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.



Projections in a W*-algebra as a continuous lattice?
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 $I$.


