# Spinor representation for $\operatorname{Spin}(V \oplus V^*)$

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.

• @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? Nov 7, 2018 at 19:43
• 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? Nov 7, 2018 at 19:48
• @Bernard Yeah for a moment I was so confused. there was an extra term added which I deleted. Thanks a lot again. Nov 7, 2018 at 19:53
• You're welcome! It's a pleasure to help. Nov 7, 2018 at 20:09

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.