Here is another way of seeing that the pin cover is not orientable. It might provide a helpful perspective when searching for other covers that are orientable.

Two pin structures on a surface are related by a twist in $Hom (\pi_1(X), \pm 1) = H^1(X, \mathbb Z/2)$, and every twist by such a cohomology class gives another pin structure. I think the twist by the orientation class in $H^1(X,\mathbb Z/2)$ just gives the original pin structure, conjugated by a rotation by $\pi$ (which is central in the orthogonal group but not in the pin group). So the moduli space of pin-structure is a torsor over the moduli space by the group $H^1(X, \mathbb Z/2)$ modulo the orientation class.

For a genus $g$ surface with a cross-cut added, this group is $(\mathbb Z/2)^{2g}$ and carries a symplectic structure. Moreover, the monodromy of this group over the moduli space is $SP_{2g} (\mathbb Z/2)$, as we can see by removing the cross-cut and using the well-known fact that the monodromy of the moduli space of Riemann surfaces is surjective on cohomology.

This implies that this torsor doesn't cover the orientation torsor, or any other two-torsor. If it did, then the subset of $H^1(X,\mathbb Z/2)$ such that twisting by gives a pin structure that maps to the same branch of the orientation torsor would be monodromy invariant, and contain some nonzero vector, hence contain all nonzero vectors, and so the map would be constant.

To orient the moduli space you likely need to find a $\mathbb Z/2$-torsor, or a torsor by a group that admits a monodromy-invariant map to $\mathbb Z/2$.

This post imported from StackExchange MathOverflow at 2017-07-30 22:08 (UTC), posted by SE-user Will Sawin