Is the polar decomposition of a product the product of polar decompositions? Given the polar decompositions $A=O_1P_1$, $B=O_2P_2$, and $AB=O_3P_3$ with $O_i$ being orthogonal matrices and $P_i$ being symmetric positive definite matrices, is there a nice way to write $O_3$ in terms of $O_1$ and $O_2$ (as well as write $P_3$ in terms of $P_1$ and $P_2$)?
 A: The singular vectors of $AB$ depends on the angles between the singular vectors of $A$ and $B$; since the polar decomposition can be derived from the singular value decomposition consequently I don't believe it'll be possible to write a very "nice" expression.
But we can try the best we can, which will reveal that the relationship doesn't really have a general form. If we decompose $A,B$ into their SVDs:
$$ A = U_1 \Sigma_1 {V_1}^*, \quad B = U_2 \Sigma_2 {V_2}^*$$
then their polar decompositions are
$$ A = O_1 P_1, \quad O_1 = U_1 {V_1}^*, P_1 = V_1 \Sigma_1 {V_1}^*$$
$$ B = O_2 P_2, \quad O_2 = U_2 {V_2}^*, P_2 = V_2 \Sigma_2 {V_2}^*$$
Starting from the SVDs let's expand the matrix product $AB$:
$$AB = U_1 \Sigma_1 {V_1}^* U_2 \Sigma_2 {V_2}^* = U_1 C {V_2}^*, \quad C = \Sigma_1 {V_1}^* U_2 \Sigma_2$$
The new matrix $C$ here depends on the angles between the left singular vectors of $A$ and the right singular vectors of $B$, as shown through the product ${V_1}^* U_2$ in the expression for $C$. Assuming we know the SVD of $C$ as
$$ C = U_3 \Sigma_3 {V_3}^*,$$
then
\begin{align}
AB & = U_1 U_3 \Sigma_3 {V_3}^* {V_2}^* \\
& = (U_1 U_3 {V_3}^* {V_2}^* ) (V_2 V_3 \Sigma_3 {V_3}^* {V_2}^*) \\
& = O P
\end{align}
But of course $U_3, \Sigma_3, V_3$ come from the SVD of $C$, and from inspecting the form of $C$ we can observe that $C$ is effectively arbitrary.
