From a recent paper by Kapustin(https://arxiv.org/abs/1406.7329), he argued that for non-orientable manifold with spin structure $Pin^{\pm}$, the corresponding time reversal symmetry $T$ squares to ${\mp}1$. The reason for this, as stated in his paper, is that reflection is related to time reversal symmetry by a Wick rotation. So if his claim is correct, wouldn't that contradict some of the results obtained in a recent paper by Ashvin(http://arxiv.org/abs/1505.04193)? In that paper, the essential argument for a tighter bound of filling is to avoid Kramer's degeneracy since time reversal symmetry $T$ squares to $-1$. However, for non-orientable Bieberbach manifolds, we should not have this degeneracy since we have $Pin^{-}$ spin structure on the manifold and $T$ should instead square to $1$ according to Kapustin.

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    $\begingroup$ Your first sentence is not a full sentence, and I'm not sure what exactly it is saying. Generally, I'm having a hard time understanding your writing. I understand that you may not be a native English speaker, but please have another look at your writing and try to make it clearer what you're trying to say: Pay special attention to writing complete sentences that are grammatically correct. $\endgroup$
    – Danu
    Jun 29, 2016 at 21:37
  • $\begingroup$ Looks clear to me now, still, don't know those papers. $\endgroup$
    – Bob Bee
    Jun 29, 2016 at 23:00

1 Answer 1


The two papers talk about very different things. In Kapustin's paper, he considered non-orientable space-time manifold to classify SPT (i.e. the partition function of the phase on these manifolds). To do that, one has to first Wick rotate to Euclidean space-time, where time-reversal becomes a mirror reflection, but with a sign change.

In Watanabe et. al. they consider the spatial manifold of the system to be a Bieberbach one, and they do not want to "twist" the space-time manifold at all, since it's all in Hamiltonian formalism. So the Pin structure there just depends on how reflections act on fermions, and time-reversal remains a global symmetry of the system.


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