This is a cross-post. Let $D \subseteq \mathbb{R}^2$ be the closed unit disk, and let $f:D \to D$ be a diffeomorphism.

Does there exist a smooth $h \in C^{\infty}(D)$ such that $h\cdot f$ is an area-preserving diffeomophism of $D$?

(Clarification: by $h\cdot f$ I mean multiplication of a scalar by a vector, i.e. $\big( h\cdot f\big) (x):=h(x)\cdot f(x)$).

Clearly, such an $h$ must send the entire boundary $\partial D$ either to $1$ or to $-1$. Another necessary condition is $\det\big(d (h\cdot f)\big) = 1$. Since

$$ d (h\cdot f)=h df+dh \otimes f=h df+f \cdot (\nabla h)^T, $$

the matrix determinant lemma implies that $$ 1=\det\big(d (h\cdot f)\big)=h^2\det(df)+(\nabla h)^T \operatorname{adj}( hdf ) f. \tag{1} $$

In fact, I am not even sure whether the PDE $(1)$ always has a solution for every diffeomorphism $f$. (that is even when omitting the requirement that $h\cdot f$ would be a diffeomorphism, or even a map from $D$ into $D$-when looking only at the PDE with no other restrictions on the result-does there always exist a solution?)

  • $\begingroup$ What is meant by $h\cdot f$? (I'd read it a priori as $h\circ f$, but then the question would be trivial taking $h=f^{-1}$). Where does "conformally" intervene in the question? If I were reading the title, I'd interpret it as whether for each $f$ there exists $h$ conformal such that $h\circ f\circ h^{-1}$ is area-preserving. $\endgroup$ – YCor Mar 26 at 15:36
  • $\begingroup$ @YCor I meant for the standard (scalar) multiplication of a vector by a scalar. Your are right that the title was a bit misleading. I have edited the question to address this, and also clarified what do I mean by $h \cdot f$. Thanks for your comment. $\endgroup$ – Asaf Shachar Mar 26 at 16:01
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    $\begingroup$ Surely yes? Pull the volume form $\omega$ back through $f$. $f^\star \omega$ is another volume form. Then the question is, does there exist $h$ so $\omega=h (f^\star\omega)$ which presumably can be shown to be true. Moser's theorem? $\endgroup$ – AlexArvanitakis Mar 26 at 16:50
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    $\begingroup$ @AlexArvanitakis Hi, I don't follow your reduction. Why is the question equivalent to finding a function $h$ such that $ \omega=h (f^\star\omega)$? According to my calculation, $(hf)^*\omega \neq h (f^\star\omega)$. I also don't see the connection to Moser's theorem. If you could elaborate that would be great. $\endgroup$ – Asaf Shachar Mar 26 at 18:50

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