I have the following, probably very elementary question: Let $Cl^{p,q}$ be the Clifford algebra on generators $e_i$, $i=1, \ldots, p+q$ with $e_i e_j = -e_j e_i$ and $e_{i}^{2}=-1$ for $i=1,\ldots,p$, $e_{i}^{2} = 1 $ for $p+1 \leq i \leq p+q$.

Let $X$ be a compact space. Karoubi (in his book ''K-Theory'', section III.4) defines $K^{p,q} (X)$ as the abelian group, manufactured by the following recipe:

Consider triples $(E,\eta_0,\eta_1)$, where $E \to X$ is a finite-dimensional vector bundle with a $Cl^{p,q}$-module structure, and $\eta_i$ is a $Cl^{p,q}$-antilinear involution (i.e., it anticommutes with the generators) on $E$. Take the free abelian group generated by the isomorphism classes of these things, and divide out the following equivalence relations:

$(E,\eta_0,\eta_1) + (F,\zeta_0,\zeta_1)=(E \oplus F, \eta_0 \oplus \zeta_0, \eta_1 \oplus \zeta_1)$; $(E, \eta_0, \eta_1)=0$ if $\eta_0$ is homotopic to $\eta_1$ (as an $Cl^{p,q}$-antilinear involution).

There are several reformulations of this equivalence relation possible (loc. cit.). One of Karoubi's main results is that the group $K^{p,q}(X)$ is isomorphic to $KO^{p-q}(X)$.

On the other hand, we have real Kasparov theory. Karoubi's $K^{p,q}(X)$ is isomorphic to $KK (Cl^{p,q};C(X)) \cong KK (R;Cl^{q,p}\otimes C(X))$.

Recall that an element in $KK (Cl^{p,q};C(X))$ is represented by a Kasparov module, i.e. a triple $(H,\phi,F)$; here $H$ is a $Z/2$-graded real Hilbert bundle on $X$. $F$ is a family of Fredholm operators on $H$, which are odd (i.e., if $\iota$ defines the grading of $H$, then $F \iota = - \iota F$). $\phi$ is a graded $\ast$-homomorphism of $C^{\ast}$-algebras $Cl^{p,q}\to B (H)$ into the bounded operators on $H$. Moreover, the operators $F-F^{\ast}$, $[F,\phi(a)]$, $(F^2-1)$ are compact for all $a \in Cl^{p,q}$ (the bracket is the graded commutator; for the experts this is because $Cl^{p,q}$ is unital).

Here is my innocent question: how can I write down an isomorphism $K^{p,q}(X) \to KK (Cl^{p,q};C(X))$ \emph{explicitly}? I assume this is easy and a matter of pure linear algebra (therefore the tag), but, as often in this area, the published literature does not delve into concrete details of this sort. I am not interested in an abstract existence proof, since I know where to find it in the literature.

EDIT: Karoubi develops another model for $K^{p,q}(X)$, let me call it $F^{p,q}(X)$. This is the abelian group, generated by pairs $(H,F)$, where $H$ is a Hilbert space with $C^{p,q+1}$-action, and $F$ is a map from $X$ to Fredholm operators on $H$, such that for all $x \in X$, $F$ is $Cl^{p,q+1}$-antilinear and selfadjoint. The equivalence relation is given by homotopy and direct sum, and invertible operators are equivalent to $0$.

It is easy to map $F^{p,q}(X)$ into Kasparov theory: If $F$ is such a family, by a spetral deformation, one can achive that $F^2 -1 $ is compact. Consider the last generator $e_{p+q+1}$ as a $Z/2$-grading. Let $\phi: Cl^{p,q} \to B(H)$ be the map given by the Clifford action. Then $(H,\phi,F)$ represents the desired element in $KK$-theory. The isomorphism $F^{p,q} \cong K^{p,q}$ is implicit; using Kuipers theorem and the long exact sequence in $K$-theory.

Thus I can reformulate my question: How do I write down explicitly the isomorphism $K^{p,q}(X) \to F^{p,q}(X)$ (in this direction, not the other one).

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