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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?

j.c.
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Edward Witten
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    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. – David Roberts Aug 24 '16 at 22:55
  • 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. – Alan Jan 10 '18 at 07:23
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    @Alan There are unlikely to be more versions on the arxiv as the paper was published in 2013 https://journals.aps.org/prb/abstract/10.1103/PhysRevB.87.155114 – j.c. Apr 05 '18 at 20:50
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    Is this the real Edward Witten ? :)) – Amr Oct 22 '21 at 13:01

2 Answers2

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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.

Tsemo Aristide
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    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? – მამუკა ჯიბლაძე Aug 25 '16 at 21:34
  • I believe you talking about the dimension of the cocycle, in the question homogeneous cocycles are considered, there is an equivalence between homogeneous and inohogeneous cocycles see http://math.stackexchange.com/questions/65531/what-is-the-motivation-for-defining-both-homogeneous-and-inhomogeneous-cochains – Tsemo Aristide Aug 25 '16 at 22:27
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    Sorry, yes, I should explain, $n$ is dimension of the cocycle. In the question, to each 2-cocycle is assigned an action of $G$ on functions $\Phi:G^2\to\mathbb C$, while to each 3-cocycle is assigned an action of $G$ on functions $\Phi:G^4\to\mathbb C$, and I don't understand where in your picture such correspondences can appear, regardless whether one works with homogeneous or inhomogeneous cocycles. – მამუკა ჯიბლაძე Aug 25 '16 at 22:42
  • What I am using in my picture are Cech representations of cocycles, not chain representations. – Tsemo Aristide Aug 25 '16 at 23:34
  • My picture says that there "exists" an $n$-stack $p_nG^{U(n)}$ the $n$-objects of this "$n$-stack" are $U(1)^{G}$ bundles over open subsets of $B_nG$. $\Lambda$ can be used to deform the classifying cocycle of this "n-stack" and such a deformation may correspond to a deformation of the canonical flat "n-connective structure". – Tsemo Aristide Aug 26 '16 at 01:51
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    OK but how does this relate to functions on $G^2$ for $n=2$ and functions on $G^4$ for $n=3$? – მამუკა ჯიბლაძე Aug 26 '16 at 07:14
  • The point here is the fact that if you write the classifying cocycle for $n=2$ of the gerbe $p_2G(U(1))$, you start with a contractible cover $(U_i){i\in I}$ and takes an object $e_i$ in the fibre of $U_i$, let $e_i^j$ be the restriction of $e_i$ to $U_i\cap U_j$, you choose a morphism between $c{ij}:e_j^i\rightarrow e_i^j$ this morphism is defined by a pullback by an element of $G$. To see this, consider the construction of Eilenberg MCLane space with simplices. In this sence the gerbe $p_2(U(1))$ corresponds to $\Lambda=0$. Then you deform with various $\Lambda$ – Tsemo Aristide Aug 26 '16 at 13:07
  • That's precisely where I get confused: the $K(A,n)$ made of simplices, the way I know, gives numbers which do not explain to me how 2-cocycles should produce functions on $G^2$ and 3-cocycles functions on $G^4$. For $K(A,1)$ it starts with $0,A,A^2,A^3,...$, for $K(A,2)$ with $0,0,A,A^3,A^6,...$ and for $K(A,3)$ with $0,0,0,A,A^4,A^{10},...$ – მამუკა ჯიბლაძე Aug 26 '16 at 13:32
  • Again, in my picture I don't write the cocycle $\mu$ expicitely. The IMPORTANT point is to construct a stack which corresponds to $\Lambda=0$, then to deform its classifying cocycle with $\Lambda$ for various $\Lambda$. I read $\hat g(\Phi)$ has a deformation of the case $\Lambda=0$, – Tsemo Aristide Aug 26 '16 at 13:48
  • To be more precise, the boundary $\Lambda$ constructed for the $3$-cocycle maybe use to deform the "polynomial stack" defined by $p_4^G(U(1))$. – Tsemo Aristide Aug 26 '16 at 14:27
  • Sorry I still don't get it. Do you mean that an $n$-cocycle can be used to deform some stack defined by $p_m^G(U(1))$ for any $n$ and any $m$? – მამუკა ჯიბლაძე Aug 26 '16 at 14:38
  • The point that you have raised I believe is the fact that the boundary $\Lambda$ defined by the $3$-cocycle $\nu$ is not a $3$-boundary, but a $4$-boundary, anyway $\Lambda$ maybe used to deform a stack in the appropriate dimension. – Tsemo Aristide Aug 26 '16 at 14:40
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    That's precisely what I want to understand - what exactly is this appropriate dimension? How does it depend on the dimension of the cocycle? – მამუკა ჯიბლაძე Aug 26 '16 at 14:41
  • The dimension of the stack used will depend of the boundary obtained. The goal of my answer is to give a geometric interpretation of the deformation obtained by the cocycle. I don't generalize the formula of $\Lambda$. What I claim is once we know $\Lambda$ it can be interpretated naturally as a deformation of an polynomial stack $P_{f(n)}^{U(1)}G$. – Tsemo Aristide Aug 26 '16 at 14:51
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    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)? – მამუკა ჯიბლაძე Aug 26 '16 at 16:00
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    This is the purpose of my answer. Anyway for $n>3$, there does not exist a clear notion of gerbes, different authors have different approaches. – Tsemo Aristide Aug 26 '16 at 16:03
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    I see. Well, for me origin of the construction already in the $n=2$ case is a complete mystery. I cannot relate it with any known constructions I know. – მამუკა ჯიბლაძე Aug 26 '16 at 16:08
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    +1 for enduringly responding to questions - not all do that. – tj_ Aug 26 '16 at 17:20
