TITLE: The difference between $l^1(G)$ and the reduced group $C^*$ algebra $C_r^*(G)$
QUESTION [14 upvotes]: Let $G$ be a group and $l^2(G)$ the Hilbert space on $G$. The complex group algebra $CG$ can be imbedded in $B(l^2(G))$, the set of all bounded linear operators, by left translation. The reduced group $C^*$ algebra $C_r^*(G)$ is the operator norm completion of $CG$. It's clear that $l^1(G)$ lies in $C_r^*(G)$. 
Q1. Are there some typical elements in $C_r^*(G)\setminus l^1(G)$? 
Q2. For a normed algebra $A$, let $C$ be the completion of the subalgebra $\{ab-ba|a,b\in A\}$ and $T(A)=A/C$. It's not hard to see $T(l^1(G))$ is the $l^1$ completion of $C[conj(G)]$, the set of all finitely supported functions on the conjugacy classes $conj(G)$. How to describle $T(C_r^*(G))$? In particular, if $g$ is an element of infinite order, can $g$ have the same image in the quotient algebra as $0$ or a finite order element in $G$? Thank you very much.

REPLY [6 votes]: Q1 seems to be related to symmetry of $\ell^1(G)$ (a Banach $^*$-algebra $A$ is symmetric if, for every $a\in A$, the spectrum of $a^*a$ is contained in $\mathbb{R}^+$). If $\ell^1(G)$ is not symmetric, then for an element  $a^*a$ with non-positive $\ell^1$-spectrum, since the $C^*$-spectrum is clearly positive, the resolvent of $a^*a$ will define elements in $C^*_r(G)\backslash\ell^1(G)$. For examples of groups with non-symmetric algebras, see papers by J. Boidol, H. Leptin and D. Poguntke from the 1980's.