Karoubi versus Kasparov K-theory 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).
 A: About 15 years ago I used  Karoubi's   description of $K$-theory  to solve prove some cutting and pasting formula for  the index families  of elliptic   problems.  To do so I needed  rephrase  Karoubi's theory into something more   flexible and more computable.  Some of the interpretations I found might be relevant to your question. I will  briefly describe one such interpretation referring for proofs and many more details to the original source, my old paper Generalized symplectic geometries and the index of families of elliptic problems, Mem. A.M.S., vol. 128, no.609, 1997. I will refer to this as the old paper. $\newcommand{\bsH}{\mathscr{H}}$
To simplify  the presentation  let me define a $C^{p,q}$-module to be a Hilbert space  $H$ equipped with a $8$_morphism  of $C^*$_algebras $\phi: C^{P,q}\to B(H)$. A  graded $C^{p,q}$-module  can then be identified witha $C^{p,q+1}$-module. $\newcommand{\eF}{\mathscr{F}}$
Suppose that $H$ is a $C^{p,q+1}$-module. Define $\eF^{p,q}$ (or $\eF^{p,q}(H)$ to be the    space  of closed, densely defined, Fredholm, selfadjoint operators $T: H\to H$ that super-commute with the $C^{p,q}$-structure, i.e.,
$$ Te_k+ e_k T=0,\;\;\forall  k=1,\dotsc, p+q. $$
The space  $\eF^{p,q}$  carries a natural topology  defined by the metric
$$ d(T_1,T_2)= \Vert \Psi(T_1-\Psi(T_2)\Vert,\;\;\Psi(\lambda=\lambda(1+\lambda^2)^{\frac{1}{2}}. $$
Denote by $\newcommand{\eBF}{\mathscr{BF}}$ $\eBF^{p,q}$ the subspace of $\eF^{p,q}$ consisting of bounded operators. Then one can show  (see here) that the inclusion $\eBF^{p,q}\hookrightarrow \eF^{p,q}$ is a homotopy equivalence and that $\eBF^{p,q}$ is a classifying space  for  Karoubi's $KO^{p,q}$, which for simplicity I will denote by $K^{p,q}$.
Thus, to a compact $CW$-complex and a  continuous map $T: X\to\eF^{p,q}$ one can associate  an element $(E,\eta_0,\eta_1)\in K^{p,q}(X)$.     To    effectively describe this correspondence
$$ (X\stackrel{T}{\to}\eF^{p,q}) \to (E,\eta_0,\eta_1), $$
one needs a new, symplectic description of Karoubi's $K$-theory, and it  is through the symplectic prism that I  got to see Kasparov's KK lurking in the background.
Let $T\in \eF^{p,q}(H)$. Recall that, by construction $H$ is a $C^{p,q+1}$-mpodule. The direct sum $\hat{H}=H\oplus H $ has a richer structure of $C^{p+1,q+1}$-module (see  section 5.2  in the  old paper)
The graph $\Gamma_T$ of $T$ is a closed subspace of $\hat{H}$. Denote by $R_T:\hat{H}\to\hat{H}$ the orthogonal reflection in $\Gamma_T$.   Observe that the subspace $H\oplus 0\in \hat{H}$ can be identified with $\Gamma_0$, the graph of the trivial linear map.   We set 
$$\Gamma_\infty=0\oplus H\subset \hat{H}. $$
Then  $R_T^2=1$, and the condition $T\in\eF^{p,q}$ is equivalent  with the following requirements.


*

*$R_T$  supercommutes tith the $C^{p+1,q+1}$-structure on $\hat{H}$.

*The  pair of subspaces $(\Gamma_0,\Gamma_T)$ is a Fredholm pair.

*The subspace $\Gamma_T$ does not intersect $\Gamma_\infty$.


