Timeline for Does the mean curvature flow naturally come with less applications than intrinsic curvature flows?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 7, 2019 at 7:36 | answer | added | slcvtq | timeline score: 11 | |
| Nov 28, 2019 at 18:35 | vote | accept | Matheus Andrade | ||
| Nov 28, 2019 at 17:35 | comment | added | Terry Tao | Related: mathoverflow.net/questions/266292/… | |
| Nov 28, 2019 at 9:31 | comment | added | Jonny Evans | The following isn't about MCF (so comment rather than an answer, sorry!), but is an example where an 'extrinsic' geometric flow has topological applications. A Kaehler manifold X can't have the same \pi_1 as a hyperbolic manifold M of dimension bigger than 2. Proof: if it did, apply harmonic map flow to the map X -> M which classifies the fundamental group; as M is negatively curved, you get a harmonic map, which then has to have 2-dimensional image by a rank estimate. This argument is due to Carlson and Toledo, building on the Eells-Sampson theory of harmonic maps and associated flow. | |
| Nov 28, 2019 at 3:57 | answer | added | Robert Haslhofer | timeline score: 18 | |
| Nov 28, 2019 at 3:10 | history | edited | Matheus Andrade | CC BY-SA 4.0 |
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| Nov 28, 2019 at 2:40 | comment | added | Matheus Andrade | I know that mean curvature isn't completely far from topology, since a sphere for example can't be embedded in a way such that the mean curvature is negative everywhere. But this is a pretty weak result and I can't think of anything else like that right now | |
| Nov 28, 2019 at 2:39 | comment | added | Matheus Andrade | @JonnyEvans My reasoning is that the mean curvature flow doesn't care about topology, only about mean curvature, and surfaces can be locally isometric (so that they have the same mean curvature) yet not homeomorphic. While sectional curvature, Ricci curvature and scalar curvature are all intrinsic, so in my head it would be much easier to obtain topological information from there rather than something that depends on ambient space. Of course, I don't trust my intuition and don't know whether this is right. So I'd really like to see some topological results using MCF. | |
| Nov 27, 2019 at 22:06 | comment | added | Jonny Evans | Of course, proving the nearby Lagrangian conjecture that way is currently out of reach and may not even work, but I just wanted to illustrate that there's no a priori reason that you can't prove topological results using mean curvature flow. The Thomas-Yau/Joyce conjectures (on special Lagrangians and Bridgeland stability conditions for Fukaya categories) have a similar flavour and the expectation seems to be that a better understanding of mean curvature flow with surgeries should help resolve these conjectures. | |
| Nov 27, 2019 at 22:03 | comment | added | Jonny Evans | "mean curvature flow, just like any other extrinsic curvature flow, can't ever be used to get topological results about our manifolds, for example. Is this correct?" What makes you think this? How could one show that something "can't ever be used"? For example, you could attempt to prove the nearby Lagrangian conjecture by using Lagrangian mean curvature flow to isotope an exact Lagrangian in a cotangent bundle to the zero section. This would show (a fortiori) that nearby Lagrangians are diffeomorphic to the zero-section. | |
| Nov 27, 2019 at 3:16 | answer | added | RBega2 | timeline score: 9 | |
| Nov 26, 2019 at 22:47 | history | asked | Matheus Andrade | CC BY-SA 4.0 |