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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.

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    $\begingroup$ $KK(\mathbb C \oplus \mathbb C,\mathbb C \oplus \mathbb C) = M_2 \mathbb Z$. $\endgroup$
    – Andreas Thom
    Commented Jan 5, 2011 at 16:59
  • $\begingroup$ When I attended a course by Jacek Brodzki (toknotes.mimuw.edu.pl/sem6/index.html) he indeed said that most of the time we'd be interested not in computing KK(A,B) but rather in understanding specific cycles. But I can't say whether he meant it generally, or only for the lecture (back then I though he meant it generally). $\endgroup$
    – Łukasz Grabowski
    Commented Jan 5, 2011 at 23:26
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    $\begingroup$ @Andreas: is this more general statement true: $KK(A^m,A^n)=M_{n,m} (KK(A;A))$? $\endgroup$
    – Johannes Ebert
    Commented Jan 8, 2011 at 15:44
  • $\begingroup$ @JohannesEbert Even though the question is older, using finite additivity of KK in both variables, I guess the statement is true. Or am I wrong? $\endgroup$
    – Sabrina Gemsa
    Commented Mar 9, 2017 at 14:09

2 Answers 2

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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.

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    $\begingroup$ * gets interpreted as code for italics, so you should use \ast on MO instead. Hopefully I formatted that correctly. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 5, 2011 at 16:33
  • $\begingroup$ This is indeed the sort of thing I had in mind - thanks! $\endgroup$
    – Paul Siegel
    Commented Jan 12, 2011 at 13:02
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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.

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  • $\begingroup$ Hi Dan! This is indeed a useful answer as well, though as you feared I don't really know how to concretely describe extension groups for C* algebras, either. If you know any good tools for computing extension groups for the sorts of C* algebras which make your isomorphism work, that would interest me as well. $\endgroup$
    – Paul Siegel
    Commented Jan 12, 2011 at 13:04

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