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Let $M$ be a closed manifold, $m$ be the normalized volume measure on $M$, and $f:M\to M$ be a $C^2$ transitive Anosov diffeomorphism. Consider the pushforward $f^km$ defined by

----------$f^km(A):=m(f^{-k}A)$ for all measurable subset $A\subset M$.

Then the Birkhoff averages $\nu_k=\frac{1}{k}\sum_{j=0}^{k-1}f^jm$ are probability measures on $M$ for all $k\ge1$. The question is:

What can we say about the measure(s) in the set $\mathcal{V}(m)$ of accumulation points of $\{\nu_k:k\ge1\}$?

We know that there exists a unique SRB measure $\mu_+$ for $f$ (and a unique SRB measure $\mu_-$ for $f^{-1}$). Do we have $\mathcal{V}(m)\subset\{\mu_+,\mu_-\}$?

edited Aug 31 2011 at 15:07

Vaughn Climenhaga
4,016●1●8●25

asked Aug 31 2011 at 1:57

Pengfei
918●2●7

A quick glance doesn't show me anything in Chernov's survey of invariant measures for hyperbolic dynamical systems unless you have some invariant measure $\mu << m$, in which case the forward and backward measures coincide. – Steve Huntsman Aug 31 2011 at 6:27

Vaughn Climenhaga · Accepted Answer · 2011-08-31 15:12:28Z UTC

5

Yes. In fact, we have $\mathcal{V}(m) = \{\mu_+\}$ whenever $m$ is a probability measure on $M$ that is absolutely continuous with respect to volume. This is shown in (0.4) of "A measure associated with Axiom-A attractors" by David Ruelle, American Journal of Mathematics 98 (1976), 619--654.

answered Aug 31 2011 at 15:12

Vaughn Climenhaga
4,016●1●8●25

	Do you mean "whenever $\mu_+$ is a"? – Steve Huntsman Aug 31 2011 at 19:29
	No, $\mu_+$ is the unique SRB measure for $f$. The initial measure $m$ is an arbitrary absolutely continuous probability measure (not necessarily an invariant one!), such as normalised volume. – Vaughn Climenhaga Aug 31 2011 at 20:17
	Got it, thanks. – Steve Huntsman Aug 31 2011 at 20:22
	Thank you for the reference! – Pengfei Sep 1 2011 at 1:38

Accumulation points of the Birkhoff average of $m$

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