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I have a question about a proof in Rosenberg and Schochet's paper "the Künneth theorem and the Universal Coefficient Theorem for Kasparov's generalized K-functor", proposition 2.6. First of all, the setting:

Def.: Let $N$ be the bootstrap class of $C^*$-algebras, it's the smallest full subcategory of the seperable nuclear $C^*$-algebras which contains the separable Type I $C^*$-algebras and and is closed under strong Morita equivalence, inductive limits, extensions, and crossed products by $\mathbb{R}$ and by $\mathbb{Z}$. And if $J$ is an ideal in $A$ and $J$ and $A$ are in $N$, then so is $A/J$. And if $A$ and $A/J$ are in $N$ then so is $J$.

Theorem (2.1): Let $A\in N$ and let $B$ be a $\sigma$-unital $C^*$-algebra such that $K_*(B)$ is an injective $\mathbb{Z}$-module. Then the map $$\gamma(A,B):KK_*(A,B)\to Hom(K_*(A),K_*(B))$$ is an isomorphism.

One step to prove this theorem is the following

Proposition (2.6): If $K_*(B)$ is injective and if $\gamma(A,B)$ is an isomorphism, then $\gamma(A\rtimes_{\rho} \mathbb{R},B)$ is an isomorphism for any continuous action of $\mathbb{R}$ on $A$.

Proof: The Thom isomorphism theorems of Connes yield natural isomorphisms

$$Hom(K_i(A\rtimes_{\rho} \mathbb{R}),K_j(B))\cong Hom(K_{i+1}(A),K_j(B))$$ and $$KK_i(A\rtimes_{\rho} \mathbb{R},B)\cong KK_{i+1}(A,B).$$The proposition follows immediately. $\Box$

The argument above is the following: $$\require{AMScd}\begin{CD} KK_i(A\rtimes_{\rho} \mathbb{R},B) @>\gamma(A\rtimes_{\rho} \mathbb{R},B)>> Hom(K_i(A\rtimes_{\rho} \mathbb{R}),K_j(B)) \\ @VV \eta V @VV \sigma V \\ KK_{i+1}(A,B) @>\gamma(A,B)>> Hom(K_{i+1}(A),K_j(B)) \\ \end{CD}$$ is a commutative diagram, where $\eta , \sigma$ and $\gamma(A,B)$ are isomorphisms, therefore $\gamma(A\rtimes_{\rho} \mathbb{R},B)$ is an isomorphism.

My question is: why is this diagram commutative?

For this, I tried to figure out how to write down the maps explicitely. The map $\gamma(A,B)$ comes from the Kasparov-product

$$KK(\mathbb{C},A)\times KK(A,B)\to KK(\mathbb{C},B)\; (\epsilon_1 , \epsilon_2)\mapsto \epsilon_1 \otimes \epsilon_1 ,$$ because it is $KK_i(\mathbb{C},A)\cong K_i(A)$, $KK_i(\mathbb{C},B)\cong K_i(B)$. Hence the Kasparov-product induces a homomorphism (which should be $\gamma(A,B)$ I think) $$KK(A,B)\to Hom(K(A), K(B)),$$ $$\epsilon_2 \mapsto (\epsilon_1 \mapsto \epsilon_1 \otimes \epsilon_2).$$

The map $\sigma$ is the contravariant $Hom(-,K_j(B))$-functor apllied to the (Connes-Thom)-isomorphism $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$.

But I don't know how to write down the isomorphisms $\sigma$ and $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ explicitely such that I'm stuck to prove that the diagram commutes.

Can you help me to prove that the diagram commutes, or do you know how to write down the maps $\sigma$ and $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ explicitely?

Best

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I think the key idea is that Connes' Thom isomorphism is itself given by a $KK$-equivalence (see for example Blackadar's book "K-theory for Operator Algebras" - Theorem 19.3.6).

This means there are classes $t \in KK_1(A, A \rtimes \mathbb{R})$ and $t^{-1} \in KK_1(A \rtimes \mathbb{R},A)$, such that the corresponding intersection products $t \otimes_{A \rtimes \mathbb{R}} t^{-1}$ and $t^{-1} \otimes_A t$ agree with the identity elements in $KK(A \rtimes \mathbb{R},A \rtimes \mathbb{R})$ and $KK(A,A)$ respectively (using Bott periodicity to identify $KK_2$ with $KK$) and the intersection product with $t$ induces the Thom isomorphism $$ K_{i+1}(A) \cong KK_{i+1}(\mathbb{C}, A) \to K_{i}(A \rtimes \mathbb{R}) \cong KK_{i}(\mathbb{C}, A \rtimes \mathbb{R}) \quad ;\quad x \mapsto x \otimes_A t $$ as well as the map $\eta$ via $$ KK_i(A \rtimes \mathbb{R},B) \to KK_{i+1}(A,B) \quad ;\quad y \mapsto t \otimes_{A \rtimes \mathbb{R}} y\ . $$

Let $x \in KK_i(A \rtimes \mathbb{R},B)$ and $y \in K_{i+1}(A)$. We have to check that $$ \sigma(\gamma(A \rtimes \mathbb{R}, B)(x))(y) = \gamma(A,B)(\eta(x))(y) $$ If we evaluate the left hand side we get $$ (y \otimes_A t)\,\otimes_{A \rtimes \mathbb{R}}\,x = y \otimes_A (t \otimes_{A \rtimes \mathbb{R}}\,x) $$ where we used the associativity of the Kasparov intersection product. But now the right hand side agrees with $\gamma(A,B)(\eta(x))(y)$ and the diagram commutes.

There is an increasing amount of people with rabbit related pseudonyms/profile pictures on mathoverflow.net (I am looking at you Bugs Bunny). Is this a trend that I missed?

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  • $\begingroup$ I don't know anything about rabbit-trends ;) (I only think watching cat videos is popular) $\endgroup$ – user62639 Jun 11 '16 at 17:15
  • $\begingroup$ I have an additional question. We can skip the assumption that $K_*(B)$ is injective, am I right? $\endgroup$ – user62639 Jun 12 '16 at 8:06
  • $\begingroup$ Injectivity of $B$ is not needed for the commutativity of the diagram. Was that your question? $\endgroup$ – Ulrich Pennig Jun 12 '16 at 10:06
  • $\begingroup$ This is clear, but if I don't overlook something the assumtion that $K_*(B)$ is injective is not needed for the proposition 2.6 $\endgroup$ – user62639 Jun 13 '16 at 6:02
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    $\begingroup$ Unless I am overlooking something here, you seem to be right. $\endgroup$ – Ulrich Pennig Jun 13 '16 at 10:17

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