TITLE: Is there a $W^{2,2}$ isometric embedding of the flat torus into $\mathbb{R}^3$?
QUESTION [9 upvotes]: It is well known that there exists a  $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$, and that this embedding cannot be $C^2$.
Is there a $W^{2,2}$ isometric embedding? (i.e an isometric map $f \in W^{2,2}(\mathbb{T}^2,\mathbb{R}^3)$?)

REPLY [3 votes]: This is just an expansion on my comment to @j.c. 's answer.
We want to use Pakzad's result cited in that answer in order to prove that there is no $f$ as in the original question.
My idea:
By Sobolev embedding theorems we have that $f\in C^{0,\alpha}$ for all $\alpha\in (0,1)$.  In particular, the image $\Sigma=f(\mathbb{T}^2)$ is a compact set.  As such, there is an $R_0>0$ so that 
$$\Sigma\subset B_{R_0}(0)$$
where here $B_{R_0}(0)$ is the open euclidean ball of radius $r_0$ centered at the origin.
Now let 
$$
R=\inf\{ \rho>0: \Sigma \subset B_\rho(0)\}
$$
and observe that, as $\Sigma$ is compact $\Sigma\cap \partial B_{R}\neq \emptyset$.  In particular, there is a $p'\in \Sigma\cap \partial B_{R}$ and so, there is a $p\in \mathbb{T}^2$ so $f(p)=p'$.  
Now pick a small radius $r$ so that $\exp_p: B_r(0)\to \mathbb{T}^2$ is an isometry onto it's image (here we use that the metric on the torus is flat) and consider the map $\tilde{f}=f\circ \exp_p|_{B_r(0)}$.  Observe, that $\tilde{f}(0)=p'$.  By Pakzad's result, up to shrinking $r$ we know that either $\tilde{f}$ is affine or there is a line through $0$ on which $\tilde{f}$ is affine.  As $f$ is isometric these affine functions are non-constant and so the image of $\tilde{f}$ contains a line segment through $p'$. This segment must leave $\bar{B}_R(0)$ which contradicts the choice of $R$.