## “Understanding” Gal(\bar Q/Q)

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I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers G = Gal(Q bar/Q). What do people mean when they say this? The Kronecker-Weber theorem gives a good idea of what the abelianization of the G looks like. But in one of Richard Taylor's MSRI talks, Taylor said that he's never heard of anyone proposing a similar direct description of G and that to understand G one studies the representations of G.

I know that there is a strong interest in showing the Langlands reciprocity conjecture [Edit: What I had in mind in writing this is evidently Clozel's conjecture, not the Langlands reciprocity conjecture - see Kevin Buzzard's post below] - that L-functions attached to l-adic Galois representations coincide with L-functions attached to certain automorphic representations. And I've heard people refer to the Tannakian philosophy which I understand as (roughly speaking) asserting that G is determined by all of its finite dimensional representations.

Here is a representation of G understood not to be a representation of G as an abstract group but as a group together with a labeling of some of the conjugacy classes of G by rational primes (the Frobenius elements)?

When people talk about "understanding G" do they mean proving [Edit: Clozel's conjecture] (in view of the Tannakian philosophy)? If not, what do they mean? If so, this conceptualization seems quite abstract to me. Is this what people mean when they say "understand G"? Can [Edit: Clozel's conjecture] be used to give more tangible statements about G?

Something that I have in mind as I write this is the inverse Galois problem (does every finite group occur as a Galois group of a normal extension of Q?) and Gross' conjecture (mostly proven by now) that for each prime p there exists a nonsolvable extension of Q ramified only at p. But I am open to and interested in other senses and respects in which one might "understand" G

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Thanks to the respondees so far. I'm hoping for a more complete answer and so decided to offer a bounty. – Jonah Sinick Oct 31 2009 at 5:05
Also, sorry for the nitpick, but I changed slightly the TeXing in the title -- feel free to revert. + Here's a quote from A.N.Parshin: "Langlands program sounds nice, but then I go and ask people -- if you go through with it, would you be able to get the full structure of absolute Galois group? And they say no, it won't be still completely known. And [studying the absolute Galois group] is, I think, more interesting. – Ilya Nikokoshev Oct 31 2009 at 10:57
Ilya, there are some interesting things in your post and I thank you for writing it, the respect in which I did not feel satisfied by your response is that it didn't (directly) address any of the questions that I asked above (about what people mean when they talk about understanding G, and whether the Langlands reciprocity conjecture has more tangible/explicit interpretations concerning the structure of G such as offering the possibility of a solution to the inverse Galois problem) – Jonah Sinick Oct 31 2009 at 18:22
The conjecture that Kevin attributes to Clozel, I would in fact attribute to Langlands, and call "the Langlands reciprocity conjecture", as you did originally. (Here "reciprocity" is used in distinction to "functoriality", which is roughly what Kevin is referring to when he writes about Langlands's conjectures.) – Emerton Feb 12 2010 at 6:56

## 9 Answers

What would it mean to understand this Galois group? You could mean several things.

You could mean trying to give the group in terms of some smallish generators and relations. This would be nice, and help to answer questions like the inverse Galois problem that Greg Muller mentioned, and having a certain family of "generating" Galois automorphisms would allow you to study questions about e.g. the representation theory in quite explicit terms. However, the Galois group is an uncountable profinite group, and so to give any short description in terms of generators and relations leads you into subtle issues about which topology you want to impose.

You could also ask for a coherent system of names for all Galois automorphisms, so that you can distinguish them and talk about them on an individual basis. One system of names comes from the dessins d'enfant that Ilya mentioned: associated to a Galois automorphism we have some associated data.

• We have its image under the cyclotomic character, which tells us how it acts on roots of unity. By the Kronecker-Weber theorem this tells us about the abelianization of the Galois group.
• We also have an element in the free profinite group on two generators, which (roughly speaking) tells us something about how abysmally acting on the coefficients of a power series fails to commute with analytic continuation.

