## August 21, 2010

### What is the Langlands Programme?

#### Posted by Tom Leinster

You probably know that the 2010 Fields Medals have been announced.

One of the four medallists, Ngô Bảo Châu, works on the Langlands programme. Now, I know the Langlands programme is famous. In particular, Laurent Lafforgue won a Fields medal for his work on the Langlands programme in 2002, so there was a flurry of publicity surrounding it then.

The thing is, I’ve never succeeded in understanding the slightest thing about it. It’s as if it’s got a hard, shiny shell—it resists all attempts at explanation, even when the person listening is a trained, interested, mathematician. But I’d like to believe that isn’t so.

I’ve heard numerous attempted explanations in the past, but as soon as I run into a term like “automorphic form” or “reductive group” I’m lost. (I can go and look those up, and I have, but I still haven’t learned anything about the Langlands programme itself.) I’ve tried reading the two explanations on the ICM page, but one’s pitched too high and the other too low. I’ve tried the Wikipedia page, but it jumps suddenly from sentences such as

It is a way of organizing number theoretic data in terms of analytic objects

to sentences such as

The Artin reciprocity law applies to a Galois extension of algebraic number fields whose Galois group is abelian, assigns L-functions to the one-dimensional representations of this Galois group; and states that these L-functions are identical to certain Dirichlet L-series or more general series (that is, certain analogues of the Riemann zeta function) constructed from Hecke characters.

I’d love it if someone could explain just one thing about the Langlands programme in terms I’ll understand. If you can do that, I’ll have learned more about it from you than I ever have from anyone else.

To give an indication of where I’m at, here’s what I think I know about the Langlands programme:

• it’s a fantastically bold, fantastically sweeping set of ideas linking together many parts of mathematics
• it’s been the source of a lot of influential work
• it has something to do with representation theory, algebraic geometry and number theory.

I’m afraid that’s the sum total of my knowledge, and I’m not even sure that the last one’s quite right. To make matters worse, I know very little about representation theory, algebraic geometry or number theory. For example, that second Wikipedia sentence contains five terms whose definitions I don’t know.

I’ll be very happy if someone can tell me something I can understand. I don’t expect to understand much, but even a tiny bit would be an advance. Thanks!

Posted at August 21, 2010 12:51 AM UTC

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### Re: What is the Langlands Programme?

http://online.itp.ucsb.edu/online/duallang-m10/

You could begin by thinking categorically. The so called categorical Langlands conjecture says: the category of some-modules for Bun(G) is equivalent to the category of some-modules for Loc(G*) where G and G* are dual Langlands groups, and Bun(G) means bundles for some curve C, and Loc stands for ‘local systems’ for some curve C. See the above online talks.

So even the ‘categorical’ version involves tons of classical geometry, and is not very categorical. The most categorical part seems to be the links to Chern-Simons and Khovanov homology and other knotty theories, despite Kapranov’s attempt some years ago to talk about 2-cat structures associated to Langlands. For me, also at your level, the trickiest thing is sorting out the different kinds of Langlands: classical, geometric, bla bla.

Posted by: Kea on August 21, 2010 1:47 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks, Marni. But unfortunately I have to spell out my ignorance here. For example, I don’t know the meaning of the following terms:

• dual Langlands groups
• local system
• curve (though of course I know roughly what this means)
• bundle for a curve (vector bundle, fibre bundle, …?)
• module for $Bun(G)$ (because I don’t know what type of thing $Bun(G)$ is; is it a ring?).

So I’m afraid I’m not enlightened yet.

The talks you link to are from a conference, “Langlands-Type Dualities in Quantum Field Theory”, currently taking place. I have a sneaking suspicion that no talk at a meeting with a title like that will explain the very basics of the Langlands Programme. I’m not going to watch all of them in order to find out, but if you know one that does do this, I’d be pleased to know which.

Posted by: Tom Leinster on August 21, 2010 2:04 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

The only part of it I understand is the part Kea mentioned, where the curve is a formal disc. But I will describe things in a different way.

Start with Omega K, the space of smooth based loops into a compact group K. This has a nice K-invariant Morse-Bott function (the one Bott invented the theory for), measuring the integral of the velocity squared (using K’s metric) around the loop. The critical points are geodesics, which being based are 1-parameter subgroups in K. Up to conjugation, they are 1-parameter subgroups in a maximal torus T, i.e. elements of the “coweight lattice” ker: Lie(T) -> T. But two different coweights may be K-conjugate, so really the critical submanifolds correspond to dominant coweights.

That begins to sound like representation theory, except for the “co”. So we decide that K’s coweight lattice should be some other group’s weight lattice. This is (hardly) the definition of the Langlands dual group.

The connection is made more precise as follows. Given a dominant coweight of K, take not just the corresponding critical set for the Morse-Bott function but its (finite-dimensional) stable manifold. Then take the closure. This turns out to be a singular projective variety, and one can identify its intersection homology with the corresponding irrep for the Langlands dual group. This is called the “geometric Satake correspondence”, and is due to some combination of Lusztig, Ginzburg, Mirkovic, Vilonen, and probably others. The history seems very murky.

If one takes the union of these singular projective varieties, one gets a sort of polynomial approximation to the smooth Omega K, and is to some extent treatable as an infinite-dimensional algebraic variety. How to describe it algebraically? It’s the space of G-bundles (G the complexification of K) over a formal disc (coordinate ring = power series) trivialized on the complement of the origin (coordinate ring = Laurent series).

You could imagine extending the above geometric representation theory to other complex curves. This is the geometric Langlands program, and what Lafforgue was involved in. Or to other curves like Spec Z. Now you’re really asking for it, and my answer grinds to a halt here.

Posted by: Allen Knutson on August 21, 2010 3:30 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks, Allen. I do appreciate you taking the time to explain, but I can also see that this thread is going to involve me having to be rather open about how ignorant I am.

I believe I followed this part:

Start with Omega K, the space of smooth based loops into a compact group K. This has a nice K-invariant Morse-Bott function (the one Bott invented the theory for), measuring the integral of the velocity squared (using K’s metric) around the loop. The critical points are geodesics, which being based are 1-parameter subgroups in K.

