Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is probably well-known to the experts but I could not find any answer neither in my head nor in the literature: Is there a (unital) C*-algebra such that its projections do not form a lattice (under the usual ordering)? Certainly, this cannot be a von Neumann algebra.

share|improve this question
2  
If you look at the universal representation of the C* algebra and then consider the double commutant of its image you get the enveloping von Neumann algebra. The question of whether the projections in the C* algebra form a sublattice of the enveloping von Neumann algebra is considered here: arxiv.org/abs/math/0601003. I hope this helps! –  Jon Bannon Jan 22 '12 at 13:44
3  
Probably I should ask this under the pseudonym unknown(Google) to spare myself embarrassment, but... $$ $$ Are there examples among the commutative $C^*$ algebras; i.e., $C(K)$ spaces? –  Bill Johnson Jan 22 '12 at 15:11
    
Bill, in the arxiv paper linked by Jon it is said (and proved) that a commutative $C^*$-algebra has always the lattice property (see the beginning of Sect. 4). –  Valerio Capraro Jan 22 '12 at 17:11
2  
Thanks, Valerio. But I was just being stupid, thinking of contractive projections on $C(K)$ instead of in $C(K)$. In the commutative case projections are just indicator functions of clopen sets. –  Bill Johnson Jan 22 '12 at 19:57

2 Answers 2

up vote 8 down vote accepted

You can find examples of AF algebras without the lattice property in Section 2 of AF Algebras with a Lattice of Projections by Aldo J. Lazar here.

share|improve this answer
    
Perfect! I was just commenting that in that arxiv paper there were only examples of C^*$-algebras with that property and not without :) –  Valerio Capraro Jan 22 '12 at 14:10

Here is perhaps the simplest example. Let $A$ be the C*-algebra of all sequences of $2 \times 2$ matrices converging to a scalar multiple of diag(1,0). Let $p$ be the constant sequence diag(1,0), and $q$ a sequence of rank 1 projections converging to diag(1,0) but never exactly equal. Then $p$ and $q$ have no upper bound at all. This example can be tweaked to make it unital by allowing any limit matrix at infinity and taking $q$ to alternate diag(1,0) and nearby but unequal projections. Then $p$ and $q$ have no least upper bound.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.