I have a (maybe dumb) question about the relation between the Künneth theorem and the Theorem-universal coefficient theorem (UCT for short) in $KK$-theory (for the setting see "The Künneth theorem and the universal coefficient theorem for Kasparov’s generalized K-functor"- Rosenberg and Schochet).

Let $N$ be the bootstrap class of $C^*$-algebras (i.e. see here commutative diagram with $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ (for $C^*$-algebras) for the definition).

UCT: Let $A\in N$ and $B$ be $\sigma$-unital. Then there is a short exact sequence $$0\to Ext_{\mathbb{Z}}^1(K_*(A), K_*(B))\to KK_*(A,B)\to Hom(K_*(A), K_*(B))\to 0.$$

Künneth: Let $A\in N$, $B$ be $\sigma$-unital and $K_*(B)$ finitely generated. Then there is a short exact sequence $$0\to K^*(A)\otimes K_*(B)\to KK_*(A,B)\to Tor_1^{\mathbb{Z}}(K^*(A),K_*(B))\to 0.$$

Is it possible to proof Künneth $\Rightarrow $ UCT, and if so, how to prove it (do you have a reference)? What is already known?

Edit: In the paper the theorems are proved independently.

By definition it is $Ext_{\mathbb{Z}}^1(K_*(A), K_*(B))=H^1(Hom_{\mathbb{Z}}(P_*,K_*(B)) [\cong H^1(Hom_{\mathbb{Z}}(K_*(A),I^*))\enspace ]$, where $P_*$ is a projective resolution of $K_*(A)$ and $I^*$ an injective resolution of $K_*(B)$. $H^1$ denotes the first cohomology of the cochain complex $Hom_{\mathbb{Z}}(P_*,K_*(B))$.

Then it is $Tor_1^{\mathbb{Z}}(K^*(A),K_*(B))=H_1(P_*\otimes K_*(B)) [\cong H_1(K^*(A)\otimes D_*))]$, where $P_*$ is a projective resolution of $K^*(A)$ and $D_*$ is a projective resolution of $K_*(B)$. $H_1$ is the first homology of the chain complex $P_*\otimes K_*(B)$.

I'm not sure if in this case we can use that homology and cohomology commute with exact functors (since the tensor product is exact for $P_*$ projective and Hom is exact for $P_*$ projective and $I^*$ injective respectively). .

Edit (very simple example): If you take $A=\mathbb{C}$ and $B=K(H)$ for a separable finite dimensional complex Hilbert space $H$, it is $Ext_{\mathbb{Z}}^1(K_*(A), K_*(B))=0=Tor_1^{\mathbb{Z}}(K^*(A),K_*(B))$ and $KK_*(A,B)\cong Hom(K_*(A), K_*(B))\cong K^*(A)\otimes K_*(B)\cong \mathbb{Z} $.

  • 1
    $\begingroup$ You forgot the bootstrap class assumption in the Künneth theorem. $\endgroup$ – Rasmus Aug 31 '16 at 8:13

It is absolutely true that these two theorems are conceptually related. But instead of one being a consequence of the other, both rather follow from the same machinery being applied to slightly different setups: left deriving the covariant hom functor vs right deriving the contravariant hom functor. This is clarified in Example 4.5 in http://arxiv.org/pdf/math/0702146.pdf

  • $\begingroup$ thank you. I'm asking because recently I someone claimed that Künneth is formally weaker than the UCT and I want to understand it. $\endgroup$ – Ypsilon Sep 1 '16 at 10:54

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.