By having a time-reversal symmetry I mean that there is a local anti-unitary symmetry (representing the non-trivial element of $Z_2$) of the state-sum construction (or, if you want, of the associated Hamiltonian). In other words there is a codimension-1 anti-unitary defect, or yet in other words there is a local basis in which all tensors involved in the state-sum have real entries.

Such a symmetry often exists, as for example in the case of the group $Z_2$. However I see no reason why such a symmetry should be there in general, and it seems to me that it actually might not exist for $Z_3$ with one of the non-trivial group cocycles.

For a theory with time-reversal symmetry all invariants associated to oriented $3$-manifolds should be real. Are there manifolds to which the non-trivial $Z_3$ (or some other) Dijkgraaf-Witten theory associates a non-real number? (By construction the invariant is real on manifolds with reflection symmetry, so one would have to test oriented 3-manifolds without reflection symmetry. Guess those exist?)

The motivation why I'm asking is that in physics, models like Dijkgraaf-Witten are called "non-chiral" because they allow gapped boundaries, but on the other hand, people refer to models as "non-chiral" if they have a time-reversal symmetry. I feel that those two notions of "non-chiral" have a large overlap but are not exactly equivalent.

  • $\begingroup$ this is a really great question. A related question, which I am going pose with absolutely no explanation: Is the trace of the T-matrix for the Drinfeld center of a unitary fusion category always real? I just spent the past hour computing and it seems like the answer might be yes.... $\endgroup$ Jun 22, 2019 at 12:03
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    $\begingroup$ @ArunDebray - I would say that if $Z$ admits a unitary symmetry associated with orientation-reversal, then it can be defined on unoriented manifolds. People usual stipulate that time-reversal symmetries are anti-unitary. $\endgroup$ Jun 22, 2019 at 13:45
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    $\begingroup$ @DanielBarter Well the trace of the T-matrix is nothing but the invariant associated to a $3$-manifold, namely $T_2\times I$ (with $T_2$ the $2$-torus and $I$ the interval) where we glue the two boundary components in a Dehn-twisted manner. Don't know much about $3$-manifolds, does this one not have a reflection symmetry? $\endgroup$
    – Andi Bauer
    Jun 22, 2019 at 18:15
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    $\begingroup$ I would say it is true for $[\omega]\in H^3(G,U(1))$ if and only if there is an automorphism $\alpha$ of $G$, such that $\alpha^\ast[\omega]=[\omega^{-1}]$. $\endgroup$ Jun 22, 2019 at 21:55
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    $\begingroup$ They have different T-matrix, so the answer is yes! Explicitly, the T-matrix $T=(T_{x,y})_{0\leq x,y\leq8}$ is given by $T_{x,y} = \delta_{x,y}\exp(\pm2\pi ix^2/9)$, respectively. $\endgroup$ Jun 22, 2019 at 22:03


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