---
abstract: 'A theorem of Davis, Figiel, Johnson and Pełczyński tells us that weakly-compact operators between Banach spaces factor through reflexive Banach spaces. The machinery underlying this result is that of the real interpolation method, which has been adapted to the category of operator spaces by Xu, showing the this factorisation result also holds for completely bounded weakly-compact maps. In this note, we show that Xu's ideas can be adapted to give an intrinsic characterisation of when a completely contractive Banach algebra arises as a closed subalgebra of the algebra of completely bounded operators on a reflexive operator space. This result was shown by Young for Banach algebras, and our characterisation is a direct analogue of Young's, involving weakly almost periodic functionals.'
author:
- Matthew Daws
title: '<span style="font-variant:small-caps;">Weakly almost periodic functionals, representations, and operator spaces</span>'
---

<span>*Keywords:* operator space, weakly-compact, dual Banach algebra, completely contractive Banach algebra, weakly almost periodic</span>

<span>2000 *Mathematical Subject Classification:* 46B70, 46H05, 46H15, 46L07, 47L25 (primary), 46A25, 46A32, 43A60.</span>

Introduction
============

In @DFJP, Davis, Figiel, Johnson and Pełczyński showed that weakly-compact operators and operators which factor through a reflexive Banach space are the same class. To do this, they used the real interpolation space method, although this was not made explicit. In @Young this result was used to give a link between Banach algebras which arise as closed subalgebra of the algebra of operators on a reflexive Banach space, and Banach algebras which admit sufficiently many so called weakly almost periodic functionals. In @Kai, Kaiser noticed that this was really a result about interpolation of Banach modules. Recently, in @Daws, we argued that such results can really be thought of as results about representing dual Banach algebras.

The use of operator spaces has attracted a lot of attention in studying non-self-adjoint operator algebras (see the monograph @BLM for example). Less well studied are the much wider class of completely contractive (or quantum) Banach algebras. The most common example is the Fourier algebra $A(G)$ for a locally compact group $G$. As $A(G)$ is the predual of the group von Neumann algebra $VN(G)$, it carries a natural operator space structure. It seems that when $A(G)$ is considered as an operator space, properties of $A(G)$ better reflect properties of $G$ (see Ruan's original paper @Ruan, or the survey paper @Runde2). This current paper was motivated by concrete questions to do with $A(G)$, a matter we consider further at the end of this paper.

I have learnt that F. Uygul has independently found the main result of this paper (Theorem \[Main\_Thm\] below). See the paper @Uygul. Our approaches are rather similar; this paper proves a factorisation result for module maps, while Uygul takes an approach closer to the presentation of @Daws. However, the module result follows easily from Uygul's result. We both essentially adapt ideas of Xu contained in @Xu.

As we suspect that this paper will mostly be read by people familiar with Banach algebras, we shall follow the notation of @Dales, in particular writing $E'$ for the dual of a Banach (or operator) space $E$, and $T'$ for the adjoint of a linear map $T$. We write $\kappa_E:E
\rightarrow E''$ for the canonical map from a Banach (or operator) space to its dual. With these exceptions, we follow @ER for notation to do with operator spaces (see also @Pisier for basic details about operator spaces). A further exception is that when $T$ is an operator between operator spaces $E$ and $F$, we always use brackets, and write $(T)_n$ for the amplification map $\mathbb M_n(E) \rightarrow \mathbb M_n(F)$. We shall also frequently abuse notation, and not notationally distinguish between the norm on $\mathbb M_n(E)$ and that on $\mathbb M_m(E)$.

Completely contractive Banach algebras
======================================

Let ${\mathcal{A}}$ be a Banach algebra which is also an operator space. Let $\Delta:{\mathcal{A}}\otimes{\mathcal{A}}\rightarrow{\mathcal{A}}; a\otimes b\mapsto
ab$ be the multiplication map. We say that ${\mathcal{A}}$ is a *completely contractive (CC) Banach algebra* when $\Delta$ extends to a completely contractive map ${\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}}\rightarrow {\mathcal{A}}$. Here, of course, ${{\widehat{\otimes}}}$ denotes the operator space projective tensor product (see @ER [Chapter 7]). By using the identification ${\mathcal{CB}}({\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}},{\mathcal{A}}) = {\mathcal{CB}}({\mathcal{A}},{\mathcal{CB}}({\mathcal{A}}))$, we see that ${\mathcal{A}}$ is a CC Banach algebra if and only if the left-regular representation of ${\mathcal{A}}$ on itself maps into ${\mathcal{CB}}({\mathcal{A}})$ and is a completely contractive homomorphism.

Similarly, let ${\mathcal{A}}$ be a CC Banach algebra, and let $E$ be a left ${\mathcal{A}}$-module. Following @Dales, all our modules shall be contractive (which can be arranged by a suitable renorming). We say that $E$ is a *completely contractive (CC) left ${\mathcal{A}}$-module* when $E$ is an operator space and the module map ${\mathcal{A}}\otimes E\rightarrow E; a\otimes x\mapsto a\cdot x$ extends to a complete contraction ${\mathcal{A}}{{\widehat{\otimes}}}E\rightarrow E$. As above, this is equivalent to the induced homomorphism $\theta:{\mathcal{A}}\rightarrow
{\mathcal{B}}(E)$ actually mapping into ${\mathcal{CB}}(E)$ and being a complete contraction. Obvious definitions apply to right ${\mathcal{A}}$-modules and ${\mathcal{A}}$-bimodules.

We turn ${\mathcal{A}}'$ into an ${\mathcal{A}}$-bimodule in the obvious way $${{\langle {a\cdot \mu} , {b} \rangle}} = {{\langle {\mu} , {ba} \rangle}} = {{\langle {\mu\cdot b} , {a} \rangle}}
\qquad (a,b\in{\mathcal{A}}, \mu\in{\mathcal{A}}').$$ As $({\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}})' = {\mathcal{CB}}({\mathcal{A}},{\mathcal{A}}')$, it is easy to see that when ${\mathcal{A}}$ is a CC Banach algebra, we have that ${\mathcal{A}}'$ is a CC ${\mathcal{A}}$-bimodule, and also for ${\mathcal{A}}''$ and so forth.

The following is the operator space version of a notion first formally defined by Runde in @Runde1 (although it had been studied before).

Let $E$ be an operator space, and suppose that ${\mathcal{A}} = E'$ is a CC Banach algebra. When the product on ${\mathcal{A}}$ is separately weak$^*$-continuous, we say that ${\mathcal{A}}$ is a *completely contractive (CC) dual Banach algebra*.

It is simple to check that the product is separately weak$^*$-continuous if and only if $\kappa_E(E)$ becomes a sub-${\mathcal{A}}$-bimodule of ${\mathcal{A}}'=E''$. Notice that the operator space structure plays little role here, essentially because duality works so well, @ER [Section 3.2]. The equivalent notion, for algebras of operators on a Hilbert space, has been widely studied (see @LM1, for example, where an operator version of our result, Theorem \[Main\_Thm\] below, is proved).

We showed in @Daws that the class of dual Banach algebras coincides with the class of weak$^*$-closed subalgebras of ${\mathcal{B}}(E)$, where $E$ is a reflexive Banach space. This result follows quite easily from the work of Young and Kasier, or at least their methods. In this paper we shall prove an analogous result for CC dual Banach algebra. Notice that when $E$ is a reflexive operator space, we have that $(E {{\widehat{\otimes}}}E')' =
{\mathcal{CB}}(E)$, so that ${\mathcal{CB}}(E)$ is a dual space. It is simple to show that ${\mathcal{CB}}(E)$ is a CC Banach algebra, and that $E{{\widehat{\otimes}}}E'$ is a submodule of ${\mathcal{CB}}(E)'$. Thus ${\mathcal{CB}}(E)$ is a CC dual Banach algebra.

Interpolation spaces
====================

It seems that we cannot escape explaining a little about interpolation spaces, given the important role played by the complex interpolation method in the theory of operator spaces.

