Santalo Formula

Let $M$ be a simple Riemannian manifold with boundary $\partial M$. For $(x,v) \in SM$, let $\tau(x,v)$ denote the exit time of the geodesic starting at $x$ with tangent vector $v$, i.e. $\tau(x,v)$ is the (necessarily unique, and finite) time at which $\exp_x(tv) \in \partial M$.

We let $\partial_+(SM)$ denote the set
$$\partial_+(SM) = \{ (x,v) \in SM \ | \ x \in \partial M, \langle v, \nu\rangle > 0 \}$$
where $\nu$ denotes the inward unit normal to $\partial M$ in $M$. The exponential map identifies $SM$ with the set
$$\Omega = \{ (x,v,t) \in \partial_+(SM) \times \mathbf{R} \ | \ 0 \leq t \leq \tau(x,v)  \},$$
via $(x,v,t) \mapsto \exp_x(tv)$. Let $\Phi: \Omega \to M$ denote this diffeomorphism. Then we have, for all $f \in C^\infty(SM)$
$$\begin{align} \int_{SM} f dvol(SM) &= \int_\Omega (\Phi^\ast f) (\Phi^\ast dvol(SM)) \\ &= \int_{\partial_+(SM)} \int_0^{\tau(x,v)} f(\phi_t(x,v)) \Phi^\ast dvol(SM). \end{align}$$
Therefore, we can compute integrals of functions over $SM$ by integrating along geodesics, provided that we can cmopute $\Phi^\ast dvol(SM)$. This is the content of the Santalo formula.

Theorem (Santalo formula). For all $f \in C^\infty(SM)$, we have
$$\int_{SM} f dvol(SM) = \int_{\partial_+(SM)} \int_0^{\tau(x,v)} f(\phi_t(x,v)) \langle v, \nu\rangle dt dvol(\partial(SM))$$

Proof. Necessarily, we must have
$$\Phi^\ast(dvol(SM)) = a(x,v) dt \wedge dvol(\partial(SM))),$$
for some function $a(x,v)$. The reason we can assume that $a$ is independent of $t$ is that $\Phi$ is defined via geodesic flow, and geodesic flow preserves the volume form on $SM$. To compute the factor $a(x,v)$, we just need to compute
$$i_{\partial / \partial t} \Phi^\ast(dvol(SM)) = \Phi^\ast(i_{\Phi_\ast(\partial / \partial t)} dvol(SM))$$
From the definition of $\Phi$, we have that $\Phi_\ast(\partial / \partial t)$ is the Reeb vector field on $SM$, i.e. the vector field generating geodesic flow. Therefore, $\Phi_\ast(\partial / \partial t)$ is equal, at a point $(x,v)$ to the horizontal lift of the vector $v$. Therefore, using the definition of the induced volume form on a hypersurface of a Riemannian manifold, we find
$$i_{\partial / \partial t} \Phi^\ast(dvol(SM)) = \langle v, \nu \rangle dvol(\partial(SM))$$
where $\nu$ is the inward pointing unit normal to $\partial(SM)$ in $SM$. This shows that $a(x,v) = \langle v, \nu \rangle$ and completes the proof.

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.