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Let $G$ be a profinite group and $M$ be a finite $G$-module. I can compute the cohomology of $G$ with coefficients in $M$ either as a topological group or as a discrete group. There is an obvious map $H^p(G,M)\to H^p(G^{\delta},M)$ (where $G^\delta$ denotes the underlying discrete group of $G$) which forgets that a $p$-cochain is continuous.

I would like to know if there are conditions on $G$ that insure that these maps are isomorphisms. In my case $G$ is finitely generated (as a topological group).

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    $\begingroup$ This should be true for $H^1$ and $G$ finitely generated by a result of Nikolov and Segal that finite-index subgroups are open. $\endgroup$
    – Ian Agol
    Commented Oct 12, 2014 at 16:37

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If $G$ is finitely generated, then $G$ is isomorphic to its profinite completion by a result of Nikolov and Segal mentioned by Ian Agol. Thus, what are you asking is equivalent to the goodness introduced by Serre (see J. P. Serre, Galois cohomology, I.2.6).

The only good finitely generated profinite groups that I know are virually polycyclic (it can be proved in the same way as Theorem 2.10 in http://arxiv.org/pdf/math/0701737.pdf). In the same paper you can find many related results. For example, a non-abelian free pro-$p$ group is not good (this is a result of A.K. Bousfield, On the p-adic completions of non-nilpotent spaces, Trans. Amer. Math. Soc. 331 (1992), 335–359).

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  • $\begingroup$ Not sure I understand why this is equivalent to goodness: I thought that was an isomorphism between the cohomology of a group with finite module coefficients and its profinite completion. $\endgroup$
    – Ian Agol
    Commented Oct 14, 2014 at 22:43
  • $\begingroup$ @Ian Agol: The profinite completion of $G^\delta$ is $G$. $\endgroup$
    – Andrei Jaikin
    Commented Oct 15, 2014 at 10:30
  • $\begingroup$ Ok, I guess I had in mind the profinite completion of a finitely generated group, but clearly $G^\delta$ will be infinitely generated if infinite. There are now many finitely generated "good" groups $G$ known, but I don't know whether this implies $\hat{G}^\delta$ is good? $\endgroup$
    – Ian Agol
    Commented Oct 15, 2014 at 17:21
  • $\begingroup$ @IanAgol: I do not think this holds. A non-abelian finitely generatede free group is good. However, I would conjecture that its profinite completion is not. In fact, I beleive that there exists a non-trivial extension of a non-abelian finitely generated free profinite group by a finite group. $\endgroup$
    – Andrei Jaikin
    Commented Oct 16, 2014 at 8:24
  • $\begingroup$ The Bousfield counter-example shows that the map on cohomology should not be expected to be an isomorphism except in very restrictive cases. I have therefore decided to accept this answer. $\endgroup$
    – Geoffroy Horel
    Commented Oct 17, 2014 at 14:49

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