These two names satisfy some relations, called the 2-, 3-, and 5-cycle relation, which are conjectured to generate all relations (at least the last time I checked), but it is difficult to know whether they actually do so. If they do, then the Galois group is the so-called Grothendieck-Teichmuller group.

The problem with this perspective is that the names aren't very explicit (and we don't expect them to be: we may need the axiom of choice to show they exist, and there are only two Galois automorphisms of C that are measurable functions!) and it seems to be a difficult problem to determine whether the Grothendieck-Teichmuller group really is the whole thing. (Or it was the last time I checked.)

However, the cyclotomic character is a nice, and fairly canonical, name associated for Galois automorphisms. We could try to generalize this: there are Kummer characters telling us what a Galois automorphism does to the system of real positive roots of a positive rational number number (these determine a compatible system of roots of unity, or equivalent an element of the Tate module of the roots of unity). This points out one of the main difficulties, though: we had to make choices of roots of unity to act on, and if Galois theory taught us nothing else it is that different choices of roots of an irreducible polynomial should be viewed as indistinguishable. Different choices differ by conjugation in the Galois group.

This brings us to the point JSE was making: if we take the "symmetry" point of view seriously, we should only be interested in conjugacy-invariant information about the Galois group. Assigning names to elements or giving a presentation doesn't really mesh with the core philosophy.

So this brings us to how many people here have mentioned understanding the Galois group: you understand it by how it manifests, in terms of its representations (as permutations, or on dessins, or by representations, or by its cohomology), because this is how it's most useful. Then you can study arithmetic problems by applying knowledge about this. If I have two genus 0 curves over Q, what information distinguishes them? If I have two lifts of the same complex elliptic curve to Q, are they the same? How can I get information about a reduction of an abelian variety mod p in terms of the Galois action on its torsion points? Et cetera.

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The 2, 3 and 5-cycle relations of GT translate the action on the moduli space curbes of genus 0 with n marked points. I don't think it was seriously conjectured that these generate all the relations. Grothendieck's original conjecture is that you have to consider all the M_{g,n} for 3g-3+n < 3 to get a full set of relations. Lochack, Nakamura and Schneps have defined a subgroup of GT (containing G_Q) that acts on the fundamental groups of all the M_{g,n}'s. It is probably strictly smaller than GT but that hasn't been proved. – YBL Nov 25 2009 at 0:19
What is the most current status on the 2-, 3- and 5-cycle relations? – Dr Shello Dec 15 2010 at 9:45

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Perhaps some articles in this book and the articles here help, and the discussion on dessin d'enfants (more) on this site.

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The book link has changed.  This is the correct link: library.msri.org/books/Book41/index.html – arsmath Dec 20 2010 at 19:21
Thanks for the corrected link! – Thomas Riepe Dec 21 2010 at 5:49

One aspect of understanding the absolute Galois group I've heard of is the 'Inverse Galois Problem'. This simply askes, Is every finite group the Galois group of some extension of Q? Its known for solvable groups, but it is unsolved (and apparently very hard) for general finite groups.

The inverse Galois problem is equivalent to asking whether any finite group can be the quotient of Gal(Q^bar, Q). In this way, its a more concrete question about the absolute Galois group than Langlands reciprocity.

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Taylor said that he's never heard of anyone proposing a similar direct description of G and that to understand G one studies the representations of G.

I remember Mazur telling me this when I was a grad student. He made this point in the following way. You shouldn't really think of Gal(Qbar/Q) as a group which has elements, but as a "group up to conjugacy" -- thus, the aspects of Galois groups that really make sense to think about are the conjugacy-invariant things: conjugacy classes (like Frobenii) and representations.

To unpack this a bit more: a Galois group is a fundamental group. But to talk about a fundamental group (as opposed to a groupoid) you need to choose a basepoint. To talk about an absolute Galois group you also need to choose a basepoint, which is to say an algebraic closure Qbar/Q. (So just as one should talk about pi_1(X,*) rather than pi_1(X), one should talk about Gal(Qbar/Q) rather than Gal(Q).) But a basepoint you can just draw with a pencil. A Galois closure of Q is not so easy.

