On dilatation factors of braids on three strands

arXiv:1307.6987v1 [math.GT] 26 Jul 2013 On dilatation factors of braids on three strands Marta Aguilera∗ Universidad de Sevilla Abstract In this work we present a natural surjective map from rigid braids in B3(in Garside sense) to SL2(N). This map provides an upper and a lower bound for the dilatation factor of a pseudo-Anosov 3-strand braid. These bounds only depend on the canonical length of the classical Garside structure of B3. 1 Introduction In this paper we review some well-known results about the braid group in three strands, and we rewrite them in terms of the Garside structure. In this way, we see that the dynamic of a braid in a super summit set (i.e. with minimal length in its conjugacy class) is easy to describe. If such a braid is reducible, it must be ∆2sσk 1or ∆2sσk 2,s, k ∈Z. Their reduction systems are simple: a curve around the first two punctures for the former, and a curve around the last two punctures for the latter. In the pseudo-Anosov case, we will point out that, despite the fact that there are only two train track graphs Γ1,Γ2as in Figure 1 that carry every foliation in the 3-times punctured disc D3, these are not train tracks for every pseudo-Anosov braid. However, if a braid β∈B3is in a super summit set, it is rigid (a Garside theoretical property), and either Γ1or Γ2is a train track for β([14], [16]). Figure 1: Train tracks for rigid braids on three strands. In addition, if βis in a super summit set, it is straightforward to obtain the associated matrix, and therefore the foliation and dilatation factor, from its Garside normal form. This allows us to study the dynamic of any braid through a conjugate in the super summit set. This also gives a natural map from rigid braids to 2 ×2 matrices with non-negative entries: Theorem 1.1. There exists a 2-to-1surjective map from the set of rigid 3-braids (modulo ∆2) to SL2(N), which sends each rigid braid to the matrix associated to its corresponding train track Γi,i= 1,2. ∗Partially supported by MTM2010-19355, P09-FQM-5112 and FEDER. 1 As a consequence, given a pseudo-Anosov braid with minimal length ℓin its conjugacy class, we can give a lower and an upper bound for its dilatation factor: Theorem 1.2. Let β∈B3be a pseudo-Anosov braid in its super summit set. If βhas canonical length ℓand dilatation factor λ, then: 1 2ℓ+ 1 + p(ℓ+ 3) (ℓ−1)≤λ, and: λ≤φℓif ℓis even, λ≤Fℓ+pF2 ℓ−1<2 √5φℓ+φ−ℓif ℓis odd, where φis the golden ratio, and Fiis the ith Fibonacci number. The lower bound is a minimum, reached only by the conjugates of β= σ1σ−(ℓ−1) 2. The upper bound is a maximum if ℓis even, reached only by the conjugates of β=σ1σ−1 2ℓ 2. If ℓis odd, the maximal dilatation factor reached only by the conjugates of β=σ1σ−1 2ℓ−1 2σ1satisfies that 2Fℓ−1< λ < 2Fℓ with 2Fℓ=2 √5φℓ+φ−ℓ. Acknowledgements. I would like to thank Dan Margalit, Bert Wiest, J´erˆome Los, Eiko Kin and Spencer Dowdall for helpful discussions and comments, some of them during a stay at Centre de Recerca Mat`ematica, to whom I also want to thank for its hospitality. I am also grateful to my advisor Juan Gonz´alezMeneses for numerous useful comments, corrections and suggestions on preliminary drafts of this paper. 2 Background The braid group on nstrands Bnis isomorphic to the mapping class group of the ntimes punctured disc fixing the boundary pointwise MCG(Dn, ∂Dn). Collapsing the boundary of Dnto a point, the mapping class group of the ntimes punctured disc can be considered as a subgroup of the (n+ 1) times punctured sphere: it will be denoted MCG(Dn). This geometric approach to Bnallows the use of the Nielsen-Thurston classification theorem [12], [17], [18]. A braid βis periodic if there exist k∈N, s ∈Z, such that βk= ∆2s, where ∆2is a Dehn twist along the disc’s boundary ∂Dn.