Every $n \times n$ real matrix $A$ has a polar decomposition $A=OP$, where $O \in O_n, P$ is symmetric positive semi-definite.

$P$ is uniquely determined by $A$, by $P(A)=\sqrt{A^TA}$, and when $A$ is invertible $O$ is also unique, given by $O=A(\sqrt{A^TA})^{-1}$.

When $A$ is not invertible, then $O$ is non-unique.

**Question:**

Can we choose a **smooth** polar factor $O(A)$ for all **non-zero** matrices $A$ with non-negative determinant?

More precisely, denote by $M_n^+$ the set of matrices with non-negative determinant.

Is there a smooth function $O:M_n^+ \setminus \{0\} \to O_n $ such that for every $A \in M_n^+ \setminus \{0\}$, $A=O(A)P(A)=O(A)\sqrt{A^TA}$?

**Remarks:**

1) We know that such a function, if exists, must give to each $A \in GL_n^+$ its unique polar factor.

2) The reason we exluded the zero matrix is that continuity implies two contradictory evaluations:

$O(0)=\lim_{t \to 0^+} O(\left(\begin{matrix}t & 0 \\ 0 & t\end{matrix}\right))=I$,

$O(0)=\lim_{t \to 0^+}=O(\left(\begin{matrix}0 & -t \\ t & 0\end{matrix}\right))=\left(\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right)$

3) The reason we had to restrict to $M_n^+ \setminus \{0\}$ (instead of working with all $M_n \setminus \{0\}$) is connectedness issues.

$M_n\setminus \{0\}$ is connected (for $n >1$). However for $O|_{O_n}=Id_{O_n}$, so if we insisted to take the domain to be all $M_n \setminus \{0\}$, the image would be $O_n$ which is disconnected.

The restriction in fact implies that $O$ is a function into $SO_n$.