But I’m afraid I couldn’t get any further. When I said I knew very little representation theory, I really meant it. And I know equally little Lie theory.

So, for example, I don’t know what a maximal torus or a weight is. I can look up maximal torus and learn that it’s a compact, connected, abelian Lie subgroup that is maximal among all such subgroups. It would probably take me a while to digest that—to understand why that’s an interesting property, and what role it plays. Certainly it wouldn’t harm me to do so, but I was hoping that I might be able to understand at least one small thing about the Langlands programme without first having to digest several substantial preliminary concepts. Or maybe that’s unrealistic? I don’t know.

Posted by: Tom Leinster on August 21, 2010 3:58 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Largely, it’s unrealistic. But we can get somewhere with K=U(n), so do let’s.

The diagonal matrices T in U(n) are a “maximal torus”. Every unitary matrix is diagonalizable. So every 1-parameter subgroup in U(n) is conjugate to one inside T. But not uniquely; we can permute the diagonal entries. So instead of getting an element of Z^n (the kernel of the exponential map from diagonal imaginary matrices -> T), we should really get a weakly decreasing list of n integers.

I’m not going to try to explain the statement that irreps of a compact Lie group correspond 1:1 to dominant weights. I’m only going to rub your face in the fact that this is a hint that representation theory of compact connected Lie groups is _much, much easier and cleaner_ than that of (say) finite groups.

Posted by: Allen Knutson on August 21, 2010 4:17 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I regret the “rub your face” phrase – it doesn’t look as cutely teasy as I’d meant it to be. What I really mean to indicate is that just as the representations of S^1 are really easy to index (by Z, this being the easy part of Fourier theory), almost the same is true for any compact connected group. Topology may seem a scary thing to add to the subject, but connectedness helps a lot.

Posted by: Allen Knutson on August 21, 2010 4:23 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Also, compared to everything else, topology doesn’t scare me.

Posted by: Tom Leinster on August 21, 2010 4:26 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Hi, Allen Knutson.

Could you please say highlight the connection between this fact about the unirreps of compact groups and the langlands conjecture? What was your point?

Posted by: Rauan Akylzhanov on May 23, 2015 11:08 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

users.ictp.it/~pub_off/lectures/lns021/Harder/Harder.ps

this is a very good introduction to the number theoretic point behind the langland’s program.

http://arxiv.org/pdf/1007.4426

this is also a more informatory article by Dalawat on consequences of the langland’s program.

both present the number theoretic aspect.

Posted by: anonymous on May 24, 2011 11:43 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Hey anonymous,

Thanks for the references. The first one assumes that you know terms such as modular form, automorphic representation and L-function, but on the other hand does have some passages that don’t require such specialist knowledge.

By the way, there might not be a need to be quite so anonymous. The email address that you’re asked to enter on the comment form doesn’t get seen by anyone other than the Café hosts and administrator, and certainly won’t lead to spam. Of course, if you don’t want to enter a valid email address, that’s entirely your choice, but you don’t have to worry on the spam front.

Posted by: Tom Leinster on May 24, 2011 12:18 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Maybe this will help, I’ll go through and explain what these things are in GEOMETRIC Langlands (and over the complex numbers), which works by analogy with the arithmetic version.

dual Langlands groups: Because in complex geometry we’re using reductive Lie groups, we can just use real compact Lie groups, and then complexify. Now, with real compact Lie groups, there’s a poset, with simply connected and centerless (adjoint form) Lie groups at the ends, consisting of all Lie groups with the same Lie algebra. The Langlands dual is what you get if you switch the order of the poset around, and then match things up.

local system: a G-local system principal G-bundle with flat connection

curve: for the purposes of my response, literally just a Riemann surface.

bundle for a curve: principal G-bundle

module for Bun(G): Here’s the complication. Bun(G) is a stack, the moduli stack of principal G-bundles on the curve C. And Loc(G) is also a stack, the moduli stack of G-local systems on C.

So the geometric Langlands correspondence would be an equivalence between local systems for the Langlands dual of G on a curve and eigensheaves for a “Hecke” operator on the stack Bun(G).

I just found this paper of Witten, which seems to do a decent job explaining things (I only skimmed) and ties Langlands duality into electro-magnetic duality: http://front.math.ucdavis.edu/0906.2747

Posted by: Charles Siegel on August 21, 2010 4:10 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Hi Charles,

Thanks. I’d like to try to understand your description of dual Langlands groups. This is from a position of never having attended a course on Lie anything, but here goes.

Now, with real compact Lie groups, there’s a poset […] consisting of all Lie groups with the same Lie algebra.

As I understand it, you’re saying that for each Lie algebra $g$, there’s a poset $P(g) = \{ \text{real compact Lie groups whose Lie algebra is}  g \}.$ (I guess that should really be isomorphism classes of such Lie groups.) Is that right? If so, what’s the order relation?

I guess I need to understand that before I go any further, but to save too many back-and-forths, I’ll ask another question at the same time.

The Langlands dual is what you get if you switch the order of the poset around, and then match things up.

It sounds to me as if this means the following: the Langlands dual of a Lie algebra $g$ is a certain Lie algebra $g^*$ with the property that $P(g^*) = P(g)^{op}$. (The right-hand side is the poset $P(g)$ with its order reversed.) But I don’t think this can be right, because you and others have been talking about Langlands dual groups, not Langlands dual Lie algebras. What’s going on?

Posted by: Tom Leinster on August 21, 2010 4:50 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Tom,

It’s the same Lie algebra. The ordering is by G_1<G_2 if and only if there is a surjective homomorphism G_1->G_2 (or vice versa, the poset turns out to be symmetric, if my understanding is correct), and you can (roughly, the operation I’m describing isn’t well-defined for all groups, so be careful, but it’s simpler than saying stuff about roots, coroots, weights, coweights, characters and cocharacters, and more intuitive) flip the partial order and lay the two posets on top of each other, and Langlands dual groups will be in the same spot. Roughly you can think of this as exchanging center and fundamental group.

And yeah, I meant isomorphism classes of compact Lie groups with Lie algebra g.

Posted by: Charles Siegel on August 21, 2010 5:02 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

It’s the same Lie algebra.

