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In Blackadar's Operator Algebras, there is the following remark after the proposition II.3.3.1 :

The projections in a C*-algebra do not form a lattice in general

In the answer of this question, it is said that:

If you restrict $\leq$ to the set of self-adjoint projections in $A$, you do get a lattice, which is isomorphic to the lattice of closed subspaces of $\mathbb{C}^2$. The supremum of $a$ and $b$ in this lattice is $I$ and the infimum of $a$ and $b$ is $0$.

Does this statement really holds? And if this is the case, can we also prove that it is a continuous lattice?

Recall that a continuous lattice is a complete lattice $L$ in which every element $y$ is equal to $\bigvee \{x \in D \mid x \ll y\}$ where $x \ll y$ ("x approximates y" or "x way below y") if for any directed set $D \subseteq P$, $y \leq \bigvee D$ implies that there is a $d \in D$ such that $x \leq d$

EDIT: Given Martin Argerami's answer, I'm now wondering if projections of an arbitrary von Neumann algebra form a continuous lattice.

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Inspector Norse
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1 Answers1

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The second statement is a statement about the algebra $M_2(\mathbb C)$. This is the von Neumann algebra $B(\mathbb C^2)$. The set of selfadjoints projections in a von Neumann algebra forms a lattice (a complete one: the union of projections is the projection onto the closure of the union of the ranges; and the meet of projections is the projection onto the intersection of the ranges).

Type I von Neumann algebras (finite-dimensional, in particular) have minimal projections, so the set of selfadjoint projections would be a continuous lattice in the sense you give. I'm not so sure for arbitrary von Neumann algebras.

For a general C$^*$-algebra (not von Neumann), the set of self-adjoint projections is usually not a lattice.

The set of selfadjoint elements of a von Neumann algebra does not form a lattice (even in the finite-dimensional case, as the example shows).

Martin Argerami
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