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Jul 9, 2019 at 15:41 comment added Willie Wong @011000: there is no relation between the $C^1$ embedding (Nash-Kuiper) and the construction by Sabitov. Sabitov sets out an explicit ansatz of a generalized surface of revolution, and essentially just solved the equations by hand and checked the regularity by hand.
Jul 9, 2019 at 15:11 comment added Willie Wong As of the very comprehensive 2001 survey of Borisenko both questions seem to be open. For Q1 if you replace "embed" by "immerse" then the result holds due to Rozendorn. (The original paper is in Russian (see Borisenko for reference) but the construction is outlined in Rozendorn's 1992 survey.) And Ian Agol already mentioned the result of Sabitov for Q2 which is only $C^{0,1}$ globally but piecewise analytic.
Jul 9, 2019 at 7:33 comment added quarague The embedding of hyperbolic space in $\mathbb{R}^3$ with Minkowski metric is explained on the wikipedia page for hyperbolic space and also in most text books that introduce hyperbolic space.
Jul 9, 2019 at 4:25 comment added 011000 I care more about the smooth embedding and immersion, so I edit the question. I know the Nash embedding theorem, but I never heard how to embed hyperbolic space in R^3 with the Minkowski metric. Could you please explain it more precisely? What's more, are there any relationship between the C^1 embedding in R^3 and the piecewise analytic immersion in R^4? Thanks for every helpful answers.
Jul 9, 2019 at 4:08 history edited 011000 CC BY-SA 4.0
I ask the smooth embedding and immersion, not just C^1.
Jul 8, 2019 at 15:43 comment added Ian Agol A piecewise analytic map (but apparently not smooth) is given here: mathscinet.ams.org/mathscinet-getitem?mr=1025303
Jul 8, 2019 at 9:28 comment added Moishe Kohan See math.stackexchange.com/questions/1528046/… and references therein.
Jul 8, 2019 at 8:51 comment added quarague I suppose you already know that one can embed hyperbolic space in $\mathbb{R}^3$ with the Minkowski metric and that the Nash embedding theorem gives an embedding for any surface into Euclidean space. The dimension bound on wikipedia gives just $n \le 51$ so that is pretty far off the $4$ or $5$ dimensions you are hoping for.
Jul 8, 2019 at 4:15 review First posts
Jul 8, 2019 at 7:20
Jul 8, 2019 at 4:11 history asked 011000 CC BY-SA 4.0