Let $X = \mathbb T^2=\mathbb R^2/\mathbb Z^2$ be the torus, $\mu$ a $T$-invariant measure on $X$, $A \in M_2(\mathbb Z)$ a hyperbolic matrix, $T_A$ the associated automorphism. Let $v^-$ be an eigenvector of the smaller eigenvalue. (The contracting eigenvector.)

We want to prove:

$h_\mu(T_a)>0$ if and only if for every $B$ with $\mu(B)>0$, there are arbitrarily large $t$ for which $\mu \left( (B+tv^-) \cap B \right) >0$.

This resembles the definitions of non-wandering set and conservative action, for $\mathbb R$-actions.

Observations in the comments:

  • The translations along $v^{-}$ may not be measure-preserving, so we cannot apply Poincaré recurrence.

  • The claim is easy in case $\mu$ is the Haar measure $m_X$.

  • $\begingroup$ Surely you see that you need to prove that the slopes are irrational...In my opinion the other direction is the complicated one. $\endgroup$
    – John B
    Commented May 2, 2018 at 12:20
  • $\begingroup$ Let me add that the sentence "This is the continuous analog of being a conservative $\mathbb Z$-action." makes no sense. In this setting (let aside infinite ergodic theory) "conservative" is just being "Haar". $\endgroup$
    – John B
    Commented May 2, 2018 at 12:27
  • $\begingroup$ The question is strange. The condition $\mu((B+tv^-)\cap B)>0$ for arbitrary large $t$ is always satisfied. Indeed, the flow of translation by $v^-$ is measure-preserving, so this is only Poincaré recurrence (even if $v^-$ has rational slope). $\endgroup$
    – user120527
    Commented May 4, 2018 at 9:47
  • $\begingroup$ @user120527 Is there a version of Poincaré recurrence that gives every $B$ with $\mu(B)>0$ a time where a positive measure of it returns? The version I know just says that almost every point, separately, returns infinitely many times. In any way, when the slopes are irrational, this is an ergodic transformation so every two sets of positive measure $\mu(A)\mu(B)>0$ have a time where they intersect in positive measure: $\mu(T^{-n}(A)\cap B)>0$ as needed here. $\endgroup$
    – Pachirisu
    Commented May 4, 2018 at 11:18
  • 1
    $\begingroup$ And by the way, there is no reason that the flow along the $v^-$ direction is measure-preserving, so Poincare recurrence simply does not apply here. (As a simple example, the point measure at $0$ is $T$-invariant, but certainly not invariant under this flow.) $\endgroup$ Commented May 4, 2018 at 17:35


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