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AFFINE EQUIVALENCE AND SADDLE CONNECTION

GRAPHS OF HALF-TRANSLATION SURFACES

HUIPING PAN

Abstract. To every half-translation surface, we associate a saddle con-nection graph, which is a subgraph of the arc graph. We prove that ev-ery isomorphism between two saddle connection graphs is induced by anaffine homeomorphism between the underlying half-translation surfaces.We also investigate the automorphism group of the saddle connectiongraph, and the corresponding quotient graph.

Keywords: half-translation surfaces, affine homeomorphisms, sad-dle connection graphs, isomorphisms

AMS MSC2010: 30F60, 30F30, 54H15

1. Introduction

1.1. Arc complex. For a compact oriented topological surface with markedpoints (X,Σ), an arc a on (X,Σ) is called properly embedded if ∂a ⊂ Σ andthe interior is disjoint from Σ. The arc complex is a simplicial complexwhose k-simplices correspond to the set of k + 1 isotopy classes of properlyembedded non-trivial arcs which can be realized pairwise disjointly outsidethe marked points. (In this paper, by an isotopy we mean an isotopy relativeto the marked points.) The arc complex is an important and useful tool forthe study of mapping class group ([5, 6]). Masur and Schleimer ([17]) provedthat the arc complex is δ-hyperbolic (see also [8]). Later, Hesel-Przytycki-Webb ([7]) proved that the arc complex is uniformly 7-hyperbolic.

The mapping class group acts naturally on the arc complex. Irmak-McCarthy ([13]) proved that every injective simplicial map from the arccomplex is induced by a self-homeomorphism. Based on this, they describedcompletely the automorphism group of the arc complex. (These results alsohold for the arc complex of non-orientable surface, [11, 12].)

1.2. Saddle connection graph. A half-translation surface is a pair (X,ω)where X is a closed Riemann surface and ω is a meromorphic quadratic dif-ferential on X which contains no poles of order greater than one. In parallelwith the arc graph, which is the 1-skeleton of the arc complex for a topo-logical surface with marked points, we can consider the saddle connectiongraph for a half-translation surface with marked points.

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http://arxiv.org/abs/1809.06248v3

2 HUIPING PAN

Let (X,ω; Σ) be a half-translation surface with marked points Σ whichcontains all the zeros and poles of ω. The meromorphic quadratic differentialω induces a singular flat metric |ω| on X whose singular points are exactlythe zeros and poles of ω. A saddle connection is an |ω|-geodesic segment onX\Σ with endpoints in Σ. The saddle connection graph of (X,ω; Σ), denotedby S(X,ω; Σ), is a graph such that the vertices are saddle connections andthe edges are pairs of interiorly disjoint saddle connections. This graphhas infinite diameter (see Proposition 2.5). Moreover, it follows from anobservation due to Minsky-Taylor ([18]) that the saddle connection graphis isometrically embedded into the arc graph. In particular, it is connectedand δ-hyperbolic (see Section 2).

1.3. Statement of Results. The aim of this paper is to investigate iso-morphisms between two saddle connection graphs, and the automorphismgroup of the saddle connection graph. A homeomorphism between two half-translation surfaces with marked points is called affine if it is affine outsidethe set of the marked points with respect to the coordinates defined by inte-grating one of the square roots of the corresponding quadratic differential.Every affine homeomorphism induces an isomorphism between the corre-sponding saddle connection graphs. The main result of this paper showsthat the converse is also true.

Theorem 1.1. Let (X,ω; Σ), (X ′, ω′; Σ′) be two half-translation surfaceswith marked points. Then every isomorphism F : S(X,ω; Σ) → S(X ′, ω′; Σ′)is induced by an affine homeomorphism f : (X,ω; Σ) → (X ′, ω′; Σ′).

Remark 1. A direct consequence of Theorem 1.1 is that every saddle con-nection graph determines a Teichmuller disk. In other words, there is aone-to-one correspondence between the set of isomorphism classes of saddleconnection graphs and the set of Teichmuller disks.

Theorem 1.2. Let (X,ω; Σ) be a half-translation surface with marked points.

(i) If (X,ω; Σ) is not a torus with one marked point, then the automor-phism group of S(X,ω; Σ) is isomorphic to the group of affine homeo-morphisms of (X,ω; Σ).

(ii) If (X,ω; Σ) is a torus with one marked point, then the automorphismgroup of S(X,ω; Σ) is an index two subgroup of the group of affinehomeomorphisms of (X,ω; Σ).

Remark 2. For a generic half-translation surface, the group of affine home-omorphisms is trivial. As a consequence, the corresponding saddle connec-tion graph has no nontrivial automorphisms, which is different from the arcgraph.

Theorem 1.3. Let (X,ω; Σ) be a half-translation surface with marked points.Let G(X,ω; Σ) be the quotient of S(X,ω; Σ) by its automorphism group.

(i) If (X,ω; Σ) is a translation surface, then G(X,ω; Σ) has infinitely manyedges if and only if (X,ω; Σ) is not a torus with one marked point.

AFFINE EQUIVALENCE AND SADDLE CONNECTION GRAPHS 3

(ii) If (X,ω; Σ) is a Veech surface, then G(X,ω; Σ) has finitely many ver-tices.

Remark 3. We don’t know whether the converse to the second statementin Theorem 1.3 is true or not (see Question 1).

1.4. Related results. To each Teichmuller disk, Smillie-Weiss ([22]) intro-duced the spine graph, which is a tree in the hyperbolic plane. They provedthat the spine graph has compact quotient by the Veech group if and onlyif the Veech group is a lattice. The dual of the spine graph is a graph whosevertices are the directions of saddle connections and whose edges are pairsof directions which are the directions of the shortest saddle connections ofsome half-translation surface in the Teichmuller disk. Nguyen studied thegraph of degenerate cylinders for translation surfaces in genus two ([20])and the graph of periodic directions for translation surface satisfying theVeech dichotomy ([21]). He proved that both of them are hyperbolic, andthat every automorphism which comes from the mapping class group is in-duced by an affine self-homeomorphism. Moreover, based on the graph ofperiodic directions, Nguyen ([21]) gave an algorithm to determine a coarsefundamental domain and a generating set for the Veech group of a Veechsurface.

Valentina Disarlo, Anja Randecker, and Robert Tang proved a similarresult independently ([3]) under the condition that all marked points arezeros or simple poles. The overall strategy of Disarlo-Randecker-Tang issimilar to ours. But the ideas in proving the triangle preserving property arequite different. They prove the triangle preserving property via developingthe “triangle test”, which is a combinatorial criterion that can detect thesimplicies on the saddle connection complex that bound triangles on theunderlying half-translation surfaces.

1.5. Outline. In Section 2, we collect some basic properties of the saddleconnection graph including the connectedness, hyperbolicity and infinitediameter. In Section 3-7, we prove Theorem 1.1. The proof consists of threesteps:

Step 1. (§3-5) Based on the existence of many “admissible pen-tagons” (see §3), we prove that the isomorphism F pre-serves triangles (Theorem 5.1).

Step 2. (§6) Fix a triangulation of (X,ω; Σ). The correspon-dence between triangles obtained in step 1 induces anaffine map between triangles. We show that these affinemaps have orientation consistency (Proposition 6.5), whichallows us to glue these affine maps between triangles toobtain a homeomorphism from (X,ω; Σ) to (X ′, ω′; Σ′)(Theorem 6.1). It then follows from the connectednessof the triangulation graph that the isotopy class of the

4 HUIPING PAN

resulting homeomorphism is independent of the choicesof triangulations (Proposition 6.7).

Step 3. (§7) We prove that the induced homeomorphism ob-tained in step 2 is isotopic to an affine homeomorphism(Proposition 7.1).

Step three is a standard argument (see [4, 20, 21]). The novel part of thispaper is step one and step two. In Section 7, we also prove Theorem 1.2. InSection 8, we prove Theorem 1.3. In Section 9, we propose two questions.

Acknowledgements. We thank Duc-Manh Nguyen for pointing out a mis-take in the proof of Theorem 1.3 in an earlier version and for informing usthe references [2, 10]. We thank Robert Tang for comments for an earlierversion. We thank Kasra Rafi for informing us the reference [18]. We alsothank Lixin Liu and Weixu Su for useful discussions. We are grateful to thereferees for careful reading of the manuscript, with many corrections andcomments. This work is supported by NSFC 11901241.

2. Preliminaries

2.1. Arc graph. An arc a on (X,Σ) is called properly embedded if ∂a ⊂ Σand the interior is disjoint from Σ. A properly embedded arc is said tobe trivial if it is isotopic to a single point (i.e. the endpoint). Otherwise,it is said to be non-trivial. The arc graph of (X,Σ), denoted by A(X,Σ),is a graph such that the vertices A0(X,Σ) are isotopy classes of properlyembedded non-trivial arcs on (X,Σ), the edges A1(X,Σ) are pairs of isotopyclasses of properly embedded non-trivial arcs which can be realized interiorlydisjoint. (In this paper, by an isotopy we mean an isotopy relative to themarked points.)

Let each edge in A(X,Σ) be of length one. This induces a metric dA onA(X,Σ). Recall that a geodesic metric space (X,d) is called δ-hyperbolic iffor every geodesic triangle [xy]∪ [yz]∪ [zx], each geodesic in {[xy], [yz], [zx]}is contained in the δ-neighbourhood of the union of the other two.

Theorem 2.1 ([17, 8, 7]). The arc graph (A(X,Σ),dA) is δ-hyperbolic.

Remark 4. In fact, Hesel-Przytycki-Webb ([7]) show that (A(X,Σ),dA) is7-hyperbolic.

2.2. Half-translation surfaces.

Definition 1 (half-translation surface). A half-translation surface is a pair(X,ω) where X is a closed Riemann surface and ω is a meromorphic qua-dratic differential on X which contains no poles of order greater than one.An half-translation surface with marked points is a triple (X,ω; Σ) such that(X,ω) is a half-translation surface, and Σ is a finite subset of X which con-tains the zeros and poles of ω.


The meromorphic quadratic differential ω induces a singular flat metric|ω| on X, which is flat outside the zeroes and poles of ω. The zeros andpoles of ω are called singular points. The cone angle at each singular pointof ω is a multiple of π.

Definition 2 (Saddle connection). A saddle connection is an |ω|-geodesicsegment on X\Σ with endpoints in Σ.

For an oriented saddle connection α, the integral∫α

√ω is called the

holonomy of α, where√ω is one of the square roots of ω. Since there is

no canonical choice of square roots of ω, we see that the holonomies of(oriented) saddle connections belong to R

2/{±1}.Proposition 2.2. Let (X,ω; Σ) be a half-translation surface with markedpoints.

• The set of holonomies of saddle connections on (X,ω; Σ) is a discretesubset of R2/{±1}.

• The set of directions of saddle connections on (X,ω; Σ) is a densesubset of S1/{±1}.

Remark 5. The first statement can be found in [9, Proposition 1]. Thesecond statement is [16, Theorem 2].

Definition 3 (Cylinder). A cylinder on (X,ω; Σ) is an open subset disjointfrom Σ, which is isometric to (R/cZ) × (0, h) with c, h ∈ R>0 and notproperly contained in any other subset with the same property. A boundarycomponent of a cylinder is called simple if it consists of only one saddleconnection. A cylinder is called semisimple if it contains at least one simpleboundary component. A cylinder is called simple if both of its boundarycomponents are simple.

By a simple closed curve on (X,ω; Σ), we mean a simple closed curve onX\Σ.Definition 4 (Cylinder curve). A simple closed curve on (X,ω; Σ) is calleda cylinder curve if it is isotopic to a core curve of some cylinder on (X,ω; Σ).

For our purpose, we need the following theorem which is stated as Lemma22 in [4].

Theorem 2.3 ([4]). Let f : (X,ω; Σ) → (X ′, ω′; Σ′) be a homeomorphismpreserving marked points, such that for each simple closed curve α, α isa cylinder curve on (X,ω; Σ) if and only if f(α) is a cylinder curve on(X ′, ω′; Σ′). Then f is isotopic to an affine homeomorphism.

2.3. Saddle connection graph.

Definition 5. The saddle connection graph of (X,ω; Σ), denoted by S(X,ω; Σ),is a graph such that the vertices S0(X,ω; Σ) are saddle connections and theedges S1(X,ω; Σ) are pairs of disjoint saddle connections.

6 HUIPING PAN

Convention. In this paper, whenever we mention the intersection betweentwo saddle connections, we mean the intersection in the interior.

Notice that the saddle connection graph S(X,ω; Σ) is a subgraph ofA(X,Σ). For any topological arc a ∈ A0(X,ω), the |ω|-geodesic representa-tive consists of several saddle connections. It follows from [18, Theorem 1.4]that S(X,ω; Σ) is a geodesically connected subset of A(X,Σ), in the sensethat any two points in S(X,ω; Σ) are connected by a geodesic of A(X,Σ)that lies in S(X,ω; Σ). This implies that S(X,ω; Σ) is connected and δ-hyperbolic.

Definition 6 (Triangulation). A triangulation Γ = {γi} of (X,ω; Σ) is aset of saddle connections {γ1, γ2, · · · , γ6g−6+3p}, where p = |Σ| and g is thegenus of X, such that

• for any i 6= j, γi and γj are disjoint;• any saddle connection γ /∈ Γ intersects γi for some γi ∈ Γ.

Lemma 2.4. Let F : S(X,ω; Σ) → S(X ′, ω′; Σ′) be an isomorphism betweentwo saddle connection graphs. Let Γ := {γi} be a triangulation of (X,ω; Σ).Then F (Γ) := {F (γ) : γ ∈ Γ} is a triangulation of (X ′, ω′; Σ).

