Santalo Formula

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

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 $f \in C^\infty(SM)$

$$\begin{aligned} \int_{SM} f dvol(SM) &= \int_\Omega (\Phi^\ast f) (\Phi^\ast dvol(SM)) \cr &= \int_{\partial_+(SM)} \int_0^{\tau(x,v)} f(\phi_t(x,v)) \Phi^\ast dvol(SM). \end{aligned}$$

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.