I'm reading "Elements of Noncommutative Geometry" of Garcia-Bondía. A Clifford module was defined as the following:
A Clifford module over a compact Riemannian manifold $(M,g)$ is a finitely generated projective right $C(M)$-module $\mathfrak{E}=\Gamma(E)$, corresponding to a complex vector bundle $E\to M$, together with a $C(M)$-linear homomorphism $c:\Gamma(\mathbb{C}l(M))\to\Gamma(\mathrm{End} E)$.
Note that $\mathbb{C}l(M)$ is bundle where every fibre is a complex Clifford algebra and $\Gamma(\cdot)$ is the totality of all sections.
It was said that this implies that $\mathfrak{E}$ is a B-A-bimodule, where $A=C(M)$ and $B=\Gamma(\mathbb{C}l(M))$. How can I prove this?
I need to find an operation: $\cdot:\Gamma(\mathbb{C}l(M)) \times \Gamma(E) \to \Gamma(E)$. Therefore I was thinking about the Clifford multiplication $\phi\cdot w= \rho(\phi)(w)$, where $\rho$ is a $\mathbb C$-representation $\mathbb{C}l_x\to \mathrm{Hom}(E_x,E_x)$, $\phi\in\mathbb{C}l_x(M)$ and $w\in E_x$ (notations are from "Spin Geometry" of Lawson and Michelsohn)