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Does there exist a function $f:\Bbb{S}^2\rightarrow\Bbb{R}^2$ which preserves the length of every rectifiable curve? That is, can a sphere be crushed flat without tears? Of course, this is a Nash-Kuiper type problem, so my question is really whether such a result is already known. If so, I would love to have a reference.

Thanks in advance!

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    $\begingroup$ Will you solve this small ambiguity about "length": "The map $f$ preserves the length of rectifiable curves" : does this mean : for every (continuous) rectifiable $u:[0,1]\to \mathbb S^2$ the length of $f\circ u$ is equal to the lenght of $u$ (length=total variation). Or does it mean: "for every $\Gamma\subset \mathbb S^2$ homeomorphic to $[0,1]$ the length of $\Gamma$ is equal to the length of $f(\Gamma)$ (length=$\mathcal H^1$) ? $\endgroup$
    – Pietro Majer
    Commented Aug 28 at 16:59
  • $\begingroup$ It is defined as the supremum of lengths of inscribed polygons. For example, the simplest non-trivial length preserving map from $\Bbb{R}^2$ to itself is given by folding along the $x$-axis. This preserves length as given by the first definition, but not as given by the second. Indeed, it suffices to consider vertical curves which cross the $x$-axis. $\endgroup$
    – Graham Smith
    Commented Aug 29 at 18:13
  • $\begingroup$ Don't forget to mention what your polygons are inscribed in. $\endgroup$
    – Daniel Asimov
    Commented Aug 30 at 0:53

2 Answers 2

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There does exist such a map, if it is not required to be $C^1$. (The nice answer of Guido De Philippis gives a proof that such a map cannot be $C^1$.). See "Lecture 4" ("Gromov's Rumpling Theorem") in the lecture notes of Petrunin-Yashinski here: https://arxiv.org/pdf/1405.6606

A more general result is in section 2.4.11 of Gromov's book Partial Differential Relations.

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    $\begingroup$ Nice reference, thank you! i will edit the answer accordingly! $\endgroup$
    – Guido De Philippis
    Commented Aug 27 at 17:22
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    $\begingroup$ That's an astonishing theorem. I'm curious if there's an existing visualization of this type of map, along the lines of what the Hevea Project did for Nash embeddings. $\endgroup$
    – Martin M. W.
    Commented Aug 28 at 0:28
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According to the reference posted below there is indeed such a map. However, such a map would not exists if one assumes some (minor) smoothness. Note that this is somehow in contrast with the Nash Kupier theorem where the map is indeed $C^1$.

In particular, I will prove below that there is no $C^1$ map $f: U\subset \mathbb R^2 \to \mathbb R^2$ such that \begin{equation*} (df)^Tdf=g \, , \label{e1} \end{equation*} where $g$ is any smooth metric with non zero Gauss curvature.

I believe that the proof below is well known, but I do not have a reference.

First, note that non existence of a smooth map satisfying the above equation follows from the invariance of the Gauss curvature under isometry.

To prove the claim one thus only has to show that any $C^1$ map satisfying the above equation is indeed smooth. For, note that it implies that \begin{equation*} (df)^{-1}=g^{-1}(df)^T \qquad \det df=\sqrt{\det g} \end{equation*} and thus \begin{equation*} \begin{pmatrix} \partial_2 f^2 &-\partial_1 f^2 \\ -\partial_2 f^1 & \partial_1 f^1 \end{pmatrix} = \det df \, (df)^{-T} =\sqrt{\det g} \,df \, g^{-T}. \end{equation*} The distributional rowise divergence of the first matrix is zero, hence $f$ is satisfying the elliptic equation $$ \mathrm{div} (\sqrt{\det g} \,Df \, g^{-T})=0. $$ Classical elliptic regularity theory implies that $f$ is smooth and thus that the classical rigidity can be applied.

To put the proof in perspective, note that this is a "non constant coefficient" version of the classical argument of Reshetnyak about the rigidity of the inclusion $df \in SO(n)$, as presented, for instance, in these notes.

As a concluding remark, observe that the rigidity comes from the equality between the dimensions of the target and source domain. The Nash-Kupier needs indeed some co-dimension to add winkles to a short embedding in order to make it isometric.

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  • $\begingroup$ That's an interesting complement. Thankyou very much! $\endgroup$
    – Graham Smith
    Commented Aug 27 at 18:24
  • $\begingroup$ By the way, is it true that if a Finsler sphere was not already conformally (Alexandrov) flat, then it cannot be $C^1$ squashed? $\endgroup$
    – Denis T
    Commented Aug 27 at 20:35
  • $\begingroup$ I think this should probably follow also from the remarks in the lecture notes linked in the other answer. Specifically, they provide an argument that a length-preserving function cannot be injective on any open set, which for $C^1$ functions implies that the differential is singular everywhere, which should contradict preserving lengths (but my differential geometry is a bit rusty, so don't quote me on that). $\endgroup$
    – tomasz
    Commented Aug 29 at 19:05

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