Timeline for Is there any physical interpretation of Nash embedding theorem?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 1, 2016 at 22:51 | comment | added | WillO | The general bound is $(n(2n+1)(2n+6)/2,n(2n+1)(2n+6)/2$ for a Lorentzian manifold of dimension $n$. | |
| Aug 1, 2016 at 22:50 | comment | added | WillO | @Blazej: I haven't double checked but I'm pretty sure you can find them here: ams.org/journals/bull/1969-75-06/S0002-9904-1969-12407-9/… | |
| Aug 1, 2016 at 22:49 | comment | added | Blazej | @WillO where did you get these numbers from? | |
| Aug 1, 2016 at 22:48 | comment | added | WillO | Any Lorentzian manifold of signature $(1,3)$ can be embedded in ${\mathbb R}^{252,252}$ (that is, ${\mathbb R}^{504}$ with signature $(252,252)$). | |
| Aug 1, 2016 at 22:45 | comment | added | Physics Guy | Ok, thanks. But, in that way, the Einsteinian Manifold could be embedded in euclidian space, even if not isometrically ? | |
| Aug 1, 2016 at 22:30 | comment | added | Blazej | Yes it is. However it is not a Riemannian manifold so Nash's theorem doesn't strictly apply. There is Whitney's theorem which says that every manifold can be embedded in some $\mathbb R^n$, but it says nothing about the question whether the embedding can be made isometric. | |
| Aug 1, 2016 at 22:07 | comment | added | Physics Guy | Ok, I did not know that it would be relevant if its a Riemannian or an Einsteinian manifold. But a pseudo-riemannian-manifold is still a differentiable manifold (with a tensorial function defined over it), or isnt it ? | |
| Aug 1, 2016 at 21:59 | history | answered | Blazej | CC BY-SA 3.0 |