Can anyone calculate KK(A,B) when neither A or B are the complex numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:38:29Z http://mathoverflow.net/feeds/question/51203 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51203/can-anyone-calculate-kka-b-when-neither-a-or-b-are-the-complex-numbers Can anyone calculate KK(A,B) when neither A or B are the complex numbers? Paul Siegel 2011-01-05T14:54:42Z 2011-01-05T20:29:37Z <p>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). </p> <p>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). </p> <p>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.</p> http://mathoverflow.net/questions/51203/can-anyone-calculate-kka-b-when-neither-a-or-b-are-the-complex-numbers/51211#51211 Answer by Makoto Yamashita for Can anyone calculate KK(A,B) when neither A or B are the complex numbers? Makoto Yamashita 2011-01-05T15:49:45Z 2011-01-05T16:32:45Z <p>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 <em>The Künneth theorem and the universal coefficient theorem for Kasparov’s generalized K-functor</em>, Duke Math. J., 55(1987), 431–474.</p> http://mathoverflow.net/questions/51203/can-anyone-calculate-kka-b-when-neither-a-or-b-are-the-complex-numbers/51242#51242 Answer by Daniel Pape for Can anyone calculate KK(A,B) when neither A or B are the complex numbers? Daniel Pape 2011-01-05T20:29:37Z 2011-01-05T20:29:37Z <p>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.</p>