Proposition 45.2.3.label Let $H$ be a complex Hilbert space and $M \subset B(H)$ be a von Neumann algebra, then
- (1)
For any $p \in \text{Proj}(M)$, the commutant $(pMp)'$ in $B(pH)$ is $pM'$.
- (2)
For any $q \in \text{Proj}(M')$, the commutant $(Mq)'$ in $B(qH)$ is $qM'q$.
Proof, [Proposition 4.2.2, AP]. (1): For any $x \in M$ and $y \in M'$, $pxp \cdot yp = ypxp = yp \cdot pxp$, so $pMp \subset (M'p)'$.
For any $x \in (M'p)' \subset B(pH)$, let $\tilde x = xp = pxp \in B(H)$, then for any $y \in M'$,
\[\tilde x y = pxp \cdot y = pxp \cdot yp = yp \cdot pxp = y \tilde x\]
By the Bicommutant Theorem, $\tilde x \in M$, $x = p\tilde x p \in pMp$, and $pMp \supset (M'p)'$.$\square$
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