What can be said about the following crossed product $C^\*$-algebra?


Let $A$ be a Kirchberg algebra with $K\_0(A) = \mathbb{Q}$ and $K\_1(A) = 0$. Consider the direct sum of $n$ copies of $A$, i.e. $B = A^n$. The permutation group $\Sigma\_n$ on $n$ letters acts on $B$ by permuting the elements. 


* What is $K\_i(B \rtimes \Sigma\_n)$?
* Is the crossed product simple and / or purely infinite? Or even a Kirchberg algebra?


Addendum: A Kirchberg algebra is unital, separable, simple, nuclear and purely infinite.


Temporary Addendum: Merry Christmas! (well, nearly)



Crossed Products by Permutation Groups
$B\rtimes\Sigma\_n$ is Morita equivalent to $A\rtimes\Sigma\_{n-1}$ with $\Sigma\_{n-1}$ acting trivially. The latter is isomorphic to $A\otimes \mathbb{C}\Sigma\_{n-1}$, which shows that $B\rtimes\Sigma\_n$ is not simple. Let $c\_n$ be the number of conjugacy classes of $\Sigma\_n$. Then $\mathbb{C}\Sigma\_{n-1}$ is the direct sum of $c\_{n-1}$ matrix algebras, so by invariance of $K\_i$ under Morita invariance we have $K\_0(B\rtimes\Sigma\_n)=\mathbb{Q}^{c\_{n-1}}$