β is reducible if there exists a non-degenerate 1-manifold C⊂Dnfixed by β. Finally, βis pseudo-Anosov if there exists a pair of transverse measured foliations (Fs, µs) y (Fu, µu), and a real number λ > 1, such that β((Fs, µs)) = (Fs, λµs) and β((Fu, µu)) = (Fu, λ−1µu). The classes of reducible and periodic braids are not disjoint, so from now on we will call reducible those elements which are reducible and non-periodic. The classification problem can be solved using the train tracks techniques introduced by Bestvina-Handel in the nineties [1], [2]. These are combinatorial objects which encode the dynamics on the surface in terms of linear algebra. In the pseudo-Anosov case, they also give the structure of the unstable foliation Fu, its measure and the dilatation factor λ. The dynamic of a braid only depends on its conjugacy class, in particular so does its Nielsen-Thurston type. So we can study the geometry of any given 2 β∈Bn, through any conjugate e β=α−1βα: Periodicity is easily recognizable in braid groups [5], and βk= ∆2sif and only if e βk= ∆2s. If βis reducible, and Cis a reduction system for β(1-manifold such that β(C) = C), then α(C) is a reduction system for e β. In the pseudo-Anosov case, if Γ is a train track for β, then α(Γ) is train track for e β. Also the combinatorial maps associated to Γ and α(Γ) are the same e fβ,Γ=e fe β,α(Γ), and so are the matrices M(β, Γ) = M(e β, α(Γ)) (see next section for definitions). 2.1 Train tracks In this section we will review some basic facts about train tracks ([1], [2], [11]). Let Γ = (V, E) be a labeled graph embedded into the punctured disc Dn, such that each component of Dn\Γ is a punctured disk or a ring and π1(Γ) = π1(Dn) = Fn. In the class of a given β∈MCG(Dn) there exists a representative automorphism fβwhich maps Γ into a tubular neighborhood Uof itself. The composition of this map with a deformation retract ι:U→Γ, allows us to associate to each β, a map e fβ: Γ →Γ. Notice that we can chose fβand ιsuch that f fβ(V)⊆V, and such that ι◦fβis injective in the interior of each edge, so for any e∈E,e fβ(e) is an edge path. Thus e fβcan be seen as a combinatorial map. Removing vertices of valence 2, and contracting edges which end in a valence 1 vertex, we can suppose such a graph Γ to have all vertices of valence at least three. In addition, we will assume that at each vertex there is a well defined tangent, so we can distinguish between those edges entering from one direction and those entering from the other. The labels of the edges must satisfy the switch condition: the sum of the labels going in from one side must be equal to the sum of those going out. A combinatorial map e fbacktracks if there is an edge a∈E, and k > 0 such that e fk(a) contains the subword ee−1or e−1e, for some e∈E. A combinatorial map is said to be efficient if it does not backtrack. Notice that e fβis efficient as combinatorial map if (ι◦fβ)kis injective in the interior of all edges for all k > 0. Given β∈MCG(Dn), the graph Γ, with the properties described above, is atrain track graph for β, if the combinatorial map e fk βis efficient ∀k≥1. In this case, the tangencies at the vertices of Γ can be chosen so that ι◦fβrespect tangencies, that is the image of each edge is a smooth edge path. We can associate to a braid βand Γ a transition matrix M=M(β, Γ), where each entry mi,j is the number of times the edge ejappears in f fβ(ei). Obviously, Mhas non-negative entries. If the graph Γ is a train track for β, M(β, Γ) contains the geometric information about β. If the transition matrix Mis reducible (i.e. Mkhas at least one zero entry for all k∈N), then the element βis reducible. If the matrix is not reducible, then βis either periodic or pseudo-Anosov. The Perron-Frobenius theorem states that, in the irreducible case, the greatest eigenvalue λis real, has multiplicity 1 and λ≥1. The element βis periodic if and only if λ= 1, and it is pseudoAnosov if and only if λ > 1. In the latter case, the eigenvalue λcoincides with the dilatation factor, and the associated eigenvector yields a measure for Γ that encodes the unstable foliation Fu[12]. 