Commutative Gelfand–Naimark theorem

Remarks

If $A$ has a unit, $\Delta(A)$ is compact and $A \cong C(\Delta(A))$. Proof idea: a character takes real values on self-adjoint elements, so $\Gamma(a^*) = \overline{\Gamma(a)}$. For self-adjoint $a$ the C* identity gives $\norm{a^{2^k}} = \norm{a}^{2^k}$, so by Gelfand's spectral radius formula $\norm{\Gamma(a)}_\infty = \norm{a}$; applying this to $a^* a$ shows that $\Gamma$ is isometric for all $a$. So the range of $\Gamma$ is a closed $*$-subalgebra that separates points and vanishes at no point, and by the Stone–Weierstrass theorem it is all of $C_0(\Delta(A))$.

Example: for $A = C(X)$ with $X$ compact Hausdorff, $\Delta(A)$ is homeomorphic to $X$ through the evaluations, and $\Gamma$ is the identity map up to this homeomorphism.

In the unital case, the theorem gives a contravariant equivalence between commutative unital C*-algebras with unital $*$-homomorphisms and compact Hausdorff spaces with continuous maps.

Going deeper

For a student

This theorem is the starting point of "noncommutative topology": a general C*-algebra is thought of as the functions on a space that does not exist. It gives the continuous functional calculus. For a normal element $a$, the C*-algebra generated by $a$ and $1$ is commutative, and its character space is the spectrum of $a$. So it is isomorphic to $C(\sigma(a))$. The noncommutative Gelfand–Naimark theorem says that every C*-algebra is isometrically $*$-isomorphic to a closed $*$-subalgebra of the bounded operators on a Hilbert space.