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 $\epsilon > 0$ be 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. $\blacksquare$