Are there any classes of (Arens regular) Banach algebras that are not operator algebras whose bidual comes from a “universal representation”, as in the case of C\*-algebras?



Classes of Banach algebras (that aren't operator algebras) whose bidual comes from a "universal representation"
This sort of depends on what you mean by "universal representation". For $C^\*$-algebras, I think the statement is usually the following: Given a $C^\*$-algebra $A$ and a representation $\pi:A\rightarrow B(H)$, let $M(\pi) = \pi(A)''$ denote the von Neumann algebra generated by $\pi(A)$. There is a unique surjective normal $\*$-homomorphism $\tilde\pi:A^{\*\*}\rightarrow M(\pi)$. (Here I followed Takesaki's book.)


For Banach algebras, we need to decide what the analogue of "von Neumann algebra" is, and perhaps what "normal" should mean.




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One solution is to focus on *dual Banach algebras*, that is, Banach algebras $A$ for which there is some predual $A\_\*$ making the multiplication separately weak$^\*$-continuous. Then "normal" means weak$^\*$-continuous. Such algebras have long been studied, with Runde undertaking a lot of study. In my paper [Dual Banach algebras: representations and injectivity](https://mathscinet.ams.org/mathscinet-getitem?mr=2289356) [arXiv:math/0604372](https://arxiv.org/abs/math/0604372) I study representations of dual Banach algebras on *reflexive* Banach spaces (using ideas of Young). Typically I fail to make a clean statement, but we have:


* If $E$ is a reflexive Banach space then $B(E)$ is a dual Banach algebra with predual $E\widehat\otimes E^\*$, and so any weak$^\*$-closed subalgebra of $B(E)$ is a dual Banach algebra.
* By Corollary 3.8 every dual Banach algebra is isometrically weak$^\*$-weak$^\*$-continuously isomorphic to a weak$^\*$-closed subalgebra of $B(E)$ for some reflexive $E$.
* There is a closed submodule $\newcommand{\wap}{\operatorname{wap}}\wap(A^\*)$ of $A^\*$ which is maximal so that the quotient map $A^{\*\*}\rightarrow\wap(A^\*)^\*$ induces a single product on $\wap(A^\*)^\*$ for either Arens product. In particular, $A$ is Arens regular if and only if $\wap(A^\*)=A^\*$.
* $\wap(A^\*)^\*$ is hence a dual Banach algebra, and it satisfies the universal property we hope for: see Proposition 2.9 (due to Runde). If $B$ is any dual Banach algebra, a homomorphism $A\rightarrow B$ admits a unique extension a weak$^\*$-continuous homomorphism $\wap(A^\*)^\*\rightarrow B$.


So let $A$ be Arens regular, and consider a representation $\pi:A\rightarrow B(E)$ for some reflexive $E$. Then we obtain a unique extension $\tilde\pi:\wap(A^\*)^\* = A^{\*\*} \rightarrow \overline{\pi(A)}^{w^\*}$. If $\pi$ induces a *bounded below* map $A/\ker\pi\rightarrow B(E)$ then $\tilde\pi$ will be surjective (from Hahn-Banach). In general, I don't see that $\tilde\pi$ will be surjective.


To better answer the **original question**: "yes". If we take $\pi:A\rightarrow B(E)$ to be a "maximal" representation on a reflexive $E$ (e.g. the sum of all cyclic representations, "cyclic" chosen to obtain a set not a proper class) then $\pi$ is an isometry (follows from Theorem 3.6, also from Young's work) and $\tilde\pi:A^{\*\*}\rightarrow B(E)$ is an isometry, so $A^{\*\*}$ is the weak$^\*$-closure of $\pi(A)$.


(In the last few years, Thiel and Gardella have studied similar things, but I don't have an exact reference.)




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A second approach is to take a more permissive approach to what a "representation" is: just any bounded homomorphism $\pi:A\rightarrow B(E)$ for any Banach space $E$. This is the same notation as a (bounded) left $A$-module, and I will sometimes use the module language. Any $A$ has an isometric representation: let $E=A\oplus\_1\mathbb C$ be the unitisation, and $\pi$ the left-regular representation, so $a\cdot (b,\alpha) = (ab+\alpha a, 0)$ for $a,b\in A, \alpha\in\mathbb C$. But $B(E)$ carries no (natural) weak$^\*$-topology.


Instead, we could use $E=A^{\*\*}\oplus\_1\mathbb C$ with the predual $A^\*\oplus\_\infty\mathbb C$, and the left-regular representation. Then the predual becomes a right module for the action $(a^\*,\beta)\cdot a = (a^\*\cdot a, \langle a^\*,a \rangle)$ with the usual action of $A$ on $A^\*$. Then $B(E)$ does not become a dual Banach algebra, but it does become a one-sided dual Banach algebra (see work of Spain, and Thiel and Gardella, for more on this notion). Furthermore, there is a standard way to extend the action of $A$ on $E$ to an action of $A^{\*\*}$ which turns $E$ into a left $(A^{\*\*},\Box)$ module, where $\Box$ is the first Arens product. A check shows that this agrees with the weak$^\*$-extension of the action of $A$, and the image in $B(E)$ is $A^{\*\*}$. Thus again $A^{\*\*}$ arises as the weak$^\*$-closure of a "universal" representation of $A$.


The difference here is that this only works with the first Arens product (a similar construction uses the second Arens product) but we don't need Arens regularity.


