cellular and hence a homeomorphismif restricted to cells (and interior of cells), it follows that f |W ′ is ahomeomorphism of W ′ onto the open set W = int(X)∪ int(e)∪ int(Y ).Moreover, it is clear that f |W ′ is orientation-preserving. Since W ′

is open and contains p, the map f is a orientation-preserving localhomeomorphism near p.

Case 3. The point p is a vertex of D′. As in the proof of Lem-ma 5.12 (v) we can choose tiles X ′j ∈ D′ and edges e′j ∈ D for j ∈ Nthat contain p and satisfy X ′j 6= X ′j+1, e′j 6= e′j+1, and e′j ⊂ ∂X ′j ∩∂X ′j+1

for all j ∈ N. There exists d′ ∈ N such that X ′d′+1 = X ′1 and such thatthe tiles X ′1, . . . , X

′d′ and the edges e′1, . . . , e

′d′ are all distinct and such

that

W ′ = p ∪ int(X ′1) ∪ int(e′1) ∪ int(X ′2) ∪ · · · ∪ int(e′d′)

is an open neighborhood of p.Moreover, by the remark following Lemma 5.12, we know that X ′j =

X ′j+d′ and e′j = e′j+d′ for all j ∈ N.Define Xj = f(X ′j) and ej = f(e′j) for j ∈ N. Since f is cellu-

lar for (D′,D), the set Xj is a tile and ej an edge in D. Moreover,ej ⊂ ∂Xj ∩ ∂Xj+1 for j ∈ N. Since X ′j and X ′j+1 are distinct tilescontaining the edge e′j+1 in their boundaries, it follows by an argumentas in Case 2 above that Xj 6= Xj+1 for j ∈ N. Moreover, since e′jand e′j+1 are distinct edges in D′ contained in X ′j+1, and f |X ′j+1 is ahomeomorphism, we also have ej 6= ej+1 for j ∈ N.

As in the proof of Lemma 5.12 (v) we see that there exists a numberd ∈ N, d ≥ 2, such that Xd+1 = X1, and such the tiles X1, . . . , Xd andthe edges e1, . . . , ed are all distinct. Moreover,

W = q ∪ int(X1) ∪ int(e1) ∪ int(X2) ∪ · · · ∪ int(ed)

is an open neighborhood of q = f(p), and Xi = Xj+d and ej = ej+d forall j ∈ N.


The periodicity properties of the indexing of the tiles X ′j and Xj

imply that d ≤ d′ and that d is a divisor of d′. Hence there exists k ∈ Nsuch that d′ = kd.

We now claim that after suitable coordinate changes near p and q,the map f can be given the form z 7→ zk.

For N ∈ N, N ≥ 2, and j ∈ N define half-open line segments

LNj = re2πij/N : 0 ≤ r < 1 ⊂ D

and sectors

ΣNj = reit : 2π(j − 1)/N ≤ t ≤ 2πj/N and 0 ≤ r < 1 ⊂ D.

We then construct a homeomorphism ψ : W → D with ψ(q) = 0 asfollows. For each j = 1, . . . , d we first map the half-open arc q∪int(ej)homeomorphically to the half-open line segment Ldj . Then q is mappedto 0, so these maps are consistently defined for q. Since Xj is a Jordanregion, we can extend the homeomorphisms on q ∪ int(ej−1) ⊂ ∂Xj

and on q ∪ int(ej) ⊂ ∂Xj to a homeomorphism of

q ∪ int(ej−1) ∪ int(ej) ∪ int(Xj)

onto the sector Σdj for each j = 2, . . . , d+ 1. Since the sets

q, int(e1), . . . , int(ed), int(X2), . . . , int(Xd+1) = int(X1)

are pairwise disjoint and have W as a union, these homeomorphismspaste together to a well-defined homeomorphism ψ of W onto D. Notethat ψ(q) = 0 and ψ(Xj ∩W ) = Σd

j for each j = 1, . . . , d.A homeomorphism ϕ : D → W ′ is defined as follows. If z ∈ D is

arbitrary, then z ∈ Σd′j for some j = 1, . . . , d′. Hence zk ∈ Σd

j , and

so ψ−1(zk) ∈ Xj ∩W . Since f is a homeomorphism of X ′j ∩W ′ onto

Xj∩W , it follows that (f |X ′j)−1(ψ−1(zk)) is defined and lies in X ′j∩W ′.We set

ϕ(z) = (f |X ′j)−1(ψ−1(zk)).

It is straightforward to verify that ϕ is well-defined and a homeomor-phism of D onto W ′ with ϕ(0) = p. It follows from the definition of ϕthat (ψ f ϕ)(z) = zk for z ∈ D, and so we have the diagram (5.10).

We assume that the tiles X ′j and the edges e′j are indexed by the pro-cedure in the proof of Lemma 5.12 (v). Then each flag (p, e′j, X ′j+1)is positively-oriented (see the remark after the proof of Lemma 5.12).Since f |X ′j is orientation-preserving, this implies that the flag (q, ej, Xj+1)is also positively-oriented. We conclude that ψ is orientation-preserving,since ψ maps the positively-oriented flag (q, ej, Xj+1) in S2 to the


positively-oriented flag (0, Ldj ,Σdj+1) in C. As follows from its defini-

tion, the map ϕ is then also orientation-preserving. Hence ψ and ϕ arelocal homeomorphisms as desired.

We have shown that in all cases the map f has a local behavior asclaimed. It follows that f is a branched covering map. Moreover, wehave seen that f near each point is a local homeomorphism unless p isa vertex of D′. It follows that each critical point of f is a vertex of D′.

(ii) Suppose in addition that every vertex of D is also a vertex ofD′. Let p be a critical point of f . Then by (i) the point p is a vertexof D′. Since f is cellular for (D′,D), the point f(p) is a vertex of D.Hence f(p) is also a vertex of D′, and we can apply the argument again,to conclude that f 2(p) is a vertex of D, etc. It follows that post(f) isa subset of the set of vertices of D. In particular, post(f) is finite, andso f is postcritically-finite.

Remark 5.23. Let the map f : S2 → S2 and the cell decomposi-tions D′ and D be as in the previous lemma, and let p be a vertex in D′.Then q = f(p) is a vertex in D. If d′ and d are the lengths of the cyclesof p in D′ and q in D, respectively, then d′ = d degf (p). Moreover, thetiles and edges of the cycle of q in D are the images under f of the tilesand edges of the cycle of p in D. This was established in Case 3 of theproof of Lemma 5.22.

We can now prove the following statement which allows us the con-struction of many Thurston maps from essentially combinatorial data(see Section 12.3 for specific examples).

Proposition 5.24. Let D0 and D1 be cell decompositions of anoriented 2-sphere S2, and L : D1 → D0 be an orientation-preservinglabeling. Suppose that every vertex of D0 is also a vertex of D1. Thenthere exists a postcritically-finite branched covering map f : S2 → S2

that is cellular for (D1,D0) and is compatible with the given labeling L.The map f is unique up to Thurston equivalence.

If deg(f) ≥ 2, i.e., if f is not a homeomorphism, then f is actuallya Thurston map.

Later we will consider triples (D1,D0, L) as in the previous proposi-tion related to Thurston maps f with invariant curves C. These triplessatisfy additional and somewhat technical conditions and lead to thenotion of a two-tile subdivision rule (see Definition 12.1).

Proof of Proposition 5.24. As in the proof of Lemma 5.14 (ii),a map f as desired is obtained from successive extensions to the skeleta


of the cell decompositionD1. Indeed, let L : D1 → D0 be an orientation-preserving labeling. If v ∈ S2 is a 1-vertex (i.e., a vertex in D1), thenL(v) is a 0-vertex (i.e., a vertex in D0). Set f(v) = L(v). This definesf on the 0-skeleton of D1. To extend this to the 1-skeleton of D1, let ebe an arbitrary 1-edge. Then e′ = L(e) is a 0-edge. Moreover, if u andv are the 1-vertices that are the endpoints of e, then u′ = f(u) = L(u)and v′ = f(v) = L(v) are distinct 0-vertices contained in e′. Hencethey are the endpoints of e′. So we can extend f to e by choosing ahomeomorphism of e onto e′ that agrees with f on the endpoints of e.In this way we can continuously extend f to the 1-skeleton of D1 sothat f |τ is a homeomorphism of τ onto L(τ) whenever τ ∈ D1 is a cellof dimension ≤ 1.

If X is an arbitrary 1-tile, then ∂X is a subset of the 1-skeleton ofD1 and hence f is already defined on ∂X. Then f |∂X is a continuousmapping of ∂X into the boundary ∂X ′ of the 0-tile X ′ = L(X). Themap f |∂X is injective. Indeed, suppose that u, v ∈ ∂X and f(u) =f(v). Then there exist unique 1-cells σ, τ ⊂ ∂X of dimension ≤ 1 suchthat u ∈ int(σ) and v ∈ int(τ). Then

f(u) = f(v) ∈ int(f(σ)) ∩ int(f(τ)) = int(L(σ)) ∩ int(L(τ))

and so the 1-cells L(σ) and L(τ) must be the same. Since L is a labelingand σ, τ ⊂ X ∈ D1, it follows that σ = τ . As the map f restricted tothe 1-cell σ = τ is injective, we conclude u = v as desired.

So f |∂X is a continuous and injective map of ∂X into ∂X ′, andhence a homeomorphism between these sets. Hence f can be extendedto a homeomorphism of X onto X ′. These extensions on different 1-tiles paste together to a continuous map f : S2 → S2 that is cellular andis compatible with the given labeling. Moreover, f |X is orientation-preserving for each 1-tile X as follows from the fact that the labelingis orientation-preserving. By Lemma 5.22 (i) and (ii) the map f isa postcritically-finite branched covering map. This shows that a mapwith the stated properties exists.

To show uniqueness up to Thurston equivalence, let g : S2 → S2

be another continuous map that is cellular for (D1,D0) and compat-ible with L. Then for each cell τ ∈ D1, the maps f |τ and g|τ arehomeomorphisms of τ onto L(τ) ∈ D0. Hence ϕτ = (g|τ)−1 (f |τ) isa homeomorphism of τ onto itself. The family ϕτ , τ ∈ D1, of thesehomeomorphism is obviously compatible under inclusions: if σ, τ ∈ D1

and σ ⊂ τ , then ϕτ (p) = ϕσ(p) for all p ∈ σ.Using this, we can define a map ϕ : S2 → S2 as follows. For p ∈ S2

pick τ ∈ D1 with p ∈ τ . Then set ϕ(p) = ϕτ (p). The compatibilityproperties of the homeomorphisms ϕτ imply that ϕ is well-defined.

5.6. FLOWERS 147

Indeed, suppose that τ, τ ′ are cells in D1 with p ∈ τ ∩ τ ′. There existsa unique cell σ ∈ D1 with p ∈ int(σ). It follows from Lemma 5.3 (ii)that σ ⊂ τ ∩ τ ′. Hence

ϕτ (p) = ϕσ(p) = ϕτ ′(p).

It is clear that g ϕ = f . Moreover, ϕ|τ = ϕτ is a homeomorphismof τ onto itself whenever τ ∈ D1. By Lemma 5.14 (i) and (iii) thisimplies that ϕ is a homeomorphism of S2 onto itself that is isotopic toidS2 rel. V1.

The sets of postcritical points of f and g are contained in the setof 0-vertices and hence in V1. So Thurston equivalence of f and gimmediately follows from the facts that gϕ = f and that ϕ is isotopicto idS2 rel. V1.

5.6. Flowers

Throughout this section f : S2 → S2 is a given Thurston map with# post(f) ≥ 3. We fix a Jordan curve C ⊂ S2 with post(f) ⊂ C.As discussed in Section 5.3 we consider the cell decompositions Dn =Dn(f, C) and use the related terminology and notation.

The results in this section are based on the following concept.

Definition 5.25 (n-Flowers). Let n ∈ N0, and p ∈ S2 be an n-vertex. Then the n-flower of p is defined as

W n(p) :=⋃int(c) : c ∈ Dn, p ∈ c.

So the n-flower W n(p) of the n-vertex p is the union of the interiorsof all cells in cycle of p in Dn (see Figure 5.1 as well as Lemma 5.12 (v)and the discussion after this lemma).

The main reason why we introduce flowers is the following. Con-sider a simply connected domain U ⊂ S2 not containing a postcriticalpoint of f and branches gn of f−n defined on U . Then it may happenthat the number of n-tiles intersecting gn(U) is unbounded as n→∞,even if the diameter of U (with respect to some base metric on S2) issmall. For example, this happens when f has a periodic critical pointp (see Section 17.2), and U spirals around one of the points in the cyclegenerated by p. However if diam(U) is sufficiently small, then gn(U) isalways contained in one n-flower as we shall see.

We first prove some basic properties of flowers.

Lemma 5.26. Let n ∈ N0, and p ∈ S2 be an n-vertex. As inLemma 5.12 let e1, . . . , ed be the n-edges and X1, . . . , Xd be the n-tilesof the cycle of p, where d ∈ N, d ≥ 2, is the length of the cycle.


(i) Then d = 2 degfn(p) and the set W n(p) is homeomorphic toD, i.e., it is an open, connected, and simply connected neigh-borhood of p. It contains no other n-vertex, and we have

W n(p) = p ∪d⋃i=1

int(Xi) ∪d⋃i=1

int(ei)(5.11)

= S2 \⋃c ∈ Dn : c ∈ Dn, p /∈ c.

(ii) We have W n(p) = X1 ∪ · · · ∪ Xd, and the set ∂W n(p) is theunion of all n-edges e with p /∈ e and e ⊂ ∂Xi for some i ∈1, . . . , d.

(iii) If c is an arbitrary n-cell, then either p ∈ c and c ⊂ W n(p),or c ⊂ S2 \W n(p).

Proof. (i) By Remark 5.23 the length d of the cycle of the vertexp (in the cell decomposition Dn) is a multiple d = kd′ of the length d′ ofthe cycle of the image point q = fn(p) (in the cell decomposition D0),where k is the degree of fn at p. Since d′ = 2, we have d = 2 degfn(p)as claimed.

The first equality in (5.11) follows from Lemma 5.12 (v). Based onthis, the argument in Case 3 of the proof of Lemma 5.22 shows thatthe set W n(p) is homeomorphic to D. Hence W n(p) is open and simplyconnected, and it follows from the first equality in (5.11) that W n(p)contains no other n-vertex than p.

Let M = S2 \⋃c ∈ Dn : c ∈ Dn, p /∈ c. If x ∈ W n(p), then x

is an interior point in one of the cells τ forming the cycle of p. So if cis any n-cell with x ∈ c, then τ ⊂ c by Lemma 5.3 (ii). This impliesp ∈ c, and so x ∈M by definition of M . Hence W n(p) ⊂M .

Conversely, if x ∈ M , let τ be an n-cell of smallest dimension thatcontains x. Obviously, x ∈ int(τ). On the other hand, the definitionof M implies that p ∈ τ . Hence τ is a cell in the cycle of p, and sox ∈ W n(p). We conclude M ⊂ W n(p), and so M = W n(p) as desired.

(ii) Equation (5.11) implies W n(p) = X1 ∪ · · · ∪Xn.Every point x ∈ ∂W n(p) is contained in one of the sets ∂Xi. Since

W n(p) is open, the point x is not contained in p∪ int(ei−1)∪ int(ei) ⊂W n(p) and hence in one of the n-edges e in the boundary of Xi distinctfrom ei−1 and ei; note that there exists such an edge, because each n-tile is an m-gon, where m = # post(f) ≥ 3, and so contains more thantwo n-edges in its boundary. Then p /∈ e, and x is contained in ann-edge with the desired properties.

5.6. FLOWERS 149

Conversely, if e is an n-edge with p /∈ e and e ⊂ ∂Xi, then e ⊂S2 \W n(p) by (5.11), and e ⊂ Xi ⊂ W n(p). Hence e ⊂ ∂W n(p).

(iii) This follows from (i) and (5.11).

Note that if we color tiles as in Lemma 5.19, then the colors of thetiles X1, . . . , Xd associated to an n-flower as in the previous lemma willalternate.

Lemma 5.27. Let k, n ∈ N0.

(i) If p ∈ S2 is an (n+k)-vertex, then fk(W n+k(p)) = W n(fk(p)).

(ii) If q ∈ S2 is an n-vertex, then the connected components off−k(W n(q)) are the (n+ k)-flowers W n+k(p), p ∈ f−k(q).

(iii) A connected set K ⊂ S2 is contained in an (n + k)-flower ifand only if fk(K) is contained in an n-flower.

Proof. (i) It is clear that q = fk(p) is an n-vertex. Let e′1, . . . , e′d′

be the (n+k)-edges and X ′1, . . . , X′d′ be the (n+k)-tiles in the cycle of

p, and define ei = fk(e′i) and Xi = fk(X ′i) for i = 1, . . . , d′. Then fromRemark 5.23 it follows that e1, . . . , ed′ are the n-edges and X1, . . . , Xd′

are the n-tiles in the cycle of q . Here we may have possible repetitionsof edges and tiles. Since the map fk is cellular for (Dn+k,Dn), we havefk(int(e′i)) = int(ei) and fk(int(X ′i)) = int(Xi) for all i = 1, . . . , d′.Using this and (5.11) the statement follows.

(ii) If p ∈ f−1(q), then p is an (n+k)-vertex. By (i) the (n+k)-flowerW n+k(p) is an open and connected subset of f−k(W n(q)). Suppose thatx ∈ ∂W n+k(p). Then by Lemma 5.26 (ii) there exists an (n + k)-tileX ′, and an (n + k)-edge e′ with p ∈ X ′, p /∈ e′, and x ∈ e′ ⊂ ∂X ′.Then X = fk(X ′) is an n-tile, e = fk(e′) is an n-edge, and q ∈ X,f(x) ∈ e ⊂ ∂X. Since fk|X ′ is a homeomorphism of X ′ onto X, wealso have q /∈ X. Lemma 5.26 (ii) implies that fk(x) ∈ ∂W n(q), andso fk(x) /∈ W n(q), because flowers are open sets.

We conclude that x ∈ S2 \ f−k(W n(q)), and so ∂W n+k(p) ⊂ S2 \f−k(W n(q)). It now follows from Lemma 5.4 that W n+k(p) is a con-nected component of f−k(W n(q)).

Conversely, suppose that U is a connected component of f−k(W n(q)).Then U is an open set and so it meets the interior int(X ′) of some(n+k)-tile X ′. Then X = fk(X ′) is an n-tile that meets W n(q). Henceq ∈ X, and so there exists an (n+ k)-vertex p ∈ X ′ with fk(p) = q.

Then by the first part of the proof, the set W n+k(p) is a connectedcomponent of f−k(W n(q)). Since W n+k(p) contains the set int(X ′) andso meets U , we must have W n+k(p) = U.


(iii) Suppose K is contained in the (n + k)-flower W n+k(p). Thenby (i) the set fk(W n+k(p)) = W n(fk(p)) is an n-flower and it containsfk(K).

Conversely, if fk(K) is contained in the n-flower W n(q), then K is aconnected set in f−k(W n(q)). Hence K lies in a connected componentof f−k(W n(q)), and hence in an (n+ k)-flower by (ii).

Similarly, as we defined an n-flower for an n-vertex, one can alsodefine an edge flower for an n-edge. These sets provide “canonical”neighborhoods for n-vertices and n-edges defined in terms of n-cells.

Definition 5.28 (Edge flowers). Let n ∈ N0, and e be an n-edge.Then the n-edge flower of e is defined as

W n(e) :=⋃int(c) : c ∈ Dn, c ∩ e 6= ∅.

We list some properties of edge flowers. They correspond to similarproperties of n-flowers as in Lemma 5.26. Note that in contrast to ann-flower, an edge flower will not be simply connected in general.

Lemma 5.29. Let e be an n-edge whose boundary ∂e consists of then-vertices u and v.

(i) Then W n(e) is an open set containing e, and

(5.12) W n(e) = W n(u) ∪W n(v) = S2 \⋃c : c ∈ Dn, c ∩ e = ∅.

(ii) We have W n(e) =⋃X ∈ Xn : X ∩ e 6= ∅ and

∂W n(e) =⋃c ∈ Dn : c ∩ e = ∅ and

there exists X ∈ Xn with X ∩ e 6= ∅ and c ⊂ ∂X,where each n-cell c in the last union either consists of onen-vertex or is an n-edge.

(iii) If c is an arbitrary n-cell, then either c∩e 6= ∅ and c ⊂ W n(e),or c ⊂ S2 \W n(e).

Proof. (i) It follows from Lemma 5.3 (i) that an n-cell c meets e ifand only if it contains one of the endpoints u and v of e. HenceW n(e) =W n(u) ∪W n(v) by the definition of flowers. By Lemma 5.26 (i) thisimplies that W n(e) is open, and, since e is an edge in the cycles of uand v, we also have

e = u ∪ int(e) ∪ v ⊂ W n(u) ∪W n(v) = W n(e).

Let M = S2 \⋃c : c ∈ Dn, c ∩ e = ∅. If an n-cell c does

not meet e, then it contains neither u nor v. Hence by (5.11) we haveS2\M ⊂ (S2\W n(u))∩(S2\W n(v)) = S2\W n(e), and so W n(e) ⊂M .

5.7. JOINING OPPOSITE SIDES 151

Conversely, let x ∈ M be arbitrary, and c be the unique n-cell csuch that x ∈ int(c). Then c ∩ e 6= ∅ and so u ∈ c or v ∈ c. It followsthat x ∈ W n(u) ∪W n(v) = W n(e). We conclude that M ⊂ W n(e),and so M = W n(e) as claimed.

(ii) By Lemma 5.3 (i) an n-tile X meets e if and only if X containsu or v. Hence by (i) and Lemma 5.26 (ii) we have

W n(e) = W n(u) ∪W n(v) =⋃X ∈ Xn : X ∩ e 6= ∅

as desired.For the second claim suppose that c is an n-cell and X an n-tile

with c∩e = ∅, X∩e 6= ∅, and c ⊂ ∂X. Then c ⊂ S2\W n(e) and c must

be an n-edge or consist of an n-vertex. Moreover, c ⊂ X ⊂ W n(e). Itfollows that c ⊂ ∂W n(e).

Conversely, let x be a point in ∂W n(e). Then by (i) the point x isalso a boundary point of W n(u) or W n(v), say x ∈ ∂W n(u).

By Lemma 5.26 (ii) there exists an n-edge e′ and an n-tile X withx ∈ e′, u ∈ X, u /∈ e′ and e′ ⊂ ∂X. If x is an n-vertex, we letc = x. Then c is an n-cell and we have c ∩ e = ∅, because W n(e)is an open neighborhood of e and c lies in ∂W n(e) ⊂ S2 \ W n(e).Moreover, X ∩ e 6= ∅ and c ⊂ e′ ⊂ ∂X. So c is an n-cell with thedesired properties containing x.

If x is not a vertex we put c = e′. Again if c ∩ e = e′ ∩ e = ∅, thenc is an n-cell with the desired properties containing x.

The other case, where e′ ∩ e 6= ∅, leads to a contradiction. Indeed,then we have v ∈ e′. Moreover, since x is not a vertex, it follows thatx ∈ int(e′); but then x ∈ int(e′) ⊂ W n(v) ⊂ W n(e) which is impossible,because x ∈ ∂W n(e) ⊂ S2 \W n(e).

(iii) If c is an n-cell and c∩ e = ∅, then c ⊂ S2 \W n(e). If c∩ e 6= ∅,then c contains u or v, and so c ⊂ W n(u) ∪W n(v) = W n(e).

5.7. Joining opposite sides

In this section f : S2 → S2 will again be a given Thurston map with# post(f) ≥ 3 and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Wefix a base metric d on S2 that induces the given topology and considerthe cell decompositions Dn = Dn(f, C) as introduced in Section 5.3.

We will define a constant δ0 > 0 such that any connected set ofdiameter< δ0 (with respect to the base metric d) is contained in a single0-flower (as introduced in Section 5.6). There is a slight difference forthe cases # post(f) = 3 and # post(f) ≥ 4. In order to treat these twocases simultaneously, the following definition is useful.


Definition 5.30 (Joining opposite sides). A set K ⊂ S2 joinsopposite sides of C if # post(f) ≥ 4 and K meets two disjoint 0-edges,or if # post(f) = 3 and K meets all three 0-edges.

We then define

(5.13) δ0 = δ0(f, C) = infdiam(K) : K ⊂ S2 is a set

joining opposite sides of C.

Then δ0 > 0. For if # post(f) = 4, then δ0 is bounded below bythe positive number

mindist(e, e′) : e and e′ are disjoint 0-edges.

If # post(f) = 3 and we had δ0 = 0, then it would follow from asimple limiting argument that the three 0-edges had a common point.This is absurd.

Lemma 5.31. A connected set K ⊂ S2 joins opposite sides of C ifand only if K is not contained in a single 0-flower (of a 0-vertex).

Proof. If K is contained in a 0-flower W 0(p), where p ∈ C is a0-vertex, then K meets at most two 0-edges, namely the ones that havethe common endpoint p. So K does not join opposite sides of C.

Conversely, suppose K does not join opposite sides of C. We haveto show that K is contained in some 0-flower. Note that K cannotmeet three distinct 0-edges.

If K does not meet any 0-edge, then K is contained in every 0-flower. If K meets only one 0-edge e, then K is contained in the0-flowers W 0(u) and W 0(v), where u and v are the endpoints of e.

If K meets two edges, then these edges share a common endpointv ∈ V0 = post(f). This is always true if # post(f) = 3 and followsfrom the fact that K does not join opposite sides of C if # post(f) ≥4. Moreover, K cannot meet a third 0-edge which implies that K ⊂W 0(v).

By the previous lemma every connected setK ⊂ S2 with diam(K) <δ0 is contained in a 0-flower.

Lemma 5.32. Let n ∈ N0, and δ0 > 0 be as in (5.13).

(i) If K ⊂ S2 is a connected set with diam(K) < δ0, then eachcomponent of f−n(K) is contained in some n-flower.

(ii) If γ : [0, 1] → S2 is a path such that diam(γ) < δ0, then eachlift γ of γ by fn has an image that is contained in some n-flower.


Here by definition a lift of γ by fn is any path γ : [0, 1]→ S2 withγ = fn γ.

Proof. (i) The set K is contained in some 0-flower W 0(p), p ∈ V0,by Lemma 5.31 and the definition of δ0. So if K ′ is a component off−n(K), then K ′ is contained in a component of f−n(W 0(p)), and hencein an n-flower by Lemma 5.27 (ii).

(ii) The reasoning is exactly the same as in (i). The image of γ iscontained in some 0-flower; by Lemma 5.27 (ii) this implies that theimage of γ is contained in an n-flower.

We will often have to estimate how many tiles are needed to connectcertain points. If we have a condition that is formulated “at the toplevel”, i.e., for connecting points in C, then the map fn can be used totranslate this to n-tiles.

Lemma 5.33. Let n ∈ N0, and K ⊂ S2 be a connected set. If thereexist two disjoint n-cells σ and τ with K ∩ σ 6= ∅ and K ∩ τ 6= ∅, thenfn(K) joins opposite sides of C.

Proof. It suffices to show that K is not contained in any n-flower,because then fn(K) is not contained in any 0-flower (Lemma 5.27 (iii))and so fn(K) joins opposite sides of C (Lemma 5.31). We considerseveral cases.

Case 1. One of the cells is an n-vertex, say σ = v, where v ∈ Vn.Then v ∈ K, so the only n-flower that K could possibly be containedin is W n(v), because no other n-flower contains the n-vertex v. Butsince σ and τ are disjoint, we have v /∈ τ , and so τ ⊂ S2 \ W n(v).Hence K ∩ (S2 \W n(v)) 6= ∅, and so W n(v) does not contain K.

Case 2. Suppose one of the cells is an n-edge, say σ = e ∈ En.Then e has two endpoints u, v ∈ Vn. The only n-flowers that meet eare W n(u) and W n(v); so these n-flowers are the only ones that couldpossibly contain K. But the set W n(e) = W n(u) ∪ W n(v) does notcontain K, because K meets the set τ which lies in the complement ofW n(e).

Case 3. One of the cells in an n-tile, say σ ∈ Xn. Then K meets∂X. Since ∂X consists of n-edges, the set K meets an n-edge disjointfrom τ . So we are reduced to Case 2.


For n ∈ N0 we denote by Dn the minimal number of n-tiles requiredto form a connected set joining opposite sides of C; more precisely,

(5.14) Dn = Dn(f, C) = minN ∈ N : there exist X1, . . . , XN ∈ Xn

such that K =N⋃j=1

Xj is connected and joins opposite sides of C.

Of course, Dn depends on f and the choice of C. If we want to empha-size this dependence, we write Dn = Dn(f, C).

From Lemma 5.33 we can immediately derive the following conse-quence.

Lemma 5.34. Let n, k ∈ N0. Every set of (n+k)-tiles whose union isconnected and meets two disjoint n-cells contains at least Dk elements.

Proof. Suppose K is a union of (n+k)-tiles with the stated prop-erties. Then the images of these tiles under fn are k-tiles and fn(K)joins opposite sides of C by Lemma 5.33. Hence there exist at leastDk distinct k-tiles in the union forming fn(K) and hence at least Dk

distinct (n+ k)-tiles in K.

Lemma 5.35. There exists M ∈ N with the following property:

(i) Each n-tile, n ∈ N, can be covered by M (n− 1)-flowers.

(ii) Each n-tile, n ∈ N0, can be covered by M (n+ 1)-flowers.

For easier formulation of this lemma and the subsequent proof, weassume for simplicity that a cover by at most M elements containsprecisely M elements. This can always be achieved by repetition ofelements in the cover.

Proof. (i) Let δ0 > 0 be as in (5.13). Then there exists M ∈ Nsuch that each of the finitely many 1-tiles X is a union of M connectedsets U ⊂ X with diam(U) < δ0. If Y is an arbitrary n-tile, n ≥ 1, thenZ = fn−1(Y ) is a 1-tile and fn−1|Y a homeomorphism of Y onto Z.Hence Y is a union of M sets of the form (fn−1|Y )−1(U), where U ⊂ Zis connected and diam(U) < δ0. Each set (fn−1|Y )−1(U) is connectedand so by Lemma 5.32 (i) it lies in an (n− 1)-flower. Hence Y can becovered by M (n− 1)-flowers.

(ii) There exists M ∈ N such that each of the two 0-tiles X can becovered by M connected sets U ⊂ X with diam(f(U)) < δ0. If Y is anarbitrary n-tile, then Z = fn(Y ) is a 0-tile. By the same reasoning asabove, the set Y is a union of M sets of the form (fn|Y )−1(U), whereU ⊂ Z is connected and diam(f(U)) < δ0.


Then U ′ = (fn|Y )−1(U) is connected, and fn+1(U ′) = f(U) whichimplies diam(fn+1(U ′)) < δ0. Hence by Lemma 5.32 (i) the set U ′ iscontained in some (n + 1)-flower. Since M of the sets U ′ cover Y , itfollows that each n-tile can be covered by M (n+ 1)-flowers.

Lemma 5.36. Let C and C be two Jordan curves in S2 that bothcontain post(f). Then there exists a number M such that each n-tile

for (f, C) is covered by M n-flowers for (f, C).

Proof. The proof is very similar to the proof of Lemma 5.35.Let δ0 = δ0(f, C) > 0 be the number as defined in (5.13). There

exists a number M such that each of the two 0-tiles X for (f, C) isa union of M connected sets U ⊂ X with diam(U) < δ0. If Y is an

arbitrary n-tile for (f, C), then Z = fn(Y ) is a 0-tile for (f, C) and fn|Yis a homeomorphism of Y onto Z. Hence Y is a union of M sets of theform (fn|Y )−1(U), where U ⊂ Z is connected and diam(U) < δ0. Eachset (fn|Y )−1(U) is connected and so by Lemma 5.32 (i) it lies in ann-flower for (f, C). Hence Y can be covered by M such n-flowers.

CHAPTER 6

Expansion

In this chapter we revisit the notion of an expanding Thurston map (seeDefinition 2.2) and study it in greater depth. We will establish basicproperties of this concept. In particular, we will show that expansionis a purely topological property of a given Thurston map. In the lastsection of this chapter we define the canonical orbifold metric of arational Thurston map. The results of this section had already beenused in our characterization of expanding rational Thurston maps (seethe proof of Theorem 1.2).

6.1. Definition of expansion revisited

Let S2 be a 2-sphere. In the following, it is often convenient to formu-late some essentially topological properties in metric terms. For this wefix a base metric on S2 that induces the given topology. Notation formetric terms will refer to this base metric unless otherwise indicated.

If C ⊂ S2 is a Jordan curve with post(f) ⊂ C, then the n-tilesfor (f, C) are precisely the closures of the connected components ofS2 \ f−n(C) (see Proposition 5.17 (v)), and so

mesh(f, n, C) = maxX∈Xn

diam(X).

Thus a Thurston map f is expanding if and only if there is a Jordancurve C ⊂ S2 with post(f) ⊂ C such that

(6.1) maxX∈Xn

diam(X)→ 0 as n→∞,

where the tiles are defined with respect to (f, C). We want to convinceourselves that this is condition is independent of the choice of the curveC.

Lemma 6.1. Let f : S2 → S2 be a Thurston map and C, C ⊂ S2 be

Jordan curves with post(f) ⊂ C, C. Then

limn→∞

mesh(f, n, C) = 0

157

158 6. EXPANSION

if and only if

limn→∞

mesh(f, n, C) = 0.

Proof. Let C, C ⊂ S2 be as in the statement of the lemma. Assumethat limn→∞mesh(f, n, C) = 0. This means that f is expanding. Then

maxX∈Xn

diam(X) = mesh(f, n, C)→ 0

as n → ∞, where Xn is the set of n-tiles for (f, C). Lemma 5.26 (ii)implies that

(6.2) diam(W n(v)) ≤ 2 maxX∈Xn

diam(X)

for each n-flower for (f, C).Now we consider tiles for (f, C). Recall from Lemma 5.36 that there

exists a number M with the following property. Each n-tile for (f, C)can be covered by M n-flowers for (f, C). If a connected set is coveredby a finite union of connected sets, then its diameter is bounded bythe sum of the diameters of the sets in the union. Combining this with(6.2), we conclude that

mesh(f, n, C) = maxdiam(X) : X is an n-tile for (f, C)≤ M max

p∈Vndiam(W n(p))

≤ 2M maxX∈Xn

diam(X)

= 2M mesh(f, n, C).

Here Vn denotes the set of n-vertices for (f, C). Hence mesh(f, n, C)→0 as n→∞ as desired.

The other implication is obtained by reversing the roles of C and

C.

The lemma shows that a Thurston map f : S2 → S2 is expanding ifand only if mesh(f, n, C)→ 0 as n→∞ whenever C ⊂ S2 is arbitraryJordan curve with post(f) ⊂ C. In particular, expansion is a propertyof the map f alone and independent of the choice of the Jordan curveC.

Lemma 5.36 that was used in the previous proof admits an improve-ment for expanding Thurston maps.

Lemma 6.2. Let f : S2 → S2 be an expanding Thurston map. Sup-

pose that C and C are two Jordan curves in S2 that both contain post(f).Then there exists a number M ∈ N with the following property: if

6.1. DEFINITION OF EXPANSION REVISITED 159

n, k ∈ N0, then every (n + k)-tile for (f, C) can be covered by M n-flowers for (f, C).

Proof. The argument is a small variation of the one that we usedto establish Lemma 5.36.

Let δ0 = δ0(f, C) > 0 be the number as defined in (5.13). Since fis expanding, there exists a number M ∈ N such that each tile X for

(f, C) is a union of M connected sets U ⊂ X with diam(U) < δ0 (in theproof of Lemma 5.36 we could guarantee this only for the two 0-tiles for

(f, C)). Indeed, since f is expanding this is trivially true for all tiles Xof sufficiently high level, because then diam(X) < δ0. There are only

finitely many tiles X for (f, C) with diam(X) ≥ δ0. The existence of asuitable constant M easily follows.

Now let n, k ∈ N0 and suppose Y is an arbitrary (n + k)-tile for

(f, C). Then Z = fn(Y ) is a k-tile for (f, C) and fn|Y is a homeo-morphism of Y onto Z. Hence Y is a union of M sets of the form(fn|Y )−1(U), where U ⊂ Z is connected and diam(U) < δ0. Eachset (fn|Y )−1(U) is connected and so by Lemma 5.32 (i) it lies in ann-flower for (f, C). Hence Y can be covered by M such n-flowers.

Our definition of expansion is somewhat ad hoc, but it has theadvantage that it relates to the geometry of tiles. As we will see,equivalent and maybe more conceptual descriptions can be given interms of the behavior of open covers of S2 under pull-backs by theiterates of the map. One can formulate this in purely topological terms.This shows that expansion is a topological property of the map; ourdefinition was based on a metric concept (namely the mesh size), butthis was just for convenience.

We start with some definitions. Let U be an open cover of S2.We define mesh(U) to be the supremum of all diameters of connectedcomponents of sets in U . If g : S2 → S2 is a continuous map, then thepull-back of U by g is defined as

g−1(U) = V : V connected component of g−1(U), where U ∈ U.

Obviously, g−1(U) is also an open cover of S2. Similarly, we denote byg−n(U) the pull-back of U by gn.

Proposition 6.3. Let f : S2 → S2 be a Thurston map. Then thefollowing conditions are equivalent:

(i) The map f is expanding.

(ii) There exists δ0 > 0 with the following property: if U is a coverof S2 by open and connected sets that satisfies mesh(U) < δ0,

160 6. EXPANSION

thenlimn→∞

mesh(f−n(U)) = 0.

(iii) There exists an open cover U of S2 with

limn→∞

mesh(f−n(U)) = 0.

(iv) There exists an open cover U of S2 with the following property:for every open cover V of S2 there exists N ∈ N such thatf−n(U) is finer than V for every n ∈ N with n > N ; i.e.,for every set U ′ ∈ f−n(U) there exists a set V ∈ V such thatU ′ ⊂ V .

Condition (iii) is the notion of expansion as defined by Haıssinsky-Pilgrim (see [HP09, Sect. 2.2]). So our notion of expansion agreeswith the one in [HP09]. Condition (iv) is essentially a reformulationof (iii) in purely topological terms without reference to the base metricon S2 (which enters in the definition of the mesh of an open cover).One can reformulate (ii) in a similar spirit. If there exists a Jordancurve C ⊂ S2 with post(f) ⊂ C and f(C) ⊂ C, then expansion of themap f can be characterized in yet another way (see Lemma 12.6).

Proof. We will show (i) ⇒ (ii) ⇒ (iii) ⇒ (i) and (iii) ⇒ (iv)⇒ (iii).

(i) ⇒ (ii): Suppose f is expanding. Pick a Jordan curve C ⊂ S2 withpost(f) ⊂ C, and let δ0 > 0 be as in (5.13) (note that # post(f) ≥ 3by Corollary 7.2). Suppose U is a cover of S2 by open and connectedsets that satisfies mesh(U) < δ0. If U ∈ U , then U is connected anddiam(U) < δ0. So if V is an arbitrary connected component of f−n(U),then by Lemma 5.32 the set V is contained in an n-flower for (f, C).Hence

diam(V ) ≤ 2 mesh(f, n, C),which implies

mesh(f−n(U)) ≤ 2 mesh(f, n, C).Since f is an expanding Thurston map, we have mesh(f, n, C) → 0,and so mesh(f−n(U))→ 0 as n→∞.

(ii) ⇒ (iii): Obvious.

(iii) ⇒ (i): Suppose U is an open cover of S2 as in (iii). Pick a Jordancurve C ⊂ S2 with post(f) ⊂ C, and let δ > 0 be a Lebesgue numberfor the cover U , i.e., every set K ⊂ S2 with diam(K) < δ is containedin a set U ∈ U . We can find a number M ∈ N such that each of thetwo 0-tiles for (f, C) can be written as a union of M connected sets Vwith diam(V ) < δ. Then each such set V is contained in a set U ∈ U .

6.2. FURTHER RESULTS ON EXPANSION 161

Now if X is an arbitrary n-tile for (f, C), then Y = fn(X) is a 0-tilefor (f, C) and fn|X is a homeomorphism of X onto Y . Hence X is aunion of M connected sets of the form (fn|Y )−1(V ), where V ⊂ Yis connected and lies in a set U ∈ U . Then (fn|X)−1(V ) lies in acomponent of f−n(U), and so

diam((fn|X)−1(V )) ≤ mesh(f−n(U)).

This impliesdiam(X) ≤M mesh(f−n(U)).

Hencemesh(f, n, C) ≤M mesh(f−n(U)).

Since mesh(f−n(U)) → 0, we also have mesh(f, n, C) → 0 as n → ∞.It follows that f is expanding.

(iii) ⇒ (iv): Suppose U is an open cover of S2 as in (iii), and V is anarbitrary open cover of S2. Let δ > 0 be a Lebesgue number for thecover V , i.e., every set K ⊂ S2 with diam(K) < δ is contained in aset V ∈ V . By (iii) we can find N ∈ N such that mesh(f−n(U)) < δfor n > N . If n > N and U ′ is a set in f−n(U), then diam(U ′) < δby definition of mesh(f−n(U)). Hence there exists V ∈ V such thatU ′ ⊂ V .

(iv) ⇒ (iii): Suppose U is an open cover of S2 as in (iv). Then U alsosatisfies condition (iii); indeed, let ε > 0 be arbitrary, and let V bethe open cover of S2 consisting of all open balls of radius ε/2. Thendiam(V ) ≤ ε for all V ∈ V . Moreover, by (iv) there exists N ∈ Nsuch that for n > N every set in f−n(U) is contained in a set in V . Inparticular, mesh(f−n(U)) ≤ ε for n > N . This shows that U satisfiescondition (iii).

6.2. Further results on expansion

In this section we collect various other useful results related to expan-sion.

Lemma 6.4. Let f : S2 → S2 be a Thurston map, n ∈ N, andF = fn. Then F is a Thurston map with post(F ) = post(f). The mapf is expanding if and only if F is expanding.

Proof. Since f is a Thurston map, the map F is a branchedcovering map on S2 with post(F ) = post(f) (see Section 2.2) anddeg(F ) = deg(f)n ≥ 2. Hence F is also a Thurston map.

Fix a Jordan curve C ⊂ S2 with post(f) = post(F ) ⊂ C. It followsfrom the definitions that

mesh(F, k, C) = mesh(f, nk, C)

162 6. EXPANSION

for all k ∈ N0. If f is expanding, then by Lemma 6.1 we have

limk→∞

mesh(f, k, C) = 0

which implies that

mesh(F, k, C) = mesh(f, nk, C)→ 0

as k →∞. Hence F is expanding.

Conversely, suppose that F is expanding. Then we know that

(6.3) limk→∞

mesh(F, k, C) = limk→∞

mesh(f, nk, C) = 0.

Let the constant M ≥ 1 be as in Lemma 5.35 for the map f and theJordan curve C. By an argument similar as in the proof of Lemma 6.1one can show that

mesh(f, l + 1, C) ≤ 2M mesh(f, l, C)

for all l ∈ N0. This implies

mesh(f, l, C) ≤ (2M)n mesh(f, nbl/nc, C)

for all l ∈ N0 and so by (6.3) we have mesh(f, l, C) → 0 as l → ∞.This shows that f is expanding.

A map f : S2 → S2 is called eventually onto, if for each non-emptyopen set U ⊂ S2 there is an iterate fn such that fn(U) = S2.

Lemma 6.5. Let f : S2 → S2 be an expanding Thurston map. Thenf is eventually onto.

As we will see (Example 6.14), there are Thurston maps that areeventually onto, but not expanding.

Proof. Let f : S2 → S2 be an expanding Thurston map. Picka Jordan curve C ⊂ S2 with post(f) ⊂ C as in Definition 2.2. Asbefore, we denote the black and white 0-tiles for (f, C) by X0

b and X0w ,

respectively.Let U ⊂ S2 be an arbitrary non-empty open set, and B(a, ε) with

a ∈ U and ε > 0 be an open ball contained in U . Since f is expanding,there is a number n ∈ N such that mesh(f, n, C) < ε/4. Then eachn-tile has diameter < ε/4. Let X be an n-tile containing the centera ∈ U of B(a, ε), and Y an n-tile that shares an n-edge with X. ThenX ∪ Y ⊂ B(a, ε) ⊂ U , and so fn(U) ⊃ fn(X ∪ Y ) = X0

w ∪ X0b = S2.

The claim follows.


Recall that a metric d on a space S is a length metric if for any twopoints x, y ∈ S we have d(x, y) = inf length(γ), where the infimum istaken over all paths γ in S joining x to y. Using this concept, one canformulate a simple criterion when a Thurston map is expanding.

Lemma 6.6. Let d be a length metric on S2 that induces the giventopology on S2, and let f : S2 → S2 be a Thurston map. If f uniformlyexpands the d-length of paths, i.e., if there is a number ρ > 1 such thatfor every path γ in S2 we have

lengthd(f γ) ≥ ρ lengthd(γ),

then f is expanding.

Proof. Let d be a length metric on S2 such that the Thurstonmap f : S2 → S2 expands the d-length of paths as in the statementof the lemma. In the following, all metric notion refers to this metricd. To prove that f is expanding, we will show that condition (iii) inProposition 6.3 is satisfied for a suitable cover U of S2.

We pick a Jordan curve C ⊂ S2 with post(f) ⊂ C, and consider tilesfor (f, C). Then the corresponding 0-flowers W 0(p), p ∈ post(f), forma cover of S2. Indeed, every point in u ∈ S2 belongs to the interior ofa (unique) 0-cell. If u is 0-vertex, then it is contained in W 0(u). If uis the interior of a 0-edge e, then u belongs to the two 0-flowers W 0(p)and W 0(q), where p and q are the endpoints of e. Finally, if u is aninterior point of one of the two 0-tiles, then u belongs to every 0-flower.

In order to obtain a cover U as in in Proposition 6.3 (iii), we wantto shrink each 0-flower W 0(p) slightly to a new set U so that we havegood control for the length of paths joining points in U to p insideW 0(p). To set this up, let r > 0 and define W 0

r (p) for p ∈ post(f) tobe the set of all points u ∈ W 0(p) such that u and p can be joined bya path γ in W 0(p) with length(γ) < r.

Claim. There exists r > 0 such that each point in S2 is containedin one of the sets W 0

r (p), p ∈ post(f).

To prove this, let u ∈ S2 be arbitrary. As we have seen, the 0-flowers cover S2, and so there exists p ∈ post(f) such that u ∈ W 0(p).Since W 0(p) is a region, there exists a path γ in W 0(p) joining u and p.The path γ may have infinite length, but it is easy to modify it so thatits length becomes finite. Indeed, we can find a finite sequence of pointsalong the original path γ including p and u so that successive points areclose. We may assume that they are so close that they can be joinedby a path of finite length that stays inside W 0(p). Here it is importantthat d is a length metric. Concatenating these paths, we obtain a newpath γ in W 0(p) joining u and p with ru := length(γ) <∞.

164 6. EXPANSION

We can choose δu > 0 such that Bu := B(u, δu) ⊂ W 0(p). Sinced is a length metric, every point v in Bu can be joined with u by apath in Bu of length < δu. If we concatenate such a path with γ, thenwe obtain a path in W 0(p) that has length < Ru + δu and stays insideW 0(p). In particular, we have uniform control for the length of suchpaths for all points in Bu. Since finitely many of the balls Bu, u ∈ S2,cover S2, the claim follows.

Now pick r > 0 as in the claim, and consider the open cover U ofS2 given by the sets W 0

r (p), p ∈ post(f). Let n ∈ N0 and p ∈ post(f)be arbitrary, and consider a component V of f−n(W 0

r (p)). Then V iscontained in a component of f−n(W 0(p)), and so there exists an n-flower W n(q) such that V ⊂ W n(q) (see Proposition 1.2) . Here q ∈ S2

is an n-vertex. Then fn(q) is a 0-vertex contained in W 0(p) whichimplies that fn(q) = p.

Now let v ∈ V be arbitrary, and u := fn(v) ∈ W 0r (p). Then there

exists a path γ in W 0r (p) ⊂ W 0(p) with length(γ) < r that joins u

and p. By Lemma 1.2 there exists a lift α of γ under fn that startsat v. Then fn α = γ and α ⊂ V ⊂ W n(q). One endpoint of α is v,while the other endpoint of α is a preimage of p under fn and hencean n-vertex. Since q is the only n-vertex in V ⊂ W n(q), it follows thatα joins v and q.

By using the fact that f expands the d-length paths by the factorρ, we see that

length(α) <1

ρnlength(fn α) =

1

ρnlength(γ) <

r

ρn.

So every point in V can be joined to q by a path of length < r/ρn. Thisimplies that diam(V ) < 2r/ρn, and it follows that mesh(f−n(U)) <2r/ρn. Since ρ > 1, we conclude that mesh(f−n(U)) → 0 as n → ∞.By Proposition 6.3, the map f is expanding.

A Thurston map f : S2 → S2 is called a Thurston polynomial ifthere exists a point in S2, denoted by ∞, that is completely invariant,i.e., f−1(∞) = ∞.

Lemma 6.7. No Thurston polynomial f is expanding.

Proof. Let f be a Thurston polynomial. We can choose a point∞ ∈ S2 that is completely invariant. Then d := degf (∞) = deg(f) ≥ 2by (2.2), and so ∞ is a critical point of f . Since ∞ is a fixed point aswell, it follows that ∞ ∈ post(f).

We pick a Jordan curve C ⊂ S2 with post(f) ⊂ C, and considertiles for (f, C). Each n-tile Xn is mapped by fn homeomorphically to


a 0-tile X0 (see Proposition 5.17 (i)). Since

∞ ∈ post(f) ⊂ C = ∂X0 ⊂ X0,

it follows that Xn contains a preimage of ∞ by fn. Since ∞ is com-pletely invariant, the only such preimage is ∞ itself, and so ∞ ∈ Xn.Therefore, each n-tile contains ∞.

We pick a point p ∈ S2 distinct from ∞. Since for each n ∈ N theset of all n-tiles forms a cover of S2, there exists an n-tile Xn containingp. Then

mesh(f, n, C) ≥ diamXn ≥ dist(p,∞) > 0,

and so mesh(f, n, C) 6→ 0 as n→∞. By (6.1) this means that f is notexpanding.

Let f : S2 → S2 a Thurston map. A Levy cycle for f is a multicurveΓ = γ1, . . . , γn (see Definition 2.16 (ii)) such that

(i) for each j = 1, . . . , n the set f−1(γj+1) contains a componentγj that is isotopic to γj rel. post(f) (here we set γn+1 = γ1);

(ii) for each j = 1, . . . , n the map f |γj−1 : γj−1 → γj is a homeo-morphism.

Since f |γj−1 : γj−1 → γj is a covering map, the last condition isequivalent to the requirement that the degree of this map is equal to1.

Lemma 6.8. Let f : S2 → S2 be a Thurston map and γ and σbe Jordan curves in S2 \ post(f) that are isotopic rel. post(f). Letγ1, . . . , γk with k ∈ N be the components of f−1(γ). Then f−1(σ) hasalso k components that we can label as σ1, . . . , σk such that for j =1, . . . , k,

(i) the curves γj and σj are isotopic rel. post(f),

(ii) the degrees of f |γj : γj → γ and f |σj : σj → σ agree.

Proof. This follows immediately if we lift a suitable isotopy byf . Indeed, for every isotopy H : S2 × [0, 1] → S2 rel. post(f) with

H0 = H(·, 0) = idS2 there is an isotopy H : S2× [0, 1]→ S2 rel. post(f)

with H0 = idS2 such that

(6.4) (Ht f)(p) = (f Ht)(p)

for all p ∈ S2, t ∈ [0, 1] (see Proposition 11.1).Let H be an isotopy rel. post(f) that deforms γ to σ rel. post(f),

i.e., H0 = idS2 and H1(γ) = σ, and let H be the lift of H by f as

above. Define σj := H1(γj). Then by definition γj and σj are isotopicrel. post(f) for j = 1, . . . , n. Moreover, (6.4) implies that σ1, . . . , σk

166 6. EXPANSION

are the components of f−1(σ) and that the degrees of f |γj : γj → γ andf |σj : σj → σ agree.

Now suppose that f is a Thurston map that has a Levy cycle Γ =γ1, . . . , γn. We consider the iterate F = fn and denote γ = γ1 forconvenience. If we use the previous lemma repeatedly, then we seethat there is a component γ of F−1(γ) that is isotopic to γ rel. post(f)such that the degree of F |γ : γ → γ is 1. We remark the existenceof such an iterate F = fn and such a (non-peripheral) Jordan curveγ ⊂ S2 \ post(f) is in fact equivalent to the existence of a Levy cycle,but we will not prove this here.

By using a lifting argument as in the proof of Lema 1.2 (based onProposition ??) one can easily show Levy cycles persist under Thurstonequivalence; so if the maps f : S2 → S2 and g : S2 → S2 are Thurstonequivalent, the f has a Levy cycle if and only if g has a Levy cycle.

If a Levy cycle Γ is an invariant multicurve, then it is clearly aThurston obtruction, since the the spectral radius of the correspondingThurston matrix A(f,Γ) is equal to 1. If the Levy cycle Γ is notinvariant, then it it is not hard to show that there is an invariantmulticurve Γ′ ⊃ Γ. Then spectral radius of A(f,Γ′) is ≥ 1, and so everyLevy cycle is contained in a Thurston obstruction. Hence accordingThurston’s Theorem 1.2 if a Thurston map has a Levy cycle (and ahyperbolic orbfold), then it cannot be equivalent to a rational map.

Our next lemma shows that a Levy cycle is also an obstruction fora Thurston map to be expanding.

Lemma 6.9. Let f : S2 → S2 be a Thurston map that has a Levycycle. Then f is not expanding.

Proof. Assume f has a Levy cycle. Then there exists an iterateF = fn and a non-peripheral Jordan curve γ1 ⊂ S2 \post(f) such thata component γ2 of F−1(γ1) is isotopic to γ1 rel. post(f) and the degreeof F |γ1 : γ2 → γ1 is equal to 1. From Lemma 6.8 it follows by inductionthat there is a sequence γkk∈N of Jordan curves in S2 \ post(f) allof which are isotopic to γ1 rel. post(f) such that γk+1 is a componentof F−1(γk) and the degree of F |γk+1 : γk+1 → γk is equal to 1 forall k ∈ N. Hence γk is a component of F−k(γ1) and the degree ofF k|γk+1 : γk+1 → γ1 is equal to 1. Moreover, each curve γk is non-peripheral.

We fix a Jordan curve C ⊃ post(f), and consider tiles and flowersfor (F, C). Let δ0 be as in (5.13) for the map F . Then we can decomposeγ1 into finitely many arcs α1, . . . , αl such that diam(αj) < δ0 for j =1, . . . , l. By Lemma 5.32 every component of F−k(αj) is contained in a


k-flower W kj for j = 1, . . . , l. It follows that for every k ∈ N the curve

γk is contained in the union of l k-flowers W k1 , . . . ,W

kl .

To reach a contradiction, assume now that f , and hence also F , isexpanding. Then we have

diam(γk) ≤l∑

j=1

diam(W kj ) ≤ 2l max

X∈XkdiamX → 0

as k →∞.On the other hand, we can find a finite open cover U of S2 consisting

of simply connected regions U each of which contains at most onepostcritical point of f . By what we have just seen, we can find k ∈ Nsuch that diam(γk) is smaller than the Lebesgue number of U . Then γkis contained in a set U ∈ U . Since U is simply connected and containsat most one postcritical point of f , the curve γk is peripheral. This isa contradiction showing that f is not expanding.

Example 6.10. We now present an example of a Thurston map fwith a Levy cycle. The map f will have no periodic critical points,and so it provides an example of a Thurston map without periodiccritcal points that is not expanding, contrasting Proposition 2.3. SinceLevy cycles persist under Thurston equivalence, f is also not equivalentto any expanding Thurston map. The same is actually true of themap from Example 6.14, but the map f considered here will have ahyperbolic orbifold.

For the construction we start with a topological sphere S2 that isa pillow similarly as in Section 1.1; so the pillow is obtained by gluingtogether two unit squares together along their boundaries. As before,we color one side (i.e., one square) of the pillow white, the other black.

The black side is divided horizontally into two rectangles, one ofwhich is colored white and the other colored black. The white side ofthe pillow is subdivided into four quadrilaterals, two white and twoblack ones as indicated on the left in Figure 6.1. Here we have cut thepillow along three sides so that we obtain a rectangle as shown in thepicture. The symbols in the picture indicate which sides have to beidentified to recover the pillow.

The map f is now constructed by mapping each white quadrilateralhomeomorphically to the white face, and each black quadrilateral tothe black side of the pillow. Here f maps vertices to vertices. InFigure 6.1 we have marked two vertices of each quadrilateral (in theleft figure), as well as two vertices of the pillow (in the right picture),to indicate the correspondence of vertices under f . Finally, we requirethat f agrees on sides shared by two vertices. The map f thus defined

168 6. EXPANSION

fγ

γ

Figure 6.1. A map with a Levy cycle.

is indeed a Thurston map (see Chapter 12 for a general discussion).The postcritical points correspond to the vertices of the pillow. Themap f has two critical points c1 and c2 with the following ramificationportraits:

c13:1// p1

// q1

c23:1// p2

// q2

.

Thus f has no periodic critical points, its signature is (3, 3, 3, 3), andf has a hyperbolic orbifold.

We consider the Jordan curve γ ⊂ S2 \ post(f) as indicated on theright of Figure 6.1. One of the components γ of f−1(γ) is shown atthe left (the other components of f−1(γ) are not shown). Clearly, γis isotopic rel. post(f) to γ. Furthermore, the degree of f : γ → γ isequal to 1, and so Γ = γ is a Levy cycle.

Lemma 6.9 implies that f is not expanding, and, by our earlierdiscussion, no Thurston map equivalent to f is expanding.

6.3. Lattes-type maps and expansion

In this section we study when a Lattes-type map is expanding. Wefirst derive a property of expanding linear maps.

It is easy to decide when a Lattes-type map is expanding. Therelevant condition is based on the following definition.

Let L : R2 → R2 be a linear map. We call it expanding if |λ| > 1for each of the (possibly complex) eigenvalues λ of L.

Proposition 6.11. Let f : S2 → S2 be a Lattes-type map and L =LA be the linear part of an affine map A as in Definition 3.3. Thenf is expanding (as a Thurston map) if and only L is expanding (as alinear map).

Moreover, every Lattes map is expanding.

Lemma 6.12. Suppose L : R2 → R2 is an expanding linear map.Then there exist constants ρ > 1 and n0 ∈ N such that

(6.5) |Ln(v)| ≥ ρn|v|for all v ∈ R2 and all n ∈ N with n ≥ n0.

6.3. LATTES-TYPE MAPS AND EXPANSION 169

Here |v| =√x2 + y2 denotes the usual Euclidean norm of v =

(x, y) ∈ R2.

Proof. The eigenvalues of L are the two (possibly identical) rootsλ1, λ2 ∈ C of the characteristic polynomial P (λ) = det(L − λ idR2) ofL. We may assume |λ1| ≤ |λ2|. Since L is expanding we have |λ1| > 1.The polynomial P has real coefficients, and so λ2 = λ1 if λ1 is not real.

There exists a basis of R2 consisting of two linearly independentvectors v1, v2 ∈ R2 such that the linear map L has a matrix represen-tation with respect to this basis of one of the following forms:

M1 = |λ1|(

cos θ − sin θsin θ cos θ

),where θ ∈ R,

M2 =

(λ1 00 λ2

), or M3 =

(λ1 10 λ1

).

We can find an inner product on R2 such that v1 and v2 form anorthonormal basis of R2 with respect to this inner product. If ‖ · ‖ isthe norm induced by this product, then

‖av1 + bv2‖ =√a2 + b2

for all a, b ∈ R. If L has a matrix representation given by M1 or M2,it is clear that ‖L(v)‖ ≥ |λ1|‖v‖ and so

(6.6) ‖Ln(v)‖ ≥ |λ1|n‖v‖for all v ∈ R2 and n ∈ N. If L is represented by the matrix M3, then asimilar estimates is harder to obtain, but one can show that that

‖Ln(v)‖ ≥ |λ1|n+1√4λ2

1 + n2‖v‖.

To see this, one bounds the operator norm of the matrix M−n3 by its

Hilbert-Schmidt norm. We leave the details to the reader.Since all norms on R2 are comparable, it follows that

|Ln(v)| ≥ ρn|v|

for all sufficiently large n independent of v ∈ R2 with ρ = |λ1|1/2 > 1,say.

Lemma 6.13. Let f : S2 → S2 be a quotient of a torus endo-morphism, d be metric on S2, π and Θ be maps as in (3.13), andΘ := Θ π : R2 → S2. Then the following statements are true:

(i) For each ε > 0 there exists δ > 0 such that d(Θ(x),Θ(y)) < εwhenever x, y ∈ R2 and |x− y| < δ.

170 6. EXPANSION

(ii) For each ε > 0 there exists δ > 0 such that diam(K) < εwhenever K ⊂ R2 is a connected set with diam(Θ(K)) < δ.

In (ii) it is understood that diam(K) refers to the Euclidean diam-eter of K, and diam(Θ(K)) to the diameter of Θ(K) with respect tothe metric d on S2.

Proof. Let T , Θ, and π be as in (3.13) so that Θ = Θ π. LetΓ ⊂ R2 a rank-2 lattice such that T ∼= R2/Γ. For γ ∈ Γ we denote byτγ : R2 → R2 the lattice translation given by τγ(u) := u+γ for u ∈ R2.

(i) We have to show that Θ is uniformly continuous on R2. NowΘ is uniformly continuous as a continuos map on the compact torusT . The map π is also uniformly continuous. This follows from the factthat the group consisting of all lattice translations τγ, γ ∈ Γ, consistsof isometries and acts cocompactly on R2, and that π = πτγ wheneverγ ∈ Γ. Hence Θ = Θ π is uniformly continuous on R2.

(ii) We argue by contradiction. Then there exists ε0 > 0 and con-nected sets Kn ⊂ R2 for n ∈ N such that diam(Kn) ≥ ε0 for n ∈ N,but diam(Θ(Kn))→ 0 as n→∞.

We pick a point xn ∈ Kn for n ∈ N. If we replace each set Kn bya suitable translate K ′n = τγ(Kn) with γ ∈ Γ (note that diam(K ′n) =diam(Kn) and Θ(K ′n) = Θ(Kn)), and pass to a subsequence if nec-essary, then we may assume that the sequence xn converges, sayxn → x ∈ R2 as n→∞.

Let p := Θ(x). Since Θ−1

(p) ⊂ T is a finite set, and Θ−1(p) =

π−1(Θ−1

(p)), there exists finitely many points z1, . . . , zk ∈ R2 suchthat

Θ−1(p) =k⋃l=1

(zl + Γ).

This implies that the distance of distinct points in Θ−1(p) is boundedaway from 0; so there exists a constant m > 0 such that |u − v| ≥ mwhenever u, v ∈ Θ−1(p) and u 6= v.

Pick a constant c with 0 < c < minε0/2,m. The set Kn isconnected, and has diameter diam(Kn) ≥ ε0 > 2c. Hence Kn cannot becontained in the disk z ∈ R2 : |z−xn| < c, and so it meets the circlez ∈ R2 : |z − xn| = c. It follows that there exists a point yn ∈ Kn

with |xn − yn| = c. By passing to another subsequence if necessary,we may assume that the sequence yn converges, say yn → y ∈ R2 asn → ∞. Then |x − y| = c < m. Note that Θ(xn),Θ(yn) ∈ Θ(Kn) forn ∈ N, and diam(Θ(Kn))→ 0 as n→∞. So

p = Θ(x) = limn→∞

Θ(xn) = limn→∞

Θ(yn) = Θ(y),


and x, y ∈ Θ−1(p). Since |x−y| = c > 0, we have x 6= y. So x and y aretwo distinct points in Θ−1(p) with |x − y| = c < m. This contradictsthe choice of m, and the claim follows.

Proof of Proposition 6.11. Our proof relies on various factsthat are discussed later in Chapters 6 and 8, and the Appendix.

We may assume that the given Lattes-type map is defined on the

Riemann sphere C. Let A, Θ, G be as in Definition 3.3. Then the

orbifold (C, αf ) is parabolic, and the map Θ: R2 → C ∼= C/G is theuniversal orbifold covering map. The Euclidean metric d0 on R2 passes

to the canonical orbifold metric d on C so that the map Θ: R2 → C isa path isometry (see Section A.7 for a detailed discussion).

Let α be a path in R2. If L = LA is the linear part of A, then themap Ln is the linear part of An for each n ∈ Z. So these maps onlydiffer by a translation. Since translations are isometries, it follows that

(6.7) length(Ln α) = length(An α)

for all n ∈ Z. Here and in the following metric notions on R2 refer tothe Euclidean metric d0. If γ := Θ α, then

fn γ = fn Θ α = Θ An α

for n ∈ N. Since Θ is a path isometry, we conclude that

lengthd(fn γ) = length(An α) = length(Ln α)

for all n ∈ N.Now suppose that L is expanding. Then Lemma 6.12 implies that

there exists N ∈ N, and a constant ρ > 1 such that

length(LN α) ≥ ρ length(α)

for all paths α in R2. If γ is an arbitrary path in C, then it has a liftunder the branched covering map Θ (Lemma A.3); so it can be writtenin the form γ = Θ α, where α is a path in R2. Hence

lengthd(fN γ) = length(LN α) ≥ ρ length(α)

= ρ lengthd(Θ α) = ρ lengthd(γ).

Lemma 6.6 implies that fN is an expanding Thurston map. Hence fis expanding (see Lemma 6.4).

To prove the converse, we assume that f is expanding, but L is not.The considerations in the beginning of the proof of Lemma 6.12 showthat then L has a real eigenvalue λ with |λ| ≤ 1. Note that λ 6= 0,because L is invertible. It follows that there exists u ∈ R2 with |u| = 1

172 6. EXPANSION

such that L(u) = λu. Let α be the parametrized line segment joining0 and u, and define αn := A−n α for n ∈ N. Then

diam(αn) = diam(A−n α) = length(A−n α)(6.8)

= length(L−n α) =1

|λ|nlength(α)

≥ length(α) = 1

for all n ∈ N.Let γ := Θ α, and γn := Θ αn for n ∈ N. Then γn is a lift of γ

under fn, because

fn γn = fn Θ A−n α = Θ An A−n α = Θ α = γ.

Since f is expanding, there exists a metric on C such that the diameterof lifts of any path under fn shrink to 0 in diameter exponentially fastas n → ∞ (see Lemma 8.7). Then these diameters tend to zero with

respect to any metric on C (that induces the given topology). Weconclude that diamd(γn)→ 0 as n→∞.

Lemma 6.13 (ii) applies to f and Θ, because f is a quotient of a

torus endomorphsm and we have the diagram (3.13) (with S2 = C).So we can find δ > 0 such that diam(K) < 1 whenever K ⊂ C isa connected set with diam(Θ(K)) < δ. Since diamd(γn) → 0, weconclude that diamd(γn) < δ for large n. Now Θ(αn) = γn, and sodiam(αn) < 1 for large n. This contradicts (6.8). It follows that Lis expanding if f is. Together with the first part of the proof, weconclude that f is expanding as a Thurston map if and only if L = LAis expanding as a linear map.

If f is a Lattes map, then we know that it can be obained as aquotient of an affine map of the form A(z) = αz + β, z ∈ C, whereα, β ∈ C, and |α|2 = deg(f) ≥ 2. So |α| > 1. On the other hand, incomplex notation the associated linear map is LA(z) = αz, z ∈ C. As areal linear map L = LA has the eigenvalues α and α. So |α| = |α| > 1,and L is expanding. By what we have seen, f is expanding as well.

We finish this chapter by giving an example of a Thurston mapthat is eventually onto, but not expanding. The example is due toK. Pilgrim.

Example 6.14. As in Example ?? let G be the crystallographicgroup consisting of all maps g of the form

u ∈ R2 7→ g(u) = ±u+ γ,

where γ ∈ Z2. Let Θ: R2 → S2 = R2/G be the quotient map.


Consider the map A : R2 → R2 given by left-multiplication of u ∈R2 (written as a column vector) by the matrix

A =

(4 22 2

).

This map has the form (3.18). The linear part L = LA of A agrees withA, because A is linear. So the matrix representing L with respect to

the standard basis on R2 is equal to A. Since det(LA) = det(A) ≥ 2,the map A induces a Lattes-type map f : S2 → S2 on the quotientS2 = R2/G. We claim that f is not expanding, but eventually onto.

Recall that later property means means that for any non-emptyopen set U ⊂ S2 there is an iterate fn such that fn(U) = S2 (see alsoLemma 6.5).

The map L = A has the eigenvalues λ1 = 3−√

5 and λ2 = 3 +√

5.Since |λ1| < 1, the map f is not expanding by Proposition 6.11.

Now consider the map linear maps B and C given by left-multipli-cation of u ∈ R2 by the matrices

B =

(2 11 1

)and C =

(2 00 2

),

respectively. Then A = B C = C B. The maps B and C again havethe form (3.18) and so descend to maps g : S2 → S2 and h : S2 → S2

respectively. Note that det(B) = 1, and so g is a homeomorphism (withan inverse induced by B−1). These maps satisfy f = g h = h g, andso fn = gn hn for all n ∈ N.

Now let U ⊂ S2 be an arbitrary non-empty open set. Then V :=Θ−1(U) is a non-empty open set in R2. Since Cn(V ) = 2nV , for suf-ficently large n the set Cn(V ) will contain arbitrarily large disks. Inparticular, there exists n ∈ N such that Cn(V ) contains a translateτγ(R) with γ ∈ Z2 of the rectangle R = [0, 1] × [0, 1/2] that is a fun-damental region for the action of G (see Example ??). For such n wehave

S2 = Θ(R) = Θ(τγ(R)) = Θ(Cn(V )) = hn(Θ(V )) = hn(U),

and so, since g is a homeomorphism,

fn(U) = (gn hn)(U) = gn(S2) = S2.

This shows that f is eventually onto.

CHAPTER 7

Thurston maps with two or three postcriticalpoints

In this chapter we investigate Thurston maps f : S2 → S2 with a post-critical set consisting of two or three elements (note that # post(f) ≥ 2by Remark 5.16). The considerations here are not essential for our mainstory and may safely be skipped by the impatient reader. For this rea-son we will sometimes deviate from the logical order and rely on someresults that will only be established in later chapters.

For starters it is easy to classify all Thurston maps with two post-critical points up to Thurston equivalence.

Proposition 7.1. Suppose f : S2 → S2 is a Thurston map with# post(f) = 2. Then f is Thurston equivalent to a power map z 7→ zn

on C, where n ∈ Z \ −1, 0, 1.

This will be proved in Section 7.1. We record the following imme-diate consequence.

Corollary 7.2. If f : S2 → S2 is an expanding Thurston map,then # post(f) ≥ 3.

Proof. By Remark 5.16 we know that # post(f) ≥ 2. If # post(f) =2, then it follows from Propositions 7.1 that the second iterate f 2 is aThurston polynomial. Then Lemmas 6.7 and 6.4 imply that f is notexpanding.

Due to this corollary, we can restrict ourselves to the case # post(f) ≥3 when we consider expanding Thurston maps.

For the case when # post(f) = 3, the relation to rational Thurstonmaps is clarified by the following theorem.

Theorem 7.3. Let f : S2 → S2 be a Thurston map with # post(f) =3. Then the following statements are true:

(i) f is Thurston equivalent to a rational Thurston map;

(ii) if the map f is expanding, then f is topologically conjugateto a rational Thurston map if and only if f has no periodiccritical points.

175

176 7. THURSTON MAPS WITH TWO OR THREE POSTCRITICAL POINTS

Part (i) of this statement is essentially a trivial case of Thurs-ton characterization of rational maps given in Theorem 2.17. We willpresent a independent proof in Section 7.1 that is in the spirit of ourcombinatorial approach. Part (ii) easily follows if this is combined withsome of our other results.

In Chapter 3 we considered Lattes or Lattes-type maps. These areThurston maps f : S2 → S2 with a parabolic orbifold and no periodiccritical points. The case when f has a parabolic orbifold, but alsoperiodic critical points is very special. Then the signature of f must be(∞,∞) or (2, 2,∞) as follows from Propositions 2.13 and 2.9. Thesemaps can easily be classified up to Thurston equivalence.

Theorem 7.4. Let f : S2 → S2 be a Thurston map. Then f hassignature

(i) (∞,∞) if and only if f is Thurston equivalent to a power map

z 7→ zn on C, where n ∈ Z \ −1, 0, 1;(ii) (2, 2,∞) if and only if f is Thurston equivalent to χ or −χ,

where χ = χn is a Chebychev polynomial of degree n ∈ N\1.

The case of signature (∞,∞) is essentially already covered byProposition 7.1. The proof of Theorem 7.4 is given in Section 7.2,where we will also review the definition of Chebychev polynomials.

7.1. Thurston maps with # post(f) ∈ 2, 3 are equivalent torational maps

We begin by looking at Thurston maps f with # post(f) = 3.

Proof of Theorem 7.3. We will rely on several technical resultsthat will be proved later. Throughout the proof f : S2 → S2 is aThurston map with # post(f) = 3.

(i) The basic idea of the proof is very similar to the considerationsin Example 2.6. We want to build polyhedral spheres S0 and S1 so thatf becomes a piecewise isometry S1 → S0. These polyhedral spherescarry natural conformal structures that make f a holomorphic map.We then apply the uniformization theorem to identify each of these

surfaces with C (under a suitable normalization) and use the fact thatif two orientation-preserving homeomorphism between spheres agree ona set P with #P ≤ 3, then they are isotopic rel. P (see Lemma 11.11).

To make this outline precise, we pick a Jordan curve C ⊂ S2 withpost(f) ⊂ C and consider the cell decompositions D0 := D0(f, C) andD1 := D1(f, C). Since # post(f) = 3, the tiles in D0 and D1 are

7.1. THURSTON MAPS WITH # post(f) ∈ 2, 3 ARE RATIONAL 177

(topological) triangles, i.e., they contain three vertices and edges intheir boundaries.

Let ∆ be a fixed equilateral Euclidean triangle of side-length 1.There exists a (non-unique) path metric d0 on S2 such that each ofthe two tiles in D0 equipped with d0 is isometric to ∆ (here and belowit is understood that the vertices of the triangles correspond to eachother under the isometry). The metric space S0 = (S2, d0) is isometricto a pillow consisting of two copies of ∆ glued together along theirboundaries (equipped with the path metric).

In a similar fashion, we want to equip S2 with a path metric d1

such that each tile in D1 is isometric to ∆. Then, roughly speaking,the metric space S1 = (S2, d1) will be an (abstract) polyhedral surfaceobtained by gluing together different copies of ∆, one for each tile inD1, according to the combinatorics of D1.

We want the metric d1 to also be compatible with f and d0. Weknow that each tile X1 in D1 is mapped homeomorphically by f to atile X0 = f(X1) in D0. Thus for d1 we choose the unique path metricon S2 so that d1|X1 is the pull-back of d0 by f : X1 → X0 = f(X1) foreach tile X1 in D1. Then the map f : (S2, d1)→ (S2, d0) is an isometryon such each tile X1, and hence a local isometry away from the verticesof D1.

The surfaces S0 = (S2, d0) and S1 = (S2, d1) each carry a uniqueconformal structure compatible with the polyhedral structure. So bythe uniformization theorem there exist conformal maps h0 and h1 from(S2, d0) and (S2, d1) (considered as a Riemann surfaces) onto the Rie-

mann sphere C, respectively. By postcomposing h0 with a suitableMobius transformation if necessary, we may assume that the maps h0

and h1 agree on the 3-element set P = post(f) ⊂ S2. Since h0 andh1 are orientation-preserving, these homeomorphisms are then isotopicrel. P (see Lemma 11.11).

Away from the vertices of D1, the map f : (S2, d1) → (S2, d0) is alocal isometry and hence conformal. It it follows that the map R :=h0 f h−1

1 is a continuous map that is holomorphic in the complementof at most finitely many points. These are removable singularities and

so R is holomorphic on C and hence a rational map. Moreover, R isa Thurston map (this follows from considerations as in the proof ofLemma 2.5) that is Thurston equivalent to f . The first part of thetheorem follows.

(ii) Suppose now in addition that f is expanding and has no periodiccritical points. Since the latter condition is invariant under Thurstonequivalence, the rational Thurston map R constructed as above will


then not have periodic critical points either, and hence is expandingby Proposition 2.3. Therefore, the maps f and R are topologicallyconjugate by Theorem 11.4.

Conversely, if f is expanding and topologically conjugate to a ra-tional map R, then R is an expanding Thurston map. Hence R has noperiodic critical points by Proposition 2.3, which implies that f cannothave periodic critical points either.

To deal with the case # post(f) = 2, we first prove an auxiliarystatement.

Lemma 7.5. Let f : S2 → S2 be a Thurston map with # post(f) =2. Then

(i) the signature of f is (∞,∞);

(ii) the set post(f) is completely invariant, i.e.,

f−1(post(f)) = post(f).

Proof. Let αf : S2 → N be the ramification function and Of =(S2, αf ) the orbifold associated with f .

(i) If this were not true, then the Euler characteristic of Of wouldbe strictly positive, contradicting Proposition 2.12.

(ii) Let p ∈ f−1(post(f)). Then condition (iii) in Proposition 2.13implies that αf (p) =∞. Thus p ∈ post(f), showing that f−1(post(f)) ⊂post(f). The reverse inclusion f−1(post(f)) ⊃ post(f) is true for everyThurston map, and so the claim follows.

Proof of Proposition 7.1. Let f : S2 → S2 be a Thurston mapwith # post(f) = 2. We want to show that f is Thurston equivalent

to the map z 7→ zn on C, where n ∈ Z \ −1, 0, 1.To see this, we slightly adjust the proof of Theorem 7.3. Namely,

we pick a point q ∈ S2 \ post(f) and define P = post(f) ∪ q. Againwe choose a Jordan curve C ⊂ S2 with P ⊂ C. Treating q as anadditional 0-vertex, we obtain cell decompositions D0 and D1 as be-fore, were each tile is a topological triangle. Then as in the proofof Theorem 7.3, we can define metrics d0 and d1 so that (S2, d0) and(S2, d1) are polyhedral surfaces that carry natural conformal structuresand so that f : (S2, d0 → (S2, d1) is a piecewise isometry. As before,

let h0 : (S2, d0) → C and h1 : (S2, d1) → C be uniformizing conformalmaps, where we may assume that h0 and h1 agree on the set P . Us-ing Lemma 11.11 again, we conclude that h0 and h1 are isotopic rel.

post(f). Then map R = h0 f h−11 : C→ C is rational Thurston map

that is Thurston equivalent to f .

7.2. PARABOLIC THURSTON POLYNOMIALS 179

For the map R we also have post(R) = 2, and so may assume thatpost(R) = 0,∞ (by conjugating R with a suitable Mobius transfor-mation). Let A = R−1(0) be the set of all zeros and B = R−1(∞)be the set of poles of R. From Lemma 7.5 it follows that A ∪ B =

post(R) = 0,∞. This implies that R(z) = czn for z ∈ C, wherec ∈ C \ 0 and n ∈ Z \ 0. Here actually n ∈ Z \ −1, 0, 1, becauseR is Thurston map and hence not a homeomorphism. So in particular,n 6= 1, which implies that by conjugating R with an auxiliary map ofthe form z 7→ αz, α ∈ C \ 0, if necessary, we may assume that c = 1.The claim follows.

7.2. Parabolic Thurston polynomials

In this section we consider Thurston maps f with a parabolic orbifoldand periodic critical points. We know that then the associated orbifoldOf has signature (∞,∞) or (2, 2,∞). These maps together with Lattesmaps considered in Chapter 3 cover all cases of Thurston maps with aparabolic orbifold.

The case when the signature is (∞,∞) has already been treated inProposition 7.1 (see also Lemma 7.5). It remains to consider the caseof signature (2, 2,∞). Our presentation follows [Mi06].

Lemma 7.6. Let f : S2 → S2 be a Thurston map with signature(2, 2,∞). Then f is a Thurston polynomial.

Proof. Suppose the signature of (the orbifold of) f is (2, 2,∞)and let p ∈ S2 be the unique point with αf (p) =∞. Lemma 2.13 (iii)implies that αf (q) =∞ for each point q ∈ f−1(p). Thus, q = p and sop is completely invariant. Therefore, f is a Thurston polynomial.

By Lemma 6.7 this implies that such a map f cannot be expanding.Moreover, we conclude that if f is rational and suitably normalized,then f is a polynomial. As we will immediately see, the polynomi-als that appear here are very special, namely Chebychev polynomials.By definition the Chebychev polynomial χn for n ∈ N0 is the uniquepolynomial such that

cos(nu) = χn(cosu), u ∈ C.

We call a Thurston polynomial f with a parabolic orbifold a para-bolic Thurston polynomial. Then f has signature (∞,∞) or (2, 2,∞).Conversely, if f is a Thurston map with signature (2, 2,∞), then f isa parabolic Thurston polynomial by Lemma 7.6. If f has signature(∞,∞), then f or f 2 is a parabolic Thurston polynomial as follows


from Lemma 7.5 (ii). An immediate classification of parabolic Thurs-ton polynomials is obtained from Theorem 7.4 which we will provebelow.

As we have seen in Chapter 3, Lattes maps can be obtained fromcrystallographic groups G acting on C. Here G contains a subgroupGtr of translations isomorphic to a rank-2 lattice. We will now discusshow the maps z 7→ zn and z 7→ ±χn(z) can be constructed from groupsG of orientation-preserving isometries acting properly discontinuouslyon C. In contrast to the Lattes case, here the subgroup of translationsGtr ⊂ G is isomorphic to a rank-1 lattice, i.e., to Z.

If we omit the requirement that each g ∈ G is orientation-preserving,then such a group G is a so-called frieze group. There are seven of themup to isomorphism. Two frieze groups do not contain orientation-reversing isometries (i.e., reflections and/or glide-reflections). Up toconjugacy by a map of the form h(z) = αz + β with α ∈ C \ 0,β ∈ C, one of these two groups is given as the group G of all isometriesg on C of the form

z 7→ g(z) = z + k, where k ∈ Z;(7.1)

and the other as the group G of all isometries g of the form

z 7→ g(z) = ±z + k, where k ∈ Z.(7.2)

Note that in both cases the action of G on C is properly discontinuous,but not cocompact.

Let G be the group of translations as in (7.1). Clearly exp(2πiz) =exp(2πiw) if and only if z = g(w) for some g ∈ G. Thus Θ: C → Cgiven by Θ(z) = exp(2πiz) for z ∈ C is a covering map induced by G(as in Chapter 3). If we fix n ∈ Z, then exp(2πinz) = exp(2πiz)n forz ∈ C; so the following diagram commutes:

(7.3) C nz//

exp(2πiz)

Cexp(2πiz)

C zn// C.

This gives a description of the map z 7→ zn analogous to the descriptionof Lattes maps in Theorem 3.1. Obviously, C/G is a cylinder, and so wemay view the map z 7→ zn as a quotient of a self-map of a cylinder. It isstraightforward to show that every map that arises as such a quotientis conformally conjugate to zn. The map z 7→ Θ(z) = exp(2πiz) is theuniversal orbifold covering map of the orbifold associated to zn (seeDefinition A.18).


Chebychev polynomials can be described in a similar vein. Let Gbe the group of isometries as in (7.2). Then cos(2πz) = cos(2πw) if andonly if w = ±z + k = g(z), where g ∈ G. Thus the branched coveringmap Θ: C → C defined by Θ(z) = cos(2πz) for z ∈ C is induced bythe group G.

Suppose A : C→ C has the form A(z) = nz + l/2 for z ∈ C, wheren ∈ N and l ∈ 0, 1. Then there exists a (unique) holomorphic mapχ : C → C such that χ(cos(2πz)) = cos(2πA(z)) for z ∈ C. Indeed,if l = 0, then A(z) = nz, and so χ = χn is the n-th Chebychevpolynomial. If l = 1, then A(z) = nz + 1/2 and χ = −χn. This meansthat in all cases the following diagram commutes:

(7.4) C A//

cos(2πz)

Ccos(2πz)

Cχ// C.

If we require that χ is non-constant and not a homeomorphism,then A(z) = nz+ l/2 with n ≥ 2 (and l ∈ 0, 1). The same argumentas in Lemma 3.10 (ii) shows that the postcritical points of χ in Care the critical values of z 7→ Θ(z) = cos(2πz), i.e., −1 and 1. Wealso have αχ(1) = αχ(−1) = 2 for the ramification function αχ ofχ, because degΘ(z) = 2 for each point z ∈ Θ−1(−1, 1). The point∞ is a fixed critical point of χ, because χ is a polynomial of degreen ≥ 2. So αχ(∞) = ∞. We conclude that χ is a Thurston mapwith post(χ) = −1, 1,∞ and signature (2, 2,∞). Note that Θ is theuniversal orbifold covering map for Oχ, the orbifold associated to χ(see Definition A.18).

The postcritical points −1 and 1 are mapped as indicated in thefollowing diagrams: if χ = χn,

−1 // 1

for n even and −1

1

for n odd;(7.5)

and if χ = −χn,

1 // −1

for n even and −1""

1dd

for n odd.(7.6)

The diagrams when n is even are similar in both cases, because themaps involved are topologically conjugate. Indeed, suppose n ∈ N iseven, define τ(z) = −z for z ∈ C, and note that χn is an even functionfor even n ∈ N. So

(τ χn τ−1)(z) = −χn(−z) = −χn(z)

for z ∈ C, which implies that −χn = τ χn τ−1 for n ∈ N even.


When n ∈ N is odd, the maps χn and −χn are not topologicallyconjugate (or Thurston equivalent) as the above diagrams show: allpostcritical points are fixed points for χn, but not for −χn.

Proof of Theorem 7.4. Statement (i) immediately follows fromProposition 7.1, Lemma 7.5, and Proposition 2.14.

To prove (ii), let f : S2 → S2 be a Thurston map with signature(2, 2,∞). Then # post(f) = 3, and so by Theorem 7.3 (i) the mapf is Thurston equivalent to a rational map, necessarily with the same

signature. So we may assume that f is a rational map on S2 = C tobegin with. Moreover, by conjugating the map with a suitable Mobiustransformation, we may assume that post(f) = −1, 1,∞ and that for

the ramification function αf : C→ N of f we have αf (−1) = αf (1) = 2and αf (∞) =∞.

The map f has a parabolic orbifold, and so Proposition 2.13 impliesthat

(7.7) αf (z) · degf (z) = αf (f(z))

for all z ∈ C. It follows that f−1(∞) = ∞, and so the rational mapf must actually be a polynomial. Equation (7.7) also shows that wehave z ∈ f−1(−1, 1) if and only if z 6= ∞ and one of the factors onthe left-hand side of (7.7) is different from 1. Then one factor is equalto 1 and the other equal to 2. We conclude that

f−1(−1, 1) = −1, 1 ∪ crit(f) \ ∞

with degf (−1) = degf (1) = 1 and degf (z) = 2 for z ∈ crit(f) \ ∞.This implies that the polynomials 1 − f(z)2 and (1 − z2)f ′(z)2 havethe same zeros of exactly the same order. If we define n = deg(f) ≥ 2,then comparison of the highest-order coefficient gives

(7.8) n2(1− f(z)2) = (1− z2)f ′(z)2

for z ∈ C. It is well known and easy to prove that then f = ±χn.Indeed, consider the even entire function g defined as g(u) = f(cosu)

for u ∈ C. Then (7.8) leads to

g′(u)2 = n2(1− g(u)2)

for u ∈ C. If we differentiate this equation, then we obtain the linearordinary differential equation

g′′(u) + n2g(u) = 0, u ∈ C.

It has the general solution g(u) = c1 cos(nu) + c2 sin(nu), c1, c2 ∈ C.Since g is even, we must have c2 = 0; moreover, c1 = g(0) = f(1) ∈


−1, 1. Hence g(u) = ± cos(nu) = f(cosu) for u ∈ C. This impliesf = ±χn.

For the converse direction suppose that the Thurston map f : S2 →S2 is Thurston equivalent to χ = ±χn with n ∈ N \ 1. We have seenearlier in this section that χ has signature (2, 2,∞). Hence f has thesame signature by Proposition 2.14.

To finish this chapter, we will discuss how one can give a Euclideangeometric model for the maps ±χn similar to the models for Lattesmaps as described in Section 1.1 and Section 3.4.

Let G be the group in (7.2). Then it induces an equivalence relation∼ on C as usual: for z, w ∈ G we have z ∼ w if and only if thereexists g ∈ G such that w = g(z). We denote by [z] the equivalenceclass of a point z ∈ C/G = C/ ∼. We equip C/G with the metricω obtained pushing forward the Euclidean metric on C (see (A.10));more explicitly,

(7.9) ω([x], [y]) := inf|z − w| : z ∈ [x], w ∈ [y]

for [x], [y] ∈ C/G.The metric space (C/G, ω) is isometric to the space ∆ obtained by

gluing two copies of the half-infinite strip S = [0, 1/2]× [0,∞) ⊂ R2 'C together along their boundaries. Here the strips carry the Euclideanmetric and ∆ the induced path metric.

To see this, note that the infinite strip F = [0, 1/2]× R ⊂ R2 ∼= Cis a fundamental region of G. Under the action of G two points in theboundary of F are identified if they are complex conjugates of eachother. So the quotient C/G is obtained by folding the strip F alongthe real axis and gluing together corresponding boundary parts of thehalf-infinite strips Sw := [0, 1/2]× [0,∞) and Sb := [0, 1/2]× (−∞, 0].The metric ω is a path metric and corresponds to the Euclidean metricon Sw and Sb. This shows that (C/G, ω) and ∆ are indeed isometric.

In the following we will identify these spaces. We will also considerthe half-infinite strips Sw and Sb as subsets and sides of ∆. We colorSw white, and Sb black. Then ∆ is a locally flat surface with twoconical singularities as indicated on the right in Figure 7.1; the conicalsingularities are labeled by −1 and 1 which we will now explain.

Namely, since the map z 7→ Θ(z) = cos(2πz) is induced by the

group G, there is an associated homeomorphism Θ: C/G→ C betweenthe quotient space C/G = ∆ and C. Under this homeomorphismthe conical singularities correspond to −1 and 1 which suggested thelabeling.


h

−1 7→−17→1

7→−117→1

−1

1

∆

Figure 7.1. Model for a Chebychev polynomial.

We now fix n ∈ N and divide each side of ∆ into n strips S ′ of equalsize. Then each strip S ′ is isometric to [0, 1

2n]×[0,∞), and hence similar

to S by the scaling factor n. We color the strips S ′ in a checkerboardfashion black and white so that strips sharing an edge have differentcolors.

One can now define a map h : ∆ → ∆ as follows. We map eachwhite strip S ′ to Sw and each black strip S ′ to Sb by an orientation-preserving Euclidean similarity. Then h is well-defined, because thedefinitions for h match on the edges where two strips intersect. Anexample of such a map is indicated in Figure 7.1.

These maps h give a Euclidean model for the maps ±χn due to thefollowing fact.

Proposition 7.7. Let n ∈ N. Then every map h : ∆→ ∆ obtainedfrom the above construction is topologically conjugate to χn or −χn.Conversely, every polynomial χn and −χn is topologically conjugate tosuch a map h.

Proof. Fix n ∈ N. Then the half-strips S ′ of the form

[ k2n, k+1

2n]× [0,∞) ⊂ Sw and [ k

2n, k+1

2n]× (−∞, 0] ⊂ Sb

for k = 0, . . . , n− 1 divide the sides Sw and Sb of ∆, respectively. Thecheckerboard coloring of these strips S ′ is uniquely determined if wespecify the coloring of S ′0 := [ 0

2n, 1

2n]× [0,∞).

If S ′0 is colored white, then the map A(z) = nz passes to the quo-tient ∆ = C/G and sends white strips S ′ to Sw and black strips S ′

to Sb by a Euclidean similarity. In other words, A induces the map


h : ∆ → ∆ discussed above. On the other hand, by (7.4) this map Apasses to the quotient χn under the map z 7→ Θ(z) = cos(2πz). This

implies that the induced homeomorphism Θ : ∆ = C/G→ C gives the

conjugacy h = Θ−1 χn Θ.If S ′0 is colored black, then A(z) = nz + 1/2 passes to the quotient

∆ = C/G and a strip S ′ is send to a strip Sw or Sb of the same colorby a Euclidean similarity. So again A induces the map h and by (7.4)

we get a conjugacy h = Θ−1 χ Θ, where χ = −χn.

CHAPTER 8

Visual Metrics

In this chapter we construct a natural class of metrics for an expand-ing Thurston map that we call visual metrics. We haven chosen thisname, because there is a close relation between these metrics and vi-sual metrics on the boundary at infinity of a Gromov hyperbolic space.Indeed, for an expanding Thurston map f : S2 → S2 one can define aGromov hyperbolic tile graph whose boundary at infinity can naturallybe identified with S2. By this identification, a metric on S2 is visualin the sense of Gromov hyperbolic spaces if and only if it is visualas it will be defined in this chapter (see Chapter 10 and in particularTheorem 10.2).

The first section of this chapter gives a quick overview of the def-inition and the basic properties of visual metrics. In Sections 8.1 and8.2 we will then provide the technical details. The final Section A.10 isdevoted to a discussion of the canonical orbifold metric for a rationalThurston map f . As we will see, this is a visual metric precisely if theorbifold of f is parabolic.

Let f : S2 → S2 be an expanding Thurston map, and C ⊂ S2 be aJordan curve with post(f) ⊂ C. We consider the cell decompositionsof S2 for (f, C) as defined in Section 5.3. One can think of the setof n-tiles as a discrete approximation of the sphere S2 and measuredistances of points by a quantity related to combinatorics of n-tiles.

Indeed, let x, y ∈ S2 be two distinct points, and X and Y be n-tileswith x ∈ X, y ∈ Y . Since f is expanding, it follows that X and Ymust be disjoint for sufficiently large n (see (6.1) and Lemma 6.1). Thisleads to the following definition.

Definition 8.1. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be a Jordan curve with post(f) ⊂ C, and x, y ∈ S2. Forx 6= y we define

mf,C(x, y) := maxn ∈ N0 : there exist non-disjoint n-tiles

X and Y for (f, C) with x ∈ X, y ∈ Y .

If x = y we define mf,C(x, x) :=∞.

187

188 8. VISUAL METRICS

xy

Figure 8.1. Separating points via tiles.

8. VISUAL METRICS 189

Note that mf,C(x, y) ∈ N if x 6= y. We usually drop both sub-scripts in mf,C(x, y) if f and C are clear from the context. A similarcombinatorial quantity that is essentially equivalent to mf,C(x, y) (seeLemma 8.6 (v)) is

m′f,C(x, y) := minn ∈ N0 : there exist disjoint n-tiles(8.1)

X and Y for (f, C) with x ∈ X, y ∈ Y ,

for x 6= y and m′f,C(x, x) :=∞.These quantities are illustrated in Figure 8.1. Here we consider the

map f : C→ C given by

f(z) = iz4 − i

z4 + i

for z ∈ C (this is the same map as in Example 14.11). We havepost(f) = 1, i ,−i, and so we can choose the unit circle C = ∂D asa Jordan curve containing the postcritical set of f . In Figure 8.1 then-tiles for (f, C) are shown for n = 1, . . . , 6. For the points x and y asindicated in the figure, we have mf,C(x, y) = 3 and m′f,C(x, y) = 4.

The number mf,C(x, y) is large if x and y are close together, i.e.,if n-tiles of high order are needed to separate the points. This is thebasis of the following definition.

Definition 8.2 (Visual metrics). Let f : S2 → S2 be an expandingThurston map. A metric % on S2 is called a visual metric (for f) ifthere exists a Jordan curve C ⊂ S2 with post(f) ⊂ C, and a constantΛ > 1 such that

(8.2) %(x, y) Λ−m(x,y)

for all x, y ∈ S2, where m(x, y) = mf,C(x, y) and where the constantC() is independent of x and y.

Here we use the convention Λ−∞ = 0. The number Λ is called theexpansion factor of the metric %. It is easy to see that the expansionfactor of each visual metric is uniquely determined. Different visualmetrics may have different expansion factors.

It is possible to identify the sphere S2 with the boundary at infinityof a certain Gromov hyperbolic graph constructed from tiles. Underthis identification, the numbers mf,C(x, y) and m′f,C(x, y) are the Gro-mov product of x and y, up to some additive constants (see Section 4.2,Chapter 10, and Lemma 10.3). A metric then is visual as defined aboveif and only if it is visual in the sense of Gromov hyperbolic spaces, seeTheorem 10.2. This explains our terminology.


Obvious questions are whether visual metrics exist, and how theydepend on the chosen Jordan curve C and the expansion factor Λ. Thisis answered by the following proposition.

Proposition 8.3. For an expanding Thurston map f : S2 → S2

the following statements are true:

(i) There exist visual metrics for f .

(ii) Every visual metric induces the standard topology on S2.

(iii) Let % be a visual metric for f with expansion factor Λ, C ⊂ S2

be a Jordan curve with post(f) ⊂ C, and m = mf,C be defined

as in Definition 8.1. Then an inequality as in (8.2) is true with

the same expansion factor Λ, where the constant A = C()

depends on C.

(iv) Any two visual metrics are snowflake equivalent, and bi-Lipschitzequivalent if they have the same expansion factor Λ.

(v) A metric % is a visual metric for some iterate F = fn if andonly if it is a visual metric for f . If Λ > 1 is the expansionfactor of % for f , then ΛF = Λn is the expansion constant of %for F = fn.

(vi) If % is a visual metric for f , then f : (S2, %) → (S2, %) is aLipschitz map.

The notion of snowflake and bi-Lipschitz equivalence were definedin Section 4.1 (in particular see (4.1)). We will postpone the proof ofProposition 8.3 to Section 8.2. A stronger result on the existence ofvisual metrics is given in Theorem 15.3.

In Sections 1.3 and 4.4 we introduced visual metrics on a moreintuitive level, where we considered certain self-similar fractal spheresconstructed as the limit of polyhedral surfaces Sn. Each surface Sn wasbuilt from squares or triangles of size Λ−n for some constant Λ > 1.A similar statement holds in general for the visual metric, which is infact a characterization.

Proposition 8.4 (Characterization of visual metric). Let f : S2 →S2 be an expanding Thurston map and % be a metric on S2. Then %is a visual metric for f with expansion factor Λ > 1 if and only if thefollowing two conditions hold:

(i) dist%(σ, τ) & Λ−n, whenever σ and τ are disjoint n-cells,

(ii) diam%(τ) Λ−n for all n-edges and all n-tiles τ .

8.1. THE NUMBER m(x, y) 191

Here cells are defined in terms of some Jordan curve C ⊂ S2 withpost(f) ⊂ C, and the implicit constants C(&) and C() are indepen-dent of the cells.

We will prove this lemma in Section 8.2. In Chapter 17 we willinvestigate the geometry of the sphere S2 equipped with a visual metric% more closely, and derive relations between the geometry of (S2, %) andproperties of the Thurston map f .

8.1. The number m(x, y)

We now turn to a more detailed exposition of the basic properties ofthe quantity m(x, y) = mf,C(x, y) as in Definition 8.1.

Recall the quantity Dn as defined in (5.14) that measures distancesin terms of lengths of tile chains. We consider a slight variant here.

Let f : S2 → S2 be an expanding Thurston map and C ⊂ S2 be

a Jordan curve with post(f) ⊂ C. We define Dn = Dn(f, C) as theminimal number of tiles of order k ≥ n for (f, C) required to joinopposite sides of C, i.e., the smallest number N ∈ N for which there aretiles Xi ∈

⋃k≥n Xl, i = 1, . . . , N , such that K =

⋃Ni=1Xi is connected

and joins opposite sides of C (see Definition 5.30).While the sets K used to define Dn are unions of tiles of order n,

the sets K in the definition of Dn are unions of tiles of order k ≥ n; in

particular, Dk ≥ Dn for k ≥ n.

