Eiko Kin and Mitsuhiko Takasawa- Pseudo-Anosov braids with small entropy and the magic 3-manifold

8/3/2019 Eiko Kin and Mitsuhiko Takasawa- Pseudo-Anosov braids with small entropy and the magic 3-manifold

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Pseudo-Anosov braids with small entropy and the magic 3-manifold

Eiko Kin and Mitsuhiko Takasawa

May 12, 2009

Abstract

We consider a surface bundle over the circle, so called the magic manifold M = Mmagic.This manifold has the smallest known volume among orientable hyperbolic 3-manifolds having3 cusps. We compute the entropy function on a fiber face H2(M,M;R). We determinehomology classes whose representatives are genus 0 fiber surfaces for M, and describe their

monodromies by braids. Among such homology classes whose representatives have n puncturesfor a given n, we decide which one realizes the minimal entropy. We show that for each n 6(resp. n = 3, 4, 5), there exists a Dehn filling of the magic manifold which gives a hyperbolicfibered manifold with an n-punctured disk fiber Dn having the monodromy n : Dn Dn suchthat n has the smallest known entropy (resp. the smallest entropy).

Keywords: mapping class group, pseudo-Anosov, entropy, hyperbolic volume, magic manifold

Mathematics Subject Classification : Primary 37E30, 57M27, Secondary 57M50

1 Introduction

Let M() be the mapping class group of an orientable surface of genus g with p punctures = g,p.Assuming that 3g 3 +p 1, elements ofM() are classified into three types: periodic, pseudo-Anosov and reducible [26]. There exist two natural numerical invariants of pseudo-Anosov mappingclasses . One is the entropy ent() which is the logarithm of the dilatation (). The otherinvariant is the volume vol() which comes from the fibration theorem by Thurston [27] whichasserts that a mapping class is pseudo-Anosov if and only if its mapping torus

T() = [0, 1]/ ,

where identifies (x, 1) with (f(x), 0) for some representative f of , is hyperbolic. We denote thethe volume ofT() by vol().

LetM

pA() be the set of pseudo-Anosov mapping classes ofM

(). Fixing the surface ,the dilatation () for MpA() is known to be an algebraic integer with a bounded degreedepending only on . The number of conjugacy classes of MpA() with dilatation bounded bysome constant is finite, see [12]. In particular the set

Dil() = {() > 1 | MpA()}

achieves its infimum (). We turn to the volume. The set

{v | v is the volume of a hyperbolic 3-manifold}The author is partially supported by Grant-in-Aid for Young Scientists (B) (No. 20740031), MEXT, Japan

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is a well-ordered closed subset ofR of order type . In particular any subset achieves its infimum.Let vol() be the minimum of {vol() | MpA()}. It is of interest to compute the minimum() (resp. vol()) and to determine the mapping class realizing the minimum. Another problemrelated to the minimal dilatation and the minimal volume can be stated as follows. For a non-negative integer c, we set

(; c) = min{() | MpA(), T() has c cusps},vol(; c) = min{vol() | MpA(), T() has c cusps}.

A problem is to compute (; c) (resp. vol(; c)) and to determine which mapping class realizesthe minimum.

In [13], the authors and Kojima provided experimental results concerning the entropy and thevolume. In the case where the mapping class group M(Dn) of the n-punctured disk Dn, theyobserved that for many pairs (n, c), there exists a mapping class simultaneously reaching both(Dn; c) and vol(Dn; c). Experiments tell us that in case c = 3, the mapping torus of the mappingclass reaching both two minimums is homeomorphic to the magic manifold Mmagic which is theexterior of the 3 chain link

C3 in the three sphere S

3 illustrated in Figure 1. Moreover for c = 2, itis observed that there exists a mapping class realizing both (Dn; 2) and vol(Dn; 2) and its mappingtorus T() is obtained from Mmagic by a Dehn filling along one cusp. Venzke also studied in thisdirection. He found a family of pseudo-Anosov mapping classes of the punctured disk having thesmall dilatation. He showed that these mapping tori are obtained from Mmagic by a Dehn fillingalong one cusp, see [28]. These studies motivate the present paper which concerns the fibrations inMmagic.

The magic manifold Mmagic is an intriguing example. It has the smallest known volume amongorientable hyperbolic 3-manifolds having 3 cusps. Many manifolds having at most 2 cusps withsmall volume are obtained from Mmagic by Dehn fillings, see [17]. For example, the figure eightknot exterior and the figure eight sister knot exterior have the smallest volume among orientablehyperbolic manifolds having a cusp [3]. The Whitehead link exterior and the Whitehead sister link,i.e. (2, 3, 8)-pretzel link exterior, have the smallest volume among orientable hyperbolic manifoldshaving 2 cusps [1]. In [17], Martelli and Petronio showed that the number of the exceptionalsurgeries for Mmagic is 5. It is not known that there exists any other hyperbolic manifold with atleast 3 cusps such that the number of exceptional surgeries is greater than or equal to 5. Recallthat for a hyperbolic 3-manifold with a torus boundary component, the distance between twoexceptional Dehn fillings is defined. The magic manifold provides many examples realizing thevarious maximal volumes of . For more details, see [9].

Let M be a hyperbolic 3-manifold with boundary M (possibly empty) which fibers over thecircle. We assume that M admits infinitely many different fibrations. In [25], Thurston introducedthe norm function XT : H2(M,M;R) R. He showed that the unit ball U with respect toXT is a compact, convex polyhedron. He described the relation between the function XT andfibrations of M as follows. For a fiber F of M, the homology class [F] H2(M,M;R) liesin the open cone int(C) with the origin over a top dimensional face of U. Conversely forany (primitive) integral class a of int(C), there exists a (connected) fiber F of M representinga. Thanks to this description, one can define the entropy function ent() : int(C) R. Moreprecisely, for each primitive integral class a int(C), the monodromy : F F on a connectedsurface F representing a must be pseudo-Anosov. One defines the entropy ofa by ent(a) = ent() =log((a)). This function defined on primitive integral classes admits a unique continuous extensionto the function on int(C). One verifies that the set

{a int(C) | a fiber representing a is homeomorphic to g,p}

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is finite for each g 0 and p 0.The mapping class group M(Dn) is isomorphic to the subgroup of M(0,n+1) consisting of the

elements which fix the (n+1)st puncture of 0,n+1. By using the natural surjective homomorphism from the n-braid group Bn to M(Dn), one can represent each element ofM(Dn) by an n-braid.A braid b is called pseudo-Anosov if (b) is a pseudo-Anosov mapping class. If this is the case, the

dilatation (b) of b is defined by the dilatation of (b).Let us define an m-braid Tm,p for m 3 which is a main example in this paper as follows.

Tm,p = (2123 m1)p2m1.

This is a horseshoe braid if gcd(p,m 1) = 1 and 1 < p m12 (Proposition 4.15). Moreover ifgcd(m1, p) = 1, then the mapping torus T((Tm,p)) is homeomorphic to Mmagic (Corollary 3.27).We set

Mnmagic = { M(0,n) | T() is homeomorphic to Mmagic}.Let f(x,y,z) be an integral polynomial

tx+y

z

tx

ty

txz

tyz

+ 1.

The following our main theorem states that for all n but 6 and 8, the minimum among the dilatationsof Mnmagic is realized by (Tn1,p) for some p = p(n) and its dilatation can be computed byusing the polynomial f(x,y,z).

Theorem 1.1. For each n 4, the minimum among the dilatations of Mnmagic is realized by:(1) the mapping class (T2k,2) in case n = 2k + 1 for k 2. The dilatation (T2k,2) equals the

largest real root of f(k1,k,0)(t) = t2k1 2(tk1 + tk) + 1.(2-i) the mapping class (1

2234) in case n = 6. The dilatation (1

2234) 2.08102 equals

the largest real root of f(3,2,1)(t) = t4

t3

2t3

t + 1.(2-ii) the mapping class (T4k+1,2k1) in case n = 4k + 2 for k 2. The dilatation (T4k+1,2k1)

equals the largest real root of f(2k+1,2k1,0)(t) = t4k 2(t2k1 + t2k+1) + 1.(3a-i) the mapping class (T3,1) in case n = 4. The dilatation (T3,1) 3.73205 equals the largest

real root of f(1,1,0)(t) = t2 4t + 1.

(3a-ii) the mapping class (T8k+3,2k+1) in case n = 8k + 4 for k 1. The dilatation(T8k+3,2k+1)equals the largest real root of f(4k1,4k+3,0)(t) = t8k+2 2(t4k1 + t4k+3) + 1.

(3b-i) the mapping class (b) in case n = 8, where

b = 11 12 13 14 15 16 11 12 13 14 11 12 13 B7.The dilatation (b) 1.72208 equals the largest real root of f(5,3,2)(t) = t6 t5 2t3 t + 1.

(3b-ii) the mapping class (T8k+7,2k+1) in case n = 8k + 8 for k 1. The dilatation (T8k+7,2k+1)equals the largest real root of f(4k+5,4k+1,0)(t) = t

8k+6 2(t4k+1 + t4k+5) + 1.The above mapping class realizing the minimal dilatation amongMnmagic is unique up to conjugacy.

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Figure 1: 3 chain link C3.

Forgetting the 1st strand of the braid Tm,p, one obtains the (m 1)-braid, say Tm,p. A familyof braids Tm,p contains the braids with the smallest dilatation. In fact,

(D3) = (T4,1) (see[19]),

(D4) = (T5,1) (see[15]) and

(D5) = (T6,2) (see[10]).

It is an open problem to determine the n-braids with the smallest dilatation for n 6. In [11]Hironaka-Kin discovered candidates with the smallest dilatation (Dn) for n odd. In [28] Venzkediscovered candidates with the smallest dilatation (Dn) for n even. We point out that all thebraids in Theorem 1.1(1)(2-ii)(3a-ii)(3b-ii) relate to such candidates. The braid T2k,2 (with oddstrands) is conjugate to the braid, say (k) with the smallest known dilatation found by Hironaka-Kin. (See Theorem 1.1(1).) For the braid Tm,p (with even strands) obtained from Tm,p in (2-ii),(3a-ii) or (3b-ii) of Theorem 1.1, the mapping class (Tm,p) is conjugate to the one given by Venzke.(See Section 4.1.1.)

Let T(n1) Bn1 be the braid given in Theorem 1.1 such that the mapping class ( T(n1))reaches the minimum among the dilatations of Mnmagic. Let T(n1) Bn2 be the braid

obtained from T(n1) by forgetting the 1st strand. As a corollary of Theorem 1.1, we obtain:Theorem 1.2. For eachn 5, there exists a Dehn filling of Mmagic which gives a hyperbolic fiberedmanifold with an (n 2)-punctured disk fiber, and with a monodromy given by (T(n1)).

In [24], Penner gave the bound of the minimal entropy of the closed surface of genus g:

log2

12g 12 log((g,0)) log 11

g.

The upper bound have been improved by several authors. The best known upper bound is log(2+

3)g

given by Minakawa [20] and Hironaka-Kin [11] independently. To show this upper bound, Hironaka-

Kin constructed a pseudo-Anosov homeomorphism on the closed surface of genus g 2 which isa lift of a pseudo-Anosov homeomorphism on the (2g + 2)-punctured sphere, represented by the(2g + 1)-braid (g+1). As a result, the dilatation of such pseudo-Anosov homeomorphism on g,0is the same as the one for (g+1), and it is the largest real root of t

2g+1 2tg+1 2tg + 1. As acorollary, |(g,0)| log((g,0)) has the upper bound 2 log(2 +

3).

