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Let $M,N$ be smooth $d$-dimensional Riemannian manifolds.

Suppose $f:M \to N$ is a differentiable isometry ($df_p$ is an isometry at every $p \in M$). I do not assume $f$ is $C^1$.

Is it true that $f$ must be smooth?

This is probably true if we assume $f$ is $C^1$ (see here).

Note: This is false when assuming $f$ is only differentiable almost everywhere, as was shown by Gromov (in fact $M$ can be non-flat, and $N$ flat).

When trying to adapt the argument to this less regular case, we hit a porblem, as described below:

A "proof":

First, note that by the inverse function theorem for everywhere differentiable maps, $f$ is a local homeomorphism.

We would like to prove it's a local isometry w.r.t the intrinsic distances, then use the Myers-Steenrod theorem.

The problem lies within the fine the details: We need to choose suitable length structures on $M,N$ such that $f$ will become an arcwise isometry. We cannot use the class of $C^1$ paths, however, since $f$ does not necessarily map $C^1$ maps to $C^1$ maps.

In a similar spirit, nothing promises us that for a differentiable path $\gamma$ such that $\|\dot \gamma(t)\|$ is integrable, $\|\dot {f \circ \gamma}(t)\|$ will be integrable.

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Asaf Shachar
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    Dear Asaf: This is not a bad question, but I do not understand the context. Do you know of any naturally occurring examples where this situation arises? Differentiability a.e. appears very often in geometry and analysis, but differentiability everywhere without $C^1$, especially for infinitesimal isometries-? It seems to me more natural to require diff a.e. with "bad set" of small Hausdorff dimension. – Moishe Kohan Feb 02 '17 at 13:42
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    Thanks for your comment. Indeed, I do not know of any natural examples, but I am curious whether there exist "unnatural" examples, in a similar spirit to Gromov's pathological maps. Indeed, Gromov has a remarkable result, that for any given metric $g$ on the unit disk $D$ (in $\mathbb{R}^n$) there exist an arcwise isometry $f:(D,g) \to (\mathbb{R}^n,e)$. (even if $g$ is non-flat or as "wild" as you can imagine. It then follows that $f$ is diff a.e (Rademacher) and moreover, $df$ is an isometry a.e – Asaf Shachar Feb 02 '17 at 13:55
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    (this follows from a measure theoretic argument, you can see here: http://mathoverflow.net/a/257199/46290). These "pathological maps duw to Gromov switch orientation in an "infinite rate" (they cannot be orientation-preserving or reversing on any open subset of $D$). To summarize, I really don't have a big specific "application" for the answer to this question in mind, but I am curious how far the pathology can be "pushed". In some sense, I agree that your formulation (with the Haussdorf measure) might be better... – Asaf Shachar Feb 02 '17 at 13:56
  • but for I now I prefer to keep waiting to see if someone has somthing intelligent to say about the quesiton as it is – Asaf Shachar Feb 02 '17 at 13:59
  • @Asaf Shachar : I am curious about Gromov's example Can you describe it or tell me a reference ? – HK Lee Feb 08 '17 at 13:51
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    @HKLee Unfortunately I cannot describe it. It involves the method known as "convex integration" which I am not familiar with, but hope to study one day. Gromov's result can be found in its book "Partial Differential Relations" - in a Corollary in the middle of page 216. (Note that Gromov defines an "isometric map" to be a continuous map which preserves lengths of rectifiable curves, see page 207). – Asaf Shachar Feb 08 '17 at 14:11

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For completeness: The answer is positive. Any differentiable isometry is smooth. This is proved here.

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Asaf Shachar
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