Does the triviality of the orthonormal frame bundle imply the triviality of the spin bundle? Let $M$ be a space and time orientable spin semi-Riemannian manifold of signature $(p,q)$, ${\rm Fr}(M)$ be its bundle of space and time oriented pseudo-orthonormal frames, $\Lambda : P\rightarrow {\rm Fr}(M)$ be the covering by the ${\rm Spin}(p,q)$-principal bundle $P$ given by a spin structure. I am trying to prove that if $M$ is metric parallelizable, then $P\rightarrow M$ must be trivial. I think it is true from an indirect argument with the additional assumption that the number of spin structures is finite.
What I tried is this: let $e$ be a global section of ${\rm Fr}(M)$. There exists a lift $\tilde e$ to $P$ iff $e_*(\pi_1(M))\subset \Lambda_*(\pi_1(P))$, and I'm basically stuck there. I tried the long exact sequence of homotopy groups, but it didn't help. I would appreciate any help, or a counter-example if I am mistaken.
 A: Recall that the set of spin structures on an $SO(n)$-bundle $E$ is either empty, or is affine over $H^1(M;\Bbb Z/2)$, which may be identified with the set of double covers of $M$ (that is, principal $\Bbb Z/2$-bundles over $M$). This is proved as Theorem 2.1.7 in Lawson and Michelsohn's book "Spin geometry"; this only discusses the case of definite characteristic. In particular, this implies that the spin structure is unique if $H^1(M;\Bbb Z/2) = 0$, and is non-unique otherwise. If two spin structures are both trivializable, then they are in particular isomorphic, so the answer to your question is: $E$ is trivial implies any spin structure on $E$ is trivializable if, and only if, $H^1(M;\Bbb Z/2) = 0$. 
The proof of this runs through a certain short exact sequence coming from the Serre spectral sequence (their formula 2.1.4). I will not discuss this precisely here, but the following discusses how to change their argument so that it remains true (without mentioning the details).
Their argument works without much change for $\text{Spin}(p,q)$, which you'll recall is a double cover of $SO^+(p,q)$ (which deformation retracts onto $SO(p) \times SO(q)$.)  Unfortunately, they relied on the statement 

The spin structures on $E$ are in natural one-to-one correspondence with 2-sheeted coverings of $P_{SO}(E)$ which are non-trivial on the fibers of $\pi$.

Here $P_{SO}(E)$ is the corresponding principal $SO(n)$-bundle to the oriented vector bundle $E$. Because $SO^+(p,q)$ has three inequivalent non-trivial double covers when $p,q \geq 2$, this definition will no longer apply. Instead, one should say

The spin structures on $E$ are in natural one-to-one correspondence with 2-sheeted coverings of $P_{SO}(E)$ which restrict to the standard covering of $SO^+(p,q)$ on each fiber of $\pi$.

Here the "standard double cover" is the double cover $\text{Spin}(p,q) \to SO^+(p,q)$. You can identify it by using the fact that $$H^1(SO^+(p,q);\Bbb Z/2) = H^1(SO(p);\Bbb Z/2) \times H^1(SO(q);\Bbb Z/2),$$ and demand that the corresponding cohomology class should restrict to the standard generator of each factor (which is the unique non-trivial element when $p,q \geq 2$, but is just zero when $p = 1$ or $q = 1$).
This now pins down a specific cohomology class in $H^1(SO^+(p,q);\Bbb Z/2)$ (said differently, a specific double covering) that your double cover should restrict to on each fiber. And from here the argument proceeds as expected; in fact, the following statement is true for any $p,q$; we do not need to demand $p,q \geq 2$.

Let $P$ be a principal $\text{SO}^+(p,q)$-bundle. Then if $P$ admits any spin structure, that spin structure provides a bijection between the set of spin structures on $P$ with the set $H^1(M;\Bbb Z/2)$. In particular, a trivial bundle has $|H^1(M;\Bbb Z/2)|$-many spin structures, and so carries $|H^1(M;\Bbb Z/2)| - 1$ non-trivial spin structures.

