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 $f \in W^{2,2}(\mathbb{T}^2,\mathbb{R}^3)$?)


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 $R$.


It is a result of Pakzad's that $W^{2,2}$ isometric immersions $f$ of bounded regular convex domains with Lipschitz boundary $\Omega\subset\mathbb{R}^2$ must be developable; more precisely, for every point $x$ of $f(\Omega)$, $f$ must restrict to an affine function on either (1) a neighborhood of $x$ or (2) a line segment passing through $x$ connecting two points of $\partial\Omega$.

This should "rule" out an isometric embedding of $\mathbb{T}^2$ to $\mathbb{R}^3$ (forgive the pun).

See Theorem II of:

Pakzad, Mohammad Reza, On the Sobolev space of isometric immersions, J. Differ. Geom. 66, No. 1, 47-69 (2004). ZBL1064.58009.

What's still missing is a classification of developable surfaces in the $W^{2,2}$ setting; I suspect it should be the same as the classical one but I should think some more...

edit 2: In fact, such a classification is not necessary to show that there is no $W^{2,2}$ immersion of a flat torus into $\mathbb{R}^3$. See RBega's answer for a much nicer proof of that fact than what follows here!

edit: The following was inspired by RBega's comment below before I fully understood it. Hopefully this works:

Suppose there is a point $p$ of type (2). Let $\Omega$ be $\mathbb{T}^2$ minus an $\epsilon$-neighborhood of the cut locus at $p$, thought of as a domain in $\mathbb{R}^2$. $\mathbb{T}^2$ minus its cut locus looks like the Voronoi cell of $p$ in the universal cover so $\Omega$ satisfies the assumptions in Pakzad's theorem. Then by Pakzad's theorem, the line segment through $f(p)$ connects the images of two points $p_1,p_2$ on $\partial\Omega$. Let $r=d_{\mathbb{T}^2}(p_1,p_2)$ and $R=d_{\mathbb{R}^3}(f(p_1),f(p_2))$, so that $f(p)$ lies on a straight line segment $l\subset\mathbb{R}^3$ of length $R$ between $f(p_1),f(p_2)$. For $\epsilon$ sufficiently small, I claim that $R>r$ which leads to a contradiction. Explicitly, $l$ is a minimizer of distance in $\mathbb{R}^3$, so $R$ is a lower bound for the distance between $f(p_1)$ and $f(p_2)$ in $f$, so the true distance $r$ cannot be achieved by $f$.

I don't have a particularly clean proof of $R>r$ for all flat tori but I think that this will follow from some case analysis on the possible Voronoi cells of lattices in $\mathbb{R}^2$. For example, consider the square flat torus of area $a^2$. We can take $\Omega$ to be a square with side length $a-\epsilon$, with $p$ at its center. Then among the pairs of points $(p_1,p_2)$ on $\partial\Omega$ that lie on some geodesic through $p$, the maximal distance $r$ between $p_1$ and $p_2$ is less than $a/2$, while the minimum length $R$ of a geodesic joining $p_1$ and $p_2$ through $p$ is at least $a-2\epsilon$.

Finally, if there can be no points of type (2), then we are left with only points of type (1), but that implies that $f$ itself is affine, which leads to a contradiction as well.

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    $\begingroup$ Thanks. Can you please say why the image of the flat Torus under a $W^{2,2}$ isometric map cannot be developable? (i.e I don't see why the theorem you mentioned imply there is no such embedding). $\endgroup$ – Asaf Shachar Nov 21 '17 at 16:14
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    $\begingroup$ @AsafShachar You're right that there's a missing argument. I was hoping that the classification of developable surfaces (see e.g. onlinelibrary.wiley.com/doi/10.1002/cpa.3160140333/full ) would easily extend to this setting but I don't see it immediately. I'll edit. $\endgroup$ – j.c. Nov 21 '17 at 17:00
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    $\begingroup$ Wouldn't the following argument work given the result of Pakzad? Pick the infimum of radii $r$ so that $f(\mathbb{T}^2)\subset B_{r}$ (here $B_r$ is the euclidean ball of radius $r$ centered at $0$). This means that there is a $p\in \mathbb{T}^2$ so that $f(p)\in B_{r}$, but any line segment through $f(p)$ lying in the sphere would have to leave the ball giving a contradiction. $\endgroup$ – Rbega Nov 22 '17 at 2:32
  • $\begingroup$ @Rbega Perhaps this is trivial, but how do you know the image $f(\mathbb{T}^2)$ is bounded? $\endgroup$ – Asaf Shachar Nov 22 '17 at 5:42
  • $\begingroup$ @AsafShachar The image $f(\mathbb{T}^2)$ must be bounded, otherwise we can find pairs of points whose distance in $\mathbb{R}^3$ is arbitrarily large. Distances in the image are bounded below by distances in $\mathbb{R}^3$, and this contradicts the facts that $\mathbb{T}^2$ has a finite metric diameter and that $f$ preserves distances. $\endgroup$ – j.c. Nov 22 '17 at 13:58

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