Lemma 8.5. Let f : S2 → S2 be an expanding Thurston map, andC ⊂ S2 be a Jordan curve with post(f) ⊂ C. Let Dn = Dn(f, C) and

Dn = Dn(f, C) for n ∈ N0.

Then Dn →∞ and Dn →∞ as n→∞.

Proof. We know that Dk ≥ Dn whenever k ≥ n. So it suffices toshow Dn →∞ as n→∞.

Let δ0 > 0 be defined as in (5.13) and suppose K = X1 ∪ · · · ∪XN

is a connected union of tiles of order ≥ n that joins opposite sides ofC. Then

δ0 ≤ diam(K) ≤N∑i=1

diam(Xi)

≤ N maxi=1,...,N

diam(Xi)

≤ N supk≥n

mesh(f, k, C).

Putting cn := supk≥n mesh(f, k, C), we conclude that N ≥ δ0/cn, and

so Dn ≥ δ0/cn.


Since f is expanding we have mesh(f, n, C)→ 0 and so also cn → 0

as n→∞. This implies that Dn →∞ as desired.

If f is expanding and C is given, then in view of the last lemma, wecan find a number k0 = k0(f, C) ∈ N such that

(8.3) Dk0 = Dk0(f, C) ≥ 10.

In the next lemma we collect some of the properties of the functionmf,C. For the proof the following terminology is useful. A (finite) chainin S2 is a finite sequence A1, . . . , AN of sets in S2 such that Ai∩Ai+1 6= ∅for i = 1, . . . , N−1. It joins two points x, y ∈ S2 if x ∈ A1 and y ∈ AN .We say that this chain is simple if there is no proper subsequence ofA1, . . . , AN that is also chain joining x and y.

If K is a compact connected set in S2, U an open cover of K, andx, y ∈ K, then one can always find a simple chain of sets in U joiningx and y.

Lemma 8.6. Let f : S2 → S2 be an expanding Thurston map, C ⊂S2 be a Jordan curve with post(f) ⊂ C, and m = mf,C.

(i) There exists a number k1 > 0 such that

minm(x, z),m(y, z) ≤ m(x, y) + k1

for all x, y, z ∈ S2.

(ii) We have

m(f(x), f(y)) ≥ m(x, y)− 1

for all x, y ∈ S2.

(iii) Let C ⊂ S2 be another Jordan curve with post(f) ⊂ C. Thenthere exists a constant k2 > 0 such that

m(x, y)− k2 ≤ mf,C(x, y) ≤ m(x, y) + k2


(iv) Let F = fn be an iterate of f . Then there exists a constantk3 > 0 such that

m(x, y)− k3 ≤ n ·mF,C(x, y) ≤ m(x, y),


(v) The number m′ = m′f,C defined in (8.1) and m are comparablein the following sense. There exists a constant k4 > 0 suchthat for all x, y ∈ S2, x 6= y it holds

m(x, y)− k4 ≤ m′f,C(x, y) ≤ m(x, y) + 1.

8.1. THE NUMBER m(x, y) 193

Proof. We fix k0 = k0(f, C) as in (8.3). Let x, y ∈ S2 be arbitrary.In order to establish the desired inequalities we may always assumex 6= y. Unless otherwise stated, tiles will be for (f, C).

(i) Let m = m(x, y) ∈ N0 be as in Definition 8.1. We can pick(m+1)-tiles X0 and Y0 containing x and y, respectively. Then X0∩Y0 =∅ by definition of m.

Define n := m + k0, and let z ∈ S2 be arbitrary. We claim thatm(x, z) ≤ n or m(y, z) ≤ n.

Otherwise m(x, z) ≥ n + 1 and m(y, z) ≥ n + 1, and so by Defi-nition 8.1 there exist numbers m1,m2 ≥ n + 1 and m1-tiles X and Zwith x ∈ X, z ∈ Z and X ∩Z 6= ∅, and m2-tiles Y and Z ′ with y ∈ Y ,z ∈ Z ′ and X ∩ Z ′ 6= ∅.

Then the set K = X∪Z∪Z ′∪Y is connected and meets the disjoint(m + 1)-tiles X0 and Y0. Thus fm+1(K) joins opposite sides of C byLemma 5.33, and consists of four tiles of order ≥ n − m = k0. Thiscontradicts (8.3), proving the claim.

So we have m(x, z) ≤ m+ k0 or m(y, z) ≤ m+ k0. This implies (i)with the constant k1 = k0 which is independent of x and y.

(ii) We may assume that m = m(x, y) ≥ 1. Then there are non-disjoint m-tiles X and Y with x ∈ X and y ∈ Y . It follows thatf(X) and f(Y ) are non-disjoint (m − 1)-tiles with f(x) ∈ f(X) andf(y) ∈ f(Y ). Hence m(f(x), f(y)) ≥ m− 1 as desired.

(iii) Let m = mf,C(x, y) ∈ N0. Then there exist m-tiles X and Y

for (f, C) with x ∈ X, y ∈ Y , and X ∩ Y 6= ∅. By Lemma 5.36 the

sets X and Y are each contained in M m-flowers for (f, C), where M

is independent of X and Y . In particular, this implies that we can finda chain of at most M such m-flowers joining x and y. Since any twotiles in the closure W n(v) of an n-flower have the point v in common, itfollows that there exists a chain X1, . . . , XN of m-tiles for (f, C) joiningx and y with N ≤ 2M . Let x1 := x, xN := y, and for i = 2, . . . , N − 1,pick a point xi ∈ Xi. Then m(xi, xi+1) ≥ m for i = 1, . . . , N−1. Henceby repeated application of (i) we obtain

m ≤ minm(xi, xi+1) : i = 1, . . . , N − 1≤ m(x1, xN) +Nk1 ≤ m(x, y) + 2Mk1.

Since 2Mk1 is independent of x and y, we get an upper bound as in(iii). A lower bound is obtained by the same argument if we reverse

the roles of C and C.(iv) The map F is also an expanding Thurston map, and we have

post(f) = post(F ) (see Lemma 6.4); so the Jordan curve C contains


the set of postcritical points of F and mF,C is defined. It follows fromProposition 5.17 (v) that the m-tiles for (F, C) are precisely the (nm)-tiles for (f, C), In the ensuing proof we will only consider tiles for (f, C).

Let mF = mF,C(x, y) and m = m(x, y); then there are non-disjoint(nmF )-tiles X and Y with x ∈ X and y ∈ Y. So m ≥ nmF which givesthe desired upper bound.

We claim that on the other hand, we have m ≤ nmF + k3, wherek3 = n+ k0 − 1. To see this assume that

m ≥ nmF + k3 + 1 = n(mF + 1) + k0.

Then we can find non-disjoint m-tiles X and Y with x ∈ X, y ∈ Y .Moreover, we can pick n(mF + 1)-tiles X ′ and Y ′ with x ∈ X ′ andy ∈ Y ′. By definition of mF we know that X ′ ∩ Y ′ = ∅, so X ′ and Y ′

are disjoint n(mF + 1)-tiles joined by the connected set K = X ∪ Y .Hence by Lemma 5.34 K must consists of at least

Dm−n(mF +1) ≥ Dk0 ≥ 10

m-tiles; but K consists of only two such m-tiles. This is a contradictionshowing the desired claim.

(v) Let m′ = m′f,C(x, y) be defined as in (8.1). Then m′ ≥ 1, becausethe two 0-tiles have nonempty intersection. So m′ − 1 ≥ 0, and thereexist (m′ − 1)-tiles X and Y with x ∈ X and y ∈ Y . Then X ∩ Y 6= ∅by definition of m′, and so m(x, y) ≥ m′ − 1.

Conversely, let m = m(x, y). Suppose m′ < m − k0. Then thereexist m′-tiles X ′ and Y ′ with X ′ ∩ Y ′ = ∅, m-tiles X and Y withX ∩ Y 6= ∅, and x ∈ X ∩X ′, y ∈ Y ∩ Y ′. Hence K = X ∪ Y is a unionof two m-tiles joining the disjoint m′-tiles X ′ and Y ′; but such a unionmust consist of at least

Dm−m′ ≥ Dk0 ≥ 10

m-tiles by Lemma 5.34. This is a contradiction showing that m− k0 ≤m′. So the claim is true with k4 = k0.

8.2. Existence and basic properties of visual metrics

We are now ready to prove Proposition 8.3 and Proposition 8.4, and inparticular the existence of visual metrics. In this section we will alsocollect various somewhat more technical statements related to visualmetrics that will be useful later on.

Proof of Proposition 8.3.(i) Fix a Jordan curve C ⊂ S2 with post(f) ⊂ C. Recall that

a function q : S2 × S2 → [0,∞) is called a quasimetric if it has the

8.2. EXISTENCE AND BASIC PROPERTIES OF VISUAL METRICS 195

symmetry property q(x, y) = q(y, x), satisfies the conditions q(x, y) =0⇔ x = y, and the inequality

(8.4) q(x, y) ≤ K(q(x, z) + q(z, y)),

holds for a constant K ≥ 1 and all x, y, z ∈ S2.We now define a quasimetric q on S2. To this purpose, we fix Λ > 1

and set

(8.5) q(x, y) := Λ−m(x,y),

for x, y ∈ S2, where m(x, y) = mf,C(x, y) ∈ N0 ∪ ∞ is as in Defini-tion 8.1.

Symmetry and the property q(x, y) = 0 ⇔ x = y are clear. Thequasi-triangle inequality (8.4) follows from Lemma 8.6 (i).

It is well-known (see [He, Prop. 14.5]) that a sufficient “snowflak-ing” of a quasimetric leads to a distance function that is comparableto a metric. This means there is a metric % and 0 < ε < 1 such thatqε %. Then % is a visual metric for f (with expansion factor Λε).

(ii) Let % be a visual metric for f satisfying (8.2), and d our fixed“base metric” on S2 that induces the standard topology of S2. We haveto show that if x ∈ S2 and xi is a sequence in S2, then %(xi, x)→ 0if and only if d(xi, x)→ 0 as i→∞.

Assume first that %(xi, x)→ 0 as i→∞. By (8.2) this is obviouslyequivalent to mi := mf,C(xi, x) → ∞. This in turn implies that foreach n ∈ N0 we have mi ≥ n for sufficiently large i. For such an i thereare non-disjoint k-tiles X, Y with x ∈ X, xi ∈ Y where k = mi ≥ n bydefinition of mf,C. Thus

d(xi, x) ≤ diamX + diamY ≤ 2 supk≥n

mesh(f, k, C).

Since f is expanding, we know the latter expression goes to 0 as n→∞.Hence d(xi, x)→ 0 as i→∞.

Conversely, suppose that d(xi, x) → 0 as i → ∞. Let n ∈ N0 bearbitrary. Then x lies in some n-flower W n(v) (see Definition 5.25 andLemma 5.26). Since flowers are open sets, we have xi ∈ W n(v) forsufficiently large i. For each of these i we can find n-tiles X and Ywith x ∈ X, xi ∈ Y , and v ∈ X ∩ Y . This implies mi ≥ n. Thereforemi →∞, hence %(xi, x)→∞ as desired.

(iii) This follows from Lemma 8.6 (iii).

(iv) This follows from (iii) and the definition of a visual metric.

(v) This follows from (iii) and Lemma 8.6 (iv).

(vi) This follows from Lemma 8.6 (ii).


We now prove the geometric characterization of visual metrics.

Proof of Proposition 8.4. Let f : S2 → S2 be an expandingThurston map and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Let %be a visual metric for f . By Proposition 8.3 (iii) we may assume that% satisfies (8.2) for m = mf,C and a constant A = C().

(i) Let k0 be defined as in (8.3), and let σ and τ be disjoint n-cells.If x ∈ σ and y ∈ τ are arbitrary, then m(x, y) < n + k0. Indeed, ifthis were not the case, then we could find (n + k)-tiles X and Y withx ∈ X, y ∈ Y , X∩Y 6= ∅, and k ≥ k0. Then K = X∪Y is a connectedset meeting disjoint n-cells. Hence by Lemma 5.33 the set fn(K) joinsopposite sides of C. On the other hand, fn(K) consists of two k-tilesX ′ = fn(X) and Y ′ = fn(Y ), where k ≥ k0. This is impossible bydefinition of k0.

Therefore, %(x, y) ≥ (1/A)Λ−n−k0 , and so we get the desired bounddist(σ, τ) ≥ (1/C ′)Λ−n with the constant C ′ = AΛk0 that is indepen-dent of n, σ, and τ .

(ii) If x, y are points in some n-tile X, then m(x, y) ≥ n. Sinceevery n-edge is contained in an n-tile, this inequality is still true if xand y are contained in an n-edge. Hence %(x, y) ≤ AΛ−m(x,y) ≤ AΛ−n,and so diam%(τ) ≤ AΛ−n whenever τ is an n-tile or n-edge, where theconstant A is as in (8.2).

A similar lower bound for the diameter of an n-edge or n-tile τfollows from (i) and the fact that every n-edge or n-tile contains twodistinct n-vertices.

To prove the converse implication, suppose now that % is a metric onS2 satisfying properties (i) and (ii) from the statement of the Lemma.We want to show that % is visual for f . Let x, y ∈ S2, x 6= y, bearbitrary, and m = mf,C(x, y).

Then we can find m-tiles X and Y with x ∈ X, y ∈ Y and X∩Y 6=∅. By (ii) we have

%(x, y) ≤ diam%(X) + diam%(Y ) . Λ−m.

We can also find (m + 1)-tiles X ′ and Y ′ with x ∈ X ′, y ∈ Y ′. Bydefinition of m we then have X ′ ∩ Y ′ = ∅. Hence by (i)

%(x, y) ≥ dist%(X′, Y ′) & Λ−m.

Since the implicit multiplicative constants in the previous inequalitiesare independent of x and y, it follows that % is a visual metric for f .

It is possible to establish the phenomenon of “exponential shrink-ing” as in Proposition 8.4 (ii) also for other types of sets. For example,


we havediam%(W

n(v)) ≤ CΛ−n

for every n-flower for (f, C) where the constant C is independent of nand v. Of particular importance will be exponential shrinking for liftsof paths.

Lemma 8.7. Let f : S2 → S2 be an expanding Thurston map, and% be a visual metric for f with expansion factor Λ > 1. Then for everypath γ : [0, 1] → S2 there exists a constant A ≥ 1 with the followingproperty: if γ is any lift of γ under fn, then

diam%(γ) ≤ AΛ−n.

Proof. Pick a Jordan curve C ⊂ S2 with post(f) ⊂ C, and let δ0 >0 be as in (5.13). Then we can break up γ into a finite number of pathsγi, i = 1, . . . , N , traversed in successive order such that diam%(γi) < δ0

for all i = 1, . . . , N . By Lemma 5.32 (ii) each lift of the pieces γiis contained in one n-flower, and so the whole lift γ in N n-flowers.Hence by Proposition 8.4 we have diam%(γ) ≤ CNΛ−n with a constantC independent of n and γ.

In general the constant A in the last lemma will dependent on γ,but the proof shows that we can take the same constant A for a familyof paths if there exists N ∈ N such that each path can be broken upinto at most N subpaths of diameter < δ0.

Let f be an expanding Thurston map, and C ⊂ S2 be a Jordancurve with post(f) ⊂ C. It is useful to define neighborhoods of pointsby using the cells in our decompositions Dn = Dn(f, C). To definethem let x ∈ S2 and n ∈ N0, and set

Un(x) =⋃Y ∈ Xn : there exists an n-tile X with(8.6)

x ∈ X and X ∩ Y 6= ∅.It is convenient to define Un(x) also for negative integers n. We setUn(x) = U0(x) = S2 for n < 0. It follows from Proposition 8.4 that thesets Un(x) resemble metric balls (defined in terms of a visual metric)very closely.

Lemma 8.8. Let % be a visual metric for f with expansion factorΛ > 1. Then there are constants K ≥ 1 and n0 ∈ N0 with the followingproperties.

(i) For all x ∈ S2 and all n ∈ ZB%(x, r/K) ⊂ Un(x) ⊂ B%(x,Kr),

where r = Λ−n.


(ii) For all x ∈ S2 and all r > 0

Un+n0(x) ⊂ B%(x, r) ⊂ Un−n0(x),

where n = d− log r/ log Λe.

Proof. (i) Let m = mf,C. If y ∈ Un(x), then m(x, y) ≥ n, and so%(x, y) . Λ−n = r. This gives the inclusion Un(x) ⊂ B%(x,Kr) for asuitable constant K independent of x and n.

Conversely, suppose that y /∈ Un(x). Then n ≥ 1. If we pick anyn-tiles X and Y with x ∈ X and y ∈ Y , then X ∩ Y = ∅ by definitionof Un(x). So by Propositiona 8.4 (ii) we have

d(x, y) ≥ dist(X, Y ) & Λ−n = r.

Hence Bd(x, r/K) ⊂ Un(x) if K is suitably large independent of x andr.

(ii) Choose n0 = dlogK/ log Λe + 1, where K is as in (i). ThenΛ−n0 ≤ 1/(ΛK). Moreover, Λ−n ≤ r ≤ ΛΛ−n, and so

KΛ−n−n0 ≤ r ≤ (1/K)Λ−n+n0 .

The desired inclusion then follows from (i).

We next show that when S2 is equipped with a visual metric, thetiles are “quasi-round”. Furthermore every tile contains points that are“deep inside” the tile.

Lemma 8.9. Let f : S2 → S2 be an expanding Thurston map, C ⊂S2 be a Jordan curve with post(f) ⊂ C, and % be a visual metric for fwith expansion factor Λ > 1. Then there exists a constant C ≥ 1 withthe following property: for every n-tile X for (f, C) there exists a pointp ∈ X such that

B%(p, (1/C)Λ−n) ⊂ X ⊂ B%(p, CΛ−n).

Proof. With a suitable constant C independent of n, an inclusionof the form

X ⊂ B%(p, CΛ−n)

holds for every n-tile X and every point p ∈ X as follows from Propo-sition 8.4 (i).

The main difficulty for an inclusion in the opposite direction isto find an appropriate point p. For this purpose let k0 ∈ N be thenumber defined in (8.3), and X be an arbitrary n-tile. Since f is anexpanding Thurston map, we have post(f) ≥ 3 (see Corollary 7.2), andso ∂X contains at least three distinct n-vertices v1, v2, v3. Using thesevertices, we can find three arcs α1, α2, α3 ⊂ ∂X with pairwise disjointinterior such that ∂X = α1∪α2∪α3 and such that αi has the endpoints


vi and vi+1 for i = 1, 2, 3, where v4 = v1. In general αi will not be ann-edge, but since it lies on ∂X, and its endpoints are n-vertices, it isthe union of all the n-edges that it contains.

We now define

Ai =⋃x∈αi

Un+k0(x)

for i = 1, 2, 3, where Un+k0(x) is defined as in (8.6). Then the set Aiis the union of all (n + k0)-tiles that meet an (n + k0)-tile that hasnonempty intersection with αi. In particular, Ai is a closed set thatcontains αi.

We claim that the sets A1, A2, A3 do not form a cover of X. Toreach a contradiction suppose that X ⊂ A1 ∪ A2 ∪ A3. We can regardX as a topological simplex with the sides αi, i = 1, 2, 3. Then theclosed sets A1, A2, A3 form a cover of X such that each set Ai containsthe side αi of the simplex for i = 1, 2, 3. A well-known result due toSperner [AH, p. 378] then implies that A1 ∩ A2 ∩ A3 6= ∅.

Pick a point x ∈ A1∩A2∩A3. Then by definition of Ai, there exist(n + k0)-tiles Xi and Yi with Xi ∩ αi 6= ∅, x ∈ Yi, and Xi ∩ Yi 6= ∅,where i = 1, 2, 3. Then the set

K =3⋃i=1

(Xi ∪ Yi)

consists of at most six (n + k0)-tiles, is connected, and meets each ofthe arcs α1, α2, α3. Hence K ′ = fn(K) is a connected set that consistsof at most six k0-tiles, and meets each of the the arcs βi = fn(αi),i = 1, 2, 3. Note that each arc βi is the union of all 0-edges edges thatit contains. Hence for i = 1, 2, 3 there exists a 0-edge ei ⊂ βi withei ∩K ′ 6= ∅. Since the arcs β1, β2, β3 have pairwise disjoint interior, itfollows that the 0-edges e1, e2, e3 are all distinct. So K ′ is a connectedset that meets three distinct 0-edges. Hence it joins opposite sides of

C. So K ′ should contain at least Dk0 ≥ Dk0 ≥ 10 tiles of order k0. Thisis a contradiction, because K ′ is a union of at most six k0-tiles.

This proves the claim that the sets A1, A2, A3 do not cover X, andwe conclude that we can find a point

p ∈ X \ (A1 ∪ A2 ∪ A3).

We claim that Un+k0(p) ⊂ X. If not, we could find a point y ∈Un+k0(p) \ X, and (n + k0)-tiles U and V with p ∈ U , y ∈ V , andU ∩ V 6= ∅. Then the connected set U ∪ V must meet ∂X, and henceone of the arcs αi; but then p ∈ Ai by definition of Ai. This is acontradiction showing the desired inclusion Un+k0(p) ⊂ X.


Using Lemma 8.8 (i) it follows that B%(p, (1/C)Λ−n) ⊂ X, whereC ≥ 1 is a constant independent of n and X.

8.3. The canonical orbifold metric as visual metric

Given an expanding Thurston map f , the question arises whether otherstandard metrics on S2 are visual metrics. To have such metrics avail-able, we restrict ourselves here to rational expanding Thurston maps

f : C → C. Recall that a rational Thurston map is expanding if andonly if it does not have any critical periodic points (see Proposition 2.3).

Lemma 8.10. Let f : C→ C be a rational expanding Thurston map.

Then the chordal metric σ on C is not a visual metric for f .

Proof. Assume σ is a visual metric for the rational expanding

Thurston map f with expansion factor Λ > 1. Let c ∈ C be a criticalpoint of f and f(c) = p. Let d = deg(f, c) ≥ 2. Tiles are defined in

terms of a fixed Jordan curve C ⊂ C with post(f) ⊂ C as usual. Forany n ∈ N0 let Xn

c be an n-tile that contains c. By Proposition 8.4 (ii)it holds diamσ(Xn

c ) Λ−n. Let Y n := f(Xn+1c ) for all n ∈ N, this is

an n-tile containing p. It holds

diamσ(Y n) Λ−nd

for all n ∈ N0, where C() is a constant independent of n. Thiscontradicts Proposition 8.4 (ii). Thus σ is not a visual metric for f .

However, there is another metric on C, the canonical orbifold metric

ω, defined in terms of such a map f . Recall that Of = (C, αf ) isthe orbifold associated with f . When Of is hyperbolic the universalorbifold cover is the hyperbolic plane X = D, when Of is parabolic, theuniversal orbifold cover is the (Euclidean) plane X = C. In any case

there is a holomorphic map Θ: X → C, such that deg(Θ, x) = α(Θ(x))for all x ∈ X . This map is called the universal orbifold covering map,see Theorem A.17.

The canonical orbifold metric ω is obtained as the push forward ofthe (hyperbolic or Euclidean) metric d0 via Θ, i.e.,

ω(p, q) := infd0(z, w) : z ∈ Θ−1(p), w ∈ Θ−1(q)

for p, q ∈ C. See A.10 for more details. The following is the main resultof this section.

Proposition 8.11. Let f : C → C be a rational Thurston mapwithout periodic critical points, and ω be the canonical orbifold metricfor f . Then ω is a visual metric for f if and only if f is a Lattes map.

8.3. THE CANONICAL ORBIFOLD METRIC AS VISUAL METRIC 201

The two implications of Proposition 8.11 will be proved separately.We first prove the “only if” statement.

Lemma 8.12. Let f : C→ C be a rational expanding Thurston mapsuch that the canonical orbifold metric ω is a visual metric for f . Thenf is a Lattes map.

Proof. Let ω be the canonical orbifold metric for f , which is as-sumed to be visual. Let Λ > 1 be the expansion factor of ω. For any

p ∈ C there is a neighborhood Up ⊂ C of p such that

ω(p, p′) σ(p, p′)1/αf (p),

for all p′ ∈ Up (see A.10). Note that αf (q) <∞ by Proposition 2.9 (ii).

Fix a point q ∈ C and let p := f(q). Let Xn be an n-tile containing qfor all n ∈ N0. Let Y n = f(Xn+1), this is an n-tile containing p. Recallfrom Proposition 8.4 (ii) that diamω(Xn) diamω(Y n) Λ−n for alln ∈ N0. Thus the above implies that

diamσ(Xn) Λ−αf (q)n and diamσ(Y n) Λ−αf (p)n.

Let d := deg(f, q). Then

diamσ(Y n) (diamσ(Xn+1))d.

Which yields

Λ−αf (p)n Λ−αf (q)dn,

for all n. Since the constant C() is independent of n, this im-plies αf (p) = αf (q)d. This shows that Of is parabolic by Proposi-tion 2.13 (iii). Thus f is a Lattes map by Theorem 3.1.

We now prove the “if” implication of Proposition 8.11.

Proposition 8.13. Let f : C → C be a Lattes map, and ω be the

canonical orbifold metric of f on C. Then ω is a visual metric for fwith expansion factor Λ = deg(f)1/2 > 1. Moreover, ω is a geodesic

metric on C and for all paths γ in C we have

(8.7) lengthω(f γ) = Λ lengthω(γ).

We will see later see (in Proposition 19.1 and Theorem 15.3) thatfor given degree of an expanding Thurston map f the number Λ =deg(f)1/2 is the largest possible expansion factor of a visual metric.So Lattes maps are special as they realize this maximal factor. This isclosely related to the characterization of Lattes maps among expandingThurston maps as given later in Theorem 19.2.


Proof. Let f : C → C be a Lattes map. By Theorem 3.1 (ii)

there are holomorphic maps Θ: C → C and A : C → C such thatf Θ = Θ A. Furthermore A(z) = az + b, where a, b ∈ C withdeg(f) = |a|2 (see (3.10)). Let Λ = |a| = deg(f)1/2 > 1. We will showthat ω is a visual metric with expansion factor Λ.

In the following, C will be equipped with the Euclidean metric, andmetric notions for subsets of C will refer to this metric. Note thatAn is Euclidean a similarity on C, that scales distances by the factor|a|n = Λn (for all n ∈ N).

We equip C with the canonical orbifold metric ω. Metric terms thatare computed with respect to ω are equipped with a subscript ω. The

metric ω is geodesic, furthermore Θ: C→ (C, ω) is a path-isometry i.e.,for any path γ ⊂ C it holds length(γ) = lengthω Θ(γ), see A.10. Thisimmmediately implies that ω(Θ(x),Θ(y)) ≤ |x−y| for all x, y ∈ C, i.e.Θ is 1-Lipschitz.

We will show that ω satisfies the conditions in Proposition 8.4. Thiswill follow essentially from Θ being a path-isometry as well as the factthat An scales by the factor Λn.

Let γ be an arbitrary path in C. Then γ has a lift under thebranched covering map Θ, and so there exists a path γ in C such thatγ = Θ γ (see Lemma A.3). Then f Θ = Θ A implies

Θ An γ = fn Θ α = fn γ,

for all n ∈ N0. Since Θ is a path-isometry and An scales distances bythe factor Λn, we conclude that

lengthω(fn γ) = length(An γ)(8.8)

= Λn length(γ) = Λn lengthω(γ).

If we take n = 1 here, then (8.7) follows.

Now pick a Jordan curve C ⊂ C with post(f) ⊂ C and consider n-cells for (f, C). Suppose σ and τ are two disjoint n-cells, where n ∈ N0.

Since ω is a geodesic metric, we can find a path γ in C joining σ andτ with

(8.9) lengthω(γ) = distω(σ, τ).

By Lemma 5.33 the path fn γ joins opposite sides of C, and so

lengthω(fn γ) ≥ diamω(fn γ) ≥ δ0,

where δ0 > 0 is defined as in (5.13) for the base metric ω on C. Notethat δ0 is independent of n, σ, and τ . Combining the last inequality

8.3. THE CANONICAL ORBIFOLD METRIC AS VISUAL METRIC 203

with (8.9) and (8.8), we arrive at the inequality

(8.10) distω(σ, τ) & Λ−n.

Here and in the following, the implicit multiplicative constant C(&) isindependent of the cells and their level n. So ω satisfies condition (i)in Proposition 8.4.

Since every n-edge or n-tile τ contains two distinct n-vertices, (8.10)also implies that

(8.11) diamω(τ) & Λ−n.

To show Proposition 8.4 (ii), it remains to prove an inequality inthe opposite direction. Since Θ is uniformly continuous, there is anumber δ1 > 0 such that for every connected set K ⊂ C, the inequalitydiamω(Θ(K)) < δ1 implies that diam(K) ≤ 1 (see Lemma 6.13). Sincef is expanding, we can find n0 ∈ N such that diamω(X) ≤ δ1, wheneverX is an n-tile with n ≥ n0.

Now suppose X is an arbitrary n-tile with n ≥ n0. Define k =

n−n0 ∈ N0 and U = int(X). Then U ⊂ C is a simply connected region

that does not contain any 0-vertices and so U ⊂ C \ post(f). Since

Θ: C \ Θ−1(post(f)) → C \ post(f) is a covering map, the inclusion

map U → C \ post(f) lifts under Θ. This implies that there exists aregion V ⊂ C such that Θ(V ) = U . Note that diamω(U) ≤ diam(V ),because distances do not increase under the map Θ. Moreover,

(8.12) Θ(Ak(V )) = fk(Θ(V )) = fk(U) ⊂ fk(X).

On the other hand, fk(X) is a tile of level n+ k = n0, and so

δ1 > diamω(fk(X)) ≥ diamω(fk(U)).

Since Ak(V ) is connected, this inequality, the relation (8.12), and thedefinition of δ1 imply that

diam(Ak(V )) ≤ 1.

It follows that

diamω(X) = diamω(U) = diamω(U) ≤ diam(V )

= Λ−k diam(Ak(V )) ≤ Λ−k = Λn0Λ−n . Λ−n.

For n-tiles X with n ≤ n0 we have diamω(X) 1 Λ−n, and so thelast inequality is trivially true. Hence diamω(X) . Λ−n for tiles X onany level n ∈ N0 if we choose a suitable constant C(.). Since everyn-edge is contained in an n-tile, we have diamω(τ) . Λ−n whenever τis an n-tile or n-edge, n ∈ N0. So (8.11) implies that diamω(τ) Λ−n

whenever τ is an n-tile or n-edge, n ∈ N. This shows that ω also


satisfies condition (ii) in Proposition 8.4. It follows that ω is a visualmetric for f with expansion factor Λ.

Proposition 8.11 clearly follows from Proposition 8.13 and Lemma 8.10.

CHAPTER 9

Symbolic dynamics

If one wants to understand a dynamical system (X, f) given by theiteration of a map f on a space X, then often one tries to find a link tosymbolic dynamics and the theory of shift operators, in particular toshifts of finite type. These operators serve as an important paradigmin dynamics. In this chapter we will study this for expanding Thurstonmaps and prove the following statement.

Theorem 9.1. Let f : S2 → S2 be an expanding Thurston map.Then f is a factor of the left-shift Σ: Jω → Jω on the space Jω of allsequences in a finite set J of cardinality #J = deg(f).

This is essentially due to Kameyama (see [Ka, Thm. 3.4]). Thebasic idea seems to go back to [Jo] (see also [Prz85]). Kameyama’snotion of an expanding Thurston map is different from ours, but hisproof carries over to our setting with only minor modifications.

It is a standard fact in Complex Dynamics that the repelling pe-

riodic points of a rational map on C are dense in its Julia set. Thefollowing statement is an analog of this for expanding Thurston maps.As we will see, it easily follows from the proof of Theorem 9.1.

Corollary 9.2. Let f : S2 → S2 be an expanding Thurston map.Then the periodic points of f are dense in S2.

Before we supply the proofs of the above results, we will first reviewsome some basic definitions related to shift operators.

Let J be a finite set. We consider J as an alphabet and its elementsas letters in this alphabet. A word is a finite sequence w = i1 . . . in,where n ∈ N0 and i1, . . . , in ∈ J . For n = 0 we interpret this asthe empty word ∅. The number n is called the length of the wordw = i1i2 . . . in. The words of length n can be identified with n-tuplesin J and are elements in the Cartesian power Jn. The letters, i.e., theelements in J , are precisely the words of length 1. If w = i1i2 . . . in andw′ = j1 . . . jm, then we denote by ww′ = i1i2 . . . inj1 . . . jm the wordobtained by concatenating w and w′.

Let J∗ be the set of all words (including the empty word) in thealphabet J . The (left-)shift Σ: J∗ \ ∅ → J∗ is defined by setting

205

206 9. SYMBOLIC DYNAMICS

Σ(i1i2 . . . in) = i2 . . . in for a word w = i1i2 . . . in ∈ J∗ \ ∅. We denoteby Jω the set of all sequences ik in J , where the sequence elementsik ∈ J are indexed by k ∈ N. More informally, we consider a sequences = ik ∈ Jω as a word of infinite length and write s = i1i2 . . . .

If s = ik ∈ Jω and n ∈ N0, then we denote by [s]n ∈ J∗ theword sn = i1 . . . in consisting of the first n elements of the sequence s.The (left-)shift Σ: Jω → Jω is the map that assigns to each sequenceik ∈ Jω the sequence jk ∈ Jω with jk = ik+1 for all k ∈ N. Inour notation we do not distinguish the shifts on J∗ \ ∅ and Jω anddenote both maps by Σ. Note that [Σ(s)]n = Σ([s]n+1) for all s ∈ Jω;indeed, if s = i1i2 . . . , then we have

[Σ(s)]n = [i2i3 . . . ]n = i2 . . . in+1 = Σ(i1 . . . in+1) = Σ([s]n+1).

If we equip J with the discrete topology, then Jω carries a naturalmetrizable product topology. This topology is induced by the ultra-metric d given by d(s, s′) = 2−N for s = ik ∈ Jω and s′ = jk ∈ Jω,s 6= s′, where N = mink ∈ N : ik 6= jk. In particular, two elementss, s′ ∈ Jω are close if and only if sk = s′k for all k = 1, . . . , n, where nis large. Equipped with this topology, the space Jω is compact.

Suppose that X and X are topological spaces, and f : X → X

and f : X → X are continuous maps. We say that the dynamical

system (X, f) is a factor of the dynamical system (X, f) if there exists

a surjective continuous map ϕ : X → X such that ϕ f = f ϕ.We are now ready for the proofs.

Proof of Theorem 9.1. Let f : S2 → S2 be an expanding Thur-ston map, and k := deg(f) ≥ 2. Fix a visual metric % for f , and letΛ > 1 be its expansion factor. In the following, metric concepts referto %. We consider tiles for (f, C) and color them black and white as inLemma 5.19. Choose a basepoint p ∈ S2 \post(f) in the interior of thewhite 0-tile X0

w .

Claim 1. supx∈S2

dist(x, f−n(p)) . Λ−n,

where C(.) is independent of n ∈ N0. In other words, the set f−n(p)forms a very dense net in S2 if n is large.

To see this, let x ∈ S2 be arbitrary. Then x lies in some n-tile Xn. If Xn is white, then Xn contains a point in f−n(p) and sodist(x, f−n(p)) ≤ diam(Xn). If Xn is black, then Xn shares an edgewith a white n-tile Y n. Then Y n contains a point in f−n(p), and sodist(x, f−n(p)) ≤ diam(Xn) + diam(Y n).

From the inequalities in both cases and Proposition 8.4 we conclude

dist(x, f−n(p)) . Λ−n,

9. SYMBOLIC DYNAMICS 207

where C(.) is independent of x and n. Claim 1 follows.

None of the points in S2 \ post(f) is a critical value for any ofthe iterates fn of f . Moreover, each iterate fn is a covering mapfn : S2 \ f−n(post(f))→ S2 \ post(f). Since p ∈ S2 \ post(f), we havef−n(p) ⊂ S2 \ f−n(post(f)) and

(9.1) #f−n(p) = deg(fn) = deg(f)n = kn

for n ∈ N. In particular,

f−1(p) ⊂ S2 \ f−1(post(f)) ⊂ S2 \ post(f),

and #f−1(p) = k. Let q1, . . . , qk ∈ S2 \post(f) be the points in f−1(p).For i = 1, . . . , k we pick a path αi : [0, 1]→ S2 \post(f) with αi(0) = pand αi(1) = qi.

Let J := 1, . . . , k, and consider the shift Σ: Jω → Jω. We wantto show that f is a factor of Σ, i.e., that there exists a continuous andsurjective map ϕ : Jω → S2 with f ϕ = ϕ Σ. In order to define ϕ,we first construct a suitable map ψ that assigns to each word in J∗ apoint in S2.

Definition of ψ. The map ψ : J∗ → S2 will be defined inductively suchthat

ψ(w) ∈ f−n(p),

whenever n ∈ N0 and w ∈ Jn ⊂ J∗ is a word of length n. For the emptyword ∅ we set ψ(∅) = p, and for the word consisting of the single letteri ∈ J we set ψ(i) := qi ∈ f−1(p).

Now suppose that ψ has been defined for all words of length ≤ n,where n ∈ N. Let w be an arbitrary word of length n+1. Then w = w′i,where w′ ∈ J∗ is a word of length n and i ∈ J . So ψ(w′) ∈ f−n(p)is already defined. Since fn(ψ(w′)) = p and fn : S2 \ f−n(post(f)) →S2 \ post(f) is a covering map, the path αi has a unique lift withinitial point ψ(w′), i.e., there exists a unique path αi : [0, 1]→ S2 withαi(0) = ψ(w′) and fn αi = αi. We now define ψ(w) := αi(1). Notethat then

fn+1(ψ(w)) = fn+1(αi(1)) = f(αi(1)) = f(qi) = p.

Hence ψ(w) ∈ f−(n+1)(p). This shows that a map ψ : J∗ → S2 withthe desired properties exists.

Claim 2. f(ψ(w)) = ψ(Σ(w)) for all non-empty words w ∈ J∗.We prove this by induction on the length of the word w. If w = i ∈

J , then

f(ψ(w)) = f(ψ(i)) = f(qi) = p = ψ(∅) = ψ(Σ(i)) = ψ(Σ(w)).

So the claim is true for words of length 1.


Suppose the claim is true for words of length ≤ n, where n ∈ N.Let w be a word of length n+ 1. Then w = w′i, where w′ is a word oflength n and i ∈ J . Let αi be the path as above, used in the definition

of ψ(w). Define βi := f αi. Then βi is a lift of αi by fn−1. Byinduction hypothesis its initial point is

βi(0) = f(αi(0)) = f(ψ(w′)) = ψ(Σ(w′)).

In other words, βi is the unique path as in the definition of ψ used to

determine ψ(Σ(w′)i) from ψ(Σ(w′)), and so ψ(Σ(w′)i) = βi(1). Hence

ψ(Σ(w)) = ψ(Σ(w′)i) = βi(1) = f(αi(1)) = f(ψ(w′i)) = f(ψ(w))

as desired, and Claim 2 follows.

Claim 3. For each n ∈ N the map ψ|Jn : Jn → f−n(p) is a bijection.In other words, the map ψ provides a coding of the points in f−n(p)by words of length n.

Again we prove this by induction on n. By definition of ψ it is truefor n = 1.

Suppose it is true for some n ∈ N. It suffices to show that themap ψ|Jn+1 : Jn+1 → f−(n+1)(p) is surjective, since both sets Jn+1

and f−(n+1)(p) have the same cardinality kn+1. So let x ∈ f−(n+1)(p)be arbitrary. Then fn(x) ∈ f−1(p), and so there exists i ∈ J withfn(x) = qi. Since

x ∈ f−(n+1)(p) ⊂ S2 \ f−(n+1)(post(f)) ⊂ S2 \ f−n(post(f)),

and fn : S2 \ f−n(post(f)) → S2 \ post(f) is a covering map, we canlift the path αi by fn to a path αi : [0, 1] → S2 whose terminal pointis x (to see this, lift αi, traversed in opposite direction, so that theinitial point of the lift is x). Then fn(αi(0)) = αi(0) = p, and soαi(0) ∈ f−n(p). By induction hypothesis there exists a word w′ ∈ Jnwith ψ(w′) = αi(0). Then αi is a path as used to determine ψ(w′i)from ψ(w′). So if we set w := w′i ∈ Jn+1, then

ψ(w) = ψ(w′i) = αi(1) = x.

This shows that ψ|Jn+1 : Jn+1 → f−(n+1)(p) is surjective. Claim 3follows.

Claim 4. If s ∈ Jω, then the points ψ([s]n), n ∈ N, form a Cauchysequence in S2 (recall that [s]n is the word consisting of the first nelements of the sequence s).

By definition of ψ the points ψ([s]n) and ψ([s]n+1) are joined by alift of one of the paths α1, . . . , αk by fn. So by Lemma 8.7 we have

(9.2) %(ψ([s]n), ψ([s]n+1)) . Λ−n,

9. SYMBOLIC DYNAMICS 209

where C(.) is independent of n and s. Hence ψ([s]n) is a Cauchysequence.

Definition of ϕ. If s ∈ Jω, then by Claim 4 the limit

ϕ(s) := limn→∞

ψ([s]n)

exists. This defines a map ϕ : Jω → S2.

Claim 5. f ϕ = ϕ Σ.To see this, let s ∈ Jω be arbitrary. Note that Σ([s]n) = [Σ(s)]n−1

for n ∈ N. Hence by Claim 2 and the continuity of f we have

f(ϕ(s)) = limn→∞

f(ψ([s]n)) = limn→∞

ψ(Σ([s]n))

= limn→∞

ψ([Σ(s)]n−1) = ϕ(Σ(s)).

Claim 5 follows.

Claim 6. The map ϕ : Jω → S2 is continuous and surjective.Let s ∈ Jω and n ∈ N. Then (9.2) shows that

(9.3) %(ϕ(s), ψ([s]n)) .∞∑l=n

Λ−l . Λ−n,

where C(.) is independent of n and s. Hence if s, s′ ∈ Σ and [s]n =[s′]n, then

d(ϕ(s), ϕ(s′)) . Λ−n,

where C(.) is independent of n, s, and s′. The continuity of ϕ followsfrom this; indeed, if s and s′ are close in Jω, then [s]n = [s′]n for somelarge n, and so the image points ϕ(s) and ϕ(s′) are close in S2.

Since Jω is compact, the continuity of ϕ implies that the imageϕ(Jω) is also compact and hence closed in S2. The surjectivity of ϕwill follow, if we can show that ϕ has a dense image in S2.

To see this, let x ∈ S2 and n ∈ N be arbitrary. Then by Claim 1we can find a point y ∈ f−n(p) with %(x, y) . Λ−n, where C(.) isindependent of x and n. Moreover, by Claim 3 there exists a wordw ∈ Jn with ψ(w) = y. Pick s ∈ Jω such that [s]n = w. Then by (9.3)we have

%(x, ϕ(s)) ≤ %(x, y) + %(y, ϕ(s)) = %(x, y) + %(ψ([s]n), ϕ(s)) . Λ−n,

where C(.) is independent of the choices. Hence

supx∈S2

dist(x, ϕ(Jω)) . Λ−n

for all n, where C(.) is independent of n. This shows that ϕ(Jω) isdense in S2 and the claim follows.

The theorem now follows from Claim 5 and Claim 6.


The procedure that we employed to code the elements in f−n(p) bywords of length n is well-known [Ne, Sect. 5.2].

Proof of Corollary 9.2. We use the notation and setup of theproof of Theorem 9.1.

It suffices to show that if x ∈ S2 and n ∈ N are arbitrary, thenthere exists a point z ∈ S2 with fn(z) = z and %(x, z) . Λ−n. Hereand in the following, C(.) is independent of x and n.

To find such a point z, we apply Claim 1 in the proof of Theorem 9.1and conclude that there exists y ∈ f−n(p) with ϕ(x, y) . Λ−n. ByClaim 3 in this proof there exists a word w ∈ J∗ of length n suchthat ψ(w) = y. Let s be the unique sequence obtained by periodicrepetition of the letters in w, i.e., s ∈ Jω is the unique sequence with[s]n = w and Σn(s) = s. Put z := ϕ(s). Then Claim 5 in the proof ofTheorem 9.1 implies

fn(z) = fn(ϕ(s)) = ϕ(Σn(s)) = ϕ(s) = z.

Moreover, by (9.3) we have

%(y, z) = %(ψ(w), ϕ(s)) = %(ψ([s]n), ϕ(s)) . Λ−n,

and so%(x, z) ≤ %(x, y) + %(y, z) . Λ−n.

The statement follows.

If in the previous argument we choose a constant sequence s ∈ Jωand set z = ϕ(s), then Σ(s) = s, and so

f(z) = f(ϕ(s)) = ϕ(Σ(s)) = ϕ(s) = z.

This shows that every expanding Thurston map has a fixed point. Amore systematic investigation of fixed points and periodic points ofexpanding Thurston maps can be found in [Li1].

CHAPTER 10

Tile graphs

An interesting feature of expanding Thurston maps is that they arelinked to negatively curved spaces. Namely, if f : S2 → S2 is a sucha map and C ⊂ S2 is a Jordan curve with post(f) ⊂ C, then onecan use the associated cell decompositions to define an infinite graphG = G(f, C). The set of vertices of this graph is given by the collectionof tiles on all levels, where it is convenient to add X−1 := S2 as a tile oflevel −1 and basepoint of the graph. One connects two vertices by anedge if the corresponding tiles have non-empty intersection and theirlevels differ by at most 1. We will study the properties of this tile graphin the present chapter. The main results are based on a recent thesisby Q. Yin [Yi].

Theorem 10.1. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Then the associatedtile graph G(f, C) is Gromov hyperbolic.

For the boundary at infinity ∂∞G we have a natural identification∂∞G = S2. By this identification the class of visual metrics in thesense of Thurston maps (see Chapter 8) and in the sense of Gromovhyperbolic spaces (see Section 4.2) are the same.

Theorem 10.2. Let f : S2 → S2 be an expanding Thurston map,C ⊂ S2 a Jordan curve with post(f) ⊂ C, and G = G(f, C) be theassociated tile graph. Then ∂∞G can be identified with S2. Under thisidentification, a metric % on ∂∞G = S2 is visual in the sense of Gromovhyperbolic spaces if and only if it is visual in the sense of expandingThurston maps.

Under the identification of S2 with ∂∞G as in the previous theorem,the number mf,C (see Definition 8.1) is the Gromov product (x · y) upto some additive constant.

Lemma 10.3. In the setting of Theorem 10.2 there is a constantc ≥ 0 such that

(10.1) mf,C(x, y)− c ≤ (x · y) ≤ mf,C(x, y) + c,


211

212 10. TILE GRAPHS

Considering the sphere S2 as the boundary at infinity of the tilegraph is the approach used by Haıssinsky-Pilgrim in [HP09]. ThusTheorem 10.2 shows that their point of view is essentially equivalentto the one developed by us.

An obvious question is how the graphs G(f, C) and G(f, C) are re-

lated for different Jordan curves C, C ⊂ S2 containing post(f). Fora Cayley graph of a group a change of the generating set leads to aquasi-isometric Cayley graphs. So in the context of expanding Thurs-ton maps one may expect a similar result. Actually, a stronger state-

ment is true: the graphs G(f, C) and G(f, C) are even rough-isometric(see Section 4.2).

Theorem 10.4. Let f : S2 → S2 be an expanding Thurston map,C, C ′ ⊂ S2 Jordan curves with post(f) ⊂ C, C ′. Then the graphs G(f, C)and G(f, C ′) are rough-isometric.

For the rest of this chapter, f : S2 → S2 will be an expandingThurston map, and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Weconsider tiles in the cell decompositions Dn(f, C), n ∈ N0, and addX−1 := S2 as a tile of level −1. Let X′ be the collection of tiles onall levels n ∈ N0 ∪ −1. In X′ we consider tiles as different if theirlevels are different even if the underlying sets of the tiles are the same.If X ∈ X′, we denote by

(10.2) `(X) ∈ N0 ∪ −1the level of the tile X, so X is an `(X)-tile.

As discussed above, we define the tile graph G = G(f, C) of f withrespect to C as follows. The set of vertices of G is equal to the set X′

of all tiles. Moreover, two distinct vertices given by a k-tile Xk and ann-tile Xn are joined by an edge if

(10.3) |k − n| ≤ 1 and Xk ∩Xn 6= ∅.So we join two vertices if the levels of the corresponding tiles differs byat most 1 and if the tiles intersect. Similarly as for Cayley graphs ofgroups, the graph G is a 1-dimensional cell complex where each edge isidentified with an interval of length 1. Then the graph G is connected.

Indeed, each point in an edge in G can be joined to a vertex, andeach vertex in G, given by an n-tile Xn, can be joined to X−1 = S2 ∈ Gas follows. Pick a point p ∈ Xn, and for i = 1, . . . , n − 1 let X i bean i-tile with p ∈ X i. Then in the vertex sequence Xn, Xn−1, . . . , X−1

two consecutive elements are joined by an edge, because the levels ofthe tiles differs by 1 and all tiles contain p and hence have non-emptyintersection. So there exists a path joining Xn and X−1 in G as desired.

10. TILE GRAPHS 213

Since G is connected, this graph carries a unique path metric so thateach edge is isometric to the unit interval. If X and Y are vertices in G,i.e., tiles in X′, we denote by |X−Y | the distance of X and Y in G, andcall this quantity the combinatorial distance of the tiles. By definitionof the metric in G it is clear that |X − Y | is equal to the minimalnumber n ∈ N0 such that there exist tiles X0 = X,X1, . . . , Xn = Y inX′ satisfying

(10.4) |`(Xi−1)− `(Xi)| ≤ 1 and Xi−1 ∩Xi 6= ∅

for i = 1, . . . , n.Note that

|X − Y | ≥ |`(X)− `(Y )|.Moreover, if X ∩ Y 6= ∅, then a simple argument similar to the oneused to show connectedness of G gives that

|X − Y | ≤ |`(X)− `(Y )|+ 1.

We pick X−1 = S2 as the basepoint in G. Note that the combinatorialdistance of a vertex X ∈ G to X−1

|X −X−1| = `(X) + 1.

We denote by (X · Y ) the Gromov product of two vertices X, Y ∈ Gwith respect to the basepoint X−1; so

(X · Y ) =1

2(|X −X−1|+ |Y − Y −1| − |X − Y |)

= 1 +1

2(`(X) + `(Y )− |X − Y |).

We want to show that the graph G equipped with its path metric isa Gromov hyperbolic space. Since every point in G has distance ≤ 1/2to a vertex, the set X′ of vertices in G is cobounded in G. Hence it isenough to consider the space of vertices X′ equipped with the metricgiven by the combinatorial distance of vertices. The key for provingGromov hyperbolicity of G is to relate the Gromov product to visualmetrics for f as discussed in Chapter 8. The basic idea goes back to asimilar argument in [BP]. Our presentation mostly follows [Yi].

For the rest of this chapter we pick a fixed visual metric % for fon S2. Metric notions on S2 will usually refer to % unless otherwisestated. If Λ > 1 is the expansion factor of %, then by Proposition 8.4there exists a constant Co ≥ 1 such that

(10.5)1

C0

Λ−`(X) ≤ diam(X) ≤ C0Λ−`(X)

214 10. TILE GRAPHS

for all X ∈ X′, and

(10.6) dist(X, Y ) ≥ 1

C0

Λ−`(X)

for all X, Y ∈ X′ with X ∩ Y = ∅ and `(X) = `(Y ).We may view tiles in two distinct ways: as vertices in the graph G,

or as subsets of the sphere S2. The following lemma provides the keyto relate these two viewpoints.

Lemma 10.5. There exists a constant C ′ ≥ 1 such that for all tilesX, Y ∈ X′ ⊂ G we have

1

C ′Λ−(X·Y ) ≤ diam(X ∪ Y ) ≤ C ′Λ−(X·Y ).

See [BP, Lem. 2.2] for a similar statement in a different (but re-lated) context.

Proof. Let X, Y ∈ X′ be arbitrary, and set n = |X − Y |.To prove the upper bound, we pick a tile chain

X0 = X,X1, . . . , Xn = Y

satisfying (10.4) and realizing the combinatorial distance n between Xand Y . Let us estimate the minimal level that of tiles in this chain.Note that

`(Xi) ≥ min`(X)− i, `(Y )− (n− i)for i = 0, . . . , n. The minimum of the right hand side occurs for

l = b(`(X)− `(Y ) + n)/2c ∈ [0, n].

Then

maxl − `(X), (n− l − 1− `(Y )) ≤ (n− `(X)− `(Y ))/2.

Hence by (10.5),

diam(X ∪ Y ) ≤n∑i=0

diam(Xi) ≤ C0

n∑i=0

Λ−`(Xi)

≤ C0

l∑i=0

Λi−`(X) + C0

n∑i=l+1

Λ(n−i)−`(Y )

≤ 2C0Λ

Λ− 1Λ(n−`(X)−`(Y ))/2

= C1Λ−(X,Y ),

where

C1 :=2C0

Λ− 1.

10. TILE GRAPHS 215

For the lower bound, let

−1 ≤ m ≤ min`(X), `(Y )

be the maximal integer for which there exist m-tiles Xm and Y m withX ∩Xm 6= ∅, Y ∩ Y m 6= ∅, and Xm ∩ Y m 6= ∅. Then

|X −Xm| ≤ `(X)−m+ 1,

|Y − Y m| ≤ `(Y )−m+ 1,

|Xm − Y m| ≤ 1.

This implies

|X − Y | ≤ |X −Xm|+ |Xm − Y m|+ |Y m − Y |≤ `(X) + `(Y )− 2m+ 3,

and so

(X · Y ) =1

2(`(X) + `(Y )− |X − Y |) + 1 ≥ m− 1/2.

Now if m = min`(X), `(Y ), then by (10.5),

diam(X ∪ Y ) ≥ maxdiam(X), diam(Y )

≥ 1

C0

maxΛ−`(X),Λ−`(Y )

=1

C0

Λ−m ≥ 1

C2

Λ−(X,Y ),

where C2 := C0Λ3/2. This gives the desired lower bound in this case.In the other case, where m < min`(X), `(Y ), we pick (m+1)-tiles

Xm+1 and Y m+1 with X ∩Xm+1 6= ∅ and Y ∩ Y m+1 6= ∅. Then we canfind points x ∈ X ∩Xm+1 and y ∈ Y ∩ Y m+1. Moreover, by definitionof m we have Xm+1 ∩ Y m+1 = ∅. Hence by (10.6),

diam(X ∪ Y ) ≥ %(x, y) ≥ dist(Xm+1, Y m+1)

≥ 1

C0

Λ−(m+1) ≥ 1

C2

Λ−(X,Y ).

So we get the desired lower bound also in this case.

The following corollary to the previous lemma relates sequencesconverging to infinity in G with points in the sphere S2.

Lemma 10.6. Let Xi be a sequence of points (tiles) in X′. Then

(i) Xi converges to infinity in G if and only if there is a uniquepoint p ∈ S2 such that Xi converges to p in S2 with respectto the Hausdorff metric.

216 10. TILE GRAPHS

(ii) Another sequence Yi in X′ that converges to infinity in Gis equivalent to Xi if and only if they converge to the samesingleton p (with respect to the Hausdorff metric in S2).

Recall from (4.8) that a sequence Xi in X′ converges to infinityif and only if

(10.7) limi,j→∞

(Xi ·Xj) =∞

and from (4.9) that a sequence Yi in X′ that converges to infinity isequivalent to Xi if (and only if)

limi→∞

(Xi · Yi) =∞.

Proof. Note that (10.7) is equivalent to

(10.8) limi,j→∞

diam(Xi ∪Xj) = 0

by Lemma 10.5. This is equivalent to diam(Xi) → 0 as i → ∞, andXi is a Cauchy sequence with respect to Hausdorff convergence in S2.This in turn happens if and only if there exists a unique point p ∈ S2

such that Xi → p as i→∞ (in the sense of Hausdorff convergence).Thus (i) holds.

Let Yi be another sequence in X′ that converges to infinity. FromLemma 10.5 we see that Yi is equivalent to Xi if and only iflimi→∞ diam(Xi ∪ Yi) = 0. This happens if and only if Yi convergeswith respect to Hausdorff convergence to the same singleton p. Thus(ii) holds.

Proof of Theorem 10.1. We use notation as before. Since theset of vertices X′ is cobounded in G, it suffices to show that X′ equippedwith the combinatorial distance is Gromov hyperbolic.

Now if X, Y, Z ∈ X′ are arbitrary, then

diam(X ∪ Y ) ≤ diam(X ∪ Z) + diam(Z ∪ Y )

≤ 2 maxdiam(X ∪ Z), diam(Z ∪ Y ).Taking logarithms with base Λ in this inequality and invoking Lemma 10.5,we obtain

(X · Y ) ≥ min(X · Z), (Z · Y ) − δ,where δ ≥ 0 is a suitable constant independent of X, Y , Z. Thestatement follows.

We are now ready to prove that our notion of visual metric on S2

agrees with the standard one on ∂∞G under a suitable identification.Recall from Section 4.2 that ∂∞G is defined to be the set of all equiva-lence classes of sequences Xi in X′ converging to infinity.

10. TILE GRAPHS 217

Proof of Theorem 10.2. The identification will be given by abijection between ∂∞G and S2. Since X′ is cobounded in G, every pointx ∈ ∂∞G can be represented by a sequence of tiles Xi in X′ convergingto infinity. By Lemma 10.6 (i) there is a unique point p ∈ S2 such thatXi converges to p (with respect to Hausdorff distance). Any twosequences Xi, Yi representing x converge to the same singleton pby Lemma 10.6 (ii). Thus the map

ϕ : ∂∞G → S2 given by ϕ(x) := p

is well defined.The map ϕ is surjective. Indeed, let p ∈ S2 be arbitrary. For

each i ∈ N we pick an i-tile Xi ∈ X′ such that p ∈ Xi. Since f isexpanding, we have diam(Xi)→ 0 as i→∞. ThusXi → p as i→∞(with respect to Hausdorff convergence in S2). Thus Xi convergesto infinity by Lemma 10.6 (i) and the point x ∈ ∂∞G represented byXi is mapped to p by ϕ.

To show that ϕ is injective, consider two points x, y ∈ ∂∞G that arerepresented by sequences Xi and Yi in X′ converging to infinity.By Lemma 10.6 (ii) it holds that they converge to the same singletonif and only if they they are equivalent. Thus ϕ(x) = ϕ(y) if and onlyif x = y. Thus ϕ is injective.

Having proved that ϕ is bijective, we ignore the original distinctionbetween points in ∂∞G and in S2, and identify ∂∞G = S2 by the mapϕ.

Let x and y be arbitrary points in S2 and let Xi and Yi be twosequences in X′ representing them in ∂∞G respectively. Recall fromthe above that this means Xi → x and Yi → y as i → ∞ withrespect to Hausdorff convergence in S2. Thus

diam(Xi ∪ Yi)→ %(x, y)

as i→∞.Recall from (4.10) the definition of the Gromov product (x · y) for

x, y ∈ ∂∞G = S2. As the basepoint in G we choose X−1 = S2 forconvenience. By (4.11) there exists a constant k ≥ 0 independent of xand y, and of the choice of the sequences Xi, Yi such that

lim infi→∞

(Xi · Yi)− k ≤ (x · y) ≤ lim infi→∞

(Xi · Yi).

Combining these two estimates yields %(x, y) Λ−(x·y) by Lemma 10.5.Since % is a visual metric, by Proposition 8.3 (iii) we also have

%(x, y) Λ−m(x,y), where m = mf,C is as in Definition 8.1. We concludethat

(10.9) m(x, y)− c ≤ (x · y) ≤ m(x, y) + c,

218 10. TILE GRAPHS

where c ≥ 0 is independent of x and y.Now let % be an arbitrary metric on S2 = ∂∞G. Recall that in the

definition of visual metrics in the sense of Gromov hyperbolic spaces(4.12) we may choose any basepoint (up to an adjustment of the mul-tiplicative constant). Similarly, by Proposition 8.3 (iii) it is not a re-striction to use the curve C in Definition 8.2 of visual metrics in thesense of expanding Thurston maps.

Thus (10.9) shows that a metric % on ∂∞G = S2 is a visual metricin the sense of Gromov hyperbolic spaces if and only if % is a visualmetric in the sense of expanding Thurston maps.

Note that in (10.9) we have proved Lemma 10.3.

Proof of Theorem 10.4. It suffices to find a rough-isometry be-

tween the set of vertices in G = G(f, C) and G = G(f, C). As before, we

denote the set of tiles for (f, C) by X′, and use the notation X′ for the

set of tiles for (f, C) (including X−1 = X−1 = S2).

For each tile X ∈ X′ we pick a tile X ∈ C of the same level (i.e.,

`(X) = `(X)) with X ∩ X 6= ∅. This assignment X 7→ X gives a

level-preserving map ψ : X′ → X′. We claim that ψ is a rough-isometry

between X′ and X′, where the spaces are equipped with their respectivecombinatorial distances.

To see this, let X, Y ∈ X′ be arbitrary, and consider X := ψ(X)

and Y := ψ(Y ). Since ψ preserves levels of tiles, we have

diam(X) Λ−`(X) = Λ−`(X) diam(X),

and similarly diam(Y ) diam(Y ). Hence

diam(X ∪ Y ) ≤ diam(X) + diam(X ∪ Y ) + diam(Y )

. diam(X ∪ Y ),

and the same argument gives diam(X ∪ Y ) . diam(X ∪ Y ). In all theprevious relations the implicit multiplicative constants are independentof X and Y . Lemma 10.5 implies that there exists a constant c ≥ 0independent of X and Y such that

(X · Y )− c ≤ (X · Y ) ≤ (X · Y ) + c.

Here Gromov products are in X′ and X′, respectively, with respect to

the basepoint X−1 = S2. Since `(X) = `(X) and `(Y ) = `(Y ), thisimplies the relation

|X − Y | − k ≤ |X − Y | ≤ |X − Y |+ k,

for combinatorial distances, where k := 2c is independent of X and Y .

10. TILE GRAPHS 219

This is the first condition (4.6) (with λ = 1) for ψ to be a rough-

isometry. It remains to show that ψ(X′) is cobounded in X′. To verify

this, let Y ∈ X′ be arbitrary. Pick a tile X ∈ X′ with `(X) = `(Y )

and X ∩ Y 6= ∅. Define X := ψ(X). It suffices to produce a uniform

upper bound for the combinatorial distance |X − Y | of X and Y in X′

independent of Y . Now

diam(X) diam(X) diam(Y ) Λ−`(Y ),

and so

diam(X ∪ Y ) ≤ diam(X) + diam(X) + diam(Y ) . Λ−`(Y ),

where again all implicit multiplicative constants are independent of thechoice of the tiles. So by Lemma 10.5 we have

(X · Y ) ≥ `(Y )− c′,

where c′ ≥ 0 is independent of the choices of the tiles. Since `(X) =

`(Y ), we conclude

|X − Y | = 2 + 2`(Y )− 2(X · Y ) ≤ k′ := 2 + 2c′,

which gives the desired uniform bound.

Remark 10.7. The rough-isometry ψ between the graphs G and Gconstructed in the previous proof is compatible with the identifications

∂∞G = S2 and ∂∞G = S2. Indeed, let p ∈ S2 be arbitrary. Viewedas point in ∂∞G, p is represented by a sequence Xi in X′ such thatXi → p (with respect to Hausdorff convergence) as i → ∞ (see

Lemma 10.6). If Xi := ψ(Xi) for i ∈ N, then Xi is a sequence of

tiles in X′. By definition of ψ the levels of Xi and Xi are the same,

thus diam Xi → 0 as i→∞. Furthermore Xi ∩ Xi 6= ∅ for i ∈ N. This

implies that Xi → p as i→∞. So if Xi represents the point p ∈ S2

under the identification ∂∞G = S2, then the image sequence Xi under

ψ also represents the point p under the identification ∂∞G = S2.

CHAPTER 11

Isotopies

In this section we present some topological facts about isotopies thatwill be important throughout the paper.

Let I = [0, 1], and X and Y be topological spaces. Recall (seeSection 2) that an isotopy between X and Y is a continuous mapH : X × I → Y such that each map Ht := H(·, t) is a homeomorphismof X onto Y . If A ⊂ X, then H is an isotopy relative to A or rel. A ifHt(a) = H0(a) for all a ∈ A and t ∈ I. If ϕ : X → Y and ψ : X → Yare homeomorphisms, we say that ϕ and ψ are isotopic (rel. A ⊂ X) ifthere exists an isotopy H : X × I → Y (rel. A) such that H0 = ϕ andH1 = ψ.

When we say that a family Ht (where it is understood that t ∈ I)of homeomorphisms from X onto Y is an isotopy between X and Y ,we consider t as a variable in I and mean that the map (x, t) ∈ X ×I 7→ Ht(x) is an isotopy. This is a slightly imprecise, but convenientway of expression. If X = Y then Ht is called an isotopy on X. IfA,B,C ⊂ X, then we say that B is isotopic to C rel. A or B can beisotoped into C rel. A if there exists an isotopy H : X × I → X rel. Awith H0 = idX and H1(B) = C. Note that this notion depends on theambient space X containing the sets A, B, C.

11.1. Equivalent expanding Thurston maps are conjugate

Recall that two Thurston maps f : S2 → S2 and g : S2 → S2 on

2-spheres S2 and S2 are (Thurston) equivalent (see Definition 2.4) if

there exist homeomorphisms h0, h1 : S2 → S2 that are isotopic rel.post(f) and satisfy that h0 f = gh1. We then have the commutativediagram:

(11.1) S2 h1//

f

S2

g

S2 h0// S2.

The maps f and g are topologically conjugate if there exists a homeo-

morphism h : S2 → S2 such that h f = g h.

221

222 11. ISOTOPIES

Obviously, the notion of Thurston equivalence is weaker than topo-logical conjugacy of the maps. We will show that under the additionalassumption that the maps are expanding, we can promote an equiv-alence between two Thurston maps to a topological conjugacy. Theidea for the proof of this statement uses well-known ideas in dynam-ics. A statement very similar to Theorem 11.4 below was proved byKameyama [Ka]. Since his notion of “expanding” is different fromours, we will present the details of the proof.

First, we state a lifting theorem that will be needed (see [Ka,Lem. 4.3]).

Proposition 11.1 (Isotopy lifting for branched covers). Let f : S2 →S2 and g : S2 → S2 be Thurston maps, and h0, h0 : S2 → S2 be home-

omorphisms such that h0| post(f) = h0| post(f) and h0 f = g h0.

Suppose H : S2 × I → S2 is an isotopy rel. post(f) with H0 = h0.

Then the isotopy H uniquely lifts to an isotopy H : S2 × I → S2 rel.

f−1(post(f)) such that H0 = h0 and g Ht = Ht f for all t ∈ I.

So if we set h1 := H1 and h1 := H1, then we have the followingcommutative diagram:

S2 H : h0'h1//

f

S2

g

S2 H : h0'h1// S2.

Proof. By an argument similar to the one used to establish (??)one can show that

(11.2) h0(post(f)) = h0(post(f)) = post(g).

This implies that

Ht(post(f)) = post(g)

for all t ∈ I. Hence Ht|S2 \ post(f) is an isotopy between S2 \ post(f)

and S2 \ post(g).Moreover, it follows from (11.2) that

h0(f−1(post(f))) = g−1(post(g)).

So the map h0|S2 \ f−1(post(f)) can be considered as a lift of

H0|S2 \ post(f) = h0|S2 \ post(f)

11.1. EQUIVALENT EXPANDING MAPS ARE CONJUGATE 223

by the (non-branched) covering maps

f : S2 \ f−1(post(f))→ S2 \ post(f)

and

g : S2 \ g−1(post(g))→ S2 \ post(g).

By the usual homotopy lifting theorem for covering maps (see [Ha,p. 60, Prop. 1.30]) the isotopy Ht|S2 \ post(f) lifts to a unique isotopy

Ht between S2 \ f−1(post(f)) and S2 \ g−1(post(g)) such that

H0 = h0|S2 \ f−1(post(f))

and g Ht = Ht f on S2 \ f−1(post(f)) for all t ∈ I.

Since Ht is constant in t on post(f), each map Ht has a continuous

extension to S2, also denoted by Ht. Then Ht|f−1(post(f)) does not

depend on t. Moreover, each map Ht is a homeomorphism from S2 onto

S2, because an inverse of Ht can be obtained by lifting the isotopy H−1t .

So Ht is an isotopy between S2 and S2 rel. f−1(post(f)). It is clearthat it has the desired properties.

Note that if in the previous proposition H is an isotopy relative to

a set M ⊂ S2 with post(f) ⊂ M , then the lift H is an isotopy rel.f−1(M). Indeed, if p ∈ f−1(M), then f(p) ∈M and so

g(Ht(p)) = Ht(f(p)) = h0(f(p)) =:q

for all t ∈ I. Thus t 7→ Ht(p) is a path contained in the finite set g−1(q)and hence a constant path.

For later use we record a simple lemma about preimages of sets.

Lemma 11.2. Let f : X → X and g : Y → Y be maps defined onsome sets X and Y , and h0, h1 : X → Y be bijections with gh1 = h0f .Then for every set A ⊂ X we have

g−1(h0(A)) = h1(f−1(A)).

Proof. Since h0, h1 are bijections, g h1 = h0 f implies h−10 g =

f h−11 . Thus(h−1

0 g)−1

(A) = g−1(h0(A)) =(f h−1

1

)−1(A) = h1(f−1(A)),

as desired.

The following lemma will be of crucial importance.

224 11. ISOTOPIES

Lemma 11.3 (Exponential shrinking of tracks of isotopies).

Let f : S2 → S2 and g : S2 → S2 be Thurston maps, and Hn : S2× I →S2 be isotopies rel. post(f) satisfying g Hn+1

t = Hnt f for n ∈ N0

and t ∈ I.If g is expanding and S2 is equipped with a visual metric for g, then

the tracks of the isotopies Hn shrink exponentially as n → ∞; moreprecisely, if d is a visual metric for g with expansion factor Λ > 1, thenthere exists a constant C ≥ 1 such that

(11.3) supx∈S2

diamd(Hnt (x) : t ∈ I) ≤ CΛ−n

for all n ∈ N0.

Proof. For all n ∈ N0 and t ∈ I we have gn Hnt = H0

t fn; so for

fixed x ∈ S2 and n ∈ N0 the path t 7→ Hnt (x) in S2 is a lift of the path

t 7→ H0t (fn(x)) by the map gn. Recall that in the proof of Lemma 8.7

we had to break up the path γ into N pieces γj so that diam(γj) < δ0

(see also (5.13)). Since H0 is uniformly continuous we can choose thenumber N uniformly for all the paths t 7→ H0

t (y), y ∈ S2. Since g isexpanding, Lemma 8.7 then implies that

supx∈S2

diamd(Hnt (x) : t ∈ I) . Λ−n

for all n ∈ N, where C(.) is independent of n.

Theorem 11.4 (Thurston equivalence implies topological conju-

gacy). Let f : S2 → S2 and g : S2 → S2 be equivalent Thurston mapsthat are expanding. Then they are topologically conjugate. More pre-cisely, if we have a Thurston equivalence between f and g as in (11.1),

then there exists a homeomorphism h : S2 → S2 such that h is isotopicto h1 rel. f−1(post(f)) and satisfies h f = g h.

Since post(f) ⊂ f−1(post(f)) and h0 and h1 are isotopic rel. post(f)this implies that h is also isotopic to h0 rel. post(f).

Proof. The main idea of the proof is to lift a suitable initial iso-topy repeatedly and use the fact that by Lemma 11.3 the tracks ofthe isotopies shrink exponentially fast. The desired conjugacy is thenobtained as a limit.

By assumption there exists an isotopy H0t between S2 and S2 rel.

post(f) such that h0 f = g h1, where h0 = H00 and h1 = H0

1 . ByProposition 11.1 we can lift the isotopy H0

t between h0 and h1 to anisotopy H1

t rel. f−1(post(f)) ⊃ post(f) between h1 and h2 := H11 . Note

that the map h1 plays two roles here: it is the endpoint H01 of the initial

isotopy H0t , and also a lift of h0.


Repeating this argument we get homeomorphisms hn and isotopies

Hnt between S2 and S2 rel. f−n(post(f)) ⊃ post(f) for all n ∈ N0 such

Hnt f = g Hn+1

t , Hn0 = hn and Hn

1 = hn+1 for all n ∈ N0 and t ∈ I.So we have the following “infinite tower” of isotopies:

...

S2 H2 : h2'h3//

f

S2

g

S2 H1 : h1'h2//

f

S2

g

S2 H0 : h0'h1// S2

We want to show that for n → ∞ the maps hn converge to a homeo-morphism h∞ that gives the desired topological conjugacy between fand g.

To see this fix a visual metric d on S2, and assume that it has the

expansion factor Λ > 1. Metric concepts on S2 will refer to this metricin the following. Since g is expanding, Lemma 11.3 implies that

(11.4) supx∈S2

diam(Hnt (x) : t ∈ I) . Λ−n

for all n ∈ N, where C(.) is independent of n. In particular,

dist(hn+1, hn) := supx∈S2

dist(hn(x), hn+1(x)) . Λ−n

for all n ∈ N0, and so there is a continuous map h∞ : S2 → S2 suchthat hn → h∞ uniformly on S2 as n→∞. Since hn−1 f = g hn, wehave h∞ f = g h∞.

The map h∞ is a homeomorphism. To see this we repeat the ar-gument where we interchange the roles of f and g. More precisely, weconsider the isotopy (H0

t )−1 between h−10 and h−1

1 . The correspondingtower of repeated lifts of this initial isotopy is given by the isotopies(Hn

t )−1 between h−1n and h−1

n+1. By the argument in the first part of

226 11. ISOTOPIES

the proof we see that the maps h−1n converge to a continuous map

k∞ : S2 → S2 uniformly on S2 as n → ∞. By uniform convergencewe have k∞ h∞(x) = limn→∞ h

−1n hn(x) = x for all x ∈ S2. Hence

k∞ h∞ = idS2 . Similarly, h∞ k∞ = idS2 , and so k∞ is a continuousinverse of h∞. Hence h∞ is a homeomorphism.

The conjugating map h = h∞ is isotopic to h1 rel. f−1(post(f)).To see this we will define an isotopy rel. f−1(post(f)) that is obtainedby concatenating (with suitable time change) the isotopies H1, H2, . . .and take h = h∞ as the endpoint at time t = 1. The precise definitionis as follows. We break up the unit interval into intervals

I = [0, 1] =

[0,

1

2

]∪[

1

2,3

4

]∪ · · · ∪

[1− 2−n, 1− 2−n−1

]∪ · · · ∪ 1.

The n-th interval in this union is denoted by In = [1− 2−n, 1− 2−n−1].Let sn : In → I, sn(t) = 2n+1(t − (1 − 2−n)), for n ∈ N0. We define

H : S2 × I → S2 by

H(x, t) := Hn+1(p, sn(t))

if p ∈ S2 and t ∈ In for some n ∈ N0, and H(p, t) = h(p) for p ∈ S2

and t = 1. We claim that H is indeed an isotopy between h1 and hrel. f−1(post(f)).

Note that H is well defined, H1 = h, and H1−1/2n = hn+1 for n ∈ N0.Moreover, Ht is a homeomorphism for each t ∈ I, and Ht|f−1(post(f))does not depend on t. To establish our claim, it remains to verifythat H is continuous. It is clear that H is continuous at each point(p, t) ∈ S2 × [0, 1).

Moreover, as follows from the uniform convergence hn → h as n→∞ and inequality (11.4), we have Ht → H1 uniformly on S2 as t→ 1.This together with the continuity of h = H1 implies the continuity ofH at points (p, t) ∈ S2 × I with t = 1.

Remark 11.5. The previous proof gives a procedure for approx-imating the conjugating map h = h∞. Indeed, as follows from theremark after the proof of Proposition 11.1, the map Hn

t is constantin t on f−n(post(f)) for all n ∈ N0. This implies that hn = hn+1 =· · · = h∞ on the set f−n(post(f)), and so the map hn sends the pointsin f−n(post(f)) to the “right” points in g−n(post(g)). The isotopy Hn

t

then deforms hn to a map hn+1 such that the points in f−(n+1)(post(f))have the correct images in g−(n+1)(post(g)) as well, etc. Since by ex-pansion the union of the sets

post(f) ⊂ f−1(post(f)) ⊂ f−2(post(f)) ⊂ . . .


is dense in S2, this gives better and better approximations of limit maph∞.

To record an immediate consequence of Theorem 11.4, we introducesome terminology related to the notion of snowflake equivalent metricsdefined in Section 6. Let (X, dX) and (Y, dY ) be metric spaces. Ahomeomorphism h : X → Y is called a snowflake equivalence if thereexist constants α > 0 and C ≥ 1 such that

1

CdX(x, x′)α ≤ dY (h(x), h(x′)) ≤ CdX(x, x′)α

for all x, x′ ∈ X. The spaces X and Y are called snowflake equivalentif there exists a snowflake equivalence between X and Y . Note thattwo metrics d and d′ on a space X are snowflake equivalent as definedin Section 6 if and only if the identity map idX : (X, d) → (X, d′) is asnowflake equivalence.

Corollary 11.6. Let f : S2 → S2 and g : S2 → S2 be expand-ing Thurston maps that are Thurston equivalent. Then S2 equipped

with any visual metric with respect to f is snowflake equivalent to S2

equipped with any visual metric with respect to g. Every homeomor-

phism h : S2 → S2 satisfying h f = g h is a snowflake equivalence.

Proof. By Theorem 11.4 we know that there exists a topological

conjugacy between f and g, i.e., a homeomorphism h : S2 → S2 suchthat h f = g h. Let d be a visual metric on S2 with respect to f ,

and d be a visual metric on S2 with respect to g. Let Λ > 1 and Λ > 1

be the expansion factors of d and d, respectively. It suffices to show

that h : (S2, d)→ (S2, d) is a snowflake equivalence.To see this pick a Jordan curve C ⊂ S2 with post(f) ⊂ C. Then

C = h(C) is a Jordan curve in S2 with post(g) = h(post(f)) ⊂ C.Since h conjugates f and g, it follows from Proposition 5.17 (iii) and(v) or, alternatively, from the uniqueness statement in Lemma 5.15that for each n ∈ N0 the images of the cells in the cell decompositionDn := Dn(f, C) of S2 under the map h are precisely the cells in the cell

decomposition Dn = Dn(g, C) of S2; so we have

(11.5) Dn = h(c) : c ∈ Dn

for all n ∈ N0. This implies that

m(h(x), h(x′)) = m(x, x′)

228 11. ISOTOPIES

S1

J

i

−1

−i

1

Figure 11.1. J is not isotopic to S1 rel. 1, i ,−1,−i.

for all x, x′ ∈ S2, where m = mg,C and m = mf,C (recall Definition 8.1).

Combining this with Proposition 8.3 (iii) we see that

d(h(x), h(x′)) Λ−m(h(x),h(x′)) = Λ−m(x,x′) = Λ−αm(x,x′) d(x, x′)α

for all x, x′ ∈ S2, where α = log(Λ)/ log(Λ) and the implicit multi-plicative constants do not depend on x and x′. It follows that h is asnowflake equivalence.

11.2. Isotopies of Jordan curves

In the following S2 is a 2-sphere equipped with a fixed base metric.It will be the ambient space for all isotopies. In this subsection we studythe problem when two Jordan curves J and K on S2 passing througha given finite set P of points in the same order can be deformed intoeach other by an isotopy of S2 rel. P . If #P ≤ 3 this is always the case(see Lemma 11.10 below).

For #P ≥ 4 this is not always true as the example in Figure 11.1shows. Here K = S1 is the unit circle and P = 1, i ,−1,−i ⊂ S1.The Jordan curve J (which contains P ) is drawn with a thick line.The curves K = S1 and J are not isotopic rel. P . In fact, J may beobtained from S1 by a “Dehn twist” around the points −i and 1. Notethat in this example we can make the Hausdorff distance (see (14.3))between J and S1 arbitrarily small.

We will need the following statement.

Proposition 11.7. Suppose J is a Jordan curve in S2 and P ⊂ Ja set consisting of n ≥ 3 distinct points p1, . . . , pn, pn+1 = p1 in cyclic

11.2. ISOTOPIES OF JORDAN CURVES 229

order on J . For i = 1, . . . , n let αi be the unique arc on J with endpointspi and pi+1 such that int(αi) ⊂ J \P . Then there exists δ > 0 with thefollowing property:

Let K be another Jordan curve in S2 passing through the pointsp1, . . . , pn in cyclic order, and let βi for i = 1, . . . , n be the arc withendpoints pi and pi+1 such that int(βi) ⊂ J \ P . If

βi ⊂ N δ(αi)

for all i = 1, . . . , n, then there exists an isotopy Ht on S2 rel. P suchthat H0 = idS2 and H1(J) = K.

In other words, if the arcs βi of the Jordan curve K are containedin sufficiently small neighborhoods of the corresponding arcs αi of J ,then one can deform J into K by an isotopy of S2 that keeps the pointsin P fixed. Even though this statement seems “obvious”, a completeproof is surprisingly difficult and involved.

As we will see, the proof of this statement easily follows from twolemmas in [Bu].

Lemma 11.8. Let Ω ⊂ S2 be a simply connected region, p, q ∈ Ωdistinct points, and α and β arcs in Ω with endpoints p and q. Thenα is isotopic to β rel. p, q ∪ S2 \ Ω.

So arcs in a simply connected region with the same endpoints canbe deformed into each other so that the endpoints and the complementof the region stay fixed. The lemma follows from [Bu, p. 413, A.6Thm. (ii)]).

Lemma 11.9. Suppose we have two Jordan curves J and K as inProposition 11.7 such that for each i = 1, . . . , n the arc αi is isotopicto βi rel. P .

Then J is isotopic to K rel. P .

This is essentially [Bu, p. 411, A.5 Thm.].

Proof of Proposition 11.7. For each arc αi there exists a sim-ply connected region Ωi that contains αi but does not contain anyelement of P different from the endpoints of αi. There exists δ > 0such that N δ(αi) ⊂ Ωi for all i = 1, . . . , n. Then by Lemma 11.8 everyarc βi in N δ(αi) with the same endpoints as αi can be isotoped to αirel. P . The proposition now follows from Lemma 11.9.

If #P ≤ 3 in Proposition 11.7, then J can always be isotoped to Krel. P .

Lemma 11.10. Suppose J and K are Jordan curves in S2 and P ⊂J ∩K is a set with #P ≤ 3. Then J is isotopic to K rel. P .

230 11. ISOTOPIES

Proof. Suppose first that P consists of exactly three distinct pointsp1, p2, p3. Define the arcs αi and βi as in Proposition 11.7. Then foreach i = 1, 2, 3 the arcs αi and βi have the same endpoints pi and pi+1

and are contained the simply connected region Ωi = S2 \ pi+2, whereindices are understood modulo 3. Hence by Lemma 11.8 each arc αi isisotopic to βi rel. P . Again Lemma 11.9 implies that J is isotopic toK rel. P .

If #P ≤ 2, we may assume that S2 = C. Then by applying thefirst part of the proof (by adding auxiliary points to P ) one sees that

both J and K are isotopic to circles on C rel. P . Hence J is isotopicto K rel. P .

Lemma 11.11. Let S2 and S2 be oriented 2-spheres, and P ⊂ S2 be

a set with #P ≤ 3. If α : S2 → S2 and β : S2 → S2 are orientation-preserving homeomorphisms with α|P = β|P , then α and β are isotopicrel. P .

Proof. The statement is essentially well-known. For the sake ofcompleteness we will give a proof, but will leave some of the detailsto the reader. These details can easily be filled in by using the factsabout isotopies that will be discussed later before Proposition 5.24. Inthe proof of the uniqueness part of this proposition, we will use verysimilar arguments.

By considering αβ−1 one can reduce the lemma to the case where

S2 = S2 and β = idS2 . Then α fixes the points in P , and we have toshow that α is isotopic to idS2 rel. P . We first assume that #P = 3.

Pick a Jordan curve K ⊂ S2 with P ⊂ K, and let J = α(K).Then P ⊂ J ∩ K, and so by Lemma 11.10 the Jordan curve J canbe isotoped into K rel. P . This implies that α is isotopic rel. P to ahomeomorphism α1 on S2 with α1(K) = K. Let e be one of the threesubarcs of K determined by P . Since α1 fixes the three points in P ,this map restricts to a homeomorphism of e that does not move theendpoints of e. Hence on e the map α1 is isotopic to the identity on erel. ∂e. By pasting the isotopies on these arcs together, one can findan isotopy h : K × I → K rel. P such that h0 = α1|K and h1 = idK .One can extend h to each of the complementary components of Kto obtain an isotopy H : S2 × I → S2 rel. P such that H1 = idS2

and H(p, t) = h(p, t) for all p ∈ K and t ∈ I. Then α2 := H0 is ahomeomorphism on S2 that is isotopic to idS2 rel. P such that α1|K =α2|K. This implies that α1 and α2 are isotopic rel. K ⊃ P . If ∼indicates that two homeomorphisms on S2 are isotopic rel. P , then wehave α ∼ α1 ∼ α2 ∼ idS2 , and so α ∼ idS2 as desired.

11.2. ISOTOPIES OF JORDAN CURVES 231

If #P ≤ 2, then we pick a set P ′ ⊂ S2 with #P ′ = 3 and P ′ ⊃ P .By the first part of the proof it suffices to find an isotopy rel. P ofthe given map α to a homeomorphism α′ that fixes the points in P ′.It is clear that such an isotopy can always be found; for an explicit

construction one can assume that S2 = C and can then obtain thedesired isotopy by post-composing α by a suitable continuous familyof Mobius transformations, for example.

The following lemma will be crucial for the proof of the uniquenessstatement on invariant Jordan curves. In its proof we will use thefollowing topological fact: ifD is a two-dimensional cell and ϕ : D → S2

is a continuous map such that ϕ|∂D is injective, then the set ϕ(int(D))contains one of the two complementary components of the Jordan curveϕ(∂D). Indeed, by applying the Schonflies Theorem and using auxiliary

homomorphisms we can reduce to the case where D = D, S2 = C,ϕ|∂D = id∂D, and ∞ /∈ ϕ(D). Then D ⊂ ϕ(D). This follows from asimple degree argument and the statement can be generalized to higherdimensions; for an elementary exposition of this and related facts indimension 2 see [Bur], in particular [Bur, Cor. 3.5].

Lemma 11.12. Let D be a cell decomposition of S2 with 1-skeletonE and vertex set V, and suppose that every tile in D contains at leastthree vertices in its boundary. If J and K are Jordan curves that areboth contained in E and are isotopic rel. V, then J = K.

Proof. Let H : S2 × I → S2 be an isotopy rel. V such that H0 =idS2 and H1(J) = K.

Note that if M ⊂ S2 is a set disjoint from V, then it remains disjointfrom V during the isotopy, i.e., if M ∩V = ∅, then Ht(M)∩V = ∅ forall t ∈ I. This follows from the fact that each map Ht, t ∈ [0, 1], is ahomeomorphism on S2 with Ht|V = idV.

Let e be an edge in D. We claim that if H1(e) ⊂ E, then H1(e) = e.First note that H1(e) is an edge in D. Indeed, since ∂e ⊂ V and theisotopy H does not move vertices, the arc H1(e) has the same endpointsas e. Moreover, int(e)∩V = ∅, and so H1(int(e))∩V = ∅ by what wehave just seen. So H1(int(e)) is a connected set in the 1-skeleton E ofD disjoint from the 0-skeleton V. By Lemma 5.5 there exists an edgee′ in D with H1(int(e)) ⊂ int(e′). Since the endpoints of H1(e) lie inV, this implies that e′ = H1(e).

To show that e′ = e we argue by contradiction and assume thate 6= e′. Then e and e′ have the same endpoints, but no other pointsin common. Hence α = e ∪ e′ is a Jordan curve that contains twovertices, namely the endpoints of e and e′, but no other vertices. Let

232 11. ISOTOPIES

Ω1 and Ω2 be the two open Jordan regions that form the complementarycomponents of α. Then both regions Ω1 and Ω2 contain vertices.

To see this note that the interior of every tile X is a connected setdisjoint from the 1-skeleton E, and hence also disjoint from α. Henceint(X) is contained in Ω1 or Ω2. Moreover, since the union of theinteriors of tiles is dense in S2, both regions Ω1 and Ω2 must containthe interior of at least one tile.

Now consider Ω1, for example, and pick a tile X with int(X) ⊂ Ω1.Then by our hypotheses the set X ⊂ Ω1 = Ω1 ∪ α contains at leastthree vertices. Since only two of them can lie on α, the set Ω1 mustcontain a vertex. Similarly, Ω2 must contain at last one vertex.

A contradiction can now be obtained from the fact that during theisotopy H the set int(e) remains disjoint from the set of vertices, buton the other hand it has to sweep out one of the domains Ω1 or Ω2 andhence it meets a vertex.

To make this rigorous, we apply the topological fact mentioned be-fore the statement of the lemma. Let D be the quotient of the productspace e× I obtained by identifying all points (u, t), t ∈ I, and by iden-tifying all points (v, t), t ∈ I, where u and v are the two endpoints ofe. Then D is a two-dimensional cell. Since the isotopy H does notmove the points u and v, the map (p, t) 7→ Ht(p) on e × I induces acontinuous map ϕ : D → S2. Moreover, ϕ|∂D is a homeomorphism of∂D onto α. Hence Ω1 or Ω2 is contained in the set

ϕ(int(D)) =⋃

t∈(0,1)

Ht(int(e)).

In particular, the set ϕ(int(D)) contains a vertex. This is a contradic-tion, because we know that no set Ht(int(e)), t ∈ I, meets V. ThusH1(e) = e as desired.

Having verified the statement about edges, it is now easy to seethat J = K. Indeed, J is a union of edges in D; to see this consider thecomponents of the set J \V. If γ is such component, then γ \ γ ⊂ V.Moreover, γ is contained in the 1-skeleton E, and does not meet the0-skeleton V. Again by Lemma 5.5 the set γ must be contained in theinterior int(e) of some edge e. This is only possible if γ = int(e). Henceγ = e. Since J is the union of the closures of these components γ, itfollows that J is the union of edges e. For each such edge e we haveH1(e) ⊂ K ⊂ E and so H1(e) = e by the first part of the proof. Thisimplies J ⊂ K. Since J and K are Jordan curves, the desired identityJ = K follows.

11.3. ISOTOPIES AND CELL DECOMPOSITIONS 233

11.3. Isotopies and cell decompositions

The main result in this section is Lemma 11.17 which gives a criterionwhen a Jordan curve C in a 2-sphere S2 can be isotoped relative to afinite set P ⊂ C into the 1-skeleton of a given cell decomposition D ofS2. We first discuss some facts about graphs that are needed in theproof. Since all the graphs we consider will be embedded in a 2-sphere,we base the concept of a graph on a topological definition rather thana combinatorial one as usual.

A (finite) graph is a compact Hausdorff space G equipped with afixed cell decomposition D such that dim(c) ≤ 1 for all c ∈ D. Thecells c in D of dimension 1 are called the edges of the graph, and thepoints v ∈ G such that v is a 0-dimensional cell in D the verticesof the graph. Note that we do allow multiple edges, i.e., two or moreedges with the same endpoints v, w. Loops however, meaning edgeswhere the two endpoints agree, are not allowed.

An oriented edge e in a graph is an edge, where one of the verticesin ∂e has been chosen as the initial point and the other vertex as theterminal point of e. An edge path in G is a finite sequence α of orientededges e1, . . . , eN such that the terminal point of ei is the initial pointof ei+1 for i = 1, . . . , N − 1. We denote by |α| = e1 ∪ · · · ∪ eN theunderlying set of the edge path. The edge path α joins the verticesa, b ∈ G if the initial point of e1 is a and the terminal point of eN is b.The number N is called the length of the edge path. The edge path iscalled simple if ei and ej are disjoint for 1 ≤ i < j ≤ N and j − i ≥ 2,and ei ∩ ej consists of precisely one point (the terminal point of ei andinitial point of ej) when j = i + 1. If the edge path α is simple, then|α| is an arc. The edge path is called a loop if the terminal point of eNis the initial point of e1.

A graph is connected (as a topological space) if and only if anytwo vertices a, b ∈ G, a 6= b, can be joined by an edge path. Thecombinatorial distance of two vertices a and b in a connected graph Gis defined as the minimal length of all edge paths joining the points(interpreted as 0 if a = b). The vertices a, b ∈ G are called neighbors iftheir combinatorial distance is equal to 1, i.e., if there exists an edge ein G whose endpoints are a and b. A vertex q ∈ G is called a cut pointof G if G \ q is not connected. A vertex q ∈ G is not a cut point ifand only if all vertices a, b ∈ G \ q, a 6= b, can be joined by an edgepath α with q /∈ |α|.

Lemma 11.13. Let G be a connected graph without cut points. Thenfor all vertices a, b, p ∈ G with a 6= b there exists a simple edge path γin G with p ∈ |γ| that joins a and b.

234 11. ISOTOPIES

a

r

p

q

q′

b

α

σ

β

Figure 11.2. Constructing a path through a, b, p.

Proof. Since G is connected, there exist edge paths in G joininga and b. By removing loops from such a path if necessary, we can alsoobtain such an edge path in G that is simple. Among all such simplepaths, there is one that contains a vertex with minimal combinatorialdistance to p. More precisely, there exists a simple edge path α in Gwith endpoints a and b, and a vertex q ∈ |α| such that the combinatorialdistance k ∈ N0 of q and p is minimal among all combinatorial distancesbetween p and vertices on simple paths joining a and b. If k = 0 thenq = p and we can take γ = α.

We will show that the alternative case k ≥ 1 leads to a contradic-tion. By definition of combinatorial distance, there exists an edge pathjoining q to p consisting of k ≥ 1 edges. The second vertex q′ on thispath as travelling from q to p is a neighbor of q whose combinatorialdistance to p is k−1 and hence strictly smaller than the combinatorialdistance of q to p. In particular, q′ 6∈ |α| by choice of q and α. Wewill obtain the desired contradiction if we can show that there exists asimple edge path σ in G that joins a and b and passes through q′.

For the construction of σ we apply our assumption that G has nocut points; in particular, q is no cut point and hence there exists anedge path β with q /∈ |β| that joins q′ to a vertex in the (nonempty) setA = |α| \ q. We may assume that β is simple and that the endpointr 6= q′ of β is the only point in |β| ∩ A.

Moreover, we may assume that r lies between a and q on the path α(the argument in the other case where r lies between q and b is similar).Now let σ be the edge path obtained by traveling from a to r alongα, then from r to q′ along β, then from q′ to q along an edge (this is


possible since q and q′ are neighbors), and finally from q to b along α.See the illustration in Figure 11.2. Then σ is a simple edge path in Gthat passes through q′ and has the endpoints a and b. This gives thedesired contradiction.

Now let S2 be a 2-sphere, and D be a cell decomposition of S2.We denote the set of tiles, edges, and vertices in D by X, E, and V,respectively. In the following the term cell, tile, etc., refers to elementsin these sets.

Let M ⊂ X be a set of tiles. We denote by |M | its underlying set;so

|M | =⋃X∈M

X.

The set

(11.6) GM :=⋃X∈M

∂X

admits a natural cell decomposition consisting of all cells contained inGM . Obviously, no such cell can be a tile, so with this cell decomposi-tion GM is a graph.

Similarly as in Section 12.2, we call a sequence of tilesX = X1, . . . , XN =Y an e-chain joining X and Y if for i = 1, . . . , N there exists an edgeei with ei ⊂ ∂Xi−1 ∩ ∂Xi. A set M of tiles is e-connected if every twotiles in M can be joined by an e-chain.

Lemma 11.14. Let M ⊂ X be a set of tiles that is e-connected.Then the graph GM is connected and has no cut points.

Proof. Let a, b ∈ GM be arbitrary vertices with a 6= b. We canpick tiles X and Y in M such that a is a vertex in X and b is a vertexin Y . By assumption there exists an e-chain X1, X1, . . . , XN in Mwith X1 = X and XN = Y . The vertices of a tile Xi lie in GM ; theysubdivide the Jordan curve ∂Xi such that successive vertices on ∂Xi

are connected by an edge and are hence neighbors in GM . An edgepath α in GM joining a and b can now be obtained as follows: startingfrom a ∈ ∂X0, use edges on the boundary of X0 to find an edge pathin GM that joins p0 = a to a vertex p1 of X1. This is possible, since X0

and X1 have a common edge and hence at least two common vertices.Then run from p1 along edges on ∂X1 to a vertex p2 of X2, and so on.Once we arrived at a vertex pk of Xk, we can reach b by running frompk to pk+1 = b along edges on Xk. In this way we obtain an edge pathα in GM that joins a and b.

A slight refinement of this argument also shows that we can con-struct the path α so that it avoids any given vertex q in GM distinct

236 11. ISOTOPIES

from a and b. Indeed, choose p0 = a as before. Since X0 and X1 haveat least two vertices in common, we can pick a common vertex p1 ofX0 and X1 that is distinct from q. There exists an arc on ∂X0 (possi-bly degenerate) that does not contain q and joins p0 and p1. This arc(if non-degenerate) consists of edges and if we follow these edges, weobtain an edge path in GM that does not contain q and joins p0 andp1. In the same way we can find an edge path in GM that avoids qand joins p1 to a vertex p2 ∈ ∂X1 ∩ ∂X2, and so on. Concatenating allthese edge paths we get a path α as desired.

This shows that GM is connected and has no cut points.

Lemma 11.15. Let M ⊂ X be a set of tiles that is e-connected, andlet a, b, p ∈ |M | be distinct vertices. Then there exists a simple edgepath α in GM with p ∈ |α| that joins a and b.

In particular, this applies if M consists of a single e-chain.

Proof. This follows from Lemma 11.14 and Lemma 11.13.

Lemma 11.16. Let γ : J → S2 be a path in S2 defined on a closedinterval J ⊂ R and M = M(γ) be the set of tiles having nonemptyintersection with γ. Then M is e-connected.

Proof. We first prove the following claim. If [a, b] ⊂ R, α : [a, b]→S2 is a path, and X and Y are tiles with α(a) ∈ X and α(b) ∈ Y , thenthere exists an e-chain X1 = X,X2, . . . , XN = Y such that Xi ∩ α 6= ∅for all i = 1, . . . , N .

In the proof of this claim, we call an e-chain X1, . . . , XN admissibleif X1 = X and Xi ∩ α 6= ∅ for all i = 1, . . . , N . So we want to find anadmissible e-chain whose last tile is Y .

Let T ⊂ [a, b] be the set of all points t ∈ [a, b] for which there existsan admissible e-chain X1, . . . , XN with α(t) ∈ XN . We first want toshow that b ∈ T .

First note that the set T is closed. Indeed, suppose that (tk) isa sequence in T with tk → t∞ ∈ [a, b] as k → ∞. Then for eachk ∈ N there exists an admissible e-chain X1

1 , . . . , XkNk

with α(tk) ∈ XkNk

.

Define Zk = XkNk

to be the last tile in this chain. Since there areonly finitely many tiles, there exists one tile, say Z, among the tilesZ1, Z2, Z3, . . . that appears infinitely often in this sequence. Sinceα(tk) → α(t∞), we have α(tk) ∈ Z for infinitely many k, and tilesare closed, we conclude α(t∞) ∈ Z. By definition of Z there exists anadmissible e-chain X1, . . . , XN with XN = Z. Then α(t∞) ∈ Z = XN ,and so t∞ ∈ T .


Obviously, a ∈ T and so T is nonempty. Since T is also closed, theset T has a maximum, say m ∈ [a, b]. We have to show that m = b; wewill see that the assumption m < b leads to a contradiction.

Consider p = α(m). Then there exists an admissible e-chainX1, . . . , XN

with p ∈ Z := XN .If p ∈ int(Z), then α(t) ∈ Z and so t ∈ T for t ∈ (m, b] close to m.

