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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?

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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.


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 arXiv: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.)


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.

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  • $\begingroup$ Thanks for the detailed answer. How do you get that $\tilde{\pi} : A^{**} \to B(E)$ is an isometry? The usual proof in the C*-algebra case uses the Kaplansky Density Theorem, and the simplest alternative proof I can think of uses the polar decomposition of normal functionals. $\endgroup$ – Cameron Zwarich May 24 at 14:14
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    $\begingroup$ Does Lemma 3.4 in your Dual Banach Algebras paper suffice as a replacement for polar decomposition? $\endgroup$ – Cameron Zwarich May 24 at 14:22
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    $\begingroup$ Yes, exactly that! Lemma 3.4 gives you the norm control you need. $\endgroup$ – Matthew Daws May 24 at 15:41

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