By using the entropy function on a fiber face for Mmagic, we show:

Theorem 1.3. For each g 2, there exists a Dehn filling ofMmagic which gives a hyperbolic fiberedmanifold with a closed fiber g,0 of genus g, and with a monodromy g : g,0 g,0 such that thedilatation of g = [g] is less than or equal to the largest real root of t

2g+1 2tg+1 2tg + 1.

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We do not know that the pseudo-Anosov homeomorphism g is conjugate to the one given byMinakawa or Hironaka-Kin.

In [5], Farb, Leininger and Margalit showed that there exists a complete, noncompact, finitevolume, hyperbolic 3-manifold M with the following property: there exist Dehn fillings on Mgiving an infinite sequence of fiberings over S1, with closed fibers gi,0 of genus gi with gi ,and with monodromy i so that (gi,0) = (i). The magic manifold could be a potential examplewhich satisfies this property.

This paper is organized as follows. Section 2 reviews basic facts. Section 3 contains the proof ofTheorem 1.1. For the proof, we first compute the entropy function for Mmagic (Theorem 3.2). Thenwe determine the homology classes whose representatives are genus 0 fiber surfaces (Corollary 3.8).We study the asymptotic behaviors of the images of these homology classes under the functionP, where P = XT()ent() (Theorem 3.10). This tells us that which class realizes the minimaldilatation among homology classes whose representatives are genus 0 fiber surfaces with n punctures(Proposition 3.11). We finally describe the monodromies for such fiber surfaces by using braids(Propositions 3.28 and 3.31). In Section 4, we discuss pseudo-Anosovs with small entropy, and weshow Theorem 1.3. We find a relation between the horseshoe map and the braids Tm,p.

Acknowledgments: We would like to thank Shigeki Akiyama and Takuya Sakasai who showed usthe proof of Proposition 3.6. We also thank Sadayoshi Kojima and Makoto Sakuma for a greatdeal of encouragement.

2 Notation and basic fact

2.1 Mapping class group

The mapping class group M() is the group of isotopy classes of orientation preserving homeo-morphisms of , where the group operation is induced by composition of homeomorphisms. Anelement of the mapping class group is called a mapping class.

A homeomorphism : is pseudo-Anosov if there exists a constant = () > 1 calledthe dilatation of and there exists a pair of transverse measured foliations Fs and Fu such that

(Fs) = 1

Fs and (Fu) = Fu.

A mapping class which contains a pseudo-Anosov homeomorphism is called pseudo-Anosov. Wedefine the dilatation of a pseudo-Anosov mapping class , denoted by (), to be the dilatation ofa pseudo-Anosov homeomorphism of . One verifies that () does not depend on the choice of arepresentative.

We recall the surjective homomorphism from the n-braid group Bn to the mapping class group

M(Dn)

: Bn M(Dn)i ti

which sends the Artin generator i for i {1, , n 1} (see Figure 2(left)) to ti, where ti isthe mapping class which represents the positive half twist about the arc from the ith punctureto the (i + 1)st puncture. The kernel of is the center of Bn which is generated by a full twistbraid (12 n1)n. By replacing the boundary of Dn with the (n + 1)st puncture, the injectivehomomorphism from M(Dn) to M(0,n+1) is induced. In the rest of the paper we regard anelement of M(Dn) as an element ofM(0,n+1).

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We say that a braid b Bn is pseudo-Anosov if (b) M(Dn) is pseudo-Anosov. When this isthe case, vol((b)) equals the hyperbolic volume of the link complement S3 \ b in the 3sphere S3,where b is a union of the closed braid of b and the braid axis (see Figure 2(right)).

Figure 2: (left) generator i. (right) braided link b.

2.2 Root of polynomial

Let f(t) be an integral polynomial of degree d. The polynomial f(t) = tdf(1/t) is called thereciprocal polynomial for f(t). The maximal absolute value of roots of f(t) is denoted by (f).

Let R(t) be a monic integral polynomial and let S(t) be an integral polynomial. We set

Qn,(t) = tnR(t) S(t)for each integer n 1. In case where S(t) = R(t), we call tnR(t)R(t) the Salem-Boyd polynomialassociated to R(t).

The following lemma shows that roots ofQn,(t) outside the unit circle are determined by thoseof R(t) asymptotically.

Lemma 2.1. Suppose that R(t) has a root outside the unit circle. Then, the roots of Qn,(t)outside the unit circle converge to those of R(t) counting multiplicity as n goes to . In particular,

(R) = limn(Qn,).

The proof of Lemma 2.1 can be found in [14, Lemma 2.5].The following proposition will be used for the computation of the dilatation.

Proposition 2.2. Letf(t) = c1tn1c2tn2 cjtnjcj+1 be a polynomial with(j +1) monomials

so that j 2, n1 > n2 > > nj > 0, c1, c2, , cj+1 R and c1, , cj > 0. (cj+1 may be non-positive.) If f(q) < 0 for some q

0, then there exists a unique real root p of f such that p > q.

Proof. We prove the proposition by an induction on the number of monomials. Let j = 2. Supposethat f(t) = c1t

n1 c2tn2 c3 satisfies f(q) < 0 for some q 0. Then the derivative f(t) equalstn21(c1n1tn1n2 c2n2), and we see that f(t) has a unique real root q = n1n2

c2n2c1n1

so that

f(x) < 0 for 0 < x < q and f(x) > 0 for x > q. Hence f|(0,q) is monotone decreasing andf|(q,) is monotone increasing. Since f(q) < 0 and f() = , the polynomial f must have aunique real root p > q.

Assuming the proposition holds up to j, we consider the polynomial

f(t) = c1tn1 c2tn2 cjtnj cj+1.

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Thenf(t) = tnj1(c1n1tn1nj c2n2tn2nj cj1nj1tnj1nj cjnj).

Set the polynomial

h(t) = c1n1tn1nj

c2n2t

n2nj

cj1nj1tnj1nj

cjnj

with j monomials. Since h(0) = cjnj < 0, the polynomial h has a unique root ph > 0 by theassumption. Because h(x) < 0 for 0 < x < ph and h(x) > 0 for x > ph, f|(0,ph) is monotonedecreasing and f|(ph,) is monotone increasing. Since f(q) < 0 for q 0 and f() = , thepolynomial f must have a unique real root p > q. This completes the proof.

2.3 Hyperbolic surface bundle over the circle

Let M be a compact, irreducible and oriented 3-manifold with the boundary M (possibly empty).Thurston introduces the pseudo-norm function XT : H2(M,M;R) R in [25]. In case where Mis a surface bundle over the circle, he describes a relation between the function XT and fibrationsof M (Theorem 2.3).

2.3.1 Thurston norm

We recall the pseudo-norm function XT : H2(M,M;R) R. For a connected surface F, let usdefine

(F) = max{0, (F)},where (F) is the Euler characteristic of F. For a disconnected surface F, let us define

(F) =

(Fi),

where the sum is over all connected components Fi of F.

First, the function XT is defined on integral classes a H2(M,M;R) as follows.XT(a) = min{(F) | F is an embedded surface in M such that a = [F]},

where [F] is the homology class of F. A surface F is a minimal representative of a if [F] = a andXT(a) = (F). The following lineality and convexity hold, see [25].

(a) XT(ka) = |k|XT(a) for an integer k Z.(b) XT(a + b) XT(a) + XT(b) for integral classes a and b.The claim (b) can be shown as follows. Let A (resp. B) be an oriented embedded surface in Mwhich is a minimal representative of a homology class a (resp. b). We define an oriented surfaceA + B embedded in M such that [A + B] = a + b as in Figure 3. Then we have

XT(a + b) (A + B) = (A) + (B) = XT(a) + XT(b).For a rational class a H2(M,M;R), we take a rational number r Q so that ra is an integralclass. Then we define

XT(a) =1

|r|XT(ra).

From (a),(b), the function XT on rational classes admits a unique continuous extension to XT :H2(M,M;R) R which is linear on the ray though the origin. The function XT is indeed a

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Figure 3: surface A + B. (arrows indicate the normal direction of surfaces.)

pseudo-norm function, for M may contain essential tori representing nonzero homology classes.If M is a hyperbolic 3-manifold, then XT becomes a norm function. The unit ball U = {a H2(M,M;R) | XT(a) 1} is a compact, convex polyhedron if XT is a norm function , seeciteThurston1. In the rest of the paper, we assume that XT is a norm function.

The following notations are needed to describe how fibrations of M are related to the Thurstonnorm.

A top dimensional face in the boundary U of the unit ball U is denoted by , and its openface is denoted by int().

The open cone with the origin over is denoted by int(C). The set of integral classes ofint(C) is denoted by int(C(Z)), and the set of rational classes

of int(C) is denoted by int(C(Q)).

Theorem 2.3 ([25]). Suppose that M is a surface bundle over the circle. Let F be a fiber offibrations ofM. Then there exists a top dimensional face satisfying the following.

(1) [F] int(C(Z

)).(2) For any a int(C(Z)), a minimal representative E of a is a fiber of fibrations of M.The face as in Theorem 2.3 is called the fiber face. Theorem 2.3 tells us the followings. Supposethat M is a surface bundle over the circle. Then a int(C(Z)) is a primitive integral class if andonly if a minimal representative E of a is connected.

Remark 2.4. It is known that if a homology class a H2(M,M;Z) has a representative F whichis a fiber of fibrations of M, then any incompressible surface which represents a is isotopic to the

fiber F. In particular F is the minimal representative of a, see [25].

2.3.2 Entropy function

Let M be a surface bundle over the circle. Suppose that M is hyperbolic. We fix a fiber face for M. The entropy function ent() : int(C(Q)) R is defined as follows. If an integral classa int(C(Z)) has a connected minimal representative F, then F is a fiber of fibrations of M.Let : F F be a monodromy. Since M is a hyperbolic manifold, the isotopy class = []must be pseudo-Anosov. The entropy ent(a) is defined as the entropy of . The dilatation (a)can be defined in the same way. For a rational number r Q and an integral class a as above,the entropy ent(ra) is defined by 1|r|ent(a). Notice that the product of the Thurston norm and theentropy XT()ent() : int(C(Q)) R is constant on the ray through the origin.

The following theorem says that the entropy function has a nice property.

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Theorem 2.5 (Matsumoto [18], McMullen [21]). The function

1

ent() : int(C(Q)) R

is a strictly concave function. Therefore, ent(

) admits a unique continuous extension to ent(

) :

int(C) R.We set P= XT()ent() : int(C) R. Since ent(a) as a goes to a homology class in theboundary (see [6]), Theorem 2.5 implies:

Corollary 2.6. The function P : int(C) R has the minimum at a unique ray through theorigin.

In other words ent() has the minimum at a unique point on the open face int().Problem 2.7. On which ray does it attain the minimum of P : int(C) R? Is the minimumalways attained at a rational point?

In Section 3.2, we solve this problem for the magic manifold.

2.4 Affine lamination

We discuss how to compute the entropy for a given homology class. The entropy for a pseudo-Anosov mapping class can be computed as follows. A smooth graph , called a train track anda smooth graph map : are associated with . The edges of are classified into two types,real edges and infinitesimal edges. The transition matrix Mreal() with respect to real edges canbe computed. Then the dilatation () equals the spectral radius of Mreal(). For more details,see [2].

Fried and Oertel study the property of the entropy function on each fiber face. They show that

there exists a polynomial, depending on the fiber face, which organize the entropy of each homologyclass. The main ingredient for the study by Oertel is affine laminations which we review in thissection.

