There should be a nontrivial 1+1 D bosonic SPT with D4 symmetry, which can be obtained by breaking the SO(3) symmetry of a Haldane chain to D4.
Concretely, write D4=Z4⋊Z2, denote the generator of Z4 by r, and denote the generator of Z2 by s. Then srs=r−1. Suppose this SPT is put on a triangulated manifold X. Denote the gauge connection corresponding to the Z4 by a∈C1(X,Z4), and the gauge connection corresponding to Z2 by b∈C1(X,Z2). Here a and b are thought of as 1-cochains that represent the flat Z4 and Z2 gauge connections, respectively.
I have a couple of related questions:
- Is the partition function of this SPT something like exp(iπ∫Xa∪b)? I think it somehow makes sense, but I do not fully understand it. In fact, now that Z4 and Z2 do not commute, what does such a cup product mean?
- If the above (or something like it) is indeed the partition function, how should I obtain it from the known partition function of the parent SO(3) SPT, exp(iπ∫Xw2(SO(3))), where w2(SO(3)) is the second Stiefel-Whitney class of the SO(3) gauge bundle that the parent Haldane chain is coupled to? I understand perhaps I need to first embed this D4 group into SO(3), and then pullback w2(SO(3)). But how should this be done properly?
- Now that the Z4 and Z2 do not commute, I think there should be a twist if we perform a coboundary operation on a. But besides this, is there any other constraint or relation between a and b? For example, does a∪b have any special property?