# Gelfand Theory

In the words of Guillemin and Sternberg, the key idea in Gelfand theory can be summarized as

Maximal ideals are the underlying ‘points’ of a commutative normed ring.

Gelfand theory can be seen as the generalization of Fourier theory, in that for commutative group algebras the Gelfand transform coincides with the Fourier transform.

## Characters

Similar to the character of a group, the character for an algebra $A$ is, roughly speaking, an approximation of representations of $A$, and the character space (also called spectrum) for $A$ the archetype of the unitary dual of a $C^\ast$-algebra.

Definition. Let $A$ be an algebra. A character on $A$ is a non-zero homomorphism $\phi:A\to \mathbb{C}$.

These are also called multiplicative linear functionals. The set $\Omega(A)$ of all characters on $A$ and call it character space of $A$. An example for a character is $\operatorname{ev}_{x_0}\in \Phi_{C(\Omega)}$ where $\operatorname{ev}_{x_0} f = f(x_0)$ for $f\in A$.

Proposition. For unital algebra $A$, there is a bijection between $\Omega(A)$ and the set of maximal ideals of codimension $1$ in $A$, via $\phi\mapsto \operatorname{Ker}(\phi)$.
Proof:  Clearly for each $\phi\in\Omega(A)$, $\operatorname{Ker}(\phi)$ is a maximal ideal of codimension $1$. The map is injective, since for $\operatorname{Ker}(\phi) = \operatorname{Ker}(\psi)$, $\phi(a-\psi(a)) = 0$ (notice that $a-\psi(a) \in \operatorname{Ker}(\psi)$ by linearity) for all $a\in A$ and hence $\phi = \psi$. The map is surjective, since given a maximal ideal $M$ of codimension 1, the quotient $\phi: A\to A/M = \mathbb{C}$ is a character with $\operatorname{Ker}(\phi) = M$.
If $A$ is a commutative Banach algebra, all the maximal ideals of $A$ has codimension $1$, thus $\Omega(A)$ is also the set of maximal ideals of $A$: For $M$ a maximal ideal, $A/M$ is a field for commutative algebra $A$; together with the fact that $A/M$ is a Banach algebra since $M$ is closed, by the Gelfand-Mazur theorem $A/M = \mathbb{C}$.

For unital Banach algebras, the character space $\Omega(A)$ can be endowed with a weak $\ast$-topology, Gelfand topology, and be made into a compact topological space, also called the character space, or the spectrum of the algebra $A$, which is also denoted by $\Omega(A)$. This is possible by the following theorem

Theorem. Let $A$ be a unital Banach algebra and $\phi\in\Omega(A)$, then $\phi$ is continuous and $\|\phi\|=1$.
Proof:  $\phi(a)\in \sigma(a)$ where $\sigma(a)$ is the spectrum of $a\in A$. This means $|\phi(a)|\leq r(a) \leq \|a\|$ where $r(a)$ is the spectral radius, so $\|\phi\|\leq 1$ since $\|a\| \leq 1$ (recall that $||a|| = \lim_{n\to \infty}\|a^n\|^{1/n}$). Now $\phi(1) = \phi(1)^2$ means $\phi(1)=1$, so $\|\phi\| = 1$.
$\Omega(A)$ is thus a subset of the space $\{\lambda\in A’ | \lambda(1_A) = 1 = |\lambda|\}$, called the state space of $A$ ($A’$ is the space of linear functions on $A$, not the commutant). Endow the space $A’$ with the weak $\ast$-topology (the topology defined by $\lambda_\nu \to \lambda$ iff $\lambda_\nu(a) \to \lambda(a)$ for each $a\in A$), the Gelfand topology on $\Omega(A)$ is the relative weak $\ast$-topology from $A’$. $\Omega(A)$ is compact in this topology.

## Gelfand transform/representation

Now for genereal unital Banach algebra $A$ the spectrum $\Omega(A)$ might be empty, but for commutative Banach algebras the Gelfand representation theorem holds, of which a corollary is that any semisimple commutative unital Banach algebra is a subalgebra of $C(\Omega(A)$.

For $a$ in $A$ a commutative Banach algebra, let $\hat{a}$ denote the Gelfand transform of $a$, defined on $\Omega(A)$ by $$\hat{a}:\Omega(A) \to \mathbb{C}: \phi\mapsto\phi(a).$$ The topology on $\Omega(A)$ is the smallest one making all the $\hat{a}$ continuous. In fact $\hat{a}\in C_0(\Omega(A))$.

Theorem(Gelfand representation) . Let $A$ be a commutative unital Banach algebra, then
1. The map $A\to C_0(\Omega(A)): a\to \hat{a}$ is a norm-decreasing homomorphism,
2. $\sigma(a) = \hat{a}(\Omega(A))$,
3. $r(a) = \|\hat{a}\|_\infty$ where $\|\cdot\|$ is the sup norm,
4. $a$ is invertible iff $\hat{a}$ is invertible
5. $\operatorname{rad}(A) = \mathcal{Q}(A) = \operatorname{Ker}(A\to C_0(\Omega(A))$ where $\mathcal{Q}(A)$ is the set of quasinilpotent elements (those with spectrum $\\{0\\}$).
Thus

1. if $A$ is semisimple, $\operatorname{Ker}(A\to C_(\Omega(A)))$ is trivial, $A\to C_(\Omega(A))$ is injective, making $A$ into a subalgebra of $C_0(\Omega(A))$.
2. $\hat{a}$ should really be seen as $\hat{a}: \Omega(A) \to \sigma(a)$.

## $C^\ast$-algebras

The Gelfand transform embeds $A$ in $C_0(\Omega(A))$ for $A$ a commutative unital semisimple Banach algebra. This is a surjection for $C^\ast$-algebras.

Theorem(Gelfand-Naimark) . Let $A$ be a commutative unital $C^\ast$-algebra, then the Gelfand transform $a\mapsto \hat{a}$ is an isometric $\ast$-isomorphism.

This is a special case for the general Gelfand-Naimark theorem, which says that every $C^\ast$-algebra is isometrically $\ast$-isomorphic to a $C^\ast$-subalgebra of bounded operators on a Hilbert space.