Explicit Form for Generated C*-Algebra

Theorem

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.

Let $S \subseteq A$.

Define:

$S^\ast = \set {x^\ast : x \in A}$

Let $\map {\text C^\ast} S$ be the $\text C^\ast$-algebra generated by $S$.

Then:

$\ds \map {\text C^\ast} S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S \cup S^\ast, \, k_1, \ldots, k_n \ge 0}^-$

where $\paren {\cdot}^-$ denotes closure in $\struct {A, \norm {\, \cdot \,} }$.

Proof

Let $C$ be the subalgebra generated by $S \cup S^\ast$.

From Explicit Form for Generated Subalgebra, we have:

$\span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$

We show that $\map {\text C^\ast} S = C^-$.

Note that $S \cup S^\ast$ is self-adjoint.

From Subalgebra Generated by Self-Adjoint Set is Self-Adjoint, $C$ is self-adjoint.

From Closure of Self-Adjoint Subset of Banach *-Algebra is Self-Adjoint, $C^-$ is self-adjoint hence a $\ast$-subalgebra of $A$.

Hence from Closed *-Subalgebra of C*-Algebra is C*-Algebra, $C^-$ is a $\text C^\ast$-subalgebra of $A$ containing $S \cup S^\ast$.

So $\map {\text C^\ast} S \subseteq C^-$.

To conclude we just need to show that $C^- \subseteq \map {\text C^\ast} S$.

By definition, $C$ is the $\subseteq$-minimal subalgebra with $S \subseteq C$.

Hence $C \subseteq \map {\text C^\ast} S$.

So $C^- \subseteq \map {\text C^\ast} S^-$ from Set Closure Preserves Set Inclusion.

From Subspace of Complete Metric Space is Closed iff Complete, we have that $\map {\text C^\ast} S$ is closed in $A$.

So from Set is Closed iff Equals Topological Closure we have $C^- \subseteq \map {\text C^\ast} S$.

Hence:

$C^- = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}^- = \map {\text C^\ast} S$

$\blacksquare$