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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.
Agol-Groves-Manning’s Theorem predicts that, for every word-hyperbolic group we can easily construct, every quasiconvex subgroup is separable (otherwise, we would find a non-residually finite hyperbolic group!).
In this section, we use graphs of groups to build new hyperbolic groups:
Combination Theorem (Bestvina & Feighn): If is a malnormal subgroup of hyperbolic groups , then is hyperbolic.
Recall: is called a malnormal subgroup of if it satisfies: if , then .
For a proof, see M. Bestvina and M. Feighn, “A combination theorem for negatively curved groups”, J. Differential Geom., 35 (1992), 85–101.
Example: Let be free, not a proper power. By Lemma 11, is malnormal, so is hyperbolic. As a special case, if is closed surface of even genus , considered as the connected sum of two copies of the closed surface of genus , then by Seifert-van Kampen Theorem, for some .
Question: (a) Which subgroups of are quasiconvex? (b) Which subgroups of are separable?
We will start by trying to answer (b). The following is an outline of the argument: Let be a finite graph so that , let be two copies of . Realize as maps , where . Let be the graph of spaces with vertex spaces , edge space , and edge maps . Then clearly, , and finitely generated subgroups are in correspondence with covering spaces . We can then use similar technique to sections 27 and 28.
Let us now make a few remarks about elevations of loops. Let be a loop in some space , i.e., and . Consider an elevation of :
The conjugacy classes of subgroups of are naturally in bijection with . The degree of the elevation is equal to the degree of the covering map .
Definition: Suppose is a covering map and is an intermediate covering space, i.e., factors through , and we have a diagram
If and are elevations of and the diagram commutes, then we say that descends to .
Let be a finite graph, a finitely generated subgroup and a loop. Let be a covering space corresponding to .
Lemma 29: Consider a finite collection of elevations of to , each of infinite degree. Let be compact. Then for all sufficiently large , there exists an intermediate, finite-sheeted covering space satisfying: (a) embeds in ; (b) every descends to an elevation of degree exactly ; (c) these are pairwise distinct.
Recall that our goal is to determine, for a given map of graphs of spaces such as the one shown below, whether the map can be extended to a covering map .
Let be graphs of spaces equipped with maps , and as before. Recall that in Stallings’ proof of Hall’s Theorem, we completed an immersion to a covering map by gluing in edges. We will aim to do the same thing with graphs of spaces.
Definition: Let be a graph of spaces, and let be the map to the underlying graph. If is a subgraph, then has a graph-of-spaces structure with underlying graph . Call a subgraph of spaces of .
We’re seeking a condition on such that is realized as a subgraph of spaces of some with a covering map such that the following diagram commutes:
Definition: For each edge map of , and each a vertex of , let
For each possible degree , let be the set of elevations of degree . We will say satisfies Stallings’ condition if and only if the following two things hold:
(a) Every edge map of is an elevation of the appropriate edge map of .
(b) For each and , there is a bijection .
So in Figure 1, the graph of spaces is something you might be able to turn into a covering. In the picture, is represented by the blue circles, and is represented by the green circles. Observe that the blue circles are in bijection with the green circles.
Corollary: satisfies Stallings’ condition if and only if can be realized as a subgraph of spaces of some such that
(a) , and
(b) there is a covering map such that the following diagram commutes:
Proof of Corollary. First we’ll show that if can be extended to a covering map as described above, then satisfies Stallings’ condition. By Theorem 17, every edge map of is an elevation. So there are inclusions
Furthermore, these maps are surjective, and clearly degree-preserving.
Now assume that satisfies Stallings’ condition. Then we build as follows. Let . As above, we have degree-preserving inclusions (this time, not surjections)
Extend these inclusions to bijections . Now we set . Each of these is an elevation
This defines an edge space and an edge map . Consider the corresponding elevation in :
Because and are of the same degree, we have a covering transformation . So we can identify them, and use as the other edge map. By construction, satisfies the conditions of Theorem 17, so there is a suitable covering map .
Exercise 25: (This will be easier later, but we have the tools necessary to do this now.) Prove that if and are LERF groups, then so is .