## Archive for September, 2009

### A question by Furstenberg

September 25, 2009

Yesterday I was talking about some properties of different dimensions with Furstenburg. Somehow I mentioned Kekaya, and he told me about the following question he has been longing to solve (which is amazingly many similarities to Kekaya):

For set $S \in \mathbb{R}^2$, if $\exists \delta>0$ s.t. for all direction $\theta, \exists$ line $l$ with direction $\theta$ s.t. $\dim (l \cap S) > \delta$. Does it follow that $\dim(A) \geq 1$?

Note that a stronger conjecture would be $\dim(A)$ is at least $1+\delta$ which when taking $\delta = 1$ would give a generalization of the $2$-dimensional Kekaya. (i.e. instead of having to have a line segment, we only require a 1-dimension set in each direction)

Reviewing the proofs of the 2-dimsional Kekaya, I found they all rely on the fact that the line segment is connected…Hence it might be interesting to even find an answer to the following question:

If $A \subseteq \mathbb{R}^2$ contains a measure 1 set in every direction, does it follow that $\dim(A)=2$?

### 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?