Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

On pg. 1 of the slides of a talk, Henri Darmon wrote:

Question: What is an interesting Diophantine equation?

A “working definition”. A Diophantine equation is interesting if it reveals or suggests a rich underlying mathematical structure.

One can adopt the perspective that the interesting elliptic curves over $\mathbb{Q}$ are those that are modular and view the fact that all elliptic curves over $\mathbb{Q}$ are modular as being of minor significance (in the sense that if some weren't, they wouldn't be so interesting).

I realize that my raising this perspective may come across as an affront to some of the celebrated research of recent times, and would hasten to emphasize that I'm asking my questions here in good faith with a view toward learning more.

What would we lose if we decided to focus only on those Galois representations that are attached to automorphic forms and ignore the possibility that some do not?

One thing that one would lose is Wiles' proof of Fermat's last theorem. Until recently, my attitude had been that the Frey Curve construction is a curiosity and that a "morally right" proof would come from the $abc$-conjecture applied to sufficiently large exponents together with the theory of arithmetic of cyclotomic fields to rule out the possibility of nontrivial solutions to the equations with smaller exponents. However, recently I came across slides from a talk by Minhyong Kim in which Kim wrote (pg. 29):

The idea of encoding points into 'larger' geometry is a common one in Diophantine geometry, as when solutions

$a^n + b^n = c^n$

to the Fermat equation are encoded into the elliptic curves

$y^2 = x(x - a^n)(x + b^n)$.

The geometry of the path torsor $\pi_1(X(\mathbb{C}); b, x)$ is an extremely canonical version of this idea.

I find the idea that the Frey curve construction is canonical to be fascinating! It raises the possibility that the proof of Fermat's last theorem coming from the study of Frey equations is morally right. [Edit: As KristianJS aptly points out, I misread Kim's quote. So I'd recur to my remark above about my impression on what a "morally right" proof of Fermat's Last Theorem would look like.]

Anyway, I'd be very interested in further examples concerning the significance of all L-functions attached to Galois representations (of suitable type) being modular.

[Added: If I remember correctly, In "The Map of My Life" Shimura wrote that he was more interested in the fact that suitable cusp forms correspond to elliptic curves than in the converse. This seems relevant. However, I cannot find the quotation and may be misremembering. I would welcome a reference from anybody who remembers this.]

[Added: I just asked another question that touches on material that may provide a partial answer to this question.]

share|improve this question
1  
This seems to suggest that the only possible rich underlying mathematical structure for a Galois representation is an automorphic one, which is of course false. There are many interesting structures in Galois representations that do not involve Langlands at all. Secondly, without proving that all Galois representation are automorphic, it can be very hard to know whether a specific individual Galois representation or family of Galois representations you care about is automorphic. There are many natural problems where one runs across a specific Galois representation or family of them. –  Will Sawin Oct 4 '12 at 19:02
1  
Some people just happen to be interested in solving diophantine equations, and in particular in understanding the arithmetic of curves/arbitrary abelian varieties/you name it over arbitrary number fields. This is to name but one source of not obviously modular $L$-functions that people find interesting in their own right. –  Alex B. Oct 5 '12 at 10:12
1  
Dear Jonah, You suggest below a perspective of throwing out Kronecker--Weber. Of course you are free to do so, but proving reciprocity laws (quadratic reciprocity, on which Gauss focussed so much effort, and then higher reciprocity laws, on which Kummer focussed most of his work, and then non-abelian reciprocity laws, which is what you are asking about) has been a major focus of many number theorists for more than 200 years. Also, proving a reciprocity law (i.e. establishing that a Galois representation or Diophantine equation is modular) introduces enormous extra structure and a new set ... –  Emerton Oct 6 '12 at 21:13
3  
Dear Jonah, What I mean is that quadratic reciprocity is about splitting of primes in the field $\mathbb Q(\sqrt{\pm p}),$ and is (in my view) most naturally proved by embedding $\mathbb Q(\sqrt{\pm p})$ inside $\mathbb Q(\zeta_p)$. Of course such an embedding can be constructed by Gauss sum's, but however one constructs it, one is using the philosophy of Kronecker--Weber/automorphy: we take a field for which we don't know reciprocity (the quadratic field), and place it in a context where we do have a reciprocity law (the cyclotomic field). Modularity for elliptic curves is used in the ... –  Emerton Oct 7 '12 at 0:29
2  
... same way: we start with a Diophantine equation for which we don't know reciprocity, but by writing it as a quotient of a modular curve, for which we have rich reciprocity laws (Eichler--Shimura, Bloch--Beilinson elements, Kato's Euler systems, ...) we can prove things about the original elliptic curve. Regards, Matthew –  Emerton Oct 7 '12 at 0:31

5 Answers 5

up vote 9 down vote accepted

Your question reminds me of a current strain of research whose starting point is Serre's conjecture, now the Khare-Wintenberger Theorem:

any continuous odd irreducible two-dimensional Galois representation over a finite field arises from a modular form

The question one might ask is then "Where are the even Galois representations?"