Let M be a compact n-dimensional manifold. A lamination is a codimension 1 closed subset ofM. We say that a lamination L has a transverse affine structure if there exists a transverse measure for the lift L ofL to the universal cover M ofM such that for any covering translation , we have() = f() for some f() R, where () denotes the pull-back of. Note that f should be ahomomorphism from 1(M) to R. If one chooses a number > 0 as a base, g = log f is an elementof Hom(1(M),R) H1(M;R). An affine lamination (L, ) is a lamination L together with thetransverse affine structure represented by the measure on L. Iff is an associated homomorphismfor the affine lamination (

L, ) and if g = log f

H1(M;R), then is called the dilatation (or

stretch factor) at the cohomology class g of (L, ). We may denote an affine lamination (L, ) byL if there exists no confusion.A branched manifoldB M of dimension m is a space with smooth structure modeled locally as

shown in Figure 4 in case m = 1, 2. The branch locus B of B is the union of points of B whose anyneighborhood can not be manifold. The sectors are the completions of the components of B \ B.A fibered neighborhood N(B) of B is a closed regular neighborhood of B in M which is foliated byintervals as fibers. A lamination L is carried by B if L can be embedded in N(B) transverse tofibers of N(B). A lamination L is fully carried by B ifL intersects every fibers of N(B). A vectorwhich assigns a real weight wi to each sector Zi ofB is an invariant measure on B if all the weightssatisfy the branch equations as shown in Figure 4(right).

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Figure 4: (left) Case: m = 1. (center) Case: m = 2. (right) branch equation w1 + w2 = w3 in casem = 1.

We now define a broken invariant measure on B which provides us a method to study affinelaminations carried by B. Let {F1, F2, , Fk} be a set of transversely oriented codimension 1submanifolds representing cohomology classes in M. Let (t, w) = (t1, t2, , tk; w1, w2, , ws)be a set of non-negative numbers. We say that (t, w) is a broken (invariant) measure on B (forF1, , Fk}) if it satisfies the followings.(1) Each ti is positive and assigned to Fi.

(2) Each wj is assigned to one of the components of B \ branch locus B (ki=1Fi).(3) The vector w = (w1, , ws) satisfies the branch equations at the branch loci.(4) The weight w+ on the plus-side of Fi equals tiw, where w is the weight on the minus-side

of Fi.

Proposition 2.8 ([22]). Each broken measure on B represents an affine lamination. Conversely,if L is an affine lamination carried by B and if F1, , Fk represent a basis for H1(M;R), thenthere exists a broken measure on B for {F1, , Fk} which represents L.

Affine laminations are obtained from pseudo-Anosov mapping classes. For a pseudo-Anosovhomeomorphism : F F, let M = F I/ be the mapping torus of , where identifies (x, 1)with ((x), 0). Let (, ) be the stable lamination of . The 2-dimensional lamination L Mare constructed from I F I by gluing {1} to {0} using . We say that L is thesuspended stable lamination for (associated with [F]). We now construct a branched surface Bwhich carries L. Let be an invariant train track for which carries the stable lamination. Then(, ) yields an invariant measure v on . The image () can be isotoped so that it transversesto fibers in the fibered neighborhood N(). The measure v on pushes forward to the measurev on (), and the measure v on gives a measure 1v on , where = (). We choose afamily of train tracks {t}0t1 from 0 = () to 1 = which is obtained from () by pinchingmore and more with increasing time. The family {t} yields a branched surface B F I. Bygluing 0 = () and 1 = using , we have a branched surface B in M which carries

L. Such

construction gives a broken measure on B having the break at the fiber F with t = . We can seethat the suspended lamination L for is an affine lamination which is represented by this brokenmeasure. Notice that the dilatation of L at the class [F] equals the dilatation for .Theorem 2.9 (Fried [6], Long-Oertel [16]). Let E and F be two fibers of M such that [E], [F] int(C) for some fiber face . Then the suspended stable lamination associated with [E] is isotopicto the one associated with [F].

Topologically, the suspended stable lamination is unique on each fiber face. By using Theorem 2.9,Fried shows:

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Figure 5: axis L for periodic map

Theorem 2.10 (Fried [6]). Let M be a hyperbolic manifold which fibers over the circle. Supposethat connected, orientable surfaces F1, , Fk represent a basis over Z of the integral lattice inH1(M;R). Then a given fiber face , there exists a polynomial P = P Z[t1, . . . , tk] such thatthe dilatation of a = x1[F1] + + xk[Fk] int(C(Q)) satisfies P(x1 , , xk) = 0.Oertel gives a recipe for computing the polynomial P = P. For more details, see [22, Section 2].

3 Magic manifold

3.1 Fiber face

The 3 chain link C3 has a symmetry in the following sense. Let L be one of the four lines in S3depicted in Figure 5(1),(2),(3) and (4). For each L, there exist an integer n and a periodic mapf : (S3, C3) (S3, C3) such that f is a 2/n rotation with respect to L. Such symmetry of C3reflects the shape of the unit ball with respect to the Thurston norm for the magic manifold. LetK, K and K be the components ofC3 such that K (resp. K, K) bounds the twice-punctureddisk F (resp. F, F) embedded in the exterior E(

C3) of

C3, see Figure 6(right). Let = [F],

= [F], and = [F]. In [25], Thurston computes the unit ball U for the magic manifold which isthe the parallelepiped with vertices = (1, 0, 0), = (0, 1, 0), = (0, 0, 1), ( + + ),see Figure 6(left). Note that the set {,,} is a basis of H2(Mmagic, Mmagic;Z).

Figure 6: (left) unit ball for magic manifold. (right) F, F and F.

The magic manifold is a hyperbolic surface bundle over the circle. The symmetry of the unitball tells us that every top dimensional face is a fiber face. We pick a shaded fiber face, say asin Figure 7(left) with vertices = (1, 0, 0), + + = (1, 1, 1), = (0, 1, 0) and = (0, 0, 1).We will consider the entropy function on int(C) in Section 3.2. The open face int() is written

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byint() = {x + y + z | x + y z = 1, x > 0, y > 0, x > z, y > z}. (3.1)

Figure 7: (left) fiber face . (center) 1 . (right) C1 .

The following lemma is needed to know the topological type of the minimal representative of agiven homology class.

Lemma 3.1. Letx +y+z be an integral homology class in int(C) (not necessarily primitive).Then we have the followings.

(1) XT(x + y + z) = x + y z.(2) Let F be an embedded surface (not necessarily minimal representative) in the magic manifold

such that [F] = x + y + z int(C(Z)). Then the number of the boundary components(F) is equal to the sum of the three greatest common devisors

gcd(x, y + z) + gcd(y, z + x) + gcd(z, x + y),

where gcd(0, w) = gcd(w, 0) is defined by w.

Proof. The claim (1) follows from (3.1). It is enough to compute the number of the boundarycomponents of a representative xF + yF+ zF to show the claim (2), since it does not dependon the choice of a representative. We put F(x,y,z) = xF + yF+ zF. Suppose that z = 0. It isnot hard to see that

(F(x,y,z)) = gcd(x, y + z) + gcd(y, z + x) + gcd(z, x + y),

since x boundary components of the surface xF, each of which is parallel to the longitude of K,contributes gcd(x, y +z) boundary components ofF(x,y,z) lying on the regular neighborhood N(K)of K, etc.

Suppose that z = 0. Then we have

(F(x,y,0)) = gcd(x, y + 0) + gcd(y, x + 0) + x + y = 2 gcd(x, y) + x + y.

The reason is that x boundary components of xF, each of which is parallel to the meridian ofK, survive as x boundary components of F(x,y,0). Similarly, y boundary components of yF, eachof which is parallel to the meridian of K, survive as y boundary components of F(x,y,0). Thiscompletes the proof.

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3.2 Fried-Oertel polynomial

By using a recipe given by Oertel, we compute a polynomial P. We set

P(t1, t2, t3) = t1 t2 + t3 + t1t2 t1t3 t2t3.

Theorem 3.2. The dilatation of x + y+ z int(C(Z)) is the largest real root of P(tx, ty, tz).In particular, the dilatation of x + y int(C(Z)) is the largest real root of P(tx, ty, t0) =tx+y 2(tx + ty) + 1.Proof of Theorem 3.2. First, we find the polynomial P(t1, t2, t3) such that the dilatation =(x + y + z) satisfies P(x, y, z) = 0. As we will see in Proposition 3.28(3a), Mmagic has a3-punctured disk fiber F = F+ with a monodromy + = 2

11 2 which is conjugate to T3,1 =

2112 . Let = + be a pseudo-Anosov homeomorphism representing +. Figure 8 shows

Mmagic = FI/ and representatives F, F and F of generators , and in Mmagic = FI/.We assign a positive real number t1, (resp. t2, t3) for F, (resp. F, F).

Figure 8: (left) F. (center) F. (right) F. (Arrows indicate the normal direction for these

surfaces.)

Let B be a branched surface fully carrying the suspended stable lamination . We find the spaceof broken measures R(B) for B.

The branched surface B can be described as a 1-parameter family of train tracks i each embed-ded in a fiber Fi (see Section 2.4). Figure 9 shows train tracks i for i = 1 > i1 > i2 > i3 > i4 > 0.A broken measure (t1, t2, t3; w1, w2, , ws) is assigned to each train track in order to describe thebroken measure on B. Two weights w1, w2 are sufficient to determine the broken measure on = 1 and other weights are determined by branch equations, see Figure 10. To determine theweight on 0 = (1), we need to describe the intersection of B with F, F and F. Figure 11

shows boundaries of F, F and F (as solid lines) together with orbits (as broken lines) of pointsp1 and p2 as in Figure 9. Since the orbit of p1 crosses F once and crosses F twice in the reverseorientation, the weight w4 in the bottom face is equal to

t2t23

w1, see Figures 11 and 12. Considering

the orbit of p2, we see that the weight w3 is equal to t1w2. Figure 12 shows the broken measure on0, which must be the same as the one on 1. Equating weights on two sectors, we get the followingequations.

t1t2(1+t3)

t3 1

t1(1+t3)t3

1 t2t23

+ t2(1+t3)t3

w2w1

=

00

.

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1

p1 p2

2

i1

p1 p2

11

i2

p1p2

2

i3

p2 p1

isotopy (rotation in theneighborhood near punctures)

i4

p2 p1

isotopy

0

p2 p1

Figure 9: 1, i1, i2, i3, i4 and 0.

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(1+t3)(t2w1+t1w2)t1+t2t3

w1

w1t3

w2t3

t2

t1

t3

(1 + 1t3)w1

(1 + 1t3)w2

t1t2

(1 + 1t3)w2

w2

1t2

(1 + 1t3)w2

Figure 10: weights on 1. (Broken lines show intersections between the fiber and F, F and F.)

p1

i1

i2 i3

1

0 i4

p1p2

p2

Figure 11: F, F, F and orbits of p1 and p2.

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w4 =t2t23w1

t3

t1

t2

t2(1+t3)t3

w1

t1(1+t3)t3

w2

w3 = t1w2

Figure 12: weights on 0.

