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Over the years, I've been somewhat in the habit of asking questions in this vein to experts in the Langlands programme.

As is well known, given an algebraic number field $K$, they propose to replace the reciprocity map $$A_K^*/K^*\rightarrow Gal(K^{ab}/K)$$ of abelian class field theory by a correspondence between the $n$-dimensional representations $\rho$ of $Gal(\bar{K}/K)$ and certain automorphic representations $\pi_{\rho}$ of $GL_n(A_K)$. (We'll skip the Weil group business for this discussion.) Substantial arithmetic information is carried on either side by the $L$-functions, which are supposed to be equal.

This involves deep and beautiful mathematics whenever something can be proved, and there are many applications, such as the Sato-Tate conjecture or this recent paper of Chenevier and Clozel:

http://www.math.polytechnique.fr/~chenevier/articles/galoisQautodual2.pdf

(I mention this one because it is in some ways very close to the point of this question.)

However, there are elementary consequences of abelian class field theory that seem not to have obvious non-abelian analogues. The one I wish to mention today has to do with the fundamental group. Given a number field $K$ (assume it's totally imaginary to avoid some silly issues), how can we tell if it has non-trivial abelian unramified extensions? Class field theory says we can look at the class group, which is quite computable in principle, and even in practice for small discriminants. But now, suppose we go on to ask the non-abelian question: which number fields have $$\pi_1(Spec(O_K))=0?$$ That is to say, when does $K$ have no unramified extension at all, abelian or not? As far as I know, there is no easy answer to this question. Niranjan Ramachandran has pointed out that there are at least ten examples, $K=\mathbb{Q}$ (oops, that's real) and $K$ an imaginary quadratic field of class number one. I know of no others. Of course I would be happy to collect some more, if someone else has them lying around.

But the question I really wanted to ask is: Suppose we are in a Langlands paradise where everything reasonably conjectured by the programme is a theorem. Does this give a way to algorithmically (as we run over fields $K$) resolve this question as in the abelian case? Otherwise, is there a sensible refinement of the usual formulation that would subsume such applications?

I'm embarrassed to admit I hadn't followed the question mentioned by David Hansen (even after commenting on it). Thanks to David for pointing it out. Of course my main question still stands. I've changed the title following Andy Putman's suggestion. The original title evolved from a (humorously) provocative version that I normally use only among friends who already know I'm a Langlands fan: 'What is the Langlands programme good for?'

Regarding jnewton's very natural thought: in addition to other difficulties, one would also need to bound $n$.

Here is one more remark concerning jnewton's suggestion. Of course in the realm of classical holomorphic cusp forms, there are infinitely many of level one. More generally, it is shown in the paper

http://www.math.uchicago.edu/~swshin/Plancherel.pdf

that whenever $G$ is a split reductive group over $\mathbb{Q}$, there are infinitely many cuspidal automorphic representations that are unramified everywhere and belong to the discrete series at $\infty$. (I presume there are other results of this sort. This one I just happen to know from a talk last Fall.) According to Clozel's conjecture as you might find in

http://seven.ihes.fr/IHES/Scientifique/asie/textes/Clozel-juil06.pdf

(conjecture (2.1)), algebraic ones among them should correspond to motivic Galois representations (after we choose a representation of the dual group)*. I don't have the expertise to recognize algebraicity in such constructions, in addition to the danger that I'm misunderstanding something more elementary. But it seems to me quite a task to show directly that there are none corresponding to Artin representations. (The only case I could do myself is the classical one.)

Now, I would like very much to be corrected on all this. But such families do seem to indicate that a 'purely automorphic' approach to the the $\pi_1$ question is somewhat unlikely, at least within the current framework of the Langlands correspondence.

I suppose I'm sabotaging my own question.

*Note that in these situations, the Galois representations don't have to be unramified, since there is the choice of a coefficient field $\mathbb{Q}_p$. In general, they should only be crystalline at $p$.

Matthew: Since I didn't really expect a complete answer to my question, if you could write your extremely informative series of comments as an answer, I will accept it. (Barring the highly unlikely possibility that someone will write something better between the time you submit your answer and the time I look at it.)

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I'm embarrassed to admit I hadn't followed the question mentioned by David Hansen (even after commenting on it). Thanks to David for pointing it out. Of course my main question still stands. I I've changed the title following Andy Putman's suggestion. The original title evolved from a (humorously) provocative version that I normally use only among friends who already know I'm a Langlands fan: 'What is the Langlands programme good for?'

