David R. MacIver's Blog

Boolean to C* algebras II: The Noncommutative Version

This one is much more speculative, as I still haven’t worked out the details. I think I worked out a few more back when I was actively investigating this stuff, but if so then I’ve forgotten them all.

The question is, what are the natural analogue of Boolean algebras in the noncommutative case?

The answer is “Err, well, I’m not quite sure!”

I’ve tried to see how far I can extend the $I(A)$ construction to the noncommutative case and see what happens. The results are interesting, but perhaps not very informative.
First of all, we can’t just look at arbitrary idempotents. These turn out to not be very well behaved. Instead we’ll look at self-adjoint idempotents - the projections. (In the commutative case every idempotent is self adjoint). We’ll call the set of these $I(A)$ as before.
Second of all, we run into a problem. The product of two projections is only a projection if they commute. So, in the noncommutative case, we don’t have the multiplication operation making it into a Boolean algebra. We can however still make it into a partial order:

Define $x prec y$ if $xy = x$.

Thie definition may seem like there should also be a right-multiplication version, but in fact there isn’t. If $x prec y$ then $x = x^* = (xy)^* = y^* x^* = yx$ (one of the many reasons it was important that these things are self adjoint).

So, if $x prec y$ and $y prec x$ then $x = xy = y$. So the relation is antisymmetric. It’s evidently transitive and reflexive, so we have a partial order.

It has top and bottom elements, $0$ and $1$.

We have an operator $neg x = 1 - x$ which maps $I(A)$ into itself. It has the following properties:


Note that it is not determined by these latter two properties. $I(A)$ is not a distributive lattice!

Is $I(A)$ even a lattice? I don’t know. It looks like it.

In the case of $End(H)$ we have $I(A)$ is isomorphic to the lattice of closed subspaces of $H$, with $neg$ being orthogonal complement and $prec$ being $subseteq$. This is a lattice, with $X wedge Y = X cap Y$ and $X vee Y$ being the closed linear span of $X cup Y$.

Further, fix $x in X setminux Y$. Then for $(pi_X pi_Y)^n x to 0$. So one might expect that $(pi_X pi_Y)^n to pi_{X cap Y}$. Unfortunately this isn’t true. Usual problem with exchanging limits. It’s true in some sort of weak topology, so if $A$ can be represented as a closed subspace of $End(H)$ in this topology then this limit is in $A$ and we have the desired result. I imagine this only works if $A$ is closed in the strong operator topology. i.e. is a von Neumann algebra. I’ll need to learn more about these before I can say for sure.
I’m not yet really sure if this works. I doubt it, and even if it does then it’s probably far too much work and there’s a better proof.

Some negative results:

The norm is not uniquely determined on the projections. Rather, every projection has norm $1$, but projections $x, y$ can have $||x - y||$ taking any value in $[0, 1]$.

Also, these posets are very very large. In the simplest example, take $A = End(mathbb{C}^2)$. Then $I(A)$ is an antichain of size $2^{aleph_0}$ plus a top and bottom element.

Note this is also an example of how badly complementation fails to be unique. For any distinct $x, y$ with $x, y not= 1, 0$ we have $x vee y = 1$ and $x wedge y = 0$.

I’ll post more as I think of it.