Since B carries affine laminations, this system must have a nontrivial solution for some t =(t1, t2, t3). The condition for the existence of a solution is that the determinant of the matrix

t1t2(1+t3)

t3 1

t1(1+t3)t3

1 t2t23

+ t2(1+t3)t3

is equal to zero. Namely,

t1 t2 + t3 + t1t2 t1t3 t2t3t3 = 0

which describes the solution variety containing R(B). Since t3 is positive, the broken measure(t1, t2, t3; w1, w2) must satisfy P(t1, t2, t3) = 0, where

P(t1, t2, t3) = t1 t2 + t3 + t1t2 t1t3 t2t3.Thus, satisfies P(x, y, z) = 0.

Next, we show that is equal to the largest real root of

P(tx, ty, tz) = tz(tx+yz tx ty txz tyz + 1).

Hence must be a root of

f(x,y,z)(t) = tx+yz tx ty txz tyz + 1.

We have f(1) =

2. By Proposition 2.2, the polynomial f has a unique real root > 1 which must

be . This completes the proof.

Remark 3.3. For x y > 0 and z > 0,

f(x,y,z)(t) = f(x+z,y+z,z)(t) = tx+y+z tx+z ty+z tx ty + 1.

Thus,(x + y z) = ((x + z) + (y + z)+ z).

Theorem 3.4. The ray through the origin and + attains the minimum of P= XT()ent() :int(C) R.

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Proof. Clearly x + y + z is in int(C) if and only if y + x + z is in int(C). The imageof x + y + z under Pequals the image of y + x+ z under P, since these homology classeshave the same entropy and the same Thurston norm. Thus if the minimum of Pis realized by theray though x + y + z, then x must equal y.

On the other hand, two homology classes x + x + and (x 1) + (x 1) have thesame Thurston norm 2x 1 if x > 1. By Remark 3.3, they have the same entropy. Thus,

P(x + x+ ) = P((x 1) + (x 1) ).

Since Pis constant on each ray though the origin, we have

P(x + x+ ) = P( + + x1),P((x 1) + (x 1) ) = P( + (x 1)1).

The minimal ray does not pass through both + + x1 and + (x 1)1, because theminimum is realized by a unique ray. Since x(> 0) is arbitrarily, the minimal ray must pass through + . This completes the proof.

3.3 Fiber surface of genus 0

Let a = x + y + z be an integral homology class in int(C) and let Fa be its minimal repre-sentative. Recall that Fa is connected if and only if a is a primitive integral class. Since {,,}is a basis of H2(Mmagic, Mmagic;Z), we see that a is a primitive integral class if and only ifgcd(x,y ,z) = 1, where gcd(x,y, 0) is defined by gcd(x, y). The topological type of Fa can be deter-mined by Lemma 3.1. In this section, we determine homology classes whose minimal representativesare connected and of genus 0.

By (3.1), x + y+ z is in int(C) such that x y if and only if x y > 0 and y > z . In thissection, we consider such homology classes for simplicity.

Lemma 3.5. Letx, y and z be integers. Suppose that x y > 0, y > z and gcd(x,y ,z) = 1. Then

gcd(x, y + z) + gcd(y, z + x) + gcd(z, x + y) x y + z 2 0. (3.2)

If (x,y ,z) satisfies the equality of (3.2), then z 0.Proof. Note that

(Fa) = (2 2g (Fa)) = (Fa) = x + y z,where g denotes the genus of Fa. Hence

g =

x + y

z + 2

(Fa)

2 0. (3.3)By substituting (Fa) = gcd(x, y + z)+gcd(y, x + z)+gcd(z, x + y) for (3.3), we have the desiredinequality. Suppose that x y > 0 > z = z (z > 0). Then

gcd(x, y z) + gcd(y, z + x) + gcd(z, x + y) x y z 2 x + y + z x y z 2= 2.

This completes the proof.

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Proposition 3.6. Let x, y and z be integers. Suppose that x y > z 0 and gcd(x,y ,z) = 1.Then the equality of (3.2) holds for (x,y ,z) if and only if (x,y ,z) is either

(1) z = 0 and gcd(x, y) = 1,

(2) (x,y ,z) = (n + 1, n , n

1) for n

0 (mod 3) and n

2, or

(3) (x,y ,z) = (2n + 1, n + 1, n) for n 1.The following proof was shown to the authors by Shigeki Akiyama.

Proof. The equality of (3.2) holds for (x,y, 0) if and only if gcd(x, y) = 1, and hence we maysuppose that x y > z > 0 by Lemma 3.5. It is easy to see that if (x,y ,z) is of either type (2) ortype (3), then it satisfies the equality of (3.2). To prove the only if part, we first show:

Claim 3.7. Let x, y and z be natural numbers. Suppose that gcd(x,y ,z) = 1. Then

{gcd(N, x), gcd(N, y), gcd(N, z)}

is pairwise coprime, where N = x + y + z.

Proof of Claim 3.7. We set gcd(gcd(N, x), gcd(N, y)) = k. Then k is a divisor of three integersN, x and y. It is also a divisor of z(= N x y). Since gcd(x,y ,z) = 1, the integer k must be 1.This completes the proof the claim.

Notice that the inequality of (3.2) is equivalent to the inequality

N gcd(N, x) gcd(N, y) gcd(N, z) 2z 2. (3.4)For all N 79, one can check that the statement of Proposition 3.6 is valid. We may suppose thatN 80. Since x > N3 and z < N3 , we have x N+13 and z N13 . Hence x 27. Let p, q and rbe natural numbers so that p > q > r and

{p,q,r} = {gcd(N, x), gcd(N, y), gcd(N, z)}.By Claim 3.7, {p,q,r} is pairwise coprime, and three numbers p, q and r are divisors ofN. Therefore

pqr N. This shows thatNp q r

N pqr p q r

pqr= 1 1

qr 1

p

1q

+1

r

. (3.5)

If r 2, then q 3 and p 5. Hence

N

p

q

r

N1

1

qr

1

p1

q

+1

r =5

6

N1 1

p 2

3

N > 2z.

Thus, no (x,y ,z) satisfies the equality of (3.4) in this case.We may suppose that r = 1. Ifq 4, then

Np q rN

34

54p

by (3.5). Since z N13 , we obtain

Np q r 2z + 2 N3

4 5

4p

2N 8

3=

8

3+

N(p 15)12p

.

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If p > 15, then 83 +N(p15)

12p > 0, which implies that no (x,y ,z) satisfies the equality of (3.4) in thiscase. If p 14, then q 13. We have N 2z = x + y z > x > 26 = 1 + 14 + 13. Thus

Np q 1 N 14 13 1 > 2z 2,

which implies that no (x,y ,z) satisfies the equality of (3.4) in this case.We may suppose that q 3. It is enough to consider the equality of (3.4) in case (q, r) =

(3, 1), (2, 1) and (1, 1). Take w {x,y ,z} so that p = gcd(N, w).

(1) Case (q, r) = (2, 1) or (3, 1).

Then Ngcd(N, w)21 = 2z 2 or Ngcd(N, w)31 = 2z 2. We set gcd(N, w) = Nk(k > 1). Then N

1 1k

2(N1)3 + 2. Since we assume that N 80 > 16, k must be 2 or 3.

(i) Case k = 2.

Then x = N2 . If (q, r) = (2, 1), then (x,y ,z) =

N2 ,

N+24 ,

N24

. We set n = N24 . We

obtain (x,y ,z) = (2n+1, n+1, n), and such (x,y ,z) satisfies the equality (3.4). If (q, r) =

(3, 1), then (x,y ,z) = N2 , N+44 , N44 . In this case, gcdN, N44 = gcd4, N44 ,which is a divisor of 4. This does not occur since (q, r) = (3, 1).

(ii) Case k = 3.

If (q, r) = (2, 1), then z = N3 12 , which can not be an integer. Let (q, r) = (3, 1).Then z = N3 1. Since gcd(N, w) = N3 , we see that w = N3 or 2N3 . If w = 2N3 ,then (x,y ,z) =

2N

3 , 1,N3 1

. This is a contradiction since y > z. If w = N3 , then

(x,y ,z) =N3 + 1,

N3 ,

N3 1

. We set n = N3 . Then (x,y ,z) = (n + 1, n , n 1). If

n 0 (mod 3), then gcd(3n, n 1) = 1, which is a contradiction since (q, r) = (3, 1).Otherwise, such (x,y ,z) satisfies the equality of (3.4).

(2) Case (q, r) = (1, 1).

Then N gcd(N, w) 1 1 = 2z 2. We have N gcd(N, w) = 2z < 2N3 . This implies thatgcd(N, w) = N2 . Thus, (x,y ,z) =

N2 ,

N4 ,

N4

, which does not occur since y > z .

This completes the proof of Proposition 3.6.

By Proposition 3.6, we immediately obtain:

Corollary 3.8. Leta = x + y+ z be an integral homology class in int(C). Suppose that x yand gcd(x,y ,z) = 1. Then the genus of a minimal representative Fa of a is 0 if and only if (x,y ,z)

satisfies either

(1) z = 0 and gcd(x, y) = 1,

(2) (x,y ,z) = (n + 1, n , n 1) for n 0 (mod 3) and n 2, or(3) (x,y ,z) = (2n + 1, n + 1, n) for n 1.

We denote the monodromy for a by a : Fa Fa.Proposition 3.9.

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(1) Leta = x + y be an integral homology class in int(C) such that gcd(x, y) = 1. Then a fiberFa is an (x + y + 2)-punctured sphere. There exists a partition P = P1

P2

P3 of the setof punctures with P1 = x + y, P2 = 1, P3 = 1. The monodromy a permutes a point of Picyclically for each i.

(2) Let a = (n + 1) + n + (n

1) be an integral homology class in int(C

) such that n

0(mod 3). Then a fiber Fa is an (n + 4)-punctured sphere. There exists a partition P =P1

P2

P3 of the set of punctures with P1 = n, P2 = 3, P3 = 1. The monodromy apermutes a point of Pi cyclically for each i.

(3) Let a = (2n + 1) + (n + 1)+ n be an integral homology class in int(C(Z)). Then a fiberFa is a (2n + 4)-punctured sphere. There exists a partition P = P1

P2

P3 of the set ofpunctures with P1 = 2n + 1, P2 = 2, P3 = 1. The monodromy a permutes a point of Picyclically for each i.

Proposition 3.9 implies that the isotopy class a = [a] can be described by the braid on the disksince a fixes a puncture P3.

Proof. We show the claim (3). Other cases are similar. We see that

(F(2n+1,n+1,n)) = gcd(2n + 1, 2n + 1) + gcd(n + 1, 3n + 1) + gcd(n, 3n + 2)

= (2n + 1) + gcd(n + 1, 2) + gcd(n, 2).

Since {gcd(n + 1, 2), gcd(n, 2)} = {2, 1}, the number of the boundary components of F(2n+1,n+1,n)equals 2n+4. Because (2n+1) boundary components ofF(2n+1,n+1,n) lie on N(K), the monodromya must permute them cyclically. Similarly, a must permute the 2 boundary components cyclicallyand fix the other 1 boundary component. This completes the proof.

Theorem 3.10.

(1) limn,mP(n + m) = 2 log(2 + 3) if limn,mn

m = 1.

(2) limnP((n + 1) + n+ (n 1)) = .

(3) limnP((2n + 1) + (n + 1)+ n) = .

Proof. (1) We see that

limn,mP(n + m) = limn,mP

nn + m

+m

n + m

= P

+ 2

= P( + ) = 2ent( + ).

By Theorem 3.2, ent( + ) = log(2 +

3).

(2) We see that

limnP

(n + 1) + n+ (n 1)n

= P( + + ) = ent( + + ) =

since + + is in .

(3) The proof is similar to the one for the claim (2).

