I am mystified by formulas that I find in the condensed matter literature (see Symmetry protected topological orders and the group cohomology of their symmetry group arXiv:1106.4772v6 (pdf) by Chen, Gu, Liu, and Wen). These formulas have been used in some very interesting work in condensed matter and I would like to know how to understand them.

I begin with the simplest case. Let $G$ be a finite group. One is given an element of $H^2(G,U(1))$ that is represented by an explicit $U(1)$-valued group cocycle $\nu(a,b,c)$. This is a homogeneous cocycle, $\nu(ga,gb,gc)=\nu(a,b,c)$ and obeys the standard cocycle condition $\nu(a,b,c)\nu^{-1}(a,b,d)\nu(a,c,d)\nu^{-1}(b,c,d)=1$ for $a,b,c,d\in G$.

Let $X=G\times G$ be the Cartesian product of two copies of $G$. We consider $G$ acting on $X=G\times G$ by left multiplication on each factor. The cocycle $\nu$ is then used to define a twisted action of $G$ on the complex-valued functions on $X$. For $g\in G$ and $\Phi: X\to \mathbb{C}$, the definition (eqn. 27 of the paper) is $$\hat g(\Phi)=g^*(\Phi) \Lambda(a,b;g)$$ where $g^*(\Phi)$ is the pullback of $\Phi$ by $g$ and (with $a,b\in G$ defining a point in $X=G\times G$, and $g_*$ an arbitrary element of $G$) $$\Lambda(a,b;g)=\frac{\nu(a,g^{-1}g_*,g_*)}{\nu(b,g^{-1}g_*,g_*)}.$$ It is shown in appendix F of the paper that this does given an action of $G$ on the functions on $X=G\times G$.

The authors also describe a version in one dimension more. In this case, $\nu(a,b,c,d)$ is a homogeneous cocycle representing an element of $H^3(G,U(1))$ and satisfying the usual cocycle relation and one takes $X=G\times G\times G\times G$ to be the Cartesian product of four copies of $G$. A twisted action of $G$ on the functions on $X$ is now defined by $$\hat g(\Phi)=g^*(\Phi) \Lambda(a,b,c,d;g)$$ with $$\Lambda(a,b,c,d;g)=\frac{\nu(a,b,g^{-1}g_*,g_*)\nu(b,c,g^{-1}g_*,g_*)}{\nu(d,c,g^{-1}g_*,g_*)\nu(a,d,g^{-1}g_*,g_*)}.$$ It is shown in appendix G that this does indeed give a twisted action of $G$ on the functions on $X$.

I presume there is supposed to be an analog of this in any dimension though I cannot see this stated explicitly.

Can anyone shed light on these formulas?

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    $\begingroup$ A trivial observation. Under the assumption that the group element $g_*$ is fixed in advance, rather than being any old element whose choice doesn't matter, for fixed $g$ the functions $\Lambda$ are coboundaries. Hence, again fixing a $g_*$, $\Lambda$ is a function from $G$ to the coboundaries, so you have a kind of 'conjugation' action of $G$ on $Hom(X,U(1))$ by pre- and post-multiplication using the obvious diagonal action on $X$ and multiplication on $U(1)$. As I said, trivial observation, but this would I hope make the generalisation to higher degree cocycles obvious. $\endgroup$
    – David Roberts
    Aug 24, 2016 at 22:55
  • $\begingroup$ Why not ask the authors of this paper? unless they are as everyone else so busy with their work so that they cannot find the time to answer your question about this paper. Anyway, I wouldn't be surprised if there are more corrections to this paper; so far 6 versions to it. $\endgroup$
    – Alan
    Jan 10, 2018 at 7:23
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    $\begingroup$ @Alan There are unlikely to be more versions on the arxiv as the paper was published in 2013 journals.aps.org/prb/abstract/10.1103/PhysRevB.87.155114 $\endgroup$
    – j.c.
    Apr 5, 2018 at 20:50
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    $\begingroup$ Is this the real Edward Witten ? :)) $\endgroup$
    – Amr
    Oct 22, 2021 at 13:01

2 Answers 2


The geometric interpretation for $1$-cocyles.

Recall the following construction due to Bisson and Joyal.

Let $p:P\rightarrow B$ be a covering space over the connected manifold $B$. Suppose that the fibres of $p$ are finite. For every topological space $X$, the polynomial functor $p(X)=\{ (u,b),b\in B, u:p^{-1}(b)\rightarrow X\}$ $p(X)$ is a total space of a bundle over $B$ whose fibres are $X^{p^{-1}(b)}$.

Here we suppose $B=BG$ the classifying bundle of $G$ and $p_G:EG\rightarrow BG$ the universal cover. We suppose that $X=U(1)$. The quotient of $EG\times Hom(G,U(1))$ by the diagonal action of $G$, where $G$ acts on $Hom(G,U(1))$ by the pullback.

