Has anyone studied boolean-algebraic completions of projection lattices of non-commutative von Neumann algebras? E.g. in order to consider them as forcing notions?

T. Jech wrote a paper in the 80's (here) investigating projection lattices of operator algebras, by treating them as forcing notions. He only looked at abelian algebras, which is natural since their lattices are complete Boolean algebras (usually measure algebras).

Has this been done for NON-abelian von Neumann algebras? Their lattices aren't boolean algebras; you'd have to take the lattices' boolean-algebraic completions first if you wanted to use them in the boolean approach to forcing.**

These lattices can have uncountable antichains so their completions can't always be measure algebras. My question is whether their completions can be new/interesting boolean algebras, or whether they always end up being products of well-known algebras, like measure algebras or Cohen algebras.

Any thoughts or references greatly appreciated!

**The boolean-algebraic completion of a poset $P$ is the regular open algebra of $P$'s order topology. This is guaranteed---under certain conditions, which obtain here---to yield a complete boolean algebra into which the original poset embeds densely in an obvious way.

  • $\begingroup$ I'm told they end up being Levy collapse algebras, but I don't quite see the proof. Let $C$ be the "collapse poset" of finite sequences of ordinals $< 2^{\omega_0}$, ordered by reverse inclusion. I see how to embed it into a (non-abelian vN algebra's) projection lattice $P$ so that every unbounded descending sequence maps to an unbounded descending sequence, and every $P$-member is compatible with (images of) $C$-members of arbitrary length. But I don't see why it must be possible to make the embedding dense in $P$. Probably I myself am being dense ... $\endgroup$ Oct 7, 2016 at 20:20

1 Answer 1


I believe I can show that a partition lattice of a separably-acting type II or III factor is forcing-equivalent to the usual (Levy) continuum-collapsing poset, on the supposition that each of its nontrivial partitions has continuum cardinality. I suspect this is always the case, and maybe it can be proved in an elegant way, but for now I can only prove that a type III projection lattice has no countable nontrivial partition, so I need the continuum hypothesis. My no-countable-nontrivial-partition proof is not very elegant or edifying so I won't post it, but I mention the main idea in a comment here: When are there countable partitions in a factor's projection lattice?

$\mathcal{R}$ is a separably-acting type II or III von Neumann factor; $\mathbb{P}$ is its projection lattice,

$ \mathbb{P} = \{ P \in \mathcal{R} : P^2 = P^* = P \} , $

with the usual ordering $P \leq Q \iff PQ = P$. The greatest and least members of $\mathbb{P}$ are the identity projection $1$ and the null projection $0$. $\mathbb{P}^+$ means $\mathbb{P} \setminus \{ 0 \}$. When $A, B \in \mathbb{P}$, we write $A \perp B$ to mean they are orthogonal ($AB = 0$).

By a partition of $P \in \mathbb{P}$ we mean a lattice partition, i.e. a maximal pairwise-disjoint subset of the projections in $\mathbb{P}^+$ that are $\leq P$. A partition is nontrivial if its cardinality is $> 1$.

A state $\phi$ on $\mathcal{R}$ is a complex linear functional on $\mathcal{R}$ that is normalized in the sense that $\phi(1) = 1$. A state is faithful if $\phi(T) = 0$ holds only when $T = 0$; it is normal if it is countably additive on sets of mutually orthogonal projections, in which case it is also continuous on $\mathbb{P}$ with respect to the strong operator topology.

For cardinals $\kappa > \aleph_0$, $C_{\kappa}$ is the set of finite sequences of ordinals $< \kappa$, ordered by reverse inclusion; this is the standard poset used to force $\kappa$ to become countable.

Recall that when $A, B$ are posets, a dense embedding $\phi : A \rightarrow B$ is an order-isomorphism onto a subset of $B$ such that every $b \in B$ has some $a \in A$ satisfying $\phi(a) \leq b$.

If one separative poset embeds densely into another, then the embedding extends naturally to an isomorphism between their boolean completions. Thus to show $B(C_{\kappa})$ isomorphic to $B(\mathbb{P})$, it suffices to show that $C_{\kappa}$ embeds densely into $\mathbb{P}^+$.

Lemma 1: Suppose that for some faithful normal state $\phi$ on $\mathcal{R}$, and all $P \in \mathbb{P}^+$, and all $\epsilon > 0$, there exists a lattice partition $X$ of $P$ such that (i) $|X| = 2^{\aleph_0}$, (ii) $Q \in X \Rightarrow \phi(Q) < \epsilon$, and (iii) for all $P' \in \mathbb{P}^+$, $P' \leq P$, there exists $Q \in X$ such that either $P' \leq Q$ or $Q \leq P'$. Then there is a dense embedding of $C_{2^{\aleph_0}}$ into $\mathbb{P}^+$.