Here is one more remark concerning jnewton's suggestion. Of course in the realm of classical holomorphiccusp forms, there are infinitely many of level one. More generally, it is shown in the paper

http://www.math.uchicago.edu/~swshin/Plancherel.pdf

that whenever $G$ is a split reductive group over $\mathbb{Q}$, there are infinitely many cuspidal automorphicrepresentations that are unramified everywhere and belong to the discrete series at $\infty$. (I presume there are other results of this sort. This one I just happen to know from a talklast Fall.) According to Clozel's conjecture as you might find in

http://seven.ihes.fr/IHES/Scientifique/asie/textes/Clozel-juil06.pdf

(conjecture (2.1)), algebraic ones among them should correspond to motivic Galois representations (after we choose a representationof the dual group)*. I don't have the expertise to recognize algebraicity in such constructions, in addition to the danger that I'm misunderstanding something more elementary. But it seems to me quite a task to show directly that there are none corresponding to Artin representations. (The only case I could do myself is the classical one.)

Now, I would like very much to be corrected on all this.But such families do seem to indicate that a 'purely automorphic' approach tothe the $\pi_1$ question is somewhat unlikely, at least within the current framework of the Langlands correspondence.

I suppose I'm sabotaging my own question.

*Note that in these situations, the Galois representations don't have to be unramified, since there is the choice of a coefficient field $\mathbb{Q}_p$. In general, they should only be crystalline at $p$.

4 added 581 characters in body; edited title

# Howdoyouusenon-abelianNon-abelian class field theory ?andfundamentalgroups

Over the years, I've been somewhat in the habit of asking questions in this vein to experts in the Langlands programme.

As is well known, given an algebraic number field $K$, the Langlands programme proposes they propose to replace the reciprocity map $$A_K^*/K^*\rightarrow Gal(K^{ab}/K)$$ of abelian class field theory by a correspondence between the $n$-dimensional representations $\rho$ of $Gal(\bar{K}/K)$ and certain automorphic representations $\pi_{\rho}$ of $GL_n(A_K)$. (We'll skip the Weil group business for this discussion.) Substantial arithmetic information is carried on either side by the $L$-functions, which are supposed to be equal.

This involves deep and beautiful mathematics whenever something can be proved, and there are many applications, such as the Sato-Tate conjecture or this recent paper of Chenevier and Clozel:

http://www.math.polytechnique.fr/~chenevier/articles/galoisQautodual2.pdf

(I mention this one because it is in some ways very close to the point of this question.)

However, there are elementary consequences of abelian class field theory that seem not to have an obvious non-abelian analogueanalogues. The one I wish to mention today has to do with the fundamental group. Given a number field $K$ (assume it's totally imaginary to avoid some silly issues), how can we tell if it has non-trivial abelian unramified extensions? Class field theory says we can look at the class group, which is quite computable in principle, and even in practice for small discriminants. But now, suppose we go on to ask the non-abelian question: which number fields have $$\pi_1(Spec(O_K))=0?$$ That is to say, when does $K$ have no unramified extension at all, abelian or not? As far as I know, there is no easy answer to this question. Niranjan Ramachandran has pointed out that there are at least ten examples, $K=\mathbb{Q}$ (oops, that's real) and $K$ an imaginary quadratic field of class number one. I know of no others. Of course I would be happy to collect some more, if someone else has them lying around.

But the question I really wished wanted to ask today is: Suppose we are in a Langlands paradise where everything reasonably conjectured by the programme is a theorem. Does this give a way to algorithmically (as we run over fields $K$) resolve this question as in the abelian case(as we run over fields $K$)? ? Otherwise, what else is there a sensible refinement of the usual formulation that would have subsume such applications?

I'm embarrassed to be added admit I hadn't followed the question mentioned by David Hansen (even after commenting on it). Thanks to David for pointing it out. Of course my main question still stands. I changed the usual title following Andy Putman's suggestion. The original title evolved from a (humorously) provocative version that I normally use only among friends who already know I'm a Langlands fan: 'What is the Langlands programme good for?'

Regarding jnewton's very natural thought: in addition to allow such applications?other difficulties, one would also need to bound $n$.

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