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

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 $P=V\Sigma V^T$, so we now have $$ V\Sigma V^T Q^T=QV\Sigma V^T \Rightarrow \Sigma V^T Q^TV=V^TQV\Sigma. $$

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 $Q \in \SO$ $F(A)=Q\cdot O(A)$ 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 $F(A)=Q\cdot O(A)$. Then $ A \perp T_{F(A)}\SO=T_{Q\cdot O(A)}\SO=QT_{O(A)}\SO,$ so for $A=Id$ we have $ Id \perp Q\skew \Rightarrow Q^T \perp \skew \Rightarrow Q^T \in \sym \Rightarrow Q^2=Id$.

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  • $\begingroup$ How about the left orthogonal polar decomposition $O'$ defined by $A=P\cdot O'(A)$? I.e. $O'(A)=O(A^T)^T$ $\endgroup$ – Dap Dec 30 '18 at 7:32
  • $\begingroup$ The left orthogonal polar equals to the right orthogonal polar; this is not trivial, but true. (This can be seen for instance by writing $A=U\Sigma V^T$ , i.e. in terms of its SVD). $\endgroup$ – Asaf Shachar Dec 30 '18 at 10:42
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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 $V\Sigma V^T$ with $V\in SO_n$ (possibly a different $V$ to the one in the question), then $P(A)=V|\Sigma| V^T$ and $Q(A)=V\operatorname{sgn}(\Sigma) V^T$ where $|\cdot|$ and $\operatorname{sgn}$ are applied entrywise to the diagonal elements. This can be seen from the expression $FS=(FV\operatorname{sgn}(\Sigma)V^T)(V|\Sigma|V^T)=OP$ 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 $D=D(\epsilon)=\operatorname{diag}(1+\epsilon,1+2\epsilon,\dots,1+n\epsilon),$ 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 $i\neq j.$ Taking $\epsilon\to 0$ and applying continuity of $D\mapsto U^T Q(UDU^T) U$ 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, $U\mapsto U^T Q(I) U$ 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.

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  • $\begingroup$ wow! This is an amazing solution, really. Do you have any insight regarding how did you think about this "commutation approach"? (i.e. $P,Q,S$ commute and the exploitation of commutation with rotations). Also, two questions about your solution: (1) The reasoning that $U^TQ(UDU^T)U$ commutes with $D$ implies it's diagonal is the same as in my answer, right? i.e. it is because $D_{\epsilon}$ has distinct values? $\endgroup$ – Asaf Shachar Jan 1 at 16:17
  • $\begingroup$ (2) I think we need to address in a special manner the case $n=2$. Since then a matrix which commutes with all rotations does not need to be $\pm id$. (it could be any rotation). So, this case is still open. $\endgroup$ – Asaf Shachar Jan 1 at 16:17
  • $\begingroup$ By the way, we also need to be a bit careful about applying continuity argument on the trace here: It is true that the mp $A \to \text{trace}(Q(A))$ is continuous. However, I only proved that the eigenvalues of $Q(A)$ are $\pm1$ when $A$ has distinct singular values. So, to conclude $\text{trace}(Q(A))$ is constant, we need to know that this set is connected and dense. I am not sure about the connectedness. I asked about this here: math.stackexchange.com/questions/3058644/… $\endgroup$ – Asaf Shachar Jan 1 at 16:56
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    $\begingroup$ @AsafShachar: I've edited to clarified these points, but to summarize: (1) yes, that relies on having distinct values on the diagonal (2) $Q^2=I$ and $Q\in SO_2$ forces $Q=\pm I$ (3) I believe I can have shown $Q(A)^2=I$ for all $A$ $\endgroup$ – Dap Jan 2 at 8:07
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    $\begingroup$ I'm just using commutation as a formal way to talk about matrices that would be simultaneously diagonal in a basis where $P-I$ is diagonal - but this basis depends on $A$ (and might not be able to be chosen continuously in $A$) $\endgroup$ – Dap Jan 2 at 8:11

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