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.




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**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).




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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?




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**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](https://www.doi.org/10.2307/1994299), [Topping](https://projecteuclid.org/euclid.pjm/1102992829), and [Fillmore](https://www.doi.org/10.2307/2034659) 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?



Which complete orthomodular lattices arise from von Neumann algebras?
Question 1: Yes, if you take the von Neumann algebra morphisms to be normal $\*$-homomorphisms. Restricting any such map to the projections will preserve sups and orthocomplements.


Question 2: No, this already fails in the commutative case. Look up "abelian AW\*-algebra" or see the discussion of Stonean spaces in volume one of Kadison-Ringrose.


Question 3: (a) Any von Neumann algebra is generated by its projections, so the functor is faithful. It is not full, just look at $M\_2(\mathbb{C})$ where the lattice of projections contains a $0$ and a $1$ and $2^{\aleph\_0}$ incomparable elements between those two. There are all kinds of morphisms from this lattice to itself that don't extend linearly.


Question 3: (b) I'm weak on category theory, but I'd imagine the $M\_2(\mathbb{C})$ example falsifies both possibilities.


Question 3: (c) I don't think there's any good answer to this question, and that's the reason research in this direction petered out. The idea of asking for a full set of normal states is good, and this does answer the question in the abelian case (again, see Kadison-Ringrose), but I doubt it will work in general. There's just no way to linearize these states on lattices. But counterexamples will take some work.


