# Taking direct sums in $K$-theory in Kirchberg-Phillips classification

A theorem by Kirchberg and Phillips states that two unital separable nuclear simple purely infinite $C^*$-algebras (so called Kirchberg algebras) are isomorphic if and only if their topological $K$-theory groups are isomorphic.

Taking here direct sums of abelian groups at the $K$-theory side, what does it mean at the Kirchberg algebras side?

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You want to know which operation on Kirchberg algebras corresponds to direct sum of $K$-groups (of course direct sum of simple algebras takes you out of simple algebras). But for your question to make sense, you need to specify the range of the $K$-theory invariants (i.e. which are the pairs of abelian groups that you can get as $(K_0,K_1)$ of a Kirchberg algebra). If direct sum of $K$-groups takes you out of the range, your question may be meaningless... –  Alain Valette Mar 4 at 17:28
@Alain: By Proposition 4.3.4 in the book "Classification of Nuclear, Simple $C^*$-algebras" by Rordam, any pair $(G_0,G_1)$ of countable abelian groups can be realized as $(K_0(A), K_1(A))$ of a unital Kirchberg algebra $A$, where $A$ can even be chosen to be in the UCT class. –  Ulrich Pennig Mar 4 at 18:25
@Ulrich: yes this fact I had in mind. One can even choose to be A a graph C∗-algebra, if one likes. – –  Hans Mar 4 at 19:59
@Ulrich: OK, then I eventually made sense of the OP. But can you describe, for instance, the operation on Kirchberg algebras that flips $K_0$ and $K_1$, i.e. which has the same effect on K-theory as tensoring by $C_0(\mathbb{R})$? –  Alain Valette Mar 4 at 20:19
@Alain: My guess would be that you can take any Kirchberg algebra with K0(A)=0 and K1(A)=ℤ, form the tensor product with that and use the Künneth theorem to see that the result has switched K-groups. –  Ulrich Pennig Mar 4 at 22:17
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