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 17Let be the number of elements with with . We say that is ofexponential growthif 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 18Any 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 19We say that is ofpolynomial growthif 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 21Let 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 22Let 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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