Ytukyg

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

The orthogonal polar factor map $O:GLp to SO$, defined by requiring $A=
O(A)P$ for some symmetric positive-definite $P$, is a smooth submersion satisfying $A perp T_{O(A)}SO$.

Question: Let $F:GLp to SO$ be a smooth submersion satisfying $A perp T_{F(A)}SO$. Does $F(A)=Q cdot O(A)$ or $F(A)= O(A) cdot Q$ for some $Q in SO$?

Edit: An equivalent reformulation of the question:

$A perp T_{F(A)}SO=F(A)skew iff A in (F(A)skew)^perp=F(A)(skew)^perp=F(A)sym$. Thus, if we define $S(A)=F(A)^{-1}A$, $S:GLp to sym$ is smooth.

So, a submersion $F:GLp to SO$ satisfies the orthogonality requirement if and only if there exist a smooth map $S:GLp to sym$, satisfying $A=F(A)S(A)$.

Using the polar decomposition, we have $O(A)P(A)=A=F(A)S(A)$, so $S(A)=Q(A)P(A)$ where $Q(A)=F(A)^{-1}O(A)$. We want to prove that $Q:GLp to SO$ is constant. I will now prove that for matrices $A$ having distinct singular values, $Q(A)$ can obtain a finite number of values. (The set of admissible values depends on $A$). I am quite sure this fact can be used to force $Q$ to be constant or at least something very constrained, but I am not sure how. Here is the proof:

Since $S(A)=Q(A)P(A) in sym$ , we have $PQ^T=(QP)^T=S^T=S=QP$. By orthogonally diagonalizing $P$, we can write $P=VSigma V^T$, so we now have
$$ VSigma V^T Q^T=QVSigma V^T Rightarrow Sigma V^T Q^TV=V^TQVSigma. $$

Setting $tilde Q=V^TQV$ we thus have $ Sigma tilde Q^T= tilde Q Sigma$ where $tilde Q in SO$. Since we assumed that the singular values of $A$ are distinct (i.e. the diagonal entries of $Sigma$ are distinct), an explicit calculation now shows that $ tilde Q$ must be diagonal. Since it is also orthogonal, we must have $tilde Q_{ii}=pm 1$ for all $i$. So, $tilde Q$ can assume a finite set of values; (which implies the same thing for $ Q$).

Comment: I am not sure for which $Q in SO$ $F(A)=Qcdot O(A)$ satisfies the requirement. A necessary condition is $Q^2=Id$; I don't know if it's sufficient.

Indeed, let $Q in SO$, and set $F(A)=Qcdot O(A)$. Then $ A perp T_{F(A)}SO=T_{Qcdot O(A)}SO=QT_{O(A)}SO,$ so for $A=Id$ we have $ Id perp Qskew Rightarrow Q^T perp skew Rightarrow Q^T in sym Rightarrow Q^2=Id$.

I will use the notation from the question, $O(A)P(A)=A=F(A)S(A)$ and $S(A)=Q(A)P(A).$

The only possibilities are $Q(A)=I$ everywhere and $Q(A)=-I$ everywhere, with $-I$ only valid $n$ is even.

To show this I will first argue that $Q(I)=pm I.$

The matrices $P(A),Q(A),S(A)$ all commute: diagonalize $S$ as $VSigma V^T$ with $Vin SO_n$ (possibly a different $V$ to the one in the question), then $P(A)=V|Sigma| V^T$ and $Q(A)=Voperatorname{sgn}(Sigma) V^T$ where $|cdot|$ and $operatorname{sgn}$ are applied entrywise to the diagonal elements. This can be seen from the expression $FS=(FVoperatorname{sgn}(Sigma)V^T)(V|Sigma|V^T)=OP$ and uniqueness of orthogonal polar decomposition.

For the case $n=2,$ we can use the above expression for $Q(A)$ to get $Q(A)^2=I.$ And $Q(A)in SO_n.$ This forces $Q(A)=pm I.$

Let $D=D(epsilon)=operatorname{diag}(1+epsilon,1+2epsilon,dots,1+nepsilon),$ and let $U$ be an arbitrary element of $SO_n.$ Since $Q(UDU^T)$ commutes with $P(UDU^T)=UDU^T,$ the conjugate $U^T Q(UDU^T) U$ commutes with $D.$ Any matrix $M$ that commutes with $D$ must be diagonal because $(DM-MD)_{ij}=M_{ij}(D_{ii}-D_{jj})=0$ forces $M_{ij}=0$ for $ineq j.$ Taking $epsilonto 0$ and applying continuity of $Dmapsto U^T Q(UDU^T) U$ we get that $U^T Q(I) U$ is diagonal. This restricts $U^T Q(I) U$ to a discrete set $operatorname{diag}(pm1,dots,pm 1).$ Since $SO_n$ is connected, $Umapsto U^T Q(I) U$ is constantly $Q(I).$ In other words $Q(I)$ commutes with every special orthogonal matrix, but for $n>2$ that forces $Q(I)=pm I$ as claimed.

From the above expression for $Q(A)$ it is clear that $Q(A)$ has eigenvalues $pm 1,$ which means the trace is a discrete invariant and must therefore be constant. If $Q(I)=I$ then the trace is $n$ and $Q(A)=I$ everywhere, and if $Q(I)=-I$ then the trace is $-n$ and $Q(A)=-I$ everywhere.