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Let $M$ be a compact, finite-dimensional Riemannian manifold, let $T: M \rightarrow M$ be an Anosov diffeomorphism, and let $\mu$ be a Sinai-Ruelle-Bowen (probability) measure. Write $\mathcal{R} = \{ R_1,\dots,R_n \}$ for a Markov partition; write $p_j^{(\mathcal{R})} := \mu(R_j)$.

Question:

Does there ever/always exist $\mathcal{R}$ s.t. $p^{(\mathcal{R})}$ is a nontrivial uniform measure on $\{1,\dots,n\}$? If not, does there ever/always exist a sequence of partitions $\mathcal{R}_m$ s.t. $p^{(\mathcal{R}_m)}$ converges to a uniform measure in some nontrivial sense?

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  • $\begingroup$ Steve, are you still interested in an answer? Probably (not at all sure) I can say that it is "yes" sometimes. Only for $\mathbb T^2$ though. It's not going to work for geodesic flows. $\endgroup$
    – Andrey Gogolev
    Commented Jun 15, 2010 at 17:17
  • $\begingroup$ I'd definitely be interested. Not sure if I can actually accept a new answer as there was a bounty that expired though... $\endgroup$
    – Steve Huntsman
    Commented Jun 15, 2010 at 20:02
  • $\begingroup$ OTOH if you answer mathoverflow.net/questions/26834 I could accept there, and for $\mathbb{T}^2$ it's the same thing anyway. $\endgroup$
    – Steve Huntsman
    Commented Jun 15, 2010 at 20:13
  • $\begingroup$ Ok, I'll think if the approach I have in mind really works. It may take some time. And again, it's not going to work for flows and for diffeos in higher dimension. Don't worry about accepting, I don't care much about it. $\endgroup$
    – Andrey Gogolev
    Commented Jun 15, 2010 at 20:58

2 Answers 2

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Since (p_1....,p_n) is an eigenvector of the transition matrix, your uniformity assumption is satisfied iff the transition matrix associated to the partition is bistochastic. I guess this is rarely the case.

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  • $\begingroup$ The transition matrix has entries which are either 0 or 1... such a matrix is bistochastic if and only if it's a permutation matrix. $\endgroup$
    – Vaughn Climenhaga
    Commented May 18, 2010 at 17:09
  • $\begingroup$ Sorry, I meant the matrix of transition probability. $\endgroup$
    – coudy
    Commented May 18, 2010 at 17:10
  • $\begingroup$ Ah... understood. $\endgroup$
    – Vaughn Climenhaga
    Commented May 18, 2010 at 17:13
  • $\begingroup$ @coudy--thanks for your answer: however, several of your statements are unclear to me. For instance, I don't see why the Markov chain associated to the partition has to be Bernoulli. There are non-Bernoulli chains that have the uniform distribution as their invariant distribution (e.g., take rows of the transition matrix to be permutations of a nondegenerate tuple summing to unity). $\endgroup$
    – Steve Huntsman
    Commented May 18, 2010 at 17:15
  • $\begingroup$ Also, there is more than one Markov partition for the cat map with fewer than N rectangles. Consider the entropy on the set of such partitions, and choose a partition with the maximal entropy S(N). It seems plausible to me that S(N)/log(N) could tend to unity. $\endgroup$
    – Steve Huntsman
    Commented May 18, 2010 at 17:21
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I wanted to elaborate on coudy's original answer, but I also think something is wrong with this argument. In the absence of any other answers I will make the bounty available for an explanation of what goes wrong here...

If $\mathcal{P}$ is a partition of $M$, write $\mathcal{P}_m^\vee := \bigvee_{j=0}^m T^j \mathcal{P}$. (Because $T$ preserves the SRB measure $\mu$ we don't have to consider, e.g., $j<0$ terms.)

A Markov partition $\mathcal{R}$ is generating, so the supremum over partitions in the Kolmogorov-Sinai entropy is realized by it: the KS entropy is $h_\mu^{KS}(T) = -\lim_m m^{-1}\sum_{R \in \mathcal{R}_m^\vee} \mu(R) \log \mu(R)$. Meanwhile the topological entropy is $h(T) = \lim_m m^{-1} \log \# \mathcal{R}_m^\vee$.

So if $\mu$ is also the measure of maximal entropy (as is the case, e.g. when $T$ is a hyperbolic toral automorphism or the Poincaré map for a geodesic flow on a surface of negative curvature), then $h_\mu^{KS}(T) = h(T)$ and in the limit we have that $-\sum_{R \in \mathcal{R}_m^\vee} \mu(R) \log \mu(R) \sim \log \# \mathcal{R}_m^\vee$, so that $\mu$ is asymptotically uniform on $\mathcal{R}_m^\vee$.

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    $\begingroup$ AFAIU, your argument is perfectly correct. The point is, you don't really tell us what you mean by "asymptotically uniform". I think that what you get with your computation is a particular case of the Shannon-McMillan Breiman theorem, valid for any ergodic transform and any partition: for a.e. x, we have 1/n mu(log R_n(x)) -> h. This implies that the ratio of the log of the measures of two elements in R_n tends to 1. But be careful, it does not imply that the ratio of the measures of two elements in the partition R_n goes to 1. It just means that this ratio is subexponential. $\endgroup$
    – coudy
    Commented May 21, 2010 at 20:41
  • $\begingroup$ OK, so the argument is less relevant, though not wrong. I've been looking at some examples with hyperbolic toral automorphisms and trying to square the behavior with what I see numerically for geodesic flow. Your point about subexponential ratios makes perfect sense now...but as you said earlier, it leaves the original question still standing. $\endgroup$
    – Steve Huntsman
    Commented May 21, 2010 at 21:46
  • $\begingroup$ BTW, my statement about "asymptotically uniform" isn't quite kosher unless it's interpreted more weakly than I'd intended, i.e., more weakly than in the spirit of the original question. $\endgroup$
    – Steve Huntsman
    Commented May 21, 2010 at 21:47

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