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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 $k>1$).

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The theorem quoted below is from:

Z. Liu, J. Malý, A strictly convex Sobolev function with null Hessian minors. Calc. Var. Partial Differential Equations 55 (2016), no. 3, Art. 58, 19 pp.

This is not really a counterexample to the question above since we relax the assumption about the determinant. The determinant of the mapping is $\det df= 0$ a.e. On the other hand the mapping is a homeomorphism so intuitively it is very close to a mapping with Jacobian positive a.e.

Theorem. Given $1\leq p<k\leq d$, there is a homeomorphism $f\in W^{1,p}((0,1)^d,\mathbb{R}^d)$ that is Holder continuous with any exponent $\alpha<1$ and such $\bigwedge^k df=0$ a.e. (so $\bigwedge^k df$ is smooth).

Clearly, the condition $\bigwedge^k df=0$ implies that $\det df=0$ a.e. This homoemorphism does not belong to $W^{1,d}$ and hence it is not smooth. Moreover, and that is a really surprising fact, $f=\nabla F$ for some convex function $F\in W^{2,p}$.

This result is truly beautiful and surprising. You should look at related references to see if they can lead to a counterexample to your question. I would suggest that you contact Zhuomin Liu.

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  • $\begingroup$ Thanks. As you said, this is not directly related since my assumptions imply that the $k$-minors contain non-trivial information (in a sense they contain all the information we can dream of, as we can reconstruct from them the differential $df$ a.e). However, this paper looks very interesting indeed, and I wasn't aware of it. Thank you. $\endgroup$ – Asaf Shachar Jan 13 '19 at 13:41

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