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. (more…)

Let ${M}$ be a compact Riemannian manifold of (strictly) negative curvature, so that ${M}$ is a ${K(\pi_1 M, 1)}$. In the previous post, we saw that the group ${\pi_1 M}$ was significantly restricted: for example, every solvable subgroup of ${\pi_1 M}$ had to be infinite cyclic. The goal of this post and the next is to understand the result of Milnor that the group ${\pi_1 M}$ is of exponential growth in an arithmetic sense.

Milnor’s (wonderful) idea is to translate this into a problem in geometry: that is, to relate the growth of the group ${\pi_1 M}$ to the volume growth of expanding balls in the universal cover ${\widetilde{M}}$. As I understand, this idea has proved enormously influential on future work on the fundamental groups of Riemannian manifolds with restricted curvature. Note that Milnor’s result also highlights the difference between positive and negative curvature: in positive curvature, the fundamental group of every compact manifold is finite. Most of this material is from Chavel’s Riemannian geometry: a modern introduction.

1. Volume growth

To begin with, let’s say something about volume growth. Let ${M}$ be a complete, simply connected Riemannian manifold whose sectional curvatures are ${\leq c < 0}$. If we choose ${p \in M}$, we know that the exponential map

$\displaystyle \exp_p : T_p M \rightarrow M$

is a diffeomorphism. Note that ${\exp_p}$ sends the euclidean ball of radius ${r}$ diffeomorphically onto the (metric) ball of radius ${r}$ in ${M}$.

Our goal is to prove:

Theorem 13 The function ${r \mapsto \mathrm{vol} ( B_M(p, r))}$ which sends ${r}$ to the volume of the ball in ${M}$ of radius ${r}$ centered at ${p}$ grows exponentially.

This theorem also highlights the sense in which negative curvature corresponds to the “spreading” of geodesics: the geodesics spread so much that the volumes of linearly expanding balls actually grow exponentially. (more…)

Let ${M}$ be a complete Riemannian manifold. In the previous post, we saw that the condition that ${M}$ have nonpositive sectional curvature had a reformulation in terms of the “spreading” of geodesics: that is, nearby geodesics in nonpositive curvature spread at least as much as they would in euclidean space. One consequence of this philosophy was the Cartan-Hadamard theorem: the universal cover of ${M}$ is diffeomorphic to ${\mathbb{R}^n}$. In fact, for a ${M}$ simply connected (and complete of nonpositive curvature), the exponential map

$\displaystyle \exp_p : T_p M \rightarrow M$

is actually a diffeomorphism.

This suggests that without the simple connectivity assumption, the “only reason” for two geodesics to meet is the fundamental group. In particular, geodesics have no conjugate points, meaning that the exponential map is always nonsingular. Moreover, again by passing to the universal cover, it follows that for any ${p \in M}$ and ${\alpha \in \pi_1 (M)}$ (relative to the basepoint ${p}$), there is a unique geodesic loop at ${p}$ representing ${\alpha}$. In this post, I’d like to discuss some of the topological consequences of having a metric of nonpositive sectional curvature. Most of this material is from Do Carmo’s Riemannian Geometry.  (more…)

Let ${(M, g)}$ be a Riemannian manifold. As before, one associates to it the curvature tensor

$\displaystyle R: TM \otimes TM \otimes TM \rightarrow TM, \quad X, Y, Z \mapsto R(X, Y) Z.$

In the previous post, we saw a quantitative expression of how the curvature is a measure of the deviation from the flatness of ${M}$. Given ${M}$, one can try to choose local coordinates around a point ${p \in M}$ which make the metric look like the euclidean metric to order 2 at ${p}$, i.e. local coordinates such that the coefficients near ${p}$ are given by

$\displaystyle g_{ij} = \delta_{ij} + O(|x|^2).$

However, we saw that the quadratic terms involve precisely the values of the curvature tensor at ${p}$. Even in the best coordinates, one can’t generally make the coefficients of a metric look euclidean to order 3: the obstruction is precisely the curvature at ${p}$. Today, I’d like to describe the interpretation of curvature in terms of geodesics. Once again, the material is standard and can be found in introductory textbooks on Riemannian geometry.

