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Let $G$ be a locally compact group, $C_0(G)$ the $C^*$-algebra of continuous functions on $G$ that vanish at infinity, $C_b(G)$ the $C^*$-algebra of bounded continuous functions on $G$. We know that $C_b(G)$ is the multiplier algebra of $C_0(G)$, and we denote the strict topology on $C_b(G) = \mathcal{M}\bigl( C_0(G) \bigr)$ by $\beta$.

Now for a strongly continuous unitary representation $(\pi, H)$ of $G$, functions of the form $\omega_{\pi,\eta,\xi} : g \in G \to (\pi(g)\eta \mid \xi) \in \mathbb{C}$ are in $C_b(G)$, and we call them matrix coefficients of the representation $\pi$. Since we can form direct sum and tensor product of two strongly continuous unitary representations, as well as the contragredient representation, we see that (the linear span of) matrix coefficients of all strongly continuous unitary representations of $G$ form a $*$-subalgebra $A_0(G)$ of $C_b(G)$.

Question. Is $A_0(G)$ strictly dense in $C_b(G)$, i.e. with respect to the $\beta$-topology?

Note that in the compact case, $C_b(G) = C_0(G) = C(G)$ and the $\beta$-topology is the same as the norm topology on the $C^*$-algebra $C(G)$ of continuous functions on $G$. In this case, The answer to the question is affirmative by Peter-Weyl. In the case where $G$ is discrete, then one can check easily that all finitely supported functions on $G$ are already matrix coefficients of the left (or right) regular representation, so the answer to the question is again affirmative. Based on these considerations, here are some sub-questions with some bias on their possible answers.

Q1. Does the question have an affirmative answer for unimodular $G$?

Q2. Can we construct some counter-example for non-unimodular $G$?

Q3. Does the question have an affirmative answer if $G$ is a real Lie group? What if the Lie group $G$ is nilpotent, or solvable, or semisimple/reductive?

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    $\begingroup$ Some remarks that may help with literature-searching. $\newcommand{\fsnorm}[1]{\Vert#1\Vert_{\rm B}}$ The algebra you have denoted by $A_0(G)$ is known as the Fourier--Stieltjes algebra of $G$, and is usually denoted by $B(G)$. $B(G)$ has been much studied: it turns out to be complete in a natural submultiplicative norm $\fsnorm{\cdot}$, so it is a commutative Banach algebra of functions on $G$. This norm is dominated by the supremum norm. $\endgroup$
    – Yemon Choi
    Dec 27, 2021 at 0:38
  • $\begingroup$ The book "Fourier and Fourier-Stieltjes Algebras on Locally Compact Groups" by E. Kaniuth & A.T.M. Lau bookstore.ams.org/surv-231 is a must-have reference for anyone who's interested in AHA. $\endgroup$
    – Onur Oktay
    Jan 8 at 6:58
  • $\begingroup$ Further to Onur's comment, a large proportion of Chapter 2 of the Kaniuth-Lau book is more or less a translation of parts of Eymard's original 1964 paper numdam.org/item/?id=BSMF_1964__92__181_0 (in French) which is freely available. $\endgroup$
    – Yemon Choi
    Jan 10 at 18:02

1 Answer 1

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First, some remarks that may help with literature-searching.$\newcommand{\fsnorm}[1]{{\Vert#1\Vert}_{\rm B}}$ $\newcommand{\supnorm}[1]{{\Vert#1\Vert}_\infty}$

The algebra you have denoted by $A_0(G)$ is known as the Fourier--Stieltjes algebra of $G$, and is usually denoted by $B(G)$, so I will do that from now on. $B(G)$ has been much studied: it turns out to be complete in a natural submultiplicative norm $\fsnorm{\cdot}$, so it is a commutative Banach algebra of functions on $G$; in fact it is also a dual Banach space for this norm, and multiplication is separately weak-star continuous, so it is an example of a dual Banach algebra.

This norm dominates the supremum norm, so $\fsnorm{\cdot}$-convergence implies $\supnorm{\cdot}$-convergence. If one takes the $\fsnorm{\cdot}$-closure inside $B(G)$ of $B(G)\cap C_c(G)$, the resulting algebra is an ideal in $B(G)$ called the Fourier algebra of $G$, usually denoted by $A(G)$. One can show that $A(G)$ is a $\supnorm{\cdot}$-dense subalgebra of $C_0(G)$ that is closed under conjugation and separates points of $G$.

(Aside: in general $B(G)\cap C_0(G)$ is larger than $A(G)$.)


Given some $f\in C_b(G)$, an $\varepsilon>0$, and a finite set $E\subset C_0(G)$, it suffices to find $h\in B(G)$ such that $\supnorm{fg-hg}\leq \varepsilon$ for all $g\in E$.

Since $C_0(G)$ is $\beta$-dense in $C_b(G)$ we can find $k \in C_0(G)$ such that $\supnorm{ fg-kg } \leq \varepsilon/2$ for all $g\in E$. Since $B(G)\cap C_c(G)$ is $\supnorm{\cdot}$-dense in $C_0(G)$ we can find $h\in B(G)\cap C_c(G)$ such that $\supnorm{k-h} \max_{g\in E} \supnorm{g} \leq\varepsilon/2$. Then $$ \supnorm{ fg -hg} \leq \supnorm{fg-kg} + \supnorm{k-h}\supnorm{g} \leq \varepsilon \quad\hbox{for all $g\in E$,} $$ as required.

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  • $\begingroup$ Thanks for all the useful information. I did some search, it seems that $B(G)$ is the image under Fourier-Stieltjes transform of the complex measure algebra $M(G)$ on $G$. For the moment, I don't see how to precisely relate this to unitary representations, i.e. why $B(G)$ is a copy of $A_0(G)$ (I know unitary representations come precisely from representations of the Fourier algebra $A(G)$, or equivalently, the convolution algebra $L^1(G)$). $\endgroup$ Dec 27, 2021 at 3:47
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    $\begingroup$ For abelian groups, $B(G)$ is the image of $M(\hat G)$ under the Fourier transform (notice the dual group here!) Checking all the definitions does indeed show that $M(\hat G)$ is given by unitary representations of $G$. Some reading for general $G$ are Eymard's original paper, or the book by Kaniuth and Lau. $\endgroup$ Dec 27, 2021 at 10:39
  • $\begingroup$ @RickSternbach As Matthew said, I'm not sure what sources you found when you searched, but they seem to have misled you into some incorrect takes. I learned most of this from Eymard's 1964 paper, or from talking to people who learned from that paper, so if you can read (mathematical) French I would recommend it as a source for the relevant definitions. $\endgroup$
    – Yemon Choi
    Dec 27, 2021 at 13:43

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