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The Nash embedding theorem tells us that every smooth Riemannian m-manifold can be embedded in $R^n$ for, say, $n = m^2 + 5m + 3$ (edit: 14 is a better bound for compact 3 manifolds thanks @mme). What can we say in the special case of 3-manifolds? For example, can we always embed a 3-manifold in $R^7$? (I believe $R^5$ is the best you can do for 2-manifolds, so that would just be the pattern $n=2m+1$).

Is the bound any tighter if we have nice manifolds? Like asking the manifold to be compact? Or compact and constant curvature?

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    $\begingroup$ This is a very difficult question answer, even for 2-manifolds and 3-manifolds. There are some partial results in Gromov's book, Partial Differential Relations. $\endgroup$
    – Deane Yang
    Commented Dec 21, 2021 at 19:51
  • $\begingroup$ @DeaneYang It occurred to me the question might be too hard. Do you think I might get more useful replies if I specialize the question to really nice manifolds, for example constant curvature? $\endgroup$
    – Ian Gershon Teixeira
    Commented Dec 21, 2021 at 20:39
  • $\begingroup$ A $3$-manifold with constant sectional or Ricci curvature is flat $3$-space, a sphere, or hyperbolic space. You know the answer for the first two. I don't believe the answer is known even for hyperbolic $3$--space. The answer is unknown even for hyperbolic $2$-space. math.stackexchange.com/questions/1528046/… $\endgroup$
    – Deane Yang
    Commented Dec 21, 2021 at 22:07
  • $\begingroup$ @Deane Surely you mean "is locally..." // OP: Gunther's version of Nash embedding gives you an isometric embedding of a compact 3-manifold into R^14. $\endgroup$
    – mme
    Commented Dec 21, 2021 at 22:13
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    $\begingroup$ Deane references Gromov's book, and I followed up, in Nash embedding theorem for 2D manifolds. See also Bill Thurston's insightful posting. All this: 2D, not 3D, manifolds. $\endgroup$
    – Joseph O'Rourke
    Commented Dec 21, 2021 at 23:52

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