eriko hironaka and eiko kin- a family of pseudo-anosov braids with small dilatation

8/3/2019 Eriko Hironaka and Eiko Kin- A family of pseudo-Anosov braids with small dilatation

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Algebraic & Geometric Topology 6 (2006) 699738 699arXiv version: fonts, pagination and layout may vary from AGT published version

A family of pseudo-Anosov braids with small dilatation

ERIKO HIRONAKA

EIKO KIN

This paper describes a family of pseudo-Anosov braids with small dilatation. The

smallest dilatations occurring for braids with 3,4 and 5 strands appear in this

family. A pseudo-Anosov braid with 2g + 1 strands determines a hyperelliptic

mapping class with the same dilatation on a genus g surface. Penner showed thatlogarithms of least dilatations of pseudo-Anosov maps on a genus g surface grow

asymptotically with the genus like 1/g, and gave explicit examples of mappingclasses with dilatations bounded above by log 11/g . Bauer later improved thisbound to log 6/g . The braids in this paper give rise to mapping classes withdilatations bounded above by log(2+

3)/g. They show that least dilatations for

hyperelliptic mapping classes have the same asymptotic behavior as for general

mapping classes on genus g surfaces.

37E30, 57M50

1 Introduction

In this paper, we study a family of generalizations of these examples to arbitrary numbers

of strands. Let B(D, s) denote the braid group on D with s strands, where D denotes a2dimensional closed disk. First consider the braids m,n in B(D, m+ n+ 1) given by

m,n = 1 . . . m1m+1 . . .

1m+n.

Matsuokas example [22] appears as 1,1 , and Ko, Los and Songs example [18] as

2,1 . For any m, n 1, m,n is pseudo-Anosov (Theorem 3.9). The dilatations of m,mcoincide with those found by Brinkmann [7] (see also Section 4.2), who also shows that

the dilatations arising in this family can be made arbitrarily close to 1.

It turns out that one may find smaller dilatations by passing a strand of m,n once around

the remaining strands. As a particular example, we consider the braids m,n defined by

taking the rightmost-strand of m,n and passing it counter-clockwise once around the

remaining strands. Figure 1 gives an illustration of m,n and m,n . The braid 1,3 is

conjugate to Ham and Songs braid 123412 . For |m n| 1, we show thatm,n is periodic or reducible. Otherwise m,n is pseudo-Anosov with dilatation strictly

Published: 12 June 2006 DOI: 10.2140/agt.2006.6.699
http://www.ams.org/mathscinet/search/mscdoc.html?code=37E30,%2057M50http://dx.doi.org/10.2140/agt.2006.6.699http://dx.doi.org/10.2140/agt.2006.6.699http://www.ams.org/mathscinet/search/mscdoc.html?code=37E30,%2057M50


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700 Eriko Hironaka and Eiko Kin

m n m n

(a) (b)

Figure 1: Braids (a) m,n and (b) m,n

less than the dilatation of m,n (Theorem 3.11, Corollary 3.32). The dilatations of

g1,g+1 (g 2) satisfy the inequality(g1,g+1)

g < 2 +

3(1)

(Proposition 3.36).

Let Msg denote the set of mapping classes (or isotopy classes) of homeomorphismson the closed orientable genusg surface Fg set-wise preserving s points. We denote

M0g by Mg . For any subset Msg , define () to be the least dilatation amongpseudo-Anosov elements of, and let () be the logarithm of (). For the braid

group B(D, s), and any subset B(D, s), define () and () in a similar way. Bya result of Penner [25] (see also McMullen [23]), (Mg) 1g .An element of Mg is called hyperelliptic if it commutes with an involution on Fgsuch that the quotient of Fg by is S

2 . Let Mg,hyp Mg denote the subset ofhyperelliptic elements ofMg . Any pseudo-Anosov braid on 2g+ 1 strands determinesa hyperelliptic element of Mg with the same dilatation (Proposition 2.10). Thus, (1)implies:

Theorem 1.1 For g 2,

(

Mg)

(

Mg,hyp)

(

B(D, 2g + 1)) mm+1

, and hence also for t 1. Since Rm(1) = 2 < 0 and Rm(2) > 0, itfollows that Rm has a simple root m with 1 < m < 2. Similarly, we can show that

for t < 0, Rm has no roots for m even, and one root if m is odd. In the odd case,

Rm(1) = 0, so Rm has no real roots strictly less than 1.Lemma 3.29 The sequence (Rm) converges monotonically to 1 from above.

