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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

$$Area(\phi(\Sigma))=Area(\Sigma)$$

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

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