Let $\mathbb D^2$ be the closed unit disk in $\mathbb R^2$. Let $f:\mathbb D^2 \to \mathbb{R}^2$ be a real-analytic orientation preserving immersion, and let $\omega:\mathbb D^2 \to \mathbb{R}^2$ be the unique harmonic map satisfying $\omega|_{\partial \mathbb D^2}=f|_{\partial \mathbb D^2}$

Does $d\omega \neq 0 $ everywhere on $\mathbb D^2$?

I have two observations:

  1. There is an open neighbourhood of $\partial \mathbb D^2$ where $d\omega \neq 0 $ .
  2. $d\omega$ is invertible outside a set of Hausdorff dimension $\le 1$.

Claim $(1)$ follows from the fact that for $p \in \partial \mathbb D^2$, we have
$$ \text{rank}(d\omega_p)\ge \text{rank}\big(d(\omega|_{\partial \mathbb D^2})_p\big)= \text{rank}\big(d(f|_{\partial \mathbb D^2})_p\big)=1.$$

For point $2$, note that $d\omega$ cannot be singular everywhere, since $$ \int_{\mathbb D^2} \det d\omega = \int_{\mathbb D^2} \det df>0.$$

Thus, $\big(\det(d\omega)\big)^{-1}(0)$ is the zero-set of a real-analytic function which is not identically zero, which implies dimension $\le 1$.

  • $\begingroup$ Dear Asaf, I wonder, do you want me to clarify anything in my answer to your question? $\endgroup$ – Dmitri Panov Dec 25 '19 at 20:16

No, this is not so.

I'll explain how to construct a counter-example, though will leave some details in the form of exercises. I will also assume that we consider just smooth maps from the disk since smooth maps can be $C^{\infty}$ approximated by analytic ones, it will be obvious from the construction that there is no difference.

Let us parametrise the boundary of $\mathbb D^2$ by angle $t$, $t\in [0,2\pi]$. Then it will be enough to find an immersion $f=(f_1,f_2)$ such that when we restrict $f_1$ and $f_2$ to the unit circle, the following equalities hold:

$$\int_{0}^{2\pi} f_1(t)\cos(t)=\int_{0}^{2\pi} f_1(t)\sin(t)=\int_{0}^{2\pi} f_2(t)\cos(t)=\int_{0}^{2\pi} f_2(t)\sin(t)=0.$$

Indeed, if we construct such an immersion then the corresponding harmonic functions $(\omega_1, \omega_2)=\omega$ will satisfy $d\omega(0,0)=0$.

The existence of such $f$ is quite obvious, plenty of ways to construct it, I'll indicate one way.

First, we start with a simple exercise:

Exercise 1. Suppose we have a finite number of distinct points $(x_1,y_1),\ldots, (x_n, y_n)\in\mathbb R^2$ then for any $0<t_1<t_2<\ldots <t_n<2\pi$ there always exists an immersion (even an embedding) from a disk $\tilde f:\mathbb D\to \mathbb R^2$ such $\tilde f(t_i)=(x_i,y_i)$.

Exercise 2. There exists $n$, distinct $t_i$'s and distinct $(x_i,y_i)$'s such that

$$\sum_i \cos(t_i)x_i=\sum_i \sin(t_i)x_i=\sum_i \cos(t_i)y_i=\sum_i \sin(t_i)y_i=0.$$

There is a huge amount of flexibility in finding such $t_i, x_i, y_i$.

Finally, we consider an immersion $\tilde f$ from Exercise 1, so that the boundary of $\mathbb D$ passes through $(x_i,y_i)$ at time $t_i$, but reparametrise it in such a way that the circle $\tilde f(t)$ spends time $2\pi(\frac{1}{n}-\epsilon)$ very close to each point $(x_i, y_i)$ and then in time $2\pi\epsilon$ runs to the point $(x_{i+1},x_{i+1})$. In such case the second 4-tuple of equalities on sums will approximately guarantee the first 4 -tuple of equalities on integrals. After this a simple perturbation argument solves the problem.

Proof of $d\omega(0,0)=0$. Recall that every harmonic function $h$ can be decomposed as $h=a_0+a_1 Re(z)+b_1 Im(z)+ a_2 Re(z^2)+b_2 Im(z^2)+...$. Such a function has zero derivative at $(0,0)$ if an only if $a_1=b_1=0$, but the integrals written above give exactly the same conditions, since $\sin(mt)$, $\cos(mt)$ are orthogonal to each other for different $m$.

  • $\begingroup$ Thank you. Can you please say why $d\omega(0,0)=0$? I don't see how it follows from the integral equalities for $f$. (Since the mean value theorem holds for $\omega$, not its derivative). I guess I am missing something. $\endgroup$ – Asaf Shachar Nov 25 '19 at 19:46
  • $\begingroup$ You are welcome! I added a paragraph to the end of the answer. $\endgroup$ – Dmitri Panov Nov 25 '19 at 20:00

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