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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?

Ian Gershon Teixeira
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    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. – Deane Yang Dec 21 '21 at 19:51
  • @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? – Ian Gershon Teixeira Dec 21 '21 at 20:39
  • 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. https://math.stackexchange.com/questions/1528046/can-a-hyperbolic-surface-be-isometrically-embedded-into-mathbb-r4 – Deane Yang Dec 21 '21 at 22:07
  • @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. – mme Dec 21 '21 at 22:13
  • I meant an optimal bound. Nash'stheorem shows that any Riemannian manifold can be isometrically embedded into Euclidean space of high enough dimension. Gunther gave a simpler proof. – Deane Yang Dec 21 '21 at 22:16
  • Is there a compact 3 manifold that saturates the bound $ R^{14} $ (for example round projective plane saturates the bound $ R^5 $ for 2 manifolds)? Or is it unclear whether or not the bound is tight? Are there certain 3 manifolds that we can definitively say cannot be isometrically embedded in $ R^7 $? I'm looking for bound information like that.// I'll update the bound in the question thanks @mme – Ian Gershon Teixeira Dec 21 '21 at 22:22
  • @Deane With better dimension bounds by a factor of about 1/2, I thought, which is what I was trying to communicate: OP's bound is R^{27}. Obviously you are more expert than I; I pulled this statement and the attribution to Gunther from Thm 1.0.3 of "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces". – mme Dec 21 '21 at 22:23
  • @mme, so the question is whether there is a better but not necessarily optimal bound for nice $3$-manifolds? To be honest, I haven't looked carefully at this in a long time, so I'm not necessarily more expert. My memory is only that Gromov improved the bounds on the dimension and it might be worth looking at his book. – Deane Yang Dec 21 '21 at 22:33
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    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. – Joseph O'Rourke Dec 21 '21 at 23:52

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