This is impossible by definition of m.If p does not belong to int(Z), then p must be a boundary point

of Z. Suppose first that p is in the interior of an edge e ⊂ ∂Z. ByLemma 5.12 (iv) there exists precisely one tile Z ′ distinct from Z suchthat e ⊂ ∂Z ′. Moreover, Z ∪ Z ′ is a neighborhood of p, and so pointsα(t) with t ∈ (m, b] close to m belong to Z or Z ′. Since Z ′ containsp and hence meets α, and Z and Z ′ share an edge, X1, . . . , XN , Z

′ isan admissible e-chain. It follows that t ∈ T for t ∈ (m, b] close to m.Again this is impossible by definition of m.

If p is a boundary point of Z, but not in the interior of an edge,then p is a vertex. The tiles in the cycle of p form a neighborhood of p,and so a point α(t) for some t ∈ (m, b] close to m will belong to a tile Z ′

in the cycle of p. It follows from Lemma 5.12 that any two tiles in thecycle of a vertex can be connected by an e-chain consisting of tiles inthe cycle. Hence there exits an e-chain Z = Z1, . . . , ZK = Z ′ such thatp ∈ Zj for j = 1, . . . , K. In particular, α ∩ Zj 6= ∅ for j = 1, . . . , K,and so X1, . . . , XN = Z = Z1, . . . , ZK = Y ′ is an admissible e-chain.Since α(t) ∈ Z ′ = ZK , we have t ∈ T , again a contradiction.

We have exhausted all possibilities proving that b ∈ T as desired.So there exists an admissible e-chain X1 = X, . . . , XN with α(b) ∈ XN .If XN = Y , then we are done. If XN 6= Y , then α(b) ∈ ∂XN ∩ ∂Y ,and so α(b) is an interior point of an edge e with e ⊂ ∂XN ∩ ∂Y , orα(b) is a vertex. As in the first part of the proof, one then can extendthe admissible e-chain X1 = X, . . . , XN to obtain an admissible e-chainwhose last tile is Y . The claim follows.

The claim now easily implies the statement of the lemma. Indeed,pick a point a ∈ J and fix a tile X ∈ X with γ(a) ∈ X. ThenX ∈ M = M(γ). If Y ∈ M is arbitrary, then there exists b ∈ J withγ(b) ∈ Y . If b ≥ a, then we apply the claim to the path α = γ|[a, b],and if b ≤ a to the path α = γ|[b, a]. This shows that we can find ane-chain in M that joins Y ∈M to the fixed tile X. Hence any two tilesin M can be joined by an e-chain in M .

For the formulation of the next statement, we need a slight gener-alization of Definition 5.30. Let C ⊂ S2 be a Jordan curve, and P ⊂ Cbe a finite set with #P ≥ 3. The points in P divide C into subarcs that

238 11. ISOTOPIES

have endpoints in P , but whose interiors are disjoint from P . We saythat a set K ⊂ S2 joins opposite sides of (C, P ) if k = # post(f) ≥ 4and K meets two of these arcs that are non-adjacent (i.e., disjoint), orif k = # post(f) = 3 and K meets all of these arcs (in this case thereare three arcs).

In the following proposition and its proof, metric notions refer tosome fixed base metric on S2.

Lemma 11.17. Let C ⊂ S2 be a Jordan curve, and P ⊂ C a finiteset with k = #P ≥ 3. Then there exists ε0 > 0 with the followingproperty:

Suppose that D is a cell decomposition of S2 with vertex set V and1-skeleton E. If P ⊂ V and

maxc∈D

diam(c) < ε0,

then there exists a Jordan curve C ′ ⊂ E that is isotopic to C rel. P andso that no tile in D joins opposite sides of (C ′, P ).

Proof. Fix an orientation of C and let p1, . . . , pk be the points inP in cyclic order on C. The points in P divide C into subarcs C1, . . . , Cksuch that for i = 1, . . . , k the arc Ci has the endpoints pi and pi+1 andhas interior disjoint from P . Here and in the following the index i isunderstood modulo k, i.e., pk+1 = p1, etc. Note that Ci ∩Ci+1 = pi+1for i = 1, . . . , k. There exists a number δ0 > 0 such that no set K ⊂ S2

with diam(K) < δ0 joins opposite sides of (C, P ) (this can be seen asin the discussion after (5.13)).

Now choose δ > 0 as in Proposition 11.7 for J = C and n = k. Wemay assume that 3δ < δ0.

We break up C into subarcs

(11.7) α1, γ1, α2, γ2, . . . , γk, α1,

arranged in cyclic order on C, such that pi is an interior point of αi andwe have αi ⊂ B(pi, δ/2) for each i = 1, . . . , k. The arcs in (11.7) havedisjoint interiors, and two arcs have an endpoint in common if and onlyif there are adjacent in this cyclic order in which case they share oneendpoint. So each “middle piece” γi does not contain any point fromP and is contained in the interior of Ci.

We choose 0 < ε0 < δ/4 so small that the distance between non-adjacent arcs in (11.7) is ≥ 10ε0 and so that

dist(pi, γi−1 ∪ γi) ≥ 10ε0

for i = 1, . . . , k.


B(pi,δ/2) B(pi+1,δ/2)

pi pi+1

Ci

viv′i

v′i−1vi+1

Ai Ai+1ai ai+1

ci

Ci

Figure 11.3. Construction of the curve C ′.

Now suppose we have a cell decomposition D of S2 such that P iscontained in the vertex set V of D and

maxc∈D

diam(c) < ε0.

Our goal is to find a Jordan curve C ′ ⊂ S2 consisting of arcs C ′i thatare unions of edges, have endpoints pi and pi+1, and satisfy

C ′i ⊂ N δ(Ci)for i = 1, . . . , k.

Let Ai be the set of all tiles intersecting αi and Ci be the set of alltiles intersecting γi, for all i = 1, . . . , k. Recall that for a given set oftiles M , we denote by |M | the union of tiles in M . Let Ai := |Ai| andCi = |Ci|. Note that

Ai ⊂ N ε0(αi) and Ci ⊂ N ε0(γi).

Furthermore

Ai ∪ Ci ∪ Ai+1 ⊂ N ε0(αi) ∪N ε0(γi) ∪N ε0(αi+1)

⊂ B(pi, δ) ∪N ε0(γi) ∪B(pi+1, δ)(11.8)

⊂ N δ(Ci),and the natural cyclic order of these sets is

(11.9) A1, C1, A2, C2, . . . , Ak, Ck, A1.

By choice of ε0 we have that if two of the sets in (11.9) are not adjacentin the cyclic order, then their distance is ≥ 8ε0 and so their intersec-tion is empty. Moreover, for i = 1, . . . , k the only one of these setsthat contains pi is Ai. The construction that follows is illustrated inFigure 11.3. Here the two large dots represent two points pi, pi+1 andthe thick line the curve C.

Consider the graphs GAi, GCi

of all edges in Ai, Ci, see (11.6). Notethat there is at least one tile contained in both Ai and Ci, namely any

240 11. ISOTOPIES

tile containing the point where αi, γi intersect. Thus GAiand GCi

,similarly GCi

and GAi+1, are not disjoint.

Pick a simple edge path c′i in GCiconnecting Ai, Ai+1. Let ci be

a subarc of |c′i| whose initial point vi lies in Ai, whose terminal pointv′i lies in Ai+1, and that has no other points in common with Ai andAi+1. The points vi and v′i are vertices. To see this, suppose that vi,for example, is not a vertex. Then, since vi ∈ c′i and c′i is an edge path,there exists an edge e ⊂ ci such that vi ∈ int(e). But then necessarilye ⊂ Ai and vi cannot be the only point of ci that belongs to Ai.

Note that v′i−1 ∈ Ci−1 and vi ∈ Ci are distinct vertices in Ai. Recallthat pi ∈ Ai. Thus it follows from Lemma 11.16 and Lemma 11.15 thatthere exists an arc ai ⊂ Ai with pi ∈ ai that consists of edges and hasthe endpoints v′i−1 and vi. Since pi /∈ Ci−1 ∪ Ci, we have v′i−1, vi 6= pi,and so pi ∈ int(ai).

If we arrange the arcs ai and ci in cyclic order

a1, c1, a2, c2 . . . , ak, ck, a1,

then two of these arcs have nonempty intersection if and only if they areadjacent in the order. If two arcs are adjacent, then their intersectionconsists of a common endpoint. Therefore, the set

C ′ := a1 ∪ c1 ∪ a2 ∪ c2 ∪ · · · ∪ ak ∪ ckis a Jordan curve that passes through the points p1, . . . , pk. Moreover,C ′ consists of edges and is hence contained in the 1-skeleton E of D.

By construction each vertex pi is an interior point of the arc ai.Thus it divides ai into two subarcs a−i and a+

i consisting of edges suchthat pi is a common endpoint of a−i and a+

i , and such that a−i sharesan endpoint with ci−1 and a+

i one with ci. Then

C ′i := a+i ∪ ci ∪ a−i+1

for i = 1, . . . , k is an arc that consists of edges and has endpoints piand pi+1. The arcs C ′1, . . . , C ′k have pairwise disjoint interior. Moreover,

C ′ = C ′1 ∪ · · · ∪ C ′k,

and by (11.8) we have

C ′i ⊂ Ai ∪ Ci ∪ Ai+1 ⊂ N δ(Ci).

Hence by Proposition 11.7 and choice of δ, the curve C ′ is isotopic toC rel. P .

It remains to show that no tile in D joins opposite sides of (C ′, P ).To see this, we argue by contradiction. Suppose that there exists atile X in D that joins opposite sides of (C ′, P ). Then K := N δ(X)


joins opposite sides of (C, P ), since C ′i ⊂ N δ(Ci) for all i = 1, . . . , k. Bychoice of δ0 we then have

δ0 ≤ diam(K) ≤ 2δ + diam(X) ≤ 2δ + ε0 < 3δ < δ0,

which is impossible.

CHAPTER 12

Subdivisions

In complex dynamics the iteration of polynomials is much better under-stood than the iteration of general rational maps. One of the reasonsis that for polynomials powerful combinatorial methods are availablesuch as external rays, Hubbard trees, or Yoccoz puzzles (see [DH84]).It is desirable to develop similar concepts for other classes of maps aswell. For Thurston maps we will introduce the notion of a two-tilesubdivision rule in this chapter. It provides a useful combinatorial toolfor their investigation.

This concept can be extracted from various previous examples (seeSections 1.1 and 1.3, or Examples 2.6 and 6.10), where we have de-scribed Thurston maps by a subdivision procedure. In these exampleswe consider a topological 2-sphere obtained as a pillow by gluing two k-gons together along their boundaries. Then the two faces of the pillow(the 0-tiles) are subdivided into k-gons (the 1-tiles) and it is specifiedhow the map sends a 1-tile to one of the 0-tiles. The common boundaryof the faces of the pillow is a Jordan curve that is invariant under themap and contains its postcritical points.

More generally, suppose f : S2 → S2 is a Thurston map with k :=post(f) ≥ 3, and C ⊂ S2 is a Jordan curve with post(f) ⊂ C. If C is f -invariant (i.e., f(C) ⊂ C), then the cell decompositions Dn = Dn(f, C)introduced in Section 5.3 have nice compatibility properties (see Propo-sition 12.5). In particular, Dm is a refinement of Dn, whenever m ≥ n.On a more intuitive level this means that each cell in Dn is subdividedby the cells in Dm of higher order. It follows from Proposition 5.10(see also Remark 5.11) that the whole sequence Dn is generated by thepair (D1,D0) and the map τ ∈ D1 → f(τ) ∈ D0, i.e., the labelingL : D1 → D0 induced by f (see Section 5.4). The triples (D1,D0, L)arising in this way lead to the following definition (see the beginningof Section 12.2 for more motivation).

Definition 12.1 (Two-tile subdivision rules). Let S2 be a 2-sphere.A two-tile subdivision rule for S2 is a triple (D1,D0, L) of cell de-compositions D0 and D1 of S2 and an orientation-preserving labeling

243

244 12. SUBDIVISIONS

L : D1 → D0. We assume that the cell decompositions satisfy the fol-lowing conditions:

(i) D0 contains precisely two tiles.

(ii) D1 is a refinement of D0, and D1 contains more than two tiles.

(iii) If k is the number of vertices in D0, then k ≥ 3 and every tilein D1 is a k-gon.

(iv) Every vertex in D1 is contained in an even number of tiles inD1.

If D0 is a cell decomposition of S2 with precisely two tiles X andY , then necessarily ∂X = ∂Y . The set C := ∂X = ∂Y is a Jordancurve which we call the Jordan curve of D0. Then C is the 1-skeletonof D0 and all vertices and edges of D0 lie on C. If k is the number ofthese vertices on C and D1 is another cell decomposition of S2, thena Thurston map that is cellular for (D1,D0) can only exist if each tilein D1 is a k-gon, i.e., it contains exactly k vertices and edges in itsboundary. Since f is not a homeomorphism, D1 contains more thantwo tiles. The number of tiles in D1 that contain a given vertex v inD1 is equal to the length of the cycle of v in D1. This number hasto be even, because it must be an integer multiple of the length ofa vertex cycle in D0 which is always equal to 2. This motivated therequirements (ii)–(iv) in Definition 12.1.

We say that a continuous map f : S2 → S2 realizes the subdivisionrule (D1,D0, L) if f is cellular for (D1,D0) and f(τ) = L(τ) for eachτ ∈ D1.

The following proposition shows that two-tile subdivision rules arisefrom Thurston maps with invariant curves.

Proposition 12.2. Suppose f : S2 → S2 is a Thurston map withpost(f) ≥ 3, and C ⊂ S2 is an f -invariant Jordan curve with post(f) ⊂C. If we define D0 = D0(f, C), D1 = D1(f, C), and L : D1 → D0 by set-ting L(τ) = f(τ) for τ ∈ D1, then (D1,D0, L) is a two-tile subdivisionrule realized by f .

Theorem 14.1 implies that every expanding Thurston map f with# post(f) ≥ 3 has an iterate F = fn that realizes a two-tile subdivisionrule.

Conversely, a two-tile subdivision rule gives rise to a Thurston mapwith an invariant curve and this map is unique up to Thurston equiv-alence.

12. SUBDIVISIONS 245

Proposition 12.3. Let (D1,D0, L) be a two-tile subdivision ruleon S2. Then there exists a Thurston map f : S2 → S2 that realizes(D1,D0, L). The map f is unique up to Thurston equivalence.

Moreover, the Jordan curve C of D0 is f -invariant and contains theset post(f).

Our concept of a two-tile subdivision rule is a special case of themore general concept of a subdivision rule (see [BS, CFP01, CFP06a,CFKP, Me02] for work related to this). The reason for the name two-tile subdivision rule is that the data given by (D1,D0) determines howthe two 0-tiles are subdivided by the cells in D1, and this together withthe labeling L can be used to create a sequence of cell decompositionDnwhere each cell τ ∈ D1 is subdivided by the cells in D2 in the same wayas the cell L(τ) ∈ D0 is subdivided by the 1-cells, etc. Our definitionis tailored to generate Thurston maps, so a more accurate term wouldhave been a “two-tile subdivision rule generating a Thurston map”,but we chose the shorter term for brevity.

Since we are mostly interested in expanding Thurston maps, wewant to find a combinatorial condition on two-tile subdivision rulesthat ensures that it can be realized by an expanding Thurston map. Tomotivate a relevant condition, suppose that f : S2 → S2 is a Thurstonmap and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. We considerthe quantities Dn = Dn(f, C) defined in (5.14). If the Jordan curveC here is f -invariant, then it is easy to see that the numbers Dn arenon-decreasing as n → ∞. We will show that we actually have anexponential increase under the additional assumption that there existsn0 ∈ N with Dn0 ≥ 2 (see Lemma 12.7). This will turn out to bekey condition related to expansion of Thurston maps realizing two-tilesubdivision rules.

Definition 12.4 (Combinatorial expansion). Let f : S2 → S2 bea Thurston map. We call f combinatorially expanding if post(f) ≥ 3,and if there exists a Jordan curve C ⊂ S2 that is f -invariant, satis-fies post(f) ⊂ C, and for which there is a number n0 ∈ N such thatDn0(f, C) ≥ 2.

The condition Dn0(f, C) ≥ 2 means that no single n0-tile for (f, C)joins opposite sides of C.

If f and C are as in the previous definition, then we say that f iscombinatorially expanding for C. This condition is indeed combinato-rial in nature, because it can be verified just by knowing the combi-natorics of the cell decompositions Dn = Dn(f, C), n ∈ N0. This inturn is determined by the combinatorics of the pair (D1,D0) and thelabeling τ ∈ D1 7→ f(τ) ∈ D0 induced by f (see Remark 5.11).


If a Thurston map f : S2 → S2 is expanding and C ⊂ S2 is an f -invariant Jordan curve with post(f) ⊂ C, then f is also combinatoriallyexpanding for C, because in this case Dn(f, C) → ∞ as n → ∞ (seeLemma 8.5). The converse is not true in general, as a combinatoriallyexpanding Thurston map need not be expanding (see Example 12.18).However, in the next chapter we will see that each combinatoriallyexpanding Thurston map is equivalent to an expanding Thurston mapwith an invariant curve (Theorem 13.2).

Now let (D0,D1, L) be a two-tile subdivision rule and C be the Jor-dan curve of D0. We will show that if a Thurston map realizing thissubdivision rule is combinatorially expanding for C, then this is true forevery Thurston map realizing the subdivision rule (Lemma 12.15). Inthis case we say that the subdivision rule is combinatorially expanding(see Definition 12.16). We will later prove that under an additionalhypothesis a two-tile subdivision rule is combinatorially expanding ifand only if it can be realized by an expanding Thurston map (Theo-rem 13.1).

This chapter is organized as follows. In Section 12.1 we summarizefacts related to Thurston maps with invariant curves and the associ-ated cell decompositions. Among them is yet another characterizationof when f is expanding (Lemma 12.6) and a crucial supermultiplicativ-ity property of the quantities Dn(f, C) (Lemma 8.5). In Section 12.2 wediscuss two-tile subdivision rules and prove Propositions 12.2 and 12.3.For two-tile subdivision rules the information given by an orientation-preserving labeling L : D1 → D0 can be further compressed: for exam-ple, it is uniquely determined if one knows the image of one flag in D1

(see Lemma 12.13 which is based on Lemma 5.21). We will concludethe section with Lemma 12.15 about combinatorial expansion of mapsrealizing a subdivision rule.

Our results pave the way for a convenient construction of Thurstonmaps from a combinatorial perspective. Section 12.3 is devoted to this.We will exhibit several Thurston maps arising from two-tile subdivisionrules.

12.1. Thurston maps with invariant curves

In the following f : S2 → S2 is a Thurston map, C ⊂ S2 is a Jordancurve with post(f) ⊂ C, and Dn = Dn(f, C) for n ∈ N0 is the celldecomposition of S2 given by the n-cells for (f, C).

As usual, a set M ⊂ S2 is called f -invariant (or simply invariantif f is understood) if

(12.1) f(M) ⊂M or equivalently M ⊂ f−1(M).

12.1. THURSTON MAPS WITH INVARIANT CURVES 247

We will mostly be interested in the case when M = C is a Jordancurve with post(f) ⊂ C. The reason for this is that then the celldecomposition Dn(f, C) induced by such an f -invariant Jordan curveC is refined by each cell decomposition Dm(f, C) of higher level m ≥ n(see Proposition 12.5 below).

Since the set post(f) is f -invariant, we have

(12.2) post(f) ⊂ f−1(post(f)) ⊂ f−2(post(f)) ⊂ . . .

We know by Proposition 5.17 (iii) that Vn = Vn(f, C) = f−n(post(f))for n ∈ N0, and so (12.2) is equivalent to the inclusions

V0 ⊂ V1 ⊂ V2 ⊂ . . .

for the vertex sets of the cell decompositions Dn.In general, a similar inclusion chain will not hold for the 1-skeleta

En := f−n(C) of Dn, but if C is f -invariant, then it follows by inductionthat

C = E0 ⊂ E1 ⊂ E2 ⊂ . . .

Actually, more is true as the following statement shows.

Proposition 12.5. Let k, n ∈ N0, f : S2 → S2 be a Thurston map,and C ⊂ S2 be an f -invariant Jordan curve with post(f) ⊂ C. Thenwe have:

(i) (Dn+k,Dk) is a cellular Markov partition for fn.

(ii) Every (n+ k)-tile Xn+k is contained in a unique k-tile Xk.

(iii) Every k-tile Xk is equal to the union of all (n+ k)-tiles Xn+k

with Xn+k ⊂ Xk.

(iv) Every k-edge ek is equal to the union of all (n+ k)-edges en+k

with en+k ⊂ ek.

Proof. (i) We know that the map fn is cellular for (Dn+k,Dn)(Proposition 5.17 (i)); so we have to show that Dn+k is a refinementof Dn (see Definition 5.6). By the invariance of C we have En+k =f−(n+k)(C) ⊃ Ek = f−k(C), and so S2 \ En+k ⊂ S2 \ Ek.

To establish the first property of a refinement, we will show thatevery (n+ k)-cell is contained in some k-tile.

Let σ be an arbitrary (n+ k)-cell. If σ is a (n+ k)-tile, then int(σ)is a connected set in S2 \ En+k ⊂ S2 \ Ek and hence contained inthe interior of a k-tile τ (see Proposition 5.17 (v)). It follows that

σ = int(σ) ⊂ τ .If σ is an (n+ k)-edge or an (n+ k)-vertex, then it is contained in

an (n+ k)-tile (Lemma 5.12 (iv) and (v)), and hence in some k-tile bywhat we have just seen.


To establish the second property of a refinement, let τ be an arbi-trary k-cell. We have to show that the (n + k)-cells σ contained in τcover τ .

If τ consists of a k-vertex p, then p is also an (n + k)-vertex, andthe statement is trivial.

If τ is a k-edge, consider the points in Vn+k that lie on τ . Notethat this includes the elements of ∂τ ⊂ Vk ⊂ Vn+k. By using thesepoints to partition τ , we can find finitely many arcs α1, . . . , αN suchthat τ = α1 ∪ · · · ∪ αN , each arc αi has endpoints in Vn+k ⊃ Vk andhas interior int(αi) disjoint from Vn+k. Then for each i = 1, . . . , Nthe set int(αi) is a connected set in Ek \ Vn+k ⊂ En+k \ Vn+k. Itfollows that int(αi) and hence also αi is contained in some (n+k)-edgeσi (Proposition 5.17 (v)). Since the endpoints of αi lie in Vn+k, theycannot lie in int(σi), and so they are also endpoints of σi. This impliesthat αi = σi. In particular, the (n+ k)-edges σ1, . . . , σN are containedin τ and form a cover of τ . The statement follows in this case.

Finally, let τ be a k-tile. If p ∈ int(τ) is arbitrary, then p is con-tained in an (n+k)-tile σ. By the first part of the proof, σ is containedin a k-cell. Since τ is the only k-cell that contains p, we must haveσ ⊂ τ . This implies that the union of the (n + k)-tiles contained in τcover int(τ). On the other hand, this union consists of finitely manytiles and is hence a closed set. It follows that the union also containsint(τ) = τ .

(ii) We have just seen that every (n + k)-tile Xn+k is contained in

a k-tile Xk. This tile is unique. For suppose Xk is another k-tile with

Xn+k ⊂ Xk. Then

∅ 6= int(Xn+k) ⊂ int(Xk) ∩ int(Xk),

and soXk and Xk have common interior points. This impliesXk = Xk.

(iii)–(iv): Both statements were established in the proof of (i).

If f and C are as in the previous proposition, then by statement (i)the pair (D1,D0) is a cellular Markov partition for f , and this partitiongenerates the cell decompositions Dn as in Proposition 5.10. If Xn isany n-tile, then by (ii) there exist unique i-tiles X i for i = 0, . . . , n− 1such that

Xn ⊂ Xn−1 ⊂ . . . ⊂ X0.

We refer to the statements (iii) and (iv) informally by saying thattiles and edges are “subdivided” by tiles and edges of higher order.


Let S = S(f, C) denote the set of all sequences Xn, where Xn isan n-tile for n ∈ N0 and

X0 ⊃ X1 ⊃ X2 ⊃ . . .

Since tiles are subdivided by tiles of higher order, for each point p ∈ S2

we can find a sequence Xn ∈ S such that p ∈⋂nX

n. Here itis understood that the intersection is taken over all n ∈ N0. In thefollowing we use a similar convention for intersections of sets labeledby some index n, k, etc., if the range of the indices is clear from thecontext.

In general, a sequence Xn ∈ S that contains a given point p ∈ S2

is not unique. Moreover, the intersection⋂nX

n may contain more thanone point. It turns out that this gives a criterion when f is expanding.

Lemma 12.6. Let f : S2 → S2 be a Thurston map, and C ⊂ S2 bean f -invariant Jordan curve with post(f) ⊂ C. Then the map f is ex-panding if and only if for each sequence Xn ∈ S(f, C) the intersection⋂nX

n consists of precisely one point.

Proof. Fix a metric on S2 that induces the standard topology onS2. If f is expanding and Xn ∈ S = S(f, C), then diam(Xn) → 0as n→∞. Hence

⋂nX

n cannot contain more than one point. On theother hand, this set is an intersection of a nested sequence of nonemptycompact sets and hence nonempty. So the set

⋂nX

n contains preciselyone point.

For the converse direction suppose that⋂nX

n is a singleton set foreach sequence Xn ∈ S. To establish that f is expanding we have toshow that

limn→∞

max diam(X) : X is an n-tile = 0.

We argue by contradiction and assume that this is not the case. Thenthere exists δ > 0 such that diam(X) ≥ δ for some tiles X of arbitrarilyhigh order.

We define a descending sequence of tiles X0 ⊃ X1 ⊃ X2 ⊃ . . . asfollows. Let X0 be a 0-tile such that X0 contains tiles X of arbitrarilyhigh order with diam(X) ≥ δ. Since the tiles are subdivided by tiles ofhigher order, and so every tile is contained in one of the finitely many0-tiles (in our case there actually two 0-tiles), there exists such a 0-tile.Note that then diam(X0) ≥ δ. Moreover, among the finitely many 1-tiles into which X0 is subdivided there must be a 1-tile X1 ⊂ X0 suchthat X1 contains tiles X of arbitrarily high order with diam(X) ≥ δ.Again this implies that diam(X1) ≥ δ. Repeating this procedure weobtain a sequence Xn ∈ S such that diam(Xn) ≥ δ for all n ∈ N0.It is easy to see that this implies that the set

⋂nX

n also has diameter


≥ δ > 0, and so it contains at least two points. This is a contradictionshowing that f is expanding.

The main idea in the previous proof is essentially Konig’s infinitylemma (see for example [Di, Lemma 9.1.3]) from graph theory; it saysthat a simplicial tree with arbitrarily long branches has an infinitebranch.

Recall the definition of the numbers Dn = Dn(f, C) in (5.14). Weknow that Dn → ∞ if f is expanding (see Lemma 8.5). If f is notnecessarily expanding, but the Jordan curve C used in the definitionof Dn is invariant, then it follows from the previous discussion thatthe numbers Dn are non-decreasing, i.e., Dn+1 ≥ Dn for all n ∈ N0.The following lemma establishes a much deeper supermultiplicativityproperty for these quantities. It implies that the numbers Dn increaseexponentially fast under the additional assumption that there existsn0 ∈ N with Dn0 ≥ 2 (see Lemma 12.8).

Lemma 12.7. Let f : S2 → S2 be a Thurston map with # post(f) ≥3, C ⊂ S2 be a Jordan curve with post(f) ⊂ C, and Dn = Dn(f, C) forn ∈ N0. Suppose that C is f -invariant. Then for all n, k ∈ N0 we have

Dn+k ≥ DnDk

if # post(f) ≥ 4, and

(12.3) Dn+k ≥ Dn(Dk − 1) + 1

if # post(f) = 3.

Before we prove this lemma let us fix some terminology. An n-chain is a finite sequence of n-tiles X1, . . . , XN , where Xi ∩ Xi+1 6= ∅for i = 1, . . . , N−1. It joins two disjoint sets A and B if A∩X1 6= ∅ andB ∩XN 6= ∅. The chain joins two points x and y if it joins the sets xand y. The n-chain is called a simple chain joining A and B if thereis no proper subsequence of X1, . . . , XN that is also a chain joining Aand B. If we put X−1 := A and XN+1 := B, then this is equivalent tothe requirement that Xi ∩Xj = ∅ whenever −1 ≤ i < j ≤ N + 1 andj − i ≥ 2.

Proof of Lemma 12.7. Case 1. # post(f) ≥ 4.Let X1, . . . , XN be a set of (n+k)-tiles whose union is connected and

joins opposite sides of C. We may assume that these tiles form a chain

joining disjoint 0-edges E and E. To prove the desired inequality, wewill break this chain into M subchains Xsi , . . . , Xsi+1−1, where M ∈ N,i = 1, . . . ,M , and s1 = 1 < s2 < · · · < sM+1 = N + 1. The lengthof each subchain (i.e., the number si+1 − si) will be at least Dn. The


Ee = c1

X1

Ee = cM+1

W k(c1)

c2 Xs2−1

Xs2

Y1

XN

Figure 12.1. The proof of Lemma 12.7.

number M of subchains will be at least Dk. Thus N ≥ DnDk, andsince the minimum over all N is equal to Dn+k, the desired inequalityfollows.

To guarantee the desired lower bound on the length, we will ensurethat each subchain Xsi , . . . , Xsi+1−1 joins disjoint k-cells. Then thelength of such a subchain is at least Dn by Lemma 5.34.

To control the number of subchains, we will associate to each one a

k-tile Yi. These k-tiles Y1, . . . , YM will form a k-chain joining E and E,and hence opposite sides of C. Thus M , which is the number of k-tilesin this chain, as well as the number of subchains, is at least Dk (bydefinition of this quantity; see (5.14)).

We now provide the details of the construction, which is illustratedin Figure 12.1. We will use auxiliary k-cells c1, c2, . . . of dimension≤ 1. If ci is 0-dimensional, then ci consists of a k-vertex pi, and we letW k(ci) := W k(pi) (see Definition 5.25). If ci is 1-dimensional, then ciis a k-edge and W k(ci) is the edge flower of ci as in Definition 5.28.

Since C is f -invariant, the cell decomposition Dk is a refinement of

D0. Hence there exist disjoint k-edges e ⊂ E and e ⊂ E with X1∩e 6= ∅and XN ∩ e 6= ∅.


For some number M ∈ N we will now inductively define k-cellsc1, . . . , cM+1 of dimension ≤ 1, k-tiles Y1, . . . , YM , and indices s1 = 1 <s2 < · · · < sM+1 = N + 1 with the following properties:

(i) c1 = e, cM+1 = e, and ci ∩ ci+1 = ∅ for i = 1, . . . ,M .

(ii) ci ∩ Yi 6= ∅ for i = 1, . . . ,M , ci+1 ⊂ ∂Yi for i = 1, . . . ,M − 1,and e ∩ YM 6= ∅.

(iii) Xsi , . . . , Xsi+1−1 is an (n+ k)-chain joining ci and ci+1 for i =1, . . . ,M .

Note that (i) and (ii) imply that E ∩ Y1 ⊃ e ∩ Y1 6= ∅, E ∩ YM ⊃e∩YM 6= ∅, and Yi∩Yi+1 ⊃ ci+1∩Yi+1 6= ∅ for i = 1, . . . ,M −1. Hence

Y1, . . . , YM will be a k-chain joining the 0-edges E and E as desired.

Let s1 = 1 and c1 = e. Suppose first that e meets W k(c1). Since e

is disjoint from e = c1 and hence from W k(c1), the points in e∩W k(c1)lie in ∂W k(c1). By Lemma 5.29 (ii) there exists a k-tile Y1 that meets

both c1 and e ⊃ e ∩W k(c1). We let M = 1, set c2 = e, and stop theconstruction. We have all the desired properties (i)–(iii).

In the other case where e ∩ W k(c1) = ∅ not all the (n + k)-tiles

X1, . . . , XN are contained in W k(c1). So there exists a smallest index

s2 ≥ 1 such that Xs2 meets S2 \W k(c1). Then s2 > s1 = 1, because

X1 meets e = c1 and is hence contained in W k(c1). To see this weuse Lemma 5.29 (iii) and the fact that X1 is contained in some k-tile. Moreover, for a similar reason we have Xs2 ⊂ S2 \W k(c1). By

definition of s2 the set Xs2−1 is contained in W k(c1). Hence every pointin the nonempty intersection Xs2−1 ∩ Xs2 lies in ∂W k(c1). Note thatthe (n + k)-tiles Xs2−1 and Xs2 do not necessarily meet in a k-vertex;but by Lemma 5.29 (iii) there exists a k-cell c2 ⊂ ∂W k(c1) of dimension≤ 1 that has common points with both Xs2−1 and Xs2 , and a k-tile

Y1 ⊂ W k(c1) with c1 ∩ Y1 6= ∅ and c2 ⊂ ∂Y1. Then c1 ∩ c2 = ∅ = c2 ∩ e,and the chain Xs1 = X1, . . . , Xs2−1 joins c1 and c2.

We can now repeat the construction as in the first step by using thechain Xs2 , . . . , XN that joins the disjoint k-cells c2 and e, etc. If in theprocess one of the cells ci has dimension 0, we invoke Lemma 5.26 (ii)and (iii) instead of Lemma 5.29 (ii) and (iii) in the above construction.The construction eventually stops, and it is clear that we obtain cellsand indices with the desired properties.

Case 2. # post(f) = 3.Let E1, E2, E3 be the three 0-edges. Consider a connected union K

of (n + k)-tiles joining opposite sides of C with N = Dn+k elements.Then K meets k-edges contained in the 0-edges, say k-edges ei ⊂ Ei


for i = 1, 2, 3. From K we can extract a simple (n + k)-chain joininge1 and e2 as well as another simple chain that joins e3 to one tile Xin the chain joining e1 and e2. Starting from this “center tile” X, wecan find three simple (n+ k)-chains that join X to the edges e1, e2, e3,respectively, and have only the tile X in common.

More precisely, for i = 1, 2, 3 we can find Ni ∈ N0 and (n+k)-chainsX,X i

1, . . . , XiNi

that join X and ei. Here the first tile X is the samein all chains and it is understood that the chain consists only of X ifNi = 0. Moreover, all the (n+ k)-tiles

X,X11 , . . . , X

1N1, X2

1 , . . . , X2N2, X3

1 , . . . , X3N3

are pairwise distinct tiles from K. Thus their number is bounded bythe number of (n+ k)-tiles in K. Since they still form a connected setjoining opposite sides of C, we have N1 +N2 +N3 + 1 = N = Dn+k.

Let Y be the unique k-tile with X ⊂ Y , and consider the chainX,X1

1 , . . . , X1N1

. Suppose that Y ∩ e1 = ∅. Since X ⊂ Y we haveN1 ≥ 1 and the chain X1, . . . , XN1 joins Y and e1. Hence this chain ora subchain must also join a k-edge e ⊂ ∂Y and e1. Then e ∩ e1 = ∅.As in the first part of the proof, we can find k-tiles Y1, . . . , YM1 joininge and e1, where M1 ∈ N and N1 ≥M1Dn.

If Y ∩ e1 6= ∅, we set M1 = 0 and do not define new k-tiles. In anycase we have that Y, Y 1

1 , . . . , Y1M1

is a chain joining Y and e1 (again weuse the convention that this chain consists only of Y if M1 = 0). Wealso have N1 ≥M1Dn (which is trivial if M1 = 0).

Using a similar construction for the other indices i = 2, 3, we obtainnumbers Mi ∈ N0 for each i = 1, 2, 3 that satisfy Ni ≥ MiDn, andchains Y, Y i

1 , . . . , YiMi

of k-tiles that join Y and ei. The union of thesek-tiles is a connected set joining opposite sides of C. Hence it containsat leastDk distinct elements. On the other hand, the number of distinctk-tiles in the union is at most M1 +M2 +M3 + 1 (note that the threechains may have other k-tiles in common apart from Y ). Hence Dk ≤M1 +M2 +M3 + 1, and it follows that

Dn(Dk − 1) + 1 ≤ Dn(M1 +M2 +M3) + 1

≤ N1 +N2 +N3 + 1 ≤ Dn+k,

which is the desired inequality (12.3).

Lemma 12.8. Let f : S2 → S2 be a Thurston map with # post(f) ≥3, C ⊂ S2 be a Jordan curve with post(f) ⊂ C, and Dn = Dn(f, C) forn ∈ N0. Suppose that C is f -invariant. Then the limit

Λ0 = limn→∞

D1/nn


exists and Λ0 = supn∈N

D1/nn ≤ deg(f).

If in addition there exists n0 ∈ N with Dn0 ≥ 2, then Λ0 > 1 andDn →∞ as n→∞.

The lemma shows that if f is combinatorially expanding for C, thenΛ0 > 1. So if Λ ∈ (1,Λ0) is arbitrary, then Dn & Λn for all n.

We will see later (Proposition 15.1) that if f is expanding, but Cis not necessarily f -invariant, then the limit Λ0 = limn→∞Dn(f, C)1/n

still exists and it is independent of C.

Proof. We use Lemma 12.7. First note that inequality (12.3) isalso true if post(f) = 4. The statements now essentially follow fromFekete’s lemma (see [KH, Prop. 9.6.4]). We provide the details for theconvenience of the reader.

A simple induction argument using (12.3) shows that if DN ≥ 2 forsome N ∈ N, then

(12.4) DkN ≥ Dk−1N + 1

for all k ∈ N. For such N let k(n) = bn/Nc. Noting that the sequenceDn is non-decreasing and using (12.4), we obtain

lim infn→∞

1

nlog(Dn) ≥ lim inf

n→∞

1

nlog(Dk(n)N)

≥ lim infn→∞

k(n)− 1

nlog(DN) =

1

Nlog(DN).

This inequality is trivially true if DN = 1, and so

lim infn→∞

1

nlog(Dn) ≥ sup

n∈N

1

nlog(Dn).

On the other hand,

lim supn→∞

1

nlog(Dn) ≤ sup

n∈N

1

nlog(Dn),

and so the limit

α := limn→∞

1

nlog(Dn) exists and α = sup

n∈N

1

nlog(Dn).

Note that Dn ≤ #Xn(f, C) ≤ 2 deg(f)n which implies α ≤ log(deg(f)).The first part of the statement follows by exponentiating α and theother relevant quantities.

For the last part suppose that there exists n0 ∈ N with Dn0 ≥ 2.Then

Λ0 = supn∈N

D1/nn ≥ D1/n0

n0> 1,


and it is clear from the definition of Λ0 as a limit that Dn → ∞ asn→∞.

We conclude this section with a lemma that shows that if a Thurs-ton map is combinatorially expanding, then there are points that lie“deep inside” a given edge or tile. It is related to Lemma 8.9 whichimplies a similar statement for expanding Thurston maps.

Lemma 12.9. Let f : S2 → S2 be a Thurston map, and C ⊂ S2 be anf -invariant Jordan curve with post(f) ⊂ C. Suppose that Dn0(f, C) ≥ 2for some n0 ∈ N.

(i) If n ∈ N0 and e is an n-edge, then there exists an (n + n0)-vertex p with p ∈ int(e).

(ii) If n ∈ N0 and X is an n-tile, then there exists an (n + n0)-edge with int(e) ⊂ int(X), and an (n + 2n0)-vertex p withp ∈ int(X).

Proof. In the previous statements and the ensuing proof it is un-derstood that the term k-cell for k ∈ N0 refers to a cell in Dk =Dk(f, C).

(i) Suppose e is an n-edge that does not contain (n+n0)-vertices inits interior. By Proposition 12.5 (iv) the n-edge e is equal to the unionof all (n + n0)-edges contained in e. Thus e must be an (n + n0)-edgeitself. Let u and v be the endpoints of e, and X be an (n + n0)-tilecontaining e in its boundary. Then K = X meets the two disjointn-cells u and v. Hence by Lemma 5.34 the set K should consistof at least Dn0 = Dn0(f, C) ≥ 2 (n + n0)-tiles. This is a contradictionproving the statement.

(ii) Let X be an n-tile. By Proposition 12.5 (iii) we know that X isthe union of all (n+n0)-tiles contained in X. In particular, there existsan (n+n0)-tile Y with Y ⊂ X. We claim that there exists an (n+n0)-edge in the boundary of Y that meets int(X). Otherwise, ∂Y ∩int(X) =∅, and as Y ⊂ X, we must have ∂Y ⊂ ∂X. Since both sets ∂Y and ∂Xare Jordan curves, this is only possible if ∂Y = ∂X. Then Y meets alln-vertices contained in ∂X, and two distinct n-vertices in particular.As in the proof of (i), this leads to a contradiction.

Hence there exists an (n + n0)-edge e with e ∩ int(X) 6= ∅. Sinceint(X) is an open subset of S2, we then also have int(e) ∩ int(X) 6= ∅.Since Dn+n0 is a refinement of Dn, by Lemma 5.7 we know that thereis a unique cell τ in Dn with int(e) ⊂ int(τ). Then int(τ)∩ int(X) 6= ∅,and so X = τ by Lemma 5.2. Hence int(e) ⊂ int(X) as desired.

By (i) there exists an (n + 2n0)-vertex p with p ∈ int(e). Then wealso have p ∈ int(X) as desired.


f

f

0

1

2

3

3

0

1

2

01

2

3

D1

D0

D0

Figure 12.2. Two two-tile subdivision rules.

12.2. Two-tile subdivision rules

In this section we consider a 2-sphere S2 and two given cell decomposi-tions D0 and D1 of S2. We call the cells in D0 the 0-cells and the cellsin D1 the 1-cells. Similarly, we refer to the tiles in D0 as the 0-tiles,the edges in D1 as the 1-edges, etc. We know by Proposition 5.24 thatunder suitable additional assumptions for such a pair (D1,D0) thereexists a postcritically-finite branched covering map f : S2 → S2 that iscellular for (D1,D0). The map f is unique up to Thurston equivalenceif additional data is provided, namely a labeling L : D1 → D0. The fol-lowing simple example illustrates that the pair (D1,D0) alone withoutthe labeling is in general not enough to determine the map f .

Example 12.10. Figure 12.2 represents two-tile subdivision rules

(D1,D0, L) and (D1,D0, L) on a sphere S2 given as a pillow obtainedby gluing two copies of a square together along their boundaries. Thecell decomposition D0 is indicated twice on the right of the figure withthe two squares as the 0-tiles and their common sides and corners as 0-edges and 0-vertices, respectively. Each of the two 0-tiles is subdividedinto nine squares of equal size. From this one obtains a refinement D1

of D0 as indicated on the left of the figure. We color the tiles black andwhite so that D1 and D0 are checkerboard tilings. Here the top square

12.2. TWO-TILE SUBDIVISION RULES 257

of the pillow is white and the colors of the 1-tiles are as indicated onthe left of Figure 12.2.

There are unique orientation-preserving labelings L and L that send1-tiles to 0-tiles of the same colors and such that the 1-vertices markedas a black dot on the left are sent to the one 0-vertex marked in thesame way in the two representations of D0 on the right (one can easilysee this directly or derive it as a special case of Lemma 12.13 (ii) below).In the figure we also show additional markings of some correspondingvertices for better illustration.

In this way we get two subdivision rules (D1,D0, L) and (D1,D0, L)

realized by Thurston maps f and f as indicated. The maps are uniquelydetermined if we require in addition that they are piecewise scaling

maps on 1-tiles. The map f assigns the same 0-tile to each 1-tile as f ,followed by an additional rotation.

The maps f and f are not Thurston equivalent. In fact, everypostcritical point for f is a fixed point, whereas no postcritical point

for f is a fixed point (each postcritical point for f is periodic of order4). Note that both maps are (Thurston equivalent to) Lattes maps.

As we know from Section 5.3, every Thurston map f : S2 → S2

arises as a cellular map for a pair of cell decompositions D1 and D0 ofS2. This gives a useful description of a Thurston map in combinatorialterms. If one wants to study the dynamics of f , one is interested inthe cell decompositions Dn obtained from pulling back D0 by fn asin Lemma 5.15. In general, in order to determine the combinatoricsof the whole sequence Dn, n ∈ N0 (i.e., the inclusion and intersectionpatterns of cells on all levels), is not enough to just know the pair(D1,D0) and the labeling τ ∈ D1 7→ f(τ) ∈ D0, but one also needsspecific information on the pointwise mapping behavior of f on the cellsin D1. Indeed, suppose g is another map that is cellular for (D1,D0)and induces the same labeling as f , i.e., f(τ) = g(τ) for all τ ∈ D1.

Let Dn be the cell decomposition of S2 obtained from D0 by pullingback by gn. Then one can show (by an argument very similar to the

considerations in the proof of Lemma 12.15 below) that Dn and Dnare isomorphic cell complexes for fixed n ∈ N0 (see Definition 5.13). In

contrast, the intersection patterns of corresponding cells in Dn and Dnon distinct levels n may be quite different.

The situation changes if (D1,D0) is a cellular Markov partition forf , because then the combinatorics of Dn is completely determined by(D1,D0) and the combinatorial data given by the labeling τ ∈ D1 7→f(τ) ∈ D0 (see Remark 5.11). This suggests that if one wants to study


Thurston maps as given by Proposition 5.24 from a purely combinato-rial point of view, then one should add the additional assumption thatD1 is a refinement of D0. If we restrict ourselves to the case where D0

contains only two tiles, then we are lead to the concept of a two-tilesubdivision rule as defined in the introduction of this chapter.

The proofs of Propositions 12.2 and 12.3 are easy consequences ofour previous considerations.

Proof of Proposition 12.2. Suppose f , C, D0, D1, and L areas in the statement. Then it immediately follows from Proposition 5.17 (i),Proposition 12.5 and the discussion after Definition 12.1 that (D0,D1, L)is a two-tile subdivision rule realized by f .

Proof of Proposition 12.3. The first part is just a special caseof Proposition 5.24. Note that f is a Thurston map; indeed, the numberof 1-tiles is equal to 2 deg(f), and also > 2 by Definition 12.1 (ii). Sodeg(f) ≥ 2.

Since D1 is a refinement of D0, the 1-skeleton C of D0 is containedin the 1-skeleton of D1. Moreover, since f is cellular for (D1,D0),this map sends the 1-skeleton of D1 into the 1-skeleton of D0. Hencef(C) ⊂ C, and so C is f -invariant. Each postcritical point of f is avertex of D0 and hence contained in C.

Suppose the map f : S2 → S2 realizes the two-tile subdivision ruleas in Proposition 12.3. If V0 denotes the set of vertices of D0 and V1

the set of vertices D1, we then have crit(f) ⊂ V1 and post(f) ⊂ V0

(see Lemma 5.22). Since the length of each cycle in D0 is 2, a vertexv in V1 is a critical point of f if and only if the length of the cycleof v in D1 is ≥ 4 (see Remark 5.23). Hence if f and g both realizethe subdivision rule, then crit(f) = crit(g) ⊂ V1. Moreover, since theorbit of any point in V1 is completely determined by the labeling, wethen also have post(f) = post(g) ⊂ V0.

Later we will need a more general version of the uniqueness partof Proposition 12.3. To formulate this, we first introduce a suitablenotion of an isomorphism between two-tile subdivision rules.

Let (D1,D0, L) and (D1, D0, L) be two-tile subdivision rules on 2-

spheres S2 and S2, respectively. We say that these subdivision rules

are isomorphic if there exist cell complex isomorphisms φi : Di → Difor i = 0, 1 such that L(φ1(τ)) = φ0(L(τ)) for τ ∈ D1 and such thatσ ⊂ τ for σ ∈ D1, τ ∈ D0 if and only if φ1(σ) ⊂ φ0(τ).

If, by abuse of notation, we denote the image of a cell τ ∈ Diunder φi by τ for i = 0, 1, then the last two conditions require that

L(τ) = L(τ) for τ ∈ D1 and that σ ⊂ τ for σ ∈ D1, τ ∈ D0 if and


only if σ ⊂ τ . So (D1,D0, L) and (D1, D0, L) are isomorphic if thecombinatorics of cells under the correspondence τ ↔ τ is the same andif this correspondence is also compatible with the labelings.

Lemma 12.11. Let (D1,D0, L) and (D1, D0, L) be two-tile subdivi-

sion rules on 2-spheres S2 and S2 that are isomorphic. If the Thurston

map f : S → S2 realizes (D1,D0, L) and the Thurston map f : S2 → S2

realizes (D1, D0, L), then f and f are Thurston equivalent.

Proof. The proof uses very similar ideas as the proof of the unique-ness part of Proposition 5.24.

There exist cell complex isomorphisms φi : Di → Di for i = 0, 1

such that L(τ) = L(τ) for τ ∈ D1 and such that σ ⊂ τ for σ ∈ D1,τ ∈ D0 if and only if σ ⊂ τ . Here we denote the image of a cell τ ∈ Diunder φi by τ for i = 0, 1.

By Lemma 5.14 (ii) there exists a homeomorphism h0 : S2 → S2

such that h0(τ) = τ for each τ ∈ D0.The map f realizes the subdivision rule (D1,D0, L). So if τ ∈ D1,

then f(τ) = L(τ) ∈ D0. Since f realizes (D1, D0, L), we have

f(τ) = L(τ) = L(τ) = f(τ) = h0(f(τ)) ∈ D0

for τ ∈ D1. The map f is cellular for (D1,D0) and so for each τ ∈ D1

the f |τ is a homeomorphism of τ onto f(τ). Similarly, the map f |τ is

a homeomorphism of τ onto f(τ) = h0(f(τ)). Hence for each τ ∈ D1

the map

ϕτ := (f |τ)−1 h0 (f |τ)

is well-defined and a homeomorphism from τ onto τ . If x ∈ τ , then

y = ϕτ (x) is the unique point y ∈ τ with f(y) = h0(f(x)).If σ, τ ∈ D1 and σ ⊂ τ , then

ϕτ |σ = ϕσ.

Indeed, if x ∈ σ, then y = ϕσ(x) ∈ σ ⊂ τ and f(y) = h0(f(x)). Henceϕσ(x) = y = ϕτ (x) by the uniqueness property of ϕτ (x).

If a point x ∈ S2 lies in two cells τ, τ ′ ∈ D1, then ϕτ (x) = ϕτ ′(x).Indeed, there exists a unique cell σ ∈ D1 with x ∈ int(σ). Thenσ ⊂ τ ∩ τ ′ by Lemma 5.3 (ii), and so, by what we have just seen, weconclude

ϕτ (x) = ϕσ(x) = ϕτ ′(x).

This allows us to define a map h1 : S2 → S2 as follows. If x ∈ S2,pick a cell τ ∈ D1 with x ∈ τ , and set

h1(x) = ϕτ (x).


Then h1 is well-defined.The definitions of h1 and ϕτ imply that h0 f = f h1. Moreover,

we have h1(τ) = τ for each τ ∈ D1. So by Lemma 5.14 (ii) the map h1

is a homeomorphism of S2 onto S2.We claim that h1(τ) = τ not only for τ ∈ D1, but also for τ ∈ D0.

Indeed, let τ ∈ D0 and x ∈ τ be arbitrary. Since D1 refines D0, thereexists σ ∈ D1 such that x ∈ σ ⊂ τ . Then σ ⊂ τ and h1(x) ∈ h1(σ) =σ ⊂ τ . We conclude that h1(τ) ⊂ τ . Conversely, if y ∈ τ , then there

exists a cell in D1 that contains y and is contained in τ . This cell hasthe form σ with σ ⊂ τ . Hence y ∈ σ = h1(σ) ⊂ h1(τ). This shows thath1(τ) = τ as desired.

This implies that for the homeomorphism h−11 h0 : S2 → S2 we have

(h−11 h0)(τ) = h−1

1 (τ) = τ for each τ ∈ D0. So by Lemma 5.14 (iii) thehomeomorphism h−1

1 h0 is isotopic to idS2 rel. V0, where V0 is the setof vertices of D0. If we post-compose an isotopy rel. V0 between h−1

1 h0

and idS2 by h1, we see that h0 and h1 are isotopic rel. V0, and hence

also isotopic rel. post(f), because post(f) ⊂ V0. Since h0 f = f h1,

the maps f and f are Thurston equivalent.

Remarks 12.12. Suppose the setup is as in the previous lemmaand its proof. We record two observations that will be important laterin the proof of Theorem 14.10.

(i) Let C be the Jordan curve of D0 and C be the Jordan curve of

D0. Then C and C are the 1-skeletons of D0 and D0, respectively. Thehomeomorphisms h0 and h1 constructed in the previous proof havethe property that they send each cell τ ∈ D0 to the corresponding

cell τ ∈ D0. Since the bijection τ 7→ τ is an isomorphism of the cell

complexes D0 and D0, it preserves the dimension of a cell. This impliesthat the maps h0 and h1 sends the 1-skeleton of D0 to the 1-skeleton

of D0, and so h0(C) = C = h1(C).(ii) The cell complex isomorphisms φi : Di → Di, i = 0, 1, as in the

definition of an isomorphism between (D1,D0, L) and (D1, D0, L) send

flags in Di to flags in Di. In the definition of an isomorphism of two-tile subdivision rules one can make the stronger additional requirementthat positively-oriented flags are sent to positively-oriented flags. If oneassumes this stronger notion of isomorphism in the previous lemma,then the maps h0 and h1 will be orientation-preserving.

If one wants to discuss specific examples of Thurston maps thatrealize a given two-tile subdivision rule (D1,D0, L), then it is convenientto represent the relevant data in a compressed form. The information


on the labeling L is completely determined by a pair of correspondingpositively-oriented flags in D1 and D0.

Lemma 12.13. Let (D1,D0) be a pair of cell decompositions of S2

satisfying conditions (i)–(iv) in Definition 12.1.

(i) If (c′0, c′1, c′2) and (c0, c1, c2) are positively-oriented flags in D1

and D0, respectively, then there exists a unique orientation-preserving labeling L : D1 → D0 with (L(c′0), L(c′1), L(c′2)) =(c0, c1, c2).

(ii) Let v′ be a vertex and X ′ be a tile in D1, and v be a vertex andX be a tile in D0. If v′ ∈ X ′ and v ∈ X, then there exists aunique orientation-preserving labeling L : D1 → D0 such thatL(v′) = v and L(X ′) = X.

So in both cases, (D1,D0, L) is a two-tile subdivision rule.Note that (i) is a special case of (ii). We formulated (i) explicitly,

because this version puts the statement in a more conceptual settingand because in the proof we use Lemma 5.21 to first establish (i) andthen derive (ii).

Proof. For i = 0, 1 denote by Vi,Ei,Xi the set of vertices, edges,and tiles of Di, respectively.

(i) To describe the labeling for (D1,D0) we proceed in the mannerdiscussed after Definition 5.20 and choose a particular index set L forthe labeling of the elements in D0 and D1.

We let L be the set that consists of two disjoint copies of Zk (onewill be for the vertices, and one for the edges), and the set b, w, whereagain we think of w representing “white” and b representing “black”.

We assign to c2 ∈ X0 the color “white”, and “black” to the othertile in X0. We assign 0 ∈ Zk to the 0-vertex v0 ∈ c0. Then there is aunique way to assign labels in Zk to the other vertices on C := ∂c2 (andthe corresponding cells of dimension 0) such that if v0, v1, . . . , vk−1 arethe vertices indexed by their label, then they are in cyclic order on C asconsidered as the boundary of the white 0-tile and in anti-cyclic orderfor the black 0-tile. Each 0-edge e is an arc on C with endpoints vland vl+1 for a unique l ∈ Zk. We label e by l (where we think of l asbelonging to the second copy of Zk). Since (c0, c1, c2) is a positively-oriented flag, and v0 is the initial point of c1, the edge c1 has the label0. All this is just a special case of Lemma 5.21. If in this way we assignto each element in D0 a label in L, we get a bijection ψ : D0 → L.Note that if (τ0, τ1, τ2) is any positively-oriented flag in D0, then itsimage under ψ has the form (l, l, w) or (l, l − 1, b) for some l ∈ Zk(cf. Lemma 5.21 (v)).


For D1 we invoke Lemma 5.21 directly to set up a suitable mapϕ : D1 → L. Since D1 satisfies the conditions of Lemma 5.21, we canfind maps LV : V1 → Zk, LE : E1 → Zk, and LX : X1 → b, w withthe properties (ii)–(iv) stated in the lemma and the normalizationsLV(v′0) = 0, where c′0 = v′0, LE(c′1) = 0, and LX(c′2) = w. The mapsLV, LE, LX induce a unique map ϕ : D1 → L such that ϕ(c) = LX(c) ifc is a 1-tile, ϕ(c) = LE(c) if c is a 1-edge, and ϕ(c) = LV(v) if c = vconsists of a 1-vertex v.

Now define L := ψ−1 ϕ : D1 → D0. The map L assigns to each1-cell c the unique 0-cell that has the same dimension as c and carriesthe same label in L as c.

It follows immediately from the properties of the maps ψ and ϕthat L preserves dimensions, respects inclusions, and is injective oncells. Hence L is a labeling according to Definition 5.20. By our nor-malizations the map L sends the flag (c′0, c

′1, c′2) to (c0, c1, c2).

Moreover, L is orientation-preserving. Indeed, ϕ maps the cellsτ0, τ1, τ2 in a positively-oriented flag in D1 to l, l, w, or to l, l − 1,b, respectively, where l ∈ Zk. These triples correspond to positively-oriented flags in D0. It follows that L has the desired properties.

To show uniqueness, we reverse the process. Given L with thestated properties, we use the same map ψ : D0 → L as above anddefine maps LV : V1 → Zk, LE : E1 → Zk, LX : X1 → b, w such thatLX(c) = (ψ L)(c) if c is a 1-tile, LE(c) = (ψ L)(c) if c is a 1-edge,and such that LV(v) = (ψ L)(c) if c = v consists of a 1-vertex v.

Then we have normalizations LV(v′0) = 0, LE(c′1) = 0, and LX(c′2) =w as in Lemma 5.21 (i). If we can show that LV, LE, LX have the prop-erties (ii)–(iv) in Lemma 5.21, then the uniqueness of L will follow fromthe corresponding uniqueness statement in this lemma.

To see this let e ∈ D1 be arbitrary, and X, Y ∈ D1 be the two tilesthat contain e in its boundary. Let u, v ∈ V1 be the two endpoints ofe. We may assume that notation is chosen so that the flag (u, e,X)is positively-oriented. Then (v, e, Y ) is also positively-oriented. Itfollows that the images of these flags under L are positively-oriented.Since L is injective on cells, and so L(u) 6= L(v), this implies thatL(X) 6= L(Y ). So L(X) and L(Y ) carry different colors (given by ψ)which implies that X and Y also carry different colors by definitionof LX. Hence LX has property (ii) in Lemma 5.21. By switching thenotation for u and v and X and Y if necessary, we may assume thatX and L(X) are white tiles. Since the flag (L(u), L(e), L(X)) ispositively-oriented, and L(X) is white, it follows that for some l ∈ Zkwe have ψ(L(u)) = l and ψ(L(e)) = l. Hence LV(u) = l and LE(e) = l.


Similarly, using that L(Y ) is black and that (L(v), L(e), L(Y )) ispositively-oriented, it follows that LV(v) = l + 1.

In other words, if we run along an oriented edge e in D1 so that awhite tile lies on the left of e, then the label of the endpoints of e (givenby LV) is increased by one, and decreased by one if a black tile lies onthe left. Hence LV has the property (iii) in Lemma 5.21. Moreover, wealso see that the label LE(e) is related to the labels of its endpoints asin statement (iv) of Lemma 5.21. The uniqueness of L follows.

(ii) Since v′ ∈ X ′ and v′ ∈ V1, we have v′ ∈ ∂X ′. There areprecisely two edges in E1 that are contained in ∂X ′ and have v′ asone of their endpoints. For precisely one of these edges e′, the triple(v′, e′, X ′) is the unique positively-oriented flag in D′ that includesv′ and X ′.

Similarly, there exists a unique edge e ∈ E0 such that (v, e,X)is a positively-oriented flag in D0. By (i) there exists an orientation-preserving labeling L : D1 → D0 that sends (v′, e′, X ′) to (v, e,X).In particular, L(v′) = v and L(X ′) = X. This shows existence of alabeling as desired.

To prove uniqueness, suppose that L : D1 → D0 is an orientation-preserving labeling with L(v′) = v and L(X ′) = X. Then the im-age of (v′, e′, X ′) under L is a positively-oriented flag of the form(v, L(e′), X). Since (v, e,X) is unique positively-oriented flag in D′that includes v and X, we have L(e′) = e. Uniqueness of L nowfollows from (i).

If we are given cell decompositions D1 and D0 as in the last lemma,then by part (ii) we can specify a unique labeling so that (D1,D0, L)becomes a two-tile subdivision rule in a very condensed form: all weneed to know is the image 0-tile X of one 1-tile X ′, and the image0-vertex v ∈ X of one 1-vertex v′ ∈ X ′. In specific examples (seeSection 12.3) one usually wants to include more information on thelabeling to get a better understanding of the mapping properties of theThurston map that realizes the subdivision rule.

Let f be a map realizing a two-tile subdivision rule (D1,D0, L). Wewant to show next that the property of f being combinatorially expand-ing for the Jordan curve C of D0 is independent of the realization. Incontrast, this is not true for expansion of the map (see Example 12.18).We require a lemma.

Lemma 12.14. Let f : S2 → S2 and g : S2 → S2 be Thurston maps.Suppose that # post(f) ≥ 3, that C ⊂ S2 is an f -invariant Jordan curve

with post(f) ⊂ C, and that h0, h1 : S2 → S2 are orientation-preserving


homeomorphisms satisfying h0| post(f) = h1| post(f), h0 f = g h1,and h0(C) = h1(C).

Then f is combinatorially expanding for C if and only if g is com-

binatorially expanding for C := h0(C) = h1(C).

Proof. We have post(g) = h0(post(f)) = h1(post(f)) (see Lemma 2.5).

Hence # post(g) = # post(f) ≥ 3. Moreover, C ⊂ S2 is a Jordan curve

with post(g) ⊂ C. This curve is g-invariant, since

g(C) = g(h0(C)) = h1(f(C)) ⊂ h1(C) = C.

So the statement that g is combinatorially expanding for C is meaning-ful (see Definition 12.4).

Pick an orientation of C. By our assumptions the map ϕ := h−11 h0

fixes the elements of post(f) pointwise and the Jordan curve C setwise.Since # post(f) ≥ 3 and post(f) ⊂ C, this implies that ϕ preserves theorientation of C. Since ϕ is an orientation-preserving homeomorphismon S2, the map ϕ sends each of the complementary components of Cto itself. Thus, ϕ is cellular for (D0,D0), where D0 = D0(f, C), and wehave ϕ(c) = c for each cell c ∈ D0. As in the proof of Lemma 5.14 (iii),this implies that ϕ is isotopic to idS2 rel. post(f). Hence h0 = h1 ϕis isotopic to h1 = h1 idS2 rel. post(f), and so there exists an isotopy

H0 : S2 × I → S2 rel. post(f) with H00 = h0 and H0

1 = h1.As in the proof of Proposition 11.4, based on Proposition 11.1 we

can repeatedly lift the initial isotopy H0. In this way we can find

isotopies Hn : S2 × I → S2 rel. post(f) such that Hnt f = g Hn+1

t

and Hn+10 = Hn

1 for all n ∈ N0 and t ∈ I. Note that Hn for n ≥ 1 isactually an isotopy rel. f−1(post(f)) ⊃ post(f).

Define homeomorphisms hn := Hn0 for n ∈ N0 (note that for n = 0

and n = 1 these maps agree with our given maps h0 and h1). Thenhn f = g hn+1 , and so

(12.5) h0 fn = gn hn

for all n ∈ N0.We have hn| post(f) = h0| post(f) which implies

(12.6) hn(post(f)) = post(g)

for all n ∈ N0. Moreover, hn|f−1(post(f)) = h1|f−1(post(f)) and so

(12.7) hn(f−1(post(f))) = g−1(post(g))

for n ∈ N0 as follows from (12.6) and Lemma 11.2.


Our hypotheses imply that if c is a cell in D0(f, C), then h0(c) is a

cell in D0(g, C). Since the set

Dn := hn(c) : c ∈ Dn(f, C)

is a cell decomposition of S2, it follows from this and (12.5) that gn

is cellular for (Dn,D0(g, C)). Since gn is also cellular for the pair

(Dn(g, C),D0(g, C)), the uniqueness statement in Lemma 5.15 implies

that Dn = Dn(g, C) for all n ∈ N0. In other words, the n-cells for (g, C)are precisely the images of the n-cells for (f, C) under the homeomor-phism hn.

We also have

(12.8) hn(C) = C

for each n ∈ N0. This can be seen by induction on n as follows. Thestatement is true for n = 0 and n = 1 by our hypothesis and by the

definition of C. Assume that hn(C) = C for some n ∈ N. Then byLemma 11.2 and the induction hypotheses we have

J := hn+1(C) ⊂ hn+1(f−1(C)) = g−1(hn(C)) = g−1(C).

It follows from (12.7) that (Hnt )h−1

n is an isotopy on S2 rel. g−1(post(g)).

It isotopes C = hn(C) ⊂ g−1(C) into J = hn+1(C) rel. g−1(post(g)). So Cand J are Jordan curves contained in the 1-skeleton g−1(C) of D1(g, C)that are isotopic relative to the set g−1(post(g)) of vertices of D1(g, C).Lemma 11.12 implies that J = C, and (12.8) follows.

Now (12.8) and (12.6) imply that a chain of n-tiles for (f, C) joinsopposite sides of C if and only if their images under hn form a chain

joining opposite sides of C. Since the images of the n-tiles for (f, C)under hn are the precisely the n-tiles for (g, C), we have Dn(f, C) =

Dn(g, C) for each n ∈ N0. The statement follows.

Now we can show the desired realization independence of combina-torial expansion.

Lemma 12.15. Let (D1,D0, L) be a two-tile subdivision rule on S2

and C be the Jordan curve of D0. Suppose that the maps f : S2 → S2

and g : S2 → S2 both realize the subdivision rule and that # post(f) =# post(g) ≥ 3. Then f is combinatorially expanding for C if and onlyif g is combinatorially expanding for C.

Proof. Let V0 and V1 be the set of vertices of D0 and D1, respec-tively. Then P := post(f) = post(g) ⊂ V0 ⊂ V1.


It follows from the proof of the uniqueness part of Proposition 5.24that there exists a homeomorphism h1 : S2 → S2 isotopic to idS2 rel.V1 ⊃ post(f) = post(g) that satisfies f = g h1. Moreover, h1(e) = efor each edge e in D1. Since D1 is a refinement of D0 and so the1-skeleton C of D0 is contained in the 1-skeleton of D1, this impliesh1(C) = C. Define h0 = idS2 . Since h1 is isotopic to idS2 rel. P we haveh1|P = idS2 |P = h0|P . Moreover, h0 f = g h1, h1(C) = C = h0(C),and both h0 and h1 are orientation-preserving homeomorphisms on S2.This shows that the hypotheses of Lemma 12.14 are satisfied (with

S2 = S2), and so f is combinatorially expanding for C if and only if g

is combinatorially expanding for C = h0(C) = h1(C) = C.

The previous lemma motivates the following definition.

Definition 12.16 (Combinatorially expanding two-tile subdivisionrules). Let (D0,D1, L) be a two-tile subdivision rule, and C be the Jor-dan curve of D0. We call (D0,D1, L) combinatorially expanding if everyThurston map f realizing (D0,D1, L) is combinatorially expanding forC.

We know that if this condition is true for one Thurston map realiz-ing the subdivision rule, then it is true for all such maps by Lemma 12.15.We will see that under an additional mild technical assumption a two-tile subdivision rule can be realized by an expanding Thurston mapif and only if the subdivision rule is combinatorially expanding (seeTheorem 13.1).

12.3. Examples of two-tile subdivision rules

In this section we present some examples of two-tile subdivision rules(D1,D0, L) and maps that realize them. In most of our examples wewill represent the underlying sphere S2 as a pillow obtained by gluingtogether two isometric copies of a polygon. This gives us a naturalcell decomposition D0 of S2, where the two polygons are the 0-tiles,and the sides and corners on the common boundaries of the polygonsthe 0-edges and 0-vertices. We assign the colors “black” and “white”to the 0-tiles. In our figures the top polygon of the pillow will be thewhite 0-tile. The description of the cell decomposition D1 is usuallymore complicated and depends on the specific case.

We know that in order to uniquely specify the labeling L : D1 → D0

it is enough to know the image of a pair (v′, X ′), where X ′ is a 1-tileand v′ ∈ X ′ a 1-vertex (see Lemma 12.13 (ii)). In general, we willinclude more information on the labeling L for a better illustrationof the behavior of the map f realizing the subdivision rule. We will

12.3. EXAMPLES OF TWO-TILE SUBDIVISION RULES 267

0−1

∞

07→−1 17→0

∞7→∞

−17→0

f1

Figure 12.3. The two-tile subdivision rule for z2 − 1.

Figure 12.4. Tiles of order 7 for Example 12.17.

assign the colors “black” and “white” to the 1-tiles indicating to whichof the 0-tiles they are sent by L (and f). With these labels the celldecomposition D1 will be a checkerboard tiling of k-gons, where k isthe number of vertices in D0.

Sometimes we will introduce markings for the 0-vertices and some1-vertices, often suggested by a natural identification of the underlying

sphere S2 with the Riemann sphere C. A 1-vertex a is then marked“a 7→ b” if the labeling L sends it to the 0-vertex marked b. Similarly,“7→ b” marks a 1-vertex that is not a 0-vertex and is sent to the 0-vertexb by L.

After these preliminaries we now proceed to discussing the exam-ples. A first simple example can be obtained from the map g in Section1.1 and the subdivision rule as indicated in Figure 1.1.

Example 12.17. Our next example is as follows. The white 0-tileis the (closure of the) upper half-plane, the black 0-tile is the (closure of


−17→1 7→−1 17→1

∞7→1

7→−1 7→−1

7→∞

7→∞

Figure 12.5. The barycentric subdivision rule.

the) lower half-plane in C. The 0-vertices are the points −1, 0,∞. Thus

the 0-edges are [−∞,−1], [−1, 0], [0,∞] ⊂ R. The cell decompositionD0 is indicated on the right of Figure 12.3.

The white 1-tiles are the first and third quadrants, and the black1-tiles are the second and forth quadrants. The 1-vertices and theirlabelings are as follows: the point ∞ is the only 1-vertex labeled ∞,the 1-vertices −1 and 1 are labeled 0, the 1-vertex 0 is labeled −1. Thecell decomposition is indicated on the left in Figure 12.3.

The pair (D1,D0) together with this orientation-preserving labelingL is a two-tile subdivision rule. It is straightforward to check that themap f1(z) = z2 − 1 realizes the subdivision rule (D1,D0, L).

This two-tile subdivision rule is not combinatorially expanding for

the Jordan curve C = R ofD0. Namely, the point∞ is its only preimageunder f1. Thus every n-tile contains ∞. Since the n-tiles cover thewhole sphere, there has to be an n-tile containing both 0 and∞ and sothe subdivision rule cannot be combinatorially expanding. In fact, forall n there exist two n-tiles that contain all postcritical points −1, 0,∞.Figure 12.4 shows the tiles of order 7.


Figure 12.6. Tiles of order 4 for the barycentric sub-division rule.

Example 12.18 (The barycentric subdivision rule). We glue twoequilateral triangles together along their boundaries to form a polyhe-

dral surface S2 that is conformally equivalent to C. The two triangles

are the 0-tiles. We can find a conformal equivalence of S2 with C suchthat the triangles correspond to the upper and lower half-planes, andthe vertices to the points −1, 1,∞. For convenience we identify thevertices with −1, 1,∞; they are the 0-vertices. The 0-edges are thethree edges of the triangles. The bisectors divide each triangle (each0-tile) into 6 smaller triangles. These 12 small triangles are the 1-tiles.The labeling of the 1-vertices is indicated in Figure 12.5. Again weobtain a two-tile subdivision rule. We can realize this subdivision ruleby a map f2 that conformally maps 1-tiles to the 0-tiles. Under the

indicated identification of S2 with C, the map is given by

f2(z) = 1− 54(z2 − 1)2

(z2 + 3)3


aa7→a 7→a

7→a

7→a 7→a

γ

f3

Figure 12.7. The subdivision rule for Example 12.19.

(see [CFKP, Example 4.6]). The subdivision rule is combinatoriallyexpanding, but the map f2 is not expanding. This follows from Propo-sition 2.3, because the point 1 is both a critical and a fixed point of f2.One can show that the Julia set of f2 is a Sierpinski carpet, i.e., a sethomeomorphic to the standard Sierpinski carpet.

It is possible to choose a different realization of the two-tile subdivi-

sion indicated in Figure 12.5 by a map f2 that is expanding. Namely, weuse affine maps to map the 1-tiles (the small triangles in the barycen-tric subdivision rule of the equilateral triangles) to the 0-tiles. In thiscase the n-tiles are Euclidean triangles for each n ∈ N. The collectionof all n-tiles is obtained from the (n − 1)-tiles similarly as the 1-tileswhere constructed from the 0-tiles: one subdivides each Euclidean tri-angle representing an (n − 1)-tile by its bisectors. It is clear that the

diameters of n-tiles tend to 0 as n→∞. Hence f2 is expanding, and sothis map is an example of an expanding Thurston map with periodiccritical points.

Example 12.19. The map h constructed in Section 1.3 realizes atwo-tile subdivision rule with the cell decompositions D1 and D0 asdescribed. The relevant information is represented in Figure 12.7. Inthis figure we also indicated the labeling L by coloring the 1-tiles toobtain a checkerboard tiling and by marking the 1-vertices that aresent to a specific 0-vertex a by L.

This subdivision rule is realized by a Thurston map f3 = h thatis combinatorially expanding. It is not equivalent to a rational map,since f3 has a Thurston obstruction given by the curve γ on the rightin Figure 12.7. We have already discussed this in Example 2.18 (seeFigure 2.2).

Example 12.20 (The 2-by-3 subdivision rule). We present anotherexample of an expanding Thurston map f4 that is not (Thurston) equiv-alent to a rational map. In a sense, this is the easiest example, but ithas a parabolic orbifold in contrast to the previous example.


0

1

∞

−1

17→0

7→1

07→0

7→−1

7→0

−17→−1

7→−1

7→0

∞7→−1

7→∞

7→∞

7→1

f4

Figure 12.8. The 2-by-3 subdivision rule.

The map f4 is a Lattes-type map with signature (2, 2, 2, 2) as inSection 3.4. Here the map A : C→ C in (3.18) is given by A(x+ yi) =2x + 3yi for x, y ∈ R, and f4 : S2 → S2 is the induced map on thequotient S2 = R2/G. As discussed in Section 3.4, we can represent thequotient space by a pillow P obtained by gluing two squares of side-length 1/2 together along their boundaries. This pillow allows a natu-

ral identification with C as we have seen in Section 1.1. This explainsthe markings of the four 0-vertices in Figure 12.8 which represents thetwo-tile subdivision rule realized by f4.

Each of the two faces (i.e., squares) of the pillow is divided into 6rectangles as shown in the figure. These 12 rectangles are the 1-tiles.Their sides and vertices are the 1-edges and 1-vertices. The coloringof 1-tiles, as well as the labeling of the 1-vertices, is indicated on theleft of Figure 12.8. The map f4 sends each of the 12 rectangles affinelyto one of the two squares forming the faces of the pillow. This impliesthat each n-tile is a rectangle with side lengths 1/2n and 1/3n. Inparticular, f4 is an expanding Thurston map.

The fact that f4 is not equivalent to a rational map follows fromTheorem 3.19. In the following, we will sketch an argument for thiswithin the framework of our present work. In our outline we will relyon some results and concepts that we will be discussed later on.

To reach a contradiction, suppose that f4 is equivalent to a rational

map R : C → C. Then R is a Thurston map with no periodic criticalpoints and is hence expanding (Proposition 2.3). So by Theorem 11.4the maps f4 and R are topologically conjugate. This implies by The-orem 17.1 (ii) that if our pillow P is equipped with a visual metric


% for f4, then (P, %) is quasisymmetrically equivalent to the standard

2-sphere (i.e., C equipped with the chordal metric). In particular, ifX0 is a 0-tile (i.e., one of the faces of the pillow P ) equipped with avisual metric %, then it can be mapped into the standard 2-sphere bya quasisymmetric map.

Now there are visual metrics for f4 with expansion factor Λ = 2(it is not hard to see this directly; it also follows from the generalargument in the proof of Theorem 15.3; indeed, if C is the boundaryof the pillow (which is f4-invariant), then we have D1 = D1(f4, C) = 2in (15.4)). If % is such a metric, then (X0, %) is bi-Lipschitz equivalentto a Rickman’s rug Rα. Here by definition the Rickman’s rug Rα for0 < α < 1 is the unit square [0, 1]2 ⊂ R2 equipped with the metric dαgiven by

dα((x1, y1), (x2, y2)) = |x1 − x2|+ |y1 − y2|α

for (x1, y1), (x2, y2) ∈ [0, 1]2. In our case, (X0, %) is bi-Lipschitz equiv-alent to Rα with α = log 2/ log 3. It is well-known that no quasisym-metric map can lower the Hausdorff dimension

dimH(Rα) = 1 + log 3/ log 2 > 2

of Rα (see [He, Theorem 15.10]); in particular, Rα and hence also(X0, %), cannot be mapped into the standard 2-sphere by a quasisym-metric map. This is a contradiction showing that f4 is not Thurstonequivalent to a rational map.

Example 12.21. We conclude this section by giving a whole classof examples that are similar to Example 12.19.

The simplest two-tile subdivision rule in this class is given as fol-lows. We consider a right-angled, isosceles Euclidean triangle T (so itsangles are π/2, π/4, π/4). The perpendicular bisector of the hypotenusedivides T into two triangles similar to T (scaled by the factor 1/

√2).

We glue two copies of the triangle T together along their boundariesto form a pillow as before. The two faces of the pillow (i.e., the twocopies of T ) are the 0-tiles, and the common sides and vertices of thesefaces are the 0-edges and 0-vertices, respectively.

We divide each of the 0-tiles (i.e., each face of the pillow) alongthe perpendicular bisector of the hypotenuse. The four triangles thusobtained are the 1-tiles. Their vertices and sides are the 1-vertices and1-edges. If the labeling is as indicated on the left of Figure 12.9, thenwe obtain a two-tile subdivision rule that can be realized by the mapf5(z) = 1− 2/z2 (this is actually a Lattes map).

As in Example 12.19, we can “add a flap” to modify the subdivisionrule. More precisely, we cut the pillow along the 1-edge that is the


−1

1

∞

17→−1

∞7→1

07→∞

−17→−1

f5

Figure 12.9. The subdivision rule realized by 1− 2/z2.

ω

1

∞

17→ω

∞7→1

ω 7→ω

07→∞

ω 7→ω

f6

Figure 12.10. Adding a flap.

perpendicular bisector of the hypotenuse of the white 0-tile. We takea copy of the pillow, scale it by the factor 1/

√2, and cut it along one

leg. We then glue the two sides of the slit to corresponding sides ofthe slit on the original pillow. This is indicated on the left in Figure12.10, where we also show the coloring of 1-tiles and the labeling of the1-vertices.

The same subdivision rule is also represented by Figure 12.11. Herethe two triangles drawn with a thick line are the 0-tiles; the white one isdrawn on the right and the black one on the left. Their edges are the 0-edges and their vertices the 0-vertices. To obtain a topological sphere,we have to match the two pairs of 0-edges with the same markings.

The white 0-tile is subdivided into four 1-tiles and the black 0-tileinto two 1-tiles. The labeling of the 1-vertices is shown in Figure 12.12.As before, this yields a two-tile subdivision rule. It can be realized bythe rational map

f6(z) = 1 +ω − 1

z3,

where ω = e4πi/3. The tiles of order 4 are shown in Figure 12.12.


17→ω

∞7→1

ω 7→ω

∞7→1 07→∞ ω 7→ω

Figure 12.11. The subdivision rule realized by f6.

This map f6 will be used to illustrate the construction of an invari-ant Jordan curve C with post(f6) ⊂ C (see Example 14.6 and Figure14.1).

The previous example can be modified as follows. Instead of addingone flap to the 1-edge bisecting the white 0-tile in Figure 12.9, we canadd n flaps. Similarly, we can glue in m flaps at the 1-edge bisectingthe black 0-tile. The resulting subdivision rule can be realized by therational map

f7(z) = 1 +ω − 1

zd,

where d = n+m+ 2 and ω = e2πi(n+1)/d.

More examples can be found in [CFKP] and [Me02]. In [Me02]and [Me09a] two-tile subdivision rules realizable by rational maps wereused to show that certain self-similar surfaces are quasispheres, i.e.,quasisymmetric images of unit sphere in R3. More general examples ofsubdivisions can be found in [CFP06b].

The Figures 12.4, 12.6, and 12.12 show symmetric conformal tilings.This means that if two tiles share an edge, they are conformal reflectionsof each other along this edge. Then the tiling can be obtained bysuccessive reflections, and so each tile encodes the information for thewhole tiling.


Figure 12.12. Tiles of order 4 of the map f6.

CHAPTER 13

Combinatorially expanding Thurston maps

In the last chapter we have constructed Thurston maps in a geomet-ric way from two-tile subdivision rules. The question arises when theThurston map realizing a subdivision rule can be chosen to be expand-ing. The key concept for an answer is the notion of combinatorialexpansion.

Theorem 13.1. Let (D1,D0, L) be a two-tile subdivision rule on a2-sphere S2 that can be realized by a Thurston map f : S2 → S2 withpost(f) = V0, where V0 is the vertex set of D0. Then (D1,D0, L) canbe realized by an expanding Thurston map if and only if (D1,D0, L) iscombinatorially expanding.

Recall that combinatorial expansion (see Definitions 12.16) for atwo-tile subdivision rule means that every Thurston map f : S2 →S2 realizing the subdivision rule is combinatorially expanding for theJordan curve C of f . In this case, C is f -invariant, # post(f) ≥ 3,post(f) ⊂ C, and there exists n0 ∈ N such that no n0-tile for (f, C)joins opposite sides of C (see Definitions 12.4).

In general, one only has post(f) ⊂ V0 for a Thurston map f re-alizing a subdivision rule as in Theorem 13.1. The stronger conditionpost(f) = V0 prevents the existence of additional vertices in V0 thathave no dynamical relevance and force an additional normalization onthe Thurston map. It is not hard to see that without the conditionpost(f) = V0 Theorem 13.1 is not true in general.

If a Thurston map f : S2 → S2 with # post(f) ≥ 3 has an invariantJordan curve C with post(f) ⊂ C, then for f to be expanding it isnecessary that f is combinatorially expanding for C. The converse isnot true in general: a combinatorially expanding Thurston map neednot be expanding. One still obtains a converse if one allows a changeof the map by a suitable isotopy.

Theorem 13.2. Let f : S2 → S2 be a Thurston map that has an in-variant Jordan curve C ⊂ S2 with post(f) ⊂ C. If f is combinatorially

277

278 13. COMBINATORIALLY EXPANDING THURSTON MAPS

expanding for C, then there exists an orientation-preserving homeomor-phism φ : S2 → S2 that is isotopic to the identity on S2 rel. post(f)such that φ(C) = C and g = φ f is an expanding Thurston map.

Note that post(g) = post(f) and g(C) = C. So the theorem saysthat if a Thurston map f is combinatorially expanding for an invariantJordan curve C ⊂ post(f), then by “correcting” the map by post-composing with a suitable homeomorphism, we can obtain an expand-ing Thurston map g with the same invariant curve and the same set ofpostcritical points.

The previous two theorems are easy consequences of the followingslightly more technical result.

Proposition 13.3. Let f : S2 → S2 be a Thurston map that hasan invariant Jordan curve C ⊂ S2 with post(f) ⊂ C. If f is combi-natorially expanding for C, then there exists an expanding Thurston

map f : S2 → S2 that is Thurston equivalent to f and has an f -

invariant Jordan curve C ⊂ S2 with post(f) ⊂ C. Moreover, there

exist orientation-preserving homeomorphisms h0, h1 : S2 → S2 that are

isotopic rel. post(f) and satisfy h0 f = f h1 and h0(C) = C = h1(C).

So up to Thurston equivalence every combinatorially expandingThurston map with an invariant Jordan curve can be promoted toan expanding Thurston map with an invariant curve.

Proposition 13.3 shows that for a Thurston map with an invariantJordan curve combinatorial expansion is sufficient for the existence ofa Thurston equivalent map that is expanding. One may ask whethercombinatorial expansion is necessary for this as well. The answer isnegative, as we will see in Example 13.21. The Thurston map f in thisexample has an invariant Jordan curve C containing all its postcriti-cal points. It is not combinatorially expanding for C (and hence notexpanding), yet equivalent to an expanding Thurston map g.

On an intuitive level the assertion of Proposition 13.3 seems quite

plausible. Namely, based on Proposition 5.24 a map f as in Propo-sition 13.3 can easily be constructed if one can find cell decompo-sitions with the same combinatorics as Dn(f, C) where the cells aresmall in diameter (with respect to a given background metric) if n islarge. Since f is combinatorially expanding, Lemma 12.7 implies thatDn(f, C) → ∞ as n → ∞. So if the level n increases one needs moreand more tiles to form a connected set joining opposite sides of C, ormore generally, to join any two disjoint k-cells. Therefore, it seemsevident that one should be able to make the cells small while keepingtheir combinatorics. If one wants to implement this idea, then one

13. COMBINATORIALLY EXPANDING THURSTON MAPS 279

faces serious difficulties that make it hard to convert this idea into avalid proof (in [CFP01, Thm. 2.3] the authors claim a more generalstatement with an argument along these lines).

For this reason our approach for the proof of Proposition 13.3 isdifferent. For the map f : S2 → S2 in this proposition to be expanding,the intersection

⋂nXn of any nested sequence Xn of n-tiles should

consist of only one point (see Lemma 12.6). In order to enforce thiscondition, we introduce a suitable equivalence relation ∼ on the sphereS2 that collapses these intersections

⋂nXn to points. We then use

Moore’s Theorem (see Theorem 13.10) to show that the quotient space

S2/∼ is also a 2-sphere. The map f will then be the induced map onS2/∼. While this approach is quite natural, it is somewhat lengthy tocarry out and will occupy the bulk of this chapter.

In the following, f : S2 → S2 will be a Thurston map and C ⊂ S2

will be an f -invariant Jordan curve with post(f) ⊂ C for which f iscombinatorially expanding.

We consider the cell decompositions Dn = Dn(f, C) for n ∈ N0 asdefined in Section 5.3. As before, we denote by Xn, En and Vn theset of n-tiles, n-edges, and n-vertices for (f, C), respectively. A subsetτ ⊂ S2 is called a tile if it is an n-tile for some n ∈ N0. We use theterms edge, vertex, cell in a similar way. In particular, in this sectionthe term “cell” will always be used with this specific meaning. We willuse the term topological cell to refer to the more general notion of cellsas defined in Section 5.1.

Since C is f -invariant, Dn+k is a refinement of Dn for all n, k ∈N0. For each X ∈ Xn+k there exists a unique Y ∈ Xn with X ⊂Y . Conversely, each n-tile Y is equal to the union of all (n + k)-tilescontained in Y , and similarly each n-edge e is equal to the union of all(n + k)-edges contained in e (all this was proved in Proposition 12.5).We will use this fact that cells are subdivided by cells of the samedimension and higher level repeatedly in the following.

The equivalence relation on S2. As in the beginning of Section 12.1,we denote by S = S(f, C) the set of all sequences Xn with Xn ∈ Xn

for n ∈ N0 and

X0 ⊃ X1 ⊃ X2 ⊃ . . .

We know (see Lemma 12.6) that expansion of a Thurston map with aninvariant curve is characterized by the condition that

⋂nX

n is alwaysa singleton set if Xn ∈ S. This may not be the case for our givenmap f , and so we want to identify all points in such an intersection⋂nX

n. This will not lead to an equivalence relation, since transitivity


may fail. As we will see, this issue is resolved if we define the relationas follows.

Definition 13.4. Let x, y ∈ S2 be arbitrary. We write x ∼ y ifand only if for all Xn, Y n ∈ S with x ∈

⋂nX

n and y ∈⋂n Y

n wehave Xn ∩ Y n 6= ∅ for all n ∈ N0.

Recall from (5.14) that Dn = Dn(f, C) denotes the minimal num-ber of n-tiles forming a connected set Kn joining opposite sides of C.Since f is combinatorially expanding for C (see Definition 12.4), wehave # post(f) ≥ 3 and so the term “joining opposite sides” is mean-ingful (see Definition 5.30). Moreover, there exists n0 ∈ N such thatDn0(f, C) ≥ 2, and so by Lemma 12.7 we have Dn = Dn(f, C)→∞ asn→∞. In combination with Lemma 5.34 this implies that if τ, σ aredisjoint k-cells and Kn is a connected set of n-tiles with σ ∩ Kn 6= ∅and τ ∩Kn 6= ∅, then the number of tiles in Kn tends to infinity andso cannot stay bounded as n → ∞. We will use this fact in the proofof the following lemma.

Lemma 13.5. The relation ∼ is an equivalence relation on S2.

Proof. Reflexivity and symmetry of the relation ∼ are clear. Toshow transitivity, let x, y, z ∈ S2 be arbitrary and assume that x ∼ yand y ∼ z. Let Xn, Zn ∈ S with x ∈

⋂nX

n and z ∈⋂n Z

n bearbitrary. We have to show that Xn ∩ Zn 6= ∅ for all n ∈ N0.

If this is not the case, then there exists n0 ∈ N0 such that Xn0 ∩Zn0 = ∅. To reach a contradiction, pick a sequence Y n ∈ S withy ∈

⋂n Y

n. Since x ∼ y and y ∼ z, we have Xn ∩ Y n 6= ∅ andY n ∩ Zn 6= ∅ for all n ∈ N0. Then Xn0 ∩ Y n ⊃ Xn ∩ Y n 6= ∅ andZn0 ∩ Y n ⊃ Zn ∩ Y n 6= ∅ for all n ≥ n0. So the n-tile Y n connects thedisjoint n0-tiles Xn0 and Y n0 for all n ≥ n0. As we discussed, this isimpossible by Lemma 5.34.

The following lemma gives convenient characterizations when twopoints are equivalent.

Lemma 13.6. Let x, y ∈ S2 be arbitrary. Then the following condi-tions are equivalent:

(i) x ∼ y,

(ii) there exist Xn, Y n ∈ S with x ∈⋂nX

n, y ∈⋂n Y

n, andXn ∩ Y n 6= ∅ for all n ∈ N0,

(iii) for all cells σ, τ ⊂ S2 with x ∈ σ, y ∈ τ , we have σ ∩ τ 6= ∅.

Proof. The implication (i) ⇒ (ii) is clear.


To show the reverse implication (ii) ⇒ (i), we assume that thereexist Xn, Y n ∈ S with x ∈

⋂nX

n, y ∈⋂n Y

n, and Xn∩Y n 6= ∅ forall n ∈ N0. We claim that if Un, V n ∈ S are two other sequenceswith x ∈

⋂n U

n and y ∈⋂n V

n, then Un ∩ V n 6= ∅ for all n ∈ N0. Toreach a contradiction assume that Un0 ∩ V n0 = ∅ for some n0 ∈ N0.We then have

Un0 ∩Xn ⊃ x 6= ∅ and V n0 ∩ Y n ⊃ y 6= ∅

for all n ∈ N. Moreover, Xn ∩ Y n 6= ∅, and so for each n ∈ N0, theKn := Xn ∪ Y n is connected, consists of two n-tiles, and meets thedisjoint n0-tiles Un0 and V n0 . As before this contradicts Lemma 5.34.Hence x ∼ y as desired.

The implication (iii) ⇒ (i) is again clear. To prove (i) ⇒ (iii),suppose that x ∼ y. We argue by contradiction and assume that thereexist cells σ, τ with x ∈ σ, y ∈ τ , and σ ∩ τ = ∅. By subdividing thecells if necessary, we may assume that σ and τ are cells on the samelevel n0.

There are sequences Xn, Y n ∈ S with x ∈⋂nX

n and y ∈⋂n Y

n. Since x ∼ y, we have Xn ∩ Y n 6= ∅ for all n.This implies that for n ∈ N0 the set Kn = Xn ∪ Y n is connected

and consists of at most two n-tiles. Moreover,

Kn ∩ σ ⊃ Xn ∩ σ ⊃ x 6= ∅,

and similarly, Kn ∩ τ 6= ∅. Hence Kn connects the disjoint n0-cellsσ and τ . Since f is combinatorially expanding, this is impossible byLemma 5.34 for large n. This gives the desired contradiction.

The previous lemma implies that all points in an intersection⋂nX

n

with Xn ∈ S are equivalent. It is clear that ∼ is the “smallest”equivalence relation with this property.

If x ∈ S2 we denote by [x] ⊂ S2 the equivalence class of x withrespect to the equivalence relation ∼, and by

S2 = S2/∼ = [x] : x ∈ S2

the quotient space of S2 under ∼. So S2 consists of all equivalence

classes of ∼. Such an equivalence class is both a point in S2 and

a subset of S2. We equip S2 with the quotient topology. Then the

quotient map π : S2 → S2, x ∈ S2 7→ [x], is continuous.

In order to to prove that S2 is in fact a topological 2-sphere, wewill need a good geometric description of the equivalence classes. Toset this up, consider a point x ∈ S2 and let n ∈ N0 be arbitrary. We


define

(13.1) Ωn = Ωn(x) =⋃x∈cn

int(cn)

where the union is taken over all n-cells cn that contain x. Recall thatint(cn) = cn if cn has dimension 0.

Note that

(13.2) Ωn

=⋃x∈Xn

Xn,

where the union is taken over all n-tiles Xn that contain x. Indeed,every cell cn in the union in (13.1) is contained in an n-tile Xn thatcontains x. Therefore, Ω

n ⊂⋃x∈Xn Xn, since the right hand side is

closed. On the other hand, for each n-tile Xn containing x we haveint(Xn) ⊂ Ωn. Thus

⋃x∈Xn Xn ⊂ Ω

n, and (13.2) follows.

Lemma 13.7. The set Ωn ⊂ S2 is a simply connected region.

Proof. We have to consider three cases. When x = v is an n-vertex then Ωn(v) = W n(v) is the n-flower of v by Definition 5.25.Recall from Lemma 5.26 (i) that such a vertex flower is a simply con-nected region.

Suppose that x is not an n-vertex, but x is contained in an n-edgeen. Then x is necessarily contained in the interior int(en) of en. Thereare precisely two distinct n-tiles Xn and Y n that contain en in theirboundaries. These are all the n-tiles that contain x. Thus

Ωn = int(Xn) ∪ int(en) ∪ int(Y n).

Then Ωn is a simply connected region (see Lemma 5.12 (iv)).Finally, suppose that x is not contained in any n-edge. Then there

is a unique n-tile Xn that contains x. Then Ωn = int(Xn) is an openJordan region, and so simply connected.

Let M ⊂ S2 be an equivalence class with respect to ∼. We select apoint x ∈M as follows.

Case 1. If M contains a vertex v, then x := v.

Case 2. If M contains no vertex, but intersects an edge e, then wechoose a point in M ∩ e for x.

Case 3. If M contains no vertex and does not intersect any edge, thenwe choose an arbitrary point in M for x.

We say that M is of vertex-type in the first, of edge-type in the second,and of tile-type in the last case. We call the point x a center ofthe equivalence class M . By Lemma 13.6 an equivalence class cannot


Ωn

x=v

M

Ωn

Xn Y n

xM

en

xΩn M

Figure 13.1. Equivalence classes of vertex-, edge-, andtile-type.

contain two distinct vertices. So if M is of vertex-type, then its centeris unique, but this may not be true in the other two cases.

With such a choice of a center x for given M , we define Ωn = Ωn(x)as in (13.1) for n ∈ N0. Note that every (n+ 1)-cell cn+1 with x ∈ cn+1

is contained in an n-cell cn with x ∈ cn+1 and int(cn+1) ⊂ int(cn) (seeLemma 5.7). Thus Ωn is a decreasing sequence of sets, i.e.,

(13.3) Ω0 ⊃ Ω1 ⊃ Ω2 ⊃ . . . .

Lemma 13.8. Let M ⊂ S2 be an arbitrary equivalence class withrespect to ∼, and Ωn be defined as above for n ∈ N0. Then

(13.4) M =⋂n

Ωn =⋂n

Ωn.

The different types of equivalence classes are illustrated in Fig-ure 13.1.


Proof. Let x ∈ M be a center of M . In order to establish theinclusion

(13.5)⋂n

Ωn ⊂M,

let y ∈⋂n Ω

nbe arbitrary. We have to show that x ∼ y (which implies

y ∈M). If this is not the case, then x 6∼ y, and so there exist sequencesXn, Y n ∈ S with x ∈

⋂nX

n and y ∈⋂n Y

n, and n0 ∈ N0 suchthat Xn0 ∩ Y n0 = ∅. On the other hand, for each n ∈ N0 we havey ∈ Ω

n, and so by (13.2) we can find an n-tile Zn with x, y ∈ Zn. Then

Xn0∩Zn ⊃ x 6= ∅ and Y n0∩Zn ⊃ y 6= ∅ , i.e., the tile Zn intersectsthe disjoint tiles Xn0 and Y n0 , for all n ∈ N0. This is impossible, sincef is combinatorially expanding for C, see Lemma 5.34. We obtain acontradiction and (13.5) follows.

To finish the proof, it is enough to show that M ⊂⋂n Ωn, or

equivalently that if

y ∈ S2 \⋂n

Ωn =⋃n

(S2 \ Ωn)

is arbitrary, then y 6∼ x (and so y 6∈M). Note that the sets S2 \Ωn forn ∈ N0 form an increasing sequence, and so y 6∈ Ωn for all sufficientlylarge n. In order to show y 6∼ x, we now consider three cases accordingto the type of M .

Case 1: M is of vertex-type. In this case x is a vertex, say ann0-vertex, where n0 ∈ N0. Then x is also n-vertex for all n ≥ n0.Fix n ≥ n0 such y 6∈ Ωn. Then there exists a unique n-cell τ withy ∈ int(τ). Since y 6∈ Ωn, we have x /∈ τ . Now x is an n-vertex andso x is an n-cell, and the n-cells τ ⊃ y and x are disjoint. ByLemma 13.6 this implies y 6∼ x as desired.

Case 2: M is of edge-type. Then x is contained in an edge, say ann0-edge en0 , where n0 ∈ N0. By successive subdivisions (see Proposi-tion 12.5 (iv)) we can find n-edges en for n ≥ n0 that contain x andthat satisfy

en0 ⊃ en0+1 ⊃ . . . .

Fix n ≥ n0 such that y 6∈ Ωn. Then there exists a unique n-cell τwith y ∈ int(τ). Since y 6∈ Ωn, we have x 6∈ τ and so en 6⊂ τ . Henceτ ∩ int(en) = ∅ by Lemma 5.3 (ii). Let u and v be the endpoints of en.Since these points are vertices, they do not belong to M by assumption.So the set ⋂

n≥n0

en ⊂⋂n

Ωn ⊂M


does not contain u or v either. It follows that there exists m ≥ n suchthat u, v /∈ em. Then y ∈ τ , x ∈ em, and τ ∩ em = ∅, because

τ ∩ em ⊂ τ ∩ (en \ u, v) = τ ∩ int(en) = ∅.By Lemma 13.6 this implies y 6∼ x as desired.

Case 3: M is of tile-type. We pick sequences Xn, Y n ∈ S withx ∈

⋂nX

n and y ∈⋂n Y

n. Since M does not meet any edges andx ∈ M , for each n ∈ N0 the tile Xn is the unique n-tile with x ∈ Xn.Then x ∈ int(Xn) and Ωn = int(Xn). In order to show that y 6∼ x, weargue by contradiction and assume y ∼ x. Then Kn := Xn∩Y n 6= ∅ foreach n ∈ N0 (see Definition 13.4). The sets Kn, n ∈ N0, are nonemptynested compact sets. Hence there exists a point z ∈

⋂nK

n. Thenx ∼ z and so z ∈M .

