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Recall, that for any graph \Gamma we built a combinatorial horoball \mathcal{H}(\Gamma).  For a group G and a collection of subgroups \mathcal{P}=\{P_1,\ldots,P_n\}and a generating set S, we built the augmented Cayley graph X by gluing copies of \mathcal{H}(\mathrm{Cay}(G)).  G is hyperbolic relative to \mathcal{P} if and only if X is Gromov hyperbolic.

Exercise 28: If A and B are finitely generated, then A*B is hyperbolic relative \{A,B\}. (Hint: X is a graph of spaces with underlying graph a tree and the combinatorial horoballs for vertex spaces.)

Example: Suppose Mis a complete hyperbolic manifold of finite volume.  So, \Gamma=\pi_1M acts on \mathbb{H}^n.  Let \Lambda be a subset of \partial\mathbb{H}^n consisting of points that are the unique fixed point of some element of \Gamma.  So \Gamma acts on \Lambda, and there only finitely many orbits.  Let P_1,\ldots,P_n be stabilizers of representatives from these orbits and let \mathcal{P}=\{P_1,\ldots,P_n\}.  Then, \Gamma is hyperbolic relative to \mathcal{P}.

Example: Let G be a torsion-free word-hyperbolic group.  Then, G is clearly hyperbolic relative to \{1\}.  A collection of subgroups P_1,ldots,P_n is malnormal if for any g\in G, P_i\cap gP_jg^{-1}\neq1 implies that i=j and g\in P_i.  G is hyperbolic relative to \mathcal{P}=\{P_1,\ldots,P_n\} if and only if \mathcal{P} is malnormal.

The collection of subgroups \mathcal{P} is the collection of peripheral subgroups.

Lemma 31: If G is torsion-free and hyperbolic relative to a set of quasiconvex subgroups \mathcal{P}, then \mathcal{P} is malnormal.

Sketch of Proof: Suppose that P_1\cap gP_2g^{-1} is infinite.  Consider the following rectangles:  Note that if k=l(g), then gP_2g^{-1} is contained in a k-neighborhood of gP_2.  Now, there exists infinite sequences p_i\in P_1 and q_i\in P_2 such that d(p_i,gq_i)\leq k.  Look at the rectangles with vertices 1, g, gp_i, p_i.  The geodesics in X between 1 and p_i and g and gq_i go arbitrarily deep into the combinatorial horoballs.  Therefore, they are arbitrarily far apart.  It follows that these rectangles cannot be uniformly slim.

Let \mathcal{N}=\{N_1,\ldots,N_n\} where each N_i\lhd P_i.  Write G/\langle\langle\bigcup_iN_i\rangle\rangle=G/\mathcal{N}.  Call this the Dehn filling of G.

Note: If G is hyperbolic relative to \mathcal{P}, then G is hyperbolic.

Theorem 21: (Groves-Manning-Osin). Suppose G is hyperbolic relative to \mathcal{P}.  Then, there exists a finite set A contained in G\smallsetminus 1 such that whenever (\bigcup_i N_i)\cap A\neq\emptyset we have

  1. P_i/N_i\to G/\mathcal{N} is injective for all i, and
  2. G/\mathcal{N} is hyperbolic relative to the collection \{P_i/N_i\};

In particular, if P_i/N_i are all hyperbolic, then so is G/\mathcal{N}.

One application of this theorem is a simple proof of a theorem of Gromov, Olshanskii, and Delzant:

Theorem 22: Let G be hyperbolic and suppose \{\langle g_1\rangle,\ldots,\langle g_n\rangle\} is malnormal, with each \langle g_i\rangle infinite.  Then, there is constant K such that for all positive integers l_1,\ldots,l_n there is an epimorphism to a hyperbolic group \phi:G\to G' such that o(\phi(g_i))=Kl_i for each i.

Theorem 12 (Gromov): Let \Gamma be torsion-free \delta-hyperbolic group.  If u,v \in\Gamma such that uv\neq vu, then for all sufficiently large m,n, \langle u^m,v^n\rangle \cong F_2.

Remark: The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities.  For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of \mathbb{Z}.

For the rest of this lecture \Gamma will be a torsion-free \delta-hyperbolic group, uv\neq vu where u,v are primitive (i.e. not proper powers).

Recall that for \Gamma torsion-free \delta-hyperbolic, u primitive implies that \langle u \rangle = C(u)= C(u^m).

