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 ${\gamma}$.

This is a linear second-order ordinary differential equation, which determines ${V}$ in terms of ${V(0)}$ and the covariant derivative ${\frac{D}{dt}_{t = 0} V}$. Once again, the values of the curvature ${R}$ (now not just at ${p}$, but at nearby points) determine ${V}$ and the spreading of the one-parameter family of geodesics ${\gamma_s}$.

As an example, we could choose tangent vectors ${v, w \in T_p M}$ and consider the one-parameter family of geodesics

$\displaystyle \gamma_s(t) = \exp_p ( t ( v + sw)) ;$

this determines a Jacobi field ${V}$ along the geodesic ${\gamma(t) = \exp_p(t v)}$ with initial values

$\displaystyle V(0) = 0, \quad \frac{D}{dt} V(0) = w.$

The values ${V(t)}$ are given by the derivatives of the exponential map. That is,

$\displaystyle V(t) = \frac{d}{ds}|_{s = 0} \exp_p( t(v + sw)) = t d( \exp_p)_{tv} (w),$

and the preceding analysis shows that this is determined in terms of ${R}$.

As a result of the Jacobi equation, we saw above that the curvature tensor controls the infinitesimal spreading of a one-parameter family of geodesics ${\gamma_s}$. More precisely, though, the sign of the curvature can be used to make a qualitative assertion:

Principle: If the (sectional) curvature is negative, nearby geodesics diverge. If the (sectional) curvature is positive, then nearby geodesics converge (after a little while).

An example of this principle is the Cartan-Hadamard theorem.

Theorem 4 (Cartan-Hadamard) Let ${M}$ be a complete Riemannian manifold of nonpositive curvature. Then the universal cover of ${M}$ is diffeomorphic to ${\mathbb{R}^n}$, so ${M}$ is a ${K(\pi_1 M, 1)}$.

The idea behind the Cartan-Hadamard theorem is that if (as we can assume without loss of generality) ${M}$ is simply connected, then for any ${p \in M}$, the exponential map

$\displaystyle \exp_p: T_p M \rightarrow M$

is a diffeomorphism.

To see this, note that the Hopf-Rinow theorem implies that the exponential map is surjective: that is, any point in the manifold can be joined to ${p}$ by a geodesic. To say that it is a diffeomorphism thus amounts to saying that ${\exp_p}$ is nonsingular everywhere, as this implies (since the map is proper) that it is a covering map.

But to say that ${\exp_p}$ is nonsingular everywhere is precisely to say that nearby geodesics “diverge”: that is, for any ${v, w \in T_p M \setminus \left\{0\right\}}$,

$\displaystyle \frac{d}{ds} \exp_p( t( v + sw)) \neq 0$

for ${t \neq 0}$. In other words, given a Jacobi field $V(t)$ along the geodesic ${\gamma(t) = \exp_p(tv)}$ (coming from the infinitesimal variation of geodesics along ${w}$) with ${V(0)}$ but ${V \not\equiv 0}$, we have that ${V(t) \neq 0}$ for ${t \neq 0}$.

But in fact,

$\displaystyle \frac{d^2}{dt^2} \left \langle V(t), V(t) \right\rangle \geq 2 \left \langle \frac{D^2}{dt^2} V(t), V(t)\right\rangle,$

and this is positively proportional, by the Jacobi equation, to minus the sectional curvature of the plane spanned by ${\{V(t), \dot{\gamma(t)}\}}$. In particular, it is always nonnegative, and that means that since ${\frac{d}{dt}|_{t = 0} \left \langle V(t), V(t)\right\rangle = 0}$, this derivative is always nonnegative and ${\left \lVert V(t) \right \rVert}$ just keeps growing in ${t}$. In other words, the infinitesimal geodesic deviation just keeps growing in time.

3. Magnification of distances

But one actually has a bit more:

Corollary 5 If ${M}$ is simply connected and complete of nonpositive curvature, then the map

$\displaystyle \exp_p : T_p M \rightarrow M$

magnifies distances: that is, given tangent vectors ${v, w}$, one has

$\displaystyle \left \lVert v - w \right \rVert \leq d_M( \exp_p(v), \exp_p(w)).$

Proof: The idea is that

$\displaystyle \left \lVert (d \exp_p)_{v *}(w) \right \rVert \geq \left \lVert w \right \rVert, \quad v, w \in T_p M$

where the notation refers to the differential of the exponential map in the direction ${w}$, at ${v \in T_p M}$: in other words, infinitesimally, ${\exp_p}$ increases lengths. Since ${\exp_p: T_p M \rightarrow M}$ is a diffeomorphism, this easily implies the result.