Proof Since M(Rm) = 2, we know that m = (Rm) > 1. Take any > 0. Let D be

the disk of radius 1+ around the origin in the complex plane. Let g(t) = t1t

and

hm(t) =2

tm+1. Then for large enough m, we have

|g(t)| = t 1t

> 2tm+1

= |hm(t)|for all t on the boundary of D , and g(t) and hm(t) are holomorphic on the complement

of D in the Riemann sphere. By Rouches theorem, g(t), g(t)+ hm(t), and hence Rm(t)

have the same number of roots outside D , which is zero.

To show the monotonicity consider Rm(m+1). Note that (m+1)m+1(t 1) 2 = 0.Hence we have

Rm(m+1) = (m+1)m(t 1) 2

= ((m+1)m (m+1)m+1)(t 1)

< 0.

Since Rm(t) is an increasing function for t > 1, we conclude that m+1 < m .

Algebraic & Geometric Topology 6 (2006)


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Corollary 3.23 and Lemma 3.29 imply the following.

Corollary 3.30 Fixing m 1, the sequences (m,n) and (m,n) converge to (Rm)as sequences in n. Furthermore, we can make (m,n) and (m,n) arbitrarily close to

1 by taking m and n large enough.

We now determine the monotonicity of (m,n) and (m,n) for fixed m 1.

Proposition 3.31 Fixing m 1, the dilatations (m,n) are strictly monotone decreas-ing, and the dilatations (m,n) are strictly monotone increasing for n

m + 2.

Proof Consider f(t) = (Rm)(t) = 2tm+1 t+ 1. Then, for t > 0,f(t) = 2(m + 1)tm 1 < 0.

Also f(1) = 2 < 0. Since bm,n = (m,n) > 1, and for n m+ 2, sm,n = (m,n) >1, it follows that (Rm)(bm,n) and (Rm)(sm,n) are both negative. We have

0 = Tm,n(bm,n) = (bm,n)n+1Rm(bm,n) + (Rm)(bm,n), and

0 = Sm,n(sm,n) = (sm,n)n+1Rm(sm,n) (Rm)(sm,n),

which imply that Rm(bm,n) > 0 and Rm(sm,n) < 0. Since Rm is increasing for t > 1,

we have

(8) sm,n < m < bm,n.

Plug bm,n into Tm,n1 , and subtract Tm,n(bm,n) = 0:

Tm,n1(bm,n) = (bm,n)n1Rm(bm,n) + (Rm)(bm,n)

= ((bm,n)n1 (bm,n)n)Rm(bm,n)

< 0

Since bm,n1 is the largest real root of Tm,n1 , we have bm,n < bm,n1 .

We can show that sm,n < sm,n+1 for n m+ 2 in a similar way, by adding the formulafor Sm,n(sm,n) to Sm,n+1(sm,n).

The inequalities (8) give the following.

Corollary 3.32 For all m, n 1 with |m n| 2, (m,n) > (m,n) .

We now fix 2g = m + n (g 2), and show that among the braids m,n and m,n ,g1,g+1 has the least dilatation.



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A family of pseudo-Anosov braids with small dilatation 729

Proposition 3.33 (1) For k = 1, . . . , m 1,(m,m) < (mk,m+k), and

(m,m+1) < (mk,m+k+1).

(2) For k = 2, . . . , m 1,(m1,m+1) < (mk,m+k), and

(m1,m+2) < (mk,m+k+1).

Proof Let = (m,m). Then plugging into Tmk,m+k gives

Tmk,m+k() = m+k+1(mk( 1) 2) 2mk+1 + 1

= 2m+2 2m+1 2m+k+1 2mk+1 + 1.Subtracting

0 = Tm,m() = 2m+2 2m+1 4m+1 + 1,

we obtain

Tmk,m+k() = 4m+1 2m+k+1 2mk+1 = 2mk+1(k 1)2 < 0.

Since (mk,m+k) is the largest real root ofTmk,m+k, wehave (m,m) < (mk,m+k).

The other inequalities are proved similarly.

