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I am trying to understand higher-form symmetries in TQFT. In particular the higher-form version of Dijkgraaf-Witten Theory.

It is known that for a 0-form symmetry we can specify the principal G-bundle through homotopy classes of the classifying map $$ M \rightarrow BG = K(G,1). $$ This is known from Homotopy Theory and Eilenberg-MacLan spaces. Indeed the homotopy classes of these maps are in bijection with the first cohomology group $H^1(M,G)$ that for a finite group is isomorphic to $\operatorname{Hom}(\pi_1(M),G)$ and fit the usual gauge theory: $$ [M,K(G,1)] \simeq H^1(M,G) \simeq \operatorname{Hom}(\pi_1(M),G) $$

I cannot find any reference for a higher version of this. Should I expect a naive generalization? This is motivated by the fact that for a 1-form symmetry $H^2(M,G)$ works as a straightforward generalization to the previous case. But does homotopy theory tell me something about the classification of gerbes via classifying maps?


There is a follow-up question to this, when the symmetry structure is an honest 2-group.

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Higher-form symmetries are abelian so, with $G$ a discrete abelian group and $p\in\mathbb{Z}_{\geq 0}$ (or $G$ a discrete group, not necessarily abelian, if $p=0$): $$ [M, K(G,p+1)] \cong \mathrm{H}^{p+1}(M;G) \cong \operatorname{Hom}\left(\pi_{p+1}(M),G\right) $$ and everything works as it should. See e.g. the Wikipedia page for Eilenberg-McLane spaces or the paper From gauge to higher gauge models of topological phases by Delcamp and Tiwari$^{(*)}$.


$^{(*)}$In that paper they do not explicitly mention $K(G,p+1)$, but they define it descriptively as a higher-classifying space $B^{p+1}G$.

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  • $\begingroup$ Thanks for the reference, that article is terribly clear. $\endgroup$
    – Badillo
    Nov 6, 2022 at 13:57
  • $\begingroup$ I have added the natural next question. I’d gladly hear if you have something to say about it. $\endgroup$
    – Badillo
    Nov 14, 2022 at 10:46
  • $\begingroup$ @Badillo You should not edit the question to ask a further question in a way that invalidates the existing answer, see e.g. this physics-meta SE thread. What you should do, instead, is revert to the previous state of the question and ask your new question in a new thread (linking this question, if it's related). $\endgroup$ Nov 14, 2022 at 17:51

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