# Defining an f-invariant measure

Suppose I have a compact oriented manifold $M$ with an orientation preserving self-diffeomorphism $f$. I wish to define a volume form on $M$ which is invariant under $f$. Certainly, it is necessary that for any open $U\subset M$, its image $f(U)$ is not a proper subset of $U$. Is this sufficient? If not, what would be sufficient?

A natural thing to try is to start with a volume form on $M$ and then try to even it out by repeatedly averaging the form with its pullback by $f$. However, I don't know how to force convergence.

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This is in general a pretty hard question. Try googling for "SRB measures" and "physical measures", there are lots of results and open questions. –  Lukas Geyer Nov 1 '12 at 22:54
Ah, thanks. Is there a specific counterexample? –  Perce Nov 2 '12 at 0:12

## 2 Answers

This is a very tough question. I don't think that you are likely to find useful necessary and sufficient conditions for the existence of an invariant volume form, especially if you require your volume forms to be smooth instead of just, say, absolutely continuous. There is one pretty general theorem that comes to mind that seems related to your topological criterion.

Let $f\colon M\to M$ be a $C^k$-smooth map, with $k\geq 2$. Suppose that there exists a Riemannian metric on $M$ with respect to which $f$ is expanding. That is, there exists a $\chi>1$ such that $\|Df(v)\|\geq \chi\|v\|$ for all tangent vectors $v$ on $M$. Then $f$ has a unique absolutely continuous invariant measure $\mu$, and this measure is strictly positive. Moreover, $\mu$ is at least $C^{k-1}$-smooth. In particular, if $f$ is $C^\infty$, the measure $\mu$ is a smooth volume form.

I believe the theorem remains true when $k = 1$ as well, except that $\mu$ may no longer be strictly positive. I could be wrong about that though. The expanding condition definitely implies your topological condition, so in that way this is a weaker statement than what you are hoping for. Then again, I don't think the statement you are hoping for is actually true. For a reference to the above theorem, see the article The measures invariant under an expanding map by Richard Sacksteder.

A word about your idea for finding an invariant volume form: it can be pushed a little further. Suppose $\omega$ is the volume form you start with, normalized to give you a probability measure $\mu$. Let $\mu_n$ be the measure $$\mu_n = \frac{1}{n}\sum_{k=0}^{n-1} f^{k}_*\mu.$$ This is a probability measure for each $n\geq 1$, but, as you say, there is no guarantee that the $\mu_n$ converge to anything, much less a volume form. However, because the weak topology on the space of Radon probability measures on $M$ is compact, there is some subsequence $\mu_{n_k}$ that converges weakly to a probability measure $\nu$. This $\nu$ is easily seen to be invariant. Of course, it will usually not be a volume form.

One final comment as far as conditions on the existence of an invariant volume form go: one thing that must be true is that almost every point $x\in M$ must be recurrent. This is a consequence of the Poincaré recurrence theorem.

Edit: I missed that you are looking at a diffeomorphism. There are no expanding diffeomorphisms of compact manifolds, so the above theorem would not apply.

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In 1932 Denjoy constructed examples of $C^1$ circle diffeomorphisms that are topologically semi-conjugate to an irrational rotation, hence uniquely ergodic, for which the unique invariant probability measure is supported on a Cantor set on the circle. With higher regularity this can't happen anymore, but even for real-analytic maps the measure can be singular w.r.t. Lebesgue measure, as shown by V.I.Arnold.

For torus maps the situation gets infinitely worse...

EDIT: I just realized that the Denjoy examples actually have an open set $U$ such that $f(U)$ is a proper subset of $U$. Anyway, the Arnold examples are honestly topologically conjugate to an irrational rotation, so for them such a set $U$ does not exist.

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