Tag Archive

You are currently browsing the tag archive for the ‘Hyperbolic groups’ tag.

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.

Tag Archive

34. Combinatorial Dehn Filling

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

15. Abelian subgroups and Residual Finiteness

Recent Posts

Pages

Recent Comments

Archives

Follow “392C Geometric Group Theory”