# The Gelfand Representation of a Commutative Banach Algebra

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

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

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

*Note that for each $x \in X$ and each $f \in \Phi_X$, $\hat{x}(f)$ simply evaluates $f$ at $x$.*

Theorem 1: Let $X$ be a commutative Banach algebra over $\mathbb{C}$ WITH unit and let $\Phi_X$ be the carrier space of $X$ with Gelfand representation $x \to \hat{x}$. Then:a) The carrier space $\Phi_X$ is a compact Hausdorff space.b) The Gelfand representation $x \to \hat{x}$ is a homomorphism of $X$ to $C(\Phi_X)$.c) For each $x \in X$, the range $\hat{x}(\Phi_X) = \mathrm{Sp}(X, x)$.d) For each $x \in X$, $r(x) = \| \hat{x} \|_{\infty}$.e) $x \in \mathrm{Sing}(X)$ if and only if $\hat{x}(f) = 0$ for some $f \in \Phi_X$. |

**Proof of a)**By the proposition referenced at the top of this page we have that since $X$ is a commutative Banach algebra with unit, $\Phi_X$ is compact in the $X$-topology and Hausdorff. $\blacksquare$

**Proof of b)**It is clear that the map $x \to \hat{x}$ is linear. Let $x, y \in X$. We want to show that $\hat{xy} = \hat{x}\hat{y}$. Let $f \in \Phi_X$. 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 $\Phi_X$, so they are equal and so the Gelfand representation $x \to \hat{x}$ is a homomorphism of $X$ to $C(\Phi_X)$. $\blacksquare$

**Proof of c)**Let $x \in X$. Recall from the theorem on The Spectrum of an Element in a Commutative Banach Algebra page that if [$X$ is a commutative Banach algebra with unit and $\lambda \in \mathbb{C}$ then $\lambda \in \mathrm{Sp}(X, x)$ if and only if $\lambda = f(x)$ for some $f \in \Phi_X$. Thus:

**Proof of d)**Let $x \in X$. Recall from The Spectrum of an Element in a Normed or Banach Algebra page that if $X$ is a Banach algebra then $\mathrm{Sp}(X, x)$ is a nonempty compact subset of $\mathbb{C}$ and moreover, $r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(X, x) \}$. Using this and the theorem referenced in part (c) for commutative Banach algebras, we have that:

**Proof of e)**Since $X$ is an algebra with unit we have by definition that $x - \lambda \in \mathrm{Sing}(X)$ if and only if $\lambda \in \mathrm{Sp}(X, x)$. We use this in the proof below.

- $\Rightarrow$ Let $x \in \mathrm{Sing}(X)$. Then $0 \in \mathrm{Sp}(X, x)$. So there exists an $f \in \Phi_X$ such that $f(x) = 0$. So [[$ \hat{x}(f) = 0

- $\Leftarrow$ Suppose that there exists an $f \in \Phi_X$ with $\hat{x}(f) = 0$. Then $f(x) = 0$ where $f \in \Phi_X$, implying that $0 \in \mathrm{Sp}(X, x)$, and so $x \in \mathrm{Sing}(X)$. $\blacksquare$

Theorem 2: Let $X$ be a commutative Banach algebra over $\mathbb{C}$ WITHOUT unit and let $\Phi_X$ be the carrier space of $X$ with Gelfand representation $x \to \hat{x}$. Then:a) The carrier space $\Phi_X$ is a locally compact Hausdorff space.b) The Gelfand representation $x \to \hat{x}$ is a homomorphism of $X$ to $C_0(\Phi_X)$.c) For each $x \in X$, $\hat{x} (\Phi_X) \setminus \{ 0\} = \mathrm{Sp}(X, x) \setminus \{ 0 \}$.d) For each $x \in X$, $r(x) = \| \hat{x} \|_{\infty}$.e) $x \in \mathrm{q-Sing}(X)$ if and only if $\hat{x}(f) = 1$ for some $f \in \Phi_X$. |

*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 $X$ is a commutative Banach algebra that $\Phi_X$ is locally compact and Hausdorff. $\blacksquare$

**Proof of b)**Let $x \in X$. We claim that $\hat{x} \in C_0(\Phi_X)$. Let $\epsilon > 0$ be given and let:

- Note that $\Phi_X^{\infty} \setminus N(\epsilon, x) = \{ f \in \Phi_X : |f(x)| < \epsilon \}$ is an open subset of $\Phi_X^{\infty}$ and so $N(\epsilon, x)$ is a closed subset of $\Phi_X^{\infty}$. But $\Phi_X^{\infty}$ is compact, so $N(\epsilon, x)$ is a closed subset of a compact space $\Phi_X^{\infty}$ and does not contain the zero functional and is hence compact. Thus $N(\epsilon, x) = \{ f \in \Phi_X : |\hat{x}(f)| \geq \epsilon \}$ is compact, so $\hat{x} \in C_0(\Phi_X)$.

- So the range of $\Phi_X$ under the Gelfand representation $x \to \hat{x}$ is contained in $C_0(\Phi_X)$. 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 $X$ to $C_0(\Phi_X)$. $\blacksquare$

**Proof of c)**From the theorem referenced in theorem 1 (c) above, we have that for every $\lambda \in \mathbb{C} \setminus \{ 0 \}$ that $\lambda \in \in \mathrm{Sp}(X, x)$ if and only if $\lambda = f(x)$ for some $f \in \Phi_X$. So:

**Proof of d)**Analogous to theorem 1 (d). $\blacksquare$

**Proof of e)**Since $X$ is an algebra without unit we have that $\mathrm{Sp}(X, x) = \{ 0 \} \cup \left \{ \lambda \in \mathbb{C} \setminus 0 : \frac{1}{\lambda} x \in \mathrm{q-Sing}(X) \right \}$. We use this in the proof below.

- $\Rightarrow$ Suppose $x \in \mathrm{q-Sing}(X)$. Then $1 \in \mathrm{Sp}(X, x)$. Let $J = \{ xy - y : y \in X \}$. Since $X$ is commutative it is easy to verify that $XJ \subseteq J$ and $JX \subseteq J$ so that $J$ is a two-sided ideal of $X$. Furthermore, it is a proper two-sided ideal since $x \neq J$, otherwise, if $x = xy - y$ for some $y \in X$ 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 $X - xX \subseteq J$ and $X - Xx \subseteq J$. So $J$ is the kernel of some $f \in \Phi_X$, i.e., $x \not \in \ker (f)$ and so $f(x) \neq 0$. Since $1 \in \mathrm{Sp}(X, x)$ we have that $f(x) = 1$.

- The converse is analogous. $\blacksquare$