Let $E$ and $F$ be Banach spaces which are embedded continuously into some Hausdorff topological vector space $X$ (in applications, we shall typically have an injection of $E$ into $F$ allowing us to take $X=F$). We define norms on the subspaces $E\cap F$ and $E+F$ of $X$ by $$\|x\|_{E\cap F} = \max\big( \|x\|_E , \|y\|_F \big) , \quad
\|w\|_{E+F} = \inf\big\{ \|x\|_E + \|y\|_F : w=x+y \big\}.$$ Obviously we can at this point replace $X$ by $E+F$ if we wish. We say that $(E,F)$ is a compatible couple. Loosely, an interpolation space is a Banach space intermediate to $E+F$ and $E\cap F$, and such that certain mapping properties hold. See @BB or @BL for further details.

These ideas can be adapted, with minor tweaking, to the setting of operator spaces, see @Xu or @Pisier [Section 2.7]. The complex interpolation method gives a Banach space $(E,F)_\theta$ where $\theta$ is a parameter between $0$ and $1$. When $E$ and $F$ are operator spaces, $(E,F)_\theta$ becomes an operator space by setting $$\mathbb M_n\big( (E,F)_\theta \big) = \big( \mathbb M_n(E),
\mathbb M_n(F) \big)_\theta \qquad (n\geq1).$$ Let $(E_1,F_1)$ be another compatible couple, and let $T:E+F
\rightarrow E_1+F_1$ be a linear map such that $T$ maps $E$ into $E_1$ in a (completely) bounded fashion, and the same for $F$ and $F_1$. Then $T:(E,F)_\theta\rightarrow(E_1,F_1)_\theta$ and $\|T\|
\leq \|T\|^{1-\theta}_{E\rightarrow E_1} \|T\|^\theta_{F\rightarrow F_1}$. Let $\mu$ be a measure space and consider the Banach spaces $L_p(\mu)$. Then $(L_1(\mu),L_\infty(\mu))$ is a compatible couple, and, isometrically, $$L_p(\mu) = \big(L_1(\mu),L_\infty(\mu)\big)_\theta
\qquad (1/p=1-\theta).$$ We give $L_\infty(\mu)$ the operator space structure it has as a C$^*$-algebra, and give $L_1(\mu)$ the operator space structure it gets from embedding $L_1(\mu)$ into the dual of $L_\infty(\mu)$. Then the above identity allows us to define an operator space structure on $L_p(\mu)$. In particular, $L_2(\mu)$ gives an example of Pisier's self-dual operator Hilbert space, see @Pisier [Section 7].

Vector-valued versions of the above shall be important for us. Let $(E_i)_{i\in I}$ be a family of operator spaces. We let $\ell_\infty(E_i)$ be the usual direct sum of operator spaces, so that $$\ell_\infty(E_i) = \big\{ (x_i)_{i\in I} : x_i\in E_i \ (i\in I),
\|(x_i)\| := \sup_i \|x_i\| \big\},$$ as a Banach space. We define $\mathbb M_n(\ell_\infty(E_i)) =
\ell_\infty( \mathbb M_n(E_i) )$ for $n\geq 1$. Similarly, $\ell_1(E_i)$ can be given an operator space structure by embedding it in $\ell_\infty(E_i')'$. Alternatively, as noted in @Pisier [Section 2.6], $\ell_1(E_i)$ is characterised by the universal property that whenever $E$ is an operator space and $T_i:E_i\rightarrow E$ is a complete contraction, for each $i\in I$, then the map $T:\ell_1(E_i)\rightarrow E$, given by $T(x_i) = \sum_{i\in I} T_i(x_i)$, is a complete contraction. When $I=\{1,2\}$, we write $E_1\oplus_1 E_2$ for $\ell_1(E_i)$, and so forth. Finally, we define $$\ell_p(E_i) = \big( \ell_1(E_i), \ell_\infty(E_i) \big)_\theta
\qquad (1/p=1-\theta).$$ We shall mainly use $\ell_2(E_i)$. We note that when $E_i=\mathbb C$ for each $i$, we have that $\ell_2(E_i) = OH(I)$, Pisier's operator Hilbert space, and that there are many characterisations of the operator space structure on $OH(I)$. We are not aware of characterisations of $\ell_2(E_i)$ which do not use complex interpolation, however.

The real interpolation method is more complicated to explain, and significantly harder to adapt to the operator space setting. Fortunately, Xu has done the hard work for us in @Xu. We shall sketch a simple case of the constructions Xu considers, following the approach of Palmer in @PalBook [Section 1.7.8] and also as used by us in @Daws.

Factoring module maps
=====================

Firstly we shall consider the Banach space case. Let ${\mathcal{A}}$ be a Banach algebra, let $E$ and $F$ be left ${\mathcal{A}}$-modules, and let $T:E\rightarrow F$ be an ${\mathcal{A}}$-module homomorphism. That is, $T(a\cdot x) = a\cdot T(x)$ for $a\in{\mathcal{A}}$ and $x\in E$. For $n\in\mathbb N$, define a new norm $\|\cdot\|_n$ on $F$ by $$\|x\|_n = \inf\big\{ 2^{-n/2} \|T\|\|y\| + \|x-T(y)\| : y\in E \big\}
\qquad (x\in F).$$ We may check that $$\|x\|_n \leq \|x\| \leq 2^{n/2}\|x\|_n \qquad (x\in F),$$ so that $\|\cdot\|_n$ is an equivalent norm on $F$. We let $$G = \Big\{ x\in F : \|x\|_G := \Big(\sum_{n=1}^\infty \|x\|_n^2\Big)^{1/2}
< \infty \Big\},$$ so that for $y\in E$, as $\|T(y)\|_n \leq 2^{-n/2} \|T\| \|y\|$, we see that $\|T(y)\|_G \leq \|T\| \|y\|$, from which it follows that $T(E)\subseteq G$. Let $\iota:G\rightarrow F$ be the inclusion map, which is norm-decreasing, and let $R$ be the map $T$, treated as map from $E$ to $G$, so that $\|R\|\leq\|T\|$. Hence $T$ factors through the normed space $G$, as $\iota \circ R = T$. Notice that $R$ has dense range, and that $\iota$ is injective, so that $R$ has the same kernel as $T$.

Let $E_0 = E / \ker T$, so that $T$ becomes an injection $E_0 \rightarrow
F$, and so we can regard $(E_0,F)$ as a compatible couple. Then $G$ is a member of the equivalence class $(E_0,F)_{1/2,2}$, this being a real interpolation space. From standard results (see, for example, @BB [Section 2.3, Proposition 1]) it follows that $G$ is reflexive if and only if $T$ is weakly-compact. In fact, this is not too hard to prove directly, which we leave as an exercise for the reader.

Notice that for $x\in F$ and $a\in{\mathcal{A}}$, we see that $\|a\cdot x\|_n
\leq \|a\| \|x\|_n$ for $n\geq1$. Hence $G$ becomes a left ${\mathcal{A}}$-module, and $\iota$ and $R$ become ${\mathcal{A}}$-module homomorphisms.

We wish to carry out a similar construction for operator spaces. A first step is to consider a different way of expressing $G$. For a Banach space $E$ and $t>0$, let $tE$ be the same space with the norm multiplied by $t$ (and similarly for an operator space $E$, where we set $\mathbb M_n(tE) = t\mathbb M_n(E)$). Set $$F_n = 2^{-n/2}\|T\| E \oplus_1 F \qquad (n\geq 1),$$ let $Y = \ell_2(F_n)$, and let $$Y_0 = \{ (x_n,y_n)\in Y : T(x_n)+y_n=0 \ (n\geq 1) \}\subseteq Y.$$ It is easy to see that $Y_0$ is a closed subspace of $Y$, so we may form the quotient space $Y/Y_0$. Finally, define $$Y_1 = \{ (x_n,y_n)+Y_0 \in Y/Y_0 : (T(x_n)+y_n) \text{ constant} \},$$ so that $Y_1$ is a closed subspace of $Y/Y_0$.

