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34. Combinatorial Dehn Filling

20 April 2009 in Course notes | Tags: Hyperbolic groups, Relatively hyperbolic groups | by jmaciejewski | Leave a comment

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 $M$ is 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

$P_i/N_i\to G/\mathcal{N}$ is injective for all $i$ , and
$G/\mathcal{N}$ 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 $i$ .

16. Free Subgroups of Hyperbolic Groups (from Sam Kim)

3 March 2009 in Course notes | Tags: Free groups, Hyperbolic groups | by elandes | 2 comments

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 $\langle v \rangle$ .
Hence 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.

And 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:

Recall 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 $d_{haus}({1,v,v^2,\dots, v^n},[1,v^n]) < R_1$

By 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 $\Pi_{\langle v\rangle}$ .
In $\mathbb{H}^2$ , there is some $m$ such that $\forall x\in \mathbb{H}^2$ either $l(\Pi_{}(x)) \leq m$ or $l(\Pi_{<v>}(x)) \leq m$ .

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

Proof:

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

Now we can prove the theorem.

Proof of Theorem 12:

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

Let $X_1 = \Pi_{}^{-1}(\lbrace u^m\mid l(u^m) > M\rbrace)$ 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 $l(\Pi_{}(x_2))\leq M$ . In particular, $X_1 \cap X_2 = \emptyset$ .

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

$\Pi_{}(u^m x_2)=u^m\Pi_{} (x_2)$

for any $m$ . In particular,

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

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 $\langle u^m, v^n \rangle \cong \mathbb{F}_2$ .

15. Abelian subgroups and Residual Finiteness

3 March 2009 in Course notes | Tags: Hyperbolic groups, Residual finiteness, Separability | by kczar44 | 1 comment

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 $g\in G$ , $gHg^{-1} \cap H \neq 1$ , 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.