It is well known that [there exists a $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$](https://mathoverflow.net/questions/31222/c1-isometric-embedding-of-flat-torus-into-mathbbr3), and that this embedding cannot be $C^2$.


Is there a $W^{2,2}$ isometric embedding? (i.e an isometric map $f \in W^{2,2}(\mathbb{T}^2,\mathbb{R}^3)$?)



Is there a $W^{2,2}$ isometric embedding of the flat torus into $\mathbb{R}^3$?
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 $R$.


