3
$\begingroup$

If $(X,T)$ is a minimal system uniquely ergodic with $\mu$. Is there $p\in X$ such that $\mu(\partial B(p,t))=0$ for all $t>0$ for some metric $d$ (with the same topology)?

This question is motivated by the answer of Anthony Quas in Balls in minimal systems.

$\endgroup$
1
  • $\begingroup$ Is there a specific case you have in mind? I would not be surprised if the answer to your question is “yes” as there is a lot of freedom amongst metrics generating the same topology, but it doesn’t seem so easy to make a general argument. $\endgroup$ Commented Aug 8, 2020 at 18:03

1 Answer 1

1
$\begingroup$

This is not a complete answer, but too long for a comment.

If you remove unique ergodicity as an assumption, then I think there exist systems where it's not possible to find points where all boundaries of balls are universally null, i.e. have $0$ measure for all $T$-invariant measures. This is because in the paper "Can one always lower topological entropy?" by Shub and Weiss, they use the hypothesis "there exists aperiodic $p \in X$ so that for a dense set of $t$, $\{x : d(p, x) = t\}$ has measure $0$ for every $T$-invariant measure" to prove that $(X, T)$ has nontrivial factors of arbitrarily small entropy.

But, it's known that not every minimal system has factors of arbitrarily small entropy; for instance, there are minimal systems of infinite entropy (due to Lindenstrauss) for which every nontrivial factor has infinite entropy. So it seems that for his systems, you can't even get universally null boundaries for a dense set of radii, let alone all radii.

On the other hand, I don't think Lindenstrauss's systems are uniquely ergodic. So I guess in theory unique ergodicity could be enough. But I don't immediately see why. And I would be surprised, since if unique ergodicity did imply your property, it should imply that all uniquely ergodic systems have nontrivial factors of strictly smaller entropy, which I don't believe to be known.

EDIT: Of course I was being stupid; in fact Shub-Weiss explicitly note that their proof implies lowerability of entropy for uniquely ergodic systems (and even systems with countably many ergodic measures) for a clear reason: the boundaries of balls with fixed center are disjoint, so of course only countably many can have nonzero measure. So I guess these observations don't help much with your original question, sorry.

$\endgroup$

You must log in to answer this question.

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