C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative

Theorem

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

Let $S \subseteq A$ be a self-adjoint set such that:

$x y = y x$ for each $x, y \in S$.

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

Then $\map {\text C^\ast} S$ is commutative.

Proof

Let $C$ be the subalgebra generated by $S$.

From Subalgebra Generated by Commuting Elements is Commutative, $C$ is commutative.

From Explicit Form for Generated C*-Algebra, we have $\map {\text C^\ast} S = C^-$.

From Closure of Commutative Set in Banach Algebra is Commutative, $C^-$ is commutative.

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

$\blacksquare$