Define a map $\alpha:Y_1 \rightarrow G$ by $$\alpha( (x_n,y_n)+Y_0 ) = T(x_1)+y_1 \qquad ((x_n,y_n)+Y_0\in Y/Y_0).$$ It is easy to see that $\alpha$ is actually well-defined on the whole of $Y/Y_0$, does map into $G$, and is an injection when restricted to $Y_1$. Let $y\in G$, so for some sequence $(x_n)$ in $E$, we have that $$\sum_{n=1}^\infty \Big( 2^{-n/2}\|T\|\|x_n\| + \|y-T(x_n)\| \Big)^2
<\infty.$$ Let $y_n = y-T(x_n)$ for each $n$, so that $(x_n,y_n)\in Y$, and by definition, $\alpha( (x_n,y_n)+Y_0 )=y$. Hence $\alpha$ is a bijection. A similar calculation shows that $\alpha$ is actually an isometry.

Now let ${\mathcal{A}}$ be a CC Banach algebra, let $E$ and $F$ be CC left ${\mathcal{A}}$-modules, and let $T:E\rightarrow F$ be a completely bounded ${\mathcal{A}}$-module homomorphism. We note that, using the complex interpolation method described above, we can give each $F_n$ and $Y$ natural operator space structures. Hence $Y_0, Y/Y_0$ and $Y_1$ all gain operator space structures, and we can hence use $\alpha$ to induce an operator space structure on $G$. It follows from Xu's work that $\iota$ and $R$ are completely bounded. Furthermore, as $G$ is a left ${\mathcal{A}}$-module, we have a homomorphism $\theta_G:{\mathcal{A}}\rightarrow{\mathcal{B}}(G)$. Xu's work shows that $\theta_G$ actually maps into ${\mathcal{CB}}(G)$ and that $\theta_G$ is norm-decreasing. However, we wish to show that $\theta_G$ is a complete contraction, in order to show that $G$ is a CC left ${\mathcal{A}}$-module. This follows from Xu's methods, but not seemingly directly from the results of @Xu.

With notation as above, there exists an absolute constant $K>0$ such that $\|R\|_{cb}\leq K\|T\|_{cb}$ and $\|\iota\|_{cb}\leq K$.

We shall only sketch this, as it follow from @Xu [Theorem 2.2], and the remark thereafter. By the universal property of $\oplus_1$, the map $$\iota_n: F_n = 2^{-n/2}\|T\|_{cb} E \oplus_1 F \rightarrow F;\
(x,y) \mapsto 2^{-n/2}(T(x)+y) 
\qquad (n\geq1, x\in E, y\in F)$$ is a complete contraction. Now consider the maps, for $1\leq t\leq\infty$, $$\beta_t: \ell_t(F_n) \rightarrow F; \
(w_n) \mapsto \sum_{n=1}^\infty 2^{-n/2}\iota_n(w_n)
= \sum_{n=1}^\infty 2^{-n}(T(x_n)+y_n)
\qquad ( w_n=(x_n,y_n)\in F_n ).$$ By the universal property, as each $\iota_n$ is a complete contraction, we have that $\|\beta_1\|_{cb}
\leq 2^{-1/2}$. Let $(w_n)\in\mathbb M_k(\ell_\infty(F_n))$, so that $$\|\beta_\infty(w_n)\| \leq \sum_{n=1}^\infty
2^{-n/2}\|\iota_n(w_n)\| \leq \sum_{n=1}^\infty
2^{-n/2}\|w_n\|_{\mathbb M_k(F_n)} \leq 
\frac{\sqrt2}{2-\sqrt2} \|(w_n)\|,$$ so that $\|\beta_\infty\|_{cb}\leq2^{1/2}(2-2^{1/2})^{-1}$. By complex interpolation, $$\|\beta_2\|_{cb} \leq \big(2^{-1/2}\big)^{1/2}
\big( 2^{1/2}(2-2^{1/2})^{-1} \big)^{1/2}
= \sqrt{ \frac{1}{2-\sqrt2} } = K.$$ Notice that $\beta_2$ vanishes on $Y_0$, and so $\beta_2$ drops to a well-defined operator from $Y/Y_0$ to $F$, and hence by restriction to a map $Y_1\rightarrow F$. Fairly obviously, $\beta_2\alpha^{-1}
= \iota$, showing that $\iota$ is completely bounded. The argument for $R$ follows similarly.

Notice that $$\|\beta_2(w_n)\| \leq \sum_{n=1}^\infty 2^{-n/2} \|\iota_n(w_n)\|
\leq \Big(\sum_{n=1}^\infty \|\iota_n(w_n)\|^2 \Big)^{1/2},$$ so we see that $\|\beta_2\|\leq 1$, so that the complex interpolation estimate of $\|\beta_2\|$ is not optimal. It seems possible that if we had a more concrete description of the operator space structure on $\ell_2(F_n)$, then we could show that $\|R\|_{cb} \|\iota\|_{cb}
\leq \|T\|_{cb}$. However, the following is a suitable work around.

Let $E, F$ and $G$ be operator spaces, and let $T\in{\mathcal{CB}}(E,F),
R\in{\mathcal{CB}}(E,G)$ and $\iota\in{\mathcal{CB}}(G,F)$ be such that $T=\iota R$. We may give $G$ an equivalent operator space structure for which $\iota$ is a complete contraction, and $\|R\|_{cb}=\|T\|_{cb}$.

By replacing $G$ by $tG$ for some $t>0$, we may suppose that $\iota$ is already a complete contraction. Define a map $\phi:\|T\|_{cb}E \oplus_1 G
\rightarrow G$ by $\phi(x,y) = R(x)+y$ for $x\in E$ and $y\in G$. Clearly $\phi$ is surjective, so we may identify $G$ with $\|T\|_{cb}E \oplus_1 G / \ker\phi$. Use this to induce a new operator space structure on $G$, say with norm $\|\cdot\|_0$, giving $G_0$. Let $w\in\mathbb M_n(G)$, so that clearly $\|w\|_0 \leq \|w\|$. Conversely, let $x\in\mathbb M_n(E)$ and $y\in\mathbb M_n(G)$ be such that $w=(R)_n(x)+y$. The map $E\rightarrow G; x \mapsto
\|T\|_{cb} \|R\|_{cb}^{-1} R(x)$ is a complete contraction, so by the universal property of $\oplus_1$, $$\|w\| = \big\|(R)_n(x) + y\big\|
\leq \big\| \big( \|T\|_{cb}^{-1}\|R\|_{cb}x , y \big) \big\|.$$ As $\|T\|_{cb} \leq \|R\|_{cb}$, the map $\|T\|_{cb}E\oplus_1 G
\rightarrow \|T\|_{cb}E\oplus_1 G; (x,y) \mapsto (x,\|T\|_{cb}\|R\|_{cb}^{-1})$ is a complete contraction, so that $$\|w\| \leq \|T\|_{cb}^{-1} \|R\|_{cb} \|(x,y)\|,$$ from which we conclude that $\|w\|\leq\|T\|_{cb}^{-1} \|R\|_{cb}\|w\|_0$, showing that $\|\cdot\|_0$ is an equivalent operator space structure on $G$. Clearly we have that $\|R:E\rightarrow G_0\|_{cb} \leq \|T\|_{cb}$.

As $\phi$ is defined to be a complete quotient map, $\phi':G_0'\rightarrow
\|T\|_{cb}^{-1}E' \oplus_\infty G'$ is a complete isometry. We may check that $$\phi'(\mu) = \big( R'(\mu), \mu \big)
\qquad (\mu\in G_0').$$ Let $\lambda\in \mathbb M_n(F')$, so that $$\begin{aligned}
\|(\iota')_n(\lambda)\|_{G_0'} &= \big\| ( (R')_n(\iota')_n(\lambda) ,
   (\iota')_n(\lambda) ) \big\|_{\|T\|_{cb}^{-1}E' \oplus_\infty G'} \\
&= \max\big( \|T\|_{cb}^{-1} \|(T')_n(\lambda)\| , \|(\iota')_n(\lambda)\| \big)
\leq \|\lambda\|.\end{aligned}$$ Hence $\| \iota:G_0\rightarrow F \|_{cb}\leq 1$, as required.

