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Suppose that we have a $d$ dimensional bosonic SPT phase, protected by some $p$-form symmetry, $G^{[p]}$. Suppose also that it is classified within group cohomology, so that we don't have to run into cobordisms. It is not clear to me what type of cohomology actually classifies higher-SPTs.

In the original Chen-Gu-Liu-Wen paper [1] it is proposed that (normal, 0-form) SPTs are classified by $\mathcal{H}^d(G,U(1))=H^d(BG,U(1))$, where $\mathcal{H}^{\bullet}$ is the Borel cohomology, whereas $H^\bullet$ is the normal topological cohomology and $BG$ is the classyfying space of $G$.

Then, for example in [2], it is mentioned that SPTs protected by a $p$-form symmetry are classified by $H^d(B^{p+1}G,U(1))$, where $B^{p+1}G= K(G,p+1)$, the Eilenberg-MacLane space. I'm assuming that $H$ in [2]'s notation is again the topological cohomology. I suppose that I could write $H^d(B^{p+1}G,U(1)) = \mathcal{H}^d(B^p G, U(1))$, since I'm just removing a classifying space. Can I reduce it even further, until I'm left with just $G$, say reach $\mathscr{H}^d(G,U(1))$? What type of cohomology would $\mathscr{H}^\bullet$ be, and how would one go about computing it? Or if it's easier without doing that, how does one compute $H^d(B^{p+1}G,U(1))$?

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 15, 155114 (2013), doi:10.1103/PhysRevB.87.155114 [arXiv:1106.4772].

[2]: C. Delcamp and A. Tiwari, From gauge to higher gauge models of topological phases, JHEP 1810, 049 (2018) doi:10.1007/JHEP10(2018)049 [arXiv:1802.10104].

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  • $\begingroup$ When you go from $B^{p+1}G$ to $B^pG$, why do you obtain Borel cohomology? Shouldn't $B^pG$ be a topological group for this? $\endgroup$
    – NDewolf
    Commented Jan 30, 2020 at 15:52
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    $\begingroup$ @NDewolf Higher-form symmetries are abelian and the classifying space of an abelian group is an abelian group, so $B^pG$ is an abelian group. $\endgroup$
    – ɪdɪət strəʊlə
    Commented Jan 30, 2020 at 16:37
  • $\begingroup$ Well, you could definitely define $\mathscr{H}^d(G,U(1)):=H^d(B^{p+1}G,U(1))$, but I don't know how that would be useful. In general, and to the best of my knowledge (which is admittedly rather limited), higher-form homology cannot be reduced to regular group homology. (For example, the order of a regular homology group is bounded by the order of $G$, but that of higher-form homology is not, so there is no "simple" way the latter could be computed using the former). The tool to compute $H^d(B^{p+1}G,U(1))$ is spectral sequences, but that belongs on math.SE (or overflow), not here, right? $\endgroup$
    – AccidentalFourierTransform
    Commented Feb 1, 2020 at 22:09

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