# The relation between Dirac and Hodge-de Rham operator

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.

• According to Berline--Getzler--Vergne (p. 130), this is indeed the case. I don't have a copy of Lawson--Michelsohn at hand, but I suspect it works out this identification in some detail. Incidentally, it is perhaps a bit better to write $\Lambda T^\ast M \cong S \otimes S^\ast$ (i.e., $S$ twisted by $S^\ast$); it makes absolutely no difference in the spin case, but this is the identification that carries over to the spin$^\mathbb{C}$ case. Nov 3, 2013 at 20:40
• Thanks for the reply. I also checked Lawson's book, but I didn't find/recognize explicit(for a less-trained reader) discussion. Actually things boils down to the two Clifford multiplication: the first one induced from that of $S$ through tensor product; the second one is the canonical one on $\Lambda^\bullet T^*M$, namely $X\cdot \equiv X\wedge - \iota_X$. If the two multiplications are identical, then everything follows. Any idea how to see the identification? Nov 5, 2013 at 21:25
• I think I know the answer now, by just explicitly compute the matrices, following Salamon's note. Nov 6, 2013 at 1:44