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 $c < 0$ 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 ${G}$ be a finitely generated group, and let ${S}$ be a finite set of generators such that ${S^{-1} = S}$. We define the norm

$\displaystyle \left \lVert\cdot \right \rVert_S: G \rightarrow \mathbb{Z}_{\geq 0}$

such that ${\left \lVert g \right \rVert_S}$ is the length of the smallest word in ${S}$ that evaluates to ${g}$. We note that

$\displaystyle d(g, h) \stackrel{\mathrm{def}}{=} \left \lVert gh^{-1} \right \rVert_S$

defines a metric on ${G}$. The metric depends on the choice of ${S}$, but only up to scaling by a positive constant. That is, given ${S, S'}$, there exists a positive constant ${p}$ such that ${\left \lVert \cdot \right \rVert_S \leq p \left \lVert\cdot \right \rVert_{S'}}$. The metric space structure on ${G}$ is thus defined up to quasi-isometry.

Definition 17 Let ${n_S(\lambda)}$ be the number of elements ${g \in G}$ with ${\left \lVert g \right \rVert}$ with ${\left \lVert g \right \rVert \leq \lambda}$. We say that ${G}$ is of exponential growth if ${n_S(\lambda)}$ grows exponentially, i.e. if

$\displaystyle \lim_{\lambda \rightarrow \infty} \frac{\log n_S(\lambda)}{\lambda} >0.$

We note that the limit in the above definition actually exists. This follows from the inequality

$\displaystyle n_S(\lambda + \nu) \leq n_S(\lambda) n_S(\nu).$

Now we use an elementary lemma:

Lemma 18 Any function ${g: \mathbb{R}_{\geq 1} \rightarrow \mathbb{R}_{\geq 1}}$ (bounded on bounded intervals) such that

$\displaystyle g(\lambda + \nu) \leq g(\lambda ) + g(\nu)$

has the property that

$\displaystyle \lim_{\lambda \rightarrow \infty} \frac{g(\lambda)}{\lambda}$

exists.

Proof: To see this, we fix ${\lambda}$, and observe that any positive real number ${x}$ can be written as ${x = n \lambda + \lambda'}$ for ${n \in \mathbb{N}}$ and ${\lambda' \in (0, \lambda)}$. Then:

$\displaystyle \frac{g(x)}{x} = \frac{g ( n \lambda + \lambda')}{ n \lambda + \lambda'} \leq \frac{n g(\lambda)}{n \lambda} + o(1) \quad \mathrm{as} \ x \rightarrow \infty .$

We find that

$\displaystyle \limsup_{x \rightarrow \infty} \frac{g(x)}{x} \leq \frac{g(\lambda)}{\lambda} ,$

and letting ${\lambda \rightarrow \infty}$, we get:

$\displaystyle \limsup_{x \rightarrow \infty} \frac{g(x)}{x} \leq \liminf_{\lambda \rightarrow \infty}\frac{g(\lambda)}{\lambda} ,$

which proves the lemma. $\Box$

Moreover, the property of having exponential growth is unchanged under variation of the generating set ${S}$.

We can make a similar definition:

Definition 19 We say that ${G}$ is of polynomial growth if ${n_S(\lambda)}$ is dominated by ${C \lambda^N}$ for ${C, N \gg 0}$.

For example, finitely generated abelian groups are of polynomial growth.

The goal is to prove:

Theorem 20 (Milnor) If ${M}$ is a compact manifold of negative curvature, then ${\pi_1 M}$ has exponential growth.

2. Outline of the proof

The argument behind Milnor’s theorem is fairly short, but it will be worth outlining. Let ${M}$ be a compact manifold of negative curvature. Choose a point ${p}$ in its universal cover ${\widetilde{M}}$, which is diffeomorphic to ${\mathbb{R}^n}$ under the exponential map ${\exp_p}$. The strategy is to relate the growth in the fundamental group ${\pi_1 M}$ with the growth in ${\widetilde{M}}$.

Namely, one has:

Theorem 21 Let ${\widetilde{M}}$ be a complete Riemannian manifold, and let ${p \in \widetilde{M}}$. Let ${\Gamma}$ be a finitely generated group of isometries of ${\widetilde{M}}$ acting freely such that the manifold ${\widetilde{M}/\Gamma}$ is compact. Then if the volumes ${\mathrm{vol}(B_{\widetilde{M}}(p, r))}$ grow exponentially in ${r}$, the group ${\Gamma}$ 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 ${\delta}$ be the diameter of ${M}$ and let ${B_\delta}$ be the ball of radius ${\delta}$ around ${p}$ in ${\widetilde{M}}$. Then ${B_\delta}$ projects surjectively onto ${M}$ under the covering projection

