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.)

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.

  • $\begingroup$ I'm just being a pedant here, but... As the orthogonal projection is bounded, surely you mean "in M" not just "affiliated with M". Also, as far as I can see, R is closed because e is a closed operator; you never need to use that e is idempotent. And also Martin's (but not, I think, Jonas's) argument seems still to work without using that e is idempotent. I ask, because $e^2=e$ and $e^2 \subseteq e$ are two possible interpretations of what "idempotent" means... $\endgroup$ Aug 26, 2010 at 8:33
  • $\begingroup$ Of course, it occurs to me that under your other conditions (namely that $e\in L^0(M,\tau)$, maybe $e^2\subseteq e$ implies that $e^2=e$... $\endgroup$ Aug 26, 2010 at 8:43
  • $\begingroup$ If by $e^2\subseteq e$ you mean inclusion of ranges, then the implication of your last comment fails even in finite dimension: if you consider the unilateral shift, $s^2\subseteq s$ but $s^2\neq s$. With this example in mind, I would say that $e^2\subset e$'' is not a very good notion of idempotent''. $\endgroup$ Aug 26, 2010 at 13:23
  • $\begingroup$ (by the way, thanks for noticing the "affiliated" mistake; it's corrected in the solution now) $\endgroup$ Aug 26, 2010 at 13:24
  • $\begingroup$ Thanks for the comments. Yes, I should have said that I want to end up with $R\in M$. I'm quite prepared to believe that one doesn't need $e^2=e$ in $L^0(M,\tau)$ for the same conclusion to hold, but I thought I'd state the problem in the specific case I was looking at. $\endgroup$
    – Yemon Choi
    Aug 26, 2010 at 20:03

2 Answers 2


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 $r\in M$.

  • $\begingroup$ That seems to do the trick, thanks. I'd been trying to do something similar with $u$ just an arbitrary element of $M$, and hadn't thought of restricting attention to unitaries in order to exploit extra structure. If no one else suggests a reference then I'll accept this answer. $\endgroup$
    – Yemon Choi
    Aug 25, 2010 at 0:32
  • $\begingroup$ Neat. You must mean $u\in M'$. $\endgroup$ Aug 25, 2010 at 5:30
  • $\begingroup$ Heh, I see I reproduced the typo as well. In my (and Martin's) defence I think the intended meaning is clear. $\endgroup$
    – Yemon Choi
    Aug 25, 2010 at 5:41
  • $\begingroup$ Thinking about this a bit more, it seems we don't actually need unitaries for the argument. Using the same notation as in Martin's answer: let $c\in M'$ so that $c\xi=ce\xi=ec\xi$ and thus $cR\subseteq R$, which is equivalent to $rcr=cr$. Since this holds for any $c\in M'$ it holds if we replace $c$ by $c^*$, giving $rc^*r=c^*r$. But now taking adjoints and remembering that $r$ is self-adjoint we get $rcr=rc$, and thus $rc=cr$. Does this look right? $\endgroup$
    – Yemon Choi
    Aug 25, 2010 at 5:56
  • $\begingroup$ Looks right to me. I've seen some sources that even use unitaries in the definition of "affiliated", so even when not needed I imagine that that formulation comes more naturally to some. $\endgroup$ Aug 25, 2010 at 6:07

Disclaimer: I fear I may be missing some subtlety here, as is often the case when I think about unbounded operators. This is an attempt to generalize the result.

A closed densely defined operator $T$ on $H$ has a unique polar decomposition $T=V|T|$ with $|T|=\sqrt{T^*T}$ and $V$ a partial isometry whose initial space is the closure of the range of $|T|$ and whose final space is the closure of the range of $T$. If $T$ is affiliated with a von Neumann algebra $M$, then $V$ is in $M$ (as stated e.g. in Nelson's paper on the bottom of page 111). Thus $VV^*$, the projection onto the closure of the range of $T$, is in $M$.

  • $\begingroup$ The argument looks ok to me. Neat! $\endgroup$ Aug 25, 2010 at 8:44

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