Proposition 3.34 For m 2,(m,m) > (m1,m+1), and

(m,m+1) (m1,m+2)with equality if and only if m = 2.

Proof Let = (m1,m+1). Then Tm,m() = 2m+2 2m+1 4m+1 + 1.

Plugging in the identity

0 = Sm1,m+1() = 2m+2

2m+1

2m+2

+ 2m

+ 1,and subtracting this from Tm,m(), we have

Tm,m() = 2m+2 4m+1 2m 2 + 2 = 2m(2 2 + 1) + 2(1 ).

The roots oft2 2t+1 are 12. Since 12 < 1 < < 2 < 1+2, 2 2+1and 1 are both negative, and hence Tm,m() < 0. Since (m,m) is the largest realroot of Tm,m(t), it follows that (m1,m+1) = < (m,m).



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For the second inequality, we plug in = (m1,m+2) into Tm,m+1 . This gives

Tm,m+1() = 2m(3 2 1) 1.

Thus, 3 2 1 < 0 would imply Tm,m+1() < 0. The polynomial g(t) =t3 t2 t 1 has one real root ( 1.83929) and is increasing for t > 1. Since (Rm) isdecreasing with m, and < (R2) 1.69562 < 1.8 by (8), we see that Tm,m+1() < 0for m 3. For the remaining case, we check that T2,3 = S1,4 .

Propositions 3.33 and 3.34 show the following.

Corollary 3.35 The least dilatation among m,n and m,n for m+ n = 2g (g 2) isgiven by (g1,g+1) .

By Corollary 3.23, Lemma 3.29 and Proposition 3.31, the dilatations (m,n) for

n m+ 2 converge to 1 as m, n approach infinity. We prove the following strongerstatement, which implies Theorems 1.1 and 1.2.

Proposition 3.36 For g 2,log(2 +

3)

g + 1< log((g1,g+1)) 1, we have

2g 1 + 1

x2g 1

2(x2g 1) 2xg = 1

2(x2g 4xg + 1) =: p(x)

for x near . Thus, p(x) has a real root larger than . Using the quadratic formula

again, we see that g = 2 +

3, and hence g < g = 2 +

3.

4 Further discussion and questions

By Propositions 3.33 and 3.34, for s 5 strands, the minimal dilatations by ourconstruction come from g1,g+1 when s = 2g+ 1; and g1,g+2 when s = 2g+ 2.

For s even, there is an example of a braid with smaller dilatation than that of g1,g+2(see the end ofSection 4.1), but for s odd, we know of no such examples.

Since (B(D, 2g+ 1)) (Mg) (Proposition 2.10), Penners lower bound [25] forelements of (Mg) extend to (B(D, 2g + 1)). Hence we have

(B(D, 2g + 1)) (Mg) log212g 12 .

For g = 2, Zhirov shows [30] thatif M2 is pseudo-Anosov with orientable invariantfoliations, then () is bounded below by the largest root of x4 x3 x2 x+ 1. Fors = 5, 1,3 is pseudo-Anosov, and its lift to F2 is orientable. Our formula shows that

the dilatation of 1,3 is the largest root of Zhirovs equation, and hence 1,3 achieves the

least dilatation among orientable pseudo-Anosov maps on F2 . This yields the following

weaker version of Ham and Songs result [12], which doesnt assume any conditions on

the combinatorics of train tracks.

Corollary 4.1 The braid 1,3 is pseudo-Anosov with the least dilatation among braids

B(D; S) on 5 strands such that all singularities of S2 \ (S {p}) for the invariantfoliations associated to the pseudo-Anosov map are evenpronged.

We discuss the following general question and related work on the forcing relation in

Section 4.1.

Question 4.2 Is there a braid B(D, 2g + 1) such that () < (g1,g+1)?



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Let Ksg Msg be the subset of mapping classes that arise as the monodromy of a fiberedlink (K, F) in S3 , where the fiber F has genusg and the link K has s components.

Question 4.3 Is there a strict inequality (Msg) < (Ksg)?

In Section 4.2, we briefly discuss what is known about bounds on dilatations of pseudo-

Anosov monodromies of fibered links, and show how the braids m,n arise in this

class.

