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This question is related to this one. Choose an embedding $\overline{\mathbf{Q}}\rightarrow \mathbf{C}$ from the algebraic closure of the field of rational numbers to the field of complex numbers. Is it realistic to say some thing about $H_{cont}^{\ast}(Aut_{\overline{\mathbf{Q}}}(\mathbf{C}),A)$ the continuous cohomology of the group of field automorphisms of $\mathbf{C}$ over $\overline{\mathbf{Q}}$ ?

$A$ is a finite $Aut_{\overline{\mathbf{Q}}}(\mathbf{C})$-module with the trivial action.

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  • $\begingroup$ What do you mean by "continuous"? $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 29, 2017 at 6:56
  • $\begingroup$ @LaurentMoret-Bailly I see $Aut_{\overline{Q}}(C)$ as a profinite group... Am I wrong ? $\endgroup$
    – Ofra
    Commented Mar 29, 2017 at 7:05
  • $\begingroup$ No, this group is not profinite since complex numbers are not algebraic over rationals (and so, cannot be presented as the union of finite degree Galois extensions of $\overline{\mathbf{Q}}$). $\endgroup$
    – Mikhail Bondarko
    Commented Mar 29, 2017 at 12:18

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