How to show that $C^{\ast}(\{1, a \} \mid \{ a=a^\ast, ||a|| \leqslant 1\}) \cong C([-1,1])$

Question

To prove that $C^{\ast}(\{1, a \} \mid \{ a=a^\ast, ||a|| \leqslant 1\}) \cong C([-1,1]):$

I was thinking that if I use the map:
$f:1 \to 1, \ \quad \ f:a \to id \ $ then: $$f(\text{Span}(\{a^n \mid n \in \mathbb{N}\})) = \text{Span}(\{z^n: n \in \mathbb{N}\} )$$ Whose completion should be the entire space $C([-1,1]).$

Since the completion of $\text{Span}(\{a^n \mid n \in \mathbb{N}\})$ is precisely $C^{\ast}(\{1, a \} \mid \{ a=a^\ast, ||a|| \leqslant 1\})$, we get an isomorphism: $C^{\ast}(\{1, a \} \mid \{ a=a^\ast, ||a|| \leqslant 1\}) \cong C([-1,1])$

Answer 1

You seem to be ommiting the crucial word, which is "universal". The universal unital C$^*$-algebra generated by a selfadjoint element with norm at most 1 is obtained in the following way:

• first one considers the algebra $A_0$ generated $1$ and $a$, i.e. $\{p(a):\ p\in\mathbb C\}$.

• then one defines $$\|p(a)\|=\sup\{\|\pi(p(a))\|:\ \pi\},$$ where the supremum is taken over all homomorphisms $\pi:A_0\to B(H_\pi)$ for some Hilbert space $H$, where one requires that $\pi(a)=\pi(a)^*$ and $\|\pi(a)\|\leq1$.

• the norm is well-defined because selfadjoint contractions exist, so the set is nonempty. It is also finite, because $\|\pi(p(a))\|\leq\|p\|_1\,$ (where $\|p\|_1$ is the sum of the absolute values of the coefficients).

• the universal unital C$^*$-algebra generated by a selfadjoint element is the C$^*$-algebra $A$ obtained as the completion of $A_0$.

• by definition, any unital C$^*$-algebra generated by a selfadjoint contraction is a quotient of $A$.

• We want to show that $A\simeq C[-1,1]$. As mentioned above, there exists a $*$-homomorphism $\pi:A\to C[-1,1]$, such that $\pi(a)=x$, the identity function. For any $*$-homomorphism we have $\sigma(\pi(a))\subset\sigma(a)$. But in this particular case we have $$ [-1,1]=\sigma(\pi(a))\subset\sigma(a)\subset[-1,1]. $$ So $\sigma(a)=[-1,1]$. More generally, given any $f\in C[-1,1]$, we have $$ f([-1,1])=f(\sigma(\pi(a)))=\sigma(f(\pi(a))=\sigma(\pi(f(a))\subset\sigma(f(a))=f(\sigma(a))\subset f([-1,1]). $$ This shows that $\sigma(f(a))=f([-1,1])$ for all $f\in C[-1,1]$. In particular, $$ \|\pi(f(a))\|=\operatorname{spr}(\pi(f(a))=\operatorname{spr}(f(a))=\|f(a)\|. $$ Thus $\pi$ is an isometry, which in particular means that it is injective. As $\pi(f(a))=f$, we also have that $\pi$ is surjective, and so $A\simeq C[-1,1]$.

Answer 2

Here is an approach relying on the universal property instead of the construction of the universal $C^\ast$-algebra. The universal $C^\ast$-algebra $C^\ast(1,a\mid a=a^\ast,\,\|a\|\leq 1)$ is uniquely (up to unital $\ast$-isomorphism) determined by the property that whenever $B$ is a unital $C^\ast$-algebra and $b\in B$ a self-adjoint contraction, then there exists a unique unital $\ast$-homomorphism $\phi\colon A\to B$ with $\phi(a)=b$.

That $C([-1,1])$ with $a=\mathrm{id}_{[-1,1]}$ has this universal property is essentially the statement of continuous functional calculus: If $b\in B$ is a self-adjoint contraction, then $\sigma(b)\subset [-1,1]$. The restriction $f\mapsto f|_{\sigma(b)}$ is a unital $\ast$-homomorphism from $C([-1,1])$ to $C(\sigma(b))$. Composed with $$ C(\sigma(b))\to B,\,g\mapsto g(b) $$ we get a unital $\ast$-homorphism from $C([-1,1])$ to $C(\sigma(b))$ that maps $\mathrm {id}_{[-1,1]}$ to $b$. That it's the unique unital $\ast$-homomorphism follows from the density of polynomials in $C([-1,1])$.