Proof. By definition, we see that for any pair of different saddle connectionsγi, γj ∈ Γ, F (γi) and F (γj) are disjoint. If F (Γ) is not a triangulation of(X ′, ω′; Σ′), then there exists a saddle connection γ′ /∈ F (Γ) of (X ′, ω′; Σ′)which is disjoint from F (γi) for all γi ∈ F (Γ). Correspondingly, F−1(γ′) /∈ Γis a saddle connection of (X,ω; Σ) which is disjoint from all γi ∈ Γ. This isa contradiction which proves the lemma!

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2.4. GL(2,R) action. Let κ = (k1, · · · , kn) ∈ {−1, 0, 1, · · · , 4g − 4}n be apartition of 4g − 4, i.e. k1 + k2 · · · + kn = 4g − 4. Denote by Q(κ) themoduli space of meromorphic quadratic differentials with marked points(X,ω; Σ) which contain n zeros of multiplicities (k1, · · · kn), where a zeroof multiplicity −1 is a pole of order one and a zero of multiplicity zerois a regular marked point. There is a natural GL(2,R) action on Q(κ),which acts on (X,ω; Σ) ∈ Q(κ) by post-composition with the atlas maps of(X,ω; Σ) (see [18, §1.8]). More precisely, let {(Ui, φi), i ∈ I} be an atlas of(X,ω; Σ), covering X except the marked points of ω. Then for a ∈ GL(2,R),a · (X,ω; Σ) is defined by the atlas {(Ui, a ◦ φi), i ∈ I}. Therefore, theidentity map between (X,ω; Σ) and a · (X,ω; Σ) is isotopic to an affinehomeomorphism, which induces an isomorphism between S(X,ω; Σ) andS(a · (X,ω; Σ)).

For each θ ∈ S1/{±1}, the straight lines in that direction induce a direc-

tional flow on (X,ω; Σ). By [14, Theorem 1], for almost every θ ∈ S1/{±1},

the induced directional flow is uniquely ergodic on (X,ω; Σ).

2.5. Infinite diameter.


Proposition 2.5. Let (X,ω; Σ) be a half-translation surface with markedpoints. The saddle connection graph S(X,ω; Σ) has infinite diameter.

Proof. We use an argument of F. Luo as explained in [15, §4.3]. Suppose tothe contrary that the diameter of S(X,ω; Σ) is finite.

By [14, Theorem 1], for almost every θ ∈ S1/{±1}, the induced directional

flow is uniquely ergodic on (X,ω; Σ). After rotation, we may assume thatthe horizontal flow of (X,ω; Σ) is uniquely ergodic. Fix a non-horizontalsaddle connection α. Let {βi}i≥1 be a sequence of saddle connections, suchthat the limit of the directions is horizontal. This implies that for everynon-horizontal segment I, there exists T > 0, such that βi intersects I forall i > T .

By assumption, the diameter of S(X,ω; Σ) is finite, there is a subsequence,still denoted by {βi}i≥1 for convenience, such that dS(α, βi) = N for some

N < ∞. For each i, consider a geodesic β0i = α, β1

i , · · · , βN−1i , βN

i = βi inthe saddle connection graph from α to βi. On the other hand, for everynon-horizontal segment I, βi intersects I for all large enough i. Therefore,the limit of the directions of βN−1

i has to be horizontal. Otherwise, βi = βNi

would intersect βN−1i for large enough i. Inductively, we obtain that the limit

of the directions of β1i has to be horizontal. As a consequence, β1

i wouldintersect the non-horizontal saddle connection α = β0

i for large enough i,which contradicts that dS(α, β

1i ) = 1.

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3. Admissible polygons

Let (X,ω; Σ) be a half-translation surface with marked points. The goalof this section is to find “admissible polygons” on (X,ω; Σ). Let us startwith the following definition.

Definition 7 (admissible map). Let Q ⊂ R2 be a polygon. A continuous

map I : Q → (X,ω; Σ) is called admissible if

(1) it is a local half-translation, i.e. the derivative is ±(1 00 1

);

(2) it is an embedding in the interior of Q;(3) I(P ) is a marked point if and only if P is a vertex of Q.

A polygon is called admissible if it admits an admissible map into (X,ω; Σ).

In particular, an admissible map is a local isometry sending sides anddiagonals (interior to Q) to saddle connections.

We will use the notation (A1A2 · · ·Am) to represent a polygon Q ⊂ R2

with counterclockwise labeled vertices A1, A2, · · ·Am.

Lemma 3.1. Let I : (A1A2 · · ·Am) → (X,ω; Σ) be an admissible map, thenfor any pair of sides AiAi+1 and AjAj+1, I(AiAi+1) and I(AjAj+1) areeither equal or disjoint.

Proof. It follows directly from the definition of admissible maps. �

8 HUIPING PAN

Definition 8. A convex polygon on R2 is called strictly convex if it is strictly

convex at each vertex, i.e. the interior angle at each vertex is strictly lessthan π.

Lemma 3.2. Let P be a convex polygon in R2. Let I : P → (X,ω; Σ) be a

local half-translation such that I(P ) ∈ Σ only if P is a vertex. Let Q1, Q2 ∈P be such that int(Q1Q2) contains no vertices of P. If I(Q1) = I(Q2), thenthe holonomy corresponding to the closed curve I(Q1Q2) is a translation.

Proof. Suppose to the contrary that the holonomy is a strict half-translation,i.e. Q1 and Q2 are identified by a strict half-translation. Let Q ∈ P be themidpoint of Q1, Q2. Then I(Q) would be a marked point of cone angleπ. Hence, Q is a vertex of P. This is a contradiction which proves thelemma. �

Lemma 3.3 (Admissible extension I). Let (A1A2 · · ·Am) ⊂ R2 be a poly-

gon without self-intersections. Let I : (A1A2 · · ·An) → (X,ω; Σ) and I :(A1An · · ·Am) → (X,ω; Σ) be two admissible maps, 3 ≤ n ≤ m − 1, suchthat

• I(A1An) = I(A1An);

• I(A1An) 6= I(AiAi+1) for all 1 ≤ i ≤ n−1 and I(A1An) 6= I(AjAj+1)for all n ≤ j ≤ m;

• for ∀1 ≤ i ≤ n−1, ∀n ≤ j ≤ m, I(AiAi+1) and I(AjAj+1) are eitherequal or disjoint.

Then the map I : (A1A2 · · ·Am) → (X,ω; Σ) defined by

I|(A1A2···An) = I, I|(A1An···Am) = I

is admissible.

Proof. To show that I is admissible, it suffices to show that I is an em-bedding in the interior. Suppose to the contrary that there exist Q1, Q2 ∈int(A1A2 · · ·Am) such that I(Q1) = I(Q2). Moreover, we may assume thatQ2 ∈ int(A1An · · ·Am)∪int(A1An) and Q1 ∈ int(A1A2 · · ·An)∪int(A1An).Since (X,ω; Σ) is a half-translation surface, the identification between Q1

and Q2 by I is a half-translation of R2. Let ζ be the induced half-translationsending Q2 to Q1. Let (A′

1A′n · · ·A′

m) be the image of (A1An · · ·Am) by ζ.Then

I(P ) = I(ζ−1(P )), ∀P ∈ (A1An · · ·Am) ∩ (A′1A

′n · · ·A′

m).

Notice thatQ1 ∈ int(A1A2 · · ·An) ∩ int(A′

1A′n · · ·A′

m).

Combined with the third assumption in the setting, this implies that

(A′1A

′n · · ·A′

m) ⊂ (A1A2 · · ·An) or (A1A2 · · ·An) ⊂ (A′1A

′n · · ·A′

m).

Without loss of generality, we may assume that (A′1A

′n · · ·A′

m) ⊂ (A1A2 · · ·An).

Then A′1A

′n = AiAj for some side or diagonal of (A1A2 · · ·An) (because

both (A′1A

′n · · ·A′

m) and (A1A2 · · ·An) are admissible polygons of (X,ω; Σ)).


Hence, I(A1An) = I(A1An) = I(A′1A

′n) = I(AiAj). Since I is admissible, it

follows that AiAj can not be a diagonal. Then AiAj must be a side. It then

follows from the second assumption in the setting that AiAj = A1An. There-

fore, ζ(A1An) = A′1A

′n = A1An. Consequently, ζ is a strict half-translation,

and the midpoint of A1An is marked point of cone angle π. This contradictsto the assumption that I is admissible. �

Let H := {(x, y) ∈ R2 : −r < y < r} be a horizontal strip with upper

boundary ∂+H and lower boundary ∂−H. Let Ai = (ai, bi) ∈ H, i =1, 2, · · · , n, such that

• bn = −r ≤ bn−1 ≤ · · · ≤ b2 ≤ b1 = r;• a1 = an = 0, ai < 0 for all i = 2, · · · , n− 1.

Lemma 3.4 (Admissible extension II). Let H, A1, A2 · · ·An be as above.Let I : (A1A2 · · ·An) → (X,ω; Σ) be an admissible map such that I(A1An) 6=I(AiAi+1) for any 1 ≤ i ≤ n − 1. Then there exists An+1 ∈ H on the rightside of A1An, such that I can be admissibly extend to (A1A2 · · ·AnAn+1).Moreover, if the image of (A1A2 · · ·An) by I is not contained in any hori-zontal cylinder, then we have An+1 ∈ H.

Proof. Let Pt ∈ ∂+H andQt ∈ ∂−H such that−−−→A1Pt =

−−−→AnQt = t(1, 0), where

t > 0. For t > 0 small, by following the horizontal lines on (X,ω; Σ) throughpoints in I(A1An), we can extend I to the polygon (A1A2 · · ·AnQtPt) whichremains to be a local half-translation. In fact, the extension fails only if someinterior point of PtQt is mapped to a marked point by I. The assumptionsthat I : (A1A2 · · ·An) → (X,ω; Σ) is an admissible map and that I(A1An) 6=

A1

Ak−1

Ak

An

An+1

A1

An

An+1

Pa

Qa

Pa

Qa

U

V

S

S

(a)

A1

Ak−1

Ak

An

An+1

A1

An

An+1

Pa

Qa

Pa

Qa

U

V

S

S

(b)

Figure 1. Admissible extension of polygons.

10 HUIPING PAN

I(AiAi+1) for any 1 ≤ i ≤ n − 1 imply that for small t > 0, the extensionI : (A1A2 · · ·AnQtPt) → (X,ω; Σ) is an embedding in the interior.

Now, let us consider the horizontal rays {LP : P ∈ I(A1An)} on (X,ω; Σ)which emanate from the points of I(A1An) and which flow away from (A1A2

· · ·An). Since (X,ω; Σ) is of finite area and {I(A1), I(An)} ⊂ Σ, it followsthat some of such rays would meet marked points during the flowing process.Let LT , T ∈ I(A1An) be a ray which meet marked points first (if there aremore than one candidates, we choose the one closer to A1). Denote bya the distance it travels from T to the first marked point. Let An+1 =(an+1, bn+1) ∈ PaQa be such that I(An+1) is the marked point met by LT

(see Figure 1). Then an+1 = a > 0 and bn+1 ≥ −r.We claim that i : (A1AnAn+1) → (X,ω; Σ) is admissible. Suppose to

the contrary that there exist M = (x, y), M = (x, y) ∈ int(A1AnAn+1)

such that I(M) = I(M). By Lemma 3.2, the identification between M andM by I is a translation. Without loss of generality, we may assume thatx ≥ x. Furthermore, we assume that y ≥ y. (The proof for the case y ≤ y

is similar.) Let k and k be respectively the slopes of AnAn+1 and MM .

Then k, k ≥ 0. If k ≤ k, then ∃M ∈ int(A1AnAn+1)\{A1, An, An+1} such

that−−−−−→MAn+1 =

−−−→MM . Then I(M) = I(An+1) ∈ Σ, which contradicts to the

construction of An+1. If k > k, then ∃M ∈ int(A1AnAn+1)\{A1, An, An+1}such that

−−−→AnM =

−−−→MM . Then I(M ) = I(An) ∈ Σ, which also contradicts

to the construction of An+1.

By Lemma 3.3, to show that I : (A1A2 · · ·An+1) → (X,ω; Σ) is admissible,it suffices to show that for ∀1 ≤ i ≤ n − 1, ∀n ≤ j ≤ n + 1, I(AiAi+1) andI(AjAj+1) are either equal or disjoint. In the following, we shall consider

An+1A1. The proof for AnAn+1 is similar.Suppose to the contrary that there exists 2 ≤ k ≤ n such that I(An+1A1)

intersects I(Ak−1Ak) but I(An+1A1) 6= I(Ak−1Ak). Let S ∈ int(An+1A1)

and S ∈ int(Ak−1Ak) such that I(S) = I(S). Since (X,ω; Σ) is a half-

translation surface, the identification between S and S by I has two cases:translation or strict half-translation.

• Case 1. The identification between S and S is a translation. Let Θ :R2 → R

2 be the translation such that Θ(S) = S. Let (A1PaAn+1QaAn)be the image of (A1PaAn+1QaAn) under the translation Θ. In par-

ticular, (A1PaQaAn) is a parallelogram with A1Pa horizontal and

PaQa vertical. Moreover,

(1) int(A1An+1) ∩ int(Ak−1Ak) = {S}.

By the construction of An+1, we see that

(2) Ak−1, Ak /∈ int(A1PaAn+1QaAn) ∪ int(A1Pa) ∪ int(AnQa).


Since I(A1An) 6= I(Ak−1Ak) and I : (A1A2 · · ·An) → (X,ω; Σ) is

admissible, it follows that Ak−1Ak and A1An are disjoint. Combining(1), (2), and the assumption about the y-coordinates of Ak−1 and Ak,

we see that Ak−1Ak intersects both A1Pa and PaQa (see Figure 1(a)),

or it intersects both AnQa and PaQa (see Figure 1(b)). Without

loss of generality, we assume that Ak−1Ak intersects both A1Pa and

PaQa.Let U, V ∈ A1An such that Ak−1U and AkV are horizontal. Then

An+1 ∈ int(Ak−1AkV U) ⊂ int(A1A2 · · ·An),

which contradicts with the assumption that (A1A2 · · ·An) is an ad-missible polygon.

