5
$\begingroup$

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

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr $\endgroup$ – John Pardon Mar 20 at 14:30
  • $\begingroup$ @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). $\endgroup$ – Sean Lawton Mar 20 at 14:33
  • 1
    $\begingroup$ 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. $\endgroup$ – John Pardon Mar 20 at 14:38
  • 1
    $\begingroup$ 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) $\endgroup$ – John Pardon Mar 20 at 14:45
  • 1
    $\begingroup$ @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). $\endgroup$ – Sean Lawton Mar 20 at 14:48
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.