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I came across this while going through some basic examples of $C^*$ algebras.

If I consider $G$ as the set of cube roots of unity, what will be $L_1(G)$? I mean what will be the structure of elements of $L_1(G)$ in this case. Actually I have to prove $L_1(G)$ here is not a $C^*$ algebra, so basically that it does not satisfy $C^*$ identity. But I am not able to find structure of $L_1(G)$.

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up vote 1 down vote accepted

Usually, for a locally compact group $G$, one denotes $L^1(G)$ the Banach space of integrable functions on $G$ with respect to the left Haar measure (it is only unique up to multiplication by a positive constant). There is an involution involving the modular function (which measures how far the left Haar measure is from being right invariant). In addition, the convolution makes it a Banach algebra.

In the case of a compact group $G$, things are nicer. There is now a unique probability measure on the Borel sets which is both left and right invariant. It is called the Haar measure, and we'll denote it $ds$. The involution is simply $$ f^*(t)=\overline{f(t^{-1})}\qquad \forall t\in G. $$ And the product is (like in the locally compact case): $$ (f*g)(t)=\int_Gf(s)g(s^{-1}t)ds. $$ This involutive Banach algebra is unital if and only if $G$ is discrete.

There are three cubic roots of the unity: $1, \omega, \omega^2=\overline{\omega}$ where $\omega=e^{2i\pi/3}$. So $G=\{1,\omega, \omega^2\}=\mathbb{Z}_3$ is of course a finite group. The Haar measure is defined by its value $1/3$ on the singletons. Every complex-valued function on $G$ is measurable and integrable: $$ \|f\|=\frac{|f(1)|+|f(\omega)|+|f(\omega^2)|}{3}. $$ Let us compute $f*f^*$. We have $$ (f*f^*)(t)=\frac{1}{3}(f(1)\overline{f(t)}+f(\omega)\overline{f(\omega^2t)}+f(\omega^2)\overline{f(\omega t)}). $$ Now consider the identity function $f(t)=t$ for $t=1,\omega, \omega^2$. Using $1+\omega+\omega^2=0$, you can check that $f*f^*$ is constant equal to $0$. Hence $$ \|f\|^2=1\neq0=\|f*f^*\|. $$ So $\|f\|$ is not a $C^*$ norm on $L^1(G)$, since this would require $\|f*f^*\|=\|f\|^2$ for every $f$ in $L^1(G)$.

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When the group $G$ is a finite group, $\ell^1(G)$ ($L_1(G)$) is equal to $\mathbb{C}G$, the group algebra of $G$. About your original problem, there is a theorem which asserts that $G$ is the trivial group iff $L^1(G)$ is a $C^\ast$-algebra. For the proof, see Proposition 2.2.21 in my lecture notes on $C^\ast$-algebras available at arXiv: 1211:3404. Of course, it is quite possible you can find an easier solution for your original problem.

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As a group, $G$ must contain the identity. – Vahid Shirbisheh Mar 24 '13 at 19:10

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