The Gelfand Representation of a Commutative Banach Algebra

# The Gelfand Representation of a Commutative Banach Algebra

Recall from The Carrier Space of a Commutative Banach Algebra page that if $\mathfrak{A}$ is a commutative Banach algebra, $\Phi_{\mathfrak{A}}$ denotes the set of multiplicative linear functionals on $\mathfrak{A}$, and $\Phi_{\mathfrak{A}}^{\infty}$ denotes the set of multiplicative linear functionals on $\mathfrak{A}$ and contains the zero functional then $\Phi_{\mathfrak{A}} \subset \Phi_{\mathfrak{A}}^{\infty} \subseteq \mathfrak{A}^*$.

We equip $\mathfrak{A}^*$ with the weak-* topology and then equip both $\Phi_{\mathfrak{A}}$ and $\Phi_{\mathfrak{A}}^{\infty}$ with the corresponding subspace topology which we call the $\mathfrak{A}$-topology on $\Phi_{\mathfrak{A}}$ (or $\Phi_{\mathfrak{A}}^{\infty}$. We then call $\Phi_{\mathfrak{A}}$ with this topology the carrier space of $\mathfrak{A}$.

 Definition: Let $\mathfrak{A}$ be a commutative Banach algebra. The Gelfand Representation of $\mathfrak{A}$ is the mapping from $\mathfrak{A}$ to $C(\Phi_{\mathfrak{A}})$ defined by $x \to \hat{x}$ where for each $x \in \mathfrak{A}$, $\hat{x} : \Phi_{\mathfrak{A}} \to \mathbb{C}$ is defined for all $f \in \Phi_{\mathfrak{A}}$ by $\hat{x}(f) = f(x)$.

Note that for each $x \in \mathfrak{A}$ and each $f \in \Phi_{\mathfrak{A}}$, $\hat{x}(f)$ simply evaluates $f$ at $x$.

