GENERICITY OF INFINITE-ORDER ELEMENTS IN HYPERBOLIC GROUPS

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1 GENERICITY OF INFINITE-ORDER ELEMENTS IN HYPERBOLIC GROUPS PALLAVI DANI 1. Introduction Let Γ be a finitely generated grou and let S be a finite set of generators for Γ. This determines a word metric on Γ, given by d(g, h) = the length of the shortest word in S reresenting g 1 h, which makes Γ into a discrete, roer metric sace. For r 1, let B S (r) be the ball of radius r centred at the identity, with resect to the word metric. We say that the generic element of Γ has a articular roerty P if {elements in B S (r) with roerty P} lim = 1 r B S (r) In this note we rove that the generic element of a finitely generated, nonelementary word-hyerbolic grou has infinite order (see Section.1 below for definitions). Let E S (r) denote the set of finite-order elements in B S (r). We will actually rove the following, somewhat stronger, result. Theorem 1.1. Let Γ be a finitely generated, non-elementary, word-hyerbolic grou, and let S be any finite generating set for Γ. Then there exist ositive constants D = D(Γ) and M = M(Γ) such that In articular, E S (r) M B S ( r + D) E S (r) (1) lim r B S (r) = 0 The limit in (1) measures the density of finite-order elements in the grou Γ. Theorem 1.1 lies in shar contrast to the case of virtually nilotent grous, for which this density is often ositive. For examle, for the square and triangle reflection grous in the Euclidean lane, the densities in (1) are 1/4 and /3 resectively. (See [D] for details). In general this limit is not a quasi-isometry invariant, and in fact, even ositivity may not be reserved under a quasi-isometry: for examle, every virtually nilotent grou has a torsion-free finite-index subgrou (for which the density in (1) is 0). Theorem 1.1 is not merely a consequence of the fact that non-elementary wordhyerbolic grous have exonential growth. In fact, there exist examles (see [L]) of finitely generated infinite torsion grous with exonential growth (for which the density in (1) is 1). The roof of Theorem 1.1 relies on the existence of a generalized notion of centre for bounded sets in δ-hyerbolic saces. In Section we review definitions and establish some results for later use. In articular, we introduce the concet of quasi-centres of bounded subsets of δ-hyerbolic 1

2 PALLAVI DANI saces. In section 3, we obtain a length bound for quasi-centres of orbits of finiteorder elements and use that to finish the roof of Theorem 1.1. I would like to thank my advisor, Benson Farb, for his guidance and time. I would also like to thank Lee Mosher for his suggestions.. Hyerbolic saces and grous.1. Definitions. We first recall some basic facts about word-hyerbolic grous given, for examle in [GH]or [BH]. Let δ > 0. A geodesic triangle in a metric sace is said to be δ-slim if each of its sides is contained in the δ-neighbourhood of the other two sides. A geodesic sace X is said to be δ-hyerbolic if every triangle in X is δ-slim. Let Γ be a finitely generated grou. Its Cayley grah with resect to a finite generating set S is the grah whose vertices are elements of Γ and whose edges connect vertices reresenting grou elements that differ by a generator on the left. G is said to be word-hyerbolic (or just hyerbolic) if its Cayley grah with resect to some generating set, endowed with the word metric, is a δ-hyerbolic metric sace. The roerty of being a hyerbolic grou is indeendent of the choice of finite generating set. A hyerbolic grou is said to be elementary if it contains a cyclic subgrou of finite index... Growth. Non-elementary hyerbolic grous have exonential growth. In fact, more is true: Theorem.1. [FN] Let Γ be a non-elementary hyerbolic grou with generating set S. Then there exist λ 1, λ > 0 and λ > 1 such that for sufficiently high r, λ 1 λ r < B S (r) < λ λ r.3. Quasi-centres. Let X be a δ-hyerbolic, roer geodesic sace and Y X be a non-emty subset of bounded diameter. Y may not have a centre, i.e. a oint in X that is equidistant from every oint of Y. However, it does have a quasicentre, a set of uniformly bounded diameter, which can often fulfil the same role as a centre. For any x X, let B(x, r) denote the ball of radius r in X, centered at x. Define ρ x = inf{r > 0 Y B(x, r)} and ρ = inf x X {ρ x }. Definition.. The quasi-centre of Y is the set C(Y ) = {x X ρ x = ρ}. Note that if x 0 minimizes ρ x, then x 0 must lie in a comact ball centered at a oint in Y, so that the quasi-centre is non-emty. The diameter of C(Y ) is uniformly bounded by δ (See [B]). The quasi-centre can be thought of as a generalization of a global fixed oint of a finite grou acting on a negatively curved Riemannian manifold. Now suose that X is the Cayley grah of a hyerbolic grou and Y is a finite subgrou. C(Y ) may not contain any vertices of the Cayley grah, but C 1 (Y ), the 1-neighbourhood of C(Y ), certainly does. Vertices in C 1 (Y ) conjugate Y into a finite ball. Theorem.3. [B, BH] Let Γ be a word-hyerbolic grou with finite generating set S. Let Y be a finite subgrou of Γ. If x is a vertex in C 1 (Y ), then for every g Y, the element x 1 gx belongs to B S (4δ + ). Theorem.3 has the following consequence.