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I like this answer very much.  But I wonder: doesn't Q come with a standard algebraic closure?  Namely, the algebraic closure of Q inside C.  Why doesn't this serve as a good "canonical basepoint" for Qbar/Q?  (Okay, canonical up to Gal(C/R) ...) – Charles Rezk Oct 31 2009 at 16:46
Seconding Charles' question! (Which I think also covers mine, namely why Gal(Qbar/Q) is deep and magical but the absolute Galois group of F_p isn't so much.) – Harrison Brown Oct 31 2009 at 17:24
What people mean by "canonical" would be: for an equation `x^2 + 1 = 0` we have chosen canonically a solution `x`. The fact that you're in `C` doesn't help --- there are still 2 solutions, you need a choice.  – Ilya Nikokoshev Nov 3 2009 at 9:33
" why Gal(Qbar/Q) is deep and magical but the absolute Galois group of F_p isn't so much." --- it's much larger, to start with and contains all `Gal F_p` in some sense. – Ilya Nikokoshev Nov 3 2009 at 9:34
I feel like this is true of most groups. E.g. the symmetric group - you need to choose an initial ordering of the letters you're permuting. – David Corwin Apr 29 at 13:16

Another aspect of "what it might mean to understand Gal(Q)" -- Shafarevich conjectured that the absolute Galois group of Q^{ab} is a free profinite group. That's pretty explicit!

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does "free profinite group" mean "profinite completion of a free group"? – Ben Webster♦ Oct 27 2009 at 14:12
From one of Harbater's papers on this: "A profinite group Π is free on a generating set S that converges to 1 if every map S → G to a profinite group G that converges to 1 uniquely extends to a group homomorphism Π → G." Here "converges to 1" refers to the profinite topology. – JSE Oct 27 2009 at 14:19
A good reference for this is Ribes-Zaleski ["Profinite groups"](books.google.com/…) page 91-94. One should note that any finite set is "converging to 1" and that in the abelian case a free profinite group (free proabelian group!) on a set converging to 1 is just a product of \Z^hat's. – Lars Oct 27 2009 at 20:51
Another quite good reference is "Field Arithmetic" of Fried-Jarden. It also discusses the connection between arithmetic/geometric properties of fields and group theoretic properties of the absolute Galois groups. E.g., Ax' Theorem: if a field K is PAC ("pseudo-algebraically closed") which just means that any geometrically irreducible variety has a rational point, then its absolute Galois group is projective, and its counter part due to Lubotzky-v.d.Dries any projective profinite group can be realized as the absolute Galois group of a PAC field. – Lior Bary-Soroker Nov 25 2009 at 2:46

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Is there any serious reason to believe (or disbelieve) this conjecture? – David Hansen Jun 4 2011 at 3:37

I'm far from expect in this topic, but here's my attempt.

First, and that's something quite straightforward, people want to study `Gal Q` (this is how I will denote it; this common shortcut is defined as `Gal F := Gal \bar F/F`) because we like Galois groups. For simple fields we know their Galois groups and we know how tremendously important they have been, from solving the equations (Abel et al) to doing group cohomology in class field theory.

As algebraic geometry matured, people learned that one of the deep reasons for importance of Galois group is that it's the fundamental group `pi_1(Spec F)` (depending on the definition, sometimes this is true only after a profinite completion, I'll omit this fine point further below). This allows us to use the whole apparatus of topology as well as the geometric intuition.

Representations of `Gal F` thus have a natural geometric meaning as some bundles (local systems) on the coverings of `Spec F`. This alone would make their study pretty important. It's very reasonable to study a space in terms of geometric objects that live on it.

