# Is a Sobolev map with smooth minors smooth on the whole domain?

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 $$k>1$$).

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 $$1\leq p, there is a homeomorphism $$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.

• 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. – Asaf Shachar Jan 13 '19 at 13:41