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Perhaps this is a naive question (and certainly an idle one):

If $\phi:\mathbb{R}^3\to \mathbb{R}^3$ is a smooth diffeomorphism with the property that for any compact surface $\Sigma\subset \mathbb{R}^3$ one has


is $\phi$ necessarily an isometry? Here we are using the euclidean metric.

If the map is distance preserving then it is an isometry by a theorem of Myers and Steenrod (see here for instance). If the map is only volume preserving then it could be affine (or worse) so the result couldn't hold.

I had thought perhaps to approximate geodesics by thin tubes but couldn't see how any such argument wouldn't also hold for volume preserving maps.

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closed as too localized by Ryan Budney, Andrés E. Caicedo, José Figueroa-O'Farrill, Sergei Ivanov, Simon Thomas Apr 27 '11 at 13:37

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

This reduces to an infinitesimal problem of characterizing the linear maps $L : \mathbb R^3 \to \mathbb R^3$ which preserve cross product lengths, and it's an undergrad linear algebra-level homework problem to show those are isometries. IMO this is more appropriate for math.stackexchange. – Ryan Budney Apr 27 '11 at 5:03
Your right I was thinking about things the wrong way. I just got hung up on the tubes. – Rbega Apr 27 '11 at 5:06