# Is the orthogonal polar factor the unique retraction $\operatorname{GL}_n^+ \to \operatorname{SO}_n$?


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?

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.
• 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?) – Asaf Shachar Dec 21 '18 at 10:35
• In fact, therer are no conditions on the map $f$, so lots of such maps exist – Andreas Cap Dec 23 '18 at 8:51