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In $B(H)$ where $H$ is a Hilbert space, we have that if p and q are (orthogonal) projections that Inf{$p$, $q$} is in $B(H)$. This also holds if we replace $B(H)$ by the phrase "the strong closure of any algebra containing p and q" where I mean things literally: algebras need not have a unit, nor be closed under *, nor be closed in any topology. Nevertheless, since we can express Inf{$p$, $q$} as the strong limit of $(pq)^n$ this holds. Is there a similar expression for Sup{$p$, $q$} so that this supremum is shown to be contained in the strong closure of any algebra that contains p and q? I've toyed with expressions like $(p+q-Inf(p, q))^n$ and $(p+q-pq)^n$ to no avail, not that these things necessarily don't work. I simply don't see how to prove that they do or don't, nor how to guess the right expression.

This is related to the following question, so I cannot use the result that I wanted to obtain here:

Showing that the WO closure of a *-algebra is a Von Neumann Algebra

Edit: I now know how to prove that the notion of "strong operator closure of an algebra" is the same thing as the notion of a strong operator closed algebra. Same thing for weak. But that has not advanced me further along my way to a solution to the stated problem above. Intuitively, I view p+q as sort of like a simple function from measure theory, with value 0 some places, 1 other places, and 2 on some intersection. If this intuition was actually "correct", then the right thing to do would be to take the nth root of p+q and take that limit. But SO closed algebras need not be closed under the taking of roots. On the other hand, there is another operation under which such an algebra is closed. Namely $x$ maps to log(1+x). (Since it has a taylor series not involving a constant term which normally maps via the functional calculus to the identity which I'm not assuming is in our algebra.) Unfortunately, this operation isn't suitable cause if you try repeated application of logs, everything just goes to -infinity.

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Actually the first of the two expressions that I guessed would be insufficient because it is not clear to me yet (though it might be once I prove the linked result) that the SO or WO closure of an algebra is an algebra. Multiplication is known to be not jointly continuous in these topologies, afterall. But in the second expression, pq is honestly in the algebra, as opposed to merely in its closure. – Jeff May 30 '12 at 20:01
As far as I know, $\inf\{p,q\}=\lim\limits_{n\to\infty}(pqp)^n$ – Norbert May 30 '12 at 20:11
That is equal to the thing I suggested, since the extra copy of p gets pushed off to infinity. – Jeff May 30 '12 at 21:18
I already did. I also tried the limit of your suggestion to the nth power. Neither works as far as I can see. Let H be C^2, and think of just about any two projections 1 dimensional not orthogonal to each other. – Jeff May 30 '12 at 23:10
I understand, but my last comment disproves your conjecture. If you think about it, that object need not even be a projection. – Jeff May 30 '12 at 23:21
up vote 2 down vote accepted

$\sup(p,q) = I - \inf(I-p,I-q) = I - \lim_{n \to \infty} (I-p-q+pq)^n$. Yes, I know you want to do without $I$, but expand this out formally and the $I$'s cancel: you get $$ \sup(p,q) = \lim_{n \to \infty} \sum_{j=1}^n {n \choose j} (-1)^{j+1} (p+q-pq)^j$$

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