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Bogdan Grechuk recently asked for elementary consequences of the Langlands program. His question reminded me of something that has nagged at me for a while. Before I state my question, let me give a bit of background and motivation.

My favorite way of motivating quadratic reciprocity is to show someone the prime factorizations of $n^2 - 5$ for $n=3,4,5,\ldots$ and to ask why we never see a prime ending in $3$ or $7$. The explanation, of course, is that if $p$ is an odd prime and $p \mid (n^2-5)$ then $$1 = \biggl({5 \over p}\biggr) = \biggl({p \over 5}\biggr)$$ by quadratic reciprocity. I think it is not too hard to jazz up this example to give an elementary motivation for (abelian) class field theory, since in the abelian case, the splitting behavior of primes is controlled by congruence conditions, and congruence conditions are pretty easy to grasp.

My question is, can we come up with a similar elementary motivation for nonabelian class field theory? In his essay, Representation theory: Its rise and role in number theory, Langlands writes down the polynomial $x^5 + 10x^3 -10x^2+35x-18$ and then remarks:

It is irreducible modulo $p$ for $p = 7, 13, 19, 29, 43, 47, 59, \ldots$ and factors into linear factors modulo $p$ for $p = 2063, 2213, 2953, 3631, \ldots\,$. These lists can be continued indefinitely, but it is doubtful that even the most perspicacious and experienced mathematician would detect any regularity. It is none the less there.

Though Langlands does go on to explain some of the theory behind this example, the reader who is hoping for an elementary example of the "regularity" alluded to above will be disappointed, because nothing like that is described explicitly in the paper.

Of course, nothing like congruence conditions exist in the nonabelian case. To describe the "regularity" properly, we need to introduce modular forms, which are difficult to describe in a completely elementary manner. But I'm reluctant to give up so quickly. Modular forms enjoy symmetries that should (I think) be translatable into identities between certain infinite series. Can we, for example, take the infinite sequence $p = 2063, 2213, 2953, 3631, \ldots$ above and write down some kind of striking infinite series identity or identities that they obey? I realize I'm being a bit vague here, but my question is in the same spirit as Bogdan Grechuk's question. That is, I'd like some sort of maximally elementary statement that can give the novice some flavor of the remarkable structure that the Langlands program aims to elucidate.

EDIT: Denis T's comment points to Emerton's answer to another MO question which is very close to what I am looking for. If I understand it correctly, it says that if we define integers $a_n$ by the equation $$q \prod_{r=1}^\infty (1-q^r)(1-q^{23r}) = \sum_{n=1}^\infty a_n q^n,$$ then the polynomial $x^6 - 6x^4 + 9x^2 + 23$ splits completely into distinct linear factors mod $p$ if and only if $a_p = 2$.

One shortcoming of this example is that it is arguably still within the realm of abelian class field theory, in particular of the theory of Hilbert class fields. It turns out that the polynomial $x^6 - 6x^4 + 9x^2 + 23$ splits completely into distinct linear factors mod $p$ if and only if $p$ is an odd prime of the form $x^2 + 23y^2$ and $p\ne 23$ (see for example Theorem 5.26 of Cox's book Primes of the Form $x^2 + ny^2$). Nevertheless, it would definitely answer my question if someone could write down an analogous explicit statement (for some other polynomial) that is still conjectural. Failing that, an analogous statement for a more interesting Galois group than $\mathfrak{S}_3$ would also be a good answer.

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    $\begingroup$ mathoverflow.net/a/12382 Does this answer provide the kind of "striking identity" you want? $\endgroup$ Commented Dec 25, 2024 at 6:10
  • $\begingroup$ @DenisT Ah, that's very close! Maybe you can post a brief summary of Emerton's answer there as an answer here. If nobody comes up with something better, I can accept it. $\endgroup$ Commented Dec 25, 2024 at 12:52
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    $\begingroup$ The scope of the question is quite lost on me. What do you mean here with "non-abelian class field theory"? If you just mean Langlands program, how exactly is this different from Bogdan's question you have asked? $\endgroup$ Commented Dec 25, 2024 at 22:01
  • $\begingroup$ Possibly one can take the modular form in Emerton's example and get a suitable congruence with a higher weight modular form, which one could describe in terms of Eisenstein series/theta functions. Would this be what you want? $\endgroup$ Commented Dec 26, 2024 at 15:55
  • $\begingroup$ @Wojowu I'm trying to narrow the focus to an elementary statement about the set of primes with respect to which a polynomial with non-abelian Galois group will (for example) split completely. $\endgroup$ Commented Dec 26, 2024 at 21:42

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Well, elementary is subjective, but one prediction of non-abelian class field theory is Dedekind's conjecture that the Riemann zeta function divides all Dedekind zeta functions. In particular, every zero of $\zeta(s)$ is also a zero of $\zeta_K(s)$. Surprisingly, abelian class field theory is sufficient to verify this for all Galois number fields, but the non-Galois case seems to require non-abelian class field theory which is still in the making.

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    $\begingroup$ This is known as Dedekind's conjecture, and more generally one believes that $\zeta_L(s)/\zeta_K(s)$ is holomorphic for any number field extension $L/K$. As this quotient factors as a product of Artin L-functions, this follows from Artin's conjecture on their holomorphy, which follows from Langlands reciprocity by identifying them with automorphic L-functions. $\endgroup$ Commented Dec 25, 2024 at 22:03
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    $\begingroup$ @Wojowu Yes. The words "Dedekind's conjecture" also appear in my post. I restricted to a special case and omitted the representation theoretic details for simplicity of exposition. $\endgroup$ Commented Dec 25, 2024 at 22:17
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    $\begingroup$ Ah somehow I've missed that you calling it by name, I promise I have actually read your post :P The goal of my comment was mostly to indicate how these results follow from "non-abelian class field theory", hopefully complementing your post. $\endgroup$ Commented Dec 25, 2024 at 22:40
  • $\begingroup$ Dedekind has nothing to do with this conjecture. It was first stated as a question by Edmund Landau. $\endgroup$ Commented Apr 15 at 7:45
  • $\begingroup$ @FranzLemmermeyer Well, "nothing to do" is a bit too harsh. There was a discussion of this story at mathoverflow.net/q/221174 In particular, KConrad wrote in a comment: For completeness, the paper of van der Waall is "On a conjecture of Dedekind on zeta-functions" Indagationes Math. 78 (1975), 83-86. He writes near the start that since Dedekind showed the ratio is holomorphic for $K$ a pure cubic field, "we will equip the following conjecture with Dedekind's name". $\endgroup$ Commented Apr 16 at 0:45
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Turns out I have a recent book that does this; Hiramatsu and Saito (2017) An Introduction to Non-Abelian Class Field Theory ::subtitle:: Automorphic Forms of Weight $1$ and $2$-Dimensional Galois Representations

Chapter 5 is $2$-Dimensional Galois Representations of odd type and non-dihedral cusp forms of weight $1$ pages 93-106.

Chapter 5, section 3 is The Case of type $A_5$ pages 101-103

Chapter 5, section 3.1 is The First Example due to Buhler pages 101 and 102.

He considers $F(x) = x^5 + 10 x^3 - 10 x^2 + 35 x - 18 .$ Then $K$ is the splitting field over $\mathbb Q$ and $G$ is the Galois group.

Theorem 5.9 (Buhler) There is an icosahedral representation $$ \rho : Gal(\bar{\mathbb Q} / \mathbb Q) \rightarrow GL_2(\mathbb C) $$ of conductor 800 such that $L(s, \rho)$ is an entire function of $s.$

They outline Buhler's proof.

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The other references in this subsection : G. Frey (1994), Construction and arithmetical applications of modular forms of low weight Deligne and Serre (1974) Formes modulaires de poids $1$ they say pages 507-530

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Preview: https://www.google.com/books/edition/Introduction_To_Non_abelian_Class_Field/_E0tDQAAQBAJ?hl=en&gbpv=1&pg=PA3&printsec=frontcover

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    $\begingroup$ Did your post get cut off, or is it meant to end with "they say pages 507–530"? $\endgroup$ Commented Dec 26, 2024 at 20:59
  • $\begingroup$ @LSpice I guess cut off, in that I went to look for more bibliographic material and saved the draft. Anyway, I think I have exactly the right reference for this question. Oh, I was going to check whether Deligne and Serre is in a book or a journal, it says Ann. scient. Ec. Norm. Sup., series, t.7 (1974), 507-530. I'm thinking journal.... annales-ens.centre-mersenne.org/index.php/ASENS $\endgroup$ Commented Dec 26, 2024 at 21:10
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    $\begingroup$ Re, I find that French publications sometimes tend to blur the line between journal and book, so I dare not speculate. Here's a free Numdam link. $\endgroup$ Commented Dec 26, 2024 at 21:27
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    $\begingroup$ @WillJagy Thanks for the answer, but I don't think this yields a statement that I would consider elementary. $\endgroup$ Commented Dec 26, 2024 at 21:44
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    $\begingroup$ Also in response to a comment you repeated a couple of times, the fact that Buhler's proof was an entire Ph.D. thesis doesn't mean that an elementary statement can't be extracted, especially since I'm not asking for a complete description of what's going on, but just a striking corollary that can be understood with minimal background. $\endgroup$ Commented Dec 26, 2024 at 21:51
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Here is a surprising application of non-abelian class field theory. (As mentioned in another answer, elementary is subjective.) Let $X$ be a smooth connected variety over $\Bbb C$ and let $Y$ be a hyperbolic curve over $\Bbb C$ of genus at least $2$. Suppose that we have two dominant morphism $f,g:X \to Y$ such that their analytifications $f^{an},g^{an}:X(\Bbb C) \to Y(\Bbb C)$ are homotopic. Then $f=g$.

This doesn't seem to be related to class field theory or number theory at all, but the proof actually uses anabelian geometry (which deserves to be called non-abelian class field theory just as much as the Langlands programme does).

