An asymptotic behavior of the dilatation for a family of pseudo-Anosov braids
aa r X i v : . [ m a t h . G T ] N ov An asymptotic behavior of the dilatation for a family ofpseudo-Anosov braids
Eiko Kin ∗ and Mitsuhiko TakasawaNovember 21, 2018 Abstract. The dilatation of a pseudo-Anosov braid is a conjugacy invariant. In this paper, westudy the dilatation of a special family of pseudo-Anosov braids. We prove an inductive formulato compute their dilatation, a monotonicity and an asymptotic behavior of the dilatation for thisfamily of braids. We also give an example of a family of pseudo-Anosov braids with arbitrarilysmall dilatation such that the mapping torus obtained from such braid has 2 cusps and has anarbitrarily large volume.Keywords: mapping class group, braid, pseudo-Anosov, dilatationMathematics Subject Classification : Primary 37E30, 57M27, Secondary 57M50
Let Σ = Σ g,p be an orientable surface of genus g with p punctures, and let M (Σ) be the mappingclass group of Σ. The elements of M (Σ), called mapping classes , are classified into 3 types: periodic,reducible and pseudo-Anosov [10]. For a pseudo-Anosov mapping class φ , the dilatation λ ( φ ) isan algebraic integer strictly greater than 1. The dilatation of a pseudo-Anosov mapping class is aconjugacy invariant.Let D n be an n -punctured closed disk. The mapping class group M ( D n ) of D n is isomorphicto a subgroup of M (Σ ,n +1 ). There is a natural surjective homomorphismΓ : B n → M ( D n )from the n -braid group B n to the mapping class group M ( D n ) [2]. We say that a braid β ∈ B n is pseudo-Anosov if Γ( β ) is pseudo-Anosov, and if this is the case the dilatation λ ( β ) of β is definedequal to λ (Γ( β )). Henceforth, we shall abbreviate ‘pseudo-Anosov’ to ‘pA’.We now introduce a family of braids. Let β ( m ,m , ··· ,m k +1 ) be the braid as depicted in Figure 1,for each integer k ≥ m i ≥
1. These are all pA (Proposition 4.1). We will provea monotonicity, an inductive formula to compute their dilatation and an asymptotic behavior ofthe dilatation for this family of braids.
Proposition 1.1. (Monotonicity) For each integer i with ≤ i ≤ k + 1 , we have λ ( β ( m , ··· ,m i , ··· ,m k +1 ) ) > λ ( β ( m , ··· ,m i +1 , ··· ,m k +1 ) ) . ∗ The first author is partially supported by Grant-in-Aid for Young Scientists (B) (No. 17740094), The Ministryof Education, Culture, Sports, Science and Technology, Japan β ( m ,m , ··· ,m k +1 ) , (center) β (2 , , , (right) β (3 , . Hence if m i ≤ m ′ i for each i , then λ ( β ( m , ··· ,m k +1 ) ) ≥ λ ( β ( m ′ , ··· ,m ′ k +1 ) ) . For an integral polynomial f ( t ) of degree d , the reciprocal of f ( t ), denoted by f ∗ ( t ), is t d f (1 /t ). Theorem 1.2. (Inductive formula) The dilatation of the pA braid β ( m , ··· ,m k +1 ) is the largest rootof the polynomial t m k +1 R ( m , ··· ,m k ) ( t ) + ( − k +1 R ( m , ··· ,m k ) ∗ ( t ) , where R ( m , ··· ,m i ) ( t ) is given inductively as follows: R ( m ) ( t ) = t m +1 ( t − − t, and R ( m , ··· ,m i ) ( t ) = t m i ( t − R ( m , ··· ,m i − ) ( t ) + ( − i tR ( m , ··· ,m i − ) ∗ ( t ) for ≤ i ≤ k. Theorem 1.3. (Asymptotic behavior) We have(1) lim m , ··· ,m k +1 →∞ λ ( β ( m , ··· ,m k +1 ) ) = 1 and(2) lim m i ,m i +1 , ··· ,m k +1 →∞ λ ( β ( m , ··· ,m k +1 ) ) = λ ( R ( m , ··· ,m i − ) ( t )) > for i ≥ , where λ ( f ( t )) denotesthe maximal absolute value of the roots of f ( t ) . For a pA braid β , let φ be the pA mapping class Γ( β ). The dilatation λ ( φ ) can be computed asfollows. A smooth graph τ , called a train track and a smooth graph map b φ : τ → τ are associatedwith φ . The edges of τ are classified into real edges and infinitesimal edges, and the transitionmatrix M real ( b φ ) with respect to real edges can be defined. Then the dilatation λ ( φ ) equals thespectral radius of M real ( b φ ). For more details, see Section 2.2.For the computation of the dilatation of the braid β ( m , ··· ,m k +1 ) , we introduce combined trees and combined tree maps in Section 3. For a given ( m , · · · , m k +1 ), one can obtain the combined tree Q ( m , ··· ,m k +1 ) and the combined tree map q ( m , ··· ,m k +1 ) inductively. For example, for ( m , m , m ) =(4 , , Q ( m ,m ,m ) , depicted in Figure 2, is obtained by gluing the combinedtree Q ( m ,m ) and another tree which depends m . The combined tree map q ( m ,m ,m ) , as shownin Figure 3, is defined by the composition of an extension of the combined tree map q ( m ,m ) andanother tree map which depends on m .By the proof of Proposition 4.1, it turns out that the spectral radius of the transition matrix M ( q ( m , ··· ,m k +1 ) ) obtained from q ( m , ··· ,m k +1 ) equals that of M real ( b φ ), where φ = Γ( β ( m , ··· ,m k +1 ) ),2igure 2: Q (4 , , (right) is obtained by gluing Q (4 , (left) and another tree (center).Figure 3: (top) q (4 , , (bottom) q (4 , , .3hat is the spectral radius of M ( q ( m , ··· ,m k +1 ) ) equals the dilatation λ ( β ( m , ··· ,m k +1 ) ). Proposition 1.1and Theorems 1.2, 1.3 will be shown by using the properties of combined tree maps.In the final part, we will consider the two invariants of pA mapping classes, the dilatation andthe volume. Choosing any representative f : Σ → Σ of a mapping class φ , we form the mappingtorus T ( φ ) = Σ × [0 , / ∼ , where ∼ identifies ( x,
0) with ( f ( x ) , φ is pA if and only if T ( φ ) admits acomplete hyperbolic structure of finite volume [7]. Since such a structure is unique up to isometry,it makes sense to speak of the volume vol( φ ) of φ , the hyperbolic volume of T ( φ ). For a pA braid β , we define the volume vol( β ) as equal to vol(Γ( β )), the volume of the mapping torus T (Γ( β )).Theorem 1.3(1) tells us that dilatation of braids can be arbitrarily small. We consider whathappen for the volume of a family of pseudo-Anosov mapping classes whose dilatation is arbitrarilysmall. It is not hard to see the following. Proposition 1.4.
