Measure algebra on the Bohr compactification vs the bidual algebras

Question

The following question probably reduces to some standard abstract harmonic analysis Twister play, but I'd still welcome some comments on it.

Let $G$ be a locally compact Abelian group and let $bG$ denote its Bohr compactification (the Pontryagin dual of $\widehat{G}$ with the discrete topology). Denote by $\mathfrak{A}$ the space $L_1(G)^{**}$ furnished with either Arens product.

Is there a canonical action of $M(bG)$ (the measure algebra on $bG$) on $L_\infty(G)$ that would give rise to an isometric homomorphism $M(bG)\to \mathfrak{A}$?

Answer 1

I'm going to say no. The "canonical" pairing of $M(bG)$ with $L^\infty(G)$ is to integrate a function in $L^\infty(G)$ against the restriction to $G$ of a measure in $M(bG)$. But this is not faithful: any measure supported on $bG\setminus G$ would go to zero in $L^\infty(G)^*$. To pick up mass on this corona we would want to extend functions in $L^\infty(G)$ to $bG$. But you have to be almost periodic to canonically extend to $bG$, so it doesn't seem like there's any way to get what you want.

Answer 2

There is at least a case when it is true though. Suppose $G$ itself is compact (this argument doesn't need $G$ abelian), so that $bG = G$. The $M(G)$ is the multiplier algebra of $L^1(G)$ and as $L^1(G)$ has a contractive approximate identity, there is an isometric embedding $M(bG) = M(G) = M(L^1(G)) \rightarrow L^1(G)^{**}$.

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Let me sketch this. Let $A$ be a Banach algebra with contractive approximate identity $(e_\alpha)$. I will regard the multiplier algebra $M(A)$ as double centralisers: pairs of maps $L,R$ from $A$ to $A$ with $$ L(ab) = L(a)b, \qquad R(ab) = aR(b), \qquad aL(b) = R(a)b \qquad (a,b\in A). $$ It turns out that, using the approximate identity, one can show that $L,R$ are automatically linear, and also (closed graph theorem) that $L,R$ are bounded. (Or make this part of the definition, if you wish).

Turn $A^*$ and $A^{**}$ into $A$-bimodules in the usual way. Given $(L,R)\in M(A)$ let $x^{**}\in A^{**}$ be an accumulation point of the bounded net $(L(e_\alpha))$. For $x^*\in A^*$ and $x\in A$ compute: $$ \langle x^{**} \cdot a, x^* \rangle = \langle x^{**}, a \cdot x^* \rangle = \lim_\alpha \langle a \cdot x^*, L(e_\alpha) \rangle = \lim_\alpha \langle x^*, L(e_\alpha)a \rangle = \lim_\alpha \langle x^*, L(e_\alpha a) \rangle = \langle x^*, L(a) \rangle. $$ Thus $x^{**}\cdot a = L(a)$ (or the canonical image thereof in $A^{**}$). Similarly, $$ \langle a \cdot x^{**}, x^* \rangle = \lim_\alpha \langle x^*, a L(e_\alpha) \rangle = \lim_\alpha \langle x^*, R(a) e_\alpha \rangle = \langle x^*, R(a) \rangle, $$ so that $a\cdot x^{**} = R(a)$. This gives us the required embedding.

For those that know about Arens products, there are clear links. I believe this construction is due to McKilligan (MathSciNet or JLMS Article).

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As Nik Weaver notices, of course $C(bG) = C(G) \subseteq L^\infty(G)$ and so we obtain a quotient map $\theta : L^\infty(G)^{*} \rightarrow M(G)$. Let $\phi:M(G)\rightarrow L^1(G)^{**}$ be the map we just constructed. Let $\mu\in M(G)$ so the associated double centraliser is $L(f) =\mu * f, R(f) = f * \mu$ for $f\in L^1(G)$. Then for $F\in C(G)$ (and writing $\cdot$ for the module actions, which are related to but not quite equal to convolution), $$ \langle \theta(\phi(\mu)), F \rangle = \langle \phi(\mu), F \rangle_{L^\infty(G)^*, C(G)}. $$ Now, a bounded approximate identity argument and a calculation shows that every $F\in C(G)$ is equal to $f\cdot F'$ for some $F'\in C(G)$ and $f\in L^1(G)$. Thus $$ \langle \theta(\phi(\mu)), F \rangle = \langle F', \mu * f \rangle_{C(G), L^1(G)} = \langle \mu, F \rangle. $$ So $\theta \circ \phi$ is the identity, and hence $M(bG)$ is a complemented subspace of $L^\infty(G)^*$ in this case.

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I have had a quick think, and I cannot see how to say much in the non-compact case.

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Edit: Some comments on uniqueness, prompted by interesting questions of Nik Weaver. Given a bounded approximate identity $(e_\alpha)$ let $x_0^{**} \in A^{**}$ be some accumulation point. Then our embedding $M(A)\rightarrow A^{**}$ is $(L,R) \mapsto L^{**}(x^{**}_0)$, which is isometric if $\|x^{**}_0\|=1$. Notice that if $x^{**} = L^{**}(x^{**}_0)$ then $x^{**}\cdot a, a\cdot x^{**} \in A \subseteq A^{**}$ for each $a\in A$.

Conversely, if $x^{**}\in A^{**}$ is any element with $x^{**}\cdot a, a\cdot x^{**} \in A$ for each $a\in A$, then we can define linear maps $L,R:A\rightarrow A$ with $L(a) = x^{**}\cdot a$ etc. and then $(L,R)\in M(A)$. It is tempting, but wrong to think that we have shown that $$ M(A) \cong \{ x^{**}\in A^{**} : A\cdot x^{**}, x^{**}\cdot A \subseteq A \}. $$ What goes wrong is that we can have a non-zero $x^{**}\in A^{**}$ with $A\cdot x^{**} + x^{**}\cdot A =\{0\}$. In the example of $A=L^1(G)$ we know that $A\cdot A* + A^*\cdot A$ is the sum of the left/right uniformly continuous functions. So any $x^{**}\in L^\infty(G)^*$ which annihilates these, but is non-zero, induces the zero multiplier. Notice that this cannot happen for $C^*$-algebras for example, as here $A^*\cdot A = A\cdot A^* = A^*$.

So, the embedding of $M(A)$ into $A^{**}$ depends on the choice of $x_0^{**}$ an accumulation point of a bai of $A$. You can characterise such $x_0^{**}$ as the "mixed identities" of $A^{**}$ (a right identity for the 1st Arens product and a left identity for the 2nd Arens product). Any such mixed identity is some accumulation point of some bai of $A$.