## Billiards

September 24, 2009

Amie Wilkinson asked me the following question some time ago:

Given a smooth convex Jordan curve $J \subseteq \mathbb{R}^2$, consider the billiard map $\varphi: J \times (0, \pi) \rightarrow J \times (0, \pi)$, let $\pi_\theta: J \times (0, \pi) \rightarrow (0, \pi)$ be the projection.

a) If $\forall (p, \theta) \in J \times (0, \pi), \pi_\theta \circ \varphi (p, \theta) = \theta$, does this imply $J$ is a circle?

(Yes, in fact we only need $\varphi$ to fix the second component of points in $\{p, q \} \times (0, \pi)$ for a chosen pair of points $p, q$. Classical geometry)

b) If $\forall p \in J, \pi_\theta \circ \varphi (p, \pi/2) = \pi/2$, does this imply $J$ is a circle?

(No, my example was a cute construction that attaches six circular arcs together)

c) What’s the smallest set $S \subseteq (0, \pi)$ s.t. if $\varphi$ fixes the second component on $J \times S$ then $J$ has to be a circle?

I am still thinking about c)…My guess is that any sub interval would work, and of course any dense subset inside a given set works equally well as the whole set…

But is it possible to have only finitely many angles? Maybe even two angles?

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