3 GENERICITY OF INFINITE-ORDER ELEMENTS IN HYPERBOLIC GROUPS 3 Theorem.4. [B, BH] A hyerbolic grou has finitely many conjugacy classes of finite subgrous..4. Polygons in δ-hyerbolic saces. Let X be a δ-hyerbolic sace. A k-gon in X is a set of oints A 1,..., A k together with geodesic segments γ 1,....γ k, such that γ i has endoints A i and A i+1 for 1 i k 1 and γ k has endoints A k and A 1. The A i s and the γ i s are called the vertices and the sides of the k-gon resectively. We will be lax and denote such a k-gon by A 1... A k, even though this doesn t comletely secify the k-gon. We will denote the side joining A i and A i+1 by A i and A i+1. A regular k-gon is one whose sides have equal lengths. A k-gon in X consists of k-geodesic segments, called its sides γ 1,..., γ k such that γ i and γ i+1 share an endoint for 1 i k 1, and γ k and γ 1 share an enoint. Note: The first one definitely seems better, since in the second, there is the additional fact that any oint is the endoint of at most two of the geodesic segments and secondly, olygons are later refered to by their vertices. The slim-triangles condition can be used to rove the following slim-olygons roerty. Lemma.5. [BH] Let X be a δ-hyerbolic sace. For any k, there is an ɛ k = ɛ k (δ) such that given any k-gon in X, each of its sides is contained in the union of the ɛ k -neighbourhoods of the other k 1 sides. In fact, we can take ɛ k = (k )δ. (A better bound is obtained in [BH]). In articular, ɛ k increases as k increases. We use the slim-olygons roerty to show that olygons have short diagonals. Lemma.6. Let P be a k-gon, k > 3, in a δ-hyerbolic sace, with side-lengths bounded by some constant l. Then there exists a diagonal of P whose length is no more than l + ɛ k, where ɛ k is as in Lemma.5. Proof. Let P = A 1 A A k. Let m i be the midoint of the side A i A i+1, for i < k, and let m k be the midoint of A k A 1. From Lemma.5 we know that corresonding to the oint m 1 on A 1 A, there is PSfrag relacements a oint on another side of P such that d(m 1, ) ɛ k. There are two ossibilities for the osition of. (shown in the figure for k = 5): A 4 A 4 m 3 m 4 A 3 A 5 A 3 m 3 m4 A 5 m m 5 m m 5 m1 m 1 A A 1 A A 1 Case 1: lies on A 1 m k or A m Case : lies on m k A k, m A 3 or one of the remaining sides

4 4 PALLAVI DANI Case 1: Assume, without loss of generality, that lies on A m. Then we have: d(a 1, A 3 ) d(a 1, m 1 ) + d(m 1, ) + d(, m ) + d(m, A 3 ) d(a, m 1 ) + ɛ k + d(, m ) + l (d(a, ) + ɛ k ) + ɛ k + d(, m ) + l d(a, m ) + l + ɛ k l + ɛ k Case : Let V be the vertex closest to, i.e. d(v, ) l. (V = A 3 in the figure). At least one of A 1 V and A V is a diagonal of P. Further, for j = 1 or, we have: d(a j, V ) d(a j, m 1 ) + d(m 1, ) + d(, V ) l + ɛ k In either case we have found a diagonal whose side-length is bounded by l + ɛ k. We will use the following consequence of Lemma.6: Lemma.7. Let P = A 1 A k be a regular olygon in a δ-hyerbolic sace, with side-length s. Then there exits η k = η k (δ), such that d(a j, A j+ ) < s + η k for some 1 j k. Proof. By alying Lemma.6 at most k times, P can be cut u into triangles using diagonals of P. At least one of these diagonals is a geodesic joining A j and A j+ for some j. Since ɛ k is the largest of ɛ 1,..., ɛ k, the length of this diagonal is at most s + (k )(ɛ k ). Now set η k = (k )(ɛ k ). 3. Finishing the roof 3.1. A length bound for the quasi-centre. Let g be a finite-order element in Γ. Then vertices in C 1 ( g ), the 1-neighbourhood of the quasi-centre of g, are aroximately half as far from the identity as g: Proosition 3.1. Let Γ be a non-elementary word-hyerbolic grou, with generating set S. Let g be a finite-order element of Γ and let x C 1 ( g ). Then there exists D = D(Γ, S) such that d(x, e) + D Proof. Let g be of order k. Pick a geodesic γ in the Cayley grah of Γ, joining g to the identity, e. Let P g denote the regular k-gon with vertices e, g,..., g k 1 and sides γ 1 = γ, γ = g(γ),..., γ k = g k 1 (γ). Let m i denote the midoint of γ i. Note that for all i, d(g i, g i+1 ) = and d(g i, m i ) = d(g i+1, m i ) = / We first rove that the set {m i 1 i k} is bounded, with diameter less than a constant which deends only on k and δ. Since the set is finite, it will be enough to show that d(m i, m i+1 ) is bounded by such a constant for every i. By Lemma.7, there exists some j such that d(g j, g j+ ) < +η k. Hence, for every i, we have d(g i, g i+ ) = d(g j, g j+ ) < + η k. For each i, ick a geodesic

