2
$\begingroup$

I've been trying to understand the definition of a group C*-algebra. Given a topological group $G$ and a C*-algebra $A$, let $u: G \to A$ define a unitary representation $U(G)$ of $G$ on $U(A)$, the unitary group of $A$. Let $\mathbb{C}G$ denote the group algebra of $G$. Then $u$ induces a homomorphism $\pi_u: \mathbb{C}G \to A$. We define the group C*-algebra of $G$ to be the completion of $\mathbb{C}G$ with respect to the norm

$$\| a \| := \sup \{ \| \pi_u(a) \|: u: G \to U(A) \text{ is a homomorphism} \}$$

I have a difficult time understanding what exactly completion with respect to this norm means. After thinking about it for a day or so, the only answer I've come up with is that given any Cauchy sequence $\{ x_n \}$ in $U(A)$ (where $A$ can be any C*-algebra and $U(A)$ can be any unitary representation), there is a corresponding Cauchy sequence $\{ y_n \}$ in $\mathbb{C}G$ and the group C*-algebra $C^*(G)$. In $C^*(G)$, we are guaranteed that the limit of this Cauchy sequence exists. Is this correct?

  • 2
    First of all I'd restrict to discrete groups first for making things a bit easier (I guess you don't want a general topological group but rather a locally compact one - to guarantee that $\|a\|$ is actually a norm). Then note that $\|\pi_{u}(a)\|$ defins a semi-norm on $\mathbb{C}G$ for any $\pi$. Now it makes sense to speak of all semi-norms that arise in this way. If there are sufficiently many representations (that's why you need some sort of restriction on $G$) to send $a$ to some *nonzero* element in $\operatorname{End}(A)$ then $\|a\|$ is non-zero. This defines a *norm* on $\mathbb{C}G$2011-05-20
  • 1
    and you complete $\mathbb{C}G$ with respect to that norm, which is a straightforward process. In the locally compact case you have to work quite a bit in order to see that there are sufficiently many representations, the compact case is a bit easier because of Peter-Weyl. I think all this is quite well explained in Bekka-de la Harpe-Valette's book on property (T) in an appendix (at least for second countable groups). Otherwise the only really detailed reference I know is Dixmier's book on $C^{\ast}$-algebras. As we have some operator-theorists here, I let them jump in for more details.2011-05-20
  • 1
    Two other references: Pedersen, $C^{\ast}$-algebras and their automorphism groups and Takesaki's Theory of operator algebras but I don't know which volume, off-hand (I guess most of this is in the exercises in volume one but volume two might have more information)2011-05-20
  • 1
    Finally, as any $C^\ast$-algebra has a faithful (hence isometric) representation on some (possibly humungous) Hilbert space by the GNS construction, it's actually enough to look at unitary representations on Hilbert spaces in the first place. The size of $G$ cuts down the possible size of that Hilbert space on which G acts unitarily and non-trivially, and thus the supremum in the definition isn't really problematic.2011-05-20
  • 0
    Thanks so much for the speedy response! I'll look up some of those references and get back here if I'm still confused.2011-05-20
  • 0
    You're welcome. As I said, I'd really think about discrete groups first, because that way you can avoid many technicalities while getting the ideas straight. As soon as you feel comfortable with discrete groups, then the locally compact case isn't *that* difficult anymore, it's more a bunch of technicalities that may obscure things more than clarify. I can't recommend a good reference for the case of discrete groups, although I'm sure it's explained in many places. Please don't take the second sentence in my second comment too seriously, I mixed up some things. Sorry about that.2011-05-20
  • 0
    After thinking about it for a couple of days and learning about the continuous functional calculus, I think I have a better idea of what is going on now. Does this sound right? If we have a specific unitary representation generated by one unitary U, then by the continuous functional calculus the C*-algebra generated by this unitary consists of continuous functions on the spectrum of U (a subset of the unit circle). However, the universal C*-algebra generated by a single unitary consists of continuous functions on the entire unit circle.2011-05-26
  • 0
    Is this because when we create the universal C*-algebra generated by a single unitary, the fact that it encodes 'all possible representations' causes the spectrum of U to 'expand' to the entire unit circle to account for all possibilities? Sorry for the very loose language here - this is all still quite foggy to me - let me know if this needs clarification.2011-05-26
  • 0
    I think I have it figured out a bit more now. The universal C*-algebra generated by a unitary $U$ consists of polynomials in $U$ and $U^*$ with the relation $U^*U=UU^*=1$. By the continuous functional calculus, these correspond to polynomials in $z$ and $\overline{z}$ in the complex plane. Then, the relation $\overline{z}z = z\overline{z} = 1$ restricts these polynomials to the unit circle. We then take the closure with respect to uniform convergence on these polynomials.2011-05-26
  • 0
    Thus, the group C*-algebra of say, $\mathbb{Z}$, is precisely the continuous functions on the unit circle, since unitary representations of $\mathbb{Z}$ are generated by a single unitary. I am still a bit foggy on the connection to the norm, but I think I'm satisfied for the time being.2011-05-26

0 Answers 0