A counter example for Liouville's theorem when the Jacobian is changing signs

Question

The famous Liouville's theorem states the following:

Let $\Omega$ be a domain in $\mathbb{R}^n$, and let $f \in W_{loc}^{1,n}(\Omega,\mathbb{R}^n)$ satisfy

1. $Jf=\det df \ge 0$ a.e. on $\Omega$ or $\det df \le 0$ on $\Omega$.
2. $df^Tdf=|Jf|^{\frac{2}{n}}\text{Id}$.

Then $f$ is either constant or a restriction to $\Omega$ of a Mobius transformation of $\mathbb{R}^n$. In particular $f$ is smooth.

Question: Is there an example for a smooth conformal map whose Jacobian changes sign?

More generally, if we drop the assumption $\text{sign}(\det df) $ is constant, are there conformal maps which are only $C^k$ for some $k$, but not $C^{\infty}$? Is there a concrete simple example of a conformal Sobolev map which is not even $C^1$?

Comment: Gromov have constructed arcwise isometries between equidimensional non-flat manifolds and flat ones. In particular, for every metric $g$ on the unit $d$-dimensional disk $\mathbb{D}^d$ there is an a.e isometry $f:(\mathbb{D}^d,g) \to (\mathbb{R}^d,e)$. ($e$ is the Euclidean metric).

Now take $g=he$ where $h \in C^{\infty}(\mathbb{D}^d)$ is an arbitrary map. For a generic $h$, the gaussian curvature of the metric $g$ will be non-zero, hence Gromov's pathological maps cannot smooth.

These arcwise isometries are of course conformal in the Euclidean sense.

(And indeed they cannot be orientation-preserving or orientation-reversing on any open subset of the domain. This is because every weakly differentiable map isometry whose differential lies in $\text{SO}(n)$ a.e is smooth).

However, Gromov's result uses convex integration which (as far as I understand) not really constructive or easily visualisable. I am looking for simple "counter-examples".

Answer 1

Allowing the Jacobian to change sign does not produce new $C^1$-smooth conformal maps in dimensions $n\ge 3$.

Proof. Suppose $f:\mathbb{R}^n\to\mathbb{R}^n$ is $C^1$ smooth and conformal. Then $Jf$ is continuous. If $Jf=0$ everywhere then $df=0$ everywhere, so $f$ is constant. Excluding this case, let $U$ be a connected component of the set $\{x : Jf(x)\ne 0\}$. By Liouville's theorem, $f$ agrees with a Möbius transformation on $U$. Since the Jacobian of a Möbius transformation is nowhere zero, the continuous function $Jf$ cannot vanish on $\partial U$. This means $\partial U$ must be empty, and $U$ is all of $\mathbb{R}^n$. $\quad\Box$

Relaxing the smoothness assumption on $f$ to Lipschitz continuity we get "folding" maps like $(x_1, \dots, x_n) \to (x_1, \dots, |x_n|)$ and others (consider origami).

Dimension $n=2$

Similar story here. Again, suppose $f\in C^1$ and let $U$ be a connected component of the set $\{x : Jf(x)\ne 0\}$. Then $f$ is either holomorphic or antiholomorphic in $U$, suppose the former without loss of generality. Then the derivative $f'$ is holomorphic in $U$ and continuous on $\overline{U}$, with zero boundary values. This forces $f'\equiv 0$, a contradiction. (This part is tricky; the maximum principle is not enough when $U$ is unbounded, but the conclusion still holds because a nonzero holomorphic function cannot vanish on a nontrivial boundary arc.)