Let Hn be the set of homology classes a = x + y + z (x y) in int(C(Z)) such thattheir minimal representatives are n-punctured spheres. By Corollary 3.8, one can determine thehomology classes of Hn. In the next section, we will prove:

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Proposition 3.11. The homology class which reaches the minimal dilatation among Hn is asfollows.

(1) k + (k 1) in case n = 2k + 1 for k 2.(2) 3 + 2+ in case n = 6 and (2k + 1) + (2k

1) in case n = 4k + 2 for k

2.

(3a) + in case n = 4 and (4k + 3) + (4k 1) in case n = 8k + 4 for k 1.(3b) 5 + 3+ 2 in case n = 8 and (4k + 5) + (4k + 1) in case n = 8k + 8 for k 1.

3.4 Proof of Proposition 3.11

This section is devoted to prove Proposition 3.11.

Lemma 3.12. For m,n > 0 such that m > n, we have the followings.

(1) P((m + 1) + m+ (m 1)) > P((n + 1) + n+ (n 1)).

(2) P((2m + 1) + (m + 1)+ m) > P((2n + 1) + (n + 1)+ n).Proof. Let us consider homology classes x + y + z with (x,y ,z) = (n+1n+2 ,

nn+2 ,

n1n+2 ) for each

n > 0. These are in the open face and pass through the line x = 13 t + 1, y =23 t + 1, z = t + 1.

Note that

P

(n + 1) + n+ (n 1)

= (n + 2)ent((n + 1) + n+ (n 1))= ent

n + 1n + 2

+

nn + 2

+

n 1n + 2

and it goes to as n goes to by Proposition 3.10(2). We have1

3ent(2 + + 0) 1

2.887>

1

4ent(3 + 2+ ) 1

2.931.

Since 1ent() : int(C(Q)) R is a strictly concave function, we have for all m > n 3,

1

4ent(3 + 2+ )

>1

(n + 2)ent((n + 1) + n+ (n 1))

>1

(m + 2)ent((m + 1) + m+ (m 1)) .

This implies the proof of (1). The proof of (2) is similar.

We set:

1 = {x + y | x + y = 1, x > 0, y > 0} int() (see Figure 7(center)). C1 = {x + y | x > 0, y > 0} (see Figure 7(right)). C1(Z) = {a | a C1 is an integral class}.

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C1(Q) = {a | a C1 is a rational class}.Lemma 3.13.

(1) For x, y N such that gcd(x, y) = 1, the monodromy for x + y is conjugate to the one fory + x.

(2) For x, y > 0, we have ent(x + y) = ent(y + x).

Proof. The existence of a rotation f : (S3, C3) (S3, C3) with respect to the line L of Figure 5(2)implies that the monodromy for x + y is conjugate to the one for y x. Obviously, themonodromy for y x equals the one for y + x. In particular, ent(x + y) = ent(y + x).Since ent() : int(C(Z)) R admits a unique continuous extension to ent() : int(C) R, theclaim (1) implies the claim (2).

Fixing a natural number n, we set

n = {x + y | x > 0, y > 0, x + y = n} C1 .

By Lemma 3.1(1), the Thurston norm of a n equals n.Lemma 3.14. ent

n+n

2

= min{ent(a) | a n} for each n N.

Proof. We fix n N. Recall that the restriction of 1ent() on n is strictly concave. Notethat ent(a) as a n or n. Thus ent()|n : n R has the unique minimum. ByLemma 3.13(2), n+n2 attains the minimum.

We setn(N) = {x + y n | x, y N, gcd(x, y) = 1}

Lemma 3.15. For each integer m 3, the minimum min{ent(a) | a m1(N)} is realized by:(1) (k 1) + k and k + (k 1) in case m = 2k.(2) (2k 1) + (2k + 1) and (2k + 1) + (2k 1) in case m = 4k + 1.(3a) + in case m = 3, and(4k 1) + (4k + 3) and (4k + 3) + (4k 1) in case m = 8k + 3

for k 1.(3b) (4k + 1) + (4k + 5) and (4k + 5) + (4k + 1) in case m = 8k + 7.

Proof. The concavity of 1ent() |n : n R and Lemma 3.14 tell us that if two homology classesx + y and x + y satisfying |x y| < |x y| are in n, then

ent(x + y) < ent(x + y). (3.6)

This implies that the minimal entropy among homology classes in m1(N) is realized by (k 1) + k or k + (k 1) if m = 2k. (See Figure 13.) The proof for other cases can be shown ina similar way.

Lemma 3.16. For k > k > 0, we have the followings.

(1) P(k + (k + c)) > P(k + (k + c)).(2) ent(k + (k + c)) > ent(k + (k + c)).

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Figure 13: first quadrant of plane. (homologies in (1) (resp. (2), (3a, 3b)) of Prop. 3.28 lie on

the lines(1) (resp. (2) and (3ab))).

Proof. For any k, we have

Pk + (k + c)

2k + c

= ent

k + (k + c)2k + c

= (2k + c)ent(k + (k + c)).

If 0 < k < k, then c2k+c >c

2k+c . By (3.6), we see that

entk + (k + c)

2k + c

> ent

k + (k + c)2k + c

.

This implies the claim (1). Therefore,

entk + (k + c)

2k + c

= (2k + c)ent(k + (k + c))

> entk + (k + c)

2k + c

= (2k + c)ent(k + (k + c))

> (2k + c)ent(k + (k + c)).

Thus, ent(k + (k + c)) > ent(k + (k + c)). This completes the proof.

Proof of Proposition 3.11. (1) We consider the case n = 2k+1. For k = 2, we see that H5 = {2+}.Ifk

= 2 and 2k

0 (mod 3),

H2k+1 is the set of homology classes of type (1) of Corollary 3.8, that

is H2k+1 = {x + y | x + y 2k1(N), x y}.In this case, k+(k1)reaches the minimal dilatation among H2k+1 by Lemma 3.15(1). Otherwise(i.e, 2k 0 (mod 3)), H2k+1 is the union of homology classes of type (1) and (2) of Corollary 3.8:

H2k+1 = {(2k 2) + (2k 3)+ (2k 4)} {x + y | x + y 2k1(N), x y}.

One needs to compare the dilatation for (2k2)+(2k3)+(2k4)with the one for k+(k1).In case k = 4,

(4 + 3) 1.46557 < (6 + 5+ 4) 1.72208.

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By Lemmas 3.16 and 3.12, for k > 4, we have

(2k 1)ent(k + (k 1)) 7ent(4 + 3)< 7ent(6 + 5+ 4)

< (2k 1)ent((2k 2) + (2k 3)+ (2k 4)).

Thus, ent(k + (k 1)) < ent((2k 2) + (2k 3)+ (2k 4)). This completes the proof incase n = 2k + 1.(2) Let us consider the case n = 4k + 2. For k = 1, we see that H6 = {3 + , 3 + 2+ }. Wehave

(3 + 2+ ) 2.08102 < (3 + ) 2.29663.For k = 2, we see that H10 = {7 + 4+ 3, 5 + 3, 7 + }. We have inequalities

(5 + 3) 1.41345 < (7 + 4+ 3) 1.55603, and (5 + 3) < (7 + )

by Lemma 3.15(2). For k = 3, we see that H14 = {11 + 6+ 5, 7 + 5, 11 + , 11 + 10+ 9}.We have (7 + 5) < (11 + ) by Lemma 3.15(2) and

(7 + 5) 1.25141 < (11 + 6+ 5) 1.39241 < (11 + 10+ 9) 1.62913.

By using the same arguments as in the case n = 2k + 1, we have for all k > 3,

(4k)ent((2k + 1) + (2k 1)) < 12ent(7 + 5),12ent(11 + 6+ 5) < (4k)ent((4k 1) + 2k + (2k 1)),

12ent(11 + 10+ 9) < (4k)ent((4k 1) + (4k 2)+ (4k 3)).

Thus, for all k 2, the homology class (2k + 1) + (2k 1), which realizes the minimal dilatationamong 4k(N), reaches the minimal dilatation among H4k+2.(3b) Let us consider the case n = 8k + 8. For k = 0, we see that H8 = {5 + 3+ 2, 5 + , 5 +4+ 3}, and

(5 + 3+ 2) 1.72208 < (5 + 4+ 3) 1.78164 < (5 + ) 2.08102.

For k 1, one can show that (4k + 5) + (4k + 1) reaches the minimal dilatation among H8k+4.(3a) The proof for the other case n = 8k + 4 is shown in a similar way.

3.5 Monodromy

Recall that Tm,p is the m-braid such that

Tm,p = (2123 m1)p2m1.

For example, T6,2 = (212345)

225 = 212345

21234

15 . Since

= m = (2123 m1)m1 = (12 m1)m Bm

is the full twist, we have:

Lemma 3.17. If p p (mod m 1), then there exists an integer k such that Tm,p = Tm,pk.

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Figure 14: (left) braid T6,2. (right) braid T5,2.

By Lemma 3.17, we have (Tm,p) = (Tm,p) M(Dm) in this case.Let us consider the braid Tm,p when gcd(m 1, p) = 1. For example, T5,2 is a reducible braid,

since a disjoint union of two simple closed curves in the 5-punctured disk D5 is invariant under themapping class (T5,2), see Figure 14(right). It is not hard to see the following.

Lemma 3.18. If gcd(m 1, p) = 1, then Tm,p is a reducible braid.Theorem 3.19. Suppose that a homology class a = x + y C1(Z) satisfies gcd(x, y) = 1.Then the isotopy class a of the monodromy a : Fa Fa can be described by the braid Tx+y+1,p

for some p = p(x, y).

The rest of section is devoted to show Theorem 3.19 and how to compute p = p(x, y).

3.5.1 Fiber surface

Let (E1, E2) be a pair of embedded surfaces in the exterior E(L) of a link L in S3 as in Figure 15.

(In particular, E1 is a punctured disk.) Depending on the orientation of E1 and E2, the surfaceE1 + E2 is either the one as in Figure 15(3) or the one as in Figure 15(4).

Lemma 3.20. Suppose that E1 + E2 is a surface of the type as in Figure 15(3). Cut E1 in E(L),and letE andE be the resulting punctured disks after cutting E1. Reglue E andE after twistingthe neighborhood of E by 360 N degrees in the clockwise direction for some natural number N.Then we obtain a new link, say L whose exterior E(L) is homeomorphic to the original linkexterior E(L). Let (E1, E

2) be a pair of embedded surfaces in E(L

) induced from (E1, E2) inE(L) (see Figure 15(2)). Let x and y be a natural numbers such that y = xN + r for some0 r < x. Then there exists an orientation preserving homeomorphism f : E(L) E(L) suchthat f(yE1 + xE2) = rE

1 + xE

2 and f(yE1) = yE

1. In particular, for (x,y ,r) = (1, N, 0), we have

f(N E1 + E2) = E2

.

In case where E1 + E2 is a surface of the type as in Figure 15(4), a similar claim in Lemma 3.20holds by replacing clockwise direction with counterclockwise direction.

Proof. By the definition of L, there exists a natural homeomorphism f : (S3, L) (S3, L). It isstraightforward to see that the induced homeomorphism f : E(L) E(L) sends N E1 + E2 to E2.More generally, one can see that f(yE1 + xE2) = rE

1 + xE

2 and f(yE1) = yE

1.