3 2.2 Garside structure Braid groups have a well-known presentation [3]: Bn=σ1, σ2,...,σn−1:σiσj=σjσi|i−j|>1 σiσjσi=σjσiσj|i−j|= 1 . These groups can be endowed with the classic Garside structure [8], [13], that is a triple (Bn, B+ n,∆), where B+ nis the monoid generated by the positive crossings σ1,...,σn−1(see Figure 2), and ∆ ∈B+ nis called the Garside element, ∆ = (σ1···σn−1) (σ1···σn−2)···(σ1σ2)σ1. For any γ, β ∈Bn, we will say that γis a prefix of β,γ4β, if γ−1β∈B+ n. This is a partial order that endows Bnwith a lattice structure (with well-defined gcd and lcm), used to define normal forms. Figure 2: Positive and negative crossings in Bn. The mapping class ∆ is represented by a half Dehn twist along the disc’s boundary. Considering braids up to Dehn twists along the boundary is equivalent to collapse the boundary to one point, so MCG(Dn) = Bn/h∆2i. From the geometric point of view βhas the same properties as ∆2iβ,∀i∈Z, so we will usually consider braids up to multiplication by ∆2. It is easy to see that ∆2is central, actually it generates the center Z(Bn) = h∆2i, [7]. The Garside element ∆ also satisfies: •[1,∆] = {s∈Bn: 1 4s4∆}, the set of simple elements, generates Bn. •The inner automorphism τcorresponding to conjugation by ∆ preserves the lattice structure. Equivalently: τ(B+ n) = ∆−1B+ n∆ = B+ n. Definition 2.1. [9] Given a braid β∈Bn, the decomposition β= ∆ss1s2···sk is its left-normal form if 1≺si≺∆∀iand siis the greatest simple prefix of sisi+1 ···sk,∀i= 1 . . . k. The integer kis the canonical length of β. The computational complexity of the calculation of the normal form of a braid on nstrands written as a product of ℓgenerators σ±1 iis O(ℓ2nlog n) for n > 3 [10], and linear in ℓif n= 3. Definition 2.2. [4] Let βbe a braid and β= ∆ss1···skits normal form. We will say the braid βis rigid if k≥1and the product skτ−s(s1)is in left normal form, or if β= ∆s,s∈Z. If a braid βhas normal form as above, then τ(β) = ∆sτ(s1). . . τ(sk) is in normal form. Therefore, βis rigid if and only if τ(β) is rigid. 4 Theorem 2.3. [4] For every pseudo-Anosov β∈Bn, there exists 0< k ≤n 2 such that βkis conjugate to a rigid braid. In B3, one can take k= 1, that is, every pseudo-Anosov braid is conjugate to a rigid braid. In the braid group B3, we will see how to extract geometric information of a pseudo-Anosov braid via a rigid conjugate. 3 Braid group in three strands 3.1 Normal forms in B3 The braid group in three strands has specially nice properties. Normal forms are easily computable and rigidity is directly recognizable. The Garside element is ∆ = σ1σ2σ1=σ2σ1σ2, and the simple elements are {1, σ1, σ2, σ1σ2, σ2σ1,∆}. Given a braid as a concatenation of σ±1 i, it is easy to rewrite it as the product of ∆s,s∈Z, and a positive word in {σ1, σ2}, using the equalities: σ−1 1= ∆−1σ1σ2σ1∆±1= ∆±1σ2 σ−1 2= ∆−1σ2σ1σ2∆±1= ∆±1σ1 Notice that each simple element, except ∆, can be written in a unique way as a word in {σ1, σ2}. The product of two proper simple elements s1s2is in normal form if and only if the last letter of s1equals the first letter of s2. Therefore, if sisi+1 is in normal form, the last crossing in siand the length of si+1 characterize the factor si+1. Repeating this process, a product of proper simple elements in normal form s1···skis determined by giving s1and the length of the factors s2,...,sk. So, the normal form of any braid β= ∆ss1···sk,k > 0, can be codified by the tuple (s;i;p1, q1,...,pr, qr)∈Z× {1,2}× N2r,r > 0. The integer sis the exponent of ∆. The first crossing in s1is σi. The other elements in the tuple indicate the length of the factors s1,...,skin the following way. The first p1simple elements in the normal form have length one (each one consists of one crossing). Then, they are followed by q1elements of length two, and after those there are p2elements of length one, etc. That is, for any j= 1,...,r, the simple factors from position Pi<j (pi+qi) + 1 to position Pi<j (pi+qi) + pj have length one, and those from position Pi<j (pi+qi) + pj+ 1 to position Pi<j+1 (pi+qi) have length two. For a coherent notation only p1and qrcould be zero. We will codify the braid ∆sby (s; 1; 0,0) = (s; 2; 0,0). Example. (−4; 1; 0,1,3,2) corresponds to the braid β= ∆−4σ1σ2·σ2·σ2·σ2·σ2σ1·σ1σ2. Recall that a braid β6= ∆s, with normal form β= ∆ss1···skis rigid if skτ−s(s1) is in normal form. That is, the normal form of ∆−sβmust start and finish with the same letter σiif sis even, and it must start and finish with different letters if sis odd. The reader can check that a braid βis rigid, in terms of the associated tuple, if and only if s+Pqjis even. It is easy to check that τ((s; 1; p1, q1,...,pr, qr)) = (s; 2; p1, q1,...,pr, qr). If a braid β= ∆ss1···skwith k > 1 is not rigid, then its conjugate by τ−s(s1) either is rigid, or its canonical