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Let $G$ be a profinite group and $k$ a field. For any finite dimensional $k$-representation $V$ of $G$, we denote its dual by $V^*$. I found the statement saying that there is an ismorphism

$\mathrm{Ext}^1(V_1,V_2)\simeq H^1(G,V_1^*\otimes V_2)$

where $V_i$ are finite dimensional $k$-representations of $G$. Could I get a reference for this isomorphism? - I am wondering how the explicit map goes.

User0829
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  • You can find this in the book by Cartan-Eilenberg. The isomorphism is completely explicit at the level of chain complexes. – Mariano Suárez-Álvarez Aug 16 '17 at 13:25
  • @MarianoSuárez-Álvarez I am really sorry, but at the moment I have no access to that book. If possible, could you give some explanation? – User0829 Aug 16 '17 at 13:31
  • Using this question, one has a functorial isomorphism $X \otimes_k Y^* \cong \mathrm{Hom}_F(Y, X)$, where $Y^* = \mathrm{Hom}_k(Y,k)$.

    When $X, Y$ are finite dimensional representations of $G$ over $k$, this corestricts to a functorial isomorphism $$ \mathrm{Hom}{k[G]}(k, X \otimes_k Y^) = (X \otimes_k Y^)^G \cong \mathrm{Hom}{k[G]}(Y, X)$$

    Taking the first derived functor should give you $$H^1(G, X \otimes_k Y^) = \mathrm{Ext}^1_{k[G]}(k, X \otimes_k Y^) \cong \mathrm{Ext}^1_{k[G]}(Y,X) $$ as desired.

     – Watson Sep 26 '18 at 14:49
  • Notice that in an abelian category $\mathscr A$, like $\mathsf{Mod}{k[G]}$, the abelian group $\mathrm{Ext}^1{\mathscr A}(Y,X)$ classifies extensions (short exact sequences) of the form $$ 0 \to X \to E \to Y \to 0. $$ (Be careful about the position of $X$ and $Y$ !) – Watson Sep 26 '18 at 14:53

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