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 $mathcal{R}_m^vee$.
No comments:
Post a Comment