There exists a family of pA mapping classes φ n of M ( D n ) such that lim n →∞ λ ( φ n ) = 1 and lim n →∞ vol( φ n ) = ∞ and such that the number of the cusps of the mapping torus T ( φ n ) goes to ∞ as n goes to ∞ . Proposition 1.4 is not so surprising, because the volume of each cusp is bounded below uniformly.We show the following.
Proposition 1.5.
There exist a family of pA mapping classes φ n of M ( D n ) such that lim n →∞ λ ( φ n ) = 1 and lim n →∞ vol( φ n ) = ∞ and such that the number of the cusps of the mapping torus T ( φ n ) is for each n . Proposition 1.5 is a corollary of the following theorem.
Theorem 1.6.
For any real number λ > and any real number v > , there exist an integer k ≥ and an integer m ≥ such that for any integer m i ≥ m with ≤ i ≤ k + 1 , we have λ ( β ( m , ··· ,m k +1 ) ) < λ and vol( β ( m , ··· ,m k +1 ) ) > v . Here we note that for a braid b , the mapping torus T (Γ( b )) is homeomorphic to the link complement S \ b in the 3 sphere S , where b is a union of the closed braid of b and the braid axis (Figure 4).When b is a braid β ( m , ··· ,m k +1 ) , the link b has 2 components, and hence the number of cusps of T (Γ( b )) is 2. A homeomorphism Φ : Σ → Σ is pseudo-Anosov ( pA ) if there exists a constant λ = λ (Φ) > dilatation of Φ, and there exists a pair of transverse measured foliations F s and F u suchthat Φ( F s ) = 1 λ F s and Φ( F u ) = λ F u . b .A mapping class φ ∈ M (Σ) is said to be pseudo-Anosov ( pA ) if φ contains a pA homeomorphism.We define the dilatation of a pA mapping class φ , denoted by λ ( φ ), to be the dilatation of a pAhomeomorphism of φ .Let G be a graph. We denote the set of vertices by V ( G ) and denote the set of edges by E ( G ).A continuous map g : G → G ′ from G into another graph G ′ is said to be a graph map . When G and G ′ are trees, a graph map g : G → G ′ is said to be a tree map . A graph map g is called Markov if g ( V ( G )) ⊂ V ( G ′ ) and for each point x ∈ G such that g ( x ) / ∈ V ( G ′ ), g is locally injective at x (thatis g has no ‘back track’ at x ). In the rest of the paper we assume that all graph maps are Markov.For a graph map g , we define the transition matrix M ( g ) = ( m i,j ) such that the i th edge e ′ i or the same edge with opposite orientation ( e ′ i ) − of G ′ appears m i,j -times in the edge path g ( e j )for the j th edge e j of G . If G = G ′ , then M ( g ) is a square matrix, and it makes sense to considerthe spectral radius, λ ( g ) = λ ( M ( g )), called the growth rate for g . The topological entropy of g isknown to be equal to log λ ( g ).In Section 2.1 we recall results regarding Perron-Frobenius matrices. In Section 2.2 we quicklyreview a result from the train track theory which tells us that if a given mapping class φ induces acertain graph map, called train track map , whose transition matrix is Perron-Frobenius, then φ ispA and λ ( φ ) equals the growth rate of the train track map. In Section 2.3 we consider roots of afamily of polynomials to study the dilatation of pA mapping classes and give some results regardingthe asymptotic behavior of roots of this family. Let M = ( m i,j ) and N = ( n i,j ) be matrices with the same size. We shall write M ≥ N (resp. M > N ) whenever m i,j ≥ n i,j (resp. m i,j > n i,j ) for each i, j . We say that M is positive (resp. non-negative ) if M > (resp. M ≥ ), where is the zero matrix.For a square and non-negative matrix T , let λ ( T ) be its spectral radius, that is the maximalabsolute value of eigenvalues of T . We say that T is irreducible if for every pair of indices i and j ,there exists an integer k = k i,j > i, j ) entry of M k is strictly positive. The matrix T is primitive if there exists an integer k > T k is positive. By definition,a primitive matrix is irreducible. A primitive matrix T is Perron-Frobenius , abbreviated to PF, if T is an integral matrix. For M ≥ T , if T is irreducible then M is also irreducible. The followingtheorem is commonly referred to as the Perron-Frobenius theorem. Theorem 2.1. [8] Let T be a primitive matrix. Then, there exists an eigenvalue λ > of T suchthat λ has strictly positive left and right eigenvectors b x and y respectively, and(2) λ > | λ ′ | for any eigenvalue λ ′ = λ of T . If T is a PF matrix, the largest eigenvalue λ in the sense of Theorem 2.1 is strictly greater than1, and it is called the PF eigenvalue . The corresponding positive eigenvector is called the
PFeigenvector .The following will be useful.
