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I have been reading the appendix in Serre's Local fields, to do with explicit computations of cup products (pg 176), but I'm stuck on one bit of lemma 4. It goes as follows

Let B be a $G$-module, $u: G \times G \rightarrow B$ a 2-cocycle and $\bar{u} \in H^2(G,B)$. Then for all $s \in G$ with $\bar{s} \in \widehat{H}^{-2}(G,\mathbb{Z})$. Then $$\bar{s} \cup \bar{u} = \bar{a}, \qquad \text{with } a=\sum_{t \in G} u(t,s)$$

Now to prove it he begins by taking a sequence $0 \rightarrow B \rightarrow B' \rightarrow B'' \rightarrow 0$ with $B'$ induced. Now since $H^2(G,B')=0$, there is a 1-cochain $f': G \rightarrow B$ such that $$u(x,y)=xf'(y)-f'(xy) + f(x), \qquad x,y \in G.$$ Then composing $f'$ with $B' \rightarrow B''$ we get a 1-cocycle $f'':G \rightarrow B$, with $d(\bar{f''})=\bar{u}$. From which he deduces that $$\bar{s} \cup \bar{u} = \bar{s} \cup d(\bar{f''})=d(\bar{s} \cup \bar{f''})$$

Now by the previous lemma we know $\bar{s} \cup \bar{f''}= \bar{f''(s)}_0$ (here the subscript 0 denotes its class in $H^{-1}(G,B''))$. Now this is where I'm stuck he then says that $d(\bar{s} \cup \bar{f''})=d( \overline{f''(s)}_0)= \overline{N(f'(s))}^0$ (where the superscript 0 denotes its class in $\widehat{H}^0$), my question is why does the $f''$ change to $f'$? I cant see why this is so, he also does something similar in lemma 2 of this same appendix, but this is the lemma I need to use.

Thank you

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It follows from the definition of the norm $N$, and is what he calls his element "a", which is $N(f') = \sum u(t,s)$. –  Chris Gerig Feb 7 '13 at 23:54
Thats the thing I though since $d$ was given by a norm, it would be that $d(f'')=N(f'')=\sum_{g \in G} g.f''(s)$? –  Chris Birkbeck Feb 8 '13 at 0:33
but it's not just that, $d(f'') = \bar{u}$, and you can see its relation to $N$ on the previous page. –  Chris Gerig Feb 8 '13 at 0:43

1 Answer 1

up vote 2 down vote accepted

The critical point with your problem is to understand how the connecting homomorphism is computed. Let $P$ be a complete resolution . Then Tate cohomology is the cohomology of the cocomplex $Hom_G(P,-)$. Let $$0 \to B \xrightarrow{i}B' \xrightarrow{\rho}B''\to 0$$ be exact. We have a commutative diagramm with exact rows: $$\begin{array}{ccccc} Hom_G(P_{-1},B) & \overset{i_\ast}{\hookrightarrow} & Hom_G(P_{-1},B') & \overset{\rho_\ast}{\twoheadrightarrow} & Hom_G(P_{-1},B'') \newline \delta \downarrow & & \downarrow\delta' & & \downarrow\delta'' \newline Hom_G(P_0,B) & \overset{i_\ast}{\hookrightarrow} & Hom_G(P_0,B') & \overset{\rho_\ast}{\twoheadrightarrow} & Hom_G(P_0,B'') \end{array}\tag{1}$$ By using $P_0=\mathbb{Z}G$ and $P_{-1}=Hom_G(P_0,\mathbb{Z}G)$, diagram $(1)$ becomes

$$\begin{array}{ccccc} B_G & \overset{i}{\hookrightarrow} & B_G^' & \overset{\rho}{\twoheadrightarrow} & B_G^''\newline \delta \downarrow & & \downarrow\delta' & & \downarrow\delta'' \newline B^G & \overset{i_\ast}{\hookrightarrow} & B^{'G} & \overset{\rho}{\twoheadrightarrow} & B^{''G} \end{array}$$

where $B_G = B/C$ with $C := \langle (g-1)b\mid g\in G,b\in B\rangle$ are the coinvariants and $\delta=N$ is multiplication with the norm of $G$ (this is shown at the beginning of VI,§ 4 in Brown's book).

Now $\overline{f''(s)} := f''(s) + C^'' =\rho(f'(s)) + C^''\in B_G^''$ represents a cohomology class. Hence $f'(s) + C^' \in B^'_G$ is a lift. By definition, $d(\overline{f''(s)}_0)$ is represented by $\delta'(f'(s) + C^')=N\cdot f'(s)$. Therefore, it suffices to show $Nf'(s)=a$:

$$Nf'(s)=\sum_{g \in G}gf'(s)=\sum_g f'(s) - \sum_g f'(gs) + \sum_g f'(g) = \sum_g u(g,s)=a.$$

Feel free to ask if you have any questions to my computations.

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Thank you very much, this is a great answer :) –  Chris Birkbeck Feb 14 '13 at 18:37

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