Do you mean that the Lie algebra I called $g^*$ is the same as the Lie algebra $g$?

So let’s see if I understand. You start with a real compact Lie group $G$. Write $g$ for its Lie algebra. There is a poset $P(g) = \{ \text{real compact Lie groups whose Lie algebra is}  g \}/\cong,$ ordered in the way that you describe. That poset comes equipped with a canonical isomorphism $i: P(g) \to P(g)^{op}$. Among the elements of $P(g)$ is $G$. So, we obtain a new Lie group $i(G)$, with the same Lie algebra as $G$. And $i(G)$ is the Langlands dual of $G$. Is that right?

(I don’t mind if there are lots of missing hypotheses; I’m only after the rough idea.)

Posted by: Tom Leinster on August 21, 2010 5:54 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Umh, no.

The Langlands Dual of $so(2n+1)$ is $sp(2n)$. These have the same rank, but are different Lie Algebras.

Posted by: Jacques Distler on August 21, 2010 8:33 AM | Permalink | PGP Sig | Reply to this

### Re: What is the Langlands Programme?

Ah, yeah, I’d forgotten completely, I’ve only thought about the ADE types, and they don’t switch, so I misremembered the statements.

Posted by: Charles Siegel on August 21, 2010 12:03 PM | Permalink | Reply to this

### One small thing

I succeeded in understanding one small thing about the Langlands programme, from Ian Grojnowski’s article on representation theory in the Princeton Companion to Mathematics. He says that

The Langlands program describes the representation theory of reductive groups.

(That’s not quite a direct quotation.) Now, I didn’t and still don’t know what a reductive group is, but I do know now that they’re algebraic groups, and that many important algebraic groups are reductive. So, I can understand the following statement:

The Langlands program describes the representation theory of many important algebraic groups.

That was something I didn’t know.

Grojnowski indicates that, according to Langlands, the representations of a reductive group are described in terms of (a) the Langlands dual group and (b) some Galois group. I guess (b) has something to do with the connections with number theory.

However, I’m still missing a sense of why the Langlands programme has such a broad sweep. A couple of paragraphs later, Grojnowski says:

It precisely unifies and generalizes harmonic analysis and number theory.

How does harmonic analysis come into it?

And… that’s an enormous claim. What does this unified, generalized, harmonic-analysis-number-theory look like? Is that what all these black-belt Langlands dudes are busy finding out?

Posted by: Tom Leinster on August 21, 2010 6:11 AM | Permalink | Reply to this

### Re: One small thing

Given how much these black belt Langlands dudes are talking about physics these days, I get the impression that they are way too bogged down in classical geometry.

Posted by: Kea on August 21, 2010 6:17 AM | Permalink | Reply to this

### Re: One small thing

If you watch Witten’s 2nd talk from last week, he only talks about physics and he concludes that (categorical) Langlands is really about S duality, but to understand it you need to go outside the stringy theories associated to the S duality (and mirror symmetry etc) and discuss M theory. And he defines a new set of theories, including a 4d TQFT (using a 5d setup) that is supposed to be the Khovanov homology analogue of Chern-Simons field theory.

But after all this, one cannot help wondering if the true categorical bones of Langlands are like the categorical bones of the ‘right’ physical theory, which may not exactly resemble the usual M theory. That is, if we asked for the physics of harmonic analysis come number theory, then I immediately think of Galois groups, automorphic forms and the Fourier transform in non-commutative geometry and quantum information theory. So the old complex curves are not primary to the discussion at all, but something that must be carefully generated from the more fundamental quantum degrees of freedom. Even string theorists talk this way these days, as if they knew that all along, lol.

Posted by: Kea on August 21, 2010 6:38 AM | Permalink | Reply to this

### Re: One small thing

Tom,

I’ve slowly been learning about the Langlands program over many years, motivated by fascination with the relations between geometric Langlands and quantum field theory. My partial understanding of the subject now is that it’s fundamentally a story about representation theory, as the quote you have from Grojnowski indicates.

The most well-known part of the Langlands story is the story about global fields (like Q, the rational numbers). But in number theory you learn that the way to think about Q is to think of it as functions on Spec (Z), points of which are primes (and “infinite primes”). The lesson is to start by looking at what happens locally, then later try and put this information together to get global information (this involves the technology of adeles).

So, it may be best to start with the local Langlands story first. This story is purely about representation theory, and instead of general reductive groups, it’s probably best to stick to the case of GL(n) (as far as I know, local Langlands is not proved and very far from understood for general reductive groups). What it says is that, for a local field F (e.g. Q_p for a prime p, R or C for an “infinite prime”), there’s a highly non-obvious and non-trivial bijection between (equivalence classes of):

1. n-dimensional representations of the Weil group of F (an arithmetic gadget associated to F, generalizing the Galois group).

2. representations of GL(n,F)

You can read this in both directions. In one direction, if you know 1, you can think of its elements as “Langlands parameters” classifying the (infinite dimensional) representations of GL(n,F) and you’ve solved a big problem in representation theory. In the other direction, what you know about representations of GL(n,F) (often from solving an analysis problem) tells you about representations of the Weil or Galois group, solving an arithmetic problem.

This matching of representations is subtle. Irreducible representations in 1 don’t just go to irreducibles in 2, they go to very special irreducibles (“supercuspidals”), from which you get other irreducibles by parabolic induction. The matching is done explicitly using the action of Frobenius in 1 and the action of Hecke algebras in 2, information sometimes packaged as L-functions.

In the case n=1, where all groups are Abelian, you get the much older and better understood (but still very non-trivial) subject of (local) class field theory.

Hope that somewhere in there there’s one piece of information that’s helpful, perhaps at least in explaining a bit of what Grojnowski is referring to.

Posted by: Peter Woit on August 21, 2010 5:53 PM | Permalink | Reply to this

### Re: One small thing

The Langlands program describes the representation theory of many important algebraic groups.