$\displaystyle \widetilde{M} \rightarrow \widetilde{M}/\Gamma = M.$

Therefore, any point in ${\widetilde{M}}$ is conjugate under ${\Gamma}$ to something in ${B_\delta}$. So ${B_\delta}$ is almost a fundamental domain for the action of ${\Gamma}$ (except that ${\Gamma}$ might take one point in ${B_\delta}$ to another in it). In other words,

$\displaystyle \widetilde{M} = \bigcup_{\gamma \in \Gamma} \gamma B_\delta.$

What does this mean? If ${n \in \mathbb{N}}$, then it follows that the ball ${B_{n\delta}}$ (of radius ${n \delta}$) is contained in the union of the various ${\gamma B_\delta}$. The key point is the following:

Key point: ${B_{n \delta}}$ is contained in the union of various ${\gamma B_{\delta}}$ for ${\left \lVert \gamma \right \rVert = O(n)}$ (with respect to any fixed system of generators).

In other words, if ${p' \in \widetilde{M}}$ is of distance ${O(n)}$ from ${p}$, then this is “witnessed” by the existence of ${\gamma \in \Gamma}$ with ${\left \lVert \gamma \right \rVert = O(n)}$ such that ${d_{\widetilde{M}}( \gamma p, p') = O(1)}$.

As a result, it will follow that

$\displaystyle \mathrm{vol}( B_{n \delta}) \leq \sum_{\gamma: \left \lVert \gamma \right \rVert \leq n} \mathrm{vol}(B_\delta) = (\mathrm{const}) \left\lvert \{ \gamma: \left \lVert \gamma \right \rVert \leq n \} \right\rvert.$

Since ${\mathrm{vol}(B_{n \delta})}$ grows exponentially, it will follow that the right-hand-side also grows exponentially, which gives the exponential growth rate of ${\Gamma}$.

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 ${\widetilde{M}, \Gamma}$ be as in the above theorem. Then there exists ${C > 0}$ such that for any ${n \in \mathbb{N}}$,

$\displaystyle B_{n \delta} \subset \bigcup_{\gamma: \left \lVert \gamma \right \rVert \leq Cn} \gamma B_{\delta},$

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

$\displaystyle S = \left\{\gamma: \gamma B_\delta \cap B_\delta \neq \emptyset\right\} .$

This is a finite set. In fact, if ${\epsilon}$ is sufficiently small, then the sets

$\displaystyle \gamma B_\epsilon(p), \quad \gamma \in \Gamma$

are all disjoint balls in ${\widetilde{M}}$ of radius ${\epsilon}$ (and the same volume). Only finitely many of them can be contained in any bounded region, so that means that ${S}$ is finite. If ${\gamma \notin S}$, then ${\gamma B_\delta }$ and ${B_\delta}$ are disjoint, and a similar argument shows that

$\displaystyle d( \gamma B_\delta, B_\delta) > \epsilon, \quad \forall \gamma \notin S,$

for some ${\epsilon > 0}$. Either ${\gamma B_\delta}$ and ${B_\delta}$ intersect, or they have to be separated by some ${\epsilon}$.

We now suppose that ${q \in \widetilde{M}}$ is a point such that

$\displaystyle d_{\widetilde{M}}(q. p) \leq n \delta, \quad \mathrm{i.e.} \ q \in B_{n \delta}.$

We would like to show that ${q}$ is in ${\gamma B_\delta}$ for a ${\gamma}$ of controlled norm.

In fact, choose a path from ${q}$ to ${p}$ of length ${\leq n \delta}$, and choose points along it

$\displaystyle q_0 = q, q_1, \dots , q_M = p$

where ${d( q_i, q_{i+1}) < \epsilon}$ and ${M}$ can be taken to be ${\leq 1 + \frac{n \delta}{\epsilon}}$. It follows that

$\displaystyle q_i \in \gamma_i B_\delta \quad \text{for some} \ \gamma_i \in \Gamma$

where ${d( \gamma_i B_\delta, \gamma_{i-1} B_\delta) < \epsilon}$ and so

$\displaystyle \gamma_{i-1}^{-1} \gamma_{i} \in S.$

It follows that the group-theoretic distance from ${\gamma_i}$ to ${\gamma_{i-1}}$ is at most ${1}$ for each ${i}$, so that the group-theoretic distance from ${\gamma_0}$ to ${\gamma_M}$ (which we can take as ${1}$) is at most ${M}$.

It follows that

$\displaystyle q \in \gamma_0 B_\delta,$

where ${\left \lVert \gamma_0 \right \rVert \leq M = O(n)}$. This completes the proof.