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 Švarc-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:
(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 .
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.
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 .