Crossed product of a rational rotation on $S^1$ Let $n \in \mathbb{N}$, and consider the action $\alpha$ of $\mathbb{Z}_n = \mathbb{Z} / n \mathbb{Z}$ on $C( S^1 )$ by
$$\alpha_k : t + \mathbb{Z} \mapsto t + \frac{k}{n} + \mathbb{Z} .$$
Consider the corresponding C*-dynamical system $\left( C( S^1) , \mathbb{Z}_n , \alpha \right)$. This book I'm reading says that the crossed product is of the form $C( S^1) \rtimes_\alpha \mathbb{Z}_n \cong M_n( C ( S^1 ))$.
I already know that for a finite group $G$, the crossed product is of the form
$$A \rtimes_\alpha G \cong \left\{ \left[ \alpha_{i}^{-1}( f ( i j^{-1} ) ) \right]_{i, j \in G} \in M_{|G|}(A) : f \in C_c (G, A) \right\} \subseteq M_{|G|}(A) ,$$
where $M_{|G|}(A)$ means the $G$-indexed square matrices in $A$. But I'm having trouble seeing how this makes for an isomorphism of $C ( S^1 ) \rtimes_\alpha \mathbb{Z}_n$ onto $M_{n}( C ( S^1 ) )$. It even seems strange, because my intuition is that it should be impossible two map a "$C ( S^1 )$-module of dimension $n$" (i.e. $C_c (G, C ( S^1 ) )$) onto a "$C ( S^1 )$-module of dimension $n^2$" (i.e. $M_n ( C ( S^1 ) )$). I don't know if I can apply this kind of rank-nullity thinking to a setting like this, but it does seem like $C ( S^1 ) \rtimes_\alpha \mathbb{Z}_n$ should be "too small" to map onto $M_n ( C ( S^1 ) )$.
Can anyone help me out here? The book I'm working through is Nikolaevs' Noncommutative Geometry: a functorial approach, if that helps.
Thanks for your time!
 A: TL;DR: the “missing” factor of $n$ in the isomorphism $C(S^1) \rtimes_\alpha \mathbb{Z}_n \cong M_n(C(S^1))$ is supplied by the fact that $S^1$ on the left-hand side is an $n$-fold cover of $S^1 \cong S^1/\mathbb{Z}_n$ on the right-hand side.
Here’s a more general picture. Suppose that $X$ and $Y$ are compact Hausdorff spaces and that $p : X \to Y$ is a Galois cover with finite group of deck transformations $G$, so that $p$ descends to a homeomorphism $X/G \to Y$. This yields not only a $C^\ast$-dynamical system $(C(X),G,\alpha)$, but also a canonical $\ast$-isomorphism $\phi : C(X) \rtimes_\alpha G \to C(X,B(\ell^2(G)))^G$ given by
$$
 \forall f \in c_c(G,C(X)), \, \forall x \in X, \, \forall \gamma \in G, \quad \phi(f)(x) \cdot \delta_\gamma := \sum_{\eta \in G} f_\eta(x\cdot\gamma^{-1}) \,\delta_{\eta \cdot \gamma};
$$
up to the $\ast$_isomorphism $B(\ell^2(G)) \cong M_{\vert G\rvert}(\mathbb{C})$ induced by the standard orthonormal basis $\{\delta_\gamma\}_{\gamma\in G}$ of $\ell^2(G)$, the map $\phi$ is precisely your injective (but not surjective) $\ast$-homomorphism $C(X) \rtimes_\alpha G \to C(X) \otimes M_{\vert G\rvert}(\mathbb{C}))$ up to the isomorphism .