The answer given (mostly by F. Calegari) is that they just don't exist when you put certain additional restrictions on your Galois representation. Suppose then that you somehow have an even two-dimensional Galois representation in your hands. Well then this Galois representation is very special in some ways that might not be apparent! You can then ask: where did this representation come from? Is it modular in some non-obvious way?

So if I were to answer your question "What would we lose if we decided to focus only on those Galois representations that are attached to automorphic forms and ignore the possibility that some do not?" I would say that you lose a lot of knowledge about what makes modular Galois representations special. In your terminology, you might lose the scope of just how interesting certain representations can be!

share|improve this answer
    
Thanks for highlighting Calegari's work. My main takeaway from your comment in conjunction with the introduction to Calegari's paper is that one can rule out the existence of even two dimensional Galois representations (satisfying suitable restrictions) by proving the potential modularity of Galois representations satisfying these restrictions. –  Jonah Sinick Oct 5 '12 at 7:06
    
I hope that somebody will eventually offer a more comprehensive answer, but I've accepted this one for now. –  Jonah Sinick Oct 5 '12 at 23:27
1  
there are lots of even mod p galois reps (just take a totally real galois number field and an irred. 2-dimensional mod p representation of its galois group over Q), calegari's theorem is for p-adic galois reps. thus it would be better to talk about the (odd) fontaine-mazur (for GL2/Q), essentially proved by kisin and emerton, rather than serre's conjecture above. –  fherzig Oct 6 '12 at 18:57
    
i don't know of any equally satisfying answer as serre's conjecture for even mod p galois reps. they are not all explained by mass forms as the projective image could be very big (like PSL2(Fp)), preventing a lift to char 0. one can try to relate them to automorphic reps, or more generally mod p cohomology of arithmetic groups (which doesn't always lift to char 0), in a more indirect way by restricting to the galois group of an imaginary quadratic extension, or (as does ash et al) by adding a character. –  fherzig Oct 6 '12 at 19:05

I am not sure I understand your question. Or rather, as I understand your question, you have already answered it. Well, you asked "what would we loose if we..." and you answered "Wiles' proof of FLT". Isn't that enough? We have developed more complicated theories for much less venerable questions!

Admitting rhetorically that it isn't enough, one can develop your answer. If we want to restrict our consideration to representation that are a priori modular, we would loose not only Wiles' proof of FLT, but also Wiles and others's proof of the Taniyama-Shimura-Weil conjecture (and this one has nothing to do with the Frey's curve), so we would lose the corollary that L-functions of elliptic curves satisfy a functional equation and have an analytic extension... We would then not even be able to formulate the beautiful conjecture of Birch and Swinnerton-Dyer! We would also loose that the symmetric powers of the representation attached to an elliptic curve are modular. So we would lose the Sato-Tate conjecture as well.

In a word, we would loose all the Galois representations coming from algebraic geometry, because we don't know a priori that they are modular/automorphic. And the fact is that those representations contain a lot of deep arithmetic and geometric (if you believe to Tate's conjecture) information.

Added: The above is about the fact that without talking of representations that are not modular/automorphic we would lose the ability even to formulate a theorem of the form "such and such representation is modular". Here's another, different, point. We also need Galois representations that are not modular, in order to prove things about representations that are modular. I am refereeing to all the arguments of p-adic families (Hida's family, eigencurve, eigenvarieties) who carry at each point a Galois representation which most of the time is not modular. but we need those point to give flesh to the eigenvariety whose "skeleton" is made of the modular ones. To give an early example of those techniques, Wiles proved that a Galois representation attached to an ordinary form is ordinary (a statement purely about modular galois representation) using Hida's families, hence non-modular representations. Nowadays those techniques are ubiquitous...

share|improve this answer
    
Joël, Birch and Swinnerton-Dyer. –  Chandan Singh Dalawat Oct 5 '12 at 14:19
    
Sure! Edited... –  Joël Oct 5 '12 at 14:21
1  
(Potentially modular, I guess, for the symmetric powers of the Tate module.) –  D. Savitt Oct 5 '12 at 15:51
    
Thanks! Regarding FLT, my suggestion was that there's another, more natural proof. Regarding BSD, if one would like, one can restrict oneself to the consideration of modular elliptic curves and their arithmetic, then BSD is well formulated. If one restricts oneself to modular elliptic curves, my impression is that the analytic continuation of the symmetric power L-functions is expected to be provable by analytic means (Kim, Shahidi, etc.)... –  Jonah Sinick Oct 5 '12 at 17:31
2  
Dear Jonah, The Kim--Shahidi approach (which builds on an idea of Langlands) has intrinsic limitations, as far as I know (in that it can only access certain symmetric powers, not all of them). Langlands envisions a proof of functoriality for symmetric powers (and hence analytic continuation of symmetric power $L$-functions) via trace formula methods, but the trace formula arguments (especially endoscopy) involve, as well as a lot of harmonic analysis, also the full force of class field theory (via Galois cohomology calculations). It is hard to remove class field theory from the arguments! –  Emerton Oct 7 '12 at 1:19

