help showing a property for a weak operator closed $^*$-subalgebra of the bounded operators of a Hilbert space.

Question

Let $A$ be a weak operator closed $^*$-subalgebra of the bounded operators of a Hilbert space. If $T \in A$, then I am trying to show that $P_{(\ker T)^\perp} \in A$(projection for $(\ker T)^\perp$. I have noticed that from polar decomposition the unique partial isometry $U$ satisfies $U^*U = P_{(\ker T)^\perp}$. I am not so sure how to proceed. Would using borel or continuous functional calculus do anything?

Answer 1

As mentioned by David, there is a rather direct argument using Borel functional calculus. But the result can be proven with only the Double Commutant Theorem.

You have $T=U|T|$ from the polar decomposition. You also have $|T|\in A$. Let $S\in A'$. For any $x\in H$ we have $$ SU|T|x=STx=TSx=U|T|Sx=US|T|x. $$ The above shows that $SU=US$ on $\def\ran{\operatorname{ran}}$ $\ran |T|$. Since $S|T|=|T|S$, $S(\ran |T|)\subset\ran|T|$. As $S^*\in A'$, if $y\in(\ran |T|)^\perp$, $$ \langle Sy,|T|x\rangle=\langle y,S^*|T|x\rangle=\langle y,|T|S^*x\rangle=0. $$ So $Sy\in(\ran |T|)^\perp=\ker |T|$. Then $SUy=0$ and $SUy=0$. So $US=SU$ also on $(\ran |T|)^\perp$ and so $SU=US$. This means that $U\in A''=A$.