Ricci-invariant class of metrics - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:25:17Z http://mathoverflow.net/feeds/question/105573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105573/ricci-invariant-class-of-metrics Ricci-invariant class of metrics ε-δ 2012-08-26T20:24:20Z 2012-08-28T08:29:56Z <p>Imagine that there is a class of Riemannian metrics $\mathcal{R}$ on 3-dimensional manifolds such that</p> <ol> <li>$\mathcal{R}$ is locally finite dimensional; i.e., there are finite number of real parameters which describe the metric locally at any point;</li> <li>Any closed 3-dimensional manifold admits a metric from $\mathcal{R}$;</li> <li>$\mathcal{R}$ is is invariant with respect to Ricci flow.</li> </ol> <p>Likely <em>such $\mathcal{R}$ does not exist</em>. Otherwise there is a good chance to simplify the Perelman's proof.</p> <blockquote> <p>Can it be proved that such $\mathcal{R}$ does not exist?</p> </blockquote> <p>UPDATE: You need to assume that metric depend continuously on the real parameters and $\mathcal{R}$ is closed in a reasonable topology. </p> http://mathoverflow.net/questions/105573/ricci-invariant-class-of-metrics/105652#105652 Answer by Robert Haslhofer for Ricci-invariant class of metrics Robert Haslhofer 2012-08-27T19:21:55Z 2012-08-28T08:29:56Z <p>EDIT: As Misha and Deane pointed out, the question is not terribly well posed. So I will interpret it somewhat broadly to better reflect our actual understanding of canonical metrics on 3-manifolds and the Ricci flow. </p> <p>By Perelman's proof of Thurston's geometrization conjecture, you can decompose any closed 3-manifold into geometric pieces. There are 8 types of geometric pieces, and they behave very simply under Ricci flow (they move just by scaling, etc). So the geometrization gives you essentially the class of metrics that you are asking for, with the constraint that you have to decompose your manifold into pieces first.</p> <p>I don't think that there is a natural class of metrics, without decomposing the manifold into pieces first. In any case, it seems unlikely that the existence of such a class of metrics would yield to a simplification of Perelman's proof: E.g. a posteriori we know that every simply connected closed 3-manifold admits a positive Einstein metric. But of course, proving that means exactly proving the Poincare conjecture (as Perelman did).</p>