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$\newcommand{\psym}{\text{Psym}_n}$ $\newcommand{\sym}{\text{sym}}$ $\newcommand{\Sym}{\operatorname{Sym}}$ $\newcommand{\Skew}{\operatorname{Skew}}$ $\renewcommand{\skew}{\operatorname{skew}}$ $\newcommand{\GLp}{\operatorname{GL}_n^+}$ $\newcommand{\SO}{\operatorname{SO}_n}$

This might be silly, but I wonder:

Let $F:\GLp \to \SO$ be a continuous retract. Is it true that $F$ must be the orthogonal polar factor, i.e. $F(A)=O$, where $A=OP,O \in \SO,P\in\psym$. Does anything changes if we assume $F$ is a deformation retract? Or if it is a smooth deformation retract?

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Retractions are a very "soft" concept so there is a huge number of those. Take any continuous function $f:PSym_n\to SO_n$ and define $F_f(A):=O\cdot f(P)$ for $A=OP$. Then this is clearly a retraction and it even is a defomormation retraction, which is smooth if $f$ is smooth.

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  • $\begingroup$ Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= O\det A$ or $F(A)= O\frac{\| A\|}{\sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ \to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym \to SO$ which is the identity on $SO$. Is it trivial such a map exists?) $\endgroup$ – Asaf Shachar Dec 21 '18 at 10:35
  • $\begingroup$ In fact, therer are no conditions on the map $f$, so lots of such maps exist $\endgroup$ – Andreas Cap Dec 23 '18 at 8:51

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