• Case 2. The identification between S and S is a strict half-translation.Let Θ be the half-translation of R

2 such that Θ(S) = S. Let

(A1PaAn+1QaAn) be the image of (A1PaAn+1QaAn) by Θ. Then

(A1PaAn+1QqAn) ∩H ⊂ {(x, y) ∈ H : x ≥ 0}.(Otherwise, some ray in {LP : P ∈ A1An} would meet the marked

point I(An) before LT meet I(An+1).) In particular, we have S ∈(A1An+1An). It then follows from Lemma 3.2 that the identification

between S and S by I is a translation, which contradicts to theassumption of this case.

If in addition, I(int(A1A2 · · ·An)) is not contained in any horizontal cylin-der, we consider the family of rays {LP : P ∈ int(I(A1An))} instead of{LP : P ∈ A1An)}. It then follows that An+1 ∈ H. �

Let H1 := {(x, y) ∈ R2 : 0 < y < d1}, H2 := {(x, y) ∈ R

2 : −d2 < y < 0}.Let ∂+ = {(x, y) ∈ R

2 : y = d1}, ∂0 = {(x, y) ∈ R2 : y = 0}, ∂− = {(x, y) ∈

R2 : y = −d2}. Let (A1A2A3A4A5) be a strictly convex pentagon such that

A1 ∈ ∂+, A2 ∈ H1 ∪ ∂+, A3A5 ⊂ ∂0, A4 ∈ ∂−.

Lemma 3.5. Let (A1A2A3A4A5), H1,H2, ∂+, ∂0, ∂− be as above. Let I :(A1A2A3A4A5) → (X,ω; Σ) be an admissible map. Suppose that I(A2A3) 6=I(A5A1).

• If d1 ≥ d2, then there exists A6 ∈ H1 such that I can be admissiblyextended to (A1A2A3A4A5A6). If in addition, I(int(A1A2A3A5)) isnot contained in any horizontal cylinder, then A6 ∈ H1.

• If d1 ≤ d2, then there exist A7 ∈ H2 such that I can be admissiblyextended to (A1A2A3A4A7A5), and A8 ∈ H2 such that I can beadmissibly extended to (A1A2A3A8A4A5).

Proof. The proof is similar to that of Lemma 3.4. �

12 HUIPING PAN

Corollary 3.6. Let γ1, γ2, γ3 be three saddle connections bounding a triangle∆ on (X,ω; Σ) which is not contained in any cylinder. Then there exist aconvex pentagon P = (P1P2P3P4P5) ⊂ R

2 which is strictly convex at eachvertex, and an admissible map I : P → (X,ω; Σ) (see Figure 2), such that

I(P1P3) = γ1, I(P1P4) = γ2, I(P3P4) = γ3.

Proof. Suppose that γ1, γ2, γ3 are arranged in the counterclockwise orderwith respect to ∆. After applying the GL(2, R) action, we may assume thatγ2 is horizontal. Let (A1A3A4) be a triangle in R

2 and I : (A1A3A4) →(X,ω; Σ) an admissible map such that I(A1A3) = γ1, I(A3A4) = γ2 andI(A4A1) = γ3.

Let M ⊂ R2 be the horizontal strip which contains A1, A3, A4 in the

boundary. Since ∆ is not contained in any cylinder, by Lemma 3.4, thereexists A2 ∈ int(M) on the left side of A1A3 such that I can be admis-sibly extended to (A1A2A3A4) → (X,ω; Σ). Consider the quadrilateral(A1A2A3A4). It is not contained in any cylinder. Again, it follows fromLemma 3.4 that there exists A5 ∈ int(M) on the right side of A1A4 suchthat I can be admissibly extended to I : (A1A2A3A4A5) → (X,ω; Σ). SinceA1, A3, A4 belong to the boundary of M while A2 and A5 belong to interior,we see that (A1A2A3A4A5) is strictly convex at each vertex. �

Lemma 3.7. Let γ1, γ2, γ3 be three saddle connections bounding a trian-gle ∆ on (X,ω; Σ) which is contained in a non-simple cylinder C. Supposethat γ3 is contained in the boundary of C. Then there exist a convex pen-tagon (A1A2A3A4A5) ⊂ R

2 and an admissible map I : (A1A2A3A4A5) →(X,ω; Σ), such that

• I(A1A3) = γ1, I(A3A4) = γ2, I(A1A4) = γ3;• the interior angle at A1 is π while the other interior angles are lessthan π (see Figure 3(a)), or the interior angle of at A4 is π whilethe other interior angles are less than π (see Figure 3(b)).

Proof. There exist a parallelogram (Q1Q2Q3Q4) on R2, and an admissible

map J : (Q1M1 · · ·MnQ2Q3N1 · · ·NkQ4) → (X,ω; Σ) whose image is C,

A5

A1

A2

A3 A4

γ3γ1

γ2

M

Figure 2. Pentagon-extension of triangles I: the triangle(A1A3A4) is not contained in any cylinder.


A2 A1 (M1)

A4

A5(Q1)

A3(Q3)

γ1

γ2

γ3

(a)

A2(M1) A1(Q1)

A5A4A3(Q3)

γ1

γ2

γ3

(b)

Figure 3. Pentagon-extension of triangles II: the triangle(A1A3A4) is contained in some non-simple cylinder.

where M1, · · · ,Mn ∈ Q1Q2 and N1, · · · , Nk ∈ Q3Q4. By assumption, C isnot simple. It follows that max{n, k} ≥ 1.

• If n ≥ 1, we may choose the polygon (Q1M1 · · ·MnQ2Q3N1 · · ·NkQ4)such that I(M1Q3) = γ1. Let A1 = M1, A3 = Q3, A5 = Q1,A2 = M2 if n > 1 or A2 = Q2 if n = 1, and A4 = Q4 if k = 0 orA4 = N1 if k ≥ 1.

• If k ≥ 1, we may choose the polygon (Q1M1 · · ·MnQ2Q3N1 · · ·NkQ4)such that I(Q1Q3) = γ1. Let A1 = Q1, A3 = Q3. A4 = N1, A2 = Q2

if n = 0 or A2 = M1 if n ≥ 1, and A5 = N2 if k > 1 or A5 = Q4 ifk = 1.

Then the resulting pentagon (A1A2A3A4A5) and the associate admissiblemap J : (A1A2A3A4A5) → (X,ω; Σ) satisfy all the properties we want. �

Lemma 3.8. Let I : (A1A2A3A4) → (X,ω; Σ) be an admissible map, where(A1A2A3A4) ⊂ R

2 is strictly convex at A2, A3, A4. If (A1A2A3A4) is not

strictly convex at A1, then there exists a sequence of admissible maps Ik :(A1Ak−1AkA3A4) → (X,ω; Σ), where k = 1, 2, · · · ,m and A0 = A2, suchthat

(i) I and Ik coincide on the triangle (A1A3A4);

(ii) (A1Ak−1AkA3) is strictly convex at each vertex for k = 1, 2, · · · ,m;

(iii) (A1AmA3A4) is strictly convex at each vertex.

Proof. After apply an action of GL(2,R), we may assume that (A1A3A4)is an equilateral triangle (see Figure 4(a)). We shall proceed by induction.The algorithm is as following.

Consider the strip H1 directed by−−−→A4A3 which contains A2,

A3, A4 in the boundary (see Figure 4(a)). Since (A1A2A3A4)is not strictly convex at A1, it follows that A1 ∈ int(H1) andthat I(int(A1A2A3A4)) by I is not contained in any cylinder

in the direction of−−−→A4A3. By Lemma 3.4, there exist A1 ∈

int(H1) below A2A3 and an admissible I1 : (A1A2A1A3A4) →(X,ω; Σ) which coincides with I on (A1A3A4), such that

14 HUIPING PAN

A4

A1 A3

A2

A1

(a)

H1

A4

A1 A3

A2

A1 A2

(b)

H2

Figure 4. Constructing a strictly convex admissible quadrilat-eral (A1AmA3A4) from an arbitrary admissible quadrilateral

(A1A2A3A4).

(A1A0A1A3) is strictly convex at each vertex, where A0 =

A2. Let θ13 be the interior angle of (A1A0A1A3) at A3. Then

θ03 < θ13 < 2π/3,

where θ03 is the interior angle of (A1A2A3) at A3 and 2π/3is the exterior angle of (A1A3A4) at A3.

If (A1A1A3A4) is strictly convex at A1, the algorithm ter-minates. Otherwise, we repeat the construction above for(A1A1A3A4).

We now show that the algorithm will terminate after finitely many steps.Suppose to the contrary that the algorithm will never terminate. In each

step, we construct an admissible map Ik : (A1Ak−1AkA3A4) → (X,ω; Σ),

where k = 1, 2, · · · ,m and A0 = A2, such that

(i) I and Ik coincide on the triangle (A1A3A4);

(ii) (A1Ak−1AkA3) is strictly convex at each vertex for k = 1, 2, · · · ,m;

(iii) (A1AmA3A4) is not strictly convex at A1.

Let θk3 be the interior angle of (A1Ak−1AkA3) at A3. Then θ03 < θk3 <

θk+1,3 < 2π/3 for all k ≥ 1. In particular, I(A3Ak) 6= I(A3Aj) for all

k > j > 1. Since I : (A1Ak−1AkA3) → (X,ω; Σ) is an admissible map, wehave

Area(A1AkA3) =1

2|A1A3| · |A3Ak| sin θk3 < Area(X,ω),


which implies that

|A3Ak| < 2Area(X,ω)|A1A3|−1(min{| sin θ03|,√3/2})−1.

LetT := 2Area(X,ω)|A1A4|−1(min{| sin θ03|,√3/2})−1.On the other hand,

there are only finitely many saddle connections on (X,ω; Σ) whose lengthare less than T. This is a contradiction which proves the lemma.

�

4. Simple cylinder preserving

In the remaining of this paper, we assume that (X,ω; Σ) and (X ′, ω′; Σ′)are half-translation surfaces with marked points, and that

F : S(X,ω; Σ) → S(X ′, ω′; Σ′)

is an isomorphism. The goal of this section is to prove that F preservessimple cylinders (see Proposition 4.3). Let us start with the following twolemmas, which will be frequently used in the sequel.

Lemma 4.1 (Triangle lemma). Let I : (A1A2A3) → (X,ω; Σ) be an ad-missible map, where (A1A2A3) is a triangle. Then I(A1A2), I(A2A3) andI(A3A1) are pairwise different.

Proof. Suppose to the contrary that I(Ai−1Ai) = I(AiAi+1) for some i ∈{1, 2, 3}. Since (X,ω; Σ) is a half-translation surface, it then follows thatthe interior angle of (A1A2A3) at Ai is π. This is a contradiction whichproves the lemma. �

Lemma 4.2 (Quadrilateral lemma ). Let I : (A1A2A3A4) → (X,ω; Σ) be anadmissible map, where (A1A2A3A4) is a strictly convex quadrilateral. Thenthere exist a strictly convex quadrilateral (A′

1A′2A

′3A

′4) , and an admissible

map

I′ : (A′1A

′2A

′3A

′4) → (X ′, ω′; Σ′)

such that

(i) F ◦ I(A1A3) = I′(A′1A

′3) and F ◦ I(A2A4) = I′(A′

2A′4);

(ii) for any triangulation Γ of (X,ω; Σ) which contains I(A1A2), I(A2A3),I(A3A4), I(A4A1), F (Γ) is a triangulation of (X ′, ω′; Σ′) which con-

tains I′(A′1A

′2), I

′(A′2A

′3), I

′(A′3A

′4), I(A

′4A

′1).

If in addition, I can be admissibly extended to a pentagon (A1A2A3A4A5)

which contains a diagonal A5A2, then F ◦ I(A1A4) = I′(A′iA

′i+1) for some

i = 1, 2, 3, 4, where A′5 = A′

1.

Proof. By Lemma 2.4, F (Γ) is a triangulation of (X ′, ω′; Σ′) which containsF ◦ I(A1A3). Therefore, there exists an admissible map I′ : (A′

1A′2A

′3A

′4) →

(X ′, ω′; Σ′) such that

• F ◦ I(A1A3) = I′(A′1A

′3) and

• I′(A′i−1A

′i) ∈ F (Γ) for each i = 1, 2, 3, 4, where A′

0 = A′4.

16 HUIPING PAN

Notice that I(A2A4) intersects I(A1A3) and intersects no other saddle

connections from Γ. Therefore, F ◦ I(A2A4) intersects I′(A′1A

′3) and inter-

sects no other saddle connections from F (Γ). Then (A′1A

′2A

′3A

′4) is strictly

convex, and F ◦ I(A2A4) = I′(A′2A

′4).

If in addition, I can be admissibly extended to a pentagon (A1A2A3A4A5)which contains a diagonal A5A2, let Γ1 be a triangulation of (X,ω; Σ) whichcontains I(A1A4), I(A1A3), and all images of sides of (A1A2A3A4A5) by I.Then I(A5A2) intersects both I(A1A4) and I(A1A3) but intersects no othersaddle connections from Γ1. Correspondingly, F ◦ I(A5A2) intersects both

F ◦ I(A1A4) and I′(A′1A

′3) but intersects no other saddle connections from

F (Γ1). Consequently, F ◦I(A1A4) = I′(A′iA

′i+1) for some i = 1, 2, 3, 4, where

A′5 = A′

1.�

Proposition 4.3 (Simple cylinder). Let I : (Q1Q2Q3Q4) → (X,ω; Σ) bean admissible map such that I(Q1Q2Q3Q4) is a simple cylinder which con-tains I(Q1Q2) and I(Q3Q4) as boundary components. Then there exists anadmissible map I′ : (Q′

1Q′2Q

′3Q

′4) → (X ′, ω′; Σ′) such that

• F ◦ I(QiQj) = I′(Q′iQ

′j) for all diagonals and sides

• I′(Q′1Q

′2Q

′3Q

′4) is a simple cylinder which contains I′(Q′

1Q′2) and

I′(Q′3Q

′4) as boundary components.

