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Suppose I have a finite group $G$, and a group cocycle $\varphi\in Z^n(G,U(1))$ (trivial action on $U(1)$) evaluated at $k$ distinct values of the inputs $g_1,...,g_n$. That is, I am given a set of values $\varphi(g_{1,1},...,g_{1,n}),...,\varphi(g_{k,1},...,g_{k,n})$. What is the minimal $k$ such that the cohomology class $[\varphi]$ can be determined uniquely?

As an example, I know that for $n=3,G=\mathbb{Z}_2$, if $$ \varphi(g_1,g_2,g_3)=\nu(g_2,g_3)\nu(g_1g_2,g_3)\nu^{-1}(g_1,g_2g_3)\nu^{-1}(g_1,g_2) $$ is a coboundary, then $$\varphi(1,0,1)\varphi(1,1,1)=1$$ and the invariant above gives $-1$ for the nontrivial cocycle. So in that case $k=2$. However, if there is a generalization to that formula I am missing it.

The case I care about the most is $n=3, G=\mathbb{Z_m}$, but if there is a general answer I'd appreciate that. I see this answer for $n=2$, is there a known generalization?

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    $\begingroup$ By $\varphi\in H^n(G,U(1))$, do you mean a cocycle or a cohomology class? $\endgroup$
    – Mikhail Borovoi
    Commented Sep 11 at 13:52
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    $\begingroup$ And by $U(1)$, do you mean $\{z\in {\mathbb C}^*\ |\ z\bar z=1\}$ ? $\endgroup$
    – Mikhail Borovoi
    Commented Sep 11 at 13:55
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    $\begingroup$ @Mikhail Borovoi might have been a slight abuse of notation. I mean that $\varphi$ is a cocycle, belonging to some unknown cohomology class $[\varphi]$. Added a small edit, hope it clarifies. For the second Q, yes, with the usual group action (I guess the question makes sense for any $G$-module) $\endgroup$
    – Yarden Sheffer
    Commented Sep 12 at 6:15
  • $\begingroup$ My memory is that the Handbook of Computational Group Theory suggests that one can restrict to a generating set in one of the variables of the cocycle, but otherwise needs to know all the values for the other input. $\endgroup$
    – Eric S.
    Commented Sep 12 at 10:45
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    $\begingroup$ You can also restrict to values from a $p$-Sylow subgroup to determine the $p$-local part of the cohomology class, by Cartan-Eilenbergs stable element formula (and do this for every $p$ to fully determine it). $\endgroup$
    – Achim Krause
    Commented Sep 13 at 11:23

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