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Some intuition: Recall that if $M$ is a closed hyperbolic manifold
then $\pi_1(M)$ is word-hyperbolic. However, a lot of interesting hyperbolic manifolds are not closed.

Example: Let $K\subset S^3$ be the figure 8 knot.

Then the complement $M_{8}=S^{3}$ $K$ admits a complete hyperbolic metric and is of finite volume.

So, here we have an example of a hyperbolic manifold which is not compact but is of finite volume. This is almost as which is almost as natural as being closed.

$M_{8}$ is homotopy equivalent to $M_{8}'$, the complement of a thickened $K$ in $S^{3}$.

$M_8'$ is a compact manifold with boundary and its interior admits a hyperbolic metric. The boundary of $M_8'$ is homeomorphic to a 2-torus, so $\partial M_8' \hookrightarrow M_8'$ induces a map $\mathbb{Z}^2\hookrightarrow\pi_1M_8'$. By Dehn’s lemma, the map is injective so $\pi_1M_8'$ cannot be word hyperbolic. The point is that $\pi_1M_8$ acts nicely on $\mathbb{H}^2$ but no cocompactly so the Svarc=Milnor lemma does not apply.

The torus boundary component of $M_8'$ corresponds to a cusp of $M_8$.

The point is that we can use cusped manifolds like $M_8'$ to build a lot of manifolds and in particular a lot of hyperbolic manifolds.

Take $M_8'$ and a solid Torus $T$ .

Choose a homeomorphism $\phi: \partial M_8' \hookrightarrow\partial T$

Definition: The manifold $M_{\phi}=M_{8}'\cup_{\phi}T$ is obtained from $M_{8}'$ by Dehn filling .

We now want to understand what we have done to $\pi_{1}M_{8}$. The map $\phi$ induces a map $\phi_{*}$:

The surjectivity of $\phi_{*}$ follows from the fact that $\phi$ is a homeomorphism. The Seifert Van Kampen theorem implies that $\pi_{1}M_{\phi}=\pi_{1}M_{8}\langle\langle \ker(\phi_{*})\rangle\rangle$, where $\langle\langle\ker(\phi_{*}) \rangle\rangle$ denotes the normal closure of $\ker(\phi_{*})$ .

Gromov-Thurston $2\pi$ theorem: Let M be any compact hyperbolic manifold and $\partial_{0}M$ be a component of $\partial M$ homeomorphic to a 2-torus for all but finitely many choices of

the Dehn filling $M_{\phi}$ is hyperbolic.

Note: by finitely many we mean finitely many maps up to homotopy.

This is a very fruitful way of building hyperbolic manifolds. The next question to ask is whether we can do the same thing for groups. So, now we will try to develop a group theoretic version of this picture.

Let $\Gamma$ be a group theoretic graph with the induced length metric. Construct a new graph $\mathcal{H}(\Gamma)$ called the combinatorial horoball on $\Gamma$ as follows: Define the vertices $V(\mathcal{H})=V(\Gamma)\times \mathbb{N}$. There are two sorts of edges in ${E}(\mathcal{H})$. We say that $(u,k)$ and $(v,k)$ are joined by a (horizontal) edge if $d_{\Gamma}(u,v) \leq 2^{k}$ and $u\neq v$. We say that $(v,k)$ and $(v,k+1)$ are joined by a (vertical) edge for all $k$.

For $k$ large enough $u'$ and $v'$ will have distance one and $L\leq 1$ iff $2^{k} \ge d_{\Gamma}(u,v)$ iff $k\leq \log_{2}d_{\Gamma}(u,v)$.

Exercise 27:
(A). For $u,vin V(\Gamma)$, $d_{\mathcal{H}}((u,0),(v,0))\approx \log_{2}d_{\Gamma}(u,v)$.

(B). For any connected $\Gamma$, $\mathcal{H}(\Gamma)$ is Gromov hyperbolic .

Let $G$ be a group and let $\mathcal{P}=\{ P_{1},\ldots, P_{n} \}$ be a finite set of finitely generated subgroups of $G$. Choose a finite generating set $S$ for $G$ such that for each $i$, $s_i=S \cap P_i$ generate $P_i$. Then $\mathrm{Cay}(G,S)$ contains natural copies of $\mathrm{Cay}(P_{i},S_{i})$.

Construct the augmented Cayley graph $X=X(G,\mathcal{P},S)$ by gluing on combinatorial horoballs equivariantly.

$X(G,\mathcal{P},S) = \mathrm{Cay}(G,S) \cup \bigcup_{i} \lbrack \mathcal{H}(\mathrm{Cay}(P_{i},S_{i})) \times G/P_{i} \rbrack / \sim$ where for each $i$
and each $gP_{i}\in G$ /$P_{i}$, $\mathcal{H}(\mathrm{Cay}(P_{i},S_{i}) \times \{ gP_{i}\}$ is glued to $g\mathrm{Cay}(P_{i},S_{i})$ along $\mathrm{Cay}(P_{i},S_ {i}) \times \{ 0 \}$.

Definition: G is hyperbolic rel $\mathcal{P}$ if and only if $X(G,P,S)$ is Gromov hyperbolic for some (any) choice of $S$.

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