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

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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.