# Is a Sobolev map with invertible smooth minors smooth?

$$\newcommand{\Cof}{\text{cof}}$$ 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).