On the other hand, if n0 ∈ N0 is large enough, then y 6∈ Ωn0 =int(Xn0). Since ∂Xn0 = Xn0 \ int(Xn0) consists of n0-edges, y ∼ x andso y ∈M , and M does not meet edges, we then actually have y 6∈ Xn0 .Thus Xn0 6= Y n0 . This means that the intersection Kn0 = Xn0 ∩ Y n0

consists of n0-cells on the boundary of Xn0 and of Y n0 , and is hencecontained in a union of n0-edges. Since z ∈M ∩Kn0 , this implies thatM meets an edge, contradicting our assumption in this case. So indeedy 6∼ x as desired.

The following consequence of the previous lemma will be one the

essential ingredients in the proof that S2 is a topological 2-sphere.

Corollary 13.9. Each equivalence class M of ∼ is a compactconnected set with connected complement S2 \M .

Proof. Let M be an arbitrary equivalence class of ∼. Then Lem-ma 13.8 and (13.3) imply that the set M is the intersection of a nestedsequence of the compact sets Ω

n, n ∈ N0. It follows from (13.2) that

each set Ωn

is connected. Hence M is also compact and connected.The complement of an open simply connected set in S2 is connected.

So Lemma 13.8 also shows that the complement S2 \ M of M is aunion of an increasing sequence of connected sets. Hence S2 \ M isconnected.

The quotient space S2 is a topological 2-sphere. After these

preparations we are ready to show that S2 is a topological 2-sphere.For the moment we consider an arbitrary equivalence relation ∼ on thesphere S2. We call it closed if (x, y) ∈ S2×S2 : x ∼ y is a closed sub-set of S2×S2 (in the older literature the term “upper semicontinuous”is often used instead). This is equivalent to the following condition: If


xn and yn are arbitrary convergent sequences in S2 with xn → xand yn → y as n→∞, and xn ∼ yn for all n ∈ N, then x ∼ y. Severalother equivalent conditions can be found in [MP, Lemma 2.2]. Theequivalence classes of a closed equivalence relation on S2 are closed andhence compact subsets of S2.

We need the following key theorem (see [Moo] for the original proof,[Da, p. 187, Thm. 1] for a stronger statement, and [Ca, Supplement 1]for a general discussion on the 2-sphere recognition problem).

Theorem 13.10 (Moore 1925). Let ∼ be an equivalence relationon a 2-sphere S2. Suppose that

(i) the equivalence relation ∼ is closed,

(ii) each equivalence class of ∼ is a connected subset of S2,

(iii) the complement of each equivalence class of ∼ is a connectedsubset of S2,

(iv) there are at least two distinct equivalence classes.

Then the quotient space S2/∼ is homeomorphic to S2.

Here it is understood that S2/∼ is equipped with the quotienttopology.

We now return to our previous considerations and consider theequivalence relation ∼ given as in Definition 13.4.

Lemma 13.11. Let ∼ be the equivalence relation on S2 as in Def-

inition 13.4. Then the quotient space S2 = S2/∼ is homeomorphic toS2.

Proof. By Lemma 13.5 our relation ∼ is indeed an equivalencerelation. It remains to verify the conditions (i)–(iv) in Theorem 13.10.

Conditions (ii) and (iii) were already proved in Corollary 13.9.

Condition (iv): There are at least two equivalence classes, becauseno two distinct vertices are equivalent by Lemma 13.6, and each post-critical point of f is a vertex (there are at least three such points).

Condition (i): Let xn and yn be convergent sequences in S2

with xn → x and yn → y as n → ∞, and suppose that xn ∼ yn forall n ∈ N. We have to show that x ∼ y. Suppose this is not the case.Then the equivalence classes [x] and [y] are disjoint. By Lemma 13.7and Lemma 13.8 there exist simply connected nested regions Ωn

x andΩny for n ∈ N0 such that

[x] =⋂n

Ωnx =

⋂n

Ωn

x and [y] =⋂n

Ωny =

⋂n

Ωn

y .


Since [x] and [y] are disjoint, the sets Ωn

x and Ωn

y will also be disjoint

for sufficiently large n, say Ωn0

x ∩ Ωn0

y = ∅. On the other hand, sinceΩn0x ⊃ [x] and Ωn0

y ⊃ [y] are open, there exists n1 ∈ N such that

xn1 ∈ Ωn0x and yn1 ∈ Ωn0

y . Since Ωn0

x and Ωn0

y consist of n0-tiles andare disjoint, this means that there exist n0-tiles σ and τ with xn1 ∈ σ,yn1 ∈ τ and σ ∩ τ = ∅. Hence xn1 6∼ yn1 by Lemma 13.6. This is acontradiction. It follows that ∼ is closed.

This shows that the conditions in Moore’s theorem are satisfied,

and so S2 is indeed a 2-sphere.

Quotients of cells and the induced cell decompositions on S2.We now study what happens to our cells under the quotient map

π : S2 → S2. If A ⊂ S2 is an arbitrary set, we denote by A its im-

age under the projection map π. So A = π(A) = [x] : x ∈ A ⊂ S2.We will see that if σ is an arbitrary cell (i.e., an element of Dn forsome n ∈ N0), then σ is a topological cell of the same dimension(Lemma 13.16). Moreover, the images σ of the n-cells σ ∈ Dn form a

cell decomposition of S2 (Lemma 13.17).

Lemma 13.12. Let M be an arbitrary equivalence class with centerx ∈M . If τ is an arbitrary cell, then

τ ∩M 6= ∅ if and only if x ∈ τ.

Proof. The “if”-implication is obvious.To show the other implication, assume that τ is a cell of level n

and x /∈ τ . Consider an n-cell σ ⊂ τ . Then x /∈ σ and so int(σ)is disjoint from Ωn = Ωn(x) by (13.1), because distinct n-cells havedisjoint interiors. Recall from Lemma 5.2 that τ is the disjoint unionof the interiors of all n-cells σ ⊂ τ . Thus τ ∩Ωn = ∅, and so τ ∩M = ∅by Lemma 13.8.

The following lemma states that if we pass to the quotient space

S2 = S2/∼, then intersection and inclusion relations of cells are pre-served. In particular we do not create “new” intersections or inclusionsbetween cells.

Lemma 13.13. If σ and τ are cells, then σ ∩ τ = σ ∩ τ . Moreover,we have σ ⊂ τ if and only if σ ⊂ τ .

Proof. The inclusion σ ∩ τ ⊂ σ ∩ τ is trivial.For the other inclusion consider an arbitrary point [x] ∈ σ∩ τ ⊂ S2.

We can assume that x is a center of M = [x] ⊂ S2. Then M meets


both cells σ and τ , and so x ∈ σ∩τ by Lemma 13.12. Thus [x] ∈ σ ∩ τ .We have proved σ ∩ τ ⊂ σ ∩ τ as desired.

In the second statement the implication σ ⊂ τ ⇒ σ ⊂ τ is trivial.For the other implication assume that σ ⊂ τ . Let k and n be thelevels of σ and τ , respectively. For the moment we make the additionalassumption that k ≥ n.

By Lemma 12.9 there exists a vertex v such that v ∈ int(σ) (notethat this is trivial if σ is a 0-dimensional cell). Then [v] ∈ σ ⊂ τ , andso [v] ∩ τ 6= ∅. Lemma 13.6 implies that v ∈ τ showing that

(13.6) int(σ) ∩ τ 6= ∅.

Since k ≥ n, the cell decomposition Dk containing σ is a refinement ofthe cell decomposition Dn containing τ . Therefore, as we have seen inthe first part of the proof of Lemma 5.7, the relation (13.6) forces theinclusion σ ⊂ τ .

If k < n, we subdivide σ into cells of level n. By the previousargument, τ will contain each of these cells, and so we always havehave σ ⊂ τ as desired.

Lemma 13.14. Let M ⊂ S2 be an equivalence class and E ⊂ S2 bea finite union of edges. Then E ∩M is connected.

Proof. Let x be a center of M . By subdividing the edges in E, wecan assume that E consists of n0-edges, where n0 ∈ N0 is large enough.For n ≥ n0 let

En :=⋃e ∈ En : e ⊂ E, x ∈ e.

Clearly Enn≥n0 is a decreasing sequence of compact connected sets.Moreover, y ∈M ∩E if and only if y ∈

⋂n≥n0

En by Lemma 13.12 andLemma 13.6. Since the latter set is connected, the claim follows.

We will need a 1-dimensional version of Moore’s Theorem. It caneasily be derived from the topological characterization of arcs and topo-logical circles (in equivalent form this as stated as two exercises in [Da,p. 21, Ex. 2 and 3]). For the convenience of the reader we will providea simple direct proof in Section A.13.

Lemma 13.15. Let J be an arc or a topological circle, and ∼ be anequivalence relation on J . Suppose that

(i) each equivalence class of ∼ is a compact and connected subsetof J ,

(ii) there are at least two distinct equivalence classes.


Then the quotient space J = J/∼ is an arc or a topological circle,respectively.

We will use this in the proof of the following lemma.

Lemma 13.16. Let τ be an edge or a tile. Then τ is an arc or a

closed Jordan region, respectively. Moreover, ∂τ = ∂τ .

Here ∂τ (and similarly ∂τ) refers as usual to the boundary of a cellτ as defined in Section 5.1. So ∂τ is the topological boundary of τ in S2

if τ is a tile, and equal to the set consisting of the two endpoints of τ ifτ is an edge. If τ is 0-dimensional cell, i.e., a singleton set consisting ofa vertex, then ∂τ = ∅, and the statement in the lemma is trivially true.So the lemma can be formulated in an equivalent form by saying that

if τ ⊂ S2 is a cell (in one of the cell decompositions D), then τ ⊂ S2 isa cell (in the general topological sense) of the same dimension, and theboundary of τ is the image of the boundary of τ under the quotientmap.

Proof. Suppose first that τ is an edge. Then our equivalencerelation∼ on S2 restricts to an equivalence relation on τ whose quotient

space can be identified with the subset τ of S2. The equivalence classeson τ have the form τ ∩M , where M ⊂ S2 is an equivalence class withrespect to ∼.

Each set τ ∩ M is compact, as ∼ is closed, and connected byLemma 13.14. Moreover, τ meets a least two distinct equivalenceclasses, as its endpoints are distinct vertices and hence not equivalent.

Lemma 13.15 implies that τ ⊂ S2 is indeed an arc.Let u and v be the two endpoints of τ . Then [u] ∩ τ is a compact

connected subset of τ containing u. Hence this set is a subarc of τ withone endpoint equal to u. This implies that the set τ \ [u] is connected,and so the set π(τ \ [u]) = τ \π(u) is also connected. Therefore, π(u)is an endpoint of τ . By the same reasoning we see that π(v) is also anendpoint of τ . Since u and v are distinct vertices, we have u 6∼ v and

so π(u) 6= π(v). Hence ∂τ = π(u), π(v) = π(u, v) = ∂τ .

If τ is a tile, say an n-tile, then τ is a closed Jordan region whoseboundary J = ∂τ is a topological circle consisting of finitely manyedges. By the Schonflies Theorem we can write S2 as a disjoint unionS2 = Ω1 ∪ J ∪ Ω2, where Ω1 and Ω2 are open Jordan regions boundedby J . Then τ coincides with one of the sets Ω1 or Ω2, say τ = Ω1.

The set J ⊂ S2 is also a topological circle as follows from the factthat ∼ is closed, Lemma 13.14, and Lemma 13.15. So we can also write

S2 as a disjoint union S2 = D1 ∪ J ∪ D2, where D1 and D2 are open


Jordan regions in S2 bounded by J . If we take preimages under the

quotient map π : S2 → S2, we get the disjoint union S2 = π−1(D1) ∪π−1(J) ∪ π−1(D2). Since point preimages under π, i.e., equivalenceclasses, are connected, the map π is monotone, and hence preimages ofconnected sets are connected [Da, p. 18, Prop. 1]. So the sets π−1(D1)

and π−1(D2) are connected open sets disjoint from π−1(J) ⊃ J . Itfollows that each of the sets π−1(D1) and π−1(D2) is contained in oneof the regions Ω1 and Ω2.

These sets cannot be contained in the same region Ωi. Indeed,

if π−1(D1) ∪ π−1(D2) ⊂ Ω1, for example, then Ω2 ⊂ π−1(J), and so

π(Ω2) ⊂ J . This means that every point in Ω2 is equivalent to a pointin J . This is impossible, because Ω2 contains the interior of an n-tile,and hence a k-vertex for some k > n (see Lemma 12.9 (ii)). Such avertex is not equivalent to any point in J by Lemma 13.6.

By what we have just seen, we may assume that indices are chosen

such that π−1(D1) ⊂ Ω1 and π−1(D2) ⊂ Ω2. ThenD1 = D1∪J ⊂ π(Ω1)and π(Ω1)∩D2 = ∅. Hence τ = π(Ω1) = D1 is a closed Jordan region.

Moreover, ∂τ = ∂D1 = J = ∂τ .

We now come to the main result of this subsection, which says thatthe quotient map π maps the cell decompositions Dn to cell decompo-

sitions Dn such that “all combinatorial properties are preserved”.

Lemma 13.17. Let n, k ∈ N0. Then we have:

(i) For each τ ∈ Dn the set τ is a topological cell in S2 of thesame dimension as τ .

(ii) For each cell τ ∈ Dn we have ∂τ = ∂τ .

(iii) For σ, τ ∈ Dn, we have σ = τ if and only if σ = τ .

(iv) Dn := τ : τ ∈ Dn is a cell decomposition of S2.

(v) The map τ ∈ Dn 7→ τ ∈ Dn is an isomorphism between the

cell complexes Dn and Dn.

(vi) Dn+k is a refinement of Dn. Moreover, for all σ ∈ Dn+k andτ ∈ Dn we have

σ ⊂ τ if and only if σ ⊂ τ .

Proof. (i) and (ii) follow from Lemma 13.16.

(iii) Let σ and τ be arbitrary n-cells, and suppose that int(σ) ∩int(τ) 6= ∅. Pick a point p ∈ int(σ)∩int(τ). Then p ∈ σ∩τ = σ ∩ τ (seeLemma 13.13), and so there exists x ∈ σ ∩ τ with π(x) = p. Then x ∈int(σ), for otherwise x ∈ ∂σ and so p = π(x) ∈ ∂σ by (ii), contradicting


the choice of p. Similarly, x ∈ int(τ). So x ∈ int(σ) ∩ int(τ) whichimplies that σ = τ . Statement (iii) follows.

(iv) From what we have seen it follows that the topological cells τfor τ ∈ Dn are all distinct, and no two have a common interior point.

Moreover, there are finitely many of these cells, and they cover S2,

because the cells in Dn cover S2. Finally, for a cell τ we have ∂τ = ∂τby (ii). Since ∂τ is a union of cells in Dn, it follows that ∂τ is a union

of cells in Dn. This shows that Dn is a cell decomposition of S2.

(v) By (i) and (iii) the map τ ∈ Dn 7→ τ ∈ Dn is a bijection

betweenDn and Dn that preserves dimensions of cells. By Lemma 13.13the map also satisfies condition (ii) in Definition 5.13. Hence it is an

isomorphism between the cell complexes Dn and Dn.

(vi) It follows immediately from the definitions and the fact that

Dn+k is a refinement of Dn that Dn+k is a refinement of Dn. The secondstatement was proved in Lemma 13.13.

The induced map f on S2. We will now show that f induces a

map f on the sphere S2. We will see later that f is a Thurston mapequivalent to f .

Lemma 13.18. Suppose that x, y ∈ S2 and x ∼ y. Then f(x) ∼f(y).

Proof. Let x, y ∈ S2 with x ∼ y be arbitrary. Pick Xn, Y n ∈S with x ∈

⋂Xn and y ∈

⋂n Y

n. Define Un = f(Xn+1) and V n =f(Y n+1) for n ∈ N0. Then Un and V n are n-tiles, and so Un, V n ∈S. Moreover, f(x) ∈

⋂n U

n and f(y) ∈⋂n V

n. Since x ∼ y we haveXn ∩ Y n 6= ∅ for all n ∈ N. Hence

Un ∩ V n = f(Xn+1) ∩ f(Y n+1) ⊃ f(Xn+1 ∩ Y n+1) 6= ∅for all n ∈ N0. Lemma 13.6 now implies that f(x) ∼ f(y) as desired.

By the previous lemma the map f : S2 → S2 given by

f([x]) = [f(x)] for x ∈ S2

is well-defined. Then f π = π f , and it follows from the properties

of the quotient topology that f is continuous (see Lemma A.9).

In the following Dn = τ : τ ∈ Dn will denote the cell decompo-

sition of S2 as provided by Lemma 13.17 (for n ∈ N0). As the next

lemma shows, the map fn has injectivity properties similar to fn.


Lemma 13.19. Let τ be an n-cell, n ∈ N. Then fn is a homeomor-

phism of τ onto σ, where σ = fn(τ). In particular, fn is cellular for

(Dn, D0).

Proof. Note that fn is a homeomorphism of τ onto σ. Hence

fn(τ) = (fn π)(τ) = (π fn)(τ) = σ

showing that fn maps τ onto σ.

So it remains to show the injectivity of fn on τ , or equivalently,that if x, y ∈ τ and fn(x) ∼ fn(y), then x ∼ y. Since every vertexand every edge is contained in a tile, we may also assume that τ is ann-tile.

If x, y ∈ τ and fn(x) ∼ fn(y), then we can pick sequences Xk andY k in S such that Xn = Y n = τ and x ∈

⋂kX

k, y ∈⋂k Y

k. Thenfn(Xk+n) and fn(Y k+n) are k-tiles for k ∈ N0. Moreover, the sequencesfn(Xk+n) and fn(Y k+n) are in S, and fn(x) ∈

⋂k f

n(Xk+n)and fn(y) ∈

⋂k f

n(Y k+n). Since fn(x) ∼ fn(y), this implies thatfn(Xk+n) ∩ fn(Y k+n) 6= ∅ for all k ∈ N0. Since Xk+n, Y k+n ⊂ τ fork ≥ 0 and fn|τ is injective, we conclude that Xk+n ∩ Y k+n 6= ∅ fork ≥ 0. Since Xn = Y n = τ , we also have Xk = Y k for k = 0, . . . , n−1.Hence Xk ∩ Y k 6= ∅ for all k ≥ 0. Lemma 13.6 then shows that x ∼ yas desired.

The fact that fn is cellular for (Dn, D0) follows from the first partof the proof and the fact that fn is cellular for (Dn,D0).

The auxiliary homeomorphisms h0 and h1. To prove that f isThurston equivalent to f , we need to define homeomorphisms h0, h1 : S2 →S2 that make the diagram

(13.7) S2 h1//

f

S2

f

S2 h0// S2

commutative and are isotopic rel. V0. The construction of these mapsfollows ideas in the proof of Lemma 12.11.

For the definition of h0 recall that S2 is the union of two 0-tilesX0

b and X0w with common boundary C. The Jordan curve C is further

decomposed into k = #V0 ≥ 3 0-edges and 0-vertices. The cell de-

composition D0 of S2 contains two tiles X0b and X0

b . Lemma 13.17 (ii)

and Lemma 13.16 show that the common boundary of X0w and X0

b is


C = π(C), which is a Jordan curve. There are k distinct vertices and

edges on C. There are no other cells in D0.We know by Lemma 13.17 (v) that the map τ ∈ D0 7→ τ ∈

D0 is an isomorphism between the cell complexes D0 and D0. SoLemma 5.14 (ii) implies that there exists a homeomorphism h0 : S2 →S2 such that h0(τ) = τ for all cells τ ∈ D0.

Now let τ ∈ D1 be arbitrary. Then f(τ) ∈ D0, and by Lemma 13.19

the map f |τ is a homeomorphism of τ onto f(τ) = h0(f(τ)). Hencethe map

ϕτ := (f |τ)−1 h0 (f |τ)

is well-defined and a homeomorphism from τ onto τ . If x ∈ τ , then

y = ϕτ (x) is the unique point y ∈ τ with f(y) = h0(f(x)). As inthe proof of Lemma 12.11, this uniqueness property implies that ifσ, τ ∈ D1 and σ ⊂ τ , then ϕτ |σ = ϕσ. From this in turn one can deducethat if a point x ∈ S2 lies in two cells τ, τ ′ ∈ D1, then ϕτ (x) = ϕτ ′(x).

This allows us to define a map h1 : S2 → S2 as follows. If x ∈ S2, we

pick τ ∈ D1 with x ∈ τ and set h1(x) := ϕτ (x). Then h1 : S2 → S2 iswell-defined.

Lemma 13.20. The map h1 : S2 → S2 is a homeomorphism of S2

onto S2 satisfying h0f = f h1. Moreover, we have h0(C) = C = h1(C)and the homeomorphisms h0 and h1 are isotopic rel. V0.

Proof. We have h1|τ = ϕτ for each cell τ ∈ D1. So the definitions

of h1 and ϕτ show that h0 f = f h1 and that h1(τ) = ϕτ (τ) = τ

for each τ ∈ D1. Since τ ∈ D1 7→ D1 is an isomorphism of cellcomplexes by Lemma 13.17 (v), the last statement implies that h1 is a

homeomorphism of S2 onto S2 (Lemma 5.14 (i)).Note that h1(τ) = τ also for each τ ∈ D0. Indeed, suppose τ ∈ D0

is arbitrary. Since D1 is a refinement of D0, for each x ∈ τ thereexists σ ∈ D1 such that x ∈ σ ⊂ τ . Then h1(x) ∈ h1(σ) = σ ⊂ τ .

So h1(τ) ⊂ τ . Conversely, let y ∈ τ be arbitrary. Since D1 is a

refinement of D0 (Lemma 13.17 (vi)), there exists a cell σ ∈ D1 suchthat y ∈ σ ⊂ τ . By Lemma 13.13 we then have σ ⊂ τ , and soy ∈ σ = h1(σ) ⊂ h1(τ). We conclude that h1(τ) = τ for each τ ∈ D0

as claimed.The Jordan curve C is the 1-skeleton of D0 and so equal to the

union of all edges e ∈ D0. We know that h0(e) = h1(e) = e = π(e) for

each such edge e. Hence h0(C) = h1(C) = π(C) = C.


If τ ∈ D0 is arbitrary, then (h−11 h0)(τ) = h−1

1 (τ) = τ . So byLemma 5.14 (iii) the homeomorphism h−1

1 h0 is isotopic to idS2 rel.V0. Hence h0 and h1 are isotopic rel. V0.

We are now ready to prove the main results of this chapter.

Proof of Proposition 13.3. If f is a Thurston map that iscombinatorially expanding for a Jordan curve C as in the statement,then we can apply all the previous considerations.

We orient the 2-sphere S2 so that the homeomorphism h0 : S2 → S2

is orientation-preserving. Then the homeomorphism h1 : S2 → S2 isalso orientation-preserving, because it is isotopic to h0.

Consider the map f : S2 → S2 defined earlier. Since f = h0 f h−1

1 , the map f is a branched covering map on S2, and h1 and h0 are

orientation-preserving homeomorphisms, it follows that the map f is a

branched covering map on S2. Moreover, we have crit(f) = h1(crit(f)).Since the maps h0 and h1 are isotopic rel. V0 = post(f), we con-

clude exactly as in the proof of Lemma 2.5 that post(f) = h0(post(f)) =

π(post(f)). In particular, post(f) is a finite set which implies that fis a Thurston map.

The Jordan curve C = π(C) satisfies

post(f) = π(post(f)) ⊂ π(C) = C.Since f(C) ⊂ C, we also have

f(C) = (f π)(C) = (π f)(C) ⊂ π(C) = C.

This shows that C is invariant with respect to f and contains the set

of postcritical points of f .In the proof of the previous lemma we have seen that h0(C) =

h1(C) = C. We also know that h0 and h1 are isotopic rel. post(f). Since

h0 f = f h1, this implies that f and f are Thurston equivalent.

It remains to show that f is expanding. Since C ⊂ S2 is an f -

invariant Jordan curve with post(f) ⊂ C, we can do this by verifying

the condition in Lemma 12.6 for C.Note thatD0(f , C) = D0. Moreover, since fn is cellular for (Dn, D0),

it follows from Lemma 5.15 that Dn(f , C) = Dn for all n ∈ N0. So the

n-tiles for (f , C) are precisely the sets X, where X is an n-tile on S2

for (f, C).Let X0 ⊃ X1 ⊃ X2 ⊃ . . . be a nested sequence of n-tiles for (f , C).

Clearly,⋂n X

n is non-empty. We have to show that this intersectiondoes not contain more than one point. From Lemma 13.17 (vi) it


follows that the corresponding sequence Xn of n-tiles for (f, C) is

nested, and so Xn ∈ S = S(f, C). To see that⋂n X

n consists of

precisely one point, we argue by contradiction and assume that⋂n X

n

contains more that one point, or equivalently, there exist two distinct(and hence disjoint) equivalence classes M and N with respect to ∼such that Mn := M ∩ Xn 6= ∅ and Nn := N ∩ Xn 6= ∅ for all n ∈N0. Since equivalence classes and tiles are compact, in this way weget descending sequences M0 ⊃ M1 ⊃ . . . and N0 ⊃ N1 ⊃ . . . ofnonempty and compact sets. Hence the sets

⋂nM

n = M ∩⋂nX

n

and⋂nN

n = N ∩⋂nX

n are nonempty. So there exists points x ∈M∩

⋂nX

n and y ∈ N∩⋂nX

n. Since x and y lie in different equivalenceclasses, they are not equivalent. On there other hand, we have x, y ∈⋂nX

n and Xn ∈ S. Hence x ∼ y by Lemma 13.6. This is a

contradiction and we conclude that f is indeed expanding.

The proofs of Theorems 13.2 and 13.1 are now immediate.

Proof of Theorem 13.2. Using the notation of Proposition 13.3and its proof, we define φ = h−1

1 h0. Then φ is isotopic to the identityon S2 rel. post(f) = V0 and φ(C) = C. Moreover,

g = φ f = h−11 h0 f = h−1

1 f h1,

and so g is topologically conjugate to the expanding Thurston map f ,and hence itself an expanding Thurston map.

Proof of Theorem 13.1. Let (D1,D0, L) be a two-tile subdivi-sion rule on S2 as in the statement, C be the Jordan curve and V0 bethe vertex set of D0.

If (D1,D0, L) can be realized by an expanding Thurston map f : S2 →S2, then C is f -invariant, post(f) ⊂ C, and # post(f) ≥ 3. SoLemma 8.5 implies that f is combinatorially expanding for C, and so(D1,D0, L) is combinatorially expanding according to Definition 12.16and the discussion following this definition.

Conversely, suppose (D1,D0, L) is combinatorially expanding. Thenthis subdivision rule can be realized by a Thurston map f : S2 → S2

that is combinatorially for C. Moreover, our hypotheses implies thatpost(f) = V0. Note that then D0 = D0(f, C). Moreover, f is cellularfor (D1,D0) and so necessarily D1 = D1(f, C) (see Lemma 5.15).

We again use the notation of Proposition 12.3 and its proof. Wedefine φ = h−1

1 h0. Then φ is a homeomorphism on S2 is isotopicto the identify on S2 rel. V0 = post(f), and is orientation-preserving.

As in the proof of Theorem 13.2 let g = φ f = h−11 f h1. Then

g : S2 → S2 is an expanding Thurston map.


f

g

C

C

Ch1 h0

Figure 13.2. The map f is not combinatorially expan-ding, but equivalent to the expanding map g.

Since φ(C) = C and φ is orientation-preserving and the identity onV0, we have φ(c) = c for each c ∈ D0. Since f is cellular for (D1,D0),the map g = φ f is cellular for (D1,D0) and we have g(c) = f(c) foreach cell c ∈ D1. Since f realizes the subdivision rule with the givenlabeling, this shows that g is also a realization. Now g is expanding,and so the claim follows.

Example 13.21. The map f : S2 → S2 is illustrated in the top ofFigure 13.2. It has four postcritical points, which are the vertices ofthe pillow shown in the top right. The front shows the white 0-tile, theback is the black 0-tile. The subdivision of the 0-tiles is indicated inthe top left of the figure. Here we have cut the pillow along three 0-edges. The left shows the subdivision of the white 0-tile, and the rightthe subdivision of the black 0-tile. On the left the preimages of onepostcritical point (a 0-vertex) are given. So we only show the labels of1-vertices that the labeling sends to the 0-vertex on the bottom left ofthe pillow.

The boundary of the pillow is the Jordan curve C containing allpostcritical points of f . It is invariant under f , but the map f isnot combinatorially expanding for C: for arbitrarily large n there will


always be an n-tile (contained in the white 0-tile) joining the 0-edgeson the left and on the right.

We will see that f is Thurston equivalent to the map g : S2 → S2,indicated in the bottom of Figure 13.2. The map g is a Lattes map,obtained similarly as the example in Section 1.1. More precisely it isobtained as a quotient of A : C → C, A(z) = 3z by a crystallographicgroup of type (2222) (see Chapter 3, in particular Theorem 3.1). It isclear that g is expanding.

We fix a homeomorphism h0 : S2 → S2 that maps the boundary ofthe pillow of f (i.e., the curve C) to the boundary of the pillow of g,and the postcritical points of f to the postcritical points of g. Let D1

and D1 denote the cell decompositions of S2 and S2, respectively, into1-cells as indicated on the top left of Figure 13.2. As in the proof of

Proposition 5.24 there is a homeomorphism h1 : S2 → S2 that maps

the curve C ⊂ S2 to the curve C ⊂ S2 indicated in the figure, is cellular

with respect to (D1, D1), and satisfies g h1 = h0 f.Obviously, the map h1 is isotopic to h0 rel. post(f). Thus f and g

are Thurston equivalent.

Note that g has an invariant curve C ′ (different from C of course)that contains all its postcritical points.

CHAPTER 14

Invariant curves

This chapter is central for this work. We will prove existence anduniqueness results for invariant curves C of expanding Thurston mapsf . We will also show that if an invariant curve exists, then it can beobtained from an iterative procedure, and that it is a quasicircle. Wealways require that C is a Jordan curve and that post(f) ⊂ C, but inthe following discussion we will often refer to such C simply as invariantcurves for brevity.

One of our main results can be formulated as follows.

Theorem 14.1. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Then for each

sufficiently large n ∈ N there exists a Jordan curve C that is invariantfor fn and isotopic to C rel. post(f).

This existence result has the following important implication.

Corollary 14.2. Let f : S2 → S2 be an expanding Thurston map.Then for each sufficiently large n there exists a two-tile subdivision rulethat is realized by F = fn.

This justifies our approach of studying expanding Thurston mapsfrom a combinatorial perspective based on cellular Markov partitions.

Invariant curves are quasicircles if the underlying metric is visual.

Theorem 14.3. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. If C is f -invariant,then C equipped with (the restriction of) a visual metric for f is aquasicircle.

This applies also to invariant curves for iterates, because if f : S2 →S2 is an expanding Thurston map, then the same is true for each iterateF = fn, n ∈ N.

If one studies rational Thurston maps f that are expanding, then

the underlying 2-sphere is the Riemann sphere C, and it is natural toequip it with the chordal metric σ. Then an invariant C curve as in theprevious theorem is a also quasicircle with respect to σ. This can be de-duced from Theorem 14.3 once we know that for such maps the chordal

299

300 14. INVARIANT CURVES

metric is quasisymmetrically equivalent to each visual metric. This willbe proved in Chapter 17. See Corollary 17.11 and Theorem 1.1, whichcontains Theorems 14.1 and 14.3 in the rational case.

As we discussed in Section 12.1, if C an f -invariant Jordan curvewith post(f) ⊂ C, then we get a sequence of cell decompositionsDn = Dn(f, C), n ∈ N0, so that each cell decomposition is refined bythe cell decompositions of higher level. We will see that Theorem 14.3implies that we get good control for the geometry of edges and tilesin these cell decompositions. Namely, the family of edges consists ofuniform quasiarcs and the boundary of tiles are uniform quasicircles.See Section 14.3 for an explanation of this terminology and Proposi-tion 14.25 for a precise statement.

In Theorem 14.1 it is necessary to pass to an iterate of the mapto guarantee existence of an invariant Jordan curve, because there areexamples of maps for which an invariant curve does not exist (see Ex-ample 14.11). One can formulate a necessary and sufficient criterionfor the existence of invariant curves.

Theorem 14.4 (Existence of invariant curves). Let f : S2 → S2 bean expanding Thurston map. Then the following conditions are equiv-alent:

(i) There exists an f -invariant Jordan curve C ⊂ S2 with post(f) ⊂C.

(ii) There exist Jordan curves C, C ′ ⊂ S2 with post(f) ⊂ C, C ′ andC ′ ⊂ f−1(C), and an isotopy H : S2× I → S2 rel. post(f) withH0 = idS2 and H1(C) = C ′ such that the map

f := H1 f is combinatorially expanding for C ′.Moreover, if (ii) is true, then there exists an f -invariant Jordan curve

C ⊂ S2 with post(f) ⊂ C that is isotopic to C rel. post(f) and isotopicto C ′ rel. f−1(post(f)).

The first condition in (ii) says that there exists a Jordan curve Cwith post(f) ⊂ C that can be isotoped rel. post(f) into its preimageunder f . This condition alone ensures that an associated f -invariant

set C with post(f) ⊂ C exists, but in general C need not be a Jor-

dan curve (see Lemma 14.17 (viii) and Example 14.22). If the map fis combinatorially expanding as stipulated in (ii), then one obtains a

Jordan curve C.Invariant curves can be constructed by an iterative procedure that

will be described in Section 14.2. In the situation of Theorem 14.4one lifts the isotopy H by the map f repeatedly to obtain a sequence

14. INVARIANT CURVES 301

of isotopies Hn : S2 × I → S2, n ∈ N0, with H0 = H such thatHn

0 = Hn(·, 0) = idS2 . One sets C0 = C and defines inductivelyCn+1 = Hn

1 (Cn) = Hn(Cn, 1) for n ∈ N0. One can then show that

the sequence Cn Hausdorff converges to the desired invariant curve C(see Proposition 14.19). An explicit knowledge of the isotopies is notreally necessary, because one can interpret this an an edge replacementprocedure (see Remark 14.21). In Section 14.2 we will discuss this andseveral examples that illustrate various phenomena in this context.

Our basic existence result Theorem 14.1 for invariant curves is com-plemented by the following uniqueness statement.

Theorem 14.5 (Uniqueness of invariant curves). Let f : S2 → S2

be an expanding Thurston map and C and C ′ be f -invariant Jordancurves in S2 that both contain the set post(f). Then C = C ′ if and onlyif C and C ′ are isotopic rel. f−1(post(f)).

This implies that in a given isotopy class rel. post(f) there areonly finitely invariant curves C (Corollary 14.8). It follows that anexpanding Thurston map f with # post(f) = 3 can have only finitelymany invariant curves C (Corollary 14.8).

The situation changes if one does not restrict the isotopy class ofC. In general an expanding Thurston maps f can have infinitely manyinvariant curves C (see Example 14.9). If, in addition, the map isrational and has a hyperbolic orbifold, then this cannot happen and fcan have only finitely many invariant curves C (see Theorem 14.10; weare grateful to K. Pilgrim for pointing this out to us).

The chapter is organized as follows. Section 14.1 is devoted to ex-istence and uniqueness results, where we provide proofs for the state-ments discussed above. The iterative procedure for the construction ofinvariant curves is explained in Section 14.2. In the final Section 14.3we discuss the quasiconformal geometry of invariant curves. Here weprove Theorem 14.3 and related results.

Much of our discussion in this chapter is quite technical. Before wego into the details, we start by looking at a specific example that willillustrate some of the main ideas.

Example 14.6. Let S2 = C and f : S2 → S2 be the map definedby f(z) = 1+(ω−1)/z3 for z ∈ S2, where ω = e4πi/3. This map was al-ready considered in Examples 2.6 and 12.21. It realizes the subdivisionrule shown in Figure 12.10 and Figure 12.11.

Note that f(z) = τ(z3), where τ(w) = 1 + (ω − 1)/w is a Mobiustransformation that maps the upper half-plane to the half-plane above


ω

1

C0

ω

1

C1

ω

1

C2

ω

1

C3

ω

1

C4

ω

1

C

Figure 14.1. The invariant curve for Example 14.6.

14. INVARIANT CURVES 303

the line through the points ω and 1 (indeed, τ maps 0, 1,∞ to∞, ω, 1,respectively). We have crit(f) = 0,∞ and post(f) = ω, 1,∞.

One can obtain an f -invariant Jordan curve C ⊂ S2 with post(f) ⊂C as follows. We start by choosing a Jordan curve C0 ⊂ S2 containingall postcritical points of f . Here we let C0 be the (extended) line

through ω and 1 (i.e., the circle on C through ω, 1,∞).Now consider f−1(C0) =

⋃k=0,...,5Rk, where

Rk = reikπ/3 : 0 ≤ r ≤ ∞is the ray from 0 through the sixth root of unity eikπ/3; see the topright in Figure 14.1. We choose a Jordan curve C1 ⊂ S2 such that

C1 ⊂ f−1(C0), post(f) ⊂ C1, and C1 is isotopic to C0 rel. post(f).

For general Thurston maps a similar choice is not always possible, butin our specific case there is a unique Jordan curve C1 ⊂ f−1(C0) withpost(f) ⊂ C1, namely C1 = R0∪R4, the union of the two rays through ωand through 1. Since # post(f) = 3, the requirement that C1 is isotopicto C0 rel. post(f) is automatic for our specific map f by Lemma 11.10.Let H : S2 × I → S2 be an isotopy rel. post(f) that deforms C0 to C1,i.e., H0 = idS2 and H1(C0) = C1.

Given the data C0, C1, and H, there are two (essentially equiva-lent) ways to obtain an f -invariant Jordan curve isotopic to C1 rel.f−1(post(f)) and hence also isotopic to C0 rel. post(f).

For the first approach we consider the Thurston map f := H1 f .Since C1 ⊂ f−1(C0) we have f(C1) ⊂ C0, and so

f(C1) = (H1 f)(C1) ⊂ H1(C0) = C1.

Thus C1 is f -invariant. The two-tile subdivision rule given by D1 =

D1(f , C1), D0 = D0(f , C1), and the labeling induced by f is as in

Figure 12.11. The map f is combinatorially expanding for C1; indeed,

no 2-tile for (f , C1) joins opposite sides of C1. Thus by Theorem 13.2there is a homeomorphism φ : S2 → S2 isotopic to the identity on

S2 rel. post(f) = post(f) such that φ(C1) = C1 and g = φ f isexpanding. Since f is also expanding (as follows from Proposition 2.3)and g is Thurston equivalent to f , there is a homeomorphism h : S2 →S2 such that h f = g h (Theorem 11.4). Then C := h−1(C1) isan f -invariant Jordan curve containing post(f). The general existenceresult for invariant curves given by Theorem 14.4 is proved in the samefashion.

For the second approach we use Proposition 11.1 to lift H = H0

by the map f to an isotopy H1 with H10 = idS2 . Then we lift H1 to


an isotopy H2 with H20 = idS2 , etc. In this way, we find a sequence

of isotopies Hn and inductively define Cn+1 := Hn1 (Cn). We will see in

Proposition 14.19 that the sequence Cn of Jordan curves converges

in the Hausdorff sense to an f -invariant Jordan curve C containing allpostcritical points of f as desired. This is illustrated in Figure 14.1.

In our example the f -invariant Jordan curve C ⊂ C with post(f) ⊂C is in fact unique. To see this, note that since # post(f) = 3, every

such curve C is isotopic rel. post(f) to the curve C0 chosen above. Sowe can find an isotopy K : S2 × I → S2 rel. post(f) with K0 = idS2

and K1(C) = C0. By Proposition 11.1 we can lift K to an isotopy

K : S2 × I → S2 rel. f−1(post(f)) with K0 = idS2 and Kt f = f Kt

for t ∈ I. Then by Lemma 11.2 we have

C ′ := K1(C) ⊂ K1(f−1(C)) = f−1(K1(C)) = f−1(C0).

So C ′ is a Jordan curve in S2 with C ′ ⊂ f−1(C0) and post(f) ⊂ C ′.Since C1 is the unique such curve, we conclude C ′ = K1(C) = C1. In

particular, C is isotopic to C1 rel. f−1(post(f)) by the isotopy K. So

every f -invariant Jordan curve C with post(f) ⊂ C lies in the sameisotopy class rel. f−1(post(f)) as C1. Hence by Theorem 14.5 (which

we will prove momentarily) there is at most one such Jordan curve C.The uniqueness of C follows.

14.1. Existence and uniqueness of invariant curves

We now turn to establishing existence and uniqueness results for gen-eral Thurston maps. We start with uniqueness results.

Proof of Theorem 14.5. Let f : S2 → S2 be an expanding Thur-ston map, and suppose that C and C ′ are f -invariant Jordan curves inS2 that both contain the set post(f) and are isotopic rel. f−1(post(f)).We have to show that C = C ′.

Under the given assumptions there exists an isotopy H0 : S2 ×I → S2 rel. f−1(post(f)) with H0

0 = idS2 and H01 (C) = C ′. Since

post(f) ⊂ f−1(post(f)), the map H0 is also an isotopy rel. post(f).Hence by Proposition 11.1 we can find an isotopy H1 : S2 × I → S2

rel. f−1(post(f)) with H10 = idS2 and f H1

t = H0t f for all t ∈ I.

Repeating this argument, we obtain isotopies Hn : S2 × I → S2 rel.f−1(post(f)) with Hn

0 = idS2 and f Hn+1t = Hn

t f for all t ∈ I andn ∈ N0.

Claim: Hn1 (C) = C ′ for n ∈ N0. To see this, we use induction on n.

For n = 0 the claim is true by choice of H0.

14.1. EXISTENCE AND UNIQUENESS OF INVARIANT CURVES 305

Suppose that Hn1 (C) = C ′ for some n ∈ N0. Then Lemma 11.2 and

the identity f Hn+11 = Hn

1 f imply that

Hn+11 (f−1(C)) = f−1(Hn

1 (C)) = f−1(C ′).Since C and C ′ are f -invariant, we have the inclusions C ⊂ f−1(C) andC ′ ⊂ f−1(C ′). In particular,

C := Hn+11 (C) ⊂ Hn+1

1 (f−1(C)) = f−1(C ′)

is a Jordan curve contained in f−1(C ′). Moreover, C and C are isotopicrel. f−1(post(f)) (by the isotopy Hn+1). Since C and C ′ are isotopic

rel. f−1(post(f)) by our hypotheses, it follows that C ′ and C are alsoisotopic rel. f−1(post(f)). Both sets are contained in f−1(C ′).

Now f−1(C ′) is the 1-skeleton of the cell decomposition D1(f, C ′).This cell decomposition has the vertex set f−1(post(f)). Moreover,since f is expanding, post(f) ≥ 3, and so every tile in D1(f, C ′) hasat least three vertices. So the hypotheses of Lemma 11.12 are satisfied

and we conclude that C ′ = C = Hn+11 (C). The claim above follows.

Fix a visual metric on S2. Then the tracks of the isotopiesHn shrinkat an exponential rate as n → ∞ (Lemma 11.3). Since Hn

0 = idS2 , itfollows that Hn

1 → idS2 uniformly as n→∞. Since Hn1 (C) = C ′ for all

n ∈ N0 by the claim, we conclude C = C ′ as desired.

Corollary 14.7 (Inv. curves in a given isotopy class rel. post(f)).Let f : S2 → S2 be an expanding Thurston map and C ⊂ S2 be aJordan curve with post(f) ⊂ C. Then there are at most finitely many

f -invariant Jordan curves C ⊂ S2 with post(f) ⊂ C that are isotopicto C rel. post(f).

Proof. Let C be such an f -invariant Jordan curve. Then there ex-

ists an isotopy H : S2×I → S2 rel. post(f) with H0 = idS2 and H1(C) =

C. Lifting H we get an isotopy H : S2× I → S2 rel. f−1(post(f)) such

that H0 = idS2 and f Ht = Ht f for all t ∈ I. Since C is f -invariant,

we have C ⊂ f−1(C). So Lemma 11.2 implies that

H1(C) ⊂ H1(f−1(C)) = f−1(H1(C)) = f−1(C).

Hence C is isotopic rel. f−1(post(f)) to the Jordan curve H1(C) thatis contained in f−1(C). Any such Jordan curve is a union of edgesin the cell decomposition D1(f, C) (see the last part of the proof ofLemma 11.12). In particular, there are only finitely many distinctJordan curves contained in f−1(C). This implies that there are onlyfinitely many isotopy classes rel. f−1(post(f)) represented by curves

C satisfying the assumptions of the corollary. Since an f -invariant


Θ

R2

GM

CM

Figure 14.2. Invariant curves for the Lattes map g.

Jordan curve C ⊂ S2 with post(f) ⊂ C is unique in its isotopy classrel. f−1(post(f)) by Theorem 14.5, the statement follows.

Corollary 14.8. Let f : S2 → S2 be an expanding Thurston mapwith post(f) = 3. Then there are at most finitely many f -invariant

Jordan curves C ⊂ S2 with post(f) ⊂ C.

Proof. Pick a Jordan curve C ⊂ S2 with post(f) ⊂ C. Since we

have # post(f) = 3, by Lemma 11.10 every Jordan curve C ⊂ S2 with

post(f) ⊂ C is isotopic to C rel. post(f). The statement now followsfrom Corollary 14.7.

In contrast to the case # post(f) = 3, expanding Thurston maps fwith # post(f) ≥ 4 can have infinitely many distinct invariant curves.

Example 14.9 (Infinitely many invariant curves). Let g be theLattes map from Section 1.1. In the following it is advantageous to usereal notation and consider the maps A and Θ used in the definition ofg as in (1.1) as maps on R2. Then A(u) = 2u for u ∈ R2. Moreover, foru1, u2 ∈ R2 we have Θ(u1) = Θ(u2) if and only if u2 = ±u1 + γ for γ ∈Γ := Z2. Recall that the extended real line R (which is the boundary ofthe pillow) is g-invariant and contains −1, 0, 1,∞ = post(g) = Θ(Γ).Let S = ∂Q be the boundary of the square Q = [0, 1/2]2 ⊂ R2. Wedenote by G the square grid obtained as the union of the horizontaland vertical lines in R2 that pass through a point in 1

2Γ = 1

2Z2. Then

Θ|S is injective and Θ(S) = Θ(G) = R. So the g-invariant curve R isobtained as the image of G under Θ. One can obtain other g-invariantJordan curves by mapping other grids by Θ.


To explain this, we consider a (2 × 2)-matrix M ∈ SL2(Z) (hereSL2(Z) denotes as usual the set of 2 × 2-matrices with integer entriesand determinant 1). We identify M with the linear map u 7→ Mu onR2 induced by left-multiplication of u ∈ R2 (considered as a columnvector) by the matrix M . Define QM = M(Q), SM := ∂QM = M(S),and the corresponding grid GM = M(G). Since M is a linear mapand M(Γ) = Γ, it follows that for u1, u2 ∈ R2 we have Θ(M(u1)) =Θ(M(u2)) if and only if u1 = ±u2 + γ, where γ ∈ Γ. This implies

that Θ|SM is injective, and so CM := Θ(SM) ⊂ C is a Jordan curve.Moreover, Θ(GM) = Θ(SM). Since A M = M A, we have

A(GM) = A(M(G)) = M(A(G)) ⊂M(G) = GM ,

and so

g(CM) = g(Θ(GM)) = Θ(A(GM)) ⊂ Θ(GM) = CM .

Hence CM is g-invariant. Since Γ = M(Γ) ⊂ GM , we also havepost(g) = Θ(Γ) ⊂ Θ(GM) = CM . So CM is an g-invariant Jordancurve that contains the set post(g). An example of this construction isindicated in Figure 14.2. The curve CM is drawn in thick on the right.

The curve CM determines the grid GM uniquely; indeed, one obtainsgenerating vectors of the two lines in GM through 0 by locally liftingCM near Θ(0) = 0 ∈ post(g) ⊂ CM to 0 by the map Θ. The whole gridGM is obtained by translating these two lines by vectors in Γ.

This implies that the mapM ∈ SL2(Z) 7→ CM is four-to-one; indeed,if M,N ∈ SL2(Z), then, as we have seen, CM = CN if and only if GM =GN . On the other hand, GM = GN if and only if M−1 N ∈ SL2(Z)is one of the four rotations around 0 (by integer multiples of π/2)that preserve the grid G. In particular, there exist infinitely many

g-invariant Jordan curves C ⊂ C with post(g) ⊂ C.

Since the map g in the previous example is a flexible Lattes map,the behavior exhibited is far from generic. In fact we have the followingresult.

Theorem 14.10. Let f : C→ C be a rational Thurston map. Sup-pose that f is expanding and has a hyperbolic orbifold. Then there are at

most finitely many f -invariant Jordan curves C ⊂ C with post(f) ⊂ C.

Proof. If C ⊂ C is an f -invariant Jordan curve with post(f) ⊂ C,then we get an associated two-tile subdivision rule (D1,D0, L), whereD1 = D1(f, C), D0 = D0(f, C) and L : D1 → D0 is the labeling inducedby f (i.e., L(τ) = f(τ) ∈ D0 for τ ∈ D1). The number of cells in D1

and D0 can be bounded only depending on deg(f) and # post(f). In


particular, we have a uniform bound independent of C. This impliesthat among the two-tile subdivision rules obtained in such a way fromf -invariant curves C, there are only finitely many up to isomorphism(this was defined in the discussion before Lemma 12.11). Here we usethe strong notion of isomorphism where we require that the cell complexisomorphisms as in the definition of an isomorphism between two-tilesubdivision rules send positively-oriented flags to positively-orientedflags (see Remark 12.12 (ii)).

This allows us to pick a finite family F of such curves C such that the

associated two-tile subdivision rule of any f -invariant Jordan curve C ⊂C with post(f) ⊂ C is isomorphic to one associated two-tile subdivisionrules for a curve in F .

Let G be the family of all Mobius transformations ϕ : C→ C withϕ f = f ϕ. If ϕ ∈ G, then ϕ(post(f)) = post(f). Now f isexpanding and so # post(f) ≥ 3. Since Mobius transformation are

uniquely determined by images of three distinct points in C, this impliesthat each ϕ ∈ G is uniquely determined by the bijection it induces onpost(f). Since there are only finitely many such bijections, G consistsof finitely many elements.

Now let C ⊂ C be an arbitrary f -invariant Jordan curve with

post(f) ⊂ C. We claim that C is isotopic rel. post(f) to one of thefinitely many Jordan curves ϕ(C), where C ∈ F and ϕ ∈ G. Sinceeach isotopy class rel. post(f) contains only finitely many f -invariant

Jordan curves C ⊂ C with post(f) ⊂ C (Corollary 14.7), this claimimplies the theorem.

To prove the claim, we use that the two-tile subdivision rule associ-

ated with C is isomorphic to one of the two-tile subdivision rules asso-ciated with a curve C ∈ F . So by Lemma 12.11 and the Remarks 12.12

there exist orientation-preserving homeomorphism h0, h1 : C→ C that

are isotopic rel. post(f) such h0 f = f h1 and h0(C) = h1(C) = C.By Thurston’s Uniqueness Theorem (see Theorem 2.19) there exists a

Mobius transformation ϕ : C → C that is isotopic to h0 rel. post(f)

such that ϕ f = f ϕ. Then ϕ ∈ G, and ϕ(C) is isotopic to h0(C) = Crel. post(f) as desired.

We now turn to existence results. As the following example shows,for a Thurston map f : S2 → S2 an f -invariant Jordan curve C ⊂ S2

with post(f) ⊂ C need not exist.


10 ∞

−i

i

R0

R2

R6

Figure 14.3. No invariant Jordan curve C ⊃ post(f).

Example 14.11. Consider the map f : C→ C defined by

f(z) = iz4 − i

z4 + i

for z ∈ C. The critical points of f are 0 and ∞, and the map has thefollowing ramification portrait:

(14.1) 04:1// −i

)) 1

∞ 4:1// i

55

So the set of postcritical points of f is given by post(f) = −i , 1, i,and f is a Thurston map. This map is also expanding as follows fromProposition 2.3.

Lemma 14.12. Let f be the map from Example 14.11. Then there

is no f -invariant Jordan curve C ⊂ C with post(f) ⊂ C.

Proof. We have f(z) = ϕ(z4) for z ∈ C, where

(14.2) ϕ(w) = iw − i

w + i, w ∈ C,

is a Mobius transformation that maps the upper half-plane to the unitdisk (note that ϕ maps 0, 1,∞ to −i , 1, i , respectively). Let C := ∂Dbe the unit circle. Then

f−1(C) =⋃

k=0,...,7

Rk, where Rk = reikπ/4 : 0 ≤ r ≤ ∞.

The postcritical points −i , 1, i lie on distinct rays Rk. Two such rayshave the points 0 and∞ in common and no other points. Thus there isno Jordan curve in f−1(C) containing all postcritical points, see Figure14.3. As we will see in Theorem 14.4, the existence of such a Jordancurve is a necessary condition for the existence of an f -invariant Jordan


curve C ⊂ C with post(f) ⊂ C (in our specific case where # post(f) = 3the choice of C does not matter since all Jordan curves that containpost(f) are isotopic rel. post(f)). Hence there is no f -invariant Jordan

curve C ⊂ C with post(f) ⊂ C. One can also see this by a simpleargument directly.

Indeed, suppose that C ⊂ C is a Jordan curve with post(f) ⊂ C and

f(C) ⊂ C. The unit circle C = ∂D is also a Jordan curve containing the

set post(f). Hence by Lemma 11.10 there exists an isotopy H : C×I →C rel. post(f) such that H0 = idC and H1(C) = C.

By Proposition 11.1 the isotopy H can be lifted to an isotopy

H : C× I → C rel. post(f) such that H0 = idC and Ht f = f Ht forall t ∈ I.

Since C ⊂ f−1(C), it follows from Lemma 11.2 that

H1(C) ⊂ H1(f−1(C)) = f−1(H1(C)) = f−1(C).

This means that the Jordan curve C ′ := H1(C) is contained in f−1(C).Moreover, it contains all postcritical points, since C does, and the points

in post(f) stay fixed under the isotopy H. As we have seen above, nosuch Jordan curve exists and we get a contradiction as desired.

By a similar (though somewhat lengthier) argument one can showthat the Lattes map f(z) = i

2(z + 1/z) does not have an f -invariant

Jordan curve C with post(f) ⊂ C (this map was considered in Exam-ple 3.24). Another such example can be found in [CFP10, Section4].

We now turn to the proof for the necessary and sufficient criterionfor the existence of an invariant Jordan curve as formulated in Theo-rem 14.4. Note that in condition (ii) of this theorem the requirement

on f is meaningful. Indeed, f = H1 f is a branched covering map of

S2 with crit(f) = crit(f). Since H1 is isotopic to idS2 rel. post(f) we

have post(f) = post(f), and so f is a Thurston map.

Furthermore, C ′ is a Jordan curve with post(f) = post(f) ⊂ C ′.Since C ′ = H1(C) ⊂ f−1(C), we have that f(C ′) = (H1 f)(C ′) ⊂H1(C) = C ′. Hence C ′ is invariant with respect to f , and it makes sense

to require that f is combinatorially expanding for C ′.

Proof of Theorem 14.4. (i) ⇒(ii) Suppose C is as in (i). Then

in (ii) we let C = C ′ = C, and the isotopy H be such that Ht =

idS2 for all t ∈ I. Then C ′ = C ⊂ f−1(C) = f−1(C), and f = f is


combinatorially expanding for the invariant curve C ′ = C, since f isexpanding.

(ii)⇒(i) Let C, C ′, H, and f be as in (ii), and define χ = H1. As we

have seen in the discussion before the proof, f is a Thurston map with

post(f) = post(f), and C ′ is an f -invariant Jordan curve containing

the set post(f) = post(f).

Since f is combinatorially expanding for C ′, Theorem 13.2 impliesthat there exists a homeomorphism φ : S2 → S2 that is isotopic to

the identity rel. post(f) = post(f) such that φ(C ′) = C ′ and g =

φ f is an expanding Thurston map. Since g = (φ χ) f , andφ χ is isotopic to the identity on S2 rel. post(f), the maps f and

g are Thurston equivalent. If notation is as in (2.7) (with S2 = S2),then we can take h0 = φ χ and h1 = idS2 . By Theorem 11.4 wecan find a homeomorphism h : S2 → S2 that is isotopic to h1 = idS2

rel. f−1(post(f)) with hf = gh. Note that then the homeomorphismh−1 is also isotopic to idS2 rel. f−1(post(f)) and we have f h−1 =h−1 g.

Let C = h−1(C ′). Then C is a Jordan curve in S2 that is isotopic toC ′ rel. f−1(post(f)), and hence isotopic to C rel. post(f); in particular,

C contains the set post(f). Moreover, C is f -invariant, because we have

f(C) = f(h−1(C ′)) = h−1(g(C ′)) = h−1(φ(f(C ′)))⊂ h−1(φ(C ′)) = h−1(C ′) = C.

The proof is complete.

Remarks 14.13. (i) The condition of combinatorial expansion incondition (ii) of Theorem 14.4 is combinatorial in nature and can easilybe checked in principle. A simple sufficient criterion for this can beformulated as follows: if no 1-tile for (f, C) joins opposite sides of C ′,then f is combinatorially expanding for C ′.

To see this, note that

f−1(C ′) = f−1(H−11 (C ′)) = f−1(C).

By Proposition 5.17 (v) this implies that the 1-tiles for (f , C ′) are pre-cisely the 1-tiles for (f, C). Hence if no 1-tile for (f, C) joins opposite

sides of C ′, then D1(f , C ′) ≥ 2 and so f is combinatorially expandingfor C ′. We will later formulate a necessary and sufficient condition for

combinatorial expansion of f (see Proposition 14.18).

(ii) The condition of combinatorial expansion in condition (ii) ofTheorem 14.4 is independent of the chosen isotopy H. More precisely,


suppose H1, H2 : S2 × I → S2 are two isotopies with H10 = H2

0 =

idS2 and H11 (C) = H2

1 (C) = C ′. Then f1 = H11 f is combinatorially

expanding for C ′ if and only if f2 = H21 f is combinatorially expanding

for C ′. This follows immediately from Lemma 12.14 (with f = f1,

g = f2, h0 = H21 (H1

1 )−1, h1 = idS2 , and C = C ′).(iii) Theorem 14.4 can be slightly modified to give necessary and

sufficient conditions for the existence of an invariant curve in a givenisotopy class rel. post(f) or rel. f−1(post(f)). An existence statementfor a given isotopy class rel. f−1(post(f)) is especially relevant in viewof the complementary uniqueness statement given by Theorem 14.5.

To formulate this precisely, let C ⊂ S2 be a given Jordan curve

with post(f) ⊂ C. Then an f -invariant Jordan curve C ⊂ S2 isotopic

to C rel. post(f) exists if and only if condition (ii) in Theorem 14.4 is

true for a Jordan curve C isotopic to C rel. post(f). This immediatelyfollows from the proof of this theorem.

Similarly, an f -invariant Jordan curve C ⊂ S2 isotopic to C rel.f−1(post(f)) exists if and only if condition (ii) in Theorem 14.4 is true

with the additional requirement that C ′ is isotopic to C rel. f−1(post(f)).

For the proof of Theorem 14.1 we require the following auxiliaryresult.

Lemma 14.14. Let f : S2 → S2 be an expanding Thurston map, andC ⊂ S2 be a Jordan curve with post(f) ⊂ C. Then for all sufficientlylarge n there exists a Jordan curve C ′ ⊂ f−n(C) that is isotopic to Crel. post(f). Moreover, C ′ can be chosen so that no n-tile for (f, C)joins opposite sides of C ′.

Proof. We fix some base metric on S2. Let P := post(f). Sincef is expanding, we have k := #P = # post(f) ≥ 3 by Corollary 7.2.Pick ε0 > 0 as in Lemma 11.17. Since f is expanding, for large enoughn we have

mesh(f, n, C) = maxc∈Dn(f,C)

diam(c) < ε0.

For such n consider the cell decomposition D = Dn(f, C) of S2. Itsvertex set is the set f−n(post(f)) ⊃ post(f) = P of n-vertices andits 1-skeleton is the set f−n(C). Hence by Lemma 11.17 there exists aJordan curve C ′ ⊂ f−n(C) that is isotopic to C rel. P = post(f) and sothat no tile in D, i.e., no n-tile for (f, C), joins opposite side of C ′.

Proof of Theorem 14.1. Let f and C be as in the statementof the theorem. By Lemma 14.14 for sufficiently large n ∈ N thereexists an isotopy H : S2× I → S2 rel. post(f) such that H0 = idS2 and


C ′ := H1(C) ⊂ f−n(C) and such that no n-tile for (f, C) joins oppositesides of C ′.

If we define F = fn for such n, then the map F is an expandingThurston map with post(F ) = post(f). The sets C and C ′ are Jordancurves with post(F ) ⊂ C, C ′, and H is an isotopy rel. post(F ) thatdeforms C into C ′ ⊂ f−n(C) = F−1(C). Moreover, F = fn is cellularfor (Dn(f, C),D0(f, C)). Hence by Lemma 5.15 we have D1(F, C) =Dn(f, C), and so the 1-cells for (F, C) are precisely the n-cells for (f, C).So no 1-tile for (F, C) joins opposite side of C ′ and by Remark 14.13 (i)the map H1 F is combinatorially expanding for C ′. This shows thatcondition (ii) in Theorem 14.4 is satisfied. Hence there exists a Jordan

curve C ⊂ S2 that is F -invariant and is isotopic to C rel. post(F ) =post(f) as desired.

Remark 14.15. In general, the fn-invariant Jordan curve C as in

Theorem 14.1 will depend on n, and one cannot expect that C is in-variant for all sufficiently high iterates of f . To illustrate this, considerthe map f from Example 14.11 (see also Lemma 14.12). Recall that

f(z) = ϕ(z4) for z ∈ C, where ϕ is as in (14.2).

The Mobius transformation ϕ maps the extended real line R to the

unit circle ∂D, and ∂D to R. This implies the unit circle C := ∂Dsatisfies f 2n(C) ⊂ C for every n ∈ N. Note that post(f) = −i , 1, i ⊂C. Thus C is a Jordan curve with post(f) ⊂ C that is invariant forevery even iterate f 2n.

On the other hand, for n ∈ N0 we have f 2n+1(∂D) ⊂ R. Since f 2n+1

is a finite-to-one map, the set f 2n+1(∂D) is infinite, and so we cannot

have f 2n+1(∂D) ⊂ ∂D (for otherwise, f 2n+1(∂D) ⊂ ∂D∩ R = −1, 1).Thus the unit circle ∂D = C is not invariant for any odd iterate of f .

Proof of Corollary 14.2. Let f : S2 → S2 be an expandingThurston map. It follows from Theorem 14.1 that for each sufficiently

large n ∈ N there exists an fn-invariant Jordan curve C ⊂ S2 with

post(f) = post(fn) ⊂ C. For such n let F = fn, D0 = D0(F, C), and

D1(F, C). Define the orientation-preserving labeling L : D1 → D0 byL(c) = F (c) for c ∈ D1. Then it is clear that (D1,D0, L) is a two-tile subdivision rule that is realized by F (see Definition 12.1 and thefollowing discussion).


14.2. Iterative construction of invariant curves

Given data as in Theorem 14.4 (ii), the f -invariant curve C can beobtained by an iterative procedure. To explain this, we first recall thedefinition of Hausdorff convergence of sets.

Let (X, d) be a compact metric space. If A,B ⊂ X are subsets ofX, then their Hausdorff distance is defined as

(14.3) distHd (A,B) = infδ > 0 : A ⊂ N δd (B) and B ⊂ N δ

d (A).Assume A and An for n ∈ N are closed subsets of X. We say thatAn → A as n→∞ in the sense of Hausdorff convergence if

limn→∞

distHd (An, A) = 0.

Note that in this case a point x ∈ X lies in A if and only if thereexists a sequence xn of points in X such that xn ∈ An for n ∈ N andxn → x as n→∞.

Now let f : S2 → S2 be an expanding Thurston map, and assume asin Theorem 14.4 (ii) that C, C ′ ⊂ S2 are Jordan curves with post(f) ⊂C, C ′ and C ′ ⊂ f−1(C), and that H : S2 × I → S2 is an isotopy rel.post(f) that deforms C to C ′, i.e., H0 = idS2 and H1(C) = C ′. For the

moment we do not assume that the map f = H1 f is combinatoriallyexpanding for C ′.

Let H0 := H. Using Proposition 11.1 repeatedly, we can find iso-topies Hn : S2 × I → S2 rel. f−1(post(f)) such that Hn

0 = idS2 andf Hn

t = Hn−1t f for all n ∈ N, t ∈ I. Now define Jordan curves

inductively by setting C0 := C, and Cn+1 := Hn1 (Cn) for n ∈ N0. Note

that then C1 = C ′.To summarize, we start with the following data for our given map

f :

(i) a Jordan curve C0 = C ⊂ S2 with post(f) ⊂ C0,

(ii) a Jordan curve C1 = C ′ ⊂ S2 isotopic to C0 ⊂ S2 rel. post(f)with C1 ⊂ f−1(C0),

(iii) an isotopy H0 : S2 × I → S2 rel. post(f) such that H00 = idS2

and H01 (C0) = C1.

We then define inductively:

(i) isotopies Hn : S2× I → S2 such that Hn0 = idS2 and f Hn

t =Hn−1t f for all n ∈ N, t ∈ I,

(ii) Jordan curves Cn+1 := Hn1 (Cn) for n ∈ N0.

Figure 14.1 illustrates this procedure for Example 14.6. Since thisexample is rather complicated and it is hard to grasp the involvedisotopies, we present a simpler example for the construction.

14.2. ITERATIVE CONSTRUCTION OF INVARIANT CURVES 315

H01

H11

. . .

C0 C1

C1 C2

C2 CFigure 14.4. Iterative construction of an invariant curve.


Example 14.16. Let f be a Lattes map obtained as in (1.1) (withg = f), where we choose

A : C→ C, u 7→ A(u) := 5u.

It is straightforward to check that the extended real line C := R =R ∪ ∞ is f -invariant and contains all postcritical points 0, 1,∞,−1of f .

As in Figure 1.1 we represent the sphere as a pillow, i.e., two squaresglued together along their boundary. The boundary of the pillow (i.e.,the boundary along which the two squares were glued together) repre-sents the curve C, the two squares represent the 0-tiles, one of which iscolored white, the other black.

The map f can then be described as follows. Each of the twosides of the pillow is divided into 5 × 5 squares, which are colored ina checkerboard fashion. The map f acts by mapping each small whitesquare to the white side of the pillow, and each small black square tothe black side. The two sides of the pillow are the 0-tiles with respect toC; the 4 vertices of the pillow are the postcritical points in this model.The small squares are the 1-tiles (for (f, C)). The coloring of the 0-and 1-tiles corresponds to a labeling map LX as in Lemma 5.21.

There exist f -invariant Jordan curves that are isotopic to C rel.post(f), but distinct from C. The construction of one such curve isillustrated in Figure 14.4. Namely, we set C0 := C. The Jordan curveC1 is shown in the top right, as well as in the middle left picture. Inthe latter picture, we see that C1 consists of 1-edges, i.e., C1 ⊂ f−1(C0).Moreover, there exists an isotopy H0 : S2 × I → S2 rel. post(f) thatdeforms C0 to C1 (i.e., H0

0 = idS2 and H01 (C0) = C1). We also see here

how the black and the white 0-tile are deformed by H01 ; namely, the

four small black squares in the top right of Figure 14.4 are part of theimage of the black 0-tile (which is at the back of the pillow) under H0

1 .The Jordan curve C2 := H1

1 (C1) consists of 2-edges, i.e., C2 ⊂f−2(C0), see the bottom left.

The two pictures in the middle of Figure 14.4 indicate how H1

deforms 1-tiles. Roughly speaking H1 deforms each black/white 1-tile“in the same fashion” as H0 deforms the black/white 0-tiles.

The curves Cn Hausdorff converge to C, which is a f -invariant Jor-

dan curve with post(f) ⊂ C (see Lemma 14.17 (viii) and Proposition14.19).

There is a conceptually different way to obtain Cn+1 from Cn, whichwill be explained in detail in Remark 14.21. Loosely speaking, wereplace each n-edge αn ⊂ Cn by (n + 1)-edges “in the same fashion”


as the 0-edge α0 := fn(αn) ⊂ C0 is replaced by the arc β1 ⊂ C1 withthe same endpoints as α0 (which are postcritical points). Note thatβ1 = H0

1 (α0), and that β1 consists of 1-edges.

To prepare the proof that under suitable conditions our iterationprocess has an invariant Jordan curve as a Hausdorff limit, we summa-rize some properties of the Jordan curves Cn.

Lemma 14.17. Let f : S2 → S2 be an expanding Thurston map,and the Jordan curves Cn for n ∈ N0 be defined as above. Then thefollowing statements are true:

(i) Cn+k ⊂ f−k(Cn) for n, k ∈ N0.

(ii) Cn+k is isotopic to Cn rel. f−n(post(f)) for n, k ∈ N0.

(iii) Cn+k ∩ f−n(post(f)) = Cn ∩ f−n(post(f)) for n, k ∈ N0.

(iv) post(f) ⊂ Cn for n ∈ N0.

(v) For n, k ∈ N0 the curve Cn+k consists of n-edges for (f, Ck).

(vi) For n ∈ N the curve Cn is the unique Jordan curve in S2 withCn ⊂ f−1(Cn−1) that is isotopic to C1 rel. f−1(post(f)).

(vii) The sequence Cn, n ∈ N0, only depends on C0 and C1 andnot of the choice of the initial isotopy H = H0 used in thedefinition of the sequence.

(viii) As n → ∞ the sets Cn Hausdorff converge to a closed f -

invariant set C ⊂ S2 with post(f) ⊂ C.

Proof. In the following we use the isotopies Hn as in the definitionof the sequence Cn, and set hn = Hn

1 for n ∈ N0.

(i) It suffices to show that Cn ⊂ f−1(Cn−1) for n ∈ N. We provethis by induction on n; this is clear for n = 1. Assume that thestatement holds for some n ∈ N; so Cn ⊂ f−1(Cn−1). Since hn = Hn

1

and hn−1 = Hn−11 are homeomorphisms with f hn = hn−1 f , we have

hn(f−1(Cn)) = f−1(hn−1(Cn)) by Lemma 11.2. Thus

Cn+1 = hn(Cn) ⊂ hn(f−1(Cn−1)) = f−1(hn−1(Cn−1)) = f−1(Cn),

and (i) follows.

(ii)–(iv) From the definition of Hn, the remark after the proof ofProposition 11.1, and induction on n we conclude that Hn is an isotopyrel. f−n(post(f)). Since Hn

0 = idS2 and

f−n(post(f)) ⊂ f−(n+k)(post(f))

for n, k ∈ N0, statements (ii) and (iii) immediately follow from this byinduction on k for fixed n. Statement (iv) follows from (iii) (with n = 0and k ∈ N0 arbitrary) and the fact that post(f) ⊂ C0.


(v) By (iv) we have #(f−n(post(f)) ∩ Cn+k) ≥ # post(f) ≥ 3. Inparticular, the points in f−n(post(f)) that lie on Cn+k subdivide thiscurve into arcs whose endpoints lie in f−n(post(f)) and whose inte-riors are disjoint from f−n(post(f)). Let α ⊂ Cn+k be one of thesearcs. Then we have int(α) ⊂ f−n(Ck) \ f−n(post(f)) by (i), and∂α ⊂ f−n(post(f)). Since by Proposition 5.17 (iii) the set f−n(Ck)is the 1-skeleton and the set f−n(post(f)) the 0-skeleton of the cell de-composition Dn(f, Ck), we conclude from Lemmas 5.4 and 5.5 that αis an edge in Dn(f, Ck), i.e., an n-edge for (f, Ck). Hence Cn+k consistsof n-edges for (f, Ck).

(vi) By (i) and (ii) we know that Cn for n ∈ N is a Jordan curvewith Cn ⊂ f−1(Cn−1) that is isotopic to C1 rel. f−1(post(f)). Let

C ⊂ f−1(Cn−1) be another Jordan curve isotopic to C1 rel. f−1(post(f)).

Then Cn and C are isotopic to each other rel. f−1(post(f)). Note thatf−1(Cn−1) is the 1-skeleton of the cell decomposition D1(f, Cn−1) and

f−1(post(f)) its set of vertices. Hence by Lemma 11.12 we have C = Cn,and the uniqueness statement for Cn follows.

(vii) It follows from (vi) and induction on n that Cn is uniquelydetermined by C0 and C1.

(viii) Pick a visual metric % for f , and let Λ > 1 be the expansionfactor of %. By Lemma 11.3 the diameters of the tracks of the isotopyHn are bounded by CΛ−n, where C is a fixed constant. Since Hn

0 = idS2

and Cn+1 = Hn1 (Cn) for n ∈ N0 this implies that distH% (Cn, Cn+1) ≤

CΛ−n for n ∈ N0. It follows that the sequence Cn is a Cauchysequence with respect to Hausdorff distance. Now the space of all non-empty closed subsets of a compact metric space is complete if it isequipped with the Hausdorff distance. Hence there exists a non-empty

closed set C ⊂ S2 such that Cn → C as n→∞ in the Hausdorff sense.Since post(f) ⊂ Cn for all n ∈ N0 by (iv), we have post(f) ⊂ C.

It remains to show that C is f -invariant. To see this, let p ∈ C bearbitrary. Then there exists a sequence pn of points in S2 such thatpn ∈ Cn for n ∈ N0 and pn → p as n→∞. By continuity of f we havef(pn) → f(p) as n → ∞. Moreover, (i) implies that f(pn) ∈ Cn−1 for

n ∈ N. Hence f(p) ∈ C, and so the set C is indeed f -invariant.

As an application of the preceding setup we prove a statement that

gives a necessary and sufficient condition for the map f in Theorem 14.4to be combinatorially expanding.


Proposition 14.18. Let f : S2 → S2 be an expanding Thurstonmap and the isotopy H0 : S2×I → S2 and Jordan curves Cn for n ∈ N0

be defined as above.

Then f = H01 f is combinatorially expanding for C1 = C ′ if and

only if there exists n ∈ N such that no n-tile for (f, C0) joins oppositesides of Cn.

Proof. Let Hn for n ∈ N0 be the isotopies used in the definition

of the curves Cn. Set hn := Hn1 . Then f = h0 f , Cn+1 = hn(Cn), and

hn f = f hn+1 for n ∈ N0. It follows by induction that for n ∈ N wehave

fn = h0 f · · · h0 f = h0 fn hn−1 · · · h1,

and so

h0 fn = fn h−11 . . . h−1

n−1.

Hence

f−n(C0) = f−n(h−10 (C1)) = (hn−1 · · · h1)(f−n(C1)).

Recall that the n-tiles for (f, C0) are the closures of the complemen-

tary components of f−n(C0), and the n-tiles for (f , C1) the closures of

the complementary components of f−n(C1) (Proposition 5.17 (v)). Sofrom the previous identity we conclude that the n-tiles for (f, C0) are

precisely the images of the n-tiles for (f , C1) under the homeomorphismhn−1 · · · h1. Note that this homeomorphism is isotopic to idS2 rel.

post(f) = post(f) and maps C1 to Cn. Thus no n-tile for (f , C1) joinsopposite sides of C1 if and only if no n-tile for (f, C0) joins oppositesides of Cn.

Now f is combinatorially expanding for C1 if and only if there exists

n ∈ N such that no n-tile for (f , C1) joins opposite sides of C1. By whatwe have seen, this is the case if and only if there exists n ∈ N such thatno n-tile for (f, C0) joins opposite sides of Cn.

The Hausdorff limit C of the curves Cn as provided by Lemma 14.17 (viii)will not be a Jordan curve in general, see Example 14.22. The follow-

ing proposition shows that C is a Jordan curve if the map f = H01 f

is combinatorially expanding for C1. Actually, one can also show that

this condition is also necessary for C to be a Jordan curve, but we willnot present the proof for this statement as it is somewhat involved.

Proposition 14.19 (Iterative procedure to obtain an invariantcurve). Let f : S2 → S2 be an expanding Thurston map and the isotopyH0 : S2× I → S2 and Jordan curves Cn for n ∈ N0 be defined as above.


If f = H01 f is combinatorially expanding for C1 = C ′, then Cn

Hausdorff converges to a Jordan curve C ⊂ S2 as n→∞. In this case,

the curve C is f -invariant and post(f) ⊂ C. Moreover, C is isotopic toC1 rel. f−1(post(f)).

By Theorem 14.5 the curve C will be the unique Jordan curve withthe given properties.

Proof. Suppose that f = H01 f is combinatorially expanding

for C1. From Theorem 14.4 it follows that there exists an f -invariant

Jordan curve C ⊂ S2 with post(f) ⊂ C that is isotopic to C0 rel. post(f)and isotopic to C1 rel. f−1(post(f)). Let K0 : S2×I → S2 be an isotopy

rel. post(f) that deforms C to C0; so K00 = idS2 and K0

1(C) = C0. UsingProposition 11.1 repeatedly, we can find isotopies Kn : S2×I → S2 rel.f−1(post(f)) with Kn

0 = idS2 such that f Kn1 = Kn−1

1 f for n ∈ N.

Claim. Cn := Kn1 (C) = Cn for all n ∈ N0.

We prove this claim by induction on n; it follows from the choiceof K0 for n = 0. Assume that the statement is true for some n ∈ N0.Then Kn

1 (C) = Cn, and so by Lemma 11.2 we have

Cn+1 = Kn+11 (C) ⊂ Kn+1

1 (f−1(C)) = f−1(Kn1 (C)) = f−1(Cn).

Since Kn+1 is an isotopy rel. f−1(post(f)), the curve Cn+1 is isotopic

to C and hence to C1 rel. f−1(post(f)). So Lemma 14.17 (vi) implies

that Cn+1 = Cn+1 proving the claim.By Lemma 11.3 the maps Kn

1 converge uniformly to the identity on

S2. Hence Cn = Kn1 (C) Hausdorff converges to the Jordan curve C as

n→∞. The statement follows.

Remark 14.20. If f : S2 → S2 is an expanding Thurston map, then

every f -invariant Jordan curve C with post(f) ⊂ C can be obtained

by our iterative procedure. Indeed, suppose that C is such a curve.

Trivially, we can then take C = C0 = C, C ′ = C1 = C, and H0t = idS2

for t ∈ I. Then Cn = C for all n ∈ N0 and so Cn → C as n→∞.Actually, a much stronger statement is true. Namely, we can start

with any Jordan curve C in the same isotopy class rel. post(f) as C.Suppose that C is such a curve. First, we claim that then there exists a

unique Jordan curve C ′ ⊂ f−1(C) that is isotopic to C rel. f−1(post(f)).To see this, let K0 : S2 × I → S2 be an isotopy rel. post(f) with

K00 = idS2 and K0

1(C) = C. By Proposition 11.1 we can lift K0 by f toan isotopy K1 rel. f−1(post(f)) with K1

0 = idS2 and K0t f = f K1

t


for all t ∈ I. Then the Jordan curve C ′ := K11(C) satisfies

C ′ = K11(C) ⊂ K1

1(f−1(C)) = f−1(K01(C)) = f−1(C).

Here we used C ⊂ f−1(C) and Lemma 11.2. This shows existence ofa curve C ′ with the desired properties. Uniqueness of C ′ follows fromLemma 11.12.

Define H : S2 × I → S2 by setting Ht = K1t (K0

t )−1 for t ∈ I.Then H is an isotopy rel. post(f) that deforms C0 := C into C1 := C ′.Indeed, we have H0 = idS2 and

H1(C0) = K11((K0

1)−1(C)) = K11(C) = C ′ = C1.

Moreover,

f := H1 f = K11 (K0

1)−1 f = K11 f (K1

1)−1.

Thus it follows from Lemma 12.14 that f is combinatorially expanding

for C1 = C ′ = K11(C).

Define the sequence Cn starting from C0 and C1 as before. FromProposition 14.19 it follows that as n → ∞ the curves Cn Hausdorffconverge to an f -invariant Jordan curve that is isotopic to C1, and

hence isotopic to C, rel. f−1(post(f)). From Theorem 14.5 it follows

that the unique such curve is C. Thus Cn → C in the Hausdorff senseas n→∞.

Remark 14.21. In the inductive definition of Cn+1 = Hn1 (Cn) one

can construct Cn+1 from Cn by an edge replacement procedure withoutexplicitly knowing the isotopy Hn. To explain this, suppose that n ∈ N,and that Cn has already been constructed (starting from given curvesC0 and C1). We know by Lemma 14.17 (v) that Cn consists of n-edgesαn for (f, C0). Then Cn+1 is obtained from Cn by replacing each n-edgeαn ⊂ Cn by a certain arc βn+1 with the same endpoints as αn.

Indeed, we can set βn+1 := Hn1 (αn) ⊂ Cn+1. Then the union of

these arcs βn+1 is equal to Cn+1. Moreover, since Hn is an isotopyrelative to the set f−n(post(f)) of n-vertices, and αn is an n-edge for(f, C0) and so has n-vertices as endpoints, the arcs αn and βn+1 havethe same endpoints.

Now the arc βn+1 is the unique arc in f−n(C1) that is isotopic to αn

rel. f−n(post(f)). This property often allows one to determine βn+1

from αn without knowing Hn explicitly.To see that this characterization of βn+1 holds, note that by Lem-

ma 14.17 (i) we have βn+1 ⊂ Cn+1 ⊂ f−n(C1). Moreover, βn+1 =Hn

1 (αn) is isotopic to αn rel. f−n(post(f)).


α0 ⊂ C0

H01

Hn1

fn|αn

αn ⊂ Cn βn+1 ⊂ Cn+1

fn|βn+1

β1 ⊂ C1

Figure 14.5. Iterative construction by replacing edges.

Suppose βn+1 ⊂ f−n(C1) is another arc that is isotopic to αn

rel. f−n(post(f)). Then the arcs βn+1 and βn+1 have endpoints inf−n(post(f)), but contain no other points in this set, since this is true

for αn. This and the inclusions βn+1, βn+1 ⊂ f−n(C1) imply that βn+1

and βn+1 are n-edges for (f, C1) (see the argument in the proof of

Lemma 14.17 (v)). Since βn+1 and βn+1 are isotopic relative to the setf−n(post(f)), which is the 0-skeleton of Dn(f, C1), it follows from the

first part of the proof of Lemma 11.12 that βn+1 = βn+1 as desired.

As we have just seen, βn+1 is an n-edge for (f, C1). Since βn+1

has endpoints in the set f−n(post(f)) ⊂ f−(n+1)(post(f)) and βn+1 ⊂Cn+1 ⊂ f−(n+1)(C0), a similar argument also shows that βn+1 consistsof (n+ 1)-edges for (f, C0).

One can look at the arc replacement procedure αn → βn+1 fromyet another point of view. Since αn is an n-edge for (f, C0), the mapfn|αn is a homeomorphism of αn onto the 0-edge α0 := fn(αn) ⊂ C0 for(f, C0) (Proposition 5.17 (i)). The endpoints of α0 lie in post(f). Thenβ1 := H0

1 (α0) is the unique subarc of C1 that has the same endpointsas α0, but contains no other points in post(f) (here it is importantthat #(C1 ∩ post(f)) = # post(f) ≥ 3). Since fn Hn

1 = H01 fn

and βn+1 = Hn1 (αn), the map fn|βn+1 is thus a homeomorphism of

βn+1 onto β1. Often, this information (together with the fact that αn

and βn+1 share endpoints) is enough to determine βn+1 uniquely. Weillustrate this procedure in Figure 14.5. Here the map f (as well asthe curves C0, C1, . . . and the isotopies H0, H1, . . . ) are as in Example14.16, see also Figure 14.4.

For example, suppose that β1 lies in a single 0-tile X0 for (f, C0),i.e., in one of the Jordan regions bounded by C0. This is not always


true, but in Example 14.16 as well as the Examples 14.22 and 14.23discussed below this is the case. Then there exists a unique n-tile Xn

for (f, C0) with αn ⊂ ∂Xn and fn(Xn) = X0; if we assign colors to tilesfor (f, C0) as in Lemma 5.19, then Xn is the unique n-tile for (f, C0)that contains αn in its boundary and has the same color as X0.

Consider the arc βn+1 := (fn|Xn)−1(β1) ⊂ Xn. Then βn+1 has thesame endpoints as (fn|Xn)−1(α0) = αn and is contained in f−n(C1).

Moreover, βn+1 is isotopic to αn rel. f−n(post(f)); this easily followsfrom Lemma 11.8, since our assumptions imply that one can find a

suitable simply connected domain Ω ⊂ S2 that contains βn+1 and αn

and no point in f−n(post(f)) except the endpoints of βn+1 and αn. By

what we have seen above, we conclude βn+1 = βn+1, and so

(14.4) βn+1 = (fn|Xn)−1(β1).

In the special case under consideration, this leads to a very conve-nient edge replacement procedure that can be summarized as follows:Suppose the arc β1 ⊂ C1 corresponding to α0 = fn(αn) ⊂ C0 lies in asingle 0-tile X0, and let Xn be the n-tile that contains αn its boundaryand has the same color as X0 (so that fn(Xn) = X0). Then αn isreplaced by the arc βn+1 in Xn that corresponds to β1 ⊂ X0 under thehomeomorphism fn|Xn of Xn onto X0.

The next example illustrates what happens if the map f in Propo-sition 14.19 is not combinatorially expanding.

Example 14.22. Let g be a Lattes map obtained as in (1.1), where

A : C→ C, u 7→ A(u) := 3u.

This map was already considered in Example 13.21. See also the bot-tom of Figure 13.2. We let C0 be the boundary of the pillow. Thecurve C1 ⊂ g−1(C0) is drawn with a thick line in the top left of Figure14.6. Clearly there is an isotopy H0 rel. post(g) (the four vertices ofthe pillow) that deforms C0 to C1. Note that g = H0

1 g is not com-binatorially expanding for C1, see Figure 13.2. Starting with the dataC0, C1, H0, we inductively define Jordan curves Cn as described before.

It is slightly more difficult than in Example 14.16 to see here howCn+1 evolves from Cn, since different 0-edges are replaced by differentarcs (consisting of 1-edges). Each 0-edge that is drawn horizontallyin Figure 14.6 is replaced by itself. Since every horizontal 1-edge ismapped by g to a horizontal 0-edge, it is also replaced by itself inthe construction of C2 from C1; in general, if in one step we obtain ahorizontal edge, then it remains unchanged in subsequent steps (seeRemark 14.21).


C1

C0

C

Figure 14.6. Since g is not combinatorially expanding,

C is not a Jordan curve.

C1 C

Figure 14.7. A non-trivial rectifiable invariant Jordan curve.

Then Cn → C as n → ∞ in the Hausdorff sense, where the set

C is indicated on the right of Figure 14.6. In this case, C is not a

Jordan curve and S2 \ C has three components. Of course, the “self-

intersections” of the limit set C can be more complicated in general.

We close this section with one more example. It gives an exampleof a non-trivial invariant curve that is rectifiable.

Example 14.23. Let f be the map from Example 14.16, i.e., theLattes map obtained as in (1.1), where we choose A : C → C, u 7→A(u) := 5u.

The curve C = C0 is the boundary of the pillow as before. Thecurve C1 (which is isotopic rel. post(f) by an isotopy H0) is the thickcurve indicated in the left of Figure 14.7. Note that no 1-tile for (f, C0)joins opposite sides of C1. Thus the sequence of curves Cn, defined

14.3. INVARIANT CURVES ARE QUASICIRCLES 325

as before, Hausdorff converges to an f -invariant Jordan curve C byProposition 14.18 and Proposition 14.19.

We briefly explain the iterative construction of the curves Cn. Notethat the three 0-edges on the top, bottom, and right side of the pilloware deformed by H0 to themselves. Recall that f maps the lower left1-tile to the white 0-tile by the map u 7→ 5u and “extends to other 1-tiles by reflection”. Note that f maps all 1-edges in C1 that are not onthe left 0-edge, to one of the 0-edges on the top, right side, or bottomof the pillow. Thus they are replaced by themselves when constructingC2 from C1, see Remark 14.21.

The resulting f -invariant Jordan curve C is shown on the right. It isnot hard to see that if the pillow is equipped with the flat metric, then

the f -invariant curve C is rectifiable. Indeed, if as before we identifythe top square of the pillow with [0, 1/2]2, then the n-edges have length5−n/2. The curve Cn contains 2n “alive” n-edges that will not remainunchanged in subsequent steps. In Cn+1 each of them is replaced by 11edges of level n+ 1. A simple computation gives

length(Cn+1) = length(Cn) + 35(2/5)n,

which implies that indeed length(C) <∞.

14.3. Invariant curves are quasicircles

Recall from Section 4.1 that a metric circle (S, d) is called a quasicircleif it is quasisymmetrically equivalent to the unit circle in R2 (equippedwith the Euclidean metric). We know (see Theorem 4.1) that (S, d) is aquasicircle if and only if it is doubling and of bounded turning. We willnow verify that this is true for an invariant curve as in Theorem 14.3.

Proof of Theorem 14.3. Suppose C is an f -invariant Jordancurve as in the statement, and let % be a visual metric on S2 withexpansion factor Λ > 1. Metric notions will be with respect to thismetric in the following.

In the ensuing proof, we will consider edges for (f, C). Since C isf -invariant, edges are subdivided by edges of higher level (see Proposi-tion 12.5 (iv)). The Jordan curve C is the union of all 0-edges, so thisimplies that C is a union of n-edges for all n ∈ N0. If n, k ∈ N0 ande is an arbitrary (n + k)-edge with e ⊂ C, then there exists a uniquen-edge e′ with e ⊂ e′ ⊂ C.

If e′ is an n-edge, then the number of (n + k)-edges e containedin e′ is bounded by # post(f) deg(f)k. Indeed, map fn|e′ is injective;so the images of these (n + k)-edges e under the map fn are distinct


k-edges, and the number of k-edges is equal to # post(f) deg(f)k (seeLemma 5.3).

After these preliminaries, we are ready to show that C equippedwith (the restriction of) % is a quasicircle. We first establish that C isdoubling. Note that in contrast (S2, %) is not doubling in general (seeTheorem 17.1 (i)).

Let x ∈ C, and 0 < r ≤ 2 diam(C). In order to show that C isdoubling, it suffices to cover B(x, r)∩C by a controlled number of setsof diameter < r/4.

It follows from Proposition 8.4 that we can find n ∈ N0 dependingon r, and k0 ∈ N0 independent of r and x with the following properties:

(i) r Λ−n,

(ii) diam(e) < r/4 whenever e is an (k0 + n)-edge,

(iii) dist(e, e′) ≥ r if n− k0 ≥ 0 and e and e′ are disjoint (n− k0)-edges.

Let E be the set of all (n + k0)-edges contained in C that meetB(x, r). Then the collection E forms a cover of C∩B(x, r) and consistsof sets of diameter < r/4 by (ii). Hence it suffices to find a uniformupper bound on #E. If n < k0, then #E ≤ # post(f) deg(f)2k0 .

Otherwise, n − k0 ≥ 0. Then we can find an (n − k0)-edge e ⊂ Cwith x ∈ e. Let e be an arbitrary (n + k0)-edge in E. Again we canfind an (n− k0)-edge e′ ⊂ C that contains e.

There exists a point y ∈ e∩B(x, r). Hence dist(e, e′) ≤ %(x, y) < r.This implies e ∩ e′ 6= ∅ by (iii). So whatever e ∈ E is, the (n − k0)-edge e′ ⊂ C meets the fixed (n− k0)-edge e. This leaves at most threepossibilities for e′, namely e, and the two “neighbors“ of e on C. Sothere are three or less (n − k0)-edges that contain all the edges in E.Since each (n− k0)-edge contains at most # post(f) deg(f)2k0 edges oforder (n+ k0), it follows that #E ≤ 3# post(f) deg(f)2k0 . In any casewe get the desired bound for #E.

It remains to show that C is of bounded turning. Let x, y ∈ Cwith x 6= y be arbitrary, and let n0 ≥ 0 be the smallest integer forwhich there exist n0-edges ex ⊂ C and ey ⊂ C with x ∈ ex, y ∈ ey andex ∩ ey = ∅. Note that n0 is well-defined, because f is expanding andso the diameter of n-edges approaches 0 uniformly as n→∞.

Then by Proposition 8.4 (i),

%(x, y) & Λ−n0 .

If n0 = 0, then

diam(C) . %(x, y)


and there is nothing to prove. If n0 ≥ 1, we can find (n0 − 1)-edgese′x ⊂ C and e′y ⊂ C with x ∈ e′x, y ∈ e′y, and e′x ∩ e′y 6= ∅. Then e′x ∪ e′ymust contain one of the subarcs γ of C with endpoints x and y. Hence

diam(γ) ≤ diam(e′x) + diam(e′y) . Λ−n0 . %(x, y).

Since the implicit multiplicative constants in the previous inequalitiesdo not depend on x and y, we get a bound as desired.

Since the class of visual metrics for f and any of its iterates coincide(see Proposition 8.3 (v)), we may apply this theorem also to any iterateof f with an invariant Jordan curve C ⊃ post(f). In particular, theJordan curve in Theorem 14.1 is a quasicircle if equipped with a visualmetric for f .

A family of quasisymmetries (possibly defined on different spaces)is called uniformly quasisymmetric if there exists a homeomorphismη : [0,∞) → [0,∞) such that each map in the family is an η-quasi-symmetry. Obviously, each finite family of quasisymmetries is uni-formly quasisymmetric. If h is an η-quasisymmetry, then h−1 is anη-quasisymmetry, where η only depends on η; actually, one can takeη : [0,∞)→ [0,∞) defined by η(0) = 0 and η(t) = 1/η−1(1/t) for t > 0.This implies that if a family of maps is uniformly quasisymmetric, thenthe family of inverse maps is also uniformly quasisymmetric.

If X, Y, Z are metric spaces, and h1 : X → Y is η1-quasisymmetricand h2 : Y → Z is η2-quasisymmetric, then h2 h1 is η-quasisymmetric,where η = η2 η1. Hence the family of all compatible compositionsof maps in two uniformly quasisymmetric families is again uniformlyquasisymmetric.

Recall (see Section 4.1) that an arc α equipped with some metric dis called a quasiarc if there exits a quasisymmetry of the unit interval[0, 1] onto (α, d). This is true and only if (α, d) is doubling and thereexists a constant K ≥ 1 such that diamd(γ) ≤ Kd(x, y), wheneverx, y ∈ α, x 6= y, and γ is the subarc of α with endpoints x and y (seeTheorem 4.1).

A family of arcs is said to consist of uniform quasiarcs if thereexists a homeomorphism η : [0,∞) → [0,∞) such that for each arc αin the family there exists an η-quasisymmetry h : [0, 1]→ α. Similarly,a family of quasicircles is said to consist of uniform quasicircles if thereexists a homeomorphism η : [0,∞)→ [0,∞) that for each quasicircle Sin the family there exists an η-quasisymmetry h : ∂D→ S. A family ofquasicircles consists of uniform quasicircles if and only if the geometricconditions characterizing quasicircles, i.e., the doubling condition and


the bounded turning condition, holds with uniform parameters. Asimilar statement is true for families of quasiarcs [TuV].

We want to show that if the assumptions are as in Theorem 14.3,then all boundaries of tiles for (f, C) are quasicircles and all edges for(f, C) are quasiarcs. Actually, the family of all boundaries of tiles con-sists of uniform quasicircles and the family of all edges consists of uni-form quasiarcs. One way to do this is to repeat the proof Theorem 14.3and show that the geometric conditions characterizing quasiarcs andquasicircles are true for the edges and boundaries of tiles with uni-form constants. We choose a different approach that is based on thefollowing lemma which is of independent interest.

Lemma 14.24. Let f : S2 → S2 be an expanding Thurston map,and C ⊂ S2 be an f -invariant Jordan curve with post(f) ⊂ C. Supposethat S2 is equipped with a visual metric % for f with expansion factorΛ > 1, and denote by Xn for n ∈ N0 the set of n-tiles for (f, C). Thenthere exists a constant C ≥ 1 with the following property:

If k, n ∈ N0, Xn+k ∈ Xn+k, and x, y ∈ Xn+k, then

(14.5)1

C%(x, y) ≤ %(fn(x), fn(y))

Λn≤ C%(x, y).

In particular, the family F = fn|Xn+k : k, n ∈ N0, Xn+k ∈ Xn+k

is uniformly quasisymmetric.

The distortion estimate (14.5) is closely related to the concept ofa conformal elevator as introduced by Haıssinsky and Pilgrim [HP09,Thm. 2.2]. See also (15.1) in Theorem 15.3 for a related statement.

Proof. In the following all cells will be for (f, C). Let m = mf,C beas in Definition 8.1. We know by Definition 8.2 and by Lemma 8.6 (iii)that %(x, y) Λ−m(x,y), whenever x, y ∈ S2. If n ∈ N0, then Lemma 8.6 (ii)implies that

m(fn(x), fn(y)) ≥ m(x, y)− n,and so

%(fn(x), fn(y)) . Λn%(x, y).

Here the implicit multiplicative constant is independent of x, y, and n.To obtain an inequality in the other direction, let x, y ∈ Xn+k ∈

Xn+k, where n, k ∈ N0. We may assume that x 6= y. Then by definitionof m(x, y) we have n + k ≤ m(x, y) < ∞. Let l := m(x, y) + 1 ∈ N.Since l > n + k, the (n + k)-tile Xn+k is subdivided by tiles of levell (Proposition 12.5 (iii)). Hence there exist l-tiles X, Y ⊂ Xn+k withx ∈ X and y ∈ Y . Then X ∩Y = ∅ by definition of m(x, y). Let X ′ :=fn(X) and Y ′ := fn(Y ). Then by Proposition 5.17 (i) the sets X ′ and


Y ′ are (l − n)-tiles. Since fn|Xn+k is injective, these tiles are disjoint,and we have fn(x) ∈ X ′ and fn(y) ∈ Y ′. So from Proposition 8.4 (i)we conclude that

%(fn(x), fn(y)) ≥ dist%(X′, Y ′) & Λ−(l−n) ΛnΛ−m(x,y) Λn%(x, y).

Here the implicit multiplicative constants are again independent of x,y, and n. The other desired inequality follows.

Inequality (14.5) immediately implies that the family F is uniformlyquasisymmetric. To see this, let k, n ∈ N0 and Xn+k ∈ Xn+k. Thenfn|Xn+k is a homeomorphism onto its image (see Proposition 5.17 (i)).Moreover, if u, v, w ∈ Xn+k, u 6= w, then by (14.5) we have

%(fn(u), fn(v))

%(fn(u), fn(w))≤ C2 %(u, v)

%(u,w).

Hence fn|Xn+k is η-quasisymmetric, where η(t) = C2t for t ≥ 0. Sinceη is independent of the chosen map, the family F is uniformly qua-sisymmetric.

Proposition 14.25. Let f : S2 → S2 be an expanding Thurstonmap and C ⊂ S2 be an f -invariant Jordan curve with post(f) ⊂ C.Suppose that S2 is equipped with a visual metric for f , and for n ∈ N0

denote by Xn the set of n-tiles and by En the set of n-edges for (f, C).Then the family ∂X : n ∈ N0 and X ∈ Xn consists of uni-

form quasicircles and the family e : n ∈ N0 and e ∈ En of uniformquasiarcs.

In particular, edges for (f, C) are quasiarcs and the boundaries ofall tiles are quasicircles.

Proof. By Theorem 14.3 there exists a quasisymmetric map h : ∂D→C. Let X be an arbitrary tile for (f, C), say an n-tile, where n ∈ N0.Then fn|X is a homeomorphism of X onto the 0-tile fn(X) (Proposi-tion 5.17 (i)), and so

fn(∂X) = ∂fn(X) = C.By Lemma 14.24 the map fn|X, and hence also the map (fn|X)−1, isa quasisymmetry. It follows that (fn|X)−1 h is a quasisymmetric mapfrom ∂D onto ∂X. Hence ∂X is a quasicircle. Actually, the family ofthese quasicircles ∂X is uniform, since the family of all relevant maps(fn|X)−1h is uniformly quasisymmetric as follows from Lemma 14.24.

