When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:31:13Z http://mathoverflow.net/feeds/question/27511 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27511/when-is-there-a-natural-riemannian-metric-whose-measure-preserves-a-self-diffeomo When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? Steve Huntsman 2010-06-08T19:52:32Z 2010-10-13T15:18:50Z <p>Let $M$ be a compact Riemannian manifold with metric $g$ and let $f \in Diff(M)$. </p> <p>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?</p> http://mathoverflow.net/questions/27511/when-is-there-a-natural-riemannian-metric-whose-measure-preserves-a-self-diffeomo/27517#27517 Answer by rpotrie for When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? rpotrie 2010-06-08T20:48:25Z 2010-06-08T20:48:25Z <p>An important necessary condition is that on periodic points, the determinant in the period must be one (that is not hard).</p> <p>I believe that what you ask has to do with <a href="http://books.google.com/books?id=9nL7ZX8Djp4C&amp;pg=PA189&amp;dq=katok+hasselblat+theorem+5.1.13&amp;hl=es&amp;ei=FKsOTPDFPJmisQaW14nyCA&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCoQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">Theorem 5.1.13</a> 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).</p> <p>In some cases (as for example expanding maps, or Anosov diffeos with the condition on periodic points), smooth measures can be constructed. </p> <p>Hope it helps. </p> http://mathoverflow.net/questions/27511/when-is-there-a-natural-riemannian-metric-whose-measure-preserves-a-self-diffeomo/27520#27520 Answer by coudy for When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? coudy 2010-06-08T21:04:28Z 2010-06-09T07:35:53Z <p>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 $x\in Fix(f^n)$. </p> <p>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).</p> <p>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.</p> <p>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 $g^*\Omega_0=\Omega_1$.</p> http://mathoverflow.net/questions/27511/when-is-there-a-natural-riemannian-metric-whose-measure-preserves-a-self-diffeomo/38129#38129 Answer by Martin for When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? Martin 2010-09-09T01:39:44Z 2010-09-09T01:39:44Z <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/27511/when-is-there-a-natural-riemannian-metric-whose-measure-preserves-a-self-diffeomo/42028#42028 Answer by Stephen Miller for When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? Stephen Miller 2010-10-13T15:18:50Z 2010-10-13T15:18:50Z <p>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).</p> <p>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 $F$.</p> <p>A good reference is Bredon: Compact Tranformation Groups</p>