Let $M$ be a finite von Neumann algebra, $B$ a von Neumann subalgebra of $M$, and $p$ be a projection in $B$. Given that $B'\cap M$ is diffuse, is it true that the relative commutant in the corner $(pBp)'\cap pMp=B'p \cap pMp$ is also diffuse?
I am reading Chapter 19.2 of the book "An introduction to $II_1$ factors" by Anantharaman-Delaroche and Popa, and though they wrote this as if it is trivial, I couldn't come up with a proof.
OK (I think) I just solved it on my own, here's just a memorandum for people:
We prove $(B'\cap M)p =B'p \cap pMp$, from which the result immediately follows.
$LHS\subset RHS$ is trivial. We prove the converse. Let $x\in B'p \cap pMp$. Then $x=yp$ for some $y\in B'$. If $y\in B'\cap M$ then we are done. Take $z\in Z(B)=Z(B')$ as the minimal projection in $Z(B)$ satisfying $zp=p$. Then, the kernel of the $\ast$-hom $B'\to B'p:x\mapsto xp$ equals $B'(1-z)$, and $x=yp=yzp$.
Then, for every $w\in M'\subset B'$, one gets $ywzp=yzwp=yzpw=xw=wx=wyzp$.
Then we get $(yw-wy)z\in B'(1-z)$, therefore $yzw=wyz$ and $yz \in (M')'=M$.
Please notice me if this is wrong.
