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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?

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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.

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

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    $\begingroup$ 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! $\endgroup$ – Berrick Caleb Fillmore Apr 27 '15 at 23:09

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