Let $\Omega \subseteq \mathbb{R}^4$ be open subset. Does there exist a non-smooth Sobolev map $f \in W^{1,p}(\Omega,\mathbb{R}^4)$ such that $df \in \{A,-A\}$ almost everywhere, where $A \in \text{GL}(\mathbb{R}^4)$ is fixed?
I am trying to see if there is a non-trivial Sobolev map whose invertible differential is doing "zig-zag" between some value and its negative. Of course, if $df=A$ is constant a.e., then $f$ must be affine.
In one-dimension, we have such a phenomena:
Just take $f(x)=|x|$ on $(-1,1)$; then $f' \in \{1,-1 \}$. It seems to me that in higher dimensions, the constraint is harder to satisfy, though if we relax the invertibility condition, it's easy: Take
$$ f(x_1,x_2,x_3,x_4)=(|x_1|,|x_1|,|x_1|,|x_1|).$$
Then $df(x)= \left( \begin{matrix} \text{sign}(x_1)&0&0&0\\\text{sign}(x_1)&0&0&0\\ \text{sign}(x_1)&0&0&0\\ \text{sign}(x_1)&0&0&0 \end{matrix} \right)$.
Trying to make $df$ invertible, we can modify $f$ to
$$ f(x_1,x_2,x_3,x_4)=(|x_1|,|x_2|,|x_3|,|x_4|),$$
but now $$df(x)=\left( \begin{matrix} \text{sign}(x_1)&0&0&0\\0&\text{sign}(x_2)&0&0\\ 0&0&\text{sign}(x_3)&0\\ 0&0&0&\text{sign}(x_4) \end{matrix} \right)$$ switches between $2^4=16$ different values, since we can change each variable separately from negative to positive.
Comment: The reason I chose $4$ dimensions, is that I am interested in even dimensions greater than $2$, and this was the smallest one. I don't know the answer even for mappings $\mathbb{R}^2 \to \mathbb{R}^2$.
Edit:
In dimension $2$, we can try do the following:
Write $f(x,y)=(|x|,y \text{sign}(x))$. Then, if $f$ were Sobolev, then clearly we would have $df =\text{sign}(x) \cdot \text{Id}$. However, I don't think the map $(x,y) \to y \text{sign}(x)$ is Sobolev.
No such a function does not exist. There is a theorem due to Ball and James that says that if you have a Sobolev function whose gradient takes the values $A$ and $B$, then necessarily $A-B$ is a rank one matrix. See Proposition 2.1 in Mueller's lecture notes. In your case $B=-A$ so $A-B=2A$ would have to be rank 1 which is not possible since you are assuming that the matrix is invertible.