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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 $p_0-q_0\equiv p_1-q_1\bmod 8$.

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 $p_0-q_0\equiv p_1-q_1\bmod 8$.

Remark. I need to give some more intuition behid the symplectic point of view to appreciate its versatility. 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 we let $R_L$ denote the orthogonal reflection in the subspace $L$, then $L$ is Lagragian iff $R_L$ anticommutes with $J$,

In algebraic terms $R_L$ defines a $\newcommand{\bZ}{\mathbb{Z}}$ $\bZ/2$-grading of the $C^{0,1}$-module $H$. However the symplectic point of view has more flexibility because it leads to certain operations which algebraically are far from intuitive. (I'm thinking here of symplectic reduction.)

where $H$ is a finite dimensional $C^{p,q}$-module. Thus an element in $K^{p,q}$ is a pair of continuous families of lagargian subspaces in a $C^{p,q}$-module. Moreover, we it suffices consider only the case when one of the families $L_0(x)$ is constant.

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. Note that only the domain of $T_{L_0,L_1}$ depends on $L_0,L_1$.

we can associate an element $T_\alpha\in [X,\eF^{p,q}]$ given by $x\mapsto T_{L_0(x), L_1(x)}$.

As mentioned before, $\eF^{p',q'}$ classifies $K^{p',q'}$ and thus the map $T_\alpha$ defines an element ${\rm ind}\; T_\alpha\in K^{p',q'}(X)$. I proved in that old paper that ${\rm ind}\; T_\alpha $ 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.

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 $J$,

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.)

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.

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 $L_0,L_1$>

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.

$$ T_{L_0,L-1} u(s)=J\frac{du}{ds},\;\;s\in (0,1). $$

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. Note that only the domain of $T_{L_0,L_1}$ depends on $L_0,L_1$.

The familly $ T_{L(x)}$ can be given a Kasparov descriptition as explained above.

$$ T_{L_0,L_1} u(s)=J\frac{du}{ds},\;\;s\in (0,1). $$

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. Note that only the domain of $T_{L_0,L_1}$ depends on $L_0,L_1$.

The familly $ T_{\alpha}$ can be given a Kasparov descriptition as explained above.