# When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism?

Let $M$ be a compact Riemannian manifold with metric $g$ and let $f \in Diff(M)$.

Under what circumstances is there a natural metric $g_f$ s.t. the associated smooth measure $\nu_f$ is preserved by $f$, and how can such a $g_f$ be obtained?

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What is the role of the original metric $g$ in your question? And what is the meaning of "natural"? –  Deane Yang Jun 8 '10 at 19:56
@Deane--The original metric is just mentioned to distinguish it from the putative $g_f$. –  Steve Huntsman Jun 8 '10 at 20:36
@Will--For background: arxiv.org/pdf/0804.0167 –  Steve Huntsman Jun 8 '10 at 20:38
Steve, as I comment below, there are many different metrics with a given measure. All you have to do is start with arbitrary metric $g$ with volume measure $\mu$ and define $g_f = g(\nu/\mu)^{2/n}$. Again, what do you mean by "natural"? –  Deane Yang Jun 9 '10 at 12:00
Why is pulling back $g$ using a diffeomorphism constructed using Moser's trick more natural or better than just changing $g$ by a conformal factor? Given the original metric $g$, the conformal factor is uniquely determined (which fits my definition of "natural"), but the diffeomorphism that pulls back $\nu$ to $\nu_f$ is far from unique. –  Deane Yang Jun 9 '10 at 16:39

Let $\Omega$ be the standard volume on your Riemannian manifold, and $\phi$ a smooth function on M. A quick computation shows that $e^\phi \Omega$ is invariant by f if and only if the following cohomological equation is satisfied: $$\phi(f^{-1}(x))-\phi(x)=log\ Jf(x)$$ where Jf is the jacobian of f. This implies for example that $Jf^n(x)=1$ for all $x\in Fix(f^n)$.

This later condition is in fact sufficient for C2 transitive Anosov diffeomorphisms (see e.g. Katok-Hasselblatt th 19.2.7). For these diffeos, this is also equivalent to saying that the SRB measure for f and the SRB measure for the inverse of f are equal (this is interesting because transitive Anosov diffeos always admit SRB measures, but of course not always smooth invariant measures).

As pointed out by Deane, any volume form can be realised as the volume associated to a Riemannian metric. Embed your manifold M in R^n, extend your volume form $f dvol_{eucl}$ to a neighborhood of the manifold, then take the restriction of the metric $f^{2/m}g_{eucl}$ to M . In fact, it is even possible to find a smooth conjuguate of f that preserves any given riemannian volume: this is the Moser trick.

Let M be a compact riemannian manifold, $\Omega_0$ and $\Omega_1$ two volume forms with the same volume. Then there is a diffeo g such that $g^*\Omega_0=\Omega_1$.

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Although as I've said I am principally concerned with the metric, you've hit upon my underlying physical motivation, viz. time-reversibility. Gallavotti has a series of papers in which he motivates the SRB and Liouville measures simultaneously, and I'm working in that context. –  Steve Huntsman Jun 8 '10 at 21:18
Then I think you are looking for the Moser trick (th 5.1.27 in KH). Let me update my answer. –  user6129 Jun 8 '10 at 21:35
Cool, sounds like: given $\nu$ and $\nu_f$, use Moser to get $h_f$ s.t. $\nu_f = h_f^*\nu$. Then set $g_f = h_f^*g$, right? –  Steve Huntsman Jun 8 '10 at 22:18
@Steve. Oups, as pointed out by Deane, I was wrong by assuming that there are volume forms that are not related to a Riemannian metric. I edited my answer. –  user6129 Jun 9 '10 at 7:37

An important necessary condition is that on periodic points, the determinant in the period must be one (that is not hard).

I believe that what you ask has to do with Theorem 5.1.13 of Katok-Hasselblatt's book. There it defines the Jacobian with respect to the original $g$ and shows that if $Jf^n(x)$ is bounded uniformly on $n$ and $x$ then there is an invariant absolutely continuous measure (however, the density needs not be smooth).

In some cases (as for example expanding maps, or Anosov diffeos with the condition on periodic points), smooth measures can be constructed.

Hope it helps.

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Thanks, it's nice to see this. But I'm really wondering about the metric more than the measure. –  Steve Huntsman Jun 8 '10 at 21:14
Steve, if you want $f$ to be only measure-preserving and not an isometry, then once you have found an invariant measure, any metric whose volume form is equal to that measure meets your criteria. And it is easy to construct lots of metrics with a given volume form. –  Deane Yang Jun 9 '10 at 3:10
@Deane: You are right. The problem here is that one does not know if the density is even continuous, so, the metric can not be changed as easily. There is however a theorem by Oxtoby-Ulam (which can be seen as a continuous counterpart to moser´s theorem) which states that if a homeomorphism preserves a non atomic probability measure which is positive in open sets, then it is conjugated to a homeomorphism preserving Lebesgue measure. –  rpotrie Jun 9 '10 at 17:55
@rpotrie: You're right about the regularity. I honestly know nothing about how to get an invariant measure with any given regularity (continuous, differentiable, or something else). I'm just observing that once you know what measure you want, the construction of a metric with the same regularity as the measure is completely straightforward. –  Deane Yang Jun 9 '10 at 18:11

Right, so your question really is: what diffeos preserve a measure equivalent to the volume (ok, with smooth Radon Nikodym derivative, if you're talking smooth metrics). This is because any volume form can be obtained from another by multiplying it with a constant.

Now, what is known in general is that a C^1 generic diffeo does not admit any invariant measure abs. cont. w.r.t. the volume (Avila-Bochi). In particular there is no smooth invariant volume measure. I have thought a LOT about how to do the C^r argument, but without a bit of success.

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A similar question occurs in the theory of topological groups. Given a measure $\mu$ on a $G$-space X, can one use $\mu$ to construct an $equivariant$ measure? The answer is yes if $G$ is compact: you define $\bar{\mu}(U)$ by integrating $\mu(gU)$ over $g\in G$, with respect to the (essentially unique) equivariant measure on $G$. If $G$ is finite, that's just the average over $g\in G$ (you usually assume the measure on $G$ integrates to one).
In your case, you could consider the subgroup of $Diff(M)$ generated by your diffeomorphism $f$. If this is compact, or more likely if $f$ lies in any compact $F\subset Diff(M)$, you should be able to define an integral of your metric $g$ over $F$.