# Are periodic billiard trajectories stable on a manifold with strictly convex boundary?

Let $(M,g)$ be a compact Riemannian manifold with strictly convex boundary. Let $\gamma:S^1\to M$ be a periodic billiard trajectory (geodesic in the interior and reflects specularly at the boundary).

For some small $\epsilon>0$, let $g_s$, $s\in(-\epsilon,\epsilon)$, be a family of metrics on $M$ so that $g_0=g$. Assume that all metrics and their dependence on the parameter $s$ is smooth, but I don't want to constrain the variation of the metric otherwise.

Is there a (smooth) family of closed curves $\gamma_s:S^1\to M$ (with some $\epsilon>0$) so that each $\gamma_s$ is a periodic billiard trajectory with respect to the metric $g_s$ and $\gamma_0=\gamma$? (With a suitable notion of smoothness at reflection points.) In some cases a perturbation of the metric can destroy the periodic trajectory (consider the 2-periodic trajectory on a square), but it seems to me that strict convexity of the boundary should add stability.

The most interesting case to me is when $M$ is three dimensional, in particular when $M$ is the unit ball in $\mathbb R^3$ and the metric $g_0$ is conformally Euclidean, but I would like to understand the problem in greater generality. Section 7.1 in this chapter (for example) discusses this question for planar polygons, but that is quite far from my goal.

In principle one could glue two copies of $M$ together along the boundary and study the stability of periodic geodesics on the closed manifold. The problem is that the metrics on the doubled manifold are not even $C^1$ (only Lipschitz) because $\partial M$ is strictly convex, so I fear that many tools are not applicable. The normal derivative of the metric tensor is the second fundamental form, so the following are equivalent: (1) the metric is $C^1$ (2) the metric is $C^2$ (3) the second fundamental form vanishes.

Edit: Carl's answer below tells that most periodic orbits are not stable if the unperturbed manifold $(M,g_0)$ is too symmetric. This is a useful answer but only to a special case. Are the orbits stable for generic metrics or under some additional assumption?

The result in the paper Spectrum of the Poincare map for periodic reflecting rays in generic domains of Petkov--Stojanov may be related.

Setting: Let $X\subset \mathbb{R}^n$ be a closed $(n-1)$-dimensional manifold (as the boundary of a connected $n$-dimensional subset $D_X$). Consider the set of the embeddings $f:X\to \mathbb{R}^n$ (denote the corresponding $n$-set by $D_f$).

The result in the above paper is very general. A very special version states that for a generic embedding $f:X\to\mathbb{R}^n$, every periodic point of the induced billiard system on $D_f$ is nondegenerate, and hence persists under any types of small perturbations of the induced billiard map. In particular, each pre-fixed periodic orbit of on $D_f$ for a generic $f$ persists under any small perturbations of the Euclidean metric of $D_f$.

• Thanks! The article looks relevant and I will look it up. – Joonas Ilmavirta Jul 30 '15 at 21:27

Not necessarily... In the example you give - a ball in three dimensions with Euclidean metric - there are continuous families of periodic orbits of all numbers of reflections $n\geq 2$ (polygons and stars on planes passing through the origin). But a typical perturbation of the metric would break the symmetry and leave only a finite number of periodic orbits for each $n$. So most of the original periodic orbits are unstable in this sense.

Edit: In response to the question edit: Persistence of a periodic orbit is a local question, depending only on the metric and boundary properties on/near the orbit. It cannot "see" the global symmetry if any. So, the unperturbed manifold need not be symmetric away from the periodic orbit. Furthermore, an isolated non-hyperbolic periodic orbit may also vanish under perturbation of Hamiltonian systems, in for example a saddle-center bifurcation. Such isolated non-hyperbolic orbits may be obtained easily in convex billiards by tuning the local boundary curvatures. The generic situation should be that each orbit is stable to perturbations of the metric, but that any given perturbation destroys infinitely many long periodic orbits.

• Why do you expect to have a finite number of periodic orbits for a fixed number of reflections in a generic metric? This is not true if the metric is radially symmetric, but radial symmetry is admittedly not a generic property. Sounds plausible, but I don't see a simple argument. – Joonas Ilmavirta Jul 28 '15 at 13:20
• According to the usual variational principle, periodic orbits are stationary points of the total length of the orbit according to the metric, fixing the $n$ points on the boundary. Generic smooth functions in a compact space have only a finite number of critical points. – user25199 Jul 28 '15 at 13:50