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Lemma 29: Suppose is a finite set of infinite degree elvations and is compact. Then for all sufficiently large , there exists an intermediate covering such that
(a) embeds in
(b) every descends to an elevation of degree
(c) the are pairwise distinct
Proof: We claim that the images of never share an infinite ray (a ray is an isometric embedding of ). Neither do two ends of the same elevation . Let’s claim by passing to the universal cover of , a tree .
For each , lift to a map . If and share an infinite ray then there exists such that and overlay in an infinite ray. The point is that correspond to cosets and . But this implies that
This implies that . So . A similar argument implies that the two ends of do not overlap in an infinite ray. This proves the claim.
Let be the core of . Enlarging if necessary, we can assume that
(ii) is a connected subgraph;
(iii) for each , for some , ;
(iv) for each , .
For each identifying with so that is identified with and is identified with . Let
For all sufficiently large ,
Now, the restriction of factors through , where . This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required.
Theorem 19: is LERF.
Recall the set-up from the previous lecture. We built a graph of spaces for .
Proof: Let be finitely generated. Let be the corresponding covering space of and let be compact. Because is finitely generated, there exists a subgraph of spaces such that . We can take large enough so that . We can enlarge so that it contains every finite-degree edge space of . Also enlarge so that
for any . For each let and let incident edge map of infinite degree .
Applying lemma 29 to , for some large , set . (Here we use the fact that vertex groups of are finitely generated)
Define as follows:
For each , the edge space is the that the lemma produced from the corresponding .
Now, by construction, can be completed to a graph of spaces so that the map
factors through and embeds. Let be identical to except with +’s and -‘s exchanged. Clearly satisfies Stallings condition, as required.
Theorem 8: Let be a -hyperbolic group with respect to . If are conjugate then there exists such that
where depends only on .
Proof: We work in . Let be such that . Let be such that . We want to find a bound on .
Let . By Lemma 9,
So . Thus . Suppose that . By the Pigeonhole Principle there exist integers such that . It follows that one can find a shorter conjugating element by cutting out the section of between and .
Recall, for , is the centralizer of .
Theorem 9: If is -hyperbolic with respect to and , then is quasi-convex in .
Proof: Again we work in . Let , . We need to prove that is in a bounded neighborhood .
Just as in the proof of Theorem 8,
Well, and are conjugate. By Theorem 8 there exists such that
But so that and .
Exercise 15: Prove that
is not hyperbolic for any Anosov .