# Is there any physical interpretation of Nash embedding theorem?

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.

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.
• 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 1 '16 at 22:30
• 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 1 '16 at 22:48