My expertise is in the local Langlands correspondence (or, at least, in local harmonic analysis), and my comment should be taken as reflecting that point of view; but I think that this is misleading in a subtle way. The algebraic representation theory of algebraic groups (at least in characteristic 0) is (I think!) much like that of compact or complex groups. The sort of representation theory with which the (local) Langlands correspondence is concerned is with the complex representation theory of rational points of reductive groups. That is, we consider not algebraic homomorphisms $\mathbb{G} \to GL_V$ (which is to say certain kinds of maps $k[GL_V] \to k[\mathbb{G}]$), where $V$ is a $k$-vector space, but rather continuous homomorphisms $\mathbb{G}(k) \to GL_V(\mathbb{C})$, where $V$ is a $\mathbb{C}$-vector space. (There are $p$-adic versions of the (local) Langlands correspondence, but my understanding is that they are less well understood.) This apparently small change in perspective makes a huge difference to the representation theory.

Posted by: L Spice on May 28, 2015 8:09 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Here’s my submission, which may be redundant.

There’s a special class of groups, called connected reductive groups, that are parameterized by combinatorial data (root data). This combinatorics is pretty simple, not too much more complicated than the classification of Lie algebras by graphs.

(I am trying not to burden anyone with any information about what a reductive group is. I hope this doesn’t count: reductive groups are groups of matrices, like GL(n) or SO(n). In three paragraphs I will say something about what kind of numbers to fill the matrices with.)

Grothendieck observed that root data come in dual pairs. At the time it might have looked only like a curiosity: without explaining anything, I can tell you that a root datum is a four-tuple (X,Y,R,C), and it turns out that (Y,X,C,R) is another root datum.

Since root data come in dual pairs, reductive groups also come in dual pairs. The partner G’ of a reductive group G is called its “Langlands dual.” There are by now startlingly intricate geometric explanations of how to build G’ directly out of G, like in Allen’s comments.

The Langlands program starts (and I will almost stop) with the following claim: homomorphisms into G (Galois representations) should be related to homomorphisms out of G’ (automorphic forms).

Actually the claim should have some number theory in it: the short story I told about G and G’ was basically true over the complex numbers (or another algebraically closed field), but in the Langlands program a number field F is involved. F plays a different role for G than it does for G’. A Galois representation is a homomorphism from Gal(F) to G-with-complex-entries. But an automorphic representation is an action of G’-with-adeles-of-F-entries on a topological vector space. (That’s where the harmonic analysis is.)

(When F is the rational numbers, an “adele” is a sequence [real number, 2-adic number, 3-adic number, 5-adic number, …] satisfying a condition.)

One sentence to try to connect this with what Charles is saying about geometric Langlands: using the dictionary between number fields and function fields, Weil suggested that G’-with-adelic-entries is analogous to the group of gauge transformations of a principal G’-bundle over a Riemann surface.

Posted by: DT on August 21, 2010 6:54 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

DT – Do you have a reference where Weil’s analogy you aluded to is worked out in detail? Actually, even the simplest case would be nice: what is the correspondence between “gauge transformations” (whatever those are) of a vector bundle (algebraic?) on the projective line over a field F, and the points of GL_n valued in the adeles of the rational function field F(t)? Does it matter whether F is the complex numbers, rather than, say, a finite field?

Posted by: SL on August 22, 2010 1:29 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

DT wrote:

Here’s my submission, which may be redundant.

Absolutely not! That’s marvellous, just the kind of thing I wanted. Thanks!

Posted by: Tom Leinster on August 22, 2010 7:59 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Try listening to David Ben-Zvi in this lecture on the Fundamental Lemma.

Posted by: David Corfield on August 21, 2010 7:56 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks. For technological reasons I can only listen to the audio, not watch the video.

You might think that listening to a talk without being able to see the board would be useless. But David’s, at least, is surprisingly informative.

Posted by: Tom Leinster on August 22, 2010 8:41 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Looking back I realize this is somewhat hidden by the format of the webpage, but there are somewhat detailed pdf notes for the talk at the same site that might help…

Posted by: David Ben-Zvi on August 22, 2010 10:48 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Typing this at a shell prompt might help.

mplayer -fs -rtsp-stream-over-tcp \
rtsp://media.cit.utexas.edu:554/math-grasp/GRASP_02_01_10.mov

it should be easy to install.)

Posted by: wb on August 25, 2010 1:37 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks very much! That works perfectly for me.

Posted by: Tom Leinster on August 25, 2010 1:51 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Gowers’s Weblog has a link to a well-written (and simple– I hope not too simple) summary of the Langlands program that also discusses Chau’s work.

Posted by: Steven H. Cullinane on August 21, 2010 6:09 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Sorry, I didn’t realize this link was to one of the two “explanations on the ICM page” you said you’d read.

Posted by: Steven H. Cullinane on August 22, 2010 2:16 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Ah, but I appreciate your comment, because you (and Jonathan vos Post, below) got me looking at a different page on Tim Gowers’s blog, about Châu’s work and the Langlands programme.

As Todd Trimble approvingly points out there, a striking positive feature of Gowers’s blog is the

absolute uninhibited intellectual honesty, not only in confessing ignorance and confusion (as in this post), but also thinking hard and out loud about ostensibly “elementary” topics on occasion.

That honesty of Gowers is particularly noticeable in that post.

Incidentally, I’ve been referring to Ngô Bảo Châu as Châu where others have been referring to him as Ngô. His Wikipedia entry says:

In this Vietnamese name, the family name is Ngô. According to Vietnamese custom, this person should properly be referred to by the given name Châu or Bảo Châu.

But I seem to be in a minority in following this advice.

Posted by: Tom Leinster on August 22, 2010 8:16 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

As Todd Trimble approvingly points out

Words failed me in conveying just how much I “approve”; in fact I am awed by the depth and probity of his posts. It takes a lot of guts (I’m tempted to insert another pluralized body part here) to be quite so unflinchingly honest and thoughtful.

It reminds me of what Rota said about John Kemeny (Indiscrete Thoughts, page 7):

Kemeny’s seminar in the philosophy of science (which that year attracted as many as six students, a record) was refreshing training in basic reasoning. Kemeny was not afraid to appear pedestrian, trivial, or stupid; what mattered was to respect the facts, to draw distinctions even when they clashed with our prejudices, and to avoid black-and-white oversimplifications. Mathematicians have always found Kemeny’s common sense revolting.

