Do there exist Markov partitions with (nearly) uniform Riemannian measures? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:20:32Z http://mathoverflow.net/feeds/question/26834 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26834/do-there-exist-markov-partitions-with-nearly-uniform-riemannian-measures Do there exist Markov partitions with (nearly) uniform Riemannian measures? Steve Huntsman 2010-06-02T15:36:03Z 2010-06-02T15:36:03Z <p>This question complements <a href="http://mathoverflow.net/questions/24813/" rel="nofollow">this one</a>; the difference is in considering Riemannian versus SRB measures.</p> <p>Let $M$ be a compact, finite-dimensional Riemannian manifold, let $T: M \rightarrow M$ be an <a href="http://en.wikipedia.org/wiki/Anosov_diffeomorphism" rel="nofollow">Anosov diffeomorphism</a>, and let $v$ be the normalized Riemannian measure. Write <code>$\mathcal{R} = \{ R_1,\dots,R_n \}$</code> for a <a href="http://en.wikipedia.org/wiki/Markov_partition" rel="nofollow">Markov partition</a>; write $v_j^{(\mathcal{R})} := v(R_j)$.</p> <blockquote> <p><strong>Question:</strong></p> <p>Does there ever/always exist $\mathcal{R}$ s.t. $v^{(\mathcal{R})}$ is a nontrivial uniform measure on <code>$\{1,\dots,n\}$</code>? If not, does there ever/always exist a sequence of partitions $\mathcal{R}_m$ s.t. $v^{(\mathcal{R}_m)}$ converges to a uniform measure in some nontrivial sense?</p> </blockquote> <p>Of course, for hyperbolic toral automorphisms the two questions are equivalent. However I would guess that this question might be easier in general, since the classical construction of Markov partitions uses an open cover of $M$ by balls of fixed radius to form an initial (typically non-Markovian) partition that is subsequently perturbed iteratively into a Markov partition.</p>