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Suppose $X$ is a smooth projective variety over a field $k$, with separable closure $\overline{k}$, Galois group $G$, and let $\overline{X}$ be $X_{\overline{k}}$.

Do we have

$$(H^j(\overline{X},\mathbf{Z}_{\ell}(1)))^G = \varprojlim_{n\ge 0}(H^j(\overline{X},\mu_{\ell^n})^G)$$

ie. does formation of Galois invariants on mod $\ell^n$ cohomology commute with inverse limits on coefficients?

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    $\begingroup$ I'm probably missing something here, but doesn't the desired equality follows from the fact that limits commute with limits, and the fact that $H^j(\overline{X}, \mathbb{Z}_{\ell}(1))$ by definition is $\varprojlim_n H^j(\overline{X}, \mu_{\ell^n})$? $\endgroup$
    – user84144
    Commented Mar 2, 2018 at 15:30
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    $\begingroup$ How is formation of Galois invariants a limit? There should be some content in here. It's a Thm of Tate that for a profinite group $G$ acting compatibly and continuously on an inverse system of discrete $G$-modules $\{T_n\}$, if each $T_n$ is finite then $H^k_{\rm cont}(G, \lim T_n) = \lim H^k(G, T_n)$. Since $X$ is smooth projective, $T_n := H^j(\overline{X},\mu_{\ell^n})$ satisfies the assumptions, and for $k = 0$ the desired commutation follows. $\endgroup$
    – user113453
    Commented Mar 2, 2018 at 16:09
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    $\begingroup$ @user113453: The $G$-invariants of a $G$-module $M$ are the limit of the diagram $$M \underset{\vdots}{\to} M$$ whose arrows are multiplication by $g$ for all $g \in G$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 2, 2018 at 16:51
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    $\begingroup$ @user113453 I think the more general statement you are claiming is not actually true. You need finiteness not just of $T_n$, but also of $H^{k-1}(G, T_n)$ for all $n$. (The point is that you have to make sure that the derived functor $\varprojlim^1 H^{k-1}(G, T_n)$ vanishes.) So everything is fine for $k = 0$ or $k = 1$, but for larger $k$ it can really go wrong. See Appendix B of Rubin's "Euler Systems", or Jannsen's "Continuous etale cohomology". In any case using this for $k = 0$ is a sledgehammer to crack a nut. $\endgroup$
    – David Loeffler
    Commented Mar 2, 2018 at 21:28

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