If $G$ is a group and $M$ a $G$-module then for $n\geq 0$ we have an action of $G$ on the cochains from $C^n(G,M)$. If $s\in G$, $a\in C^n(G,M)$ then $(sa)_{s_1,\ldots,s_n}=sa_{s^{-1}s_1s,\ldots,s^{-1}s_ns}$, $\forall s_1,\ldots,s_n\in G$.

This action induces an action of $G$ on $H^n(G,M)$, which is known to be trivial. (See, e.g., Brown, Chapter III, Proposition 8.3.)

I'm interested in the following identity.

Let $a\in C^n(G,M)$ and $s\in G$ and let $b\in C^{n-1}(G,M)$, where $$b_{s_1,\ldots,s_{n-1}}=\sum_{k=0}^{n-1}(-1)^ka_{s_1,\ldots,s_k,s,s^{-1}s_{k+1}s,\ldots,s^{-1}s_{n-1}s}~\forall s_1,\ldots,s_{n-1}\in G.$$ Then we have $sa-a-db=c$, where $$c_{s_1,\ldots,s_n}=\sum_{k=0}^n(-1)^kda_{s_1,\ldots,s_k,s,s^{-1}s_{k+1}s,\ldots,s^{-1}s_ns}~\forall s_1,\ldots,s_n\in G.$$

As a consequence, if $a\in Z^n(G,M)$ then $da=0$ so $c=0$ so in $H^n(G,M)$ we have $[sa-a]=[sa-a-db]=[c]=[0]$ so $s[a]=[a]$ so the action of $G$ on $H^n(G,M)$ is trivial.

The proof is straightforward, but it's quite annoying to write it down. I checked it for $n\leq 3$ and it is quite easy to see the pattern. My question is whether somebody saw this somewhere. I looked for it in a couple of books, but didn't find it.

In the paper I'm writing I only need the cases $n=1$ and $2$, with $G$ commutative, so those annoying $s^{-1}s_is$ are replaced by $s_i$. This is easy to write down, only a few lines, but I would rather quote the general result, provided it is written somewhere.

  • $\begingroup$ Please could you clarify what your question is: it must be to prove an identity in cocycles (modulo coboundaries?). But on my reading, you seem to be asking for a proof that $G$ acts trivially on $H^n(G,M)$, which as you say, is in Brown. $\endgroup$ Oct 12, 2019 at 11:59
  • $\begingroup$ This identity holds for arbitrary cochains, not only for cocycles. (As I wrote, $a\in C^n(G,M)$.) As a consequence, in the particular case when $a\in Z^n(G,M)$, we get $sa-a-db=c=0$ so $sa=a+db$, which implies $s[a]=[a]$, so we get an alternative proof of the fact that $G$ acts trivially on $H^n(G,M)$. However, this is not the reason why I need this result. I need it for arbitrary cochains, not for cocycles. (And only for $n=1$ or $2$ and $G$ commutative.) I know how to prove it, but if it is already somewhere, I would rather quote it than write it down. (It's a reference request.) $\endgroup$ Oct 12, 2019 at 21:39


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