Let $X$ be a compact Hausdorff space. It is well-known that every homomorphism $F : \mathcal{C}(X) \to \mathbb{R}$ is the evaluation $f \mapsto f(x)$ at some point $x \in X$. The usual proof is not really constructive, but for $X=[0,1]$ there is a constructive one. For details see my crosspost on math.SE. Feel free to replace $\mathbb{R}$ by $\mathbb{C}$.

**Question.** Is there an explicit example of $X$ and $F : \mathcal{C}(X) \to \mathbb{R}$ as above such that 1) syntactically $F$ is not defined as an evaluation, 2) one does not see directly that $F$ is an evaluation, 3) some computation has to be done to find the point $x \in X$ such that $F$ is the evaluation at $x$?

Background: Gelfand duality states that the adjunction between $\mathrm{Spec}$ and $\mathcal{C}$ is actually an equivalence, which means that A) for every compact Hausdorff space $X$ the unit $X \to \mathrm{Spec}(\mathcal{C}(X))$ is an isomorphism and B) for every commutative unital $C^*$-algebra $A$ the counit $A \to \mathcal{C}(\mathrm{Spec}(A))$ is an isomorphism. There are many important applications for B), for example the existence of the functional calculus, but I don't know of any specific application for A) (as a result independent from this duality). It would be nice to have some computational example which shows the relevance of A). Actually I'm after another duality, where A) is already proven but its significance is unclear.

isan answer?) $\endgroup$ – Qfwfq Jan 21 '12 at 13:35