 Theorem 1: Let $\mathfrak{A}$ be a commutative Banach algebra over $\mathbb{C}$ WITH unit and let $\Phi_{\mathfrak{A}}$ be the carrier space of $\mathfrak{A}$ with Gelfand representation $x \to \hat{x}$. Then: a) The carrier space $\Phi_{\mathfrak{A}}$ is a compact Hausdorff space. b) The Gelfand representation $x \to \hat{x}$ is a homomorphism of $\mathfrak{A}$ to $C(\Phi_{\mathfrak{A}})$. c) For each $x \in \mathfrak{A}$, the range $\hat{x}(\Phi_{\mathfrak{A}}) = \mathrm{Sp}(\mathfrak{A}, x)$. d) For each $x \in \mathfrak{A}$, $r(x) = \| \hat{x} \|_{\infty}$. e) $x \in \mathrm{Sing}(\mathfrak{A})$ if and only if $\hat{x}(f) = 0$ for some $f \in \Phi_{\mathfrak{A}}$.
• Proof of a) By the proposition referenced at the top of this page we have that since $\mathfrak{A}$ is a commutative Banach algebra with unit, $\Phi_{\mathfrak{A}}$ is compact in the $\mathfrak{A}$-topology and Hausdorff. $\blacksquare$
• Proof of b) It is clear that the map $x \to \hat{x}$ is linear. Let $x, y \in \mathfrak{A}$. We want to show that $\hat{xy} = \hat{x}\hat{y}$. Let $f \in \Phi_{\mathfrak{A}}$. Then since $f$ is multiplicative we have that $\hat{xy}(f) = f(xy) = f(x)f(y) = \hat{x}(f)\hat{y}(f)$. So $\hat{xy}$ and $\hat{x}\hat{y}$ agree on $\Phi_{\mathfrak{A}}$, so they are equal and so the Gelfand representation $x \to \hat{x}$ is a homomorphism of $\mathfrak{A}$ to $C(\Phi_{\mathfrak{A}})$. $\blacksquare$
• Proof of c) Let $x \in \mathfrak{A}$. Recall from the theorem on The Spectrum of an Element in a Commutative Banach Algebra over C page that if [$\mathfrak{A}$ is a commutative Banach algebra over $\mathbb{C}$ with unit and $\lambda \in \mathbb{C}$ then $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$ if and only if $\lambda = f(x)$ for some $f \in \Phi_{\mathfrak{A}}$. Thus:
(1)
\begin{align} \quad \hat{x}(\Phi_{\mathfrak{A}}) = \{ f(x) : f \in \Phi_{\mathfrak{A}} \} = \mathrm{Sp}(\mathfrak{A}, x) \quad \blacksquare \end{align}
• Proof of d) Let $x \in \mathfrak{A}$. Recall from The Spectrum of an Element in a Normed or Banach Algebra over C page that if $\mathfrak{A}$ is a Banach algebra then $\mathrm{Sp}(\mathfrak{A}, x)$ is a nonempty compact subset of $\mathbb{C}$ and moreover, $r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, x) \}$. Using this and the theorem referenced in part (c) for commutative Banach algebras, we have that:
(2)
\begin{align} \quad r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, x) \} = \max \{ |\lambda| : \lambda = f(x), \: f \in \Phi_{\mathfrak{A}} \} = \sup_{f \in \Phi_{\mathfrak{A}}} \{ |f(x)| \} = \sup_{f \in \Phi_{\mathfrak{A}}} \{ |\hat{x}(f)| \} = \| \hat{x} \|_{\infty} \end{align}
• Proof of e) Since $\mathfrak{A}$ is an algebra with unit we have by definition that $x - \lambda \in \mathrm{Sing}(\mathfrak{A})$ if and only if $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$. We use this in the proof below.
• $\Rightarrow$ Let $x \in \mathrm{Sing}(\mathfrak{A})$. Then $0 \in \mathrm{Sp}(\mathfrak{A}, x)$. So there exists an $f \in \Phi_{\mathfrak{A}}$ such that $f(x) = 0$. So [[$\hat{x}(f) = 0 •$\Leftarrow$Suppose that there exists an$f \in \Phi_{\mathfrak{A}}$with$\hat{x}(f) = 0$. Then$f(x) = 0$where$f \in \Phi_{\mathfrak{A}}$, implying that$0 \in \mathrm{Sp}(\mathfrak{A}, x)$, and so$x \in \mathrm{Sing}(\mathfrak{A})$.