$\newcommand{\eFL}{\mathscr{FL}}$.  We denote by $\eFL^{p+1,q+1}(\hat{H})$ the  set of closed subspaces  $L$ of $\hat{H}$  such that the reflection $R_L$ in $L$ satisfies    the conditions $1$ and $2$ above.  (In the old paper I called these spaces generalized lagrangian spaces of type $(p+1,q+1)$. We have thus produced  an inclusion
$$\eF^{p,q}(H)\to \eFL^{p+1,q+1}(\hat{H}). $$
One can show two things. First, the space $\eFL^{p+1,q+1}$  classifies $K^{p+1,q+1}$ and second, the above  inclusion is a homotopy equivalence. (The proof uses Bott periodicity.)  One can use a process of symplectic reduction  to canonically associate to continuous family $L: X\to \eFL^{p+1,q+1}(X)$ an  element 
$$ (E,\eta_0,\eta_1)\in K^{p+1,q+1}(X)\cong K^{p,q}(X). $$ 
Observe that the elements of $\eFL^{p+1,q+1}$  can be identified with  selfajoint operators $R:\hat{H}\to \hat{H}$ such that $R^2=1$  and  super-commute with the $C^{p+1,q+1}$-structure and they satisfy condition 2.  This almost looks like a Kasparov element.
To  actually get a Kasparov element  consider a smooth, odd, nondecreasing function $\newcommand{\bR}{\mathbb{R}}$
$$\beta : \bR\to \bR,\;\;\beta (t)=\pm 1\;\;\mbox{if $\pm t>1$}. $$
For $\newcommand{\ve}{{\varepsilon}}$ $\ve>0$ we set $\beta_\ve(t)=\beta(t/\ve)$.
Then for any $T\in\eF^{p,q}$, $\beta_\ve (F)$ defines a Kasparov element for $\ve>0$ sufficiently small.
Remark.  I will describe below an explicit map
$$K^{p,q}(X)\to [X, \eF^{p+1,q+1}]. $$
This is more or less what you need since $K^{p,q}(X)\cong K^{p+1,q+1}(X)$ and more generally, $K^{p_0,q_0}(X)\cong K^{p_1,q_1}(X)$ if $p_0-q_0\equiv p_1-q_1\bmod 8$.    
The symplectic point of view   is crucial since the above map is inspired by Floer's work on    symplectic (Floer) homology. To justify the  symplectic terminology let me discuss a simple example.     
Let us  look at a $C^{1,0}$-module. This is a real Hilbert space  equipped with a   orthogonal operator $J: H\to H$ such that $J^2=-1$, $J^*=-J$.  In the finite dimensional case think $H=\bR^n\oplus \bR^n$, $J(x,y)=-(y,x)$.)    A subspace $L\subset H$ is called Lagrangian if $JL=L^\perp$. 
In the  case of $\bR^n\oplus \bR^n= T^*\bR^n$ equipped with the canonical symplectic structure we obtain in this fashion the classical notion of Lagrangian subspace. If  $P_L$ denotes the orthogonal projection onto $L$ and $R_L=2P_L-1$ denotes the orthogonal reflection in the subspace $L$,   then  $L$ is Lagragian iff $R_L$ anticommutes with $J$,
$$R_LJ+JR_L=0. $$
In algebraic terms  $R_L$ defines a $\newcommand{\bZ}{\mathbb{Z}}$ $\bZ/2$-grading  of the $C^{1,0}$-module $H$.  However the symplectic  point of view  has more   flexibility because it leads to  certain operations which algebraically  do not seem natural.  (I'm thinking here of the process  symplectic reduction.)
In general   given  a $C^{p,q}$-module $H$, we define a  $(p,q)$-Lagrangian in $H$ to be a subspace  $L\subset H$   such that $R_L$ supercommutes with the $C^{p,q}$-structure, i.e., $R_L$ is  a $\bZ/2$-grading of  the $C^{p,q}$-module $H$. Denote by  $\DeclareMathOperator{\Lag}{Lag}$ $\Lag^{p,q}(H)$ the space of $(p,q)$-lagragians in $H$.   Observe that  and element in $K^{p,q}(X)$ is defined by a continuous map
