Does Ricci flow with surgery come from sections of a smooth Riemannian manifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:20:05Z http://mathoverflow.net/feeds/question/35584 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35584/does-ricci-flow-with-surgery-come-from-sections-of-a-smooth-riemannian-manifold Does Ricci flow with surgery come from sections of a smooth Riemannian manifold? Richard Borcherds 2010-08-14T17:03:55Z 2010-08-14T18:34:45Z <p>More precisely, is Ricci flow with surgery on a 3-dimensional Riemannian manifold M given by the "constant-time" sections of some canonical smooth 4-dimensional Riemannian manifold? </p> <p>There would be a discrete set of times corresponding to the surgeries, but the 4-dimensional manifold might still be smooth at these points even though its sections would have singularities. The existence of such a 4-manifold is well known if there are no singularities: the problem is whether one can still construct it in the presence of singularities. </p> <p>Background: Ricci flow on M in general has finite time singularities. These are usually dealt with by a rather complicated procedure, where one stops the flow just before the singularities, then carefully cuts up M into smaller pieces and caps off the holes, and constructs a Riemannian metric on each of these pieces by modifying the metric on M, and then restarts the Ricci flow. This seems rather a mess: my impression is that it involves making several choices so is not really canonical, and has a discontinuity in the metric and topology on M. If the flow were given by sections of a canonical smooth 4-manifold as in the question this would give a cleaner way to look at the surgeries of Ricci flow. </p> <p>(Presumably if the answer to the question is "yes" this is not easy to show, otherwise people would not spend so much time on the complicated surgery procedure. But maybe Ricci flow experts know some reason why this does not work.)</p> http://mathoverflow.net/questions/35584/does-ricci-flow-with-surgery-come-from-sections-of-a-smooth-riemannian-manifold/35586#35586 Answer by Agol for Does Ricci flow with surgery come from sections of a smooth Riemannian manifold? Agol 2010-08-14T17:59:34Z 2010-08-14T17:59:34Z <p>This is an open question, as far as I know. Perelman makes a comment in one of his papers to the effect that he would like to achieve some sort of canonical Ricci flow-with-surgery in space-time (see section 13.2 of <a href="http://front.math.ucdavis.edu/0211.5159" rel="nofollow">his first paper</a>). There are several unresolved issues having to do with the formation of the singularities in Ricci flow that make this question difficult. On the other hand, recently Angenent, Knopf, and Caputo have shown that one may do a canonical surgery in the rotationally symmetric case: <a href="http://www.ma.utexas.edu/users/danknopf/" rel="nofollow">Minimally invasive surgery for Ricci flow singularities.</a> There are several simplifications in the rotationally symmetric case which makes their approach possible: the Ricci flow is reduced to a (coupled) ODE, so one may apply the maximum principle to show that there are finitely many singularities, and analyze the asymptotics. Since the Ricci flow becomes rotationally symmetric near a singularity in dimension 3 to first order, there is some hope that their approach should work without the rotationally symmetric hypothesis. </p> http://mathoverflow.net/questions/35584/does-ricci-flow-with-surgery-come-from-sections-of-a-smooth-riemannian-manifold/35588#35588 Answer by Rbega for Does Ricci flow with surgery come from sections of a smooth Riemannian manifold? Rbega 2010-08-14T18:24:07Z 2010-08-14T18:34:45Z <p>To build a little on what Agol said:</p> <p>For mean curvature flow of hypersurfaces, the analogous question is at least partially known to be true (sort of). The advantage of the mean curvature flow over the Ricci flow in this case is that there are already are good notions of weak solution--that is solutions defined through singularities but that agree with classical solutions when the latter exist. The two most important are the Brakke flow and the level set flow. In principal the latter does exactly what you want, namely given a $\Sigma_0$ a closed hypersurface in $\mathbb{R}^{n+1}_x$ there is a hypersurface'' $\mathcal{M}$ in $\mathbb{R}^{n+1}\times \mathbb{R}^{\geq 0}$. So that $\partial \mathcal{M}=\Sigma_0 \subset \mathbb{R}^{n+1} \times 0$ and each level set $\lbrace x_{n+2}=t \rbrace \cap \mathcal{M}$ can be interpreted as the flow of $\Sigma_0$ at time $t$ (indeed if $\Sigma_0$ is smooth these level sets agree with the usual flow up to the first singular time, thereafter they continue to exist while the classical flow ceases to make sense). It should be pointed out that in principal $\mathcal{M}$ is quite singular and there are some subtleties about when the level sets have non-empty interior. If you are willing to start with a mean convex $\Sigma_0$, White has shown that the latter never occurs and that (I believe) the levels are almost everywhere smooth.</p> <p>On the other hand the surgery question for mean curvature flow is a bit trickier. Huisken and Sinestrari give such a surgery when (if I recall correctly) $n\geq 3$ and the initial surface is 2-convex (i.e. the sum of the lowest two principal curvatures is positive...a stronger condition than mean convex). Recently, there were a couple of papers on the arxiv where it was shown that if you took the surgery times (of the H-S surgery procedure) closer and closer to the singular time then you you would limit to the level set flow. </p> <p>This seems to be the flavor of what you are looking for, though it should be pointed out that the regularity of the higher dimensional guy $\mathcal{M}$ is somewhat unclear in general. </p>