Sorry for the vague title, I can't think of a better one that isn't overly long.

Suppose that $S$ is a commuting set of projection operators on a Hilbert space. I'll introduce the following notation: if $p \in S$, let $p^+ \equiv p$ and $p^- \equiv 1 - p$. Let $I \equiv ${$+, -$}. The projections are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$; this makes the set of all projections into a complete lattice. Is the following identity true?

$\sup_{f \in I^S} \inf_{p \in S} p^{f(p)} = 1$

In the case where $S$ is finite with elements $p_1, p_2, \ldots p_n$, the left hand side of this equation is simply the product over $i$ of $p_i + (1 - p_i)$, so I'm interested in whether this can be generalised to the infinite case. It's easy to see that the following two statements are equivalent to the above:

If $\inf_p p^{f(p)} x = 0$ for all $f$, then $x = 0$

If $\sup_p p^{f(p)} x = x$ for all $f$, then $x = 0$

but I have no idea how to prove either of these.

My reason for asking is that I'm trying to show that, if $\mathcal{H}_1$ and $\mathcal{H}_2$ are Hilbert spaces, then if a projection on $\mathcal{H}_1 \otimes \mathcal{H}_2$ is of the form $\sup_i p_i \otimes q_i$, with $p_i$ and $q_i$ drawn from some complete Boolean algebras of projections on $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively, then the $q_i$ may be chosen to satisfy $q_i q_j = 0$ when ever $i \neq j$. So if anybody knows of an alternative way to prove that, or knows that it's false, then by all means say so.

`$_{p \in S} p^{f(p)}$`

will be a minimal projection if it is not 0, so the sup will be 1 iff S generates a completely atomic von Neumann algebra. – Jesse Peterson Jul 26 '10 at 4:29