Example: G is quasi-isometric to 1 if and only if G is finite.

Definition: A metric space X is proper if closed balls of finite radius in X are compact.  The action of a group \Gamma on a metric space X is cocompact if X/\Gamma is compact in the quotient topology.

The Švarc-Milnor Lemma: Let X be a proper geodesic metric space.  Let \Gamma act cocompactly and properly discontinuously on X.  (Properly discontinuously means that for all compact K\subseteq X, |\{\gamma\in \Gamma | \gamma K\cap K \neq \emptyset \}| < \infty.)  Then \Gamma is finitely generated and, for any x_0\in X, the map

\Gamma \rightarrow X
\gamma \mapsto \gamma.\ x_0

is a quasi-isometry (where \Gamma is equipped with the word metric).

Proof: We may assume that \Gamma is infinite and X is non-compact.  Let R be large enough that the \Gamma-translates of B=B(X, R) cover X.  Set

S=\{s\in \Gamma\setminus 1 | s\bar{B}\cap \bar{B} \neq \emptyset \}

Let r=\inf\{d(\bar{B},\gamma\bar{B}) | \gamma\in\Gamma , \gamma \not\in S \cup \{1\} \}.  Let \lambda=\max_{s\in S} d(x_0, sx_0).   We want to prove that:
(a) S generates,
(b) \forall \gamma\in\Gamma,

\underbrace{ \lambda^{-1}d(x_0,\gamma x_0)}_{i} \leq \ell_S(\gamma)\leq \underbrace{ \frac{1}{r}d(x_0,\gamma x_0) + 1}_{ii}
(c) \forall x\in X, there exists \gamma\in\Gamma such that

d(x,\gamma x_0) \leq R

Note: d_S(1,\gamma)=\ell_S(\gamma) and d_S(\gamma, \delta)=\ell_S(\gamma^{-1}\delta).

(c) is obvious.
(b-i) is also obvious.
To complete the proof we need to show (a) and (b-ii).

Assume \gamma\not\in S\cup\{1\}.  Let k be such that

R + (k-1)r \leq d(x_0, \gamma x_0) < R + kr

As \gamma\not\in S\cup\{1\}, k>1.  Choose x_1, \cdots, x_{k+1} =\gamma x_0, such that d(x_0, x_1)<R and d(x_i, x_{i+1})<r for each i>0.   Choose 1=\gamma_0,\gamma_1,\cdots, \gamma_{k-1},\gamma_k=\gamma such that x_{i+1}\in\gamma_i\bar{B} for each i.  Let s_i=\gamma_{i-1}^{-1}\gamma_i, so \gamma=s_1\cdots s_k.  Now

d(\bar{B}, s_i\bar{B}) \leq d(\gamma_{i-1}^{-1} x_i, s_i\gamma_i^{-1}x_{i+1}) = d(x_i, x_{i+1}) <r

So, s_i\in S\cup\{1\}.  Therefore S generates \Gamma.


\ell_S(\gamma) \leq k \leq \frac{1}{r}d(x_0, \gamma x_0) + \frac{r-R}{r} < \frac{1}{r}d(x_0,\gamma x_0)+1

as required.

Corollary: If K\subseteq \Gamma is a finite index subgroup of a finitely generated group then K is quasi-isometric to \Gamma.

Two groups G_1 and G_2 are commensurable if they have isomorphic subgroups of finite index.  Clearly, if G_1 and G_2 are commensurable then they are quasi-isometric.

Example: Sol = \mathbb{R}^2 \rtimes_E \mathbb{R}.
Semidirect product is taken over the matrix eqlatex This means that Sol=\{(x,y,t) | x,y,t\in\mathbb{R} \}, but

(x,y,t)(x',y',t')=(x+e^tx', y+e^{-1}y', t+t')

Let A\in SL_2(\mathbb{Z}) with eigenvalues \lambda, \lambda^{-1} with \lambda >1.  Let \Gamma=\mathbb{Z}^2 \rtimes_A \mathbb{Z}.

\Gamma_A sits inside Sol as a uniform lattice, meaning Sol/\Gamma_A is a compact space.

Exercise 11: What is this quotient?

So, \Gamma_A is a quasi-isomorphic to Sol  But, Bridson-Gersten showed that \Gamma_A and \Gamma_{A'} are commensurable if and only if the corresponding eigenvalues \lambda, \lambda' have a common power.

Exercise 12: Let T_k be the infinite regular 2k valent tree.  Prove that for all k, k' \geq 2, T_k is quasi-isometric to T_{k'}.