4
$\begingroup$

This question appears for first time (to my knowledge) in

×2 and ×3 invariant measures and entropy Daniel J. Rudolph Ergodic Theory and Dynamical Systems / Volume 10 / Issue 02 / June 1990, pp 395 - 406

I want to know what progress has been done in this specific direction?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

The question of the existence of a nonatomic measure which is jointly invariant under these two maps, and is not equal to Lebesgue measure, is called the Furstenberg $\times 2 \times 3$ problem and is a notorious open problem in ergodic theory which I understand to have been posed by Hillel Furstenberg in 1967. There have been numerous extensions of, and variations upon, Rudolph's theorem (to commuting pairs of Anosov automorphisms, for example, in which project Anatoly Katok has been active; or to the Hausdorff dimension of closed sets which are invariant under both maps, such as in the recent Annals paper by Hochman and Shmerkin) but my understanding is that no fundamental progress on the original statement of the problem has been made since 1990.

Some other related results have been discussed in these questions.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Do you believe the answer is yes or no? $\endgroup$
    – user39115
    Commented Jul 22, 2014 at 9:30
  • $\begingroup$ I am not sure. I am not an expert on this question, but I am surprised by the fact that the positive-entropy case is so much easier than the zero-entropy case (but having said that, in a natural sense there are "more" zero entropy measures than positive entropy ones). I suppose if pressed I believe that no non-atomic jointly invariant measures exist. $\endgroup$
    – Ian Morris
    Commented Jul 22, 2014 at 9:45
  • 3
    $\begingroup$ Furstenberg was asked this question directly at a conference at the University of Maryland a couple of years ago. He responded that he wanted to believe in the beauty of mathematics, which many in the audience (including me) took to mean he thought that there were no other jointly invariant continuous measures other than Lebesgue. But his response was a best ambiguous. $\endgroup$
    – Douglas Lind
    Commented Oct 25, 2014 at 18:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .