Polar decomposition of tensor product of operators in von Neumann algebra

Question

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?

Answer 1

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. 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.

Answer 2

To answer the second question, let $\mathcal{M}_1$ and $\mathcal{M}_2$ be the von Neumann subalgebras generated by $T$ and $S$, respectively. Then $\mathcal{M}_1\otimes\mathcal{M}_2$ embeds in $\mathcal{M}\otimes\mathcal{M}$, and as $\mathcal{M}_1$ and $\mathcal{M}_2$ are abelian we have $\mathcal{M}_1\cong L^\infty(X)$ and $\mathcal{M}_2\cong L^\infty(Y)$ for some $X$ and $Y$, with $T$ corresponding to $g \in L^\infty(X)$ and $S$ corresponding to $h \in L^\infty(Y)$. The tensor product of $g$ and $h$ is just the function $f(x)g(y)$ and functional calculus works by composition.