If u and v do not commute we can show there is some point u^p on \langle u \rangle arbitrarily far from \langle v \rangle .
pic11Hence we have the following lemma.

Lemma 13: d_{haus}(\langle u \rangle , \langle v \rangle )=\infty

If u and v do not commute there is some point u^p on \langle u\rangle arbitrarily far from \langle v\rangle .

Proof: Suppose not. That means \exists R_0 > 0 such that \forall u^p \in \langle u \rangle \exists v^q \in \langle v \rangle such that  d(u^p,v^q) = d(1,u^{-p}v^q) < R_0.  So u^{-p}v^q is in B(1,R_0).  But the Cayley graph is locally finite so B(1,R_0) has finitely many elements.  By the Pigeonhole Principle \exists p\neq r such that u^{-p}v^q=u^{-r}v^s for some q, s.  Then \langle u \rangle = C(u) =C(u^{p-r})=C(v^{q-s})=C(v)=\langle v \rangle .  But then uv=vu. \Rightarrow\Leftarrow .

For a moment view \langle u \rangle and \langle v \rangle as the horizontal and vertical geodesics in \mathbb{H}.  For two points x on \langle u \rangle and y on \langle v \rangle , we can argue that the geodesic between them curves toward the origin.

pic2And so we have Lemma 14.

Lemma 14: There exists R > 0 such that \forall m,n, [u^m,v^n]\cap B(1,R)\neq \emptyset .

Proof:

pic3Recall that \phi : \mathbb{Z}\to \Gamma by \phi (i)= u^i is a quasi-isometric embedding.  So by Theorem 6, d_{haus}({1,u,u^2,\dots, u^m},[1,u^m]) < R_1 and d_{haus}({1,v,v^2,\dots, v^n},[1,v^n]) < R_1

pic4By Lemma 13 choose u^p \in \langle u \rangle such that
d(u^p,\langle v \rangle) > 2R_1 + \delta .  Choose u_p \in [1,u^m] such that d(u_p,u^p) < R_1 .  Now, u_p must be \delta-close to [u^m,v^n] so for some point x on the geodesic between v^n and u^m, d(u_p, x) < \delta .  Then d(1,[u^m,v^n])\leq d(1,u^p) + d(u^p,u_p) + d(u_p, x) \leq l(u^p)+R_1 + \delta \Box

For a subgroup H \subseteq \Gamma, one can choose a closest point projection \Pi_H : \Gamma \to H which is H-equivariant. (Write \Gamma = \cup H{g_i}.  Choose \Pi_H(g_i)=h_i where h_i and g_i are close and declare \Pi_H to be H-equivariant.)  \Pi_{H} is typically not a group homomorphism.

We’re interested in \Pi_{\langle u\rangle} and \Pi_{\langle v\rangle}.
pic5In \mathbb{H}^2, there is some m such that \forall x\in \mathbb{H}^2 either l(\Pi_{<u>}(x)) \leq m or l(\Pi_{<v>}(x)) \leq m.

pic6

Lemma 15: \exists M such that \forall x\in Cay(\Gamma), l(\Pi_{<u>}(x)) \leq M or l(\Pi_{<v>}(x)) \leq M .

Proof:
pic7

Let y\in[\Pi_{<u>}(x), \Pi_{<v>}(x)]\cap B(1,R).  WLOG, y is \delta-close to p\in[x,\Pi_{<u>}(x)] and d(1, \Pi_{<u>}(x) \leq d(1,p)+d(p,\Pi_{<u>}(x)) \leq d(1,p) +d(p,1) since \Pi_{<u>}(x) is the closest point to x (in particular compared to u^0=1).  So d(1,\Pi_{<u>}(x)) \leq 2d(1,p)\leq 2(R+\delta)\Box .

Now we can prove the theorem.

Proof of Theorem 12:

The idea is to use the Ping-Pong Lemma on the Cayley graph.

pic8Let X_1 = \Pi_{<u>}^{-1}(\lbrace u^m\mid l(u^m) > M\rbrace) and let X_2= \Pi_{<v>}^{-1}(\lbrace v^n\mid l(v^n) > M\rbrace), where M is provided by Lemma 15.  For all x_1\in X_1 we have l(\Pi_{<v>}(x_1))\leq M and likewise for all x_2\in X_2 we have l(\Pi_{<u>}(x_2))\leq M.  In particular,  X_1 \cap X_2 = \emptyset.

