Can anyone calculate KK(A,B) when neither A or B are the complex numbers? - MathOverflow

Can anyone calculate KK(A,B) when neither A or B are the complex numbers?

2011-01-05T14:54:42Z

Here I am referring to Kasparov's KK-theory, a bivariant functor on the category of separable C* algebras. It is well known that $KK(A, \mathbb{C})$ is K-homology and $KK(\mathbb{C}, B)$ is K-theory, both of which can be calculated for a huge collection of C* algebras (often by topological methods).

I am wondering if anybody knows how to actually calculate $KK(A,B)$ when it is not simply isomorphic to K-theory or K-homology (so I guess I also have to exclude $C_0(\mathbb{R})$. It seems that most of the time the interest in KK groups is not actually calculating them but in constructing specific elements (such as the Dirac / dual Dirac elements in proofs of the Baum-Connes conjecture).

It occurred to me that I have never actually seen anybody explicitly calculate $KK(A, B)$ in nontrivial cases. I am left wondering if this is because nobody knows how to do it or if it just isn't particularly useful to do so. Explanations or references are both appreciated.

Answer by Makoto Yamashita for Can anyone calculate KK(A,B) when neither A or B are the complex numbers?

2011-01-05T15:49:45Z

There is Rosenberg-Schochet universal coefficient theorem, which says $KK(A,B)\simeq Ext(K_{\ast}(A),K_{\ast+1}(B))\oplus Hom(K_{\ast}(A),K_{\ast}(B))$ (not canonically) when $A$ is $KK$-equivalent to a commutative algebra. It was proved in The Künneth theorem and the universal coefficient theorem for Kasparov’s generalized K-functor, Duke Math. J., 55(1987), 431–474.

Answer by Daniel Pape for Can anyone calculate KK(A,B) when neither A or B are the complex numbers?

2011-01-05T20:29:37Z

Hi Paul, I do not know if this is interesting (enough) for you, but in 'The operator K-functor and extensions of C*-algebras' Kasparov shows that for suitable A and B (I guess A,B seperable, A nuclear) there exists an isomorphism from $\mathrm{KK}(A,B)$ onto a group $\mathrm{Ext}(A,C_0(\mathbb{R})\otimes B)$ of 'stable equivalence classes of extensions' $$0\xrightarrow{}{} C_0(\mathbb{R})\otimes B\otimes\mathcal{K}\xrightarrow{}{} C \xrightarrow{}{} A \xrightarrow{}{} 0\text{ .}$$. Maybe the RHS is not concrete enough for this to qualify as an answer.