Proof. Let Γ be triangulation of (X,ω; Σ) which contains I(Q1Q2), I(Q2Q3) =I(Q4Q1), I(Q3Q4), I(Q3Q1). By Lemma 4.2, there exist a strictly convexquadrilateral (Q′

1Q′2Q

′3Q

′4), and an admissible map

I′ : (Q′1Q

′2Q

′3Q

′4) → (X ′, ω′; Σ′)

such that

• F ◦ I(Q1Q3) = I′(Q′1Q

′3) and F ◦ I(Q2Q4) = I′(Q′

2Q′4);

• F (Γ) is a triangulation of (X ′, ω′; Σ′) which contains I′(Q′1Q

′2), I

′(Q′2Q

′3),

I′(Q′3Q

′4), I(Q

′4Q

′1).

Q2

Q5

Q4

Q1

Q3

(a)

Q′

2 Q′

1

Q′

4

Q′

5

Q′

3

∆′

(b)

Figure 5. Correspondence between Simple cylinders.


If (X,ω; Σ) is a torus with one marked point, then Γ1 := {I(Q1Q2), I(Q2Q3),I(Q1Q3} is a triangulation. Therefore, F (Γ1) := {F◦I(Q1Q2), F◦I(Q2Q3), F◦I(Q1Q3)} is a triangulation of (X ′, ω′; Σ′). Then I′(Q′

1Q′2) = I′(Q′

3Q′4) and

I′(Q′2Q

′3) = I′(Q′

4Q′1). The proposition follows.

In the following, we assume that (X,ω; Σ) is not a torus with one markedpoint. Let η1, η2, η3 be three different interior saddle connections of C suchthat each of them intersects both I(Q1Q3) and I(Q2Q3). Then each of

{F (η1), F (η2), F (η3)} intersects both I′(Q′1Q

′3) and F ◦ I(Q2Q3) while dis-

joint with other saddle connections from F (Γ). Therefore, F ◦ I(Q2Q3) =

I′(Q′i−1Q

′i) for some i = 1, 2, 3, 4, where Q′

0 = Q′4. Up to relabelling, we may

suppose that F ◦ I(Q2Q3) = I′(Q′2Q

′3). By Lemma 4.1, we see that

I′(Q′2Q

′3) 6= I′(Q′

1Q′2), I′(Q′

2Q′3) 6= I′(Q′

3Q′4).

If I′(Q′2Q

′3) 6= I′(Q′

1Q′4), consider the triangles from the complement of F (Γ)

on (X ′, ω′; Σ′). Let ∆′ be the one which contains I′(Q′2Q

′3) as a boundary

and which is not contained in the image of (Q′1Q

′2Q

′3Q

′4) under the admissible

map I′ (see Figure 5(b) ). The other two boundary saddle connections of ∆′

can not be I′(Q′2Q

′3). In this case, there are at most two saddle connections

of (X ′, ω′; Σ′) which intersects both I′(Q′2Q

′3) and I′(Q′

1Q′3) while disjoint

with other saddle connections from F (Γ), where the maximum holds only if∆′ ∪ (Q′

1Q′2Q

′3Q

′4) is a pentagon strictly convex at each vertex. Therefore,

I′(Q′1Q

′4) = I′(Q′

2Q′3). In particular, the image of (Q′

1Q′2Q

′3Q

′4) by I′ is a

simple cylinder on (X ′, ω′; Σ′).Next, we consider F ◦ I(Q1Q2) and F ◦ I(Q3Q4). Since (X,ω; Σ) is not a

torus with one marked point, it follows that I(Q1Q2) 6= I(Q3Q4). By Lemma3.4, we can admissibly extend I to a convex pentagon (Q1Q5Q2Q3Q4) whichis strictly convex at Q5 (see Figure 5(a)). By Lemma 4.2, this implies that

F (Q1Q2) is I′(Q′

1Q′2) or I

′(Q′3Q

′4), say I′(Q′

1Q′2). Similarly, it can be shown

that F ◦ I(Q3Q4) = I′(Q′3Q

′4). �

5. Triangle preserving

The goal of this section is to prove the following theorem.

Theorem 5.1. Let γ1, γ2, γ3 be three saddle connections on (X,ω; Σ) whichbound a triangle. Then F (γ1), F (γ2), F (γ3) also bound a triangle on (X ′, ω′; Σ′).

Our strategy is to show that F preserves convex pentagons.

Theorem 5.2 (Pentagon preserving). Let I : (A1A2A3A4A5) → (X,ω; Σ)be an admissible map, where (A1A2A3A4A5) is a convex pentagon strictlyconvex at A2, A3, A4, A5. Then there exists an admissible map

I′ : (A′1A

′2A

′3A

′4A

′5) → (X ′, ω′; Σ′),

where (A′1A

′2A

′3A

′4A

′5) is strictly convex at A′

2, A′3, A

′4, A

′5, such that

F ◦ I(AiAj) = I′(A′iA

′j)

18 HUIPING PAN

A2 A1

A4

A5

A3

(a)

B′

2(A′

4)B′

1(A′

1)

B′

4(A′

2)

B′

5(A′

5)

B′

3(A′

3)

(b)

A2

A1

A4

A5

A3

(c)

B′

2(A′

4)B′

1(A′

1)

B′

4(A′

2)

B′

5(A′

5)

B′

3(A′

3)

(d)

Figure 6. Correspondence between pentagons. In (a) and(b), the pentagons are not strictly convex at A1 and A′

1. In(c) and (d), the pentagons are strictly convex at each vertex.

for all sides and diagonals.

Proof of Theorem 5.1 assuming Theorem 5.2. Let ∆ be a triangle boundedby γ1, γ2, γ3. If ∆ is contained in some simple cylinder, the theorem followsfrom Proposition 4.3. If ∆ is not contained in any simple cylinder, it thenfollows from Lemma 3.7 and Corollary 3.6 that there exists an admissiblemap I : (A1A2A3A4A5) → (X,ω; Σ), where (A1A2A3A4A5) is a convexpentagon strictly convex at A2, A3, A4, A5, such that ∆ is contained in theimage of (A1A2A3A4A5) by I. By Theorem 5.2, there exists an admissiblemap I′ : (A′

1A′2A

′3A

′4A

′5) → (X ′, ω′; Σ′), where (A′

1A′2A

′3A

′4A

′5) is a convex

pentagon, strictly convex at A′2, A

′3, A

′4, A

′5, such that F ◦I(AiAj) = I′(A′

iA′j)

for all diagonals and sides AiAj . Since γ1, γ2, γ3 bound a triangle which iscontained in the image (A1A2A3A4A5) by I, then F (γ1), F (γ2), F (γ3) alsobound a triangle which is contained in the image of (A′

1A′2A

′3A

′4A

′5) by I′.

In particular, F (γ1), F (γ2), F (γ3) bound a triangle on (X ′, ω′; Σ′). �

The reminder of this section is to prove Theorem 5.2.

Lemma 5.3 (Pentagon lemma I). Let I : (A1A2A3A4A5) → (X,ω; Σ) be anadmissible map, where (A1A2A3A4A5) is a convex pentagon strictly convexat A2, A3, A4, A5. Then there exists an admissible map

I′ : (A′1A

′2A

′3A

′4A

′5) → (X ′, ω′; Σ′),


where (A′1A

′2A

′3A

′4A


2, A′3, A

′4, A

′5 satisfy the follow-

ing.

(i) If (A1A2A3A4A5) is strictly convex at each vertex, so is (A′1A

′2A

′3A

′4A

′5).

(ii) F ◦ I(AiAj) = I′(A′iA

′j) for all diagonals AiAj .

(iii) For any triangulation Γ of (X,ω; Σ) which contains I(A1A2), I(A2A3),I(A3A4), I(A4A5) and I(A5A1), F (Γ) is a triangulation of (X ′, ω′; Σ′)

which contains I′(A′1A

′2), I

′(A′2A

′3), I

′(A′3A

′4), I(A

′4A

′5), I

′(A′5A

′1).

Proof. Let Γ be a triangulation of (X,ω; Σ) which contains I(A2A4), I(A1A4)and I(Ai−1Ai) for all i = 1, 2, 3, 4, 5, where A0 = A5. Then by Lemma 2.4,F (Γ) is a triangulation of (X ′, ω′; Σ′) which contains F ◦I(A2A4), F ◦I(A1A4)and F ◦ I(Ai−1Ai) for all i = 1, 2, 3, 4, 5, where A0 = A5.

Consider the admissible quadrilateral (A1A2A3A4). By Lemma 4.2, thereexists a strictly convex quadrilateral (B′

1B′2B

′3B

′4), and an admissible map

I′ : (B′1B

′2B

′3B

′4) → (X ′, ω′; Σ′) (see Figure 6), such that

• F ◦ I(A1A3) = I′(B′1B

′3) and F ◦ I(A2A4) = I′(B′

2B′4);

• F ◦ I(A1A4) = I′(B′i−1B

′i) for some i ∈ {1, 2, 3, 4}, say B′

1B′2;

• I′(B′i−1B

′i) ∈ F (Γ) for each i = 1, 2, 3, 4, where B′

0 = B′4;

Next, we consider the admissible pentagon (A1A2A3A4A5). Let ∆′ be the

triangle in (X ′, ω′; Σ′)\F (Γ) which contains I′(B′1B

′2) in the boundary and

which is not contained in I′(B′1B

′2B

′3B

′4). Then we can admissibly extend I′

to the pentagon (B′1B

′4B

′3B

′2B

′5) such that

• I′(B′1B

′2B

′5) = ∆′;

• F ◦ I(A3A5) = I′(B′3B

′5), F ◦ I(A1A4) = I′(B′

1B′2);

• I′(B′5B

′2), I

′(B′5B

′1) ∈ F (Γ),

where for the second claim we use the fact that I(A3A5) intersects bothI(A2A4) and I(A1A4) while disjoint from other saddle connections in F (Γ),for the third claim we apply Lemma 4.2 for the convex quadrilateral (A1A3A4A5).

In particular, (B′1B

′4B

′3B

′2B

′5) is strictly convex at B′

5, B′2, B

′3, B

′4. By rela-

beling A′1 = B′

1, A′2 = B′

4, A′3 = B′

3, A′4 = B′

2, A′5 = B′

5 (see Figure 6(b)(d)),we get the desired admissible map I′ : (A′

1A′2A

′3A

′4A

′5) → (X ′, ω′; Σ′).

If in addition, (A1A2A3A4A5) is also strictly convex at A1, then F ◦I(A2A5) intersects only I′(A′

1A′3) and I′(A′

1A′4) and disjoint with any other

saddle connections in I′(int(A′1A

′2A

′3A

′4A

′5)). Therefore, (A′

1A′2A

′3A

′4A

′5) is

also strictly convex at A′1 and F ◦ I(A2A5) = I′(A′

2A′5). �

Next, we deal with the sides of (A1A2A3A4A5). For each diagonal Ai−1Ai+1,let

D(Ai−1Ai+1):= max{d(Ai, Ai−1Ai+1),d(Ai−2, Ai−1Ai+1),d(Ai+2, Ai−1Ai+1)}

where d(·, ·) represents the Euclidean distance on R2, and where A−1 = A4,

A0 = A5, A6 = A1, and A7 = A2.

20 HUIPING PAN

Lemma 5.4 (Pentagon lemma II). Let I : (A1A2A3A4A5) → (X,ω; Σ) andI′ : (A′

1A′2A

′3A

′4A

′5) → (X ′, ω′; Σ′) be as in Lemma 5.3. Suppose further that

(A1A2A3A4A5) is also strictly convex at A1.

(i) If d(A3, A2A4) = D(A2A4), then

F ({I(A2A3), I(A3A4)}) = {I′(A′2A

′3), I

′(A′3A

′4)}.

(ii) If d(A5, A2A4) = d(A1, A2A4) = D(A2A4) > d(A3, A2A4) andI(A1A2A4A5) is contained in a cylinder C whose boundary contains

I(A2A4),then F ◦ I(A1A2) = I′(A′1A

′2), F ◦ I(A4A5) = I′(A′

4A′5).

(iii) If d(A1, A2A4) = D(A2A4) and I(A1A2A4A5) is not contained inany cylinder whose boundary contains I(A2A4), then

F ◦ I(A1A2) = I′(A′1A

′2).

(iv) If d(A5, A2A4) = D(A2A4) and I(A1A2A4A5) is not contained inany cylinder whose boundary contains I(A2A4), then

F ◦ I(A4A5) = I′(A′4A

′5).

Similar results also hold for A3A5, A1A3, A1A4, and A2A5.

Proof. The overall strategy is to admissibly extend I and I′ to hexagons byusing Lemma 3.3 and Lemma 3.5. Since (A1A2A3A4A5) is strictly convexat each vertex, by relabeling the vertices of (A1A2A3A4A5), we see that thecases of A3A5, A1A3, A1A4, and A2A5 are all equivalent to the case of A2A4.In the following, we prove the case of A2A4.

(i). By Lemma 3.5, we see that I can be admissibly extend a hexagon(A1A2A6A3A4A5) such that (A2A6A3A4) is a strictly convex quadrilateral(see Figure 7(a)). Consider (A1A2A3A4). It follows from Lemma 4.2 that

(3) F ◦ I(A2A3) ∈ {I′(A′1A

′2), I

′(A′2A

′3), I

′(A′3A

′4), I

′(A′4A

′1)}.

