There are a whole bunch of theorems in Riemannian geometry to the effect that “if the Riemannian manifold ${M}$ has property A of the curvature, then it has the topological property B.” Over the rest of MaBloWriMo and in the following weeks, I aim to talk about a few such results. The first one characterizes manifolds of negative curvature.

Negative curvature

Let ${M}$ be a Riemannian manifold with Riemannian metric ${g}$ Say that ${M}$ has negative curvature if for all ${p \in M, X,Y \in T_p(M)}$,

$\displaystyle g( R(X,Y)Y, X) \leq 0.$

(Later I will interpret this in terms of the sectional curvature, which I have not yet defined.)

Statements

Theorem 1 (Cartan-Hadamard) Let ${M}$ be a complete Riemannian manifold of negative curvature. Then for ${p \in M}$, the map ${\exp_p: T_p(M) \rightarrow M}$ is a covering map. In particular, if ${M}$ is simply connected, then it is diffeomorphic to ${\mathbb{R}^n}$.

Of course, the diffeomorphism doesn’t have to preserve the Riemannian metric.

The strategy of the proof is as follows. First, we will show that the map ${\exp_p}$ is an immersion (though in general not injective), using the discussion yesterday about how Jacobi fields determine the differential of the exponential map. Then we will invoke

Proposition 2 Let ${M}$ be a complete Riemannian manifold. Suppose ${p \in M}$ and the map ${\exp_p}$ is an immersion. Then ${\exp_p}$ is a covering map.

The condition of the result is often stated to the effect that “${M}$ has no conjugate points to ${p}$.”

To see this, we will appeal to yet another result:

Theorem 3 (Ambrose) Let ${f: M \rightarrow N}$ be a surjective morphisms of Riemannian manifolds with ${M}$ complete. Suppose ${f}$ preserves the metric on the tangent spaces. Then ${f}$ is a covering map.

I will work backwards to prove these three results. (more…)