I edited this post to fix some sign issues.  (11/29)

I now want to discuss a result of Myers, which I can summarize as follows:

If ${M}$ is a complete Riemannian manifold with positive, bounded-below curvature, then ${M}$ is compact.

This is a very loose summary—Myers’ theorem actually gives a lower bound for the diameter of ${M}$. Moreover, I haven’t explained what “bounded below curvature” actually means. To say that the sectional curvature is bounded below is sufficient, but we can do better.

I will now outline how the proof works.

The first thing to notice is that any two points ${p,q \in M}$ can be joined by a length-minimizing geodesic ${\gamma}$, by the Hopf-Rinow theorems. In particular, if we can show that every sufficiently long geodesic (of length ${>L}$, say) doesn’t minimize length, then ${M}$ is necessarily of diameter at most ${L}$.

All the same, the length function as a map ${\mathrm{Curves} \rightarrow \mathbb{R}}$ is not so easy to work with; the energy integral is much more convenient. Moreover, we know that if ${\gamma}$ minimizes length and is a geodesic, it also minimizes the energy integral.

If ${\gamma}$ is a geodesic that minimizes the energy integral (among curves with fixed endpoints), then in particular we can consider a variation ${\gamma_u}$ of ${\gamma}$, and consider the function ${E(u):=E(\gamma_u)}$; this necessarily has a minimum at ${u=0}$. It follows that ${E''(0) \leq 0}$. If we apply the second variation formula, we find something involving the curvature tensor that looks a lot like sectional curvature.

Before turning to the details, I will now define the refinement of sectional curvature we can use:

Ricci curvature

Given a Riemannian manifold ${M}$ with curvature tensor ${R}$ and ${p \in M}$, we can define a linear map ${T_p(M) \rightarrow T_p(M)}$,

$\displaystyle X \rightarrow -R(X,Y)Z$

that depends on ${Y,Z \in T_p(M)}$. The trace of this linear map is defined to be the Ricci tensor ${\rho(Y,Z)}$. This is an invariant definition, so we do not have to do any checking of transformation laws.

A convenient way to express this is the following: If ${E_1, \dots, E_n}$ is an orthonormal basis for ${T_p(M)}$, then by linear algebra and skew-symmetry

$\displaystyle \boxed{\rho(Y,Z) = \sum R(E_i, Y, E_i, Z) .}$

The Ricci curvature has many uses. Considered as a ${(1,1)}$ tensor (by the functorial isomorphism ${\hom(V \otimes V, \mathbb{R}) \simeq \hom(V, V)}$ for any real vector space ${V}$ and the isomorphism ${T(M) \rightarrow T^*(M)}$ induced by the Riemannian metric), its trace yields the scalar curvature, which is just a real-valued function on a Riemannian manifold. It is also used in defining the Ricci flow, which led to the recent solution of the Poincaré conjecture.  I may talk about these more advanced topics (much) later if I end up learning about them–I am finding this an interesting field, and may wish to pursue geometry further.

Statement of Myers’ theorem

Theorem 1 Let ${M}$ be a complete Riemannian manifold whose Ricci tensor ${\rho}$ satisfies $\displaystyle \rho(X,X) \geq C|X|^2$

for all ${X \in T_p(M), p \in M}$. Then

$\displaystyle \mathrm{diam}(M) \leq \pi \sqrt{ \frac{n-1}{C}}.$

A corollary of this is that if the sectional curvature is bounded below by ${\delta}$—which implies by the boxed equation above that ${\rho(X,X) \geq ( n - 1) \delta|X|^2}$ for all ${X}$, because we can choose the orthonormal basis ${E_1, \dots, E_n}$ above such that ${E_2, \dots, E_n}$ are orthogonal to ${X}$—then

$\displaystyle \mathrm{diam}(M) \leq \frac{\pi}{ \sqrt{ \delta}}.$

Also, the finite diameter implies compactness since ${M}$ is complete as a metric space.

The fundamental group

Theorem 2 Let ${M}$ be a compact Riemannian manifold with positive Ricci curvature; then its fundamental group is finite.

I think this is especially interesting because it relates the Ricci curvature of the manifold, which is defined heavily in terms of the Riemannian metric and not in terms of ${M}$ itself, with the bare-bones topology of ${M}$ expressed in the fundamental group.

The proof, however, is fairly straightforward from what we have already done. It will be sufficient to show that the universal covering space ${\tilde{M}}$ of ${M}$ is compact, because if ${f: \tilde{M} \rightarrow M}$ is the covering map, then ${f^{-1}(x)}$ for ${x \in M}$ is discrete and closed, hence finite. So ${\tilde{M}}$ is finitely-sheeted over ${M}$, which implies the theorem.

I will now show ${\tilde{M}}$ is compact.

Now it is a basic fact that any covering space of a smooth manifold can be made into a smooth manifold (this is also true with “smooth” replaced by real-analytic, complex, etc.). Moreover, we can pull back the Riemannian metric ${g}$ on ${M}$ via ${f}$ to get ${f^*g}$ on ${\tilde{M}}$. Since ${f}$ is locally an isometry, it preserves curvature; in particular, ${\tilde{M}}$ has bounded-below Ricci curvature because ${M}$ does.

Moreover, ${\tilde{M}}$ is a complete Riemannian manifold. Indeed, more generally, any covering space of a complete Riemannian manifold—with the pulled-back Riemannian metric–is complete. The reason is that if ${f}$ is the covering map, a curve ${\gamma}$ is a geodesic iff ${f \circ \gamma}$ is one; this too follows from the local isometry property of ${f}$. Therefore, if we start at a point ${\tilde{p} \in \tilde{M}}$ and start a geodesic from ${\tilde{p}}$, we can project it to ${M}$ via ${f}$, extend it to a geodesic on ${(-\infty,\infty)}$ on ${M}$ by completeness, and use the covering space property to lift it upward to ${\tilde{M}}$ (uniquely since the starting point is fixed) to get a geodesic in ${\tilde{M}}$. So geodesics in ${\tilde{M}}$ are infinitely extendable, which implies completeness.

Now the Myers theorem applied to ${\tilde{M}}$ shows compactness and establishes the second theorem.