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

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

**The Š****varc-Milnor Lemma:** Let be a proper geodesic metric space. Let act cocompactly and properly discontinuously on . (Properly discontinuously means that for all compact .) Then is finitely generated and, for any , the map

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

**Proof:** We may assume that is infinite and is non-compact. Let be large enough that the -translates of cover . Set

Let . Let . We want to prove that:

(a) generates,

(b) ,

(c) , there exists such that

**Note:** and .

(c) is obvious.

(b-i) is also obvious.

To complete the proof we need to show (a) and (b-ii).

Assume . Let be such that

As , . Choose , such that and for each . Choose such that for each . Let , so . Now

So, . Therefore generates .

Also,

as required.

**Corollary:** If is a finite index subgroup of a finitely generated group then is quasi-isometric to .

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

**Example:** .

Semidirect product is taken over the matrix This means that , but

Let with eigenvalues with . Let .

sits inside as a *uniform lattice*, meaning is a compact space.

**Exercise 11:** What is this quotient?

So, is a quasi-isomorphic to But, Bridson-Gersten showed that and are commensurable if and only if the corresponding eigenvalues have a common power.

**Exercise 12:** Let be the infinite regular valent tree. Prove that for all , is quasi-isometric to .

## 3 comments

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9 February 2009 at 9.08 pm

Henry WiltonThanks, Allison! I edited a couple of things, including giving Švarc his hacek.

9 February 2009 at 9.18 pm

metaficionadoThank you for fixing the hacek and the matrix.

10 February 2009 at 9.19 am

Henry WiltonI should also just mention that the proof of the Švarc-Milnor Lemma is the one given by de la Harpe in his

Topics in Geometric Group Theory.