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Hi all, sorry if this is a dumb question, I don't know much about von Neumann algebras except the definition and a few relevant facts I've managed to prove by myself so I expect the answer will turn out to be well known. Anyway, let $\mathcal{H}$ be a Hilbert space, and suppose that $P$ is a commuting set of self-adjoint projections on $\mathcal{H}$, with the additional two properties:

1) $P$ is closed under complements, i.e. if $p \in P$ then so is $1 - p$.

2) $P$ is closed under suprema of arbitrary subsets, i.e. if $S \subseteq P$ then $\sup S \in P$ (here the projections on $\mathcal{H}$ are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$).

Now let $V$ denote the smallest von Neumann algebra containing $P$. Suppose that $p \in V$ is a self-adjoint projection. Is $p \in P$?

I know that $p$ is necessarily in the closure (relative to the weak operator topology) of the set of finite sums $\sum_i \lambda_i p_i$, where $p_i \in P$ and $\lambda_i \in \mathbb{R}$. It seems like it may be possible to derive a contradiction from the assumption that $q$ has a strictly smaller range than $p$, where $q \equiv \sup ${$ r \in P | r \leq p $}. But I don't know how to proceed.

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2 Answers 2

up vote 5 down vote accepted

The answer "yes" follows from Theorem 2.8 of Bade's "On Boolean algebras of projections and algebras of operators," 1955, which is in the more general context of algebras of operators on a Banach space.

Bade had previously proven a less general result that still covers your case, dealing with algebras of operators on reflexive spaces, in Theorem 3.4 of "Weak and strong limits of spectral operators," 1954.

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That's great, thanks. It will take me a while to digest the papers but it's good to know the answer. –  Phil Wild Jul 24 '10 at 2:13
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This is off the cuff, not a complete answer, and I haven't checked through it carefully, but it was getting too long for a comment. I'm making it community wiki in case someone sees a way to iron out the wrinkles, although if they have a better approach they should feel free to write it up as a separate answer.

Anyway. Since $V$ is an abelian von Neumann algebra, it has the form $C(\Omega)$ for some compact, extremally disconnected toppological space. Then your original set $P$ can be regarded as the set of indicator functions of a family $\mathcal F$ of clopen subsets; $\mathcal F$ is closed under taking complements and under formation of arbitrary unions.

Here is the part which I haven't quite nailed down at time of writing: we note/claim that any element of $V$ is the pointwise limit of some bounded net of finite linear combinations of characteristic functions of sets in ${\mathcal F}$, in particular $p$, which must be the indicator function of some clopen $X\subseteq\Omega$, must be approximable in this way. Now we consider the family ${\mathcal C}$ of all $A\in{\mathcal F}$ which have non-empty intersection with $X$. The approximability assumption should force $\bigcup_{A\in\mathcal C} A$ to be $X$, and thus $X\in\mathcal F$, which is the same as saying that $p\in P$ as required.

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Thanks for your reply, though there are a couple of points I don't get. Am I right in thinking that $\Omega$ can be taken to be the maximal ideal space of $V$? If so then can't we only assume that $\mathcal{F}$ is closed under closures of arbitrary unions, rather than the unions themselves? If so I don't think this really matters, see below. I don't think your choice of $\mathcal{C}$ works - if $X$ is non-empty then surely the whole of $\Omega$ is in $\mathcal{C}$? Suppose instead we let $\mathcal{C}$ consist of all $A \in \mathcal{F}$ with $A \subseteq X$. [contd.] –  Phil Wild Jul 24 '10 at 2:17
    
Then if $B \equiv \overline{\cup \mathcal{C}}$ then clearly $B \subseteq X$. To show the reverse inclusion, let $x \in X$ . By your statement about pointwise limits, for each $y$ not in $X$ there exists $A \in \mathcal{F}$ such that $x \in A$ , and some neighbourhood of $y$ is disjoint from $A$. By compactness, a finite set of such neighbourhoods covers the complement of $X$ ; the intersection of the corresponding $A$'s is in $\mathcal{F}$, contains $x$ and is contained in $X$, so $x \in B$. [contd.] –  Phil Wild Jul 24 '10 at 2:31
    
This leaves your claim that every element of $V$ is the pointwise limit of a net of etc. I don't see why this is so, is it possible to explain to a know-nothing like me? I haven't been able to figure out any "nice" description of the topology on $C(\Omega)$ which makes the natural isomorphism $C(\Omega) \to V$ into a homeomorphism. By the way, can anybody reading this recommend a good introductory text book where I can read about Von Neumann algebras? I've read Rudin's "Functional Analysis" and that's pretty much the limit of my knowledge. –  Phil Wild Jul 24 '10 at 2:38
    
typo $\mathcal V=\mathcal F$ (fixed) –  Rasmus Bentmann Jul 24 '10 at 9:46
    
Phil: sorry for the incomplete nature of my "answer" - I'm afraid I have v. limited internet access for the next few days so probably won't be able to respond properly until then. But I'll try to think a bit more about how this can be patched up. –  Yemon Choi Jul 24 '10 at 22:56
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