Any commutative AW$^*$-algebra has the form $C(X)$ where $X$ is a Stonean space.

In Takesaki's book Theory of Operator Algebra's, he shows that any Stonean space can be decomposed into three parts: a hyperstonean part, a part with a dense meagre set and a part where every regular positive measure has a nowhere dense support. Let's call these Type A,B and C.

Type A is a von Neumann algebra, while type B and C allow no normal states, so they are in a certain sense maximally non-von Neumann.

A Type C Stonean space doesn't allow a faithful state, since all states/measures have nowhere dense support.

My question is: Is there a Type B commutative AW$^*$-algebra that has a faithful state? Or in other words: Does there exist a Stonean space with a dense meagre set that has a measure on it that is nonzero on every open set.

edit: Just want to add that the question can be reframed in another way: Is there a commutative AW$^*$-algebra with a faithful state that isn't a von Neumann algebra? There is a result that says that any AW$^*$ II-factor with a faithful state is a von Neumann algebra, while there are III-factor AW$^*$-algebra's that aren't von Neumann, so I wouldn't be surprised if there was some result about the commutative case in either direction.


Found an answer myself: We can take any Type B commutative AW*-algebra $C(X)$ such that $X$ is separable, i.e. has a dense countable subset. Then there is a faithful positive unital map $F:C(X)\rightarrow L^\infty(\mathbb{N})$. Since $L^\infty(\mathbb{N})$ definitely has faithful states $C(X)$ then has those as well.

That there exist such $X$'s such that $C(X)$ isn't a von Neumann algebra is shown in for instance On classifying monotone complete algebras of operators by Saito and Wright.


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