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.

| cite | improve this answer | |
  • $\begingroup$ In $C_{c}(\Gamma, *)$, the * should be a C*-algebra. But $B\odot A$ is only a vector space with out norm. $\endgroup$ – Yan kai Aug 24 '14 at 12:12
  • 1
    $\begingroup$ 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)) $. $\endgroup$ – Martin Argerami Aug 24 '14 at 12:30
  • $\begingroup$ Oh, yes, I understand wrong. Thanks, Martin. $\endgroup$ – Yan kai Aug 24 '14 at 17:20
  • $\begingroup$ You're welcome. This works because "reduced" is spatial. I wonder if it is true at the max level, but I wouldn't even know how to think about it. $\endgroup$ – Martin Argerami Aug 24 '14 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.