# Can harmonic maps with immersive boundary conditions have singular points?

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 $$\le 1$$.

• Dear Asaf, I wonder, do you want me to clarify anything in my answer to your question? – 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.

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 $$0 there always exists an immersion (even an embedding) from a disk $$\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 $$m$$.

• 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. – Asaf Shachar Nov 25 '19 at 19:46
• You are welcome! I added a paragraph to the end of the answer. – Dmitri Panov Nov 25 '19 at 20:00