C star algebra generated by a normal operator

Assume Countable Choice and let $H$ be a nonzero complex Hilbert space with $T\in \mathcal{B}(H)$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators). The unital $\ast$-algebra generated by $T$ is the smallest subset $P(T)\subseteq \mathcal{B}(H)$ containing the identity $I$ and $T$ that is closed under addition, scalar multiplication, multiplication and the adjoint. The C*-algebra generated by $T$ is its norm closure

$C^{\ast }(I,T):={\overline{P(T)}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}\subseteq \mathcal{B}(H)\mathrm{.}$

For an arbitrary $T$, the concrete elements of $P(T)$ are finite linear combinations of finite words in the two letters $T$ and $T^{\ast }$. When $T$ is normal, the two letters commute and every such word can be reordered, so in that case

$P(T)=\{{\sum }_{j,k}{c}_{jk}{T}^j(T^{\ast }{)}^k:\text{only finitely many }{c}_{jk}\mathrm{\ne }0\}\mathrm{.}$

Well-definedness and the algebra structure. The set $P(T)$ is a complex $\ast$-subalgebra of $\mathcal{B}(H)$ containing $I$, and the closure of a $\ast$-subalgebra of a C*-algebra is again a unital $\ast$-subalgebra: sums, products and adjoints of limits are the limits of the corresponding sums, products and adjoints, because the algebra operations and the adjoint are continuous (Hilbert-adjoint identities, Bounded Hilbert operators form a C star algebra, C star algebra). A closed subset of the complete space $\mathcal{B}(H)$ is complete, so $C^{\ast }(I,T)$ with the inherited norm, multiplication, unit and adjoint is itself a unital complex C*-algebra, a unital C*-subalgebra of $\mathcal{B}(H)$ with the same identity $I$.

Commutativity is exactly normality. If $T$ is normal (Self-adjoint, positive, unitary and normal operators), then $T$ commutes with $T^{\ast }$, hence any two words in $T$ and $T^{\ast }$ commute, hence any two $\ast$-polynomials commute, so $P(T)$ is commutative; commutativity passes to the closure because if $A_n{B}_n=B_n{A}_n$ and $A_n\to A$, $B_n\to B$, then $AB=BA$ by continuity of multiplication. Conversely, if $C^{\ast }(I,T)$ is commutative, then the elements $T$ and $T^{\ast }$ of it commute, that is $T{T}^{\ast }=T^{\ast }T$ and $T$ is normal. In particular $C^{\ast }(I,T)$ is a nonzero commutative unital C*-algebra exactly when $T$ is normal, and only that case is fed to the Gelfand theory later on this page.

Minimality convention. $C^{\ast }(I,T)$ is the smallest closed unital $\ast$-subalgebra of $\mathcal{B}(H)$ containing $T$: every closed unital $\ast$-subalgebra contains $P(T)$ and hence its closure. No generator other than $I$ and $T$ is adjoined, and the definition does not presuppose any particular representation of the generated algebra (The operator norm as the least bound and as the unit-sphere or unit-ball supremum for the norm used in the closure).