Is an infinite topological direct sum of amenable Banach algebras amenable again?
Can you give me a good reference about this notion?


Thanks



Infinite topological direct sum of amenable Banach algebra
The OP has now clarified that he or she is asking about $c\_0$-direct sums. To answer the question we need the notion of the amenability constant of a Banach algebra. Everything that follows is probably "folklore", in the sense that


(i) Johnson knew how to do it in 1972 but, due to the different culture at the time of mathematical publication, did not bother formally writing it down in his 1972 monograph *Cohomology in Banach Algebras* or his Amer J. Math article of the same year;


(ii) successive generations of researchers, who take seriously the study of amenability of Banach algebras, will *almost surely* have worked these results out for themselves during their own studies.


Actually, this result might also be somewhere in Runde's book, but I haven't checked. (**UPDATE 2015-11-10:** Tomasz Kania has pointed out that this is Corollary 2.3.19 in Runde's *Lectures on Amenability*.)


**Definition.** Given an amenable Banach algebra $B$, the amenability constant of $B$ is defined to be 
$${\rm AM}(B):=\inf \{ \Vert M \Vert\_{(B\widehat\otimes B)^{\*\*}}\;\; \colon \;\;\hbox{$M$ is a virtual diagonal for $B$}\}. $$
(By weak-star compactness, the infimum is actually attained.) We adopt the convention that if $B$ is a non-amenable Banach algebra, ${\rm AM}(B)=+\infty$.


*Remark 1.*
It is an easy exercise to show that if $q:A\to B$ is a surjective homomorphism of Banach algebras then ${\rm AM}(B)\leq \Vert q\Vert^2{\rm AM}(A)$.


**Lemma 2.** Suppose $A$ is a Banach algebra and $(B\_i)\_{i\in \Lambda}$ is a family of closed subalgebras such that $\bigcup\_{i\in\Lambda} B\_i$ is dense in $A$. Then ${\rm AM}(A)\leq \sup\_{i\in\Lambda} {\rm AM}(B\_i)$.


Note that since closed subalgebras of ${\bf M}\_2({\mathbb C})$ can be non-amenable, the inequality in this lemma is usually not an equality.


*Outline of the proof of Lemma 2.* We may assume the RHS of the inequality is finite, bounded above by $C$ say; otherwise there is nothing to prove. Consider $(M\_i)\_{i\in\Lambda}$ where $M\_i$ is a virtual diagonal for $B\_i$ such that ${\rm AM}(B\_i)=\Vert M\_i \Vert$. Then $(M\_i)$ is a net in $(A\widehat{\otimes} A)^{\*\*}$, bounded by $C$; take any weak-star cluster point $M$. Then $\Vert M\Vert\leq C$ and one can check $M$ is indeed a virtual diagonal for $A$. **QED.**


**Lemma 3.**
Let $A\_1$ and $A\_2$ be Banach algebras, and let $B=A\_1\oplus\_\infty A\_2$ i.e. with the norm $\Vert (a\_1,a\_2)\Vert := \max (\Vert a\_1\Vert, \Vert a\_2\Vert)$.
Then ${\rm AM}(B)= \max \{{\rm AM}(A\_1), {\rm AM}(A\_2)\}$.


*Proof of lemma 3.* Since there are contractive, surjective algebra homomorphisms $q\_j:B\to A\_j$ ($j=1,2$) we have ${\rm AM}(B)\geq \max \{{\rm AM}(A\_1), {\rm AM}(A\_2)\}$ by Remark 1. Conversely, using the contractive embedding
$$ (A\_1\widehat\otimes A\_1) \oplus\_\infty (A\_2\widehat\otimes A\_2) \to (A\_1\oplus\_\infty A\_2)\widehat\otimes(A\_1\oplus\_\infty A\_2) $$
and the corresponding embedding at the level of biduals, we can combine a virtual diagonal $M\_1$ for $A\_1$ and a virtual diagonal $M\_2$ for $A\_2$ to get a virtual diagonal $M$ for $A$, which satisfies $\vert M \Vert \leq \max( \Vert M\_1\Vert, \Vert M\_2\Vert)$. **QED.**


**Proposition.** Let $(A\_n)\_{n=1}^\infty$ be a sequence of Banach algebras, and let $A=c\_0{\rm-}\bigoplus\_{n=1}^\infty A\_n$. Then 
${\rm AM}(A) = \sup\_n {\rm AM}(A\_n) <\infty$.


*Proof that LHS ≥ RHS.* This is like the proof of Lemma 2, using the contractive surjective homomorphisms $A\to A\_n$ for $1\leq n\leq\infty$, and I omit the details.


*Proof that LHS ≤ RHS.* Let $C=\sup\_n {\rm AM}(A\_n)$; we may assume $C<\infty$, otherwise there is nothing to prove. Now let $B\_n = A\_1 \oplus \dots \oplus A\_n$. By Lemma 3 and induction we know that ${\rm AM}(B\_n)\leq C$, for each $n$. Since $\bigcup\_{n\geq 1} B\_n$ is dense in $A$, applying Lemma 2 completes the proof. **QED.**




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**OLD answer, before the question was clarified**


Tomek Kania has already pointed out the key "counterexample" (assuming that by infinite topological direct sum you mean what I would call the $\ell^\infty$-direct sum).


I should also point out that there are commutative counterexamples. Indeed, let $A={\ell^\infty}{\rm-}\bigoplus \ell^1({\bf T}\_d)$ where ${\bf T}\_d$ denotes the unit circle equipped with the discrete topology, and the multiplication on $\ell^1({\bf T}\_d)$ is given by convolution. Then, as observed in e.g. Proposition 5.8 of <http://arxiv.org/abs/0708.4195> $A$ quotients onto $M(G\_0)$ where $G\_0$ is the Bohr compactificaton of ${\bf T}\_d$.


If $A$ were weakly amenable, then $M(G\_0)$ would be weakly amenable; but by results of Brown and Moran $M(G\_0)$ has non-zero point derivations, hence is not weakly amenable. Therefore $A$ is not weakly amenable, and in particular cannot be amenable.


