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Sep 12, 2020 at 18:01 comment added Igor Khavkine @TimCampion For the record, your (2) need not presume that the fermion field is a spinor. The point is to capture the anti-commutativity, not the half-integer spin property. If you want your fermion to also be a spinor, a spin structure has to be provided just as in (1) and you have to parity shift $\Pi\mathcal{V}$ the appropriate spinor bundle, as indicated in this answer. Only when $\mathcal{V}$ is locally trivialized to such spinor fermion fields look like simple superfunctions on $M$.
Sep 12, 2020 at 15:47 comment added user1504 Yes, the structure sheaf is locally isomorphic to $C^\infty \otimes \wedge E$, and the transition functions for E give you the spin structure. But E has rank 0 in the special case of real manifolds, so there's no transition data.
Sep 12, 2020 at 15:35 comment added Tim Campion But now I think maybe I see -- when you expand a superfunction in local coordinates, it looks a lot like a section of a bundle of Clifford algebras. So maybe the spin structure is obtained as a subbundle of this associated Clifford bundle (via the multiplicative embedding of $Spin(p,q)$ into $Cl(p,q)$)? Part of what makes it confusing is that ordinary manifolds are special cases of supermanifolds, and not every ordinary manifold admits a spin structure...
Sep 12, 2020 at 15:35 comment added Tim Campion Thanks -- this is exactly the sort of thing I was hoping for, right from the very first sentence! I guess one aspect that puzzles me is that when things are presented as in (2), it's not clear how extract a spin structure from the supermanifold data. And in this respect, the appeal to Batchelor's theorem confuses me because Batchelor relates relates the "super" structure to a plain vector bundle -- with structure group $GL(n)$ -- so where does the spin come from?
Sep 12, 2020 at 15:07 history answered user1504 CC BY-SA 4.0