Let $A$ be a von Neumann algebra. Then a classic observation is that the set of projections $\Pi(A)$ is naturally a complete orthomodular lattice.
Question 1: Is the construction $A \mapsto \Pi(A)$ a functor from von Neumann algebras to complete orthomodular lattices?
For this to make sense, I should say what a morphism of von Neumann algebras is -- but I'm not sure what the appropriate choice of morphism is. I should also say what a morphism of complete orthomodular lattices is, and here there is a natural guess -- a morphism $f: L \to M$ should be a function which preserves $(-)^\perp$ and sups (equivalently, infs).
The construction $A \mapsto \Pi(A)$ makes sense even when $A$ is just a $C^\ast$-algebra, except that we only know that $\Pi(A)$ is an orthoposet (maybe an ortholattice?).
Question 2: Let $A$ be a $C^\ast$-algebra, and suppose that $\Pi(A)$ is a complete orthomodular lattice. Does it follow that $A$ is a von Neumann algebra?
Question 3: Assuming the answer to Question 1 is "yes", let $\Pi: vNA \to COML$ be the above functor from von Neumann algebras to complete orthomodular lattices.
(a.) Is the functor $\Pi$ fully faithful?
(b.) Does the functor $\Pi$ have a left or right adjoint?
(c.) What is the essential image of the functor $\Pi$?
(3c) is the title question, of course. It seems there was substantial interest in this question in the '60's and '70's. I've come across work by Holland, Topping, and Fillmore identifying various properties of orthomodular lattices of the form $\Pi(A)$ not shared by all complete orthomodular lattices. I believe there is a characterization those lattices of the form $\Pi(A)$ where $A = B(H)$ is the algebra of all bounded operators on a Hilbert space $H$ (though I'm having trouble tracking down where I read this). But the trail seems to go cold after this period. Did the question just go out of fashion?
I do have the sense that there might be a characterization which has something to do with "having a full set of states". For instance, suppose we define a state on a complete orthomodular lattice $L$ to be a function $f: L \to [0,1]$ which preserves order and directed sups, is additive ($f(0) = 0$, $f(a \vee b) = f(a) + f(b)$ if $a \perp b$), and is normalized so that $f(1) = 1$ (I think maybe this should really be called a "normal state" or something like that?). Say that $L$ has a full set of states if for $a, b \in L$ we have $a \leq b \Leftrightarrow f(a) \leq f(b)$ for all states $f$. Then I believe that if $L = \Pi(A)$, then $L$ has a full set of states. Is the converse true?