The proof that the family e : n ∈ N0 and e ∈ En consists of uni-form quasiarcs runs along the same lines. First note that each 0-edgeis a subarc of C, and hence corresponds to a subarc of ∂D under thequasisymmetry h. Since this subarc can be mapped to the unit interval


[0, 1] by a bi-Lipschitz homeomorphism, each 0-edge is quasisymmetri-cally equivalent to [0, 1] and hence a quasiarc.

Now let e be an arbitrary edge for (f, C), say an n-edge, wheren ∈ N0. Then fn|e is a homeomorphism of e onto the 0-edge fn(e)(Proposition 5.17 (i)). Moreover, there exists an n-tile X with e ⊂ X(Lemma 5.12). Then fn|e is the restriction of the map fn|X to e, andit follows from Lemma 14.24 that fn|e is a quasisymmetry. Hence e isquasisymmetrically equivalent to a 0-edge and hence a quasiarc.

Lemma 14.24 actually implies that the family fn|e : n ∈ N0 and e ∈En is uniformly quasisymmetric. So each edge is quasisymmetricallyequivalent to a 0-edge by a quasisymmetry in a uniformly quasisym-metric family. Since there are only finitely many 0-edges, this impliesthat the family of all edges for (f, C) consists of uniform quasiarcs.

A quasidisk is a closed topological disk (i.e., a topological cell ofdimension 2) that is quasisymmetrically equivalent to the closed unitdisk D. A family of closed topological disks is said to consist of uni-form quasidisks if each disk X in the family can be mapped to D by anη-quasisymmetry, where η is independent of X. It is a natural ques-tion whether the family X : n ∈ N0, X ∈ Xn of tiles obtained froman invariant curve as in the previous theorem actually consists of uni-form quasidisks. This is true if and only if the expanding Thurstonmap f is topologically conjugate to a rational map without periodiccritical points. One direction easily follows from Theorem 1.1 (iii) andCorollary 17.11 proved later.

For the other direction suppose that f and C are as in Proposi-tion 14.25 and that the two 0-tiles equipped with a visual metric %are quasidisks. Then one can show that (S2, %) is a quasisphere (thisrequires the solution of a so-called welding problem). So by Theo-rem 17.1 (ii) the map f is topologically conjugate to a rational mapwithout periodic critical points. We skip the details for this implicationas we will not use the result.

CHAPTER 15

The combinatorial expansion factor

Suppose f : S2 → S2 is a Thurston map with post(f) ≥ 3 and C ⊂ S2

is a Jordan curve with post(f) ⊂ C. In Section 5.6 we introduced thequantity Dn(f, C) as the minimal number of n-tiles for (f, C) to form aconnected set that joins opposite sides of C (see Definition (5.30) and(5.14)). In this chapter we study the asymptotic behavior of Dn(f, C)as n→∞. We will see that for an expanding Thurston map, Dn(f, C)grows at an exponential rate independent of C.

Proposition 15.1. Let f : S2 → S2 be an expanding Thurstonmap, and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Then the limit

Λ0(f) := limn→∞

Dn(f, C)1/n

exists. Moreover, this limit is independent of C and we have 1 <Λ0(f) <∞.

We call Λ0(f) the combinatorial expansion factor of f . Later wewill see that Λ0(f) ≤ deg(f)1/2 (Proposition 19.1).

The combinatorial expansion factor is well-behaved under takingiterates and invariant under topological conjugacy.

Proposition 15.2. Let f : S2 → S2 be an expanding Thurstonmap. The the following statements are true:

(i) Λ0(fn) = Λ0(f)n for n ∈ N.

(ii) Suppose g : S2 → S2 is an expanding Thurston map that istopologically conjugate to f . Then Λ0(g) = Λ0(f).

The main result of this chapter relates the combinatorial expansionfactor with expansion factors of visual metrics.

Theorem 15.3. Let f : S2 → S2 be an expanding Thurston map,and Λ0(f) ∈ (1,∞) be its combinatorial expansion factor. Then thefollows statements are true:

(i) If Λ is the expansion factor of a visual metric for f , then1 < Λ ≤ Λ0(f).

331

332 15. THE COMBINATORIAL EXPANSION FACTOR

(ii) Conversely, if 1 < Λ < Λ0(f), then there exists a visual metric% for f with expansion factor Λ. Moreover, the visual metric% can be chosen to have the following additional property:

For every x ∈ S2 there exists a neighborhood Ux of x suchthat

(15.1) %(f(x), f(y)) = Λ%(x, y) for all y ∈ Ux.

This statement shows that if Λ is the expansion factor of a visualmetric for f , then 1 < Λ ≤ Λ0(f), but conversely, the existence of avisual metric with expansion factor Λ is only guaranteed for 1 < Λ <Λ0(f). This statement is optimal, since a visual metric with expansionfactor Λ = Λ0(f) need not exist in general. We will discuss an exampleat the end of this chapter (see Example 15.8).

We now proceed to supply the proofs. We need some preparationand start with some lemmas.

Lemma 15.4. Let n ∈ N0, f : S2 → S2 be a Thurston map with# post(f) ≥ 3, and C ⊂ S2 a Jordan curve with post(f) ⊂ C. If thereexists a connected set K ⊂ S2 that joins opposite sides of C and thatcan be covered by M ∈ N n-flowers for (f, C), then Dn(f, C) ≤ 4M .

Proof. We first assume that # post(f) = 3. Let K be as in thestatement. By picking a point from the intersection of K with each ofthe three 0-edges, we can find a set x, y, z ⊂ K such that x, y, zmeets opposite sides of C. Since K is connected and can be coveredby M n-flowers, we can find n-vertices v1, . . . , vM ∈ S2 such that x ∈W n(v1), y ∈ W n(vM) and W n(vi)∩W n(vi+1) 6= ∅ for i = 1, . . . ,M − 1.Then it follows from Lemma 5.26 (i) that there exists a chain consistingof n-tiles X1, . . . , X2M joining x and y (recall the terminology discussedbefore Lemma 8.6). Similarly, there exists a chain X ′1, . . . , X

′2M of n-

tiles joining x and z. The union K ′ of the n-tiles in these two chainsis a connected set consisting of at most 4M n-tiles. It contains the setx, y, z and hence joins opposite sides of C. Thus Dn(f, C) ≤ 4M .

If # post(f) ≥ 4, the proof is similar and easier. In this case wecan find a set x, y ⊂ K that joins opposite sides of C. By the sameargument as before, we get the bound Dn(f, C) ≤ 2M .

Lemma 15.5. Let f : S2 → S2 be an expanding Thurston map, and

C, C ⊂ S2 be Jordan curves with post(f) ⊂ C, C. Then there existconstants C1, C2 ≥ 1 such that

(15.2)1

C1

Dn(f, C) ≤ Dn+1(f, C) ≤ C1Dn(f, C)

15. THE COMBINATORIAL EXPANSION FACTOR 333

and

(15.3)1

C2

Dn(f, C) ≤ Dn(f, C) ≤ C2Dn(f, C)

for all n ∈ N0.

Proof. Set Dn = Dn(f, C) and Dn = Dn(f, C) for n ∈ N0.To show (15.2) for a suitable constant C1, we fix n ∈ N0 and pick a

connected set K joining opposite sides of C that consists of Dn n-tilesfor (f, C). By Lemma 5.35 (ii) we can cover K by MDn (n+1)-flowers,where M ∈ N is independent of n. Hence by Lemma 15.4 we haveDn+1 ≤ CDn, where C = 4M . An inequality in the opposite directionfollows from a similar argument again based on Lemma 5.35 (i) andLemma 15.4.

To establish (15.3), we consider δ0 = δ0(f, C) > 0 defined as in

(5.13) for f , C, and a base metric d on S2. Since f is expanding, there

exists n0 ∈ N0 such that diamd(X) < δ0/2, whenever X is n0-tile for(f, C).

We can find a compact connected set K joining opposite sides of

C that consists of Dn n-tiles for (f, C). Then diamd(K) ≥ δ0 and so

K contains two points with d(x, y) ≥ δ0. There exist n0-tiles X andY for (f, C) such that x ∈ X and y ∈ Y . By choice of n0 we have

X ∩ Y = ∅, and so K joins n0-tiles for (f, C) that are disjoint. Hence

fn0(K) joins opposite sides of C by Lemma 5.33. Every n-tile for (f, C)can be covered by M n-flowers for (f, C), where M only depends on Cand C (Lemma 5.36). This and Lemma 5.27 imply that if n ≥ n0, then

we can cover fn0(K) by MDk (n− n0)-flowers for (f, C).So by Lemma 15.4 we have

Dn−n0 ≤ 4MDn,

and the first part of the proof implies

Dn ≤ Cn01 Dn−n0 ≤ 2MCn0

1 Dn.

If n ≤ n0 we get a similar bound from the inequalities Dn ≤ 2 deg(f)n0

and Dn ≥ 1. It follows that there exists a constant C independent ofn such that

Dn ≤ CDn

for all n ∈ N0. An inequality in the opposite direction is obtained by

reversing the roles of C and C and using an inequality similar to (15.2)

for Dn.


Proof of Proposition 15.1. A consequence of (15.3) is that if

C, C ⊂ S2 are Jordan curves with post(f) ⊂ C, C and the sequence

Dn(f, C)1/n converges as n → ∞, then Dn(f, C)1/n also convergesand has the same limit. So if the limit exists, it does not depend on C.

To show existence, we may impose additional assumptions on C;namely by Theorem 14.1, we may assume that C is invariant for someiterate F = fN of f . Since F is also an expanding Thurston map(Lemma 6.4), it follows from Lemma 8.5 and Lemma 12.8 that thelimit

Λ0(F, C) := limn→∞

Dn(F, C)1/n

exists and that Λ0(F, C) ∈ (1,∞).Since the n-tiles for (F, C) are precisely the (nN)-tiles for (f, C), we

have DnN(f, C) = Dn(F, C) for all n ∈ N0, and so

DnN(f, C)1/(nN) = Dn(F, C)1/(nN) → Λ0(f) := Λ0(F, C)1/N ∈ (1,∞)

as n → ∞. Invoking (15.2) we conclude that Dn(f, C)1/n → Λ0(f) asn→∞. The statement follows.

Proof of Proposition 15.2. (i) If F = fn is an iterate of f ,then, as was pointed out in the previous proof, we have

Dk(F, C) = Dnk(f, C)whenever k ∈ N0 and C is a Jordan curve with post(f) ⊂ C. Thisimplies

Λ0(fn) = Λ0(F ) = limk→∞

Dk(F, C)1/k = limk→∞

Dnk(f, C)1/k = Λ0(f)n.

(ii) By assumption there exists a homeomorphism h : S2 → S2 suchthat h f = g h. Pick a Jordan curve C ⊂ S2 with post(f) ⊂ C and

let C = h(C). Then C is a Jordan curve that contains post(g), and, asin the proof of Corollary 11.6, we have

Dn(g, C) = h(c) : c ∈ Dn(f, C)

for n ∈ N0. This implies that Dn(f, C) = Dn(g, C) for all n ∈ N0 andso

Λ0(g) = limk→∞

Dn(g, C)1/n = limn→∞

Dn(f, C)1/n = Λ0(f)

as desired.

We now proceed to proving Theorem 15.3, the main result of thischapter. So let f : S2 → S2 be an expanding Thurston map. We fixa Jordan curve C ⊂ S2 with post(f) ⊂ C, and let Dk = Dk(f, C) fork ∈ N0. In the following cells will be for (f, C). In Proposition 15.1 the


combinatorial expansion factor Λ0(f) was defined, and we proved that1 < Λ0(f) <∞.

The proof of the first part of Theorem 15.3 is easy.

Proof of Theorem 15.3 (i). Suppose % is a visual metric for fwith expansion factor Λ. Then there exists a constant C ≥ 1 such that

diam(X) ≤ CΛ−k

for all k-tiles (Proposition 8.4 (ii)). Let δ0 = δ0(f, C) > 0 be defined asin (5.13) for f , C, and the metric %.

For each k ∈ N0 there exists a connected set joining opposite sidesof C that consists of Dk k-tiles. Hence

δ0 ≤ diam(K) ≤ CDkΛ−k.

Taking the kth roots here and letting k → ∞, we conclude that Λ ≤Λ0(f) as desired.

It remains to prove part (ii). For a given expansion factor Λ ∈(1,Λ0(f)) we have to construct a visual metric that satisfies (15.1). Inthe example of Section 1.3 we have already encountered visual metricswith this local expansion property. The construction in the generalcase is involved and much more difficult than the general constructionof visual metrics as in Section 8.2. We will first do this under additionalassumptions on the map and then for the general case.

Construction of the metric under additional assumptions. Weassume in addition to our standing assumptions that the Jordan curveC is f -invariant and that Λ ∈ (1,Λ0(f)) satisfies

(15.4) Λ ≤ D1 = D1(f, C).In this case, we will now construct a visual metric % with expansion

factor Λ that satisfies (15.1). We first introduce some terminology.

A tile chain P is a finite sequence of tiles X1, . . . , XN , where Xj ∩Xj+1 6= ∅ for j = 1, . . . , N−1. Here we do not require the tiles to be ofthe same level. A subchain of P is a tile chain of the form Xj1 , . . . , Xjs ,where 1 ≤ j1 < · · · < js ≤ N . The tile chain joins the sets A,B ⊂ S2

if A ∩X1 6= ∅ and B ∩XN 6= ∅. It joins the points x, y ∈ S2 if it joinsx and y. Every chain joining two sets A and B contains a simplesubchain joining A and B, i.e., a chain that does not contain a propersubchain joining the sets. A chain X1, . . . , XN joining two disjoint setsA and B is simple if and only if Xi∩Xj = ∅ whenever 0 ≤ i, j ≤ N + 1and |i− j| ≥ 2, where X0 = A and XN+1 = B.

Define the weight of a k-tile Xk to be

(15.5) w(Xk) := Λ−k,


and the w-length of a tile chain P consisting of the tiles X1, . . . , XN as

lengthw(P ) :=N∑j=1

w(Xj).

Now for x, y ∈ S2 we define

(15.6) %(x, y) := infP

lengthw(P ),

where the infimum is taken over all tile chains P joining x and y.Obviously, such tile chains exist and the infimum can be taken oversimple tile chains P .

Lemma 15.6. The distance function % defined in (15.6) is a visualmetric with expansion factor Λ.

Proof. Symmetry and the triangle inequality immediately followfrom the definition of %. Since f is expanding, we also have %(x, x) = 0for x ∈ S2 .

Let x, y ∈ S2 with x 6= y be arbitrary, and define m = m(x, y) =mf,C(x, y) (see Definition 8.1). Then there exist m-tiles X and Y withx ∈ X, y ∈ Y and X ∩ Y 6= ∅. So X, Y is a tile chain joining x and y,and thus

%(x, y) ≤ w(X) + w(Y ) = 2Λ−m.

To prove the claim, it remains to establish a lower bound %(x, y) ≥(1/C)Λ−m for a suitable constant C independent of x and y.

Pick (m + 1)-tiles X ′ and Y ′ with x ∈ X ′ and y ∈ Y ′. ThenX ′ ∩ Y ′ = ∅ by definition of m. Every tile chain joining x and ycontains a simple tile chain joining X ′ and Y ′.

So let P be a simple tile chain joining X ′ and Y ′, and suppose itconsists of the tiles X1, . . . , XN . Let k ∈ N0 be the largest level of anytile in P . If k ≤ m+ 1, then we get the favorable estimate

(15.7) lengthw(P ) ≥ Λ−k ≥ Λ−m−1.

Otherwise, k > m + 1. We want to show that then we can replacethe k-tiles in P by (k − 1)-tiles without increasing the w-length of thetile chain. The construction is illustrated in Figure 15.1.

To see this, set X0 = X ′, XN+1 = Y ′, and let Xi, where 1 ≤ i ≤ N ,be the first k-tile in P . Since P is a simple chain joining X ′ and Y ′,the tile Xi is not contained in Xi−1 and so it has to meet ∂Xi−1. Sincethe level of Xi−1 is < k, we can find a (k − 1)-edge e ⊂ ∂Xi−1 withe ∩ Xi 6= ∅. Here and below we use the fact that C is f -invariant,and so cells of any level are subdivided by cells of higher level. Every(k − 1)-tile meets e or is contained in the complement of the edge


x

y

X′=X0 X1

Xi−1 Xi

e Z

Wk−1(e)

Xj

XN

Y ′=XN+1

Figure 15.1. Replacing k-tiles by (k − 1)-tiles.

flower W k−1(e) (see Lemma 5.29 (iii)). Since tiles of level ≤ k − 1 aresubdivided into tiles of level k − 1, this implies also that every tile oflevel ≤ k − 1 meets e or is contained in the complement of W k−1(e).

Now P is simple and so no tile in the “tail” Xi+1, . . . , XN , XN+1

meets e. Let j′ ∈ N be the largest number such that i ≤ j′ ≤ N andall tiles Xi, . . . , Xj′ are k-tiles. Then Xj′+1 has level ≤ k − 1. Sincethis tile does not meet e, it is contained in S2 \W k−1(e), and so thetiles Xi, . . . , Xj′ form a chain of k-tiles joining e and S2 \W k−1(e). Letj ∈ N be the smallest number with i ≤ j ≤ j′ such that Xj meets thecomplement ofW k−1(e). ThenXi, . . . , Xj is a chain P k of k-tiles joininge and S2 \W k−1(e). In particular, P k joins two disjoint (k − 1)-cellsas follows from the definition of an edge flower (see Definition 5.28).Moreover, Xj is the only tile in the chain P k that meets the complementof W k−1(e).

Since P k joins disjoint (k−1)-cells, it follows from Lemma 5.34 thatP k has at least D1 elements, and so by (15.4),

lengthw(P k) ≥ D1Λ−k ≥ Λ−k+1.

Let Z be the unique (k−1)-tile with Z ⊃ Xj. Then Z ∩Xj+1 6= ∅. Wealso have Z ∩ e 6= ∅; for otherwise Xj ⊂ Z ⊂ S2 \W k−1(e). Then j > iand Xj−1 meets Xj and so the complement of W k−1(e) contradictingthe definition of j. So Z ∩ Xi−1 ⊃ Z ∩ e 6= ∅. Thus we can replacethe subchain P k of P by the single (k − 1)-tile Z to obtain a chain P ′

joining X ′ and Y ′. It satisfies

lengthw(P ′) = lengthw(P )− lengthw(P k) + w(Z) ≤ lengthw(P ).

By passing to a subchain of P ′ we can find a simple chain P ′′ joining X ′

and Y ′ that contains fewer k-tiles than P and satisfies lengthw(P ′′) ≤lengthw(P ).


Continuing this process, we can remove all k-tiles from the chain tilejoining X ′ and Y ′ without increasing its w-length. If k−1 > m+1, wecan repeat the process and remove the (k− 1)-tiles without increasing

the w-length, etc. In the end we obtain a tile chain P joining X ′ and

Y ′ that contains no tiles of level > m + 1 and satisfies lengthw(P ) ≤lengthw(P ). Thus

lengthw(P ) ≥ lengthw(P ) ≥ Λ−m−1.

This together with the previous estimate (15.7) implies

%(x, y) ≥ Λ−m−1.

This is an inequality as desired, and so % is indeed a visual metric withexpansion factor Λ. In particular, this implies that % induces the giventopology on S2 (see Proposition 8.3 (ii)).

Lemma 15.7. The visual metric % as defined in (15.6) has the ex-pansion property (15.1).

Proof. We first show that

(15.8) %(f(x), f(y)) ≤ Λ%(x, y),

for all x, y ∈ S2 with %(x, y) < 1.Indeed, suppose x, y ∈ S2 are arbitrary points with %(x, y) < 1.

Let P be an arbitrary tile chain that joins x and y and suppose thatit consists of the tiles X1, . . . , XN . We may assume in addition that Psatisfies lengthw(P ) < 1. Then P does not contain 0-tiles and hencef(X1), . . . , f(XN) is a tile chain joining f(x) and f(y). Calling thelatter chain f(P ), we have

lengthw(f(P )) = Λ lengthw(P ).

Taking the infimum over all such chains P , we obtain the desired in-equality (15.8).

For an inequality in the other direction we now consider two casesfor x ∈ S2.

Case 1. x /∈ crit(f).Then we can find an open neighborhood U of x on which f is a homeo-morphism. Then U ′ = f(U) is an open set containing f(x). We canchoose ε > 0 and δ ∈ (0, 1) such that B%(x, δ) ⊂ U , B%(f(x), ε) ⊂ U ′,and f(B%(x, δ)) ⊂ B%(f(x), ε).

Define Ux = B%(x, δ), and let y ∈ Ux be arbitrary. Then we have%(f(x), f(y)) < ε. Consider a tile chain P ′ joining f(x) and f(y) whosew-length is close enough to %(f(x), f(y)) so that lengthw(P ′) < ε. Bydefinition of the metric %, for every point z that lies on a tile in P ′, wehave %(f(x), z) ≤ lengthw(P ′) < ε. Hence P ′ lies in B%(f(x), ε) ⊂ U ′.


It follows that (f |U)−1 is defined on every tile X ′ in P ′, and so byLemma 5.18 (i) the Jordan region X = (f |U)−1(X ′) is a tile containedin U . If k is the level of X ′, then k+1 is the level of X. By consideringthese images of tiles in P ′ under (f |U)−1, we get a tile chain P joiningx and y with lengthw(P ) = (1/Λ) lengthw(P ′). Taking the infimumover P ′, we obtain

(15.9) %(x, y) ≤ (1/Λ)%(f(x), f(y)),

as desired.Case 2. x ∈ crit(f).Then x ∈ f−1(post(f)), and so x is a 1-vertex. Consider the flower

U = W 1(x), and its image U ′ = f(W 1(x)) = W 0(f(x)). These are openneighborhoods of x and f(x), respectively, and the map f |U \ x is a(non-branched) covering map of U\x onto U ′\f(x) (all this followsfrom the considerations in the proof of Lemma 5.22). Again we canchoose ε > 0 and δ ∈ (0, 1) such that B%(x, δ) ⊂ U , B%(f(x), ε) ⊂ U ′,and f(B%(x, δ)) ⊂ B%(f(x), ε).

Define Ux = B%(x, δ), and let y ∈ Ux be arbitrary. In order toshow (15.9), we may assume x 6= y. Then %(f(x), f(y)) < ε andf(x) 6= f(y). Consider a tile chain P ′ joining f(x) and f(y) consistingof tiles X ′1, . . . , X

′N . We can make the further assumptions that X ′1 is

the only tile in this chain that contains f(x) and that lengthw(P ′) isclose enough to %(f(x), f(y)) such that lengthw(P ′) < ε. As before thisimplies that P ′ lies in U ′. We now choose a path γ : [0, N ] → U ′ withthe following properties:

(i) γ(0) = f(x), γ(N) = f(y) and γ(t) 6= f(x) for t 6= 0,

(ii) γ([i− 1, i]) ⊂ X ′i for i = 1, . . . , N ,

(iii) γ(i− 1/2) ∈ int(X ′i) for i = 1, . . . , N .

Since the tiles X ′i are Jordan regions, such a path γ can easily beobtained by first running from f(x) in X ′1 to an interior point of X ′1,then to a point in X ′1 ∩ X ′2, then to an interior point of X ′2, etc., andfinally to f(y) 6= f(x) in X ′N . Since X ′1 is the only tile in P ′ containingf(x), this can be done so that the path never meets f(x) except in itsinitial point.

There exists a lift α of this path (under f) with endpoints x and y,i.e., a path α : [0, N ]→ U with α(0) = x, α(N) = y, and f α = γ. Toobtain α, lift γ|(0, N ] under the covering map f |U \ x such that thelift ends at y, and note that the lift has a unique continuous extensionto [0, N ] by choosing x to be its initial point.

Using this lift α, we can construct a lift of our tile chain P ′ asfollows. Consider a tile X ′i in P ′ and let ki be its level. Set pi :=


α(i − 1/2) and p′i := γ(i − 1/2). Then f(pi) = p′i ∈ int(X ′i). ByLemma 5.18 (ii) there exists a unique (ki + 1)-tile Xi with pi ∈ Xi andf(Xi) = X ′i.

Note that

γ((0, N ]) ⊂ U ′ \ f(x) = W 0(f(x)) \ f(x) ⊂ S2 \ post(f)

and that the map

f : S2 \ f−1(post(f))→ S2 \ post(f)

is a covering map. This implies that α|[i − 1, i] is the unique lift ofγ|[i − 1, i] with α(i − 1/2) = pi. On the other hand, the path βi =(f |Xi)

−1(γ|[i−1, i]) is also a lift of γ|[i−1, i] under f with βi(i−1/2) =pi by definition of Xi. Hence βi = α|[i− 1, i] and so α([i− 1, i]) ⊂ Xi.It follows that x = α(0) ∈ X1, y = α(N) ∈ XN , and Xi ∩ Xi+1 ⊃α(i) 6= ∅ for i = 1, . . . , N − 1.

Therefore, the tiles X1, . . . , XN form a tile chain P joining x and y.The level of each tile in P exceeds the level of the corresponding tile inP ′ by exactly 1. Hence lengthw(P ) = (1/Λ) lengthw(P ′). Taking theinfimum over P ′ , we again obtain inequality (15.9)

Combining (15.8) and (15.9) we see that every point x ∈ S2 has aneighborhood Ux such that (15.1) holds.

This concludes the proof for the existence of the visual metric %with the desired properties as in Theorem 15.3 (ii) under the addi-tional assumptions that C is f -invariant and that (15.4) holds. Wenow consider the general case.

Proof of Theorem 15.3 (ii). We can choose an iterate F =fn of f such that F has an F -invariant Jordan curve C ⊂ S2 withpost(f) = post(F ) ⊂ C (Theorem 14.1). Note that Dk(f, C)1/k →Λ0(f) > Λ. Hence if n is sufficiently large, which we may assume bypassing to an iterate of F , we also have

D1(F, C) = Dn(f, C) ≥ Λn.

This means that F is an expanding Thurston map that satisfies theadditional assumptions that guarantee the existence of a metric withthe desired properties by earlier considerations. We call this metricd in order to distinguish it from the metric % that we are trying toconstruct. Then d is a visual metric for F with expansion factor Λn,and for each x ∈ S2 there exists an open neighborhood Ux of x suchthat

(15.10) d(F (x), F (y)) = d(fn(x), fn(y)) = Λnd(x, y)

for all y ∈ Ux.


We now define % as

(15.11) %(x, y) =1

n

n−1∑i=0

Λ−id(f i(x), f i(y))

for x, y ∈ S2. It is clear that % is a metric on S2.Property (15.1) for % follows from the corresponding property (15.10)

for d with the same sets Ux, x ∈ S2; indeed, if x ∈ S2 and y ∈ Ux thenby (15.10) we have

%(f(x), f(y)) =1

n

n−1∑i=0

Λ−id(f i+1(x), f i+1(y))

=1

n

(Λn−2∑i=0

Λ−(i+1)d(f i+1(x), f i+1(y)) + Λd(x, y)

)

= Λ1

n

n−1∑i=0

Λ−id(f i(x), f i(y)) = Λ%(x, y).

It remains to show that % is a visual metric for f with expansionfactor Λ. Let m = mf,C and mF = mF,C be as in Definition 8.1. Sinced is a visual metric for F with expansion factor Λn, we have

d(x, y) Λ−nmF (x,y) Λ−m(x,y)

for all x, y ∈ S2 by Lemma 8.6 (iv). Hence

%(x, y) ≥ 1

nd(x, y) & Λ−m(x,y).

Moreover, by Lemma 8.6 (ii) we have

m(f i(x), f i(y)) ≥ m(x, y)− iand so

d(f i(x), f i(y)) Λ−m(f i(x),f i(y)) ≤ ΛiΛ−m(x,y)

for all i ∈ N0. Hence

%(x, y) .1

n

n−1∑i=0

Λ−m(x,y) = Λ−m(x,y).

It follows that %(x, y) Λ−m(x,y) for all x, y ∈ S2 where C() isindependent of x and y. This shows that % is a visual metric for f withexpansion factor Λ.

We conclude the chapter with an example showing that for an ex-panding Thurston map f one can in general not expect the existenceof a visual metric with expansion factor Λ = Λ0(f).


Example 15.8. The example is a Lattes-type map as in Section 3.3.We consider the crystallographic group G consisting of all isometries gon R2 of the form u 7→ g(u) = ±u+γ, where γ ∈ Z2. Then the quotientspace S2 := R2/G is a 2-sphere. Let Θ: R2 → S2 = R2/G be thequotient map. We know that Θ is induced by G and so Θ(u1) = Θ(u2)for u1, u2 ∈ R2 if and only if there exists g ∈ G such that u2 = g(u1).

As described in Section 3.3, one may view S2 = R2/G as a pillowobtained by folding the rectangle R = [0, 1] × [0, 1/2] along the line` = (x, y) ∈ R2 : x = 1/2 and identifying the boundaries of thesquares Q = [0, 1/2] × [0, 1/2] and Q′ = [1/2, 1] × [0, 1/2] under thisoperation. In particular, C := Θ(∂Q) = Θ(∂Q′) is a Jordan curvecontaining the four critical values of Θ which are the four corners ofthe pillow.

Let

M =

(2 20 2

).

Note that then for n ∈ N we have

(15.12) Mn =

(2n n2n

0 2n

)and M−n =

(2−n −n2−n

0 2−n

).

Now define A : R2 → R2 as A(u) = Mu for u ∈ R2, where u isconsidered as a column vector. Then there exists a unique Lattes-typemap f : S2 → S2 such that the diagram

R2 A//

Θ

R2

Θ

S2 f// S2

commutes. The map f has signature (2, 2, 2, 2) and its four postcrit-ical points are the critical values of Θ. In particular, post(f) ⊂ C.Proposition 6.11 implies that the Thurston map f is an expanding.

Since A induces a map on the quotient R2/G, we have AgA−1 ∈ Gwhenever g ∈ G (Lemma A.31). This implies that if n ∈ N0 andα ∈ 1

2Z2, then Θ is injective on the parallelogram A−n(α+Q). Indeed,

if u1, u2 ∈ α + Q and Θ(A−n(u1)) = Θ(A−n(u2)), then there existsg ∈ G such that A−n(u2) = g(A−n(u1)). This implies that u2 = h(u1)with h := An g A−n ∈ G. Then Θ(u2) = Θ(u1); since Θ is injectiveon the square α+Q,we conclude that u2 = u1. The injectivity of Θ onA−n(α +Q) follows.

This implies that for n ∈ N0 the set Xn = Θ(A−n(α + Q)) is aJordan region whenever α ∈ 1

2Z2. The n-tiles for (f, C) are precisely

the sets Xn of this form. This is clear for n = 0. If n ∈ N0 is arbitrary,


then fn Θ = Θ An. Since Θ|(α+Q) is a homeomorphism of α+Qonto the 0-tile Θ(α+Q), it follows that fn|Xn is a homeomorphism ofthe Jordan region Xn onto a 0-tile. So by Lemma 5.18 (i), the set Xn

is indeed an n-tile. Since these sets Xn cover Θ(R2) = S2, there are noother n-tiles.

If a point u = (x, y) ∈ R2 (now considered as a row vector) lies inA−n(Q), then −n2−n−1 ≤ x ≤ 2−n−1 and 0 ≤ y ≤ 2−n−1 as followsfrom (15.12). This implies that if we equip S2 with the flat metricobtained by pushing the Euclidean metric forward by Θ, then for eachn-tile Xn we have diam(Xn) n2−n (note that this implies that theflat metric on S2 is not a visual metric for f ; see Proposition 8.4 (i)).We conclude that Dn := Dn(f, C) & 2n/n for n ∈ N.

If Xi := Θ(A−n(αi + Q)) with αi = (0,−i/2) for i = 1, . . . , N :=d2n/ne, then X1, . . . , XN is a chain of n-tiles joining opposite sides ofC. Hence Dn ≤ N . 2−n/n for n ∈ N. It follows that Dn 2n/n, andso

Λ0(f) = limn→∞

D1/nn = 2.

If Λ is the expansion factor of a visual metric, then, as we have seenin the proof Theorem 15.3 (i), we must have 1 . DnΛ−n 2nΛ−n/n. Itfollows that a visual metric for f with expansion factor Λ = Λ0(f) = 2does not exist.

CHAPTER 16

Measure-theoretic dynamics

In this chapter we investigate measure-theoretic properties of the dy-namical system given by iteration of an expanding Thurston map. Wewill first review some definitions and the necessary background frommeasure-theoretic dynamics in Section 16.1. Our goal in the otherSection 16.2 is then to prove the following statement.

Theorem 16.1. Let f : S2 → S2 be an expanding Thurston map.Then there exists a unique measure νf of maximal entropy for f . Themap f is mixing for νf .

Since mixing implies ergodicity, it follows that f is ergodic withrespect to νf .

On our way to prove the previous theorem, we will be able to com-pute the topological entropy htop(f) of f .

Corollary 16.2. Let f : S2 → S2 be an expanding Thurston map.Then htop(f) = log(deg(f)).

For expanding Thurston maps without periodic critical points, The-orem 16.1 can be derived from general results due to Haıssinsky andPilgrim [HP09]. We will present a different approach. We consideran iterate F = fn of our given expanding Thurston map f that hasan F -invariant Jordan curve C with post(F ) ⊂ C. Then in the celldecomposition Dk(F, C), k ∈ N0, generated by F and C each cell issubdivided by cells of higher order. This will allow us to construct aspecific F -invariant probability measure νF that assigns to each k-tilemass proportional to deg(F )−k, where the proportionality factor onlydepends on the color of the tile (see Proposition 16.10).

By using coverings by tiles, it is also easy to obtain the estimatehtop(F ) ≤ log(deg(F )) (see the proof of Lemma 16.7). On the otherhand, the 1-tiles in D1(F, C) form a measurable partition of S2 thatgenerates (in a suitable sense) the Borel σ-algebra on S2 (see Lem-ma 16.5). This allows us to explicitly compute the measure-theoreticentropy hνF (F ) of F with respect to νF as hνF (F ) = log(deg(F )) (seethe proof of Proposition 16.10). It follows that νF is a measure of max-imal entropy for F . One then shows that this measure νF is actually

345

346 16. MEASURE-THEORETIC DYNAMICS

f -invariant and the unique measure of maximal entropy for f ; this isformulated in Theorem 16.11 which immediately implies Theorem 16.1

Our main point here is to give a rather concrete and elementarydescription of the measure of maximal entropy of an expanding Thurs-ton maps and to establish some of its basic properties. We will nottouch upon many interesting related questions such as equidistributionof preimages and periodic points or more general measures such as equi-librium measures for suitable potentials. These subjects are thoroughlyinvestigated in [Li1, Li2].

16.1. Review of measure-theoretic dynamics

In this section we briefly discuss the necessary concepts related to topo-logical and measure-theoretic entropy. For more background on thesetopics see [KH, Wa].

In the following, (X, d) is a compact metric space, and g : X → Xa continuous map. For n ∈ N0, and x, y ∈ X we define

(16.1) dng (x, y) = maxd(gk(x), gk(y)) : k = 0, . . . , n− 1.

Then dng is a metric on X. Let D(g, ε, n) be the minimal number ofsets whose dng -diameter is at most ε > 0 and whose union covers X.One can show that the limit

h(g, ε) := limn→∞

1

nlog(D(g, ε, n))

exists [KH, p. 109, Lem. 3.1.5]. Obviously, the quantity h(g, ε) is non-increasing in ε. One defines the topological entropy of g (see [KH,Sect. 3.1.b]) as

htop(g) := limε→0

h(g, ε) ∈ [0,∞].

If one uses another metric d′ on X, then one obtains the same quantityfor htop(g) if d′ induces the same topology on X as d [KH, p. 109,Prop. 3.1.2]. The topological entropy is also well-behaved under it-eration. Indeed, if n ∈ N, then htop(g

n) = nhtop(g) [KH, p. 111,Prop. 3.1.7 (3)].

We denote by B the σ-algebra of all Borel sets on X. A measureon X is understood to be a Borel measure, i.e., one defined on B. If Xis compact and A measure µ is called g-invariant if

(16.2) µ(g−1(A)) = µ(A)

for all A ∈ B. Note that by continuity of g, we have g−1(A) ∈ Bwhenever A ∈ B. We denote by M(X, g) the set of all g-invariantBorel probability measures on X.

16.1. REVIEW OF MEASURE-THEORETIC DYNAMICS 347

If µ is a probability measure on a compact metric space X, thenit is regular. This means that for every ε > 0 and every Borel setA ⊂ X there exists a compact set K ⊂ A with µ(A \ K) < ε (innerregularity) and an open set U ⊂ X with A ⊂ U and µ(U \ A) < ε(outer regularity). See [Ru, Thm. 2.18] for a more general result thatcontains this statement as a special case.

A semi-algebra S is a system of sets in X satisfying the followingproperties: (i) ∅ ∈ S, (ii) A∩B ∈ S, whenever A,B ∈ S, and (iii) X\Ais a finite union of disjoint sets in S, whenever A ∈ S. A semi-algebragenerates B if B is the smallest σ-algebra containing S.

Let S be a semi-algebra generating B. If µ and ν are two measureson X and µ(A) = ν(A) for all A ∈ S, then µ = ν. Similarly, in orderto show that a measure µ is g-invariant it is enough to verify (16.2)for all sets A in S (see [Wa, p. 20, proof of Thm. 1.1] for the simpleargument on how to verify these statements).

Let µ ∈ M(X, g). Then g is called ergodic for µ if for each setA ∈ B with g−1(A) = A we have µ(A) = 0 or µ(A) = 1. The map g iscalled mixing for µ if

(16.3) limn→∞

µ(g−n(A) ∩B) = µ(A)µ(B)

for all A,B ∈ B. It is easy to see that if g is mixing for µ, then g isalso ergodic.

To establish mixing, one only has to verify (16.3) for sets A and Bin a semi-algebra generating B ([Wa, p. 41, Thm. 1.17 (iii)]; note thatthe terminology in [Wa] slightly differs from ours). If µ, ν ∈M(X, g),g is ergodic for µ, and and ν is absolutely continuous with respect toµ, then ν = µ [Wa, p. 153, Rems. (1)].

Our next goal is to define the metric entropy of g for a measure µ.We will follow [KH, Sect. 4.3] with slight differences in notation andterminology (see also [Wa, Ch. 4]).

Let µ ∈M(X, g). A measurable partition ξ for (X,µ) is a countablecollection ξ = Ai : i ∈ I of sets in B such that µ(Ai ∩ Aj) = 0 fori, j ∈ I, i 6= j and

µ

(X \

⋃i∈I

Ai

)= 0.

Here I is a countable (i.e., finite or countably infinite) index set. Thesymmetric difference of two sets A,B ⊂ X is defined as

A∆B = (A \B) ∪ (B \ A).

Two measurable partitions ξ and η for (X,µ) are called equivalent ifthere exists a bijection between the sets of positive measure in ξ and


the sets of positive measure in η such that corresponding sets have asymmetric difference of vanishing µ measure. Roughly speaking, thismeans that the partitions are the same up to sets of measure zero.

Let ξ = Ai : i ∈ I and η = Bj : j ∈ J be measurable partitionsof (X,µ). Then

ξ ∨ η := Ai ∩Bj : i ∈ I, j ∈ Jis also a measurable partition, called the join of ξ and η. The join offinitely many measurable partitions is defined similarly.

Letg−1(ξ) := g−1(Ai) : i ∈ I

and for n ∈ N define

(16.4) ξng = ξ ∨ g−1(ξ) ∨ · · · ∨ g−(n−1)(ξ).

The entropy of ξ (for given g) is

Hµ(g, ξ) =∑i∈I

µ(Ai) log(1/µ(Ai)) ∈ [0,∞].

Here it is understood that the function φ(x) = x log(1/x) is continu-ously extended to 0 by setting φ(0) = 0.

One can show that if Hµ(g, ξ) < ∞, then the quantities Hµ(g, ξng ),n ∈ N0, are subadditive in the sense that

Hµ(g, ξn+kg ) ≤ Hµ(g, ξng ) +Hµ(g, ξkg )

for all k, n ∈ N0 [KH, p. 168, Prop. 4.3.6]. This implies ([Wa, p. 87,Thm. 4.9]; see also the argument in the last part of the proof ofLemma 12.7) that

hµ(g, ξ) := limn→∞

1

nHµ(g, ξng ) ∈ [0,∞)

exists and we have

hµ(g, ξ) = infn∈N

1

nHµ(g, ξng ).

The quantity hµ(g, ξ) is called the (metric) entropy of g relative to ξ.The (metric) entropy of g for µ is defined as

(16.5) hµ(g) = suphµ(g, ξ) : ξ is a measurable

partition of (X,µ) with Hµ(g, ξ) <∞.In this definition it is actually enough to take the supremum over allfinite measurable partitions ξ (this easily follows from “Rokhlin’s in-equality” [KH, p. 169, Prop. 4.3.10 (4)]).

We call a finite measurable partition ξ a generator for (g, µ) if thefollowing condition is true: Let A be the smallest σ-algebra containing

16.2. THE MEASURE OF MAXIMAL ENTROPY 349

all sets in the partitions ξng , n ∈ N. Then for each Borel set B ∈ Bthere exists a set A ∈ A such that µ(A∆B) = 0. If for every set B ∈ Band for every ε > 0, there exists n ∈ N and a union A of sets in ξngwith µ(A∆B) < ε, then ξ is a generator for (g, µ). If ξ is a generator,then hµ(g) = hµ(g, ξ) by the Kolmogorov-Sinai Theorem [Wa, p. 95,Thm. 4.17].

If µ ∈M(g,X) and n ∈ N, then [KH, pp. 171–172, Prop. 4.3.16 (4)]

(16.6) hµ(gn) = nhµ(g).

If α ∈ [0, 1] and ν ∈M(g,X) is another measure, then [Wa, p. 183,Thm. 8.1]

hαµ+(1−α)ν(g) = αhµ(g) + (1− α)hν(g).

The topological entropy is related to the metric entropy by theso-called variational principle. It states that [Wa, p. 188, Thm. 8.6]

htop(g) = suphµ(g) : µ ∈M(g,X).A measure µ ∈ M(g,X) for which htop(g) = hµ(g) is called a measureof maximal entropy.

Let X be another compact metric space. If µ is a measure on X

and ϕ : X → X is continuous, then the push-forward ϕ∗µ of µ by ϕ is

the measure given by ϕ∗µ(A) = µ(ϕ−1(A)) for all Borel sets A ⊂ X.

Suppose g : X → X is a continuous map, and µ ∈M(X, g) and µ ∈M(X, g). Then the dynamical system (X, g, µ) is called a (topological)

factor of (X, g, µ) if there exists a continuous map ϕ : X → X such thatϕ∗µ = µ and g ϕ = ϕ g. Then we have the following commutativediagram:

(X,µ)g//

ϕ

(X,µ)

ϕ

(X, µ)g// (X, µ).

In this case, hµ(g) ≤ hµ(g) [KH, p. 171, Prop. 4.3.16].

16.2. The measure of maximal entropy

In this section we fix an expanding Thurston map f : S2 → S2. Ourgoal is to describe a measure of maximal entropy for f and show itsuniqueness. We will freely use the notation and the results discussedin the previous section.

By Theorem 14.1 we can find a sufficiently high iterate F = fn

of f that has an F -invariant Jordan curve C ⊂ S2 with post(f) =post(F ) ⊂ C. Then F is also an expanding Thurston map (Lemma 6.4).


In the following we consider the cell decompositions Dk = Dk(F, C) fork ∈ N0. A cell is a cell in any of the cell decompositions Dk, k ∈ N0,and the terms tiles, edges, and vertices are used in a similar way. Asusual, we denote by Xk and Ek the set of k-tiles and k-edges for (F, C),respectively. By Proposition 12.5 the cell decomposition Dm+k is arefinement of Dk for m, k ∈ N0, and so cells are subdivided by cells ofhigher order.

We denote by X0w and X0

b the two 0-tiles, and color the tiles for(F, C) as in Lemma 5.19. In particular, X0

w is colored white and X0b is

colored black.For k ∈ N0 let wk be the number of white and bk be the number

of black k-tiles contained in X0w , and similarly let w′k and b′k be the

number of white and black k-tiles contained in X0b . Then it follows

from the discussion after Lemma 5.19 that

(16.7) wk + w′k = bk + b′k = deg(F )k.

Note that b1, w′1 6= 0. Indeed, suppose that b1 = 0, for example.

Then X0w contains only white 1-tiles. Let X ⊂ X0

w be such a 1-tile,e ⊂ X be a 1-edge with e ⊂ ∂X, and Y be the other 1-tile containinge. Then Y is black and so Y ⊂ X0

b . Hence

e ⊂ X ∩ Y ⊂ X0w ∩X0

b = ∂X0w .

Since ∂X is a union of 1-edges, it follows that ∂X ⊂ ∂X0w . As X0

w andX are Jordan regions and X ⊂ X0

w , this is only possible if X = X0w .

Hence X0w is a 1-tile and F |X0

w is a homeomorphism of X0w onto itself.

Applying Lemma 5.18 (i) repeatedly, we see that X0w is a k-tile for

each k ∈ N0. This is impossible, because F is expanding and so thediameters of k-tiles approach 0 as k →∞.

Define

(16.8) w :=b1

b1 + w′1, b :=

w′1b1 + w′1

.

Then w, b > 0 and w + b = 1. It follows from (16.7) for k = 1 that thematrix

(16.9) A =

(w1 b1

w′1 b′1

)has the eigenvalues λ1 = deg(F ) and λ2 = w1 − b1 with respectiveeigenvectors

v1 =

(wb

)and v2 =

(1−1

).

Here |λ2| = |w1 − b1| < λ1 = deg(F ). Indeed, since 1 ≤ b1 ≤ deg(F )and 0 ≤ w1 ≤ deg(F ), we otherwise had w1 = 0 and b1 = deg(F ) ≥ 2.


Then the white 0-tile contains only black 1-tiles. Arguing similarly asin the discussion before the lemma, we see that then there can be onlyone such tile, and so b1 = 1. This is a contradiction.

The existence of a largest positive eigenvalue λ1 for A with a corre-sponding eigenvector with all positive coordinates is an instance of thePerron-Frobenius Theorem ([KH, p. 52, Thm. 1.9.11]).

Let k, l,m ∈ N0 withm ≥ l ≥ k be arbitrary. The map F k preservescolors of tiles, i.e., if Xm is an m-tile, then F k(Xm) is an (m − k)-tile with the same color as Xm. Moreover, if Y l is an l-tile, then itfollows from Lemma 5.18 (i) that the map F k|Y l induces a bijectionXm 7→ F k(Xm) between the m-tiles contained in Y l and the (m− k)-tiles contained in the (l − k)-tile Y l−k := F k(Y l).

If we use this for m = k + 1 and l = k, then we see that a whitek-tile contains w1 white and b1 black (k + 1)-tiles, and similarly eachblack k-tile contains w′1 white and b′1 black (k + 1)-tiles. This leads tothe identity

(16.10)

(wk+1 bk+1

w′k+1 b′k+1

)=

(wk bkw′k b′k

)(w1 b1

w′1 b′1

),

for k ∈ N0. This implies

Ak =

(wk bkw′k b′k

)for k ∈ N0. The following lemma is another consequence of (16.10).

Lemma 16.3. For all k ∈ N we have

wk = w deg(F )k + b(w1 − b1)k, bk = w deg(F )k − w(w1 − b1)k,

w′k = b deg(F )k − b(w1 − b1)k, b′k = b deg(F )k + w(w1 − b1)k.

Since |w1−b1| < deg(F ), the terms with deg(F )k in these identitiesare the main terms for large k.

Proof. This follows from (16.7), (16.8), and (16.10) by induction.

The next lemma provides an important connection between cellsand general Borel sets.

Lemma 16.4. Let S be the collection of all sets consisting of theempty set and the interiors of all cells. Then S is a semi-algebra gen-erating the Borel σ-algebra B on S2.

Proof. We first show that S is a semi-algebra by verifying condi-tions (i)–(iii) of a semi-algebra.

Condition (i): By definition of S we have ∅ ∈ S.


Condition (ii): Let A,B ∈ S. In order to show that A∩B ∈ S, we mayassume that A = int(σ) and B = int(τ), where σ is a k-cell and τ is anl-cell, and k ≥ l. Let p ∈ int(τ) be arbitrary. Then by Lemma 5.2 thereexists a unique k-cell c with p ∈ int(c). Since Dk is a refinement of Dl,there exists a unique l-cell τ ′ with int(c) ⊂ int(τ ′) (see Lemma 5.7).Then τ and τ ′ are both l-cells containing the point p in their interiors.This implies that τ ′ = τ , and so int(c) ⊂ int(τ).

It follows that int(τ) can be written as a disjoint union of interiorsof k-cells. This implies that either A ∩ B = int(σ) or A ∩ B = ∅. Inboth cases, A ∩B ∈ S.

Condition (iii): Let A ∈ S be arbitrary. If A = ∅, then S2 \ A = S2,and so S2\A is equal to the disjoint union of the interiors of the 0-cells,and so a disjoint union of elements in S.

If A = int(τ) where τ is a k-cell, then S2 \ A is the disjoint unionof the interiors of all k-cells distinct from τ . Again S2 \A is a disjointunion of sets in S.

So S is indeed a semi-algebra.

S generates B: Let A be the smallest σ-algebra containing S. Since Sconsists of Borel sets, we have A ⊂ B. So in order to show that A = Bis suffices to establish that U ∈ A for each open subset U of S2.

Let p ∈ U be arbitrary. Then for each k ∈ N0 the point p iscontained in the interior of some k-cell. Since F is expanding, thediameter of k-cells approaches 0 as k → ∞. Hence there exists a cellc with p ∈ int(c) ⊂ U . This implies that U as a union of elementsin S. Since for each k ∈ N0 there are only finitely many k-cells, thecollections S is countable, and so U is a countable union of elementsin S. Hence U ∈ A as desired.

In the following we set

E∞ =⋃k∈N0

F−k(C).

Then E∞ is a Borel set. Proposition 5.17 (iii) (applied to the map F )implies that E∞ is equal to union of all edges. Since every vertex iscontained in an edge, the set E∞ also contains all vertices. Moreover,we have

(16.11) F−1(E∞) = E∞.

Indeed, since the preimage of every edge is a union of edges, it is clearthat F−1(E∞) ⊂ E∞. For the other inclusion note that F−1(C) ⊃ C


and so F−(k+1)(C) ⊃ F−k(C) for k ∈ N0. Thus

F−1(E∞) = F−1

( ⋃k∈N0

F−k(C))

=⋃k∈N0

F−(k+1)(C)

⊃⋃k∈N0

F−k(C) = E∞,

as desired.

Lemma 16.5. Let µ be an F -invariant probability measure on S2

with µ(E∞) = 0. Then for each k ∈ N the set Xk of k-tiles forms ameasurable partition of S2. It is equivalent to the partition ξkF whereξ = X1. Moreover, ξ = X1 is a generator for (F, µ).

Proof. Note that µ(E∞) = 0 implies that all edges are sets of µ-measure zero. Since every vertex is contained in an edge, we also haveµ(v) = 0 for all vertices v. The k-tiles cover S2 and two distinct k-tiles have only edges or vertices, i.e., a set of µ-measure zero in common.Hence Xk is a measurable partition of S2.

Let X ∈ Xk be arbitrary. Then for i = 1, . . . , k there exist uniquei-tiles X i with X = Xk ⊂ Xk−1 ⊂ . . . ⊂ X1. Put Yi = F i−1(X i) fori = 1, . . . , k. Then Y1, . . . , Yk are 1-tiles. We claim that

(16.12) X = Y1 ∩ F−1(Y2) ∩ · · · ∩ F−(k−1)(Yk).

To verify this, denote the right hand side in this equation by X. Then

it is clear that X ⊂ X. We verify X = X by inductively showing

that for any point x ∈ X we have x ∈ X i for i = 1, . . . , k, and sox ∈ Xk = X.

Indeed, since X ⊂ Y1 = X1 this is clear for i = 1. Suppose x ∈ X i−1

for some i with 2 ≤ i ≤ k. To complete the inductive step, we have to

show x ∈ X i. Note that x ∈ X ⊂ F−(i−1)(Yi) and so F i−1(x) ∈ Yi. Themap F i−1|X i−1 is a homeomorphism of X i−1 onto the 0-tile F i−1(X i−1).Moreover, x ∈ X i−1, X i ⊂ X i−1, and F i−1(x) ∈ Yi = F i−1(X i). Henceby injectivity of F i−1 on X i−1 we have x ∈ X i as desired.

Equation (16.12) shows that every element in Xk belongs to ξkF .This implies that the measurable partitions Xk and ξkF are equivalent(ξkF may contain additional sets, but they have to be of measure zero).

To establish that ξ is a generator, let B ⊂ S2 be an arbitrary Borelset and ε > 0. By what we have seen, it is enough to show that thereexists k ∈ N and a union A of k-tiles such that µ(A∆B) < ε.

By regularity of µ there exists a compact set K ⊂ A and an openset U ⊂ S2 with K ⊂ A ⊂ U and µ(U \ K) < ε. Since the diameterof tiles goes to 0 uniformly with the order of the tiles, we can choose


k ∈ N so large that every k-tile that meets K is contained in the openneighborhood U of K. Define

A =⋃X ∈ Xk : X ∩K 6= ∅.

Then K ⊂ A ⊂ U . This implies A∆B ⊂ U \K, and so

µ(A∆B) ≤ µ(U \K) < ε

as desired. The proof is complete.

The last lemma allows us to easily compute the entropy of F -invariant measures µ once we know that µ(E∞) = 0. The followingfact will be useful for verifying this.

Lemma 16.6. There exists 1 ≤ L < deg(F ) such that for all k,m ∈N0 and each m-edge e there exists a collection M of (m+ k)-tiles with#M ≤ CLk such that e is contained in the interior of the set

⋃X∈M X.

Here C is independent of k.

The total number of (m + k)-tiles is 2 deg(F )m+k. So the lemmasays that for large k, the m-edge e can be covered by a substantiallysmaller number of (m+ k)-tiles.

Proof. It follows from Lemma 8.9 and Proposition 8.4 that onecan find k0 ∈ N such that for every s-tile X, s ∈ N0, there exist two(s+ k0)-tiles Y and Z, one white and one black, with Y ⊂ int(X) andZ ⊂ int(X).

Every white s-tile contains wk0 white and bk0 black (s + k0)-tiles,and every black s-tile contains w′k0 white and b′k0 black (s + k0)-tiles.

By (16.7) we also know that wk0 + w′k0 = bk0 + b′k0 = deg(F )k0 .Now let e be an arbitrary m-edge. For each l ∈ N0 we will define

certain collections Tl of (m + lk0)-tiles whose union contains e in itsinterior. We will denote the number of white tiles in Tl by Nw

l , thenumber of black tiles in Tl by Nb

l , and define Nl = maxNwl , N

bl .

Then the number of tiles in Tl is bounded by 2Nl.Let T0 be the set of all m-tiles that meet e. Then the union of the

tiles in T0 is the closure of the edge flower of e and so it contains e inits interior.

Suppose the collection Tl has been constructed. Then we subdivideeach of the tiles U in Tl into (m+ (l+ 1)k0)-tiles and remove one whiteand one black tile (m+ (l+ 1)k0)-tile from the interior of U . We defineTl+1 as the collection of all tiles obtained in this way from tiles in Tl.Since int(U) ∩ e = ∅ for each U ∈ Tl, the union of the tiles in Tl+1 stillcontains e in its interior. Then for the number of white tiles in Tl+1 we


have the estimate

Nwl+1 = Nw

l (wk0 − 1) +Nbl (w′k0 − 1)

≤ Nl(wk0 + w′k0 − 2) = Nl(deg(F )k0 − 2).

Similarly,

Nbl+1 ≤ Nl(deg(F )k0 − 2),

and so

Nl+1 ≤ Nl(deg(F )k0 − 2).

Let

L := (deg(F )k0 − 2)1/k0 < deg(F ).

Then

#Tl ≤ 2Nl ≤ 2N0Lk0l

is a bound for the total number of tiles in Tl.Now let k ∈ N be arbitrary, and l ∈ N0 be the smallest number

with lk0 ≥ k. For each (m+ lk0)-tile U in Tl we can pick a (m+ k)-tilethat contains U . Let M be that collection of all (m+ k)-tiles obtainedin this way. Then the union of all tiles in M contains e in its interiorand we have

#M ≤ #Tl ≤ 2N0Lk0l ≤ 2N0L

k0Lk = CLk,

where C = 2N0Lk0 . The claim follows.

The constant C in the previous lemma depends on e. If we requirethe weaker property that the collection M of (m+k)-tiles only covers e,then we can choose the collection so that #M ≤ CLk with a constantC independent of e. Indeed, in this case, we can choose T0 to consistof the two m-tiles X and Y , one white and one black, that contain ein their boundary. Then N0 = 1 and this leads to an inequality of thedesired type with a constant C independent of e.

In the next lemma we obtain an upper bound for the topologicalentropy of f .

Lemma 16.7. htop(f) ≤ log(deg(f)).

We will verify that actually htop(f) = log(deg(f)) (see the proof ofCorollary 16.2).

Proof. Since htop(F ) = nhtop(f) and deg(F ) = deg(f)n, it sufficesto show that htop(F ) ≤ log(deg(F )).

To show that htop(F ) ≤ log(deg(F )), we fix a base metric d on S2

and let ε > 0 be arbitrary. Since F is expanding, we can find k0 ∈ N0

such that diam(X) ≤ ε whenever X ∈ Xk for k ≥ k0.


Now if k ∈ N0 and X ∈ Xk+k0 is arbitrary, then F i(X) is a (k −i + k0)-tile for i = 0, 1 . . . , k, and so diam(F i(X)) < ε. This impliesthat the diameter of X with respect to the metric dkF (see (16.1)) is≤ ε. Since the number of (k + k0)-tiles is equal to 2 deg(F )k+k0 andthese tiles form a cover of S2, it follows that D(F, ε, k) ≤ 2 deg(F )k+k0 ,and so h(ε, F ) ≤ log(deg(F )). Letting ε → 0 we conclude htop(F ) ≤log(deg(F )) as desired.

Since the curve C is F -invariant, we can restrict F to C to obtaina map F |C : C → C. The following lemma shows that the topologicalentropy of this restriction is strictly smaller than log(deg(F )).

Lemma 16.8. htop(F |C) < log(deg(F )).

Proof. The proof is very similar to the first part of the proof ofLemma 16.7. Again let d be a base metric on S2.

Since C consists of # post(f) 0-edges, by Lemma 16.6 we can cover Cby a collection Mk of k-tiles, where #Mk ≤ CLk. Here 1 ≤ L < deg(F )and C is independent of k. The k-edges in the boundaries of the k-tilesin Mk then form a cover of C. It is clear that each k-edge contained inC must belong to this collection. Hence if Ek is the set of all k-edgescontained in C, we have #Ek ≤ C ′Lk with a constant C ′ independentof k.

Now let ε > 0 be arbitrary. Since F is expanding, we can findk0 ∈ N0 such that diam(X) ≤ ε whenever X ∈ Xk for k ≥ k0. Sinceevery k-edge is contained in a k-tile, we also have diam(e) ≤ ε whenevere ∈ Ek for k ≥ k0.

If k ∈ N0 and e ∈ Ek+k0 is arbitrary, then F i(e) is a (k − i + k0)-edge for i = 0, 1 . . . , k, and so diam(F i(e)) < ε. This implies that thediameter of e with respect to the metric dkF is ≤ ε.

It follows that D(F |C, ε, k) ≤ #Ek+k0 ≤ C ′Lk0+k, and so h(ε, F ) ≤log(L). Letting ε → 0 we conclude htop(F |C) ≤ log(L) < log(deg(F ))as desired.

Remark 16.9. The topological entropy of F |C can in fact be com-puted explicitly. Since we do not need this in the following, we willonly give an outline of the procedure without proof.

As described in Section 5.4, we label the 0-edges E1, . . . , Ek so thatthey are in cyclic order on the boundary of the white 0-tile X0

w . Herek = # post(F ). If en is an n-edge, then F n maps it homeomorphicallyto a 0-edge. We say that en is of type i if F n(en) = Ei.

Each 0-edge is subdivided into 1-edges. Let aij be the number of1-edges of type j into which Ei is subdivided, and A = (aij)i,j=1,...,k bethe k × k-matrix with entries aij. It is easy to see that for n ∈ N the


entry anij of the n-th power An = (anij) of A gives the number of n-edgesof type j into which the 0-edge Ei is subdivided. Based on this one canshow that for the topological entropy of F |C we have htop(F |C) = ρ(A),where ρ(A) is the spectral radius of A.

The spectral radius ρ(A) in turn can easily be computed from anymatrix norm. If for an k × k-matrix M = (mij) we set

‖M‖ =k∑

i,j=1

|mij|

for example, then htop(F |C) = ρ(A) = limn→∞‖An‖1/n. Note that for this

matrix norm, ‖An‖ is equal to the number of n-edges contained in C.

After this preparation we are now ready to construct a measure νFthat will turn out to be the unique measure of maximal entropy for Fand f .

Proposition 16.10. There exists a unique probability measure νFon S2 such that for each X ∈ Xk, k ∈ N0, we have

(16.13) νF (X) =

w deg(F )−k

b deg(F )−kif

X is white,X is black.

Then νF (E∞) = 0. Moreover, the measure νF is F -invariant, F ismixing for νF , and hνF (F ) = log(deg(F )).

Here w and b are as in (16.8). The proposition implies in particularthat edges and vertices are sets of νF -measure zero, and that F isergodic for νF .

Proof. The proof proceeds in several steps.

Construction of νF and νF (E∞) = 0: For each k-tile X, k ∈ N0, putw(X) = w(degF )−k if X is white and w(X) = b(degF )−k if X is black.

If X ∈ Xk is white, then w1 is the number of white (k + 1)-tilescontained in X, and b1 the number of black (k + 1)-tiles contained inX. Since w1 + w′1 = deg(F ), we have∑

Y ∈Xk+1,Y⊂X

w(Y ) =w1w + b1b

deg(F )k+1=

w1b1 + b1w′1

(b1 + w′1) deg(F )k+1

=b1

(b1 + w′1) deg(F )k= w(X).


A similar equation is also true for black k-tiles. If we iterate theseidentities, we get

(16.14)∑

Y ∈Xk+m,Y⊂X

w(Y ) = w(X)

for all k,m ∈ N0 and all X ∈ Xk.For A ⊂ S2 we now define

(16.15) ν∗(A) = infU

∑X∈U

w(X),

where the infimum is taken over all covers U of A by tiles (not necessar-ily of the same order). By subdividing tiles into tiles of high order andusing (16.14), one sees that in the infimum in the definition of ν∗(A) itis enough to only consider covers by tiles whose orders exceed a givennumber k. Based on this and the fact that maxx∈Xk diam(X) → 0 ask →∞, it is clear that ν∗ is a metric outer measure, i.e., if A,B ⊂ S2

are sets with dist(A,B) > 0 (with respect to some base metric on S2),then

ν∗(A ∪B) = ν∗(A) + ν∗(B).

It is a known fact that the restriction of a metric outer measure to theσ-algebra of Borel sets is a measure. We denote this restriction of ν∗

by νF .If A ⊂ S2 is compact, then it is enough to consider only finite covers

by tiles in (16.15). Indeed, suppose U = Xi : i ∈ N is an infinitecover of the compact set A ⊂ S2 by tiles, and let ε > 0 be arbitrary.It follows from Lemma 16.6 that for each i ∈ N we can find a finitecollection Ui of tiles such that Xi ⊂ int(X ′i), where

X ′i =⋃X∈Ui

X

and ∑X∈Ui

w(X) ≤ w(Xi) + ε/2i.

Finitely many sets int(X ′i1), . . . , int(X ′im) will cover A. Then

U ′ = Ui1 ∪ · · · ∪ Uimis a finite collection of tiles that covers A, and we have∑

X∈U ′w(X) ≤

∑X∈U

w(X) + ε.

Since ε > 0 was arbitrary, we conclude that for compact sets A we getthe same infimum in (16.15) if we restrict ourselves to finite covers bytiles.


A consequence of this is that νF (X) = w(X) for each X ∈ Xk,k ∈ N0. Indeed, by definition of νF we obviously have νF (X) ≤ w(X).

For an inequality in the opposite direction, it is enough to consideran arbitrary finite cover U of X by tiles. By subdividing the tiles in U ifnecessary, we may also assume that they all have the same order l andthat l ≥ k. Since U is a cover of X and l-tiles have pairwise disjointinteriors, this implies that Y ∈ U whenever Y ∈ Xl and Y ⊂ X. Hence

w(X) =∑

Y ∈Xl,Y⊂X

w(Y ) ≤∑Y ∈U

w(Y ).

Taking the infimum over all U we get w(X) ≤ νF (X) as desired.Since νF (X) = w(X) for all tiles X, we have (16.13).

It follows from Lemma 16.6 and the definition of νF , that if e is anedge, then νF (e) = 0. Since E∞ is the (countable) union of all edges,we have νF (E∞) = 0. This also shows that

νF (S2) =∑X∈X0

νF (X) =∑X∈X0

w(X) = w + b = 1,

and so νF is a probability measure.

Uniqueness of νF : Suppose that ν is another probability measure onS2 satisfying the analog of (16.13). Then from Lemma 16.6 it followsthat each edge is a set of ν-measure zero. Hence νF (int(c)) = ν(int(c))for all cells c. Since the empty set together with the interiors of allcells form a semi-algebra S generating the Borel σ-algebra on S2 (seeLemma 16.4), we conclude that ν = νF .

νF is F -invariant: To show that νF is F -invariant, it is enough to verifythat

(16.16) νF (F−1(A)) = νF (A)

for all sets A in the semi-algebra S. This is true if A = ∅.Edges are sets of measure zero, and the preimage of an edge is

a finite union of edges (see Proposition 5.17 (ii)). This implies that(16.16) holds if A = int(e) for an edge e, or if A = v for a vertex v(since every vertex is contained in an edge).

Moreover, if X is a tile, then X \ int(X) is a union of edges, and sowe have νF (F−1(int(X))) = νF (F−1(X)) and νF (int(X)) = νF (X). Soin order to establish (16.16), it remains to show that

νF (F−1(X)) = νF (X)

for all tiles X. To see this, note that if X is a k-tile, then F−1(X) isa union of deg(F ) (k + 1)-tiles that have the same color as X. Sincethe intersection of any two distinct (k+ 1)-tiles is contained in a union


Zm+k

Zm

Y l

Fm

Xk

X0

Figure 16.1. Bijection of tiles.

of edges, and hence a set of measure zero, it follows that from (16.13)that

νF (F−1(X)) = deg(F )νF (X)

deg(F )= νF (X).

F is mixing for νF : It suffices to show that for all sets A and B in thesemi-algebra S we have

νF (F−m(A) ∩B)→ νF (A)νF (B)

as m→∞. Based on the fact that edges are sets of measure zero andthat the preimage of each edge under F is a finite union of edges, forthis it suffices to show that for all tiles X and Y we have

νF (F−m(X) ∩ Y )→ νF (X)νF (Y )

as m→∞.So let k, l,m ∈ N0, X = Xk ∈ Xk and Y = Y l ∈ Xl be arbitrary.

We may assume that m ≥ l. Then F−m(Xk) is a union of (m+k)-tilesthat have the same color as Xk. Since edges are sets of measure zeroand the (m+ k)-tiles subdivide the l-tile Y l, it follows that

(16.17) νF (F−m(Xk) ∩ Y l) =

νF (Xk)

deg(F )m·#Zm+k ∈ Xm+k : Zm+k ⊂ Y l, Fm(Zm+k) = Xk.

Let X0 ∈ X0 be the unique tile with Xk ⊂ X0, and Y 0 := F l(Y l) ∈X0. We assume that X0 and Y 0 are both white; the other cases aresimilar. Then Y = Y l is also white.

Every (m+k)-tile Zm+k lies in a unique “parent” m-tile Zm. SinceY l is an l-tile and m ≥ l, we have Zm+k ⊂ Y l if and only if Zm ⊂ Y l.If Fm(Zm+k) = Xk, then Fm(Zm) is a 0-tile containing Xk, and soFm(Zm) = X0. Conversely, if Zm is an m-tile and Fm(Zm) = X0,then it follows from Lemma 5.18 (i) that Zm+k := (Fm|Zm)−1(Xk) isthe unique (m + k)-tile with Zm+k ⊂ Zm and Fm(Zm+k) = Xk. Thesituation is illustrated in Figure 16.1. These statements imply that the


map Zm+k 7→ Zm that assigns to each (m + k)-tile Zm+k its uniqueparent m-tile Zm induces a bijection between the sets Z ∈ Xm+k :Zm+k ⊂ Y l, Fm(Zm+k) = Xk and Zm ∈ Xm : Zm ⊂ Y l, Fm(Zm) =X0. Hence

(16.18) #Zm+k ∈ Xm+k : Zm+k ⊂ Y l, Fm(Zm+k) = Xk =

#Zm ∈ Xm : Zm ⊂ Y l, Fm(Zm) = X0.

Since X0 is white, the last quantity is equal to the number of whitem-tiles contained in Y l. Applying the homeomorphism F l|Y l, we seethat this number is equal to wm−l, the number of white (m − l)-tilescontained in the white 0-tile Y 0 = F l(Y l). Combing this with (16.17),(16.18), and Lemma 16.3, we get

νF (F−m(X) ∩ Y ) =νF (X)

deg(F )m· wm−l →

νF (X)

deg(F )l· w = νF (X)νF (Y )

as m→∞.

The identity hνF (F ) = log(deg(F )): According to Lemma 16.5 themeasurable partition ξ = X1 is a generator for (F, νF ), and so hνF (F ) =hνF (F, ξ). Moreover, for each k ∈ N, the measurable partition ξkF isequivalent to the measurable partition Xk given by the k-tiles. Hence

HνF (F, ξkF ) = HνF (F,Xk) =∑X∈Xk

νF (X) log(1/νF (X))

= w log(deg(F )k/w) + b log(deg(F )k/b)

= k log(deg(F )) + w log(1/w) + b log(1/b).

This implies

hνF (F ) = hνF (F, ξ) = limk→∞

1

kHνF (F, ξkF ) = log(deg(F )).

The proof is complete.

We can now identify the topological entropy of f .

Proof of Corollary 16.2. We know that htop(F ) ≤ log(deg(F )),as we have seen in the proof of Lemma 16.7. Moreover, we havehνF (F ) = log(deg(F )) by Proposition 16.10, and so htop(F ) ≥ log(deg(F ))by the variational principle. It follows that htop(F ) = log(deg(F )).Since F = fn and so deg(F ) = deg(f)n and htop(F ) = nhtop(f), theclaim follows.

Theorem 16.11. The measure νF is the unique measure of maximalentropy for f , i.e., the unique f -invariant probability measure νF withhνF (f) = htop(f). Moreover, f is mixing for νF .


Proof. We first show uniqueness. So let µ be a probability mea-sure that is f -invariant and satisfies hµ(f) = htop(f). Then µ is F -invariant and satisfies

(16.19) hµ(F ) = nhµ(f) = nhtop(f) = htop(F ) = log(deg(F )).

We will show that this implies that µ = νF . The proof proceeds inseveral steps.

We can write µ as a convex combination µ = βµs+(1−β)µa, whereµs is a probability measure that is singular with respect to νF and µais a probability measure that is absolutely continuous with respect toνF . Since νF and µ are F -invariant, it follows from the uniqueness ofthe decomposition of a measure into singular and absolutely continuousparts that µs and µa are also F -invariant. Since F is ergodic for νFand µa is F -invariant and absolutely continuous with respect to µF , itfollows that µa = νF .

If β = 0, then µ = νF and we are done. If β > 0, then we can usethe equation

log(deg(F )) = hµ(F ) = βhµs(F ) + (1− β)hνF (F )

= βhµs(F ) + (1− β) log(deg(F )),

to conclude thathµs(F ) = log(deg(F )).

We will show that this is impossible by proving that for every F -invariant probability measure µ that is singular with respect to νF wemust have

hµ(F ) < log(deg(F )).

The uniqueness of νF will then follow.So let µ be such a measure and consider the union E∞ of all edges.

Assume first that µ(E∞) > 0. By (16.11) we can then write µ as aconvex combination µ = αµ1 +(1−α)µ2 of two F -invariant probabilitymeasures µ1 and µ2, where α = µ(E∞), µ1 is concentrated on E∞, andµ2 on S2 \ E∞.

Since µ1 is F -invariant, we have µ1(F−k(C)) = µ1(C) for all k ∈ N0.On the other hand, C ⊂ F−k(C), and so µ1(F−k(C) \ C) = 0. Thisimplies that µ1(E∞ \ C) = 0. So µ1 is actually concentrated on C.Therefore, by the variational principle and by Lemma 16.8 we have

hµ1(F ) = hµ1(F |C) ≤ htop(F |C) < log(deg(F )).

We also have hµ2(F ) ≤ htop(F ) = log(deg(F )), and so

hµ(F ) = αhµ1(F ) + (1− α)hµ2(F ) < log(deg(F )).

In this case we are done.


In the remaining case we have µ(E∞) = 0. Then by Lemma 16.5ξ = X1 is a generator for (F, µ), and so hµ(F ) = hµ(F, µ). In particular,

hµ(F ) = limk→∞

1

k

∑X∈Xk

µ(X) log(1/µ(X)),

and the limit is given by the infimum of the sequence elements.Since µ and νF are mutually singular, we can find a Borel set A ⊂ S2

with µ(A) = 1 and νF (A) = 0. Using inner regularity of µ and outerregularity of νF , for each ε > 0 we can find a compact set K ⊂ S2 andan open set U ⊂ S2 with K ⊂ A ⊂ U , µ(K) > 1− ε, and νF (U) < ε. Ifk is sufficiently large, then we can cover the set K by k-tiles containedin U .

Using this for smaller and smaller ε > 0, we conclude that for eachk ∈ N we can find a set Mk ⊂ Xk such that for Ak :=

⋃X ∈ Mk

we have µ(Ak) → 1 and νF (Ak) → 0 as k → ∞. Note that νF (X) ≥c deg(F )−k for each X ∈ Xk, where c > 0 is independent of k and X.Hence

#Mk ≤ νF (Ak) deg(F )k/c.

We also have #Xk = 2 deg(F )k.The function x 7→ φ(x) = x log(1/x) is concave and has a maximum

equal to 1/e on [0, 1]. This implies that if M ⊂ Xk is arbitrary andA =

⋃X∈M X, then∑

X∈M

µ(X) log(1/µ(X)) ≤ #M · φ(

1

#M

∑X∈M

µ(X)

)= µ(A) log(#M/µ(A)) ≤ µ(A) log(#M) + 1/e.

Hence

k log(deg(F )) = khµ(F )

≤∑X∈Xk

µ(X) log(1/µ(X))

=∑X∈Mk

µ(X) log(1/µ(X)) +∑

X∈Xk\Mk

µ(X) log(1/µ(X))

≤ µ(Ak) log(#Mk) + µ(S2 \ Ak) log(#Xk) + C1

≤ µ(Ak) log(νF (Ak)) + (µ(Ak) + µ(S2 \ Ak)) log(deg(F )k) + C2

= µ(Ak) log(νF (Ak)) + k log(deg(F )) + C2.

Here the constants C1 and C2 do not depend on k. An inequality ofthis type is impossible as

µ(Ak) log(νF (Ak))→ −∞


for k →∞.This shows that if there is a measure of maximal entropy for f ,

then it has to agree with νF . Actually, we proved that νF is the uniquemeasure of maximal entropy for F .

We now show that νF is f -invariant and a measure of maximalentropy for f . Indeed, the measure f∗νF is F -invariant and the triple(S2, F, f∗νF ) is a factor of (S2, F, νF ) by the map f . It follows thathf∗νF (F ) ≤ hνF (F ). Iterating this and noting that fn∗ νF = F∗νF = νFby F -invariance of νF , we obtain

hνF (F ) = hfn∗ νF (F ) ≤ hfn−1∗ νF

(F ) ≤ · · · ≤ hf∗νF (F ) ≤ hνF (F ).

Hence hf∗νF (F ) = hνF (F ), and so f∗νF is a measure of maximal entropyfor F . By uniqueness of νF we have f∗νF = νF showing that νF is f -invariant. Moreover,

hνF (f) = hνF (F )/n = log(deg(F ))/n = log(deg(f)) = htop(f),

and so νF is a measure of maximal entropy for f . By the first part ofthe proof we know that it is the unique such measure.

It remains to show that f is mixing for νF . Indeed, since F ismixing for νF (Proposition 16.10) and νF is f -invariant, we have thatfor all m ∈ 0, . . . , n− 1, and all Borel sets A,B ⊂ S2,

νF (f−(nl+m)(A) ∩B) = νF (F−l(f−m(A)) ∩B)→νF (f−m(A))µF (B) = νF (A)νF (B)

as l→∞. Hence we get the desired relation

νF (f−k(A) ∩B)→ νF (A)νF (B)

as k →∞. The proof is complete.

Proof of Theorem 16.1. The statement follows immediately fromTheorem 16.11.

CHAPTER 17

The geometry of the visual sphere

If f : S2 → S2 is an expanding Thurston map and % a visual metric,then we call the metric space (S2, %) the visual sphere for f . Of course,(S2, %) depends on the choice of the visual metric %, but by Proposi-tion 8.3 (iv) any two such choices for a given Thurston map f producesnowflake equivalent metric spaces. Accordingly, we think of the vi-sual sphere of a Thurston map as uniquely determined up to snowflakeequivalence. In this chapter we investigate geometric features of the vi-sual sphere that are invariant under such equivalences and relate themto the dynamics of f .

The following statement is one of the main results here.

Theorem 17.1. Let f : S2 → S2 be an expanding Thurston mapand % be a visual metric for f . Then the following statements are true:

(i) (S2, %) is doubling if and only if f has no periodic criticalpoints.

(ii) (S2, %) is quasisymmetrically equivalent to C if and only if fis topologically conjugate to a rational map.

(iii) (S2, %) is snowflake equivalent to C if and only if f is topolog-ically conjugate to a Lattes map.

Here it is understood that C is equipped with the chordal metric.Statement (ii) characterizes the visual spheres that are quasispheres.

It provides an interesting analog of Cannon’s conjecture (see Section 4.3).We know that two expanding Thurston maps are Thurston equiv-

alent if and only if they are topologically conjugate (Theorem 11.4).Thus (ii) and (iii) can be reformulated as follows:

(ii’) (S2, %) is a quasisphere if and only if f is Thurston equivalentto a rational Thurston map with no periodic critical points.

(iii’) (S2, %) is snowflake equivalent to the standard unit sphere ifand only if f is Thurston equivalent to a Lattes map.

For the “only if” implication in (ii’) one has to assume that the rationalmap has no periodic critical points (or impose an equivalent condition);see Example 17.12.

365

366 17. THE GEOMETRY OF THE VISUAL SPHERE

Much more can be said in the case (i) of the previous theorem.

Proposition 17.2. Let f : S2 → S2 be a an expanding Thurstonmap without periodic critical points, % be a visual metric for f withexpansion factor Λ > 1, and νf be the measure of maximal entropy off . Then the metric measure space (S2, %, νf ) is Ahlfors Q-regular with

Q :=log(deg(f))

log(Λ).

In particular, (S2, %) has Hausdorff dimension Q and

0 < HQ% (S2) <∞.

Here HQ% is Q-dimensional Hausdorff measure on (S2, %). For the

definition of the measure of maximal entropy νf see Chapter 16. Thestatement implies that under the given assumptions we have HQ

% (M) νf (M) for each Borel set M ⊂ S2 with C() independent of M .

Since the Hausdorff dimension of (S2, %) must be ≥ 2, it also followsthat Λ ≤ deg(f)1/2. Combining this with Theorem 15.3 we obtain theupper bound Λ0(f) ≤ deg(f)1/2 for the combinatorial expansion factorof f . We will later see that this is true for every expanding Thurstonmap without the additional assumption that f has no periodic criticalpoints (see Proposition 19.1).

It is worth recording some of the consequences of our results forrational Thurston maps explicitly.

Theorem 17.3. Let f : C→ C be a rational Thurston map with noperiodic critical points. Then the following statements are true:

(i) For each sufficiently large n ∈ N there exists a quasicircle

C ⊂ C with post(f) ⊂ C that is fn-invariant (i.e., fn(C) ⊂ C).

(ii) Each fn-invariant Jordan curve C ⊂ C with post(f) ⊂ C is aquasicircle.

(iii) Suppose C is an f -invariant Jordan curve C ⊂ C with post(f) ⊂C. Let En the set of n-edges and Xn be the set of n-tiles for(f, C). Then the family e : n ∈ N0, e ∈ En consists of uni-form quasiarcs, and the family X : n ∈ N0, X ∈ Xn ofuniform quasidisks.

Here the underlying metric is the chordal metric on C. For theconcepts of uniformity used here see Section 14.3.

An immediate consequence of (iii) is that the family ∂X : n ∈N0, X ∈ Xn consists of uniform quasicircles. Note that this andthe statement about the arcs in (iii) does not a priori follow from

17.1. LINEAR LOCAL CONNECTEDNESS 367

Proposition 14.25, because we use different underlying metrics. Wewill prove though that for a rational Thurston map f without peri-odic critical points the chordal metric is quasisymmetrically equivalentto each visual metric for f (see Lemma 17.11). Once we know this,Theorem 17.3 (iii) can easily be deduced from Proposition 14.25.

A consequence of Theorem 17.3 is that each sufficiently high iter-

ate of an rational Thurston map f : C → C that is expanding has aparticularly nice Markov partition, where the tiles are quasidisks.

Another important property of visual spheres from the viewpointof quasiconformal geometry is that they are linearly locally connected.This and related properties will be discussed in Section 17.1.

Theorem 17.1 (i) and Proposition 17.2 will be proved in Section 17.2.In Section 17.3 we will establish Theorem 17.1 (ii). We postpone theproof of Theorem 17.1 (iii) to Section 18.5.

17.1. Linear local connectedness

Recall (see Section 4.1) that a metric space (X, d) is said to be linearlylocally connected (often abbreviated as LLC) if there exists a constantC ≥ 1 such that the following two conditions are satisfied:

(LLC 1) If p ∈ X, r > 0, and x, y ∈ Bd(x, r), x 6= y, then there existsa continuum E ⊂ X with x, y ∈ E and E ⊂ Bd(p, Cr).

(LLC 2) If p ∈ X, r > 0, and x, y ∈ X \Bd(x, r), x 6= y, then there ex-ists a continuum E ⊂ X with x, y ∈ E and E ⊂ X\Bd(p, r/C).

It is easy to see that first condition (i) is satisfied if and only if Xis of bounded turning (as defined in Section 4.1).

The space (X, d) is called annularly linearly locally connected (ab-breviated as ALLC) if there exists a constant C ≥ 1 with the followingproperty: If p ∈ X, r > 0, and x, y ∈ Bd(p, 2r) \ Bd(p, r), then thereexists a path γ in X joining x and y with

γ ⊂ Bd(p, Cr) \Bd(p, r/C).

The following proposition shows that the visual sphere for an ex-panding Thurston map is linearly locally connected and annularly lin-early locally connected.

Proposition 17.4. Let f : S2 → S2 be an expanding Thurstonmap, and suppose that S2 is equipped with a visual metric % for f .Then the following statements are true:

(i) There exists a constant C ′ ≥ 1 such that any two points x, y ∈S2 can be joined by a path α in S2 with

diam(α) ≤ C ′%(x, y).


(ii) (S2, %) is annularly linearly locally connected.

(iii) (S2, %) is linearly locally connected.

In this statement and in its proof below all metric concepts refer tothe visual metric %.

The statements (i)–(iii) are not logically independent, but one canshow the implications (ii) ⇒ (iii) ⇒ (i) for quite general spaces. Theensuing proof will not rely on this directly.

Proof. Let Λ > 1 be the expansion factor of %. Then for someJordan curve C ⊂ S2 with post(f) ⊂ C we have %(u, v) Λ−m(u,v) foru, v ∈ S2, where m(u, v) = mf,C(u, v) (see Definitions 8.1 and 8.2). Inthe following, all cells will be for (f, C).

(i) Let x, y ∈ S2, x 6= y, be arbitrary, and n = m(x, y) ∈ N0. Thenthere exist n-tiles X and Y with x ∈ X, y ∈ Y , and X ∩ Y 6= ∅. SinceX and Y are Jordan regions, we can find a path α in X ∪ Y that joinsx and y. Then by Proposition 8.4 we have

diam(α) ≤ diam(X) + diam(Y ) . Λ−n %(x, y),

where the implicit multiplicative constants are independent of x andy. Statement (i) follows.

(ii) Let p ∈ S2, r > 0, and x, y ∈ B(p, 2r)\B(p, r). In the followingall implicit multiplicative constants will be independent of these initialchoices of p, r, x, and y. We have to find a path γ joining x and y suchthat

(17.1) γ ⊂ B(p, Cr) \B(p, r/C),

where C ≥ 1 is a suitable constant. The construction of γ is illustratedin Figure 17.1.

Definen := maxm(p, x),m(p, y)+ 1.

ThenΛ−n min%(p, x), %(p, y) r.

Let X, Y, Z be n-tiles with x ∈ X, y ∈ Y , and p ∈ Z. Then bydefinition of n we have X ∩ Z = ∅ and Y ∩ Z = ∅.

Since f is expanding, we can choose k0 ∈ N0 as in (8.3). In par-ticular, every connected set of k0-tiles joining opposite sides of C mustcontain at least 10 k0-tiles.

Consider the set Un+k0(p) as defined in (8.6). This is the set ofall (n + k0)-tiles that intersect an (n + k0)-tile containing p. Thenfn(Un+k0(p)) is connected, and consists of k0-tiles. This set cannotjoin opposite sides of C; for otherwise, we could find a connected set of

17.1. LINEAR LOCAL CONNECTEDNESS 369

x

x′

X

y

y′

Y

β⊂γ

γγα

v

p

Z

Un+k0 (p)

Wn(v)

Figure 17.1. Construction of γ.

six k0-tiles with this property (see the proof Lemma 8.9 for a similarreasoning). This is impossible by definition of k0. Hence fn(Un+k0(p))is contained in a 0-flower (Lemma 5.31) which implies that Un+k0(p) iscontained in an n-flower (Lemma 5.27 (iii)). So there exists an n-vertexv with p ∈ Un+k0(p) ⊂ W n(v). Since Z contains p, this tile must be oneof the n-tiles forming the cycle of v. So v ∈ Z, and v /∈ X, Y . This inturn implies that X and Y do not meet W n(v) (see Lemma 5.26 (iii)).

Pick a path α in S2 that joins x and y and satisfies (i). ByLemma 11.16 we can find a set M of n-tiles that forms an e-chainjoining X and Y so that each tile in M has non-empty intersectionwith α. Pick n-vertices x′ ∈ ∂X, y′ ∈ ∂Y . Since X and Y do not con-tain v, we have x′, y′ 6= v. Consider the graph GM = ∂U : U ∈ M.It consists of n-edges, is connected, has no cut points (Lemma 11.14),and contains x′ and y′ as vertices. Hence there exists an edge pathin GM joining x′ and y′ whose underlying set β does not contain v.Then this edge path does not contain any edge in the cycle of v and soβ ∩W n(v) = ∅. Let γ be the path in S2 that is obtained by runningfrom x to x′ along some path in X, then from x′ to y′ along β, andthen from y′ to y along some path in Y . Then γ joins x and y.

Since the sets X, Y, β have empty intersection with W n(v) andhence with Un+k0(p), it follows that γ∩Un+k0(p) = ∅. So by Lemma 8.8we have

dist(p, γ) & Λ−(n+k0) Λ−n r.

Hence there exits a constant C1 ≥ 1 independent of the initial choicessuch that

γ ∩B(p, r/C1) = ∅.


The set γ can be covered by n-tiles that meet α. Since

diam(α) ≤ C ′%(x, y) ≤ 4C ′r . r,

and

maxdiam(U) : U is an n-tile . Λ−n r,

we conclude that

diam(γ) ≤ diam(α) + 2 maxdiam(U) : U is an n-tile . r.

Since the initial point x of γ has distance ≤ 2r from p, it follows thatthere exists a constant C2 ≥ 1 independent of the initial choices suchthat γ ⊂ B(p, C2r). If we set C = maxC1, C2, then (17.1) follows.

(iii) To show that (S2, %) is linearly locally connected, we verify thetwo relevant conditions; here we can use possibly different constants Cin each of the conditions.

Let p ∈ X, r > 0, and x, y ∈ B(p, r), x 6= y, be arbitrary. Choose apath α as in (i), and define E := α and C = 2C ′ + 1. Then x, y ∈ E,and, since diam(α) ≤ C ′%(x, y) ≤ 2C ′r, we have

E ⊂ B(p, r + diam(α)) ⊂ B(p, Cr).

The first of the LLC-conditions follows.For the second condition, let p ∈ X, r > 0, and x, y ∈ X \ B(p, r)

with x 6= y be arbitrary. Let α be a path in S2 joining x and y. Ifα ∩ B(p, r) = ∅, define E := α. Then E is a continuum with x, y ∈ Eand E ⊂ X \B(p, r).

If α meets B(p, r), then, as we travel from x to y along α, thereexists a first point with x′ ∈ B(p, 2r). Note that if x ∈ B(p, 2r), thenx′ = x, and d(p, x′) = 2r otherwise. In any case, x′ ∈ B(p, 2r)\B(p, r).Let αx be the subpath of α obtained by traveling along α starting fromx until we reach x′. Then αx ⊂ X \B(p, r).

Traveling along α in the opposite direction starting from y, wedefine a point y′ ∈ B(p, 2r) \ B(p, r) and a subpath αy ⊂ X \ B(p, r)of α joining y and y′ similarly. Then x′, y′ ∈ B(p, 2r) \ B(p, r). Henceby (ii) there exists a path γ in S2 that joins x′ and y′ and satisfiesγ ⊂ X \B(p, r/C). Here C ≥ 1 is a constant independent of the initialchoices. Now define E = αx ∪ γ ∪ αy. Then E is a continuum withx, y ∈ E and E ⊂ X \ B(p, r/C). It follows that the second LLCcondition is satisfied as well.

17.2. Doubling and Ahlfors regularity

Here part (i) of Theorem 17.1 as well as Proposition 17.2 are proved.We first need some preparation.

17.2. DOUBLING AND AHLFORS REGULARITY 371

Let f : S2 → S2 be a branched covering map on a 2-sphere S2.Recall that a point p ∈ S2 is called periodic if there exists n ∈ N suchthat fn(p) = p. The smallest n for which this is true is called the periodof the periodic point. The point p is called pre-periodic if there existsn ∈ N0 such that fn(p) is periodic.

The following lemma is well-known.

Lemma 17.5. Let f : S2 → S2 be a branched covering map. Thenf has no periodic critical points if and only if there exists N ∈ N suchthat

deg(fn, p) ≤ N

for all p ∈ S2 and all n ∈ N.

Proof. Note that for p ∈ S2 and n ∈ N we have

(17.2) deg(fn, p) =n−1∏k=0

deg(f, fk(p)).

So if p is a periodic critical point of period l, say, and d = deg(f, p) ≥ 2,then

deg(fn, p) ≥ ddn/le ≥ 2n/l →∞as n→∞. Hence deg(fn, p) is not uniformly bounded.

If f has no periodic critical point, then the orbit p, f(p), f 2(p), . . .of a point p ∈ S2 can contain each critical point at most once. Henceby (17.2) we have

deg(fn, p) ≤ N :=∏

c∈crit(f)

deg(f, c).

Note that the last product is finite, because f has only finitely manycritical points.

Corollary 17.6. Let f : S2 → S2 be a Thurston map with no peri-odic critical points. Then there is a constant N ∈ N with the followingproperty: if C ⊂ S2 is a Jordan curve with post(f) ⊂ C, then for eachn ∈ N0 and each vertex v in Dn(f, C) the cycle of v has length at mostN .

In other words, the closure of each n-flower W n(v) contains at mostN tiles of level n, where N only depends on f .

Proof. By Lemma 17.5, there existsN ′ ∈ N such that deg(fn, p) ≤N ′ for all p ∈ S2 and n ∈ N0. This implies that the cycle of each n-vertex (defined with respect to any Jordan curve C ⊂ S2 with post(f) ⊂C) has length at most N := 2N ′ (see Lemma 5.26).


We are now ready to prove the first part of Theorem 17.1.

Proof of Theorem 17.1 (i). Assume first that f has no peri-odic critical points. By Theorem 14.1 there exists an iterate F = fn

and an F -invariant Jordan curve C ⊂ S2 with post(f) = post(F ) ⊂ C.Then F is also an expanding Thurston map (Lemma 6.4) and it hasno periodic critical points as easily follows from (2.5). So by Corol-lary 17.6 there exists a number N ∈ N such that the cycle for eachvertex in Dn(F, C), n ∈ N0, has length ≤ N .

It suffices to show that S2 is doubling equipped with a visual metricfor F , because the class of visual metrics for f and F agree (Proposi-tion 8.3 (v)).

Fix such a visual metric % for F , and denote by Λ > 1 its expansionfactor. In the following, all cells are for (F, C). To establish that S2 isdoubling, we now proceed similarly as in the proof of Theorem 14.3.

Let x ∈ S2 and 0 < r ≤ 2 diam(S2) be arbitrary. We have to coverB(x, r) by a controlled number of sets of diameter < r/4. Again, usingProposition 8.4, we can find n ∈ N0 depending on r, and k0 ∈ N0

independent of r and x with the following properties:

(i) r Λ−n,

(ii) diam(X) ≤ r/4 whenever X is an (n+ k0)-tile,

(iii) dist(X, Y ) ≥ r whenever n− k0 ≥ 0 and X and Y are disjoint(n− k0)-tiles.

Let T be the set of all (n + k0)-tiles that meet B(x, r). Then thecollection T forms a cover of B(x, r) and consists of sets of diameter< r/4 by (ii). Hence it suffices to find a uniform upper bound on#T , independent of x and r. If n < k0, then #T ≤ 2 deg(F )2k0 (seeProposition 5.17 (iv)) and we have such a bound.

Otherwise, n− k0 ≥ 0. Pick an (n− k0)-tile X with x ∈ X. If Z isan arbitrary (n+ k0)-tile in T , then we can find a unique (n− k0)-tileY that contains Z (here we use that C is F -invariant and so each tileis subdivided by tiles of higher level).

There exists a point y ∈ Z ∩B(x, r). Hence dist(X, Y ) ≤ %(x, y) <r. This implies X ∩ Y 6= ∅ by (iii). So whatever Z ∈ T is, thecorresponding (n−k0)-tile Y ⊃ Z meets the fixed (n−k0)-tile X. Hence

Y must share an (n−k0)-vertex v with X which implies Y ⊂ W n−k0(v).

Since by choice of N the set W n−k0(v) contains at most N tiles of level(n−k0), and the number of (n−k0)-vertices in X is equal to # post(F ),this leaves at most N# post(f) possibilities for Y .

17.2. DOUBLING AND AHLFORS REGULARITY 373

Since every (n− k0)-tile contains at most 2 deg(f)2k0 tiles of order(n + k0), it follows that #T ≤ 2N# post(f) deg(f)2k0 . So we get auniform bound as desired, which shows that (S2, %) is doubling.

To show the other implication, we use the following fact aboutdoubling spaces, which is easy to show: in every ball there cannot betoo many pairwise disjoint smaller balls that all have the same radius.More precisely, for every η ∈ (0, 1) there is a number K such that everyball open ball of radius r contains at most K pairwise disjoint openballs of radius ηr.

Now suppose f : S2 → S2 is an expanding Thurston map suchthat S2 equipped with some visual metric is doubling. Pick a Jordancurve C ⊂ S2 with post(f) ⊂ C. In the following, cells will be for(f, C). Let p ∈ S2 and n ∈ N. In order to show that f has no periodiccritical points, it suffices to give a uniform bound on d = deg(fn, p) (seeLemma 17.5). For this we may assume that deg(fn, p) ≥ 2. Then p isan n-vertex and the closure of the n-flower W n(p) consists of precisely2 deg(fn, p) tiles of level n. These n-tiles have pairwise disjoint interiorand each interior contains a ball of radius r Λ−n (see Lemma 8.9).

On the other hand, diam(W n(p)) . Λ−n. Since S2 is doubling, itfollows that the number of these tiles and hence deg(fn, p) is uniformlybounded from above by a constant independent of n and p. Hence fhas no periodic critical points.

We now prove the Ahlfors regularity of (S2, %) in the case when fhas no periodic critical points.

Proof of Proposition 17.2. As in the statement, let f : S2 →S2 be an expanding Thurston map without periodic critical points, %be a visual metric, and ν = νf be the measure of maximal entropy forf . Assume that Λ > 1 is the expansion factor of %.

By Theorem 14.1 we can fix an iterate F = fn and an F -invariantJordan curve C ⊂ S2 with post(f) ⊂ C. The map F is expanding byLemma 6.4 and % is a visual metric for F with expansion factor Λ′ := Λn

by Proposition 8.3 (v). In the following, cells are defined for (F, C) andmetric notions refer to %. The measure ν is also the measure of maximalentropy νF for F (see Proposition 16.10 and Theorem 16.11).

Consider an arbitrary closed ball B(x,R), where x ∈ S2 and 0 <R ≤ diam(S2). We use the sets Um(x) as defined in (8.6) for themap F . Since F does not have periodic critical points, the length ofthe cycle of each vertex is uniformly bounded (Corollary 17.6). Thisimplies that each set Um(x) consists of a uniformly bounded numberof tiles. In particular, these sets are closed. Moreover, by Lemma 8.8


(applied to F ) we have

Um+n0(x) ⊂ B(x,R) ⊂ Um−n0(x),

where m = d− log(R)/ log(Λ′)e and n0 ∈ N0 is a constant independentof the ball.

Noting that

Q =log(deg(F ))

log(Λ′)

and using Proposition 16.10, we obtain

νF (Um+n0(x)) νF (Um−n0(x)) deg(F )−m

exp(log(R) log(deg(F ))/ log(Λ′)) = RQ,

and so ν(B(x,R)) = νF (B(x,R)) RQ. Here the constants C() areindependent of the ball. The Ahlfors Q-regularity of (S2, %, ν) follows.

This in turn implies that ν(M) HQ% (M) for every Borel set M ⊂

S2, where C() is independent of M . Since ν is a probability measure,we conclude that 0 < HQ

% (S2) < ∞. It follows that the Hausdorff

dimension of (S2, %) is equal to Q.

17.3. Quasisymmetry and rational Thurston maps

In this section we prove Theorem 17.1 (ii). We will also derive The-orem 17.3 as a consequence of this and other previous results. Theproof of Theorem 17.1 (ii) mostly follows [Me02] and [Me10]. It wasindependently established in [HP09] by a different method.

The more difficult implication in Theorem 17.1 (ii) amounts to prov-

ing that if a rational Thurston map f : C → C is expanding, then itsvisual sphere is a quasisphere. For this one wants to show that the

chordal metric σ on C is quasisymmetrically equivalent to each visualmetric % for f . The key for this is to analyze the metric properties ofthe cell decompositions Dn(f, C) with respect to the chordal metric σ.This is of independent interest and we record the relevant facts in aseparate statement.

Proposition 17.7 (Tiles in the chordal metric). Let f : C→ C be

a rational Thurston map without periodic critical points and C ⊂ C be

a Jordan curve with post(f) ⊂ C. We equip C with the chordal metricσ and denote by Xn for n ∈ N0 the collection of n-tiles for (f, C). Thenthe following statements are true:

(i) If X, Y ∈ Xn, n ∈ N0, and X ∩ Y 6= ∅, then

diam(X) diam(Y ).

17.3. QUASISYMMETRY AND RATIONAL THURSTON MAPS 375

(ii) If X, Y ∈ Xn, n ∈ N0, and X ∩ Y = ∅, then

dist(X, Y ) & diam(X).

