# Is there a volume-preserving diffeomorphism of the disk with prescribed singular values?

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...

• 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? – Adrián González-Pérez 2 days ago