If $T=V|T|\text { and } S=W|S|$ is the polar decomposition of $T$. Is it true that the polar decomposition of $T\otimes S$ is $T\otimes S=(V\otimes W)(|T| \otimes |S|)$. If $T$ and $S$ are self-adjoint what is $f(T\otimes S)$ where f is continous function appropriately defined on the domain?



Polar decomposition of tensor product of operators in von Neumann algebra
For the 1st part, the answer is "yes".


Let $T,S$ be bounded operators on $H$ and $K$ respectively. As $(T\otimes S)^\*(T\otimes S) = |T|^2\otimes|S|^2$ it follows that $|T\otimes S| = |T|\otimes |S|$. With polar decompositions $T=V|T|, S=W|S|$ we have that $(V\otimes W)|T\otimes S| = T\otimes S$. Clearly $V\otimes W$ is a [partial isometry](https://en.wikipedia.org/wiki/Partial_isometry). If $T\otimes S = U(|T|\otimes |S|)$ is the polar decomposition, then $U$ equals $V\otimes W$ on the image of $|T|\otimes |S|$. The question is whether $(V\otimes W)(\tau)=0$ for any $\tau$ in $\operatorname{Im}(|T|\otimes |S|)^\perp$; as this, by construction, is true of $U$, and would imply that $U=V\otimes W$.


Now, $p=V^\*V$ is the projection onto the closure of the image of $|T|$, and similarly $q=W^\*W$ is the projection onto the closure of the image of $|S|$. We know that
$$ (\tau|(|T|\otimes|S|)(x\otimes y))=0 \qquad (x\in H, y\in K). $$
Which implies that
$$ (\tau|pz \otimes qw)=0 \qquad (z\in H, w\in K), $$
which implies
$$ ((p\otimes q)\tau|z\otimes w)=0 \qquad (z\in H, w\in K), $$
which implies $(p\otimes q)\tau=0$. Thus
$$ 0 = (V\otimes W)(p\otimes q)\tau = (VV^\*V\otimes WW^\*W)\tau
= (V\otimes W)\tau, $$
as we wanted to prove.


