Suppose $L$ is a completely positive contraction of $B(H)$ into itself which is idempotent (i.e. $L^2 = L$) and suppose the range of $L$ is a von Neumann factor $M$. Suppose further that $L$ is normal, so $$ L(A) = \sum_{i=1}^\infty S_i A S_i^* $$ for $A\in B(H)$. Then $M$ is a type I factor. Has anyone seen a proof of this? I believe, if the range of $L$ is a von Neumann algebra $R$, then $R$ is atomic. The key is $L$ is normal.
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2$\begingroup$ Dear Bob: if Nik's answer is satisfactory, you can mark it as "officially accepted", which will take your question off the list of "questions still requiring an answer" $\endgroup$– Yemon ChoiMay 7, 2018 at 1:04
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This is in Bruce Blackadar's book, Operator Algebras: Theory of C*-Algebras and Von Neumann Algebras. See Theorem IV.2.2.3.
answered May 6, 2018 at 3:38