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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 $M$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

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 $\langle v \rangle$.
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 $l(\Pi_{}(x)) \leq m$.

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

Proof:

Let $y\in[\Pi_{}(x), \Pi_{}(x)]\cap B(1,R)$.  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 $X_2= \Pi_{}^{-1}(\lbrace v^n\mid l(v^n) > M\rbrace)$, where $M$ is provided by Lemma 15.  For all $x_1\in X_1$ we have $l(\Pi_{}(x_1))\leq M$ 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,

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

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 $d(g, \gamma)$.

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, then $d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma)$, a contradiction. Similarly $t'. 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.