 Solution

A simple Riemannian manifold has strictly convex boundary and a unique geodesic between each pair of points, smoothly dependent on its endpoints; there are no conjugate points and no trapped geodesics. Let $\nu$ denote the outward unit normal. For a boundary unit vector with $⟨v,\nu ⟩<0$, follow its unit-speed geodesic ${\gamma }_{x,v}$ until the next boundary intersection at time $\tau (x,v)>0$. Its geodesic scattering relation isThus it maps incoming boundary unit vectors to outgoing ones, recording both exit point and direction.

If the scattering relation is defined on the full boundary unit bundle, extend it on outgoing vectors by following the same oriented geodesic backwards to its previous boundary intersection, and fix tangential vectors. This extension is an involution, because the second application follows the same segment back to its starting point with the same velocity. The exit time is not a component of the scattering relation itself; together with it one obtains lens data. For a simple Riemannian metric, $\tau$ is the boundary distance between the endpoints of the segment.