Let $d\ge 3$ and $2 \le k \le d-1$ be integers, where **at least one of $k,d$ is odd**. 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 $ is smooth. Is $f$ smooth?

$f$ must be smooth on $\Omega_0=\{ x \in \Omega \, | \, \bigwedge^k df_x \in \text{GL}(\bigwedge^{k}\mathbb{R}^d) \}$, which is an open subset of full measure on $\Omega$. I ask if $f$ is smooth on all $\Omega$.

*A proof sketch for regularity on $\Omega_0$:* (for the full details see theorems 1.1 and 1.2 here).

On $\Omega_0$, we can smoothly reconstruct $df$ from $\, \,\bigwedge^k df$: 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.

Without the invertibility assumption $\bigwedge^k df \in \text{GL}(\bigwedge^{k}\mathbb{R}^d)$ **everywhere on $\Omega$**, we cannot inverse the map $df \to \bigwedge^k df$ on all $\Omega$ (as the rank can fall on a subset of measure zero). Thus, I don't see how to advance beyond the "good" set $\Omega_0$.

**Edit:**

If $k$ is odd, I can say a bit more under additional assumptions: Suppose that $f \in W^{1,\infty}(\Omega,\mathbb{R}^d)$, and that $\text{rank}(\bigwedge^k df)>k$ everywhere on $\Omega$ and that $\bigwedge^k df$ is smooth. Then $f$ is smooth.

*Proof sketch:*

Since $\text{rank}(\bigwedge^k df)=\binom{\text{rank}(df)}{k}$, $\text{rank}(\bigwedge^k df)>k$ implies $\text{rank}(df) > k$ a.e., and in this regime the map $df \to \bigwedge^k df$ is smoothly invertible. (This relies upon the pointwise algebraic fact that the exterior power map $A \to \bigwedge^k A$ is invertible when restricting the domain to the set of matrices of rank larger than $k$).

For full details, see theorem 5.10 here.

*Comment:*

I don't think that (naive) quantitative estimates are possible here: Small $\|\bigwedge^k df\|$ does not imply small $\|df\|$; if one singular value of $df$ is $n$, and all the rest are $\frac{1}{n}$, then all the singular values $\bigwedge^k df$ are bounded below by $1$ (If $k>1$).