By a non-orientable Riemann surface ${\cal C}$, I mean a compact non-orientable two-manifold without boundary that is endowed with a conformal structure.
Such objects have a moduli space that is somewhat like the more familiar moduli space of oriented Riemann surfaces, but this moduli space is itself non-orientable. (See corollary 2.3 of https://arxiv.org/abs/1309.0383, where this is proved with a simple explicit example.)
My question is this: Suppose that ${\cal C}$ is endowed with a $pin^+$ structure. (This is one of the two possible analogs of a spin structure in the non-orientable case, the other being a $pin^-$ structure.) I've come to suspect that the moduli space of non-orientable Riemann surfaces with a $pin^+$ structure is itself orientable. I wonder if this is a known result and where it might be found.
Best Answer
I think the answer to my question is no, this moduli space is not orientable. This can be proved via the same example used in corollary 2.3 of https://arxiv.org/abs/1309.0383, which I cited in the question.
A Klein bottle can be constructed as a cylinder whose boundaries are replaced by crosscaps or one-sided circles. Remove a disc from the Klein bottle; we get a pair of pants or three-holed sphere, except it has one true boundary and two ``boundaries'' that are really one-sided circles. This is an unorientable manifold with boundary and it can be glued onto the boundary of any possibly orientable Riemann surface with one boundary circle to make a compact example without boundary.
The pair of pants variant described in the last paragraph has a diffeomorphism that acts trivially on the true boundary, exchanges the two one-sided circle ''boundaries'', and preserves the orientation that one would have if one omits the two one-sided circles. (This diffeomorphism acts as a Dehn twist by pi near the true boundary.) One can pick a pin^+ structure on the pair of pants variant that is preserved by the diffeomorphism. This diffeomorphism reverses the orientation of the moduli space of conformal structures on the Riemann surfaces in question, basically because it exchanges the two ''length'' coordinates associated to the one-sided circle ''boundaries.''