(This appears in Rota’s reminiscences of the logician Alonzo Church. Rota relates what Kemeny, a former student of Church, thought of great mathematicians as ‘great people’: “There is no reason why a great mathematician should not also be a great bigot. Look at your teachers in Fine Hall, at how they treat one of the greatest living mathematicians, Alonzo Church.”)

Posted by: Todd Trimble on August 22, 2010 10:06 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Let us not forget that John was, I believe, the first mathematician prepared to reveal his miserable ignorance online.

Posted by: David Corfield on August 23, 2010 9:51 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

But David was the first to publicly announce that my ignorance was ‘miserable’.

Posted by: John Baez on August 27, 2010 7:12 AM | Permalink | Reply to this

### Gowers writes; Re: What is the Langlands Programme?

ICM2010 — Ngo laudatio

“… The conversation turned to the topic of how many Fields medals could in theory be given for advances in the Langlands programme. My view was I suppose the official line, which is that it is such a deep and difficult area that any major advance is huge news, though I couldn’t resist a joke comparison to pole vault records, where people who are in a position to beat them deliberately don’t beat them by much because you get big money for beating world records. (I’m not seriously suggesting that somebody who had a proof of all the Langlands conjectures would sit on it, or release it only gradually.) Assaf (and I hope he won’t mind my making his views public) was more sceptical, maintaining that all mathematicians have their icebergs to explore and that the Langlands programme was not as unusual in this respect as perhaps it is sometimes conveyed as being. He said that he likes to ask the experts whether if they could assume all the results they wanted that are currently conjectural, they would know more about any concrete Diophantine equations. Apparently they don’t particularly like this question. Whether it is an appropriate criterion to judge the area is of course a matter for debate. In fact, that is what prompted me to say that perhaps the iceberg was the true and fascinating object of study in that area. I didn’t think of saying it at the time, but after a while there is not much interest in solving more and more Diophantine equations (not that Assaf was claiming that there was), and attention must turn to more global phenomena somehow. Perhaps that is what algebraic number theory is….”

Posted by: Jonathan Vos Post on August 21, 2010 7:56 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

As Peter points out it’s useful to separate the local and global parts of the story. The local Langlands program is a general principle that tells us what representations of groups like GLn, SLn, SOn, Spn (all your favorite matrix groups) look like, if you take the entries of the matrices to be real, complex or p-adic numbers (though the same principles work also for finite fields and other settings). Namely there’s a notion of “nice” representation – these include all finite dim reps but also all reasonable infinite dimensional ones…

The local Langlands program says these representations are classified by “Langlands parameters”. What are these? remember for a finite group there is the same number of conjugacy classes and irreducible reps, but no canonical way to match them up. The idea very roughly is that reps should match CANONICALLY with conjugacy classes in ANOTHER group, the dual group. Something like this is the content of the Langlands program over finite fields (due to Lusztig). For the other fields it’s still generally the idea, except rather than looking at conjugacy classes of ELEMENTS (aka homomorphisms from Z) in the dual group, we look at conjugacy classes of homomorphisms from a Galois group attached to the situation. Depending on the field then you may decide where the complexity lies — for p-adic groups the Galois group is super intricate, while over R or C it’s easy, so in the former you think of this really as Galois data while in the latter you think of it basically as conjugacy classes in this dual group (for which there’s always an easy combinatorical prescription). That’s it in a nutshell.

Then there’s the global Langlands program which says what happens when you look at fields like rational (meromorphic) functions on a curve (Riemann surface), or more importantly, number fields. In this case we don’t describe ALL representations of our matrix groups, but the “deep” ones, aka automorphic. These are by definition the representations that arise when we study harmonic analysis on special locally symmetric spaces, like the modular curve (upper half plane mod SL2Z). Again the classification of these is by conjugacy classes of reps of a Galois group, which is now a super hard and subtle Galois group (eg that of the rational numbers). This correspondence of reps and Galois data ends up being the most powerful tool we have to understand these Galois groups, hence a huge variety of problems in number theory (such as Fermat’s last theorem etc etc…)

Posted by: David Ben-Zvi on August 21, 2010 9:35 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks a lot, David, and Peter W. That’s a really nice description.

How literally is the following true?

remember for a finite group there is the same number of conjugacy classes and irreducible reps, but no canonical way to match them up. The idea very roughly is that reps should match CANONICALLY with conjugacy classes in ANOTHER group, the dual group.

Is it literally true that given a finite group $G$, there is another group $G'$ such that the irreducible reps of $G$ correspond canonically with the conjugacy classes of $G'$? If so, is $G'$ a finite group?

Something like this is the content of the Langlands program over finite fields (due to Lusztig)

which makes me think that it can’t be true in quite that simplicity, because what I wrote didn’t even mention fields.

Posted by: Tom Leinster on August 22, 2010 9:29 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

It’s true for a large class of finite groups, the finite groups of Lie type - eg matrix groups GL,SL,SO,Sp etc over finite fields.. this class of groups contains almost all finite simple groups (other than cyclic, where the same holds since it’s easy for abelian groups, symmetric groups where analogous stories hold – the “F_1 case” – and some of the exceptional groups). Lusztig classified the reps of this enormous class of finite groups (that certainly should have earned him a Fields medal!), and they’re classified by conjugacy classes in the dual groups. I won’t give the precise statement but I think this is not a particularly misleading simplification.. it’s a pretty amazing piece of math!

Posted by: David Ben-Zvi on August 22, 2010 10:46 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Wow!

Posted by: Tom Leinster on August 23, 2010 1:09 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

A thought after all these months: if symmetric groups fall into this picture by being $GL(n, \mathbb{F}_1)$, and if GLs are Langlands self-dual, this explains how Young diagrams parameterize conjugacy classes and at the same time irreducible representations.

Posted by: David Corfield on July 10, 2012 5:23 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

You may find it helpful to have a look at the paper “An elementary introduction to the Langlands program”, by Stephen Gelbart (AMS Bulletin, vol.10, number 2, 1984). It is very clearly written and is aimed at non-experts.

Posted by: Bisi Agboola on August 22, 2010 12:57 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks—that looks good. Here’s the link.

