# C*-algebras with bizzarre structure of projections

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.

• 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! Jan 22, 2012 at 13:44
• 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? Jan 22, 2012 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). Jan 22, 2012 at 17:11
• 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. Jan 22, 2012 at 19:57

• Perfect! I was just commenting that in that arxiv paper there were only examples of C^*$-algebras with that property and not without :) Jan 22, 2012 at 14:10 Here is perhaps the simplest example. Let$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.