5 GENERICITY OF INFINITE-ORDER ELEMENTS IN HYPERBOLIC GROUPS 5 joining g i and g i+, and consider the triangle g i g i+1 g i+. Since this triangle is δ- slim, each of the oints m i and m i+1 is δ-close to a side not containing it. Without loss of generality we may assume that one of the following three cases occurs: PSfrag relacements g i+1 g i+1 m i m i+1 m i m i+1 g i g i+ g i+1 g i g i+ Case 1: d(m i, ) < δ Case : d(m i, ) < δ ase 3: d(m i, 1 ) < δ; d(m i+1, ) < δ g i m i 1 m i+1 g i+ Case 1: We show that d(, m i+1 ) < δ as well. By the triangle inequality we have: d(, m i+1 ) + d(m i+1, g i+1 ) d(g i+1, m i ) + d(m i, ) = d(, m i+1 ) + < + δ = d(, m i+1 ) < δ So d(m i, m i+1 ) < δ. Case : Once again d(m i, m i+1 ) < δ. The roof is similar to that of Case 1. Case 3: The triangle inequality imlies that d(g i, 1 ) > d(g,e) δ and d(g i+, ) > d(g,e) δ (since d(g i, m i ) = d(g i+, m i+1 ) = d(g,e) ), so that d( 1, ) = d(g i, g i+ ) d(g i, 1 ) d(g i+, ) < + η k = η k + δ + δ So d(m i, m i+1 ) < η k + 4δ. Thus the diameter of the set {m i 1 i k} is at most k(η k + 4δ). This means each of its elements (in articular, m 1 ) is aroximately equidistant from the vertices e, g,... g k 1. More recisely, for all i, we have d(g i, m 1 ) d(g i, m i ) + d(m i, m 1 ) + δ + k(η k + 4δ) This imlies ρ ρ m1 d(e, g)/ + k(η k + 4δ). Now by Lemma.4 there are only finitely many conjugacy classes of finite-order elements. Let D = 1 + max{k(η k + 4δ) k is the order of an element in Γ} Note that D deends only on Γ and S. If x C 1 ( g ), then d(x, e) ρ k(η k + 4δ) D

6 6 PALLAVI DANI 3.. Proof of Theorem 1.1. For every g E S (r), ick any x g C 1 ( g ). Let D be as in Proosition 3.1, i.e. d(x g, e) d(g,e) + D. We now have a ma φ : E S (r) B S ( r + D) g x g Given x B S ( r + D), φ 1 (x) consists of elements h in Γ, such that x C 1 ( h ). By Lemma.3 above, x 1 hx is an element of B S (4δ + ), a finite ball. Then if M = B S (4δ + ), there are at most M choices for x 1 hx, and hence for h, so that φ 1 (x) M. It follows that E S (r) M B S ( r + D) Note that M deends only on Γ and S. Now since Γ is non-elementary, Lemma.1 guarantees the existence of λ > 1 and λ 1, λ > 0, such that for sufficiently high r, λ 1 λ r < B S (r) < λ λ r. So E S (r) F (Γ, S) = lim r B S (r) lim M B S ( r + D) λ λ r +D lim r B S (r) r λ 1 λ r = 0 References [B] N. Brady, Finite subgrous of hyerbolic grous Internat. J. Algebra Comut. 10 (000), no. 4, [BH] M. Bridson; A. Haefliger, Metric saces of non-ositive curvature, Grundlehren der Mathematischen Wissenschaften [Fundamental Princiles of Mathematical Sciences], 319. Sringer-Verlag, Berlin, [D] P. Dani, Finding the density of finite-order elements in virtually nilotent grous, in rearation. [GH] [FN] K. Fujiwara; A. Nevo, Maximal and ointwise ergodic theorems for word-hyerbolic grous, Ergodic Theory Dynam. Systems 18 (1998), no. 4, [L] I.G. Lysnok, Infinite Burnside grous of even eriod, (Russian) Izv. Ross. Akad. Nauk Ser. Mat. 60 (1996), 3 4; translation in Izv. Math. 60 (1996),