A question about the definition of group $C^\ast$-algebra

Question

Let $G$ be a local compact group, then group $C^\ast$-algebra of $G$ is defined as the completion of $C_c(G)$ with respect to some norm.

By now, I have seen three norms.

1. $\|f\|=\sup\|\pi(f)\|$, where the supremum is taken over all cyclic $\ast$-representation of $C_c(G)$.
2. $\|f\|=\sup\|\pi(f)\|$, where the supremum is taken over all non-degenerate $\ast$-representation of $C_c(G)$.
3. $|f|=\sup\|f\|$, where the supremum is taken over all $C^\ast$-norm $\|\cdot\|$.

Are they all same one?

Who can tell me which norm is the right norm in the definition of group $C^\ast$-algebra?

Thanks!

Answer 1

It is a well-known fact that every representation of a $*$-algebra decomposes in the direct sum of cyclic representations. The way the group C*-algebra is defined is by taking the supremum over all the representations of $C_c(G)$, but this procedure inevitably leads to Russell's paradox. One way to avoid this is to consider the family of all C*-seminorms on $C_c(G)$, take the supremum, and construct the enveloping C*-algebra of $C_c(G)$ the usual way: from the supremum you actually hit the maximum, which is the maximal C*-seminorm. If this is not a C*-norm then you quotient by $N=\{f\in C_c(G)\ |\ \Vert f\Vert_{\text{max}} = 0\}$ and then take the completion w.r.t. $\Vert\ \cdot\ \Vert_{\text{max}}$, otherwise you can skip the step of quotenting out and go directly to the completion.

Answer 2

There is no set-theoretic issue here. You never define the set of all representations, you define the set of real numbers $$\{ x \in \mathbb{R} : x = \pi(a), \pi : A \to B(H) \text{ is a representation } \}$$ and then take the supremum. The quantification over all representations occurs in the application of the Axiom (Schema) of Replacement to $\mathbb{R}$.

However, if you do want to take a set of representations for some reason, then just restricting to cyclic representations isn't good enough; there can be no set of all cyclic representations, simply because the class of separable Hilbert spaces is not a set. In the case of a locally compact group you know that $C_c(G)$ has a bounded approximate identity of norm at most $1$ and enough representations to separate points. The GNS construction implies that every cyclic representation is unitarily equivalent to some GNS representation, and thus it suffices to take the set of GNS representations corresponding to states on $C_c(G)$.