(iii) If k, n ∈ N0, X ∈ Xn, Y ∈ Xn+k, and X ∩ Y 6= ∅, then

diam(X) & diam(Y ).

(iv) If k, n ∈ N0, X ∈ Xn, Y ∈ Xn+k, and X ∩ Y 6= ∅, then

diam(X) . diam(Y ),

where C() = C(k).

(v) Let C ⊂ C be another Jordan curve with post(f) ⊂ C. If

n ∈ N0, X is an n-tile for (f, C), X is an n-tile for (f, C), and

X ∩ X 6= ∅, then

diam(X) diam(X).

(vi) If x, y ∈ C and x 6= y, then

σ(x, y) diam(X),

whenever X ∈ Xn contains x, where n = mf,C(x, y).

The implicit multiplicative constants in (i), (ii), (iii), (v) are indepen-dent of tiles involved in the inequalities and their levels, and in (vi)independent of x, y, and X.

So in the previous inequalities we can choose all implicit multiplica-tive constants only depending on f and C with the exception of (iv),where we also have dependence on the level difference k of the tiles (butnot on the specific tiles). Statements (iii) and (iv) essentially say thatif two tiles for (f, C) have non-empty intersection, then their diametersare comparable in the following way: the diameter of the lower-leveltile bounds the diameter of the higher-level tile up to a uniform con-stant, while in the converse inequality the constant depends only onthe level difference.

The proof of the previous proposition will essentially follow fromtwo observations. First, two sets of tiles with the same combinatoricsare conformally equivalent. Second, there are only finitely many dif-ferent combinatorial types, because the local degrees of all iterates ofthe map are uniformly bounded. The statements can then be derivedfrom distortion estimates for conformal maps.

To fill in the details of this outline, we first remind the reader of theclassical Koebe distortion theorem, see [Po, Thm. 1.3 and Cor. 1.4]. Inits formulation we use the Euclidean metric and the notationBC(a, ρ) =z ∈ C : |z − a| < ρ for Euclidean disks.


Theorem 17.8 (Koebe’s distortion theorem). Let R > r > 0,z0 ∈ C, B:= BC(z0, R), and g : B → C be conformal map. Then for allz, w ∈ BC(z0, r) we have

(17.3) |g′(z)| |g′(w)| and

(17.4) |g(z)− g(w)| |g′(z)||z − w|.Here C() = C(r/R), and C(r/R)→ 1 as r/R→ 0. Moreover,

(17.5) dist(g(z0), ∂g(B)) ≥ 1

4|g′(z0)|R.

The following lemma provides a version of Koebe’s distortion the-

orem for conformal maps on multiply connected subregions of C. Herewe want to formulate the distortion estimates in terms of the chordalmetric. If h is a conformal map on a subregion Ω of C, we define thederivative of h with respect to the chordal metric as

‖h′(z)‖σ =(1 + |z|2)|h′(z)|

1 + |h(z)|2for z ∈ Ω.

This has to be interpreted as a suitable limit if z = ∞ or h(z) = ∞.To get uniform distortion estimates, one has to make the additionalassumption that the image h(Ω) of the conformal map is sufficientlysmall. We will require that the image h(Ω) is contained in a hemisphere(i.e., a chordal disk of radius

√2).

Lemma 17.9. Let Ω ⊂ C be a region, and A,B ⊂ Ω continua. Thenfor all conformal maps h : Ω→ Ω′ := h(Ω) whose image Ω′ is contained

in a hemisphere of C we have

diamσ(h(A)) ‖h′(a)‖σ and(17.6)

distσ(h(A), ∂Ω′) & ‖h′(a)‖σ for each a ∈ A,(17.7)

where C() = C(A,Ω) and C(&) = C(A,Ω). Moreover,

diamσ(h(A)) diamσ(h(B)),(17.8)

where C() = C(A,B,Ω).

It is easy to see that without the requirement that Ω′ is small enoughthe lemma is not true in general.

Proof. Note that necessarily Ω 6= C. So by using auxiliary rota-tions, we may assume that Ω ⊂ C and that Ω′ ⊂ D. Then the chordaland the Euclidean metric on Ω′ are bi-Lipschitz equivalent. Moreover,we also have ‖h′(a)‖σ |h′(a)| for each a ∈ A, where C() = C(A).


So it suffices to prove the desired inequalities for the Euclidean metricand the usual derivative |h′(a)|.

In the following, all metric notions refer to the Euclidean metric. IfD = BC(z0, r) is a Euclidean disk, we use the notation 2D = BC(z0, 2r)for the disk with the same center and twice the radius. A Harnack chain(in Ω) is a sequence D1, . . . , Dn of Euclidean disks with 2Di ⊂ Ω fori = 1, . . . , n and Di∩Di+1 6= ∅ for i = 1, . . . , n−1. We call n the lengthof the Harnack chain, and say that it connects two points u, v ∈ Ω ifu ∈ D1 and v ∈ Dn. Note that if u and v are two points in Ω that canbe connected by a Harnack chain of length n, then repeated applicationof (17.3) leads to |h′(u)| |h′(v)| with C() = Cn

0 , where C0 ≥ 1 is auniversal constant.

Now any two points in a compact subset K of Ω can be joined by aHarnack chain whose length is uniformly bounded only depending onK and Ω. This implies that

(17.9) |h′(u)| |h′(v)| for all u, v ∈ A,

where C() = C(A,Ω) is independent of u, u, and h.Let a, u, v ∈ A be arbitrary. Then there exists a Harnack chain in Ω

that connects u and v and has length uniformly bounded from above.Then (17.4), the triangle inequality, and (17.9) give

|h(u)− h(v)| . |h′(a)|with C(.) = C(A,Ω). Hence diam(h(A)) . |h′(a)| with an implicitmultiplicative constant only depending on A and Ω. This gives one ofthe estimates in (17.6).

To show the other estimate in (17.6), we pick a Euclidean disk Dwith 2D ⊂ Ω that contains two distinct points u0, u1 ∈ D ∩ A. Thenby (17.4) and (17.9) we have

diam(h(A)) ≥ |h(u0)− h(u1)| & |h′(u0)||u0 − u1| |h′(u0)| |h′(a)|for each a ∈ A with implicit multiplicative constants only dependingon A and Ω.

To prove (17.7), we note that by (17.5) and (17.6) we have

dist(h(a), ∂Ω′) ≥ 1

4|h′(a)| dist(a, ∂Ω) |h′(a)| diam(h(A))

for all a ∈ Ω, where C() = C(A,Ω). Inequality (17.7) follows.Finally, in order to establish (17.8) pick a ∈ A and b ∈ B. By

similar arguments as above one sees that diam(h(B)) |h′(b)| withC() = C(B,Ω), and that |h′(a)| |h′(b)| with C() = C(A,B,Ω).Hence

diam(h(A)) |h′(a)| |h′(b)| diam(h(B))


with C() = C(A,B,Ω) as desired.

Let D be a cell complex. A subset D′ ⊂ D is called a subcomplexof D if the following condition is true: if τ ∈ D′, σ ∈ D, and σ ⊂ τ ,then σ ∈ D′. If D′ is a subcomplex of D, then the cells in D′ form acell decomposition of the underlying set

|D′| :=⋃c : c ∈ D′.

Now suppose that f : C→ C is a rational Thurston map, and C ⊂ Cis a Jordan curve with post(f) ⊂ C. Consider the cell decompositions

Dn = Dn(f, C) of C. If τ ∈ Dn, then fn|τ is a homeomorphism of the n-cell τ onto the 0-cell fn(τ). So the map τ 7→ fn(τ) induces a labelingDn → D0 (see Definition 5.20). We call this the natural labeling onDn. Similarly, the map τ 7→ fn(τ) induces a natural labeling on everysubcomplex of Dn.

Lemma 17.10. Let n,m ∈ N0, D be a subcomplex of Dn, and D′ bea subcomplex of Dm equipped with the natural labelings. If ψ : D → D′is a label-preserving isomorphism, then there exists a homeomorphismh : |D| → |D′| such that

(i) h(τ) = ψ(τ) for each τ ∈ D,

(ii) h maps int(|D|) conformally onto int(|D′|).

See Definition 5.13 for the notion of a label-preserving isomorphismof cell complexes. Roughly speaking, the lemma says that combinato-rial equivalence of two subcomplexes D and D′ gives conformal equiv-alence of their underlying sets.

Note that since f and all of its iterates are orientation-preserving,fn maps positively oriented flags in Dn to positively oriented flagsin D0. A similar statement is true for fm and Dm. Hence ψ mapspositively oriented flags in D to positively oriented flags in D′. Thisimplies that homeomorphism h in the lemma is necessarily orientation-preserving. Of course, this is also follows from conformality statement(ii) if int(|D|) is dense in |D|.

Proof. If τ ∈ D, then fn(τ) = fm(ψ(τ)), because ψ is label-preserving. Hence we can define a homeomorphism hτ := (fm|ψ(τ))−1(fn|τ) of τ onto ψ(τ). It is clear that the maps hτ are compatibleunder inclusions of cells: if σ, τ ∈ D and σ ⊂ τ , then hτ |σ = hσ.We now define a map h : |D| → |D′| as follows. For p ∈ |D| pickτ ∈ D with p ∈ τ . Set h(p) := hτ (p). By an argument as in theproof of of Proposition 5.24 one can show that h is well-defined and ahomeomorphism of |D| onto |D′|. Obviously, h has property (i).


To establish property (ii) it suffices to show that h is holomorphic

(as a map on a subset of C) near each interior point p of |D|. Weconsider several cases. If p is an interior point of a tile X ∈ D, thenthis is clear by definition of h. Suppose p is an interior point of ann-edge e. Then the two n-tiles X, Y ∈ Dn that contain e in theirboundaries are in D. Similarly, if X ′ = ψ(X), Y ′ = ψ(Y ), e′ = ψ(e),then X ′ and Y ′ are the two m-tiles that contain the m-edge e′ in theirboundaries. Let X0 = fn(X) = fm(X ′), Y 0 = fn(Y ) = fm(Y ′) ande0 = fn(e) = fm(e′), and set U = int(X) ∪ int(e) ∪ int(Y ), U ′ =int(X ′) ∪ int(e′) ∪ int(Y ′), and U0 = int(X0) ∪ int(e0) ∪ int(Y 0). Itfollows from the considerations in the proof of Lemma 5.15 that fn|Uis a conformal map of U onto U0 and that fm is a conformal map of U ′

onto U0. Hence g = (fm|U ′)−1 (fn|U) is a conformal map of U ontoU ′. Obviously, g = h|U , and so h is holomorphic near p ∈ int(e) ⊂ U .

Finally, if p is an n-vertex, then there is an open neighborhoodU ⊂ |D| of p that contains no other n-vertex. By what we have seen,h is holomorphic on U \ p. Since h is continuous in p, the point p isa removable singularity, and hence h is holomorphic near p.

Proof of Proposition 17.7. Unless otherwise stated, in the fol-lowing all cells are for (f, C).

(i) For n ∈ N0 and X, Y ∈ Xn with X ∩ Y 6= ∅, we considerthe complex D(X, Y ), equipped with the natural labeling, consistingof all n-cells c for which there exists an n-tile Z with c ⊂ Z andZ ∩ (X ∪ Y ) 6= ∅. Obviously,

(17.10) |D(X, Y )| =⋃Z ∈ Xn : Z ∩ (X ∪ Y ) 6= ∅.

Let Ω(X, Y ) be the interior of |D(X, Y )|. Then Ω(X, Y ) is a regioncontaining X and Y .

Suppose that X ′, Y ′ is a pair of non-disjoint m-tiles, m ∈ N0. Wecall D(X, Y ) and D(X ′, Y ′) equivalent if there exists a label-preservingisomorphism ψ : D(X, Y ) → D(X ′, Y ′) with ψ(X) = X ′ and ψ(Y ) =Y ′. If D(X, Y ) and D(X ′, Y ′) are equivalent, then by Lemma 17.10there exists a conformal map h : Ω(X, Y )→ Ω(X ′, Y ′) with h(X) = X ′

and h(Y ) = Y .Since f has no periodic critical points, the length of the cycle of

each vertex is uniformly bounded (Corollary 17.6). So each n-vertexis contained in a uniformly bounded number of n-tiles independentof n. This implies that the number of n-tiles, and hence the num-ber of n-cells, in D(X, Y ) is uniformly bounded by a number inde-pendent of X, Y , and n. Therefore, among the complexes D(X, Y )


there are only finitely many equivalence classes. Since f is expand-ing, there are also only finitely many complexes D(X, Y ) such thatΩ(X, Y ) is not contained in a hemisphere. Hence we can find finitelymany complexes D(X1, Y1), . . . ,D(XN , YN) such that each complexD(X, Y ) not in this list is equivalent to one complex D(Xi, Yi) andsuch that Ω(X, Y ) is contained in a hemisphere. It follows from (17.8)in Lemma 17.9 (applied to A = Xi, B = Yi, Ω = Ω(Xi, Yi), and theconformal map h : Ω(Xi, Yi) → Ω(X, Y ) produced by Lemma 17.10)that diam(X) diam(Y ) with C() independent of X, Y , and n.

(ii) The argument is very similar to the previous one. For n ∈ N0

and X ∈ Xn, we consider the cell complex D(X), equipped with thenatural labeling, consisting of all n-cells c for which there exists ann-tile Z with c ⊂ Z and Z ∩X 6= ∅. Then

(17.11) |D(X)| =⋃Z ∈ Xn : X ∩ Z 6= ∅.

If we define Ω(X) to be the interior of |D(X)|, then |D(X)| is a regionthat contains X. Moreover, if Y is an n-tile with X ∩ Y = ∅, thenΩ(X) is disjoint from Y .

If m ∈ N0 and X ′ is an m-tile, then we call the complexes D(X)and D(X ′) equivalent if there exists a label-preserving isomorphismψ : D(X)→ D(X ′) with ψ(X) = X ′.

Again there only finitely many equivalence classes of the complexesD(X). Based on Lemma 17.10 and (17.6) and (17.7) in Lemma 17.9,we conclude that for each n-tile Y with X ∩ Y = ∅, we have

dist(X, Y ) ≥ dist(X, ∂Ω(X)) & diam(X),

where C(&) does not depend on X and Y .

(iii)–(iv) Let n, k ∈ N0, X ∈ Xn, Y ∈ Xn+k, and X ∩ Y 6= ∅.If W is any n-flower, then any two n-tiles contained in W have

an n-vertex in common, and hence have comparable diameter by (i).This implies that diam(Z) diam(W ) whenever Z is an n-tile withZ ∩W 6= ∅. We also see that diam(W ) diam(W ′), whenever W andW ′ are n-flowers with W ∩W ′ 6= ∅.

Now by Lemma 6.2 (applied for C = C) we can cover the (n+k)-tileY with M n-flowers W1, . . . ,WM , where M is independent of Y . SinceX and Y have a point in common, we may assume that X ∩W1 6= ∅.Moreover, since Y is connected, we may assume that each flower in thelist meets one of the previous ones. Then

diam(X) diam(W1) diam(Wi)


for i = 1, . . . ,M . This implies

diam(Y ) ≤M∑i=1

diam(Wi) diam(W1) diam(X).

Since in the previous inequalities all implicit multiplicative constantsonly depended on f and C, claim (iii) follows.

For (iv) note that by choosing tiles that contain a point in X ∩ Y ,we can find tiles X i of level i = n, . . . , n + k such that X = Xn,Y = Xn+k, and X i ∩ X i+1 6= ∅ for i = n, . . . , n + k − 1. If we canshow that diam(X i) . diam(X i+1) with a uniform constant C(.) onlydepending on f and C, then we conclude that diam(X) . diam(Y )with a constant C(.) = C(k) as desired.

In other words, we are reduced to proving the inequality diam(X) .diam(Y ) under the additional assumption that Y ∈ Xn+1, where theimplicit multiplicative constant is supposed to depend only on f andC.

By Lemma 5.35 (ii) and (i) the n-tile X can be covered by a con-trolled number of (n + 1)-flowers whose diameter is comparable todiam(Y ). If one combines this with similar estimates as in the pre-vious argument, then (iv) immediately follows.

(v) If X and X are as in the statement, then by Lemma 5.36

and (i) we can cover X by M n-flowers W1, . . . ,WM for (f, C) withdiam(Wi) diam(X) for i = 1, . . . ,M . Here C() and the number

M are independent of n, X, and X. By an estimate as in the proof

of (iii), we conclude diam(X) . diam(X). For the other inequality we

reverse the roles of X and X and use an analog of (i) for the curve C.(vi) Let x, y ∈ C with x 6= y be arbitrary, n := mf,C(x, y), and X

be an n-tile with x ∈ X. By Definition 8.1 there are n-tiles Xn andY n with x ∈ Xn, y ∈ Y n and Xn ∩ Y n 6= ∅. Then by (i) we have

diam(X) diam(Xn) diam(Y n),

and so

σ(x, y) ≤ diam(Xn) + diam(Y n) diam(X).

On the other hand, pick (n+1)-tiles Xn+1 and Y n+1 with x ∈ Xn+1

and y ∈ Y n+1. Then Xn+1 ∩ Y n+1 = ∅ by definition of n. We havex ∈ X ∩Xn+1, and so diam(Xn+1) diam(X) by (iii) and (iv). Then(ii) implies that

σ(x, y) ≥ dist(Xn+1, Y n+1) & diam(Xn+1) diam(X).


Since in the previous inequalities all the implicit multiplicative con-stants can be chosen independent of x, y, and X, the statement fol-lows.

Lemma 17.11. Let f : C → C be a rational Thurston map with noperiodic critical points. Then each visual metric % for f is quasisym-metrically equivalent to the chordal metric σ.

Proof. By Proposition 2.3 the map f is expanding. Let % be a vi-

sual metric for f . We have to show that the identity map idC : (C, %)→(C, σ) is a quasisymmetry. We will actually prove that this map isweakly quasisymmetric: there exists a constant H ≥ 1 such that wehave the implication

(17.12) %(x, z) ≤ %(x, y)⇒ σ(x, z) ≤ Hσ(x, y)

for all x, y, z ∈ C. A weak quasisymmetry from a connected dou-bling space to a doubling space is quasisymmetric (see Proposition 4.4).

We may apply this statement, because (C, σ) and (C, %) are doubling.Since f has no periodic critical points, the last fact follows from The-orem 17.1 (i).

We pick a Jordan curve C ⊂ C with post(f) ⊂ C and denote byΛ > 1 the expansion factor of %. In the following, we will consider tilesfor (f, C)

Now suppose that x, y, z ∈ C are points with %(x, z) ≤ %(x, y).In the following estimates, all implicit multiplicative constants can bechosen independent of x, y, z.

Define n = mf,C(x, y) and l = mf,C(x, z), where mf,C is as in Defi-nition 8.1. Then

Λ−l %(x, z) ≤ %(x, y) Λ−n.

Hence there exists a constant k0 ∈ N0 independent of x, y, z such n ≤l + k0. Now we pick an n-tile Xn, an l-tile X l, and an (l + k0)-tileXn+k0 that all contain x. Then

σ(x, z) diamσ(X l) by Proposition 17.7 (vi),

diamσ(X l+k0) by Proposition 17.7 (iii) and (iv),

. diamσ(Xn) σ(x, y) by Proposition 17.7 (iii) and (vi).

Since the implicit constants here are independent of x, y, z, the map

idC : (C, %) → (C, σ) is indeed weakly quasisymmetric. The statementfollows.

We are now ready to prove the second part in Theorem 17.1.


Proof of Theorem 17.1 (ii). Let f : S2 → S2 be an expandingThurston map, and % be a visual metric for f .

Suppose first that f is topologically conjugate to a rational map.Obviously, this rational map is then itself an expanding Thurstonmap. Moreover, by Corollary 11.6 the conjugating homeomorphismis a snowflake equivalence with respect to visual metrics, and in par-ticular a quasisymmetry. This implies that in order to show that thevisual sphere of f is a quasisphere, we may actually assume that f itselfis a rational Thurston map that is expanding.

Then f : C→ C has no periodic critical points by Proposition 2.3.So by Lemma 17.11 the visual metric % is quasisymmetrically equivalent

to the chordal metric σ. Hence the identity map idC : (C, %) → (C, σ)

is a quasisymmetry, and so (C, %) is a quasisphere. This proves the firstimplication of the theorem.

For the converse direction suppose that f : S2 → S2 is an expand-ing Thurston map, % is a visual metric for f on S2, and that there

exists a quasisymmetry h : (S2, %) → (C, σ). Since all visual metricsare snowflake and hence also quasisymmetrically equivalent, we mayalso assume that % is a visual metric for f satisfying (15.1) in Theo-rem 15.3.

The map h−1 is also a quasisymmetry; so h and h−1 are η-quasi-symmetric for some distortion function η. We consider the conjugate

g = h f h−1 : C→ C of f by h.We claim that the family of iterates gn : n ∈ N is uniformly

quasiregular, i.e., each map gn is K-quasiregular with K independentof n (see [Ri, Ch.1, Sec. 2] for the definition of a K-quasiregular map).The reason is that with the metric % satisfying (15.1), the map f islocally “conformal”, and so the dilatation of gn = h fn h−1 canbe bounded by the dilatations of h and h−1, and hence by a constantindependent of n.

To be more precise, let n ∈ N, u ∈ C, and for small ε > 0 consider

points v, w ∈ C with σ(u, v) = σ(u,w) = ε. Define x = h−1(u),y = h−1(v), z = h−1(w). By Theorem 15.3 (ii) we have that if ε > 0 issufficiently small (depending on u and n), then

σ(gn(u), gn(v))

σ(gn(u), gn(w))=

σ(h(fn(x)), h(fn(y)))

σ(h(fn(x)), h(fn(z)))

≤ η

(%(fn(x), fn(y))

%(fn(x), fn(z))

)= η

(%(x, y)

%(x, z)

)= η

(%(h−1(u), h−1(v))

%(h−1(u), h−1(w))

)≤ H := η(η(1)).


Hence

H(gn, u) :=

lim supε→0

max

σ(gn(u), gn(v))

σ(gn(u), gn(w)): v, w ∈ C, σ(u, v) = σ(u,w) = ε

≤ H

for all u ∈ C and n ∈ N. This inequality implies that gn is locally

H-quasiconformal on the set C \ crit(gn) (according to the so-called“metric” definition of quasiconformality; see [Va1, Sect. 34]).

In particular, this implies that gn|C\crit(gn) is K-quasiregular withK = K(H) independent of n. Since the finite set crit(gn) is removablefor quasiregularity (see [Ri, Ch. 7, Sect. 1]), we conclude that gn isK-quasiregular with K independent of n.

So the family of iterates gn : n ∈ N of g is uniformly quasiregular.This implies that g is topologically conjugate to a rational map (see[IM, p. 508, Thm. 21.5.2]). Hence f is also topologically conjugate toa rational map.

Proof of Theorem 17.3. By Proposition 2.3 the map f is anexpanding Thurston map (and hence also every iterate of f). So by

Theorem 14.1 there exists a Jordan curve C ⊂ C with post(f) ⊂ Cthat is fn-invariant for suitable n ∈ N. Equipped with the chordalmetric, every such curve is a quasicircle as follows from Theorem 14.3and Lemma 17.11 (note that the class of visual metrics for f and forany iterate of f are the same). Statements (i) and (ii) follow.

Suppose the curve C is actually f -invariant, and consider cells for(f, C). Then by Proposition 14.25 the family of all edges consists of

uniform quasiarcs if the underlying metric on C is a visual metric forf . Again by Lemma 17.11 we can switch to the chordal metric σ inthis statement.

Similarly, the family of the boundaries of all tiles consists of uniformquasicircles for σ. Now it is a standard fact that a closed Jordan region

X ⊂ C bounded by a quasicircle ∂X is a quasidisk. More precisely,if h : ∂D → ∂X is an η-quasisymmetry, then it can be extended to anη-quasisymmetry H : D→ X. Here η-depends not only η, but also on

a lower bound for the diameter of C \X (see [Bo2, Prop. 5.3 (ii)]). Inparticular, if we have a family of such Jordan regions whose boundariesform a family of uniform quasicircles, then the family will consist ofuniform quasidisks, if there is a positive uniform lower bound for thediameters of the complements of the regions. Since f is expanding,there are only finitely many tiles not contained in hemispheres, and so


we have such a uniform lower bound for the family of all tiles for (f, C).Statement (iii) follows.

Example 17.12. The “if” part of Theorem 17.1 (ii) is not true if oneonly requires that f is Thurston equivalent to a rational map. Indeed,

in Example 12.18 we considered two maps f2 and f2 and on a triangular

pillow identified with the Riemann sphere C. Here f2 : C→ C is ratio-

nal and not expanding (it has a critical fixed point), while f2 : C→ Cis an expanding (non-rational) Thurston map. Both maps realize thebarycentric subdivision rule and are hence Thurston equivalent (Propo-

sition 12.3). So f2 is Thurston equivalent to a rational map.

On the other hand, f2 has a critical fixed point. So if % is a visual

metric for f2, then (C, %) is not doubling by Theorem 17.1 (i). Hence

(C, %) is not a quasisphere.

CHAPTER 18

Rational expanding Thurston maps:measure-theoretic properties

In this chapter we consider rational expanding Thurston maps f : C→C. Our main objective is to finish the proof of Theorem 17.1, moreprecisely we will prove part (iii) of this theorem here.

In order to use ergodic methods, we first need an f -invariant mea-sure that is absolutely continuous with respect to Lebesgue measure

on C.

Theorem 18.1. Let f : C → C be a rational expanding Thurs-ton map. Then there exists a unique f -invariant ergodic probabil-

ity measure λ on C that is absolutely continuous with respect to (2-dimensional) Lebesgue measure.

It will be convenient to start the construction of λ from the canon-ical orbifold measure Ω. This measure is induced by the universal orb-

ifold covering map Θ: X → C. Recall that X = D if the orbifold Ofassociated to f is hyperbolic, and X = C if Of is parabolic. Roughlyspeaking, Ω is the push-forward of (hyperbolic, respectively Euclidean)area on X, see Section A.11 for details. In the case of Lattes maps,this measure is already f -invariant, as well as (essentially) the measureof maximal entropy.

Theorem 18.2. Let f : C→ C be a Lattes map. Then the canonicalorbifold measure Ω for f is f -invariant and there is a constant c > 0such that

cΩ = νf = λ.

Here λ is the measure from Theorem 18.1. In general the followingholds.

Theorem 18.3. Let f : C → C be a rational expanding Thurston.Its measure of maximal entropy is absolutely continuous with respect toLebesgue measure if and only if f is a Lattes map.

The above is a weak form of a theorem by A. Zdunik, which saysthat the result is true for any rational map, not only postcritically finiteones, see [Zd].

387

388 18. RATIONAL EXPANDING THURSTON MAPS

**** Old stuff ****The results are mostly from [Me09a], which we follow closely. See

also [Zd] for closely related results.

Definition 18.4 (Lyapunov exponent). Let f : C → C be a ra-

tional map, and µ be a Borel probability measure on C. Then theLyapunov exponent (of f with respect to µ in the chordal metric) is

Lµ :=

∫log f ] dµ.

Thus Lµ measures the average (with respect to µ) expansion off with respect to the chordal metric. The Lyapunov exponent isnon-negative for any measure µ supported on the Julia set of f (see[Prz93]). If µ is ergodic and invariant for f it follows from Birkhoff’sergodic theorem together with the chain rule that for µ-almost every

point x ∈ C

(18.1) limk

1

klog((fk)](x)

)= lim

k

1

k

k−1∑j=0

log(f ](f j(x))

)= Lµ.

Here f will be a rational expanding Thurston map. We will considerLyapunov exponents Lν , Lω with respect to two measures. Namely withrespect to the measure of maximal entropy ν = νf for f (see Chap-ter 16), and with respect to Lebesgue measure. More precisely withrespect to the measure ω which is the (unique) ergodic f -invariant prob-ability measure that is absolutely continuous with respect to Lebesgue

measure on the Riemann sphere C. This measure will be defined in thenext section.

Theorem 18.5. Let f be a rational expanding Thurston map, andthe Lyapunov exponents Lν , Lω be defined as above. Then

Lω ≤1

2log(deg f) ≤ Lν .

The first inequality is an equality if and only if f is a Lattes map. If fis a Lattes map there is equality in the second inequality as well.

We believe that equality holds for both inequalities if and only if fis a Lattes map, but we do not have a proof for the “only if” part forthe second inequality.

We now fix a visual metric d for f with expansion factor Λ > 1.To compare the expansion rates of f with respect to d and the chordalmetric σ we define

(18.2) αν :=log Λ

Lναω :=

log Λ

Lω.

18. RATIONAL EXPANDING THURSTON MAPS 389

Theorem 18.5 may then be rewritten as follows.

Corollary 18.6. Let f : C→ C be a rational expanding Thurstonmap, and αν , αω be defined as above. Then

αν ≤2 log Λ

log(deg f)=

2

dimH(C, d)≤ αω.

The first inequality is an equality if f is a Lattes map. The secondinequality is an equality if and only if f is a Lattes map.

The 2 in the above expression should be viewed as the Hausdorff

dimension of (C, σ). Thus the expression 2/ dimH(C, d) compares theHausdorff dimensions with respect to σ and d.

Theorem 18.7. Let f : C → C be a rational expanding Thurstonmap, αν, αω, and d be as above. Then

(i) for Lebesgue almost every point x ∈ C

limy→x

log d(x, y)

log σ(x, y)= αω;

(ii) for ν-almost every point x ∈ C

limy→x

log d(x, y)

log σ(x, y)= αν .

The dimension of a probability measure µ on a set S equipped witha metric d is defined as

(18.3) dim(µ, d) := infdimH(A, d) | A ⊂ S measurable, µ(A) = 1.

The normalized Lebesgue measure on the sphere C is denoted by λ

(i.e., λ(C) = 1).

Theorem 18.8. Let d be a visual metric for a rational expanding

Thurston map f : C→ C, α as above. Then

(i)

dim(λ, d) =2

αω;

(ii)

dim(ν, σ) =αν log(deg f)

log Λ=

log(deg f)

Lν.

Note that (i) implies by Corollary 18.6 that

dim(λ, d) ≤ dimH(C, d),


where equality occurs if and only if f is a Lattes map. Thus if f is not

a Lattes map, there is a set A ⊂ C of full Lebesgue measure, whoseHausdorff dimension with respect to the visual metric d is strictly less

than dimH(C, d).

18.1. Ergodicity of Lebesgue measure for rational maps

In this section we prove the following.

Theorem 18.9. Let f : C → C be a rational expanding Thurston

map. Then f is ergodic for the Lebesgue measure on C.

This result is well-known, we follow the argument from [McM94a,Theorem 3.9] closely.

To prepare the proof, we first show the following lemma. Let V∞ :=⋃∞n=0 f

−n(post(f)).

Lemma 18.10. There exists a neighborhood U ⊂ C of post(f), such

that for every z0 ∈ C \V∞ the sequence of iterates zn := fn(z0) has a

subsequence znk∈ C \ U (for all k ∈ N).

Proof. For convenience we use a visual metric % for f as in The-orem 15.3 (ii) with expansion factor Λ > 1. Then there is a k0 ∈ N,such that for any p ∈ post(f) the ball Up = B%(p,Λ

−k0) satisfies (15.1),i.e.,

%(f(x), f(p)) = Λ%(x, p), for all x ∈ Up

furthermore

%(p, p′) ≥ 2Λ−k0+1 for distinct points p, q ∈ post(f).

In particular, the balls Up and Uq are disjoint for distinct p, q ∈ post(f).Now let U :=

⋃p∈post(f) Up.

Consider now an arbitrary point z0 ∈ C \V∞. Let zn = fn(z0) bean iterate for some n ∈ N. We want to show that there is an m ≥ n,

such that zm ∈ C \ U . If zn ∈ C \ U we are done. Assume now thatzn ∈ U , thus zn ∈ Up for a p ∈ post(f).

Case 1.Λ−k0−1 ≤ %(x, p) < Λ−k0 .

ThenΛ−k0 ≤ %(f(x), f(p)) < Λ−k0+1.

Thus for q = f(p) ∈ post(f) it holds zn+1 = f(zn) /∈ Uq. Furthermorefor any p′ ∈ post(f) \ q if holds

%(zn+1, p′) ≥ %(p′, q)− %(zn+1, q) ≥ 2Λ−k0+1 − Λ−k0+1 ≥ Λ−k0 ,

18.1. ERGODICITY OF LEBESGUE MEASURE FOR RATIONAL MAPS 391

meaning that zn+1 /∈ Up′ . Thus zn+1 /∈ U as desired.

Case 2.Λ−k−1 ≤ %(x, p) < Λ−k,

for some k > k0. Then the point fk−k0(zn) satisfies the assumptions of

Case 1 (in the ball Bp′ , where p′ = fk−k0(p)). Thus zk−k0+1 ∈ C \ U asdesired.

Before proceeding to the proof of Theorem 18.9, let us remind thereader of the Lebesgue density theorem, which we state only in R2 forsimplicity. Let A ⊂ R2 be a Borel set. Then (Lebesgue) almost everypoint z0 ∈ A is a Lebesgue density point of A, meaning that

(18.4) limε→0+

area(B(z0, ε) ∩ A)

area(B(z0, ε))= 1.

Here “area” denotes the Lebesgue measure on R2 and balls are takenwith respect to the Euclidean metric. In the above we may take qua-siballs instead.

Lemma 18.11. Let A ⊂ R2 be a Borel set, and z0 ∈ A be a Lebesguedensity point of A as above. Let Vn be a sequence of sets such thatz0 ∈ Vn for all n ∈ N, and diamVn → 0 as n → ∞. Furthermore weassume that there is a constant K <∞ such that for rn := diamVn itholds

B(z0, rn/K) ⊂ Vn ⊂ B(z0, rn)

for all n ∈ N. Then

limn→∞

area(Vn ∩ A)

area(Vn)= 1.

Proof. Note that (??) is equivalent to

area(B(z0, ε)) \ A)

area(B(z0)), ε)→ 0 as ε→ 0 + .

Let Vn be a sequence of sets as in the statement, with correspond-ing constant K < ∞. Let rn = diamVn, Bn := B(z0, rn), andB′n(zn, rn/K). Then

area(Vn \ A) ≤ area(Bn \ A)

As well as

area(Vn) ≥ area(B′n) =1

K2area(Bn).

Thusarea(Vn \ A)

area(Vn)≤ K2 area(Bn \ A)

area(Bn)→ 0

as n→∞, which yields the statement.


Proof of Theorem 18.9. Let A ⊂ C be a fully f -invariant set,meaning that f−1(A) = A = f(A), such that area(A) > 0. To provethat f is ergodic for Lebesgue measure, we need to show that area(A) =1.

Let z0 ∈ C \ V ∞ be a Lebesgue density point of A. Now let U ⊂ Cbe a neighborhood of post(f) as in Lemma 18.10. Then there is asubsequence znk

of the sequence of iterates zn = fn(z0), such that

znk∈ C \ U for all k ∈ N.

Let δ0 := dist(post(f), C\U). This means that every ball B(znk, δ0)

is a simply connected domain contained in C \ post(f). It follows thatthere is a univalent branch gk of f−nk defined on B(znk

, δ0) that mapsznk

to z0. Let Bk := B(znk, δ0/2) and Vk := gk(Bk). We note that

the diameter of Vk → 0 as k → ∞. This is most easily seen using thecanonical orbifold metric ω of f . Namely, by Proposition A.25 there isa constant ρ > 1 such that

diamω(Vk) ≤ δ0 ρ−nk ,

for all k ∈ N. Recall that the chordal metric σ is quasisymmetricallycomparable to ω (see Lemma A.24 (i)), thus it follows that diamσ Vk →0 as k →∞.

Note that by Koebe distortion (see Theorem 17.8), the sets Vn sat-isfy the assumption of Lemma ??. Thus

area(Vk ∩ A)

area(Vk)→ 1

as k → ∞. Since A is f -invariant, it follows using Koebe distortionthat

area(Vk \ A) area(Bk \ A)‖g′k(z0)‖2σ area(Bk \ A) area(Vk),

with constants C() that are independent of k. Hence

area(Bk ∩ A)

area(Bk)→ 1

as k →∞.By compactness there is a convergent subsequence of znk

, whichwe still denote by znk

for convenience. Let z∞ := limk→∞ znkand

B := B(z∞, δ0/4).For sufficiently large k it holds B ⊂ Bk. Then area(B \ A) ≤

area(Bk \ A) as well as area(B) ≥ 14

area(Bk), meaning that

area(B \ A)

area(B)≤ 4

area(Bk \ A)

area(Bk).

18.2. THE JACOBIAN OF A MEASURE 393

Since the right hand side goes to 0 as k →∞ it follows that

area(B \ A) = 0.

Recall from Lemma 6.5 that f is eventually onto, meaning there is

an n ∈ N such that fn(B) = C. Since A is f -invariant it follows thatfn(B∩A) = A. Since area(B\A) = 0 it follows that area(fn(B\A)) =0. Thus area(A) = area(fn(B∩A)) = 1. Thus f is ergodic with respectto Lebesgue measure as desired.

18.2. The Jacobian of a measure

Definition 18.12 (Jacobian). Let (X,F , µ), (X ′,F ′, µ′) be mea-sure spaces, i.e., F is a σ-algebra on the set X, and µ is a measureon (X,F); analog statements hold for (X ′,F ′, µ′). Let T : X → X ′ bea measurable map. Then a µ-measurable function J : X → [0,∞) iscalled a Jacobian of T if for any F -measurable set A ⊂ X on which Tis injective, the set T (A) is F ′-measurable and

µ′(T (A)) =

∫A

J dµ.

We will sometimes write J = JT,µ,µ′ if we want to emphasize theinvolved map and measures. Most of the time our measure spaces willbe identical, i.e., (X,F , µ) = (X ′,F ′, µ′). We then write J = Jµ =Jf,µ = Jf depending on what dependence we want to emphasize.

If A ∈ F is a set on which T is injective, and ρ′ : X ′ → R is afunction in L1(µ′) it holds

(18.5)

∫T (A)

ρ′ dµ′ =

∫A

ρ′ T · JT dµ.

Let (X ′′,F ′′, µ′′) be a third measure space, and S : X ′ → X ′′ be ameasurable map. Then

JST (x) = JS(T (x))JT (x),

if the Jacobians JT and JS exist.A necessary and sufficient condition for the existence of the Ja-

cobian is that for any set A on which T is injective the measure(T |A)∗µ′ := µ′ T |A on A is absolutely continuous with respect toµ. In this case J is the Radon-Nikodym derivative d(T |A)∗µ′/dµ.

We will however be only interested in the case where X and X ′ will

be D,C, or C. The measures µ, µ′ will be absolutely continuous with

respect to Lebesgue measures on D,C, and C. Furthermore T = f willbe a holomorphic function. In this case the existence of the Jacobianfollows from elementary calculus.


Indeed let dA denote Lebesgue measure on X and X ′. Let κ andκ′ be the densities of µ and µ′, i.e., dµ = κdA and dµ′ = κ′dA. Then

(18.6) Jf,µ,µ′(z) =|f ′(z)|2 κ′(f(z))

κ(z).

Assume now that X = X ′ = C and µ = µ′ is an f -invariant proba-

bility measure on C absolutely continuous to (2-dimensional) Lebesgue

measure on C. Define tiles in terms of a Jordan curve C ⊂ C withpost(f) ⊂ C, that has 0 area (i.e., µ(C) = 0). Then f is injective onany 1-tile X ∈ X1. This shows that

(18.7)

∫J dµ =

∑X∈X1

∫Jdµ =

∑X∈X1

µ(f(X)) = deg f.

The (measure theoretic) entropy of f : (C, µ) → (C, µ), where µ is

a Borel measure on C, may be expressed via the Jacobian

(18.8) h = hµ =

∫log Jµ dµ.

This is known as Rohlin’s formula (see for example [PU], Chapter 1).

***** old stuff *****

We remind the reader of the following (well-known) fact. A Borel

measure µ on C is invariant for f if and only if for all ϕ ∈ L1(µ)

(18.9)

∫ϕ f dµ =

∫ϕdµ.

This is immediate for characteristic functions and follows for L1(µ)functions by the usual approximation process. Thus, using the above

hω =

∫log(f ])2 dω +

∫log κ(f(z)) dω −

∫log κ(z) dω = 2Lω.

(18.10)

Using Jensen’s inequality we obtain

(18.11) 2Lω = hω =

∫log J dω ≤ log

∫J dω = log deg f,

where equality occurs if and only if J = const = deg f (ω almosteverywhere).

It was shown in [Zd] that Lebesgue measure is not absolutely con-tinuous to the measure of maximal entropy unless f is a Lattes map(has parabolic orbifold). Thus for the measure ω (which is absolutely

18.3. THE INVARIANT ABSOLUTELY CONTINUOUS MEASURE 395

continuous with respect to λ), there is inequality in the above. We willhowever prove this directly here.

18.3. The invariant absolutely continuous measure

In the following, f : C→ C is a rational Thurston map with no periodiccritical points. In this section we construct an f -invariant probability

measure λ on C that is absolutely continuous with respect to Lebesgue

measure on C.For simplicity let dA be (2-dimensional) Lebesgue measure on the

plane C ⊂ C. Then the Jacobian with respect to dA is J = |f ′|2.

Consider a probability measure µ on C that is absolutely continuous

with respect to dA, i.e., dµ = κdA, where κ ∈ L1(C, dA) with∫dµ = 1.

The pushforward of µ by f , i.e., f∗µ = µ f−1 is given as follows. Let

z ∈ C \ post(f) arbitrary. To simplify the discussion, we furthermoreassume that z as well as f−1(z) are contained in C (i.e., none of thepoints considered here is ∞). Let f−1

j (j = 1, . . . , d = deg(f)) bebranches of the inverse of f defined in a suitable small neighborhoodof z. Then

df∗µ(z) =∑

j=1,...,d

κ(f−1j (z))dA(f−1

j (z))

=∑

w∈f−1(z)

κ(w)|f ′(w)|−2dA(z) =∑

w∈f−1(z)

κ(w)J−1(w)dA(z).

Clearly µ is f -invariant if and only if f∗µ = µ, which by the aboveis the case if and only if

κ(z) =∑

w∈f−1(z)

κ(w)J−1(w)

for Lebesgue almost every z ∈ C. Here, we could have used otherbackground measures instead, the Jacobian then has to be computedwith respect to this measure.

This motivates the following definition. Let P = post(f) and CP :=

C\P . To simplify (at least conceptually) the discussion, we will restrict

ourselves to non-negative functions κ that are continuous on CP . To

this end let C(CP ) be the space of continuous functions ρ : C→ [0,∞).We will later see that we can in fact restrict ourselves to the set ofbounded continuous functions ρ : CP → [0,∞) denoted by Cb(C). The

Ruelle or transfer operator R : C(CP ) → C(CP ) is defined as follows.


For ρ ∈ C(CP ) let R(ρ) be the function defined as

(18.12) R(ρ)(x) =∑

y∈f−1(x)

ρ(y)Jf (y)−1

for x ∈ CP .It will be convenient to compute the Jacobian Jf with respect to

the orbifold measure Ω, which is the measure induced on C via theuniversal orbifold cover. The construction of Ω is given in Section A.11,it is reviewed here for the convenience of the reader.

Let αf be the ramification function of f , so that Of = (C, αf )is the orbifold associated to f . Recall that by assumption f has no

periodic critical points, thus αf (p) 6= ∞ for all p ∈ C (see Proposi-tion 2.9 (ii)). Recall from Section A.8 that the universal orbifold cov-

ering map Θ: X → C is a holomorphic branched covering map withdeg(Θ, u) = αf (f(u)) for all u ∈ X. Here X = D if Of is hyperbolic,and X = C if Of is parabolic. If the case of X = C, we always assumethat C is equipped with Lebesgue measure dA of the plane. WhenX = D, we assume that D is equipped with hyperbolic area measuredA = dAh. Recall that a map τ : X → X such that Θ = Θ τ is calleda deck transformation of Θ. X. Conversely, for any u, v ∈ X with

Θ(u) = Θ(v) ∈ CP there is a unique deck transformation τ : X → Xwith τ(u) = v. See Proposition A.21 and Corollary A.20.

Every deck transformation τ is a Mobius transformation, in fact an(hyperbolic or Euclidean) isometry. Hence, it preserves (hyperbolic orEuclidean) area. We denote by G the group of all deck transformationsof Θ. G acts cocompactly on X (see Proposition A.21 (iv)), so thereis a fundamental domain FG ⊂ X of G, such that the closure FG iscompact. See Definition A.26 as well as Theorem A.27.

The canonical orbifold measure Ω (of Of ) on C is defined as

(18.13) Ω(V ) := area(Θ−1(V ) ∩ FG).

for any Borel set V ⊂ C. Here “area” denotes hyperbolic area in thecase when Of is hyperbolic, or the Euclidean area in the case when Ofis parabolic. The canonical orbifold measure Ω is well-defined, satisfies

0 < Ω(C) <∞, and is mutuallly absolutely continuous with respect to

Lebesgue measure on C, see Lemma A.28.From now on, the Jacobian Jf of f refers to this measure Ω. Fur-

thermore, with respect to these measures on X and C, the Jacobian of

the map Θ: X → C is JΘ ≡ 1.


Let d = deg(f), then on a suffiently small neighborhood V of a

given point x ∈ CP , there are d inverse branches g1, . . . , gd : V → CP

of f defined on V that give the d distinct preimages of a point v ∈ V .Then

R(ρ)(v) =d∑

k=1

ρ(gk(v))Jf (gk(v))−1

for v ∈ V . Together with (18.5) this shows that R(ρ) ∈ C(CP ) for ρ ∈C(CP ), and we also see thatR is a continuous linear operator on C(CP )

if we equip C(CP ) with the topology of locally uniform convergence.

More specifically, if ρn is a sequence in C(CP ) with ρn → ρ ∈ C(CP )

locally uniformly on CP , then R(ρn)→ R(ρ) locally uniformly on CP .

Lemma 18.13. The Ruelle operator satisfies

(i) R : C(CP ) ∩ L1(C,Ω)→ C(CP ) ∩ L1(C,Ω). Furthermore∫Cρ dΩ =

∫CR(ρ) dΩ,

for all ρ ∈ C(CP ) ∩ L1(C,Ω).

(ii) A measure λ of the form dλ = ρ dΩ where ρ ∈ C(CP ) ∩L1(C,Ω) is f -invariant if and only if ρ is a fixed point of R.

Both statements follow directly from the fact that the pushforward

of a measure λ on C of the form dλ = ρ dΩ where ρ ∈ C(CP )∩L1(C,Ω)is given by df∗λ = R(ρ) dΩ as we saw in the beginning of this section.

It will be convenient to lift the Ruelle operator to the universalorbifold cover of Of , i.e., to X = D in the hyperbolic case or X = Cin the parabolic case.

Let XP = Θ−1(CP ) = X \ crit(Θ). Given a continuous function

ρ : CP → [0,∞) the lift of ρ by Θ is the continuous function ρ : XP →[0,∞) given by

(18.14) ρ(z) = ρ(Θ(z))

for all z ∈ XP . Note that ρ is equivariant with respect to the groupof deck transformations G, meaning that ρ = ρ g for all g ∈ G.Denote by C(X,G) the set of all G-equivariant continuous functionsρ : XP → [0,∞).

Lemma 18.14.

(i) For any ρ ∈ C(X,G) there is a unique function ρ ∈ C(CP )satisfying (18.13).


(ii) It holds ρ ∈ L1(C,Ω) if and only if ρ ∈ L1(FG). In this case∫Cρ dΩ =

∫FG

ρ dA.

Proof. (i) Given ρ ∈ C(X,G) we define ρ ∈ CP by

ρ(x) := ρ(z),

where z ∈ Θ−1(x) is arbitrary. Clearly this is well defined and ρ is thelift of ρ as in (18.13), meaning that ρ = ρ Θ. Since ρ(x) = ρ(Θ−1(x))for a suitable branch of the inverse of Θ defined on some small ballin CP , it follows that ρ ∈ C(CP ). That ρ is the unique continuousfunction satisfying (18.13) is obvious.

(ii) This follows immediately from (18.4) and the fact that JΘ ≡1.

Recall from Lemma A.23 that there are d = deg(f) nonequivalentlifts of inverse branches of f to the orbifold cover, i.e., holomorphicfunctions A1, . . . , Ad : X → X with the following properties:

f Θ Ak = Θ for k = 1, . . . , d, and(18.15)

f−1(x) = Θ(A1(z)), . . . ,Θ(Ad(z)),

whenever x ∈ C and z ∈ Θ−1(x).These functions Ak allow us to lift the Ruelle operator to the orb-

ifold cover X in an elegant form. Namely, for a G-equivariant functionρ : XP → [0,∞) we define

(18.16) R(ρ)(z) =d∑

k=1

ρ(Ak(z))JAk(z)

for all z ∈ XP . Here the Jacobian JAkof Ak is computed with respect

to the hyperbolic area in the case when X = D, and with respect to (2-dimensional) Lebesgue measure when X = C. Let us record properties

of R.

Lemma 18.15 (Lift of the Ruelle operator). The operator R is welldefined and satisfies the following.

(i) R ∈ C(X,G) for any ρ ∈ C(X,G), i.e.,

R : C(X,G)→ C(X,G).

(ii) For any ρ ∈ C(CP ) it holds

R(ρ) = R(ρ).


(iii) For any ρ ∈ C(X,G)∫FG

R(ρ) dA =

∫FG

ρ dA.

(iv) R has a fixed point ρ ∈ C(CP ) if and only if R has a fixedpoint ρ ∈ C(X,G). In this case ρ = ρ Θ.

That R is well defined means that it is independent of the choiceof the lifts A1, . . . , Ad. The second property is the reason that we call

R the lift of R to the universal orbifold cover.

Proof. Let A′1, . . . , A′d : X → X be other nonequivalent lifts of in-

verse branches of f to the universal orbifold cover, i.e., maps satisfyingthe two properties in (18.14). Relabeling these maps if necessary, thereare deck transformations g1, . . . , gd ∈ G such that A′k = gk Ak, fork = 1, . . . , d (see Lemma A.23). Consider now a function ρ ∈ C(X,G).Since ρ is G-equivariant it follows that ρ(A′k(z)) = ρ(gk Ak(z)) =ρ(Ak(z)). Note that Jgk ≡ 1, since gk is an isometry. Thus

JA′k(z) = JgkAk= Jgk(Ak(z))JAk

(z) = JAk(z).

This shows thatd∑

k=1

ρ(A′k(z))JA′k(z) =d∑

k=1

ρ(Ak(z))JAk(z).

Thus R is well defined.

(i) Consider a ρ ∈ C(X,G). Clearly R(ρ) is a continuous functionXP → [0,∞). To show that it is G-equivariant, let z ∈ XP and g ∈ Gbe arbitrary. Let w = g(z). As before Jg ≡ 1, since g is an isometry.Thus

JAkg(z) = JAk(g(z))Jg(z) = JAk

(g(z)).

We remind the reader that Ak 6= Ak g in general, but A1 g, . . . , Ad g are again d non-equivalent lifts of f−1 by Θ, meaning they satisfy

(18.14), see Lemma A.23. We have already seen that R is well defined,

i.e., the maps Ak g could be used in the definition of R. Thus

R(ρ)(w) = R(ρ(g(z)) =d∑

k=1

ρ(Ak g(z))JAk(g(z))

=d∑

k=1

ρ(Ak g(z))JAkg(z) = R(ρ)(z).

This shows that R(ρ) isG-equivariant, thus R(ρ) ∈ C(X,G) as desired.


(ii) Recall that we equip X with the hyperbolic area in the casewhen X = D and with the 2-dimensional Lebesgue measure in the case

when X = C. Furthermore, C is equipped with the canonical orbifold

measure. The Jacobian of Θ: X → C with respect to these measuresis JΘ ≡ 1. Recall that the lifts Ak satisfy Θ = f Θ Ak, therefore

JΘ = 1 = Jf (Θ Ak(z)) JΘ(Ak(z))︸︷︷︸=1

JAk(z) = Jf (Θ Ak(z))JAk

(z),

which yields

(18.17) Jf (Θ(Ak(z))−1 = JAk(z)

for all z ∈ XP . Consider now arbitrary ρ ∈ C(CP ) and z ∈ Xp. Thenusing (18.13)

R(ρ)(z) = R(ρ)(Θ(z)) =∑

y∈f−1(Θ(z))

ρ(y)Jf (y)−1

=d∑

k=1

ρ(Θ(Ak(z)))Jf (Θ(Ak(z))−1 =d∑

k=1

ρ(Ak(z))JAk(z)

= R(ρ)(z),

as desired.

(iii) This follows immediately from Lemma 18.12 (i) together withLemma 18.13 (ii).

(iv) Let ρ ∈ C(CP ) be a fixed point of R. Then by (ii) R(ρ) =

R(ρ) = ρ, i.e., ρ = ρ Θ ∈ C(X,G) is a fixed point of R.

Now let ρ ∈ C(X,G) be a fixed point of R. According to Lemma 18.13 (i)

there is a unique function ρ ∈ C(CP ) with ρ = ρ Θ. Again by (ii)

it follows that R(ρ) = R(ρ) = ρ. The uniqueness of ρ shows thatR(ρ) = ρ as desired.

It will be enough for our purposes to consider bounded continuousfunctions. Let Cb(X,G) ⊂ C(X,G) be the set of all bounded contin-

uous G-equivariant functions ρ : XP → [0,∞) and Cb(CP ) ⊂ C(C) be

the set of all bounded continuous functions ρ : CP → [0,∞).

Lemma 18.16. The operators R and R map bounded functions tobounded functions, i.e.,

R : Cb(X,G)→ Cb(X,G)

R : Cb(CP )→ Cb(C).


Proof. Recall that FG ⊂ X is a fundamental domain of the ac-tion of G on X, and that FG is compact. Let the nonequivalent liftsA1, . . . , Ad of f−1 by Θ be given as before. Then JAk

assumes its max-imum on FG for every k = 1, . . . , d. Let ρ ∈ Cb(X,G) be arbitrary.

It follows that R(ρ) is bounded on FG by definition (18.15), hence

bounded on X, since R(ρ) is G-equivariant by Lemma 18.14 (i).

To see the second statement we note that ρ ∈ Cb(CP ) if and only if

ρ = ρΘ ∈ Cb(X,G). Lemma 18.14 (ii) thus shows that R : Cb(CP )→Cb(CP ).

The previous statement is one of the main reasons to use the orbifold

measure and the orbifold lift R, since it is not easy to see that R mapsbounded functions to bounded functions directly from the definition in

(18.11). Clearly any function ρ ∈ Cb(CP ) is integrable. Thus using

the lift R instead of R allows us to avoid questions about integrability,particularly near postcritical points.

From the previous lemma it is obvious that Lemma 18.14 remains

true if we replace any C(X,G) by Cb(X,G) and any C(CP ) by Cb(CP )in the statement.

Thus using Lemma 18.14 together with the previous remark as well

as Lemma 18.12, it is enough to find a fixed point ρ ∈ Cb(X,G) of Rto obtain the f -invariant measure that is absolutely continuous with

respect to Lebesgue measure on C.

Consider first the parabolic case, i.e., when the universal orbifoldcover is the plane X = C. Since our map f has no periodic criticalpoints by assumption, this means that f is a Lattes map by Theo-rem 3.1 (i). From Theorem 3.1 (ii) we see that the lifts Ak : C → Cof f−1 by Θ are of the form Ak(z) = z/αk + βk, where αk, βk ∈ C and|αk|2 = d = deg(f). Thus JAk

≡ 1/|αk|2 = 1/d. It follows that ρ ≡ 1

is a fixed point of R. Thus the canonical orbifold measure Ω is in fact

f -invariant. Let c = 1/Ω(C) = 1/ area(FG), then

(18.18) λ = cΩ

is our desired probability measure that is absolutely continuous withrespect to Lebesgue measure. We also note that in this case the Ja-cobian with respect to λ agrees with the Jacobian with respect to Ω.This yields together with (18.16) that for any Lattes map

(18.19) Jλ(z) = d = deg(f),

for all z ∈ C \ post(f).


Having done the parabolic case, we assume from now on that theorbifold Of of f is hyperbolic, meaning that the orbifold cover is X =D. Note however, that the following argument actually works equallywell in the parabolic case.

To obtain a fixed point of the lift of the Ruelle operator R inCb(X,G), we will show that it leaves a certain convex and compactsubset K of Cb(X,G) invariant.

Lemma 18.17 (Fixed point of R). Assume there is a compact setK ⊂ Cb(D, G) that is convex, bounded, contains a constant function

ρc = const = c > 0, and satisfies R(K) ⊂ K. Then R has a fixed pointin K.

Note that the functions in Cb(D, G) are bounded below (by 0)according to the definition. Thus “K is bounded” means that it isbounded above, i.e., there is a constant N < ∞ such that ρ(z) ≤ Nfor all ρ ∈ K and z ∈ DP .

Proof. Let ρ1 := ρc ≡ c > 0. Define

ρn :=1

n

n−1∑i=0

Ri(ρ1) ∈ K

for n ∈ N (here Ri denotes the i-th iterate of Ri). By compactness ofK , there exists a convergent subsequence ρnk

that converges locally

uniformly on CP to a function ρ ∈ K. We claim that R(ρ) = ρ. Indeedfor any x ∈ DP and n ∈ N

|R(ρn)(x)− ρn(x)| = 1

n

∣∣Rn(ρ1)(x)− ρ1(x)∣∣ ≤ N

n→ 0 as n→∞.

Here N is the upper bound of the functions in K. Hence

R(ρ)(x) = limk→∞R(ρnk

)(x) = limk→∞

ρnk(x) = ρ(x)

for all x ∈ CP . The claim follows.

The remainder of this section deals with the construction of a setK ⊂ Cb(D, G) as in the previous lemma.

LetG∼ be the equivalence relation on D induced by G. This means

zG∼ w if and only if there exists g ∈ G with w = g(z), which is the

case if and only if Θ(z) = Θ(w) (for all z, w ∈ D). Put differently, the

equivalence classes ofG∼ are exactly the orbits of G.

Let A be the class of all holomorphic functions ϕ : D→ D such that

(18.20) ϕ(z)G∼ ϕ(w)⇒ z

G∼ w


for all z, w ∈ D. Recall from (A.20) that the lifts Aj of f−1 by Θsatisfy this implication. It follows that arbitrary compositions Aj1 · · · Ajn (where j1, . . . , jn ∈ 1, . . . , d) satisfy the same condition,meaning they are contained in A. More generally, A is closed undercompositions, i.e., ϕ, ψ ∈ A ⇒ ϕ ψ ∈ A.

Define for all z, w ∈ DP

M(z, w) := supJϕ(z)/Jϕ(w) : ϕ ∈ A.

Recall that Jϕ denotes the Jacobian of ϕ ∈ A with respect to thehyperbolic metric on D as before.

Lemma 18.18. Let M be defined as above. Then

M : DP × DP → [1,∞)

is continuous and satisfies the following.

(i) For all z, w ∈ DP , and A ∈ A

M(A(z), A(w))JA(z) ≤M(z, w)JA(w).

(ii) Fix an arbitrary p ∈ crit(Θ) and a ball B = B(p, δ) ⊂ Daround p. Then for any fixed w0 ∈ DP the map

z 7→M(z, w0) is bounded on B \ p.

Proof. We first note that idD ∈ A, thus M(z, w) ≥ 1 for allz, w ∈ D.

If z0 ∈ CP , then there exists a small open disk B = B(z0, δ) ⊂ DP

such that no two distinct points in B are G-equivalent. It follows thateach function ϕ ∈ A is injective on B. Hence finiteness and positivity ofM onB×B follow from Koebe’s distortion theorem (see Theorem 17.8).These properties can easily be promoted to uniform upper bounds forarbitrary pairs in any compact subset of DP by using the (obvious)inequality

(18.21) M(u,w) ≤M(u, v)M(v, w)

for u, v, w ∈ DP in combination with a chaining argument.

(i) Let z0, w0 ∈ DP be arbitrary. Let B = B(z0, δ) ⊂ DP be thesame disk as above, and similarly B′ = B(w0, δ

′) ⊂ DP be a diskaround w0 such that no two points in B′ are equivalent. Again byKoebe distortion there is a function C : [0, δ) → [1,∞) such that forall z ∈ B

1

C(|z − z0|)Jϕ(z) ≤ Jϕ(z0) ≤ C(|z − z0|)Jϕ(z)


where C(|z−z0|)→ 1 as z → z0. Here the function C does not dependon ϕ. A similar estimate holds in the ball B′ with a different functionC ′. Then

1

C(|z − z0|)C ′(|w − w0|)≤∣∣∣∣ Jϕ(z)

Jϕ(w)− Jϕ(z0)

Jϕ(w0)

∣∣∣∣ ≤ C(|z−z0|)C ′(|w−w0|),

for z ∈ B, w ∈ B′. This implies the continuity of M on DP × DP .

Let ϕ,A ∈ A, recall that ϕ A ∈ A. Furthermore

JϕA(u) = Jϕ(A(u)) · JA(u)

for u ∈ D.If we use this for points z, w ∈ U ′, then we conclude that

Jϕ(A(z))

Jϕ(A(w))JA(z) =

JϕA(z)

JϕA(w)JA(w) ≤M(z, w)JA(w).

Taking the supremum over all ϕ ∈ A yields (i).

(ii) We now consider a point p ∈ crit(Θ) ⊂ D, as well as a mapA ∈ A. Roughly speaking, this map behaves as w 7→ wk near p insuitable coordinates. This shows that the Jacobian of A is boundednear p. Using Koebe distortion, we will show that these bounds areindependent of the chosen map A.

Here are the details. Note first that p is a point with non-trivialstabilizer subgroup Gp (see Proposition A.21 (ii)). Recall that anyg ∈ Gp is an isometry that fixes p, meaning it is a rotation around p.Let m = #Gp.

Consider the distance of p to other points in its G-orbit, i.e.,

dist(p,G(p) \ p) = infdh(p, (g(p)) : g ∈ G \Gp.

Since G acts properly discontinuously, this infimum is attained anddist(p,G(p) \ p) > 0.

Let δ = 13

dist(p,G(p) \ p), and B = B(p, δ) be a (hyperbolic)ball around p. Then g(B) ∩B = ∅ for g ∈ G \Gp.

Consider now an arbitrary map A ∈ A. Let B′ := A(B). From theprevious considerations together with (18.19), it follows that the com-ponent of A−1(B′) containing p is B. Thus the restriction A|B : B →B′ := A(B) is a proper map such that its degree k = deg(A|B) dividesm, and p is its only critical point. Thus A|B is conformally conjugateto wk : D→ D. More precisely, there are disks D1 = B(0, r1) ⊂ D andD2 = B(0, r2) ⊂ D, where 0 < r1, r2 < 1, as well as conformal mapsh : B → D1, and ψ : D2 ⊂ D→ B′ such that

A = ψ(h(z)k)


for all z ∈ D1. Put differently, the diagram

B′ BA

oo

h

D2 ⊂ D

ψ

OO

D1 ⊂ Dwkoo

commutes. Note that h is independent of the chosen map A, whereasψ depends on A. We have

A′(z) = ψ′(h(z)k)kh(z)k−1h′(z).

Fix a w0 ∈ B with dh(p, w0) = 12δ. Since |kh(w0)k−1h′(w0)| is

a constant, it follows that |A′(w0)| |ψ′(h(w0)k)|, hence JA(w0) |ψ′(h(w0)k)|2, where the constant C() is independent of A.

Consider now an arbitrary point z ∈ B(p, 12δ). By Koebe distortion

it follows that

|h′(z)| 1 as well as |ψ′(h(z)k)| |ψ′(h(w0)k)|

where the constants are independent of A and z. Clearly |h(z)k−1| isbounded on B(p, 1

2δ). Thus

|A′(z)| . |A′(w0)|,

which implies

JA(z) . JA(w0).

Thus M(z, w0) is bounded for z ∈ B(p, 12δ). We have proved the claim

using this point w0. Now let w0 ∈ DP be arbitrary. From (18.20)we obtain that M(z, w0) ≤ M(z, w0)M(w0, w0) is bounded for z ∈B(p, 1

2δ). The claim follows.

We now define the set K ⊂ Cb(D, G) as the set of all functionsρ ∈ Cb(D, G) satisfying

ρ(z) ≤M(z, w)ρ(w) for all z, w ∈ DP ,(18.22)

as well as ∫FG

ρ dAh = 1.(18.23)

Recall that FG ⊂ D was a fundamental domain of G.

Lemma 18.19 (Properties of K). The set K satisfies the following.

(i) K is bounded, meaning there is a constant N <∞ such that

ρ(z) ≤ N for all ρ ∈ K and z ∈ DP .


(ii) K is locally bounded below, meaning for any compact set C ⊂DP there is a constant c > 0 such that

ρ(z) ≥ c for all ρ ∈ K and z ∈ C.

(iii) K is compact, convex, and contains a nonnegative constantfunction.

(iv) K is invariant by R in the sense that R(K) ⊂ K.

Proof. (i) Consider a fixed w0 ∈ DP . Then there is a ball B =B(w0, δ) ⊂ DP such that g(B) ∩B = ∅ for all g ∈ G \ idD. Then forany function ρ ∈ K it holds∫

B

ρ dAh ≤∫FG

ρ dAh.

Note that trivially M(w0, w0) = 1. If ρ(z0) is large, continuity of Mtogether with (18.21) thus implies that

∫FGρ dAh is large. Thus ρ does

not satisfy the normalization (18.22).

Therefore ρ(w0) ≤ N <∞ for all ρ ∈ K. Since ρ(z) ≤M(z, w0)ρ(w0)for all z ∈ DP , it follows from Lemma 18.17 (ii) that ρ is uniformlybounded on FG∩DP . Since ρ is G-equivariant, and FG is a fundamentaldomain of G, it follows that ρ is uniformly bounded on DP ,

(ii) The argument is very similar the the previous one. Namelylet w0 ∈ DP be arbitrary. If ρ(w0) is small it follows from ρ(z) ≤M(z, w0)ρ(w0) together with Lemma 18.17 (ii) that ρ is small on FG ∩DP . Thus ρ does not satisfy the normalization (18.22).

(iii) Since M(z, w) ≥ 1 for all z, w ∈ DP it is clear that K containsthe nonnegative constant function ρc ≡ c = 1/ area(FG). The convexityof K is clear as well. Let ρn be a sequence in K that converges locallyuniformly to ρ. Clearly ρ is contained in Cb(D, G), satisfies (18.21), aswell as the normalization (18.22) (here we are using (i)). Thus K iscompact.

(iv) It remains to show that R(K) ⊂ K. Let ρ ∈ K be arbitrary.

We first note that Lemma 18.14 (iii) shows that R(ρ) again satisfiesthe normalization in (18.22). It remains to show that

R(ρ)(z) ≤M(z, w)R(ρ)(w)

for all z, w ∈ DP . Recall from Lemma 18.17 (i) that

M(Ak(z), Ak(w))JAk(z) ≤M(z, w)JAk

(w)


for z, w ∈ DP and k = 1, . . . , d. Hence (see (18.15))

R(ρ)(z) =d∑

k=1

ρ(Ak(z))JAk(z)

≤d∑

k=1

ρ(Ak(w))M(Ak(z), Ak(w))JAk(z) by (18.21)

≤d∑

k=1

ρ(Ak(w))M(z, w)JAk(w) = M(z, w)R(ρ)(w)

for all ρ ∈ Cb(D, G) and z, w ∈ DP . The claim follows.

We are now ready to prove the existence of the f -invariant abso-lutely continuous probability measure λ.

Proof of Theorem 18.1.By Lemma 18.18 and Lemma 18.16 the lift of the Ruelle operator Rhas is a fixed point ρ ∈ K. Thus R has a fixed point ρ where ρ = ρΘby Lemma 18.14 (iv). By Lemma 18.12 (ii) it follows that λ = ρ dΩ isf -invariant.

Since ρ ∈ K the normalization (18.22), together with Lemma 18.13 (ii)yield ∫

Cρ dΩ =

∫FG

ρ dA = 1,

thus λ is a probability measure on C.

Since ρ : CP → (0,∞) is a bounded continuous function Clearly, λis mutually absolutely continuous with respect to the canonical orbifoldmeasure Ω, which in turn is mutually absolutely continuous with re-

spect to Lebesgue measure on C, see Lemma A.28 (ii). Thus, it followsfrom Theorem 18.9 that f is ergodic with respect to λ.

The uniqueness of λ follows, since two ergodic f -invariant proba-bility measures are identical or singular.

Let us record the properties of the density ρ of λ.

Lemma 18.20. Let λ be the f -invariant probability measure andρ be its density with respect to the canonical orbifold measure, i.e.,dλ = ρ dΩ. Then

(i) ρ : C \ post(f)→ (0,∞) is a bounded continuous function.

If κ is the density of λ with respect to Lebesgue measure dA on the

plane C ⊂ C, i.e., dλ = κ dA it follows using Lemma A.28 (ii) that

(18.24) κ : C \ post(f)→ (0,∞) is continuous.


18.4. Lattes maps, the measure of maximal entropy, andLebesgue measure

We now prove Zdunik’s theorem. Let f : C → C be a rationalexpanding Thurston map and µ be an f -invariant probability measure

on C. Recall from Chapter 16 that the (metric) entropy hµ of µ satisfieshµ ≤ log(deg(f)). Furthermore equality is attained for µ = νf , i.e., themeasure of maximal entropy constructed in Proposition 16.10.

Theorem 18.21. Let f : C → C be a rational expanding Thurstonmap and λ be the f -invariant probability measure that is absolutelycontinuous with respect to Lebesgue measure according to Theorem 18.1.Then for the entropy hλ it holds

hλ = log deg(f) if and only if f is a Lattes map.

We first prove the following lemma.

Lemma 18.22. Let f : C → C be a rational expanding Thurstonmap. Then f is a Lattes map if and only if for all p ∈ post(f), n ∈ N,and q, q′ ∈ f−n(p) \ post(f) it holds

(18.25) deg(fn, q) = deg(fn, q′).

Proof. Let f : C → C be a rational expanding Thurston map.

Assume first that f is a Lattes map. Since the orbifold Of = (C, αf )associated to f is parabolic (see Theorem 3.1) it follows from Proposi-tion 2.13 (iii) by induction that for any n ∈ N and q ∈ f−n(p)\post(f)

αf (p) = deg(fn, q).

This proves (18.24).

To show the other implication, assume now that (18.24) holds. Fixa p ∈ post(f).

Claim 1. It holds

deg(fn, q) = deg(fm, q′)

for all q ∈ f−n(p)\post(f), and q′ ∈ f−m(p)\post(f), where n,m ∈ N.Indeed, given such q, q′, let u ∈ f−k(q) (where k ∈ N). Then

deg(fk, u) = 1, so

deg(fn+k, u) = deg(fn, q) deg(fk, u) = deg(fn, q).

Similarly for u′ ∈ f−j(q′) (where j ∈ N) it holds deg(fm+j, u′) =deg(fm, q′). Choose k, j ∈ N such that n + k = m + j. Then u, u′ ∈f−(n+k)(p) \ post(f) and (18.24) implies

deg(fn, q) = deg(fn+k, u) = deg(fm+j, u′) = deg(fm, q′),

18.4. LATTES MAPS, ENTROPY, AND LEBESGUE MEASURE 409

which is the claim.

Claim 2. The ramification function of f is given by

αf (p) = deg(fn, q)

for all p ∈ C, where q ∈ f−n(p) \ post(f).Indeed, αf (p) is (by Definition 2.7) the lowest common multiple of

all numbers deg(fk, u), where k ∈ N and u ∈ f−k(p).Let u ∈ f−k(p) be arbitrary. Note that for j ∈ N sufficiently

large, there is a q ∈ f−j(u) that is not a postcritical point. Herewe are using the fact that f does not have periodic critical points, seeProposition 2.3. Then deg(fk+j, q) = deg(fk, u) deg(f j, q) is a multipleof deg(fk, u). Claim 2 now follows from Claim 1.

After these preparations let p, q ∈ C with f(q) = p be arbitrary.Since f has no periodic critical points, there is a sufficiently large n ∈ Nand a u ∈ f−n(q) that is not a postcritical point. Then fn(u) = q andfn+1(u) = p. Thus αf (q) = deg(fn, u) and αf (p) = deg(fn+1, u) byClaim 2. Therefore

αf (p) = deg(fn+1, u) = deg(fn, u) deg(f, q) = αf (q) deg(f, q).

This shows that f has parabolic orbifold by Proposition 2.13 (iii). Thusf is a Lattes map by Theorem 3.1 (i).

Proof of Theorem 18.20. Let f : C→ C be a rational expand-ing Thurston map, and λ be the f -invariant, absolutely continuous

measure on C given by Theorem 18.1.By Rohlin’s formula (18.7) we have

hλ =

∫log Jλ dλ.

Combined with Jensen’s inequality this yields

hλ ≤ log

∫Jλ dλ = log(deg(f)),

where equality is achieved is and only if Jλ = const = deg(f) λ-almost

everywhere (i.e., Lebesgue almost everywhere) on C.

Assume first that f is a Lattes map. Recall from (18.18) thatJλ ≡ deg(f) λ-almost everywhere. Thus hλ = log(deg(f)) by theabove.

Assume now that hλ = log(deg(f)), so Jλ = const = log(deg(f)

λ-a.e. on C.


Now let p ∈ post(f) be arbitrary. For an n ∈ N we consider pointsq, q′ ∈ f−n(p) \post(f). Without loss of generality we can assume thatp = 0 and that q, q′ ∈ C.

To avoid having to compute the derivative of f with respect to thecanonical orbifold metric, we consider the density κ of λ with respect

to the 2-dimensional Lebesgue measure on the plane C ⊂ C. Thus

dλ = κ dA,

where κ : C → [0,∞) is the Radon-Nikodym derivative dλ/dA. Notethat the Jacobian is given by

Jf,λ(z) = |f ′(z)|2κ(f(z))/κ(z).

This yields by induction

Jfn,λ(z) = |(fn)′(z)|2κ(fn(z))/κ(z),

for all n ∈ N.Recall from (18.23) that κ(z) 1 in a neighborhood of q as well as

in a neighborhood of q′. Let d = deg(fn, q) and d′ = deg(fn, q′). Then|(fn)′(z)| |z − q|d−1 in a neighborhood of q, as well as |(fn)′(z)| |z − q′|d′−1 in a neighborhood of q′. Thus

const = Jfn,λ(z) = |(fn)′(z)|2κ(fn(z))/κ(z) |z − q|2(d−1)κ(fn(z)).

Since fn(z) = a(z− q)d +O(|z− q|d+1) near q (for some a ∈ C \ 0, itfollows that κ is comparable to a radial function, i.e., there is a functionk : (0,∞)→ (0,∞) such that

κ((z − q)d) k(|z − q|d) |z − q|−2(d−1).

Setting w = (z − q)d yields

κ(w) k(|w|) |w|−2(1−1/d).

Applying exactly the same argument to fn in a neighborhood of q′

yields

κ(w) |w|−2(1−1/d′).

Combining these two estimates results in d = d′. This shows that f isa Lattes map by Lemma 18.21.

The previous result implies Theorem 18.3 (which is Zdunik’s theo-rem in our setting) relatively easily.

Proof of Theorem 18.3. Let f be a rational expanding Thurs-ton map, νf be its measure of maximal entropy, and λ be the f -invariantmeasure that is absolutely continuous with respect to Lebesgue mea-sure.

18.4. LATTES MAPS, ENTROPY, AND LEBESGUE MEASURE 411

Since νf and λ are both ergodic f -invariant probability measures

on C they are either identical or singular to each other. Recall that νfis the unique measure of maximal entropy, and hνf = log deg(f), seeTheorem 16.1 and Corollary 16.2. It follows from Theorem 18.20 thatνf and λ are identical if f is a Lattes map and singular otherwise.

Proof of Theorem 18.2. Let f : C → C be a Lattes map. We

have seen in (18.17) that for c = 1/Ω(C) the measure λ = cΩ is an

f -invariant probability measure on C. In the previous proof we haveseen that λ and the measure of maximal entropy νf are identical, thuscΩ = λ = νf as desired.

We are ready to prove Theorem 17.1 (iii). We first need an elemen-tary lemma.

Lemma 18.23. Let ϕ : X → Y be a snowflake homeomorphism be-tween metric spaces (X, dX) and (Y, dY ), meaning there is a β > 0 suchthat

dY (ϕ(x), ϕ(y)) dX(x, y)β

for all x, y ∈ X, where C() is independent of x and Y . Then thefollowing statements are true.

(i) If (X, dX) is Ahlfors QX-regular, it follows that (Y, dY ) isAhlfors QY -regular, where QY = QX/β.

(ii) If (X, dX) is Ahlfors QX-regular and (Y, dY ) is Ahlfors QY -regular, it follows that β = QX/QY .

(iii) Assume (X, dX) is Ahlfors QA-regular and (Y, dY ) is AhlforsQY -regular. Then the QX-dimensional Hausdorff measure of(X, dX) is finite and non-zero if and only if the QY -dimensionalHausdorff measure is finite and non-zero,

HQX (X) 1 ⇔ HQY (Y ) 1.

The proof of this lemma follows directly from the fact that ϕ yieldsa bijection between sets in X of diameter D to sets in Y of diameterDβ.

Proof of Theorem 17.1 (iii). Let % be a visual metric for theexpanding Thurston map f : S2 → S2.

Assume first that f is topologically conjugate to a Lattes map.Since such a topological conjugacy is in fact a snowflake homeomor-phism with respect to visual metrics (see Corollary 11.6), we can as-

sume that f : C→ C is a Lattes map.Recall from Proposition 8.11 that the canonical orbifold metric

ω is a visual metric for f (see Section A.10 for the definition of ω).


Since two visual metrics are snowflake equivalent according to Propo-

sition 8.3 (iv), we may assume that % = ω. The spaces (C, ω) and (C, σ)however, are bi-Lipschitz equivalent, see Lemma A.24 (ii). Recall that

σ is the chordal metric on C. It follows that (S2, %) is snowflake equiv-

alent to (C, σ) as desired.

To prove the other implication, assume now that (S2, %) is snowflake

equivalent to (C, σ). In particular they are quasisymmetric. Thus, byTheorem 17.1 (ii), the map f is topologically conjugate to a rational

map. As before, we can assume that f : C → C is in fact a rationalmap (which is postcritically finite and has no periodic critical points).

Let ϕ : (C, %)→ (C, σ) be a snowflake homeomorphism, i.e.,

σ(ϕ(x), ϕ(y))β σ(x, y),

where β > 0, for all x, y ∈ C (here C() is a constant independent

of x, y). Recall from Proposition 17.2 that (C, %) is Ahlfors Q-regular,

where Q = log(deg(f))/ log(Λ). Clearly (C, σ) is Ahlfors 2-regular.Thus by Lemma 18.22 (ii) β = Q/2.

Assume that f is not a Lattes map. Then by Theorem 18.3 there is

a set E ⊂ C that has full measure for the measure of maximal entropy

νf , but is a zero-set for the Lebesgue measure on C. Since (E, %) isAhlfors Q-regular, it follows by Lemma 18.22 (i) that ϕ(E) viewed as

a set in (C, σ) is Ahlfors 2-regular, more precisely by Lemma 18.22 (iii)its 2-dimensional Hausdorff measure is non-zero.

Recall from Lemma 17.11 that the map id: (C, σ) → (C, %) is aquasisymmetry. Thus the composition

(C, σ)id−→ (C, %)

ϕ−→ (C, σ),

is a quasisymmetric map that maps the Lebesgue null-set E to a setof non-zero Lebesgue measure; hence is not absolutely continuous. Weremind the reader of the well-known fact that a quasisymmetric map

(C, σ)→ (C, σ) is quasiconformal, see [AIM, Theorem 3.4.1]. Further-

more, any quasiconformal map (C, σ) → (C, σ) is absolutely continu-ous, see [AIM, Theorem 3.1.2].

This contradiction shows that f is indeed a Lattes map as desired.

18.5. Chordal versus visual metric: a.e. comparisons

Here we finish the proof of Theorem 18.7, Theorem 18.8, Theorem 18.5,as well as Theorem 17.1 (iii).

Lemma 18.24. here Xk is (any) k-tile containing x.

18.5. CHORDAL VERSUS VISUAL METRIC: A.E. COMPARISONS 413

In the following tiles will be defined in terms of any Jordan curve

C ⊂ C with post(f) ⊂ C.

Proposition 18.25. Let f : C→ C be a rational expanding Thurs-ton map, ω the absolutely continuous (with respect to Lebesgue mea-sure) invariant measure from Section ??, and ν = νf be the measureof maximal entropy for f . Then

(i) for Lebesgue almost every point x ∈ C

limk−1

klog(diamXk) = Lω;

(ii) for ν almost every point x ∈ C it holds

limk−1

klog(diamXk) = Lν .

Here Xk is a (any) k-tile containing x, for any k ∈ N.

We will prove the first statement first. It is essentially a consequenceof the Shannon-McMillan-Breiman theorem. The reader should reviewthe notation from Chapter 16.

Let (X,B, µ) be a probability space, and g : X → X be an ergodicmeasure preserving map. Let ξ be a countable measurable partitionfor (X,B, µ). We define ξ(x) to be the element of ξ which contains x.This is defined for µ-almost every x ∈ X. The information function(to ξ and µ) is

Iξ(x) := − log µ(ξ(x)),

for µ-almost every x ∈ X. Thus Iξ assumes the constant value− log µ(A)on A ∈ ξ.

The Shannon-McMillan-Breiman theorem is now the following, see[Pet, Section 6.2] for a proof. Recall the definition of ξng from (16.4),hµ is the measure theoretic entropy of µ.

Theorem 18.26 (Shannon-McMillan-Breiman). Let g : X → X bean ergodic measure preserving map on (X,B, µ), and ξ be a countablemeasurable partition that is a generator for (X,B, µ). Then

limn

1

nIξng (x) = h = hµ,

for µ-almost every x ∈ X.

Proof of Proposition 18.24 (i). From Proposition 17.7 (v) itfollows that to prove the statement we may define tiles in terms of any

Jordan curve C ⊂ C with post(f) ⊂ C.


Thus we assume that C is invariant with respect to some sufficientiterate F = fn, see Theorem 1.1. Recall that ω is the f -invariant

probability measure on C, that is absolutely continuous with respectto Lebesgue measure λ. Clearly ω is invariant for F as well. Since C isa quasicircle it follows that λ(C) = ω(C) = 0, since ω is invariant for Fit follows that for E∞ :=

⋃k∈N0

F−k(C) it holds ω(E∞) = 0. Let XmF

denote the set of m-tiles defined in terms of (F, C).From Lemma 16.5 it follows that ξ := X1

F is a generator for (F, µ)and the partition ξmF is equivalent to Xm

F . Thus for Lebesgue almost

every point x ∈ C

(18.26) − 1

mlog(ω(Xk

F ))→ hω(F ) = nhω(f) = 2nLω,

as m → ∞ by Theorem 18.25 (see also (16.6) and (18.9)). Here eachXmF ∈ Xm

F contains x. Recall that hω(F ), hω(f) are the (measure the-oretic) entropies for F , respectively f , with respect to ω.

Let x ∈ C\post(f) be arbitrary and δ := dist(x, post(f)). Let x becontained in Xm

F ∈ XmF for some m ∈ N. Since f (hence F ) is expand-

ing we can assume that for sufficiently large m we have diamXmF < δ/2.

Then it follows from Proposition ?? ?? that ω(XmF ) area(Xm

F ), whereC() = C(δ). Recall from Proposition 17.7 ?? that area(Xm

F ) (diamXm

F )2. This yields with (18.25)

limm− 1

mlog(diamXm

F ) = nLω.

Thus the statement is proved for the iterate F = fn. It remains toprove it for the map itself, i.e., for tiles defined in terms of (f, C).

To this end let x ∈ C be a point where (18.25) holds, Xk be ak-tile defined in terms of (f, C) containing x. Let k = mn + j, wherej = 0, . . . , n− 1. Recall from Proposition 17.7 ?? that

diamXmn+j diamXmn = XmF ∈ Xm

F .

Thus

limk−1

klog(diamXk) = lim

m− 1

mnlog(diamXmn) = Lω.

We now prepare to prove the second part in Proposition 18.23.

Lemma 18.27. For any ε > 0 there is a set Aε ⊂ C with ν(Aε) ≤ ε

such that for any x ∈ C \ Aε it holds

diamXk [(fk)](x)]−1,


here Xk is any k-tile containing x. The constant C() = C(ε) isindependent of k.

Proof. To prove the claim, we first have to define the excluded set,

i.e., Aε. Clearly points x ∈ C where (fk)](x) = 0 for some k ≥ 1 haveto be excluded (i.e., any k-vertex). The set Aε will be a neighborhoodof the Jordan curve C (thus a neighborhood of the postcritical points)together with all preimages.

Recall from Proposition 17.7 (v) that another Jordan curve C ⊂ Cwith post(f) ⊂ C yields tiles with comparable diameter. More precisely,

if X is an n-tile for (f, C) and X is an n-tile for (f, C) with X ∩ X 6= ∅then diamX diam X (where C() is independent of X, X). Thus itis enough to prove the statement for tiles defined with respect to any

Jordan curve C ⊂ C with post(f) ⊂ C.Thus we assume that C is invariant for a suitable iterate F = fn

(see Theorem 14.1). Let X1C be the set of all 1-tiles (defined in terms

of (F, C)) that intersect C. Fix a constant 0 < τ < 1. By considering ahigher iterate (i.e., F = fnm) we can assume that the measure of theirunion, i.e., A1 :=

⋃X1C is small, more precisely

ν(A1) ≤ τν(X0w ) and ν(A1) ≤ τν(X0

b ),

see Lemma 16.6 and Proposition 16.10. Recall that X0w , X

0b denote the

two 0-tiles. Let Ak+1 := F−k(A1) for any k ≥ 0. The set

A = A(τ) :=⋃k∈N

Ak

will be the exceptional set. We need to show that it has small measure,and that points not in this set satisfy the statement of the claim above.

Consider a 1-tile X1 that is not contained in X1C. Note that A2 :=

F−1(A1) is a union of 2-tiles. Recall that F maps each n-tile Xn to an(n − 1)-tile and ν(F (Xn)) = deg(F )ν(Xn). Thus it follows from theabove that

ν(A2 ∩X1) ≤ τν(X1).

This means that ν(A1 ∪ A2) ≤ τ + τ 2.In the same fashion it follows that for any 2-tile X2 not contained

in A1 ∪ A2 it follows that for ν(A3 ∩X2) ≤ τν(X2) and ν(A1 ∪ A2) ≤τ + τ 2 + τ 3.

Continuing inductively we obtain

ν(A) = τ + τ 2 + · · · = τ

1− τ.

Clearly we can make ν(A) ≤ ε by choosing τ sufficiently small.


Let x ∈ C \ A and k ≥ 0 be arbitrary. We want to show thatthe statement of the lemma is satisfied. Let Xk be the k-tile contain-ing x, which we assume without loss of generality to be white. ThenF k : Xk → X0

w is a homeomorphism (see Proposition 5.17 (i)). Thusthe inverse G : X0

w → Xk is well defined, it is furthermore holomorphic(hence conformal) on int(X0

w ). The tile Xk is not contained in Ak, andx is not contained in Ak+1. Thus y := F k(x) /∈ A1.

This means that y is contained in an interior 1-tileX1, i.e., in a 1-tileX1 that is not contained in A1. Then Xk+1 := G(X1) is a (k + 1)-tilecontained in Xk containing x. Koebe distortion (i.e., (17.6)) togetherwith Proposition 17.7 ?? yields

(18.27) diamXk diamXk+1 G](y) = [(F k)](x)]−1,

Here C() is independent of x. A completely analog argument holds for

black k-tiles, meaning that the estimate above holds for any x ∈ C\A1

and k-tile Xk containing x.

Thus the statement is proved for the iterate F = fn. To prove itfor the map f itself we first note that A contains a neighborhood of

the critical points of f . Thus for all x ∈ C \ A it holds f ](x) 1.

Furthermore each iterate of x is in C \ A, thus f ](f j(x)) 1 for eachj ≥ 0. This yields by the chain rule

(fmn+j)](x) (fmn)](x).

for each j = 0, . . . , n− 1. Here C() is independent of x, j and m.

We are now ready to finnish the proof. Let x ∈ C \ A be arbitraryand Xk be a k-tile containing x. Let k = mn+j, where j = 0, . . . , n−1.Then it follows that

diamXk diamXmn by Proposition 17.7 ??

[(Fm)](x)]−1 = [(fmn)](x)]−1 by (18.26)

[(fk)](x)]−1.

The constant C() is independent of x and k as desired.

Proof of Proposition 18.24 (ii).Let A1/n be a set as in Lemma 18.26. Then

⋂nA1/n is a zero

set for ν. Thus ν-almost every point x ∈ C satisfies the estimatefrom Lemma 18.26 as well as (18.1). Combining those expressionsimmediately yields the statement.


Proof of Theorem 18.7. We fix a Jordan curve C ⊂ C withpost(f) ⊂ C. Tiles will be defined in terms of (f, C). In the following,we write a = b± C if b− C ≤ a ≤ b+ C.

(i) Let x ∈ C be a point satisfying Proposition 18.24 (i), Xm ∈ Xm

a tile containing x for each m ∈ N. Let m = mf,C(x, y) (see Definition8.1), then y → x⇔ m→∞. Thus,

limy→x

log d(x, y)

log σ(x, y)= lim

m

−m (log Λ± C/m)

log(diamXm)± C

= limm

log Λ± C/m− 1m

log diamXm ± C/m=

log Λ

Lω= αω.

The proof of part (ii) is almost identical. More precisely we haveto replace ω by ν in the above (and use Proposition 18.24 (ii)).

We now prepare the proof of Theorem 18.8.

Lemma 18.28. Let ϕ : X → Y be a homeomorphism between metricspaces such that there is an α ≥ 0 satisfying

limy→x

log|ϕ(x)− ϕ(y)|log|x− y|

= α,

for all x ∈ X. Then the Hausdorff dimension of the spaces satisfies

dimH(Y ) = dimH(X)/α.

Here the metrics on X, Y are again denoted by the Polish notation(i.e., by |x− x′|, |y − y′|).

Proof. Let ε > 0 be arbitrary, then for any x ∈ X there is an = n(x, ε) ∈ N, such that for all y ∈ B(x, 1

n) ⊂ X it holds

α− ε ≤ log|ϕ(x)− ϕ(y)|log|x− y|

≤ α + ε

⇔ |x− y|α+ε ≤ |ϕ(x)− ϕ(y)| ≤ |x− y|α−ε.Let Xn be the set of points x ∈ X where the above holds (in the

ball B(x, 1n)). Clearly this implies

dimH(Xn)/(α + ε) ≤ dimH(f(Xn)) ≤ dimH(Xn)/(α− ε).Taking the (countable) union of all Xn yields

dimH(X)/(α + ε) ≤ dimH(Y ) ≤ dimH(X)/(α− ε).This finishes the the proof, since ε > 0 was arbitrary.


Proof of Theorem 18.8.(i) Let A ⊂ C be a Borel set with λ(A) = ω(A) = 1, such that

for all x ∈ A the conclusion in Theorem 18.7 (i) holds. Note thatdimH(A, σ) = 2, since Lebesgue measure λ is Ahlfors 2-regular on(A, σ). Applying Lemma 18.27 to id : (A, σ)→ (A, d) yields

dimH(A, d) = 2/αω,

finishing the proof.

(ii) Now let A ⊂ C be a Borel set with ν(A) = 1 such thatfor all x ∈ A the conclusion in Theorem 18.7 (ii) holds. Note that

dimH(A, d) = dimH(C, d) = log(deg f)/ log Λ = Q, since ν is AhlforsQ-regular on (A, d) (see Proposition 17.2 ??). Applying Lemma 18.27to id : (A, σ)→ (A, d) yields

dimH(A, d) = dimH(C, d) =log(deg f)

log Λ=

dimH(A, σ)

αν.

This finishes the proof.

Now we can finish the proof of Theorem 18.5.

Proof of Theorem 18.5, second part.Clearly it holds dim(ν, σ) ≤ 2. This yields with Theorem 18.8 (ii)

dim(ν, σ) =log(deg f)

Lν≤ 2.

Thus 12

log(deg f) ≤ Lν .We now assume that f is a Lattes map. Recall from ... that if

x ∈ C \ post(f) is a fixed point for fn, then (fn)](x) = (deg f)n/2. Itis well-known that the measure of maximal entropy is the weak limitof 1

#x=fn(x)∑x=fn(x) δx, where the sum is taken over all fixed points

of fn.Assume now that x0 is fixed point of fn. Thus periodic with period

n. Considering the points x0, x1 = f(x0), . . . , xn−1, xn = fn(x0) = x0

it holds by the chain rule∑j

log f ](xj) = log((fn)](x0)

)= (deg f)n/2.

It follows that

Lν =

∫log f ] dν = lim

n

1

#x = fn(x)∑

#x=fn(x)

log f ](x)

=1

2log(deg f),


as desired.

We can now finish the proof of Theorem 17.1.

Proof of Theorem 17.1 (iii). Let d be a visual metric for theexpanding Thurston map f : S2 → S2. Assume (S2, d) is snowflakeequivalent to S2 equipped with the standard metric. In particularthey are quasisymmetric. Thus, by Theorem 17.1 (ii), the map f istopologically conjugate to a rational map. So we can assume that

f : C → C is in fact a rational map (which is postcritically finite andhas no periodic critical points).

Let ϕ : (C, d) → (C, σ) be the snowflake equivalence. This meansthat there is a β > 0, such that

σ(ϕ(x), ϕ(y))β d(x, y),

for all x, y ∈ C (where C() is a constant independent of x, y). Clearlyϕ changes the (Hausdorff) dimension by the factor β. Thus we have

β = 2/Q, where Q = dimH(C, d). The map ϕ maps Q-dimensionalHausdorff measure to (2-dimensional) Lebesgue measure λ (up to mul-tiplicative constants).

Assume now that f is not a Lattes map. Then by Theorem 18.8and Corollary 18.6 there is a set A ⊂ S2 that has full (2-dimensional)Lebesgue measure λ, yet is a (Hausdorff) Q-dimensional zero set in

(C, d).

The map id: (C, σ)→ (C, d) is a quasisymmetry (see Corollary 17.11).We now get a contradiction since the composition

(C, σ)id−→ (C, d)

ϕ−→ (C, σ),

is a quasiconformal map that maps A to a zero set (with respect toLebesgue measure λ); hence is not absolutely continuous. This contra-diction shows that f is indeed a Lattes map.

Assume now that f is topologically conjugate to a Lattes map. Soassume that f is a Lattes map. Recall from ... that the flat metric

on C is a visual metric. Clearly C equipped with this flat metric is

bi-Lipschitz equivalent to the standard sphere (C, σ). Since two visualmetrics are snowflake equivalent it follows that every visual metric is

snowflake equivalent to (C, σ) as desired.

Note that id : (C, d)→ (C, σ) is not bi-Lipschitz.

CHAPTER 19

A combinatorial characterization of Lattes maps

We now revisit Lattes maps. We will give a characterization of Lattesmaps in terms of their combinatorial expansion behavior. This is basedon a recent thesis by Qian Yin.

Let f : S2 → S2 be an expanding Thurston map, and C ⊂ S2 bea Jordan curve with post(f) ⊂ C. We defined the quantity Dn(f, C)for n ∈ N0 as the minimal cardinality of a set of n-tiles for (f, C)whose union is connected and joins opposite sides of C (see (5.14) andDefinition 5.30). The combinatorial expansion factor Λ0(f) of f asintroduced in Chapter 15 is related to the growth rate of Dn(f, C), andgiven by

Λ0(f) = limn→∞

Dn(f, C)1/n.

We have already seen that Λ0(f) ≤ deg(f)1/2 in the case that f hasno periodic critical points; see the discussion after Proposition 17.2).This implies that Dn(f, C) cannot grow much faster than deg(f)n/2

as n → ∞. As the following statement shows, up to a multiplicativeconstant this is actually a precise upper bound for the growth rate ofDn(f, C).

Proposition 19.1. Let f : S2 → S2 be an expanding Thurstonmap, and C ⊂ S2 be a Jordan curve with post(f) ⊂ C. Then thereexists a constant c > 0 such that

(19.1) Dn(f, C) ≤ c deg(f)n/2

for all n ∈ N. Moreover, we have Λ0(f) ≤ deg(f)1/2.

We will prove this later in this section.In view of Proposition 19.1 one can ask whether there are maps for

which Dn(f, C) actually grows as fast as deg(f)n/2 as n→∞. It turnsout that essentially Lattes maps are characterized by this property.This is the main result of this section.

Theorem 19.2. Let f : S2 → S2 be an expanding Thurston map.Then f is topologically conjugate to a Lattes map if and only if thefollowing conditions are true:

(i) f has no periodic critical points.

421

422 19. A COMBINATORIAL CHARACTERIZATION OF LATTES MAPS

(ii) There exists c > 0, and a Jordan curve C ⊂ S2 with post(f) ⊂C such that

(19.2) Dn(f, C) ≥ c deg(f)n/2

for all n ∈ N0.

This theorem is due to Qian Yin [Yi]. The proof will occupy the restof the section. We will follow Yin’s approach with some modificationsand simplifications.

If f is a rational map, then we get a slightly stronger statement.

Corollary 19.3. Let f : C→ C be a rational expanding Thurstonmap. Then f is a Lattes map if and only if condition (ii) in Theo-rem 19.2 is satisfied.

If (19.2) is true for some Jordan curve C, then the same conditionholds for each Jordan curve C ′ ⊂ S2 with post(f) ⊂ C ′ (in general witha different constant c′ > 0 depending on C ′). This follows from thefact that Dn(f, C) Dn(f, C ′) which was shown in Lemma 15.5 (see(15.3)).

By Proposition 19.1 the inequality Dn(f, C) . deg(f)n/2 is alwaystrue, so (19.2) says that Dn(f, C) deg(f)n/2 as n → ∞. This as-ymptotic behavior of Dn(f, C) implies that Λ0(f) = deg(f)1/2. Thisequality is slightly weaker than the requirement (19.2). It would bevery interesting to characterize the expanding Thurston maps whosecombinatorial expansion factor is maximal in this sense. Besides forexpanding Thurston that are topologically conjugate to Lattes maps, itis also satisfied for certain Lattes-type maps (similiar to Example 15.8where the associated linear map on R2 has a shear component).

19.1. Visual metrics, 2-regularity, and Lattes maps

Here we begin the proof of Theorem 19.2. Some parts are somewhattechnical. We delegate those to the next section, and finish the proofassuming these technical results.

The main implication (i.e., the sufficiency) in Theorem 19.2 will bea consequence of the following.

Theorem 19.4. Let f : S2 → S2 be an expanding Thurston map.Then f is topologically conjugate to a Lattes map if and only if

(i) f has no periodic critical points, and

(ii) there is a visual metric % for f , such that (S2, %) is Ahlfors2-regular.

19.1. VISUAL METRICS, 2-REGULARITY, AND LATTES MAPS 423

Proof. Assume first that f is topologically conjugate to a Lattes

map. So we can in fact assume that f : C→ C is a Lattes map. Thenf has no periodic critical point (see Theorem 3.1 (i)). Furthermorethe canonical orbifold metric ω for f is a visual metric with expansionfactor Λ = deg(f)1/2, see Proposition 8.11. By Proposition 17.2 the

space (C, ω) is Ahlfors 2-regular.

Assume now that f : S2 → S2 is an expanding Thurston map with-out periodic critical points, and % is a visual metric such that (S2, %)is Ahlfors 2-regular. Here we may use the measure of maximal entropyνf or the 2-dimensional Hausdorff measure H2

% by Proposition 17.2.

The metric space (S2, %) is linearly locally connected by Proposi-tion 17.4 (iii). Hence Theorem 4.2 applies and (S2, %) is quasisymmet-

rically equivalent to the Riemann sphere C equipped with the chordalmetric σ. This in turn implies by Theorem 17.1 (ii) that f is conjugateto a rational map. Since our assumptions are preserved under such a

conjugacy, we are reduced to the case where f : C → C is a rationalmap.

