On a spin manifold $M$, we can define Dirac operator \begin{equation} D: \Gamma(M,S) \to \Gamma(M,S) \end{equation} and in particular $D^-: \Gamma(M,S_-)\to \Gamma(M,S_+)$. Let us consider twist the spin bundle by vector bundle $E$ wit connection, and therefore we obtain twisted Dirac operator $D_E$. Now consider special case where $E = S$ or $E = S_+$.
My question is:
Is the twisted Dirac operator $D_{E = S}: \Gamma(S\otimes S) \to \Gamma(S\otimes S)$ equal to the usual Hodge-de Rham operator $d + d^*$? (upon identifying $S \otimes S = {\Lambda ^ \bullet }{T^*}M$. Note that if we view $\Lambda^{\bullet}T^*M$ as a Clifford-module, there is a canonical Dirac operator $D|_{\Lambda^\bullet T^*M}$ which is $D|_{\Lambda^\bullet T^*M} = d+d^*$. However, in my limited understanding, I don't know if $D|_{\Lambda^\bullet T^*M} = D_{E = S}$)
Is the twisted ${D^ - }:\Gamma ({S_ - } \otimes {S_ + }) = \Gamma \left( {{T^*}M} \right) \to \Gamma (M,{S_ + } \otimes {S_ + }) = \Gamma \left( {\Lambda _ + ^2{T^*}M} \right) \oplus {C^\infty }\left( M \right)$ the same as ${d^*} + {d^ + }$?
I tried to prove they are equal. And to do so I "guessed" the explicit correspondence between bi-spinors and forms, but the result I obtained turns out to be a mess. Thank you!
Edit: Actually the problem boils down to identifying two Clifford multiplications acting on $\wedge^\bullet T^*M\otimes \mathbb{C}$: the 1st one induced from that on $S$ through tensor product, the 2nd one is the canonical multiplication, namely $X \cdot \equiv \sqrt(2)(X\wedge\;\; - \iota_X)$. But I could not find explicit identification.