Lemma 2.2. [8, Theorem 1.6, Exercise 1.17] Let T be a primitive matrix, and let s be a positivenumber. Suppose that a non-zero vector y ≥ satisfies T y ≥ s y . Then,(1) λ ≥ s , where λ is the largest eigenvalue of T in the sense of Theorem 2.1, and(2) s = λ if and only if T y = s y .Proof. (1) Let b x be a positive left eigenvector of T . Then, b x T y = λ b xy ≥ s b xy . Hence we have λ ≥ s .(2) (‘Only if’ part) Suppose that s = λ , and suppose that T y ≥ λ y and T y = λ y . Premultiplyingthis inequality by a positive left eigenvector b x of T , we have b x T y (= λ b xy ) > λ b xy . Hence λ > λ , which is a contradiction.(‘If’ part) Suppose that T y = s y . Premultiplying this equality by a positive left eigenvector b x of T , we obtain λ = s . (cid:3) For a non-negative k × k matrix T , one can associate a directed graph G T as follows. The graph G T has vertices numbered 1 , , · · · , k and an edge from the j th vertex to the i th vertex if and onlyif the ( i, j ) entry T i,j = 0. By the definition of G T , one easily verifies the following. Lemma 2.3.
Let T be a non-negative square matrix.(1) T is irreducible if and only if for each i, j , there exists an integer n i,j > such that the directedgraph G T has an edge path of length n i,j from the j th vertex to the i th vertex.(2) T is primitive if and only if there exists an integer n > such that for each i, j , the directedgraph G T has an edge path of length n from the j th vertex to the i th vertex. A smooth branched 1-manifold τ embedded in D n is a train track if each component of D n \ τ iseither a non-punctured k -gon ( k ≥ k -gon ( k ≥
1) or an annulus such that aboundary component of the annulus coincides with the boundary of D n and the other componenthas at least 1 prong. A smooth map from a train track into itself is called a train track map .Let f : D n → D n be a homeomorphism. A train track τ is invariant under f if f ( τ ) can becollapsed smoothly onto τ in D n . In this case f induces a train track map b f : τ → τ . An edge of6 is called infinitesimal if there exists an integer N > b f N ( τ ) is a periodic edge under b f . An edge of τ is called real if it is not infinitesimal. The transition matrix of b f is of the form: M ( b f ) = M real ( b f ) A M inf ( b f ) ! , where M real ( b f ) (resp. M inf ( b f )) is the transition matrix with respect to real (resp. infinitesimal)edges. The following is a consequence of [1]. Proposition 2.4.
A mapping class φ ∈ M ( D n ) is pA if and only if there exists a homeomorphism f : D n → D n of φ and there exists a train track τ such that τ is invariant under f , and for theinduced train track map b f : τ → τ , the matrix M real ( b f ) is PF. When φ is a pA mapping class, wehave λ ( φ ) = λ ( M real ( b f )) . For an integral polynomial S ( t ), let λ ( S ( t )) be the maximal absolute value of roots of S ( t ). For amonic integral polynomial R ( t ), we set Q n, ± ( t ) = t n R ( t ) ± S ( t )for each integer n ≥
1. The polynomial R ( t ) (resp. S ( t )) is called dominant (resp. recessive )for a family of polynomials { Q n, ± ( t ) } n ≥ . In case where S ( t ) = R ∗ ( t ), we call t n R ( t ) ± R ∗ ( t ) the Salem-Boyd polynomial associated to R ( t ). E. Hironaka shows that such polynomials have severalnice properties [3, Section 3]. The following lemma shows that roots of Q n, ± ( t ) lying outside theunit circle are determined by those of R ( t ) asymptotically. Lemma 2.5.
Suppose that R ( t ) has a root outside the unit circle. Then, the roots of Q n, ± ( t ) outside the unit circle converge to those of R ( t ) counting multiplicity as n goes to ∞ . In particular, λ ( R ( t )) = lim n →∞ λ ( Q n, ± ( t )) . The proof can be found in [3]. We recall a proof here for completeness.
Proof.
Consider the rational function Q n, ± ( t ) t n = R ( t ) ± S ( t ) t n . Let θ be a root of R ( t ) with multiplicity m outside the unit circle. Let D θ be any small diskcentered at θ that is strictly outside of the unit circle and that contains no roots of R ( t ) other than θ . Then, | R ( t ) | has a lower bound on the boundary ∂D θ by compactness. Hence there exists anumber n θ > θ such that | R ( t ) | > | S ( t ) t n | on ∂D θ for any n > n θ . By the Rouch´e’stheorem, it follows that R ( t ) and R ( t ) ± S ( t ) t n (hence R ( t ) and Q n, ± ( t )) have the same m roots in D θ . Since D θ can be made arbitrarily small and there exist only finitely many roots of R ( t ), theproof of Lemma 2.5 is complete. (cid:3) Lemma 2.6.
