13
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ 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! $\endgroup$
    – Jon Bannon
    Commented Jan 22, 2012 at 13:44
  • 6
    $\begingroup$ 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? $\endgroup$
    – Bill Johnson
    Commented Jan 22, 2012 at 15:11
  • 1
    $\begingroup$ 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). $\endgroup$
    – Valerio Capraro
    Commented Jan 22, 2012 at 17:11
  • 2
    $\begingroup$ 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. $\endgroup$
    – Bill Johnson
    Commented Jan 22, 2012 at 19:57

2 Answers 2

Reset to default
11
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Perfect! I was just commenting that in that arxiv paper there were only examples of C^*$-algebras with that property and not without :) $\endgroup$
    – Valerio Capraro
    Commented Jan 22, 2012 at 14:10
19
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .