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It is well known that $3$-cocycles in $H^3(\mathbb{Z}/n\mathbb{Z};U(1))$ are of the form $$w(a,b,c)=\exp(\frac{2\pi ik}{n^2}a(b+c-[b+c])),$$ for $k\in\mathbb{Z}_n$, $a,b,c\in\mathbb{Z}_n$ and $[b+c]=b+c\mod{n}$.
(See https://mathoverflow.net/questions/121065/explicit-3-cocycle-of-a-cyclic-group)

How do I show that this is indeed a $3$-cocycle?

According to me, the identity that must be shown is $(\delta^3w)(a,b,c,d)=0$.
Here I use $\delta^nf(x_1,...,x_{n+1})=f(x_2,...,x_{n+1})(\Pi_{i=1}^nf(x_1,...,x_ix_{i+1},...,x_{n+1})^{(-1)^i})f(x_1,...,x_n)^{(-1)^{n+1}}$.

Since we have a product of exponentials, we can only regard the sum of the terms between brackets. Thus we will only look at $w'(a,b,c)=a(b+c-[b+c])$ and we want $(\delta^3w')(a,b,c,d)=0\mod{n^2}$.

We then obtain $$\begin{align} (\delta^3w')(a,b,c,d)=&w'(b,c,d)w'(a+b,c,d)^{-1}w'(a,b+c,d)w'(a,b,c+d)^{-1}w'(a,b,c)\\ \end{align}$$ $$\begin{align} =b&(c+d-[c+d])+\\ -(a+b)&(c+d-[c+d])+\\ a&(b+c+d-[b+c+d])+\\ -a&(b+c+d-[b+c+d])+\\ a&(b+c-[b+c])\\ =a&(b-d+[c+d]-[b+c]). \end{align}$$

How is $a(b-d+[c+d]-[b+c])$ equal to $0\mod{n^2}$?

Frime Kemic
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  • Can you expand on the $[b+c]$ notation ? How is it defined in term of $b$ and $c$ ? Of course, if one takes $[b+c]=b+c$ it satisfies the congruence but it is not interesting... If it is really the reminder of the division, then the last equality may fail. – Roland May 18 '17 at 12:12
  • @wonderich In your answer to the question linked, you stated that it is easy to see that the formula satisfies the cocycle condition. Could you maybe expand on that? As you can see above, my computations do not work out – Frime Kemic Jun 07 '17 at 13:11
  • Can you please answer my question @FrimeKemic ? Maybe I can help... – Roland Jun 07 '17 at 16:14
  • @Roland I think it is the remainder of the division, but indeed you can find examples where the statement will not hold. I'm not sure what else it could be though... – Frime Kemic Jun 07 '17 at 20:52
  • Ok this is what I thought and there are indeed counter-example of the statement. (Still this is a weird notation, it also suggest we should take $a,b,c$ and $d$ in $[0;n-1]$ otherwise it will depend on the choice of the representative). What is weird is that $a$ might be invertible, so the question is really if $(b-d+[c+d]-[b+c])=0$ mod $n^2$. But if $n>2$, for the congruence to hold, we really need $b-c+[c+d]-[b+c]=0$... – Roland Jun 07 '17 at 21:43
  • In the linked mathoverflow thread, there is a comment giving an article on Arxiv in which you can get another description of the cocycle. Did you take a look ? – Roland Jun 07 '17 at 21:51

2 Answers2

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I just came across this question. Using Roland's hint about looking at the article on arxiv, I found that the correct formula is in prop 2.3

$$w(a,b,c)=exp(\frac{2πik}{n} a[\frac{b+c}{n}]),$$ where $[x]$ denotes the greater integer less than or equal to x. It is very fast to verify this is a 3-cocycle using the trick in prop 2.1 in the braided monoidal gr-categories article.

Nathan VanDer Werf
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The mistake in the original question is the following: $w'(a+b, c, d) = [a + b]\cdot (c + d - [c + d])$ instead of $(a + b)\cdot (c + d - [c + d])$. So, you should make each argument modulo $n$ before using the formula of the 3-cocycle. With this changes the formula is correct. Indeed: \begin{multline} w'(b, c, d) - w'(a+b, c, d) + w'(a, b+c, d) - w'(a, b, c+d) + w'(a, b, c) = \\ = b\cdot(c + d - [c+d]) - [a+b]\cdot(c+d-[c+d]) + a\cdot([b+c] + d - [b+c+d]) - \\ - a\cdot(b + [c+d] - [b+c+d]) + a(b+c - [b+c]) = \\ = (a+b)\cdot (c+d) + [a+b]\cdot [c+d] - (a + b)\cdot [c + d] - [a + b]\cdot (c + d) \end{multline}

This is quite symmetric expression. Let $a+b = \delta_{12} + [a+b]$ and $c+d = \delta_{34} + [c + d]$. Here $\delta_{12}, \delta_{34}\in\{0, n\}$, because the value $a + b$ can be less than $n$ (and in this case $\delta_{12} = 0$) or greater than $n$ (and in this case $\delta_{12} = n$). The situation with $c + d$ is the similar. Finally: \begin{multline} \delta^3(w')(a, b, c, d) = \delta_{12}\cdot \delta_{34} + \delta_{12}\cdot[c+d]+[a+b]\cdot \delta_{34} + [a+b]\cdot[c+d] - \\ - \delta_{12}\cdot[c+d] - [a+b]\cdot[c+d] - \delta_{34}\cdot[a+b] - [c+d]\cdot[a+b] + \\ + [a+b]\cdot[c+d] = \delta_{12}\cdot\delta_{34} = 0 (mod\ n^2). \end{multline}

user1043756
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