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 {<r}, and {d(m, S)<r}. 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.

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