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Let $M$ be a compact smooth manifold and $m$ be a normalized volume induced by some Riemannian metric on $M$. Let $f\in\mathrm{Diff}^1(M)$ and $R_f$ be the set of recurrent points of $f$ (A point $x$ is recurrent if $x\in\omega(f,x)\cap \alpha(f,x)$, or equally $\liminf_{n\to\pm\infty}d(f^nx,x)=0$).

The Hopf decomposition of $(M,f)$ is a partition $M=C_f\sqcup D_f$ such that

  • the restriction $f|_{C_f}$ is conservative. That is, if $E\subset {C_f}$ is a measurable subset with $[f^nE:n\in\mathbb{Z}]$ mutually disjoint, then $m(E)=0$;

  • the restriction $f|_{D_f}$ is dissipative. In fact ${D_f}$ is the disjoint union of $[f^nF:n\in\mathbb{Z}]$ for some measurable subset $F\subset M$.

See here (Jon Aaronson, An introduction to infinite ergodic theory, Page 15).


By Poincare Recurrence theorem (Page 17, same book) we have $m(C_f\backslash R_f)=0$.

My question is about the converse direction:

  1. If $m(R_f)>0$, will we have $m(C_f)>0$?

  2. When will the following be true: $m(R_f\backslash C_f)=0$?

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It should be pointed out that conservativity here is weaker than the usual measure--preserving requirement. –  Pengfei Sep 4 '11 at 3:44

1 Answer 1

up vote 2 down vote accepted

Let me first notice that the proof of the Poincare recurrence theorem you are referring to is actually valid for any action of a countable group with an invariant measure. Now, if instead of $\mathbb Z$ we are talking about general groups, then there are, for instance, examples of Fuchsian groups such that their action on the boundary circle is minimal (any orbit is dense), but nonetheless completely dissipative with respect to the Lebesgue measure (see this paper and the references therein). Roughly speaking, it is possible that no open set is wandering, but the action is still completely dissipative. The point is that dissipativity is a measure theoretical property, whereas the recurrent set is defined in topological terms.

In your question the group, the state space and the measure are all more specific, and I don't think this particular question has ever been addressed. Still, a priori I don't see any reason why topological recurrence on a set of positive measure would prevent the action from being completely dissipative.

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Thank you for the reference! What do you mean for a measure being quasi-invariant? Poincare recurrence theorem requires more than preserving the measure class (take the north-south map on S^2 for example). –  Pengfei Sep 4 '11 at 14:29
    
Sorry - that's my typo. Of course, I wanted to say that the Poincare recurrence theorem is valid for any action with an INVARIANT measure. I've corrected my answer. –  R W Sep 4 '11 at 16:45

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