The Index Form

Let $f: [0,T] \times (-\epsilon, \epsilon) \to M$ be a family of parametrized curves in a Riemannian manifold $(M, g)$. To simplify this calculation, we assume that $f(0,s) = p, f(T, s) = q$ for some $p,q \in M$ and all $s \in (-\epsilon, \epsilon)$. (This assumption is not necessary, but without it our variational formulae will have additional boundary terms.)

For convenience, set $\dot f = \partial f / \partial t$ and $f' = \partial f / \partial s$. For each $s \in (-\epsilon, \epsilon)$ we define the energy functional $E = E(s)$ to be
\[ E(s) = \frac{1}{2} \int_0^T |\dot f|^2 dt. \]
The first variation is
\begin{align}
\frac{dE}{ds} &= \int_0^T \langle \nabla_{f'} \dot f, \dot f \rangle dt \\\
&= \int_0^T \langle \nabla_{\dot f} f', \dot f \rangle dt \\\
&= -\int_0^T \langle f', \nabla_{\dot f}\dot f \rangle dt
\end{align}

Set $\gamma(t) := f(t,0)$ and $X(t) = f'(t)$ (thought of as a vector field supported on $\gamma$). Evaluating the above at $s=0$ we obtain
\[ \left.\frac{dE}{ds}\right|_{s=0} = -\int_0^T \langle X, \nabla_{\dot \gamma} \dot \gamma \rangle dt, \]
which shows immediately that

Theorem. $\gamma$ is a critical point of the energy functional if and only if $\nabla_{\dot \gamma} \dot \gamma = 0$.


The second variation is
\begin{align}
\frac{d^2 E}{ds^2}
&= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle
 + \langle f', \nabla_{f'}\nabla_{\dot f}\dot f \rangle dt \\\
&= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle
 + \langle f', \nabla_{\dot f}\nabla_{f'}\dot f \rangle dt
 + \langle f', R(f', \dot f)\dot f \rangle dt \\\
&= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle
 - \langle \nabla_{\dot f}f', \nabla_{f'}\dot f \rangle dt
 + \langle f', R(f', \dot f)\dot f \rangle dt \\\ &= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle
 - \langle \nabla_{\dot f} f', \nabla_{\dot f} f'\rangle dt
 + \langle f', R(f', \dot f)\dot f \rangle dt
\end{align}

Assume now that $\gamma$ is a geodesic, i.e. $\nabla_{\dot \gamma} \dot \gamma = 0$. Then evaluating the above at $s=0$, we obtain
\[ \frac{d^2 E}{ds^2} = \int_0^T |\nabla_{\dot \gamma} X|^2 - \langle X, R(X, \dot \gamma) \dot \gamma \rangle dt. \]

Definition. Let $\gamma$ be a geodesic. The index form associated to variations $X,Y$ of $\gamma$ is
\begin{align} I(X,Y) &= \int_0^T \langle \nabla_{\dot \gamma} X, \nabla_{\dot \gamma} Y \rangle dt
 - \langle Y, R(X, \dot \gamma) \dot \gamma \rangle \\\
&= -\int_0^T \langle Y, \nabla_{\dot \gamma}^2 X + R(X, \dot\gamma)\dot \gamma \rangle
\end{align}
It follows from symmetries of the Riemann tensor that $I(X,Y) = I(Y, X)$ and also $I(X,X) = E''$ as above.

Theorem. Suppose that $X$ is the infinitesimal variation of a family of affine geodesics about a fixed geodesic $\gamma$. Then
\[ \nabla_{\dot \gamma}^2 X + R(X, \dot\gamma)\dot\gamma = 0. \]
In particular, $I(X, -) = 0$.

Proof. Let $f(t,s)$ denote the family as above. By hypothesis, we have that $\nabla_{\dot f} \dot f = 0$ for all $s$, so that
\[ \nabla_{f'} \nabla_{\dot f} \dot f = 0. \]
Commuting the derivatives using the curvature tensor, we have
\[ 0 = \nabla_{\dot f} \nabla_{f'} \dot f + R(f', \dot f) \dot f. \]
Now use $\nabla_{\dot f} f' = \nabla_{f'} \dot f$ and evaluate at $s=0$ to obtain
\[ 0 = \nabla_{\dot \gamma}^2 X + R(X, \dot \gamma)\dot\gamma. \]

Boundary Distance

Recently, I've been learning some topics related to machine learning, and especially manifold learning. These both fall under the general notion of inverse problems: given some mathematical object $X$ (it could be a function $f: A \to B$, or a Riemannian manifold $(M,g)$, or a probability measure $d\mu$ on a space $X$, etc.), can we effectively reconstruct $X$ given only the information of some auxiliary measurements? What if we can only perform finitely many measurements? What if the measurements are noisy? Can we reconstruct $X$ at least approximately? Can we measure in some precise way, how close our approximate reconstruction is to the unknown object $X$? And so on, and so forth.

Anyway, this post is about a cute observation, which I was reminded of while reading a paper on the inverse Gel'fand problem. Let $M$ be a compact manifold with smooth boundary $\partial M$. Then with no additional data required, we have a Banach space $L^\infty(\partial M)$ consisting of the essentially bounded measureable functions on the boundary. Since it is a Banach space, it comes with a complete metric $d_\infty(f,g) := \|f-g\|_{L^\infty(\partial M)}$.

Now, suppose that $g$ is a Riemannian metric on $M$. Then we have the Riemannian distance function $d_g(x,y)$ which is defined to be the infimum of arclengths of all smooth paths connecting $x$ and $y$. For any $x \in M$, we obtain a function $r_x \in L^\infty(\partial M)$ defined by
\[ r_x(z) = d_g(x,z), \forall z \in \partial M. \]
This gives a map $\phi_g: M \to L^\infty(\partial M)$, defined by $x \mapsto r_x$.

Theorem. Suppose that for any two distinct $x,y \in M$, there is a unique length-minimizing geodesic connecting $x$ and $y$. Then $\phi_g: M \to L^\infty(\partial M)$ is an isometric embedding, i.e. $d_g(x,y) = d_\infty(r_x, r_y)$ for all $x,y \in M$.

Proof. Let $x,y$ be distinct and let $\gamma$ be the unique geodesic from $x$ to $y$. For any point $z$ on the boundary, we have
\[ |d_g(x,z) - d_g(y,z)| \leq d_g(x,y). \]
which is the triangle inequality. Now let $\gamma$ be the unique geodesic from $x$ to $y$, and extend $\gamma$ until it hits some boundary point $z_\ast$. Then since $x,y,z_\ast$ all lie on a length-minimizing geodesic, we have
\[ d_g(x,z_\ast) - d_g(y,z_\ast) = d_g(x,y). \]
Therefore, the bound above is always saturated, and we find
\[ \sup_{z \in \partial M} |d_g(x,z) - d_g(y,z)| = d_g(x,y). \]
But the expression on the left is nothing but the $L^\infty(\partial M)$-norm of $r_x-r_y$, so the theorem is proved.

Hamilton-Jacobi equation and Riemannian distance

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
\[ u' \geq 1 + \kappa u^2. \]

Proof. From preceding formulas, $|Hr|^2$ can be expressed in terms of the Ricci curvature and the radial derivative of $\Delta r$. On the other hand, $|\Delta|^2$ is bounded above by $(n-1) |Hr|^2$. The claimed inequality then follows from simple rearrangement.

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.