Suppose that R ( t ) has no roots outside the unit circle, and suppose that Q n, ± ( t ) hasa real root µ n greater than for sufficiently large n . Then, lim n →∞ µ n = 1 . roof. For any ε >
0, let D ε be the disk of radius 1 + ε around the origin in the complex plane.Then, for any sufficiently large n , we have | R ( t ) | > | S ( t ) t n | for all t on ∂D ε . Moreover, R ( t ) and ± S ( t ) t n are holomorphic on the complement of D ε in the Riemann sphere. By Rouch´e’s theorem, R ( t ) and R ( t ) ± S ( t ) t n (hence R ( t ) and Q n, ± ( t )) have no roots outside D ε . Hence µ n converges to 1as n goes to ∞ . (cid:3) For an n × n matrix M , let M ( t ) be the characteristic polynomial | tI − M | of M , where I = I n isthe n × n identity matrix. Let M ∗ ( t ) be the reciprocal polynomial of M ( t ). Then, M ∗ ( t ) = t n (cid:12)(cid:12)(cid:12) t I − M (cid:12)(cid:12)(cid:12) = | I − tM | , that is M ∗ ( t ) equals the determinant of the matrix I − tM .This section introduces combined tree maps . Given two trees we combine these trees with anothertree of star type having the valence n + 1 vertex and define a new tree, say Q n . When two treemaps on Q n satisfy certain conditions ( L1 , L2 , L3 and R1 , R2 , R3 ), we can define the combinedtree map q n on Q n and obtain a family of tree maps { q n : Q n → Q n } n ≥ . In Section 3.1 wegive a sufficient condition that guarantees M ( q n ) is PF. In Section 3.2 we consider combined treemaps in a particular setting. Then, we give a formula for M ( q n )( t ) and M ( q n ) ∗ ( t ) and analyze theasymptotic behavior of the growth rate for q n . This analysis will be applied to train track maps inSection 4. We assume that all trees are embedded in the disk D . By the trivial tree T , we mean the treewith only one vertex. Let G n, + and G n, − be trees of star type as in Figure 5, having one vertex ofvalence n + 1.Figure 5: trees (left) G n, + and (right) G n, − having one vertex of valence n + 1.Let G L (resp. G R ) be a tree (possibly a trivial tree) with a valence 1 vertex, say v L (resp. v R ).Let w L and w R be vertices of G n, + as in Figure 5, and glue G L , G n, + and G R together so thatfor S ∈ { L, R } , v S and w S become one vertex (Figure 6). The resulting tree Q n, + is called the combined tree , obtained from the triple ( G L , G n, + , G R ). We define the combined tree Q n, − , obtainedfrom the triple ( G L , G n, − , G R ) in the same manner.Before we define combined tree maps on Q n, + / − , we label the edges of Q n, + / − . Let ℓ be thenumber of edges of G L , and let r be the number of edges of G R plus 1. Note that the number ofedges of Q n, + / − is ℓ + n + r . 8igure 6: combined trees Q n, + : (left) general case, (right) case where G R is a trivial tree. • The edges of G n, + / − are numbered ℓ + 1 to ℓ + n + 1 in the clockwise/counterclockwise directionas in Figure 7. • The edge of G L sharing a vertex with the ( ℓ + 1) st edge is numbered ℓ and the remaining edgesof G L are numbered 1 to ℓ − • The edge of G R sharing a vertex with the ( ℓ + n + 1) st edge is numbered ℓ + n + 2 and the remainingedges of G R are numbered ℓ + n + 3 to ℓ + n + r arbitrarily.The edge numbered i is denoted by e i .Now we take a tree map g L : Q n, + / − → Q n, + / − satisfying the following conditions. L1 The map g L restricted to the set of vertices of E ( Q n, + / − ) \ ( E ( G L ) ∪ { e ℓ +1 } ) is the identity. L2 g L ( G L ) ⊂ G L . L3 The edge path g L ( e ℓ +1 ) passes through e ℓ +1 only once and passes through e ℓ .Next, we take a tree map g R : Q n, + / − → Q n, + / − satisfying the following conditions. R1 The map g R restricted to the set of vertices of E ( Q n, + / − ) \ ( E ( G R ) ∪ { e ℓ + n +1 } ) is the identity. R2 g R ( G R ) ⊂ G R . R3 The edge path g R ( e ℓ + n +1 ) passes through e ℓ + n +1 only once and passes through e ℓ + n +2 .Finally, we define the tree map g n : Q n, + / − → Q n, + / − satisfying the following conditions. n1 The map g n restricted to the set of vertices of E ( Q n, + / − ) \ ( E ( G n, + / − ) ∪ { e ℓ , e ℓ + n +2 } ) is theidentity. n2 g n rotates the subtree G n, + / − as in Figure 7. n3 The image of each e ∈ { e ℓ , e ℓ + n +2 } is as in Figure 7. The length of the edge path g n ( e ) is 3.The composition q n = g R g n g L : Q n, + / − → Q n, + / − is called the combined tree map , obtained from the triple ( g L , g n , g R ). It makes sense to considerthe transition matrices M ( g S ) of g S | G S : G S → G S , S ∈ { L, R } and M ( g n ) of g n | G n, + / − : G n, + / − →G n, + / − . The transition matrix M ( q n ) has the following form: M ( q n ) = M L A B M n CD E M R , where M n = ∗ ∗ (3.1)9igure 7: (top) g n rotates G n, + , (bottom) g n rotates G n, − . The edges e ℓ and e ℓ + n +2 and theirimages are drawn in bold.(each empty space in M n represents the number 0), and the block matrices satisfy M L ≥ M ( g L ), M n ≥ M ( g n ) and M R ≥ M ( g R ). (In fact M L = M ( g L ), although we will not be using this fact.)Throughout this subsection, we assume that the trees G L and G R are not trivial. It is straight-forward to see the following from the defining conditions of g L , g n and g R . Lemma 3.1.
Let m i,j be the ( i, j ) entry of M ( q n ) . We have(1) m ℓ + n,ℓ + n +2 = 1 and m ℓ + n +1 ,ℓ + n +2 = 1 , and(2) m ℓ,ℓ +1 > , m ℓ +1 ,ℓ +1 > and m ℓ + n +1 ,ℓ +1 > . Moreover, m ℓ +1 ,j = m ℓ + n +1 ,j for each j with ≤ j ≤ ℓ and m ℓ +1 ,j > for some ≤ j ≤ ℓ , and(3) m ℓ + n +2 ,ℓ +1 > . An important feature is that the growth rate of q n is always greater than 1 if M ( g L ) and M ( g R )are irreducible, which will be shown in Proposition 3.3. We first show that M ( q n ) is irreduciblein this case. Notice that M ( g n ) is always irreducible, and since M n ≥ M ( q n ) so M n must beirreducible as well. Lemma 3.2.
