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For two unital (separable) $C^*$-algebras $A$ and $B$, let $(H,\rho,F)$ be a $KK$-cycle in the sense of Kasparov, or in the sense of Wikipedia :)

I wonder if there us a natural way to "forget" the right Hilbert $B$-module structure, and "project" in some sense to a Fredholm module over $A$.

This is inspired by the following construction: Looking at the KK-theory level instead, we have the Kasparov product $$ KK(A,B) \times K(B,\mathbb{C}) \to K(A,C). $$ Which by pairing with a $K$-theory class, produces a $K$-homology class.

QUESTION: Is there some natural way to lift this Kasparav contraction to a simple operation to turn $KK$-cycles into $K$-homology classes?

Edit: As Ulrich points out in the comments below, one way to do this might be to use a trace, or state, on $B$.

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  • $\begingroup$ What definition of K-homology are you using? Perhaps the most common definition in the operator algebras literature uses the language of Fredholm modules, and these (and the relations between them) are defined exactly the same as $KK$-cycles if you replace "Hilbert module over $\mathbb{C}$" with "Hilbert space". $\endgroup$
    – Paul Siegel
    Commented Apr 18, 2019 at 20:54
  • $\begingroup$ @Paul: Yes, of course - just plug in $\mathbb{C}$ to $KK$ and get $K$-homology. But for a given $KK$-cycle, I'm asking for a way to produce from it a $K$-homology cycle. I've given a naive guess above, but I'm not sure it makes sense. $\endgroup$
    – Max Schattman
    Commented Apr 18, 2019 at 21:05
  • $\begingroup$ The problem is that the $B$-valued inner product will not give you a $\mathbb{C}$-valued inner product. Therefore you can not turn your right Hilbert $B$-module $H$ into a Hilbert space. If $B$ has a trace, this might work. $\endgroup$
    – Ulrich Pennig
    Commented Apr 18, 2019 at 21:35
  • $\begingroup$ Let me elaborate on my previous comment further. A $KK(A, \mathbb{C})$ - cycle is a triple $(H, \rho, F)$ where $H$ is a countably generated Hilbert module over $\mathbb{C}$, $\rho$ is a $*$-representation of $A$ on $H$ as even bounded operators which commute with $\mathbb{C}$, and $F$ is a bounded odd operator which commutes with $\mathbb{C}$ and satisfies three identities modulo $\mathbb{C}$-compact operators. On the other hand a Fredholm module over $A$ is a triple $(H, \rho, F)$ where $H$ is a graded separable Hilbert space, $\rho$ is a $*$-representation of $A$ on $H$... $\endgroup$
    – Paul Siegel
    Commented Apr 18, 2019 at 23:35
  • $\begingroup$ ... as even bounded operators, and $F$ is a bounded odd operator which satisfies the same three identities as in KK-theory modulo compact operators. The relations which define the group $KK(A, \mathbb{C})$ and the K-homology of $A$ are homotopy, unitary equivalence, and direct sum in both cases. Note that a Hilbert $\mathbb{C}$-module is just a Hilbert space, all bounded operators commute with $\mathbb{C}$, and a $\mathbb{C}$-compact operator is just a compact operator. So how do we get from a $KK$-cycle to a Fredholm module? A $KK$-cycle is a Fredholm module! $\endgroup$
    – Paul Siegel
    Commented Apr 18, 2019 at 23:41

1 Answer 1

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Here are just some trivial observations that came to mind after thinking about this a little longer: You are essentially asking for a canonical class in the $K$-homology group $K^0(B) = KK(B,\mathbb{C})$. In general this does not exist except in very special situations. For example, if $\text{Ext}^1_{\mathbb{Z}}(B,\mathbb{Z}) = 0$, then $K^0(B) \cong \hom(K_0(B),\mathbb{Z})$, ie. any group homomorphism $K_0(B) \to \mathbb{Z}$ lifts uniquely to a $K$-homology class. A trace on $B$ produces such a group homomorphism.

An interesting situation, which produces a class in $K^0(B)$ is when $B$ is the completion of an algebra $\mathcal{B}$ that is part of a spectral triple $(\mathcal{B}, \mathcal{H}, D)$. In case $B$ is commutative, this boils down to the statement that $K$-homology classes arise from Dirac operators on spin-manifolds. For details about the $K$-homology classes associated to spectral triples see this article (in particular Prop. 4.4)

https://ro.uow.edu.au/cgi/viewcontent.cgi?referer=https://www.google.com/&httpsredir=1&article=1745&context=eispapers

by Carey, Phillips and Rennie.

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