Consider a bosonic system with time reversal symmetry T and a unitary on-site Z2 symmetry. Suppose the symmetry is realized in a special way such that T2=(−1)B acting on any physical local operator, where B=0 if the operator is \mathbb{Z}_2 even, and B=1 if the operator is \mathbb{Z}_2 odd. How many SPT phases are there, in 1+1 and 3+1 dimensions?
1 Answer
\newcommand{\Z}{\mathbb{Z}}
I'm not entirely sure that this is what you're looking for, but it seems like the total symmetry is a G=\Z_2\rtimes \Z_2^T, with antiunitary \Z_2^T, so the number of SPT phases is inside [1] \mathrm{H}^{d+1}(G,U(1)) = \begin{cases} \Z_2^{(d+2)/2} & d = 0 \mod 2 \\ \Z^{(d+1)/2}_{2} & d = 1 \mod 2 \end{cases} So it would be two SPT phases in (1+1)D and four SPT phases in (3+1)D.
References
[1] X. Chen, Z. C. Gu, Z. X. Liu and X. G. Wen, Symmetry protected topological orders and the group cohomology of their symmetry group, Phys. Rev. B 87, no.15, 155114 (2013) [arXiv:1106.4772].
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\begingroup thanks! I think you mean G= Z_2\rtimes Z_2^T, and the cohomology is in Eq.(J122) of the paper cited above? If so, there are certain typo above, because d/2 is fractional for d=1 mod 4 etc. \endgroup Commented Jul 9, 2020 at 14:23
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\begingroup Yes, you're right, sorry about the typos. I corrected both now. \endgroup Commented Jul 9, 2020 at 14:33