Proof of Theorem 6: Let be a quasi-geodesic in a
-hyperbolic space
. We can replace
by
as in Lemma 8. Let
,
, and
such that
is maximal. We need to bound
. Let
be such that
, and
such that
. Let
such that
, and
such that
. Finally, let
be a path in
obtained by concatenating
, the section of the image of
with endpoints
and
, and
.
By construction, does not intersect
, and part 3 of Lemma 8 shows that
.
Lemma 7 gives a bound on the length of paths based on the distance between their endpoints:
.
The left hand side of this inequality increases linearly in , while the right increases logarithmically, so that
must be bounded above for the equality to hold, and clearly this upper bounded depends only on the constants
,
and
. QED.
It now makes sense to make the following definition.
Definition: A finitely generated group is called (word-) hyperbolic if some (any) Cayley graph for
is Gromov-hyperbolic. Equivalently, a group
is hyperbolic if it acts properly discontinuously and cocompactly by isometries on a proper Gromov-hyperbolic metric space.
Examples:
a) Free groups
b) is not hyperbolic for
.
c) Let be any closed hyperbolic manifold. Then
is word-hyperbolic.
d) More generally, is word-hyperbolic for any
with negative sectional curvature bounded away from
.
Without getting into too much detail, we briefly mention the following theorem of Gromov as an indication of how general the class of hyperbolic groups really is.
Theorem (Gromov): A “randomly chosen” finitely presented group is “almost surely” word-hyperbolic.
Definition: A subspace of a geodesic metric space
is quasiconvex if there exists a
such that, for all
and for all
,
.
Example: Consider with the
-metric. Then the diagonal subgroup
is not quasiconvex (though it is quasi-embedded).
Theorem 6 implies that this kind of poor behavior does not occur in hyperbolic space.
Corollary: Suppose that is a word-hyperbolic group and
is a subgroup. Then
is quasiconvex in some (any) Cayley graph of
if and only if
is finitely generated and
is a quasi-embedding.
Proof: is immediate from Theorem 6. For the other direction, fix a generating set
for
, assume
is quasiconvex in the Cayley graph of
with constant
, and let
be in
. Consider a geodesic in the Cayley path of
from
to
, which we can take to be of the form
for
in
. Let
be the vertices of this geodesic, so
.
By quasiconvexity, for each there exists
in
such that
. Take
and
. Let
, so
. Let
. Note that
. For each
, we have that
and so
. Therefore,
is generated by
, a finite set. Furthermore, we have shown that
.
But it’s clear that (as each element of
has
-length at most
) so the inclusion of
into
is a quasi-isometric embedding. QED.
In light of this, the following definition makes sense
Definition: A subgroup of a hyperbolic group
is called quasiconvex if it is a quasiconvex space of some (any) Cayley graph of
.
Exercise 14: If is a retraction and
is finitely generated, then the inclusion
is a quasi-isometric embedding. (Hint: you can choose a generating set
for
such that
or
for all
.
Example: Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a retract of a finite-index subgroup, so every finitely generated subgroup of a free group is quasiconvex.



3 comments
Comments feed for this article
19 February 2009 at 1.07 pm
Henry Wilton
Thanks for the pictures, David – definitely beyond the call of duty.
13 June 2012 at 9.55 am
benjamin apra
It seems me that the proof for theorem 6 is not complete: it only proves that the geodesic is in a D-neigboorhood of the quasi geodesic, but doesn’t prove the other inclusion, so that the the hausdorff distance is not bounded.
Am i clear enough?
thank you
8 July 2012 at 1.58 pm
Henry Wilton
Benjamin, thanks for the comment, and sorry for being so slow to reply. I think you’re right. I haven’t had a chance to think hard about it, but I don’t think there’s any difficulty with the other direction – a similar argument should work.