# Is the orthogonal polar factor the unique submersion satisfying an orthogonality relation?


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 $$Id \perp Q\skew \Rightarrow Q^T \perp \skew \Rightarrow Q^T \in \sym \Rightarrow Q^2=Id$$.

• How about the left orthogonal polar decomposition $O'$ defined by $A=P\cdot O'(A)$? I.e. $O'(A)=O(A^T)^T$ – Dap Dec 30 '18 at 7:32
• 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). – Asaf Shachar Dec 30 '18 at 10:42

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.

• 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? – Asaf Shachar Jan 1 at 16:17
• (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. – Asaf Shachar Jan 1 at 16:17
• 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/… – Asaf Shachar Jan 1 at 16:56
• @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 $A$ – Dap Jan 2 at 8:07
• 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 $A$) – Dap Jan 2 at 8:11