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Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $\mathbb{R}^4?$

Sean Lawton
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GAUTAM NEELAKANTAN
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2 Answers2

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The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $g\geq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^\infty$-isometric embedding into $\mathbb{R}^4$).

Since the smallest known $C^\infty$-embedding for the hyperbolic plane is $\mathbb{R}^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $\mathbb{R}^3$ for $r\geq 2$. Later Efimov generalized this to closed hyperbolic surfaces.

I believe these facts and references may be found in:

Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.

Sean Lawton
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    There are visualizations of the Nash--Kuiper embeddings here: http://hevea-project.fr/ – John Pardon Mar 20 '19 at 14:30
  • @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $g\geq 2$ that are not presently visualized (as far as I know). – Sean Lawton Mar 20 '19 at 14:33
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    That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all. – John Pardon Mar 20 '19 at 14:38
  • @JohnPardon I was trying to contrast the situation for $g=1$ and $g\geq 2$ (minimal smooth embedding vs visualization of $C^1$-embedding). Sorry if it confused you. I made an edit that hopefully makes my intent more clear. Thanks for the comment. – Sean Lawton Mar 20 '19 at 14:40
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    There isn't really any difference between the $g=1$ and $g\geq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw) – John Pardon Mar 20 '19 at 14:45
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    @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done). – Sean Lawton Mar 20 '19 at 14:48
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I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^{\infty}$ isometric embedding $V \to \mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303.

Michael Albanese
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