$\newcommand{\Cof}{\text{cof}}$ Let $k,d$ be even integers, such that $d\ge3$ and $2 \le k \le d-1$. Let $\Omega \subseteq \mathbb{R}^d$ be open, and let $f \in W^{1,p}(\Omega,\mathbb{R}^d)$, for some $p \ge 1$.

Question: Suppose that $\det df>0$ a.e. and that $\bigwedge^k df \in \text{GL}(\bigwedge^{k}\mathbb{R}^d)$ is smooth. Is $f$ smooth?

When $k,d$ are not both even, the answer is positive. Here is the idea: (for the full details see theorem 1.1 here).

$\bigwedge^k df$ uniquely determines $df$ (assuming $\det df>0$) in a way that makes the inverse map smooth: If $A,B \in \text{GL}^+(\mathbb{R}^d)$ and $\bigwedge^k A=\bigwedge^k B$, then $A=B$. The exterior power map $\psi: A \to \bigwedge^k A$ is a smooth embedding when considered as a map $\text{GL}^+(\mathbb{R}^d) \to \text{GL}(\bigwedge^{k}\mathbb{R}^d)$; $\text{Image} (\psi)$ is a closed embedded submanifold of $\text{GL}(\bigwedge^{k}\mathbb{R}^d)$, which makes $\psi:\text{GL}^+(\mathbb{R}^d) \to \text{Image} (\psi)$ a diffeomorphism. Composing $x \to \bigwedge^k df_x$ with the smooth inverse of $\psi$ finishes the job.

In the case when $k,d$ are both even, $\bigwedge^k A=\bigwedge^k (-A)$ and both $A,-A \in \text{GL}^+(\mathbb{R}^d)$. Thus the $k$-minors cannot distinguish beween a map and its negative, so theoretically $df$ could "switch" between "something" and its negative, thus violating smoothness.

Edit 1:

A naive approach would be to try to construct a (non-smooth) Sobolev map $f$ whose differential $df$ "zig-zags" between a given fixed invertible matrix $A$ and its negative. However, such a map does not exist; Indeed, if the gradient of a non-affine Sobolev map takes only the values $A$ and $B$, then necessarily $A − B$ is a rank one matrix. (This is Proposition 2.1 in 5).

Comment: Every possible counter-example for the case where $k,d$ are both even must have non-continuous weak derivatives. (Indeed, if the weak derivatives are continuous, we can inverse the map $A \to \bigwedge^k A$ locally, since we "know which branch to choose"- $\bigwedge^k A=\bigwedge^k B$ implies $A=\pm B$.)

Edit 2:

I can reduce the question to the case where $k$ is a power of $2$. The idea is that if $k=2^rm$ where $m$ is odd, then if the $k$-minors $\bigwedge^k df$ are smooth, then so is $\bigwedge^{2^r} df$. (essentially because the $k$-minors determine the $2^r$-minors, for full details see section 6.2.1 here).


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