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Let $\Gamma$ be an infinite discrete abelian group and $A=\ell^1(\Gamma)$ denote its group algebra. Clearly, $A_*=c_0(\Gamma)$ is a predual of $\ell^1(\Gamma)$ for which $(A,A_*)$ is a dual Banach algebra.

By Matthew Daws' Representation Theorem, there exists a reflexive Banach space $E$ and a weak$^*$ continuous, isometric Banach algebra isomorphism $\phi:A\to B(E)$ onto a subalgebra of $B(E)$ (Corollary 3.8 in [Daws2007]).

Question: Do we have a simple, explicit relationship between $\Gamma$ and $E$?

In the same article, $E$ was given as an $\ell^2$-sum of the interpolation spaces between $A_*$ and $\mu\cdot A$ for unit-norm $\mu\in A_*$. Here $\cdot$ denotes the convolution.

Perhaps obvious to experts: $E$ does not admit an equivalent uniformly convex norm, since otherwise $B(E)$ (and so all of its subalgebras) would be Arens regular. However, $\ell^1(\Gamma)$ is Arens regular iff $\Gamma$ is finite.

As a side question

Question: Given $\Gamma$, do we have a characterization of reflexive $E$ for which $B(E)$ contains no subalgebra that is isometrically & weak$^*$ isomorphic to $\ell^1(\Gamma)$ ?

LSpice
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Onur Oktay
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    A short (but not uninformed) answer to your first question: NO. (Matt is welcome to correct me, but we have discussed this in the distant past.) – Yemon Choi Mar 04 '23 at 01:05
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    Regarding the second question: an even earlier result of Daws establishes that B(E) is Arens regular for every super-reflexive E. So certainly if one is given an infinite group Gamma, one knows that the class of E in your question cannot contain any super-reflexive members. – Yemon Choi Mar 04 '23 at 01:08
  • @YemonChoi Thank you for this short but valuable information. – Onur Oktay Mar 04 '23 at 17:11

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