Let q n = g R g n g L : Q n, + / − → Q n, + / − be the combined tree map. Assume that both M ( g L ) and M ( g R ) are irreducible. Then, M ( q n ) is irreducible.Proof. Note that M L M R and M n are irreducible. Let G q n be the directed graph of M ( q n ). Weidentify vertices of G q n with edges of Q n, + / − . Let V L (resp. V R , V n ) be the set of vertices of G q n coming from the set of edges of the subtree G L (resp. G R , G n, + / − ) of Q n, + / − . Lemma 3.1(2) showsthat there exists an edge connecting the set V L to the set V n , and there exists an edge connectingthe set V n to the set V L . This is also true between V n and V R by Lemma 3.1(1,3). Thus, one canfind an edge path between any two vertices of G q n . (cid:3) Proposition 3.3.
Under the assumptions of Lemma 3.2, M ( q n ) is PF.Proof. Lemma 3.1(2) says that the directed graph G q n has an edge from the vertex v ℓ +1 to itself,and we denote such edge by e . Since M ( q n ) is irreducible, for any vertex v of G q n there exists anedge path E = e e · · · e n ( v ) from v ℓ +1 to v . Thus, for any n ≥ n ( v ) we have an edge path e · · · eE of length n from v ℓ +1 to v . Since the number of vertices is finite, there exists an integer N > w of G q n and any integer n ≥ N we have an edge path of length n from v ℓ +1 w . Since there exists an edge path from any vertex x of G q n to v ℓ +1 , we can find a sufficientlylarge integer N ′ such that for any pair of vertices x and w there exists an edge path of length N ′ from x to w . Thus, M ( q n ) is PF. (cid:3) The following property is crucial in proving Proposition 1.1 and Theorem 1.3.
Proposition 3.4.
Under the assumptions of Lemma 3.2, we have λ ( M ( q n )) > λ ( M ( q n +1 )) > .Proof. To compare M ( q n +1 ) with M ( q n ) we introduce a new labeling of edges of Q n +1 , + / − .The trees G L and G R are the common subtrees for both trees Q n, + / − and Q n +1 , + / − . Edges of thesubtrees G L and G R of Q n +1 , + / − are numbered in the same manner as those of Q n, + / − , and edgesof G n +1 are numbered ℓ + 1 , ℓ + n + r + 1 , ℓ + 2 , ℓ + 3 , · · · , ℓ + n + 1in the clockwise/counterclockwise direction. Here the edge sharing a vertex with the ℓ th edge isnumbered ℓ + 1.Let M ( q n ) = ( m i,j ) ≤ i,j ≤ ℓ + n + r be the matrix given in (3.1). Then, M ( q n +1 ) = ( m ′ i,j ) ≤ i,j ≤ ℓ + n + r +1 with new labeling has the following form: M ( q n +1 ) = M L A ∗ B . .. C ∗ D E M R . Put s = λ ( M ( q n +1 )) > y = t ( y , · · · , y ℓ + n + r +1 ) be the PF eigenvector for M ( q n +1 ).Then, ℓ + n + r +1 X j =1 m ′ i,j y j = sy j for i with 1 ≤ i ≤ ℓ + n + r + 1 . (3.2)For i = ℓ + 1 and i = ℓ + n + r + 1 of (3.2) we have ℓ + n + r +1 X j =1 m ′ ℓ +1 ,j y j = ℓ +1 X j =1 m ℓ +1 ,j y j + y ℓ + n + r +1 = sy ℓ +1 and y ℓ +2 = sy ℓ + n + r +1 . These two equalities together with s > ℓ +1 X j =1 m ℓ +1 ,j y j + y ℓ +2 > sy ℓ +1 . (3.3)The equalities (3.2) for all i = ℓ + 1 , ℓ + n + r + 1 together with the inequality (3.3) imply M ( q n )ˆ y ≥ s ˆ y , where ˆ y = t ( y , · · · , y ℓ + n + r ) . By Lemma 2.2(1), we have λ ( M ( q n )) ≥ s = λ ( M ( q n +1 )). By Lemma 2.2(2) together with (3.3),we have λ ( M ( q n )) > s . (cid:3) .2 Asymptotic behavior of growth rate In this section we concentrate on the combined tree obtained from the triple ( G L , G n, + / − , T ). Weassume that g L ( G L ) = G L and study the combined tree map q n = g n g L .Let R be the subtree of Q n, + / − such that R is obtained from the subtree G L together with the( ℓ + 1) st edge. (Hence E ( R ) = { e , e , · · · , e ℓ +1 } .) For an example of R , see Figure 8.By the assumption g L ( G L ) = G L , we have q n ( R ) ⊃ R and hence the following tree map r : R → R is well defined: for each e ∈ E ( R ), the edge path r ( e ) is given by the edge path q n ( e ) byeliminating edges which do not belong to E ( R ). The tree map r does not depend on the choice of n . The transition matrix M ( r ) is given by the upper-left ( ℓ + 1) × ( ℓ + 1) submatrix of M ( q n ). Wecall R the dominant tree and r the dominant tree map for a family of combined tree maps { q n } n ≥ .Figure 8: (left) Q n, + , (right) its subtree R We now define a polynomial S ( t ) (resp. U ( t )) as follows: Consider the matrix tI − M ( q n ) (resp. I − tM ( q n )) and replace the ( ℓ + 1) st row by the last row. Take the upper-left ( ℓ + 1) × ( ℓ + 1)submatrix of the resulting matrix, denoted by S (resp. U ), and then S ( t ) (resp. U ( t )) is definedequal to the determinant of S (resp. U ). It is not hard to see that the matrices S and U do notdepend on n .The following statement, which will be crucial later, tells us that M ( r )( t ) is the dominantpolynomial and S ( t ) is the recessive polynomial for a family of polynomials { M ( q n )( t ) } n ≥ . Proposition 3.5.