The whole Galois group is very complicated. Fortunately, since the Galois groups of `F_q` and `Q_p` (`p`-adic numbers) are known, we know lots of factorgroups of it. This is the standard topic of algebraic number theory courses.

Moreover, if we restrict ourselves to the *Abelian part of Galois group, its structure has been completely established by the class field theory. In other words, all one-dimensional representations of `Gal Q` are known. That begs a natural question about higher-dimensional reps — indeed this goes on by the name of Langlands program.

The Langlands program is such a huge topic that I don't feel able even starting to talk about it. It might be a good idea to post questions here if you'll feel brave enough to learn it :)

One more topic in the discussion of the structure of `Gal F` is about the specific (conjugacy classes) of operators that live there, called Frobenius operators. The amazing thing about them is that they behave, formally, in a way similar to the knot operators in the fundamental group of a threefold without knots. This image is reinforced by the computations of the dimension of `Spec Z` that give an answer of 3, providing fruitful connections to the theory of real 3-manifolds, knots and their L-functions.

Finally, here's the direction that was explored by many people and was probably made famous through Grothendieck's work. Consider a variety `Z := P^1-(0, 1, \infty)`, the projective plane without three points. It is really interesting as it makes sense over any field. Now suppose we consider it as a scheme over Q and ask for its fundamental group. Then one would have an exact sequence, if I'm not mistaken,

````0  -->  pi_1(Z/\bar Q)  -->  pi_1(Z/Q)  -->  Gal Q --> 0
````

which implies, by standard reasoning, that `Gal Q` acts on `pi_1(Z/\bar Q)`, which is a very straightforward group — it's simply a topological fundamental group of a plane without two points, so it's freely generated by two loops. Grothendieck called it a cartographic group and related it to dessin's d'enfants. The standard references are very well-written in T.'s answers and there are hopefully more questions on this topic coming to MathOverflow ;)

As you see, this is a very interesting direction that we should continue to explore. One more thing I like about it is that it's also very suitable to MathOverflow format, since there are both many specific questions as well as opportunities to write reviews and to produce new results by collaboration.

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Conc. primes as knots and Spec Z as 3-manifold: Fits that to the Poincare conjecture? Topologists view 3-manifolds as Kirby-equivalence classes of framed links. How would that be with Spec Z? Then, topologists have things like virtual 3-manifolds, has that analogies in arithmetics?      – Thomas Riepe Nov 3 2009 at 14:01
I'm not a specialist in this topic, and it's a different from the original question here. Do you mind posting a separate question about this? – Ilya Nikokoshev Nov 3 2009 at 14:59

Khare-Larsen-Savin have used functoriality (ie the Langlands program) to help solve the inverse Galois problem, see for example link text. They show that some finite groups of Lie type occur as Galois groups.

As for the Langlands program in general, solving it may not give an answer to "What is Gal(Qbar/Q)?", however it can certainly help. For example, the use of automorphic forms in Iwasawa theory has been quite fruitful such as Ribet's proof of the converse to Herbrand's theorem, the proof of the Main conjecture by Mazur-Wiles and Wiles, etc. For example, Eisenstein series, their congruences with cusp forms and the associated Galois representations are used in Ribet's paper (Inv. Math. 34 (1976)) to construct unramified p-extensions. It can be easier to hide an automorphic form up your sleeve.

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$\newcommand{\bb}{\mathbb}$ This is a response to Charles' remark to JSE's answer, why doesn't $\bar{\bb Q}$ come with a standard algebraic closure inside the complex numbers $\bb C$?

First, if one considers an abstract extension $K/\bb Q$, then $K$ has $d = [K:\bb Q]$ embeddings into the complex numbers, which can not, a priori, be distinguished in any way. (e.g. maps from $K = \bb Q[x]/(x^3-2)$ to $\bb C$ require a "choice" of $2^{1/3}$ in $\bb C$.

