Question about the covering space of a spin manifold

Question

I'm a little confused about how to prove that a covering space $\widetilde{M}$ of a spin manifold $M$ is also a spin manifold. If not, is there some counterexample?

Could you please give me some help with the details? Thanks a lot.

Answer 1

Suppose $\pi:\widetilde{M}\rightarrow M$ is a covering map of smooth manifolds and that $M$ is spin. This means what $w_1(TM) = w_2(TM) = 0$ where $w_i(TM)\in H^i(M;\mathbb{Z}/2\mathbb{Z})$ denote the Stiefel-Whitney classes and $TM$ denotes the tangent bundle to $M$

Proposition: The pull back bundle $\pi^\ast(TM)$ is isomorphic to $T\widetilde{M}$.

Proof: Recall that by definition, $\pi^\ast(TM) = \{(\tilde{p},(p,v))\in \widetilde{M}\times TM: \pi(\tilde{p}) = p\}$, where the projectoin $\pi^\ast(TM)\rightarrow \widetilde{M}$ maps $(\tilde{p},(p,v))$ to $\tilde{p}$.

Define a map $\phi:T\widetilde{M}\rightarrow \pi^\ast TM$ by $\phi(\tilde{p}, w) = (\tilde{p}, (\pi(\tilde{p}), \pi_\ast w))$. I will leave it to you to verify that $\phi$ is a bundle isomorphism. $\square$

Now, using naturality of Stiefel-Whitney classes, we find that for $i=1,2$ that $$w_i(T\widetilde{M}) = w_i(\pi^\ast(TM)) = \pi^\ast(w_i(TM)) = \pi^\ast(0) = 0.$$ Thus, $\widetilde{M}$ is also spin.