Tools for the Langlands Program? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:35:26Z http://mathoverflow.net/feeds/question/15370 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15370/tools-for-the-langlands-program Tools for the Langlands Program? Ben 2010-02-15T21:32:29Z 2010-07-23T21:43:16Z <p>Hi,</p> <p>I know this might be a bit vague, but I was wondering what are the hypothetical tools necessary to solve the Langlands conjectures (the original statments or the "geometic" analogue). What I mean by this is the following: for the Weil Conjectures it became clear that, in order to prove them, one needed to develop a marvelous cohomology theory that would explain Weil's observations. Of course, we all know that etale cohomology is that marvelous tool. By analogy, what "black box" tools are necessary for the Langlands program? Broadly speaking what tools do we need for the Langlands program?</p> <p>Curious grad student, Ben</p> http://mathoverflow.net/questions/15370/tools-for-the-langlands-program/15388#15388 Answer by Kevin Buzzard for Tools for the Langlands Program? Kevin Buzzard 2010-02-15T23:36:17Z 2010-02-15T23:36:17Z <p>There are all sorts of problems with the Langlands conjectures that we (as far as I know) have no idea at all how to approach. As a very simple example of an issue for $GL(2)$ over $\mathbf{Q}$ that we cannot do, consider this: there should be a canonical bijection between continuous even (i.e. det(complex conj)=+1) irreducible 2-dimensional representations $Gal(\overline{\mathbf{Q}}/\mathbf{Q})\to GL(2,\mathbf{C})$ and normalised algebraic cuspidal Maass new eigenforms on the upper half plane. This is a sort of non-holomorphic analogue of the Deligne-Serre theorem which relates the odd irreducible Galois representations to holomorphic weight 1 newforms. One way of nailing this bijection is that given a Maass newform, then for all primes $p$ not dividing the level, the eigenvalue of $T_p$ (suitably normalised) should be the trace of the representation evaluated at the Frobenius element in the Galois group.</p> <p>You want a black box which will solve all of Langlands---then you need a black box which will solve this. Unfortunately it seems to me that firstly you'll need several good new ideas to resolve even this simple case, and secondly there is more than one strategy and it's not clear what will work first. As examples of the problems one faces: given the Galois representation, that's just a lump of algebra---a finite amount of data. However is one going to construct a bunch of analysis from it?? One way might be via the theory of base change, which works a treat for cyclic extensions, and just enough has been developed in order to resolve the problem for Galois representations with solvable image (one uses a lot more than the statement that the group is solvable---one uses that it is also "small"---this is not just a formal consequence of cyclic base change). This is the Langlands-Tunnell theorem, which gives the Maass form from the Galois representation if it has solvable image. In the non-solvable case one can dream of non-solvable base change, but non-solvable base change is really nothing but a dream at this point. So there's one big black box but that will only resolve one direction of one small fragment of the Langlands conjectures.</p> <p>Now what about the other way? Well here we're even more in the dark. Given an algebraic Maass form, we can't even prove that its Hecke eigenvalues are algebraic numbers, let alone the sum of two roots of unity. In the holomorphic modular form case we can get bases of the spaces of forms using e.g. coherent cohomology of the modular curve considered as an algebraic curve over $\mathbf{Q}$, or (in weights 2 or more) singular cohomology of a (typically non-trivial) local system on the curve. Both these machines produce $\mathbf{Q}$-vector spaces with Hecke actions, and hence char polys are in $\mathbf{Q}[x]$ and so eigenvalues are algebraic. But with algebraic Maass forms we have no such luxury. They are not cohomological, so we can't expect to see them in singular cohomology of a local system, and they are not holomorphic, so we can't expect to see them in coherent cohomology either. So we, vaguely speaking, need a black box which, given certain finite-dimensional complex vector spaces with Hecke actions, produces finite-dimensional $\mathbf{Q}$-vector spaces out of thin air, which when tensored up to the complexes give us back our groups. People have tried using base change to do this, or other known instances of functoriality, but everything so far has failed and it's not clear to me that one even has a conjectural approach for doing this direction. And I'm only talking about proving that the eigenvalues are algebraic---not even coming close to attaching the Galois representation!</p> <p>So one vague black box "non-abelian base change", and one hard problem that as far as I know no-one has ideas about, and, if you put these together, you would solve one teeny tiny insy winsy little part of the Langlands programme. Makes the Weil conjectures look like a walk in the park!</p> http://mathoverflow.net/questions/15370/tools-for-the-langlands-program/15394#15394 Answer by JS Milne for Tools for the Langlands Program? JS Milne 2010-02-16T00:44:11Z 2010-02-16T00:44:11Z <p>Hey! We're making progress. It used to be called the Langlands philosopy. [Oops, this was meant to be a comment on fpqc's comment.]</p> http://mathoverflow.net/questions/15370/tools-for-the-langlands-program/15412#15412 Answer by Emerton for Tools for the Langlands Program? Emerton 2010-02-16T03:02:47Z 2010-07-23T14:28:03Z <p>This answer deals with the classical Langlands program (if you like, the Langlands program for number fields).