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Nash embedding theorem says that every Riemannian manifold can be (isometrically) embedded into $R^n$. That means that every $RM$ is a sub-manifold to $R^n$.

Since General Relativity is defined on a pseudo-Riemannian manifold and classical theories are defined on a "simple" Euclidean space, I want to ask what the embedding theorem means for the relation between GR and classical physics.

Qmechanic
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Physics Guy
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  • The embedding is not unique, though, and it would change dynamically (including the necessary embedding dimension!) as the manifold changes. How do you interpret that? – CuriousOne Aug 01 '16 at 21:59
  • Related: https://physics.stackexchange.com/q/8932/2451 , https://physics.stackexchange.com/q/267916/2451 and links therein. – Qmechanic Feb 08 '19 at 09:36

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Spacetime in General Relativity is not Riemannian so surely it can't be embedded isometrically in $\mathbb R^n$. I suppose it might be possible to embed it in some $\mathbb R^n$ with different signature. However, I don't see any relevance. This is merely a mathemathical fact. Our world is not a differential manifold after all.

Blazej
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  • 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 ? – Physics Guy Aug 01 '16 at 22:07
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    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. – Blazej Aug 01 '16 at 22:30
  • Ok, thanks. But, in that way, the Einsteinian Manifold could be embedded in euclidian space, even if not isometrically ? – Physics Guy Aug 01 '16 at 22:45
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    Any Lorentzian manifold of signature $(1,3)$ can be embedded in ${\mathbb R}^{252,252}$ (that is, ${\mathbb R}^{504}$ with signature $(252,252)$). – WillO Aug 01 '16 at 22:48
  • @WillO where did you get these numbers from? – Blazej Aug 01 '16 at 22:49
  • @Blazej: I haven't double checked but I'm pretty sure you can find them here: http://www.ams.org/journals/bull/1969-75-06/S0002-9904-1969-12407-9/S0002-9904-1969-12407-9.pdf – WillO Aug 01 '16 at 22:50
  • The general bound is $(n(2n+1)(2n+6)/2,n(2n+1)(2n+6)/2$ for a Lorentzian manifold of dimension $n$. – WillO Aug 01 '16 at 22:51