# Unitary representation of a group of automorphism on an abelian algebra

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

$$U_t$$ 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 $$f \circ \phi$$.
• Thank you @NikWeaver for your reply! With a general C*-algebra, how to understand if $U_t$ is an element or not of the algebra? Because, if we for example look at CCR algebra with finite degrees of freedom, the automorphism is generally representated as unitary operator where the unitary operator is an element of the algebra. Or am I missing something? Commented Oct 25, 2023 at 17:12
• If $A$ is abelian then the only way for all automorphisms to be inner is for there to be no non-identity automorphisms. If $A=C(X)$ then this happens when $X$ has no non-trivial self-homeomorphisms. Commented Oct 26, 2023 at 9:10
• The first place I'd go is $C^*$-Algebras and Their Automorphism Groups by Pedersen. Commented Oct 26, 2023 at 15:36