Radicals of Biduals of Beurling Algebras Can Be Different for the Two Arens Products
Jared T. White
TL;DR
The paper resolves a question of Dales and Lau by constructing a Beurling algebra on the free group $ ext{F}_3$ with a non-symmetric weight $\omega$ from an infinite generating set $X$, for which the Jacobson radical of the bidual differs between the two Arens products $oxed{oxempty}$ and $oxed{oxempty}$. The method builds $ ext{A}= ext{l}^1( ext{F}_3,\omega)$ and introduces weak*-limit functionals $\Phi_0$ and $\\Psi_0$ so that $\, ext{I}= ext{A}^{**} oxed{ ext{rad}}$ under $oxempty$ contains a nonzero element with nilpotent left action, while the analogous right action under $ riangle$ is not radical; a second functional $\, extPsi_0$ ensures non-quasi-nilpotence of $\, extPhi_0 riangle extPsi_0$. The result demonstrates that $ ext{rad}( ext{A}^{**}, oxempty) e ext{rad}( ext{A}^{**}, riangle)$ and provides a nontrivial example beyond the commutative/operator-algebra cases, enriching the understanding of bidual radicals and Arens regularity.
Abstract
Let $\operatorname{rad}$ denote the Jacobson radical of a Banach algebra, and let $\Box$ and $\Diamond$ denote the two Arens products on its bidual. We give an example of a Beurling algebra $\mathcal{A}$ for which $\operatorname{rad}(\mathcal{A}^{**}, \Box) \neq \operatorname{rad}(\mathcal{A}^{**}, \Diamond)$, answering a question of Dales and Lau. The underlying group in our example is the free group on three generators.
