Lemma 59.57.4. Let $G$ be a topological group. Let $R$ be a ring. Let $M$, $N$ be $R\text{-}G$-modules. If $M$ is finite projective as an $R$-module, then $\text{Ext}^ i(M, N) = H^ i(G, M^\vee \otimes _ R N)$ (for notation see proof).

Proof. The module $M^\vee = \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ endowed with the contragredient action of $G$. Namely $(g \cdot \lambda )(m) = \lambda (g^{-1} \cdot m)$ for $g \in G$, $\lambda \in M^\vee$, $m \in M$. The action of $G$ on $M^\vee \otimes _ R N$ is the diagonal one, i.e., given by $g \cdot (\lambda \otimes n) = g \cdot \lambda \otimes g \cdot n$. Note that for a third $R\text{-}G$-module $E$ we have $\mathop{\mathrm{Hom}}\nolimits (E, M^\vee \otimes _ R N) = \mathop{\mathrm{Hom}}\nolimits (M \otimes _ R E, N)$. Namely, this is true on the level of $R$-modules by Algebra, Lemmas 10.12.8 and 10.78.9 and the definitions of $G$-actions are chosen such that it remains true for $R\text{-}G$-modules. It follows that $M^\vee \otimes _ R N$ is an injective $R\text{-}G$-module if $N$ is an injective $R\text{-}G$-module. Hence if $N \to N^\bullet$ is an injective resolution, then $M^\vee \otimes _ R N \to M^\vee \otimes _ R N^\bullet$ is an injective resolution. Then

$\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ) = \mathop{\mathrm{Hom}}\nolimits (R, M^\vee \otimes _ R N^\bullet ) = (M^\vee \otimes _ R N^\bullet )^ G$

Since the left hand side computes $\text{Ext}^ i(M, N)$ and the right hand side computes $H^ i(G, M^\vee \otimes _ R N)$ the proof is complete. $\square$

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