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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.

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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – Physics Guy
    Commented Aug 1, 2016 at 22:07
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    $\begingroup$ 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. $\endgroup$
    – Blazej
    Commented Aug 1, 2016 at 22:30
  • $\begingroup$ Ok, thanks. But, in that way, the Einsteinian Manifold could be embedded in euclidian space, even if not isometrically ? $\endgroup$
    – Physics Guy
    Commented Aug 1, 2016 at 22:45
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    $\begingroup$ Any Lorentzian manifold of signature $(1,3)$ can be embedded in ${\mathbb R}^{252,252}$ (that is, ${\mathbb R}^{504}$ with signature $(252,252)$). $\endgroup$
    – WillO
    Commented Aug 1, 2016 at 22:48
  • $\begingroup$ @WillO where did you get these numbers from? $\endgroup$
    – Blazej
    Commented Aug 1, 2016 at 22:49

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