Is there a non-affine harmonic map with constant determinant?

Question

Does there exist a smooth harmonic map $f:\mathbb{R}^2 \to \mathbb{R}^2$ such that:

1. $\det(df)$ is constant.
2. $f$ is not affine.

Is there such a map with $\det(df) \neq 0$?

($f$ is harmonic if each of its two components is a harmonic function).

Answer 1

There is no non-affine harmonic self-map of the plane with constant Jacobian determinant, and here is a nice proof using complex analysis. Assume that there is such a function, and without loss of generality assume that $\det (df) = 1$. Every harmonic function in the plane is the sum of a holomorphic and an anti-holomorphic function, so there exist entire functions $g$ and $h$ with $f = g + \bar{h}$ (where the bar denotes complex conjugation.) Using Wirtinger derivatives (just for notational convenience) we get $$ 1 = \det (df) = \left| \frac{\partial f}{\partial z} \right|^2 - \left| \frac{\partial f}{\partial \bar{z}} \right|^2 = |g'|^2 - |h'|^2, $$ so $$ |g'|^2 = |h'|^2 + 1 \ge 1. $$ This shows that $1/g'$ is a bounded entire function, so by Liouville's theorem it is constant, as is $g'$. This immediately implies that $|h'|$ and thus $h'$ is also constant, so $g$ and $h$ are affine, which implies that $f$ is affine, too.

EDIT: One can actually avoid the use of Liouville's theorem to show that this is really a local result. Writing $G = (g')^2$ and $H = (h')^2$, the equation above is $|G| = |H|+1$. At all points where $H$ does not vanish, both $|G|$ and $|H|$ are smooth with $|G'|=\|\nabla |G|\|=\| \nabla |H| \| = |H'|$. This shows that $G' = \lambda H'$ with some constant $|\lambda|=1$, so $G = \lambda H + \mu$ with another constant $\mu$. Since $|G| \ne |H|$, we also get $\mu \ne 0$. Since the image of any concentric circle under the map $r \mapsto \lambda r + \mu$ intersects any other concentric circle in at most two points, the map $H$ must be constant, again giving that $f$ is affine. (This is a little exercise in complex analysis, in a way more elementary, but a little less elegant than the argument above using Liouville.)