12
$\begingroup$

The Gelfand-Naimark theorem establishes that a complex commutative Banach algebra $A$ with an identity and an involution $x\to x^*$ satisfying $\|x x^*\|=\|x\|^2$ is (isometrically isomorphic to) a $C(K)$ space where $K$ is the compact space consisting of the maximal ideals in $A$.

Is there an analogous characterization of the complex commutative Banach algebras with an identity and an involution which are a convolution algebra $L_1(G)$ with $G$ a locally compact abelian group?

$\endgroup$

2 Answers 2

5
$\begingroup$

Rieffel has given in his Ph.D. thesis a fairly satisfactory characterisation of commutative $L_1(G)$-algebras. This requires some terminology.

Let $A$ be a commutative (possibly non-unital) Banach algebra and let $f\in A^*$ be a (norm-one) character. Then $f$ is termed $L^\prime$-inducing, when $A$ is an abstract $L$-space under the ordering given by the cone

$$\{x\in A\colon \|x\| = f(x)\}$$

and $| x\cdot y | \leqslant |x| \cdot |y|$ in the sense of the above-introduced ordering (here $|x|$ stands for the Banach-lattice modulus). Rieffel then offers the following characterisation:

Theorem. Let $A$ be a commutative, semi-simple Banach algebra. Suppose that every norm-one character on $A$ is $L^\prime$-inducing and $A$ is Tauberian. Then $A$ is Banach-algebra isometrically isomorphic to $L_1(G)$ for some locally compact group $G$.

It is to be noted that there exist characterisations of $L_1(S)$-algebras, where $S$ is a locally compact semigroup, that are in a spirit similar to the above result. There is a recent (2016) result by Lau and Hung extending this further to general Fourier-Stieltjes algebras.

$\endgroup$
1
$\begingroup$

This is a partial answer for the case where $G$ is discrete (equivalently, $L^1(G)$ is unital). First note that if $G$ is a locally compact group then $G$ is discrete if and only if its dual group $\widehat{G}$ is compact. Furthermore, in this case $L^1(G)$ is isomorphic to $C(\widehat{G})$.

Edit: This last assertion is wrong. However, $\widehat{G}$ is still the spectrum of $L^1(G)$, and so the Gelfand transform $L^1(G) \longrightarrow C(\widehat{G})$ exhibits $C(\widehat{G})$ as the $\mathrm{C}^*$-algebra generated from $L^1(G)$ (i.e, every $*$-algebra map $L^1(G) \longrightarrow A$ to a $\mathrm{C}^*$-algebra $A$ factors through a unique $\mathrm{C}^*$-algebra map $C(\widehat{G}) \longrightarrow A$). Consequently, the observations below (which I still believe are correct), do not seem to be very relevant to the question.

Now commutative unital $\mathrm{C}^*$-algebras of the form $C(\widehat{G})$ for $\widehat{G}$ a compact (Hausdorff) abelian group are exactly those commutative unital $\mathrm{C}^*$-algebras which carry a compatible Hopf algebra structure (where the tensor product of $\mathrm{C}^*$-algebras is the completed tensor product). This just follows from the fact that the functor $K \mapsto C(K)$ is a contravariant equivalence, and that the coproduct in the category of commutative unital $\mathrm{C}^*$-algebras coincides with the completed tensor product.

$\endgroup$
8
  • 3
    $\begingroup$ Can $L^1(G)$ be isomorphic to $C(\hat G)$ when $G$ is infinite discrete? $\endgroup$ Sep 15, 2015 at 19:39
  • 1
    $\begingroup$ The third sentence is, for infinite G, incorrect. As noted many times on MO, $L^1(\Omega)$ cannot be isomorphic as a Banach space to $C(K)$ if $K$ is infinite. $\endgroup$
    – Yemon Choi
    Sep 15, 2015 at 19:44
  • $\begingroup$ Talk of Hopf algebras etc should instead characterize $L^1(G)$ for G locally compact (not nec abelian) as the predual of a commutative Kac algebra. In my view this is really not in the spirit of the original question, but perhaps @M.González can comment on this? $\endgroup$
    – Yemon Choi
    Sep 15, 2015 at 19:47
  • 1
    $\begingroup$ I was thinking about a practical criterium. For example, it obviously follows from the Gelfand-Naimark theorem that the space $Bo[0,1]$ of bounded Borel functions on $[0,1]$ is a $C(K)$ space. Moreover I am not familiar with Hopf algebras and Kac algebras. $\endgroup$ Sep 16, 2015 at 7:13
  • 3
    $\begingroup$ Unfortunately your guess is still incorrect, in general, since $L^1(G)$ always has trivial Jacobson radical (this works for all locally compact $G$; for LCA $G$ this comes out of Fourier analysis applied to the Gelfand transform, as shown in e.g. Rudin's book Fourier Analysis on Groups). So when $G$ is infinite, $L^1(G)\to C_0(\widehat{G})$ is never surjective. In the case $G={\bf Z}$ this boils down to the concrete statement: there exist continuous functions on the circle whose Fourier series are not absolutely summable $\endgroup$
    – Yemon Choi
    Sep 16, 2015 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.