Now, observe that $p : X \to Y$ defines a principal $G$-bundle and that $G \leq U(B(\ell^2(G)))$ acts by conjugation on the finite-dimensional $C^\ast$-algebra $B(\ell^2(G))$. Thus, we get an associated locally trivial bundle of finite-dimensional $C^\ast$-algebras $\mathcal{A} := X \times_{G} B(\ell^2(G))$ over $Y \cong X/G$ with typical fibre $B(\ell^2(G)) \cong M_{\lvert G \rvert}(\mathbb{C})$ and a canonical isomorphism $p^\ast : \Gamma(Y,\mathcal{A}) \to C(X,B(\ell^2(G)))^G$ given by
$$
 \forall a \in \Gamma(Y,\mathcal{A}), \quad p^\ast (a) := a \circ p,
$$
where $\Gamma(Y,\mathcal{A})$ is the $C^\ast$-algebra of continuous global sections of $\mathcal{A} \to Y$. Putting everything together, then, you therefore have a $\ast$-isomorphism $(p^\ast)^{-1} \circ \phi : C(X) \rtimes_\alpha G \to \Gamma(Y,\mathcal{A})$.
At last, if $\mathcal{A} \to Y$ is trivial as a locally trivial bundle of finite-dimensional $C^\ast$-algebras, i.e., if there exists an isomorphism of locally trivial bundles of finite-dimensional $C^\ast$-algebras $\mathcal{A} \cong Y \times B(\ell^2(G))$, then each such isomorphism yields a corresponding isomorphism of $C^\ast$-algebras
$$
 G(Y,\mathcal{A}) \to \Gamma(Y,Y \times B(\ell^2(G)) \cong C(Y) \otimes M_{\lvert G\rvert}(\mathbb{C});
$$
hence, you get the desired chain of $\ast$-isomorphisms
$$
 C(X) \rtimes_\alpha G \cong C(X,B(\ell^2(G)))^G \cong \Gamma(Y,\mathcal{A})  \cong C(Y) \otimes M_{\lvert G\rvert}(\mathbb{C}).
$$
How does this play out in your case? You have an $n$-fold Galois cover $p : S^1 \to S^1$ given by
$$
 \forall t \in \mathbb{R}, \quad p(t+\mathbb{Z}) := nt+\mathbb{Z},
$$
whose group of deck transformations is precisely $\mathbb{Z}_n$ with the action you specify. Moreover, the $\mathbb{Z}_n$-action by conjugation on $B(\ell^2(\mathbb{Z}_n))$ can be identified with the $\mathbb{Z}_n$-action on $M_n(\mathbb{C})$ generated by conjugation by the appropriate $n \times n$ shift matrix. Your injective $\ast$-homomorphism
$$
 C(S^1)\rtimes_\alpha\mathbb{Z}_n \to C(S^1,M_n(\mathbb{C}))
$$
is not surjective but rather maps onto the proper $\ast$-subalgebra
$
 C(S^1,M_n(\mathbb{C}))^{\mathbb{Z}_n};
$
in turn, the $n$-fold covering map $p : S^1 \to S^1$ induces a $\ast$-isomorphism $C(S^1,M_n(\mathbb{C}))^{\mathbb{Z}_n} \to \Gamma(S^1,S^1 \times_{\mathbb{Z}_n} M_n(\mathbb{C}))$. But now, since $S^1 \setminus \{0+\mathbb{Z}\} \cong (0,1)$ is contractible and since $U_n(\mathbb{C})$ is path connected, it follows that $S^1 \times_{\mathbb{Z}_n} M_n(\mathbb{C}) \to S^1$ is trivial, and hence that $\Gamma(S^1,S^1 \times_{\mathbb{Z}_n} M_n(\mathbb{C})) \cong C(S^1) \otimes M_n(\mathbb{C})$. Thus, putting everything together, you get a chain of $\ast$-isomorphisms
$$
 C(S^1) \rtimes_\alpha \mathbb{Z}_n \cong C(S^1,M_n(\mathbb{C}))^{\mathbb{Z}_n} \cong \Gamma(S^1,S^1 \times_{\mathbb{Z}_n} M_n(\mathbb{C})) \cong C(S^1) \otimes M_n(\mathbb{C}),
$$
where the $S^1$ in $C(S^1) \rtimes_\alpha \mathbb{Z}_n$ is an $n$-fold cover of the $S^1$ in $C(S^1) \otimes M_n(\mathbb{C})$. An explicit construction of the composition of all these $\ast$-isomorphisms can be found in Example 10.9 of these notes by Chris Phillips.
