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Jan 1, 2023 at 22:10 answer added Anton Petrunin timeline score: 2
Mar 28, 2020 at 6:23 history edited Asaf Shachar CC BY-SA 4.0
I have cleaned up the question and pointed on a purely PDE-type subquestion.
Mar 26, 2020 at 18:50 comment added Asaf Shachar @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.
Mar 26, 2020 at 16:50 comment added AlexArvanitakis 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?
Mar 26, 2020 at 16:01 comment added Asaf Shachar @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.
Mar 26, 2020 at 15:59 history edited Asaf Shachar CC BY-SA 4.0
added 131 characters in body; edited title
Mar 26, 2020 at 15:36 comment added YCor 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.
Mar 26, 2020 at 15:28 history asked Asaf Shachar CC BY-SA 4.0