# Harmonic maps are light

Assume $f\colon \mathbb{D}\to\mathbb{R}^2$ is a harmonic map and $x\notin f(\partial\mathbb{D})$. Is it true that $f^{-1}\{x\}$ is totally disconnected?

I hope that the answer is yes. But actually I need a "yes", with a CAT(0) space instead of $\mathbb{R}^2$.

So, I need a generalizable proof.

P.S. For CAT(0) target the answer is "no" --- if the target is a cone with large total angle then the tip of the cone might have a tree as an inverse image for harmonic map; this example is constructed in "Harmonic maps between flat surfaces with conical singularities" by Ernst Kuwert.

As your question operates with $f(\partial D)$, I assume that $f$ is continuous in $\overline{D}$, though you do not mention this explicitly. Then the answer is yes, the zero set of $f$ is discrete (not just totally disconnected).
Suppose wlog that $0\not\in f(\partial D)$. Suppose by contradiction that zeros of $f$ have an accumulation point $z^*$ in $D$. Then, according to a theorem of Wilmshurst, there is an open analytic arc $\gamma$ which has $z^*$ in its interior, and such that $f(z)=0$ on $\gamma$. Consider now this whole arc $\gamma$ (the maximal arc on which $f(z)=0$). It cannot reach the boundary $\partial D$ because on $\partial D$ we have $|f(z)|>\delta>0$ by assumption. So $\gamma$ must contain a loop, but then $f\equiv 0$ inside this loop and thus everywhere.
• Thank you, it seems to be generalizable --- all you need is a local formula $$f(z)-f(0)=a\cdot z^k+b\cdot \bar z^k+o(|z|^k)$$ for an integer $k$ and $|a|+|b|\ne 0$. I am sure it is still true for the surfaces, altho I do not know why. – Anton Petrunin Jun 13 '17 at 3:05