Why does free probability use Von Neumann algebras instead of C star algebras?

Question

My level of understanding is very rudimentary--basically I understand that if you only let your variables live in a star-algebra, you can't take limits. But why is it better to take the weak star closure instead of the norm closure?

Answer 1

I'm not yet an expert in the field, so maybe someone more experienced can add to this later.

In my understanding, Voiculescu was originally interested in the free group factor problem, which concerns specifically the von Neumann algebras coming from the free groups on $n$ generators. For this reason, much of the focus of free probability theory has been on the von Neumann algebraic level, hoping to obtain some nice invariants for these von Neumann algebras.

That said, free probabilistic techniques are still interesting on the C* level. For a reference on free probability that focuses only on C* aspects and not at all on the von Neumann algebraic aspects, try the nice book "Lectures on the Combinatorics of Free Probability" by Nica and Speicher.

Answer 2

Two things:

1. One constructs the free product by fixing a state on each algebra, and considering the GNS representation given by such state. So, there is no real advantage to restricting to C$^*$-algebras, as the free product as constructed by Voiculescu will embed naturally in the free product of the weak operator closures.

2. Even in the classical case (i.e., independence) one also considers von Neumann algebras: one uses $L^\infty(X)$ and not $C(X)$. I'm definitely no probabilist, but I think that even basic objects like conditional expectations would not be defined over the continuous functions (and you wouldn't have characteristic functions).

3. What Josh said.