I've recently encountered the following statement:- "The map $r:Spin(p,q)\rightarrow SO(p,q)$ is a double cover". However in the proof there's no explicit construction of that map. Does someone know how it's defined?
EDIT:I was wrong, I think that the explicit map is just (given $g\in\Gamma(p,q)$) $$r_{g}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$$ $$x\mapsto \alpha(g)xg^{-1}\equiv-gxg^{-1}$$ However is still not clear to me how can I regard $r_{g}(x)$ as an element of $SO(p,q)$.
We can regard $Pin(p,q) $ as the group generated by the elements of $\mathbb{R}^n$ with $||g||^2 = 1$, and $r_g$ is the reflection through the hyperplane perpendicular to $g/||g||$: notice that $O(p,q)$ is generated by such transformations. When we restrict to the Spin group, we obtain the composition of an even number of reflections, hence the double cover restricts to $SO(p,q)$ (because reflections have determinant equal to $-1$).