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Let $0<\sigma_1<\sigma_2$ such that $\sigma_1\sigma_2=1$, and let $D_1 \subseteq \mathbb{R}^2$ be the closed unit disk.

Does there exist a smooth map $f:D_1 \to D_1$ such that $df$ has everywhere the fixed singular values $\sigma_1,\sigma_2$? Is there such a diffeomorphism of $D_1$?

The linear map $x \to \begin{pmatrix} \sigma_1 & 0 \\\ 0 & \sigma_2 \end{pmatrix}x$ does not satisfy the requirement, since it gets outside of $D_1$. (as $ \sigma_2 > 1$).

If we exclude a ray from $D_1$, then there exist a map satisfying the requirements, given by $re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$.

Edit:

$f$ must be measure-preserving. Can we use some ergodic theory here to deduce an obstruction? I tried using bounds on singular values of matrix products but so far without success.

I think that by considering iterates $f^n$ of $f$, we might get a contradiction...

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  • $\begingroup$ I have trouble parsing your example in $D_1$ minus a ray, in particular is not surjective or injective, don't you want it to be a diffeomorphism? $\endgroup$ – Adrián González-Pérez 2 days ago

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