Timeline for Quantitative upper bound on mean curvature of an isometric embedding
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 5, 2019 at 8:55 | history | edited | Raziel | CC BY-SA 4.0 |
added 205 characters in body
|
| Jan 5, 2019 at 8:30 | comment | added | Raziel | This was specified in the second comment above: by mean curvature I mean the norm of the trace of the second fundamental form of the embedding. I will edit my post to state it clearly. | |
| Jan 5, 2019 at 0:34 | comment | added | Deane Yang | The wording of your question seems to assume that mean curvature is a scalar function. Note that in fact the mean curvature of a submanifold with codimension greater than $1$ is a normal vector field. | |
| Jan 5, 2019 at 0:05 | comment | added | Deane Yang | I think the answer is already provided in previous MathOverflow discussion. What's not mentioned there is that the second deformation step is probably best addressed using Gunther's proof, which is far simpler than previously known proofs. Tao (terrytao.wordpress.com/2016/05/11/…) has a nice description of the entire proof. You just need to verify that at each step, you can bound the second fundamental form in terms of the curvature, injectivity radius, and any prior bound on the second fundamental form. This is messy but should be straightforward. | |
| Jan 4, 2019 at 21:59 | answer | added | RBega2 | timeline score: 2 | |
| Jan 4, 2019 at 21:26 | comment | added | Deane Yang | Have you looked at Gromov’s book Partial Differential Relations? It’s possible that it addresses this or similar questions. | |
| Jan 4, 2019 at 21:02 | history | edited | Raziel | CC BY-SA 4.0 |
deleted 2 characters in body
|
| Jan 4, 2019 at 20:06 | comment | added | RBega2 | This is not my area of expertise, but it seems that the fact that you can isometrically embedded $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ in a way that has arbitrarily large mean curvature suggests you are going to have to essentially reproduce Nash's proof, but with quantitative estimates. | |
| Jan 4, 2019 at 20:00 | comment | added | Raziel | The second fundamental form of the embedding is exactly what I would like to control (more precisely, the norm of its trace). I tried to clarify a bit the question in light of your comment. | |
| Jan 4, 2019 at 20:00 | history | edited | Raziel | CC BY-SA 4.0 |
Clarified the question after Daene comment
|
| Jan 4, 2019 at 19:49 | history | edited | Raziel | CC BY-SA 4.0 |
Clarified the question after Daene comment
|
| Jan 4, 2019 at 19:16 | comment | added | Deane Yang | Could you say why a bound on the second fundamental form isn’t sufficient for your needs? | |
| Jan 4, 2019 at 18:33 | history | edited | Raziel | CC BY-SA 4.0 |
deleted 7 characters in body
|
| Jan 4, 2019 at 17:46 | history | edited | Raziel | CC BY-SA 4.0 |
edited body
|
| Jan 4, 2019 at 17:37 | history | asked | Raziel | CC BY-SA 4.0 |