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Let $M$ be a von Neumann algebra on $B(H)$. and let $p \in M$ be a projection How can we prove that the compression $pMp$ is a von Neumann algebra on $B(pH)$?

I saw a result that if $p \in M$ is a projection, then $(pMp)'=pM'p$. Using this can we conclude that

$(pMp)''=pM''p=pMp$ since $M$ is a von neumann algebra? I know if $M$ is a von Neumann algebra so is $M'$. But to apply the result $(pMp)'=pM'p$ once more we need $p\in M'$ right?

Please help me.

budi
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1 Answers1

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One cannot expect $p\in M'$ to hold in general. (Take for example $M=M_2$ and $p=\bigl(\begin{smallmatrix}1&0 \\ 0&0\end{smallmatrix}\bigr)$.)

But why not instead prove that $pMp$ is a von Neumann algebra directly by showing that $pMp$ is a weakly closed $C^*$-subalgebra of $B(pH)$?

Interestingly, showing that $pMp$ is a von Neumann algebra by proving that it is the commutant in $B(pH)$ of the von Neumann algebra $pM'$ is more difficult but possible, see Corollary 5.5.7 of Kadison and Ringrose, and has the advantage of giving an expression of the commutant of $pMp$ in $B(pH)$.

Bram Westerbaan
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