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I'm studding Hitchin's Generalized Calabi-Yau Manifolds https://arxiv.org/abs/math/0209099 and I've stuck here:

Suppose that $V$ is a vector space and denote its dual by $V^*$. Now we know that the $\bigwedge^\bullet(V^*)$ that is the exterior algebra over the dual space is a representation for Clifford algebra $CL(V \oplus V^*)$ by the action $$(v,\xi).\varphi=i_v\varphi+\xi\wedge\varphi , (v,\xi)\in CL(V \oplus V^*)$$

we are mainly interested in those representations of Spin group $\operatorname{Spin}(V \oplus V^*)$ that is not a representation of $SO(V \oplus V^*)$ and we call them Spinor representation. I need to find out if the restriction of this representation to the subgroup $\operatorname{Spin}(V \oplus V^*)$ of $CL(V \oplus V^*)$ is one of these representations. I see that if we take $$\rho:CL(V \oplus V^*)\rightarrow \operatorname{End}(\bigwedge^\bullet(V^*))$$ by restricting the representation we have
$$\rho:\operatorname{Spin}(V \oplus V^*)\rightarrow GL(\bigwedge^\bullet(V^*))$$ and we have $$\rho(-1)=-\operatorname{id}$$ so why it cant be a representation of $SO(V \oplus V^*)$? I know that $\operatorname{Spin}(V \oplus V^*)$ is a double cover of $SO(V \oplus V^*)$ but can't see how its relevant. That would be perfect if after figuring this out I get to understand how tensoring $\bigwedge^\bullet(V^*)$ in the space of top forms of $V$ $$\bigwedge^\bullet(V^*)\otimes (\bigwedge^n V)^\frac{1}{2}$$ will contruct another Spinor representation and in what aspects this will arise more useful constructions than $\bigwedge^\bullet(V^*)$ so Hitchin prefered this one.

Any help would be a lot appreciated.

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  • $\begingroup$ @bernard Thanks a lot for correcting my awful mistakes but I can't find the reason of changing the title? can you make some clarifications about that? $\endgroup$ Nov 7, 2018 at 19:43
  • $\begingroup$ I only changed the formatting of Spin, to make it mathematically correct. At least that's what I intended to do. Did I change something else inadvertently? $\endgroup$
    – Bernard
    Nov 7, 2018 at 19:48
  • $\begingroup$ @Bernard Yeah for a moment I was so confused. there was an extra term added which I deleted. Thanks a lot again. $\endgroup$ Nov 7, 2018 at 19:53
  • $\begingroup$ You're welcome! It's a pleasure to help. $\endgroup$
    – Bernard
    Nov 7, 2018 at 20:09

1 Answer 1

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To be able to get a well defined representation of the quotient you would want $\rho(-1)=id$, since if the representation passed to the quotient elements of the same equivalence class would have to give the same transformation.

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  • $\begingroup$ Thanks a lot, This helped me with the first part of question, do you have any ideals about the alternative representation? $\endgroup$ Nov 9, 2018 at 9:47

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