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Hello MO World

I'm working on a paper involving embedding your favourite measure-preserving transformation into a topological model (think Krieger generator theorem: embedding in a full shift) and have run into a question about entropy and Bowen balls.

Setup: $T$ is a homeomorphism from a compact metric space $X$ to itself. Definition: $B(x,n,\delta)=\lbrace y\colon d(T^ix,T^iy) < \delta\text{ for $0\le i < n$}\rbrace$ is a Bowen $(n,\delta)$-ball around $x$. Let $\mu$ be an ergodic invariant measure for $T$ and let $A$ be a set of positive measure.

What can be said about the functions $$ f_n(x)=\frac{\mu(A\cap B(x,n,\delta))}{\mu(B(x,n,\delta))} $$ as $n\to\infty$ for fixed $\delta$?

Notice that if $X$ is a shift space, then the $B(x,n,\delta)$ partition the space and the $f_n$ are just conditional expectations with respect to that partition. In that case, $\int f_n \ d\mu(x)$ is equal to $\mu(A)$ for all $n$. This is the kind of conclusion I'd like to find in the general case.

One interpretation of the question is that I'm asking: if you pick a point $x$ according to the measure $\mu$ and then pick a second point $y$ according to the restriction of $\mu$ to the Bowen ball around $x$, then is the distribution of $y$ `similar' to $\mu$?

For a non-dynamical example where the resampled distribution is not the same, consider the set $\lbrace 1,2,3\rbrace$ with the usual distance in $\mathbb R$. If you pick a point $x$, and then sample a point $y$ in the 1.5 ball around $x$, you're more likely to get to 2 than you are to 1 or 3.

If anyone has ideas, or has seen something similar, I'd really like to hear about it...

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Perhaps you are not interested anymore on the question, but let me note that one should really not expect a positive answer. Of course, when there are different Lyapunov exponents (allow me to think of differentiable maps), the Bowen balls will be aligned more and more with certain invariant directions as $n$ grows, and so one should expect instead do obtain densities of an induced measure along these invariant directions (or invariant foliations). Notice also that the measures induced in this manner (with these conditionals) need not be invariant under the original dynamics.

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  • $\begingroup$ So I no longer need the result for the paper that I was working on, but I'm still interested in the question. I believe that, for example if $T$ is an Anosov automorphism of $\mathbb T^d$ (with a variety of Lyapunov exponents), then the answer to the question is `yes' (as the Bowen balls are translates of small ellipsoids around $x$). $\endgroup$
    – Anthony Quas
    Commented Dec 22, 2015 at 23:25
  • $\begingroup$ Sure, since in that case strong stable and strong unstable foliations are collections of subspaces. But the example to have in mind should better be your (favorite) nonuniformly hyperbolic system possibly with holonomies that are at most Hölder continuous. The densities will be sitting on the strong stable leaves. The same goes for a noninvertible dynamics with some lower dimensional invariant foliation. $\endgroup$
    – John B
    Commented Dec 22, 2015 at 23:34
  • $\begingroup$ Unfortunately I'm not used to thinking about this kind of system, and would certainly not be in a position to construct an example based on this description. $\endgroup$
    – Anthony Quas
    Commented Dec 22, 2015 at 23:46
  • $\begingroup$ In other words, if you let $n\to\infty$ in the case of a toral automorphism, the density will be that of the marginal on the family of subspaces that are really the strong unstable foliation. The Lebesgue density lemma applies to the integration of the marginals, and so by an almost everywhere argument also to the strong unstable foliation and its densities (we have a Fubini theorem in this case!). But in general we would need absolute continuity and this spoils having in general Lebesgue density theorems on the invariant foliations. This is the reason why the the answer can be negative. $\endgroup$
    – John B
    Commented Dec 22, 2015 at 23:56
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    $\begingroup$ Do you think you could construct an explicit counterexample, or is this more of a meta-argument for why you think this shouldn't work? $\endgroup$
    – Anthony Quas
    Commented Dec 23, 2015 at 6:04

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