To be slightly more precise: let $M\subset B(H)$ be a finite von Neumann algebra equipped with a faithful normal trace $\tau$, and let $L^0(M,\tau)$ be the completion of $M$ in the measure topology; this is an algebra, whose elements can be identified with those densely-defined and closed operators on $H$ that are affiliated with $M$. (See e.g. E. Nelson, *Notes on noncommutative integration*, JFA 1974). Let $e$ be an idempotent in $L^0(M,\tau)$, not necessarily self-adjoint; then it is not hard to show that $R=\{ \xi\in H : e\xi=\xi\}$ is a closed subspace of $H$.


**Question:** is the orthogonal projection onto $R$ affiliated with $M$?


I suspect the answer is yes (and would like it to be, for some calculations I'm doing at the moment) but am having difficulties nailing the argument down. Given that this should, if true, be a pretty basic bit of operator algebra theory, and standard knowledge, I'd be grateful if someone could point me to a reference. (I currently have somewhat limited library access, but this might well be covered in Kadison & Ringrose for instance.)




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**Edit/update:** both Martin Argerami and Jonas Meyer have given straightforward proofs of the desired result, and a quick check in Kadison & Ringrose vol. 1 has not turned up any explicit statement (probably because the result turns out to be so basic). Since I can't accept both their answers, I'm accepting Martin's on grounds of personal preference.



range projection of an unbounded idempotent affiliated to a finite von Neumann algebra
I couldn't find it in Kadison & Ringrose. But what about this: let $\xi\in R$ and let $u$ be a unitary in $M'$. Since $e$ is affiliated with $M$, $ue=eu$. So $u\xi=ue\xi=eu\xi\in R$. This shows that $uR\subset R$ for any unitary $u$, and so $uR=R$ for any unitary $u$ in $M'$. This in turn is equivalent to $ur=ru$, where $r$ is the orthogonal projection onto $R$. As $u$ was any unitary in $M'$, we conclude that $r\in M$.