1. Curvature and geodesic deviation

There’s another way to think of curvature, which also leads to this: curvature measures how nearby geodesics spread. To think about this, suppose we have a one-parameter family ${\gamma_s}$ of geodesics in ${M}$, where ${\gamma = \gamma_0}$ is the starting point of the variation. One then has a vector field

$\displaystyle V = \left( \frac{d}{ds} \gamma_s\right)_{s = 0}$

along the curve ${\gamma}$, which measures the infinitesimal “spreading” of the one-parameter family ${\gamma_s}$. Now, a computation shows that ${V}$ satisfies the equation

$\displaystyle \frac{D^2}{dt^2} V(t) + R( V, \dot{\gamma}(t)) \dot{\gamma(t)} = 0,$

in other words that ${V}$ is a Jacobi field. Here ${\frac{D}{dt}}$ is covariant differentiation along the curve ${\gamma}$. (more…)

Today we will prove the fixed point theorem, which I restate here for convenience:

Theorem 1 (Elie Cartan) Let ${K}$ be a compact Lie group acting by isometries on a simply connected, complete Riemannian manifold ${M}$ of negative curvature. Then there is a common fixed point of all ${k \in K}$.

There is a Haar measure on ${K}$. In fact, we could even construct this by picking a nonzero alternating ${n}$-tensor (where ${n=\dim K}$) at ${T_e(K)}$, and choosing the corresponding ${K}$-invariant ${n}$-form on ${K}$. This yields a functional ${C(K) \rightarrow \mathbb{R}}$, which we can assume positive by choosing the orientation appropriately. This yields the Haar measure ${d \mu}$ by the Riesz representation theorem.

Now define ${J(q) := \int_K d^2(q,kp) d \mu(k).}$ This is a continuous function ${M \rightarrow \mathbb{R}}$ which has a minimum, because ${J(q)>J(p)}$ for ${q}$ outside some compact set containing ${p}$. Let the minimum occur at ${q_0}$. I claim that the minimum is unique, which will imply that it is a fixed point of ${K}$.

It can be checked that ${J}$ is continuously differentiable; indeed, let ${q_t}$ be a curve. Then ${d^2(q_t,kp)}$ can be computed as in yesterday when ${kp \neq q_t}$; when they are equal, it is still differentiable with zero derivative because of the ${d^2}$. (I am sketching things here because I don’t currently want to dive into the technical details; see Helgason’s book for them.)

So now take ${q_t}$ to be a geodesic joining the minimal point ${q_0}$ to some other point ${q_1}$. Now

$\displaystyle \frac{d}{dt} J(q_t)|_{t=0} = \int_K \frac{d}{dt} d^2(q_t, kp) |_{t=0} d\mu(k) = 0.$

Then we get

$\displaystyle \int_K d(q_0,kp) \cos \alpha d\mu(k) = 0$

where ${\alpha}$ is an appropriate angle as in yesterday’s post. When ${q_0=kp}$, this ${\alpha}$ is not well-defined, but ${d(q_0,kp)=0}$, so it is ok. Now

$\displaystyle \int_K d^2(q_1, k.p) d\mu \geq \int_K \left( d^2(q_0, kp) + d^2(q_0,q_1) - 2 d(q_0, kp) \cos \alpha \right) d \mu(k) .$

This is because of the cosine inequality. But the cosine part vanishes, so this is strictly greater than ${J(q_0)}$. In particular, since ${q_1}$ was arbitrary, ${q_0}$ was a global minimum for ${J}$—and it is thus a fixed point.

I am now aiming to prove an important fixed point theorem:

Theorem 1 (Elie Cartan) Let ${K}$ be a compact Lie group acting by isometries on a simply connected, complete Riemannian manifold ${M}$ of negative curvature. Then there is a common fixed point of all ${k \in K}$.

There are several ingredients in the proof of this result. These will provide examples of the techniques that I have discussed in past posts.

Geodesic triangles

Let ${M}$ be a manifold of negative curvature, and let ${V}$ be a normal neighborhood of ${p \in M}$; this means that ${\exp_p}$ is a diffeomorphism of some neighborhood of ${0 \in T_p(M)}$ onto ${V}$, and any two points in ${V}$ are connected by a unique geodesic. (This always exists by the normal neighborhood theorem, which I never proved. However, in the case of Cartan’s fixed point theorem, we can take ${V=M}$ by Cartan-Hadamard.)

So take ${a=p,b,c\in V}$. Draw the geodesics ${\gamma_{ab}, \gamma_{ac}, \gamma_{bc}}$ between the respective pairs of points, and let ${\Gamma_{ab}, \Gamma_{bc}}$ be the inverse images in ${T_p(M) = T_a(M)}$ under ${\exp_p}$. Note that ${\Gamma_{ab}, \Gamma_{ac}}$ are straight lines, but ${\Gamma_{bc}}$ is not in general. Let ${a',b',c'}$ be the points in ${T_p(M)}$ corresponding to ${a,b,c}$ respectively. Let ${A}$ be the angle between ${\gamma_{ab}, \gamma_{ac}}$; it is equivalently the angle at the origin between the lines ${\Gamma_{ab}, \Gamma_{ac}}$, which is measured through the inner product structure.

