Semi-Simple Commutative Banach Algebras
Semi-Simple Commutative Banach Algebras
Recall from The Radical of a Commutative Banach Algebra page that if $\mathfrak{A}$ is a commutative Banach algebra then the radical of $\mathfrak{A}$ is defined to be the set:
(1)\begin{align} \quad \mathrm{Rad}(\mathfrak{A}) = \bigcap_{f \in \Phi_{\mathfrak{A}}} \ker (f) \end{align}
We proved that $\mathrm{Rad}(\mathfrak{A})$ is nonempty (as it contains $0$) and that:
(2)\begin{align} \quad \mathrm{Rad}(\mathfrak{A}) = \{ x \in \mathfrak{A}: r(x) = 0 \} \end{align}
We give a special name to commutative Banach algebras $\mathfrak{A}$ over $\mathbb{C}$ whose radical contains ONLY the $0$.
| Definition: Let $\mathfrak{A}$ be a commutative Banach algebra over $\mathbb{C}$. $\mathfrak{A}$ is said to be Semi-Simple if $\mathrm{Rad}(\mathfrak{A}) = \{ 0 \}$. |
The following proposition gives criteria for when a commutative Banach algebra $\mathfrak{A}$ over $\mathbb{C}$ is semi-simple.
| Proposition 1: Let $\mathfrak{A}$ be a commutative Banach algebra over $\mathbb{C}$. The following statements are equivalent: 1) $\mathfrak{A}$ is semi-simple. 2) $\Phi_{\mathfrak{A}}$ separates the points of $\mathfrak{A}$. 3) The Gelfand representation $x \to \hat{x}$ is a monomorphism of $\mathfrak{A}$ into $C_0(\Phi_{\mathfrak{A}})$. 4) The spectral radius $r$ defined on $\mathfrak{A}$ is an algebra norm on $\mathfrak{A}$. |
Recall that if $\mathfrak{A}$ is a commutative Banach algebra over $\mathbb{C}$ then $r(x) = \| \hat{x} \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)|$. We use this throughout the proof below.
- Proof: $1) \Rightarrow 2)$ Let $\mathfrak{A}$ be semi-simple. Then $\mathrm{Rad}(\mathfrak{A}) = \{ 0 \}$. Let $x, y \in \mathfrak{A}$ be such that $x \neq y$. Suppose instead that for every $f \in \Phi_{\mathfrak{A}}$ we have that $f(x) = f(y)$. Then by linearity of $f$ we have that $f(x - y) = 0$, so $x - y \in \ker (f)$ for every $f \in \Phi_{\mathfrak{A}}$, implying that $x - y \in \mathrm{Rad}(\mathfrak{A})$. But $\mathrm{Rad}(X) = \{ 0 \}$, so $x - y = 0$, a contradiction. Thus there must exist an $f \in \Phi_{\mathfrak{A}}$ such that $f(x) \neq f(y)$, i.e., $\Phi_{\mathfrak{A}}$ separates the points of $\mathfrak{A}$.
- $2) \Rightarrow 3)$ Let $\Phi_{\mathfrak{A}}$ separate the points of $\mathfrak{A}$. We have already proven that the Gelfand representation $x \to \hat{x}$ is a homomorphism of $\mathfrak{A}$ to $C_0(\Phi_{\mathfrak{A}}) =$ when $\mathfrak{A}$ has a unit. When $\mathfrak{A}$ does not have a unit, we have proven already that $\Phi_{\mathfrak{A}}$ is compact and that the Gelfand representation is a homomorphism of $\mathfrak{A}$ to $C(\Phi_{\mathfrak{A}})$. But since $\Phi_{\mathfrak{A}}$ is compact, $C(\Phi_{\mathfrak{A}}) = C_0(\Phi_{\mathfrak{A}})$.
