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.



Dualizing the trivial action on a $C^*$-algebra
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 $C(G)$.


