Amenable Groups and Amenable Banach Algebras

Amenable Groups

Definition: Let $G$ be a group. An Invariant Mean of $\ell^{\infty}(G)$ is a positive linear functional $\mu : \ell^(G) \to \mathbf{R}$ with the following properties: a) $\mu (1) = 1$. b) If for each $h \in G$, $T_h : \ell^{\infty} (G) \to \ell^{\infty}$ denotes the left translation operator defined for each $m \in \ell^{\infty}(G)$ by $T_h(m) \in \ell^{\infty}$ being defined for all $g \in G$ by $[T_h(m)](g) = m(h^{-1}g)$, then we have the property that $\mu (T_h(m)) = \mu(m)$. A group $G$ is said to be an Amenable Group if there exists an invariant mean $\mu$ on $\ell^{\infty} (G)$.

For each $h \in G$, the left translation operator $T_h : \ell^{\infty} (G) \to \ell^{\infty}(G)$ takes each $m \in \ell^{\infty}(G)$ and left translates it $T_h(m) \in \ell^{\infty}(G)$ via the formula defined for all $g \in G$ by $[T_h(m)](g) = m(h^{-1}g)$. Note that $m(g) = [T_h(m)](hg)$ for all $g \in G$.

Amenable Banach Algebras

Definition: Let $\mathfrak{A}$ be a Banach algebra and let $X$ be a Banach $\mathfrak{A}$-bimodule. The First Cohomology Group of $\mathfrak{A}$ and $X$ is denoted $H^1(\mathfrak{A}, X)$ and is defined to be the quotient space $Z^1(\mathfrak{A}, X) / B^1(\mathfrak{A}, X)$.

Here, $Z^1(\mathfrak{A}, X)$ denotes the space of all bounded $X$-derivations, which we have already proven is a subspace of $\mathrm{BL}(\mathfrak{A}, X)$, and, $B^1(\mathfrak{A}, X)$ denotes the space of all inner bounded $X$-derivations, which we have already proven is a subspace of $Z^1(\mathfrak{A}, X)$.

Let $\mathfrak{A}$ be a Banach algebra. Let $X$ be a Banach $\mathfrak{A}$-bimodule. Let $X^*$ denote the dual space of $X$, and regard it as a Banach $\mathfrak{A}$-bimodule as follows. For all $a \in \mathfrak{A}$ and $f \in X^*$ define the multiplication $af$ to be the functional $af : X \to \mathbf{F}$ defined for all $x \in X$ by:

(1)

\begin{align} \quad (af)(x) = f(xa) \end{align}

Similarly, for all $a \in \mathfrak{A}$ and $f \in X^*$ define the multiplication $fa$ to be the functional $fa : X \to \mathbf{F}$ defined for all $x \in X$ by:

(2)

\begin{align} \quad (fa)(x) = f(ax) \end{align}

Definition: Let $\mathfrak{A}$ be a Banach algebra. Then $\mathfrak{A}$ is said to be Amenable if $H^1(\mathfrak{A}, X^*) = \{ 0 \}$ for all Banach $\mathfrak{A}$-bimodules $X$ (where each $X^*$ is regarded as a Banach $\mathfrak{A}$-bimodule as above).

Equivalently, $\mathfrak{A}$ is amenable if whenever $X$ is a Banach $\mathfrak{A}$-bimodule we have that every bounded $X^*$-derivation is an inner bounded $X^*$-derivation.

Proposition 1: Let $\mathfrak{A}$ be amenable. If $\sigma$ is a multiplicative linear functional on $\mathfrak{A}$ then there exists an $F \in A^{**} \setminus \{ 0 \}$ such that $F(\sigma) = 1$ and $F(fa) = \sigma(a) F(f)$ for all $a \in \mathfrak{A}$ and for all $f \in \mathfrak{A}^*$.

Recall that if $X$ and $Y$ are normed spaces and $T : X \to Y$ then the adjoint of $T$ is the mapping $T^* : Y^* \to X^*$ defined for all $f \in Y^*$ by $T(f) = f \circ T$ and furthermore, $\| T \| = \| T^* \|$.

• Proof: Let $\mathfrak{A}^*$ be a Banach $\mathfrak{A}$-bimodule with multiplicative defined as follows. Define the left module multiplication for all $a \in \mathfrak{A}$ and all $f \in \mathfrak{A}^*$ by:

(3)

\begin{align} \quad af = \sigma (a) f \end{align}

• Define the right module multiplication for all $a \in \mathfrak{A}$ and all $f \in \mathfrak{A}^*$ by $fa$ being the function defined for all $b \in \mathfrak{A}$ by $(fa)(b) = f(ab)$.

• First it is important to note that $af, fa \in \mathfrak{A}^*$ for all $a \in \mathfrak{A}$ and all $f \in \mathfrak{A}^*$. To see why, fix $a \in \mathfrak{A}$ and fix $f \in \mathfrak{A}^*$. Then for all $b \in \mathfrak{A}$ we have that:

(4)

\begin{align} \quad \| (af)(b) \| = \| \sigma (a) f(b) \| \leq \| \sigma(a) \| \| f(b) \| \leq \underbrace{[\| \sigma \| \| a \| \| f \|]}_{\mathrm{fixed}} \| b \| \end{align}

• (Where $\| \sigma (a) \| \leq \| \sigma \| \| a \|$ since $\sigma$ is a multiplicative linear funtional, and every multiplicative linear functional is bounded). And:

(5)

\begin{align} \quad \| (fa)(b) \| = \| f(ab) \| \leq \| f \| \| ab \| \leq \underbrace{\| f \| \| a \|}_{\mathrm{fixed}} \| b \| \end{align}

• Now, since $\sigma$ is a multiplicative linear functional we have that $\sigma \in \mathfrak{A}^*$. So by the definition of left module multiplication, for every $a \in \mathfrak{A}$ we have that $a \sigma = \sigma (a) \sigma$. And by the definition of right module multiplication, for every $a \in \mathfrak{A}$ we have that $\sigma a$ is the function defined for all $b \in \mathfrak{A}$ by $(\sigma a)(b) = \sigma (ab) = \sigma (a) \sigma (b)$, i.e., $\sigma a = \sigma (a) \sigma$. Thus, for all $a \in \mathfrak{A}$:

(6)

\begin{align} \quad a \sigma = \sigma a = \sigma (a) \sigma \end{align}

• Thus $\mathfrak{A}(\mathbb{C} \sigma)= (\mathbb{C} \sigma)\mathfrak{A} \subseteq \mathbb{C} \sigma$, showing that $\mathbb{C} \sigma$ is an $\mathfrak{A}$-bimodule. And further, since $\mathfrak{A}^*$ is a Banach $\mathfrak{A}$-bimodule, from above we see that $\mathbb{C} \sigma$ is a $\mathfrak{A}$-bimodule that is a closed (since $\mathbb{C} \sigma$ is finite-dimensional) submodule of $\mathfrak{A}^*$. Let $X = \mathfrak{A}^* / \mathbb{C} \sigma$. Let $q : \mathfrak{A}^* \to X$ be the quotient mapping.

• Let $q^*: X^* \to \mathfrak{A}^{**}$ be the adjoint mapping.

• Let $v \in \mathfrak{A}^{**}$ be such that $v(\sigma) = 1$. Let $\delta$ be the inner bounded $\mathfrak{A}^{**}$-derivation defined for all $a \in \mathfrak{A}$ by:

(7)

\begin{align} \quad \delta(a) = av - va \end{align}