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.

  • 1
    $\begingroup$ 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 '16 at 21:37
  • $\begingroup$ Looks clear to me now, still, don't know those papers. $\endgroup$ – Bob Bee Jun 29 '16 at 23:00

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.