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.

1. Torsion-freeness of the fundamental group

Let’s explore some consequences of these assertions.

Theorem 8 If ${M}$ is complete of nonpositive curvature, then any ${\alpha \in \pi_1(M) \setminus \left\{1\right\}}$ has infinite order.

To see this, note that if ${\alpha}$ had finite order, it would define a free action of a finite group ${\mathbb{Z}/n}$ on the universal cover ${\widetilde{M}}$, which is diffeomorphic to ${\mathbb{R}^n}$. But there is a homotopical obstruction to this: no nontrivial finite group ${G}$ can act freely on ${\mathbb{R}^n}$. In fact, then ${\mathbb{R}^n/ G}$ would be a finite-dimensional model for a ${K(G, 1)}$ (see for instance this MO thread). It is known that no nontrivial finite group ${G}$ can admit a finite-dimensional ${K(G, 1)}$: in fact, the group cohomology

$\displaystyle H^*(G; \mathbb{Z}) \simeq H^*( K(G, 1); \mathbb{Z})$

is nonzero in infinitely many dimensions.

In particular, the fundamental group of any nonpositively curved manifold (say, a compact Riemann surface of positive genus) can have no torsion (although it is usually highly non-abelian).

It is also possible to prove the above theorem by geometric rather than topological means. In fact, one can prove the stronger result:

Theorem 9 (Cartan) Let ${M}$ be a simply connected complete manifold of nonpositive curvature. Let ${G}$ be a compact group acting on ${M}$ by isometries. Then ${G}$ has a fixed point.

The proof of this result uses the notion of the center of mass of a compact subset ${K \subset M}$, which is an extension of the notion in euclidean space to any Riemannian structure on ${\mathbb{R}^n}$ of nonpositive curvature. Given ${K \subset M}$, there is a unique ${p \in M}$ which minimizes

$\displaystyle \int_{K} d_M(x, p)^2;$

this is called the center of mass. (The measure on ${K}$ is the restriction of the Riemannian measure on ${M}$.) More generally, this notion makes sense whenever one has a compact measure space ${K}$ with a continuous map

$\displaystyle f: K \rightarrow M.$

If ${G}$ acts on ${M}$, then for any ${m \in M}$, the center of mass of the map

$\displaystyle G \rightarrow M, \quad g \mapsto gm$

is a fixed point for the ${G}$-action.

2. Preissman’s theorem

In strictly negative curvature, one can say a little more about the fundamental group: it is highly non-abelian.

Theorem 10 Let ${M}$ be a compact Riemannian manifold of strictly negative sectional curvature. Then any abelian subgroup of ${\pi_1(M)}$ is cyclic.

In particular, this provides a strong topological restriction on negatively curved manifolds, which rules out negatively curved metrics on tori. Note that ${\pi_1(M)}$ itself can’t be cyclic, because then ${M}$ would have the homotopy type of ${S^1}$.

Example 1 No compact product manifold can have negative curvature. In fact, if ${M \times N}$ is a product of two compact manifolds, then any ${\alpha \in \pi_1(M), \beta \in \pi_1(N)}$ commute with each other in ${\pi_1(M \times N)}$. Therefore if ${M \times N}$ had negative curvature, one of ${M, N}$ would have to be contractible, which is a contradiction unless one of ${M, N}$ is a point.

Let ${\widetilde{M}}$ be the universal cover of ${M}$, so that ${\widetilde{M}}$ is diffeomorphic to ${\mathbb{R}^n}$ via the exponential map at any point. The proof of Preissman’s theorem is based on the following fact: given a geodesic triangle ${ABC}$ in ${\widetilde{M}}$ between points ${A, B, C \in \widetilde{M}}$, the sum of the angles is strictly less than ${\pi}$. This is a consequence of the fact that the corresponding triangle in euclidean space of lengths ${d_{\widetilde{M}}(A, B), d_{\widetilde{M}}(B, C), d_{\widetilde{M}}(A, C)}$ has larger angles. In particular, for any geodesic quadrilateral in ${\widetilde{M}}$, the sum of the angles must be less than ${2\pi}$.

With this in mind, we will prove that any fixed-point-free isometry of ${\widetilde{M}}$ has a unique invariant geodesic, called an axis.

Lemma 11 Given an isometry ${\alpha: \widetilde{M} \rightarrow \widetilde{M}}$ without fixed points, there is a unique geodesic of ${\widetilde{M}}$ left invariant by ${\alpha}$.

