Where does the double commutant theorem fails for $AW^*$-algebras?

Question

Commutative $AW^*$-algebra are characterized as those $C^*$-algebras such that their space of projections is a complete boolean algebra (see http://en.wikipedia.org/wiki/AW*-algebra).

Von Neumann algebras are characterized as those $C^*$-subalgebras $A$ of $B(H)$ satisfying any of these three equivalent conditions - see section 4.6 in Pedersen's book "Analysis Now" (specifically Theorem 4.6.7 and Proposition 4.6.15):

• $A$ overlaps with its double commutant $A''$ (where $A'=\{T:ST=TS \, for \, all \, S\in A\}$).

• $A$ is weakly closed ($A$ is closed in the topology generated by the seminorms $T\mapsto |(Tx,y)|$ for $x,y\in H$).

• $A$ is $\sigma$-weakly closed ($A$ is closed in the topology generated by by the seminorms $T\mapsto |\sum_n(Tx_n,y_n)|$ for $x_n,y_n\in H$ such that $\sum \|x_n \|,\sum\|y_n\|<\infty$).

Let now for example $A$ be the commutative $AW^*$-algebra given by $[f]_I$ with $f:\mathbb{R}\to\mathbb{C}$ Borel and bounded and $I$ the ideal generated by functions with meager support. It is shown in Tristan Bice's answer to Examples of hyperstonean space that this $AW^*$-algebra does not carry normal positive functionals and thus it is not a Von Neumann algebra. However $A$ is a $C^*$-algebra so (By the GNS-construction) it has an isomorphic copy as a $C^*$-subalgebra of $B(H)$ (It may be the case that $H$ is a non-separable Hilbert space though, but I don't think this matters in what follows). Where does the double commutant Theorem (4.6.7) fails for this $AW^*$-subalgebra of $B(H)$? I'm a bit puzzled - I can expect that there is a problem with $\sigma$-weakly continuous functionals, but I don't see where the problem can arise with weakly continuous functionals.

Answer 1

The first statement is not quite right - $C([0,1])$ is a commutative C*-algebra but not an AW*-algebra, even though its only projections are the constant 0 and 1 functions, which certainly constitutes a complete Boolean algebra. However, it is true that AW*-algebras are precisely the real rank zero C*-algebras whose projections form a complete lattice. In the commutative case this corresponds to the fact that a topological space is extremally disconnected iff it is totally disconnected and its clopen sets form a complete lattice. I should have mentioned this myself in the previous post you refer to.

To represent an abstract C*-algebra $A$ concretely on a Hilbert space $H$, we usually consider the universal representation $\pi=\bigoplus\pi_\phi$ where $\phi$ ranges over all states on $A$ and $\pi_\phi$ comes from the GNS construction, as you mention. I'm not sure if this answers your question, but you can show that $\pi[A]$ is not a von Neumann subalgebra of $B(H)$, i.e. $\pi[A]\neq\pi[A]''$, not just for the AW*-algebra you had in mind, but in fact for any infinite dimensional C*-algebra $A$. This is because you can identify $\pi[A]\subseteq\pi[A]''$ with $A\subseteq A^{**}$ and C*-algebras are reflexive if and only if they are finite dimensional (see Finite dimensional $C^*$-algebras). So, strange as it may seem, even a von Neumann algebra $A$ can be represented faithfully as a non-weakly closed subalgebra of $\mathcal{B}(H)$, as long as $A$ is infinite dimensional.

Alternatively, you could consider the atomic representation, where $\phi$ above ranges only over pure states, which is still faithful on $A$. But here too we can show that $\pi[A]\neq\pi[A]''$ whenever $A$ is infinite dimensional, by even constructing a projection $p\in\pi[A]''\setminus\pi[A]$ (which is perhaps more what you are looking for?). To see this, first note that it suffices to consider unital $A$ - otherwise $1\in\pi[A]''\setminus\pi[A]$. Now take a maximal commutative C*-subalgebra $B$ of $A$, which necessarily contains the unit and is also infinite dimensional. Identify $B$ with $C(X)$, where $X$ is the space of all pure states on $B$ with the weak*-topology. As $B$ is unital, $X$ is compact, and as $B$ is infinite dimensional, $X$ is infinite. But infinite discrete spaces are not compact so $X$ must contain a non-isolated point $\phi$, which can be extended to a pure state on all of $A$. Thus we have $v\in H$ with $\phi(a)=\langle\pi(a)v,v\rangle$, for all $a\in A$ (and $T\mapsto\langle Tv,v\rangle$ is certainly a weakly continuous functional on $\mathcal{B}(H)$). As $\{\pi(b):b\in B^1_+\text{ and }\phi(b)=0\}$ is directed, it has a supremum $p\in\pi[B]''$ with $\langle pv,v\rangle=0$. But $B$ is commutative so $\pi[B]\subseteq\pi[B]'$ and hence $\pi[B]''\subseteq\pi[B]'$. Thus if we had $q\in A$ with $\pi(q)=p$ then we would have $q\in B'$ and hence $q\in B$ by maximality. As $b\leq q$, for all $b\in B^1_+$ with $\phi(b)=0$, we must have $q=1$, as $\phi$ is not isolated. But $1=\phi(1)=\langle\pi(1)v,v\rangle=\langle\pi(q)v,v\rangle=\langle pv,v\rangle=0$, a contradiction. Thus $p\notin\pi[A]$.

To ensure that $\pi[A]=\pi[A]''$, we must restrict to a different class of states on $A$, namely the normal states. And if there are not enough of these (e.g. none in Dixmier's example) then $\pi$ will no longer be faithful.