TITLE: Can $ {L^{1}}(G) $ be a $ C^{*} $-algebra?
QUESTION [10 upvotes]: Let $ G $ be a locally compact abelian group. Then $ {L^{1}}(G) $ is a commutative algebra when equipped with convolution. Is there an involution $ ^{*} $ on $ {L^{1}}(G) $ so that it becomes a $ C^{*} $-algebra? We can show that the map $ f \mapsto \overline{f} $ is an involution, but with this involution, $ {L^{1}}(G) $ is not a $ C^{*} $-algebra. I believe the answer is negative, but I can’t prove it. If this is the case, can we inject $ {L^{1}}(G) $ into a larger algebra which is a $ C^{*} $-algebra?

REPLY [3 votes]: A $C^*$-algebra $A$ is isometric to an $L_1$-space (even as Banach space!) iff it is one dimensional.
Assume $A$ is isometric to $L_1$ space and $\operatorname{dim}(A)>1$, then $A$ is weakly sequentially complete. By result of Sakai (proposition 2), this is possible only if $A$ is finite dimensional. By classification theorem for $C^*$ algebras we know that $A$ is finite $\ell_\infty$-sum of finite dimensional matrix algebras: $A=M_{n_1}\oplus_\infty\ldots\oplus_\infty M_{n_k}$. Since $\operatorname{dim}(A)>1$, then either $n_i\geq 2$ for some $i$ or $k\geq 2$. In both cases we see that $A$ contains a copy of $\ell_\infty^2$. Thus we have an embedding of $\ell_\infty^2$ into $A$ which is finite dimensional $\ell_1$-space. The latter is impossible by result of Lyubich (theorem 1).
Therefore $\dim(A)=1$, that is $A=\mathbb{C}=L_1(G)$, where $G$ is unique group consisting of one element - its identity.