We have(1) M ( q n )( t ) = t n M ( r )( t ) + S ( t ) , and(2) M ( q n ) ∗ ( t ) = t n U ( t ) + M ( r ) ∗ ( t ) .Proof. The transition matrix M ( q n ) = ( m i,j ) is of the form M ( q n ) = M ( r ) 1 . .. 1 m ℓ + n +1 , · · · m ℓ + n +1 ,ℓ +1 , and it is easy to see that m ℓ +1 ,j = m ℓ + n +1 ,j for 1 ≤ j ≤ ℓ . For the proof of (1) (resp. (2)), applythe determinant expansion with respect to the last row of tI − M ( q n ) (resp. I − tM ( q n )). (cid:3) Proposition 3.6.
Suppose that M ( g L ) is irreducible. Then, we have(1) M ( q n ) is PF for each n and λ ( M ( q n )) > λ ( M ( q n +1 )) , and(2) λ ( M ( r )) = lim n →∞ λ ( M ( q n )) . roof. (1) The proof is parallel to the proofs of Propositions 3.3 and 3.4.(2) Apply Lemma 2.5 with Proposition 3.5(1). (cid:3) This section is devoted to proving Proposition 1.1 and Theorems 1.2, 1.3, 1.6.
Proposition 4.1.
The braids β ( m , ··· ,m k +1 ) are pA. By a result of W. Menasco’s [6, Corollary 2], if L is a non-split prime alternating link which is not atorus link, then S \ L has a complete hyperbolic structure of finite volume. Since β ( m , ··· ,m k +1 ) is a2 bridge link as depicted in Figure 13, his result tells us that β ( m , ··· ,m k +1 ) is pA. Here we will showProposition 4.1 by using Proposition 2.4. As a result, we will find the polynomial whose largestroot equals the dilatation of β ( m , ··· ,m k +1 ) . Proof of Proposition 4.1.
To begin with, we define a tree Q ( m , ··· ,m k +1 ) and a tree map q ( m , ··· ,m k +1 ) on the tree Q ( m , ··· ,m k +1 ) inductively.For k = 1 let Q ( m ,m ) be the combined tree obtained from the triple ( G m , + , G m , − , T ). Takethe tree maps g m and g m with conditions n1 , n2 , n3 and let us define q ( m ,m ) as the combinedtree map q ( m ,m ) = g m g m : Q ( m ,m ) → Q ( m ,m ) (Figure 9) . Figure 9: case ( m , m ) = (4 , k . Let Q ( m , ··· ,m k +1 ) be the combined tree ob-tained from the triple ( Q ( m , ··· ,m k ) , G m k +1 , + / − , T ) in case k + 1 odd/even. We extend q ( m , ··· ,m k ) : Q ( m , ··· ,m k ) → Q ( m , ··· ,m k ) to a tree map b q : Q ( m , ··· ,m k +1 ) → Q ( m , ··· ,m k +1 ) satisfying L1 , L2 , L3 so that for the edge e of G m k +1 , + / − sharing a vertex with the edge of Q ( m , ··· ,m k ) ,the length of the edge path b q ( e ) is 3. Let us define the combined tree map q ( m , ··· ,m k +1 ) = g m k +1 b q : Q ( m , ··· ,m k +1 ) → Q ( m , ··· ,m k +1 ) .
13y Proposition 3.6(1), the transition matrix M ( q ( m , ··· ,m k +1 ) ) is PF.We now deform Q = Q ( m , ··· ,m k +1 ) into a train track τ ( m , ··· ,m k +1 ) (as in Figure 10):1. Puncture a disk near each valence 1 vertex of Q , and connect a 1-gon at the vertex whichcontains the puncture.2. Deform a neighborhood of a valence m i + 1 vertex of the subtree G m i , + / − of Q into an ( m i + 1)-gon for each i .3. For each i + 1 odd/even, puncture above/below the vertex of valence 2 which connects the twosubtrees G m i , + / − and G m i +1 , − / + . Deform a neighborhood of the vertex and connect a 1-gon whichcontains the puncture.Figure 10: numbers 1 , , q = q ( m , ··· ,m k +1 ) induces the graph map b q on τ = τ ( m , ··· ,m k +1 ) into itself. Since b q rotates apart of the train track smoothly (Figure 11), it turns out that b q is a smooth map. It is easy to showthe existence of a representative homeomorphism f of φ = Γ( β ( m , ··· ,m k +1 ) ) such that τ is invariantunder f and such that b q is the train track map induced by f . The transition matrix of b q withrespect to real edges of τ is exactly equal to the PF matrix M ( q ( m , ··· ,m k +1 ) ). By Proposition 2.4,the braid β ( m , ··· ,m k +1 ) is pA. This completes the proof of Proposition 4.1. (cid:3) Figure 11: b q rotates a part of the train track smoothly. Example 4.2.
Let us express the formula to compute the dilatation of the braids β (4 ,m ) for m ≥ .We know that the dilatation of the braid β (4 ,m ) is the largest root of the polynomial M ( q (4 ,m ) )( t ) forthe combined tree map q (4 ,m ) . By using q (4 , shown in Figure 9, the transition matrix for q (4 , is M ( q (4 , ) = . he transition matrix M ( r ) of the dominant tree map r (Figure 12) for a family of combinedtree maps { q (4 ,m ) } m ≥ is the upper-left × matrix. Hence the dominant polynomial M ( r )( t ) for { q (4 ,m ) } m ≥ equals t − t − t with the largest root ≈ . . In this case the recessive polynomial S ( t ) is S ( t ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − t − t − t − − − t − − − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − t − t + 1 . By Proposition 3.5(1), the dilatation of β (4 ,m ) is the largest root of M ( q (4 ,m ) )( t ) = t m M ( r )( t ) + S ( t ) = t m ( t − t − t ) + ( − t − t + 1) , and Proposition 3.6(2) says that lim m →∞ λ ( β (4 ,m ) ) ≈ . . Figure 12: dominant tree map r : R → R for { q (4 ,m ) } m ≥ .We are now ready to show Proposition 1.1. Proof of Proposition 1.1.