Of course, this doesn't answer Charles' question, which, I imagine, is more along the following lines. Why doesn't one simply start with the complex numbers, and then consider the set of algebraic numbers inside $\bb C$? The resulting field is clearly isomorphic to $\bar{\bb Q}$, and, moreover, comes with a canonical embedding into $\bb C$.

The problem arises when one wants to define Frobenius elements. Defining such elements amounts to giving a choice of embedding from $\bar{\bb Q}$ into $\bar{\bb Q}_p$. So there is a choice to be made for every $p$! Thinking of $\bar{\bb Q}$ inside $\bb C$ fixes this choice for "$p=0$" only.

To make this completely explicit, consider the splitting field $K$ of $x^3 - 2$. In the "fields live inside $\bb C$" optic, $K$ is the field $\bb Q(2^{1/3},\sqrt{-3})$ where $2^(1/3)$ is real and the imaginary part of $\sqrt{-3}$ is positive. Clearly $Gal(K/\bb Q) = S_3$, where we can think of $(123)$ as sending $2^{1/3}\mapsto e^{2\pi i/3} 2^{1/3}$. If $p = 7$, then $Frob_p$ has order 3 in $S_3$. We can ask, does $Frob_p = (123)$ or $(132)$? We find that $$Frob_p = (123) \quad\mbox{if } \sqrt{-3}\equiv 2\mod 7$$ $$Frob_p = (132) \quad\mbox{if } \sqrt{-3}\equiv 5\mod 7$$ and knowing that the imaginary part of $\sqrt{-3}$ is positive does not allow us to determine $Frob_p$ without choosing an embedding of $K$ into $\bar{\bb Q}_7$.

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Thanks, this is a point that I have long wondered about and your remark clarifies the matter completely. – Jonah Sinick Nov 1 2009 at 18:44
Thanks! So what I take from this is that the only canonical constructions are the completions Q_p (with Q_&infin;=R). So if X is the algebraic space associated to Q, then for each p we get a covering space X_p->X, associated to the field of algebraic numbers in Q_p. If X had a distinguised base point, we could name the X_p's using subgroups of &pi;_1X. Since we can't, all we know is the conjugacy class of &pi;_1X_p inside &pi;_1X. For rational primes p, this is the congugacy class of the subgroup "generated" by Frob_p. – Charles Rezk Nov 1 2009 at 20:12
I hate the fact that you can't even use html characters in comments. – Charles Rezk Nov 1 2009 at 20:13

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Charles, yes, that's correct, although literally to have a Frobenius at p one should work with extensions unramified at p, which amounts to working with pi_1(Z[1/N]) for some set N, and taking p not dividing N. (For p|n, one has to work with the still complicated group Gal(Qbar_p/Q_p), rather than Gal(Fbar_p/F_p), which is topologically cyclic and generated by Frobenius. – Lavender Honey Nov 1 2009 at 23:29
Very very nice!!! – Filippo Alberto Edoardo Nov 6 at 23:06

I am pretty sure that when different number theorists say "one of the main goals of number theory is to understand Gal(Q-bar/Q)" they may well mean different things.

One example of what someone might mean, which has been touched upon above, but which I'd like to stress (because it's what I mean whenever I start seminars with this generic-sounding statement!) is Clozel's conjectures in his Ann Arbor paper.

Langlands would conjecture that for any automorphic representation of GL_n over a number field K, there should be an associated n-dimensional complex representation of what people might now call the Langlands group of K (Langlands really conjectured something a little more precise: he conjectured that a certain category that he could not quite define, but whose objects he understood, should have the structure of a Tannakian category; I've tried to make this statement "concrete" in the above).

One of the problems with Langlands' conjecture was simply that it's not the sort of thing that can be checked by example, because no-one really knows an intrinsic construction of the Langlands group of K.

However, there are theorems in the literature about constructing Galois representations from automorphic representations, for example class field theory (which tells us what the abelianisation of this Langlands group should look like: ((adeles of K)^* / K_^* ) ) and Deligne-et-al's theorem attaching p-adic Galois representations to holomorphic cusp forms for GL_2 over Q.

