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.

  • $\begingroup$ 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 Aug 1 '16 at 22:07
  • 1
    $\begingroup$ 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 Aug 1 '16 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 Aug 1 '16 at 22:45
  • 1
    $\begingroup$ 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 Aug 1 '16 at 22:48
  • $\begingroup$ @WillO where did you get these numbers from? $\endgroup$ – Blazej Aug 1 '16 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.