$\blacksquare$ Theorem 2: Let$\mathfrak{A}$be a commutative Banach algebra over$\mathbb{C}$WITHOUT unit and let$\Phi_{\mathfrak{A}}$be the carrier space of$\mathfrak{A}$with Gelfand representation$x \to \hat{x}$. Then: a) The carrier space$\Phi_{\mathfrak{A}}$is a locally compact Hausdorff space. b) The Gelfand representation$x \to \hat{x}$is a homomorphism of$\mathfrak{A}$to$C_0(\Phi_{\mathfrak{A}})$. c) For each$x \in \mathfrak{A}$,$\hat{x} (\Phi_{\mathfrak{A}}) \setminus \{ 0\} = \mathrm{Sp}(\mathfrak{A}, x) \setminus \{ 0 \}$. d) For each$x \in \mathfrak{A}$,$r(x) = \| \hat{x} \|_{\infty}$. e)$x \in \mathrm{q-Sing}(\mathfrak{A})$if and only if$\hat{x}(f) = 1$for some$f \in \Phi_{\mathfrak{A}}$. If$A$is a locally compact Hausdorff topological space, then$C_0(A)$is the algebra of all continuous complex-valued functions$f$on$A$such that$f$vanishes at infinity. That is,$f \in C_0(E)$if and only if for all$\epsilon > 0$we have that$\{ x \in E : |f(x)| \geq \epsilon \}$is compact in$E$. If$E$is a compact space then$C_0(E) = C(E)$. • Proof of a) By the proposition represented at top of this page we have that since$\mathfrak{A}$is a commutative Banach algebra that$\Phi_{\mathfrak{A}}$is locally compact and Hausdorff.$\blacksquare$• Proof of b) Let$x \in \mathfrak{A}$. We claim that$\hat{x} \in C_0(\Phi_{\mathfrak{A}})$. Let$\epsilon > 0be given and let: (3) \begin{align} \quad N(\epsilon, x) = \{ f \in \Phi_{\mathfrak{A}} : |f(x)| \geq \epsilon \} \end{align} • Note that\Phi_{\mathfrak{A}}^{\infty} \setminus N(\epsilon, x) = \{ f \in \Phi_{\mathfrak{A}} : |f(x)| < \epsilon \}$is an open subset of$\Phi_{\mathfrak{A}}^{\infty}$and so$N(\epsilon, x)$is a closed subset of$\Phi_{\mathfrak{A}}^{\infty}$. But$\Phi_{\mathfrak{A}}^{\infty}$is compact, so$N(\epsilon, x)$is a closed subset of a compact space$\Phi_{\mathfrak{A}}^{\infty}$and does not contain the zero functional and is hence compact. Thus$N(\epsilon, x) = \{ f \in \Phi_{\mathfrak{A}} : |\hat{x}(f)| \geq \epsilon \}$is compact, so$\hat{x} \in C_0(\Phi_{\mathfrak{A}})$. • So the range of$\Phi_{\mathfrak{A}}$under the Gelfand representation$x \to \hat{x}$is contained in$C_0(\Phi_{\mathfrak{A}})$. Proving that$x \to \hat{x}$is a homomorphism was shown in Theorem 1, and thus, the Gelfand representation$x \to \hat{x}$is a homomorphism of$\mathfrak{A}$to$C_0(\Phi_{\mathfrak{A}})$.$\blacksquare$• Proof of c) From the result referenced in Theorem 1 (c) above, we have that for every$\lambda \in \mathbb{C} \setminus \{ 0 \}$that$\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$if and only if$\lambda = f(x)$for some$f \in \Phi_{\mathfrak{A}}. So: (4) \begin{align} \quad \hat{x}(\Phi_{\mathfrak{A}}) \setminus \{ 0 \} = \{ \hat{x}(f) : f \in \Phi_{\mathfrak{A}} \} \setminus \{ 0 \} = \{ f(x) : f \in \Phi_{\mathfrak{A}} \} \setminus \{ 0 \} = \mathrm{Sp}(\mathfrak{A}, x) \setminus \{ 0 \} \quad \blacksquare \end{align} • Proof of d) Analogous to Theorem 1 (d).\blacksquare$• Proof of e) Since$\mathfrak{A}$is an algebra without unit we have that$\mathrm{Sp}(\mathfrak{A}, x) = \{ 0 \} \cup \left \{ \lambda \in \mathbb{C} \setminus 0 : \frac{1}{\lambda} x \in \mathrm{q-Sing}(\mathfrak{A}) \right \}$. We use this in the proof below. •$\Rightarrow$Suppose$x \in \mathrm{q-Sing}(\mathfrak{A})$. Then$1 \in \mathrm{Sp}(\mathfrak{A}, x)$. Let$J = \{ xy - y : y \in \mathfrak{A} \}$. Since$\mathfrak{A}$is commutative it is easy to verify that$\mathfrak{A}J \subseteq J$and$J\mathfrak{A} \subseteq J$so that$J$is a two-sided ideal of$\mathfrak{A}$. Furthermore, it is a proper two-sided ideal since$x \neq J$, otherwise, if$x = xy - y$for some$y \in \mathfrak{A}$then rearranging this equation gives us that$x + y - xy = 0$, i.e.,$x \circ y = 0 = y \circ x$(again using the commutativity hypothesis) so that$x$is quasi-invertible, a contradiction. Lastly, it is a proper modular two-sided ideal with modular unit$x$since$(1 - x)\mathfrak{A} \subseteq J$and$\mathfrak{A}(1 - x) \subseteq J$. So$J$is the kernel of some$f \in \Phi_{\mathfrak{A}}$, i.e.,$x \not \in \ker (f)$and so$f(x) \neq 0$. Since$1 \in \mathrm{Sp}(\mathfrak{A}, x)$we have that$f(x) = 1$. • The converse is analogous.$\blacksquare\$