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

This question has now been cross-posted at mathoverflow.

While working on a variational problem, I have reached to the following question.

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

Does there exist a smooth map $$f:D \to D$$ such that $$df$$ has everywhere the fixed singular values $$\sigma_1,\sigma_2$$ and $$\det(df)=1$$? Is there such a diffeomorphism of $$D$$?

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

If we exclude a ray from $$D$$, then there is such a map, given by $$re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$$.

Furthermore, when $$\frac{\sigma_2}{\sigma_1}=n$$ is an integer, this map is given by $$z \to \frac{z^n}{|z|^{n-1}}$$ which is in $$W^{1,\infty}(D,\mathbb{R}^2)$$. Thus, if we could approximate such a map by smooth maps having fixed singular values, we would finish.

Here is an argument (given by a colleague) showing that $$f$$ cannot be the gradient of a function:

Suppose that $$f=\nabla u$$. Then $$df=\operatorname{Hess}u$$ is symmetric and has real eigenvalues. Since $$\det(df)=1$$, at every point of $$D$$ both eigenvalues are positive or both are negative. Thus $$\operatorname{tr}(df) \neq 0$$ has a definite sign.

By composing $$f$$ with the map $$x \to -x$$ we can assume the eigenvalues are always positive. Now, $$\int_{D} \operatorname{div} f = \int_{\partial D} \langle f, n \rangle \le \operatorname{Vol}(\partial D) =2\operatorname{Vol}( D),$$

where in the inequality we have used the fact that $$|f| \le 1$$. We showed that $$\operatorname{div} f \le 2$$ on average, so there exist a point $$x \in D$$ where $$\operatorname{div}f (x)=\lambda_1(df_x) + \lambda_2(df_x) \le 2.$$

Since the eigenvalues are positive, and $$df=\operatorname{Hess}u$$ is symmetric, we have $$\lambda_i=\sigma_i$$, so $$\sigma_1+\sigma_2=\sigma_1(df_x) + \sigma_2(df_x) \le 2$$.

This contradicts the AM-GM inequality $$\frac{\sigma_1+\sigma_2}{2} > \sqrt{\sigma_1 \sigma_2}=1$$, which is strict here since $$\sigma_1 \neq \sigma_2$$.

I tried using Helmholtz decomposition to treat the general case, but that doesn't seem to lead anywhere.

• 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 Feb 14 at 14:00
• You are right. However, I am also interested in non-diffeomorphic examples. (I have edited the question to make it clearer). – Asaf Shachar Feb 18 at 14:41