Where can I find the proof that any block (maximal boolean subalgebra) $\mathbf{B}$ of the orthomodular lattice $\mathcal{L}$ of closed subspaces of a separable Hilbert space $\mathcal{H}$ is atomic?
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I don't think this is true.
Firstly, remark that the lattice $L$ of closed subspaces of a Hilbert space $H$ is isomorphic with the lattice $P$ of projections on $H$ via the map $P\mapsto L$ given by $p\mapsto pH$. The order on $P$ then satisfies $p\leq q$ if and only if $pH\subseteq qH$.
Next, the algebra $B(H)$ of bounded operators (which contains $P$ as subset) is a von Neumann algebra on $H$. Let $M$ be a maximal commutative *-subalgebra of $B(H)$. Then $M''$ is a commutative von Neumann algebra on $H$ containing $M$, and by maximality of $M$, we find $M=M''$, so $M$ is a commutative von Neumann algebra. It follows that $M$ is generated as von Neumann algebra by its projections $B$, i.e., $B''=M$.
$B$ must be a block. Indeed, if there is some Boolean subalgebra $A\subseteq P$ containing $B$ as a subset, then all elements of $A$ commute as projections on $H$, hence $A''$ is a commutative von Neumann algebra on $H$ containing $M=B''\subseteq A''$. By maximality of $M$ it follows that $M=A''$, which means that $A$ is a set of projections in $M$. Since $B$ is the set of all projections in $M$, it follows that $B=A$.
Now, in the book Foundation of Quantum Theory by Klaas Landsman, which is open source (follow the link), we find on on page 601 a classification of maximal commutative *-subalgebras of $B(H)$, as well as the assertion that there are no atomic projections (which are precisely the atoms in the block corresponding to the maximal commutative *-subalgebra) in one type of maximal commutative *-subalgebras, namely the one of algebras isomorphic to $L^\infty(0,1)$.
Hence $P$, and so also $L$ has a block that is not atomic.
answered Aug 15, 2017 at 21:23
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$\begingroup$ Many thanks for your replay and for the reference. My question was motivated from reading different statements identifying blocks with orthogonal basis of atoms. One for all, in a answer to a question in this site, “Any Boolean subalgebra of P(H) is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis.” Your answer is exhaustive and incontestable but I have yet confusing ideas because of something not well understood. $\endgroup$– dioxoidCommented Aug 16, 2017 at 15:20
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$\begingroup$ From your answer I deduce that for any boolean subalgebra B of P(H) may not exist a maximal orthogonal set of atoms E, such that any element of B is generated from a subset of E. Otherwise any boolean subalgebra of P(H) must be a subalgebra of the block generated from the powerset of some E and all blocks would be atomic! From the other side if B is a boolean subalgebra of P(H), any projection in B is generated from different sets of orthogonal atoms and because of commutativity in B, it seems (?) that B must be constructed by one (if maximal) or more maximal orthogonal sets of atoms. $\endgroup$– dioxoidCommented Aug 16, 2017 at 15:21
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$\begingroup$ I guess the argument in my reply shows that the remark in the answer to that other question is not completely correct. You are correct that for every projection $p$, there is some collection $C_p$ of pairwise othogonal atoms whose supremum is $p$. However, apparently, there is a Boolean subalgebra $B$ such that $\bigcup\{C_p:p\in B\}$ contains two elements that do not commute. $\endgroup$ Commented Aug 16, 2017 at 18:05