$$ X\to {\Lag}^{p,q}(H)\times {\Lag}^{p,q}(H), \;\;x\mapsto (L_0(x), L_1(x))$$
where $H$ is a finite dimensional $C^{p,q}$-module. Thus an element  in $K^{p,q}$ is a pair of continuous families of lagrangian subspaces in a $C^{p,q}$-module. Moreover, it suffices  consider only the case  when  one of the families  $L_0(x)$ is constant. 
Fix a finite dimensional $C^{p',q'}$-module  $H$, $p'=p+1$, $q'=q+1$. Denote by $J$ the operator on $H$  defined by the  multiplication by $e_{p+1}$ so that $J^*=-J$, $J^2 =-1$. 
To a pair of  lagrangians $L_0, L_1\in \Lag^{p',q'}(H)$  we associate an operator  $T_{L_0,L_1}\in\eF^{p,q}(\bsH)$ where 
$$\bsH=L^2(0,1, H), $$
and $T_{L_0,L_1}$ is the closed, Fredholm  selfadjoint  unbounded operator on $\bsH$ with domain
$$ D(T_{L_0,L_1})=\bigl\lbrace u\in L^{1,2}(0,1; H);\; u(0)\in L_0,\;\;u(1)\in L_1\bigr\rbrace, $$
such that 
$$ T_{L_0,L_1} u(s)=J\frac{du}{ds},\;\;s\in (0,1). $$
Above, $L^{1,2}$ denotes the Sobolev spaces of functions with  first order derivative in $L^2$. 
The motivation for the operator $T_{L_0,L_1}$ comes from symplectic Floer homology. In his papers on the (Floer) homology  of a pair of lagrangian submanifolds A. Floer  investigated the operators $T_{L_0,L_1}$ in the case $p=1, q=0$ and the indices of one-parameter families of such operators.   Note that only the domain of $T_{L_0,L_1}$ depends on $L_0,L_1$.     The action of $_{L_0,L_1}$ is independent of the lagrangians $L_0,L_1$>
To an element  $\alpha\in K^{p',q'}(X)$   represented by a continuous map 
$$ X\ni x \to (L_0(x), L_1(x))\in {\Lag}^{p',q'}(H)\times {\Lag}^{p',q'}(H) $$
we can associate  an element $T_\alpha\in [X,\eF^{p,q}]$ given by the continuous map 
$$ X\ni x\mapsto T_{L_0(x), L_1(x)}\in\eF^{p,q}(\bsH). $$  
As mentioned before, $\eF^{p,q}(\bsH)$ classifies  $K^{p,q}$ and thus the    map $T_\alpha$ defines an element ${\rm ind}\; T_\alpha\in K^{p,q}(X)$.  In that old paper I proved  that ${\rm ind}\; T_\alpha \in K^{p,q}(X)$ coincides with $\alpha\in K^{p',q'}(X)$ via the canonical isomorphism $K^{p,q}(X)\to K^{p',q'}(X)$. The proof uses  crucially the process of symplectic reduction. For details see Thm. 5.5 in the old paper.
The familly $ T_{\alpha}$ can be given a Kasparov descriptition as explained above.
A: To get such an isomorphism one needs to require Karoubi's triples $(E,\eta_0,\eta_1)$ to consist of Euclidean vector bundles, and gradings of such (as well as the $Cl^{p,q}$-module on $E$). Given such a triple, one obtains a Kasparov $Cl^{p,q},C_0(X)$-module $(H,\varphi,F)$ as follows. $H$ is the completion of $C_c(X,E)$ with respect to the obvious $C_0(X)$-inner-product, $\varphi: Cl^{p,q}\longrightarrow \mathcal{L}(H)\cong C_b(X,E)$ is induced by the $Cl^{p,q}$-module $\rho:Cl^{p,q}\longrightarrow \mathcal{L}(E)$, and $F:=F_{\frac{1}{2}(\eta_0+\eta_1^*)}$; where for $f\in \mathcal{L}(E)$, $F_f(\phi)(x):=f(\phi(x)), \phi\in C_c(X,E)$. The converse is an application of Serre-Swan Theorem (and then the operator $F$ in the Kasparov module, viewed as a linear endomorphism of a hermitian vector bundle $E$, provides the gradings $\eta_0=F$ and $\eta_1=F^*$ ) ... 
