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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):


          GeodesicLoopCube
    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.

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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.

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  • $\begingroup$ Thanks, Anton, for the point about a limit set of simple closed geodesics. $\endgroup$ – Joseph O'Rourke Nov 2 '15 at 18:11

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