2
$\begingroup$

Which von Neumann algebras acting on separable Hilbert space $H$ have uncountable antichains of projections? ("Antichain" meaning a set of projections any pair of which has no shared nonzero subprojection.)

I'm pretty sure $\mathcal{B}(H)$ (the algebra of all bounded operators on $H$) does have one. For instance, for any orthogonal $x, y \in H$ and all real $c$, if $P_c$ denotes the unique one-dimensional projection whose range includes $x + cy$, then $\{P_c : c \in \mathbb{R} \}$ is an uncountable antichain.

I'm equally sure that an abelian algebra without minimal projections cannot have one, since its lattice of projections is basically a measure algebra with a countable dense subset.

But with non-abelian type II and type III algebras, I'm not sure where to begin ... even defining what the projections are in concrete cases seems hard; all the examples I've found are defined with intricately-constructed unitary operators rather than with projections. Can anyone help me out here, or point me to a good discussion of the properties of projection lattices of the various types of vN algebras? Thanks!

$\endgroup$
5
  • $\begingroup$ How do projections "share subprojections"? Does this mean they are pairwise orthogonal? $\endgroup$
    – rschwieb
    Commented Jan 12, 2016 at 3:26
  • $\begingroup$ I probably should have defined it more formally ... By calling $S$ an "antichain of projections" I mean that for any distinct $P, Q \in S$, there exists no nonzero projection that is both a subprojection of $P$ and a subprojection of $Q$. This doesn't imply that $P$ and $Q$ are orthogonal unless you're in an abelian algebra, I think -- unless I'm mistaken, the $P_c$ I define form an antichain but are not pairwise orthogonal. $\endgroup$ Commented Jan 12, 2016 at 3:41
  • $\begingroup$ I guess I wasn't clear because that isn't helping. What is a subprojection? $\endgroup$
    – rschwieb
    Commented Jan 12, 2016 at 3:57
  • $\begingroup$ Projection $P$ is a subprojection of projection $Q$ iff $P$'s range is $\subseteq$ $Q$'s range. Equivalently, iff $PQ = P$. (The term is from Kadison & Ringrose's book.) Also thanks @rschwieb for the Goodearl reference -- that led me to some references to some lattice theory papers; it looks like a lot was figured out about the structure of these projection lattices pre-1980. $\endgroup$ Commented Jan 12, 2016 at 4:25
  • $\begingroup$ I was sure it would be useful. Also FFR, you can thank posts that are useful with upvotes... That's the norm here. $\endgroup$
    – rschwieb
    Commented Jan 12, 2016 at 11:48

2 Answers 2

2
$\begingroup$

Unless I'm misreading your definition, this is an antichain of projections in $M_2(\mathbb C)$: $$ P_t=\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix},\ \ t\in[0,1]. $$

This idea can be made to work easily in any type I von Neumann algebra.

For types II and III, a deeper idea is needed, namely halving: given any projection $P_0\in\mathcal M$, there exists $P\leq P_0$ with $P\sim P_0-P$. So there exists $V$ with $V^*V=P$ and $VV^*=P_0-P$. Now, with $Q=P_0-P$, $$ \mathcal M_0=\text{span}\{P,V^*,V,Q\} $$ is a subalgebra of $\mathcal M$, isomorphic to $M_2(\mathbb C)$. Now we can construct the $P_t$ as above, namely $$ P_t=t\,P+\sqrt{t-t^2}\,(V^*+V)+(1-t)\,Q,\ \ \ t\in[0,1]. $$

$\endgroup$
4
  • $\begingroup$ Yes, I think that's the best example for type I von Neumann algebras. Can type II and III algebras have uncountable antichains? $\endgroup$ Commented Jan 13, 2016 at 16:19
  • $\begingroup$ The idea of the $P_t$ works in any non-commutative von Neumann algebra, as they all contain a copy of $M_2(\mathbb C)$. Please see the edit. $\endgroup$ Commented Jan 14, 2016 at 4:54
  • $\begingroup$ Brilliant, this is so helpful. I am still trying to nail down formally why the span of those operators is isomorphic to $M_2(\mathbb{C})$ but I see intuitively that it is. Thanks again. $\endgroup$ Commented Jan 14, 2016 at 16:47
  • $\begingroup$ I had tried to provide a simple argument by it was wrong, as the $P$ and $Q$ were not orthogonal to each other. I have edited the answer. To understand the isomorphism, just notice that $P,V^*,V,Q$ play the roles of $E_{11}, E_{12}, E_{21}, E_{22}$. $\endgroup$ Commented Jan 14, 2016 at 18:50
1
$\begingroup$

Without being too familiar with the material, I am not 100% certain, but I think that Goodearl's von Neumann regular rings discusses what you want. It definitely covers the types of VNR rings and their idempotents.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .