Timeline for Nash embedding theorem for manifolds with boundary
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2022 at 21:59 | comment | added | Ben McKay | If I remember correctly, there was some concern with Nash's proof in recent years, but the Gunther proof (made famous partly by Deane) is widely accepted. | |
| Dec 18, 2022 at 19:15 | answer | added | Deane Yang | timeline score: 8 | |
| Dec 18, 2022 at 17:32 | comment | added | Anton Petrunin | @DeaneYang could you make an answer from your comment (so the question would disappear from unanswered). | |
| Nov 4, 2019 at 10:52 | comment | added | Sebastian Goette | For applications, one might want the manifold to lie in a halfspace with the boundary of $M$ mapping to the boundary of that halfspace. Is that possible as well? | |
| Nov 3, 2019 at 20:56 | comment | added | Deane Yang | @RyanBudney, that's a good point. You can indeed do it, but, in doing so, you're effectively showing that the standard proof of the Nash theorem for an open manifold is easily adapted to a manifold with boundary. | |
| Nov 2, 2019 at 20:02 | comment | added | Ryan Budney | Or do the reverse of what Deane suggests: remove the boundary, embed that Riemann manifold, then check to see if you can ensure that extends to the manifold with boundary. | |
| Nov 2, 2019 at 20:01 | comment | added | Deane Yang | I believe it's straightforward to embed $M$ smoothly into an open manifold $N$ (maybe by attaching an open collar to $\partial M$?). It is then straightforward to extend the Riemannian metric smoothly to $N$. Now you can apply that Nash theorem to $N$. | |
| Nov 2, 2019 at 19:15 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Nov 2, 2019 at 19:11 | history | asked | Ryan Vaughn | CC BY-SA 4.0 |