# How can I justify that the compact manifold in string theory must be orientable in one dimension

I am preparing a presentation on Calabi-Yau manifolds where I do not want to use advanced mathematics, but I would like to show that under some minimal assumptions in one (complex) dimension, the compactification of spacetime in string theory would have to be by "folding the extra dimensions into a Calabi-Yau manifold", i.e. a torus.

First, we want it to allow for a Ricci-flat metric, i.e. a flat metric. Assuming Gauss-Bonnet you see that the Euler characteristic must be 0.

For supersymmetry to survive in the compact manifold, we want a Killing spinor, or (which is a bit stronger) a covariantly constant spinor. I thought to relax that to the presence of a nonvanishing vector field, so that Poincaré-Hopf gives us an Euler characteristic of 0 again.

In any case, we see that the Euler characteristic must be 0, that a torus works, but we could still have a Klein bottle.

Can anyone think of a way to make it plausible that the manifold making up the compact dimensions in string theory must be must be orientable? Or another argument why it must admit a complex structure?

Thanks!

• This seems like a physics question, not a math one. Anyway, for mathematical comments, note that Gauss-Bonnet only applies on oriented surfaces, but of course a surface with a flat metric has an oriented double cover with a flat metric, so the argument still works. Further, spin structures are only defined on oriented manifolds, so to talk about "spinors" you are doing something like a $\text{Pin}^\pm$ structure. Maybe the fact that a Pin structure has to be flying around is enough to justify that you don't want your surface to be a Klein bottle? – user98602 Sep 6 '18 at 4:53
• Ah yes, of course we cannot integrate over the whole space if it is not oriented. And for the mere existence of a spinor, we need an $\text{SO}(2)$ bundle, which also requires orientability. That should indeed be enough. Thanks! – doetoe Sep 6 '18 at 5:02