Recall the homomorphism $\theta_G:{\mathcal{A}}\rightarrow{\mathcal{B}}(G)$, which actually maps into ${\mathcal{CB}}(G)$ by Xu's work.

\[mod\_okay\] With notation as above, $\theta_G$ is a complete contraction.

We first show that our claim holds for the original definition of $G$, and then check that the above renorming procedure does not chance our conclusions. Fix $n\geq 1$ and $a=(a_{kl})\in\mathbb M_n({\mathcal{A}})$ with $\|a\|\leq 1$. Then $A:=(\theta_G)_n(a)\in\mathbb M_n({\mathcal{CB}}(G)) =
{\mathcal{CB}}(G,\mathbb M_n(G))$. Let $m\geq 1$ and let $w\in\mathbb M_m(G)$, so that $(A)_m(w)\in\mathbb M_m(\mathbb M_n(G))
= \mathbb M_{m\times n}(G)$. We wish to show that $\|(A)_m(w)\| \leq \|w\|$, which would both demonstrate Xu's result that $\theta_G$ maps into ${\mathcal{CB}}(G)$, and would show that $\theta_G$ is a complete contraction.

Recall the definitions of $(F_k)_{k\geq 1}, Y, Y_0, Y_1$ and $\alpha$ from above. Clearly, for $k\geq 1$, $$(\theta_E)_n(a)\in{\mathcal{CB}}\big( 2^{-k/2}\|T\|_{cb}E,
\mathbb M_n(2^{-k/2}\|T\|_{cb}E) \big)$$ with norm at most $\|a\|\leq1$, as $\theta_E$ is a complete contraction. By the universal property of $\oplus_1$, it is clear that $$A_k:= (\theta_E)_n(a) \oplus (\theta_F)_n(a) : F_k \rightarrow
\mathbb M_n(F_k)$$ is a complete contraction, for each $k\geq 1$. Clearly the diagonal map $(\oplus_k A_k):\ell_\infty(F_k) \rightarrow \mathbb M_n(\ell_\infty(F_k))
= \ell_\infty(\mathbb M_n(F_k))$ is a complete contraction. Again, by the universal property of $\oplus_1$, $(\oplus_k A_k):\ell_1(F_k)
\rightarrow \mathbb M_n(\ell_1(F_k))$ is also complete contraction. Hence, by complex interpolation, $$(\oplus_k A_k): Y=\ell_2(F_k) \rightarrow \mathbb M_n(Y)$$ is a complete contraction. It is clear that as $T$ is an ${\mathcal{A}}$-module homomorphism, $(\oplus_k A_k)$ leaves $Y_0$ invariant, and hence $(\oplus_k A_k)$ drops to a complete contraction on $Y/Y_0$. Similarly, $(\oplus_k A_k)$ restricts to a complete contraction on $Y_1$.

It is a simple check that $(\alpha)_{m\times n} (\oplus_k A_k)
(\alpha^{-1})_m = (A)_m$, and so we conclude that $(A)_m$ is a contraction, as required.

Now consider the renorming. Again let $a\in\mathbb M_n({\mathcal{A}})$ and $x\in\mathbb M_m(G)$, so that $(A)_m(x)\in\mathbb M_{m\times n}(G)$. Suppose that $\|x\|_0<1$, so that $x=\phi(y,z) = (R)_m(y)+z$ for some $y\in\mathbb M_m(E)$ and $z\in\mathbb M_m(G)$ with $\|(y,z)\|_{\|T\|_{cb}E\oplus_1 G} < 1$. Let $$w = \big( (A)_m(y) , (A)_m(z) \big) \in
\mathbb M_m( \|T\|_{cb}E \oplus_1 G ),$$ so that $$\|w\|_{\|T\|_{cb}E\oplus_1 G} \leq
\|A\|_{cb}\|y,z\|_{\|T\|_{cb}E\oplus_1 G} < \|a\|.$$ We then observe that $$(A)_m(x) = ((\theta_G)_n(a))_m(R)_m(y) + A(z)
= (R)_{m\times n} ((\theta_E)_n(a))_m(y) + A(z)
= (\phi)_n(w),$$ so that $\|(A)_m(x)\| \leq \|\phi\|_{cb} \|w\| < \|a\|$, showing that $\|A:G_0\rightarrow G_0\|_{cb}\leq\|a\|$ as required.

We have hence shown the following factorisation result.

\[mod\_fac\_thm\] Let ${\mathcal{A}}$ be a CC Banach algebra, let $E$ and $F$ be CC left ${\mathcal{A}}$-modules, and let $T:E\rightarrow F$ be a completely bounded ${\mathcal{A}}$-module homomorphism. The following are equivalent:

1.  $T$ is weakly-compact;

2.  there exists a reflexive CC left ${\mathcal{A}}$-module $G$, an injective complete contraction $\iota:G\rightarrow F$, and a completely bounded map $R:E\rightarrow G$ with $\|R\|_{cb} = \|T\|_{cb}$, $T = \iota R$, and such that $R$ and $\iota$ are ${\mathcal{A}}$-module homomorphisms.

Representing CC dual Banach algebras
====================================

Theorem \[mod\_fac\_thm\] is central to proving Young's representation theorem (a fact explicitly noticed by Kaiser in @Kai). The situation for CC Banach algebras is more complicated, because we need to take account of the matrix structures $\mathbb M_n({\mathcal{A}})$. Recall that while $\mathbb M_n({\mathcal{A}})$ is obviously an algebra, the product need not be uniformly bounded in $n$. Indeed, this is equivalent to ${\mathcal{A}}$ being completely isomorphic to a subalgebra of ${\mathcal{B}}(H)$ for a Hilbert space $H$ (see @ER [Chapter 17] and @LM1 for example).

\[Main\_Thm\] Let ${\mathcal{A}}$ be a CC dual Banach algebra with predual ${\mathcal{A}}_*$. Then there exists a weak$^*$-weak$^*$-continuous completely isometric homomorphism $\theta:{\mathcal{A}}\rightarrow{\mathcal{CB}}(E)$ for some reflexive operator space $E$.

We may suppose that ${\mathcal{A}}$ is unital. If not, we replace ${\mathcal{A}}$ by ${\mathcal{A}}\oplus_1 \mathbb C$, the unitisation of ${\mathcal{A}}$, and we may check that properties of $\oplus_1$ imply that ${\mathcal{A}}\oplus_1 \mathbb C$ is a CC dual Banach algebra, with predual ${\mathcal{A}}_* \oplus_\infty \mathbb C$. Denote the unit of ${\mathcal{A}}$ by $e_{{\mathcal{A}}}$.

Let $n\geq 1$, we have that $\mathbb M_n({\mathcal{A}}_*)$ is a CC left ${\mathcal{A}}$-module. Let $\mu=(\mu_{ij})\in\mathbb M_n({\mathcal{A}}_*)$ with $\|\mu\|=1$, and let $T:{\mathcal{A}} \rightarrow \mathbb M_n({\mathcal{A}}_*)$ be the map $T(a) = a\cdot\mu$, so that $T$ is completely contractive. Recall that Gantmacher's Theorem tells us that $T$ is weakly-compact if and only if $T''({\mathcal{A}}'') \subseteq \mathbb M_n({\mathcal{A}}_*)$. Let $b=(b_{ij}) \in \mathbb
M_n({\mathcal{A}}_*)' = \mathbb T_n({\mathcal{A}})$ (see @ER [Proposition 7.1.6]). Then $${{\langle {T'(b)} , {a} \rangle}} = \sum_{i,j=1}^n {{\langle {b_{ij}} , {a\cdot\mu_{ij}} \rangle}}
= \sum_{i,j=1}^n {{\langle {\mu_{ij}\cdot b_{ij}} , {a} \rangle}},$$ so that $T'(b) = \sum_{i,j} \mu_{ij}\cdot b_{ij} \in{\mathcal{A}}_* \subseteq{\mathcal{A}}'$. For $\Phi\in{\mathcal{A}}''$, let $c\in{\mathcal{A}}$ be such that ${{\langle {c} , {\mu} \rangle}} =
{{\langle {\Phi} , {\mu} \rangle}}$ for $\mu\in{\mathcal{A}}_*$. Then we have that $${{\langle {T''(\Phi)} , {b} \rangle}} = \sum_{i,j=1}^n {{\langle {\Phi} , {\mu_{ij}\cdot b_{ij}} \rangle}}
= \sum_{i,j=1}^n {{\langle {c} , {\mu_{ij}\cdot b_{ij}} \rangle}}
= \sum_{i,j=1}^n {{\langle {b_{ij}} , {c\cdot\mu_{ij}} \rangle}}.$$ Thus $T''(\Phi) = T(c)\in\mathbb M_n({\mathcal{A}}_*)$, and we conclude that $T$ is indeed weakly-compact.

