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Given a locally compact Hausdorff group $G$, one can construct several Banach star-algebras using $G$ (and its associated Haar measure): $L^1 (G)$, $M(G)$ (regular complex measures on $G$), $L^{\infty} (G)$, $C^* (G)$, $C^*_r (G)$, $W^* (G)$, etc. (see this Wiki article for some details). Then one can ask, for instance, which Banach *-algebras can be represented (i.e. are isometrically *-isomorphic) to/as $L^{\infty} (G)$. In this case every such algebra is an abelian Von-Neumann algebra, and every abelian Von-Neumann algebra can be represented as $L^ \infty (X,\mu)$ for some set $X$ with measure $\mu$`, so the question reduces to the question "which measures are Haar measures (on some LCH group)?". However, I'm mostly interested in this type of question for the algebras $L^1 (G)$. It is known that such a Banach *-algebra is semisimple and symmetric for any abelian LCH group $G$, so obviously one can't represent all commutative Banach *-algebras as $L^1 (G)$. So, to summarize, I have two questions:

  1. Is there an operator-theoretic characterization of the Banach algebras which are isometrically *-isomorphic to $L^1 (G)$ for some $G$? In particular, is there an abelian, symmetric, semisimple Banach *-algebra which isn't $L^1 (G)$ for some $G$?

  2. Are there classes of Banach *-algebras to which one can associate an LCH group $G$ in a canonical way (up to an isomorphism of topological groups), so that the algebra is naturally isomorphic to a group algebra $L^1 (G)$?

Also, if there is some known reference for this kind of problems, I'll be glad to know of it.

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2 Answers 2

up vote 7 down vote accepted

A characterization of the Banach algebras which are isometrically *-isomorphic to $L^1(G)$ for some $G$ was given by P. L. Patterson, Characterization of algebras arizing from locally compact groups, Trans. Amer. Math. Soc. 329 (1992), 489-506. This paper contains references to earlier ones dealing with special classes of groups, in particular Abelian and compact ones.

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The MR-Review of that paper is here: ams.org/mathscinet-getitem?mr=1043862 –  Theo Buehler Apr 2 '11 at 18:30
    
Great!! Thanks. –  Mark Apr 3 '11 at 19:33
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[This is too long to be a comment, but is at best a partial answer; I just thought I'd write it down while I have spare time.]

Regarding the 2nd part of Question 1: you need to impose some extra hypotheses to capture the "real" question I suspect you are interested in. Otherwise, just take $C[0,1]$, which is certainly abelian, symmetric as a *-algebra, and semisimple. This cannot be isomorphic to $L^1$ of anything, even as a Banach space. (To give a bit more detail: every map from $C[0,1]$ to $L^1$ is weakly compact, so if the two were isomorphic as Banach spaces then the identity map on $C[0,1]$ would be weakly compact, implying $C[0,1]$ is reflexive which is not the case.) If you want non C*-examples then $C^1[0,1]$ will also do, for much the same reasons.

If you're after a counter-example where the underlying Banach space looks like $L^1$ or $\ell^1$, then I suggest the convolution algebra $\ell^1(S)$ where $S$ is an infinite semilattice (=commutative semigroup in which all elements are idempotent); the involution is just complex conjugation. Semisimplicity is proved in an old paper of Hewitt and Zuckerman (I am not sure of the spelling of the 2nd author's name, and as I am out of the office can't check right now). Off the top of my head, the quickest, if not the most direct, way to show that such an algebra is not isomorphic to an abelian group algebra, is to use the fact (Duncan & Namioka, late 1970s) that $\ell^1(S)$ for such $S$ is never amenable, while every abelian $L^1$-group algebra is amenable (Johnson, 1972). There ought to be a simper way to distinguish the two classes of algebra, though.

Update. If you are interested only in isometric isomorphism of Banach algebras then there is a much easier way to show $\ell^1(S)$ is not a group algebra. For if $\theta: \ell^1(S)\to L^1(G)$ is an isometric algebra isomorphism, then by examining where extreme points of the unit ball go, you find that $G$ must be discrete and for each $x\in S$, $\theta(\delta_x)=\delta_{\phi(x)}$ where $\phi:S \to G$ is an isomorphism of semigroups. Now $S$ has lots of idempotents, $G$ has only one, and we have reduced to absurdity as they say.

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If you want yet another argument that $C[0,1]$ cannot be isomorphic as a Banach algebra to an abelian group algebra, then: there are two canonical products one can introduce on the double dual of a Banach algebra, turning it into a Banach algebra (these are called the Arens products on the second dual). For $C[0,1]$ the two Arens products on $C[0,1]^{**}$ coincide; for $L^1(G), the two Arens products on $L^1(G)^{**}$ are different. –  Yemon Choi Apr 3 '11 at 19:06
    
Well, I didn't expect all Banach -algebras to be isomorphic to an L^1, or even the commutative, semisimple, symmetric ones. Your remarks do point out, though, that the answer is probably no for most C-algebras (the commutative unital ones are $C(X)$ for $X$ compact), since $L^1 (G)$ is rarely a C*-algebra. –  Mark Apr 3 '11 at 19:40
    
Right, which is why I was surprised at your Question 1, and thought that perhaps you'd meant to ask something different. By the way, the answer is definitely no for every infinite-dimensional C*-algebra, whether commutative or not -- you can either use the reflexivity argument in my main answer or an appeal to Arens regularity as in my comment. If you're looking at finite-dimensional C*-algebras, and allowing non-isometric isomorphism, then I suspect you'd need to get into the representation theory of finite groups to see what kinds of matrix algebra can arise. –  Yemon Choi Apr 3 '11 at 20:28
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