$\hat g(\Phi)=g^*(\Phi)$

is the polynomial construction $p_G(X)$. It corresponds to $\Lambda=0$.

Remark that we can define non zero $\Lambda$ and the definition:

$\hat g(\Phi)(a)=g^*(\Phi)\Lambda(a)$

defines a $U(1)^G$ bundle isomorphic to $p_G(X)$ and we can see these bundle as a deformation of the canonical flat connection of $p_G(U(1))$.

Interpretation of n-cocycles, n>1

2-cocycles classify gerbes or stacks. There is a notion of classifying space for gerbes. If $G$ is a commutative group, the classifying spaces of a $G$-gerbe is $K(G,2)$. Let $B_2G$ be the classifying space of the $G$-gerbes. The universal gerbe $p_G$ is a functor $:E_2G\rightarrow Ouv(B_2G)$ where $Ouv(B_2G)$ is the category of open subsets of $B_2G$. For every open subset $U$ of $B_2G$, an object of the fibre of $U$ is a $G$-bundle. We can generalize the Bisson Joyal construction here:

If $p_U:T_U\rightarrow U$ is an object of ${E_2G}_U$ the fibre of $U$, we define $p_U(X)$ the polynomial functor associated to $p_U$, we obtain a gerbe $E_2^XG$ such that for every open subset $U$ of $B_2G$, the fibre of $U$ are the bundles $p_U(X)$. Its classifying cocyle is defined by a covering $(U_i)_{i\in I}$ of $B_2G$ and $c_{ijk}: U_{ijk}\rightarrow U(1)^G$. Remark that if $\mu$ is a $U(1)$ valued $2$-cocycle, we can express $\Lambda$ with Cech cohomology and obtain a $2$-boundary $d_{ijk}$.

There exists a notion of connective structure on gerbes, a notion which represents a generalization of the notion of connection. The cocyle $c_{ijk}d_{ijk}$ is a deformation of the canonical flat connective structure defined on $E_2^{U(1)}G$.

For higher dimensional cocyles, there is a notion of $3$-gerbe, but for $n>3$, the notion of $n$-gerbes is not well understood since the notion of $n$-category which must be used to buil such a theory is not well-known also.

Bisson, T., Joyal, A. (1995). The Dyer-Lashof algebra in bordism. CR Math. Rep. Acad. Sci. Canada, 17(4), 135-140.

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    $\begingroup$ In the question, $\Phi$ is a function on $G^2$ for $n=2$ and on $G^4$ for $n=3$, could you explain how does this fit into your picture? $\endgroup$ Aug 25, 2016 at 21:34
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    $\begingroup$ OK but how does this relate to functions on $G^2$ for $n=2$ and functions on $G^4$ for $n=3$? $\endgroup$ Aug 26, 2016 at 7:14
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    $\begingroup$ That's precisely what I want to understand - what exactly is this appropriate dimension? How does it depend on the dimension of the cocycle? $\endgroup$ Aug 26, 2016 at 14:41
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    $\begingroup$ Thanks a lot, now I think I finally understand. So you only provide a geometric interpretation of what $\Lambda$ does, not a conceptual explanation of why $\Lambda$ is of this particular form, and why $f(2)=2$ and $f(3)=4$ (if I read these numbers correctly)? $\endgroup$ Aug 26, 2016 at 16:00
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    $\begingroup$ +1 for enduringly responding to questions - not all do that. $\endgroup$
    – tj_
    Aug 26, 2016 at 17:20

Maybe formula (27) is some kind of re-indexed loop transgression map in the case $n=2$. In general, loop transgression is a map

$ \tau : H^n(BG) \rightarrow H^{n-1}(L(BG))$

where $L(X) = Maps(S^1, X)$ is the free loop space of a topological space $X$. Here and elsewhere, I'll take the coefficient group to be $U(1)$.

There should be a continuous map

$ \psi : B(Y_{G \times G}) \rightarrow L(BG) $.

Here, $Y_{G \times G}$ is the groupoid (the "action groupoid") corresponding to the diagonal action of $G$ on $G \times G$ from the left, and $B(Y_{G \times G})$ is its classifying space. The map $\psi$ should be
induced by the equivariant map $G \times G \rightarrow G$ given by $(a,b) \mapsto ab^{-1}$.

Now, suppose you start with a 2-cocycle $\nu \in H^2(BG)$, and transgress it to $\tau(\nu) \in H^1(L(BG))$ and then pull it back to get $\psi^*(\tau(\nu)) \in H^1(B(Y_{G \times G}))$. Maybe one can interpret their $\Lambda (a,b; g)$ function as an element in $H^1(B(Y_{G \times G}))$, and that in fact

$ \Lambda = \psi^*(\tau(\nu)). $

However, I haven't been able to get that to work. If true, it may suggest a similar story for higher $n$.


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