Gelbart definitely starts well:

Herein lies the agony as well as the ecstasy of Langlands’ program. To merely state the conjectures requires much of the machinery of class field theory, the structure theory of algebraic groups, the representation theory of real and $p$-adic groups, and (at least) the language of algebraic geometry. In other words, though the promised rewards are great, the initiation process is forbidding.

I hope the rest of it displays this much awareness too.

Posted by: Tom Leinster on August 22, 2010 8:26 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

This might be more your style: a recent Kapustin paper on TQFTs.

Posted by: Kea on August 22, 2010 4:02 AM | Permalink | Reply to this
Weblog:
Excerpt:
Tracked: August 22, 2010 3:23 PM

### Re: What is the Langlands Programme?

If I were to recommend just one article that attempts to provide a large-scale view of the program, it would be Langlands’ lecture at the Helsinki congress.

In brief, he explains:

(1) The arithmetic of varieties are expected to be reflected in special values of their $L$-functions, or natural factors of their $L$-functions (that is, motivic $L$-functions), as in the class number formula and the conjecture of Birch and Swinnerton-Dyer (generalized by Deligne, Bloch, Beilinson, and Kato over the years subsequent to this lecture).

(2) The special values in question often lie outside the domains of convergence of the initial definitions as Euler products, and hence, call for analytic continuation, as in the conjecture of Hasse and Weil. (In B-S-D, one must take the value at 1, where the original Euler product doesn’t converge.)

(3) All known cases of this analytic continuation rely on the identification of the arithmetic-geometric $L$-function with an automorphic $L$-function, i.e., an $L$-function associated to an automorphic representation of the adelic points $G(A_F)$ of a reductive group $G$. These usually have nice properties coming from harmonic analysis on the group.

These considerations leads to the conjecture

*Every motivic $L$-function is an automorphic $L$-function.*

Whether or not the words therein makes sense at yet, this sentence is really the core of the Langlands program. In a real sense, the rest are details. I know I’m not explaining anything in this post. But it might still be useful to take home that sentence as something to mull over and keep coming back to, at least if you’re serious about understanding Langlands.

By the way, even though it’s important at some point, for the big picture, you lose nothing by assuming $G$ to be $GL_n$. In fact, the so-called functoriality conjecture implies that we can always take $G$ to be $GL_n$. In short, don’t worry too much at the outset about the dual of a group.

Posted by: Minhyong Kim on August 23, 2010 4:25 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

To add one more remark to the last paragraph, Langlands’ also puts the punchline more or less in this form:

*All $L$-functions of interest (motivic or automorphic) are standard $L$-functions.*

A standard $L$-function refers to one associated to an automorphic representation of some $GL_n(A_F)$. For these, all nice analytic properties are known.

Posted by: Minhyong Kim on August 23, 2010 4:37 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Minhyong wrote:

at least if you’re serious about understanding Langlands.

Personally, yes and no.

“No” in the sense that if I were truly serious about it, I’d totally retrain as a mathematician. It would transform my life.

“Yes” in the sense that I might sit in a colloquium seriously listening and seriously trying to understand, even if the subject appears to have no direct bearing on my own research, and even knowing that I’m unlikely to spend very much time outside the colloquium thinking about it.

And it’s a serious business in the sense that for experts to convey something of their subject to non-experts takes serious time and thought. I value that effort, and I think that expository work (e.g. your own Why everyone should know number theory) is an important and underrated part of mathematical culture. No one can be an expert on everything; and for the subjects on which we’re not expert, we rely on expository work. Good exposition combats the fragmentation of mathematics.

I’m miles from understanding your comments above. As entertainment (or, more pompously, as a sociological experiment), let me explain what goes through my mind when I read your conjectural slogan:

Every motivic $L$-function is an automorphic $L$-function.

Here we go.

• Motive.  Something in algebraic geometry. Grand dream of Grothendieck. Should unify lots of cohomology theories. Lots of people have tried to make the dreams precise, with partial success.
• $L$-function.  Dirichlet series with some further properties. Riemann’s $\zeta$ is one. Number theorists are always talking about them.
• Automorphic.  Presumably refers to automorphic forms, not automorphisms. Automorphic forms are some mega souped-up version of modular forms. Modular forms are something to do with holomorphic functions on the upper half-plane, and something to do with lattices in $\mathbf{C}$. A famous theorem connects modular forms with elliptic curves.

So a motivic $L$-function sounds like something from the world of algebraic geometry and number theory, whereas an automorphic $L$-function sounds like something from the world of complex analysis and number theory (and probably algebraic geometry too, because that’s always working its way in). Hence, the impression your slogan leaves with me is

Every one of these important objects from arithmetic geometry has some kind of analytical content.

Does that bear any resemblance to the truth?

Posted by: Tom Leinster on August 24, 2010 6:47 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I threw out the slogan exactly because it’s the kind of thing that might well be emphasized at a colloquium as compared, for example, to the subtle relation between a group and its dual. It’s a sentence that’s fairly easy to remember, with the potential to provide some non-trivial immediate inspiration, and then to gradually flesh out over the years.

In any case, your summary is quite accurate. Your last sentence might be elaborated upon in two parts:

(1) $L$-functions coming from arithmetic-geometric objects are analytically *natural*.

Note that this is far from obvious given their definition as a big Euler product, amalgamating very local arithmetic data. More precisely, this naturality is expressed as the Hasse-Weil conjecture that these $L$-functions have analytic continuation (with a few well-understood poles) to the whole plane and satisfy a functional equation. Note that if you write down a random Euler product convergent on a half-plane, nothing like this would happen. The Hasse-Weil conjecture says the Euler product for an arithmetic-geometric $L$-function has much more structure than such a random product.

(2) They are analytically natural because they can be identified with automorphic $L$-functions.

This is the main content of the Langlands program. The reason that automorphic $L$-functions can be expected to be natural is because they should reflect group-theoretic symmetries of the automorphic forms from which they are built, even though the general definition is still an Euler product. To get some sense of this phenomenon, it is probably good to work out the case of the $L$-function of a modular form $f$ of weight $k$. This has the right sort of beginner-intermediate level of complexity to convey some sense of the general phenomenon.