length is strictly smaller than k(or both things happen). This 5 conjugation is known as cycling [9], and a finite number of iterations provides a rigid conjugate of any initial non-periodic braid. The Nielsen-Thurston type of a rigid braid it is easily recognizable: Proposition 3.1. [16] Every braid β∈B3is conjugate to a braid α= (s; 1; p1, q1,...,pr, qr) called Murasugi representative of βsuch that: •If βis periodic, r= 1 and –s∈Zand (p1, q1) = (0,0), or –sis odd and (p1, q1) = (1,0), or –sis even and (p1, q1) = (0,1). •If βis reducible, then r= 1,αis rigid and (p1, q1)∈ {(k, 0),(0, k) : k > 0}. •If βis pseudo-Anosov, αis rigid distinct from above. The conjugacy problem in Bncan be solved by building the finite set of braids with minimal canonical length in the conjugacy class, called super summit set [9]. Notice that the non-periodic Murasugi representatives are rigid. Hence all Murasugi representatives have minimal canonical length in their conjugacy class, so they belong to their super summit set. Conversely, any braid βin a super summit set satisfies that either βor τ(β) is a Murasugi representative. Lemma 3.2. [6] Let βbe a braid in B3with canonical length at least 2. Then the super summit set of β,SSS(β), is the set of rigid conjugates of β. Actually, SSS(β)consists of either two closed orbits under cycling, conjugate to each other by ∆, or one closed orbit under cycling, self conjugated by ∆. 3.2 Train tracks of braids in B3 Firstly, we want to point out the difference between train tracks as they have been defined in the section above, from those graphs that carry foliations. A train-track graph for a braid βand the corresponding combinatorial map provide a matrix M. This matrix Mdetermines the Nielsen-Thurston type of β, and if it is pseudo-Anosov it determines the unstable foliation. However the reciprocal does not hold: from the foliation one cannot obtain a train track for β. Actually every admissible foliation in D3(see Figure 3) is carried by one of the two graphs Γ1or Γ2in Figure 1 1with certain labels [11]. However not every pseudo-Anosov 1In literature, the graphs G1,G2appear more often than Γ1. The two foliations on the first row in Figure 3 are carried by G1and G2respectively. We will use Γ1instead of Gi, despite the fact that it is not so intuitive to see that Γ1also carries both. Notice that if x > y we could split Γ1to get G1, changing the labels v1=y u1=x−y > 0. And if x < y we can split Γ1to get G2. Similarly, Γ2can be split to get graphs G3and G4, mirror images of G1 and G2. 6 braid admits Γ1or Γ2as train track, see the example below. Later it will be shown that if the braid is rigid, then yes, it admits either Γ1or Γ2as train track. Figure 3: Foliations in B3. Example. Neither Γ1nor Γ2are train tracks for the braid β= (σ1σ1)−1σ2σ−1 1(σ1σ1). The image of Γiunder the action of βcan not be embedded into a tubular neighborhood of Γirespecting tangencies (see Figure 4). Figure 4: Γ1, Γ2are not train track graph for β= (σ1σ1)−1σ2σ−1 1(σ1σ1). We could also check the combinatorial maps e βfor Γ1and Γ2. Let’s label the edges around the punctures from left to right e1, e2, e3. We will consider ei (i= 1,2,3) oriented anticlockwise, and x, y from right to left. Γ1 e β →Γ1 x→y y→(−x)e2(−y)e3y e2(−y)e3 Γ2 e β →Γ2 x→y y→(−y) (−x)e3x e2(−x) The reader can check that in either case, e βbacktracks in the second iteration. The following theorem states that for a pseudo-Anosov rigid braid it is straightforward to get a train track and its transition matrix M. Actually, 7 Mis the product of some simple matrices associated to the simple elements of the normal form. This is not the case in general: the product of transition matrices of two braids does not give any information about the product of the braids. Theorem 3.3. [14] Let β= (2s; 1; p1, q1,...,pr, qr)or β= (2s+1; 2; p1, q1,...,pr, qr) be a rigid pseudo-Anosov braid in B3. Then, the graph Γ1(see Figure 1) is a train-track graph for β. Its transition matrix is M(β, Γ1) = Lp1Uq1···LprUqr, where L=1 0 1 1 U=1 1 0 1 . In fact this theorem can be easily extended to reducible braids. The graph Γ1is also a train tack for the braids (2s; 1; k, 0) = ∆2sσk 1, where k > 0. Its associated matrix is Lk=1 0 k1, which is obviously