Hang on a minute though---Langlands was conjecturing the existence of complex Galois representations of some group we don't have a definition of---how does Deligne's theorem fit into this? Here we have a p-adic representation (which wouldn't be continuous if you chose an isomorphism Q_p-bar = C and considered the induced complex representation) of a group that at least we have a definition of (Gal(Q-bar/Q) in Deligne's case). So in fact Deligne isn't proving a special instance of Langlands' conjecture, he's proving something else.

Clozel formulated precisely what that "something else" should be: he conjectured that to any algebraic automorphic representation of GL_n(adeles of K) there should be an attached p-adic Galois representation. I think that nowadays one would also conjecture that conversely given a geometric p-adic Galois representation (i.e. a representation Gal(K-bar/K)-->GL_n(Q_p-bar) which looks reasonable in some way that can be made precise) it should arise in this way. People like Emerton and Kisin are well on the way to proving this for K=Q and n=2.

So "proving Clozel's conjecture" would be one way of making precise "understanding Gal(Q-bar/Q)". What this would boil down to in this case would be understanding all the representations of Gal(Q-bar/Q) which were (a) taking values in Aut(V) with V a f.d. vector space over the p-adics, (b) unramified outside a finite set of primes (c) de Rham at p, and one would be understanding them in the sense that one would be parametrising the set of unramified conjugacy classes Frob_ell in terms of something totally different (automorphic representations, which are analytic gadgets).

The book of Harris and Taylor is to a large extent the state of the art nowadays (but the theory is currently moving so fast) [EDIT: see Toby Gee's comment below]. However the base field K is essentially always either totally real or CM in this book, so in some sense the general case of Clozel's conjecture is still wide wide open.

Note finally that in some sense what Langlands conjectures is still open even for modular elliptic curves. If E is a modular elliptic curve over Q then the Galois representation naturally associated to E is its Tate module. But the representation of the Langlands group attached to E would be a representation to GL2(C), whose image would land in U2(C), the trace of Frob_p would be a sub p/sqrt(p), and (in the non-CM case) the conjugacy classes of the Frob_p's would be "evenly distributed in U(2)"---a statement which turns into the Sato-Tate conjecture when you unravel it. Note that Taylor et al proved the Sato-Tate conjecture but they certainly did not prove it by constructing this representation! Indeed it's in some sense an ill-defined question to construct this representation, because we don't know what the Langlands group is.

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A much better answer than mine, +1. – Ilya Nikokoshev Nov 2 2009 at 11:34
A pedantic note: Harris-Taylor isn't quite state of the art (though I guess it is to a "large extent"): in particular the hypotheses there on the existence of a discrete series place have been completely removed by Shin and Chenevier-Harris, amongst others. – TG Nov 2 2009 at 12:50
Thanks Toby. When I wrote the comment about Harris-Taylor, what I really meant was "I know things have happened since then but don't quite know who to attribute things to so I won't go there".   Passing remark: I now don't quite understand the canonical thing to do. Am I supposed to edit my post to reflect your comments, which will then make your comment look ridiculous? I think I'll just leave it. – Kevin Buzzard Nov 2 2009 at 14:42
Kevin - Thanks for your interesting and informative post. I added an edit to my post (under the heading [EDIT] to avoid taking your response out of context just as you wish to avoid taking Toby's comment out of context) to account for your correction. – Jonah Sinick Nov 3 2009 at 4:21
I should perhaps clarify something: amongst the algebraic automorphic representations are those "of Artin type" (I think there must be a better name for them, but it escapes me)---I am talking about the ones which should correspond to finite image Galois representations. For these representations, Clozel's conjecture and Langlands' conjecture should "basically coincide" and should both predict finite image Galois reps to GL(n,complexes). Even proving this statement (attaching reps to pi's of Artin type on GL(n)) would be an extraordinary achievement. – Kevin Buzzard Nov 3 2009 at 16:20
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