5
$\begingroup$

While analyzing a variational problem, I came to the following question:

Let $\mathbb D^n \subseteq \mathbb{R}^n$ be the closed $n$-dimensional unit ball, and let $f: \mathbb D^n \to \mathbb{R}^n$ be a smooth orientation-preserving immersion. Denote by $\omega_f :\mathbb D^n \to \mathbb{R}^n$ the unique harmonic map satisfying $\omega_f|_{\partial \mathbb D^n}=f|_{\partial \mathbb D^n}$.

$d\omega_f$ must be invertible outside a set of measure zero in $\mathbb D^n$. Indeed, $\omega_f$ is real-analytic, and so is $\det d\omega_f$, which is not identically zero, since $$ \int_{\mathbb D^n} \det d\omega_f = \int_{\mathbb D^n} \det df>0.$$

Now, the zero-set of a real-analytic function which is not identically zero has measure zero.

Question: Do there exist $f_k \in C^{\infty}(\mathbb D^n, \mathbb{R}^n)$ such that $d\omega_{f_k} \in \text{GL}^+$ are everywhere and $f_k \to f$ in $W^{1,2}$?

($\omega_{f_k}$ is the harmonic map corresponding to the Dirichlet problem imposed by $f_k$.)

Note that even though $d\omega_f \in \text{GL}$ a.e., it can "spend time" in both $\text{GL}^+$ and $\text{GL}^-$. Here is an example:

Let $f : \mathbb D^2 \to \mathbb R^2$ be defined by $ f(x,y) = (x-2y^2,y). $ We have $$df=\left(\begin{matrix}1 & -4y \\ 0 & 1\end{matrix}\right)$$

and thus $f$ is an orientation-preserving immersion.

The solution to the Dirichlet problem in this case is $\omega_f(x,y) = (x^2 - y^2 + x - 1,y)$, so $$d\omega_f=\left(\begin{matrix}1+2x & -2y \\ 0 & 1\end{matrix}\right)$$ and $\det(d\omega_f)=1+2x>0 \iff x>-\frac{1}{2}$.

$\endgroup$
2
  • $\begingroup$ In the highlighted Question, do you mean $f_k\to f$? $\endgroup$ Aug 24 '19 at 10:44
  • $\begingroup$ Yes, thank you. This was a typo. $\endgroup$ Aug 24 '19 at 11:41
0
$\begingroup$

It seems that the answer is negative for dimension $n=2$ . I am not sure if higher dimensions can be reduced to the $2D$ case.

Here is the argument for $n=2$:

Suppose that there exist $f_k \in C^{\infty}(\mathbb D^2, \mathbb{R}^2)$ such that $d\omega_{f_k} \in \text{GL}^+$ everywhere and $f_k \to f$ in $W^{1,2}$.

The convergence $f_k \to f$ in $W^{1,2}$ implies that $d\omega_{f_k} \to d\omega_f$ in $L^2$. (This follows from the fact that the trace operator is a continuous surjection onto the fractional Sobolev space $W^{1/2,2}(\partial \Omega)$, see here for details).

This implies that $\det(d\omega_{f_k}) \to \det(d\omega_f)$ in $L^1$ (here use the fact that the dimension is $2$), so up to passing to a subsequence, $\det(d\omega_{f_k})$ converges pointwise a.e. to $\det(d\omega_f)$.

By our assumption, $\det(d\omega_{f_k}) > 0$ everywhere, so this implies $\det(d\omega_f) \ge 0$ a.e.

Now, taking any $f$ whose $\omega_f$ does not satisfy this, we get a contradiction. (e.g. the example described in the question).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.