reducible. If a braid βhas Γ1as train track and Mis the associated matrix, then τ(β) has Γ2= ∆(Γ1) as a train track. We order the edges by taking the one labeled by xfirst, so βand τ(β) have the same associated matrix M. Therefore the above theorem can be written for any rigid braid: Corollary 3.4. Every rigid braid in B3admits either Γ1or Γ2of Figure 1 as a train track graph. Let εbe the automorphisms of Bnthat maps σito σ−1 i. For any braid β, τ◦ε(β) is its vertical mirror image. This implies that τ◦ε(β) has the same train track graph but it switches labels, hence both rows and columns of the transition matrix exchange places, M(τ◦ε(β)) = M(ε(β)) = m2,2m2,1 m1,2m1,1. So, βand ε(β) have the same dilatation factor, and if −→ v= (v1, v2) is the M(β) eigenvector associated to λ, then −→ w= (v2, v1) is the M(ε(β)) eigenvector associated to λ. This verifies an obvious fact: βand ε(β) have analogous dynamics. Notice that the incidence matrix being a product of L’s and U’s, belongs to SL2(N). Furthermore, Land Uare very special elements of this monoid. The following result, which is well known, is implicitly shown in the subsequent discussion. Theorem 3.5. The monoid SL2(N)is freely generated by L,U. This theorem means that given any 2×2 matrix Mwith non negative entries and determinant 1, it admits a unique product decomposition in terms of Land U. We will call this the LU-decomposition of M. Because det(M) = 1 and it has nonzero entries, we can say that Mhas a biggest row R(M)and a smallest row r(M), meaning R(M) 1≥r(M) 1and R(M) 2≥r(M) 2(where at least one of the inequalities is strict). If r(M)= (m1,1, m1,2), then the first factor of the LUdecomposition of Mis L: M=r(M) R(M)⇒M=LM′,where M′=r(M) R(M)−r(M)∈SL2(N). And obviously, if the first row is R(M), then the first factor is U. This gives us an algorithm to compute the LU-decomposition of M. At each step kMk∞> 8 kM′k∞, therefore the algorithm ends. We could have defined the analogous algorithm defining the columns C(M)and c(M), obtaining the decomposition from right to left. Given M∈SL2(N), we define the length of M, and we write ℓ(M), as the length on the associated LU-decomposition. We will denote M(i) the product of the first ifactors of M, for i= 1,...,ℓ(M). With this notation M(ℓ(M)) = M, the first factor of Mis M(1) and the ith factor is M(i−1)−1M(i). 3.3 Rigid 3-braids and SL2(N) As a consequence of the results in the section above, we can prove Theorem 1.1. Proof: [of theorem1.1] The map from rigid braids to matrices in SL2(N) is explicit in both senses. Given the normal form of a rigid braid β= ∆ss1···sk, we can construct from the associated tuple (s;i;p1, q1,...,pr, qr) the LU-word Lp1Uq1···LprUqr. Reciprocally, given a 2 ×2 square matrix M, the LUdecomposition provides the values pj, qj. The value smust be chosen even or odd, so that the associated braid is rigid. The 2 ×2 identity matrix is associated to powers of ∆, so 1 and ∆ (mod ∆2) are the two preimages of the identity matrix. In the other cases, because there are always two options for i, the correspondence restricted to Bn/h∆2iis 2 to 1. Due to Lemma 3.2, it follows that two rigid conjugated braids must have conjugate matrices in SL2(N), that is cyclic permutations of L,Ufactors. Let J=0 1 1 0 . Because J=J−1and L=JUJ, we have that M(ε(β)) = JM(β)J. The LU-decomposition of a matrix Mcan be obtained from that of JMJ by exchanging L’s and U’s. However, these two matrices are not conjugated in SL(N) in general. Due to the above results, we can explicitly give the set of dilatation factors in B3: Corollary 3.6. The set of dilatation factors for pseudo-Anosov braids in three strands is: (λ=T+√T2−4 2, T ∈N, T ≥3). Proof: If the incident matrix associated to a braid is M∈SL2(N), the dilatation factor is the biggest root of the polynomial x2−T x + 1, where T= trace(M)∈N. So the dilatation factor λof a pseudo-Anosov braid only depends on T=trace(M). As det(M) = 1 and all entries of Mare non-negatives integers, we have T > 1. Only Lk, Ukhave trace equal to 2, but these matrices are reducible. Therefore, for pseudo-Anosov braids T > 2, and every value of T > 2 can be obtained at least for the matrix LUT−2=1T−2 1T−1, which corresponds, among others, to the braid σ1σ−(T−2) 2. 9