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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
(i) ;
(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,
Also
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 .


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