States given by Dirac measure

Question

Let $G$ be an infinite cyclic group. Then we have $C^*(G)\cong C^*(\mathbb{Z})\cong C(S^1)$. Now let $\delta_{x_0}$ be the state corresponding to the Dirac measure at $x_0 \in S^1$ given by $\delta_{x_0}(f)=f(x_0)$ which is evaluation at $x_0$. How can I get a closed form expression for a state on $C^*(G)$ given by the Dirac delta by the identifications. Theoretically I can write it composing with isomorphism from $C^*(G)$ to $C(S^1)$. But can we make it a more clear form?

Answer 1

The map $\delta_{x_0}$ is even a unital $\ast$-homomorphism, which makes things quite easy. An isomorphism (of course you have to fix one) from $C^\ast(G)$ to $C(S^1)$ is given by sending $u_g$ to $\mathrm{id}_{S^1}$, where $g$ is a generator of $G$. After composition with $\delta_{x_0}$, the element $u_g$ is sent to $x_0$.

In other words, your state is the unique unital $\ast$-homomorphism (given by the unital property of $C^\ast(G)$) that sends $u_g$ to $x_0$.