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?


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?