Hamilton-Jacobi equation and Riemannian distance

This post has been migrated from my old blog, the math-physics learning seminar.

Consider the cotangent bundle \(T^\ast X\) as a symplectic manifold with canonical symplectic form \(\omega\). Consider the Hamilton-Jacobi equation

\[ \frac{\partial S}{\partial t} + H(x, \nabla S) = 0, \]

for the classical Hamilton function \(S(x,t)\). Setting \(x=x(t), p(t) = (\nabla S)(x(t), t)\) one sees immediately from the method of characteristics that this PDE is solved by the classical action

\[ S(x,t) = \int_0^t (p \dot{x} - H) ds, \]

where the integral is taken over the solution \((x(s),p(s))\) of Hamilton’s equations with \(x(0)=x_0\) and \(x(t) = x\). The choice of basepoint \(x_0\) involves an overall additive constant of \(S\), and really this solution is only valid in some neighbourhood \(U\) of \(x_0\). (Reason: \(S\) is in general multivalued, as the differential “\(dS\)” is closed but not necessarily exact.)

Now consider the case where \(X\) is Riemannian, with Hamiltonian \(H(x,p) = \frac{1}{2} |p|^2\). The solutions to Hamilton’s equations are affinely parametrized geodesics, and by a simple Legendre transform we have

\[ S(x, t) = \frac{1}{2} \int_0^t |\dot x|^2 ds \]

where the integral is along the affine geodesic with \(x(0) = x_0\) and \(x(t) = x\). Since \(x(s)\) is a geodesic, \(|\dot x(s)|\) is a constant (in \(s\)) and therefore

\[ S(x, t) = \frac{t}{2} |\dot x(0)|^2. \]

Now consider the path \(\gamma(s) = x(|\dot x(0)|^{-1}s)\). This is an affine geodesic with \(\gamma(0) = x_0\), \(\gamma(|\dot x(0)|t) = x\) and \(|\dot \gamma| = 1\). Therefore, the Riemannian distance between \(x_0\) and \(x\) (provided \(x\) is sufficiently close to \(x_0\)) is

\[ d(x_0, x) = |\dot x(0)| t. \]

Combining this with the previous calculation, we see that

\[ S(x, t) = \frac{1}{2t} d(x_0, x)^2. \]

Now insert this back into the Hamilton-Jacobi equation above. With a bit of rearranging, we have the following.

Theorem. Let \(x_0\) denote a fixed basepoint of \(X\). Then for all \(x\) in a sufficiently small neighborhood \(U\) of \(x_0\), the Riemannian distance function satisfies the Eikonal equation

\[ |\nabla_x d(x_0, x)|^2 = 1. \]

Now, for convenience set \(r(x) = d(x_0, x)\). Then \(|\nabla r|^2 = 1\), from which we obtain (by differentiating twice and contracting)

\[ g^{ij} g^{kl}\left(\nabla_{lki} r \nabla_j r + \nabla_{ki}r \nabla_{lj} r\right) = 0.\]

Quick calculation shows that

\[ \nabla_{lki} r = \nabla_{ilk} r - \left.R_{li}\right.^{b}_k \nabla_b r \]

Therefore, tracing over \(l\) and \(k\) we obtain

\[ g^{lk} \nabla_{lki} r = \nabla_i ( \Delta r) + Rc(\nabla r, -) \]

Plugging this back into the equation derived above, we have

\[ \nabla r \cdot \nabla(\Delta r) + Rc(\nabla r, \nabla r) + |Hr|^2 = 0, \]

where \(Hr\) denotes the Hessian of \(r\) regarded as a 2-tensor. Now, using \(r\) as a local coordinate, it is easy to see that \(\partial_r = \nabla r\) (as vector fields). So we can rewrite this identity as

\[ \partial_r (\Delta r) + Rc(\partial_r, \partial_r) + |Hr|^2 = 0. \]

Now, we can get a nice result out of this. First, note that the Hessian \(Hr\) always has at least one eigenvalue equal to zero, because the Eikonal equation implies that \(Hr(\partial_r, -)=0\). Let \(\lambda_2, \dots, \lambda_n\) denote the non-zero eigenvalues of \(Hr\). We have

\[ |Hr|^2 = \lambda_2^2 + \dots + \lambda_n^2, \]

while on the other hand

\[ |\Delta r|^2 = (\lambda_2 + \dots + \lambda_n)^2 \]

By Cauchy-Schwarz, we have

\[ |\Delta r|^2 \leq (n-1)|Hr|^2 \]

Proposition. Suppose that the Ricci curvature of \(X\) satisfies \(Rc \geq (n-1)\kappa\), and let \(u = (n-1)(\Delta r)^{-1}\). Then

\[ u' \geq 1 + \kappa u^2. \]

Now, the amazing thing is that this deceptively simple inequality is the main ingredient of the Bishop-Gromov comparison theorem. The Bishop-Gromov comparison theorem, in turn, is the main ingredient of the proof of Gromov(-Cheeger) precompactness. I hope to discuss these topics in a future post.