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Certain non-flat, 2D metrics can be visualized as a 3D surface. The metric for the surface of the unit sphere, $$ds^2 = d\theta^2+\sin^2\theta\,d\phi^2,$$ would be the most familiar example. Others are more esoteric: in Martin's General Relativity: A Guide ..., he visualizes the following metric as a infinitely long trumpet-shaped surface: $$ds^2=\frac{1}{r^2}dr^2 + r^2d\phi^2,$$

What determines whether or not a 2D metric can be described by a 3D surface?

Doubt
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  • https://en.wikipedia.org/wiki/Nash_embedding_theorem – Kosm May 08 '17 at 22:58
  • The canonic example: a flat torus has no 3d isometric embedding. The nash embedding theorem statss that every differentiable riemannian (resp pseudoriemannian) n-dimensional manifold can be embedded isometrically into 2n dimensional eucliean (resp minkowski) space. –  May 08 '17 at 23:00
  • So which is it: according to this theorem, is there, or is there not a way to represent a 2D metric, such that its geodesics are described by geodesics on a 3D surface? – Doubt May 09 '17 at 01:58
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    @Qmechanic I should think it would. – E.P. May 15 '17 at 12:44
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    @NeuroFuzzy, you may be right on that; the embedding is "only" $C^1$. The authors say that "the normal vector exhibits a fractal behaviour" which would perhaps be undesirable for physics applications. – typesanitizer May 15 '17 at 12:46
  • It is certainly not an easy question. I would imagine that the Poincaré conjecture takes care of the surfaces with genus $0$. For higher genus surfaces I don't think that the question can be answered that simply. – asdas May 17 '17 at 00:36
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    Would [math.se] be a better home for this question? See also Phys.SE chat discussion. – Qmechanic May 09 '17 at 02:40
  • @theindigamer Interesting! I would think some physically important smoothness condition is violated in doing that though? –  May 15 '17 at 03:19
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    @NeuroFuzzy, your flat torus example doesn't seem correct: http://m.pnas.org/content/109/19/7218 – typesanitizer May 15 '17 at 02:04
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    It depends on the degree of smoothness you require: If it is $C^1$ then there are no restrictions on the metric (Nash-Kuiper's theorem). Otherwise ($C^2$-smooth), there is no known complete answer, only partial results, e.g. every positively curved metric embeds isometrically in $R^3$. – Moishe Kohan May 24 '17 at 02:36

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