# A formula that proves that $G$ acts trivially on $H^*(G,M)$

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.

• 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. Oct 12, 2019 at 11:59
• 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.) Oct 12, 2019 at 21:39