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Let $G$ be a profinite group, and $A$ a G-module. If $G$ is the projective limit of {$G_{\alpha}$}, and $A$ the direct limit of {$A_{\alpha}$}, then
$H^*(G,A)$ is isomorphic to $dir lim_{\alpha} H^*(G_{\alpha},A_{\alpha})$.
Here the cohomology groups are defined via the group of continuous functions from $G$ to $A$. At page 26 of this book it is asserted that this theorem is false on discontinuous cochains. And my question is: is there an example illustrating this statement?
Thanks in advance.

awllower
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  • So you're asking for an example where $H^_{\text{cont.}}(G,A) \not\cong H^_{\text{discont.}}(G,A)$? – Bruno Joyal Dec 04 '12 at 18:52
  • Actually I would like to see an example showing that the equality holds not, on discontinuous cochains. Thanks for your attention in any case. – awllower Dec 05 '12 at 01:17
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    Dear @awllower, I don't understand the question... What equality? Please be more precise. – Bruno Joyal Dec 05 '12 at 13:20
  • I mean this isomorphism, which I referred to earlier as the equality: $H^_{\text{cont.}}(G,A) \cong dir limH^{\text{cont.}}(G{\alpha},A_{\alpha})$ That is to say, the "theorem" alludes to that isomorphism exactly. – awllower Dec 05 '12 at 14:43

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If you take a prime $p$ and a non-commutative finitely generated free pro-$p$ group $F$, then the second continuous cohomology group with coefficients in $\mathbb Z/p$ is trivial $$H^2_{\sf cont}(F,\mathbb Z/p) =0$$ but the discrete (discontinuous) cohomology group is uncountable (in particular, it is nontrivial) $$H^2_{\sf disc}(F,\mathbb Z/p)\ne 0.$$ (see this or this ).

Since $$H^2_{\sf cont}(F,\mathbb Z/p)=\varinjlim H^2_{\sf disc}(F/U,\mathbb Z/p)$$ and $F=\varprojlim F/U,$ where $U$ runs over all open normal subgroups, this is a counterexample to your statement.

Sergei Ivanov
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