Fix m and p, and consider the braid Tm,p. Now, we shall define two embedded surfaces Fm,p andFm,p in E(Tm,p) (See Figure 16). The surface Fm,p is the m-punctured disk in E(Tm,p) which isbounded by the braid axis of Tm,p. Clearly, Fm,p is a fiber in E(Tm,p) with monodromy Tm,p. The

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Figure 15: (1) surface E1 and E2 in E(L). (2) surface E1 and E

2 in E(L

) (in case where thenumber N equals 2 in Lemma 3.20). (3, 4) two types of E1 + E2.

Figure 16: surface Fm,p and Fm,p in E(Tm,p).

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surface Fm,p is the (p + 1)-punctured disk in E(Tm,p) which is bounded by Km,p, where Km,p isthe knot obtained by the closing the 1st strand of Tm,p.

Given a pair (k, ) N N, the following method (T) enables us to visualize another fiberFm(k,),p(k,) in E(Tm,p) with the monodromy Tm(k,),p(k,).

(T) Applying Lemma 3.20 for the link Tm,p, a pair (Fm,p, Fm,p) and N = k, we obtain a pair ofembedded surfaces(Fm,k(m1)+p, Fm,k(m1)+p) = (Fm,p, kFm,p + Fm,p)

in E(Tm,k(m1)+p), since k = (2123 m1)k(m1). Apply Lemma 3.20 once more for the linkTm,k(m1)+p, a pair (Fm,k(m1)+p, Fm,k(m1)+p) and N = , we obtain a pair of embedded surfaces

(Fm(k,),p(k,), Fm(k,),p(k,)) = (Fm,k(m1)+p, Fm,k(m1)+p + Fm,k(m1)+p)in E(Tm(k,),p(k,)), where Tm(k,),p(k,) = T(k(m1)+p)+m,k(m1)+p. We have a fiber Fm(k,),p(k,) ofE(Tm,p) with the monodromy Tm(k,),p(k,) which can be easily visualized. The homology class of

Fm(k,),p(k,) equals [(kFm,p + Fm,p) + Fm,p]. For the case where Tm,p = T5,1 and (k, ) = (1, 2), wehave Tm(k,),p(k,) = T15,5, see Figure 19(2,3,4).Remark 3.21. Let x and y be natural numbers such that 0 < x < y. Suppose that

y = xk + r1 (0 < r1 < x, x,r1 N),x = r1 + r2 (0 < r2 < r1, , r2 N).

By applying Lemma 3.20 twice, we have a homeomorphism f : E(Tm,p) E(Tm(k,),p(k,)) suchthat f(xFm,p + yFm,p) = r2Fm(k,),p(k,) + r1Fm(k,),p(k,). In particular

x

Fm,p + yFm,p

=

r2

Fm(k,),p(k,) + r1Fm(k,),p(k,)

.

Let q = (q1, q2, , qj) be a sequence of natural numbers. The number j is called the lengthof q, and it is denoted by |q|. For q = (k1, 1, , kj, j) with length |q| = 2j even, let q(i) be asubsequence (k1, 1, , ki, i) for i j. For q = (0, k1, 1, , kj, j) with length |q| = 2j +1 odd,let q(i) be a subsequence (0, k1, 1, , ki, i) for i j.

We will define a fiber Fm(q),p(q) in E(C3) with monodromy Tm(q),p(q) associated to a sequenceqsuch that [Fm(q),p(q)] C1. To do so, we define a fiber Fm(i),p(i) in E(C3) (associated to q) withmonodromy Tm(i),p(i) inductively as follows. Figure 18(1) is another diagram of the 3 chain link C3.Surfaces E (resp. E) drawn in Figure 17 is a representative of the homology class (resp. ).We first consider a sequence q = (k1, 1, , kj, j) with even length. Then two cases can occur.Case 1-even: q = (k1, 1).

We apply Lemma 3.20 twice. First, for L = C3, a pair (E, E) and N = k1, apply Lemma 3.20.Let (E, k1E+ E) be a pair of embedded surface induced from (E, E) in a new link L

exteriorE(L). Second, apply Lemma 3.20 for L, a pair (k1E+ E, E) and N = 1. Then we have a pairof embedded surfaces (k1E+ E, 1(k1E+ E) + E) in a new link L

exterior E(L). We seethat the link L is a braided link of Tm(q),p(q) = T(k1+1)1+2,k1+1. Hence

(Fm(q),p(q), Fm(q),p(q)) = (k1E+ E, 1(k1E+ E) + E).Therefore Fm(q),p(q) is a fiber in E(C3) with monodromy Tm(q),p(q), and we have

Fm(q),p(q)

= 1 + (1k1 + 1) C1(Z).

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Figure 17: surface E and E in E(C3).

For example in case q = (1, 1), the braid Tm(q),p(q) equals T4,2, see Figure 18.

Case 2-even: q = (k1, 1, , kj, j) with length 2j > 2.For i = 1, we have defined a fiber Fm(1),p(1) in E(C3) E(Tm(1),p(1)) with monodromy Tm(1),p(1) asin case 1-even. Suppose that we have a fiber Fm(i),p(i) in E(C3) E(Tm(i),p(i)) with monodromyTm(i),p(i). Apply (T) for Tm(i),p(i), a pair (ki+1, i+1). Then we have a pair of embedded surfaces

(Fm(i+1),p(i+1), Fm(i+1),p(i+1)) = (Fm(i)(ki+1,i+1),p(i)(ki+1,i+1), Fm(i)(ki+1,i+1),p(i)(ki+1,i+1))such that Fm(i+1),p(i+1) is a fiber of E(C3) with monodromy

Tm(i+1),p(i+1) = Tm(i)(ki+1,i+1),p(i)(ki+1,i+1).

By induction, it is shown that

Fm(i+1),p(i+1) C1(Z).

Next, let us consider a sequence q = (0, k1, 1, , kj, j) with odd length.Case 1-odd: q = (0).

Applying Lemma 3.20 for C3, a pair (E, E) and N = 0, we obtain a pair of embedded surfaces(E, 0E + E) in a new link L exterior E(L). We see that the link L is a braided link ofTm(q),p(q) = T0+2,1. Hence

(Fm(q),p(q), Fm(q),p(q)) = (E, 0E + E).Therefore Fm(q),p(q) is a fiber in E(C3) with monodromy Tm(q),p(q), and we have

Fm(q),p(q)

= 0 + C1(Z).

Case 2-odd: q = (0

, k1

, 1

,

, kj

, j

) with length 2j + 1 > 1.

For i = 0, we have defined a fiber Fm(0),p(0) in E(C3) E(Tm(0),p(0)) with monodromy Tm(0),p(0)as in case 1-odd. For i 1, in the same manner as case 2-even, a fiber Fm(i),p(i) in E(C3) withmonodromy Tm(i),p(i) = Tm(q(i)),p(q(i)) is given inductively, and we see that

Fm(i),p(i)

C1(Z).For the case where q = (3, 1, 2), see Figure 19.

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Figure 18: Case 1-even: q = (1, 1). fiber F4,2 with monodromy T4,2.

Figure 19: construction of fiber associated to q = (3, 1, 2). (1) C3. (2) T5,1. (3) T5,5. (4) T15,5.

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3.5.2 Continued fraction

We review some results on continued fractions. Let us consider a continued fraction

w1 +1

w2 +

1

w3 + 1

wj1 +1

wj

for wi N. For simplicity, we denote such continued fraction by

w1 +1

w2 +

1

w3 + +1

wj1 +1

wj. (3.7)

The number j in (3.7) is called the length of the continued fraction of the form (3.7).We define [w1, w2, , wj] N inductively as follows.

[w1] = w1.

[w1, w2] = w1w2 + 1.

[w1, w2, , wi] = [w1, w2, , wi1] wi + [w1, w2, , wi2] .

The next lemma is elementary and well-known.

Lemma 3.22.

(1) w1 +1w2 +

1w3 ++

1wj1 +

1wj

=[w1,w2, ,wj ][w2,w3, ,wj ] .

(2) [w1, w2,

, wj] = [wj, wj

1,

, w1].

Definition 3.23. Let u and v be natural numbers such that gcd(u, v) = 1. We define two finitesequences of integers r = (r0, r1, , rj+1) and q = (q1, q2, , qj) according to the Euclideanalgorithm. We set r0 = max{u, v} and r1 = min{u, v}. Write r0 = r1q1 + r2 (0 r2 < r1).

If r2 = 0, then r1 must be 1 since gcd(u, v) = 1. In this case, we set r = (r0, r1 = 1, r2 = 0)and q = (q1).

Suppose that r2 = 0. Let us define q2, q3, and r3, r4, such that r2 > r3 > r4 > inductively as follows. Let qi be the quotient when ri1 is divided by ri, and let ri+1 be itsremainder. By the monotonicity r0 > r1 > > ri > ri+1 > 0, there exists j Nsuch that rj+1 = 0. Then rj must be 1. We set r = (r0, r1,

, rj = 1, rj+1 = 0) and

q = (q0, , qj).The fraction r0r1 can be expressed by the following two kinds of continued fraction by using the

sequence q = (q1, q2, , qj).r0r1

= q1 +1

q2 +

1

q3 + +1

qj1 +1

qj= q1 +

1

q2 +

1

q3 + +1

qj1 +1

(qj 1) +1

1(3.8)

with length j and length j + 1 respectively. We can choose the one with odd (resp. even) lengthamong two continued fractions expressing r0r1 .

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Proof of Theorem 3.19. Let x and y be natural numbers satisfying gcd(x, y) = 1. If x > y (resp.x < y ), we choose the continued fraction

max{x, y}min{x, y} = w1 +

1

w2 +

1

w3 + +1

wj1 +1

wj(3.9)

with odd length (resp. even length). Now, we define a finite sequence s = (s0, s1, , sj+1) whichis defined by

s0 = max{x, y},s1 = min{x, y},

si+1 = si1 siwi for i 1.Notice that sj = 1 and sj+1 = 0. (If the continued fraction in (3.9) is the first type in (3.8), thenthe finite sequence s equals r of Definition 3.23.)

Suppose that x < y . Note that the continued fraction of (3.9) has even length by definition. Letus write q = (w1, w2, , wj) = (k1, 1, , k j

2, j

2). A fiber Fm(q),p(q) in E(C3) with monodromy

Tm(q),p(q) associated to q is a representative of x + y. In fact by Remark 3.21,x + y

= [xE + yE] (= [s1E + s0E])

=

s3Fm(1),p(1) + s2F(m(1),p(1))...

=

s2i+1Fm(i),p(i) + s2iF(m(i),p(i))...

= sj+1Fm( j2 ),p( j2 ) + sjFm( j2 ),p( j2 ) (= 0Fm( j2 ),p( j2 ) + 1Fm( j2 ),p( j2 ))=

Fm(q),p(q)

.

Since a representative Fm(q),p(q) ofx + y is an (x + y + 2)-punctured sphere, m(q) equals x + y + 1.The argument for the case x > y is similar as above.

3.5.3 Computation ofp = p(x, y) in Theorem 3.19

In this section we give a recipe to compute the number p of Tx+y+1,p in Theorem 3.19. First, wegive an example to see how the number p comes from a pair (x, y).

Example 3.24. We explain the braid T15,5 is a monodromy for 11 + 3 and why the number

p = 5 is derived from two finite sequences q = (q1, q2, q3) = (3, 1, 2) and r = (r0, r1, r2, r3, r4) =(11, 3, 2, 1, 0) of Definition 3.23 associated to two numbers 3 and 11. By the proof of Theorem 3.19,we see that

[11E + 3E] =

2F5,1 + 3F5,1 = 2F5,5 + 1F5,5 = 0F15,5 + 1F15,5 .A simple description of these equalities is as follows.

