Is there a generalization of De Rham cohomology for spinors fields?

I can see that one can construct p form fields out of spinor field by contraction of the type $\bar{\psi} \gamma^{a_1} \gamma^{a_2}...\gamma^{a_p}\gamma$.

Now we can consider the integral of the p form fields on p-cycles. There is a natural derivative like operation acting on spinor fields $\gamma \cdot \partial$. Does this map have the the desired properties to make a cohomology in some way. If I cannot use this map can I use some other operator to suitably generalize the exterior derivative operator.

I have a background in physics and not in mathematics, please keep that in view when you write your answer.

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    $\begingroup$ What do you mean by spinor fields? You can construct a complex of forms with values in any vector bundle, and then try to compute its homology. You may also want to check the Atiah--Singer theorem concerning the relation between analytic and topological invariants. $\endgroup$ – Alex Degtyarev Jun 18 '15 at 9:05
  • $\begingroup$ @AlexDegtyarev Thanks, I will learn about the things you mentioned. I have not learnt about the subject rigorously. But I know there is a definition , A section of the spinor bundle S, is called a spinor field. en.wikipedia.org/wiki/Spinor_bundle . I have some understanding from physics,like the Dirac field. Where one considers the representation of the clifford algebra $\{\gamma^\mu,\gamma^\nu\} = g^{\mu\nu}$.say g is a minkowski metric. There is representation of size $2^{d/2}$, which may be reduced. So the what I mean by spinor field is the vectorfield on which the clifford acts. $\endgroup$ – Prathyush Jun 18 '15 at 10:23
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    $\begingroup$ Exactly! Just one spinor field is not enough to speak about $d$: you need several bundles. Usually these are $E\otimes\Omega^p$ for a fixed $E$. (A very good introduction to this is Hirzebruch's "Topological methods in algebrac geometry".) Alternatively, in the spinor case, you can speak about the Dirac operator and its index, which equals the signature of the manifold by (Hirzebruch--)Atyah--Singer. $\endgroup$ – Alex Degtyarev Jun 18 '15 at 10:28
  • $\begingroup$ Crossposted from math.stackexchange.com/q/1329146/11127 $\endgroup$ – Qmechanic Jun 19 '15 at 6:52

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