Theorem 12 (Gromov): Let be torsion-free
-hyperbolic group. If
such that
, then for all sufficiently large
,
.
Remark: The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities. For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of .
For the rest of this lecture will be a torsion-free
-hyperbolic group,
where
are primitive (i.e. not proper powers).
Recall that for torsion-free
-hyperbolic,
primitive implies that
.
If and
do not commute we can show there is some point
on
arbitrarily far from
.
Hence we have the following lemma.
Lemma 13:
If and
do not commute there is some point
on
arbitrarily far from
.
Proof: Suppose not. That means such that
such that
. So
is in
. But the Cayley graph is locally finite so
has finitely many elements. By the Pigeonhole Principle
such that
for some
. Then
. But then
.
.
For a moment view and
as the horizontal and vertical geodesics in
. For two points
on
and
on
, we can argue that the geodesic between them curves toward the origin.
And so we have Lemma 14.
Lemma 14: There exists such that
,
.
Proof:
Recall that
by
is a quasi-isometric embedding. So by Theorem 6,
and
By Lemma 13 choose
such that
. Choose
such that
. Now,
must be
-close to
so for some point
on the geodesic between
and
,
. Then
.
For a subgroup , one can choose a closest point projection
which is
-equivariant. (Write
. Choose
where
and
are close and declare
to be
-equivariant.)
is typically not a group homomorphism.
We’re interested in and
.
In
, there is some
such that
either
or
.
Lemma 15: such that
,
or
.
Proof:
Let . WLOG,
is
-close to
and
since
is the closest point to
(in particular compared to
). So
.
.
Now we can prove the theorem.
Proof of Theorem 12:
The idea is to use the Ping-Pong Lemma on the Cayley graph.
Let
and let
, where
is provided by Lemma 15. For all
we have
and likewise for all
we have
. In particular,
.
Let . By
-equivariance,
for any . In particular,
by the triangle inequality. Similarly,
for all and all
. Because
and
are quasi-isometrically embedded, it follows that
and
for
.
Therefore, by the Ping-Pong Lemma .
2 comments
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4 March 2009 at 6.49 am
Henry Wilton
Wow! That’s a lot of mathematics and a lot of pictures. Many many thanks, to both Sam and Emily. Great job. I tidied up the proof of Theorem 12 a little.
2 May 2011 at 10.55 am
The Tits alternative and non-positive curvature | Here there be dragons
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