Proof: Let $P_{ \langle \rangle }$ (indexed by the empty sequence) be $1 \in \mathbb{P}$. Let $\{ P_{\langle \alpha \rangle} : \alpha < 2^{\omega_0} \}$ be a lattice partition of $P_{\langle \rangle}$ of the kind supposed, with $\epsilon = 1/2$. Next, for each $\alpha$ obtain a lattice partition $\{ P_{\langle \alpha, \beta \rangle} : \beta < 2^{\omega_0} \}$ of $P_{\langle \alpha \rangle}$, with $\epsilon = 1/4$. Continue in this manner so that for every finite sequence $\vec{\alpha}$ of ordinals $< 2^{\omega_0}$, $\phi( P_{\vec{\alpha}} ) < 2^{ - |\vec{\alpha}|}$. Clearly $\vec{\alpha} \mapsto P_{\vec{\alpha}}$ is an embedding of $C_{2^{\aleph_0}}$ into $\mathbb{P}^+$; we must confirm that it is a dense embedding.

For $n > 0$, define $X_n := \{ P_{\vec{\alpha}} : | \vec{\alpha} | = n \}$. Each $X_n$ is then a lattice partition of $1 \in \mathbb{P}$ satisfying (i), (ii), (iii) for $\epsilon = 2^{ - n}$. Now given $P \in \mathbb{P}^+$, fix $n$ such that $2^{-n} < \phi(P)$; then by (iii), some $P_{\vec{\alpha}} \in X_n$ satisfies either $P \leq P_{\vec{\alpha}}$ or $P_{\vec{\alpha}} \leq P$; but the first alternative is excluded because each $P_{\vec{\alpha}} \in X_n$ satisfies $\phi(P_{\vec{\alpha}}) < 2^{-n} < \phi(P)$.

Lemma 2: If, for all $P \in \mathbb{P}^+$, every nontrivial lattice partition of $P$ has cardinality $2^{\aleph_0}$, then there is a dense embedding of $C_{2^{\aleph_0}}$ into $\mathbb{P}^+$.

Proof: Fix $P \in \mathbb{P}^+$, $\epsilon > 0$, and a faithful normal state $\phi$ on $\mathcal{R}$; it suffices to show there exists a lattice partition $X$ of $P$ meeting requirements (i), (ii), (iii) of Lemma 1.

Enumerate as $\{ P_\alpha : \alpha < 2^{\aleph_0} \}$ the projections in $\mathbb{P}^+$ that are $< P$.

We define $Q_\alpha$ by induction on $\alpha$, for all $\alpha < 2^{\aleph_0}$, with the induction hypothesis that the previously-defined $Q_\beta$ are pairwise-disjoint. At step $\alpha$, if $P_\alpha \leq Q_\beta$ for some $\beta < \alpha$, let $Q_\alpha = 0$. Otherwise, under the present lemma's supposition, $\{ Q_\beta \wedge P_\alpha : \beta < \alpha \}$ cannot be a partition of $P_\alpha$ (in the lattice $\mathbb{P}$), because its cardinality is less than $2^{\aleph_0}$. Thus $\{ Q_\beta \wedge P_\alpha : \beta < \alpha \}$ is a pairwise-disjoint but not maximal pairwise-disjoint subset of the projections in $\mathbb{P}^+$ below $P_\alpha$; so there exists in $\mathbb{P}^+$ a projection $< P_\alpha$ that is disjoint from all $Q_\beta$, $\beta < \alpha$. Let $Q_\alpha$ be such a projection satisfying $\phi(Q_\alpha) < \epsilon$ (it follows the fact that these projections form a downward-closed subset of $\mathbb{P}^+$, which has no minimal projections, and from the normality of $\phi$, that we can require an arbitrarily small $\phi$ value here).

Let $X$ be the set of all the nonzero $Q_\alpha$ so obtained. Clearly requirements (ii) and (iii) of Lemma 1 are met, and $X$ is pairwise-disjoint. Moreover $X$ must be a maximal pairwise-disjoint subset of the projections in $\mathbb{P}^+$ below $P$ -- i.e. a partition of $P$ -- because any such projection we might hope to add to $X$ was already enumerated as one of the $P_\alpha$, and either $P_\alpha$ was $\leq Q_\beta$ for some $\beta < \alpha$ or $P_\alpha \wedge Q_\alpha \neq 0$. Finally, by supposition, $X$ has the cardinality required by (i).

Corollary: If CH holds and $\mathbb{P}$ has no countable nontrivial partition then there is a dense embedding of $C_{2^{\aleph_0}}$ into $\mathbb{P}^+$.

Proof: CH means that $2^{\aleph_0}$ is $\aleph_1$, the smallest uncountable ordinal. This plus the supposition that $\mathbb{P}$'s nontrivial partitions are uncountable implies they all have cardinality $2^{\aleph_0}$ (note $| \mathbb{P} | \leq | \mathcal{R} | \leq | \mathcal{B}(H) |$, which is $\leq 2^{\aleph_0}$ when $H$ is separable, because $\mathcal{B}(H)$ then has a countable dense subset in the strong operator topology, from which at most $2^{\aleph_0}$ convergent nets can be chosen). And this, plus the observation that $\{ P' \in \mathbb{P} : P' \leq P \}$ is order-isomorphic to all of $\mathbb{P}$ for any $P \in \mathbb{P}^+$, implies the supposition of Lemma 2.


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