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Group cohomology and condensed matter

+ 14 like - 0 dislike
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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?


This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user Edward Witten

asked Aug 24, 2016 in Theoretical Physics by Edward Witten (180 points) [ revision history ]
edited Aug 27, 2016 by Dilaton
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.

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user David Roberts

1 Answer

+ 5 like - 1 dislike

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 if we suppose that the action of $G$ on $U(1)$ is not trivial, 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.

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user Tsemo Aristide
answered Aug 25, 2016 by Tsemo Aristide (40 points) [ no revision ]
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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?

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user მამუკა ჯიბლაძე
OK but how does this relate to functions on $G^2$ for $n=2$ and functions on $G^4$ for $n=3$?

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user მამუკა ჯიბლაძე
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 $\Lambda$

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user Tsemo Aristide
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 $0,0,0,A,A^4,A^{10},...$

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user მამუკა ჯიბლაძე
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 $\Lambda=0$,

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user Tsemo Aristide
Most recent comments show all comments
+1 for enduringly responding to questions - not all do that.

This post imported from StackExchange MathOverflow at 2016-08-27 09:42 (UTC), posted by SE-user tj_

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