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.)


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…)