# Is $KK^G(\mathbb{C}^n,B)$ countably additive in $B$ and countable?

Let $G$ be a finite discrete groupoid, $A=\mathbb{C}^n$ a finite dimensional, commutative $C^*$-algebra and assume we have given a $G$-action on $A$. Note that the action of $G$ on $\mathbb{C}^n=C_0(\{1,\ldots,n\})$ just permutes the factors $\mathbb{C}$.

Is then $KK^G(A,B)$ countably additive in $B$, that is, $$KK^G(A,\bigoplus_{n=1}^\infty B_n) \cong \bigoplus_{n=1}^\infty KK^G(A,B_n)$$ for all separable $G$-$C^*$-algebra $B_n$?

Note that this is true for $A=C_0(G^{(0)})$ because $$KK^G(C_0(G^{(0)}),B) \cong K(B \rtimes G)$$ by Green-Julg.

The question is related to the generalized Künneth theorem, which allows to compute $KK(A,B)$ from the knowledge of $K(A)$ and $K(B)$. Note, however, that the $G$-equivariant Künneth theorem is a complicated matter already, or particularly, for finite groups, already for $G$ with only two elements, by some recent publication by Rosenberg.

Another subquestion was if $KK^G(A,B)$ is a countable set.

• What action of $G$ on $\mathbb{C}^n$? – Qiaochu Yuan Apr 2 '15 at 21:19
• @Yuan: Any arbitrary action. – hans Apr 3 '15 at 10:32
• I mean, you talk about "the action" without introducing an action first. – Qiaochu Yuan Apr 3 '15 at 18:47