Recall the tree Q ( m , ··· ,m k +1 ) and the tree map q ( m , ··· ,m k +1 ) used inthe proof of Proposition 4.1. For i even, Q ( m , ··· ,m k +1 ) is also the combined tree obtained fromthe triple ( Q ( m , ··· ,m i − ) , G m i , − , Q ( m i +1 , ··· ,m k +1 ) ). Then, q ( m , ··· ,m k +1 ) is also the combined tree mapgiven by b q ( m , ··· ,m i − ) g m i b q ( m i +1 , ··· ,m k +1 ) : Q ( m , ··· ,m k +1 ) → Q ( m , ··· ,m k +1 ) , where b q ( m , ··· ,m i − ) and b q ( m i +1 , ··· ,m k +1 ) are suitable extensions of q ( m , ··· ,m i − ) and q ( m i +1 , ··· ,m k +1 ) respectively. By Proposition 3.4, the claim holds.For i odd, Q ( m , ··· ,m k +1 ) is the combined tree obtained from ( Q ( m , ··· ,m i − ) , G m i , + , Q ′ ( m i +1 , ··· ,m k +1 ) ),where Q ′ ( m i +1 , ··· ,m k +1 ) is the tree obtained from Q ( m i +1 , ··· ,m k +1 ) by the horizontal reflection. Then,the proof is similar to that for the even case. (cid:3) We turn to the proof of Theorems 1.2 and 1.3.
Proof of Theorem 1.2.
By Proposition 2.4 and by the proof of Proposition 4.1, the dilatation of β ( m , ··· ,m k +1 ) is the largest root of M ( q ( m , ··· ,m k +1 ) )( t ). Fixing m , · · · , m k ≥
1, let R ( m , ··· ,m k ) and r ( m , ··· ,m k ) be the dominant tree and the dominant tree map for { q ( m , ··· ,m k +1 ) } m k +1 ≥ , and we set R ( m , ··· ,m k ) ( t ) = M ( r ( m , ··· ,m k ) )( t ) . By Proposition 3.5, M ( q ( m , ··· ,m k +1 ) )( t ) = t m k +1 R ( m , ··· ,m k ) ( t ) + S ( m , ··· ,m k ) ( t ) and (4.1) M ( q ( m , ··· ,m k +1 ) ) ∗ ( t ) = t m k +1 U ( m , ··· ,m k ) ( t ) + R ( m , ··· ,m k ) ∗ ( t ) , (4.2)15here S ( m , ··· ,m k ) ( t ) is the recessive polynomial for { M ( q ( m , ··· ,m k +1 ) )( t ) } m k +1 ≥ and U ( m , ··· ,m k ) ( t )is the dominant polynomial for { M ( q ( m , ··· ,m k +1 ) ) ∗ ( t ) } m k +1 ≥ . Claim 4.3.
We have(1) S ( m , ··· ,m k ) ( t ) = ( − k +1 R ( m , ··· ,m k ) ∗ ( t ) and(2) U ( m , ··· ,m k ) ( t ) = ( − k +1 R ( m , ··· ,m k ) ( t ) .Proof. It is enough to show Claim 4.3(1). For if (1) holds, by (4.1) we have M ( q ( m , ··· ,m k +1 ) )( t ) = ( − k +1 M ( q ( m , ··· ,m k +1 ) ) ∗ ( t ) . This together with (4.1), (4.2) implies Claim 4.3(2).We prove Claim 4.3(1) by an induction on k . For k = 1, this holds [4, Theorem 3.20(1)]. Weassume Claim 4.3(1) up to k −
1. Then, we have S ( m , ··· ,m k − ) ( t ) = ( − k R ( m , ··· ,m k − ) ∗ ( t ), M ( q ( m , ··· ,m k ) )( t ) = ( − k M ( q ( m , ··· ,m k ) ) ∗ ( t ) and U ( m , ··· ,m k − ) ( t ) = ( − k R ( m , ··· ,m k − ) ( t ) . (4.3)For k ≥
2, the transition matrix M ( q ( m , ··· ,m k +1 ) ) has the block form: M ( q ( m , ··· ,m k +1 ) ) = n + 1 n + 1 · · · n k − + 1 n k + 1 n k +1 M ( q ( m , ··· ,m k ) ) 11 n k + 1 1 2 2 · · · n k +1 · · · , where n j = m + · · · + m j + j for 1 ≤ j ≤ k + 1. Note that the last edge of the tree R ( m , ··· ,m j ) isnumbered n j . In this case the polynomial S ( m , ··· ,m k ) ( t ) is the determinant of a matrix: tI − M ( r ( m , ··· ,m k − ) ) − t . ... .. − − − − − · · · − t − − − − · · · − − . Subtract the second last row from the last row of this matrix, and let A = ( a i,j ) be the resultingmatrix. Applying the determinant expansion with respect to the last row of A , we have S ( m , ··· ,m k ) ( t ) = | A | = n k +1 X j =1 ( − n k +1+ j a n k +1 ,j | A n k +1 ,j | A i,j is the matrix obtained by A with row i and column j removed. Since a n k +1 ,j = 0 for j = n k , n k + 1 and a n k +1 ,n k = − t , a n k +1 ,n k +1 = −
1, we have | A | = t | A n k +1 ,n k | − | A n k +1 ,n k +1 | . We note that | A n k +1 ,n k +1 | = M ( q ( m , ··· ,m k ) )( t ). For the computation of | A n k +1 ,n k | , subtract the sec-ond last row of A n k +1 ,n k +1 − from the last row, and for the resulting matrix, apply the determinantexpansion of the last column successively. Then, we obtain | A n k +1 ,n k | = − t m k − R ( m , ··· ,m k − ) ( t ) + S ( m , ··· ,m k − ) ( t ) . Thus, S ( m , ··· ,m k ) ( t ) = − M ( q ( m , ··· ,m k ) )( t ) − t m k R ( m , ··· ,m k − ) ( t ) + tS ( m , ··· ,m k − ) ( t ) . In the same manner we have R ( m , ··· ,m k ) ∗ ( t ) = M ( q ( m , ··· ,m k ) ) ∗ ( t ) + t m k U ( m , ··· ,m k − ) ( t ) − tR ( m , ··· ,m k − ) ∗ ( t ) . By using (4.3), these two equalities imply Claim 4.3(1). This completes the proof.We now turn to proving Theorem 1.2. We will prove an inductive formula for R ( m , ··· ,m k ) ( t ). Itis not hard to show that R ( m ) ( t ) = t m +1 ( t − − t .For k ≥
