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.

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