Notice that I(A6A4) intersects I(A2A3) while disjoint from I(A1A4) andI(A2A5). Correspondingly, F ◦I(A6A4) intersects F ◦I(A2A3) while disjoint

from I(A′1A

′4) and I(A′

2A′5). Therefore, (3) is reduced to

(4) F ◦ I(A2A3) ∈ {I′(A′2A

′3), I

′(A′3A

′4)}.

Similarly, we have

(5) F ◦ I(A3A4) ∈ {I′(A′2A

′3), I

′(A′3A

′4)}.

Since I(A2A3) 6= I(A3A4) by Lemma 4.1, it then follows that

F ({I(A2A3), I(A3A4)}) = {I′(A′2A

′3), I

′(A′3A

′4)}.

(ii). If I(A1A2) = I(A4A5), then C is a simple cylinder. It then follows

from Lemma 4.3 and Lemma 5.3 that F ◦ I(A1A2) = I′(A′1A

′2) and F ◦

I(A4A5) = I′(A′4A

′5). In the following, we assume that I(A1A2) 6= I(A4A5).

By Lemma 3.5, we can admissibly extend I to a hexagon (A1A7A2A3A4A5)


A1

A5

A4

A3

A2

A6

(a)

A1

A5

A4

A3

A2

A6

(b)

Figure 7. Identifying edges of strictly convex pentagons I:d(A3, A2A4) = D(A2A4)

such that (A1A7A2A4A5) is a convex pentagon which is strictly convex atA7. It then follows from Lemma 4.2 that

(6) F ◦ I(A1A2) ∈ {I′(A′4A

′5), I

′(A′5A

′1), I

′(A′1A

′2), I

′(A′2A

′4)}.

Since F ◦ I(A2A4) = I′(A′2A

′4), it then follows from Lemma 4.1 that F ◦

I(A1A2) 6= I′(A′2A

′4). Then (6) is reduced to

(7) F ◦ I(A1A2) ∈ {I′(A′4A

′5), I

′(A′5A

′1), I

′(A′1A

′2)}.

Let us consider A5A1. There are two subcases: |A5A1| > |A2A4| and|A5A1| ≤ |A2A4|.

Subcase 1. |A5A1| ≤ |A2A4|. It follows that C is not a semisimple cylin-der which contains I(A5A1) as a simple boundary component. Therefore,we can admissibly extend I|(A1A2A4A5) to a convex pentagon (A1A8A2A4A5),(see Figure 8(a)), such that

• (A1A8A2A4A5) is strictly convex at A2, and• I(A1A8A2A4A5) ⊂ C.

Recall that d(A5, A2A4) = d(A1, A2A4) = D(A2A4) > d(A3, A2A4). ThenI(int(A2A3A4)) ∩ C = ∅. Consequently, I : (A1A8A2A3A4A5) → (X,ω; Σ)is admissible (by Lemma 3.3).

• Subcase 1a. F ◦ I(A1A2) = I′(A′5A

′1). Let Γ be a triangulation

of (X,ω; Σ) which contains I(A1A2), I(A1A4), I(A2A4), and theimages of sides of (A1A8A2A3A4A5) by I. Then by Lemma 5.3,we can also admissibly extend I′|(A′

1A′

2A′

4A′

5) to a convex pentagon

(A′1A

′2A

′4A

′5A

′8) which is strictly convex at A′

5. (see Figure 8(b))such that

– F ◦ I(A8A4) = I′(A′8A

′4), and

– I′(A′1A

′8), I

′(A′5A

′8) ∈ F (Γ),

where we use the fact that I(A4A8) intersects both I(A1A2) andI(A2A5) while disjoint from any other saddle connections in Γ for the

22 HUIPING PAN

first claim. Together with Lemma 3.3 and Lemma 5.3, this impliesthat I′ : (A′

1A′2A

′3A

′4A

′5A

′8) → (X ′, ω′; Σ′) is also admissible. Then

I′(A′8A

′4) and I′(A′

1A′3) are disjoint on (X ′, ω′; Σ′), which contradicts

to that I(A8A4) intersects I(A1A3) on (X,ω; Σ).

• Subcase 1b. F ◦ I(A1A2) = I′(A′4A

′5). Similarly as in Subcase 1a,

we can admissibly extend I′ to a hexagon (A′1A

′2A

′3A

′4A

′9A

′5), (see

Figure 8(c)), such that F ◦ I(A8A4) = I′(A′9A

′1). Then I′(A′

9A′1)

and I′(A′1A

′3) are disjoint on (X ′, ω′; Σ′), which contradicts to that

I(A8A4) intersects I(A1A3) on (X,ω; Σ).

Consequently, F ◦ I(A1A2) = I′(A′1A

′2) by (7).

Subcase 2. |A5A1| > |A2A4|. Consider the strip in the direction of A4A5

which contains A1, A4, A5 in the boundary. It then follows from Lemma 3.4that I : (A1A2A3A4A5) → (X,ω; Σ) can be admissibly extended to a strictlyconvex hexagon (A1A2A3A4A5A9) (see Figure 8(d)). Then I(A9A2) inter-sects I(A1A4) while disjoint from both I(A3A5) and I(A2A4). Correspond-

ingly, F ◦ I(A9A2) intersects I′(A′

1A′4) while disjoint from both I(A′

3A′5) and

I(A′2A

′4). This implies that

(8) F ◦ I(A5A1) ∈ {I′(A′5A

′1), I

′(A′1A

′2)}.

Similarly, by considering I(A9A4) instead of I(A9A2), we see that

(9) F ◦ I(A5A1) ∈ {I′(A′5A

′1), I

′(A′4A

′5)}.

There are two more subcases depending on whether I′(A′1A

′2) = I′(A′

4A′5) or

not.

• Subcase 2a. I′(A′1A

′2) = I′(A′

4A′5). Then by Proposition 4.3, we

see that I(A1A2) = F−1 ◦ (I′(A′1A

′2)) = F−1 ◦ (I′(A′

4A′5)) = I(A4A5).

Therefore, I(A2A4) = I(A5A1), which contradicts the assumptionthat |A5A1| > |A2A4|.

• Subcase 2b. I′(A′1A

′2) 6= I′(A′

4A′5). Then

(10) F ◦ I(A5A1) = I′(A′5A

′1).

Since |A5A1| > |A2A4|, it follows that C is not a semisimple cylin-der which contains I(A2A4) as a simple boundary component. Sim-ilarly as in subcase 1a, we can admissibly extend I|(A1A2A3A4A5) toa hexagon (A1A10A2A3A4A5) such that (A1A10A2A4A5) is convexand strictly convex at A1 (see Figure 8(e)). Notice that I(A10A5)intersects I(A1A4) while disjoint from I(A2A4) and I(A3A5). Cor-

respondingly, F ◦ I(A10A5) intersects I(A′1A

′4) while disjoint from

I(A′2A

′4) and I(A′

3A′5). Therefore, (6) is reduced to

(11) F ◦ I(A1A2) ∈ {I′(A′1A

′2), I

′(A′5A

′1)}.

Combined with (10), this implies that F ◦ I(A1A2) = I′(A′1A

′2).


A1 A5

A4

A3

A2

A8

(a)

A′

1 A′

5

A′

4

A′

3

A′

2

A′

8

(b)

A′

1 A′

5

A′

4

A′

3

A′

2 A′

9

(c)

A1 A5

A4

A3

A2

A9

(d)

A1 A5

A4

A3

A2A10

(e)

Figure 8. Identifying edges of strictly convex pentagons II:I(A1A2A4A5) is contained in a cylinder whose boundary con-tains I(A2A4)

24 HUIPING PAN

Similarly, it can be shown that F ◦ I(A4A5) = I′(A′4A

′5).

(iii). By Lemma 3.5, we can admissibly extend I to a hexagon

(A1A11A2A3A4A5)

such that (A1A11A2A4A5) is a strictly convex pentagon. The remaining ofthe proof is similar to that of subcase 1a and subcase 1b in case (ii).

(iv). By interchanging the labels of A2 and A4, A1 and A5, we see thatthis case is equivalent to the case (iii).

�

If (A1A2A3A4A5) is not strictly convex at A1, we need to modify Lemma5.4. In this case, by interchanging the labels A2 and A5, A3 and A4, we seethat A1A3 and A1A4 are equivalent, A2A4 and A3A5 are equivalent. ButA1A3 and A2A4 are not equivalent in general.

Lemma 5.5 (Pentagon lemma III). Let I : (A1A2A3A4A5) → (X,ω; Σ)and I′ : (A′

1A′2A

′3A

′4A

′5) → (X ′, ω′; Σ′) be as in Lemma 5.3. Suppose that

(A1A2A3A4A5) is not strictly convex at A1.

• (A2A4 and A3A5)(i) If d(A3, A2A4) = D(A2A4), then

F ({I(A2A3), I(A3A4)}) = {I′(A′2A

′3), I

′(A′3A

′4)}.

(ii) If d(A5, A2A4) = D(A2A4), then

F ◦ I(A5A4) = I′(A′5A

′4).

Similar results also hold for A3A5.• (A1A3 and A1A4)(iii) If d(A2, A1A3) = D(A1A3), then

F ({I(A1A2), I(A2A3)}) = {I′(A′1A

′2), I

′(A′2A

′3)}.

(iv) If d(A5, A1A3) = d(A4, A1A3) = D(A1A3), then

{F ◦ I(A3A4), F ◦ I(A1A5)} ⊂ {I′(A′3A

′4), I

′(A′1A

′5), I

′(A′4A

′5)}.

(v) If d(A5, A1A3) = D(A1A3) > d(A4, A1A3), then

F ◦ I(A1A5) = I′(A′1A

′5).

(vi) If d(A4, A1A3) = D(A1A3) > d(A5, A1A3), then

F ◦ I(A3A4) = I′(A′3A

′4).

Similar results also hold for A1A4.

Proof. The proofs of cases (i) and (iii) are similar to the proof of case (i) inLemma 5.4. It remains to consider (ii), (iv), (v), and (vi).

(ii). Notice that (A′1A

′2A

′3A

′4A


2, A′3, A

′4, A

′5. If

(A′1A

′2A

′3A

′4A

′5) is also strictly convex at A′

1, then it is a strictly convexpentagon. It then follows from the first conclusion of Lemma 5.3 that(A1A2A3A4A5) is also a strictly convex pentagon, which contradicts the


A1

A2

A5

A3 A4

A7

(a)

A1

A2

A5

A3 A4

A8

(b)

A1

A2

A5

A3 A4

A9

(c)

Figure 9. Identifying edges of non-strictly convex pentagons.

assumption that (A1A2A3A4A5) is not strictly convex at A1. Consequently,(A′

1A′2A

′3A

′4A

′5) is not strictly convex at A′

1.By Lemma 3.5, we can admissibly extend I to a hexagon (A1A2A3A4A7A5)

such that (A1A2A4A7A5) is convex and strictly convex atA2, A4, A7, A5 (seeFigure 9(a)). Let Γ1 be a triangulation of (X,ω; Σ) which contains I(A2A4),I(A1A4), I(A5A4), and the images of all sides of (A1A2A3A4A7A5) by I.

By Lemma 5.3, we see that F (Γ1) contains I(A′2A

′4), I(A

′1A

′4), F ◦ I(A5A4)

and all images of sides of (A′1A

′2A

′3A

′4A

′5) by I′. Notice that I(A2A7) in-

tersects both I(A1A4) and I(A5A4) while disjoint from any other saddle

connections in Γ1. Correspondingly, F ◦ I(A2A7) intersects both I′(A′1A

′4)

and F ◦ I(A5A4) while disjoint from any other saddle connections in F (Γ1).This implies that

F ◦ I(A5A4) ∈ {I′(A′5A

′4), I

′(A′5A

′1), I

′(A′1A

′2)}.

Since I(A1A7) intersects I(A5A4) while disjoint from both I(A1A3) andI(A2A4), it follows that F ◦ I(A1A7) intersects F ◦ I(A5A4) while disjoint

from both I′(A′1A

′3) and I′(A′

2A′4). This implies that F ◦I(A5A4) 6= I′(A′

1A′2).

If F ◦ I(A5A4) = I′(A′5A

′1), then by Lemma 5.3, we can admissibly ex-

tend I′|(A′

1A′

2A′

4A′

5) to a pentagon (A′

1A′2A

′4A

′5A

′7) such that F ◦ I(A2A7) =

I′(A′2A

′7) (since I(A2A7) intersects both I(A1A4) and I(A5A4) ). This im-

plies that (A′1A

′2A

′4A


1, which contradicts to the

26 HUIPING PAN

assumption that (A′1A

′2A

′3A

′4A

′5) is also not strictly convex at A′

1. Conse-

quently, F ◦ I(A5A4) = I′(A′5A

′4).

(iv) If I(A3A4) = I(A5A1), then the claim follows from Proposition 4.3.Otherwise, it follows directly from Lemma 3.5, Lemma 4.2 and Lemma 4.1.

(v) It follows from Lemma 3.5 that we can admissibly extend I to ahexagon (A1A2A3A4A5A8) such that (A1A3A4A5A8) is a strictly pentagon(see Figure 9(b)). It then follows Lemma 4.2 and Lemma 4.1 that

F ◦ I(A5A1) ∈ {I′(A′3A

′4), I

′(A′4A

′5), I

′(A′5A

′1)}.

On the other hand, I(A4A8) intersects I(A5A1) while disjoint from I(A1A3)and I(A2A4). Therefore,

F ◦ I(A5A1) ∈ {I′(A′4A

′5), I

′(A′5A

′1)}.

