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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?

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    $\begingroup$ In the commutative case (mathoverflow.net/questions/71259/…) the answer to (3c) is not known either. In fact, this is a famous open research problem, originally formulated by von Neumann. $\endgroup$ Commented Jul 5, 2020 at 20:44
  • $\begingroup$ @DmitriPavlov Although, as Nik Weaver points out (and you point out in that question), in the abelian case, some form of answer is known (admitting a full set of normal states) -- it's just not entirely satisfactory from some perspectives. $\endgroup$ Commented Jul 5, 2020 at 20:51
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    $\begingroup$ Counterexample for question 2: $\Pi(C([0,1]))$ has two elements, the constant $0$ and $1$ functions. This is a complete Boolean algebra. But $C([0,1])$ is not a von Neumann algebra. $\endgroup$ Commented Jul 8, 2020 at 19:17

2 Answers 2

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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.

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  • $\begingroup$ Thanks! How essential is it to use $M_2(\mathbb C)$ here? After all, in light of Gleason's theorem, this is a very "exceptional" von Neumann algebra. Could one just as well use $M_3(\mathbb C)$ to show that $\Pi$ is not full? $\endgroup$ Commented Jul 5, 2020 at 20:53
  • $\begingroup$ You make a good point. $M_3(\mathbb{C})$ does seem much more rigid ... I'll have to think about this. $\endgroup$
    – Nik Weaver
    Commented Jul 5, 2020 at 21:20
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I'm actually starting to think that the situation is not as bad as indicated in Nik Weaver's answer. Some notation:

  • If $\mathcal A$ is a $C^\ast$-algebra or JB-algebra (resp. von Neumann algebra or JBW-algebra), let $\Pi(\mathcal A)$ be its ortholattice of projections. Let $State(\mathcal A)$ (resp. $State_n(\mathcal A)$) be its convex space of states (resp. normal states). If $L$ is a complete orthomodular lattice, let $State_n(L)$ be the convex space of normal states on $L$. If $K$ is a convex space, let $Aff(K)$ be the space of bounded ($\mathbb R$-valued) affine functionals on $K$.

We are led to the following considerations:

  1. By Gleason's Theorem, if $\mathcal A$ is a von Neumann algebra without a direct summand of $M_2(\mathbb C)$, the natural restriction map $State_n(\mathcal A) \to State_n(\Pi(\mathcal A))$ is an isomorphism.

So in order to recover $\mathcal A$ from $\Pi(\mathcal A)$, it typically suffices to recover it from $State_n(\mathcal A)$.

  1. Let $\mathcal A$ be a JB-algebra (resp. JBW-algebra). Then as Banach spaces, there is a natural isomorphism $Aff(State(\mathcal A)) \cong \mathcal A$ (resp. $Aff(State_n(\mathcal A)) \cong \mathcal A$). Moreover, Alfsen and Schultz have shown (see Geometry of State Spaces of Operator Algebras and predecessor works) how to recover the Jordan multiplication on $\mathcal A$ purely from the convex structure on $State(\mathcal A)$ (resp. $State_n(\mathcal A)$), including giving a characterization of those convex spaces for which this can be done. (Briefly, they identify conditions under which a functional calculus can be developed. Since a Jordan multiplication is determined by its squaring operation, this yields a candidate Jordan multiplication, and they identify further conditions ensuring that the candidate multiplication is bilinear and hence actually is a Jordan multiplication.)

Thus if $\mathcal A$ is a $C^\ast$-algebra (resp. von Neumann algebra), then the self-adjoint part of $\mathcal A$ and its Jordan multiplication can be reconstructed from $State_n(\mathcal A)$. But because $\mathcal A$ is not typically isomorphic to its opposite algebra (a fact which taking $State_n(\mathcal A)$ forgets), there is no hope in general to recover the algebra multiplication on $\mathcal A$. However,

  1. Alfsen and Schultz have shown that any JB-algebra (resp. JBW-algebra) $\mathcal A$ is "locally" the self-adjoint part of at most two different $C^\ast$-algebras (resp. von Neumann algebras), which are opposite to each other. Moreover, they have characterized in terms of "orientation" data on the convex space $State(\mathcal A)$ (resp. $State_n(\mathcal A)$) exactly when a global algebra multiplication can be found, and how to distinguish between the different possible choices.

Putting this together, we see that the functor $\mathcal A \mapsto State_n(\mathcal A)$ is "almost injective" on isomorphism classes, its essential image can be explicitly characterized, and the additional data to make the functor 1-to-1 on isomorphism classes has been described. This leaves me suspecting that with the right notion of morphism of oriented convex space, the functor becomes fully faithful.

This is not quite the same as characterizing things in terms of $\Pi(\mathcal A)$, but if we exclude von Neumann algebras with summands of $M_2(\mathbb C)$, then $State_n$ factors through $\Pi$. In particular, we have a description of the essential image of $\Pi$, in terms of the normal state space of a complete orthomodular lattice $L$. It might be nice to have a more direct description, but the state space of $L$ is a pretty natural invariant to consider, so one might also be satisfied with this.

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    $\begingroup$ @TimCampion The normal state space of $L^\infty([0,1])$ has no extreme points, so is not compact in any locally convex topology. Normal state spaces are only closed bounded convex sets. $\endgroup$ Commented Jul 8, 2020 at 19:53
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    $\begingroup$ @TimCampion For some reason I forgot to mention this earlier (maybe because it doesn't relate to the official title of the post) but it is possible to use the Alfsen--Shultz-Hanche-Olsen--Størmer results to turn the state functor on C$^*$-algebras and normal state functor on W$^*$-algebras into categorical equivalences. I worked out the details of this during my PhD. Essentially all you need is already in the two volumes on state spaces by Alfsen and Shultz. $\endgroup$ Commented Jul 15, 2020 at 14:46
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    $\begingroup$ @TimCampion The maps of state spaces that are images of Jordan homomorphisms are those preserving complementary projective faces (Lemma 5.15 of Geometry of State Spaces...). The simplest way of ensuring that you get $*$-homomorphisms was suggested to me by Bram Westerbaan - use Choi's theorem that a 2-positive Jordan homomorphism is a $*$-homomorphism, so either keep the state space as an "operator convex set", or reconstruct $M_2(\mathcal{S}(A))$ from an orientation on $\mathcal{S}(A)$. Even in the case when $A \cong A^{\mathrm{op}}$ the orientation is needed to distinguish *-homomophisms $\endgroup$ Commented Jul 15, 2020 at 14:50
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    $\begingroup$ from Jordan homomorphism that don't preserve products, so even for cases as simple as $M_n(\mathbb{C})$. One thing I'll add -- as far as I know, it is an unsolved problem to find a structure defined solely in terms of projections that acts like an orientation of the state space, and so allows us to distinguish *-homomorphisms from Jordan homomorphisms. It needs to be a structure because of course the transpose map induces an isomorphism of projection lattices. Coincidentally, I recently asked an expert about it at a conference who agreed it is a very difficult problem. $\endgroup$ Commented Jul 15, 2020 at 14:55
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    $\begingroup$ @TimCampion Regarding state spaces, there are some references in this old answer and a sentence about the nonunital case. $\endgroup$ Commented Jul 15, 2020 at 14:59

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