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edited May 21, 2013 at 16:08

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)}$$F:=F_{\frac{1}{2}(\eta_0+\eta_1^*)}$; where for $f\in \mathcal{L}(E)$, $F_f(\phi)(x):=f(\phi(x)), \phi\in C_c(X,E)$. The converse is an application of Serre-Swan Theorem (and then the operator $F$ in the Kasparov module, viewed as a linear endomorphism of a hermitian vector bundle $E$, provides the gradings $\eta_0=\eta_1=F$$\eta_0=F$ and $\eta_1=F^*$ ) ...

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 $\eta_0=\eta_1=F$) ...

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 $\eta_1=F^*$ ) ...

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created May 21, 2013 at 16:02

Mkouboi

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 $\eta_0=\eta_1=F$) ...