Consider a bosonic system with time reversal symmetry $\mathcal{T}$ and a unitary on-site $\mathbb{Z}_2$ symmetry. Suppose the symmetry is realized in a special way such that $$\mathcal{T}^2= (-1)^B$$ 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 1



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.


[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].

  • $\begingroup$ 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. $\endgroup$
    – user34104
    Jul 9, 2020 at 14:23
  • $\begingroup$ Yes, you're right, sorry about the typos. I corrected both now. $\endgroup$ Jul 9, 2020 at 14:33

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