Let x_2\in X_2.  By \langle u\rangle-equivariance,

\Pi_{<u>}(u^m x_2)=u^m\Pi_{<u>} (x_2)

for any m.  In particular,

l(\Pi_{<u>}(u^m x_2))\geq l(u^m)-l(\Pi_{<u>}(x_2))\geq l(u^m)-M

by the triangle inequality.  Similarly,

l(\Pi_{<v>}(v^n x_1))\geq l(v^n)-l(\Pi_{<v>}(x_1))\geq l(v^n)-M

for all x_1\in X_1 and all n.  Because \langle u\rangle and \langle v\rangle are quasi-isometrically embedded, it follows that u^mX_2 \subset X_1 and v^n X_1\subset X_2 for m,n >>0.

Therefore, by the Ping-Pong Lemma \langle u^m, v^n \rangle \cong \mathbb{F}_2.

Finally, we are in a position to prove that a hyperbolic group has no subgroup isomorphic to \mathbb Z^2.

Theorem 11. Let \gamma \in \Gamma with o(\gamma)=\infty. Then |C(\gamma):\langle \gamma \rangle|<\infty.

Proof. By Lemma 10, we can assume that \gamma is not conjugate to any element of length \leq 4\delta by replacing \gamma with a power of itself. Suppose g\in C(\gamma). We need to bound d(g, \gamma).

henrylec151

Replacing g with \gamma^{-r}g for some r, we may assume that d(1,g)=d(g,\langle \gamma \rangle). We will be done if we can bound l(g).

Suppose l(g)>2(l(\gamma)+2\delta). By dividing into triangles, we see that any geodesic rectangle is 2\delta-slim, in the same way that triangles are \delta-slim.

Because the rectangle with vertices 1, \gamma, g\gamma, g is 2\delta-slim, there exists g_t, g_{t'} \in [1,g] such that d(g_t, \gamma g_{t'}) \leq 2\delta.

If t<t'-2\delta, then d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma), a contradiction. Similarly t'<t-2\delta. So |t-t'|<2\delta. Therefore, d(g_t, \gamma g_t)<4\delta.

But l(g_t^{-1}\gamma g_t)=d(g_t, \gamma g_t)<4\delta. This is a contradiction since we assumed that \gamma is not conjugate to anything so short. Therefore l(g)\leq2(l(\gamma)+2\delta). Thus |C(\gamma):\langle \gamma \rangle|<\infty.

An element of a group is torsion if its order is finite.

A group is torsion if every element is torsion.

A group is  torsion-free if no nontrivial elements are torsion.

Corollary. Every non-trivial abelian subgroup of a hyperbolic group is virtually cyclic.

Lemma 11. Let \Gamma be a torsion-free hyperbolic group. Whenever \gamma \in \Gamma is not a proper power, then \langle \gamma \rangle is malnormal.

Definition. A subgroup H of a group G is malnormal if for all g\in GgHg^{-1} \cap H \neq 1 , then g\in H.

Remark. By Theorem 11, if \Gamma is hyperbolic and torsion-free, centralizers are cyclic.

Proof of Lemma. Suppose g \langle \gamma \rangle g^{-1}\cap \langle \gamma \rangle \neq 1.

Therefore for some p, q \neq 0, g\gamma^{p}g^{-1}=\gamma^q.

By Lemma 10, |p|=|q|. Therefore g^2\gamma^pg^{-2}=\gamma^p. Thus g^2\in C(\gamma^p)= \langle \gamma \rangle. Therefore g \in \langle \gamma \rangle.

Exercise 17. Prove that if x, y, z \in \Gamma where \Gamma is hyperbolic and torsion-free and xy=yx and yz=zy and y\neq 1, then xz=zx. That is, \Gamma is commutative transitive.

We now turn briefly to a fundamental open question about hyperbolic groups.  This question is a theme of the course.

Question. Is every word-hyperbolic group residually finite?

The fundamental groups of hyperbolic manifolds are linear, so residually finite by Selberg’s Lemma.

What about for negative curved manifolds?

Evidence for:

Theorem (Sela). Every torsion-free hyperbolic group is Hopfian.

Theorem (I. Kapovich-Wise). If every nontrivial hyperbolic group has a proper finite-index subgroup, then every hyperbolic group is residually finite.

Evidence against:

Theorem (Agol-Groves-Manning). If every hyperbolic group is residually finite, then every quasi-convex subgroup of every hyperbolic group is separable.

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