Let $Γ \subset SO(n + 1)$ be a finite group acting freely on the unit sphere $S^n \subset \mathbb{R}^{n+1}$. Show that the quotient $\Gamma \backslash S^n$ admits a spin structure if and only if there is a subgroup $\tilde{\Gamma} \subset \operatorname{Spin}(n+1)$ such that the covering homomorphism $\tilde{Ad}: \operatorname{Spin}(n+1)\to SO(n+1)$ restricts to an isomorphism $\tilde{\Gamma} \cong Γ$.
My question: For the "only if" part, the map $\tilde{\Gamma} \backslash \operatorname{Spin}(n+1) \to \Gamma \backslash S^n$ gives a spin structure. However I am stuck at the "if" part. How would I construct this $\tilde{\Gamma}$?
Thank you very much!
I think this argument works.
Suppose that $S^n / \Gamma$ is Spin. Let $Q$ be the $SO(n)$ frame bundle of the sphere an let $Q / \Gamma$ be the $SO(n)$ frame over $S^n / \Gamma$. Let $P$ be the $Spin(n)$ reduction of $Q$. The sphere is simply connected so $P$ is a simply connected cover of $Q$.
Now the long exact sequence of $SO(n+1) \to Q/\Gamma \to S^n / \Gamma$ gives via the long exact sequence of homotopy groups $$ \pi_2(Q) = 0 \to \mathbb{Z}_2 \to \pi_1(Q/\Gamma) \to \pi_1(S^n / \Gamma) \cong \Gamma \to 0$$ and since the latter is spin we have that the sequence splits (see Dirac operators in Riemannian Geometry by Friedrich, page 38) we have that $\pi_1(Q/\Gamma) = \Gamma \oplus \mathbb{Z}_2$, on the other hand since $P \to Q/\Gamma$ is the universal cover with fiber $Ad^{-1}(\Gamma)$ we should intuitively have $$ Ad^{-1}(\Gamma) \cong \pi_1(Q / \Gamma) \cong \mathbb{Z}_2 \oplus \Gamma $$ So that you can choose a preimage $\tilde \Gamma =\{0\} \oplus \Gamma$ of $\Gamma$ under $Ad$.
Edit: actually as $Q$ is isomorphic to $SO(n+1)$ and $P$ is probably its double cover $Spin(n+1)$, this argument migth be rigourous as all cover maps are lie group homomorphisms and fundamental groups are identified with the kernels of such maps.