(r0, r1) (r2, r1) (r2, r3) (r4, r3)

(11, 3) q1=3

(2, 3) q2=1

(2, 1) q3=2

(0, 1)(3.10)

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On the other hand, by Figure 19 (or by using the argument in Case 2-odd in Section 3.5.1), we seethat T15,5 is a monodromy for 11 + 3. Recall that Km,p is the knot obtained by the closing the1st strand of Tm,p. LetK

m,p be the knot obtained by the closing the rest of strands, i.e, K

m,p equals

the closed braid of Tm,p removing Km,p. For the braid Tm,p, we have a pair of numbers

(p,m

1) = (i(Fm,p, Km,p), i(Fm,p, Km,p)),where i( , ) is the intersection number. In the process to find the monodromy associated to q =(3, 1, 2), a finite sequence of pairs of such intersection numbers obtained from T5,1, T5,5 and T15,5is described as follows.

(1, 1) q1=3

(1, 4) q2=1

(5, 4) q3=2

(5, 14) (3.11)

We extend the sequences of (3.10) and (3.11) to the left. Then we have

(r0, r0 + r1) (r0, r1) (r2, r1) (r2, r3) (r4, r3)

(11, 14)

q0=1(11, 3)

q1=3(2, 3)

q2=1(2, 1)

q3=2(0, 1)

(0, 1) q0=1

(1, 1) q1=3

(1, 4) q2=1

(5, 4) q3=2

(5, 14)

These show that

14

11= 1 +

1

3 +

1

1 +

1

2=

[1, 3, 1, 2]

[3, 1, 2]=

[q0, q1, q2, q3]

[q1, q2, a3],

14

5= 2 +

1

1 +

1

3 +

1

1=

[2, 1, 3, 1]

[1, 3, 1]=

[q3, q2, q1, q0]

[q2, q1, q0].

Thus, the number p = 5 equals [q2, q1, q0].

Proposition 3.25. LetTx+y+1,p(x,y) be the braid in Theorem 3.19 which represents the monodromy

for x + y C1(Z).(1) Let max{x,y}min{x,y} = w1 +

1w2 +

1w3 ++

1wj1 +

1wj

be the continued fraction in (3.9). Then

p = p(x, y) = [wj1, wj2, , w1, w0 = 1] = [w0 = 1, w1, , wj1].

(2) p = p(x, y) satisfies p(max{x, y}) (1)j (mod x + y).Proof. (1) We have

x + y

max{x, y} = w0 +1

w1 +

1

w2 + +1

wj1 +1

wj,

where w0 = 1. It is not hard to show the claim (1) by using the argument of Example 3.24.

(2) By induction, one can show thatw0 11 0

w1 11 0

wj 11 0

=

[w0, w1, , wj] [w0, w1, , wj1][w1, w2, , wj] [w1, w2, , wj1]

. (3.12)

Since the determinant of each matrix

wi 11 0

is 1,

(1)j+1 [w0, w1, , wj1] [w1, w2, , wj] (mod [w0, w1, , wj]).Note that x + y = [w0, w1, , wj], p = [wj1, wj2, , w0] = [w0, w1, , wj1], and max{x, y} =[w1, w2, , wj ]. Thus, (1)j p(max{x, y}) (mod x + y).

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Theorem 3.26. Let p 1 and m 3 be integers such that gcd(p,m 1) = 1. Then there existx, y N with x + y = m 1 such that Tm,p gives a monodromy for x + y C1(Z).Proof. Let : N N be the Euler function. The number of braids Tm,p satisfying 1 p m 1and gcd(p,m 1) = 1 equals (m 1). On the other hand, the number of homology classes

x +y m1(N

) satisfying gcd(x, y) = 1 equals (m1). Let x +y and x +y are distinctelements in m1(N). It is enough to show that p(x, y) = p(x, y) since 1 p(x, y), p(x, y) m1.Suppose that (x, y) = (y, x). Then the concavity of ent()|m1 : m1 R and Lemma 3.13

imply that ent(x + y) = ent(x + y), and hence Tm,p(x,y) = Tm,p(x,y) which implies thatp(x, y) = p(x, y).

Suppose that (x, y) = (y, x). By Proposition 3.25(2), we see that

p(x, y)(max{x, y}) +p(y, x)(max{x, y}) = (p(x, y) +p(y, x))(max{x, y}) 0 (mod x + y). (3.13)

Since gcd(max{x, y}, x + y) = 1, we have p(x, y) +p(y, x) 0 (mod x + y). Thus, p(x, y) p(y, x)(mod x + y). This completes the proof.

Theorem 3.26 immediately gives:Corollary 3.27. Let p 1 and m 3 be integers such that gcd(p,m 1) = 1. Then S3 \ Tm,p ishomeomorphic to S3 \ C3.Proposition 3.28. Letm 3 be integers. The followings are homology classes realizingmin{ent(a) | a m1(N)} and their monodromies.(1) If m = 2k, then(k 1) + k and k + (k 1) realize the minimum and their monodromies

are given by T2k,2 (or T2k,2k3).

(2) If m = 4k + 1, then (2k 1) + (2k + 1) and (2k + 1) + (2k 1) realize the minimum andtheir monodromies are given by T4k+1,2k+1 (or T4k+1,2k

1).

(3a) If m = 3, then + realize the minimum and its monodromy is given by T3,1. If m = 8k + 3(k 1), then (4k 1) + (4k + 3) and (4k + 3) + (4k 1) realize the minimum and theirmonodromies are given by T8k+3,2k+1 (or T8k+3,6k+1).

(3b) If m = 8k + 7, then (4k + 1) + (4k + 5) and (4k + 5) + (4k + 1) realize the minimumand their monodromies are given by T8k+7,6k+5 (or T8k+7,2k+1).

Proof. We show the claim in case m = 2k. Other cases can be shown in a similar way. ByLemma 3.15, the homology classes a = (k 1) + k and a = k + (k 1) realize the minimum.Let us consider the monodromies Tm,p(k1,k) and Tm,p(k,k1). Let (x, y) = (k 1, k). Since x < y,the continued fraction which is chosen in (3.9) is y

x

= w1 +1

w2, where w1 = 1 and w2 = k

1. By

Proposition 3.25(1), p(k 1, k) = [w1, w0] = [1, 1] = 2. By (3.13),

p(k 1, k) +p(k, k 1) 0 (mod 2k 1).

Hence p(k, k 1) = 2k 3. By Lemma 3.13(1), T2k,2 or T2k,2k3 gives the monodromy for a anda.

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3.6 Proof of Theorem 1.1

In Propositions 3.11 and 3.28, we have proved Theorem 1.1 except n = 6, 8. To complete the proofof Theorem 1.1, we shall give monodromies for two homology classes 3 + 2+ and 5 + 3+ 2in Proposition 3.31.

Lemma 3.29.

(1) The 5-braided link 12234 and the 4-braided link T4,2 are isotopic to the (2, 4, 6)-pretzellink.

(2) The braided link b for the 7-braid b as in Theorem 1.1(3b-i) is isotopic to the 5-braided link

1223415 .

Proof. (1) This is an easy exercise and we leave the proof for the readers. (Note: T4,2 is conjugateto the 4-braid 2123

21, and it might be easier to see the latter braided link is isotopic to the

(2, 4, 6)-pretzel link.)

(2) Let be an n-braid. By deforming the axis of , the braided link can be represented by theclosed braid of Bn+2, where = 1n+12n 2n1 21 21 22 2n (1, 2 {1, 1}), seeFigure 20. By using this method, 12234

15 is represented by the closed 7-braid

a, wherea = 16 (1

2234

15 )

15

14

13

12

11

11

12

13

14

15 .

On the other hand, the braided link b (Figure 21(left)) can be represented by a closed 6-braid asin Figure 21(center) whose link type equals a closed 7-braid as in Figure 21(right). Namely, b isisotopic to the closure of the 7-braid b:

b = 6 1 2 3 4 1 2 3 1 2 54 4 3 5 4 3 2 1 1 2 3 4 5,

where i stands for 1i . We see that a is conjugate to b, since the super summit set for a isequal to the one for b. (The super summit set is a complete conjugacy invariant, see [4].) Infact, the super summit set consists of 4 elements 17 1234321543654321,

17 1213432543654321,

17 1232145432654321 and 17 1232143254654321, where i stands for i. (One can use the com-

puter program Braiding by Gonzalez-Meneses for a computation of the super summit set [8].)

Thus, the link types of b and 1223415 are the same. This completes the proof.

Figure 20: (left) braided link . (right) closed braid representing .

Lemma 3.29 together with Corollary 3.27 implies:

Corollary 3.30.

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Figure 21: (left) braided link b. (center) closed 6-braid representing b. (right) closed 7-braid brepresenting b.

(1) S3 \ 12234 is homeomorphic to S3 \ C3.(2) S3 \ b is homeomorphic to S3 \ C3.Proposition 3.31.

(1) (12234) is conjugate to the monodromy for 3 + 2+ .

(2) (b) is conjugate to the monodromy for 5 + 3+ 2.

Proof. (1) We see that H6 = {3 + 2+ , 3 + }, see the proof of Proposition 3.11. By Propo-sition 3.9, the monodromy for 3 + permutes 4 punctures cyclically and fixes two 1 punctures.On the other hand, the monodromy for 3 + 2+ permutes 3 punctures cyclically, and the map-ping class (1

2234) permutes 3 punctures cyclically. By Corollary 3.30(1), we conclude that

(12234) is conjugate to the monodromy for 3 + 2+ .

(2) We see that H8 = {5+, 5+3+2, 5+4+3}. The mapping class (b) permutes 5 punc-tures cyclically, 2 punctures cyclically and fixes the other 1 puncture. Among

H8, 5+3+2 is the

only class whose monodromy permutes 2 punctures cyclically. By Corollary 3.30(2), we concludethat (b) must be conjugate to the monodromy for 5 + 3+ 2.

4 Further discussion

4.1 Pseudo-Anosovs with small entropy

4.1.1 Punctured sphere

Forgetting the 1st strand of the m-braid Tm,p, one obtains the (m 1)-braid, say Tm,p. Forexample, T6,2 = (1234)

224 . Notice that Tm,p may not be pseudo-Anosov, even though Tm,p

is pseudo-Anosov. In the case where Tm,p is pseudo-Anosov, it is easy to see that the inequalityent(Tm,p) ent(Tm,p) holds. A question is when the equality holds. The following lemma, whichis obvious by the definition of pseudo-Anosov homeomorphisms, is an answer.

Lemma 4.1. Let m,p be a pseudo-Anosov homeomorphism which represents the mapping class(Tm,p) M(Dm). Corresponding to the 1st strand of the braid Tm,p, there exists a puncture ofthe disk, say am,p, which is fixed by m,p. Suppose that invariant foliations Fs (or Fu) associatedto m,p has no 1-pronged singularity at am,p. Then T

m,p is a pseudo-Anosov braid such that

ent(Tm,p) = ent(Tm,p).

The family of braids Tm,p contains the braids with smallest entropy. In fact,

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T4,1 = 112 B3 is the 3-braid with the smallest entropy, see Matsuoka [19].

T5,1 = 1213 B4 is the 4-braid with the smallest entropy, see Ko-Los-Song [15]. T6,2 123412 B5 is the 5-braid with the smallest entropy, see Ham-Song [10].

