# Can ${L^{1}}(G)$ be a $C^{*}$-algebra?

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?

In what follows, we assume that $G$ is a locally compact Hausdorff group that does not have to be abelian.

You can certainly define an involution $^{*}$ on ${L^{1}}(G)$ by $$\forall f \in {L^{1}}(G), ~ \forall x \in G: \quad {f^{*}}(x) \stackrel{\text{df}}{=} \overline{f(x^{-1})} \cdot \Delta(x^{-1}),$$ where $\Delta$ denotes the modular function of $G$. This makes ${L^{1}}(G)$ into a convolution $*$-algebra.

This being said, suppose that $(U,\mathcal{H})$ is a strongly continuous Hilbert-space representation of $G$. Then we can define a $*$-homomorphism $\pi_{U}: ({L^{1}}(G),\star,^{*}) \to (B(\mathcal{H}),\circ,^{*})$ by $$\forall f \in {L^{1}}(G): \quad {\pi_{U}}(f) \stackrel{\text{df}}{=} \int_{G} f(x) \cdot U_{x} ~ \mathrm{d}{x},$$ where the convergence of the integral is with respect to the strong operator topology. Evidently, $({L^{1}}(G),\star,^{*},\| \cdot \|_{1})$ is a Banach $*$-algebra and $\left( B(\mathcal{H}),\circ,^{*},\| \cdot \|_{B(\mathcal{H})} \right)$ is a $C^{*}$-algebra, so it can be deduced, via an easy argument using the concept of a ‘spectrum’, that $\pi_{U}$ is norm-decreasing (i.e., $\| {\pi_{U}}(f) \|_{B(\mathcal{H})} \leq \| f \|_{1}$ for all $f \in {L^{1}}(G)$) and thus continuous.

We can now define a universal $C^{*}$-norm $\| \cdot \|_{u}$ on $({L^{1}}(G),\star,^{*})$ subordinate to $\| \cdot \|_{1}$ by $$\| f \|_{u} \stackrel{\text{df}}{=} \sup \! \left( \left\{ \| {\pi_{U}}(f) \|_{B(\mathcal{H})} ~ \middle| ~ \text{ (U,\mathcal{H})  is a Hilbert-space representation of  G } \right\} \right).$$ Taking the completion of ${L^{1}}(G)$ with respect to $\| \cdot \|_{u}$ results in a $C^{*}$-algebra that we shall call the group $C^{*}$-algebra of $G$, denoted by ${C^{*}}(G)$.

In order for the definition of ${C^{*}}(G)$ to make any sense, we need to determine if a Hilbert-space representation of $G$ exists in the first place. Fortunately, such representations exist, and there is a well-known one, called the left-regular representation and denoted by $\lambda$, that assigns $G$ to left-translation operators on ${L^{2}}(G)$ and has the property that $\pi_{\lambda}$ is injective; if $f \not\equiv 0$, then $$\| f \|_{u} \geq \| {\pi_{\lambda}}(f) \|_{B({L^{2}}(G))} > 0.$$ We therefore have a continuous injective $*$-homomorphism from $({L^{1}}(G),\star,^{*},\| \cdot \|_{1})$ to ${C^{*}}(G)$.

To prove that $({L^{1}}(G),\star,^{*},\| \cdot \|_{1})$ itself is not necessarily a $C^{*}$-algebra, consider $G = \Bbb{Z}$. Then the $C^{*}$-identity is not satisfied because \begin{align} \| \delta_{0} + i \delta_{1} + \delta_{2} \|_{1}^{2} = 9 \quad \text{but} \quad \| (\delta_{0} + i \delta_{1} + \delta_{2})^{*} \star (\delta_{0} + i \delta_{1} + \delta_{2}) \|_{1} & = \| \delta_{-2} + 3 \delta_{0} + \delta_{2} \|_{1} \\ & = 5 \\ & \neq 9. \end{align} In fact, we have the following theorem.

Thm. If $G$ is a discrete group of order $\geq 2$, then $({\ell^{1}}(G),\star,^{*},\| \cdot \|_{1})$ is not a $C^{*}$-algebra.

Both of your questions are thereby settled.

• Your definition of $f^*$ has $x$ free on the left hand side and no free $x$ on the right hand side! – Mariano Suárez-Álvarez Apr 27 '15 at 20:14
• @Mariano: Corrected. Thanks! – Berrick Caleb Fillmore Apr 27 '15 at 20:22

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.

• I usually sort downloaded articles according to field, sub-field and sub-sub-field, but I’ll be sure to put these two papers together! – Berrick Caleb Fillmore Apr 27 '15 at 23:09