Recall from Corollary 17.11 that the identity map idC : (C, σ) →(C, %) is a quasisymmetry. Hence by Theorem 4.3 the Hausdorff 2-

measure on C with respect to the chordal metric σ (i.e., the standard

Lebesgue measure) and the Hausdorff 2-measure H2% on C with respect

to the visual metric % are absolutely continuous with respect to eachother. On the other hand, by Proposition 17.2 the measure of maximalentropy νf of f and H2

% are comparable. Thus, νf is absolutely contin-

uous with respect to Lebesgue measure on C. Zdunik’s theorem, i.e.,Theorem 18.3, implies that f is a Lattes map.

We have seen in the above proof that using Proposition 17.2 theprevious theorem immediately implies the following.

Corollary 19.5. Let f : S2 → S2 be an expanding Thurston map.Then f is topologically conjugate to a Lattes map if and only if

(i) f has no periodic critical points, and

(ii) there is a visual metric % for f with expansion factor Λ =deg(f)1/2.

Thus to prove that conditions (i) and (ii) of Theorem 19.2 implythat f is topologically conjugate to a Lattes map, it is enough to con-struct a visual metric % with expansion factor Λ = deg(f)1/2. To thisend we define for arbitrary x, y ∈ S2

(19.3)Nn(x, y) := min lengthP : P n-tile chain connecting x and y.


The following lemma will provide the main step in the proof ofTheorem 19.2.

Lemma 19.6. Let f : S2 → S2 be an expanding Thurston map that

has an invariant Jordan curve C ⊂ C with post(f) ⊂ C. Furthermorewe assume that condition (19.2) from Theorem 19.2 is satisfied for C.Then

Nn(x, y) Λn−m(x,y),

where Λ := deg(f)1/2 for all x, y ∈ S2 and n ≥ m(x, y). The constantC() is independent of x and y as well as n.

Here, m = mf,C is the number given in Definition 8.1. Recall that wedefined m(x, x) =∞. Thus, there is no n ≥ m(x, y) in this case. Theproof of this lemma will occupy the bulk of the next section. Assumingthe previous lemma, we can finish the proof of Theorem 19.2.

Lemma 19.7. Let f : S2 → S2 be an expanding Thurston map thatsatisfies condition (19.2) from Theorem 19.2. Then there is a visualmetric % for f with expansion factor Λ = deg(f)1/2.

Proof. Let f : S2 → S2 be an expanding Thurston map that sat-isfies (19.2) for some Jordan curve C, with post(f) ⊂ C.

We assume first that C is f -invariant. Let Λ := deg(f)1/2. For allx, y ∈ S2 define

(19.4) %(x, y) = lim supn→∞

Λ−nNn(x, y).

Note that clearly Nn(x, x) = 1 for all x ∈ S2 and n ∈ N. ThusLemma 19.6 shows that

%(x, y) Λ−m(x,y),

where m = mf,C and C() is independent of x and y. To prove that% is a visual metric for f it remains to show that % is indeed a metric.Note that the above shows that %(x, y) is finite and %(x, y) = 0 if andonly if x = y (for all x, y ∈ S2). The symmetry of % is obvious as well.Finally, the triangle inequality follows from the inequality

Nn(x, z) ≤ Nn(x, y) +Nn(z, y)

valid for all n ∈ N0 and x, y, z ∈ S2. Thus % is a visual metric withexpansion factor Λ = deg(f)1/2.

After having done the special case when C if f -invariant, let us nowconsider the general case.

By Theorem 14.1 we can pick an iterate F = fk of f that has anF -invariant Jordan curve C ⊂ S2 with post(f) = post(F ) ⊂ C ′. Since


f has no periodic critical points, it follows that F has no periodic criti-cal points. Recall from Lemma 15.5 that condition (19.2) is essentiallyindependent of the chosen Jordan curve. More precisely, since we as-sume that f satisfies this condition, we can assume it is satisfied forthe F -invariant curve C. Since every n-tile for (F, C) is an nk-tile for(f, C) it follows

Dn(F, C) = DnN(f, C) & deg(f)nN/2 = deg(F )n/2,

for all n ∈ N. Thus, using the argument in the first part of the proof,there is a visual metric % for F with expansion factor ΛF = deg(F )1/2.From Proposition 8.3 (v) it follows that % is a visual metric for f withexpansion factor Λ = (ΛF )1/k = deg(F )1/2k = deg(f)1/2.

Let us now finish the proof of Theorem 19.2, still assuming thatLemma 19.7 has been proved.

Proof of Theorem 19.2. We first prove the sufficiency part ofthe conditions in Theorem 19.2. To this end let f : S2 → S2 be an

expanding Thurston map without periodic critical points and C ⊂ Ca Jordan curve with post(f) ⊂ C such that (19.2) is satisfied. ByLemma 19.7 there is a visual metric % for f with expansion factorΛ = deg(f)1/2. Corollary 19.5 shows that f is topologically conjugateto a Lattes map.

To prove the reverse implication, let f : S2 → S2 be a Lattes map.We want to prove that f satisfies the conditions (i) and (ii) in Theo-rem 19.2. Since these conditions are invariant under topological con-jugacy (in a suitable sense), we may assume that f is a Lattes map

on the Riemann sphere C. Then f has no periodic critical points (seeTheorem 3.1 (i)), and so (i) is true.

We know that f has a parabolic orbifold. Let ω be the canonical

orbifold metric of f on C, see Section A.10. Then by Proposition 8.13this metric is a visual metric for f with expansion factor Λ = deg(f)1/2.

Now we pick a Jordan curve C ⊂ C with post(f) ⊂ C and con-sider tiles for (f, C). Fix n ∈ N. Then, according to the definitionof Dn(f, C), we can find a connected union K of n-tiles that joins op-posite sides of C and consists precisely of Dn(f, C) such tiles. Thendiamω(K) ≥ δ0, where δ0 > 0 is the constant in (5.13) for the under-

lying base metric ω on C. Among the n-tiles that form K we can finda chain X1, . . . , XN of distinct n-tiles that joins two points x, y ∈ K


with ω(x, y) = diamω(K). Then N ≤ Dn(f, C), and

N∑i=1

diamω(Xi) ≥ ω(x, y) ≥ δ0.

Since ω is a visual metric for f with expansion factor Λ = deg(f)1/2,we have diamω(Xi) ≤ CΛ−n with C > 0 independent of n, by Propo-sition 8.4 (ii). Hence CN deg(f)−n/2 ≥ δ0, and so

Dn(f, C) ≥ N & deg(f)n/2,

where C(&) is independent of n. This shows that (ii) is also true.

It suffices to produce a visual metric for F with expansion factordeg(F )1/2 = deg(f)k/2, because this will be visual metric for f withexpansion factor deg(f)1/2 (see Proposition 8.3 (v)). Since the growthbehavior of Dn(f, C) as n → ∞ is essentially independent of C up tomultiplicative constants, we may assume that (ii) in Theorem 19.2 issatisfied for the F -invariant curve C. Then

This implies that

Nn(x, y) ≤ #P . Λn−k

for n ≥ k, and soThis is the upper bound for %(x, y) in (19.11).To prove the corresponding lower bound, let Xk+1 and Y k+1 be

(k + 1)-tiles with x ∈ Xk+1 and y ∈ Y k+1. Then Xk+1 ∩ Y k+1 = ∅ bydefinition of k = m(x, y). If n ≥ k + 1 and P is an arbitrary chain ofn-tiles joining x and y, then P also joins Xk+1 and Y k+1. Since theseare two disjoint (k + 1)-cells, it follows from Lemma 5.34 that

length(P ) ≥ Dn−k−1(f, C).So by using hypothesis (ii) of Theorem 19.2, we have

Nn(x, y) ≥ Dn−k−1(f, C) & Λn−k−1.

Hence we get the desired lower bound

%(x, y) = lim supn→∞

Nn(x, y)Λ−n & Λ−k−1 Λ−k.

The relation (19.11) follows.Now we can show that % is a metric. Namely, by (19.11) the quan-

tity %(x, y) is finite for all x, y ∈ S2. Symmetry of % and the relation%(x, y) = 0 ⇔ x = y are clear (for the the implication ⇒ we use thelower bound in (19.11)). Finally, the triangle inequality follows fromthe inequality



valid for all n ∈ N0 and x, y, z ∈ S2.Moreover, % is a visual metric for f by (19.11). Its expansion factor

is equal to Λ = deg(f)1/2. The first part of the statement follows.Now suppose in addition that f has no periodic critical points.

Then Proposition 17.2 implies that S2 equipped with this metric % isAhlfors 2-regular.

is as follows. The necessity of the conditions (i) and (ii) is fairlyeasy to derive from properties of Lattes maps. The hard part of thetheorem is to show that (i) and (ii) are sufficient for an expandingThurston map to be topologically conjugate to a Lattes map. For thisone first shows that under these conditions there exists a visual metricfor f that is Ahlfors 2-regular. This is the most involved part of theargument. Then one invokes Theorem 4.2 and Theorem 17.1 (ii) toreduce to the case that f is a rational map. Moreover, one can thenapply Proposition 4.3 to argue that the measure of maximal entropyof f is absolutely continuous with respect to Lebesgue measure on the

Riemann sphere C. Invoking Theorem 18.3 one then concludes thatf is a Lattes map. The argument actually shows that if we add theassumption that f is a rational map in Theorem 19.2, then condition(ii) is satisfied if and only if f is a Lattes map.

We now turn to the details. We first record the relatively easynecessity part of Theorem 19.2.

Proof of Theorem 19.2 (Necessity part). Since the condi-tions (i) and (ii) in the statement are invariant under topological con-jugacy (in a suitable sense), we may assume that f is a Lattes map on

the Riemann sphere C. Then f has no periodic critical points (Theo-rem 3.1), and so (i) is true.

We know that f has a parabolic orbifold. Let ω be the canonical

orbifold metric of f on C, see Section A.10. Then by Proposition 8.13this metric is a visual metric for f with expansion factor Λ = deg(f)1/2.

Now we pick a Jordan curve C ⊂ C with post(f) ⊂ C and con-sider tiles for (f, C). Fix n ∈ N. Then, according to the definitionof Dn(f, C), we can find a connected union K of n-tiles that joins op-posite sides of C and consists precisely of Dn(f, C) such tiles. Thendiamω(K) ≥ δ0, where δ0 > 0 is the constant in (5.13) for the under-

lying base metric ω on C. Among the n-tiles that form K we can finda chain X1, . . . , XN of distinct n-tiles that joins two points x, y ∈ Kwith ω(x, y) = diamω(K). Then N ≤ Dn(f, C), and

N∑i=1

diamω(Xi) ≥ ω(x, y) ≥ δ0.


Since ω is a visual metric for f with expansion factor Λ = deg(f)1/2,we have diamω(Xi) ≤ CΛ−n with C > 0 independent of n. HenceCN deg(f)−n/2 ≥ δ0, and so

Dn(f, C) ≥ N & deg(f)n/2,

where C(&) is independent of n. This shows that (ii) is also true.

For the proof of the sufficiency part of Theorem 19.2 we need somepreparation; Proposition 19.1 will be proved along the way. Our mainintermediate goal is to establish Proposition 19.14 below that guar-antees the existence of an Ahlfors 2-regular visual metric for a mapsatisfying the conditions in Theorem 19.2. This relies on the fact thatin this case any two points in the underlying 2-sphere can be joined by arelatively short chain of n-tiles. To obtain such chains we use a dualityargument that will give the existence of short chains provided certainseparating sets of tiles do not have too small cardinality. The keyingredient of this duality argument is a well-known graph-theoreticalstatement, namely Menger’s Theorem. Before we formulate this result,we first record some definitions.

We consider a finite graph G. Here we take the combinatorial pointof view; so G is just a pair (V,E), where V is a finite set called theset of vertices of G and E ⊂ V × V is a symmetric subset of the set ofpairs in V called the set of edges of G. For x, y ∈ V we write x ∼ y if(x, y) ∈ E. In this case we say that x and y are joined by an edge (inG). The set E is symmetric, and so we consider edges as unoriented.

A path in G is a finite sequence x1, . . . , xn of vertices in G so thatsuccessive vertices are joined by an edge. If A,B ⊂ V , then a path inG joins A and B, if the first vertex of the path lies in A, and its lastvertex lies in B. A path joining A and B in G is called an A-B-path.A set A ⊂ V is connected if for all a, a′ ∈ A there exists a path inA joining a and a′. A set K ⊂ V separates A and B if every A-B-path contains an element in K. Given these definitions, we have thefollowing statement.

Theorem 19.8 (Menger’s Theorem). Let G be a finite graph withvertex set V , and A,B ⊂ V . Then the minimal cardinality of a setseparating A and B in G is equal to the maximal number of pairwisedisjoint A-B-paths in G.

For the proof and more background see [Di, Sect. 3.3]. This theoremcan be seen as a special case of the max-flow min-cut theorem (see [Di,Sect. 6.2]).


An important consequence for us will be that if M is a lower boundfor the cardinality of a set separating A and B, then there exists anA-B-path containing at most #V/M vertices.

The graphs G that we will consider in our context will have setsof tiles as vertex sets. In this situation we want to use the topologyof the underlying 2-sphere to obtain information on separating sets inthe graph. This is based on some topological facts that are essentiallywell-known and that we will discuss next. In the following S2 will bea fixed topological 2-sphere as usual.

If Ω ⊂ S2 is a region, and A,B ⊂ Ω, then a set K ⊂ Ω separatesthe sets A and B in Ω if for every path γ in Ω joining A and B (i.e.,γ has one endpoint in A and one in B), we have γ ∩K 6= ∅. Note thatthis is meaningful even if K is not disjoint from A or B. The set Kseparates x ∈ Ω and y ∈ Ω (or x ∈ Ω and a set B ⊂ Ω) if K separatesA = x and B = y (or x and B) in Ω. We omit the phrase “inΩ” if Ω is understood.

The following well-known fact is very useful.

Lemma 19.9 (Janiszewski’s lemma). Let K,L ⊂ R2 be two closedsets with K ∩ L = ∅. If two points x, y ∈ R2 are separated by K ∪ L,then they are separated by K or by L.

A version of this can be found in [Bi, Thm. III.4.A]; exactly thesame statement is true if R2 is replaced by a 2-sphere S2.

We need a slightly more sophisticated lemma in the same spirit.

Lemma 19.10. Let Ω ⊂ S2 be a simply connected region, A,B ⊂ Ωconnected sets, and K ⊂ Ω a set that is relatively closed in Ω and hasfinitely many connected components. If K separates A and B in Ω,then one of the components of K separates A and B in Ω.

This is a rather straightforward consequence of Janiszewski’s lemmaif we make the additional assumption that the sets A and B do not meetK. This assumption is not convenient in our later applications of thelemma though. Possible crossings of the sets A or B with K complicatethe situation and require a somewhat more involved argument.

Proof. If Ω 6= S2, then Ω is homeomorphic to R2, and, as wehave a purely topological statement, we may assume Ω = R2. We willfirst present the proof in this case, and comment on the minor changesnecessary for Ω = S2 after the argument.

We first establish the statement under an additional hypothesis.

Special Case: The set B is a singleton set, i.e., B = b, whereb ∈ R2.


We run an induction on the number of components of K. Theinduction beginning is clear. For the induction step, we assume thatthe statement is true (for given A and B) for sets K with at mostn ∈ N components. Now let K be a closed set in R2 with n + 1components that separates A and B. Then we can decompose K asK = K ′ ∪ C, where C is a component of K, and K ′ consists of theother n components of K. Since K is closed, the sets K ′ and C arealso closed.

Let S ⊂ A be the set of all points a ∈ A that are separated fromB by C. If S = A, then C separates A and B, and we are done. IfS = ∅, then Janiszewski’s lemma implies that K ′ separates every pointin a ∈ A from B = b (here we are using the assumption that B is asingleton). Hence K ′ separates A and B. Since K ′ has n components,our induction hypothesis applies; so there exists a component of K ′,and hence also a component of K, that separates A and B. Again weare done.

So we are reduced to the case where S 6= ∅ and S 6= A. The setS is relatively closed in A. Indeed, suppose that an is a sequence ofpoints in S with an → a ∈ A as n → ∞. To reach a contradiction,assume that a /∈ S. Then C does not separate a and B, and so thereis path γ in R2 joining a and B that does not meet C. In particular,a /∈ C, and so there exists a small path-connected neighborhood of adisjoint from C; but then by traveling first to a in this neighborhoodand then along γ, for large n we can join an and B by a path thatavoids C, contradicting our assumption that an ∈ S.

Since the set S 6= ∅, A is relatively closed in A, and A is connected,the set S cannot be relatively open in A. Hence there exists a pointa ∈ S and a sequence an of points in A \ S with an → a as n→∞.Then necessarily a ∈ C; for otherwise, we can again find a small path-connected neighborhood of a disjoint from C. Then we could travelfrom a to an for large n, and then, since an ∈ A\S, from an to B alonga path disjoint from C. This contradicts the fact a ∈ S.

Since an ∈ A\S and K = K ′∪C separates an and B, Janiszewski’slemma implies that K ′ separates an and B = b. The same argumentfor the proof that S is relatively closed in A, also shows that the setof points in A separated by K ′ from B is relatively closed in A. Sincean → a as n → ∞, we conclude that K ′ separates a and B. Nowa ∈ C, and so there exists a point in C that K ′ separates from B.

This in turn implies that K ′ separates C and B. Indeed, if this isnot the case, we can find a path α that avoids K ′ and joins a pointc ∈ C to B. Now K ′ is closed and so R2 \K ′ is a union of open regions.Since C ⊂ R2 \K ′ is connected, the set C is contained in one of these


A B

a

c

C

α

β

K ′

γ

S

Figure 19.1. Proof of Lemma 19.10.

regions. Since regions are path-connected, we can join c ∈ C and a ∈ Cby a path β that avoids K ′. Concatenating α and β gives a path inR2 \K ′ joining a and B. This is impossible as K ′ separates a and B.The situation is illustrated in Figure 19.1.

Since K ′ separates C and B, the set K ′ also separates A and B.Indeed, suppose γ is a path joining A and B. We claim that it meetsK ′. Since K = K ′ ∪C separates A and B, it must meet K ′ or C. If itmeets C, then it also meets K ′ as K ′ separates C and B. So γ meetsK ′ in any case.

We can now apply the induction hypotheses to K ′. Since K ′ sepa-rates A and B, and has only n components, there exists a componentof K ′, and hence also a component of K, that separates A and B.

The general case: A,B ⊂ R2 are arbitrary connected sets.The statement in the above special case (with the roles of A and B

reversed) gives the following version of Janiszewski’s lemma: Let A bea singelton set in R2, B ⊂ R2 be connected, and K,L ⊂ R2 be closedsets with finitely many components. If K ∩L = ∅ and K ∪L separatesA and B, then K or L separate A and B.

Indeed, by what we have seen, one of the components of K ∪ Lseparates A and B, which implies that K or L separates A and B.


The proof in the general case is now a repetition of the proof in thespecial case. The only difference is that we apply the above versionof Janiszewski’s lemma instead of the original version. We used thistwice: to reduce to the case S 6= ∅, and to argue that if an ∈ S \ A,then K ′ separates an and B. In both cases, when B is not necessarily asingelton set, we can apply the modified version of Janiszewski’s lemmawith the desired conclusion.

This concludes the proof if Ω is homeomorphic to R2. The proof inthe case Ω = S2 is the same. The only difference is that we apply theS2-version of Janiszewski’s lemma mentioned after the formulation ofthe R2-case.

In the following technical lemma, we consider an expanding Thur-ston map with an f -invariant Jordan curve C ⊂ S2 with post(f) ⊂ C.Cells will be for (f, C). We consider n, k ∈ N0 with n ≥ k, a collec-tion Z0, . . . , ZN of distinct k-tiles, where N ∈ N0, and we suppose thatthere exist k-edges E1, . . . , EN with Ei ⊂ Zi−1 ∩ Zi for i = 1, . . . , N .Then according to the terminology introduced in Section 11.3, the tilesZ0, . . . , ZN form a simple e-chain. Since n ≥ k, the k-cells are subdi-vided into n-cells. We form a graph G whose vertex set consists of alln-tiles contained in any of the k-tiles Zi. If

(19.5) Ω :=N⋃i=0

int(Zi) ∪N⋃i=1

int(Ei),

then V is the set of all n-tiles X with int(X) ⊂ Ω. In G we connecttwo distinct vertices in V given by n-tiles X and Y by an edge if thereexists an n-edge e ⊂ X ∩ Y with int(e) ⊂ Ω.

Note that if X, Y ∈ V , X 6= Y , and X and Y share an n-edge e,then there are two possibilities. Namely, X and Y may lie in the samek-tile Zi. Then necessarily int(e) ⊂ int(Zi) ⊂ Ω and so the vertices Xand Y are joined by an edge in G. If X and Y lie in different k-tiles Zi,then the only situation where X and Y are joined by an edge in G iswhen X and Y lies in subsequent k-tiles of the e-chain, say X ⊂ Zi−1

and Y ⊂ Zi, and int(e) ⊂ int(Ei) ⊂ Ω. So we only join “across”the k-edges E1, . . . , EN , but not across any other k-edge contained inZ0 ∪ · · · ∪ ZN .

Given a set of n-tiles M we use the notation |M | for the underlyingpoint set of a set M of n-tiles; i.e.,

|M | :=⋃X∈M

X.


Such a set M may be viewed as a set of vertices in G, i.e., a subsetof V . The next lemma relates these two viewpoints, in particular weaddress connectedness of n-tiles in G and in Ω.

Lemma 19.11. With the given assumptions, the following state-ments are true:

(i) The set Ω defined in (19.5) is a simply connected region in S2.

(ii) Let γ be a path in Ω and M ⊂ V be the set of all n-tiles Xwith int(X) ⊂ Ω and X ∩ γ 6= ∅. Then M is connected in G,and M ′ = |M | ∩ Ω a connected subset of Ω.

Similarly if γ be a path in the boundary of one of the k-tiles∂Zi, and M ⊂ V be the set of all n-tiles X with X ⊂ Zi andX ∩ γ 6= ∅. Then M is connected in G, and M ′ = |M | ∩ Ω aconnected subset of Ω.

(iii) If A,B,K ⊂ V , then K separates A and B in G if and onlyif K ′ = |K| ∩Ω separates A′ = |A| ∩Ω and B′ = |B| ∩Ω in Ω.

(iv) Let A,B ⊂ V , and suppose that each of the sets |A| and |B|contains at least two k-vertices and that A′ = |A| ∩ Ω andB′ = |B| ∩ Ω are connected. Let K ⊂ V be a set of mini-mal cardinality that separates A and B in G. Then |K| is aconnected set not contained in a k-flower. In particular,

(19.6) #K ≥ Dn−k(f, C).

Proof. In the formulation of the lemma and its ensuing proof itis understood that all cells are for (f, C).

(i) For each i = 1, . . . , N the set

Ui := int(Zi−1) ∪ int(Ei) ∪ int(Zi)

is an open region (see Lemma 5.12 (iv)). This implies that Ω is openand connected, and hence a region.

The additional statement that Ω is simply connected follows fromthe fact that every loop γ in Ω can be homotoped to a constant loopinside Ω. To see this, define

Ωl :=l⋃

i=0

int(Zi) ∪l+1⋃i=1

int(Ei) ⊂ Ω

for l = 0, . . . , N , where we set EN+1 = ∅. For each i = 1, . . . , N − 1there exits a deformation retraction of the set

int(Ei) ∪ int(Zi) ∪ int(Ei+1)


onto int(Ei) which implies that there exists a deformation retractionof Ωi onto Ωi−1. Here it is important that

(int(Zi) ∪ int(Ei+1)) ∩ Ωi−1 = ∅

which follows from the facts that the k-tiles Z0, . . . , ZN are all distinctand that the set int(Ei) only meets the k-tiles Zi−1 and Zi for i =1, . . . , N (this follows from Lemma 5.12 (iv)).

By using these deformation retractions successively, every closedloop in Ω = ΩN can be homotoped inside Ω to a loop in Ω0 ⊂ Ω. Theset Ω0 is an open Jordan domain with an open arc on its boundary ifN ≥ 1 and an open Jordan region if N = 0. Hence Ω0 is contractible,and the simple connectivity of Ω follows.

(ii) Paths in G precisely correspond to e-chains consisting of n-tiles X0, . . . , Xl ∈ V for which there exist n-edges e1, . . . , el with ei ⊂Xi−1 ∩ Xi and int(ei) ⊂ Ω for i = 1, . . . , l − 1, where l ∈ N0. For therest of the proof of the present lemma we call such an e-chain of n-tilesadmissible.

Let M ⊂ V be the set associated to a path γ as in the statement.The proof that M is connected in G amounts to showing that every twon-tiles in M can be joined by an e-chain of n-tiles that is admissiblein this sense. The argument for this is along the lines of the proof ofLemma 11.16 with small modifications which we now outline.

It is enough to consider paths γ : [a, b]→ S2 defined on a compactinterval [a, b] ⊂ R, and show that if γ ⊂ Ω or γ ⊂ ∂Zi, and if X, Y ∈ V ,and γ(a) ∈ X, γ(b) ∈ Y , then there is an admissible e-chain of n-tilesjoining X and Y . As in the proof of Lemma 11.16 one can find amaximal number m ∈ [a, b] such that there is an admissible e-chain Pof n-tiles X1 . . . , XM with X1 = X and p := γ(m) ∈ U := XM .

We claim that m = b; again one argues by contradiction and as-sumes that a ≤ m < b. The case p ∈ int(U) immediately gives acontradiction. So p must be contained in the boundary of U , and thereare two cases for p depending on whether p belongs to the interior ofan n-edge, or whether p is an n-vertex.

Suppose first that p ∈ int(e), where e ⊂ ∂U is an n-edge. Thenthere exists a unique n-tile U ′ 6= U with e ⊂ U ′. If γ is a path inΩ, then p ∈ U ′, and so U ′ ∈ V . This follows from the fact that eachn-cell that meets Ω has interior contained in Ω. We can extend ouradmissible n-chain P by the n-tile U ′ which gives a contradiction to themaximality of m as in the proof of Lemma 11.16. The other possibilityis if γ is a path in the boundary ∂Zi of one of the k-tiles Zi. Thene ⊂ ∂Zi, and so int(e) is a neighborhood of p in ∂Zi. This implies that


γ(t) ∈ int(e) ⊂ U for t > m close to m. Again this contradicts themaximality of m.

The other case for p is that p is an n-vertex. If γ is a path in Ω, thenall the n-tiles in the cycle of p belong to V , and we get a contradictionas in the proof of Lemma 11.16. If γ is a path in the boundary ∂Zi ofone of the k-tiles Zi, then p ∈ ∂Zi. Exactly two n-edges belonging tothe cycle of p are contained in ∂Zi. This implies that the Jordan curve∂Zi splits the n-tiles in the cycle of p into two groups, namely thosewith interior contained in Zi and those with interior disjoint from Zi.Moreover, the n-tiles in the first group Vp all belong to V , their unionforms a relative neighborhood of p in Zi, and every two of them can bejoined by an admissible e-chain of n-tiles in Vp. From these facts oneagain gets a contradiction to maximality of m.

We conclude that m = b. So there exists an admisssible e-chain Pof n-tiles whose first tile is X and whose last tile is an n-tile Y ′ ∈ Vwith γ(b) ∈ Y ′. Using considerations very similar to the above, onecan show that P extends to an admissible e-chain whose last tile is Y .

The connectedness of M in G follows. The connectivity of the setM ′ = |M | ∩ Ω can easily be derived from this. Indeed, to connect twopoints p, q ∈M ′, one chooses n-tiles X, Y ∈M with p ∈ X and q ∈ Y .Then by the first part of the statement, one can find an admissible e-chain consisting of tiles X0 = X, . . . , Xl = Y in M . There are n-edgese1, . . . , el with ei ⊂ Xi−1 ∩Xi and int(ei) ⊂ Ω for i = 1, . . . , l. To finda path γ that stays in |M | ∩Ω and joins p and q, we travel from p to apoint in int(e1) along an arc whose interior stays in int(X0). We crossover to the interior of the next tile X1 and travel along an arc whoseinterior stays in int(X1) to a point in int(e2), etc., until we finally joina suitable point in int(el) to q by an arc whose interior stays in int(Xl).The concatenation of these arcs gives a path in M ′ = |M | ∩ Ω joiningp and q.

(iii) “⇐” With the given setup suppose that K ′ = |K|∩Ω separatesA′ = |A| ∩ Ω and B′ = |B| ∩ Ω in Ω. A path in G joining A and Bgives an adimissible e-chain P consisting of n-tiles X0, . . . , XM ∈ Vwith X0 ∈ A and XM ∈ B for which there exist n-edges e1, . . . , eMwith ei ⊂ Xi−1∩Xi and int(ei) ⊂ Ω for i = 1, . . . ,M . We have to showthat P ∩K 6= ∅.

Similarly as in the last part of the proof of (ii), one can find a pathγ that lies in the set

M⋃i=0

int(Xi) ∪M⋃i=1

int(ei) ⊂ Ω,


and has one endpoint in int(X1) and one in int(XM) Then γ does notmeet any tile in V that does not belong to P . The path γ joins A′ andB′ in Ω and hence meets K ′ ⊂ |K| by our assumptions. By choice ofγ we then must have P ∩K 6= ∅ as desired.

(iii) “⇒” Suppose K separates A and B in G, and suppose γ is apath in Ω joining A′ = |A|∩Ω and B′ = |B|∩Ω. Let M be the set of alln-tiles that meet γ. Since γ ⊂ Ω we have M ⊂ V . By (ii) the set M isconnected in G, and so there exists a path P in G that starts in A andends in B and consists of n-tiles in M ⊂ V . By our assumptions wehave P ∩K 6= ∅. In particular, M and K have a tile in common whichimplies that γ ∩K ′ 6= ∅, where K ′ = |K| ∩ Ω. Hence K ′ separates A′

and B′ in Ω.

(iv) Let K ⊂ V be a set of minimal cardinality that separates setsA and B in G that satisfy the hypotheses. The existence of such aset follows from the fact that the whole vertex set V separates A andB according to our definition. Then by (iii) the set K ′ = |K| ∩ Ωseparates A′ = |A| ∩ Ω and B′ = |B| ∩ Ω in Ω. By Lemma 19.10 acomponent of K ′ separates A′ and B′ in Ω. Since the interiors of tilesin V are connected subsets of Ω, this component of K ′ is of the formL′ = |L| ∩ Ω with L ⊂ K. By (iii) the set L separates A and B in G.Hence L = K by minimality of K. This shows that K ′ = |K| ∩Ω, andhence also |K| ⊂ K ′, is connected.

To show that |K| is not contained in a k-flower we argue by con-tradiction, and assume that |K| ⊂ W k(p) for some k-vertex p. Eachof the sets |A| and |B| contain at least two k-vertices. So we can pickk-vertices a ∈ |A| and b ∈ |B| with a, b 6= p. There are correspondingn-tiles X ∈ A and Y ∈ B with a ∈ X and b ∈ Y . Since X, Y ∈ V , eachof these n-tiles must be contained in one of the k-tiles Zi. To keep thenotation in the following argument simple, let us assume that X ⊂ Z0

and Y ⊂ ZN (the general case requires inessential modifications). Thena ∈ ∂Z0 and b ∈ ∂ZN .

By the same argument as in the proof of Lemma 11.14, we can nowfind a path α in the (topological) graph G′ = ∂Z0∪· · ·∪∂ZM that joinsa and b and avoids p. Namely, we can join a to one of the endpointsof the k-edge E1 ⊂ Z0 ∩ Z1 by a (possibly degenerate) path α0 ⊂ ∂Z0

consisting of k-edges that avoid p. Then we join the endpoint of α0 inE1 to one of the endpoints of E2 by a path α1 ⊂ ∂Z1 of k-edges thatavoid p, etc. In this way, we obtain paths αi ⊂ ∂Zi for i = 0, . . . , Nwhose concatenation gives a path α of k-edges in G′ that joins a and band does not contain p. Then α consists of k-cells that do not containp, and so α ∩ W k(p) = ∅. It follows that the compact sets α and


|K| ⊂ W k(p) are disjoint, and so these sets have positive distance withrespect to some base metric on S2.

By a small modification we can push α inside Ω to obtain a pathβ in Ω that is disjoint from |K| and joins A′ and B′. To do this,we slightly move the initial point a of α0 into int(X) ⊂ int(Z0) ⊂ Ωand the other endpoint of α0 to int(E1) ⊂ Ω. We can join these newendpoints by a path β0 that follows the original path α0 closely andstays inside int(Z0) ∪ int(E1) ⊂ Ω. We slightly move the last point ofα1 into int(E1), and join the endpoint of β0 to this point by a path β1

that follows α1 closely and stays inside int(E1)∪ int(Z1)∪ int(E2) ⊂ Ω.Continuing in this way, we get a collection of paths β0, . . . , βN whoseconcatenation is a path β in Ω that joins A′ and B′ and stays so closeto the original path α that β ∩ |K| = ∅.

This contradicts the fact that K ′ ⊂ |K| separates A′ and B′ in Ω.So |K| is not contained in a k-flower.

Inequality (19.6) now easily follows. Indeed, M = fk(|K|) is aconnected union of (n− k)-tiles. This set joins opposite sides of C. Forotherwise, M is contained in a 0-flower (Lemma 5.31) which in turnimplies by Lemma 5.27 (iii) that the connected set |K| ⊂ f−k(M) iscontained in a k-flower; but we have just seen that this is not the case.Since M joins opposite sides of C, the number of (n − k)-tiles in thisset must be at least Dn−k(f, C). The set K contains at least as manyn-tiles as M contains (n− k)-tiles; inequality (19.6) follows.

The previous lemma allows us to give a proof of Proposition 19.1.

Proof of Proposition 19.1. We first assume C is f -invariant.We want to apply Lemma 19.11 with k = 0 and N = 1 where

Z0 = X0w and Z1 = X0

b are the two 0-tiles and E1 is some 0-edge(necessarily on the boundary of both Z0 and Z1). In this case Ω isS2 with the union of the 0-edges distinct from E1 removed. We canpick 0-edges E0 ⊂ Z0 and E2 ⊂ Z1, so that E0, E1, E2 are distinct.Moreover, if # post(f) ≥ 4, we may assume that E0 ∩ E2 = ∅.

Now let n ∈ N0 and consider the graphG as defined in Lemma 19.11.In the present situation, the vertex set V actually is equal to the setof all n-tiles. So #V = 2 deg(f)n. Let A ⊂ V be the set of all n-tilesthat are contained in Z0 and meet E0, and B ⊂ V be the set of alln-tiles that are contained in Z1 and meet E2. Both |A| and |B| con-tain two 0-vertices, namely the endpoints of E0 and E2, respectively.Moreover, A′ = |A|∩Ω and B′ = |B|∩Ω are connected as follows fromLemma 19.11 (ii) applied to the path γ given by parametrizations ofthe arcs E0 and E2, respectively.


If K ⊂ V separates A and B in G, then #K ≥ Dn(f, C) byLemma 19.11 (iv). So by Menger’s Theorem there are at least Dn(f, C)pairwise disjoint A-B-paths in G. Then one of them must have car-dinality ≤ #V/Dn(f, C). This path gives an e-chain P consisting ofn-tiles X1, . . . , XM with X1 ∈ A, XM ∈ B, and

#P = M ≤ #V/Dn(f, C) = 2 deg(f)n/Dn(f, C).Then X1 meets E0 and XM meets E2. If # post(f) ≥ 4, then E0

and E2 are disjoint 0-edges which implies that |P | joins opposite sidesof C. Hence M ≥ Dn(f, C), and so

Dn(f, C) ≤M ≤ 2 deg(f)n/Dn(f, C),which gives

(19.7) Dn(f, C) ≤√

2 deg(f)n/2.

This is an upper bound for Dn(f, C) as desired.The other case, where # post(f) = 3, is along the same lines, but

slightly more subtle. In this case, E0, E1, E2 are the three 0-edges. Theunderlying set |P | of our e-chain P meets E0 and E2. We want to showthat it also meets E1. To see this we choose a path γ that starts inan interior point of X1 ends in an interior point XM and stays inside|P | ∩ Ω. This is possible, since P forms an e-chain where successivetiles have a common n-edge e whose interior is contained in Ω.

Now E1 splits our simply connected region Ω into the complemen-tary components int(Z0) and int(Z1). Since γ stays in Ω, starts inint(Z0), and ends in int(Z1), it must meet E1. Hence |P | ⊃ γ alsomeets E1. So |P | meets all three 0-edges, which means that this setjoins opposite sides of C (according to the definition of this terminologyfor # post(f) = 3). Again we have M ≥ Dn(f, C) and get (19.7). Thiscompletes the proof of the statement when C is invariant.

In the general case, when C is not necessarily f -invariant, we canfind an iterate F = fN of f that has an F -invariant Jordan curveC ′ ⊂ S2 with C ′ ⊃ post(F ) = post(f) (Theorem 14.1). Then by thefirst part of the proof

Dk(F, C ′) . deg(F )k/2 = deg(f)kN/2.

Since k-tiles for (F, C ′) are (kN)-tiles for (f, C ′), this means

DkN(f, C ′) . deg(f)kN/2.

If n ∈ N0 is arbitrary, we can write it as n = kN + l, where k ∈ N0 andl ∈ 0, . . . , N − 1. Now we know by Lemma 15.5 (see (15.2)) that

Dn+1(f, C ′) . Dn(f, C ′).


Applying this inequality at most (N − 1)-times, we are led to

Dn(f, C ′) = DkN+l(f, C ′) . DkN(f, C ′). deg(f)kN/2 ≤ deg(f)n/2.

We also know by Lemma 15.5 (see (15.3)) that

Dn(f, C) Dn(f, C ′).So

Dn(f, C) Dn(f, C ′) . deg(f)n/2.

Since in all these inequalities the implicit multiplicative constants areindependent of the index n (or k as in some of the previous inqualities),the statement follows.

In the following two lemmas we make the assumption that f : S2 →S2 is an expanding Thurston map with an f -invariant Jordan curveC ⊂ S2 with post(f) ⊂ C that satisfies condition (ii) in Theorem 19.2.We allow the possibility that f has periodic critical points. Tiles inthese statements will be for (f, C).

Lemma 19.12. Let n, k ∈ N, n ≥ k, and suppose that Xk and Y k

are two k-tiles for which there exists an (k−1)-tile Uk−1 with Xk, Y k ⊂Uk−1.

Then there exists an e-chain P consisting of n-tiles that starts in ann-tile contained in Xk, ends in an n-tile contained in Y k, and satisfies

(19.8) #P ≤ c′ deg(f)(n−k)/2,

where c′ > 0 only depends on f and C.

The lemma essentially says that under our assumptions, k-tiles witha common parent can be joined by an e-chain of n-tiles with controlledlength.

Proof. Since fk−1|Uk−1 is a homeomorphism that maps k-tiles to1-tiles, the number of k-tiles contained in Uk−1 is bounded by N0 :=2 deg(f), the number of 1-tiles.

We can pick a path γ ⊂ int(Uk−1) that joins Xk and Y k. Then γonly meets k-tiles contained in Uk−1. By Lemma 19.11 (ii) this impliesthat there exists a simple e-chain consisting of k-tiles

Z0 = Xk, Z1, . . . , ZN = Y k

with N ≤ N0. We are now in the situation of Lemma 19.11. Let Gbe the graph as defined in this lemma with vertex set V , and A ⊂ Vand B ⊂ V consist of all n-tiles contained in Z0 = Xk and ZN = Y k,respectively. Then |A| = Xk and |B| = Y k contain the k-vertices on


the boundary of Xk and Y k, respectively, and hence at least two k-vertices. Moreover, if we define Ω as in Lemma 19.11, and A′ = |A|∩Ωand B′ = |B| ∩ Ω, then int(Xk) ⊂ A′ ⊂ Xk and int(Y k) ⊂ B′ ⊂ Y k.This implies that A′ and B′ are connected.

So Lemma 19.11 (iv) applies and shows in combination with Menger’sTheorem that the number of disjoint A-B-paths in G is bounded belowby Dn−k(f, C). Let P such a path of minimal length. Note that P formsan e-chain of n-tiles whose first n-tile is contained in Xk and its last inY k. The number #V of vertices in G is equal to the number of n-tilescontained in any of k-tiles Zi, and hence bounded by N1 deg(f)n−k,where N1 = 2(N0 + 1). It follows that

#P ·Dn−k(f, C) ≤ #V ≤ N1 deg(f)n−k.

Using the lower bound (19.2) we conclude

#P ≤ c′ deg(f)(n−k)/2,

where c′ > 0 only depends on f and C.

Lemma 19.13. Let n, k ∈ N0, n ≥ k, Xn and Y n be two n-tiles,and suppose that there exists a k-tile Uk with Xn, Y n ⊂ Uk. Thenthere exists an e-chain P of n-tiles that starts in Xn, ends in Y n, andsatisfies

(19.9) #P ≤ c′n−k∑i=0

2n−k−i deg(f)i/2,

where c′ ≥ 1 is a constant only depending on f and C.

As we will see in the proof, we can take the same constant c′ in(19.9) as in (19.8) if c′ ≥ 1 as we may assume.

We will later apply this lemma when deg(f)1/2 > 2. In this case,the sum on the right hand side of (19.9) has geometrically increasingterms, and so (19.9) says that #P . deg(f)(n−k)/2 with C(.) onlydepending on f and C.

Proof. We prove this for fixed n ∈ N (and arbitrary tiles Xn andY n) by downward induction on k = n, n− 1, . . . , 0.

The statement is true for k = n. Indeed, in this case Xn = Y n =Uk, and so the single tile Xn = Y n forms a suitable chain P . We have#P ≤ 1 which gives a bound as in (19.9) for n = k if c′ ≥ 1.

Now we assume that the statement is true for some number 0 <k ≤ n. Then it is also true for k − 1. To see this, suppose that wehave n-tiles Xn and Y n, and a (k − 1)-tile Uk−1 with Xn, Y n ⊂ Uk−1.Then there exist unique k-tiles Xk and Y k with Xn ⊂ Xk ⊂ Uk−1 and


Y n ⊂ Y k ⊂ Uk−1. By Lemma 19.12 there exists an e-chain P1 of n-tileswith

#P1 ≤ c′ deg(f)(n−k)/2

that starts in an n-tile Xn ⊂ Xk and ends in an n-tile Y n ⊂ Y k. Wecan now apply the induction hypothesis to the n-tiles Xn, Xn ⊂ Xk to

find an e-chain P0 of n-tiles that starts in Xn, ends in Xn, and satisfies

#P0 ≤ c′n−k∑i=0

2n−k−i deg(f)i/2.

Similarly, we can find an e-chain P2 of n-tiles that starts in Y n, ends

in Y n, and satisfies

#P2 ≤ c′n−k∑i=0

2n−k−i deg(f)i/2.

Concatenating P0, P1, P2, leads to an e-chain P of n-tiles that startsin Xn, ends in Y n, and satisfies

#P ≤ #P0 + #P1 + #P2

≤ 2c′n−k∑i=0

2n−k−i deg(f)i/2 + c′ deg(f)(n−k)/2

≤ c′n−k+1∑i=0

2n−k+1−i deg(f)i/2.

The statement follows.

Proposition 19.14. Let f : S2 → S2 be an expanding Thurstonmap satisfying condition (ii) in Theorem 19.2. Then there exists avisual metric % for f with expansion factor Λ = deg(f)1/2. Moreover,if f has no periodic points in addition, then the metric space (S2, %) isAhlfors 2-regular.

Proof. By Theorem 14.1 we can pick an iterate F = fN of f thathas an F -invariant Jordan curve C ⊂ S2 with post(f) = post(F ) ⊂C ′. It suffices to produce a visual metric for F with expansion factordeg(F )1/2 = deg(f)N/2, because this will be visual metric for f withexpansion factor deg(f)1/2 (see Proposition 8.3 (v)). Since the growthbehavior of Dn(f, C) as n → ∞ is essentially independent of C up tomultiplicative constants, we may assume that (ii) in Theorem 19.2 issatisfied for the F -invariant curve C. Then

Dn(F, C) = DnN(f, C) & deg(f)nN/2 = deg(F )n/2,


Let f : S2 → S2 be an expanding Thurston map satisfying conditions(i) and (ii) of Theorem 19.2. Then by Proposition 19.14 there exists avisual metric % for f such that (S2, %) is Ahlfors 2-regular. By Propo-sition 17.4 (iii) the metric space (S2, %) is linearly locally connected.Hence Theorem 4.2 applies and (S2, %) is quasisymmetrically equivalent

to the Riemann sphere C equipped with the chordal metric σ. This inturn implies by Theorem 17.1 (ii) that f is conjugate to a rational map.Since our hypotheses (i) and (ii) in Theorem 19.2 are preserved undersuch a conjugacy, we are reduced to the case where f is a rational map.

The proof will be complete if we can show that an expanding Thurs-

ton map f : C → C is a Lattes map if it is rational and satisfies thehypotheses (i) and (ii) of Theorem 19.2. By what we have just seen,

there exists a visual metric % for f on C such that (C, %) is Ahlfors andso condition (ii) in Theorem 19.2 is also true for the iterate F = fN off .

In other words, by replacing our original map f by a suitable iterate(which we will now also call f), we have reduced the proof of the firstpart of the statement to the case where the Jordan curve C ⊂ S2 in(ii) in Theorem 19.2 is f -invariant. Passing to an even higher iterate ifnecessary, we may also assume that

(19.10) Λ := deg(f)1/2 > 2.

For n ∈ N0 and x, y ∈ S2 we now let Nn(x, y) ∈ N be the minimalcardinality of a chain of n-tiles whose first tile contains x and whoselast contains y. We define

%(x, y) := lim supn→∞

Λ−nNn(x, y) ∈ [0,∞].

We first claim that

(19.11) %(x, y) Λ−m(x,y),

where m(x, y) = mf,C(x, y) ∈ N0∪∞ is as in Definition 8.1. Here andin the following all implicit multiplicative constants are independent ofx and y.

To see this, we may assume that x 6= y (if x = y both sides in(19.11) are equal to 0), and set k := m(x, y) ∈ N0. Then by definitionof k = m(x, y) there exist k-tiles Xk and Y k with x ∈ Xk, y ∈ Y k, and

Xk ∩ Y k 6= ∅. For each n ≥ k we can pick n-tiles Xn, Xn, Y n, Y n with

x ∈ Xn, y ∈ Y n, Xn, Xn ⊂ Xk, Y n, Y n ⊂ Y k, and Xn ∩ Y n 6= ∅. Thenby (19.10) and Lemma 19.13 there exists an e-chain P1 of n-tiles with

P1 . Λn−k whose first tile is Xn and whose last tile is Xn. Similarly,there exists an e-chain P2 of n-tiles with #P2 . Λn−k whose first tile is


Y n and whose last tile is Y n. Since Xn∩ Y n 6= ∅, the union P = P1∪P2

is a chain of n-tiles (not necessarily an e-chain) with #P . Λn−k whosefirst tile is Xn and whose last tile is Y n.

This implies that

Nn(x, y) ≤ #P . Λn−k

for n ≥ k, and so


Λ−nNn(x, y) . Λ−k.

This is the upper bound for %(x, y) in (19.11).To prove the corresponding lower bound, let Xk+1 and Y k+1 be

(k + 1)-tiles with x ∈ Xk+1 and y ∈ Y k+1. Then Xk+1 ∩ Y k+1 = ∅ bydefinition of k = m(x, y). If n ≥ k + 1 and P is an arbitrary chain ofn-tiles joining x and y, then P also joins Xk+1 and Y k+1. Since theseare two disjoint (k + 1)-cells, it follows from Lemma 5.34 that

length(P ) ≥ Dn−k−1(f, C).So by using hypothesis (ii) of Theorem 19.2, we have

Nn(x, y) ≥ Dn−k−1(f, C) & Λn−k−1.

Hence we get the desired lower bound


Nn(x, y)Λ−n & Λ−k−1 Λ−k.

The relation (19.11) follows.Now we can show that % is a metric. Namely, by (19.11) the quan-

tity %(x, y) is finite for all x, y ∈ S2. Symmetry of % and the relation%(x, y) = 0 ⇔ x = y are clear (for the the implication ⇒ we use thelower bound in (19.11)). Finally, the triangle inequality follows fromthe inequality


valid for all n ∈ N0 and x, y, z ∈ S2.Moreover, % is a visual metric for f by (19.11). Its expansion factor

is equal to Λ = deg(f)1/2. The first part of the statement follows.Now suppose in addition that f has no periodic critical points.

Then Proposition 17.2 implies that S2 equipped with this metric % isAhlfors 2-regular.

We are now ready to prove the other half of Theorem 19.2.

Proof of Theorem 19.2 (Sufficiency part). 2-regular. We

know (see Corollary 17.11) that the identity map idC : (C, σ)→ (C, %)is a quasisymmetry. Hence by Theorem 4.3 the Hausdorff 2-measure

on C with respect to the chordal metric σ (i.e., the standard Lebesgue


measure) and the Hausdorff 2-measure H2% on C with respect to the

visual metric % are absolutely continuous with respect to each other.On the other hand, by Proposition 17.2 the measure of maximal entropyµ of f and H2

% are comparable. Thus, µ is absolutely continuous with

respect to Lebesgue measure on C. Theorem 18.3 implies that f is aLattes map.

Proof of Corollary 19.3. Suppose f is a rational Thurstonmap that is expanding. If f is a Lattes map, then condition (ii) inTheorem 19.2 is true as we have seen in the proof of the necessity partof this theorem. Conversely, if the condition is satisfied, then f is aLattes map, as we have seen in the proof of the sufficiency part ofTheorem 19.2.

CHAPTER 20

Outlook

20.1. Invariant Peano curves, matings, and fractal tilingsassociated to expanding Thurston maps

In Chapter 14 invariant Jordan curves were constructed for an ex-panding Thurston maps. The following related result is obtained in[Me13].

Theorem 20.1. Let f be an expanding Thurston map. Then foreach sufficiently high iterate F = fn there is a Peano curve γ : S1 → S2

(onto) such that F (γ(z)) = γ(zd) (for all z ∈ S1). Here d = degF .This means that the following diagram commutes.

S1 zd//

γ

S1

γ

S2

F// S2

Furthermore, we can approximate the Peano curve γ as follows. Thereis a homotopy Γ: S2 × [0, 1]→ S2, with Γ(z, 0) = z, such that

Γ(z, 1) = γ(z) for all z ∈ S1.

Here we view S1 ⊂ S2 as the equator.

In fact Γ may be chosen to be a pseudo-isotopy, meaning it is anisotopy on [0, 1).

The result may be paraphrased as follows. Via γ we can view thesphere S2 as a parametrized circle S1. Wrapping this parametrizedcircle (which is S2) around itself d times yields the map F .

According to Sullivan’s dictionary there is a close correspondencebetween the dynamics of rational maps and of Kleinian groups [Sul85].Cannon-Thurston construct (in [CT07]) an invariant Peano curve γ : S1 →S2 for the fundamental group of a (hyperbolic) 3-manifold M3 thatfibers over the circle. Theorem 20.1 may be viewed as the correspond-ing result in the case of rational maps. Thus it provides another entryin Sullivan’s dictionary.

There is a converse to Theorem 20.1, see [Me13].

445

446 20. OUTLOOK

H0

H1

H2

γ0 γ1

γ1 γ2

γ2 γ3

Figure 20.1. Construction of γ for the map g.

Theorem 20.2. Let f : S2 → S2 be a Thurston map, F = fn aniterate (for some n ∈ N), and γ : S1 → S2 be a Peano curve (onto)such that F (γ(z)) = γ(zd) for all z ∈ S1. Then f is expanding.

The construction of the invariant Peano curve γ in Theorem 20.1is very close to the construction of an invariant Jordan curve fromSection 14.2.

In both cases we start with a Jordan curve C containing all post-critical points of f . In Section 14.2 we deform C to a Jordan curveC ′ ⊂ F−1(C). In the construction of the invariant Peano curve howeverwe deform C to all of F−1(C). Since this set is not a Jordan curve, thisis achieved by a pseudo-isotopy H rel. post(f). Recall that a pseudo-isotopy on S2 is a homotopy H : S2 × [0, 1] → S2 with H(·, 0) = idS2

that is an isotopy on [0, 1).The 0-th approximation γ0 of γ is the curve C. Its deformation by

H is the 1-st approximation γ1. The pseudo-isotopy H0 = H can belifted by F n to pseudo-isotopies Hn for each n ∈ N. Deforming γn byHn yields the next approximation γn+1. The invariant Peano curve γis the limit of the approximations γn.

20.1. INVARIANT PEANO CURVES, MATINGS, FRACTAL TILINGS 447

The construction is illustrated in Figure 20.1 for the Lattes map gfrom Section 1.1. Here C = γ0 is the boundary of the pillow (which

is actually the extended real line R = R ∪ ∞). The large dotsindicate the postcritical points. Furthermore we have chosen to drawthe approximations γn not on the sphere but in the orbifold covering,i.e., in C. This just means here that the pillow has been cut in asuitable way and folded into the plane.

Douady and Hubbard observed in the early 80’s that the Julia set ofcertain rational maps “contains” the Julia set of certain polynomials.This prompted them to introduce the notion of mating of polynomi-als. This is an operation that geometrically combines two polynomials.Often this yields a rational map. In fact Thurston’s classification ofrational maps (Theorem 2.17) was to a large extend motivated by thequestion when a map arising as a mating “is” a rational map. Alsothe notion of Thurston equivalence naturally appears in this context.An excellent introduction to matings can be found in [Mi04]. Thereare many different variants of matings. We will only define the mostimportant one (for us) here, an overview of the different constructionscan be found in [MP].

Theorem 20.3. Let P = zn + an−1zn−1 + · · ·+ a0 be a monic (i.e.,

the leading coefficient is 1) polynomial. Then

(i) P is conformally conjugate to zn in a neighborhood of ∞, i.e.,there is a conformal map ϕ : U → V , where U , V are neigh-borhoods of ∞, such that

(20.1) ϕ(zn) = P (ϕ(z)),

for all z ∈ U . This is Bottcher’s theorem, see [Mi, §9] or[CG, II.4].

(ii) If the filled Julia K set of P is connected this conjugacy extends

conformally to φ : C \ D → C \ K. So φ is the Riemann mapand satisfies (20.1), i.e., φ(zn) = P (φ(z)).

(iii) If K is furthermore locally connected, the map φ extends tothe unit circle S1 = ∂D, by Caratheodorys theorem (see forexample [Mi, Theorem 17.14]). We call the restriction of thisextension to the circle

σ : S1 → J

the Caratheodory loop. Here J = ∂K is the Julia set ofP . Since σ is the extension of φ it follows from ((ii)) that

448 20. OUTLOOK

σ(zd) = f(σ(z)) for all z ∈ S1, i.e., the following diagramcommutes

(20.2) S1 zd//

σ

S1

σ

J f// J .

Note however that σ is not injective in general, i.e., it is a semi-conjugacy. The previous theorem is to a large extend the reason thatthe dynamics of polynomials is much better understood than the dy-namics of rational maps. Namely it enables one to describe Julia setsof polynomials in combinatorial terms.

The (topological) mating is defined as follows. Let Pw, Pb be twomonic (i.e., their leading coefficient is 1) polynomials of the same de-gree. Futhermore their Julia sets, equivalently their filled Julia setsKw,Kb, are connected and locally connected. Let σw : S1 → Kw andσb : S1 → Kb be their Caratheodory loops. Consider the disjoint unionKwtKb. The map PwtPb is naturally defined on this set (i.e., it is themap that restricts to Pj on Kj ⊂ Kw t Kb for j = w, b). We considerthe equivalence relation ∼ on Kw t Kb generated by (i.e., the smallestequivalence relation satisfying)

(20.3) σw(z) ∼ σb(z)

for each z ∈ S1 = ∂D. The mating of Kw and Kb is defined asKw⊥⊥Kb := Kw t Kb/∼.

From (20.2) it follows that for all xw ∈ Kw, xb ∈ Kb it holds

xw ∼ xb ⇒ Pw(xw) ∼ Pb(xb).

This implies that the map PwtPb : KwtKb → KwtKb descends to thequotient by ∼, i.e., yields a map

Pw⊥⊥Pb : Kw⊥⊥Kb → Kw⊥⊥Kb.

This map is called the (topological) mating of Pw and Pb. Identifyingσw(z) with σb(z) in (20.3) is customary, but not essential. Identifyingσw(z) with σb(z) would result in the mating of Pw with the polynomial

Pb(z).The space Kw⊥⊥Kb may not be a “nice” topological space. In fact

it is not known (not even expected) that it is always Hausdorff. Some-what surprisingly though, mating “often” results in maps Pw⊥⊥Pb thatare topologically conjugate to a rational map. Perhaps even more sur-prisingly this even happens in cases where Kw and Kb are dendrites,i.e., have no interior. The following is obtained in [Me09c].

20.1. INVARIANT PEANO CURVES, MATINGS, FRACTAL TILINGS 449

Theorem 20.4. Let f : S2 → S2 be an expanding Thurston mapwithout periodic critical points. Then every sufficiently high iterateF = fn is topologically conjugate to the mating of two polynomialsPw, Pb.

The polynomials Pw and Pb are postcritically finite as well as monicand of the same degree d as F . The Julia sets of Pw and Pb are dendrites.

Theorem 20.1 and Theorem 20.4 are closely related. Let us denoteJ := Jw/ ∼ = Jb/ ∼ ⊂ Kw⊥⊥Kb. In the case when the mating

Pw⊥⊥Pb is topologically conjugate to a rational map R : C → C by

the homeomorphism h : Kw⊥⊥Kb → C, the Julia set JR of R satisfiesJR = h(J ). When the Julia sets of Pw and Pb are dendrites it followsthat J = Kw⊥⊥Kb.

The Caratheodory loop σw : S1 → Jw descends to the quotient, i.e.,to Kw⊥⊥Kb. From (20.2) it follows that this quotient map σ = σw/∼satisfies

(20.4) S1 zd//

σ

S1

σ

J Pw⊥⊥Pb// J .

Thus Theorem 20.4 implies Theorem 20.1. However, Theorem 20.4is proved via Theorem 20.1. More precisely the equivalence relationon S1 induced by the invariant Peano curve γ : S1 → S2 from Theo-rem 20.1, i.e.,

s ' t :⇔ γ(s) = γ(t),

for all s, t ∈ S1 is considered. Is is proved that this equivalence relationagrees with the equivalence relation induced on S1 by σ from (20.4)(i.e., s ≈ t⇔ σ(s) = σ(t) for all s, t ∈ S1) for suitable polynomials Pw

and Pb.

Let γ : S1 → S2 be an invariant Peano curve for an expandingThurston map F as in Theorem 20.1. Let d = degF . Divide the circleS1 into d arcs Aj := z ∈ S1 : 2πj/d ≤ arg(z) ≤ 2π(j + 1)/d forj = 0, . . . d − 1. Since γ(zd) = F (γ(z)) (for all z ∈ S1) it follows thatF maps each set Bj := γ(Aj) to the whole sphere S2. Thus the setsBj form a Markov partition for F . This Markov partition howeveris geometrically not “as nice” as the one obtained from an invariantJordan curves C as in Chapter 14. Since C is a Jordan curve, the setsfrom the partition are Jordan domains. Their boundaries are uniformquasicircles (with respect to any visual metric) by Proposition 14.25.

On the other hand, the sets Bj are not Jordan domains in general,i.e., they may have local cutpoints. Figure 20.2 and Figure 20.3 show

450 20. OUTLOOK

Figure 20.2. Fractal tiling obtained from invariantPeano curve.

two examples of fractal tilings obtained in this way. Both examplescome from Lattes maps (it is much harder to explicitly compute theinvaraint Peano curves in the general setting).

20.2. Open problems

The authors hope that the results in the present paper will serve as afoundation for future work in the area. Our line of investigation hasbeen continued in [Me13, Me09c, Yi]. In the following we will discussa selection of questions and open problems that are worth pursuing.

The basis of our combinatorial approach is the existence of cellularMarkov partitions for Thurston maps. We know by Corollary 14.2 thatsuch a cellular Markov partition (related to a two-tile subdivision rule)exists for every sufficiently high iterate of an expanding Thurston map.This leads to the following question:

20.2. OPEN PROBLEMS 451

Figure 20.3. Another fractal tiling obtained from aninvariant Peano curve.

Problem 1. Does every expanding Thurston map f : S2 → S2 havea cellular Markov partition?

In the introduction we conjectured that the answer should be af-firmative. To prove the conjecture, one essentially has to construct aconnected graph G ⊂ S2 with post(f) ⊂ G that is f -invariant, i.e.,f(G) ⊂ G. A different way to phrase the problem is to ask whetherf (and not some iterate fn) can be described by a (suitably defined)k-tile subdivision rule. Here k would be the number of components ofS2 \G.

As we have seen, the dynamics of an expanding Thurston map fgenerates a class of visual metrics, and so a fractal geometry on the2-sphere on which it acts. This does not rule out that the map f canactually be described by a smooth model.

Problem 2. Is every expanding Thurston map f : S2 → S2 topo-

logically conjugate to a smooth expanding Thurston map g : C→ C?

This question was raised by K. Pilgrim. We expect that this is truefor at least every sufficiently high iterate of f .

We know that every expanding Thurston map f has iterates fn

with invariant curves as in Theorem 14.1. This naturally leads to the

452 20. OUTLOOK

question whether one can bound the order of this iterate in terms ofsome natural invariants of the map.

Problem 3. Let f : S2 → S2 be an expanding Thurston map. Isthere a number N0 ∈ N, depending on some natural data such asdeg(f), # post(f), and Λ0(f) such that for all n ≥ N0 there existsan fn-invariant Jordan curve C ⊂ S2 with post(f) ⊂ C?

Recall that Λ0(f) denotes the combinatorial expansion factor of f(see Section 15).

According to one of our main results every expanding Thurstonmap has an iterate that can be described by a two-tile subdivisionrule (Corollary 14.2). If a map realizes a subdivision rule, it shouldbe possible to extract all relevant information about the map from thecombinatorial data. This underlying principle is the common threadfor the next three problems.

Every two-tile subdivision rule is realized by a Thurston map thatis unique up to Thurston equivalence. In contrast, a Thurston mapmay realize combinatorially different two-tile subdivision rules. Thisleads to the following question:

Problem 4. Suppose two Thurston maps f and g realize differenttwo-tile subdivision rules. How can one decide from combinatorial datawhether the maps are Thurston equivalent?

Of course, there are several simple necessary conditions, such asdeg(f) = deg(g) and # post(f) = # post(g), whose validity can easilybe checked from the subdivision rules. In addition, the maps have tohave the same critical portrait (see Example 14.11). A related question,namely when polynomials with the same critical portrait are Thurstonequivalent, is answered in [BN].

Problem 5. Let f be an expanding Thurston map that realizes atwo-tile subdivision rule. Is there an effective way to compute the com-binatorial expansion factor Λ0(f) from the combinatorial description?

Since Λ0(f) is defined as a limit, a priori one cannot expect to findΛ0(f) by a finite procedure. However, if f is a Lattes or Lattes-typemap (see Example 15.8), then Λ0(f) is the smallest absolute value ofany eigenvalue of the matrix describing the underlying torus endomor-phism [Yi]. In general, one may speculate that if f realizes a two-tilesubdivision rule with underlying cell decompositions D and D′, thenΛ0(f) is related to the spectral radius of a matrix that is obtained fromthe incidence relations of the cells in D′ and their images under f inD.


Problem 6. Let f be an expanding Thurston map that realizes atwo-tile subdivision rule. Is it possible to decide from the subdivisionrule whether f is Thurston equivalent to a rational map?

As discussed in the introduction, Thurston proved a necessary andsufficient condition for a Thurston map to be equivalent to a rationalmap [DH]. In principle, it is possible to check this criterion if themap is given by a subdivision rule. In practice, the sufficiency part ishard to verify, since it involves the behavior of the map on infinitelymany homotopy classes of curves. So in Problem 6 we really ask for asubstantially simpler criterion, preferably based on the verification of afinite condition. As we have seen (Theorem 7.3), the case # post(f) = 3is completely understood. The case # post(f) = 4 should also berather accessible. For understanding the underlying issues it mightbe worthwhile to reprove Thurston’s result by using a combinatorialapproach as in the present paper.

From Theorem ?? it follows that for the solution of Problem 6one may equivalently ask if one can decide from the subdivision rulewhether S2 equipped with a visual metric d is quasisymmetricallyequivalent to the standard 2-sphere (i.e., the unit sphere in R3). Thisis closely related to Cannon’s conjecture in Geometric Group Theorywhich can be reformulated as a similar quasisymmetric equivalenceproblem (see [Bo1] for more background, and [Ca94] for a differentapproach). The general problem when a metric 2-sphere (S2, d) is qua-sisymmetrically equivalent to the standard 2-sphere is a hard problemthat is not completely understood (for results in this direction, see[BK]).

It is well-known that there are many analogies between ComplexDynamics and the theory of Kleinian groups (informally refered to asSullivan’s dictionary). It seems very fruitful to explore this systemati-cally for Thurston maps and translate statements for these maps intocorresponding group theoretic analogs. Expanding Thurston maps cor-respond to Gromov hyperbolic groups whose boundary at infinity is a2-sphere. This is suggested, for example, by Theorem ??, which can beseen as an analog of Cannon’s Conjecture, or by Remark ??. For Gro-mov hyperbolic groups whose boundary at infinity is a Sierpinski car-pet there is an analog of Cannon’s conjecture—the Kapovich-Kleinerconjecture. It predicts that these groups arise from some standard sit-uation in hyperbolic geometry. It would be interesting to formulate ananalog of this for Thurston maps in the spirit of Theorem ??.

If f : S2 → S2 is an expanding Thurston map without periodic crit-ical points, then S2 equipped with a visual metric d is Ahlfors regular

454 20. OUTLOOK

(Proposition 17.2). The infimal exponent Q such that (S2, d) is qua-sisymmetrically equivalent to an Ahlfors Q-regular space is called the(Ahlfors regular) conformal dimension of (S2, d). This is an importantnumerical invariant of the fractal geometry of (S2, d). Actually, thisconformal dimension only depends on f and not on the choice of thevisual metric d, since all visual metrics are quasisymmetrically equiva-lent.

Problem 7. Is it possible to determine the conformal dimensionof a 2-sphere equipped with a visual metric of an expanding Thurstonmap f in terms of dynamical data of f?

Bonk-Geyer-Pilgrim [Bo1] stated a conjecture expressing this con-formal dimension in terms of eigenvalues of certain matrices related tothe dynamics of f . One half of this conjecture has recently been provedHaıssinsky-Pilgrim [HP08].

To prove his characterization of rational maps among Thurstonmaps [DH], Thurston used an iteration procedure on a suitable Te-ichmuller space. This idea may also be fruitful for studying someproblems related to our existence criteria for invariant Jordan curves.Namely, in view of Theorem 14.4 one is led to the general question howto find non-trivial sets K ⊂ S2 with post(f) ⊂ K (and maybe addi-tional geometric features) for which there exists an isotopy rel. post(f)that deforms the set into a subset of its preimage f−1(K). It may bepossible to reformulate this as a fixed point problem under iteration ofa pull-back operation on a suitable space.

As we discussed above, Problem 1 is equivalent to finding an f -invariant graph G with suitable properties. If one does not insist ondescribing the map by a cellular Markov partition, then other invariantsets K may lead to other natural combinatorial descriptions of the mapor its iterates.

Actually, in many cases one can show the existence of an fn-invariantarc A ⊂ S2 with post(f) ⊂ A. Slightly more general, one can ask if aninvariant tree T ⊂ S2 with post(f) ⊂ T exists, i.e., a graph T ⊂ S2 (inthe sense of Section 11.3) such that S2 \ T is connected.

Problem 8. Let f : S2 → S2 be expanding Thurston map. Doesthere exists a tree T ⊂ S2 with post(f) ⊂ T that is invariant under asufficiently high iterate F = fn?

Is there a necessary and sufficient condition for the existence of anf -invariant tree T with post(f) ⊂ T?

If such an invariant tree T exists, then Ω0 = S2 \ T is a simplyconnected region that is subdivided by the complementary components


Ω1 of F−1(T ). Moreover, on each component Ω1 the map F |Ω1 is ahomeomorphism onto Ω0. A similar statement then also holds for theiterates of F . Since there is only one open “0-tile” Ω0, the dynamics ofF is described by a one-tile subdivision rule. This would simplify thecombinatorial description of F ; on the other hand, open tiles then areonly simply connected regions and not necessarily open Jordan regions.So the simplified combinatorial description comes at the price of a morecomplicated geometry of tiles.

For postcritically finite polynomials such an invariant tree alwaysexists, namely the Hubbard tree introduced by Douady and Hubbard,see [DH84]. In fact the Hubbard tree determines the polynomialuniquely up to conjugation by a map z 7→ az + b, a ∈ C \ 0, b ∈ C,see [Poi].

Cannon-Floyd-Parry show in [CFP10, Theorem 3.1] that a suffi-ciently high iterate of each Lattes map has an invariant tree. Notehowever that “Lattes map” is used there in a more restrictive sense,i.e., it denotes a quotient of a double cover of a conformal torus en-domorphism. They also show in [CFP10, Section 4] that it is indeednecessary to take an iterate in Problem 8. This means they show thatfor a certain Lattes map f no f -invariant tree T with post(f) ⊂ Texists. Indeed the authors have been informed by Bill Floyd that fora sufficiently high iterate F = fn such an F -invariant tree T withpost(f) ⊂ T always exists (Cannon-Floyd-Parry).

If one further relaxes the geometric requirements on “tiles”, thenother combinatorial descriptions of (expanding) Thurston maps exist.The result presented in Section 9 can be reinterpreted in this vein.This is related to the description of f via “limit spaces” of its iteratedmonodromy group (see [Ne] and the discussion below).

Yet another point of view to obtain combinatorial descriptions ofThurston maps is to consider sets that are “dual” to invariant setscontaining the postcritical set. To illustrate this, we consider the Lattesmap g from Section 1.1, and again represent the sphere S2 as a pillowas in Figure 1.1. Let K be the subset of the pillow that is of the unionof four Jordan curves, namely the two curves consisting of the pointsat distance 1/3 and at 2/3, respectively, from the bottom edge of thepillow, and the two closed curves of the points at distance 1/3 and 2/3from the left edge. Then K is g-invariant and for each n ∈ N0 the setg−n(K) is a graph that is the 1-skeleton of a cell decomposition of thepillow. Moreover, if C is the (g-invariant) boundary of the pillow, theng−n(K) is a dual graph to g−n(C).

456 20. OUTLOOK

In this paper we mostly considered expanding Thurston maps. Onemay ask whether combinatorial descriptions exist for more generaltypes of maps.

Problem 9. Let f : S2 → S2 be a Thurston map (not necessarilyexpanding). Is there a Jordan curve C ⊂ S2 with post(f) ⊂ C that isinvariant for some iterate fn? Are there other natural decompositionsof the sphere S2 that are invariant (in a suitable sense) under f orsome iterate fn?

When f is a polynomial natural decompositions are obtained viaexternal rays (see [DH84]). Closely related are Yoccoz puzzles (see[Hu93] and [Mi00]). A lack of such a description for rational maps isone of the main reasons why their study is harder than the study ofpolynomials.

Similarly, one can ask if and how our results extend to maps thatare not postcritically-finite. The theory of coarse expanding dynamicalsystems developed in [HP09] should be relevant here.

Problem 10. Let R : C → C be a rational map (not necessarily

postcritically-finite) whose Julia set is the whole Riemann sphere C.Does there exist a natural combinatorial description of R or some iter-ate Rn?

The rational maps R : C → C of a given degree d ≥ 2 form acomplex manifold Rd of dimension 2d + 1. M. Rees has shown [Re]that the set of points inRd where the corresponding rational map R hasas Julia set equal to the whole sphere has positive measure with respectto the natural measure class on Rd. These points/rational maps areobtained by a slight perturbation of a certain expanding Thurston mapR0 ∈ Rd. It would be interesting to find combinatorial descriptions ofexpanding Thurston maps that change under such perturbations in acontrolled manner.

Our combinatorial approach seems quite relevant in the study of theiterated monodromy group of a Thurston map f : S2 → S2. To state thebasic definitions, let G = π1(S2 \ post(f), p) be the fundamental groupof S2 \post(f) represented by homotopy classes of loops in S2 \post(f)with initial point p ∈ S2 \ post(f). Then there exists a natural actionof G on the disjoint union T of the sets f−n(p), n ∈ N0. Namely, ifx ∈ f−n(p) ⊂ T , and g ∈ G is represented by a loop α starting at p,then we can lift α by the map fn to a path α with initial point x. If yis the endpoint of α, then y ∈ f−n(p), and we define g(x) := y.

The action G y T is not effective in general, i.e., there may beelements g ∈ G that act as the identity on T . Let H be the ineffective


kernel of this action, i.e., the normal subgroup of G consisting of allelements inG acting as the identity on T . Then the iterated monodromygroup of f is defined as the quotient group Γ = G/H. The action ofG on T induces a natural action Γ y T . By the same method as inthe proof of Theorem 9.1 one can find a coding of the elements in T bywords in a finite alphabet. Then it is not hard to write down explicitrecursive relations for the action of suitably chosen generators of Γ onthese words. In principle, this describes the group Γ completely, butit seems hard in specific cases to get a complete understanding of thestructure of the group.

Problem 11. What is the iterated monodromy group of an expand-ing Thurston map? In particular, what is the growth behavior of sucha group?

The last question is interesting, because there is an example of a(non-expanding) Thurston map (namely, the polynomial f(z) = z2 +i)whose iterated monodromy group is a group of intermediate growth[BP]. Formerly, these groups had been regarded as rather exotic ob-jects.

We conclude our discussion with some remarks on higher dimen-sions. The concept of postcritically-finite maps has been generalizedto several complex variables and the dynamics of these maps has beenstudied (see, for example, [FS92, FS94, Jo]). A possible different

direction is the theory of quasiregular mappings on Rn = Rn ∪ ∞.

Problem 12. Develop a theory of quasiregular maps in higher di-mensions that generalizes the theory of Thurston maps in dimension2.

It is not even clear what the basic definitions should be. It seemsreasonable to require that a quasiregular Thurston map f : Rn → Rn

is uniformly quasiregular (i.e., there exists K ≥ 1 such that fn is K-quasiregular for each n ∈ N) and that the postcritical set post(f) off (i.e., the forward orbit of the branch set of f) is “small” and “non-exotic”. For example, one could require that post(f) is an embedded

cell complex in Rn of codimension 2.

APPENDIX A

In this appendix we collect various facts whose discussion would haveinterrupted the main flow of our presentation.

A.1. Metric space terminology

Let (X, d) be a metric space. A path γ in X is a continuous mapγ : [a, b] → X defined on some interval I = [a, b] ⊂ R. Sometimesconsiders also paths defined on half open or open intervals I ⊂ R. Apath γ joins two points x, y ∈ X if γ(a) = x and γ(b) = y. As usualwe define the length of γ as

lengthd(γ) := supn∑k=1

d(γ(tk−1), γ(tk)) ∈ [0,∞],

where the supremum is taken over all n ∈ N and all points t0 = a <t1 < · · · < tn = b. The path γ is called rectifiable (with respect to d) ifL := lengthd(γ) < ∞. In this case, we define L(t) := lengthd(γ|[a, t])for t ∈ [a, b]. Then there is a unique path γ : [0, L] → X, called thearclength parametrization of γ, such that γ(L(t)) = γ(t) for all t ∈ [a, b].If ρ : X → [0,∞] is a Borel function we define the path integral of ρalong the rectifiable path γ as∫

γ

ρ ds :=

∫ L

0

ρ(γ(s)) ds.

The metric d is called a length metric if

d(x, y) := infγ

lengthd(γ)

for x, y ∈ X, where the infimum is taken over all paths γ in X joiningx and y. The metric is a geodesic metric if this infimum is attained asa minimum. A path realizing the infimum of the length of all pathsjoining two points x, y ∈ X is called a geodesic segment joining x andy.

The Riemann sphere C can be identified with the unit sphere in R3

via stereographic projection. The chordal metric σ on C is the metric

459

460 A

that corresponds the Euclidean metric in R3 under this identification.More explicitly,

(A.1) σ(z, w) =2|z − w|√

1 + |z|2√

1 + |w|2

for z, w ∈ C, and

σ(∞, z) = σ(z,∞) = limw→∞

σ(z, w) =2√

1 + |z|2

for z ∈ C.The spherical derivative measures the expansion of a rational map

R with respect to the chordal metric. It is given by

(A.2) R](z) = limw→z

σ(R(w), R(z))

σ(w, z)=

1 + |z|2

1 + |R(z)|2|R′(z)|

for all z ∈ C. If z = ∞ or R(z) = ∞, the last expression has to beunderstood in a suitable limiting sense.

Let D := z ∈ C : |z| < 1 be the unit disk in the complex planeC. Then the hyperbolic metric d0 on D is defined as

(A.3) d0(u, v) = infγ

∫γ

2|dz|1− |z|2

u, v ∈ D,

where the infimum is taken over all paths γ in D that join u and v andare rectifiable with respect to the Euclidean metric on D. The space(D, d0) is geodesic and a model of the hyperbolic plane. The conformalautomorphisms of D are precisely the orientation-preserving isometriesof (D, d0).

By definition a Euclidean or hyperbolic cone C is a metric spaceisometric to a disk D = u ∈ C : |u| < r, where r > 0, equipped witha conformal metric with length element

(1 + |χ|)β|u|β−1 |du|

1 + χ|u|2β.

Here χ = 0 or χ = −1 depending on whether the cone is Euclideanor hyperbolic, and β > 0 (note that this length element is obtainedby pulling back the hyperbolic or Euclidean length element by themap u 7→ z = uβ). The number 2πβ is called the cone angle of C(the geometric significance of the cone angle is more apparent in othergeometric models for C, but we will not discuss this).

A.2. ORIENTATIONS ON SURFACES 461

A.2. Orientations on surfaces

Orientation is a subject that is easy to grasp on an intuitive level, butis notoriously difficult to discuss rigorously without some sophisticatedmathematical concepts or facts. We first recall the fairly standard wayof introducing orientation for surfaces by using homology groups, andthen discuss an alternative and very intuitive approach to orientationbased on the concept of a flag.

Let M be a compact and connected n-dimensional topological man-ifold (without boundary). If the singular homology group Hn(M) (withcoefficients in Z) is isomorphic to Z, then we call M orientable. Thisis true, for example, if M is a 2-sphere or a 2-dimensional torus (theonly cases we are interested in).

We say that M is oriented if one of the two generators of Hn(M) ∼=Z has been chosen as the fundamental class [M ] of M . If f : M →N is a homeomorphism between compact and connected oriented n-dimensional topological manifold M and N , then f induces an isomor-phism f∗ : Hn(M)→ Hn(M), and so f∗([M ]) = [N ] or f∗([M ]) = −[N ].In the first case we say that f is orientation-preserving, and in the sec-ond that f is orientation-reversing.

We mention in passing that in this framework we can also definethe degree of a continuous map. Namely, if M and N are oriented n-dimensional topological manifold M and N with fundamental classes[M ] and [N ], respectively, and f : M → N a continuous map, then itsdegree deg(f) ∈ Z is the unique integer such that f∗([M ]) = deg(f)[N ],where f∗ : Hn(M) → Hn(N) is the map between homology groups in-duced by f . Note that the sign of deg(f) depends on the orientationschosen on M and N . If M = N and we choose the same orientation insource and target, then deg(f) is independent of this choice.

For open manifolds or manifolds with boundary one has to resort tosuitable relative homology groups, if wants to give precise definitionsfor concepts related to orientation. This is somewhat technical, andwill only discuss this in a simple relevant case to give the general idea.

Let M be a compact and connected oriented surface. Then theorientation on S2 induces an orientation on every Jordan region X ⊂ S2

which in turn induces an orientation on ∂X and on every arc α ⊂∂X. These orientations are represented by generators in the homologygroups H2(X, ∂X), H1(∂X), and H1(α, ∂α), respectively.

To see how to get canonical generators in these groups from thefundamental class of M , first note that we have a natural isomorphisms

H2(M) ∼= H2(M,M \ int(X)) ∼= H2(X, ∂X) ∼= Z

462 A

induced by the inclusion map and excision. Hence we get an inducedorientation on X as represented by a generator of H2(X, ∂X) obtainedas the image of [M ].

Similarly, we have natural isomorphisms H2(X, ∂X) ∼= H1(∂X)(from the long exact sequence of relative homology) and H1(∂X) ∼=H1(α, ∂α). They give us canonical generators of the relevant homologygroups once we have an orientation on M .

On a more intuitive level, an orientation of an arc is just a selectionof one of the endpoints as the initial point and the other endpoint asthe terminal point. This can easily be reconciled with the homological

viewpoint if one uses isomorphism H1(α, ∂α) ∼= H0(∂α) for reducedhomology.

Suppose the Jordan region X ⊂ M in the oriented surface M isequipped with the induced orientation. If α ⊂ ∂X is an arc with agiven orientation, then we say that X lies to the left or to the right of αdepending on whether the orientation on α induced by the orientationof X agrees with the given orientation on α or not. Similarly, we saythat with a given orientation of ∂X the Jordan region X lies to the leftor right of ∂X.

An alternative way to introduce orientation is by using the notionof a flag. We will outline this only for surfaces. Let M be a connected(possibly open) surface. By definition a (topological) flag on a M isa triple (c0, c1, c2), where ci is an i-dimensional cell for i = 0, 1, 2,c0 ⊂ ∂c1, and c1 ⊂ ∂c2. So a flag in M is a closed Jordan region c2

with an arc c1 contained in its boundary, where the point in c0 is one ofthe endpoints of c1. We orient the arc c1 so that the point in c0 is theinitial point in c1. If we already have an orientation on M , then theflag is called positively- or negatively-oriented (for the given orientationon M) depending on whether c2 lies to the left or to the right of theoriented arc c1.

One can turn this around to give an alternative definition of ori-entation. Namely, we call to flags (c0, c1, c2) and (c′0, c

′1, c′2) equivalent

if there exists a homeomorphim f : M → M that is isotopic to idMand satisfies f(ci) = c′i for i = 0, 1, 2. On every connected surface Mthere are at most two equivalence classes of flags. This can easily bederived from that fact that if X, Y ⊂M are Jordan regions, then thereexists a homeomorphim f : M → M that is isotopic to idM and sat-isfies f(X) = Y . To get such a homeomorphism, one shrinks X andY by isotopies on M into small neighborhoods of points x ∈ int(X)and y ∈ int(Y ), and moves the neighborhood of x to the neighborhoodof y by an isotopy. In the shrinking process it is important that for

A.2. ORIENTATIONS ON SURFACES 463

every Jordan region Z ⊂ M there is a Jordan region Z ′ ⊂ M suchthat Z ⊂ int(Z ′). This easily follows from the fact that the topologicalcircle ∂Z is “tame” and so has a neighborhood that is homeomorphicto an annulus.

This outline of the argument also makes it obvious that the home-omorphism f on M that is isotopic to idM and satisfies f(X) = Y canbe constructed so that it agrees with idM outside a suitable compactsubset of M .

We call M orientable if there exist precisely two such equivalencesclasses of flags in M . An orientation on M is a choice of one of theequivalence as a distinguished flags. We say that the flags in this classare positively oriented and the flags in the other class are negativelyoriented.

Any positively-oriented flag determines the orientation uniquely. Soon orientable surfaces such as the plane C or a 2-sphere we can thinkof an orientation just as a choice of some flag as positively oriented.

The standard orientation on C or on C is the one for which thestandard flag (c0, c1, c2) is positively-oriented, where c0 = 0, c1 =[0, 1] ⊂ R, and

c2 = z ∈ C : 0 ≤ Re(z) ≤ 1, 0 ≤ Im(z) ≤ Re(z).Let M be an oriented surface, and Ω be a region in M . Then Ω

is orientable. (In the proof of this fact it is important that relevantisotopies on Ω between flags in Ω can be made to fix points outside asufficiently large compact subset on Ω. This allows one to extend theisotopy to M). We can represent the orientation on M by a flag inΩ. This flag represents a unique orientation on Ω, called the inducedorientation on Ω.

An orientation on a not necessarily connected surface is a choice onan orientation on each of its connected components (if each of thesecomponents is orientable). If U is an arbitrary open subset of an ori-ented surface M , then each of the components of U is contained in acomponent of M . We equip each of these components of U with theinduced orientation from the corresponding component of M . Thisdefine the induced orientation on U .

If f : M → N is a homeomorphism between connected and orientedsurfaces M and N , then either f maps all positively-oriented flags in Mto positively-oriented flags in N , or all positively-oriented flags in M tonegatively-oriented flags in N . We say that f is orientation-preservingin the first case, and orientation-reversing in the second.

Since edges and tiles in a cell decomposition D of S2 are arcs andclosed Jordan regions, respectively, it makes sense to speak of oriented

464 A

edges and tiles in D. A flag in D is a flag (c0, c1, c2), where c0, c1, c2

are cells in D. If ci are i-dimensional cells in D for i = 0, 1, 2, then(c0, c1, c2) is a flag in D if and only if c0 ⊂ c1 ⊂ c2.

A.3. Covering maps

Before we turn to branched covering maps, we remind the reader ofsome well-known facts about covering maps. They are true in greatgenerality, but we restrict ourselves mostly to covering maps betweenorientable surfaces.

Let X and Y be oriented surfaces, and π : X → Y be continuousand surjective map. Then π is called a covering map if every pointp ∈ Y has an open neighborhood U ⊂ Y such that π−1(U) can bewritten as a disjoint union

π−1(U) =⋃i∈I

Ui

of open sets Ui ⊂ X such that π|Ui is a orientation-preserving homeo-morphism of Ui onto U for each i ∈ I. Here I is some index set. Wesay that a set U as in this definition is evenly covered by π. Usually,one does not insist on the maps π|Ui to be orientation-preserving; thisadditional requirement is motivated by our definition of a branched cov-ering map: without it such a map covering map would not necessarilybe a branched covering map.

A covering map π : X → Y is a local homeomorphism, i.e., an openand continuous map such that for every point x ∈ X there exist anopen neighborhood V such that f |V is a homeomorphism of V ontof(V ). Moreover, π is orientation-preserving in addition. Conversely, ifX and Y are compact, then every orientation-preserving local homeo-morphism π : X → Y is a covering map.

Let π : X → Y be a covering map, Z a topological space, andf : Z → Y is a continuous map. A continuous map g : Z → X is calleda lift of f if π g = f . In this case, we have the commutative diagram:

(A.4) X

π

Z

g>>

f// Y.

The next lemma is a standard fact about existence and uniquenessof lifts.

Lemma A.1 (Existence and uniqueness of lifts). Let X and Y beoriented surfaces, π : X → Y be a covering map, and Z be a path-connected and locally path-connected topological space.

A.3. COVERING MAPS 465

(i) Suppose g1, g2 : Z → Y are two continuous maps such thatπ g1 = π g2. If there exists z0 ∈ Z with g1(z0) = g2(z0),then g1 = g2.

(ii) Suppose Z is simply connected, f : Z → Y is a continuousmap, and z0 ∈ Z and y0 ∈ Y are points such that f(z0) =π(y0). Then there exists a continuous map g : Z → X suchthat g(z0) = x0 and f = π g.

In (i) the maps g1 and g2 are lifts of f := π g1 = π g2. So thestatement says that lifts of maps are uniquely determined by the imageof one point.

Statement (ii) guarantees the existence of a lift g of f with g(z0) =x0. By (i) this lift g of f satisfying g(z0) = x0is unique.

A special and important case is if Z = [0, 1], and z0 = 0. Then f isa path in Y , and the statement says that we can lift it to a unique pathg in X if we prescribe any point in the fiber π−1(f(0)) as the initialpoint of the lift.

Let X be a path-connected and locally path-connected topologicalspace, and x0 ∈ X be a basepoint in X. Recall that the fundamentalgroup π1(X, x0) of X with respect to x0 consists of all homotopy classesall loops in X based (i.e, starting and ending) at x0 (see [Ha] for precisedefinitions). Simple connectivity of X means that π1(X, x0) = 0, andso π1(X, x0) is the trivial group only consisting of the unit element.

Suppose Y is another path-connected and locally path-connectedtopological space with basepoint y0, and f : X → Y a continuous mapthat is basepoint-preserving in the sense that f(x0) = y0. If we assignto each class [γ] ∈ π1(X, x0) represented by a loop γ in X based atx0, the class [f γ] represented by the image loop f γ, then we get awell-defined induced group homomorphism f∗ : π1(X, x0)→ π1(Y, y0).

Using these concepts, one can formulate a lifting criterion that ismore general than Lemma A.1 (ii).

Lemma A.2 (General lifting criterion). Let X and Y be orientedsurfaces, π : X → Y be a covering map, and Z be a path-connectedand locally path-connected topological space. Let x0 ∈ X, y0 ∈ Y ,z0 ∈ Z, and suppose f : Z → Y is a continuous map and π : X → Yis a covering map with π(x0) = y0 = f(z0). Then there exists a mapg : Z → X with π g = f and g(z0) = x0 if and only if f∗(π1(Z, z0)) ⊂π∗(π1(X, x0)).

For the proof see [Ha].

466 A

A.4. Branched covering maps

Let X and Y be compact oriented surfaces, and π : X → Y be acontinuous and surjective map. Recall from Section 1.2 that π is abranched covering map if for each point p ∈ X there exists d ∈ N, openneighborhoods U of p and U ′ of q := π(p), open neighborhoods V andV ′ of 0 ∈ C, and orientation-preserving homeomorphisms ϕ : U → Vand ψ : U ′ → V ′ with ϕ(p) = 0 and ψ(q) = 0 such that

(ψ π ϕ−1)(z) = zd

for all z ∈ V .So branched covering maps are modeled on non-constant holomor-

phic map between Riemann surfaces. Every such map is a branchedcovering map.

For maps between surfaces that are not necessarily compact, onehas to adjust the definition of a branched covering map, because itis desirable to have a condition that corresponds to the condition forcovering maps that points in the target have neighborhoods that areevenly covered. This condition is automatic in the case of compactsurface. Accordingly, we make the following definition.

Let X and Y be oriented surfaces, and π : X → Y be a continuousand surjective map. Then π is a branched covering map if for eachpoint q ∈ Y there exists an open neighborhood U ′ with the followingproperty: for some index set I 6= ∅ we can write π−1(U ′) as a disjointunion

π−1(U ′) =⋃i∈I

Ui

of open sets Ui ⊂ X such that Ui contains precisely one point pi ∈π−1(q). Moreover, we require that for each i ∈ I there exists di ∈ N,open neighborhoods Vi and V ′i of 0 ∈ C, and orientation-preservinghomeomorphisms ϕi : Ui → Vi and ψi : U

′ → V ′ with ϕ(pi) = 0 andψ(q) = 0 such that

(ψ π ϕ−1)(z) = zdi

for all z ∈ Vi.This definition is equivalent to our original definition of a branched

covering map if X and Y are compact in addition. For given π thenumber di is uniquely determined by p = pi and called the local degreeof π at p, denoted by degf (p). Note that our definition allows differentlocal degrees at points in the same fiber π−1(q).

Every branched covering map is surjective, open (images of opensets are open), and discrete (the preimage set of every point is a discreteset).

A.4. BRANCHED COVERING MAPS 467

A ramified point of a branched covering map f : X → Y is a pointq ∈ Y , such that at least one point p ∈ π−1(q) has local degreedegf (p) ≥ 2. Note that the set of ramified points is discrete. In-deed, if U ′ ⊂ X is a neighborhood of q as above, it follows that U ′

contains no ramified point distinct from q.

We next discuss existence and uniqueness for lifts under branchedcovering maps. The basic diagram is again (A.4), where π is now agiven branched covering map. A map g as in (A.4), is called a lift of f(under π). We first record an important special case of this situation.

Lemma A.3 (Lifting paths under branched covering maps). Let Xand Y be surfaces, π : X → Y be a branched covering map, γ : [0, 1]→Y be a path in Y , and x0 ∈ π−1(γ(0)). Then there exists a pathα : [0, 1]→ X with α(0) = x0 and π α = γ.

So every path can be lifted under a branched covering map. More-over, any point in the fiber over the initial point of the original pathcan be prescribed as the initial point of the lift. In general, the liftis not uniquely determined, because it can make various “turns” as itruns through critical points of π.

Proof. If we break γ up into small subpaths, we can reduce to thesituation where γ runs in an open subset of X that is evenly covered byπ. Using local coordinates, we can further make the assumption thatX and Y are equal to the open unit disk D in C, and π(z) = zn forz ∈ Y = D, where n ∈ N.

The set [0, 1] \ γ−1(0) can be written as a disjoint union

[0, 1] \ γ−1(0) =⋃k∈A

Ik

of its countably many connected components Ik. Here A is an countable(possibly empty) index set, and we may assume that A = 1, . . . , n ifthere are n ∈ N distinct components, and A = N if there are infinitelymany components.

The components Ik, k ∈ A, are pairwise disjoint non-empty inter-vals Ik that are relatively open in [0, 1]. Each path γk = γ|Ik, k ∈ A,is contained in D \ 0, and so it has a lift αk under the covering mapπ : D \ 0 → D \ 0. If 0 is contained in one of these intervals (whichhappens precisely if γ(0) 6= 0), we may assume that 0 ∈ I1. Then wecan choose the corresponding lift α1 so that α1(0) = x0.

Let a be an endpoint of one of the intervals Ik, and assume thata 6∈ Ik. Then γ(a) = 0, which implies

limt∈Ik→a

αk(t) = 0.

468 A

So we find a unique continuous extension of αk to a by setting αk(a) =0. We conclude that αk has a unique extension to Ik, which we alsodenote by αk.

Now define α : [0, 1]→ Y = D by setting

α(t) =

0, if t ∈ γ−1(0),αk(t), if t ∈ Ik for some k ∈ A.

Clearly, π α = γ. Moreover, α(0) = x0. Indeed, this is true if0 ∈

⋃k∈A Ik, because then 0 ∈ I1, and α(0) = α1(0) = x0 by our

previous choices. If 0 6∈⋃k∈A Ik, then 0 ∈ γ−1(0), and so γ(0) = 0.

Then π−1(γ(0)) = π−1(0) = 0, and so necessarily x0 = 0; but thenα(0) = 0 = x0.

It remains to show that α is continuous on [0, 1]. We have α|Ik = αkfor k ∈ A, and α|γ−1(0) ≡ 0. So α is continuous if we restrict it to anyof the closed sets Ik, k ∈ A, or γ−1(0). Since the union of these sets is[0, 1], the the continuity of α follows if A is a finite set. If A is infinite,a more careful argument is need. In the case, we have diam(Ik) → 0and so diam(γ|Ik) → 0 as k → ∞. This implies diam(α|Ik) → 0 ask →∞. Using this, it is not hard to verify the continuity of α at eachpoint t0 ∈ [0, 1]. We leave the details to the reader.

The following statement is the analog of Lemma 1.2 for branchedcovering maps.

Lemma A.4 (Lifting branched covering maps). Let X, Y, Z be sur-faces, and π : X → Y be a branched covering map.

(i) Suppose g1, g2 : Z → X are two continuous and discrete mapssuch that π g1 = π g2. If there exists a point z0 ∈ Z suchthat g1(z0) = g2(z0) =: x0, and π(x0) ∈ Y \ π(crit(π)), theng1 = g2.

(ii) Suppose Z is simply connected, f : Z → Y is a branched cov-ering map, and z0 ∈ Z and x0 ∈ X are points such thatf(z0) = π(x0).

If for all y ∈ Y , z ∈ f−1(y) and x ∈ π−1(y), we have

degπ(x)| degf (z),

then there exists a branched covering map g : Z → X such thatg(z0) = x0 and f = π g.

A.4. BRANCHED COVERING MAPS 469

The situation is illustrated in the following commutative diagram:

X

π

Z

g>>

f// Y.

If π is a covering map as in the previous statement, then we call a fiberπ−1(y0), y0 ∈ Y , clean if it does not contain critical points of π. Themaps g1 and g2 in (i) are lifts of f := π g1 = π g2 under π. So (i)says that lifts are uniquely determined by the image of a point thatmaps into a clean fiber of π.

Proof. (i) The maps g1 and g2 are lifts of f := π g1 = π g2

under the branched covering map π : Z → Y . We are claiming thatwith the given normalization such a lift is unique.

To see this define PY = π(crit(π)), PZ = π−1(PY ), PX = f−1(PY ) =g−1

1 (PZ) = g−12 (PZ). Note that these are discrete subsets of Y , Z, and

X, respectively. Let X ′ = X \ PX , Y ′ = Y \ PY , and Z ′ = Z \ PZ .Then x0 ∈ X ′, f(X ′) ⊂ Y ′ = π(Z ′), and g1(X ′) = g2(X ′) = Y ′.Moreover, if we restrict π to Z ′, then we obtain a covering map in theusual sense. Now a lift of a continuous map on a connected set undera covering map is uniquely determined by the image of one point (see[Ha, Proposition 1.34]). Since PX is a discrete set, X ′ is connected.Hence g1|X ′ = g2|X ′. Moreover, X ′ is dense in X, and so it followsthat g1 = g2 as desired.

(ii) This is an existence statement for lifts under branched coveringmaps. It follows from a topological analog for branched covering mapsof the classical monodromy theorem in complex analysis. This gener-alization is fairly straightforward, and we will only give an outline ofthe argument.

A germ (of a lift of f) at a point x ∈ X is a branched coveringmap gx : Ux → Z defined on an an open and connected neighborhoodUx ⊂ X of x such that f |Ux = π gx. We consider two such germs at xas equivalent if they agree on a neighborhood of x (actually one shoulddefine a germ as an equivalence class of such local lifts at x, but forthe sake of easier exposition it is convenient to ignore the distinctionbetween a local lift and the equivalence class that it represents.)

The value gx(x) does not determine gx uniquely in general, but itfollows from (i) that the value gx(x

′) for a point x′ 6= x sufficiently closeto x gives a uniqueness statement for gx.

This fact allows one to define a notion of an analytic continuationof a germ gx along a path α : [0, 1] → X with α(0) = x. Such a

470 A

continuation is given by germs gα(t) at α(t) for t ∈ [0, 1] such that gα(0)

is equivalent with the given germ gx; moreover, we require that thesegerms are compatible in the following sense: if t0 ∈ [0, 1] is arbitrary,then each germ gα(t) for t ∈ [0, 1] close enough to t0 is equivalent to agerm obtained by restricting gα(t0) to a suitable neighborhood of α(t).

As in the classical monodromy theorem, one can show that if ananalytic continuation of gx along a path α : [0, 1]→ X exists, then upto equivalence the germ gα(0) uniquely determines gα(1). In this casewe say that the (equivalence class of the) germ gα(1) is obtained byanalytic continuation of gα(0) along α.

Analytic continuation of a germ along homotopic paths with thesame endpoints leads to equivalent germs. More precisely, let x, x′ ∈ X,H : [0, 1] × [0, 1] → X be continuous, and assume that x = H(s, 0)and x′ = H(s, 1) for all s ∈ [0, 1]. Suppose that we have an analyticcontinuation of a germ at x along every path t 7→ αs(t) := H(s, t),s ∈ [0, 1]. Let g0

x′ and g1x′ be the two germs at x′ obtained by analytic

continuation of the given germ along α0 and α1, respectively. Then g0x′

and g1x′ are equivalent.

Using these facts one can now give the proof of (ii) as follows.Our hypotheses imply that degz0(g)| degf (x0). Hence there exists alocal germ gx0 of a lift of f at x0 (to check this one writes π near z0

and f near x0 as power functions in suitable local coordinates as in(2.1)). Moreove, since degg(z)| degf (x) whenever x ∈ f−1(y) and z ∈π−1(y), the analytic continuation of gx0 along any path α : [0, 1] → Xwith α(0) = x0 exists, because this condition ensures that we neverencounter any “singularities” in the analytic continuation along α.

