3
$\begingroup$

Suppose that we have a morphism between profinite groups $f: G_{1}\rightarrow G_{2}$ such that $f^{\ast}:H_{cont}^{\ast}(G_{2},A)\rightarrow H_{cont}^{\ast}(G_{1},A) $ is an isomorphism for any finite trivial $G_{i}$-module $A$. What can we say in general about $f$ ? and in particular when $G_{i}=\mathrm{Gal}(\overline{k_{i}}|k_{i})$, $i=1,2$.

$\endgroup$
4
  • 1
    $\begingroup$ By "trivial $G_i$-module" you mean the module has the trivial action, right? And presumably $f$ is continuous? So it looks to me like this says no more and no less than the statement that the map from the abelianisation of $G_1$ to the abelianisation of $G_2$ induced by $f$ is an isomorphism. By the way, how will continuous cohomology differ from profinite cohomology if $A$ is finite? Maybe they're the same? Or have I missed a subtlety? $\endgroup$
    – znt
    Dec 20 '16 at 16:02
  • $\begingroup$ @znt Yes I do mean trivial action, and f is continuous with respect to the profinite topology. I do not know what is the profinite cohomology. I suspect that it is the same notion, no ? $\endgroup$ Dec 20 '16 at 16:34
  • $\begingroup$ By profinite cohomology I meant the direct limit of group cohomology $H^*(G/N,A^N)$ as $N$ runs through the normal subgroups -- the usual way one defines cohomology of a profinite group acting discretely (stabilisers are open) on an abelian group. It's an exact functor (unlike continuous cohomology, for which there are traps). $\endgroup$
    – znt
    Dec 20 '16 at 20:29
  • $\begingroup$ @znt that is fine! we are talking about the same thing in this case I guess. $\endgroup$ Dec 20 '16 at 20:59
5
$\begingroup$

Here is one recent result in this direction, taken from I. Efrat and J. Minac, Galois groups and cohomological functors, Trans. of the AMS, http://www.ams.org/journals/tran/0000-000-00/S0002-9947-2016-06724-0/home.html :

Let $q=p^s$ be a prime power. Suppose that $G_i=\mathrm{Gal}(\overline{K_i}/K_i)$ for fields $K_i$, both containing a root of unity of order $q$, and take the trivial $G_i$-module $A=\mathbb{Z}/q\mathbb{Z}$ ($i=1,2$). For a profinite group $G$ consider the closed subgroups $G^{(2)}=G^q[G,G]$ and $G^{(3)}=(G^{(2)})^q[G^{(2)},G]$ (resp., $G^{(3)}=(G^{(2)})^{2q}[G^{(2)},G]$) when $p>2$ (resp., $p=2$). Then the map $f^*$ is an isomorphism if and only if $f$ induces an isomorphism $G_1/G_1^{(3)}\cong G_2/G_2^{(3)}$ of profinite groups.

$\endgroup$
1
  • $\begingroup$ Thank you very much! I have finally found the full version! very impressive and in the same time it is great that the question was solved! $\endgroup$ Dec 21 '16 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.