Profinite groups as absolute Galois groups

It is a well-known result that all profinite groups arise as the Galois group of some field extension.

What profinite groups are the absolute Galois group $\mathrm{Gal}(\overline{K}|K)$ of some extension $K$ over $\mathbb{Q}$?

The answer is simple enough in the finite case:

• (Artin-Schreier) Only the trivial one and $C_2$.

This might tempt us to think that absolute Galois groups are not that diverse. But 0-dimensional anabelian geometry shows us differently:

• (Neukirch-Uchida) There's as many different absolute Galois groups as non-isomorphic number fields.

The answer might still turn out to be boring, but I haven't seen this discussed anywhere. The closest is Szamuely in his book "Galois Groups and Fundamental Groups", where he seems delighted by the fact that the absolute Galois group of $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{q})$ are non-isomorphic for primes $p\neq q$.

• do you have a reference for your first claim? Thank you. – Ofra Mar 6 '16 at 16:13
• Artin-Schreier tells you more generally that the only possible finite subgroups are cyclic of order $2$. This rules out many infinite examples, e.g. infinite products of finite groups. – Qiaochu Yuan Mar 6 '16 at 16:16
• @Ofra Sure. Williiam C. Waterhouse "Profinite groups are Galois groups" (1974) Proceedings of the AMS Vol. 42 Number 2. – Myshkin Mar 6 '16 at 16:17
• @QiaochuYuan Indeed! I'll edit the question and mention it. – Myshkin Mar 6 '16 at 16:20
• I got confused by "some extension $K$ over $\mathbb{Q}$". It just means "some field $K$ of characteristic zero". – YCor Mar 6 '16 at 17:47

This is a very good question which is a big open problem. There are a number of theorems, some of them easy and some very difficult, and also a number of conjectures, restricting the class of groups which may turn out to be absolute Galois groups. But it seems that nobody has any idea about how a precise description of the class of absolute Galois groups might look like.

One general point-of-view position on which the experts seem to agree is that the right object to deal with is not just the Galois group $G_K=\operatorname{Gal}(\overline{K}|K)$ considered as an abstract profinite group, but the pair $(G_K,\chi_K)$, where $\chi_K\colon G_K\to \widehat{\mathbb Z}^*$ is the cyclotomic character of the group $G_K$ (describing its action in the group of roots of unity in $\overline K$). The question about being absolute Galois should be properly asked about pairs $(G,\chi)$, where $G$ is a profinite group and $\chi\colon G\to \widehat{\mathbb Z}^*$ is a continuous homomorphism, rather than just about the groups $G$.

For example, here is an important and difficult theorem restricting the class of absolute Galois groups: for any field $K$, the cohomology algebra $H^*(G_K,\mathbb F_2)$ is a quadratic algebra over the field $\mathbb F_2$. This is (one of the formulations of) the Milnor conjecture, proven by Rost and Voevodsky. More generally, for any field $K$ and integer $m\ge 2$, the cohomology algebra $\bigoplus_n H^n(G_K,\mu_m^{\otimes n})$ is quadratic, too, where $\mu_m$ denotes the group of $m$-roots of unity in $\overline K$ (so $G_K$ acts in $\mu_m^{\otimes n}$ by the character $\chi_K^n$). This is the Bloch-Kato conjecture, also proven by Rost and Voevodsky.

Here is a quite elementary general theorem restricting the class of absolute Galois groups: for any field $K$ of at most countable transcendence degree over $\mathbb Q$ or $\mathbb F_p$, the group $G_K$ has a decreasing filtration $G_K\supset G_K^0\supset G_K^1\supset G_K^2\supset\dotsb$ by closed subgroups normal in $G_K$ such that $G_K=\varprojlim_n G_K/G_K^n$ and $G_K/G_K^0$ is either the trivial group or $C_2$, while $G_K^n/G_K^{n+1}$, $n\ge0$ are closed subgroups in free profinite groups. (Groups of the latter kind are called "projective profinite groups" or "profinite groups of cohomological dimension $\le1$".) One can get rid of the assumption of countability of the transcendence degree by considering filtrations indexed by well-ordered sets rather than just by the integers.

Here is a conjecture about Galois groups of arbitrary fields (called "the generalized/strengthened version of Bogomolov's freeness conjecture). For any field $K$, consider the field $L=K[\sqrt[\infty]K]$ obtained by adjoining to $K$ all the roots of all the polynomials $x^n-a$, where $n\ge2$ and $a\in K$. In particular, when the field $K$ contains all the roots of unity, the field $L$ is (the maximal purely inseparable extension of) the maximal abelian extension of $K$. Otherwise, you may want to call $L$ "the maximal radical extension of $K$". The conjecture claims that the absolute Galois group $G_L=\operatorname{Gal}(\overline{L}|L)$ is a projective profinite group.

