# What information about a locally compact group $G$ is encoded in $C_r^\ast(G)$ which is not in $L^1(G)$?

Let $G$ be a locally compact group and let $C_r^\ast(G)$ denote its reduced group $C^\ast$-algebra. Many features of a $G$ can be realized from $L^1(G)$ or $C_r^\ast(G)$. For example, $G$ is discrete iff $L^1(G)$ (resp. $C_r^\ast(G)$) is unital, or $G$ is abelian iff $L^1(G)$ (resp. $C_r^\ast(G)$) is commutative.

I was wondering what kind of information about the group $G$ is encoded in $C_r^\ast(G)$ which is not in $L^1(G)$?

Is it correct if we say $C_r^\ast(G)$ is just a $C^\ast$-completion of $L^1(G)$ and in fact we lose some information about $G$ by considering $C_r^\ast(G)$ instead of $L^1(G)$?

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The Banach algebra $L^1(G)$ determines $G$ and its topology by Wendel's theorem: math.stackexchange.com/q/328443. This is not true for $C_{r}^\ast(G)$, so it seems that the question in the title is the wrong way around and the last sentence is the right way of looking at it. – Martin Mar 24 '13 at 17:39
@Martin: That's exactly my point. I'd like to know what makes mathematicians to consider $C_r^\ast(G)$ instead of $L^1(G)$. When $G$ is abelian, $C_r^\ast(G)$ appears in the Pontryagin duality, $C_r^\ast(G)$ is a $C^\ast$-algebra and therefore easier to work with. What else can be said about the benefits of $C_r^\ast(G)$. For example, I am tempted to say the computation of $K$-theory of $C_r^\ast(G)$ is easier, but I am not sure about it. – Vahid Shirbisheh Mar 24 '13 at 18:05
The natural reason to look at the reduced $C^*$-algebra is that it encodes information about the unitary representations of $G$, and more precisely about the representations weakly contained in the regular representation (aka tempered representations). – YCor Mar 24 '13 at 19:13
The other reason to consider the reduced $C*$-algebra is that $C^*$-algebras are Good and Tasteful Things while Banach algebras are Old and Pointless Things. (There might conceivably be some sarcasm here.) – Yemon Choi Mar 24 '13 at 22:18
I personally prefer $C_r^*(G)$ because there is a natural assembly map from the K-homology of $BG$ to the K-theory of $C_r^*(G)$ which is an isomorphism in many cases (conjecturally in all cases). Thus the K-theory of $C_r^*(G)$ can be studied using techniques from topology. – Paul Siegel Mar 25 '13 at 3:12