Applying Theorem \[mod\_fac\_thm\], we find a reflexive operator space $G_\mu$ and complete contractions $\iota_\mu:G_\mu\rightarrow
\mathbb M_n({\mathcal{A}}_*)$ and $R_\mu:{\mathcal{A}}\rightarrow
G_\mu$, such that $\iota_\mu R_\mu = T$, and with $\iota_\mu$ and $R_\mu$ being ${\mathcal{A}}$-module homomorphisms. Let $\theta_\mu:{\mathcal{A}}\rightarrow
{\mathcal{CB}}(G_\mu)$ be induced by the module action. Let $x_\mu=R_\mu(e_{{\mathcal{A}}})
\in G_\mu$, so that $$T(a) = \iota_\mu R_\mu(a\cdot e_{{\mathcal{A}}}) = 
\iota_mu \theta_\mu(a) R_\mu(e_{{\mathcal{A}}}) =
\iota_\mu \theta_\mu(a)(x_\mu)
\qquad (a\in{\mathcal{A}}).$$ Hence, for $a=(a_{kl})\in\mathbb M_n({\mathcal{A}})$, $$(T)_n(a) = \big( a_{kl}\cdot\mu \big)
= \big( \iota_\mu \theta_\mu(a_{kl})(x_\mu) \big)
= (\iota_\mu)_n \big( \theta_\mu(a_{kl})(x_\mu) \big)
= (\iota_\mu)_n (\theta_\mu)_n(a)(x_\mu),$$ noting that $(\theta_\mu)_n(a) \in {\mathcal{CB}}(G,\mathbb M_n(G))$. Then note that $$\begin{aligned}
 \langle{{\langle {a} , {\mu} \rangle}}\rangle
&= \big( {{\langle {a_{kl}} , {\mu_{ij}} \rangle}} \big)
= \big( {{\langle {e_{{\mathcal{A}}}} , {a_{kl}\cdot\mu_{ij}} \rangle}} \big)
= \big((T)_n(a)\big)(e_{{\mathcal{A}}})
= \big( (\iota_\mu)_n (\theta_\mu)_n(a)(x_\mu) \big) (e_{{\mathcal{A}}}),\end{aligned}$$ so that, as $\|x_\mu\| \leq 1$, $$\|(\theta_\mu)_n(a)\| \geq \| (\theta_\mu)_n(a)(x_\mu) \|
\geq \big\| (\iota_\mu)_n \big( (\theta_\mu)_n(a)(x_\mu) \big) (e_{{\mathcal{A}}}) \big\|
\geq \| \langle {{\langle {\mu} , {a} \rangle}} \rangle \|.$$

Let $\Lambda = \{ \mu\in\mathbb M_n({\mathcal{A}}_*) : n\geq 1, \|\mu\|=1 \}$, and let $E = \ell_2(\{G_\mu : \mu\in\Lambda\})$. Using the same argument as in the proof of Proposition \[mod\_okay\], we see that the homomorphism $\theta = \oplus_{\mu\in\Lambda} \theta_\mu:{\mathcal{A}}\rightarrow{\mathcal{B}}(E)$ maps into ${\mathcal{CB}}(E)$, and is a complete contraction. We can treat $x_\mu$ as a member of $E$ for each $\mu$, so we see that for $a\in\mathbb M_n({\mathcal{A}})$, $$\begin{aligned}
 \|a\| &=
\sup\{ \| \langle {{\langle {\mu} , {a} \rangle}} \rangle \| : \mu\in\mathbb M_n({\mathcal{A}}_*), \|mu\|=1 \} \\
&\leq \sup\{ \| (\theta_\mu)_n(a) \| : \mu\in\mathbb M_n({\mathcal{A}}_*), \|\mu\|=1 \}
\leq \|(\theta)_n(a)\| \leq \|a\|, \end{aligned}$$ and so we must have that $\theta$ is a complete isometry, as required.

Finally, we wish to show that $\theta$ is weak$^*$-weak$^*$-continuous. That is, we wish to show that there exists a complete contraction $\theta_*:E'{{\widehat{\otimes}}}E\rightarrow{\mathcal{A}}_*$ such that $\theta_*'=\theta$. As $\theta$ is a diagonal map on $E = \ell_2(G_\mu)$, it is sufficient to check that each $\theta_\mu$ is weak$^*$-continuous. Fix $\mu=(\mu_{ij})
\in\mathbb M_n({\mathcal{A}}_*)$. Recall that $T_\mu$ has dense range, and that $\iota_\mu$ is injective. We have that $\iota_\mu':\mathbb T_n({\mathcal{A}})
\rightarrow G_\mu'$ has dense range if and only if $\iota_\mu''$ is injective. Now, as $G_\mu$ is reflexive, we identify $G_\mu$ with $G_\mu''$, and we see that $\iota_\mu'' = \kappa_{\mathbb M_n({\mathcal{A}}_*)}\iota_\mu$, which is injective, showing that $\iota_\mu'$ does indeed have dense range. Let $x\in G_\mu$ and $\lambda\in G_\mu'$. By density, we may suppose that $x = R_\mu(b)$ and that $\lambda=\iota_\mu'\kappa_{{\mathcal{A}}}(c)$ for some $b\in{\mathcal{A}}$ and $c=(c_{ij})\in\mathbb T_n({\mathcal{A}})$. Then $$\begin{aligned}
{{\langle {\theta_\mu'\kappa_{E_\mu'{{\widehat{\otimes}}}E_\mu}(\lambda\otimes x)} , {a} \rangle}}
&= {{\langle {\lambda} , {\theta_\mu(a)(x)} \rangle}}
= {{\langle {\iota_\mu'\kappa_{{\mathcal{A}}}(c)} , {\theta_\mu(a)R_\mu(b)} \rangle}}
= {{\langle {\iota_\mu R_\mu(ab)} , {c} \rangle}} \\ &= {{\langle {ab\cdot\mu} , {c} \rangle}}
= \sum_{i,j=1}^n {{\langle {ab\cdot\mu_{ij}} , {c_{ij}} \rangle}}
= \sum_{i,j=1}^n {{\langle {a} , {b\cdot\mu_{ij}\cdot c_{ij}} \rangle}},\end{aligned}$$ so that $\theta_\mu'\kappa_{E_\mu'{{\widehat{\otimes}}}E_\mu}(\lambda\otimes x) \in {\mathcal{A}}_*$. Hence $\theta_\mu$ is weak$^*$-continuous, as required.

Representing general CC Banach algebras
---------------------------------------

Let ${\mathcal{A}}$ be a Banach algebra, let $E$ be a left ${\mathcal{A}}$-module, and define $${\operatorname{WAP}}(E,{\mathcal{A}}) = {\operatorname{WAP}}(E) = \{ x\in E : {\mathcal{A}}\rightarrow E;
a\mapsto a\cdot x \text{ is weakly-compact} \},$$ and similarly for CC Banach algebras ${\mathcal{A}}$ and CC left ${\mathcal{A}}$-modules. In particular, we say that ${\operatorname{WAP}}({\mathcal{A}}')\subseteq {\mathcal{A}}'$ is the space of *weakly almost periodic functionals*, the term coming from abstract harmonic analysis. We note that some authors write ${\operatorname{WAP}}({\mathcal{A}})$ for this space.

