Gelfand Theory | Itinerarium Mentis

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

The map $A\to C_0(\Omega(A)): a\to \hat{a}$ is a norm-decreasing homomorphism,
$\sigma(a) = \hat{a}(\Omega(A))$,
$r(a) = \|\hat{a}\|_\infty$ where $\|\cdot\|$ is the sup norm,
$a$ is invertible iff $\hat{a}$ is invertible
$\operatorname{rad}(A) = \mathcal{Q}(A) = \operatorname{Ker}(A\to C_0(\Omega(A))$ where $\mathcal{Q}(A)$ is the set of quasinilpotent elements (those with spectrum $\\{0\\}$).

Thus

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 $C_0(\Omega(A))$.
$\hat{a}$ 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.