Now ${d(a,b) = l(\gamma_{ab}) = l(\Gamma_{ab}) = d(a',b')}$ from the figure and since geodesics travel at unit speed, and similarly for ${d(a,c)}$. Moreover, we have ${d(b,c) = l(\gamma_{bc}) \geq l(\Gamma_{bc}) \leq d(b',c')}$, where the first inequality comes from the fact that ${M}$ has negative curvature and ${\exp_p}$ then increases the lengths of curves; this was established in the proof of the Cartan-Hadamard theorem.

We have evidently by the left-hand-side of the figure

$\displaystyle d(b',c')^2 = d(a',c')^2 + d(a',b')^2 - 2d(a',b')d(a',c') \cos A.$

In particular, all this yields

$\displaystyle \boxed{d(b,c)^2 \geq d(a,c)^2 + d(a,b)^2 - 2d(a,b)d(a,c) \cos A.}$

So we have a cosine inequality.

There is in fact an ordinary plane triangle with sides ${d(a,b), d(b,c),d(a,c)}$, since these satisfy the appropriate inequalities (unless ${a,b,c}$ lie on the same geodesic, which case we exclude). The angles ${A',B',C'}$ of this plane triangle satisfy

$\displaystyle A \leq A'$

by the boxed equality. In particular, if we let ${B}$ (resp. ${C}$) be the angles between the geodesics ${\gamma_{ab}, \gamma_{bc}}$ (resp. ${\gamma_{ac}, \gamma_{bc}}$), then by symmetry and ${A'+B'+C=\pi}$

$\displaystyle \boxed{A + B+ C \leq \pi.}$

This is a fact which I vaguely recall from popular-math books many years back.   The rest is below the fold. (more…)

Apologies for some initial bugs in the formulas–I have now corrected them.

Today I will prove the Cartan-Hadamard theorem.

Nonvanishing of Jacobi fields

The key lemma is that (nontrivial) Jacobi fields do not vanish.

Lemma 1 Let ${M}$ be a Riemannian manifold of negative curvature, ${\gamma: [0,M]}$ a geodesic on ${M}$, and ${J}$ a Jacobi field along ${\gamma}$ with ${J(0)=0}$. If ${\frac{D}{dt}V(t)|_{t=0} \neq 0}$, then ${J(t) \neq 0}$ for all ${t > 0}$.

Indeed, we consider ${ \frac{d^2}{dt^2} \left \langle J(t), J(t)\right \rangle}$, which equals

$\displaystyle 2\frac{d}{dt} \left \langle \frac{D}{dt} J(t), J(t) \right \rangle = 2\left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle + 2 \left| \frac{D}{dt} V(t) \right|^2.$

I claim that this second derivative is negative, which will follow if we show that

$\displaystyle \left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle \geq 0.$

But here we can use the Jacobi equation and the antisymmetry of the curvature tensor to turn ${\left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle }$ into

$\displaystyle \left \langle R(\dot{\gamma}(t), J(t)) \dot{\gamma(t)}, J(t) \right \rangle = -\left \langle R( J(t),\dot{\gamma}(t) ) \dot{\gamma(t)}, J(t) \right \rangle \geq 0.$

(The last inequality is from the assumption of negative curvature.)

This proves the claim.

Now there are arbitrarily small ${t}$ with ${ \left \langle J(t), J(t) \right \rangle \neq 0}$ because ${\frac{D}{dt} J(t)|_{t=0} \neq 0}$, so in particular there must be arbitrarily small ${t}$ with ${ \frac{d}{dt} \left \langle J(t), J(t) \right \rangle > 0}$. In particular, this derivative is always positive. This proves the claim.

I followed Wilkins in the proof of this lemma.

By yesterday’s post, it’s only necessary to show that ${\exp_p}$ is a regular map. Now if ${X,Y \in T_p(M)}$
$\displaystyle d(\exp_p)_X(Y) = J(1)$
where ${J}$ is the Jacobi field along the geodesic ${\gamma(t) = \exp_p(tX)}$ with ${J(0)=0, \frac{D}{dt} J(t)|_{t=0} = Y}$. This is nonzero by what has just been proved, which establishes the claim and the Cartan-Hadamard theorem.