2, one can verify R ( m , ··· ,m k ) ( t ) = tM ( q ( m , ··· ,m k ) )( t ) − t m k R ( m , ··· ,m k − ) ( t ) + tS ( m , ··· ,m k − ) ( t ) . (4.4)Substitute the two equalities M ( q ( m , ··· ,m k ) )( t ) = t m k R ( m , ··· ,m k − ) ( t ) + ( − k R ( m , ··· ,m k − ) ∗ ( t ) and S ( m , ··· ,m k − ) ( t ) = ( − k R ( m , ··· ,m k − ) ∗ ( t )into (4.4), then we find the inductive formula R ( m , ··· ,m k ) ( t ) = t m k ( t − R ( m , ··· ,m k − ) ( t ) + ( − k tR ( m , ··· ,m k − ) ∗ ( t ) . This completes the proof of Theorem 1.2. (cid:3)
Proof of Theorem 1.3.
To begin with, we show
Claim 4.4.
We have(1) lim m i →∞ lim m i +1 →∞ · · · lim m k +1 →∞ λ ( β ( m , ··· ,m k +1 ) ) = 1 for i = 1 and(2) lim m i →∞ lim m i +1 →∞ · · · lim m k +1 →∞ λ ( β ( m , ··· ,m k +1 ) ) = λ ( R ( m , ··· ,m i − ) ( t )) > for i ≥ .Proof. (1) By Theorem 1.2 and by Lemma 2.5lim m k +1 →∞ λ ( β ( m , ··· ,m k +1 ) ) = λ ( R ( m , ··· ,m k ) ( t )) . R ( m , ··· ,m i ) ( t ) = M ( r ( m , ··· ,m i ) )( t ). It is not hard to see that the matrix M ( r ( m , ··· ,m i ) )for i ≥ R ( m , ··· ,m i ) ( t ) is greater than 1. Then, by using theinductive formula of R ( m , ··· ,m i ) ( t ) in Theorem 1.2 together with Lemma 2.5, we havelim m i →∞ λ ( R ( m , ··· ,m i ) ( t )) = λ ( R ( m , ··· ,m i − ) ( t ))for i ≥
2. Since R ( m ) ( t ) = t m +1 ( t − − t , we have lim m →∞ λ ( R ( m ) ( t )) = 1 by Lemma 2.6. Thiscompletes the proof of (1).(2) The proof is identical to that of (1). This completes the proof of Claim 4.4.Claim 4.4(1) says that for any λ > m i ( λ ) for each i with 1 ≤ i ≤ k + 1such that λ ( β m ( λ ) , ··· ,m k +1 ( λ ) ) < λ . Set m = max { m i ( λ ) | i = 1 , · · · k + 1 } . By Proposition 1.1 λ ( β ( m , ··· ,m k +1 ) ) < λ whenever m i > m . This completes the proof of Theorem 1.3(1).The proof of Theorem 1.3(2) is identical to that of (1), but using Claim 4.4(2) instead ofClaim 4.4(1). (cid:3) We show the existence of two kinds of families of pA mapping classes with arbitrarily smalldilatation and with arbitrarily large volume.
Proof of Proposition 1.4.
There exists a family of pseudo-Anosov mapping classes ψ n of M ( D n )such that lim n →∞ λ ( ψ n ) = 1 . It suffices to show that for any pA mapping class φ ∈ M (Σ g,p ), there exists a family of pA mappingclasses ˆ φ n ∈ M (Σ g,p ( n ) ) such that the dilatation of ˆ φ n is same as φ and the volume of ˆ φ n goes to ∞ as n goes to ∞ .Let Φ ∈ φ be a pA homeomorphism. Since the set of periodic orbits of Φ is dense on Σ g,p ,one can find a periodic orbit of Φ, say Q = { q , · · · , q s } . Now puncture each point of Q , thenthe pA mapping class φ ′ ∈ M (Σ g,p + s ) induced by φ satisfies λ ( φ ′ ) = λ ( φ ). On the other hand,vol( φ ′ ) > vol( φ ) since T ( φ ) is a complete hyperbolic manifold obtained topologically by filling acusp of T ( φ ′ ) with a solid torus [9, Section 6]. The volume of any cusp is bounded below uniformly.Thus, if we puncture periodic orbits of Φ ∈ φ successively, we obtain a family of pA mapping classwith the desired property. (cid:3) Finally, we show Theorem 1.6.
Proof of Theorem 1.6 . By Theorem 1.3(1), for each integer k ≥
1, the dilatation of β ( m , ··· ,m k +1 ) goes to 1 as m , · · · , m k +1 all go to ∞ . Thus, it suffices to show that the volume of β ( m , ··· ,m k +1 ) goes to ∞ as k goes to ∞ .One verifies that β ( m , ··· ,m k +1 ) is a 2 bridge link as in Figure 13. In particular it is an alternatinglink with twist number k + 1. Theorem 1 in [5] tells us that for each m , · · · , m k +1 ≥ β ( m , ··· ,m k +1 ) ) >
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