Let us quickly recall the Arens products (see @Daws [Section 2] for further details about the following ideas, although we note that most of this is folklore). Let ${\mathcal{A}}$ be a Banach algebra. We define bilinear actions ${\mathcal{A}}''\times
{\mathcal{A}}', {\mathcal{A}}'\times{\mathcal{A}}'' \rightarrow {\mathcal{A}}'$ by $${{\langle {\Phi\cdot\mu} , {a} \rangle}} = {{\langle {\Phi} , {\mu\cdot a} \rangle}}, \quad
{{\langle {\mu\cdot\Phi} , {a} \rangle}} = {{\langle {\Phi} , {a\cdot\mu} \rangle}}
\qquad (\Phi\in{\mathcal{A}}'', \mu\in{\mathcal{A}}', a\in{\mathcal{A}}).$$ Then we define bilinear maps ${\Box},{\Diamond}:{\mathcal{A}}''\times{\mathcal{A}}''
\rightarrow{\mathcal{A}}''$ by $${{\langle {\Phi{\Box}\Psi} , {\mu} \rangle}} = {{\langle {\Phi} , {\Psi\cdot\mu} \rangle}},\quad
{{\langle {\Phi{\Diamond}\Psi} , {\mu} \rangle}} = {{\langle {\Psi} , {\mu\cdot\Phi} \rangle}}
\qquad (\Phi,\Psi\in{\mathcal{A}}'',\mu\in{\mathcal{A}}').$$ It can be shown that ${\Box}$ and ${\Diamond}$ are Banach algebra products on ${\mathcal{A}}''$, called the *first* and *second Arens products*, and that $a\cdot\Phi = \kappa_{{\mathcal{A}}}(a){\Box}\Phi =
\kappa_{{\mathcal{A}}}(a){\Diamond}\Phi$ for $a\in{\mathcal{A}},\Phi\in{\mathcal{A}}''$, and similarly on the right. When ${\Box}={\Diamond}$, we say that ${\mathcal{A}}$ is *Arens regular*. This is equivalent to ${\operatorname{WAP}}({\mathcal{A}}')={\mathcal{A}}'$. Indeed, more is true, as a shall see shortly.

We now sketch how to apply these ideas to a CC Banach algebra ${\mathcal{A}}$. As before, this idea has been studied for operator algebras (see @Ruan1 for example) but we have not been able to find a good source for CC Banach algebras; no doubt the following is known to experts though. Let $\Theta:{\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}}\rightarrow{\mathcal{A}}$ be the completely contractive multiplication map, so that $\Theta':{\mathcal{A}}'
\rightarrow{\mathcal{CB}}({\mathcal{A}},{\mathcal{A}}')$ is also a complete contraction. With the convention that $${{\langle {T} , {a\otimes b} \rangle}} = {{\langle {T(a)} , {b} \rangle}}
\qquad (T\in({\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}})'={\mathcal{CB}}({\mathcal{A}},{\mathcal{A}}'),
a\otimes b\in{\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}}),$$ we may check that $\Theta'(\mu)(a) = \mu\cdot a$ for $a\in{\mathcal{A}}$ and $\mu\in{\mathcal{A}}'$. Define a map $\alpha:{\mathcal{A}}''{{\widehat{\otimes}}}{\mathcal{A}}''
\rightarrow({\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}})''={\mathcal{CB}}({\mathcal{A}},{\mathcal{A}}')'$ by $${{\langle {\alpha(\Phi\otimes\Psi)} , {T} \rangle}} = {{\langle {\Phi} , {T'(\Psi)} \rangle}} 
= {{\langle {T''(\Phi)} , {\Psi} \rangle}}
\qquad (\Phi,\Psi\in{\mathcal{A}}'', T\in{\mathcal{CB}}({\mathcal{A}},{\mathcal{A}}')),$$ so that as $({\mathcal{A}}''{{\widehat{\otimes}}}{\mathcal{A}}'')'={\mathcal{CB}}({\mathcal{A}}'',{\mathcal{A}}''')$ and the map ${\mathcal{CB}}({\mathcal{A}},{\mathcal{A}}')\rightarrow {\mathcal{CB}}({\mathcal{A}}'',{\mathcal{A}}''');
T\mapsto T''$ is a complete isometry, we see that $\alpha$ is a complete contraction. Finally, we see that $${{\langle {\Theta''\alpha(\Phi\otimes\Psi)} , {\mu} \rangle}}
= {{\langle {\Phi} , {\Theta'(\mu)'(\Psi)} \rangle}} = {{\langle {\Phi} , {\Psi\cdot\mu} \rangle}}
= {{\langle {\Phi{\Box}\Psi} , {\mu} \rangle}}
\qquad (\Phi,\Psi\in{\mathcal{A}}'', \mu\in{\mathcal{A}}'),$$ which shows that $({\mathcal{A}}'',{\Box})$ is a CC Banach algebra. If we choose the other convention for identifying ${\mathcal{CB}}({\mathcal{A}},
{\mathcal{A}}')$ with the dual of ${\mathcal{A}}{{\widehat{\otimes}}}{\mathcal{A}}$, we will find a proof that $({\mathcal{A}}'',{\Diamond})$ is a CC Banach algebra.

Let ${\mathcal{A}}$ be a Banach algebra, and let $X\subseteq{\mathcal{A}}'$ be a closed submodule. The following are equivalent:

1.  $X\subseteq{\operatorname{WAP}}({\mathcal{A}}')$;

2.  the first (or, equivalently, second) Arens product drops to a well-defined product on $X' = {\mathcal{A}}''/X^\perp$ turning $(X',X)$ into a dual Banach algebra.

See, for example, @Daws [Proposition 2.4] or compare with @Lau [Theorem 5.6].

Loosely, we can say that ${\mathcal{A}}''/{\operatorname{WAP}}({\mathcal{A}}')^\perp$ is the largest quotient of ${\mathcal{A}}''$ on which the Arens products agree. The above will clearly still hold for CC Banach algebras. Combining these observations with our factorisation theorem, we have the following.

\[wap\_thm\] Let ${\mathcal{A}}$ be a CC Banach algebra, and let $q:{\mathcal{A}}''
\rightarrow {\operatorname{WAP}}({\mathcal{A}}')'={\mathcal{A}}''/{\operatorname{WAP}}({\mathcal{A}}')^\perp$ be the quotient map. The following are equivalent:

1.  \[cond\_one\] the map $q\kappa_{{\mathcal{A}}}:{\mathcal{A}}\rightarrow{\operatorname{WAP}}({\mathcal{A}}')'$ is a complete isometry;

2.  \[cond\_two\] there is a homomorphism $\theta:{\mathcal{A}}\rightarrow {\mathcal{CB}}(E)$, which is a complete isometry, for some reflexive Banach space $E$.

We may replace the word "isometry" by "isomorphism onto its range" above. Furthermore, we may also replace the phrase "a complete isometry" by "an injection" above.

We shall show the isometric version; the isomorphic and injective versions are similar. If (\[cond\_one\]) holds, then ${\mathcal{A}}$ is completely isometric to a subalgebra of some CC dual Banach algebra, and so (\[cond\_two\]) holds by Theorem \[Main\_Thm\].

