# $C^2$ isometric embedding of the flat torus into $\mathbb{R}^3$

What is the reason that there is no $C^2$ isometric embedding of the flat torus inside $\mathbb{R}^3$? Is there an explicit proof of this fact anywhere? The flat metric must violate some condition for the $C^2$ isometric embedding. And as we know, the condition for a local $C^2$ isometric embedding is given by the Gauss and Codazzi-Mainardi relations. I do not understand how these relations are getting violated by the flat metric on torus. Kindly cite some reference. Thanks in advance.

Edit 1:

I am following this paper: Han and Lin, On the isometric embedding of torus in $\mathbb{R}^3$, Methods and Applications of Analysis 15, pp. 197-204, 2008.

Here, the sufficient conditions for the existence of a global smooth isometric embedding of the torus of genus 1 $\mathbb{T}$ with a Riemannian metric $a$ $(\mathbb{T}, a)$ are given. But these conditions, as can be clearly seen, are given for the existence of the standard embedded torus in $\mathbb{R}^3$, which is too strict. The original question is about the existence of an isometrically embedded torus in $\mathbb{R}^3$, not necessarily the tadard torus. So there must be a different set of sufficient conditions than the ones given in the above reference. Can one of these sufficient conditions be the requirement that the the subset of $\mathbb{T}$ where the Gaussian curvature $K$ of $a$ is positive is non-empty?

• As far as I know, the flat torus can be isometrically embedded in $\mathbb{R}^3$ by a $C^1$ mapping. You probably want to add you are asking for smooth embedding or at least $C^2$. – C. Falcon May 22 '17 at 2:40
• Thanks for pointing this out. I will edit the question! – Ayan May 22 '17 at 2:43
• As C. Falcon, says, the flat torus can be $C^1$-embedded in $\mathbb{R}^3$. But this is the best you can get... It is a cute fact that, every compact $C^2$ surface in $\mathbb{R}^3$ has positive curvature somewhere. – Amitai Yuval May 22 '17 at 2:44
• So, the condition that "every compact $C^2$ surface in $\mathbb{R}^3$ has positive curvature somewhere" is a sufficient condition for the $C^2$ global isometric embedding of a compact 2-dimensional manifold, besides the local Gauss and Codazzi-Mainardi conditions? Also, how to check the Gauss-Codazzi-Mainardi relations in the case of flat torus, when no second fundamental form $b_{\alpha\beta}$ is specified a priori? – Ayan May 22 '17 at 2:47
• Can I use the Gauss formula for that? Because, if there is a isometric immersion of Torus into $\mathbb{R}^3$ then second fundamental form is null, so every geodesic of the immersion is a geodesic of $\mathbb{R}^3$. That argument is possibly wrong, I know. Someone can say where? – Irddo May 22 '17 at 2:49

Let $S$ be a compact immersed surface in the euclidan space. By some standard argument, you can find a system of linear Euclidian coordinates $x,y,z$ so that the restriction of the last coordinate, say $z$, is a Morse function when restricted to your surface. Near a point $(x_0,y_0,z_0)$ where $z$ is maximal, the surface is described by the graph a function $z=f(x,y)$, with $f'(x_0,y_0)=(0,0)$. The second fundamental quadratic form at this point is $1/2f"(x_0,y_0)$, and by the Taylor formula, it must be negative definite ($f$ has a maximum). Hence its determinant , which is the Gaussian curvature must be strictly positive.
• (+1) Or even more low-tech: Fix a point $O$ of $\mathbf{R}^{3}$ arbitrarily, pick a point $p$ of $S$ farthest from $O$ (so that $S$ lies inside the sphere centered at $O$ of radius $\|p - O\|$), and show the product of the principal curvatures at $p$ is positive. – Andrew D. Hwang May 23 '17 at 10:57