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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.

figure 1

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}.

fig2

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_{*}:

lecture_4_17_09_xymatrix1

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

lecture_4_17_09_xymatrix2

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.

fig6
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 .

fig7

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 .
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.

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