If F ◦ I(A5A1) = I′(A′4A

′5), it follows from Lemma 5.3 and Lemma 3.3 that

I′ can be admissibly extended to a hexagon (A′1A

′2A

′3A

′4A

′8A

′5) such that

(A′1A

′3A

′4A

′8A

′5) is a strictly convex pentagon. Therefore, (A′

1A′2A

′3A

′4A

′8A

′5)

is a convex hexagon strictly convex at A′2, A

′3, A

′4, A

′8, A

′5. Let Γ2 be a

triangulation of (X,ω; Σ) which contains I(A1A3), I(A1A4), I(A1A5) andall images of sides of (A1A2A3A4A5A8) by I. Then by Lemma 2.4 and

Lemma 5.3, F (Γ2) is a triangulation of (X ′, ω′; Σ′) which contains I′(A′1A

′3),

I′(A′1A

′4), I

′(A′4A

′5) and all images of sides of (A′

1A′2A

′3A

′4A

′8A

′5) by I′. Notice

that I′(A′2A

′8) intersects each of {I′(A′

1A′3), I

′(A′1A

′4), I

′(A′4A

′5)} while dis-

joint from any other saddle connection in F (Γ2). Therefore, F−1 ◦ I′(A′

2A′8)

intersects each of {I(A1A3), I(A1A4), I(A4A5)} while disjoint from any othersaddle connection in Γ2. This implies that (A1A2A4A5A8) is strictly convexat A1. In particular, (A1A2A4A5) is strictly convex at A1, which contradictsthe assumption that (A1A2A)3A4A5) is not strictly convex at A1.

(vi) It follows from Lemma 3.5 that we can admissibly extend I to ahexagon (A1A2A3A9A4A5) such that (A1A3A9A4A5) is a strictly convexpentagon. It then follows Lemma 4.2 and Lemma 4.1 that

F ◦ I(A3A4) ∈ {I′(A′3A

′4), I

′(A′4A

′5), I

′(A′5A

′1)}.

Notice that I(A1A9) intersects each of I(A2A4), I(A3A5), I(A3A4). There-

fore, F ◦ I(A3A4) = I′(A′3A

′4). Otherwise, F ◦ I(A1A9) would disjoint from

I′(A′2A

′4) or I

′(A′3A

′5). �

Based on Lemma 5.4 and Lemma 5.5, we now prove the following.

Lemma 5.6 (Pentagon lemma IV). Let I : (A1A2A3A4A5) → (X,ω; Σ)and I′ : (A′

1A′2A

′3A

′4A

′5) → (X ′, ω′; Σ′) be as in Lemma 5.3. Then there exist

two adjacent sides of (A1A2A3A4A5), say Ak−1Ak and AkAk+1, such that

F ◦ I(Ak−1Ak) = I′(A′k−1A

′k) and F ◦ I(AkAk+1) = I′(A′

kA′k+1). Moreover,

if (A1A2A3A4A5) is not strictly convex at A1, then k can be chosen to bedifferent from 1.


Diagonals Case 1 Case 2 Case 3

A2A4 F (γ1) = γ′1 F (γ4) = γ′4 F ({γ2, γ3}) = {γ′2, γ′3}A3A5 F (γ2) = γ′2 F (γ5) = γ′5 F ({γ3, γ4}) = {γ′3, γ′4}A1A3 F (γ5) = γ′5 F (γ3) = γ′3 F ({γ1, γ2}) = {γ′1, γ′2}A4A1 F (γ3) = γ′3 F (γ1) = γ′1 F ({γ4, γ5}) = {γ′4, γ′5}A5A2 F (γ4) = γ′4 F (γ2) = γ′2 F ({γ1, γ5}) = {γ′1, γ′5}

(a)

CaseSides

CaseSides

CaseSides

Identified Identified Identified

(111) γ1, γ2, γ5 (112) γ1, γ2, γ3 (113) γ1, γ2(121) γ5, γ1 (122) γ5, γ1, γ3 (123) γ5, γ1, γ2(131) γ5, γ1 (132) γ1, γ3, γ4 (133) γ1, γ2(211) γ2, γ4, γ5 (212) γ2, γ3, γ4 (213) γ1,γ2,γ4(221) γ4, γ5 (222) γ3, γ4, γ5 (223) γ4, γ5(231) γ3, γ4, γ5 (232) γ3, γ4 (233) γ3, γ4(311) γ2, γ3, γ5 (312) γ2, γ3 (313) γ1, γ2, γ3(321) γ5 (322) γ2, γ3, γ5 (323) γ1, γ2, γ3, γ5(331) γ2, γ3, γ4, γ5 (332) γ2, γ3, γ4 (333) γ1, γ2, γ3, γ4

(b)Table 1. In table (a), we list all three possibilities for each diag-

onal. In table (b), we list all 27 possibilities for A2A4, A3A5 and

A1A3, where the triple (ijk) represents the possibility correspond-

ing to case i for A2A4, case j for A3A5, and case k for A1A3.

Proof. For convenience, let γi := I(AiAi+1), γ′i := I(A′

iA′i+1), i = 1, 2, 3, 4, 5,

where A6 = A1 and A′6 = A′

1. γi is said to be identified if F (γi) = γ′i. Theproof will be split into two cases depending on whether (A1A2A3A4A5) isstrictly convex at A1.

Case I: (A1A2A3A4A5) is strictly convex at A1.

Consider the diagonals A2A4, A3A5 and A1A3. By Lemma 5.4, there arethree cases for each diagonal as listed in Table 1(a). Therefore, there are27 possibilities in total. Let us denote by the triple (ijk) the possibilitycorresponding to case i for A2A4, case j for A3A5, and case k for A1A3. Welist all possibilities in Table 1(b), where γi is said to be identified by F ifF (γi) = γ′i. (For cases (331), (332) and (333), we have F ({γ2, γ3}) = {γ′2, γ′3}and F ({γ3, γ4}) = {γ′3, γ′4}. To show that all of γ2, γ3, γ4 are identified, itsuffices to show that γ3 is identified. Suppose to the contrary that γ3 isnot identified, then F (γ2) = F (γ4) = γ′3. Therefore γ2 = γ4. It then

28 HUIPING PAN

follows from Proposition 4.3 that I′(A′2A

′3A

′4A

′5) is a simple cylinder which

contains γ′3 as an interior saddle connection, which contradicts the fact that(A′

1A′2A

′3A

′4A

′5) is an admissible strictly convex pentagon.)

We see that for each possibility (ijk) 6= (321), there exists some m suchthat F (γm) = γ′m and F (γm+1) = γ′m+1), where γ6 = γ1 and γ′6 = γ′1. Tofinish the proof of this case, it remains to consider the possibility (321).Notice that in this possibility, we have

(12) F (γ5) = γ′5, F ({γ2, γ3}) = {γ′2, γ′3}.Let us consider one more diagonal A1A4. There are three subcases.

• If F (γ3) = γ′3, then by Equation (12), we also have F (γ2) = γ′2. Theproof completes.

• If F (γ1) = γ′1, the proof completes.• If F ({γ4, γ5}) = {γ′4, γ′5}, then by Equation (12), we also have F (γ4) =γ′4. The proof completes.

Diagonals Case 1 Case 2

A2A4 F (γ4) = γ′4 F ({γ2, γ3}) = {γ′2, γ′3}A3A5 F (γ2) = γ′2 F ({γ3, γ4}) = {γ′3, γ′4}

(a)

CaseSides

CaseSides

Identified Identified

(11) γ2, γ4 (12) γ3, γ4(21) γ2, γ3 (22) γ2, γ3, γ4

(b)

Table 2. In table (a), we list all possibilities for diagonals A2A4

and A3A5. In table (b), we list all four possibilities for the combi-

nation of A2A4, A3A5, where the pair (ij) represents the possibility

corresponding to case i for A2A4 and case j for A3A5.

Case II: (A1A2A3A4A5) is not strictly convex at A1.

In this case, for each diagonal, there are two or four cases by Lemma5.5 (see Table 2(a)) Consider the diagonals A2A4 and A3A5. There are4 possibilities in total. Let us denote by the triple (ij) the possibilitycorresponding to case i for A2A4 and case j for A3A5. We list all possi-bilities in Table 2(c). For case (22), we have F ({γ2, γ3}) = {γ′2, γ′3} andF ({γ3, γ4}) = {γ′3, γ′4}. To show that all of γ2, γ3, γ4 are identified, it suf-fices to show that γ3 is identified. Suppose to the contrary that γ3 is notidentified, then F (γ2) = F (γ4) = γ′3. Hence d(A5, A2A4) = D(A2A4). It


then follows from the second statement of Lemma 5.5 that γ4 is identified.This is a contradiction which proves that all of γ2, γ3, γ4 are identified.

We see that for each possibility (ij) 6= (11), there exists some m such thatF (γm) = γ′m and F (γm+1) = γ′m+1), where γ6 = γ1 and γ′6 = γ′1. To finishthe proof of this case, it remains to consider the possibility (11). Notice thatin this possibility, we have

(13) F (γ2) = γ′2, F (γ4) = γ′4.

Let us consider the diagonal A1A3. By Lemma 5.5,there are four cases.

(C1) F (γ5) = γ′5;(C2) F (γ3) = γ′3;(C3) F ({γ1, γ2}) = {γ′1, γ′2};(C4) {F (γ3), F (γ5)} ⊂ {γ′3, γ′4, γ′5}, A1A3 and A4A5 are parallel.

If F (γ3) = γ′3 or F (γ5) = γ′5, then the proof completes. If F ({γ1, γ2}) ={γ′1, γ′2}, then by Equation (13), it follows that F (γ1) = γ′1. The proof alsocompletes. If {F (γ3), F (γ5)} ⊂ {γ′3, γ′4, γ′5}, together with (13), this reducesto

{F (γ3), F (γ5)} = {γ′3, γ′5}.If F (γ3) = γ′3 and F (γ5) = γ′5, then the proof completes. It remains toconsider the case

(14) F (γ3) = γ′5, F (γ5) = γ′3.

Similarly, let us consider the diagonal A1A4. By Lemma 5.5, there arefour more possibilities.

(C5) F (γ3) = γ′3;(C6) F (γ1) = γ′1;(C7) F ({γ4, γ5}) = {γ′4, γ′5};(C8) {F (γ1), F (γ3)} ⊂ {γ′1, γ′2, γ′3}, A1A4 and A2A3 are parallel.

If F (γ3) = γ′3, F (γ1) = γ′1, or F ({γ4, γ5}) = {γ′4, γ′5}, then the proof com-pletes as discussed above. If {F (γ1), F (γ3)} ⊂ {γ′1, γ′2, γ′3}, together with(13), this reduces to

{F (γ1), F (γ3)} = {γ′1, γ′3}.If F (γ3) = γ′3 and F (γ1) = γ′1, the proof completes. It remains to considerthe case

(15) F (γ1) = γ′3, F (γ3) = γ′1.

Next, let us consider the case where both (C4) and (C8) hold. In thiscase, we have γ5 = F−1(γ′3) = γ1. Recall that A1A3 and A4A5 are parallelfor (C4), and that A1A4 and A2A3 are parallel for (C8). Then

d(A2, A1A3) = d(A5, A1A3) = D(A1A3),

which reduces this case to case (C3). The proof completes.�

30 HUIPING PAN

(A11)A1

(A41)A5

(A31)A3

A4

(A21)A2

(a)

(A11)A′

1

(A41)A′

5

A′

3(A31)

A′

4

A′

2(A21)

(b)

Figure 10. Caption to Figure 10: Identifying edges ofstrictly convex quadrilaterals.

5.1. Quadrilaterals. A direct consequence of Lemma 5.6 is that we canimprove Lemma 4.2.

Lemma 5.7. Let I : (A1A2A3A4) → (X,ω; Σ) and I′ : (A′1A

′2A

′3A

′4) →

(X ′, ω′; Σ′) be two admissible maps, where (A1A2A3A4) and (A′1A

′2A

′3A

′4)

are strictly convex quadrilaterals. Suppose that

• F ◦ I(A1A3) = I′(A′1A

′3), F ◦ I(A2A4) = I′(A′

2A′4),

• F ◦ I(A1A2) = I′(A′1A

′2), F ◦ I(A1A4) = I′(A′

1A′4),

F ◦ I(A3A4) = I′(A′3A

′4).

Then F ◦ I(A2A3) = I′(A′2A

′3).

Proof. If the image of (A1A2A3A4) by I is a simple cylinder, then the lemmafollows from Proposition 4.3. In the following, we assume that the image of(A1A2A3A4) by I is not a simple cylinder. In particular, I(A1A2) 6= I(A3A4)and I(A1A4) 6= I(A2A3).

Let θi be the interior angle of (A1A2A3A4) at Ai, i = 1, 2, 3, 4. If θ2+θ3 ≤π, it then follows from Lemma 3.4 and Lemma 4.2 that F ◦ I(A2A3) =

I′(A′i−1A

′i) for some i = 1, 2, 3, 4, where A′

0 = A′4. Since I(A1A4) 6= I(A2A3),

it follows that F ◦I(A2A3) 6= I′(A′1A

′4). It then follows from Lemma 4.1 that

F ◦ I(A2A3) = I′(A′2A

′3).

Suppose that θ2 + θ3 > π. Consider the algorithm below.

Since θ2 + θ3 > π, then θ1 + θ4 < π. By Lemma 3.4,we can admissibly extend I to a strictly convex pentagon(A1A2A3A4A5). It then follows from Lemma 5.3 that wecan also admissibly extend I′ to a strictly convex pentagon(A′

1A′2A

′3A

′4A

′5) such that

F ◦ I(A2A5) = I′(A′2A

′5), F ◦ I(A3A5) = I′(A′

3A′5).


By Lemma 5.6, there exists

AmAn ∈ {A1A5, A4A5, A2A3}such that F ◦ I(AmAn) = I′(A′

mA′n).

• If F ◦ I(A2A3) = I′(A′2A

′3). The algorithm terminates.

