Galois groups are nice compact Hausdorff groups, and therefore possess a bounded Haar measure, unique if we insist that the total volume be $1$. What is the Haar measure on the absolute Galois group of $\mathbb{Q}$ (or of any other number field)? Is there any arithmetic meaning for a subgroup being "big" or "small" with respect to this measure?

Two remarks:

    1. In the book Field Arithmetic by M. Fried and M. Jarden there is a small treatment of Haar measures on Galois groups in $\S 18.5$, but I did not find much a part from general statements which, in case the base field is $\mathbb{Q}$, do not look that much sexy (to me).

    2. Following Tate's thesis, it seems natural to put on the abelianization $$ G_\mathbb{Q}^\text{ab}=\mathrm{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})=G_\mathbb{Q}/[G_\mathbb{Q},G_\mathbb{Q}] $$ the quotient measure coming from first putting on the ideles group $\mathbb{A}_\mathbb{Q}^\times$ the "product Haar measure" coming from each local factor – this is what Tate does, indeed (observe that the normalization is not trivial, here, since the ideles are only locally compact); and, then, identifying the group $G_\mathbb{Q}^\text{ab}$ with a quotient of $\mathbb{A}_\mathbb{Q}^\times$ via global Class Field Theory. But then

      i) Does the Haar measure on $\mathbb{A}_\mathbb{Q}^\times$ really descend to a Haar measure on its quotient $G_\mathbb{Q}^\text{ab}$?

      ii) if i) holds, we get a Haar measure on $G_\mathbb{Q}^\text{ab}$, with respect to which it has some volume $vol_{ideles}$. Similarly, we can hope that the Haar measure on the absolute Galois group induces a Haar measure on its abelianization, the latter being a quotient of the former: this will give $G_\mathbb{Q}^\text{ab}$ another volume $vol_{galois}$. Are the values $vol_{galois}$ and $vol_{ideles}$ related (and interesting)?

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    $\begingroup$ Haar measure on a profinite group is uniquely determined by the fact that it pushes forward to Haar measure on all finite quotients. Do you want a more explicit description than this? $\endgroup$ – Qiaochu Yuan Nov 20 '12 at 9:31
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    $\begingroup$ To be explicit, a subgroup of volume $1/n$ has index $n$, and if it is open and normal, then it can be identified with the absolute Galois group of a Galois extension of degree $n$. By uniqueness up to normalization, there really isn't much more to say. $\endgroup$ – S. Carnahan Nov 20 '12 at 9:39
  • $\begingroup$ Ah ok, I see. Indeed, I agree that there is no much more to say. Thank you! $\endgroup$ – Filippo Alberto Edoardo Nov 20 '12 at 9:41
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    $\begingroup$ Both your question (up to normalization) can be answered with yes. These procedures are connected to the quotient measure (pretty wellknown). The more interesting question is what are the normalizations and how to justify them. $\endgroup$ – Marc Palm Nov 20 '12 at 10:37

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