- In either case, all that remains to show is that the Gelfand representation is injective. Let $x, y \in \mathfrak{A}$ and suppose that $\hat{x} = \hat{y}$. Then for all $f \in \Phi_{\mathfrak{A}}$ we have that $\hat{x}(f) = \hat{y}(f)$, i.e., for all $f \in \Phi_{\mathfrak{A}}$ we have that $f(x) = f(y)$. If $x \neq y$ then $\Phi_{\mathfrak{A}}$ cannot separate the points of $\mathfrak{A}$. Thus we must have that $x = y$, and so the Gelfand representation is a monomorphism of $\mathfrak{A}$ to $C_0(\Phi_{\mathfrak{A}})$.
- $3) \Rightarrow 4)$ Suppose the Gelfand representation $x \to \hat{x}$ is a monomorphism of $\mathfrak{A}$ into $C_0(\Phi_{\mathfrak{A}})$. We want to show that the spectral radius is a norm on $\mathfrak{A}$. First observe $r : \mathfrak{A} \to [0, \infty)$.
- 1. Showing that $r(x) = 0$ if and only if $x = 0$: Suppose $r(x) = 0$. Then $0 = r(x) = \|\hat{x} \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)|$, so that $\hat{x}(f) = 0$ for all $f \in \Phi_{\mathfrak{A}}$. But also $\hat{0}(f) = 0$ for all $f \in \Phi_{\mathfrak{A}}$. Since the Gelfand representation is a monomorphism (and hence injective) and $\hat{x} = \hat{0}$ we have that $x = 0$. On the other hand if we suppose that $x =0$ then $r(0) = \| \hat{0} \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{0}(f)| = \sup_{f \in \Phi_{\mathfrak{A}}} |f(0)| = 0$. So we conclude that $r(x) = 0$ if and only if $x = 0$.
- 2. Showing that $r(\alpha x) = |\alpha| r(x)$: Let $\alpha \in \mathbb{C}$, $x \in \mathfrak{A}$. Then:
\begin{align} \quad r(\alpha x) = \| \alpha x \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{\alpha x} (f)| = \sup_{f \in \Phi_{\mathfrak{A}}} |f(\alpha x)| = |\alpha| \sup_{f \in \Phi_{\mathfrak{A}}} |f(x)| = |\alpha| \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)| = |\alpha| \| x \|_{\infty} = |\alpha| r(x) \end{align}
- 3. Showing that $r(x + y) \leq r(x) + r(y)$: Let $x, y \in \mathfrak{A}$. Then:
\begin{align} \quad \quad r(x + y) = \| x + y \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x + y}(f) | = \sup_{f \in \Phi_{\mathfrak{A}}} |f(x + y)| \leq \sup_{f \in \Phi_{\mathfrak{A}}} |f(x)| + \sup_{f \in \Phi_{\mathfrak{A}}} |f(y)| = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)| + \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{y}(f)| = \| x \|_{\infty} + \| y \|_{\infty} = r(x) + r(y) \end{align}
- 4. Showing that $r(xy) \leq r(x)r(y)$: Let $x, y \in \mathfrak{A}$. Then:
\begin{align} \quad \quad r(xy) = \| xy \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} | \hat{xy}(f)| = \sup_{f \in \Phi_{\mathfrak{A}}} |f(xy)| \leq \sup_{f \in \Phi_{\mathfrak{A}}} |f(x)| \cdot \sup_{f \in \Phi_{\mathfrak{A}}} |f(y)| = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)| \cdot \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{y}(f)| = \| x \|_{\infty} \| y \|_{\infty} = r(x)r(y) \end{align}
- Thus the spectral radius $r$ is an algebra norm on $\mathfrak{A}$.
- $4) \Rightarrow 1)$ Suppose that the spectral radius $r$ is an algebra norm on $\mathfrak{A}$. Then in particular, $r(x) = 0$ if and only if $x = 0$. We have proven already that $\mathrm{Rad}(\mathfrak{A}) = \{ x \in X : r(x) = 0 \}$ at the proposition referenced at the top of this page. Therefore we see that $\mathrm{Rad}(\mathfrak{A}) = \{ 0 \}$, i.e., $\mathfrak{A}$ is semi-simple. $\blacksquare$