4.1 The forcing relation on the braid types

The existence of periodic orbits of dynamical systems can imply the existence of

other periodic orbits. Continuous maps of the interval give typical examples for such

phenomena. Boyland introduced the notion of braid types, and defined a relation on the

set of braid types to study an analogous phenomena in the 2dimensional case. Recall

that there is an isomorphism

B(D; S)/Z(B(D; S)) M(D; S).Let f: D D be an orientation preserving homeomorphism with a single periodicorbit

S. The isotopy class of f relative to

Sis represented by Z(

B(D;

S)) for some

braid B(D; S) by using the isomorphism above. The braid type ofSfor f, denotedby bt(S,f), is the conjugacy class [Z(B(D; S))] in the group B(D; S)/Z(B(D; S)). Tosimplify the notation, we will write [] for [Z(B(D; S))]. Let

bt(f) = {bt(P,f) | P is a single periodic orbit for f},and BT the set of all braid types for all homeomorphisms of D. A relation on BT isdefined as follows: For bi BT (i = 1, 2),

b2 b1 (For any f : D D, if b2 bt(f), then b1 bt(f)).We say that b2 forces b1 if b2 b1 . It is known that gives a partial order on BT(see Boyland [6] and Los [21]), and we call the relation the forcing relation.

The topological entropy gives a measure of orbits complexity for a continuous map of

the compact space (see Walters [29]). Let h(f) 0 be the topological entropy of f.For a pseudo-Anosov braid B(D; S), log(()) is equal to h(), which in turnis the least h(f) among all f with an invariant set S such that bt(S,f) = [] (seeFathiLaudenbachPoenaru [10, Expose 10]). One of the relations between the forcing

relation and the dilatations is as follows.



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Theorem 4.4 (Los [21]) Let 1 and 2 be pseudo-Anosov braids. If [2] [1]and [2] = [1], then (2) > (1) .

The forcing relation on braids m,n and m,n was studied by Kin [17].

Theorem 4.5 For any m, n 1,(1) [m,n] [m,n+1] ,(2) [m,n] [m+1,n] ,(3) [m,n]

[m,] if

m + 2, and

(4) [m,n] [m,] if n m + 2.

S1 R1 R R0 S0H(R0)

H(R1)

H(S0)

H(S1)

Figure 26: Smalehorseshoe map

The Smalehorseshoe map H : D D is a diffeomorphism such that the action ofHon three rectangles R0,R1 , and R and two half disks S0, S1 is given in Figure 26. Therestriction ofH to Ri (i = 0, 1) is an affine map such that H contracts Ri vertically and

stretches horizontally. The restriction ofH to Si (i = 0, 1) is a contraction map. Katok

showed [15] that any C1+ surface diffeomorphism ( > 0) with positive topological

entropy has a horseshoe in some iterate. This suggests that the Smalehorseshoe map is

a fundamental model for chaotic dynamics.

The set

=

nZ

Hn(R0 R1)

is invariant under H. Let 2 = {0, 1}Z , and

: 2 2(. . . w1 w0w1 . . .) (. . . w1w0 w1 . . .), wi {0, 1}

the shift map. There is a conjugacy K : 2 between the two maps H| : and : 2 2 as follows:

K : 2x (. . . K1(x)K0(x)K1(x) . . .),



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where

Ki(x) =

0 if Hi(x) R0,1 if Hi(x) R1.

If x is a period k periodic point, then the finite word (K0(x)K1(x) . . . Kk1(x)) is calledthe code for x. We say that a braid is a horseshoe braidif there is a periodic orbit for

the Smalehorseshoe map whose braid type is []. We define a horseshoe braid type in

a similar manner. For the study of the restricted forcing relation on the set of horseshoe

braid types, see the papers [8, 11] by de Carvalho and Hall.

R1 R0

ab

c

d

e

a b c d e

1 0

Figure 27: Periodic orbit with the code 10010 and its braid representative

The result by Katok together with Theorem 4.4 implies that horseshoe braids are relevant

candidates realizing the least dilatation. It is not hard to see that the braid type of the

periodic orbit with the code

1 0 . . . 0n1

1 0 . . . 0m

or 1 0 . . . 0n1

1 0 . . . 0 m1

1 (n m + 2)

is represented by [m,n](= [m,n]) (For the definition ofm,n , see the end ofSection 3.2).

