# Can every diffeomorphism be rescaled into a volume preserving one?

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?)

• 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. – YCor Mar 26 at 15:36
• @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. – Asaf Shachar Mar 26 at 16:01
• 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? – AlexArvanitakis Mar 26 at 16:50
• @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. – Asaf Shachar Mar 26 at 18:50