# Is $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ a commutative algebra?

Consider $l^\infty(\mathbb{Z})$ the Banach space of bounded complex valued functions on the abelian group $\mathbb{Z}$ with the supremum norm. It has a natural action by $\mathbb{Z}$ given by $(zf)(g):=f(z^{-1}g)$. Then let $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ be the algebra (multiplication by composition) consisting of all bounded $\mathbb{Z}$-equivariant operators $l^\infty(\mathbb{Z})\rightarrow l^\infty(\mathbb{Z})$.

Question: Is this a commutative algebra?

If $l^\infty$ is replaced by $l^2$ the algebra is the von Neumann algebra $L(\mathbb{Z})$ and in this cases $\mathbb{C}(\mathbb{Z})$ lies densely (wrt WOT or SOT) in $L(\mathbb{Z})$ as a subalgebra and therefore $L(\mathbb{Z})$ is commutative. But I think $\mathbb{C}(\mathbb{Z})$ does not lie densely in $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$, so the situation is a bit different.

Question: What happens if $\mathbb{Z}$ is replaced by an arbitrary abelian group?

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The following is an abstract (Banach) algebraic take on Werner's construction. Let $A=\ell^1(\mathbb Z)$ with convolution (but in general $A$ is any Banach algebra). We turn the dual space $A^*$ into an $A$-bimodule (though in our example, $A$ is commutative) by dualising the actions: $$(a\cdot f)(b) = f(ba), \quad (f\cdot a)(b) = f(ab) \qquad (f\in A^*, a,b\in A).$$ Then you can check that $T:A^*\rightarrow A^*$ is $\mathbb Z$-equivariant if and only if $T$ is a module homomorphism: I'll write $T\in\hom_A(A^*)$ to mean $T(f\cdot a) = T(f)\cdot a$ (the right action is more appropriate in the non-commutative group setting). If $e\in A$ is the unit then $$T(f)(a) = T(f)(ae) = (T(f)\cdot a)(e) = T(f\cdot a)(e) \qquad (T\in \hom_A(A^*), f\in A^*, a\in A).$$ So again $T$ is determined uniquely by $m\in A^{**}, m(f) = T(f)(e)$. Conversely, given $m\in A^{**}$ we can define $T\in\hom_A(A^*)$ by this relation. So $\hom_A(A^*)\cong A^{**}$. A similar result holds if $A$ only has a bounded approximate identity.

So $A^{**}$ becomes an algebra for the product induced from $\hom_A(A^*)$. Define $$(m\cdot f)(a) = m(f\cdot a) \qquad (m\in A^{**}, f\in A^*,a\in A).$$ Then if $m_i$ is associated to $T_i$ for $i=1,2$, $$(m_1m_2)(f) = (T_1\circ T_2)(f)(e) = T_1(T_2(f))(e) = m_1(T_2(f)).$$ However, for $a\in A$, $$T_2(f)(a) = m_2(f\cdot a) = (m_2\cdot f)(a)$$ and so $$(m_1m_2)(f) = m_1(m_2\cdot f)$$ This is precisely the "1st Arens product", say $\Box$, as studied in Banach algebra theory. By making the symmetric choices we get the "2nd Arens product", say $\diamond$, which might differ. If $A$ is commutative then $m_1\Box m_2 = m_2\diamond m_1$ and so $\hom_A(A^*)$ is commutative precisely when the Arens products agree.

However, Young showed in "The irregularity of multiplication in group algebras" Quart J. Math. Oxford Ser. (2) 24 (1973), 59–62, MathSciNet that for any locally compact group $G$, we have that $L^1(G)$ is not Arens regular. So Werner's question has a negative answer for all Abelian groups (with the axiom of choice!)

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Delurking to say: very nice! –  Yemon Choi Oct 29 '13 at 16:08
Minor nitpick: your last paragraph is missing the word "infinite" in both sentences. (Please don't thump me for pedantry!) Also, just to make explicit something which I think is tacit in your final sentence: without choice, we could have $hom_A(A^*)=A$ for $A=\ell^1(G)$ for every $G$, in which case Werner's question has a positive answer for all Abelian groups. –  Yemon Choi Oct 29 '13 at 17:14

I might have an answer to the problem which depends on the set theoretic axioms used. I would be glad if someone could double check this.

Let $G=\mathbb{Z}$. First observe the following: Let $\chi$ be a functional in the dual space $l^\infty(G)'$. A $G$-equivariant linear bounded operator $A:l^\infty(G)\rightarrow l^\infty(G)$ is uniquely determined by the value $A(f)(1)$ for each $f\in l^\infty(G)$. So define $\varphi(\chi)$ to be the operator $A$ with $A(f)(1)=\chi(f)$. One has to check that this gives a bounded operator wrt to the sup-norm, but that is easy. We have defined a map $\varphi:\mathcal{B}^G(l^\infty(G))\rightarrow l^\infty(G)'$. One can easily find an inverse to this map, so it is bijective. Now one can calculate a formula for the product on $l^\infty(G)'$ which respects $\varphi$. It can be written as $$(\chi*\tilde{\chi})(f)=\chi\big(g\mapsto\tilde{\chi}(f(g\_))\big)$$ We will give two elements in $l^\infty(G)'$ which do not commute wrt to this product. This will crucially use the axiom of choice (see also below). Denote by $c$ the Banach space of functions on $G$ such that the limit for $x\rightarrow -\infty$ and also the limit for $x\rightarrow\infty$ exists and take the sup-norm as norm. Define two elements $\chi$ and $\tilde{\chi}$ in $c'$ as follows: $$\chi(f)=\lim_+f+\lim_-f$$ $$\tilde{\chi}(f)=\lim_+f-\lim_-f$$ where $\lim_-$ means the limit for $x\rightarrow -\infty$ and $\lim_+$ means the limit for $x\rightarrow\infty$. Using Hahn-Banach extend these functionals to $l^\infty(G)$ (axiom of choice!). Now choose any bounded function $F:G\rightarrow\mathbb{R}$ which converges to $0$ in the negative and to $1$ in the positive. It is now easy to see that $$(\chi*\tilde{\chi})(F)=2$$ but $$(\tilde{\chi}*\chi)(F)=0$$

Weakening the axiom of choice AC to dependent choice DC and adding another axiom yields that $l^\infty(G)'=l^1(G)$, see: Martin Väth, The dual space of L∞ is L1, Indag. Mathem., N.S., 9 (4), 1998, 619–625. That is, every functional on $l^\infty(G)$ is given by pointwise multiplication with a function in $l^1(G)$. It is easy to see that the product above is commutative in this case.

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