Here for the braids b and b with the same strands, b b means that b is conjugate to b. It is anopen problem to determine the braids of Bn with the smallest entropy for n 6.

All the braids in Proposition 3.28 have been studied from the view point of their entropies.Hironaka-Kin studies a family of braids

(k) = 12 2k212 2k4 B2k1 (k 3)

with odd strands [11]. It is easy to see that (k) T2k,2 (cf. Proposition 3.28(1)). Each braid (k)has the smallest known entropy. Venzke finds a family of braids with small entropy [28]. Let usintroduce his family of braids n. Set Ln = n1n2 1 Bn. He defines the braid n Bnas follows.

n = L2n

11

12 if n = 2k 1 (k 3),

n = L2k+1n

11

12 if n = 4k (k 2),

n = L2k+1n

11

12 if n = 8k + 2 (k 1),

n = L6k+5n

11

12 if n = 8k + 6 (k 1),

6 = 5432154354.

It is not hard to see that 2k1 T2k,2, 4k T4k+1,2k+1, 8k+2 T8k+3,2k+1, 8k+6 T8k+7,6k+5,and 6 T6,2 (cf. Proposition 3.28(2)(3a)(3b)). By using Lemma 4.1, one can show that

ent(2k

1) = ent(T2k,2),

ent(4k) = ent(T4k+1,2k+1),

ent(8k+2) = ent(T8k+3,2k+1),

ent(8k+6) = ent(T8k+7,6k+5).

It is surprising that each braid with the smallest known entropy of n 6 strands can be derivedfrom the braid Tn+1,p for some p with the smallest entropy among a family of braids {Tn+1,p},where p is any integer such that 1 p m 1 and gcd(p, m 1) = 1.

Let T(m) Bm be one of the two braids which realizes the minimum in Proposition 3.28. Forexample, T(2k) = T2k,2 or T2k,2k3. Let T(m) Bm1 be the braid obtained from T(m) by forgettingthe 1st strand ofT(m). One can apply Lemma 4.1 to show that (T(m)) = (T

(m)). By Theorem 3.2

and Proposition 3.28, we have the followings.

Corollary 4.2.

(1) (T(2k)) equals the largest real root of P(tk1, tk, t0) = t2k1 2(tk1 + tk) + 1.

(2) (T(4k+1)) equals the largest real root of P(t2k1, t2k+1, t0) = t4k 2(t2k1 + t2k+1) + 1.

(3a) (T(8k+3)) equals the largest real root of P(t4k1, t4k+1, t0) = t8k+2 2(t4k1 + t4k+3) + 1.

(3b) (T(8k+7)) equals the largest real root of P(t4k+1, t4k+5, t0) = t8k+6 2(t4k+1 + t4k+5) + 1.

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We now discuss the monotonicity of the entropy of braids T(m). The following proposition is acorollary of Lemma 3.16 and Proposition 3.28.

Proposition 4.3.

(1) ent(T(2k)) > ent(T(2(k+1))).

(2) ent(T(4k+1)) > ent(T(4(k+1)+1)).

(3a) ent(T(8k+3)) > ent(T(8(k+1)+3)).

(3b) ent(T(8k+7)) > ent(T(8(k+1)+7)).

Next, we consider an inequality between ent(T(m)) and ent(T(m+1)).

Lemma 4.4. We have ent(T(2k1)) > ent(T(2k)).

Proof. One can prove the claim by using the argument in the proof of Lemma 3.16.

In contrast to Lemma 4.4, it is not true that ent(T(2k)) > ent(T(2k+1)) for all k. For example,

ent(T(6)) < ent(T(7)),

ent(T(10)) < ent(T(11)).

Our computer experiment shows that it seems that ent(T(2k)) > ent(T(2k+1)) for all k but k = 3, 5.See the computation of ent(T(m)) below.

Lemma 4.4 tells us that we have an inequality ent(T(2k)) < ent(T(2k1)) for two (2k 1)-braids.The above example shows that we have inequalities ent(T(6)) < ent(T

(7)) for two 6-braids and

ent(T(10)) < ent(T(11)) for two 10-braids.

m ent(T(m)) m ent(T(m)) m ent(T(m)) m ent(T(m))

3 1.31696 4 0.962424 5 0.831443 6 0.543535

7 0.732858 8 0.382245 9 0.346031 10 0.295442

11 0.302271 12 0.240965 13 0.224273 14 0.203526

15 0.200958 16 0.176191 17 0.166609 18 0.155345

19 0.152136 20 0.13892 21 0.132708 22 0.125641

23 0.122845 24 0.114683 25 0.11033 26 0.105485

We turn to the asymptotic behavior of the entropy ofT(m). In [11, Proposition 3.36], it is shownthat for m = 2k,

limm |(0,m)|ent(T

(m))(= limm(m 2)ent(T(m))) = 2 log(2 +

3).

Hence limm(m 1)ent(T(m)) = 2log(2 + 3) since ent(T(m)) tends to 0. We give alternativeproof for this claim from the view point of the geometry of the magic manifold.

Corollary 4.5. We have

limm |(0,m+1)|ent(T(m)) = limm(m 1)ent(T(m)) = 2 log(2 +

3).

Proof. Recall that T(m) is a monodromy for some homology class whose minimal representative isan (m + 1)-punctured sphere 0,m+1. By Theorem 3.10(1) and Proposition 3.28, we have

limm |2 (m + 1)|ent(T(m)) = 2 log(2 +

3).

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Remark 4.6. Venzke shows that each mapping torus T(n) except n = 6 is obtained from Mmagicby a Dehn surgery along a single cusp. He also computes the surgery coefficient, see [28].

Remark 4.7. Using train track maps, Hironaka-Kin computes the polynomial whose largest realroot equals the dilatation of (k), see [11]. Venzke also computes such polynomial for braids n byusing the same technique, see [28].

Finally, we propose the following conjecture.

Conjecture 4.8. The braid T(m) or the braid T(m+1) realize the minimal entropy among all braids

of Bm.

4.1.2 Closed surface of genus g

In the rest of this section, we prove Theorem 1.3. By Lemma 3.1 and Theorem 3.2, it is easy tosee the following lemma.

Lemma 4.9. For each g N, let a = (g + 1) + (g + 1) + be an integral homology class inint(C).

(1) A minimal representative Fa is a 3-punctured surface of genus g.

(2) The dilatation (a) is the largest real root of

f(g+1,g+1,1) = t2g+1 2tg+1 2tg + 1.

(3) limg P(a) = limg(2g + 1)ent((g + 1) + (g + 1)+ ) = 2 log(2 +

3).

One verifies the claim (3) in the above lemma by using limg P(a) = P( + ).Let a : g,0 g,0 be a homeomorphism on the closed surface of genus g obtained from the

monodromy a : g,3 g,3 (for the homology class a) by forgetting all 3 punctures. Recall thatin case the mapping class a = [a] is pseudo-Anosov, the dilatation of a is less than or equal tothe one for the mapping class a = [a]. For the proof of Theorem 1.3, it is enough to show:

Lemma 4.10. The mapping class a is pseudo-Anosov.

Proof. Recall that K, K and K are components of the 3 chain link C3. Let us consider a surfaceF(g+1,g+1,1) = (g + 1)F + (g + 1)F+ F which is a representative of a = (g + 1) + (g + 1)+ .We see that the boundary component of F(g+1,g+1,1) which lies on the regular neighborhood N(K)

has the boundary slope (g+1)g . The boundary slopes for the other two boundary components of

F(g+1,g+1,1) which lie on N(K) or N(K) are(g+1)g and 2g. (Notice that the boundary slopes

do not depend on the choice of a representative of the homology class.)

A result in [17] tells us that the Dehn filling Mmagic(g) = Mmagic((g+1)g , (g+1)g , 2g) alongslopes (g+1)g ,

(g+1)g and 2g is a hyperbolic manifold. Hence a is pseudo-Anosov since its

mapping torus T(a) is homeomorphic to Mmagic(g). This completes the proof.

4.2 Asymptotic behavior of entropy function

We consider asymptotic behaviors of the entropy function for a family of homology classes inProposition 3.6.

Theorem 4.11. Letx + y be a homology class in C1.

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(1) limx,y ent(x + y) = 0.

(2) limy ent(x + y) =

log2

x.

Of course, limx ent(x + y) =log2

y by symmetry.

Proof. (1) We may suppose that x y.By [16, Theorem 3.5], we have an inequality

ent(a + b) min{ent(a), ent(b)}

for a, b int(C). Hence for all > 0 so that x > 0 and for all > 0,

ent(x + (x + )) min{ent((x ) + x), ent( + )}.

Notice that ent( + ) goes to

as goes to 0. If one takes > 0 sufficiently small, then one

may assume thatent(x + (x + )) ent((x ) + x).

Since ent() is continuous, we have ent(x + (x + )) ent(x + x). Thus,

limx ent(x + (x + )) limx ent(x + x) = limx

1

xent( + ) = 0.

Since > 0 is arbitrary, the proof is completed.

(2) By Theorem 3.2, the dilatation of x + y + 0 C1(Z) is the largest real root of

P(tx, ty, t0) = P(tx, ty, 1) = tyRx(t) + (Rx)(t),

where Rx(t) = tx 2. By Lemma 2.1, the largest real root of P(tx, ty, 1) converges to 21/x, which

is the unique real root of Rx(t), as y . This claim can be extended to homology classes ofC1(Q), that is the dilatation of x + y C1(Q) converges to 21/x as y . Since the entropyfunction on C1(Q) can be extended to C1 uniquely, the proof is completed.

Proposition 4.12. The entropy of the homology classes (n + 1) + n + (n 1) int(C)converges to the logarithm of the golden mean 1+

5

2 as n goes to .Proof. We have

P(tn+1, tn, tn1) = tn1

tn(t2 t 1) + (t2 t 1)

.

If (n + 1) + n+ (n 1) is an integral homology class, then its dilatation n is the largest realroot of tn(t2 t 1) + (t2 t 1). The polynomial t2 t 1 has the real root 1+

5

2 > 1. By

Lemma 2.1, n converges to1+

52 as n goes to . Since ent() is continuous on int(C), the proof

is completed.

4.3 Relation between horseshoe braid and braid Tm,p

The horseshoe map was discovered by S. Smale around 1960. For the study of the dynamicalsystems, this map is significant since it is a simple factor possessing the chaotic dynamics. In thissection, we relate monodromies for homology classes in C1(Z) to the horseshoe map.

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The horseshoe map H : D D is a diffeomorphism of the disk D such that the action of H onthe rectangle R and two half disks S0, S1 is given as in Figure 22. More precisely, the restrictionH|Ri for i {0, 1} is an affine map such that H contracts Ri vertically and stretches horizontally,and H|S0S1 : S0 S1 S0 S1 is a contraction map. The set =

jZ

Hj(R) is invariant under

H. Notice that the map H| : can be described by using the symbolic dynamics as follows.We set S = {0, 1}Z, that is the set of all two sided infinite sequences s = ( s1s0|s1 ) of 0and 1, where we put the symbol | between the 0th element and the 1st element. We introduce themetric on Sas follows.

d(s, t) =iZ

|si ti|2|i|

,

where s = ( s1s0|s1s2 ) and t = ( t1t0|t1t2 ).Theorem 4.13 (Smale). Let s : S S be the shift map, i.e, s is a homeomorphism which isdefined by

s(

s0

|s1s2

) = (

s0s1

|s2

Documents

Eiko Kin and Mitsuhiko Takasawa- Pseudo-Anosov braids with small entropy and the magic 3-manifold