If $M$ is non-orientable, then it has a finite cover which is orientable (in particular, the orientable double cover).


If $M$ is non-spin, then it does not necessarily have a finite cover which is spin, e.g. $M = \mathbb{CP}^2$. As a cover of a spin manifold is spin, a necessary condition for $M$ to admit such a finite cover is that its universal cover is spin (which is not the case in the previous example). In analogy with the first sentence, note that the universal cover is always orientable.



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> Let $M$ be a closed smooth manifold whose universal cover is spin. Is there a finite cover of $M$ which is spin?
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This question is partially motivated by the study of positive scalar curvature. The Dirac operator on a spin manifold can be used to obtain obstructions to positive scalar curvature à la Lichnerowicz, Hitchin, Gromov & Lawson, Rosenberg, etc. More generally, these techniques can be applied to manifolds which admit a spin cover. Things are generally more difficult in the non-compact case than in the compact case, so if the answer to my question were 'yes', we could just pass to a compact cover which is spin and apply the techniques there (as opposed to passing to the potentially non-compact universal cover).



If the universal cover of a manifold is spin, must it admit a finite cover which is spin?
No, this is not true: for each dimension $d \geq 4$, there is a closed, oriented $d$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.


The reason is simply that there are finitely presented groups which have no nontrivial finite quotient.
One example is Higman's group $H$, see
<https://en.wikipedia.org/wiki/Higman_group>.


The key features of $H$ are:


1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ has a classifying space $BH$ which is a finite $2$-dimensional CW-complex.


The proof of 1,2 can be found in Tao's blog <https://terrytao.wordpress.com/2008/10/06/finite-subsets-of-groups-with-no-finite-models/>,
and the proof of 3,4 in ''The topology of discrete groups'' by Baumslag, Dyer, Heller).


Now pick an element $1 \neq x \in H$, which induces an injective homomorphism $\mathbb{Z} \to H$ and form
the amalgamated product $G=H \ast\_{\mathbb{Z}} H$. The group $G$ is infinite and has no nontrivial homomorphism to a finite group $F$, since any homomorphism
$G \to F$ must vanish on the two copies of $H$.


The pushout $BH \cup\_{S^1} BH$ is aspherical by Whiteheads asphericity theorem and hence a model for $BG$.
I have designed things so that $H\_2(BG) \cong \mathbb{Z}$ and all other homology groups are trivial.
In particular, $G$ is perfect, and the Quillen plus construction $BG^+$ must be homotopy equivalent to $S^2$, so that there is a homology equivalence
$f:BG \to S^2$. Now let $V \to S^2$ be the nontrivial oriented vector bundle of rank $d$, which has $w\_2 (V) \neq 0$. It follows that the vector bundle
$f^\ast V \to BG$ is not spin. $BG$ has no nontrivial cover, and $BG$ is aspherical, so the pullback of $f^\ast V$ to the universal cover is trivial.


Now there exists, when $d \geq 4$, a closed $d$-manifold $M$ with a $2$-connected map $\ell: M \to BG$ and a bundle isomorphism $TM\oplus \mathbb{R}\cong \ell^\ast f^\ast V \oplus \mathbb{R}$.
This is achieved by surgery below the middle dimension. In particular, $\pi\_1 (M)\cong G$. Hence $\pi\_1(M)$ has no nontrivial finite index normal subgroup, and therefore no nontrivial finite index subgroup at all. It follows that $M$ does not have a nontrivial finite cover.


By construction, $w\_2 (TM) \neq 0$, but the universal cover of $M$ is stably parallelizable.


