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I am working on a paper of R.P Langlands called "Representations of abelian algebraic groups", available here: http://www.sunsite.ubc.ca/DigitalMathArchive/Langlands/pdf/AbelianAlg-ps.pdf

Now on page 13-14 he makes the following claim, which I have summarized.

Let $G$ be a group, $L$ a $G$-module and let $G$ act trivially on $\mathbb{Q} / \mathbb{Z}$.

If we start with the natural paring $$L \otimes Hom(L,\mathbb{Q} / \mathbb{Z}) \rightarrow \mathbb{Q} / \mathbb{Z},$$ we get a cup product paring $$\nu: \widehat{H}^{-3}(G,L) \otimes \widehat{H}^{2}(G, Hom(L,\mathbb{Q} / \mathbb{Z})) \rightarrow \widehat{H}^{-1}(G,\mathbb{Q} / \mathbb{Z}).$$

Now he then goes on to claim the following. If we take $\alpha \in \widehat{H}^{2}(G, Hom(L,\mathbb{Q} / \mathbb{Z}))$, then $\alpha =0 $ if and only if for all $\beta \in \widehat{H}^{-3}(G,L)$ we have $\nu(\beta \otimes \alpha)=0$. Now why is this true?

Thank you

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up vote 2 down vote accepted

This is explained completely in Ken Brown's Cohomology of Groups (always check here!), with a section devoted to it: A Duality Theorem, Section VI.7.

Ultimately, we have ordinary composition products from homological algebra, and using the evaluation map $Hom(M,\mathbb{Q}/\mathbb{Z})\otimes M\to \mathbb{Q}/\mathbb{Z}$ for a duality pairing gives us the desired result. Alternatively, you can exhibit a fundamental class and use the cap product to establish the pairing.

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