Let ${(M, g)}$ be a Riemannian manifold. As before, one associates to it the curvature tensor

$\displaystyle R: TM \otimes TM \otimes TM \rightarrow TM, \quad X, Y, Z \mapsto R(X, Y) Z.$

In the previous post, we saw a quantitative expression of how the curvature is a measure of the deviation from the flatness of ${M}$. Given ${M}$, one can try to choose local coordinates around a point ${p \in M}$ which make the metric look like the euclidean metric to order 2 at ${p}$, i.e. local coordinates such that the coefficients near ${p}$ are given by

$\displaystyle g_{ij} = \delta_{ij} + O(|x|^2).$

However, we saw that the quadratic terms involve precisely the values of the curvature tensor at ${p}$. Even in the best coordinates, one can’t generally make the coefficients of a metric look euclidean to order 3: the obstruction is precisely the curvature at ${p}$. Today, I’d like to describe the interpretation of curvature in terms of geodesics. Once again, the material is standard and can be found in introductory textbooks on Riemannian geometry.

1. Curvature and geodesic deviation

There’s another way to think of curvature, which also leads to this: curvature measures how nearby geodesics spread. To think about this, suppose we have a one-parameter family ${\gamma_s}$ of geodesics in ${M}$, where ${\gamma = \gamma_0}$ is the starting point of the variation. One then has a vector field

$\displaystyle V = \left( \frac{d}{ds} \gamma_s\right)_{s = 0}$

along the curve ${\gamma}$, which measures the infinitesimal “spreading” of the one-parameter family ${\gamma_s}$. Now, a computation shows that ${V}$ satisfies the equation

$\displaystyle \frac{D^2}{dt^2} V(t) + R( V, \dot{\gamma}(t)) \dot{\gamma(t)} = 0,$

in other words that ${V}$ is a Jacobi field. Here ${\frac{D}{dt}}$ is covariant differentiation along the curve ${\gamma}$. (more…)

Apologies for some initial bugs in the formulas–I have now corrected them.

Today I will prove the Cartan-Hadamard theorem.

Nonvanishing of Jacobi fields

The key lemma is that (nontrivial) Jacobi fields do not vanish.

Lemma 1 Let ${M}$ be a Riemannian manifold of negative curvature, ${\gamma: [0,M]}$ a geodesic on ${M}$, and ${J}$ a Jacobi field along ${\gamma}$ with ${J(0)=0}$. If ${\frac{D}{dt}V(t)|_{t=0} \neq 0}$, then ${J(t) \neq 0}$ for all ${t > 0}$.

Indeed, we consider ${ \frac{d^2}{dt^2} \left \langle J(t), J(t)\right \rangle}$, which equals

$\displaystyle 2\frac{d}{dt} \left \langle \frac{D}{dt} J(t), J(t) \right \rangle = 2\left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle + 2 \left| \frac{D}{dt} V(t) \right|^2.$

I claim that this second derivative is negative, which will follow if we show that

$\displaystyle \left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle \geq 0.$

But here we can use the Jacobi equation and the antisymmetry of the curvature tensor to turn ${\left \langle \frac{ D^2}{dt^2} J(t), J(t)\right \rangle }$ into

$\displaystyle \left \langle R(\dot{\gamma}(t), J(t)) \dot{\gamma(t)}, J(t) \right \rangle = -\left \langle R( J(t),\dot{\gamma}(t) ) \dot{\gamma(t)}, J(t) \right \rangle \geq 0.$

(The last inequality is from the assumption of negative curvature.)

This proves the claim.

Now there are arbitrarily small ${t}$ with ${ \left \langle J(t), J(t) \right \rangle \neq 0}$ because ${\frac{D}{dt} J(t)|_{t=0} \neq 0}$, so in particular there must be arbitrarily small ${t}$ with ${ \frac{d}{dt} \left \langle J(t), J(t) \right \rangle > 0}$. In particular, this derivative is always positive. This proves the claim.

I followed Wilkins in the proof of this lemma.

By yesterday’s post, it’s only necessary to show that ${\exp_p}$ is a regular map. Now if ${X,Y \in T_p(M)}$

$\displaystyle d(\exp_p)_X(Y) = J(1)$

where ${J}$ is the Jacobi field along the geodesic ${\gamma(t) = \exp_p(tX)}$ with ${J(0)=0, \frac{D}{dt} J(t)|_{t=0} = Y}$. This is nonzero by what has just been proved, which establishes the claim and the Cartan-Hadamard theorem.

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…)