[Paper Review] Module super-amenability for semigroup algebras

Abasalt Bodaghi, Massoud Amini|ArXiv.org|Dec 23, 2009

Advanced Operator Algebra Research9 references3 citations

TL;DR

This paper introduces module super-amenability for Banach algebras acting on another Banach algebra, and establishes that for an inverse semigroup $S$ with an upward directed set of idempotents $E$, the semigroup algebra $\ell^1(S)$ is $\ell^1(E)$-module super-amenable if and only if the group homomorphic image $S/\approx$ is finite. The result extends Selivanov’s theorem on super-amenability of group algebras to inverse semigroups via module structure.

ABSTRACT

Let $S$ be an inverse semigroup with the set of idempotents $E$. In this paper we define the module super-amenability of a Banach algebra which is a Banach module over another Banach algebra with compatible actions, and show that when $E$ is upward directed and acts on $S$ trivially from left and by multiplication from right, the semigroup algebra $ \ell ^{1}(S)$ is $\ell^{1}(E)$-module super-amenable if and only if an appropriate group homomorphic image of $S$ is finite.

Motivation & Objective

• To extend the concept of super-amenability to module structures in Banach algebras.
• To characterize module super-amenability for semigroup algebras of inverse semigroups with upward directed idempotent sets.
• To establish a semigroup analog of Selivanov’s theorem on super-amenability of group algebras.
• To show that $\ell^1(S)$ is $\ell^1(E)$-module super-amenable precisely when the group image $S/\approx$ is finite.
• To demonstrate that module super-amenability does not imply standard super-amenability, even when the algebra has an identity.

Proposed method

• Define module super-amenability for a Banach algebra $\mathcal{A}$ that is a Banach module over another Banach algebra $\mathfrak{A}$ with compatible actions.
• Introduce the notion of a module diagonal, showing that module super-amenability is equivalent to the existence of such a diagonal in $\mathcal{A} \widehat{\otimes} \mathcal{A}$.
• Use the quotient algebra $\ell^1(S)/J \cong \ell^1(S/\approx)$ to reduce the problem to group algebras.
• Leverage Selivanov’s theorem, which states that $L^1(G)$ is super-amenable iff $G$ is finite, to analyze the group image $S/\approx$.
• Apply results on bounded approximate identities for $\ell^1(E)$ acting on $\ell^1(S)$ when $E$ satisfies condition $D_1$, ensuring module super-amenability is preserved under quotienting.
• Construct explicit module diagonals in matrix algebras $M_n(\mathfrak{G})$ to illustrate non-super-amenable but module super-amenable examples.

Experimental results

Research questions

• RQ1 When is the semigroup algebra $\ell^1(S)$ of an inverse semigroup $S$ module super-amenable over $\ell^1(E)$, where $E$ is the set of idempotents in $S$?
• RQ2 What structural conditions on $S$ ensure that $\ell^1(S)$ is $\ell^1(E)$-module super-amenable?
• RQ3 How does the finiteness of the group image $S/\approx$ relate to module super-amenability of $\ell^1(S)$?
• RQ4 Can a Banach algebra be module super-amenable without being super-amenable, even with an identity?
• RQ5 What role does the upward directedness of $E$ play in the module super-amenability of $\ell^1(S)$?

Key findings

• The semigroup algebra $\ell^1(S)$ is $\ell^1(E)$-module super-amenable if and only if the group homomorphic image $S/\approx$ is finite.
• The quotient algebra $\ell^1(S)/J$ is isomorphic to $\ell^1(S/\approx)$, and this isomorphism preserves the module structure.
• Module super-amenability of $\ell^1(S)$ implies that $\ell^1(S)/J$ is super-amenable, which forces $S/\approx$ to be finite by Selivanov’s theorem.
• There exist examples of $\ell^1(S)$ that are $\ell^1(E)$-module super-amenable but not super-amenable, such as $\ell^1(\mathbb{N})$ under the max operation.
• The matrix algebra $M_n(\mathfrak{G})$ is $\mathfrak{G}$-module super-amenable even when $\mathfrak{G} = \ell^1(S)$ is not weakly amenable, showing that module super-amenability is strictly weaker than super-amenability.
• The identity element in $\ell^1(\mathbb{N})$ does not yield a diagonal, proving that $\ell^1(\mathbb{N})$ is not super-amenable despite being module super-amenable.