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 $M(bG)\to \mathfrak{A}$?

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    $\begingroup$ But $C(b G)$ isometrically embeds in $L^1(G)^* \cong L^\infty(G)$, so wouldn't you expect $M(b G) \cong C(b G)^*$ to be a quotient of $L^\infty(G)^*$? $\endgroup$ – Nik Weaver Apr 17 at 20:36

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.

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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 $M(bG) = M(G) = M(L^1(G)) \rightarrow L^1(G)^{**}$.

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).

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.

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

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 $A$.

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  • $\begingroup$ But in this case, when $G$ is compact, can't we trivially embed $M(G)$ in $L^\infty(G)^*$ by taking $\mu$ to integration against $\mu$? Seems like this is a homomorphism for either Arena product. $\endgroup$ – Nik Weaver Apr 18 at 20:17
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    $\begingroup$ @NikWeaver Nope: what about a point mass? $L^\infty(G)$ is only equivalence classes for Haar measure, so singular measures can't be integrated against. $\endgroup$ – Matthew Daws Apr 18 at 20:35
  • $\begingroup$ So how does a point mass act on $L^\infty(G)$ in your construction? For that matter, how does it correspond to a multiplier of $L^1(G)$? $\endgroup$ – Nik Weaver Apr 18 at 23:07
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    $\begingroup$ There is another way to view the construction (maybe a better one). If we convolve an $L^1$ and an $L^\infty$ function then we get a uniformly continuous function, which we can integrate against. Then take a limit along a bai in $L^1(G)$. So a point mass $\delta_s$ say "integrates" against an $L^\infty(G)$ function $F$ say by integrating $F$ against $L^1(G)$ functions which ever more closely approximate a point mass at $s$: so sort of average of the value of $F$ near $s$. I guess it's all rather non-constructive... $\endgroup$ – Matthew Daws Apr 19 at 7:51
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    $\begingroup$ Yes... Well, just some accumulation point. I think this slightly subtle point is lost in usual arguments in some phrase like "moving to a subnet if necessary, we may assume that..." $\endgroup$ – Matthew Daws Apr 20 at 7:58

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