</p> <p>There are (at least) two aspects to this program: </p> <p>(a) functoriality: this is Langlands original conjecture, explained in the letter to Weil, and further developed in "Problems in the theory of automorphic forms" and later writing. It is a conjecture purely about automorphic forms. Langlands has outlined an approach to proving it in general is his papers on the topic of "Beyond endoscopy" (available online at his collected works). </p> <p>A proof of functoriality would imply, among other things, the non-solvable base-change discussed in Kevin's answer.</p> <p>It seems that for the "beyond endoscopy" program to work as Langlands envisages it, one would need unknown (and seemingly out of reach) results in the analytic number theory of $L$-functions. </p> <p>(b) reciprocity: this is the conjectured relationship between automorphic forms and Galois representations/motives. It has two steps: attaching Galois representations, or even motives, to (certain) automorphic forms, and, conversely, showing that all Galois representations of motives arise in this way. (This converse direction typically incorporates the Fontaine--Mazur conjecture as well, which posits a purely Galois-theoretic criterion for when a Galois representation should arise from a motive.)</p> <p>If one is given the direction automorphic to Galois, then there are some techniques for deducing the converse direction, namely the Taylor--Wiles method. However this method is not a machine that automatically applies whenever one has the automorphic to Galois direction available; in particular, it doesn't seem to apply in any straightforward way to Galois representations/motives for which some $h^{p,q}$ is greater than 1 (in more Galois-theoretic terms, which have irregular Hodge--Tate weights). Thus in particular, even if one could attach Galois representations to (certain) Maass forms, one would still have the problem of proving that every even 2-dimensional Artin representation of $G_{\mathbb Q}$ arose in this way.</p> <p>As to constructing Galois representations attached to automorphic forms, here the idea is to use Shimura varieties, and one can hope that, with the fundamental lemma now proved, one will be able to get a pretty comprehensive description of the Galois representations that appear in the cohomology of Shimura varieties. (Here one will also be able to take advantage of recent progress in the understanding of integral models of Shimura varieties, due to people like Harris and Taylor, Mantovan, Shin, Morel, and Kisin, in various different contexts.)</p> <p>The overarching problem here is that, not only do not all automorphic forms contribute to cohomology (e.g. Maass forms, as discussed in Kevin's answer), but also, not all automorphic forms appear in any Shimura variety context at all. Since Shimura varieties are currently the only game in town for passing from automorphic forms to Galois representations, people are thinking a lot about how to move from any given context to a Shimura variety context, by applying functoriality (e.g. Taylor's construction of Galois reps. attached to certain cuspforms on $GL_2$ of a quadratic imaginary field), or trying to develop new ideas such as $p$-adic functoriality. While there are certainly ideas here, and one can hope for some progress, the questions seem to be hard, and there is no one black box that will solve everything.</p> <p>In particular, one could imagine having functoriality as a black box, and asking if one can then derive reciprocity. (Think of the way that Langlands--Tunnell played a crucial role in the proof of modularity of elliptic curves.) Langlands has asked this on various occasions. The answer doesn't seem to be any kind of easy yes.</p> http://mathoverflow.net/questions/15370/tools-for-the-langlands-program/33142#33142 Answer by Greg Kuperberg for Tools for the Langlands Program? Greg Kuperberg 2010-07-23T21:43:16Z 2010-07-23T21:43:16Z <p>I don't know a whole lot about the Langlands program, but if there is one tool that seems to come up a lot in geometric Langlands, it's perverse sheaves. You see a lot of singular algebraic varieties in geometric Langlands, and perverse sheaves are meant as a singular generalization of a vector bundle with a flat connection. Ordinary sheaves are already a singular generalization of vector bundles, but not the relevant one. Perverse sheaves (which are made from sheaves but not sheaves themselves) are a more apropos generalization that incorporates and sort-of just is intersection (co)homology.</p> <p>I can also say that I wasn't going to learn about perverse sheaves until I had to. However, I have now seen several important papers, in the related categorification program, that read this way: "Perverse sheaves + necessary restrictions = a good solution". So now I might be slowly getting used to them. I can also see that even the formalism perverse sheaves or intersection homology is sort-of inevitable. In some of the simpler constructions, the varieties (over $\mathbb{C}$, say) are non-singular and certain answers arise as ordinary cohomology products or intersection products. For instance, the Schubert calculus in a Grassmannian manifold. What choice do you have if the Grassmannian is replaced by a singular variety $X$? For some of these categorification/Langlands questions, you can either propose wrong answers, or ad hoc answers, or you can automatically get the right answer by using intersection homology on $X$. (With middle perversity, as they say.)</p>