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    For those curious, I presume the paper by Wen et al. implies that the general formula is $f(n) = 2^{n-1}$. I don't have a conceptual explanation on the purely mathematical side, but the way these two numbers enter on the physics side is as follows: $n-1$ corresponds to the number of spatial dimensions we are considering, and $2^{n-1}$ then naturally counts the number of nearest neighbours. (More correctly, the latter counts the number of sites involved in the smallest entangled quantum state that builds up the topological phase of matter associated to a given $n$-cocycle.) – Ruben Verresen Aug 30 '16 at 18:54
  • @RubenVerresen Well in principle $2^{n-1}$ might appear in the case of square/cubical/hypercubical lattice and other kinds of hypertilings might produce other kind of $f(n)$. There are three possibilities here - for a given cocycle, either all choices of hypertilings produce equivalent/isomorphic twisted actions; or they produce nonequivalent twisted actions; or (less probable, but still possible) only hypercubical tilings produce correct formula for $\Lambda$ that might be used to twist the action. I don't see which is true. – მამუკა ჯიბლაძე Aug 31 '16 at 17:24
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    Let me add that although hypercubical tiling is the only type of regular tiling which exists in every dimension, there still are some "less regular" ones, as well as a finite amount of low-dimensional regular exceptions. – მამუკა ჯიბლაძე Aug 31 '16 at 17:25
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    I can verify that given a homogeneous $n$-cocycle of $G$ with coefficients in the abelian group $A$, there's a closed "simplicial" formula for twisted actions on functions $G^{n} \to A$ and a closed "orthoplicial" formula for twisted actions on functions $G^{2(n-1)} \to A$. Here I mean closed as a function of the cocycle and the dimension. I didn't derive it explicitly yet, but there should also be "cubical" formulas for twisted actions on functions $G^{2^{n-1}} \to A$. These correspond to the only three infinite families of regular polyhedra. – Eric S. Sep 01 '16 at 00:47
  • ... For what it's worth, after finding these formulas, it appears there ought to be lots more of them (although not closed as a function of dimension). There's probably a more general way to put this, but at least for any polytope with $v$ vertices homeomorphic to $S^{n-1}$, there will be a corresponding way to twist the action of $G$ on functions $G^v \to A$ using an $n$-cocycle. Even having observed this, I still don't really know what's going on! – Eric S. Sep 01 '16 at 00:54
  • Nice Eric, can you show us your formulas ? – Tsemo Aristide Sep 01 '16 at 02:43
  • @EricS. How do you obtain a tesselation from a regular polytope? Afaik the only tesselation by regular polytopes which exists in every dimension is the one by hypercubes. – მამუკა ჯიბლაძე Sep 01 '16 at 22:18
  • @მამუკა ჯიბლაძე You don't. Tessellations don't have anything to do with the construction of Wen et al. Rather, they give a Hamiltonian prescription for what the phase ought to look like on a small piece P of (n-1)-dimensional space (small meaning homeomorphic to a ball). You can then put the phase on an (n-1)-dimensional (say, PL) manifold M by decomposing M into a union of copies of P glued together along their boundaries. It's unclear how things behave vis-a-vis smooth vs PL, but the physicists don't really seem to care because smooth = PL in dimensions 2, 3 and 4. – Eric S. Sep 02 '16 at 01:16
  • @Tsemo. Since the formulas are too long for comments, I will post an answer containing them soon. Before I do this though, I want to stare at them a little longer, in order to see if the generalizations shed any light on the OP's original formulas. – Eric S. Sep 02 '16 at 01:22
  • @EricS. Seems like we are reading the paper in different ways. According to my interpretation of it, the number $v$ in your $G^v\to A$ is not the number of vertices of $P$ but the number of copies of $P$ meeting at each vertex when decomposing the manifold into a union of glued copies of $P$. In particular, their formulæ require that at each vertex of the decomposition the same number of copies of $P$ meet, so that this is some kind of regular tesselation. Maybe our points of view are dual to each other, I don't know. But in any case what I call tesselation is what you call decomposition. – მამუკა ჯიბლაძე Sep 02 '16 at 06:42
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    I'd be very surprised if it's actually necessary to have the same number of $P$ around each vertex. Whatever is going on in the CGLW Hamiltonians (which CGLW never really write down....), my understanding is that after gauging them, one should get Hamiltonians for the twisted Dijkgraaf-Witten theory. And the latter type of Hamiltonians can be constructed given any cell-decomposition, cf. http://mathoverflow.net/a/194633/68910 Basically, the correspondence in Walker's third example ought to lift to the "pre-gauging" SPT phase. – Eric S. Sep 02 '16 at 20:26
  • @EricS. I believe that's another question. Independently of it, the $v$ in $G^v\to A$ is, in your reading, the number of vertices of $P$ and in my reading, the number of $P$s meeting at a vertex. In both cases, if you want a constant $v$, you need some regularity restriction - either constancy of numbers of vertices of the cells or of the numbers of cells meeting at each vertex. If you want arbitrary cell decompositions, you have to deal with functions on $G^v$ with varying $v$, what kind of structure must it be? In any case, this is beyond constructions of the paper in question I believe. – მამუკა ჯიბლაძე Sep 03 '16 at 22:41
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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 $n$.

Bruce Bartlett
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