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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.

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

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