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Let $\Omega\subset\mathbb{R}^d$ be an open regular domain and let $f\in W^{1,\infty}(\Omega;\mathbb{R}^d)$ satisfy that $df\in\operatorname{SO}(d)$ almost-everywhere. It was proved by Reshetnyak (in a wider Riemannian context) that $f$ is in fact smooth, hence an affine Euclidean isometry.

Question: Does a similar assertion in Minkowski space $\mathbb{R}^{1,d}$? The Euclidean proof relies on ellipticity, hence doesn't seem generalizable.

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  • $\begingroup$ Can you provide a citation for Reshetnyak's result? (I am having difficulty seeing how ellipticity is involved.) $\endgroup$
    – Willie Wong
    Commented Jun 22, 2023 at 17:58
  • $\begingroup$ @WillieWong, here is one paper about this: arxiv.org/pdf/1701.08892.pdf $\endgroup$
    – Deane Yang
    Commented Jun 22, 2023 at 22:18
  • $\begingroup$ @WillieWong Just as a sketch of the proof, for every Sobolev map between manifolds of equal dimensions, div(cof df))=0 (with the appropriate interpretation)). If df is in SO(), then cof(df) = df, i.e., the map is (weakly) harmonic, hence smooth. (The complete proof involves more technical details, but that's the gist of it.) $\endgroup$
    – Raz Kupferman
    Commented Jun 23, 2023 at 19:26
  • $\begingroup$ Thanks @DeaneYang and Raz. $\endgroup$
    – Willie Wong
    Commented Jun 23, 2023 at 23:09
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    $\begingroup$ @WillieWong If the map is everywhere orientation preserving, then no problem. But if you allow the map to change orientation then the projection on O(d) is no longer a Sobolev map $\endgroup$
    – Raz Kupferman
    Commented Jul 7, 2023 at 5:50

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