Do there exist Markov partitions with (nearly) uniform Riemannian measures?

2010-06-02T15:36:03Z

This question complements this one; the difference is in considering Riemannian versus SRB measures.

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

Question:

Does there ever/always exist $\mathcal{R}$ s.t. $v^{(\mathcal{R})}$ 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. $v^{(\mathcal{R}_m)}$ converges to a uniform measure in some nontrivial sense?

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.

Do there exist Markov partitions with (nearly) uniform Riemannian measures? - MathOverflow

Do there exist Markov partitions with (nearly) uniform Riemannian measures?