The way that anabelian geometry appears in the proof is that $X$, $Y$, $f$ and $g$ may actually be defined over a finitely generated (though generally transcendental) extension of $\Bbb Q$ and there one can use the injectivity of the Hom-conjecture in anabelian geometry for hyperbolic curves over finitely generated fields.

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Let me explain the difficulty with your request for someone to write down an analogous explicit statement that is still conjectural.

The Langlands program concerns a correspondence between Galois representations and automorphic forms. The main partial results are Galois-to-automorphic results of the form "For each Galois representation such that ... there exists an automorphic form such that ..." and automorphic-to-Galois results that are the other way around.

The explicit example you give involves writing down both the Galois representation and the automorphic form explicitly, of course with the Galois representation described elementarily by a field extension and the automorphic form described elementarily by a product formula. The problem with this is that if we have either a Galois-to-automorphic result or an automorphic-to-Galois result that applies to the example, it will no longer be conjectural. (Once the existence of a Galois representation or automorphic form is proven, finding the right one is "just" a computation, which for equations as simple as the ones you wrote down is, as far as I know, easy.)

So the key difficulty is that automorphic-to-Galois results are known in such great generality that it may not be possible to find another sufficiently explicit example. Emerton's example is a holomorphic modular form, and automorphic-to-Galois results are known for all holomorphic modular forms. In fact, anything you write down with an explicit power series is going to be holomorphic, so you seem to be in a lot of trouble here.

Automorphic-to-Galois results are expected, but not known for Maass forms with Laplacian eigenvalue 1/4. Galois-to-automorphic results are known by Langlands-Tunnell when the Galois representation has solvable image. But this is fine, we may want non-solvable image anyways to avoid the applicability of ordinary class field theory. Maass forms with eigenvalue 1/4 are perfectly concrete, one can compute with them, for example showing that the first two nonsolvable ones with squarefree conductor have level 1951 and 2141 and finding the associated polynomials of degree 5, as in the paper Numerical computations with the trace formula and the Selberg eigenvalue conjecture by Booker and Strömbergsson. But they're not given by any kind of snazzy q-series formula. Instead one uses the trace formula to compute with them.

For automorphic forms on other groups I believe the situation is similar: The automorphic-to-Galois theorems are very powerful, handling the holomorphic cases and non-holomorphic forms are even harder to write down. I'm not truly an expert on this and it's possible you could find some useful gap in the automorphic-to-Galois theorems (maybe something to do with exceptional groups?) But even if you had such a scenario where the automorphic forms was holomorphic you could actually write down a formula for it in a reasonable amount of space, and extract the Hecke eigenvalues in a nice way (since they're no longer simply the $p$th coefficient of the $q$-series).

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  • $\begingroup$ Thanks; this is very illuminating! Probably, then, I should back off from asking for something conjectural, and settle for something that definitely goes beyond abelian class field theory. I took another look at Chandan Singh Dalawat's answer to his own question and it is very close to what I'm looking for. The only slightly annoying thing is that explicitly writing down a polynomial that generates the number field $\mathbb{Q}(E[l])$ for $l=7$ or even $l=3$ is painful. I asked Magma to compute one and the coefficients were huge. But maybe that's unavoidable. $\endgroup$ Commented Jan 9 at 19:42
  • $\begingroup$ @TimothyChow I think there is some hope. Ideally one wants to consider a modular form of weight $1$ whose associated Galois representation has Galois group $A_5$ which then corresponds to a degree $5$ polynomial. The issue is then how to write down the modular form of weight $1$. There are eta quotients, but these probably can't produce any modular forms of high enough level. But there are other ways to explicitly write down modular forms, such as theta functions. Theta functions of weight $1$ only correspond to dihedral modular forms, $\endgroup$ Commented Jan 9 at 19:46
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    $\begingroup$ Maybe something can be done with the example from Joe Buhler's Ph.D. thesis. As I mentioned in a separate comment, Buhler didn't have an immediate answer to my question; "his" modular form has no slick formula. But again, if all I'm after is an attractive corollary rather than a complete characterization, I feel that there is still some hope. $\endgroup$ Commented Jan 10 at 14:39
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    $\begingroup$ @TimothyChow Buhler's modular form has weight 1 and level 800. There are other modular forms of weight 1 and level 800 that do have slick formulas. If we multiply one of them by Buhler's form, we get a modular form of weight 2 and level 800, and then we can divide to get a formula for Buhler's. The advantage of weight 2 is we have access to many more theta functions. If we choose the multiplier to be an eigenform then the product will have a fixed character and we can restrict attention to spaces of modular forms with that character as nebentypus. $\endgroup$ Commented Jan 13 at 11:48
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    $\begingroup$ @TimothyChow However, these spaces seem to be 100-200 dimensional, so even if a formula in terms of theta function exists, it will likely be a linear combination of 100-200 theta functions. One can try to go to spaces of even higher level but this just makes the situation worse. $\endgroup$ Commented Jan 13 at 11:48

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