Conversely, suppose there exists a complete isometry $\theta:{\mathcal{A}}\rightarrow{\mathcal{CB}}(E)$. For $n,m\geq 1$, $x\in\mathbb M_m(E)$ and $\mu=(\mu_{ij})\in\mathbb M_{n\times m}(E')$ with $\|x\|=\|\mu\|=1$, consider the map $$\Lambda_{x,\mu} = \Lambda:\mathbb M_n({\mathcal{A}})\rightarrow\mathbb M_{(n\times m)^2}; \
a=(a_{kl})\mapsto \langle{{\langle {\mu} , {((\theta)_n(a))_m(x)} \rangle}}\rangle.$$ Then notice that for $a=(a_{kl})\in\mathbb M_n({\mathcal{A}})$, $$\Lambda(a) = \big( {{\langle {\mu_{ij}} , {\theta(a_{kl})(x_{rs})} \rangle}} \big)
= \big( {{\langle {\theta'(\mu_{ij}\otimes x_{rs})} , {a_{kl}} \rangle}} \big)
= \langle{{\langle {\big(\theta'(\mu_{ij}\otimes x_{rs})\big)} , {a} \rangle}}\rangle
= \langle{{\langle {\lambda} , {a} \rangle}}\rangle,$$ say, for some $\lambda\in\mathbb M_{n\times m\times m}({\mathcal{A}}')$. We claim that actually $\lambda\in\mathbb M_{n\times m\times m}({\operatorname{WAP}}({\mathcal{A}}'))$, which is equivalent to $\theta'(\mu_{ij}\otimes x_{rs}) \in {\operatorname{WAP}}({\mathcal{A}}')$ for each $i,j,r,s$. The claim follows by the observation that the map ${\mathcal{A}}\rightarrow{\mathcal{A}}'; a\mapsto a\cdot\theta'(\phi\otimes y)$ factors through the reflexive Banach space $E$, for any $\phi\in E'$ and $y\in E$. For $b\in\mathbb M_{n\times m\times m}({\mathcal{A}})$, we have that $$\langle{{\langle {\lambda} , {b} \rangle}}\rangle
= \big({{\langle {\theta'(\mu_{ij}\otimes x_{rs})} , {b_{kl}} \rangle}}\big)
= \big( {{\langle {\mu_{ij}} , {\theta(b_{kl})(x_{rs})} \rangle}} \big)
= \langle{{\langle {\mu} , {((\theta)_{n\times m\times m}(b))_m(x)} \rangle}}\rangle,$$ We thus see that $$\|\lambda\| = \sup\{ \| \langle{{\langle {\lambda} , {b} \rangle}}\rangle \| :
b\in\mathbb M_{n\times m\times m}({\mathcal{A}}), \|b\|\leq 1 \}
\leq \|\mu\| \|x\| = 1.$$ We conclude that for $a\in\mathbb M_n({\mathcal{A}})$, $$\begin{aligned}
 \| (q\kappa_{{\mathcal{A}}})_n(a) \|
&= \sup\{ \|\langle{{\langle {\lambda} , {a} \rangle}}\rangle\| : \lambda\in
   \mathbb M_{n\times m\times m}({\operatorname{WAP}}({\mathcal{A}}')),
   \|\lambda\|\leq 1 \} \\
&\geq \sup\{ \|\Lambda_{x,\mu}(a)\| : \|x\|=\|\mu\|=1 \}
= \|(\theta)_n(a)\|_{cb} = \|a\|,\end{aligned}$$ while clearly $\| (q\kappa_{{\mathcal{A}}})_n(a) \| \leq \|a\|$, showing that $q\kappa_{{\mathcal{A}}}$ is a complete isometry, as required.

In the case of Banach algebras, the above is due to Young, @Young; our proof is closer in nature to Kaiser's presentation in @Kai.

Notice that $q\kappa_{{\mathcal{A}}}$ being an injection is independent of the particular operator space structure on ${\mathcal{A}}$; this can be restated by saying that a CC Banach algebra admits an injective representation on a reflexive operator space if and only if the underlying Banach algebra admits an injective representation on a reflexive Banach space. In contrast, it seems possible that $q\kappa_{{\mathcal{A}}}$ might be an isomorphism, while ${\mathcal{A}}$ admits some operator space structure turning it into a CC Banach algebra for which $q\kappa_{{\mathcal{A}}}$ is not a *complete* isomorphism. We consider this question for Fourier algebras below.

Fourier algebras
================

Let $G$ be a locally compact group, and consider the convolution algebra $L_1(G)$. The space ${\operatorname{WAP}}(L_1(G)')$ is a classical object, which has been widely studied (along with generalisations for semigroups, see @BJM). In particular, ${\operatorname{WAP}}(L_1(G)')$ is a sub-C$^*$-algebra of $L_\infty(G)$, say with (compact) character space $G^{{\operatorname{WAP}}}$. Then $G$ naturally embeds densely into $G^{{\operatorname{WAP}}}$, and $G^{{\operatorname{WAP}}}$ inherits a semigroup structure (in fact induced by the Arens products, see @Daws [Section 7] for example). We can abstractly characterise $G^{{\operatorname{WAP}}}$ as a certain semigroup compactification of $G$.

Now consider instead the Fourier algebra $A(G)$. When $G$ is abelian, $A(G) = L_1(\hat G)$ where $\hat G$ is the dual group of $G$. Hence ${\operatorname{WAP}}(A(G)')$ has an interpretation in terms of $\hat G$. In particular, ${\operatorname{WAP}}(A(G)')$ is a sub-C$^*$-algebra of $VN(G)$, and certainly ${\operatorname{WAP}}(A(G)')$ is not all of $VN(G)$ (as $L_1(G)$ is only Arens regular when $G$ is finite, indeed, see @LL which proves much more). For non-abelian groups, ${\operatorname{WAP}}(A(G)')$ was first studied in @Gran and @DR. Surprisingly, when $G$ is not abelian, it is, in general, unknown if ${\operatorname{WAP}}(A(G)')$ is a sub-C$^*$-algebra of $VN(G)$ (see, for example, @Hu for recent work on this problem). Indeed, it is not even known if ${\operatorname{WAP}}(A(G)')=VN(G)$ can occur for infinite $G$ (see @Forrest1 for partial results). However, when $G$ is amenable and discrete, for example, it is known that ${\operatorname{WAP}}(A(G)') = C^*_r(G)$, the reduced C$^*$-algebra of $G$, as we would expect by analogy with the abelian case.

In the context of Kac algebras, see @ES, or Locally Compact Quantum Groups, see @KV, we view $L_1(G)$ and $A(G)$ as being dual to each other, in some technical sense. As noted in the introduction, it seems to be necessary to use the operator space structure on $A(G)$ to fully realise this idea. Indeed, one could also argue that one should think of $L_1(G)$ as an operator space, but as its dual is a commutative C$^*$-algebra, $L_1(G)$ gets the max quantisation, and so in this case we actually do not gain any new structure over viewing $L_1(G)$ as simply a Banach space.

Actually, this is not quite true. For example, let ${\mathcal{A}}$ be a closed subalgebra of ${\mathcal{B}}(E)$ for some reflexive Banach space $E$. It would seem to be a reasonable conjecture that if we give ${\mathcal{A}}$ the max quantisation (see @ER [Section 3.3]) then ${\mathcal{A}}$ becomes a closed subalgebra of ${\mathcal{CB}}(F)$ for a suitable reflexive operator space $F$. We have, however, been unable to prove this[^1].

In the case of $L_1(G)$, we can argue as follows, however. Let $M(G)$ be the Banach algebra of measures on $G$, with convolution product. Then $M(G) = C_0(G)'$ is a dual Banach algebra, and when we give $M(G)$ the natural operator space structure this induces, we see that $M(G)$ gets the max quantisation, and is hence a CC Banach algebra. Then $L_1(G)$ is completely isometrically a subspace of $M(G)$, and by Theorem \[Main\_Thm\], $M(G)$ is completely isometrically a subspace of ${\mathcal{CB}}(E)$ for some reflexive operator space $E$. The same hence applies to $L_1(G)$, so by Theorem \[wap\_thm\], ${\operatorname{WAP}}(L_1(G)')$ induces the operator space structure on $L_1(G)$.

Similarly, let $B(G)$ be the Fourier-Stieltjes algebra of $G$, so that $B(G) = C^*(G)'$, and hence inherits an operator space structure turning it into a CC dual Banach algebra (see @Runde2 [Section 6] for example). Then $A(G)$ is a closed ideal in $B(G)$, and the operator space structures agree, so we again see that the canonical map $A(G) \rightarrow {\operatorname{WAP}}(A(G)')'$ is a complete isometry onto its range. Hence ${\operatorname{WAP}}(A(G)')$ cannot be too "badly behaved".