The fact that X is simply connected and the homotopy invarianceof analytic continuation as discussed guarantees that up to equivalencethe initial choice of the germ gx0 leads to a unique germ gx at everypoint x ∈ X. If we define g(x) = gx(x), then g is a branched coveringmap from X to Z such that g(x0) = z0 and f = π g.

Lemma A.5 (Compositions of branched coverings). Let X, Y, Z besurfaces and f : X → Y , g : Y → Z, h : X → Z be continuous mapssuch that h = g f . Assume two of the maps f, g, h are branchedcovering maps, then the third one is a branched covering map as well.

STILL TO BE DONE!

A.5. Quotients of maps

Let ∼ be an equivalence relation on a set X. We denote by X/∼ thequotient set consisting of all equivalence classes [x] := y ∈ X : x ∼ y,

A.5. QUOTIENTS OF MAPS 471

x ∈ X, and by π : X → X/∼ the natural projection that sends eachpoint x ∈ X to its equivalence class [x]. If X is a topological space,then we equip X/∼ with the quotient topology. In this topology a setU ⊂ X/∼ is open if and only if π−1(U) is open. Then π is a continuousmap. Moreover, a map f : X/∼ → Z into another topological space Zis continuous if and only if f π : X → Z is continuous. Actually, thisfunctorial property characterizes the quotient topology on X/∼.

A set A ⊂ X is called saturated (for ∼) if it is a union of equivalenceclasses. Obviously, the set A is saturated if and only if A = π−1(π(A)).

Let Θ: X → Y be a surjective map between sets X and Y . Theequivalence relation induced by Θ on X is given by

(A.5) x ∼ y :⇐⇒ Θ(x) = Θ(y),

for x, y ∈ X. Note that if A ⊂ X is saturated for ∼, then Θ(X \A) =Y \Θ(A). There is a natural map h : X/∼→ Y given by

(A.6) h : [x] 7→ Θ(x)

which clearly is well defined and bijective. The following lemma is wellknown.

Lemma A.6. Let Θ: X → Y be a continuous and surjective mapbetween topological spaces X and Y , and ∼ be the equivalence relationon X induced by Θ.

The map h : X/∼→ Y from (A.6) is a homeomorphism if and onlyif Θ(U) ⊂ Y is open for every saturated open set U ⊂ X.

Proof. Let h be defined as in (A.6). It is bijective. Note thatΘ = h π. Since Θ is continuous, h is continuous by the functorialproperty of the quotient topology.

Assume first that Θ(U) ⊂ Y is open for every saturated open setU ⊂ X. To show that h is a homeomorphism it remains to show thath−1 is continuous, or equivalently that h is open. So let V ⊂ X/∼ beopen. Then U := π−1(V ) is a saturated open set with π(U) = V , andso h(V ) = (h π)(U) = Θ(U) is open as desired.

Assume now that h is not homeomorphism. Then there exists anopen set V ⊂ X/ ∼ such that h(V ) ⊂ Y is not open. Note thatU := π−1(V ) ⊂ X is open and saturated, but Θ(U) = h π(U) = h(V )is not open.

It is easy to verify the condition of the previous lemma in the casesthat are relevant for us.

Corollary A.7. Let X, Y be topological spaces, Θ: X → Y bya continuous surjective map, and ∼ be the equivalence relation on X

472 A

induced by Θ. Then the map h : X/∼ → Y given by h([x]) = Θ(x) isa homeomorphism in the following cases.

(i) X is compact and Y is Hausdorff.

(ii) Θ: X → Y is an open map.

(iii) Θ is a branched covering between surfaces X, Y .

Proof. The statement holds in case (ii), since the condition fromLemma A.6 is trivially satisfied when Θ is open.

This immediately implies case (iii), since every branched cover isopen.

(i) Let X and Y be as in the statement, and U ⊂ X be saturatedand open. Then X \ U is also saturated. This set is also closed, andhence compact. Thus, Θ(X \ U) = Y \ Θ(U) is compact. Since Y isHausdorff, this implies that this set is closed. So Θ(U) is open. Weconclude from Lemma A.6 that X/∼ and Y are homeomorphic.

We now consider equivalence relations that are induced by groupactions. Statements that are similar in spirit to the ones above holdin this case. To state these precisely, we will first discuss some relatedterminology that is also relevant elsewhere.

Let X be a topological space and G be a group of homeomorphismsacting on G. The equivalence relation ∼ on X induced by G is definedby

(A.7) x ∼ y :⇐⇒ there exists g ∈ G such that y = g(x)

for x, y ∈ X. The equivalence class [x] of a point x ∈ X with respectto G is equal to its orbit Gx := g(x) : g ∈ G under G. We denotethe quotient space by X/G, and equip it with the quotient topology.As before, we denote by π the quotient map π : X → X/G given byπ(x) = [x] = Gx for x ∈ X.

A fundamental domain for the action of G is a set F ⊂ X suchthat F is equal to the closure of its interior int(F ), every orbit of Ghas at least one point in F , and at most one point in int(F ). If F isa fundamental domain for G, then, at least on an intuitive level, oneobtains the quotient space X/G from F by identifying the points onthe boundary of F that lie in the same orbit.

Suppose Y is another topological space and Θ: X → Y is a contin-uous and surjective map. We say that Θ is induced by the action of agroup G acting on X if Θ(x) = Θ(y) for x, y ∈ X is and only if thereexists g ∈ G such y = g(x). In this case, the equivalence relation on Xinduced by Θ is the same as the equivalence relation induced by G.

A.5. QUOTIENTS OF MAPS 473

The stabilizer Gx of a point x ∈ X is the subgroup of G consistingof all elements g ∈ G with g(x) = x. We call the action cocompact ifthere exists a compact set K ⊂ X such that X =

⋃g∈G g(K). So then

the translates of K under G cover X, and X/G = π(K). In particular,the quotient space X/G is compact as the continuous image π(K) ofthe compact set K.

Suppose in addition that X is a proper metric space, i.e., a spacewhere all closed balls are compact. The action of G on X is calledproperly discontinuous if for each compact set K ⊂ X the set g ∈ G :g(K) ∩K 6= ∅ is finite. We call the action of G on X geometric if itis cocompact and properly discontinuous, and each element of G is anisometry on X.

Lemma A.8. Let X be a topological space, Y be a Hausdorff space,G be a group of homeomorphisms acting on cocompactly X, and Θ: X →Y be a surjective and continuous map induced by G. Then there existsa unique homeomorphism ϕ : X/G→ Y such that Θ = ϕ π.

So by the homeomorphism ϕ one can identify Y with X/G and Θwith the quotient map π : X → X/G.

Proof. We define ϕ([x]) = Θ(x) for [x] ∈ X/G. Since Θ is in-duced by G, is clear that ϕ : X/G→ Y is a well-defined map satisfyingΘ = ϕ π. This last identity determines ϕ uniquely. The map ϕ isinjective, as follows from the fact that Θ is induced by G. It is alsosurjective, because Θ is, and continuous by the functorial property ofthe quotient topology. So ϕ is a continuous bijection. To show thatϕ is a homeomorphism, it suffices to prove that ϕ(A) ⊂ Y is closedwhenever A ⊂ X/G is a closed. Now since the action of G on X iscocompact, the quotient X/G is compact. So if A ⊂ X/G is closed,then A, and hence also ϕ(A), are compact. Since Y is a Hausdorffspace, it follows that ϕ(A) is closed as desired.

Let ∼ be an equivalence relation on a set X, and f : X → X be amap. We say that f descends to the quotient X/ ∼ if there exists a

map f : X/∼→ X/∼ such that f π = π f . In this case, we havethe following commutative diagram:

Xf

//

π

X

π

X/∼ f// X/∼ .

474 A

Here necessarily f([x]) = [f(x)] for x ∈ X. This shows that if f

descends to X/∼, then f is uniquely determinedIt is easy to characterize when a map f descends. The relevant con-

dition is that ∼ should be invariant for f , i.e., we have the implication

x ∼ y ⇒ f(x) ∼ f(y)

for all x, y ∈ X, or equivalently, that for every saturated set A ⊂ Xthe set f−1(A) is saturated.

Lemma A.9 (Quotients of maps). Let ∼ be an equivalence relationon a set X, and f : X → X be a map.

(i) The map f descends to a map f on X/∼ if and only if ∼ isinvariant for f .

(ii) Suppose that X is a topological space, and f is a continuous

map that descends to the map f : X/ ∼→ X/ ∼. Then f iscontinuous.

Proof. (i) Suppose ∼ is invariant for f . Then the map f : X/∼→ X/ ∼ given by f([x]) := [f(x)] for [x] ∈ X/ ∼ is well-defined.

Obviously, f π = π f which shows that f descends to X/∼.

Conversely, suppose that f descends to the map f on X/∼. Letx, y ∈ X with x ∼ y be arbitrary. Then [x] = [y], and so

[f(x)] = (π f)(x) = (f π)(x) = f([x]) = f([y]) = [f(y)].

Hence f(x) ∼ f(y), and we see that ∼ is invariant.

(ii) Suppose the continuous map f : X → X descends to the map

f : X/ ∼→ X/ ∼. Then f π = π f is continuous. Hence f iscontinuous by the functorial property of the quotient topology.

Lemma A.10. Let Θ: X → Y be a surjective map that induces anequivalence relation ∼ on X. Let f : X → X be a map such that ∼is invariant for f . Then there is a unique map g : Y → Y such thatΘ f = g Θ, i.e., the following diagram commutes

Xf//

Θ

X

Θ

Yg// Y.

(i) If X is compact and Y is a Hausdorff space it holds that g iscontinuous.

(ii) If f , Θ are branched covering maps between surfaces it holdsthat g is a branched covering.

A.6. TORI AND LATTICES 475

(iii) If X, Y are Riemann surfaces and Θ, f are both holomorphicit follows that the map g is holomorphic.

In each of these three cases g is topologically conjugate to the the de-

scended map f : X/∼ → X/∼.

Proof. For any y ∈ Y , consider a x ∈ X with Θ(x) = y. Defineg(y) := Θ f(x). It is elementary to check that g is well-defined,satisfies Θ f = g Θ, and is the unique such map.

(i) Recall from part (i) of Corollary A.7 that h : X/∼ → Y givenby h([x]) = Θ(x) is a homeomorphism and from Lemma A.9 that f

descends to a continuous map f : X/∼ → X/∼. Let [x] ∈ X/∼ then

g h([x]) = g Θ(x) = Θ f(x) = h([f(x)]) = h f([x]),

i.e., h is a topological conjugacy between f and g. This implies thecontinuity of g.

(ii) Using Corollary A.7 (iii) it follows as in (i) that g is continuous

and topologically conjugate to f . Since Θ f = g Θ it follows fromLemma A.5 that g is a branched covering map.

(iii) We can assume that X (hence Y ) is connected. Since Θ issurjective it follows that it is not constant, hence open. Using Corol-lary A.7 (ii) it follows as in the proof of (i) that g is continuous and

topologically conjugate to f .Consider first a point x ∈ X that is not a critical point of Θ.

Thus there is a branch of Θ−1 defined in a neighborhood Vy ⊂ Y ofy = Θ(x), such that Θ−1(y) = x (which is holomorphic). It followsthat g = Θ f Θ−1, thus holomorphic, on Vy.

Assume now that x ∈ X is a critical point of Θ. Let Ux ⊂ Xbe a neighborhood of x that contains no other critical point of Θ.Then Vy = Θ(Ux) is a neighborhood of y = Θ(x). From the previousargument it follows that g is holomorphic at any w ∈ Vy \ y. Thusy is an isolated singularity. Since g is continuous at y it follows fromRiemann’s removability theorem that g is holomorphic at y.

A.6. Tori and lattices

In this section we review some facts about tori and lattices. For someof the related terminology see also the beginning of Chapter 1.2.

A lattice Γ ⊂ R2 is a non-trivial discrete subgroup of R2 (consid-ered as a group with vector addition). Often it is more convenient toconsider a lattice as a discrete subgroup of C ∼= R2. In the following,

476 A

we will freely switch back and forth between these different viewpointsand real and complex notation.

The rank of a lattice is the dimension of the subspace of R2 (consid-ered as a real vector space) spanned by the elements in Γ. The latticecan be a rank-1 lattice. Then, using complex notation, we can writeit in the form Γ = ωZ, where ω ∈ C \ 0. The other case is that Γis a rank-2 lattice. Then there exist generators ω1, ω2 ∈ C \ 0 withIm(ω2/ω1) > 0 such that

Γ = ω1Z⊕ ω2Z := mω1 + nω2 : m,n ∈ Z.If γ ⊂ R2 ∼= C is a rank-2 lattice, then we have an equivalence

relation ∼ on R2 given by

x ∼ y :⇐⇒ x− y ∈ Γ

for x, y ∈ R2. We denote the quotient space R2/∼ by R2/Γ, and byπ : R2 → R2/Γ the quotient map that sends a point x ∈ R2 to itsequivalence class [x]. In complex notation, we use C/Γ to denote thequotient space; the quotient map is then a map π : C→ C/Γ.

We equip R2/Γ with the quotient topology. Then T = R2/Γ is a 2-dimensional torus, and π : R2 → T = R2/Γ is a covering map; actually,since we insist on covering maps to be orientation-preserving, we haveto equip T with a suitable orientation here. The lattice translations

τγ : R2 → R2, u ∈ R2 7→ τγ(u) := u+ γ, γ ∈ Γ,

are deck transformations for π; so π = π τγ for γ ∈ Γ. We will seemomentarily that actually every deck transformation of π has this form(Lemma A.11 (i)).

The conformal structure on C induces a unique conformal structureon T = R2/Γ = C/Γ. It is represented by a complex atlas on C/Γgiven by suitable inverse branches of π. Then T is a complex torus andπ : C→ T a holomorphic map.

Every torus can be repesented in the form R2/Γ up to orientation-preserving homeomorphisms. Actually, we can choose Γ = Z2 if thisis convenient. Up to conformal equivalence every complex torus hasa representation of the form C/Γ. Here two complex tori T = C/Γand T ′ = C/Γ′ obtained from rank-2 lattices Γ,Γ′ ⊂ C are conformallyequivalent if and only if there exists α ∈ C \ 0 such that Γ′ = αΓ.

Lemma A.11. Let Γ ⊂ R2 be a rank-2 lattice, T = R2/Γ, andπ : R2 → T = R2/Γ be the quotient map.

(i) For a continuous map ϕ : R2 → R2 we have π ϕ = π if andonly if there exists γ ∈ Γ such that ϕ = τγ.


(ii) If A : T → T is a torus endomorphism, then A can be lifted toa homeomorphism on R2, i.e., there exists a homeomorphismA : R2 → R2 such that A π = π A. The homeomorphism Ais orientation-preserving, and unique up to post-compositionby a translation τγ, γ ∈ Γ.

(iii) If A : T → T is a torus endomorphism, then there exists aunique R-linear map L : R2 → R2 with L(Γ) ⊂ Γ such that forevery lift A as in (ii) we have

(A.8) A τγ A−1 = τL(γ) = L τγ L−1

for all γ ∈ Γ.

(iv) If A : T → T is a torus endomorphism and L the map as in(iii), then deg(A) = det(L).

Statement (i) implies that the deck transformations for π are pre-cisely the lattice translations τγ, γ ∈ Γ. Note that from (ii) we get acommutative diagram of the form

(A.9) R2 A//

π

R2

π

TA// T.

The map L in (ii) can be viewed as the map induced by A onthe fundamental group of T = R2/Γ. More precisely, if x0 ∈ T andy0 := A(x0), then A induces a map A∗ : π1(T, x0) → π1(T, y0) (seeSection 1.2). Moreover, we have a natural isomorphism π1(T, x0) ∼=Γ. Namely, if [`] ∈ π1(T, x0) is an element of the fundamental group

represented by a loop ` based at x0, then we can lift ` to a path ˜

on R2 under π. In general, ˜ is not a loop; if u0 ∈ R2 is the initialpoint of ˜ and v0 ∈ R2 the other endpoint of ˜, then γ := v0 − u0 ∈ Γ,and one can show that the map [`] 7→ γ gives a well-defined groupisomorphism π1(T, x0)→ Γ. Similarly, we have a natural isomorphismπ1(T, y0) → Γ. Under these isomorphisms, the map A∗ : π1(T, x0) →π1(T, y0) corresponds to the homomorphism L : Γ → Γ. As we onlymention this to put the map L in familiar context, we will omit adetailed justification of the claims in the preceding discussion.

Proof of Lemma A.11. (i) Let ϕ : R2 → R2 be a continuousmap with π ϕ = π. Then we can consider ϕ as a lift of π : R2 → Tunder the covering map π : R2 → T , because we have the commutative

478 A

diagram

R2

π

R2

ϕ>>

π// T.

Fix u0 ∈ R2, and define v0 = ϕ(u0). Then π(v0) = π(ϕ(u0)) =π(u0). This means that u0 ∼ v0 are equivalent with respect to theequivalence realation ∼ induced by Γ. Thus, there exists γ ∈ Γ suchthat v0 = u0 + γ. Consider the corresponding lattice translation τγ.Then π τγ = π, and so τγ is also a lift of π under the covering map π.Since τγ(u0) = v0 = ϕ(u0), Lemma 1.2 implies that ϕ = τγ.

(ii) We can lift the map A π : R2 → T under the covering mapπ to get a continuous map A : R2 → R2 satisfying π A = A π(Lemma A.4 (ii)); then we have the commutative diagram

R2

π

R2

A

>>

Aπ// T.

(Note that this is essentially the same as (A.9)).Fix u0 ∈ R2, and define v0 := A(u0) ∈ R2. Note that ψ := A

π : R2 → T is also a covering map, and π(v0) = π(A(u0)) = A(π(u0)) =ψ(u0). So by lifting π : R2 → T under this covering map we can finda continuous map B : R2 → R2 with B(v0) = u0 and π = =ψ B =A π B, and obtain the commutative diagram

u0 ∈ R2

ψ=Aπ

v0 ∈ R2

B99

π// T,

Then (A B)(v0) = A(u0) = v0 and

π (A B) = A π B = π = π idR2 .

So we conclude A B = idR2 by Lemma A.4 (i) applied to the coveringmap π. Similarly, (B A)(u0) = B(v0) = u0 and

ψ (B A) = A π B A = π A = A π = ψ = ψ idR2 ,

which implies B A = idR2 by the same lemma applied to the coveringmap ψ. It follows that A is a homeomorphism on R2 with inverse B.

The relation π A = A π shows that locally A can be writtenas a composition of A π and a branch of π−1. Since these maps are


orientation-preserving, A has this property as well. These considera-tions show that A has a lift A as desired.

Suppose A : R2 → R2 is another continuous map with πA = Aπ.Then

π (A A−1) = π A B = A π B = ψ B = π.

By (i) there exists τ ∈ Γ such that AA−1 = τγ. Then A = τγ A. Thisshows that A is unique up to post-composition by a lattice translation.Note that clearly every map for the form τγ A, τ ∈ Γ, is actually alift of A.

(iii) Let A be homeomorphic lift of A as is (ii). If γ ∈ Γ is arbitrary,then

π A τγ A−1 = A π τγ A−1

= A π A−1 = π A A−1 = π.

So A τγ A−1 is a deck transformation of the covering map π. By(i)this implies that there exists a unique γ′ ∈ Γ such that

A τγ A−1 = τγ′ .

Note that for γ, σ ∈ Γ we have

A τγ+σ A−1 = A τγ τσ A−1 = τγ′ A τσ A−1

= τγ′ τσ′ = τγ′+σ′ .

Moreover, if

τγ′ = A τγ A−1 = τ0 = idR2 ,

then

τγ = A−1 idR2 A = idR2 ,

and so γ = 0. This implies that the map γ 7→ γ′ gives an injectivegroup homomorphism L : Γ → Γ such that A τγ A−1 = τL(γ) for allγ ∈ Γ. We can uniquely extend L to an invertible R-linear map, onR2, also denoted by L. Then L(Γ) ⊂ Γ and the first equality in (A.8)is true by definition of L. The second equation in (A.8) is actually truefor every invertible linear map L.

Uniqueness of L is clear.

(iv) Let A : R2 → R2 be a lift of A as in (iii). We define a homotopyH : R2 × [0, 1]→ R2 as

H(u, t) = (1− t)A(u) + tL(u), u ∈ R2, t ∈ [0, 1].

480 A

Let Ht := H(·, t) for t ∈ [0, 1]. Then H0 = A and H1 = L. For allu ∈ R2, t ∈ [0, 1], and γ ∈ Γ we have

(Ht τγ)(u) = (1− t)A(τγ(u)) + tL(τγ(u))

= (1− t)τL(γ)(A(u)) + tτL(γ)(L(u))

= (1− t)A(u) + tL(u) + L(γ) = (τL(γ) Ht)(u).

Here we used (A.8). Hence Ht τγ = τL(γ) Ht for all t ∈ [0, 1] andγ ∈ Γ. This implies that the equivalence relation ∼ induced by Γ ininvariant under Ht, and so Ht descends to a continuous map H t onT = R2/Γ (see Lemma A.9). Then π Ht = H t π for all t ∈ [0, 1]. Inparticular,

A π = π A = π H0 = H0 π,

which implies H0 = A.Define H : T × [0, 1]→ T as H(x, t) = H t(x) for (x, t) ∈ T × [0, 1].

Then H is continuous. Indeed, if (xn, tn) is a sequence in T × [0, 1]with (xn, tn) → (x, t) ∈ T × [0, 1] as n → ∞, then by considering alocal inverse of π near (x, t), we can find points u ∈ R2 and un ∈ R2

for n ∈ N with π(u) = x, π(un) = xn and un → u as n→∞. Then asn→∞,

H(xn, tn) = H tn(π(un)) = π(Htn(un)) = π(H(un, tn))→ π(H(u, t))

= π(Ht(u)) = H t(π(u)) = H(x, t)

by the continuity of H and π.The map H is a homotopy on T with the time-t maps H t. In

particular, the maps A = H0 and L := H1 on T are homotopic.Since degrees of maps are invariant under homotopies, it follows thatdeg(A) = deg(L). Note that L π = H1 π = H1 π = L π. So L isa map on T induced by the linear map L.

It is a standard fact that then deg(L) = det(L). One can seethis as follows by using some basic concepts from differential geometry.Namely, we can consider T = R2/Γ as a complex torus and hence as asmooth manifold. Then L is a smooth map on T . There exists a unique2-form α on T whose pull-back π∗(α) is equal to the standard volumeform on R2 given by ω = dx ∧ dy, where x and y are the standardcoordinates on R2. If we pull ω back by L, we get L∗(ω) = det(L)ω.

This implies that L∗(α) = det(L)α. On the other hand,∫

T

L∗(α) = deg(L)

∫T

α.

A.7. ORBIFOLDS AND GEOMETRIC STRUCTURES 481

Hence

det(L)

∫T

α =

∫T

L∗(α) = deg(L)

∫T

α.

Since

∫T

α 6= 0 this implies that deg(L) = det(L). We conclude that

deg(A) = deg(L) = det(L)

as desired.

In Lemma A.11 we can consider T = C/Γ a complex torus. Thenthe quotient map π : C → T = C/Γ is holomorphic. Suppose A : T →T is holomorphic torus endmorphism, and A : C → C is a homeomor-phic lift of A as in Lemma A.11 (ii). Then π A = A π. This showsthat locally the map A is given as a composition of the holomorphicmap A π with a branch of π−1, which is also holomorphic. We con-clude that A : C → C is holomorphic itself. Since this map is also ahomeomorphism on C it must be of the form A(z) = αz+ β for z ∈ C,where α, β ∈ C, α 6= 0. The associated linear map L : C → C as inLemma A.11 (iii) is then given by L(z) = αz for z ∈ C. By part (iv) ofthis lemma we have deg(A) = det(L) = |α|2 (in this computation wehave to consider L as an R-linear map).

A.7. Orbifolds and geometric structures

In this section we discuss how to obtain an associated geometric struc-ture from orbifold data given by a ramification function on a surfaceS. We will restrict ourselves to the only case that will be important

for us, namely when S is the Riemann sphere C. We start by settingup and reviewing some relevant terminology.

Now let (C, α) be an orbifold with the Riemann sphere as the under-

lying surface, and a ramification function α : C → N. In the followingwe set P := p ∈ S2 : α(p) ≥ 2. This finite set is the support of theramification function α. We will make the additional assumptions that

α(p) < ∞ for each p ∈ C and that (C, α) is hyperbolic or parabolic.These will be the only relevant cases for us.

One can find a Riemann surface X that serves as a model space

for the orbifold (C, α), and is equipped with a suitable metric d0. If

the orbifold (C, α) is hyperbolic, then X = D and d0 is the hyperbolic

metric on D. If (C, α) is parabolic, then X = C and d0 is the Eu-clidean metric. In both cases, (X, d0) is a geodesic metric space, andthe orientation-preserving isometries of (X, d0) are given by conformalautomorphisms of X.

482 A

By the universal orbifold covering theorem (see [?]) one can find

a surjective holomorphic map Θ: X → C and a group of orientation-preserving isometries Γ of (X, d0) that acts geometrically such that thefollowing properties are true:

(i) if p ∈ C and x ∈ Θ−1(p), then degΘ(x) = α(p),

(ii) for x, y ∈ X we have Θ(x) = Θ(y) if and only if there existsγ ∈ Γ such that γ(x) = y.

We call Θ the universal orbifold covering map of (C, α), and Γ theassociated group of deck transformations.

Condition (i) says that the local degree of Θ is constant on eachfiber Θ−1(p) and is given by the value α(p) of the ramification functionα at p (this fact explains the term chosen to designate α). The map

Θ is a covering map over C \ P . Condition (ii) implies that Γ is thegroup of deck transformation of this cover, i.e., Γ consists precisely ofall homeomorphisms γ of X such that Θ = Θ γ. The stabilizer of a

point x in a fiber Θ−1(p), p ∈ C, is a cyclic group of order α(p). Thisjustifies a remark in the previous section that α(p) should be thoughtof as the order of a cyclic group.

We get a canonical metric structure on (C, α) by identifying thisspace with the metric quotient X/Γ. More explicitly, we define the

canonical orbifold metric d on (C, α) by

(A.10) d(p, q) := infd0(z, w) : z ∈ Θ−1(p), w ∈ Θ−1(q)

for p, q ∈ C. Since Γ acts geometrically on (C, d0) and transitivelyon Θ−1(p), for each z0 ∈ Θ−1(p) there exists w0 ∈ Θ−1(q) such thatd(p, q) = d0(z0, w0). In particular, the infimum in (A.10) is attained asa minimum. This implies that d(p, q) > 0 if p 6= q, and it follows that

d is indeed a metric on C (the other properties of a metric immediatelyfollow from the definition of d).

We have d(Θ(z),Θ(w)) ≤ d0(z, w) for all z, w ∈ D. It is easy to seethat a stronger condition is true locally: for each z ∈ X there exists aneighborhood Nz of z in X such that

(A.11) d(Θ(z),Θ(w)) = d0(z, w)

for all w ∈ Nz. This together with a simple covering argument impliesthat the map Θ is a path isometry: if γ is a path in X, then

lengthd(Θ γ) = lengthd0(γ).

The metric space (C, d) is geodesic. Indeed, a geodesic joining

two points p, q ∈ C can be obtained as follows. We can pick points

A.7. ORBIFOLDS AND GEOMETRIC STRUCTURES 483

z ∈ Θ−1(p) and w ∈ Θ−1(q) such that d(Θ(z),Θ(w)) = d0(z, w). Since(X, d0) is geodesic, we can find a geodesic segment joining z and w.Since Θ is a path isometry, the path Θ γ must then be a geodesic

segment in (C, d) joining p and q.It follows from the local behavior of Θ and a compactness argument

that there exists a constant C ≥ 1 such that

(A.12)1

Cσ(p, q) ≤ d(p, q) ≤ Cσ(p, q)ε0

for all p, q ∈ C. Here ε0 := 1/maxα(p) : p ∈ C. In particular, d

induces the standard topology on C.

If (C, α) is an orbifold and d the canonical orbifold metric, then

near each point p ∈ C, the space (C, d) is isometric to a cone with coneangle 2π/α(p). Here the cone is hyperbolic or Euclidean depending

on whether the orbifold is hyperbolic or parabolic. The points p ∈ Cwhere α(p) ≥ 2 are precisely the conical singularities of (C, d), where

the cone angle differs from 2π. Near all other points, (C, d) is locallyisometric to the model space (X, d0).

Consider a sector, i.e., a set of the form

S = z ∈ C : |z| < r, 0 ≤ arg(z) ≤ θ,

for some r > 0, θ > 0. If we identify the line segments arg(z) = 0and arg(z) = θ in S we obtain a cone of cone angle θ. The (imageof) the point 0 (under this identification) is called the tip of the cone.Such a cone is equipped with the induced length metric.

Cones with cone angle θ > 2π are possible, in this case one hasto replace C in the above definition by the Riemann surface of thelogarithm.

We will be mostly interested in cones of cone angle θ = 2π/n forsome n ∈ N. Let us identify Zn with the n roots of unity, i.e., Zn =e2πi/k : k = 0, . . . , n−1. This group acts on C by multiplication (i.e.,as rotations). Consider now the disk Dr = z ∈ C : |z| < r equippedwith the metric d given by

d(x, y) = min|x− gy| : g ∈ Zn,

for all x, y ∈ Dr. Then (Dr, d) is a cone with cone angle 2π/n.

A locally flat surface with cone points is a metric space S that is asurface (i.e., a two dimensional manifold) that is locally isometric to anopen set in the plane R2 ∼= C or a cone. More precisely for any pointx ∈ S there is a neighborhood U ⊂ S of x and an isometry ϕ : U → V ,where either V ⊂ R2 is open or V = C is a cone and ϕ(x) is the tip of

484 A

the cone C. In the latter case we call x a cone point of S. It followsthat cone points are isolated. In the first case we call ϕ : U → V ⊂ R2

a flat chart of S.Clearly every polyhedral surface is a locally flat surface with cone

points when equipped with the induced length metric.

Lemma A.12. Let G be a group that acts properly discontinuous onC by orientation preserving isometries. The equivalence relation ∼ onC is defined by

x ∼ y if and only if x = g(y) for a g ∈ G.

We define a metric d on C/G := C/∼ by

d(x, y) := min|x− g(y)| : g ∈ G.

Then (C/G, d) is a locally flat surface with conical singularities.

Proof.

Note that a cone C (with cone angle θ) may be mapped to thedisk D(z, r′) = |z| ≤ r′ by the map z 7→ zα, where α = 2π/θ (andr′ = rα).

Every locally flat surface S with cone points may be naturallyequipped with a conformal structure as follows. Every flat chart (i.e.,isometry) ϕ : U → V ⊂ R2, is a chart.

Consider now a cone point x ∈ S, and an isometry ϕ : U → C. HereU is a neighborhood of x and C is a cone (with cone angle θ > 0). Thecomposition of ϕ : U → C and zα : C → D(0, r′) then is a chart. Clearlychanges of coordinates are conformal, so X equipped with these chartsis a Riemann surface, i.e., these charts define a conformal structure onS.

In the case our locally flat surface with cone points is a polyhedron,where each face is a convex polygon we can map each the interior of twofaces intersecting in an edge isometrically to the union of two polygonsin the plane.

Lemma A.13. Let S be a locally flat surface with cone points thatis homeomorphic to the sphere S2. Let f : S → S be a map that isholomorphic with respect to flat charts.

Then there is a map h : S → C that is conformal with respect to the

conformal structure on S given above, and a rational map g : C → C

A.8. ORBIFOLDS AND COVERINGS 485

such that g h = h f , i.e., the following diagram commutes

Sf//

h

S

h

Cg// C.

Proof. The conformal map h : S → C exists by the uniformization

theorem. The map g := h f h−1 : C → C is holomorphic on thecomplement of the preimages of the cone points. Thus g is holomorphicby the Riemann removability theorem, i.e., g is a rational map.

***

A.8. Orbifolds and coverings

Orbifolds may be defined in much greater generality, we however re-strict ourselves to the case of 2-dimensional orbifolds such that all itssingular points are cone points. Such an orbifold O is a pair (S, α),

where S is a 2-dimensional manifold, and α : N → S is a map, suchthat the set of points in S points x ∈ S with α(x) ≥ 2 is discrete. Apoint x ∈ S with α(x) ≥ 2 is called a cone point of the orbifold O.A point x ∈ S with α(x) = ∞ is called a puncture of the orbifold.Punctures should be thought of as being removed from the orbifold. Ifα ≡ 1 we identify the orbifold (S, α) with the surface S, conversely anysurface S may be viewed as such an orbifold (abusing notation).

The Euler characteristic of an orbifold O = (S, α) is defined as

(A.13) χ(O) := χ(S)−∑x∈S

(1− 1

α(x)

),

where χ(S) is the Euler characteristic of the underlying manifold S.Here and in the following we use the convention 1/∞ = 0. If S isa compact surface this may be computed as follows. Let D be a celldecomposition of S (see Section 5.1) such that each cone point x of isa vertex of D (i.e., x is a 0-cell of D). Each cone point x counts as1/α(x) of a point, more precisely we define

(A.14) #V =∑ 1

α(x),

where the sum is taken over all vertices of D. Then

χ(O) = #F −#E + #V,

where #F is the number of faces (i.e., of 2-cells) and #E is the numberof edges (i.e., 1-cells) of D. Thus with the slight modification (A.14)

486 A

the Euler characteristic of an orbifold is given in the usual way. Notethat punctures do not contribute to #V in (A.14). We will sometimesremove the punctures from the orbifold O = (S, α). This means weconsider

(A.15) S0 = S \ x ∈ S : α(x) =∞.

Consider now two orbifolds O = (S, α) and O′ = (S ′, α′). Assumefirst that O and O′ do not contain punctures. Let f : S ′ → S be abranched covering map. Then f viewed as a map on the orbifolds,written as f : O′ → O, is called an orbifold covering map if

(A.16) α(f(x′)) = deg(f, x′)α′(x′),

for all x′ ∈ S ′. The degree of the orbifold covering map is defined tobe the degree of the underlying branched covering map.

Assume now that O and O′ have punctures. In this case an orbifoldcovering map does not need to be defined on punctures, though itmay be. The formal definition is as follows. Let S0 ⊂ S∗ ⊂ S andS ′0 ⊂ S ′∗ ⊂ S ′, so S∗ and S ′∗ are obtained by removing some puncturesfrom S and S ′. An orbifold covering map f : O′ → O is a branchedcovering map f : S ′∗ → S∗, satisfying (A.16) for all x ∈ S ′∗. Again thedegree of this orbifold covering map is defined to be the degree of theunderlying branched covering map.

We have the following variant of the Riemann-Hurwitz formula (2.3).

Lemma A.14. Let O′ = (S ′, α′) and O = (S, α) be orbifolds andf : S ′ → S be a branched covering map of degree d. Then

χ(O′) = dχ(O)−∑x′∈S′

(deg(f, x′)

α(f(x′))− 1

α′(x′)

).

Proof. Note that∑

x′∈S′ deg(f, x′) = d. It holds

χ(O′) = χ(S ′)−∑x′∈S′

(1− 1

α′(x′)

)= dχ(S)−

∑x′∈S′

(deg(f, x′)− 1)−∑x′∈S′

(1− 1

α′(x′)

)


where the (standard) Riemann-Hurwitz (2.3) formula was used

= dχ(S)−∑x∈S

∑x′∈f−1(x)

deg(f, x′)

(1− 1

α(f(x′))

)

−∑x′∈S′

(deg(f, x′)

α(f(x′))− 1

α′(x′)

)= dχ(S)− d

∑x∈S

(1− 1

α(x)

)−∑x′∈S′

(deg(f, x′)

α(f(x′))− 1

α′(x′)

)where

∑x′∈S′ deg(f, x′) = d was used

= dχ(O)−∑x′∈S′

(deg(f, x′)

α(f(x′))− 1

α′(x′)

).

For coverings of orbifolds the second term vanishes (see (A.16)),thus the following holds.

Corollary A.15. Let O′ = (S ′, α′) and O = (S, α) be orbifoldsand f : O′ → O be an orbifold covering map of degree d. Then

χ(O′) = dχ(O).

The following is a reformulation of Proposition 2.13 (iii).

Lemma A.16. Let f : S2 → S2 be a Thurston map, αf : S2 → N beits ramification function (see Definition 2.7), and Of = (S2, αf ) be thecorresponding orbifold. Then f is parabolic if and only if f : Of → Ofis an orbifold covering map.

The following is the existence theorem for the universal orbifoldcovering. Proofs can be found in [Lam, 4.8.2], [Th, Proposition 13.2.4and Theorem 13.3.6] and (in much greater generality) in [BH].

Theorem A.17. Let O = (S, α) be an orbifold that is parabolic orhyperbolic. Then the following holds.

(i) There exists a simply connected surface X and a branched cov-ering map Θ: X → S0 such that

deg(Θ, x) = α(Θ(x),

for all x ∈ X;

488 A

(ii) if S is a Riemann surface and O is hyperbolic, then X = D isthe hyperbolic plane and Θ is holomorphic;

(iii) if S is a Riemann surface and O is parabolic, then X = C isthe complex plane and Θ is holomorphic.

The first property implies that deg(Θ, x) = deg(Θ, y) for all x, y ∈X with Θ(x) = Θ(y), i.e., the degree of Θ is constant in each fiber.Note that Θ does not cover the punctures of O. If we identify X withthe orbifold O′ = (X,αX), where αX ≡ 1, the map Θ: X → O is anorbifold covering map.

Definition A.18. For a hyperbolic or parabolic orbifold O =(S, α) the map Θ: X → O from Theorem A.17 is called the universalorbifold covering map of O.

An orientation preserving homeomorphism g : X → X is called adeck transformation of the universal orbifold cover if

Θ g = Θ.

The group of all such deck transformations is called the fundamentalgroup π1(O) of the orbifold O. Is is possible to define this fundamentalgroup abstractly and construct the universal orbifold cover from this.Note that the above implies that in the case when Θ is holomorphic,any g ∈ π1(O) is holomorphic as well.

The universal orbifold cover satisfies the following universal prop-erty.

Theorem A.19. Let O = (S, α) and O′ = (S ′, α′) be orbifolds andΘ: X → O be the universal orbifold covering map and Φ: O′ → Obe an orbifold covering. Furthermore let x ∈ X, s ∈ S and s′ ∈ S ′

be points such that Θ(x) = Φ(s′) = s and α(s) = 1. Then there is a

unique orbifold covering map Θ : X → O′ such that f(x) = s′ and

Θ = Φ Θ.

So the following diagram commutes:

XΘ

Θ

O′

Φ !!

O.The theorem above is [Th, Proposition 13.2.4], however the proof fol-lows directly from Lemma A.4.


Corollary A.20 (Uniqueness of the universal orbifold cover). LetO = (S, α) be an orbifold that is parabolic or hyperbolic and Θ: X → Oas well as Θ′ : X ′ → O be universal orbifold coverings. Let x ∈ Xand x′ ∈ X ′ such that Θ(x) = Θ′(x) and deg(Θ, x) = deg(Θ′, x) =α(Θ(x)) = 1. Then there is a unique homeomorphism A : X ′ → Xwith A(x′) = x and Θ A = Θ′.

If Θ and Θ′ are holomorphic is follows that A is biholomorphic.

Proof. Apply Theorem A.19 twice to Θ: X → O and Θ′ : X ′ →O. Here we first view Θ as the universal covering map and Θ′ asa covering map, then Θ′ as the universal covering map and Θ as acovering map. This yields maps A : X ′ → X and B : X → X ′ withA(x′) = x, B(x) = x′ and

Θ A = Θ′

Θ = Θ′ B

Thus Θ = Θ A B as well as Θ′ = Θ′ B A. From the uniquenessstatement in Theorem A.19 it follows that A B = idX as well asB A = idX′ , thus A is a homeomorphism.

It is immediate that A is holomorphic in the case when Θ and Θ′

are.

In the following we collect properties of the group of deck transfor-mations. We will mainly consider the case when the underlying surface

is the Riemann sphere C, which is the case most relevant for us.

Proposition A.21 (Deck transformations of the universal orbifoldcover). Let O = (S, α) be an orbifold that is hyperbolic or parabolic, andΘ: X → O be the universal orbifold covering map. The group of decktranformations G = π1(O) corresponding to the universal orbifold coversatisfies the following.

(i) The map Θ is induced by the group G, meaning that for allx, y ∈ X:

Θ(x) = Θ(y) if and only if g(x) = y for an g ∈ G.

Thus G acts transitively on fibers of Θ.

(ii) For any x ∈ X it holds

deg(Θ, x) = #Gx.

Assume now further that O = (C, α), then Θ is holomorphic. Thenthe universal orbifold cover is X = D, i.e., the hyperbolic plane whenO is hyperbolic; or X = C is the Euclidean plane when O is parabolic.

490 A

In the first case X = D is equipped with the hyperbolic metric, in thesecond case X = C is equipped with the Euclidean metric. Then

(iii) G acts by isometries and properly discontinuously on X.

If additionally α(p) 6= ∞ for all p ∈ C (O has no punctures) itholds

(iv) G acts cocompactly on X.

Thus G acts geometrically on X in this case.

It follows that in the holomorphic case, every g ∈ Gx is an isometryof D or C with a fixed point, hence a rotation (for any x ∈ X).

A.9. Orbifold lifts of Thurston maps

We now consider the orbifoldO = (S2, αf ) of a Thurston maps f : S2 →S2. We can lift f (or rather branches of its inverse) to the universalorbifold cover Of .

Lemma A.22 (Existence of lifts to the universal orbifold cover).Let f : S2 → S2 be a Thurston map and O = (S2, αf ) be the associatedorbifold. Let Θ: X → Of be the universal canonical orbifold coveringmap. Let z0, w0 ∈ X be points such that (f Θ)(w0) = Θ(z0). Then

(i) there exists a branched covering map A : X → X with A(z0) =w0 and

f Θ A = Θ.

(ii) If z0 /∈ crit(Θ) the map A is unique.

(iii) If f : C → C (as well as Θ) is holomorphic, the map A isholomorphic as well.

So in this setting, we have the following commutative diagram:

(A.17) w0 ∈ XΘ

z0 ∈ XAoo

Θ

S2 f// S2.

One should think of A as a lift of a suitable inverse branch of f−1 underthe branched covering map Θ.

Proof. We apply the lifting criterion giving by Lemma A.4 (ii)for the branched covering maps f Θ: X → S2 and Θ: X → S2

(corresponding to π and f in the lemma, respectively). To see thatthe assumptions of this lemma are satisfied, let z, w ∈ X with Θ(z) =(f Θ)(w) be arbitrary. Define p = Θ(z) and q = Θ(w). Then p =f(q), and so deg(f, q)αf (q) divides αf (p) (see Lemma 2.8 (i)). Since

A.9. ORBIFOLD LIFTS OF THURSTON MAPS 491

deg(Θ, w) = αf (q) and deg(Θ, z) = αf (p) by definition of the universalcovering map Θ, this implies that

deg(f, q)αf (q) = deg(f, q) deg(Θ, w) = deg(f Θ, w)

divides deg(Θ, z) = αf (p). So by Lemma A.4 (ii) there exists a branchedcovering map A : X → X with A(z0) = w0 and f ΘA = Θ, finishing(i).

(ii) Assume now that z0 /∈ crit(Θ). Recall that the degree of theuniversal orbifold covering map Θ is constant in each fiber. Thus p0 =Θ(z0) is not a critical value of Θ.

Since f ΘA = Θ it follows that p0 is not a critical value of f Θ.Note that for w0 = A(z0) we have f Θ(w0) = Θ(z0) = p0. Thus theuniqueness of A follows from Lemma A.4 (i).

When f , as well as Θ, are holomorphic is follows immediately thatA is holomorphic.

A map A : X → X satisfying f Θ A = Θ as in the previouslemma is called a lift of a branch of the inverse of f to the orbifoldcover, or a lift of f−1 by Θ for short.

Consider now y ∈ S2 \ post(f) and x ∈ f−1(y). Let z ∈ Θ−1(y)and w,w′ ∈ Θ−1(x) be arbitrary. Since y is not a postcritical points,i.e., αf (y) = 1, it follows that the points z /∈ crit(Θ). By the previouslemma there are unique maps A : X → X and A′ : X → X with f Θ A = f Θ A′ = Θ and A(z) = w as well as A′(z) = w′. ByCorollary A.20 there is a deck transformations g ∈ G = π(Of ), i.e.,maps g : X → X and h : X → X, of Θ such that g(w) = w′. In thiscase we have

(A.18) g A = A′.

Indeed A = g A is a map A : X → X with A(z′) = w′. Furthermore

f Θ A = f Θ g A = f Θ A = Θ.

The uniqueness in Lemma A.22 (ii) shows (A.18). Two lifts A,A′ as in(A.18) are called equivalent lifts of f−1. Clearly equivalent lifts map agiven point z to points in the same G-orbit.

It should be pointed out that for a given lift A of f−1 by Θ anda deck transformation g ∈ G = π(Of ) the maps A and A g are notequivalent in general. Put differently, points z, z′ ∈ X that are in thesame G-orbit will not be mapped to points in the same G-orbit by Ain general.

Lemma A.23 (d lifts of f−1). Let f : S2 → S2 be a Thurston mapwith d = deg(f). Then

492 A

(i) there are d pairwise non-equivalent lifts A1, . . . , Ad of f−1 byΘ;

(ii) these lifts satisfy that the following. For any z ∈ X let y =Θ(y) ∈ S2, then

f−1(y) = Θ(A1(z)), . . . ,Θ(Ad(z));(iii) for any g ∈ G the maps A1 g, . . . , Ad g are non-equivalent

lifts of f−1 by Θ. In particular

(A.19) Θ(A1 g(z)), . . . ,Θ(Ad g(z)) = Θ(A1(z)), . . . ,Θ(Ad(z)),for all z ∈ X.

Note that the expression in (ii) is not true for points y ∈ S2 withαf (y) =∞, since in this case y is not in the image of Θ.

Proof. (i) Let y ∈ S2 \post(f) be arbitrary. Let x1, . . . , xd be thedistinct points in f−1(y). Let z ∈ Θ−1(y) and wj ∈ Θ−1(xj) for j =1, . . . , d be arbitrary. Let Aj be the lift of f−1 by Θ with Aj(z) = wj.Note that the points w1, . . . , wd are mapped by Θ to distinct points,meaning they are in distinct G-orbits. Thus the lifts A1, . . . , Ad are notequivalent, since they map the point z to points in distinct G-orbits.

On the other hand, let A : X → X be a lift of f−1 by Θ. Sincef Θ A(z) = Θ(z) = y it follows that Θ A(z) ∈ f−1(y), i.e.,ΘA(z) = xj for some j = 1, . . . , d. Then A(z) ∈ Θ−1(xj). Thus thereis a g ∈ G with g(A(z)) = wj. As in the discussion preceeding thislemma this implies that g A = Aj.

(ii) Consider z ∈ X and let y = Θ(z). Then any lift A of f−1

satisfies f Θ A(z) = Θ(z) = y. Thus Θ(A(z)) ∈ f−1(y). There-fore Θ(A1(z), . . . ,Θ(Ad(z)) ⊂ f−1(y). Conversely let x ∈ f−1(y) bearbitrary. Note that αf (y) < ∞, since it is in the image of Θ. Sinceαf (x) divides αf (y) it follows that αf (x) <∞. Thus x is in the imageof Θ. Let w ∈ Θ−1(x). By Lemma A.22 (i) there is a lift A of f−1

by Θ such that A(z) = w. We have already shown that A is equiva-lent to one of the lifts Aj. Thus x = Θ(A(z)) = Θ(Aj(z)). Thereforef−1(y) ⊂ Θ(A1(z)), . . . ,Θ(Ad(z)), finishing the proof.

(iii) Let A1, . . . , Ad be non-equivalent lifts of f−1 by Θ as in (i) andg ∈ G be a deck transformation. Clearly f Θ Aj g = Θ g = Θ,thus Aj g is a lift of f−1 by Θ for j = 1, . . . , d.

Let z ∈ X \ crit(Θ). Then w = g(z) /∈ crit(Θ). It follows thatthe points A1(w), . . . , Ad(w) (i.e., the points A1 g(z), . . . , Ad g(z))are contained in pairwise distinct orbits of G. Thus the maps A1 g, . . . , Ad g are pairwise not equivalent.

Since Θ(g(z)) = Θ(z) = y, (A.19) follows from (ii).

A.10. THE CANONICAL ORBIFOLD METRIC 493

We have seen that a lift A : X → X of f−1 by Θ is not G-equivariantin the sense that it does not necessarily map points in the same G-orbitto points in the same G-orbit. However, the following is true. Letz, z′, w, w′ ∈ X be points with A(z) = w and A(z′) = w′ then

(A.20) wG∼ w′ ⇒ z

G∼ z′.

HereG∼ denotes the equivalence relation of being contained in the same

G-orbit. Indeed wG∼ w′ implies

Θ(z) = f Θ A(z) = f Θ(w) = f Θ(w′) = f Θ A(z′) = Θ(z′),

as desired.

A.10. The canonical orbifold metric

Assume an orbifold O and its universal orbifold cover Θ: X →O is given. Using Θ to push the metric on X to O results in thecanonical orbifold metric. Note however, that such a metric is onlywell defined if X is equipped with a natural metric and the group ofdeck transformations act by isometries.

We restrict ourselves to the case where the unlying surface is the

Riemann sphere, i.e., to orbifolds of the form O = (C, α). Furthermorewe assume that O is parabolic our hyperbolic. Then the universalorbifold cover is X = D in the hyperbolic case, or X = C in theparabolic case. In the first case X = D is equipped with the hyperbolicmetric d0 from (A.3), in the second case X = C is equipped with theEuclidean metric d0. We will remove the punctures from the sphere,meaning we will consider

C0 := C \ p ∈ C : α(p) =∞.

The universal orbifold covering map then is a holomorphic map Θ: X →C0. The associated group of deck transformations G = π1(O) acts ge-ometrically on X.

The canonical orbifold metric ω on O is defined by

(A.21) ω(p, q) := infd0(z, w) : z ∈ Θ−1(p), w ∈ Θ−1(q)

for p, q ∈ C0. Since G acts geometrically on (X, d0) and transitivelyon Θ−1(p), for each z0 ∈ Θ−1(p) there exists w0 ∈ Θ−1(q) such thatω(p, q) = d0(z0, w0). In particular, the infimum in (A.21) is attained asa minimum. This implies that ω(p, q) > 0 if p 6= q, and it follows that ω

is actually a metric on C0 (the other properties of a metric immediatelyfollow from the definition of ω).

494 A

From the uniqueness of the universal orbifold covering map Θ (Corol-lary A.20), it follows that the metric ω is uniquely determined if O ishyperbolic and unique up to a scaling factor if O is parabolic. There-fore, in the following we will refer to ω as the canonical orbifold metricon O.

An equivalent way to define this metric is as follows. Recall that

X/G is homeomorphic to C0 = Θ(X), the homeomorphism is given by

ϕ : X/G→ C0, ϕ : [x] 7→ Θ(x),

see Corollary A.7. Define on X/G

ω([x], [y]) := infd0(g(x), h(y)) : g, h ∈ G,

for all [x], [y] ∈ X/G. Clearly the map ϕ : (X, ω)→ (C0, ω) then is anisometry.

Fix an x ∈ X and let p = Θ(x) ∈ C. Let Bx ⊂ X be a ball aroundx (with respect to d0). Note that every (rotation) g ∈ Gx fixes Bx.Then Bx/Gx is a cone with cone angle 2π/#Gx = 2π/α(p). Here themetric on Bx/Gx is defined analog as above. The cone is hyperbolic orEuclidean depending on whether the orbifold is hyperbolic or parabolic.

From the above it follows that there is a neighborhood Up ⊂ C of p that

is isometric with respect to ω to such a cone. Thus points p ∈ C where2 ≤ α(p) < ∞ are precisely the conical singularities of the orbifold

O = (C, α), Near all other points, (C0, ω) is locally isometric to themodel space (X, d0).

We have ω(Θ(z),Θ(w)) ≤ d0(z, w) for all z, w ∈ D. It is easy to seethat a stronger condition is true locally: for each z ∈ X there exists aneighborhood Nz of z in X such that

(A.22) ω(Θ(z),Θ(w)) = d0(z, w)

for all w ∈ Nz. This together with a simple covering argument impliesthat the map Θ is a path isometry: if γ is a path in X, then

(A.23) lengthω(Θ γ) = lengthd0(γ).

The metric space (C0, ω) is geodesic. Indeed, a geodesic joining two

points p, q ∈ C0 can be obtained as follows. We can pick points z ∈Θ−1(p) and w ∈ Θ−1(q) such that ω(Θ(z),Θ(w)) = d0(z, w). Since(X, d0) is geodesic, we can find a geodesic segment joining z and w.Since Θ is a path isometry, the path Θ γ must then be a geodesic

segment in (C0, ω) joining p and q.Let us compare the canonical orbifold metric ω with the chordal

metric σ. Fix a point p ∈ C0. Since deg(Θ, x) = α(p) for any x ∈


Θ−1(p) it follows that there is a neighborhood Up ⊂ C0 of p such that

(A.24) ω(p, q) σ(p, q)1/α(p),

for all q ∈ Up. Here C() = C(p).

Consider now a puncture p ∈ C of O (i.e., α(p) = ∞). Let

γ : [0, 1) → C0 be a curve that is rectifiable with respect to σ andends at p. Any lift γ by Θ leaves any compact set of X, thus hasinfinite length. From (A.23) it follows that γ has infinite length withrespect to ω.

Note that (A.24) shows that when O has no punctures, ω induces

the standard topology on C.

Lemma A.24. Let O = (C, α) be an orbifold without punctures.Then

(i) id : (C, σ)→ (C, ω) is quasisymmetric;

(ii) (C, σ) and (C, ω) are bi-Lipschitz equivalent.

Let us remark that (A.24) shows that id : (C, σ) → (C, ω) is neverbi-Lipschitz (the inverse is Lipschitz though). Furthermore when O has

punctures, (C0, σ) and (C0, ω) are never quasisymmetrically equivalent,since a quasisymmetry maps a bounded set to a bounded set.

Proof. We first compare σ and ω locally in a more precise fashion.Consider points z1 = r1e

iθ1 , z2 = r2eiθ2 ∈ C, where r1, r2 > 0, |θ2−θ1| ≤

π. It is easy to verify that the Euclidean metric satisfies

(A.25) |r2eiθ2 − r1e

iθ1| |r2 − r1|+ r1|θ2 − θ1|.

Consider a point p ∈ C, for convenience we assume p = 0, so thatthe chordal metric is bi-Lipschitz equivalent to the Euclidean metric insome neighborhood Up. Let α = α(p), then it holds

ω(z1, z2) |r1/α2 eiθ2/α − r1/α

1 eiθ1/α|(A.26)

|r1/α2 − r1/α

1 |+ r1/α1 |θ2 − θ1|/α

This implies that in Up it holds

σ(z1, z2) . ω(z1, z2) . σ(z1, z2)1/α.

A compactness argument implies that there is a constant C ≥ 1 suchthat

(A.27)1

Cσ(z, w) ≤ ω(z, w) ≤ Cσ(z, w)ε0

for all z, w ∈ C. Here ε0 := 1/maxα(p) : p ∈ C.

496 A

(i) Recall that we need to show that there is a homeomorphismη : [0,∞)→ [0,∞) such that

σ(u, v)

σ(u,w)≤ t⇒ ω(u, v)

ω(u,w)≤ η(t)

for all u, v, w ∈ C with u 6= w and t > 0.Fix a constant δ > 0, which will be determined later. If σ(u,w) ≥ δ

or if σ(u, v) ≥ δ it follows immediately from (A.27) that this implicationis satisfied.

It remains to consider the case when σ(u, v) ≤ δ and σ(u,w) ≤ δ.One of the prototypical examples of a quasisymmetric map in the planeis the radial stretch which is the map R1/α : C→ C given by

R1/α : reiθ 7→ r1/αeiθ,

where α > 0. The map id: (C, σ) → (C, ω) behaves locally as the

radial stretch, up to bi-Lipschitz equivalence. Consider a point p ∈ C.As before we assume for convenience that p = 0. Using (A.26) and(A.25) it holds for z1 = r1e

iθ1 and z2 = r2eiθ2 in a neighborhood Up of

p (where |θ2 − θ1| ≤ π)

ω(r1eiθ1 , r2e

iθ2) |r1/α2 − r1/α

1 |+ r1/α1 |θ2 − θ1|/α

|r1/α2 − r1/α

1 |+ r1/α1 |θ2 − θ1|

|R1/α(r1eiθ1)−R1/α(r2e

iθ2)|.

Here the constants C() only depend on the neighborhood Up. Since

the radial stretch is quasisymmetric it follows that id : (C, σ)→ (C, ω)is quasisymmetric in Up.

Cover C with finitely many sets Up in which id : (C, σ)→ (C, ω) isquasisymmetric by the above. Let δ > 0 be the Lebesgue number ofthis covering, i.e., every ball Bσ(q, δ) is contained in one of the sets Up.

Combined with the argument above, quasisymmetry on C follows.

(ii) Recall that every point p in (C, ω) has a neighborhood thatis isometric to a (hyperbolic or parabolic) cone. Each such cone is bi-Lipschitz to a Euclidean disk. Using compactness it is possible to show,

that this implies that (C, ω) is bi-Lipschitz to (C, σ). Alternatively,

one may define a map ϕ : C → C that is the radial stretch Rα in theneighborhood of each point p with α = α(p) ≥ 2, and the identity

elsewhere. From the above it follows that ϕ : (C, ω) → (C, σ) is bi-Lipschitz, but we do not work out the details here.


We can now prove the important expansion property of the canon-ical orbifold metric.

Proposition A.25. Let f : C → C be a rational Thurston mapwithout periodic critical points, and let ω be the canonical orbifold met-ric of f . Then there exists a constant ρ > 1 such that

(A.28) lengthω(f γ) ≥ ρ lengthω(γ)

for all paths γ in C.

This lemma is essentially well-known; see [Mi, Thm. 19.6] or [CG,V.4.3.1], for example. As we will see, if f has a hyperbolic orbifold,then the main idea for the proof is to lift inverse branches of f−1 to D,and use the fact that holomorphic maps of D into itself are contractingin the hyperbolic metric.

Proof. Let α = αf be the ramification function of f . If Of =

(C, α) is parabolic, then f is a Lattes map (see Theorem 3.1) and thestatement follows from Proposition 8.13. So we may assume that Ofis hyperbolic. Then its universal orbifold cover is the hyperbolic plane

represented by the open unit disk D. Let Θ: D → C be the universalorbifold covering map and ω be the canonical orbifold metric of Of .As we know, it is obtained by pushing the hyperbolic metric d0 on Dforward by Θ.

We will show that ω has the property that there is a constant ρ > 1such that

(A.29) ‖f ′(q)‖ω := lim infq′→q

ω(f(q′), f(q))

ω(q′, q)≥ ρ

for all q ∈ C. This expression should be viewed as the derivative of fwith respect to the metric ω. To show this inequality, it is enough to

show that for each point q0 ∈ C there exists an open neighborhood Nof q0 and a constant ρ′ > 1 such that ‖f ′(q)‖ > ρ′ for all q ∈ N . Then

the estimate (A.29) follows by covering C with finitely many such setsN .

So let q0 ∈ C be arbitrary, and set p0 := f(q0). We can find pointsz0, w0 ∈ D with Θ(w0) = q0 and Θ(z0) = p0. Then by Lemma A.22 wecan find a holomorphic map A : D→ D such that A(z0) = w0 and

(A.30) f Θ A = Θ.

498 A

So in this setting, we have the following commutative diagram:

(A.31) w0 ∈ D

Θ

z0 ∈ DAoo

Θ

Cf

// C.

One should think of A as a lift of a suitable inverse branch of f−1

under the branched covering map Θ. By the Schwarz-Pick lemma thederivative of A with respect to the hyperbolic metric d0 satisfies

‖A′(z)‖ :=(1− |z|2)|A′(z)|

1− |A(z)|2= lim

z′→z

d0(A(z′), A(z))

d0(z′, z)≤ 1

for all z ∈ D.Let us show that in fact ‖A′(z)‖ < 1 for all z ∈ D. Assume not, then

A is an automorphism of D. Let u ∈ D be arbitrary and v = f(u). Letz ∈ Θ−1(v) and w = A(z). Since Θ is the universal orbifold coveringmap we obtain from (A.30)

α(v) = deg(Θ, z) = deg(f Θ A, z) = deg(f, u) deg(Θ, w) deg(z)︸︷︷︸=1

= deg(f, u)α(u).

This implies that Of is parabolic by Proposition 2.13 (iii), which is acontradiction.

So in particular, ‖A′(z0)‖ < 1. Since the map z 7→ ‖A′(z)‖ iscontinuous, we can find a neighborhood U of z0 and a constant k < 1such that ‖A′(z)‖ ≤ k < 1 for all z ∈ U . The set N := Θ(A(U)) isneighborhood of Θ(A(z0)) = q0. If q ∈ N is arbitrary, then we can picka point w ∈ U with Θ(A(w)) = q. Moreover, if qn is any sequencecontained in N \ q with qn → q as n → ∞, then there exists asequence wn in U with Θ(A(wn)) = qn for all n ∈ N and wn → w asn→∞. Hence

lim infn→∞

ω(f(qn), f(q))

ω(qn, q)= lim inf

n→∞

ω((f Θ A)(wn), (f Θ A)(w))

ω(Θ(A(wn),Θ(A(w)))

= lim infn→∞

ω(Θ(wn),Θ(w))

ω(Θ(A(wn),Θ(A(w))

= lim infn→∞

d0(wn, w)

d0(A(wn), A(w))

=1

‖A′(w)‖≥ 1/k > 1.

A.11. THE CANONICAL ORBIFOLD MEASURE 499

In the third equality we used the fact that the map Θ is locally a“radial isometry” (see A.22). We conclude that ‖f ′(q)‖ ≥ ρ′ := 1/k forall q belonging to the neighborhood N of q0. Since q0 was arbitrary,inequality (A.29) follows.

Now let γ be a path in C. Inequality (A.29) in combination witha covering argument implies that lengthω(f γ) ≥ ρ lengthω(γ). So fexpands the lengths of paths with respect to the metric ω by the factorρ > 1.

Note that in the case when f has periodic critical points, an esti-mate as (A.28) holds in a neighborhood of the Julia set, by the exactsame argument as above.

A.11. The canonical orbifold measure

Given an orbifold with a universal covering map, we may pushthe measure on the universal orbifold cover to the orbifold to obtainthe canonical orbifold measure. We restrict ourselves to the situationwe will be needing. More precisely we will be interested in orbifoldsthat arise from rational expanding Thurston maps. These have nopunctures, the group of deck transformationsG then acts geometrically,see Proposition A.21.

Definition A.26. Let G be a group of orientation-preserving isom-etries acting on X = C or X = D properly discontinuously and cocom-pactly. A fundamental domain of G is a connected open set FG ⊂ Xsatisfying the following.

(i) FG contains at most one point from each G-orbit, meaningthat if x ∈ FG and y = g(x) 6= x for a g ∈ G it follows thaty /∈ FG.

(ii) ⋃g∈G

g(FG) = X

(iii)area(∂FG) = 0.

(iv) FG is compact.

As mentioned above, our definition of a fundamental domain ismore restrictive than usual. We refer the reader to [Bea, Chapter 9]as well as [Ra, § 6.6] for more general settings.

Theorem A.27. For any group G of orientation-preserving isonetriesof X = C or X = D acting properly discontinuously and cocompactlythere exists a fundamental domain FG as above.

500 A

Outline of proof. The standard fundamental domain is the Dirich-let domain given as follows. Fix a point z ∈ X, such that its stabilizerGz is trivial. In the case when G is the group of deck transformations

of the universal orbifold covering map Θ: X → C this means thatz /∈ crit(Θ). For any g ∈ G \ idX let

Hg(z0) := x ∈ X : d(x, z) < d(x, g(z).

This is a half-plane in X (bounded by the line x ∈ X : d(x, z) =d(x, g(z))). The Dirichlet domain is

(A.32) D(z) :=⋂

g∈G\idX

Hg(z).

It is well-known that D(z) is in fact a fundamental domain in the senseof Definition A.26. A proof can be found in [Ra, Theorem 6.6.13 andTheorem 6.6.9], see also [Bea, Theorem 9.4.2].

The (only) case that will be relevant for us, is when G is the group of

deck transformations of the universal orbifold covering map Θ: X → Cof the orbifold associated to a rational expanding Thurston map f .Then, we can construct a fundamental domain by elementary means.

Let C ⊂ C be a Jordan curve with post(f) ⊂ C such that area(C) = 0

(here area denotes the Lebesgue measure on the sphere C). We consider

D = D0(f, C) the cell decomposition of C given by C, see Section 5.3.This contains two 0-tiles X0

w and X0b , which are colored white and

black respectively. Exactly as in Lemma 5.15, we can define a cell

decomposition D of X such that Θ is cellular. A tile Y in D is coloredwhite if Θ(Y ) = X0

w and black if Θ(Y ) = X0b . Exactly as in Lemma 5.19

every edge e ∈ D is contained in a white tile Yw ∈ D and a black tile

Yb ∈ D. Then

FG := int(Yw) ∪ int(Yb) ∪ int(e)

is a fundamental domain of G. Indeed Θ is injective on FG, which showsthat property (i) of Definition A.26 is satisfied. Furthermore (FG) =

Θ(Yw∪Yb) = C, which shows property (ii) of Definition A.26. Note thatΘ: Yw → X0

w as well as Θ: Yb → X0b are homeomorphisms. It follows

that FG = Yw ∪ Yb is compact. Thus property (iv) of Definition A.26is satisfied. Finally since Θ is holomorphic, it is absolutely continuous.Therefore areaX(∂Yw) = areaX(∂Yb) = areaC(C) = 0. Since ∂FG ⊂(∂Yw ∪ ∂Yb) property (iii) of Definition A.26 follows.

We now fix a fundamental domain FG for the group of deck trans-

formations G of Θ. The canonical orbifold measure Ω (of Of ) on C is

A.11. THE CANONICAL ORBIFOLD MEASURE 501

defined as

(A.33) Ω(V ) := area(Θ−1(V ) ∩ FG).

for any Borel set V ⊂ C. Here “area” denotes hyperbolic area in thecase when Of is hyperbolic, or the Euclidean area in the case whenOf is parabolic. Put differently, Ω is the push forward of (hyper-

bolic or Euclidean) area by the bijective map (Θ|FG) : FG → C. Sincearea(∂FG) = 0, we note that

Ω(V ) = area(Θ−1(V ) ∩ FG).

Thus 0 < Ω(C) <∞.

Lemma A.28 (Canonical orbifold measure). The canonical orbifoldmeasure Ω is well-defined. Furthermore

(i) if V ⊂ C is a connected open set on which a branch Ψ of Θ−1

is defined. Then

Ω(V ) = area(Ψ(V )).

(ii) Let dA denote Lebesgue measure on C ⊂ C, and κ : CP →(0,∞) be the density of Ω with respect to Lebesgue measure,

then the density is given as follow. If w = Θ(z) ∈ CP then

κ(w) = ‖Θ′(z)‖−2.

Here ‖Θ′(z)‖ = |Θ′(z)| in the case when X = C and ‖Θ′(z)‖ =

|Θ′(z)|1−|z|2

2in the case when X = D. In particular Ω and Lebesgue

measure on C are mutually absolutely continuous.

Proof. To show that the canonical orbifold measure is well-defined,we need to show that it is independent of the particular fundamentaldomain chosen. We repeat the argument from [Bea, Theorem 9.1.3]for the convenience of the reader.

To this end let F1 and F2 be two fundamental domains of G and Ω1

and Ω2 be the canonical orbifold measures defined in terms of F1 and

502 A

F2. Then for any Borel set V ⊂ CΩ1(V ) = area(Θ−1(V ) ∩ F1)

= area(Θ−1(V ) ∩ F1 ∩

⋃g∈G

g(F 2)︸︷︷︸=X

)

=∑g∈G

area(Θ−1(V ) ∩ F1 ∩ g(F 2))

=∑g∈G

area(Θ−1(V ) ∩ F 1 ∩ g(F2))

here we are the fact that area(F ) = area(F ) = g(F ) = g(F ) as well asg(Θ−1(V )) = Θ−1(V ), for any fundamental domain and g ∈ G,

=∑g∈G

area(Θ−1(V ) ∩ g−1(F 1) ∩ F2)

= area(Θ−1(V ) ∩ F2 ∩

⋃g∈G

g(F 1))

= area(Θ−1(V ) ∩ F2) = Ω2(V ).

(ii) Let w0 ∈ C be an arbitrary point that is not a critical valueof Θ. Pick a z0 ∈ Θ−1(w0). Note that the stabilizer Gz0 is trivial, seeProposition A.21 (ii). Let B = B(z0, δ) ⊂ X be a sufficiently smallball such that B ∩ g(B) = ∅ for all g ∈ G \ idX. Using the Dirichletdomain D(z0) =: FG (see (A.32)) as fundamental domain, we see that

B ⊂ FG. Let B′ := Θ(B) ⊂ C. Then Θ: B → B′ is a homemorphism,the inverse of which we denote by Ψ. Then

Ω(B′) = area(B) =

∫B=Ψ(B′)

dA =

∫B′‖Ψ′(w)‖2dA(w).

For Ψ(w) = z it follows that ‖Ψ′(w)‖ = ‖Θ′(z)‖−1. This shows (ii). Inturn, this implies property (i) immediately.

A.12. Planar crystallographic groups

Here we collect some facts about planar crystallographic or wallpapergroups. All of this material is fairly standard, and we mostly omitthe proofs. For general background see [Ra, Section 7.4]. Recall (seethe introduction of Chapter 3) that we call a group G of orientation-preserving Euclidean isometries on R2 a (planar) crystallographic groupif it acts properly discontinously and cocompactly on R2. We will

A.12. PLANAR CRYSTALLOGRAPHIC GROUPS 503

briefly outline how one can associate a canonical orbifold with a crys-tallographic group, and how this leads to a classification of such groups.

In the following G is a fixed planar crystallographic group. Theaction of G on R2 induces an equivalence relation ∼ for points in R2:for u, v ∈ R2 we have u ∼ v if there exists g ∈ G such that v = g(u).We denote the corresponding quotient space by R2/G, and by Θ: R2 →R2/G the quotient map.

The stabilizer Gu of a point u ∈ R2 (consisting of all g ∈ G suchthat g(u) = u) is a finite subgroup of G, because G acts properlydiscontinuously. Since G consists of orientation-preserving isometries,Gu is a finite cyclic group. Moreover, the set u ∈ R2 : #Gu ≥ 2cannot have limit points in R2, and so it is a closed and discrete subsetof R2.

The Euclidean metric d0 on R2 descends to a metric d on the quo-tient Z := R2/G as in (A.10). The map Θ preserves small “radial”distances similarly as in (A.22). This implies that Θ: (R2, d0)→ (Z, d)is a path isometry. Locally near a point p ∈ Z, the space Z = R2/Gequipped with the metric d is isometric to a Euclidean cone of coneangle 2π/n, where n = #Gu, u ∈ Θ−1(p). Note that if u, v ∈ Θ−1(p),then the groups Gu and Gv are conjugate in G and hence have thesame cardinality.

It follows that every point in Z has a neighborhood homeomorphicto an open disk, and so Z is a surface. It is connected as the continu-ous image of R2. Moreover, since G consists of orientation-preservingisometries and acts cocompactly, this surface is compact and orientable.The quotient map Θ: R2 → Z = R2/G is a branched covering map,and we have degΘ(u) = #Gu for all u ∈ R2. If p ∈ Z and u, v ∈ Θ−1(p),then

(A.34) deg(Θ, u) = #Gu = #Gv = deg(Θ, v),

and so the local degree of Θ is constant in each fiber Θ−1(p).It follows from the uniqueness statement for lifts under branched

covering maps (see Lemma A.4 (i)) that every homeomorphism ϕ : R2 →R2 satisfying Θ = Θ ϕ belongs to G. So G is the group of deck trans-formations of Θ. The branched covering map Θ is induced by G (inthe sense discussed after Theorem 3.1); namely, for u, v ∈ R2 we haveΘ(u) = Θ(v) if and only if there exists g ∈ G such that g(u) = v.

If we set α(p) := deg(Θ, u), where u ∈ Θ−1(p), then α : Z → N iswell-defined by (A.34). Since the every compact subset of R2 can onlycontain finitely many points u ∈ G where #Gu ≥ 2, there are at mostfinitely many points p ∈ Z such that α(p) ≥ 2. So α is a ramificationfunction on Z, and (Z, α) is an orbifold. The conformal structure

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on C ∼= R2 descends to a natural conformal structure on Z so thatΘ: R2 ∼= C→ Z becomes a holomorphic map. The map Θ: C→ (Z, α)is then the universal orbifold covering map (see Section A.7).

For every crystallographic group G the subgroup Gtr consisting ofall translations in G is a normal subgroup of finite index (see [Ra,Thm. 7.4.2 and the subsequent discussion]). This group Gtr is isomor-phic to a rank-2 lattice in R2, and so the quotient space T := R2/Gtr isa (2-dimensional) torus. The quotient map R2 → R2/Gtr is a coveringmap π : R2 → T .

Since Gtr ⊂ G, there is a natural quotient map Θ: T = R2/Gtr →Z = R2/G such that Θ = Θ π. Then we have the following commu-tative diagram:

(A.35) R2 π//

Θ

''

T = R2/GtrΘ// Z = R2/G.

Since π is a covering map, and Θ is a branched covering map, Θ is abranched covering map as well.

Proposition A.29. Let G be a planar crystallographic group (con-sisting of orientation-preserving isometries). Then the quotient spaceZ = R2/G is a 2-dimensional torus or 2-dimensional sphere. Thesecases occur depending on whether G consists only of translations ornot.

Proof. As before, let Gtr be the subgroup of G consisting of alltranslations in G, and T = R2/Gtr and Z = R2/G. As we discussed,the quotient T is a torus, and Z is a compact, connected, and orientablesurface. If G consists only of translations, then G = Gtr and Z = T ,and so Z is a torus.

Suppose that G 6= Gtr, and consider the maps in (A.35). Everyorientation-preserving isometry on R2 different from a translation hasa unique fixed point. Since G contains such elements by assumption,it follows that there exists u ∈ R2 with #Gu ≥ 2. Then u := π(u) ∈ Tis a critical point of Θ, because deg(π, u) = 1 and so

(A.36) 2 ≤ #Gu = deg(Θ, u) = deg(Θ, u) · deg(π, u) = deg(Θ, u).

If χ(Z) is the Euler characteristic Z, then the Riemann-Hurwitz for-mula (2.3) implies that

deg(Θ)χ(Z) =∑

c∈crit(Θ)

(deg(Θ, c)− 1) > 0,


and so χ(Z) > 0. On the other hand, χ(Z) = 2− 2g, where g ∈ N0 isthe genus of Z. Hence g = 0 and Z is a 2-sphere.

The argument in the previous proof actually allows us to concludethat the orbifold (Z, α) is parabolic in case Z = S2 is a 2-sphere. To

see this, define d := deg(Θ) ∈ N. If p ∈ Z and q ∈ Θ−1

(p), we can picku ∈ R2 with π(u) = q. Then Θ(u) = (Θ π)(u) = Θ(q) = p, and asimilar computation as in (A.36) shows that

deg(Θ, q) = deg(Θ, u) = #Gu = α(p).

So the local degree of Θ is constant equal to α(p) in the fiber Θ−1

(p).It follows that

d =∑

q∈Θ−1

(p)

deg(Θ, q) = α(p) ·#Θ−1

(p).

Since Z is a 2-sphere and so χ(Z) = 2, the Riemann-Hurwitz formulashows that

2d =∑

c∈crit(Θ)

(deg(Θ, c)− 1) =∑p∈S2

∑q∈Θ

−1(p)

(deg(Θ, q)− 1)

=∑p∈S2

(d−#Θ−1

(p)) = d∑p∈S2

(1− 1

α(p)

).

If we divide both sides by d, then we see that the orbifold (Z, α) isindeed parabolic.

Since α(p) 6=∞ for p ∈ Z, the orbifold (Z, α) can only have one ofthe following signatures: (2, 2, 2, 2), (2, 4, 4), (3, 3, 3), (2, 4, 6). Thesesignatures correspond to the signatures of Thurston maps with par-abolic orbifolds and no periodic critical points (see Propositions 2.13and 2.9).

If G and G′ are two planar crystallographic groups, then we saythat they are conjugate (by an orientation-preserving Euclidean simi-larity) if there exists an an orientation-preserving Euclidean similarityh : R2 → R2 such that G′ = h G h−1 : g ∈ G. If we use complexnotation, then every orientation-preserving Euclidean similarity h onC ∼= R2 can be written in the form z ∈ C 7→ h(z) = az + b, wherea, b ∈ C, a 6= 0. Note that if two plane crystallographic groups are con-jugate, then they are isomorphic as groups. The following propositiongives a classification of such groups up to conjugacy.

Proposition A.30. Let G be a plane crystallographic group (con-sisting of orientation-preserving isometries). Then G is conjugate by

506 A

an orientation-preserving Euclidean similarity to one of the followinggroups G′ consisting of isometries on C of the form

(i) z 7→ g(z) = z +m+ nτ, m, n ∈ Z,where τ ∈ C, Im τ > 0, is fixed (torus-case);

(ii) z 7→ g(z) = ±z +m+ nτ, m, n ∈ Z,where τ ∈ C, Im τ > 0 is fixed ((2, 2, 2, 2) case);

(iii) z 7→ g(z) = ikz +m+ ni, m, n ∈ Z, k = 0, 1, 2, 3,

((2, 4, 4)-case);

(iv) z 7→ g(z) = ω2kz +m+ nω, m, n ∈ Z, k = 0, 1, 2,

where ω = eiπ/3 ((3, 3, 3)-case);

(v) z 7→ g(z) = ωkz +m+ nω, m, n ∈ Z, k = 0, . . . , 5,

where ω = eiπ/3 ((2, 3, 6)-case).

These different cases are distinguished by the quotient Z = R2/G.In case (i) the space Z is a torus, while in cases (ii)–(v) the space Z isa 2-sphere and the orbifold (Z, α) has a signature as indicated in eachcase.

Outline of proof. One possibility to justify this statement isto invoke the well-known classification of planar crystallographic orwallpaper groups. There are 17 such groups if we allow them to con-tain isometries that reverse orientation (i.e., reflections and glide reflec-tions). Among these groups there are 5 that contain only orientation-preserving isometries on C. These groups are sometimes denoted byp1, p2, p3, p4, p6, where the number indicates the highest order ofa stabilizer subgroup contained in G. Groups of type p1 are groupsof lattice translations isomorphic to Z2. Then Z = R2/G is a torus,and up to conjugacy (by an orientation-preserving Euclidean similar-ity) one obtains one of the groups in (i). Similarly, the cases p2, p4,p3, p6 correspond to the cases (ii), (iii), (iv), (v), respectively.

Alternatively, one can also prove the statement directly based onour previous considerations. Namely, if G = Gtr, then G containsonly translations and we are again in case (i). Otherwise, by Propo-sition A.29 and the subsequent considerations Z = C/G = R2G is a2-sphere, and the associated orbifold (Z, α) has one of the signatures(2, 2, 2, 2), (3, 3, 3), (2, 4, 4), (2, 3, 6). From the signatures we can readoff the different possibilities for the cardinalities of the stabilizers Gu,u ∈ C. In each of these cases we can find u ∈ C so that the stabilizerGu has maximal cardinality n, where n ∈ 2, 3, 4, 6. Moreover, thecardinality of any other stabilizer divides n.


Up to conjugacy, we may assume that u = 0. The group G0 = Gu

is then a cyclic group of order n generated by a rotation g0 ∈ G0 ⊂ Gof the form z 7→ ζz, where ζ is a primitive n-th root of 1. For Gtr wecan find a rank-2 lattice Λ ⊂ C such that Gtr consists precisely of thetranslations z 7→ z + γ, where γ ∈ Λ.

If g ∈ G is arbitrary, then g has the form z 7→ g(z) = αz + β,where α, β ∈ C, |α| = 1. If α = 1, then g is a translation and g ∈ Gtr.If α 6= 1, then g has precisely one fixed point v ∈ C. Moreover, theorder of α in the multiplicative group C∗ is equal to the order of g inG. Hence this order divides #Gv. On the other hand, #Gv|n, and soαn = 1. This implies that α is an n-th root of 1, and hence a powerof ζ. This in turn implies that g can be written as a composition ofa power of g0 and a translation, which necessarily belongs to Gtr. Weconclude that g0 and Gtr generate G.

If n = 2, then g0(z) = −z, and we are in case (ii). In the other cases,where n ∈ 3, 4, 6, we can say more about Gtr and the correspondingrank-2 lattice Λ. Namely, since Gtr is a normal subgroup of G, we haveg0 Gtr g−1

0 = Gtr which is equivalent to ζΛ = Λ. This additionalrotational symmetry of Λ determines Λ uniquely up to scaling by afactor a ∈ C \ 0, and Gtr up to conjugacy by the correspondingsimilarity z 7→ az (which does not change g0).

Namely, up to such a scaling Λ = Z ⊕ iZ = m + ni : m,n ∈ Zif n = 4, and Λ = Z ⊕ ωZ = m + nω : m,n ∈ Z if n = 3, 6, whereω = eiπ/3. Since g0 and Gtr generate G, this leads to the cases (iii),(iv), (v).

The groups G′ from cases (iii), (iv), and (v) are indicated in Fig-ure 3.1, Figure 3.2, and Figure 3.3, respectively. In each case, G′ is thegroup of all orientation-preserving isometries of the plane C that keepthe tiling invariant, i.e., a group element g ∈ G′ maps each triangle inthe tiling to another one of the same color. Furthermore, g maps eachpoint marked by a dot to another such point. The dots indicate theunderlying lattice Λ, i.e., the orbit of 0 ∈ C by the group of transla-tions G′tr ⊂ G′. Actually, the point 0 has the property that its G′-orbitequals its G′tr-orbit. As we have seen, the lattice Λ is equal to Z⊕ iZin case (iii), and to Z⊕ ωZ in cases (iv) and (v).

Let G be a planar crystallographic group. For our study of Lattesand and Lattes-type maps we are interested in when an orientation-preserving homeomorphism A : R2 → R2 induces a well-defined contin-uous map on the quotient R2/G. The following lemma gives a necessaryand sufficient criterion.

508 A

Figure A.1. Group associated to signature (2, 4, 4).




Lemma A.31. Let G be a planar crystallographic group, and A : R2 →R2 be an orientation-preserving homeomorphism. Then A descends toa well-defined map f on Z = R2/G, i.e., f : Z → Z is the unique mapwith f Θ = Θ A, if and only if

(A.37) A g A−1 ∈ G for all g ∈ G.

Moreover, if this condition is true, then f is branched covering map,and A also descends to well-defined covering map A on T = R2/Gtr.

Proof. Firts that if a map f : Z → Z with f Θ = Θ A exists,then it is uniquely determined as follows from the surjectivity of thequotient map Θ: R2 → Z = R2/G.⇐ Suppose that there exists a map f : Z → Z such that f Θ =

Θ A. For g ∈ G we consider A g A−1. This is an orientation-preserving homeomorphism on R2 satisfying

Θ A g A−1 = f Θ g A−1

= f Θ A−1 = Θ A A−1 = Θ.

Hence AgA−1 is a deck transformation of Θ, and so AgA−1 ∈ G.

⇒ Conversely, suppose condition (A.37) is satisfied. Let∼ be equiv-alence relation induced by G (as discussed in the beginning of thissection). Then we have have the implication

(A.38) u ∼ v ⇒ A(u) ∼ A(v) for u, v ∈ R2.

Indeed, if u ∼ v for u, v ∈ R2, then there exists g ∈ G such thatv = g(u). By (A.37) we have h := A g A−1 ∈ G, and

h(A(u)) = (A g A−1)(A(u)) = A(g(u)) = A(v).

Hence A(u) ∼ A(v).Now if p ∈ Z = R2/G is arbitrary, pick u ∈ R2 such that Θ(u) = p

and define f(p) := Θ(A(u)) ∈ Z. Then by (A.38) the map f is welldefined, and we have f Θ = Θ A.

If (A.37) is true, then we can say more about the induced map f .Namely, it follows from the basic properties of quotient topologies thatthen f is continuous on Z. Locally we can write f as a composition ofΘ A with an inverse branch of Θ−1. This implies that f is actually abranched covering map on the compact surface Z.

If g ∈ Gtr ⊂ G and g 6= idR2 , then g has no fixed points. Thisimplies that h := A g A−1 cannot have any fixed points either. Onthe other hand, h ∈ G is an orientation-preserving isometry, and so h

510 A

must be a translation. Hence h = A g A−1 ∈ Gtr. This is triviallyalso true if g = idR2 , and we see that

A g A−1 ∈ Gtr for all g ∈ Gtr .

An argument similar to the previous considerations implies that Adescends to a well-defined map on the torus T = R2/Gtr, i.e., thereexists a map A : T → T such that A π = π A, where π : R2 → T =R2/Gtr is the quotient map. Moreover, A is a branched covering map,and hence a torus endomorphism.

If Θ is as in (A.35), then the maps from the previous lemma satisfy

Θ A π = Θ π A = Θ A = f Θ = f Θ π.Since π is surjective, this implies that Θ A = f Θ. In particular, fis the quotient of the torus endomorphism A : T → T ; so we have thefollowing commutative diagram:

(A.39) R2 A//

π

R2

π

TA//

Θ

T

Θ

S2 f// S2.

The set of points p ∈ C with non-trivial stabilizer Gp = g ∈ G :g(p) = p is discrete. Indeed, let p, q ∈ C be two such points and

g(z) = a(z − p) + p ∈ Gp, h(z) = b(z − q) + q ∈ Gq,

where g, h 6= idC. Thus a, b ∈ C \ 1 with |a| = |b| = 1). Then

(A.40) h−1 g−1 h g(z) = z + (q − p)(1− a−1)(1− b−1),

is a translation. Since G acts discretely it follows that p = q or |p−q| ≥const > 0.

A.13. Quotients of circles and arcs

The purpose of this section is to provide a proof of Lemma 13.15 thatgave a criterion when a quotient J/∼ of a circle or an arc J is again acircle or an arc.

Proof of Lemma 13.15. We may assume that J is equal to theunit interval [0, 1] or the unit circle ∂D ⊂ C. We denote by [u] ⊂ J theequivalence class of a point u ∈ J , and by π : J → J/∼ the quotientmap that sends each point u ∈ J to [u] (considered as a point in J/∼).By our assumptions, in both cases J = [0, 1] and J = ∂D each set [u] is

A.13. QUOTIENTS OF CIRCLES AND ARCS 511

a subarc of J (possibly degenerate). This implies that the equivalencerelation ∼ is closed. Indeed, suppose un and vn are two sequencesin J with un ∼ vn for n ∈ N, un → u ∈ J and vn → v ∈ J as n→∞.Let αn := [un] = [vn] for n ∈ N0. If in the sequence α1, α2, . . . one ofthe sets αn appears infinitely often, then, by passing to a subsequenceif necessary, we may assume that α := α1 = α2 = . . . . Then un andvn are sequences in α. Since α is a closed set, we conclude u, v ∈ α,and so, since α is also an equivalence class, we have u ∼ v as desired.

If none of the sets appears αn appears infinitely often in the se-quence, then necessarily diam(αn)→ 0 as n→∞, as J does not carryinfinitely many pairwise disjoint arcs with diameter bounded away from0. This implies |un − vn| → 0 as n→∞, and so u = v; again we haveu ∼ v showing that ∼ is closed.

Since ∼ is closed, the quotient space J/∼ is Hausdorff; this followsfrom the fact that the complement of the diagonal in J × J is anopen set, and hence a neighborhood for each of the points it contains.Moreover, the space J/∼ is also compact, since it is the continuousimage of J under the continuous map π. So J/∼ is a compact Hausdorffspace in both cases J = [0, 1] and J = ∂D.

Now we restrict ourselves to the case J = [0, 1]. For u, v ∈ [0, 1]we define [u] < [v] iff [u] 6= [v] and u < v. This is well-defined, and atotal order for the equivalence classes of ∼, i.e., for u, v ∈ J we haveprecisely one of the cases [u] < [v], [v] < [u], or [u] = [v].

We want to show that J/∼ is homeomorphic to [0, 1]. For this itsuffices to find a surjective and continuous map h : [0, 1] → [0, 1] suchthat h(u) = h(v) for u, v ∈ [0, 1] if and only if u ∼ v. For then h will

descend to a continuous bijection h of the compact Hausdorff space

J/∼ onto [0, 1]. Then h is a homeomorphism as desired.We now construct h. By enumerating the sets [u] for u ∈ [0, 1] ∩Q

we can find a sequence of closed subintervals I0, I1, . . . of [0, 1] withI0 = [0], I1 = [1] such that the intervals I0, I1, . . . are pairwise distinctequivalence classes for ∼ and such that

D :=⋃n

In

is dense in [0, 1]. Note that by our hypotheses we have I0 6= I1. More-over, whenever u, v ∈ [0, 1] and [u] < [v], then there exists n ∈ N0

with [u] < In < [v]; otherwise, each q ∈ [u, v] ∩ Q, and hence eachpoint in [u, v], would be contained in one of the sets [u] and [v]; thisis impossible since [u] and [v] are closed disjoint sets with non-emptyintersection with [u, v], and [u, v] is connected. So between any two

512 A

distinct equivalence classes there is always one of the sets In; this andI0 6= I1 imply that the list I0, I1, . . . is infinite.

We now inductively pick a number yn ∈ [0, 1] for each set In (notnecessarily an element of In) so that In < Ik if and only if yn < yk.This is done as follows. Let y0 = 0 and y1 = 1, and suppose numbersy0, . . . , yn with the desired properties have been chosen for the setsI0, . . . , In, where n ≥ 1. For some bijection φ : 0, . . . , n → 0, . . . , nwith φ(0) = 0 and φ(n) = 1, we have

Iφ(0) = I0 < Iφ(1) < · · · < Iφ(n) = I1.

Then there exists a unique number l ∈ 0, . . . , n− 1 such that

Iφ(l) < In+1 < Iφ(l+1).

Define yn+1 := 12(yφ(l) + yφ(l+1)). It is clear that this gives a correspon-

dence In ↔ yn for n ∈ N0 as desired.If In < Ik, then there exists m > maxn, k such that In < Im < Ik.

This and the definition of yn imply that the set D′ := yn : n ∈ N0 isdense in [0, 1]; indeed, if we use the relation < to successively order thevalues y0 = 0, y1 = 1, y2, . . . , and yn < yk are immediate neighbors inthis ordering up to some point, then eventually another value ym willfall in the “gap” (yn, yk) cutting it in half, and so all gaps between thevalues yn will become small.

There exists a unique function h : D → D′ such that h|In ≡ ynfor all n ∈ N0. Then h is non-decreasing, and we can extend it to anon-decreasing function on [0, 1], also denoted h, by setting

h(u) := suph(t) : t ∈ D ∩ [0, u] for u ∈ [0, 1].

Then h is a non-decreasing function on [0, 1] with a dense image in[0, 1]. Hence h is continuous (at each point left-hand and right-handlimit exist and agree with the function value), and surjective.

It remains to verify that h(u) = h(v) if and only if u ∼ v. Supposefirst that u, v ∈ [0, 1] and u ∼ v. If u = v, then h(u) = h(v). Ifu 6= v, then [u] = [v] is a non-degenerate interval, and so it must beamong the intervals I0, I1, . . . by density of D, say [u] = [v] = In. Thenh(u) = yn = h(v) by definition of h.

Conversely, suppose that u, v ∈ [0, 1] and u 6∼ v. Without loss ofgenerality u < v. Then [u] < [v] and we can choose intervals In and Ikwith k, n ∈ N0 such that [u] < In < Ik < [v]. Then the properties of himply

h(u) ≤ h|In = yn < h|Ik = yk ≤ h(v),

and so h(u) 6= h(v) as desired. This completes the proof that J/∼=[0, 1]/∼ is homeomorphic to [0, 1], and hence an arc.

A.13. QUOTIENTS OF CIRCLES AND ARCS 513

Note that the images of the endpoints of J , i.e., 0 and 1, under thequotient map π are the endpoints of the quotient arc J/∼. Indeed, theendpoints of an arc α are precisely the points p ∈ α for which α\p isconnected. Note that [0, 1] \ [0] (the unit interval with the equivalenceclass of [0] removed) is a half-open interval and hence connected. Hence

(J/∼) \ π(0) = π([0, 1] \ [0])

is connected as the continuous image of the connected set [0, 1] \ [0].So π(0) is one of the endpoints of J/∼, and a similar argument showsthat π(1) is the other endpoint.

We now turn to the case J = ∂D. Then by our hypotheses we canpick points u, v ∈ ∂D with u 6∼ v. In particular, u 6= v. Let α and βbe the two subarcs of ∂D with endpoints u and v. Define p = π(u),

q = π(v), α = π(α), β = π(β). Then p 6= q, and our hypotheses imply

that α ∩ β = p, q (a connected set in ∂D that meets both α and βmust meet u or v). By the first part of the proof (consider ∼ restricted

to α and β), the sets α and β are arcs, and they have the endpointsp and q. So they have precisely their endpoints in common. Everycompact Hausdorff space that can be represented as a union of twosuch arcs is homeomorphic to ∂D. Hence J = ∂D/∼ is homeomorphicto ∂D and thus a topological circle.

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Index

‖h′‖σ, 376

αf , 42Ahlfors regular, 20, 94, 366annularly linearly locally connected

(ALLC), 367Aut(C), 58

bi-Lipschitz, 91bi-Lipschitz equivalent, 91Birkhoff’s ergodic theorem, 388boundary at infinity, 97bounded turning, 93branched covering map, 31

Cannon’s conjecture, 18, 101canonical orbifold metric, 491cell, 107

decomposition, 108of 2-sphere, 115induced by Thurston map, 124refinement, 110

cell complex, 108isomorphism, 121natural labeling of, 378

cellular map, 111cellular Markov partition, 111center of equivalence class, 282Chebychev polynomial, 176, 179, 184chordal metric, 200, 374, 457closed equivalence relation, 285cocompact, 58combinatorial expansion factor, 331combinatorially expanding, 245,

255, 263, 265, 277, 278two-tile subdivision, 246, 266

cone, 492

conjugacy, 36convergence to infinity, 97, 215crit(f), 32critical cycle, 43critical point, 32

periodic, 35critical value, 32crystallographic group, 58, 59cycle of vertex, 117

∂∞G, 100∂∞X, 97δ0, 152degf (p), 32dimension of measure, 389distortion, 376Dn, 154, 250, 254doubling, 18, 93, 365doubling measure

metric, 95

e-chain, 137, 235e-connected, 235edge flower, 150edge path, 233edge-type, 282En, 129entropy

metric, 348of partition, 348topological, 346

equivalent partition, 347equivalent sequences, 97equivariant, 54, 72ergodic, 347eventually onto, 35, 162

521

522 INDEX

expanding, 34, 35, 157–159, 161,249, 277, 278

combinatorially, 245, 246, 255,263, 265, 277, 278

expansion factor, 189

f -invariant, 246Jordan curve, 247, 299–301, 305,

307, 366measure, 346

factor of a dynamical system, 206factor of dynamical system, 349flag, 116, 460flexible Lattes map, 86flower, 147

G(f, C), 212G-equivariant, 54, 72generator, 348graph, 233Gromov

hyperbolic, 96, 211group, 100

product, 96, 211, 213group action

cocompact, 58properly discontinuous, 58

Hausdorffconvergence, 314dimension, 366distance, 314

holomorphic torus endomorphism, 55hyperbolic metric, 458

information function, 412invariant

equivalence relation, 472Jordan curve, 247, 299–301, 305,

307measure, 346multicurve, 48set, 246

Isom(C), 58isomorphism

of cell complexes, 121of subdivisions, 258

isotopy, 36, 221lift, 222of Jordan curve, 228

iterated monodromy group, 454

Jacobian, 392Jensen’s inequality, 393joining opposite sides, 152, 238Jordan curvef -invariant, 247, 299–301, 305,

307, 366of D0, 244

Koebe distortion, 376Konig’s infinity lemma, 250

Λ, 189Λ0, 254, 331`(X), 212labeling, 134

compatible, 134natural, 378orientation-preserving, 134subdivisions, 261

Lattes map, 2, 53, 55, 200, 201, 365flexible, 86

Lattes-type map, 56, 73Lebesgue number, 160left-shift, 205length of cycle, 117level of cell, 129Levy cycle, 165linearly locally connected (LLC), 93,

367local degree, 32Lyapunov exponent, 388

M(X, g), 346mf,C , 187, 192, 211m′f,C , 189

map realizing subdivision, 244mating, 446measurable partition, 347measure

invariant, 346of maximal entropy, 345, 349, 357,

361, 366regular, 347

mesh, 34, 157metric

canonical orbifold, 491chordal, 200, 374, 457hyperbolic, 458

INDEX 523

visual, 17, 187, 189, 190, 197, 200,332, 365–367

metric doubling measure, 95metric entropy, 348mixing, 347Moore’s Theorem, 286multicurve, 48

invariant, 48non-peripheral, 48

n-cell, 129chain, 250edge, 129edge flower, 150flower, 147skeleton, 108tile, 129vertex, 129

natural labeling, 378non-peripheral, 48

Of , 44orbifold, 41, 44

Euler characteristic, 45hyperbolic, 45parabolic, 45, 47, 55, 176signature, 46

orbit, 33, 58orientation-preserving Thurston

equivalent, 37

parabolicorbifold, 47, 55, 176

partitionequivalent, 347measurable, 347

periodic critical point, 35periodic point, 35, 205peripheral, 48planar crystallographic group, 59post(f), 33postcritical point, 33postcritically-finite, 33properly discontinuous, 58

quasi-isometry, 97quasi-Mobius, 96quasiarc, 93

uniform, 327, 366

quasicircle, 93, 299, 366uniform, 327, 366

quasidisk, 330uniform, 366

quasimetric, 194quasisphere, 102, 365Quasisymmetric uniformization

problem, 92quasisymmetrically equivalent, 92quasisymmetry, 91

uniform, 327, 328weak, 95, 382

quotient of torus endomorphism, 57,65

%, 189R], 458ramification function, 42, 44ramification portrait, 34refinement of cell decomposition, 110regular measure, 347Rickman’s rug, 272Riemann-Hurwitz formula, 32rough-isometry, 97Ruelle operator, 395

σ, 457S(f, C), 249Shannon-McMillan-Breiman, 412shift, 205signature, 46skeleton, 108snowball, 102snowflake equivalent, 18, 92, 365snowflake homeomorphism, 91snowsphere, 102spherical derivative, 458stabilizer subgroup, 58subadditive, 348subcomplex, 378subdivision, 243, 299

isomorphic, 258symbolic dynamics, 205

Thurstonmatrix, 49obstruction, 49, 79polynomial, 164

Thurston equivalent, 37, 224, 259

524 INDEX

orientation-preserving, 37Thurston map, 33

expanding, 34, 35, 157, 159, 161,249, 277, 278

obstructed, 270orbifold of, 44parabolic, 47, 55, 176rational, 33, 35, 49, 55, 175, 307,

365, 366signature, 46

tile, 374tile chain, 335tile graph, 212tile-type, 282topological entropy, 346topologically conjugate, 36, 224torus endomorphism, 56

holomorphic, 55quotient of, 57, 65

transfer operator, 395two-tile subdivision, 243

combinatorially expanding, 246,266

isomorphism of, 258realization of, 244, 299

type of equivalence classes, 282

uniformquasiarcs, 327, 366quasicircles, 327, 366quasidisks, 366

uniformly quasisymmetric, 327, 328

vertex-type, 282visual metric, 17, 187, 189, 190, 197,

200, 211, 332, 365–367on ∂∞X, 98, 211

visual sphere, 365, 367Vn, 129

weak quasisymmetry, 382weakly quasisymmetric, 95

X0w , X

0b , 128

Xn, 129

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