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Sep 10, 2018 at 17:46 vote accept Simon Lentner
Jun 20, 2016 at 22:29 history edited Qiaochu Yuan CC BY-SA 3.0
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Jun 20, 2016 at 22:03 comment added Branimir Ćaćić Ah, OK, that makes sense. And even in the $\mathbb{C}$-algebra case, given that $\mathbb{CP}^{2k}$ is spin$^\mathbb{C}$ but not spin, I suppose you'd might as well work with something more precisely akin to spin$^\mathbb{C}$?
Jun 20, 2016 at 21:55 comment added Qiaochu Yuan @Branimir: I guess it depends on whether you have in mind the case that $k$ is a $\mathbb{C}$-algebra or an $\mathbb{R}$-algebra.
Jun 20, 2016 at 21:47 comment added Branimir Ćaćić Unless I'm missing something, what you're describing is more like a spin$^\mathbb{C}$ structure, no? Following Plymen, to get something like a spin structure, you'd presumably need $M$ isomorphic to $\operatorname{Hom}_k(M,k)$ as $\operatorname{Cl}(V,q)$-modules, where $\operatorname{Hom}_k(M,k)$ is given the $\operatorname{Cl}(V,q)$-module structure induced by the order-reversing anti-automorphism on $\operatorname{Cl}(V,q)$.
Jun 20, 2016 at 20:18 history answered Qiaochu Yuan CC BY-SA 3.0