• Otherwise, we may assume that F ◦I(A1A5) = I′(A′1A

′5).

Let us relabel the quadrilaterals (A1A2A3A5) and (A′1A

′2A

′3A

′5)

by setting A11 = A1, A21 = A2, A31 = A3, A41 = A5 andA′

11 = A′1, A

′21 = A′

2, A′31 = A′

3, A′41 = A′

5. Then

I(A21A31) = I(A2A3), I(A′21A

′31) = I′(A′

2A′3)

F ◦ I(A11A31) = I′(A′11A

′31), F ◦ I(A21A41) = I′(A′

21A′41),

F ◦ I(A11A21) = I′(A′11A

′21), F ◦ I(A11A41) = I′(A′

11A′41),

F ◦ I(A31A41) = I′(A′31A

′41).

Let θ21 and θ31 be respectively the interior angles of(A11A21A31A41) at A21 and A31. Then

0 < θ21 ≤ θ2, 0 < θ31 < θ3

0 < θ21 + θ31 < θ2 + θ3 < 2π.

If θ21 + θ31 ≤ π, the algorithm terminates. Otherwise,we repeat the construction above for (A11A21A31A41).

We now show that the algorithm will terminate after finitely many steps.Suppose to the contrary that the algorithm will never stop. Then in eachstep, we construct two admissible maps I : (A1iA2iA3iA4i) → (X,ω; Σ) andI′ : (A′

1iA′2iA

′3iA

′4i) → (X ′, ω′; Σ′) such that

I(A2iA3i) = I(A2A3), I(A′2iA

′3i) = I′(A′

2A′3)

F ◦ I(A1iA3i) = I′(A′1iA

′3i), F ◦ I(A2iA4i) = I′(A′

2iA′4i),


′2i), F ◦ I(A1iA4i) = I′(A′

1iA′4i),


′4i).

and

0 < θ2,i+1 ≤ θ2i ≤ θ2, 0 < θ3,i+1 ≤ θ3i ≤ θ2(16)

π < θ2,i+1 + θ3,i+1 < θ2i + θ3i < 2π,(17)

where for the third inequality, we use the observation that either θ2,i+1 < θ2ior θ3,i+1 < θ3i for each i ≥ 1. Consequently, either {θ2i}i≥1 or {θ3i}i≥1 hasa strictly decreasing subsequence, say {θ2i}i≥1. For convenience, we alsodenote by {θ2i}i≥1 the decreasing subsequence. In particular, the corre-

sponding saddle connections {I(A1iA2i)}i≥1 are pairwise different. On theother hand, it follows from (16) and (17) that

(18) π − θ3 < θ2i ≤ θ2, π − θ2 < θ3i ≤ θ3.

32 HUIPING PAN

Therefore,

I(|A1iA2i|) = |A1iA2i| =2Area(A1iA2iA3i)

|A2iA3i| sin θ2i≤ 2Area(X,ω; Σ)

|A2A3|min{sin θ2, sin θ3}:= L.

Recall that there are finitely many saddle connections on (X,ω; Σ) withlength at most L. This is a contradiction which proves the lemma.

�

Proof of Theorem 5.2. By Lemma 5.3, there exists an admissible map

I′ : (A′1A

′2A

′3A

′4A

′5) → (X ′, ω′; Σ′)

such that F ◦ I(AiAj) = I′(A′iA

′j) for all diagonals AiAj. By Lemma 5.6,

we may assume that F ◦ (A1A2) = I′(A′1A

′2) and F ◦ (A2A3) = I′(A′

2A′3).

Consider the quadrilateral (A1A2A3A4), it follows from Lemma 5.7 that F ◦I(A3A4) = I′(A′

3A′4). Similarly, by considering (A1A2A3A5) and (A2A3A4A5),

we see that F ◦ I(A4A5) = I′(A′4A

′5) and F ◦ I(A5A1) = I′(A′

5A′1). �

6. Homeomorphism

The goal of this section is to prove the following theorem.

Theorem 6.1. Every isomorphism F : S(X,ω; Σ) → S(X ′, ω′; Σ′) is inducedby a homeomorphism f : (X,ω; Σ) → (X ′, ω′; Σ′), such that

(i) f(Σ) = Σ′, and(ii) for any saddle connection γ on (X,ω; Σ), f(γ) is isotopic to F (γ).

Moreover, if (X,ω; Σ) is not a torus with one marked point, then such ahomeomorphism is unique up to isotopy. If (X,ω; Σ) is a torus with onemarked point, then there are two such homeomorphisms up to isotopy.

6.1. Triangulation graphs.

Definition 9. Two triangulations Γ1 and Γ2 of (X,ω; Σ) differ by an ele-mentary move if

• there exist β1 ∈ Γ1 and β2 ∈ Γ2 such that Γ1\β1 = Γ2\β2, and• there exist γ1, γ2, γ3, γ4 ∈ Γ1\{β1} = Γ2\{β2} such that they bounda strictly convex quadrilateral on (X,ω; Σ) which contains β1, β2 asdiagonals.

Definition 10. The triangulation graph of (X,ω; Σ), denoted by T(X,ω; Σ),is a graph whose vertices are triangulations of (X,ω; Σ), and whose edgesare pairs of triangulations which differ by an elementary move.

Proposition 6.2 ([2, 10, 19, 23]). For any half-translation surface withmarked points (X,ω; Σ), the triangulation graph T(X,ω) is connected.

Remark 6. The above proposition holds for general flat surfaces (simpli-cial surfaces) (see [2, Proposition 11,12],[10, Theorem 1],[23, Theorem 1.5]).Nguyen ([19, Theorem 6.2]) provides an elementary proof for the case ofhalf-translation surfaces.


6.2. Orientation consistency. By Theorem 5.1, we know that if the sad-dle connections γ1, γ2, γ3 bound a triangle ∆ on (X,ω; Σ), their imagesF (γ1), F (γ2), F (γ3) also bound a triangle on (X ′, ω′; Σ′), which is denotedby ∆′. This correspondence induces an affine homeomorphism between ∆and ∆′, which is called the F -induced affine homeomorphism and denoted byf∆. Our goal is to “glue” these F -induced affine homeomorphisms betweentriangles according to some triangulation of (X,ω; Σ) to obtain a globallywell defined homeomorphism from (X,ω; Σ) to (X ′, ω′; Σ′). To do this, weneed to clarify the orientation consistency among affine homeomorphismsbetween triangles.

Definition 11. Two triangles ∆1 and ∆2 on (X,ω; Σ) are called coconvexif there exists an admissible map I : (A1A2A3A4) → (X,ω; Σ) such that

• (A1A2A3A4) is strictly convex at each vertex;• both ∆1 and ∆2 are contained in the image of (A1A2A3A4) by I.

Lemma 6.3. Let ∆ and ∆ be two coconvex triangles on (X,ω; Σ). Supposethat the F -induced affine homeomorphism for ∆ is orientation preserving.Then the F -induced affine homeomorphism for ∆ is also orientation pre-serving.

Proof. Let I : (A1A2A3A4) → (X,ω; Σ) be an admissible map whose image

contains both ∆ and ∆. In particular, (A1A2A3A4) is strictly convex ateach vertex. Suppose that I(A1A2A4) = ∆. By Lemma 4.2 and Theorem5.1 there exists an admissible map I′ : (A′

1A′2A

′3A

′4) → (X ′, ω′; Σ′), where

(A′1A

′2A

′3A

′4) is a strictly convex quadrilateral, such that F ◦ I(AiAj) =

I′(A′iA

′j) for all i 6= j. Suppose that the vertices of (A1A2A3A4) are la-

beled in the counterclockwise order. Then the vertices of (A1A2A4) is alsolabeled in the counterclockwise order. Since the F -induced affine homeo-morphism on I(A1A2A4) is orientation preserving, we see that the verticesof (A′

1A′2A

′4) is also labeled in the counterclockwise order. Therefore, the

vertices of (A′1A

′2A

′3A

′4) are labeled in the counterclockwise order. (Other-

wise, the vertices of (A′1A

′2A

′4) would be labeled in the clockwise order.) As

a consequence, the F -induced affine homeomorphism on the image of anytriangle in (A1A2A3A4) is also orientation preserving. In particular, the

F -induced affine homeomorphism for ∆ is also orientation preserving. �

Lemma 6.4. For any two triangles ∆1 and ∆2 on (X,ω; Σ), there exists asequence of triangles ∆0 = ∆,∆1, · · · ,∆m+1 = ∆2 such that ∆k and ∆k+1

are coconvex for each 0 ≤ k ≤ m.

Proof. Notice that each triangle is a connected component of (X,ω; Σ)\Γ forsome triangulation Γ. To prove the lemma, by Proposition 6.2, it suffices toconsider the case that

∆1 ⊂ (X,ω; Σ)\Γ1, ∆2 ⊂ (X,ω; Σ)\Γ2,

where Γ1 and Γ2 are two triangulations differing by an elementary move.

34 HUIPING PAN

By definition, there exists an admissible map I : (A1A2A3A4) → (X,ω; Σ),where (A1A2A3A4) is a strictly convex quadrilateral, such that

I(A1A3) ∈ Γ1, I(A2A4) ∈ Γ2, Γ1\{I(A1A3)} = Γ2\{I(A2A4)}.Let ∆1 = I(A1A2A3) and ∆2 = I(A2A3A4). Then ∆1 and ∆2 are coconvex.

Consider the triangles in (X,ω; Σ)\Γ1 and (X,ω; Σ)\Γ2. There exist tri-

angles ∆1,j (0 ≤ j ≤ m1) and ∆2,j (0 ≤ j ≤ m2), such that

• ∆1,0 = ∆1, ∆1,m1= ∆1, ∆2,0 = ∆2, ∆2,m2

= ∆2;

• ∆i,j, ∆i,j+1 ⊂ (X,ω; Σ)\Γi, ∀i = 1, 2,∀0 ≤ j ≤ mi − 1;

• ∆i,j and ∆i,j+1 share a common boundary, ∀i = 1, 2,∀0 ≤ j ≤ mi−1.

Next, consider the pair (∆i,j, ∆i,j+1). By Lemma 3.8, there exists a se-

quence of triangles ∆i,j,0 = ∆i,j, ∆i,j,1, · · · , ∆i,j,nij= ∆i,j+1, such that each

pair of adjacent triangles are coconvex.Now, let us replace each pair (∆i,j, ∆i,j+1) by the sequence ∆i,j,0 = ∆i,j,

∆i,j,1, · · · , ∆i,j,nij= ∆i,j+1. Then for the new sequence which starts at ∆1

and ends at ∆2, any adjacent pair of triangles are coconvex.�

Combining Lemma 6.3 and Lemma 6.4, we have the following proposition.

Proposition 6.5. Let F : S(X,ω; Σ) → S(X ′, ω′; Σ′) be an isomorphism.Then either

(i) for every triangle on (X,ω; Σ), the induced affine homeomorphism be-tween triangles is orientation preserving, or

(ii) for every triangle on (X,ω; Σ), the induced affine homeomorphism be-tween triangles is orientation reversing.

Corollary 6.6. Let F : S(X,ω; Σ) → S(X ′, ω′; Σ′) be an isomorphism.Then any triangulation Γ of (X,ω; Σ) induces a homeomorphism fΓ between(X,ω; Σ) and (X ′, ω′; Σ′), such that

(1) fΓ(Σ) = Σ′, and(2) for any saddle connection γ ∈ Γ, fΓ(γ) = F (γ).

Proof. It follows from Theorem 5.1 and Proposition 6.5. �

The homeomorphism fΓ obtained in Corollary 6.6 is called the F -inducedhomeomorphism with respect to Γ. In the following, we prove that theisotopy class of fΓ relative to Σ and Σ′ is independent of the choices oftriangulations.

Proposition 6.7. Let F : S(X,ω; Σ) → S(X ′, ω′; Σ′) be an isomorphism.For any two triangulations Γ1 and Γ2, the F -induced homeomorphisms fΓ1

and fΓ2are isotopic.

Proof. By Proposition 6.2, it suffices to prove it for the case that Γ1 andΓ2 differ by an elementary move. Let Γ1 = {α1, γ2, γ3, γ4, γ5, · · · , γk} and


Γ2 = {β1, γ2, γ3, γ4, γ5, · · · , γk} such that γ2, γ3, γ4, γ5 bound a strictly con-vex quadrilateral Q on (X,ω; Σ) whose diagonals are α1, β1. Correspond-ingly, F (γ2), F (γ3), F (γ4), F (γ5) bound a strictly convex quadrilateral Q′ on(X ′, ω′; Σ′) whose diagonals are F (α1), F (β1). By construction, fΓ1

|X1\Q =fΓ2

|X1\Q. Notice that α1 and β1 divide Q into four triangles. Let fQ be the

piecewisely affine map from Q to Q′ whose restriction to each of these fourtriangles is affine. Let f12 : (X,ω; Σ) → (X ′, ω′; Σ′) be a homeomorphismsuch that f12|X1\Q = fΓ1

|X1\Q = fΓ2|X1\Q and f12|Q = fQ. Then both fΓ1

and fΓ2are isotopic to f12. Therefore, fΓ1

and fΓ2are isotopic.

�

Proof of Theorem 6.1. If (X,ω; Σ) is not a torus with one marked point,they every triple of saddle connections bound at most one triangle. Thenthe theorem follows from Corollary 6.6 and Proposition 6.7.

If (X,ω; Σ) is a tours with one marked point. Then every triangula-tion Γ consists of three saddle connections, which bound two triangles on(X,ω; Σ). Therefore, the F -induced homeomorphisms with respect to Γhas two choices, which results in two isotopy classes of homeomorphismssatisfying the condition in the theorem. �

7. Affine Homeomorphism

Our goal in this section is to prove the following proposition, which is thelast piece for proving the main theorem.