Proof: For uniqueness, the idea is that if there were two axes ${\gamma, \gamma'}$ for ${\alpha}$, we would be able to construct a quadrilateral of total angle ${2\pi}$. Namely, observe first that ${\gamma, \gamma'}$ don’t intersect: if they intersected at ${p}$, they’d intersect at ${\alpha(p)}$ as well and two geodesics in ${\widetilde{M}}$ that intersect twice are the same. Thus, choose a point ${A}$ on ${\gamma}$, and draw a geodesic ${\Gamma}$ from ${p}$ to some point ${B}$ on ${\gamma'}$. Then one has a quadrilateral ${AB \alpha(B) \alpha(A)}$ (based on the geodesics ${\gamma, \gamma', \Gamma, \alpha(\Gamma)}$) whose angles sum to ${2\pi}$ from the construction, a contradiction.

For existence, the idea is to choose ${p \in \widetilde{M}}$ such that ${d_{\widetilde{M}}(p, \alpha(p))}$ is minimized. We can do this for the following reason: the function

$\displaystyle f(p) \stackrel{\mathrm{def}}{=} d_{\widetilde{M}}(p, \alpha(p))$

is invariant under ${\pi_1}$ and descends to a function on the compact manifold ${M}$. Indeed, if ${\pi: \widetilde{M} \rightarrow M}$ is the projection, then ${f(p)}$ is the length of the shortest loop at ${p}$ in the homotopy class ${\alpha}$. So there is a point ${p}$ which minimizes this distance.

Now let ${\gamma}$ be the geodesic segment from ${p}$ to ${\alpha(p)}$. The claim is that ${\alpha}$ preserves the entire geodesic defined by ${\gamma}$ and that ${\gamma}$ is therefore an axis. To see this, it suffices to show that the broken geodesic path from ${p}$ to ${\alpha^2(p)}$ given by ${\gamma}$ and ${\alpha(\gamma)}$ is actually a geodesic segment. But if it is not a segment — if there is a twist in the path — then for ${x}$ close to ${p}$, we would get a path from ${x}$ to ${\alpha(x)}$ of length ${d(p, \alpha(p))}$ which is not a geodesic (it has a twist) and which can be shortened, a contradiction.

$\Box$

With the lemma proved, the proof of Preissman’s theorem is now rather straightforward. Consider ${\alpha, \beta \in \pi_1(M)}$, acting on the universal cover ${\widetilde{M}}$, and suppose that they commute. We know that ${\alpha}$ has an axis ${\gamma \subset \widetilde{M}}$ such that ${\alpha(\gamma) = \gamma}$. Since ${\left\{\alpha, \beta\right\}}$ commute, it follows that ${\beta(\gamma)}$ is also an axis for ${\alpha}$ and therefore ${\beta(\gamma ) = \gamma}$. More generally, given any abelian subgroup ${G \subset \pi_1(M)}$, this reasoning shows that there is a single axis for all of ${G}$.

But if ${G \neq \left\{1\right\}}$ is an abelian subgroup of ${\pi_1 M}$ with axis ${\gamma}$, then there is an injective homomorphism

$\displaystyle G \rightarrow \mathbb{R}$

because any ${G}$ must act on ${\gamma}$ by translation by a constant along the geodesic. However, the only finitely generated abelian group that imbeds in ${\mathbb{R}}$ is ${\mathbb{Z}}$, so ${G= \mathbb{Z}}$.

Corollary 12 (Byers) Under the above hypotheses, any solvable subgroup of ${\pi_1 M}$ is infinite cyclic.

Proof: It suffices to show that if ${G \subset \pi_1 M}$ is a subgroup with the property that ${G}$ has a subgroup ${H}$ with ${H}$ abelian and ${[G, G] \subset H}$, then ${G \simeq \mathbb{Z}}$.

To see this, choose a geodesic ${\gamma}$ in ${\widetilde{M}}$ (as in the previous proof) invariant under ${H}$. For ${b \in G, a \in H}$, one has

$\displaystyle bab^{-1}a^{-1} \gamma = \gamma,$

so that, since ${a}$ leaves ${\gamma}$ invariant,

$\displaystyle bab^{-1} \gamma = \gamma, \quad \text{or} \quad ab^{-1} \gamma = b^{-1}\gamma.$

It follows that ${b^{-1} \gamma = \gamma}$ since axes are unique, and we find that ${G}$ itself leaves the geodesic ${\gamma}$ invariant. The same reasoning as above now shows that ${G}$ has to be infinite cyclic. $\Box$

A somewhat related manifestation of the high degree of non-abelianness of ${\pi_1 M}$ comes from the theorem (of Milnor) that ${\pi_1 M}$ is a group of exponential growth under these hypotheses.

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