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.



Question about projections on a Hilbert space
Here's a counterexample. For each natural number $n$, let $D\_n$ be the set of those $x\in[0,1]$ whose binary expansion has a 1 in the $n$-th place. Then the operation of multiplication by the characteristic function of $D\_n$ is a projection operator $p\_n$ on $L^2[0,1]$. Let $S$ be the set of these operators $p\_n$; they commute. For any $f:S\to I$, the infimum of the operators $p^{f(p)}$ is zero. For example, if $f$ were identically $+$, then the range of $p\_n$ consists of functions supported on $D\_n$ (up to measure 0, of course), and so the infimum would project to functions supported on the intersection of the $D\_n$, i.e., a one-point set (up to measure 0 again); such functions are 0 in $L^2$. For the general case, where some of the values of $f$ are $-$, you can use the same argument with some of the $D\_n$ replaced by their complements.


