Free product of $C^*$-algebras

Question

i want to figure out what the "free product of two $C^*$-algebras" precisely is. Thus suppose $A$ and $B$ are $C^*$-algebras. How to find or construct $A\ast B$? I readed something about the $C^*$-algebra generated by $A$ and $B$. Is $A\ast B$ therefore the C^*-algebra of all psooible products $c_1c_2...c_n$ for $c_i\in A\cup B$? What is the norm on such a free product? Can someone explain this explicit because i don't find good literature about this?! Thank you very much.

Answer 1

It is too long to write details here (you can see in them in chapter 1 of Voiculescu-Dykema-Nica, for instance).

You first construct the vector space spanned by $$\{a_1a_2\cdots a_n: \ n\in\mathbb N, a_j\in A\cup B,\ \text{no consecutive }a_j,a_{j+1}\text{ belong to the same algebra} \}. $$ Then you quotient by the obvious linearity relations $$a_1a_2(\alpha a_3 +\beta a_3')a_4\cdots a_n=\alpha\,a_1a_2a_3\cdots a_n+\beta\,a_1a_2a_3'\cdots a_n,$$ etc.

This quotient is a $*$-algebra. Then you show that this algebra has many representations, by constructing explicit representations in terms of the individual representations of $A$ and $B$.

And then you define $A*B$ as the enveloping C$^*$-algebra.