Unitary representation of a group of automorphism on an abelian algebra

Question

Given an abelian C*-algebra $\mathcal{A}$, a state $\omega$, a strongly continuous group of *-automorphism $\{\tau_t : t \in \mathcal{R}\}$, and given a representation $ (\pi(\mathcal{A}), \mid \Omega \rangle)$, where $\pi: \mathcal{A} \to \mathcal{B}(H)$ and $H$ is a Hilbert space, such that $\forall a \in \mathcal{A}, \, \omega (a) = \langle \Omega \mid \pi(a) \Omega \rangle$, is it true that $\pi(\tau_t(a)) = U_t \pi(a) U_{t}^{\dagger} \iff \omega(\tau_t(a)) =\omega(a) \, \forall a \in \mathcal{A}, \, t \in \mathcal{R} $ ?

I was trivially thinking that, since the algebra is abelian, then

$\omega(\tau_t(a)) = \langle \Omega \mid U_t \pi(a) U_{t}^{\dagger} \Omega \rangle = \langle \Omega \mid U_t U_{t}^{\dagger}\pi(a) \Omega \rangle = \langle \Omega \mid \pi(a) \Omega \rangle = \omega(a)$,

but I am afraid I'm missing something, so any help is very welcome.

I have already posted this question on MathStackExchange, but without any luck: https://math.stackexchange.com/questions/4790179/covariant-representation-of-abelian-algebra

Answer 1

$U_t$ doesn't belong to the algebra, so you can't assume it commutes with every $\pi(a)$. In fact, no non-trivial automorphism can be implemented by a unitary belonging to the algebra, if the algebra is abelian.

It's so easy to give counterexamples to your question that I think it's better if I leave that as an exercise. Let $X$ be a compact Hausdorff space (could even be finite) with a non-trivial self-homeomorphism $\phi$, then take a nonconstant continuous function $f$ on $X$ and find a state that takes different values on $f$ and $f \circ \phi$.