In order to see this, we observe that the previous formula

$\displaystyle V(t) = \frac{d}{ds}|_{s =0 } \exp( t( v + sw))$

is a Jacobi field along the geodesic ${\exp(tv)}$, and that it satisfies ${V(0) = 0, V'(0) \stackrel{\mathrm{def}}{=} \frac{D}{dt}|_{t=0} V(t) = w}$. We saw previously that it is never zero after ${t =0}$. We now want to bound from below the growth rate. Let ${f(t) = \left \langle V(t), V(t)\right\rangle}$. Then,

$\displaystyle f(0) = f'(0) = 0, \quad f''(0) = 2\left \langle V'(0), V'(0)\right\rangle = 2 \left \langle w, w\right\rangle ,$

and ${f''(t) \geq 0}$ for all ${t}$. By integrating twice, this implies that

$\displaystyle \left \langle V(t), V(t)\right\rangle = f(t) \geq t^2 \left \langle w,w\right\rangle$

for all ${t}$. Since ${V(t) = t d \mathrm{exp}_{* tv}(w)}$, this provides the desired bound on differential of the exponential map.

$\Box$

Let’s now drop the assumption that ${M}$ is simply connected and complete. We can then use the above analysis to give a complete characterization of when ${M}$ has negative curvature.

Theorem 6 A Riemannian manifold ${M}$ has nonpositive curvature if and only if for every ${p \in M}$, there is a neighborhood ${U}$ of zero in ${T_p M}$ such that the exponential map is defined on ${U}$ and

$\displaystyle \exp_p : U \rightarrow M$

increases distances.

Geometrically, this means that given a small triangle in ${M}$, the length of the third side will be larger than expected from euclidean space (in terms of the other two sides and the angle between them).

Proof: We’ve already more or less proved that if ${M}$ has negative curvature, then the exponential map increases distances. Now assume that the exponential map increases distances.

Choose ${v, w \in T_p M}$ orthonormal, and consider the Jacobi field ${V(t) = \frac{d}{ds}|_{s = 0} \exp_p( t( v + sw))}$. We’ll look at the behavior of ${f(t) = \left \langle V(t), V(t)\right\rangle}$ for ${t}$ small, which requires a short computation.

Namely, we know that ${f(0) = f'(0) = 0}$, while

$\displaystyle f''(t) = 2 \left \langle V''(t), V(t)\right\rangle + 2 \left \langle V'(t), V'(t)\right\rangle,$

and

$\displaystyle f'''(t) = 6 \left \langle V''(t), V'(t)\right\rangle + 2 \left \langle V'''(t), V(t)\right\rangle ,$

and

$\displaystyle f^{(4)}(t) = 8 \left \langle V'''(t), V'(t)\right\rangle + 6 \left \langle V''(t), V''(t)\right\rangle + 2 \left \langle V^{(4)}(t), V(t)\right\rangle$

Therefore:

$\displaystyle f''(0) = 2 \left \langle v, v\right\rangle ,$

and

$\displaystyle f'''(0) = 6\left \langle V''(0), w\right\rangle = 0,$

because ${V''(0) = 0}$ by the Jacobi equation. Finally,

$\displaystyle f^{(4)}(0) = 8 \left \langle V'''(0), V'(0) \right\rangle = 8 \left \langle - R(w, v) v, w\right\rangle.$

This follows by ${V'''(0) = -R(w, v) v}$ as one sees by differentiating the Jacobi equation. Observe that the sectional curvature through the plane spanned by ${v,w}$ pops up.

Anyway, it follows that

$\displaystyle f(t) = t^2 \left \langle v, v\right\rangle - \frac{t^4}{3}\left \langle R( w, v) v, w\right\rangle + O(t^5) .$

If ${\exp_p}$ magnifies distances, then this is true infinitesimally: the norm of a Jacobi field has to grow more quickly than ${t^2 \left \langle v, v\right\rangle}$, which implies by the above asymptotic formula that the sectional curvatures ${\left \langle R(w, v) v, w\right\rangle }$ are nonpositive. $\Box$