Polar decomposition in a von Neumann algebra

Question

Let $M \subseteq B(H)$ be a von Neumann algebra and $T \in M$. If $T=U|T|$ is the polar decomposition of T, why is $U \in M$? I'm thinking it's because $M$ is SOT-closed, but I'm not entirely sure.

Answer 1

The polar decomposition of $T \in M$ is given by $T=U|T|$, where U is a partial isometry, $|T|=(T^{*}T)^{\frac{1}{2}}$ and $\text{ker }T= \text{ker }|T|=\text{ker }U$. Clearly, $|T| \in M$ since it is an SOT closed subalgebra of $B(H)$. To show that $U \in M$, it suffices to show $U \in M''$ by the double commutant theorem. Let $S\in M'$.

$TSx=STx=SU|T|x$ and $TSx=U|T|Sx=US|T|x$. Hence $SU$ and $US$ agree on $\overline{Ran(|T|)}$. By the self adjointness of $|T|$, $\overline{Ran(|T|)}^{\perp}=\text{ker } |T|= \text{ker }U$.

Let $x\in \text{ker }|T|$.

$|T|Sx=S|T|x=0$. Hence $S(\text{ker }|T|)\subset\text{ker }|T|=\text{ker } U$ so that $US=0$ on ker $|T|$. $SUx=0$ clearly, if $x \in \text{ker } |T|=\text{ker }U$.

So $US=SU$ whence $U\in M''=M$.