The following situation is given: Let $A$ be a unital, separable, nuclear $C^*$-Algebra, $i:\mathbb{C}\to A$ the unital embedding. All $C^*$-algebras are considered as trivially graded. Consider the induced map of $i$ in $KK$-theory: $i^0:KK^0(A,\mathbb{C})\to KK^0(\mathbb{C},\mathbb{C}),\; [E,\phi, T]\mapsto [E,\phi\circ i, T]$, where $[E,\phi, T]$ denotes a Kasparov $A-\mathbb{C}$-module. We denote with $\mathcal{O}_\infty$ the cuntz algebra in infinitely many generators. In the proof I'm trying to understand one multiplies the map $i^0$ with the $KK$-equivalence class given by the unital embedding $\iota:\mathbb{C}\to \mathcal{O}_\infty$ to obtain a map $v^0:KK^0(A,\mathcal{O}_\infty)\to KK^0(\mathbb{C},\mathcal{O}_\infty)$. Identifying $KK^0(\mathbb{C},\mathcal{O}_\infty)$ with $K_0(\mathcal{O}_\infty)$, the author claimes that the map $v^0$ sends a *-homomorphism $\varphi:A\to \mathcal{O}_\infty\otimes \mathcal{K}$ to $[\varphi(1_A)]_0$.

Why is $v^0$ sending a *-homomorphism $\varphi:A\to \mathcal{O}_\infty\otimes \mathcal{K}$ to $[\varphi(1_A)]_0$?

I'm trying to reconstruct $v^0$ using definitions of induced maps in $KK$-theory, multiplication of maps in $KK$-theory and so on. But I still don't get it:

First of all, $\iota$ induces a $KK$-equivalence $[\iota]\in KK^0(\mathbb{C},\mathcal{O}_\infty)$, thus there exists an element $[\gamma]\in KK^0(\mathcal{O}_\infty, \mathbb{C})$ such that $[\iota]\otimes [\gamma]=1\in KK^0(\mathcal{O}_\infty,\mathcal{O}_\infty)$ and such that $[\gamma]\otimes [\gamma]=1 \in KK^0(\mathbb{C},\mathbb{C})$. Furthermore, there are isomorphisms $$-\otimes [\iota]: KK^0(\mathbb{C},\mathbb{C})\to KK^0(\mathbb{C},\mathcal{O}_\infty)$$, and $$-\otimes [\gamma]: KK^0(A,\mathcal{O}_\infty)\to KK^0(A,\mathbb{C})$$ induced by the Kasparov-product $\otimes$. Now, my guess is that $v^0$ is the composition $(-\otimes [\iota])\circ i^0 \circ (-\otimes [\gamma])$. And then we compose $v^0$ with the isomorphism $KK^0(\mathbb{C},\mathcal{O}_\infty)\to K_0(\mathcal{O}_\infty)$. But I don't know an explicit formula for the last isomorphism, which fits in this situation.

I appreciate any help.


I don't think that there is anything special about $\mathcal O_\infty$ being used here. One observation that might help is that the map $KK(A, \mathcal O_\infty)\to KK(\mathbb C, \mathcal O_\infty)$ given by Kasparov product with the $KK$ class of $i_0$ is the same as the map induced by $i_0$ using the functoriality of $KK$: this is a general property of the Kasparov product, holding for all $*$-homomorphisms. So the map $v_0$ sends the class of $\varphi$ to the class of $\varphi\circ i_0$. Now the standard identification of $KK(\mathbb C, \mathcal O_\infty)$ with $K(\mathcal O_\infty)$ sends the class of $\varphi\circ i_0$ to the class of the idempotent $\varphi\circ i_0(1_{\mathbb C})= \varphi(1_A)$.


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.