$C^1$ isometric embedding of flat torus into $\mathbb{R}^3$ - MathOverflow

$C^1$ isometric embedding of flat torus into $\mathbb{R}^3$

2010-07-09T17:46:19Z

I read (in a paper by Emil Saucan) that the flat torus may be isometrically embedded in $\mathbb{R}^3$ with a $C^1$ map by the Kuiper extension of the Nash Embedding Theorem, a claim repeated in this Wikipedia entry. I have been unsuccessful in finding a description of such a mapping, or an image of what the embedding looks like. I'd be grateful to any pointers on this topic. Thanks!

Addendum. It seems Benoît Kloeckner's answer below is definitive. What I asked for apparently does not yet exist, but is "in process" and will soon be available through the work of the Hévéa project.

[23Apr2012] This is taken from the link in DamienC's comment and Benoît's update in the latter's answer below:

Answer by Benoît Kloeckner for $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$

2010-07-09T18:35:45Z

A group of french mathematicians and computer scientists are currently working on this. The project is named Hévéa, and has already produced a few images. Edit: a few images and the PNAS paper have been released, see http://math.univ-lyon1.fr/~borrelli/Hevea/Presse/index-en.html

Just a few word to explain what I understood of their method (which is by using h-principle) from the few image I saw in preview. Start with a revolution torus. The meridians are cool, because they all have the same length, as expected from those of a flat torus. But the parallels are totally uncool, because their lengths differ greatly: they witness the non-flatness of the revolution torus.

Now perturb your torus by adding waves in the direction of the meridians (like an accordion), with large amplitude on the inside and small amplitude on the outside. If you design this perturbation well, you can manage so that the parallels now all have the same length. Of course, the perturbed meridian have now varying lengths! So you do the same thing by adding small waves in another direction, getting all meridians to have the same length again. You can iterate this procedure in a way so that the embedding converges in the $C^1$ topology to a flat embedded torus. But to prove that the precise perturbation you chose in order to get a nice image does converge, and that your maps are embeddings needs work (getting an immersion is easier if I remember well).

Also, the Hévéa project plans to draw images of Nash spheres, that is $C^1$ isometric embeddings of spheres of radius $>1$ inside a ball of unit radius.

Answer by Dick Palais for $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$

2010-07-09T20:36:34Z

On the other hand, if you are willing to settle for conformally flat, there is a beautiful theory of these. (The idea is to consider flat embeddings in the three-sphere, and then "project them into $R^3$ using stereographic projection.) The classification of flat embeddings of the torus in the three-sphere goes back to Bianchi in the 1800's, and Ulrich Pinkall recently found some particularly nice ones (the so-called "Bianchi-Pinkall Tori) by taking inverse images of a simple-closed curve under the Hopf fibration (so one set of circles are Hopf fibers). If you would like to see some example images, some applets to play with and morph them, and an explanatory pdf file, have a look here:

http://virtualmathmuseum.org/Surface/bianchi-pinkall_tori/bianchi-pinkall_tori.html

Answer by Joseph O'Rourke for $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$

2010-08-03T12:21:54Z

I would like to mention something I learned from Igor Pak subsequent to posing this question: there is a piecewise-linear embedding of the flat torus! In the paper by V. A. Zalgaller, "Some bendings of a long cylinder," Journal of Mathematical Sciences, 100(3):2228--2238, 2000 (translated from a 1997 article in the Russian journal Zapiski Nauchnykh Seminarov POMI), he proves this theorem:

"Theorem 1. A direct flat torus can be isometrically embedded in $\mathbb{R}^3$ 'in the origami style' if its development is a rectangle sufficiently large compared to its altitude."

He defines a direct flat torus as the result of identifying the opposite sides of a rectangle. (I have never seen the term "direct flat torus," and I don't know what role the modifier "direct" plays.) "In the origami style" describes how he bends a triangular prism such that "its middle part is broken to a complicated form." The embedding is a triangular prism bent into the shape of a regular hexagon. The "bendings" are piecewise-linear crinklings of the surface.