Let's say we have a $C^{*}$-algebra $A$, a map $\alpha \in Aut(A)$ with $\alpha ^{n}=id$ and $n\in \mathbb{N}$. Of what structure is the crossed product $A \rtimes_{\alpha}\mathbb{Z}_{n}$ then?

The construction I know begins with the space $C_c(\mathbb{Z}_n, A)$ of formal polynomials $\sum_{k\in\mathbb{Z}_{n}}t^kx_k$ where $x_0, ..., x_{n-1} \in A$ on which one can define a suitable addition, multiplication, involution and a $C^*$-norm. The crossed product $A \rtimes_{\alpha}\mathbb{Z}_{n}$ is the completion of $C_c(\mathbb{Z}_n, A)$ by this norm, then.

My question: Since the definition of the norm for crossed products looks rather complicated, what do the elements of $C_c(\mathbb{Z}_n, A)$ look like? Isn't $C_c(\mathbb{Z}_n, A)$ already complete since $\mathbb{Z}_n$ is finite or do I miss something?


In the construction you mention, I don't know what the "suitable" norm is.

Remember that the goal of the cross product is to implement the automorphisms by unitary conjugation.

The way I know the construction is, you embed $A\subset B(H)$ and $C_c(\mathbb Z_n,A)\subset B(H^n)$ in the following way. We actually consider $$C_c(\mathbb Z_n,A)\subset M_n(A),$$ by embedding $A\hookrightarrow M_n(A)$ via $$a\longmapsto \pi(a)=\sum_{k=1}^n\alpha^{k}(a)\otimes E_{kk},$$ and $\mathbb Z_n\hookrightarrow M_n(A)$ by $$m\longmapsto U_m=\sum_{k=1}^nE_{k,k+m-1},$$ where all the sums in the coordinates are interpreted in $\mathbb Z_n$.

One can show that $$\tag{1}U_m\pi(a)U^*_m=\pi(\alpha^m(a)),$$ and that $$\tag{2} C_c(\mathbb Z_n,A)=C^*(\{\pi(A),U_1,\ldots,U_n\})=\{X\in M_n(A):\ X_{kj}=\alpha^{k-1}(X_{1,j-k+1})\}. $$ Now we can assign to $C_c(\mathbb Z_n, A)$ its natural operator norm in $B(H^n)$, which agrees with norm pointwise convergence of the coordinates. With this, it is easy to check that $C_c(\mathbb Z_n, A)$ is closed. So $(1)$ gives a concrete expression for $A\rtimes_\alpha\mathbb Z_n$.

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  • $\begingroup$ Hello Martin, thanks for your answer. I guess you mean $C_c(\mathbb{Z}_n, A)$ and $A \rtimes_\alpha \mathbb{Z}_n$ instead of $C_c(\mathbb{Z}, A)$ and $A \rtimes_\alpha \mathbb{Z}$ at the end? And if I understand you correctly we have $A \rtimes_\alpha \mathbb{Z}_n = C_c(\mathbb{Z}, A)$ since the right hand side is closed, then? $\endgroup$ – worldreporter14 Aug 19 '17 at 19:44
  • $\begingroup$ Yes, absolutely. With $\mathbb Z$ the ideas are the same, but you don't get a closed algebra. $\endgroup$ – Martin Argerami Aug 19 '17 at 20:20

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