This is the second post intended to understand some of the ideas in Milnor’s “Note on curvature and the fundamental group.” This is the paper that introduces the idea of growth rates for groups and proves that the fundamental group in negative curvature has exponential growth (as well as a dual result on polynomial growth in nonnegative curvature). In the previous post, we described volume comparison results in negative curvature: we showed in particular that a curvature bounded above by meant that the volumes of expanding balls grow exponentially in the radius. In this post, we’ll explain how this translates into a result about the fundamental group.
1. Growth rates of groups
Let be a finitely generated group, and let
be a finite set of generators such that
. We define the norm
such that is the length of the smallest word in
that evaluates to
. We note that
defines a metric on . The metric depends on the choice of
, but only up to scaling by a positive constant. That is, given
, there exists a positive constant
such that
. The metric space structure on
is thus defined up to quasi-isometry.
Definition 17 Let
be the number of elements
with
with
. We say that
is of exponential growth if
grows exponentially, i.e. if
We note that the limit in the above definition actually exists. This follows from the inequality
Now we use an elementary lemma:
Lemma 18 Any function
(bounded on bounded intervals) such that
has the property that
exists.
Proof: To see this, we fix , and observe that any positive real number
can be written as
for
and
. Then:
We find that
and letting , we get:
which proves the lemma.
Moreover, the property of having exponential growth is unchanged under variation of the generating set .
We can make a similar definition:
Definition 19 We say that
is of polynomial growth if
is dominated by
for
.
For example, finitely generated abelian groups are of polynomial growth.
The goal is to prove:
Theorem 20 (Milnor) If
is a compact manifold of negative curvature, then
has exponential growth.
2. Outline of the proof
The argument behind Milnor’s theorem is fairly short, but it will be worth outlining. Let be a compact manifold of negative curvature. Choose a point
in its universal cover
, which is diffeomorphic to
under the exponential map
. The strategy is to relate the growth in the fundamental group
with the growth in
.
Namely, one has:
Theorem 21 Let
be a complete Riemannian manifold, and let
. Let
be a finitely generated group of isometries of
acting freely such that the manifold
is compact. Then if the volumes
grow exponentially in
, the group
has exponential growth.
By the analysis in the previous post, the volumes of balls in a complete, simply connected Riemannian manifold of bounded-above negative curvature grow exponentially. So this theorem will be sufficient.
Let’s now try to outline how this result, in turn, is proved. Let be the diameter of
and let
be the ball of radius
around
in
. Then
projects surjectively onto
under the covering projection
Therefore, any point in is conjugate under
to something in
. So
is almost a fundamental domain for the action of
(except that
might take one point in
to another in it). In other words,
What does this mean? If , then it follows that the ball
(of radius
) is contained in the union of the various
. The key point is the following:
Key point: is contained in the union of various
for
(with respect to any fixed system of generators).
In other words, if is of distance
from
, then this is “witnessed” by the existence of
with
such that
.
As a result, it will follow that
Since grows exponentially, it will follow that the right-hand-side also grows exponentially, which gives the exponential growth rate of
.
3. Completion of the proof
The previous section contained the main ideas of the proof of Milnor’s theorem, and reduced it to the following:
Lemma 22 Let
be as in the above theorem. Then there exists
such that for any
,
where the norm is with respect to a system of generators to be specified.
To see this, we’re going to take as system of generators the set
This is a finite set. In fact, if is sufficiently small, then the sets
are all disjoint balls in of radius
(and the same volume). Only finitely many of them can be contained in any bounded region, so that means that
is finite. If
, then
and
are disjoint, and a similar argument shows that
for some . Either
and
intersect, or they have to be separated by some
.
We now suppose that is a point such that
We would like to show that is in
for a
of controlled norm.
In fact, choose a path from to
of length
, and choose points along it
where and
can be taken to be
. It follows that
where and so
It follows that the group-theoretic distance from to
is at most
for each
, so that the group-theoretic distance from
to
(which we can take as
) is at most
.
It follows that
where . This completes the proof.
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