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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)

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$B\rtimes\Sigma_n$ 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 $K_0(B\rtimes\Sigma_n)=\mathbb{Q}^{c_{n-1}}$

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  • $\begingroup$ Do you mean $K_0$ instead of $K_1$? $\endgroup$ Dec 23 '13 at 22:04

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