This paper grew out of an attempt to use the operator space structure to study weakly almost periodic functionals on $A(G)$. Our hope was that using the factorisation definition of weakly compact, we might find a new definition for operator spaces. However, as we have seen above, this is not the case. Of course, it remains possible that ${\operatorname{WAP}}(A(G)')$ is a well-behaved space, and that simply further work is required. Alternatively, maybe we need to use the operator space structure on $A(G)$ in another way.

Two possibilities come to mind. Let $T:E\rightarrow F$ be a weakly-compact, completely bounded map. Then $(T)_n$ is also weakly-compact. To follow the analogy with complete boundedness, we would want to attach some value, corresponding somehow to a "measure of weak compactness", to each $(T)_n$, and define $T$ to be "completely weakly-compact" if these values remained bounded. Perhaps we could use the factorisation definition of $T$, and use some invariant of the arising reflexive operator space.

We can define $T:E\rightarrow F$ to be completely bounded if and only if the map $$I\otimes T : {\mathcal{K}}(\ell_2) \otimes_{\min} E \rightarrow
{\mathcal{K}}(\ell_2) \otimes_{\min} F$$ is bounded (this approach is taken in @Pisier). Obviously $I\otimes T$ is never weakly-compact (even if $E=F=\mathbb C$, we just get the identity map on ${\mathcal{K}}(\ell_2)$, which is not a reflexive space). Can we find some property of $I\otimes T$ which implies that $T$ is weakly-compact?

<span>9</span>

<span><span style="font-variant:small-caps;">B. Beauzamy</span>, *Espaces d'Interpolation Réels: Topologie et Géométrie*, (Springer-Verlag, Berlin, 1978).</span>

<span><span style="font-variant:small-caps;">J. Bergh, J. L<span>ö</span>fstr<span>ö</span>m</span>, *Interpolation spaces. An introduction*, (Springer-Verlag, Berlin-New York, 1976).</span>

<span><span style="font-variant:small-caps;">J.F. Berglund, H.D. Junghenn, P. Milnes</span>, *Analysis on semigroups; Function spaces, compactifications, representations*, (John Wiley & Sons, Inc., New York, 1989).</span>

<span><span style="font-variant:small-caps;">D.P. Blecher, C. Le Merdy</span>, *Operator algebras and their modules—an operator space approach*, (Oxford University Press, Oxford, 2004).</span>

<span><span style="font-variant:small-caps;">H.G. Dales</span>, *Banach algebras and automatic continuity*, (Clarendon Press, Oxford, 2000).</span>

<span><span style="font-variant:small-caps;">W.J. Davis, T. Figiel, W.B. Johnson, A. Pełczyński</span>, 'Factoring weakly compact operators', *J. Functional Analysis* 17 (1974) 311–327.</span>

<span style="font-variant:small-caps;">M. Daws</span>, 'Dual Banach algebras: representations and injectivity', *Studia Math.* to appear (see `arXiv:math.FA/0604372`).

<span><span style="font-variant:small-caps;">C.F. Dunkl, D.E. Ramirez</span>, 'Weakly almost periodic functionals on the Fourier algebra', *Trans. Amer. Math. Soc.* 185 (1973) 501–514.</span>

<span><span style="font-variant:small-caps;">M. Enock, J.-M. Schwartz</span>, *Kac algebras and duality of locally compact groups*, (Springer-Verlag, Berlin, 1992).</span>

<span><span style="font-variant:small-caps;">E.G. Effros, Z.-J. Ruan</span>, *Operator spaces*, (Oxford University Press, New York, 2000).</span>

<span><span style="font-variant:small-caps;">B. Forrest</span>, 'Arens regularity and discrete groups', *Pacific J. Math.* 151 (1991) 217–227.</span>

<span><span style="font-variant:small-caps;">E.E. Granirer</span>, 'Weakly almost periodic and uniformly continuous functionals on the Fourier algebra of any locally compact group', *Trans. Amer. Math. Soc.* 189 (1974) 371–382.</span>

<span><span style="font-variant:small-caps;">Z. Hu</span>, 'Open subgroups of $G$ and almost periodic functionals on $A(G)$', *Proc. Amer. Math. Soc.* 128 (2000) 2473–2478.</span>

<span><span style="font-variant:small-caps;">S. Kaijser</span>, 'On Banach modules. I.', *Math. Proc. Cambridge Philos. Soc.* 90 (1981) 423–444.</span>

<span><span style="font-variant:small-caps;">J. Kustermans; S. Vaes</span>, 'A simple definition for locally compact quantum groups', *C. R. Acad. Sci. Paris Sér. I Math.* 328 (1999) 871–876.</span>

<span><span style="font-variant:small-caps;">A.T.-M. Lau</span>, 'Uniformly continuous functionals on the Fourier algebra of any locally compact group', *Trans. Amer. Math. Soc.* 251 (1979) 39–59.</span>

<span><span style="font-variant:small-caps;">A.T.-M. Lau, V. Losert</span>, 'On the second conjugate algebra of $L\sb 1(G)$ of a locally compact group', *J. London Math. Soc. (2)* 37 (1988) 464–470.</span>

<span><span style="font-variant:small-caps;">C. Le Merdy</span>, 'An operator space characterization of dual operator algebras', *Amer. J. Math.* 121 (1999) 55–63.</span>

<span>Banach algebras and the general theory of $^*$-algebras, Vol 1</span> <span>Cambridge University Press, Cambridge, 1994</span>

<span><span style="font-variant:small-caps;">G. Pisier</span>, *Introduction to operator space theory*, (London Mathematical Society Lecture Note Series, 294. Cambridge University Press, Cambridge, 2003).</span>

<span><span style="font-variant:small-caps;">Z.-J. Ruan</span>, 'The operator amenability of $A(G)$', *Amer. J. Math.* 117 (1995) 1449–1474.</span>

<span><span style="font-variant:small-caps;">Z.-J. Ruan</span>, 'A characterization of nonunital operator algebras', *Proc. Amer. Math. Soc.* 121 (1994) 193–198.</span>

<span><span style="font-variant:small-caps;">V. Runde</span>, 'Applications of operator spaces to abstract harmonic analysis', *Expo. Math.* 22 (2004) 317–363.</span>

<span><span style="font-variant:small-caps;">V. Runde</span>, 'Amenability for dual Banach algebras', *Studia Math.* 148 (2001) 47–66.</span>

<span>Periodicity of functionals and representations of normed algebras on reflexive spaces.</span> <span>Proc. Edinburgh Math. Soc. (2)</span><span>20</span><span>1976/77</span><span>99–120</span>

<span style="font-variant:small-caps;">F. Uygul</span>, 'A Representation Theorem for Completely Contractive Dual Banach Algebras', `arXiv:math.FA/0702200`.

<span><span style="font-variant:small-caps;">Q. Xu</span>, 'Interpolation of operator spaces', *J. Funct. Anal.* 139 (1996) 500–539.</span>

*Author's Address:*

*Email:* `chelsea94@example.org`

[^1]: As Uygul implicitly points out in @Uygul, this is actually rather simple for dual Banach algebras. Let ${\mathcal{A}}$ be a dual Banach algebra with predual ${\mathcal{A}}_*$, and give ${\mathcal{A}}_*$ the min quantisation, so that ${\mathcal{A}}$ gets the max quantisation, and is hence a CC (dual) Banach alegbra. Hence ${\mathcal{A}}$ is weak$^*$-weak$^*$ isometric to a weak$^*$-closed subalgebra of ${\mathcal{CB}}(E)$ for some reflexive operator space $E$. Morally, we should now be able to draw conclusions for any Banach algebra ${\mathcal{A}}$ with the max quantisation, as in this case, ${\operatorname{WAP}}({\mathcal{A}}')'$ will also get the max quantisation. However, the max quantisation only respects quotients and not necessarily subspaces.

{'source': 'arXiv_out/math0701935'}