Proposition 7.1. Let f : (X,ω; Σ) → (X ′, ω′; Σ′) be a homeomorphismwhich induces an isomorphism f∗ : S(X,ω; Σ) → S(X ′, ω′; Σ′). Then f isisotopic to an affine homeomorphism.

Proof. Recall that a simple closed curve on a half-translation surface withmarked points is called a cylinder curve if it is isotopic (relative to markedpoints) to the core curve of some cylinder.

By Theorem 2.3, it suffices to prove that for each simple closed curve α,α is a cylinder curve on (X,ω; Σ) if and only if f(α) is a cylinder curve on(X ′, ω′; Σ′). Since f∗ is an isomorphism, it suffices to prove that if α is acylinder curve on (X,ω; Σ), then f(α) is a cylinder curve on (X ′, ω′; Σ′).

Let C be an cylinder on (X,ω; Σ) with core curve α. Let

I : (Q1Q2 · · ·QnQn+1 · · ·Qm) → (X,ω; Σ)

be an admissible map such that

• I(QnQn+1) = I(QmQ1) is an interior saddle connection of C;• I(QiQi+1) is a boundary saddle connection for 1 ≤ i ≤ n − 1 andn+ 1 ≤ i ≤ m− 1.

If C is a simple cylinder, it then follows from Proposition 4.3 that f(α) isa cylinder curve on (X ′, ω′; Σ′).

If C is not a simple cylinder, consider a triangulation of

(Q1Q2 · · ·QnQn+1 · · ·Qm).

36 HUIPING PAN

By Theorem 5.1, there exists an admissible map

I′ : (Q′1Q

′2 · · ·Q′

nQ′n+1 · · ·Q′

m) → (X ′, ω′; Σ′)

such that f∗(I(QiQj)) = I′(Q′iQ

′j) for all sides and diagonals QiQj.

To show that f(α) is a cylinder curve on (X ′, ω′; Σ′), it is equivalent to

show that Q′iQ

′i+1 and Q′

jQ′j+1 are parallel for any 1 ≤ i ≤ n − 1 and

n+ 1 ≤ j ≤ m− 1.Consider the quadrilateral (Q′

1Q′2Q

′n+1Q

′n+2). Let θ′1, θ

′2, θ′n+1, θ

′n+2 be

the interior angles at Q′1, Q

′2, Q

′n+1 and Q′

n+2 respectively. If Q′1Q

′2 and

Q′n+1Q

′n+2 are not parallel, then either θ′2 + θ′n+1 > π or θ′n+2 + θ′1 > π.

Without loss of generality, we may assume that θ′2 + θ′n+1 > π. It thenfollows from Lemma 3.4 that we can admissibly extend I′|(Q′

1Q′

2Q′

n+1Q′

n+2)

to a strictly convex pentagon (Q′1Q

′2Q

′n+1Q

′n+2B

′). Correspondingly, byTheorem 5.2, we can also admissibly extend I|(Q1Q2Qn+1Qn+2) to a strictlyconvex pentagon (Q1Q2Qn+1Qn+2B), which contradicts to the assumptionthat I(Q1Q2Qn+1Qn+2) is contained in the cylinder C whose boundary is

parallel to I(Q1Q2). Therefore, Q′1Q

′2 and Q′

n+1Q′n+2 are parallel. Similarly,

we can show that Q′iQ

′i+1 and Q′

jQ′j+1 are parallel for any 1 ≤ i ≤ n−1 and

n+ 1 ≤ j ≤ m− 1.�

Proof of Theorem 1.1. By Theorem 6.1, there exists a homeomorphism f :(X,ω; Σ) → (X ′, ω′; Σ′) such that for any saddle connection γ on (X,ω; Σ),f(γ) is isotopic to F (γ). It then follows from Proposition 7.1 that f is

isopotic to an affine homeomorphism f : (X,ω; Σ) → (X ′, ω′; Σ′). �

Proof of Theorem 1.2. Let Aut(S(X,ω; Σ)) be the automorphism group ofS(X,ω; Σ). Let Aff(X,ω; Σ) be the group of affine self-homeomorphismsof (X,ω; Σ). Since each affine self-homeomorphism of (X,ω; Σ) induces anautomorphism of S(X,ω; Σ), there exists a natural group homomorphism

F : Aff(X,ω; Σ) → Aut(S(X,ω; Σ).

By Theorem 1.1, F is surjective. Moreover, it follows from Theorem 6.1that if (X,ω; Σ) is a torus with one marked point, then F is two-to-one.Otherwise, F is injective.

�

8. Quotient graph

The goal of this section is to prove Theorem 1.3. Recall that for two

vectors ~a = (x1, y1) and ~b = (x2, y2) on R2, ~a ∧~b := x1y2 − x2y1. We start

with the following lemma.

Lemma 8.1. If (X,ω; Σ) is a translation surface which is not a torus withone marked point, then there are saddle connections α, β1, β2, · · · , such thatβi is disjoint from α for all i ≥ 1, and that limi→∞ |

∫αω ∧

∫βiω| = ∞.


γ+1 γ+2 γ+3 γ+4

γ−3 γ−1 γ−4 γ−2

Figure 11. The red region represents a simple cylinder deter-mined by γ±

1 and the saddle connection connecting the left end-points of γ±

1 .

Proof. It follows from [16, Theorem 2] that (X,ω; Σ) has infinitely manycylinders.

First, we claim that there exists a cylinder C0 such that the closure is aproper subset of (X,ω; Σ). Indeed, let C ⊂ (X,ω; Σ) be a cylinder. Withoutloss generality, we may assume that C is horizontal. If (X,ω; Σ) 6= C, theclaim follows. If (X,ω; Σ) = C, then for every saddle connection δ+ in theupper boundary component of C, there is a corresponding saddle connectionδ− in the lower boundary component, such that

∫γ+ ω =

∫γ− ω (see Figure

11). Let (γ+1 , γ−1 ) be such a pair, they determine a simple cylinder C1 as

illustrated in Figure 11. By assumption, (X,ω; Σ) is not a torus with onemarked point, each boundary component of C contains at least two saddleconnections. In particular, C1 is a proper subset of C = (X,ω; Σ).

Next, let α be a non-horizontal saddle connection on (X,ω; Σ)\C0, let{βi}i≥1 be a sequence of interior saddle connections of C0. Then βi is disjointfrom α for all i ≥ 1, and that limi→∞ |

∫αω ∧

∫βiω| = ∞.

�

Proof of Theorem 1.3. (i) Notice that every edge in the saddle connectiongraph is represented by a pair of disjoint saddle connections.

If (X,ω; Σ) is a torus with one marked point, the group Aff+(X,ω; Σ)of orientation-preserving affine homeomorphisms of (X,ω; Σ) is isomorphicto SL(2,Z), and the set of pairs of disjoint saddle connections has oneAff+(X,ω; Σ)-orbit. In particular, G(X,ω; Σ) has one vertex and one edge.

If (X,ω; Σ) is not a torus with one marked point, let {α1, β1}, {α2, β2}be two pairs of non-parallel, disjoint saddle connections on (X,ω; Σ). Theyrepresent two edges e1, e2 of S(X,ω; Σ). Suppose that there is an automor-phism F of S(X,ω; Σ) such that F (e1) = e2. By Theorem 1.1, there isan affine homeomorphism f of (X,ω; Σ) such that f({α1, β1}) = {α2, β2}.Therefore, |

∫α1

ω ∧∫β1

ω| = |∫α2

ω ∧∫β2

ω| 6= 0. On the other hand, by

Lemma 8.1, the set

{|∫

α

ω ∧∫

β

ω| : α, β are disjoint saddle connections}

is an infinite set. As a consequence, G(X,ω; Σ) has infinitely many edges.

38 HUIPING PAN

(ii) If (X,ω; Σ) is Veech surface, it follows from [24, Theorem 6.8] (seealso [22, Theorem 1.3]) that the set of triangles on (X,ω; Σ) has finitelymany Aff(X,ω; Σ)-orbits, where Aff(X,ω; Σ) is the group of affine homeo-morphisms of (X,ω; Σ). This implies that the set of saddle connections hasfinitely many Aff(X,ω; Σ)-orbits, since each saddle connection is containedin at least one triangle. In particular, G(X,ω; Σ) has finitely many vertices.

�

9. Questions

In this section, we propose two questions. The first question concernsTheorem 1.3.

Question 1. (i) Characterize those half-translation surfaces whose quotientgraph have finitely many edges. (ii) Is it true that the quotient graph hasfinitely many vertices if and only if the underlying half-translation surfaceis a Veech surface?

Remark 7. Let T(X,ω; Σ) be the spine tree defined by Smillie-Weiss (see[22, §4] for the definition). Suppose that G(X,ω; Σ) has finite vertices.Then the set of directions of saddle connections has finite Aff(X,ω; Σ)-orbits, which implies that the set of components of H2\T(X,ω; Σ) has finiteAff(X,ω; Σ)-orbits. To prove that (X,ω; Σ) is a Veech surface, it sufficesto prove that every component of H2\T(X,ω; Σ) has finite quotient area byAff(X,ω; Σ). This is equivalent to show that every saddle connection hasnon-trivial stabilizers in Aff(X,ω; Σ).

Irmak-McCarthy ([13]) proved that every injective simplicial map from anarc graph to itself is induced by some self-homeomorphism of the underlyingsurface (see also [1, 11, 12]). We may ask a similar question for the saddleconnection graph.

Question 2. Let (X,ω; Σ) be a half-translation surface with marked points.Is it true that every injective simplicial map F : S(X,ω,Σ) → S(X,ω; Σ) isinduced by some affine homeomorphism f : (X,ω; Σ) → (X,ω; Σ)?

References

[1] Aramayona J., Simplicial embeddings between pants graphs, Geom. Dedicata 144(2010), 115-128.

[2] Bobenko A. I. and B. A. Springborn, A discrete Laplace-Beltrami operator for sim-plicial surfaces, Discrete Comput. Geom. 38:4 (2007), 740-756.

[3] Disarlo V., Randecker A. and R. Tang, Rigidity of the saddle connection complex,arXiv:1810.00961.

[4] Duchin M., Leininger C. and K. Rafi, Length spectra and degeneration of flat metrics,Invent math. 182(2010), 231-277.

[5] Harer J. L., Stability of the homology of the mapping class groups of orientablesurfaces. Ann. of Math. (2) 121 (1985), no. 2, 215–249.

[6] Harer J. L., The virtual cohomological dimension of the mapping class group of anorientable surface. Invent. Math. 84 (1986), no. 1, 157–176.


[7] Hensel S., Przytycki P. and R. C. H. Webb, 1-slim triangles and uniform hyperbolicityfor arc graphs and curve graphs. J. Eur. Math. Soc. 17 (2015), no. 4, 755–762.

[8] Hilion A. and C. Horbez, The hyperbolicity of the sphere complex via surgery paths.J. Reine Angew. Math. 730 (2017), 135–161.

[9] Hubert P. and T. A. Schmidt, An introduction to Veech surfaces. Handbook of dy-namical systems. Vol. 1B, 501–526, Elsevier B. V., Amsterdam, 2006.

[10] Indermitte C., Th. M. Liebling, M. Troyanov,and H. Clemenon, Voronoi diagrams onpiecewise flat surfaces and an application to biological growth. Theor. Comput. Sci.263, 263-274 (2001).

[11] Irmak E., Injective Simplicial Maps of the Arc Complex on Nonorientable Surfaces,Algebr. Geom. Topol. 9 (2009) 2055-2077.

[12] Irmak E., Injective Simplicial Maps of the Complexes of Curves of NonorientableSurfaces, Topology Appl. 153 (2006), no. 8, 1309-1340.

[13] Irmak E. and J. D. McCarthy, Injective Simplicial Maps of the Arc Complex, TurkishJ. Math. 34 (2010), no. 3, 339-354.

[14] Kerckhoff S., Masur H. and J. Smillie, Ergodicity of Billiard Flows and QuadraticDifferentials, Ann. of Math., Vol. 124, No. 2 (1986), pp. 293-311.

[15] Masur H. and Y. Minsky, Geometry of the complex of curves I: Hyperbolicity, Invent.math. 138, 103-149 (1999).

[16] Masur H., Closed trajectories for quadratic differentials with an application to bil-liards. Duke Math. J. 53 (1986), no. 2, 307–314.

[17] Masur H. and S. Schleimer, The geometry of the disk complex. J. Amer. Math. Soc.26 (2013), no. 1, 1–62.

[18] Minsky Y. and S. J. Taylor, Fibered faces, veering triangulations, and the arc com-plex. Geom. Funct. Anal. 27 (2017), no. 6, 1450-1496.

[19] Nguyen D.-M., Triangulations and volume form on moduli space of flat surfaces,Geom. Funct. Anal. Vol. 20 (2010) 192-228.

[20] Nguyen D.-M., Translation surfaces and the curve graph in genus two. Algebr. Geom.Topol. 17 (2017), no. 4, 2177-2237.

[21] Nguyen D.-M., Veech dichotomy and tessellations of the hyperbolic plane, preprint,arXiv:1808.09329.

[22] Smillie J. and B. Weiss, Characterizations of lattice surfaces. Invent. Math. 180(2010), no. 3, 535-557.

[23] Tahar G., Geometric triangulations and flip, C. R. Math. Acad. Sci. Paris 357 (2019),no. 7, 620-623.

[24] Vorobets Y. B., Plane structures and billiards in rational polygons: the Veech al-ternative. (Russian) Uspekhi Mat. Nauk 51 (1996), no. 5(311), 3-42; translation inRussian Math. Surveys 51 (1996), no. 5, 779-817.

Huiping Pan, Department of mathematics, Jinan University, 510632, Guangzhou,

China

E-mail address: [email protected]

arxiv:1809.06248v3 [math.gt] 16 aug 2020

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