# Dualizing the trivial action on a $C^*$-algebra

Let $G$ be a finite abelian group (cosidered as a discrete topological group), $A$ a unital separable $C^*$-algebra. Let $T\colon G\to \operatorname{Aut}(A)$, $T_g(a)=a$ for all $g\in G$ the trivial action.

Let $Lt\colon G\to \operatorname{Aut}(C(G))$ the left translation, i.e. $Lt$ is given by $Lt_g(f)(h)=f(g^{-1}h)$ for a continuous function $f\in C(G)$ and $h\in G$.

My question: Is $T\otimes Lt\colon G\to \operatorname{Aut}(A\otimes C(G))$ the dual action of $T$, or is it possible to identify the dual action of $T$ with $T\otimes Lt$? ($\otimes$ here means the minimal tensor product of $C^*$-algebras, but note that $C(G)$ is nuclear)

First of all, the dual action is an action $\hat{T}\colon \hat{G}\to \operatorname{Aut}(A\rtimes_T G)$, defined by $$\hat{T}_\tau(\sum_{s\in G}a_s \delta_s)=\sum_{s\in G}a_s\tau(s) \delta_s$$ for $\tau\in\hat{G}$ and $\sum_sa_s \delta_s\in C_C(G,A)$. $\hat{G}$ denotes the dual group (the Pontryagin dual) of $G$. Since $T$ is the trivial action, there is an isomorphism $$A\rtimes_T G\to A\otimes_{max}C^*(G),$$ induced by the canonical map $\sum_sa_s\cdot \delta_s\mapsto \sum_sa_s\otimes \delta_s$ on the $*$-algebraic tensor products ($C^*(G)$ means the full group $C^*$-algebra of $G$). Since $C^*(G)$ is nuclear, $A\otimes_{max}C^*(G)\cong A\otimes C^*(G)$. Forthermore, there is an isomorphism $$F\colon C^*(G)\to C(\hat{G}),$$ given by the Fourier transform $F(f)(\chi)=\sum\limits_{g\in G}f(g)\chi(g)$, where $f\in C_C(G)$ and $\chi\in \hat{G}$.

All in all, after composing these isomorphisms, we obtain an action $\hat{T}\colon \hat{G}\to \operatorname{Aut}(A\otimes C(\hat{G}))$ which does not look like $T\otimes Lt\colon \hat{G}\to \operatorname{Aut}(A\otimes C(\hat{G}))$ to me, or does it?

The last step is to identify $\hat{G}\cong G$ (but this is not canonically isomorphic). I appreciate your help.

Using the isomorphisms you wrote down, the dual action is the left translation action, composed with the automorphism $g\mapsto g^{-1}$.
To see this, first notice that the $C^*$-algebra $A$ plays no role in any of the definitions, so it can safely be disregarded in the computation. We're thus considering the action of $\widehat{G}$ on the group algebra $C^*(G)=\mathbb{C}[G]$ given by $\tau\cdot \delta_s = \tau(s)\delta_s$ for $\tau\in \widehat{G}$ and $s\in G$. The Fourier/Gelfand transform $F:\mathbb{C}[G]\to C(\widehat{G})$ sends $\delta_s$ to the function $F(\delta_s):\chi\mapsto \chi(s)$. We have $$F(\tau\cdot \delta_s)(\chi) = \tau(s)\chi(s) = F(\delta_s)(\tau\chi),$$ showing that the action $\widehat{G}\rightarrow\operatorname{Aut}(C(\widehat{G}))$ is given by $(\tau\cdot f)(\chi) = f(\tau\chi)$. Of course, the same formula continues to hold if you choose an isomorphism $\widehat{G}\cong G$ and use that same isomorphism to identify $C(\widehat{G})$ with $C(G)$.