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From the Nash Embedding theorem e.g. see

https://en.wikipedia.org/wiki/Nash_embedding_theorem

an m dimensional Riemannian manifold has an isometric embedding into a dimension of m(3m+11)/2 or m(m+1)(3m+11)/2 depending on if its compact or not.

My questions are: 1) does there always exist an isometric embedding, in the sense of the of the Nash theorem above, of an m dimensional Riemannian manifold into m(m+1)/2 space? 2) Is the first question known to be false for any m?

As far as I'm aware the answer to the first question is probably true because there are specific known examples of m manifolds into m(m+1)/2 space for certain m.

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    $\begingroup$ it seems that the Poincare disk with constant negative curvature cannot be embedded into $\mathbb{R}^3$. $\endgroup$ – Orest Bucicovschi Oct 15 '17 at 21:50
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The answer is no, at least for $m=2$. Indeed, even the (much weaker) Whitney embedding theorem doesn't hold for $m=2$ and $n = \frac{m(m+1)}{2} = 3$. One of examples is the projective plane $\mathbb{R} P^2$, which doesn't admit a smooth embedding into $\mathbb{R}^3$ (as any closed surface in $\mathbb{R}^3$ can be oriented).

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  • $\begingroup$ Does $\mathbb{R}\mathbb{P}^2$ admits a topological embedding into $\mathbb{R}^3$? $\endgroup$ – Eduardo Longa Oct 16 '17 at 20:51
  • $\begingroup$ Also no (although the proof I had in mind takes advantage of the differential structure). I meant to use the word smooth only as an opposition to isometric. $\endgroup$ – Michał Miśkiewicz Oct 16 '17 at 20:57
  • $\begingroup$ I understood what you meant, I am just curious for a proof that no topological embedding exists. $\endgroup$ – Eduardo Longa Oct 16 '17 at 20:59
  • $\begingroup$ If (for some reason) you already know that any representation of the clique $K_6$ in $\mathbb{R}^3$ contains a nontrivial link (i.e. two disjoint subgraphs $K_3$ have nonzero linking number), this impossibility can be shown by noting that on $\mathbb{R}P^2$ one can draw $K_6$ so that any two subgraphs $K_3$ are unlinked, i.e. one of the cirles bounds a disk disjoint with the other circle. Here is a popular article about it, in Polish. $\endgroup$ – Michał Miśkiewicz Oct 16 '17 at 22:31

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