You are currently browsing allison’s articles.

**Last time**:

**Lemma 24**: Fix basepoints as usual. There is a lift of such that if and only if

.

Furthermore, if the lift exists, it is unique.

**Lemma 25**: Given a choice of , there exists an elevation of at . Furthermore, is unique in the sense that if is another elevation of and then there is a homeomorphism and the diagram

commutes.

**Proof**:

Let be defined by . By Lemma 24, the composition

lifts at , call this lift . Suppose

Then . But by Lemma 24, . This implies .

Another, more categorical construction uses the **fibre product**.

The fibre product is defined by

.

There are obvious maps

given by forgetting factors.

**Exercise**: If is a covering map then is a covering map.

**Lemma 26**: Fix . Let and let for . Let , and let be the connected component containing . Then is an elevation of at , and every elevation of arises in this way.

**Proof**: To prove that is an elevation we just observe that . Now suppose

is an elevation. Then , with .

The covering map factors trhough , and so is a covering map. Because is an elevation, is a homeomorphism onto its image, a connected component of .

What has this got to do with graphs of spaces/groups?

Let be a vector space, and let be an edge map. Define to be the mapping cylinder

comes with a map such that and . This is an inclusion , and . Let be a covering map.

Let be the fibre product. There’s a map ; . Clearly, .

Therefore, is an injection. It’s easy to see that induces a bijection at the level .

**Lemma 27**: Any covering space arises as the fibre product of a covering map .

**Proof**: Let be a covering map

let be the fibre product of and .

Define by . As before, is a covering map.

**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 .

## Recent Comments