$L(f,s)=\frac{1}{(2\pi)^s \Gamma(s)}\int_{0}^{\infty} f(iy)y^s(dy/y).$

You might be able to work out analytic continuation and functional equation using the formula $f(-1/z)=\pm i z^k f(z).$ It’s also possible and important to express (or define) this also as an Euler product over primes. However, the key point is that the $L$-function of an automorphic form (souped up to that of an automorphic representation) has many such expressions, some of which make the analytic naturality (almost) obvious. Langlands proposes that identity with such a thing is the right point of focus if we want to prove naturality for arithemtic-geometric (i.e. motivic) $L$-functions.

By the way, the Dirichlet series expression on either side is still important, but better not to focus on for now. In some sense, it’s important for applications.

Posted by: Minhyong Kim on August 25, 2010 4:12 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

In the elaboration, maybe I should add

(0) Important information about an arithmetic-geometric object can be accessed through an analytic object, its $L$-function.

That $L$-functions should have such an important role in the theory looks a priori quite mysterious, perhaps even arbitrary. Nevertheless, a complex network of theorems, conjectures, and experiments indicates that they are good things to look at.

By the way, to give some sense of this scary term ‘motivic $L$-function’, I should remind you that the $L$-function (also called the ‘zeta function’) of a variety decomposes $L(X,s)=\prod_i L(H^i(X),s)^{(-1)^i}$ as a consequence of the Lefschetz trace formula for varieties over finite fields. (This is only true up to finitely many Euler factors that you should ignore for now. In fact, the left-hand-side as written only makes sense up to finitely many Euler factors, as mentioned in the recent discussion between John and Matt. But again, please ignore this.) You lose very little by thinking of the $L(H^i(X),s)$ when someone refers to a motivic $L$-function. The reason it becomes more general is that the Galois representation on $H^i$ out of which the $L$-function is built sometimes breaks up into subrepresentions $V^i_j$, in which case, we can further write $L(H^i(X),s)=\prod_j L(V^i_j,s).$ The constituents are then the most primitive motivic $L$-functions. You can guess that in an ideal world, breaking up $L(X,s)$ in this way should help us to understand it.

Posted by: Minhyong Kim on August 25, 2010 5:59 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

“*Every motivic L-function is an automorphic L-function.*”

So, whatever is the meaning of it, if this conjecture is proven, the Langlands program will be finished. Is that it?

Posted by: Daniel de França MTd2 on August 23, 2010 7:04 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Not really. The programme has been continuously evolving. For example, there is quite likely something profound behind a categorical version, as alluded to by many people, rather than one that focuses on $L$-functions. The input from the geometric version (concerning local systems on Riemann surfaces over $\mathbb{C}$) as well as ingredients of Ngo’s proof certainly do create the impression that we are still missing something essential in the classical formulation even over number fields. Nevertheless, the conjecture recalled above really has been the center of it all so far. It’s still the main problem around which others revolve in the Langlands program.

Posted by: Minhyong Kim on August 24, 2010 12:19 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Hi Minhyong, the discussion remembers me at an old question on MO. Do you know a bit more about that? Best, Thomas

Posted by: Thomas on August 24, 2010 7:23 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Dear Thomas,

It looks interesting, but perhaps you could elaborate a bit on the question?

Best,

Minhyong

Posted by: Minhyong Kim on August 24, 2010 2:12 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Dear Minhyong,

I just had tried to fit my mental image, derived from a - on my side not very well understood - article by Rapoport (p.261-271), with Yoshida’s drawing. I thought that the passage on the left side of Yoshida’s landscape between semisimple and normal L-Fct.s is performed by the MWC and the RZ spectral sequence, the corresponding passage on the right side being the Ramanujan conjecture. My puzzlement comes from trying to locate Rapoport’s statements in Yoshida’s drawing and understanding what the other points and connections there are and how (and in which directions) those connections go.

Best,
Thomas

Posted by: Thomas on August 27, 2010 2:34 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I made an attempt to answer your question on MO. Essentially, Yoshida’s drawing is a schematic outlining both the intuitions and the main lines underlying the proof of his result with Taylor (the goal that Minhyong refers to in his reply), namely the local-global compatibility result for automorphic forms and their associated Galois representations. For more details, see my answer on MO.

As for the relationship with Rapoport’s article, what Taylor and Yoshida are doing is essentially carrying out the program that Rapoport describes, namely, they establish the monodromy weight conjecture (in their context), and they have the generalized Ramanujan conjecture already (this was proved by Harris and Taylor); this lets them deduce their desired result.

Posted by: Matthew Emerton on August 28, 2010 7:47 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Dear Thomas,

I’m afraid I don’t know. The portion to concentrate on, I think, is near the top where Yoshida lays out his ‘goal,’ which is to show that the restriction to a local Galois group of the global Galois representation (associated to a certain automorphic representation) is compatible with the local Langlands correspondence.

Minhyong

Posted by: Minhyong Kim on August 27, 2010 2:58 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Many thanks to you and Matthew! Asking you on such general questions pays off, so I’m now heartened to ask what you think about these rumors on extending the concept of motives? Would that have any implications in relation to the Langlands programme?

Posted by: Thomas on August 28, 2010 10:21 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

A coming seminar: “Quantum motives (related MO q.) : realizations, detection, applications”, incl. a lecture “Quantum motives: review (of) the classical idea of how to linearize algebraic geometry with an eye to utilizing it in the quantum setup.” and a minicourse “Geometric Langlands and quantum motives: a link”. Sounds fascinating!

Posted by: Thomas on September 5, 2010 7:28 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Dear Minhyong,

Could you tell us, also in “mysterious ways”, the categorical version, as alluded by many people?

And what is “The input from the geometric version (concerning local systems on Riemann surfaces over ℂ) as well as ingredients of Ngo’s proof certainly do create the impression that we are still missing something essential in the classical formulation even over number fields.”?

Best,

Daniel.

Posted by: Daniel de França MTd2 on August 24, 2010 3:15 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Well, there’s not much to say as of now. The situation is that one conjectures (more or less) a bijection

Motivic Galois representations $\leftrightarrow$ Algebraic automorphic representations

at the level of isomorphism classes of irreducible objects. One might hope for a correspondence at the level of categories.

