Relationship between group $C^\ast$-algebras $C^\ast(G)$ and graph $C^\ast$-algebras $C^\ast(E)$

Question

Let $E=E(G,S)$ be the graph defined by a group $G$ and a subset $S$ of $G$. What is relationship between group $C^\ast$-algebras $C^\ast(G)$ and graph $C^\ast$-algebras $C^\ast(E)$?

Answer 1

In general, the universal group $C^*$-algebra $C^*(G)$ and the graph algebra $C^*(E)$ of its Cayley graph with respect to a generating set may be very different. For example, let's consider a finite group $G$. If $S$ is a generating set, $E$ is strongly connected. If $G$ is not a cyclic group, then the adjacency matrix for $E$ is not a permutation matrix. In this case, $C^*(G)$ is isomorphic to the group algebra, but $C^*(E)$ is purely infinite.

I think the relationship you might be looking for is the relationship between $G$ and $C^*(E)$. Since $G$ acts on $E$, by the universal property of $C^*(E)$, we have that $G$ acts on $C^*(E)$. Kumjian and Pask (http://www.uow.edu.au/~dpask/index_files/papers/codgaga.pdf) showed that the reduced crossed product $C^*(E)\rtimes_\lambda G$ is isomorphic to $C^*(E/G)\otimes \mathcal{K}(\ell^2(G))$.

Answer 2

Assuming E is the Cayley digraph, not very much. Take the Cayley digraph of a finite non-cyclic group with respect to a set of generators. The graph C*-algebra is an infinite dimensional simple C*-algebra and the group algebra is finite dimensional and not simple.