Theorem 4.1 of Heat Kernels and Dirac Operators involves an isomorphism $$C(V)\otimes\mathrm{End}_{C(V)}(E)\to\mathrm{End}(E).$$ Let me call it $\Phi$. Unfortunately the definition of $\Phi$ is not mentioned in the book. I would like to explain what I believe to be the correct definition and hear your opinion:
I think that there are two equivalent ways to define the isomorphism:
- Probably the most obvious guess is $\Phi(a\otimes b)=c_E(a)\circ b$. Of course it is not clear this $\Phi$ is invertible, but this follows from the fact that this definition of $\Phi$ is equivalent to the following one:
- Let $T\in L(S\otimes W,E)$ be the isomorphism of Clifford modules defined in proposition 3.27. $S$ is the spinor module, i.e. $S$ is a Clifford module and the Clifford multiplication $$C(V)\to\mathrm{End}(S)$$ is an isomorphism of super-algebras. In addition the book mentions that $\mathrm{End}(W)$ and $\mathrm{End}_{C(V)}(E)$ are isomorphic. As explained here I believe that the isomorphism simply maps $b\in\mathrm{End}(W)$ to $T\circ(1\otimes b)\circ T^{-1}$. Let $b/S\in\mathrm{End}(W)$ be the element of $\mathrm{End}(W)$ associated to $b\in\mathrm{End}_{C(V)}(E)$, then we can define a clearly invertible $\Phi$ through the following formula: $$\Phi(a\otimes b)=T\circ (c_S(a)\otimes b/S)\circ T^{-1}$$ Let me conclude by explaining why both definitions are equivalent: $$T\circ(c_S(a)\otimes b/S)\circ T^{-1}=\underbrace{T\circ(\underbrace{c_S(a)\otimes 1}_{=c_{S\otimes W}(a)})\circ T^{-1}}_{=c_E(a)}\circ\underbrace{T\circ(1\otimes b/S)\circ T^{-1}}_{=b}=c_E(a)\circ b$$