Explicit isomorpshim between $C^*(\mathbb{Z})$ and $C(S^1)$

Question

I know that for an abelian group $G$, $C^*(G)$ is isomorphic to $C(\hat{G})$. Hence we have $C^*(\mathbb{Z})$ isomorphic to $C(\hat{\mathbb{Z}})$ which is inturn isomorphic to $C(S^1)$. Can we have an explicit isomorphism between $C^*(\mathbb{Z})$ and $C(S^1)$? We have $C^*(\mathbb{Z})$ is a universal $C^*$ generated by a unitary $u$. Mapping $u$ to the function $z \mapsto z$ will it be an isomorphism?

Answer 1

I will give a proof that avoids Fourier theory, as the usual isomorphism $C(\hat{G}) \cong C^*(G)$ for $G$ an abelian discrete group (so that $\hat{G}$ is a compact space) is simply a consequence of Fourier theory.

You already established that $$A:= C^*(u,1\mid u^*u = uu^* = 1)\cong C^*(\mathbb{Z})$$ or in other words that $C^*(\mathbb{Z})$ is the universal $C^*$-algebra generated by a unitary.

I will prove that $A \cong C(S^1)$ and you then obtain the desired isomorphism.

Indeed, let $z: S^1 \hookrightarrow \mathbb{C}$ be the inclusion map. Then $z$ is a unitary in $S^1$ so by the universal property of the universal $C^*$-algebra, there is a unique unital $*$-morphism $\sigma: A \to C(S^1)$ such that $\sigma(u)= z$. The functional calculus yields a unital isometric $*$-isomorphism $C(\sigma_A(u)) \to A$ such that the identity on $\sigma_A(u)$ is mapped to $u$. Moreover, since $u$ is a unitary in $A$ we have an inclusion map $\sigma_A(u) \hookrightarrow S^1$ which induces a unital $*$-morphism $C(S^1) \to C(\sigma_A(u))$. Composing with the functional calculus map, we obtain a unital $*$-morphism $\tau: C(S^1) \to A$ and it is readily verified that $\sigma$ and $\tau$ are inverse of each other, as they are each other inverse on the generators. Thus $\sigma$ and $\tau$ are isomorphisms.

In particular, the canonical isomorphism $$C(S^1) \cong A \cong C^*(\mathbb{Z})$$ is given by sending the identity function $z$ to $u$ and $1$ to $1$, as you correctly guessed.