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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
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 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 .