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Let $\mathbb{D}^2$ be the closed unit disk, and let $f:\mathbb{D}^2 \to \mathbb{R}^2$ be a real-analytic map, satisfying $\det df>0$ everywhere except on a set of Hausdorff dimension not greater than $1$. Define $$U:= \{ p \in \mathbb{D}^2 \, | \, \det df>0\},$$

and set $Q=Q(df)=df(\sqrt{df^Tdf})^{-1}$ to be the orthogonal polar factor of $df$; $\sqrt{}$ is the unique symmetric positive-definite matrix square root. $Q$ is well-defined on $U$-in particular it is defined a.e. on $\mathbb{D}^2$. The restriction $Q|_U$ is smooth, even real-analytic. So, we can consider the derivatives $Q_x,Q_y$ of $Q$ on $U$.

Question: Are $Q_x,Q_y$ locally integrable on $\mathbb D^2$? i.e. is it true that $Q_x,Q_y \in L^1_{loc}(\mathbb D^2)$?

The easiest case is probably when $f$ is conformal a.e. (so it's actually holomorphic), as then $Q(df)=\frac{\sqrt 2}{\|df\|}df$ is obtained from $df$ simply by normalization. I asked on this case separately here.

Comment:

The only possible problem is when we approach points where $df=0$, as the polar factor map $\text{GL}_2^+ \to \text{SO}_2$ can be extended to a smooth map $\text{GL}_2^+ \cup (\text{rank} = 1) \to \text{SO}_2$.

However, the polar factor cannot be extended continuously on all $\text{GL}_2^+ \cup \det^{-1}(0)$, as singularities occur:

  1. $\lim_{t \to 0} Q(\left(\begin{matrix}t & 0 \\ 0 & t\end{matrix}\right))=\text{sgn}(t) \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$, and

  2. $\lim_{t \to 0}Q(\left(\begin{matrix}0 & -t \\ t & 0\end{matrix}\right))=\text{sgn}(t) \left(\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right)$.


This question is a first step in understanding wether or not $Q \in W^{1,p}(\mathbb{D}^2,\mathbb{R}^{4})$ for some $p \ge 1$. (As you can see here).

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