Let $M$ be a von Neumann algebra and $p$ is a projection in $M$. Does there exist relationship between the commutant $(pMp)'$ of $pMp$ and $pM'p$.
I know the fact that if $p\in M'$, we have $(pMp)'=pM'p.$
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Let $N=M'$. Then $p\in N'$. So, by the fact you know, $$ (pM'p)'=(pNp)'=pN'p=pMp. $$ Taking commutant (and here you need to be careful when taking double commutants of degenerate things), $$\tag1 pM'p+(1-p)B(H)(1-p)=(pMp)'. $$ If you want to consider $pMp\subset B(pH)$, then $$ pM'p=(pMp)'. $$
answered Mar 16, 2021 at 16:06
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$\begingroup$ If $pMp\subseteq B(pH)$, how to conclue that $pM'p=(pMp)'$? $\endgroup$ Commented Mar 16, 2021 at 21:07
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$\begingroup$ You just compress equation $(1)$ with $p$. $\endgroup$ Commented Mar 16, 2021 at 21:30