# Isometric embedding of a genus g surface

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

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.

• There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr – John Pardon Mar 20 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 at 14:33
• 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 at 14:38
• 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 at 14:45
• @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 at 14:48

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.