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Let $a,b,c,d\in\mathbb{Z}/n\mathbb{Z}$ and $$f(a,b,c)=\exp\left(\frac{2\pi x i}{n^2}a(b+c-[b+c])\right).$$ Here $x\in\{0,...,n-1\}$ and $[b+c]=b+c\mod n$.

Now, it is given that the following holds $$f(a,b,c)f(b,c,d)f(a+b,c,d)^{-1}f(a,b+c,d)f(a,b,c+d)^{-1}=1$$

My question is:
What does $b+c\mod n$ mean here?

Thoughts:
My assumption was that it is the remainder of division by $n$. However, I calculated $a(b+c-[b+c])+b(c+d-[c+d])-(a+b)(c+d-[c+d])+a(b+c+d-[b+c+d])-a(b+c+d-[b+c+d]) =a(b-d+[c+d]-[b+c])$.
But we can find examples such that $$\exp\left(\frac{2\pi x i}{n^2}a(b-d+[c+d]-[b+c])\right)\neq 1$$ if we take $[\cdot]$ to be the remainder. So what else could it be?

Edit:
The formula given in the source in the comments is $$\delta\alpha(g_1,...,g_{k+1})=\alpha(g_1,...,g_k)^{(-1)^{k+1}}\alpha(g_2,...,g_{k+1})\prod_{i=1}^k\alpha(g_1,...,g_ig_{i+1},...,g_{k+1})^{(-1)^{i}}$$ The condition which must be satisfied is $$\delta\alpha=1$$ so in our case this would give $$f(a,b,c)f(b,c,d)f(a+b,c,d)^{-1}f(a,b+c,d)f(a,b,c+d)^{-1}=1.$$

Frime Kemic
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  • Can you mention one such example specifically? – Arthur Jun 15 '17 at 13:21
  • @skyking According to the answer of https://mathoverflow.net/questions/121065/explicit-3-cocycle-of-a-cyclic-group (and other sources I found online), $f$ should satisfy a certain condition, namely the one I state in my question. – Frime Kemic Jun 15 '17 at 13:34
  • You still haven't told us where "it is given". That equation $f(a,b,c)\ldots=1$ looks suspicious, and we would like to check that you have it right. – TonyK Jun 15 '17 at 13:43
  • You used the wrong formula in your counter example. It should be $[c+d]$ – skyking Jun 15 '17 at 13:44
  • @TonyK Sorry, I misunderstood that. The condition can be found on page 12 of http://math.ucr.edu/home/baez/qg-winter2005/group.pdf – Frime Kemic Jun 15 '17 at 14:04
  • @skyking You are right, I will search for another one – Frime Kemic Jun 15 '17 at 14:04
  • Take $n=3,a=1,b=2,c=1,d=1$ then $(b-d+[c+d]-[b+c])=2-1+2-0=3$, but $3/9$ is not an integer – Frime Kemic Jun 15 '17 at 14:17
  • "The condition can be found...": No it can't. Do you mean the page containing the heading "3. Topological Actions"? I see no such formula there. If the formula in the question is your re-phrasing of a formula from that paper, then I think you must have made a mistake. – TonyK Jun 15 '17 at 14:18
  • @TonyK See edit, please – Frime Kemic Jun 15 '17 at 14:24
  • Hmmm. I'm not going to try and decipher that.. But if you replace $f(a,b,c)$ by $f(a,c,d)$ in your product (or indeed $f(a,d,c)$), then it's easy to see that it's true; otherwise it looks false. – TonyK Jun 15 '17 at 14:41
  • @TonyK The formula is quite easy, since $k=3$ and we don't have that many terms. According to me, you should then obtain $f(a,b,c)$ and not the ones you mention – Frime Kemic Jun 15 '17 at 14:53

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