I'm crrently reading the paper about the Künneth-theorem for $C^*$-algebras: http://msp.org/pjm/1982/98-2/pjm-v98-n2-p15-s.pdf and I'm trying to understand remark 4.9. I henceforth asumme that $A$ is a separable nuclear $C^*$algebra and $B$ is a separable $C^*$algebra. Furthermore, I will adopt all notations from the paper.

We assume that:

(1). If $K_*(B)$ is free and if $f:A'\to A''$ is a $*$-homomorphism inducing an isomorphism $f_*:K_*(A')\to K_*(A'')$, then $(f\otimes 1)_*:K_*(A'\otimes B)\to K_*(A''\otimes B)$ is an isomorphism.

And we want to show that:

(2). Künnth-theorem (theorem 4.1) holds in full generality.
(As mentioned in the paper, it suffices to assume that $A$,$B$ are separable $C^*$ algebras with $A$ nuclear and $K_*(B)$ torsion free and to show that $\alpha(A,B)$ is an isomorphism).

What I have done so far: Assume (1) and that $K_*(A)$ is free. Then I obtain that (2) holds with $K_*(A \otimes B) \cong K_*(A) \otimes K_*(F)$. However, I the assumption (that $K_*(A)$ is free) is too strong and the result is somehow strange. My try was the following: I tried to prove $\operatorname{coker}(1 \otimes v_*)=\operatorname{coker} ((1 \otimes v)_*)$ and $\ker(1 \otimes v_*)=\ker ((1 \otimes v)_*)$ as in proof of thm 4.1 and for this I'm using diagram (4.5). But I used that $\alpha(A,F)$ is an isomorphism (which is correct even though if we would drop that $A$ is in the bootstrap-class) to use that $K_*(A) \otimes K_*(F) \cong K_*(A \otimes F)$.

First I considered the top-row in diagram (4.5) in the proof of theorem 4.1: Since $K_*(B)$ is free it is $\operatorname{Tor}(K_*(A),K_*(B))=0$ and $\mu_*:K_*(F)\to K_*(B)$ is an isomorphism, it follows that $1\otimes ( \mu)_*:K_*(A) \otimes K_*(F)\to K_*(A) \otimes K_*(B)$ is an isomorphism as well. Thus, $K_*(A)\otimes K_*(C)=0$ (since the sequence is exact) and $1 \otimes v_*=0$.

Now, if we assume that $A$ nuclear and that $K_*(A)$ is free, then we can apply (1) to $\mu$ as follows : since $\mu_*:K_*(F)\to K_*(B)$ is an isomorphism, $(\mu \otimes 1_A)_*:K_*(F \otimes A)\to K_*( B \otimes A)$ is an isomorphism. Applying the 6-term exact sequence, it follows that $K_*(C \otimes A) =0$. And then we have in the top row of this diagram that $(1 \otimes v)_*=0$. It follows that $\ker(1 \otimes v_*)=\ker ((1 \otimes v)_*) =0$ and: $\operatorname{coker}(1 \otimes v_*) \cong [K_*(A) \otimes K_*(F)]/Im (1 \otimes v_*)=K_*(A) \otimes K_*(F)$, $\operatorname{coker} ((1 \otimes v)_*) \cong K_*(A \otimes F)/\operatorname{Im}((1 \otimes v)_*)\cong K_*(A \otimes F)$, hence $\operatorname{coker}(1 \otimes v_*)=\operatorname{coker} ((1 \otimes v)_*)$.

Thank you for any help.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.