# An exercise about reduced crossed product of a C*-dynamical system

Here is an Exercise in a book "C*-algebras and Finite-Dimensional Approximations" by N.P.Brown and N. Ozawa.

Exercise 4.1.3. Let $A$ and $B$ be two C*-algebras and $\Gamma$ be a discrete group. If $\alpha:\Gamma\rightarrow \mathrm{Aut}(A)$ is an action and $\tau\otimes\alpha:\Gamma\rightarrow Aut(B\otimes A)$ is defined by $(\tau\otimes\alpha)_{g}=\mathrm{id}_{B}\otimes\alpha_{g}$, then $$(B\otimes A)\rtimes_{\tau\otimes\alpha, r}\Gamma\cong B\otimes(A\rtimes_{\alpha, r}\Gamma).$$

Here, the "$B\otimes A$" denote the tensor product equiped with minimal norm. And $A\rtimes_{\alpha, r}\Gamma$ denotes the reduced crossed product of a C*-dynamical system $(A, \Gamma, \alpha)$, which is the norm closure of the image of a regular representation $C_{c}(\Gamma, A)\rightarrow B(H\otimes l^{2}(\Gamma))$. While the $C_{c}(\Gamma, A)$ denotes the linear space of finitely supported functions on $\Gamma$ with values in $A$.

This looks like an "associativity" property. Both algebras live in $B(H_B\otimes H_A\otimes \ell^2(\Gamma))$, so the topology is the same.
At the pre-closure level, you should convince yourself that $C_c(\Gamma,B\odot A)$ and $B\odot C_c(\Gamma,A)$ are equal, and that their closures give the two algebras you want to consider.
• In $C_{c}(\Gamma, *)$, the * should be a C*-algebra. But $B\odot A$ is only a vector space with out norm. – Yan kai Aug 24 '14 at 12:12
• Why should it be? If I'm not wrong, $C_c (\Gamma, D)$ is the set of finitely supported functions $\Gamma\to D$. I don't see any topological structure of $D$ playing a role. Besides, $B\odot A$ is not "only a vector space without norm"; it is a $*$-algebra, and nothing prevents you to consider a norm in it, like the one induced by $B\otimes A$. But in any case, as I understand it $C_c (\Gamma,*)$ is an algebraic structure, you give it a topology by embedding it in $B (H\otimes\ell^2 (\Gamma))$. – Martin Argerami Aug 24 '14 at 12:30