I want to make a couple of comments about the premise of this question. First, the OP asks what the consequences would be if we "focus only on those Galois representations that are attached to automorphic forms and ignore the possibility that some do not". But of course it is certainly the case that there exist Galois representations that are not attached to automorphic forms (Galois representations that are ramified at infinitely many primes, for instance).

Second, the OP writes about "Galois representations that are attached to automorphic forms" as though this phrase is well-defined. But do you mean the automorphic forms that conjecturally believed to have Galois representations attached to them, or the automorphic forms that are currently known to have Galois representations attached to them? The latter is a (growing but proper!) subset of the former; and note again that the former is not "all automorphic forms" (see e.g. the paper of Buzzard-Gee in the 2011 LMS Durham Symposium).

So, there's a conjectural map from (some) automorphic forms to (some) Galois representations, and this has been understood for decades to have significant arithmetic consequences for representations in the image of the map. The most generous way I can think to interpret the question is "What would happen if we satisfied ourselves with statements of the form "If $\rho$ is in the image of this map in an instance where the map has been constructed...", without attempting to characterize the image in some easily checkable way?", and the answer is simply that we would lose most or all of the arithmetic consequences.

share|improve this answer
    
Regarding your first question - Yes, I know that not all Galois representations are attached to automorphic forms. Regarding your other two questions: perhaps I should have phrased by question more carefully. I had in mind the correspondence between elliptic curves over Q and weight 2 cusp eigenforms for congruence subgroups of SL(2, Z) in particular, but also meant to indicate that I'm interested in the question in general generality as well. –  Jonah Sinick Oct 5 '12 at 16:52
    
When you say "we would lose most or all of the arithmetic consequences," which arithmetic consequences do you have in mind? –  Jonah Sinick Oct 5 '12 at 16:52

You would lose all the p-adic families of Galois representations, that are necessary to prove properties of the modular ones. Hida's family was necessary for the proof ofMTT, for example.

share|improve this answer
    
Thanks for the response! What's MTT? –  Jonah Sinick Oct 5 '12 at 16:17
    
MTT = Mazur–Tate–Teitelbaum. A conjecture about L-invariants proved by Greenberg and Stevens using Hida families. –  Rob Harron Oct 5 '12 at 16:52

Proving modularity of finite image Galois representations seems to be the most feasible way of proving the Artin conjecture. In fact, this was one of Langlands' original motivations.

share|improve this answer
5  
I suppose you could, but I think it would not be in the spirit of non-abelian class field theory. For $GL_{1}$, this is like asking "why bother studying all abelian extensions when we can just study the cyclotomic extensions?" You'd be throwing out Kronecker-Weber! –  Kevin Ventullo Oct 5 '12 at 7:46
3  
One might as well throw out all theorems containing a universal quantifier. –  Will Sawin Oct 5 '12 at 16:50
1  
Dear Jonah, The statement that all abelian extensions are class fields is encapsulated in Artin's reciprocity law. This generalizes and synthesizes a huge amount of earlier number theory (including much of the number-theoretic work of Gauss, Eisenstein, and Kummer). It has many applications. I find your perspective interesting to think about, but I'm mentioning this just so you have a sense of what's at stake in the discussion. Incidentally, in modern proofs of class field theory, the proof that all abelian extensions are class fields comes before the proof that class field exist and ... –  Emerton Oct 7 '12 at 0:42
1  
@Jonah: I didn't mean to suggest that I find the existence of Haar measure interesting on its own. I was comparing the distinction "subfields of cyclotomic fields" vs. "finite abelian extensions of Q" with "locally compact groups admitting a Haar measure" vs. "locally compact groups" (include Hausdorff condition both times). Things of the first type are trivially of the second type, but it sure is nice that the things of the second type are also of the first type. As a simple example of what you gain, if you show a Galois extension of Q is abelian then Kron-Web theorem tells you [contd] –  KConrad Oct 7 '12 at 4:44
5  
that the splitting law in that extension will be describable by congruence conditions. Do you agree that is nice? Isn't it nice that the hard property that an odd prime has the form $x^2 + y^2$ is the same as being 1 mod 4, which is easy to check? Abelian extensions can arise in nature not directly as a subfield of some cyclotomic field, so the KW theorem is telling you something valuable about that field (e.g., by KW its zeta-function decomposes as a product of Dirichlet L-functions, which a priori is not obvious at all). –  KConrad Oct 7 '12 at 4:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.