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.

Proof of Ambrose’s theorem

The idea is to use exponential coordinates. Firstly ${f}$ takes geodesics to geodesics, at least locally. Indeed, ${l(f \circ c) = l(c)}$ if ${c}$ is a curve in ${M}$, geodesics locally minimize length, and ${f}$ is locally an isomorphism of manifolds. By splitting a geodesic in ${M}$ into small pieces, it follows that ${f}$ maps geodesics in ${M}$ to a broken geodesic in ${N}$, but since ${f}$ is smooth, ${f}$ just maps ${M}$-geodesics to ${N}$-geodesics.

So let ${n \in N}$. We will find a small neighborhood ${U}$ of ${n}$ such that ${f^{-1}(U)}$ has components diffeomorphic to ${U}$. Indeed, take ${A \subset T_n(N)}$ an open star-shaped set containing ${0}$ mapped diffeomorphically onto some open set, say ${U}$. If ${r>0}$ is taken very small and ${A}$ is chosen as ${B_r(0_n) = \{ v \in T_n(N): |v| < r\}}$, we can take ${U}$ to be ${B_r(n) = \{ n' \in N: d(n,n') < r \}}$ if ${d}$ is the metric on ${N}$ induced by the Riemannian metric.

Let ${S = f^{-1}(n)}$. Because ${f}$ is a local diffeomorphism, ${S}$ is discrete. I claim that $\displaystyle \boxed{f^{-1}(U) = \{ m \in M: d(m, S) < r \}.}$

Indeed, if ${a,b \in M}$, then ${d(a,b) \geq d(f(a),f(b))}$ because ${f}$ preserves curve lengths. This implies that the second set is contained in the first.

For the other inclusion, let us choose ${s \in S}$ and draw the commutative diagram The image in ${N}$ is ${B_r(n)}$, while by Hopf-Rinow II the image in ${M}$ is ${B_r(s)}$. From the diagram it is even clear that ${f: B_r(s) \rightarrow B_r(n)}$ is a diffeomorphism.

If ${d(f(m), n) < r}$ we can choose a geodesic between ${n}$ and ${f(m)}$ which we can locally, piece-by-piece, lift it to a piecewise-smooth curve ${c}$ starting at ${m}$ and ending at some point of ${f^{-1}(n)}$. Now ${d(c(a),c(b)) }$ is proportional to ${|a-b|}$ so ${c}$ is actually a geodesic (hence smooth) by these arguments. In particular, the length of ${c}$ is ${, and ${d(m, S). This proves the other inclusion.

So we have ${f^{-1}(U) = \bigcup_{s \in S} B_r(s)}$. Each ${B_r(s)}$ is diffeomorphic to ${B_r(n)}$. All we need to show is that ${B_r(s), B_r(s')}$ are disjoint for ${s' \neq s}$. Otherwise we could choose ${q \in B_r(s) \cap B_r(s')}$ and choose geodesics from ${s}$ to ${q}$ in ${B_r(s)}$, and ${s' }$ to ${q}$ in ${B_r(s')}$. The images give two different geodesics from ${f(q)}$ to ${n}$, contradiction. This proves Ambrose’s theorem.

Proof of the proposition

Since the map ${\exp: T_p(M) \rightarrow M}$ is surjective (H-R II) and an immersion by assumption, we can consider the pulled back metric ${\exp^*g}$ on ${T_p(M)}$ and apply Ambrose’s theorem to it. The only thing we have to check is that with the Riemannian metric ${\exp^*g}$ on ${T_p(M)}$, it is complete. But geodesics through the origin exist of arbitrary length—they are just the straight lines through the origin, since these lines are mapped onto geodesics in ${M}$. By H-R, this “completeness at a point” implies the completeness of the whole manifold.

I will actually prove the Cartan-Hadamard theorem in the next post.