# Projections in a W*-algebra as a continuous lattice?

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

It is known that the set of projections in an arbitrary W*-algebra is a complete orthomodular lattice. I would like to know for which kind of W*-algebras this lattice is also continuous.

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Is the article by Kaplansky "Any orthocomplemented complete modular lattice is a continuous geometry." useful? –  Abel Stolz Jun 7 at 7:38
Theorem 6 of this article states that any orthocomplemented modular lattice of type II is a continuous geometry. A continuous geometry is a lattice L that is complemented, modular, meet continuous, and join continuous. I don't know how it relates to continuous lattices but I know that continuous lattices can be characterized equationally : encyclopediaofmath.org/index.php/Continuous_lattice –  Rennela Jun 9 at 18:34
According to Gierz' Compendium on continuous lattices, continuous lattices form an overlapping subclass of continuous geometry. –  Rennela Jun 10 at 9:28
I'm going to say the von Neumann algebra has to be either atomic abelian or finite dimensional for this to happen. If $M$ is nonatomic then I don't think any projection besides 0 is "way below" any other projection, because if $p$ is nonzero then we can write $p = \bigvee p_\alpha$ where the $p_\alpha$ are directed and all strictly less than $p$, so if $p \leq q$ we have $q = \bigvee (p_\alpha + q-p)$ and this shows that $p$ is not way below $q$. So you need minimal projections to even have a chance. But $B(H)$ fails for infinite dimensional $H$: take $H = l^2$, let $p_1$ be the rank one projection whose range is spanned by the vector $e_1$, and for $n \geq 2$ let $p_n$ be the rank one projection whose range is spanned by the vector $e_1 + \frac{1}{n}e_n$. Then $e_1$ is not contained in the span of $p_2 \vee \cdots \vee p_n$ because if it were then $e_j$ would be too for $2 \leq j \leq n$, but the span of $p_2 \vee \cdots \vee p_n$ only has dimension $n-1$. However, the range of $\bigvee p_n$ contains vectors arbitrarily close to $e_1$, hence it contains $e_1$, hence it contains $e_n$ for $n \geq 2$, hence it is everything. Thus $I$ is the join of the directed sequence $p_2$, $p_2 \vee p_3$, $p_2 \vee p_3 \vee p_4$, $\ldots$, but $p_1$ is not less than any of these projections. This shows that $p_1$ is not way below $I$, and by symmetry no nonzero projection is way below $I$.
The lattice of projection of a finite von Neumann algebra is isomorphic to a lattice of closed subspaces $\mathbb{C}^n$. I found out that the lattice of the closed balls of a separable Hilbert space, ordered by reversed inclusion, is a continuous lattice (Edalat, Heckmann, "A computation model for metric spaces"). But if I follow your argument, for the usual order on von Neumann algebras, I can only hope to get a continuous lattice with projections in a finite-dimensional von Neumann algebra, right? –  Rennela Jun 9 at 19:21
No, I think it either has to be finite dimensional or atomic abelian, like $l^\infty$. –  Nik Weaver Jun 10 at 3:00