Hence, m,n (n m+ 2) is a horseshoe braid. Figure 27 illustrates the periodic orbitwith the code 10010 and its braid representative.

For the case of even strands, there is a horseshoe braid having dilatation less than our

examples. In fact, the braid type of period 8 periodic orbit with the code 10010110 is

given by [= (123456)37], which satisfies () = 1.4134 . . . < (2,5) =

1.5823 . . ..

4.2 Fibered links

For a fibered link (K, F) with fibering surface F, the monodromy (K,F) : F Fis the homeomorphism defined up to isotopy such that the complement of a regular

neighborhood of K in S3 is a mapping torus for (K,F) . Define (K,F) to be the



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characteristic polynomial for the monodromy (K,F) restricted to first homology

H1(F,R). If K is a fibered knot, then (K,F) is the Alexander polynomial of K (see

Kawauchi [16] and Rolfsen [26]).

The homological dilatation of a pseudo-Anosov map : F F is defined to be (f),where f is the characteristic polynomial for the restriction of to H1(F;R). Thus, if

(K, F) is a fibered link and (K,F) is the monodromy, then ((K,F)) is the homological

dilatation of(K,F) . In the case where (K,F) is a pseudo-Anosov map, ((K,F)) and

((K,F)) are equal if (K,F) is orientable (see Rykken [27]).

Any monic reciprocal integer polynomial is equal to (K,F) for some fibered link (K, F)

up to multiples of (t 1) and t (see Kanenobu [14]). In particular, any reciprocalPerron polynomial1 can be realized. On the other hand, if(K,F) is orientable, then

((K,F)) is in general strictly greater than ((K,F)).

Leininger [20] exhibited a pseudo-Anosov map L : F5 F5 with dilatation L ,where

log(L) = 0.162358.

A comparison shows that this number is strictly less than our candidate for the least

element of (B(D, 2g + 1) for g = 5:log((4,6)) = 0.240965.

The pseudo-Anosov map L is realized as the monodromy of the fibered (2, 3, 7)pretzel knot. Its dilatation L is the smallest known Mahler measure greater than 1

among monic integer polynomials (see Boyd [5] and Lehmer [19]).

In the rest of this section, we will construct fibered links whose monodromies are

obtained by lifting the spherical mapping classes associated to m,n . We set g = m+n2 .Let Sbe the set of marked points on int(D) corresponding to the strands of m,n , andF the double covering of D, branched over S. Then F has one boundary component ifm+ n is even and two boundary components if m+ n is odd. Let m,n be the lift of

the pseudo-Anosov representative m,n of m,n M(D; S) to F. Using an argumentsimilar to that in the proof of Proposition 2.10, we have

(m,n) = (m,n ) = (m,n).

Note that m,n is 1pronged near each of the boundary of F if m + n is odd.

Let Km,n be the twobridge link given in Figure 28. By viewing (S3, Km,n) as the result

of a sequence of Hopf plumbings see Hironaka [13, Section 5], one has the following.

1A monic integer polynomial f is Perron if f has a root (f) > 1 such that (f) > || forall roots = (f).



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N

(N positive half twists)

(N negative half twists)

N

n

m+1

=

=

Figure 28: Twobridge link associated to m,n

Proposition 4.6 The complement of a regular neighborhood ofKm,n in S3 is a mapping

torus for m,n .

The fibered links Km,n and the dilatations ofm,n were also studied by Brinkmann [7].

Let m,n be the Alexander polynomial for Km,n . SalemBoyd sequences for m,nwere computed by Hironaka [13]. Proposition 3.18 implies the following.

Lemma 4.7 If m and n are both odd, then (m,n) = (m,n) = (Km,n ).

Question 4.8 Let m,n be the pseudo-Anosov representative of m,n . Is there afibered link K in S3 such that the complement of a regular neighborhood of K in S3 is

a mapping torus for a lift of m,n ?

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Department of Mathematics, Florida State University

Tallahassee FL 32306-4510, USA

Department of Mathematical and Computing Sciences, Tokyo Institute of Technology

2-12-1-W8-45 Oh-okayama, Meguro-ku, Tokyo 152-8552, Japan

[email protected], [email protected]

Received: 23 July 2005 Revised: 13 April 2006
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eriko hironaka and eiko kin- a family of pseudo-anosov braids with small dilatation

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