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.


1 Answer 1


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 agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.