• Thank you, that was a very interesting answer! Can you drop a few references or names of where I can find out more about all this? – Myshkin Mar 6 '16 at 23:26
• One reference is my paper "Koszul property and Bogomolov's conjecture", published in IMRN in 2005. This covers the generalized/strengthened Bogomolov conjecture story. You may want to look into an updated version with an improved/expanded appendix, available from arxiv.org/abs/1405.0965 . This covers also the filtration-by projective profinite groups story (see Remark 2 at the end). – Leonid Positselski Mar 6 '16 at 23:32
• The references to Voevodsky's proofs are "Motivic cohomology with $\mathbf Z/2$-coefficients", Publ. Math. IHES 2003 (for the Milnor conjecture, which also implies the above formulation of the Bloch-Kato conjecture for $m$ equal to a power of $2$) and "On motivic cohomology with $\mathbf Z/l$-coefficients, Annals of Math. 2011 (for any prime number $l$ in the role of $m$, which implies the assertion for an arbitrary $m\ge2$). – Leonid Positselski Mar 6 '16 at 23:40
• Finally, another relevant name that I should mention is Ido Efrat. An important contribution of his (perhaps also done independently by Koenigsmann) was to propose a conjectural exhastive list of all the absolute Galois groups (with cyclotomic characters) that happen to be finitely generated profinite groups. So, if you do not insist on knowing all the absolute Galois groups but agree to restrict your attention to the finitely generated ones, then there is indeed a precisely formulated conjecture describing what these are. – Leonid Positselski Mar 6 '16 at 23:47
• @LeonidPositselski thanks for the great answer. Is there a statement or a reference you could provide to help me appreciate the power of the Rost-Voevodsky theorems on the cohomology ring? Also, where could I see that filtration by closed subgroups in action? – PrimeRibeyeDeal Jul 11 '17 at 12:28

In order to study absolute Galois groups of fields one may also use another consequence of the aforementioned Rost-Voevodsky theorem: in fact, for any field $K$, the cohomology algebra $H^*(G_K,\mathbb{F}_p)$ is a quadratic algebra over the field $\mathbb{F}_p$ also for $p$ odd; and this is still true for the algebra $H^*(G_K(p),\mathbb{F}_p)$, where $G_K(p)$ denotes the maximal pro-$p$ quotient of $G_K$, if $K$ contains a primitive $p$-th root of 1 - i.e., $G_K(p)$ is the Galois group of the compositum of all Galois $p$-extensions $L/K$, and it is called the maximal pro-$p$ Galois group of $K$. Note that the class of such Galois pro-$p$ groups includes all absolute Galois groups which are pro-$p$.

One reduces to pro-$p$ groups as in general they are easier to deal with than profinite groups.

Pro-$p$ groups whose $\mathbb{F}_p$-cohomology is a quadratic algebra - and thus they are «good candidates» for being realized as maximal pro-$p$ Galois groups - are studied by S. Chebolu, I. Efrat and J. Minàc in the paper Quotients of absolute Galois groups which determine the entire Galois cohomology (2012), and in my paper Bloch-Kato pro-$p$ groups and locally powerful groups (2014). Both papers are available also on the arXiv.

Also, recently I. Efrat, E. Matzri, J. Minàc and N.D. Tan proved that the $\mathbb{F}_p$-cohomology algebra of maximal pro-$p$ Galois groups of fields (containing a primitive $p$-th root of 1) satisfies another property, called the «vanishing of triple Massey products»: in the papers Triple Massey products and Galois theory and Triple Massey products vanish over all fields Minàc and Tan produce some examples of pro-$p$ groups which are not realizable as maximal pro-$p$ Galois groups (and thus as absolute Galois groups).

All the papers I mentioned are available in the arXiv: if you wish I may provide the links.

There is a paper of Jochen Königsmann which restricts the profinite groups isomorphic to absolute Galois groups. Unfortunately, I can't remember the title.

Edit: I found it: http://math.usask.ca/fvk/jkproduct.pdf "Products of absolute Galois groups"

• Thanks! I've also found interesting this other paper of him, "Relatively projective groups as absolute Galois groups". – Myshkin Mar 6 '16 at 23:36