Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 4 / c

Let $X$ be the character space of an algebra, consisting of nonzero multiplicative complex linear functionals on $A$. We construct the required Gelfand transform and prove its properties.

Characters exist. A maximal proper ideal exists by the axiom of choice and is closed: if it were dense it would contain an element within distance one of the identity, hence an invertible element, contradicting properness. The quotient is a complex unital Banach division algebra, and the Gelfand-Mazur theorem identifies it with $\mathbb{C}$, producing a character. The same argument applied to the proper ideal generated by $a-\lambda$ provesEvery character has norm one, since its value belongs to the spectrum and is bounded by $\Vert a\Vert$, while $\chi (1)=1$. The character equations are weak-star closed in the dual unit ball. Thus Banach-Alaoglu theorem makes $X$ compact Hausdorff; separability of $A$ makes it metrizable.

The algebra is inverse-closed in $B(H)$. If $a\in A$ is invertible there, $a^{\ast }a\ge cI$ for some $c>0$. For $M=\Vert a^{\ast }a\Vert$, the Neumann series for $(a^{\ast }a{)}^{-1}=M^{-1}[I-(I-a^{\ast }a/M){]}^{-1}$ converges in $A$, since the positive contraction in brackets has norm at most $1-c/M<1$. Therefore $a^{-1}=(a^{\ast }a{)}^{-1}{a}^{\ast }\in A$. This proves spectral permanence for C-star algebras in the present concrete setting. In particular a self-adjoint element has real spectrum in $A$, so characters take real values on it. Decomposing $a$ into self-adjoint real and imaginary parts gives $\chi (a^{\ast })=\overline{\chi (a)}$.

Define $\hat{a}(\chi )=\chi (a)$. This is a unital star-homomorphism to $C(X)$. Every element of the abelian star-algebra is normal, and the C-star identity gives $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral radius formula therefore gives $r_A(a)=\Vert a\Vert$. The displayed spectrum identity now yields $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$, proving isometry and injectivity. Its image is closed because $A$ is complete, contains the constants, is closed under conjugation, and separates distinct characters. The Stone-Weierstrass theorem makes that image dense in $C(X)$, hence equal to it. $\boxed{A\cong C(X)}$ isometrically as a star-algebra.