# An exercise in group cohomology

Here is an exercise from Serre's "local fiels" when he starts to do cohomology: Let G act on an abelian group A, f be an inhomogenous n cochain, i.e. $f\in C^n(G,A).$ Define an operator T on f, $Tf(g_1,g_2,\cdots,g_n)=g_1g_2\ldots g_n f(g_n^{-1},g_{n-1}^{-1},\ldots,g_1^{-1})$. It is clear that $T^2f=f$. It is also not too hard to show $T(df)=(-1)^{n+1}d(Tf)$. Thus f is a cocycle iff Tf is, and f is a coboundary iff Tf is. When n=1, it is straightforward to see -f is cohomologous to Tf. Then the exercise wants us to show when n= 0,3 mod 4, f is cohomologous to Tf, while when n=1,2 mod 4, Tf is cohomologous to -f.
Any idea will be appreciated.

-
The reference is in Serre "Local fields" p.113 –  Ying Zhang Apr 6 '10 at 14:25
Why not modify the definition of T with signs so that you get a true map of cochain complexes, not just up to sign? –  Wilberd van der Kallen Apr 6 '10 at 14:56
The exercise is at the end of Ch. VII, §4 in the 3rd French edition, on page 121. Refering to pages in such popular books tends to be not a great idea! :) –  Mariano Suárez-Alvarez Apr 6 '10 at 15:04
@Wilberd:That will be a good idea, but it seems to me the naive modification T':=(-1)^nT doesn't work, since we have T showing up on both sides of T(df)=(−1)n+1d(Tf). @Mariano:Thanks! I should have said I mean p.113 from the translated English version GTM 67. –  Ying Zhang Apr 6 '10 at 15:29
Did I suggest to do it naively? –  Wilberd van der Kallen Apr 6 '10 at 15:29

Fix the signs as Wilberd suggests in the comments, check that you get a natural automorphism of the complex computing cohomology, see that it induces in fact an automorphism of the universal $\delta$-functor $H^\bullet(G,\mathord-)$, and see what it does in degree zero.
I think I got it now, like you said, it uses the property of $H^{\bullet}(G,-)$ being a universal $\delta$ functor, thus if I have $0\to C^{\bullet}(G,A)\to C^{\bullet}(G,A)\to 0$ with the identity map, and $0\to C^{\bullet}(G,A)\to C^{\bullet}(G,A)\to 0$ with the modified T map, on the cohomology level these commute in degree 0. By the property of being a universal $\delta$ functor, I got a commutative square with vertical arrows $H^i(G,A)\to H^i(G,A)$ that is an isomorphism, with upper horizontal arrows induced by identity and lower by T. –  Ying Zhang Apr 6 '10 at 16:40
(Sorry I've never tried typing a comm diag in a comment..) So it should be a general statement that any map that induce identity between $H^0(G,A)$ actually induce the identity on any $H^i(G,A)$? Thanks again. –  Ying Zhang Apr 6 '10 at 16:40
Any map of $\delta$-functors. Yes. –  Mariano Suárez-Alvarez Apr 6 '10 at 19:00
Yes, actually it seems we don't even need to check whether it is an automorphism on the chain level. Just knowing it is a chain map, coincide with identity on $H^0$ and then can proceed by universal property that maps between universal $\delta$ functors are unique up to an iso. (actually, only one side of the $\delta$-functor being universal will suffice.) Pls excuse me for being slow on this, now everything seem to piece together. –  Ying Zhang Apr 6 '10 at 21:33