I'm trying to understand this point, though it seems rather basic. My understanding is that for a von Neumann algebra $M$ on a Hilbert space $\mathscr{H}$, the projections satisfy the property that if $\{p_i\}_{i \in I}$ is a family of projections in $M$, then we can take and infimum and a supremum by
\begin{align*}
\land_i p_i & = \left[ \bigcap_i p_i \mathscr{H} \right] , \\
\lor_i p_i & = \left[ \bigcup_i p_i \mathscr{H} \right] ,
\end{align*}
where $[K]$ denotes the orthogonal projection onto the closed span of $K \subseteq \mathscr{H}$. The infimum and supremum are both contained in $M'' = M$, and satisfy the identities $\land_i p_i = 1 – \lor_i (1 – p_i) , \lor_i p_i = 1 – \land_i (1 – p_i)$. We also have minimal element $0 \in M$ and maximal element $1 \in M$.
I'm confused as to how this doesn't say that the space of projections in $M$ is a Boolean algebra. I'm confident that I'm just missing something simple, but I am missing it all the same.
Best Answer
A Boolean algebra must also satisfy the distributive law $p\vee(q\wedge r)=(p\vee q)\wedge (p\vee r)$. This is not true for projections on a Hilbert space in general. For instance, imagine that $\mathscr{H}$ is $2$-dimensional and $p,q,$ and $r$ are three different rank $1$ projections. Then $p\vee q=p\vee r=1$ but $q\wedge r=0$ so $p\vee(q\wedge r)=p\neq 1=(p\vee q)\wedge(p\vee r)$.
(It is true for commuting projections, since if $p$ and $q$ commute then $p\wedge q=pq$ and $p\vee q=p+q-pq$, and then the distributive law above can be verified by a simple computation. As a result, the projections in a commutative von Neumann algebra form a complete Boolean algebra.)