Distributivity in boolean subalgebras of orthomodular lattice

Question

A boolean subalgebra $B$ of the orthomodular lattice $L$ of closed subspaces of a separable Hilbert space, may be defined like a sublattice with $0$ and $1$, with pairwise commuting elements. How to prove that, in this subalgebra, the distributive property holds?

Answer 1

You can find the proof of a stronger result in

Greechie, Richard J. On generating distributive sublattices of orthomodular lattices. Proc. Amer. Math. Soc. 67 (1977), no. 1, 17-22.

Namely, Greechie shows that if $L$ is an orthomodular lattice and $S\subseteq L$ is a subset with the property that, whenever $a, b, c, \in S$, at least one of them commutes with the other two, then $S$ generates a distributive sublattice of $L$.

Answer 2

Many thanks. In the meantime I found a more simple proof in my stronger hypothesis that all elements of $B$ are pairewise commuting. I use the fact that if $\breve{a}C\breve{b}$ and $\breve{a}\wedge\breve{b}=\breve{0}$, than $\breve{a}\bot\breve{b}$ and prove the only not trivial inequality $\breve{a}\wedge(\bigvee_{i\in I}\breve{b}_i)\leq\bigvee_{i\in I}(\breve{a}\wedge\breve{b}_j)$.

Since $\breve{a}\wedge(\bigvee_{i\in I}\breve{b}_i),\neg(\bigvee_{i\in I}(\breve{a}\wedge\breve{b}_i))\in B$, they are commuting, then it suffices to prove that $[\breve{a}\wedge(\bigvee_{i\in I}\breve{b}_i)]\wedge\neg[\bigvee_{i\in I}(\breve{a}\wedge\breve{b}_i)]=\breve{0}$.
If $\breve{c}=[\breve{a}\wedge(\bigvee_{i\in I}\breve{b}_i)]\wedge\neg[\bigvee_{i\in I}(\breve{a}\wedge\breve{b}_i)]$, then $\breve{c}\leq\breve{a}$, $\breve{c}\leq\bigvee_{i\in I}\breve{b}_i$ and $\breve{c}\leq\neg[\bigvee_{i\in I}(\breve{a}\wedge\breve{b}_i)] =\bigwedge_{i\in I} \neg(\breve{a}\wedge\breve{b}_i) \Rightarrow \breve{c}\leq\neg(\breve{a}\wedge\breve{b}_j)$ for all $j\in I$, from which, taking into account that $\breve{a}\wedge\neg\breve{b}_j\leq\neg\breve{b}_j\leq\neg\breve{a}\vee\neg\breve{b}_j =\neg(\breve{a}\wedge\breve{b}_i)$, because of orthomodularity $$\breve{c}=\breve{c}\wedge\breve{a}\leq\neg(\breve{a}\wedge\breve{b}_j)\wedge [(\breve{a}\wedge\breve{b}_j)\vee(\breve{a}\wedge\neg\breve{b}_j)]=\breve{a}\wedge\neg\breve{b}_j\leq\neg\breve{b}_j,\qquad \forall j\in I\qquad\Rightarrow\qquad\breve{c}\leq \bigwedge_{i\in I}\neg\breve{b}_i=\neg\bigvee_{i\in I}\breve{b}_i$$ wich together $\breve{c}\leq\bigvee_{i\in I}\breve{b}_i$, gives $\breve{c}=\breve{0}$.