Let $S$ be the surface of a convex body embedded in $\mathbb{R}^3$. For me $S$ is a convex polyhedron, but I am happy to view $S$ as a smooth body with positive Gaussian curvature at each point, or with non-negative Gaussian curvature at each point. Likely these various versions do not affect the question I am posing. Maybe even convexity is not required.

Let $x \in S$ be an arbitrary point on $S$, and $u$ an arbitrary direction vector tangent to $S$ at $x$. Shoot off a geodesic $\gamma$ from $x$ in direction $u$, and let it proceed until it intersects its own path at some point $y$. Rather than let the geodesic cross itself at $y$, have it instead reflect from $\gamma$ like a mirror: angle of incidence $=$ angle of reflection. And continue: every time $\gamma$ would cross itself, instead it reflects. Call this $\gamma$ a reflecting geodesic. ("Self-avoiding" is eye-catching but inaccurate.)

Q. For generic $x$ and $u$, is it true that for "most" surfaces $S$, a reflecting geodesic $\gamma$, emanating from $x$ in direction $u$, converges to a point?

I would prefer not to attempt to define precisely what "most surfaces" means, but in the polyhedral world, it would suffice for $S$ to be the convex hull of random points in space. (Vertices would be hit with probability zero.) I know that, without the "generic" qualifier, a reflecting geodesic might get caught in an infinite loop. For example, on a cube, $\gamma$ intersects itself at $90^\circ$ (essentially: because of the Gauss-Bonnet theorem):

    A geodesic starting at the center of the left-front face eventually intersects itself at $90^\circ$ on the bottom face.
These reflecting geodesics may seem contrived, but something close to these came up in my research, and I am hoping that an answer to Q might help.


1 Answer 1


It is true.

If the set of limit points of your trajectory is not a single point it has to be a (limit set of) simple closed geodesics.

On the other hand, from Gauss--Bonnet formula, it follows that most of convex closed polyhedral surfaces do not admit simple closed geodesic.

  • $\begingroup$ Thanks, Anton, for the point about a limit set of simple closed geodesics. $\endgroup$ Nov 2, 2015 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.