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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 $\frac{m(3m+11)}{2}$ or $\frac{m(m+1)(3m+11)}{2}$ 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 $\frac{m(m+1)}{2}$ 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 $\frac{m(m+1)}{2}$ space for certain $m$.

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    $\begingroup$ it seems that the Poincare disk with constant negative curvature cannot be embedded into $\mathbb{R}^3$. $\endgroup$
    – orangeskid
    Oct 15, 2017 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, 2017 at 20:51
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    $\begingroup$ 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, 2017 at 20:57
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    $\begingroup$ I understood what you meant, I am just curious for a proof that no topological embedding exists. $\endgroup$
    – Eduardo Longa
    Oct 16, 2017 at 20:59
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    $\begingroup$ 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, 2017 at 22:31

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