Regarding geometric Langlands and Ngo, both make use of Hitchin fibrations in different ways. One could ask if there’s something fundamental about their occurrence that needs to be built into the number theory. It’s possible to go on to wildly speculate that perverse sheaves are somehow more natural than automorphic forms, that number theory should have something like T-duality, and so on.

Posted by: Minhyong Kim on August 25, 2010 3:24 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I agree with these comments.. One can view Ngo’s work very naturally as part of the geometric Langlands program. Namely they are part of the geometric (or categorified, or motivic) form of the trace formula over function fields. One can say that much of the progress in the work on the fundamental lemma comes from the realization that various objects like p-adic orbital integrals that are crucial in the classical story are in fact shadows of motivic/geometric/categorified objects which one can study geometrically, in particular applying Hodge theory/theory of weights, on which Ngo’s proof ultimately relies. This is one of the lessons (that I think Minhyong is alluding to) from the new developments, and one can hope this categorification will have many more consequences for the classical story.

Posted by: David Ben-Zvi on August 25, 2010 8:05 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

In light of Minhyong’s very lucid answers, it seems worth mentioning (especially in the context of Ngo’s work; and it seems that in the West he is referred to, Western-style, as Ngo, rather than as Chau) that there are two aspects to what is usually called “the Langlands program”.

The one that Minhyong explained, to the effect that motivic $L$-functions should also be automorphic $L$-functions, is known as reciprocity. Roughly, one thinks of “motivic” as meaning “Diophantine”, so motivic $L$-functions encode Diophantine information (often in elaborate, and conjectural, ways, as in the Birch–Swinnerton-Dyer conjecture and its generalizations), while the automorphic $L$-functions are supposed to be thought of as accessible, understandable, perhaps even computable in some way.

A simple illustration of this kind of phenomenon is given by the famous result, due essentially to Gauss, that the Galois group of $\mathbf Q(\zeta_n)$ over $\mathbf Q$ (here $\zeta_n$ is a primitive $n$th root of 1), is canonically isomorphic to $(\mathbf Z/n)^{\times}$, where an element $a$ in this group of units acts on $\zeta_n$ just by raising to the $a$th power. The group $(\mathbf Z/n)^{\times}$ is a very simple example of an automorphic object (the underlying reductive group is just $GL_1$), and it is eminently understandable. With this isomorphism in hand, it is straightforward to solve the Diophantine problem of describing all subfields of $\mathbf Q(\zeta_n)$. For example, taking $n = p$, an odd prime, it is easy to deduce that $\mathbf Q(\zeta_p)$ has a unique quadratic subfield, equal to $\mathbf Q(\sqrt{\pm p})$ (the sign being chosen so that $\pm p \equiv 1$ mod 4), and using the description of Gal($\mathbf Q(\zeta_p)/\mathbf Q$) afforded in this way, one can easily deduce (for example) the law of quadratic reciprocity. (This is the origin of the appelation reciprocity for the general problem of relating motivic and automorphic $L$-functions.)

There is another aspect of the Langlands programme, though, in fact the one that Langlands first introduced, and in which the role of reductive groups other than $GL_n$ becomes more apparent, and this is the problem of functoriality . Very roughly, it says that if we have a map of dual groups $H^{\vee} \to G^{\vee}$, then automorphic forms on $H$ give rise to automorphic forms on $G$. This is where harmonic analysis enters the picture: describing/computing/constructing automorphic forms is a non-abelian analogue of Fourier theory (non-abelian because the groups $G$ and $H$ are typically non-abelian), and this problem is about constructing an automorphic form on $H$, given some corresponding data related to $G$.

Ngo’s theorem, the so-called Fundamental Lemma, allows one (through combining his work with earlier work of Langlands, Arthur, and many others) to solve this problem of functoriality in special cases, namely when $H$ is a so-called endoscopic group attached to $G$. (This is a very technical notion; roughly, it means that the map $H^{\vee} \to G^{\vee}$ is obtained by identifying $H^{\vee}$ with the centralizer of some element in $G^{\vee}$.)

Most instances of functoriality are not endoscopic, but nevertheless, the fact that even the endoscopic case could not be solved was a big obstruction (psychologically, as well as technically) to thinking about the general problem, so the fact that it is now essentially solved, thanks to Ngo, represents enormous progress.

Although functoriality and reciprocity are different problems, they are related heuristically and technically, and share many common features (often at a deep level, where it is not obvious that they should). Thus Ngo’s result also has implications in the theory of Shimura varieties, which is one of the basic testing grounds for reciprocity. Just to give an ideal of what I’m talking about here: Shimura varieties are certain varieties which are defined in terms of reductive groups, so that they are naturally quite close to the automorphic world; but nevertheless, proving reciprocity for them is a major task, not yet complete, and Ngo’s result has played a major role in recent advances on this front. (Let me close with a parenthetical remark directed at those readers who may know some of the theory of modular forms and modular curves, but don’t know the general theory of Shimuva varieties: as an illustration of the way the harmonic analysis problems related to functoriality and the Diophantine problems related to reciprocity intertwine, let me mention that Langlands discovered the phenomenon and problems of endoscopy in the context of trying to generalize Eichler–Shimura theory — the theory that relates cohomology of modular curves to modular forms — to the context of general Shimura varieties.)

Posted by: Matthew Emerton on August 25, 2010 6:49 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

To add even more length to an already long post: in the function field case, there is less difference between functoriality and reciprocity then in the number field case. In particular, the latter essentially implies the former, and so it it the latter on which function field work seems to focus (especially in its modern incarnation as geometric Langlands).

One already sees this in class field theory: over function fields, the idele class group and the abelianized Galois group are almost identical (it is just a question of replacing a $\mathbf Z$ by its profinite completion).

But in the number field case, the archimedean places make the idele class group quite a bit bigger and more mysterious. When one passes to the non-abelian context, this persists, so that not all automorphic forms are related to motives. Thus functoriality is its own problem, with its own implications that are of interest independent of Diophantine applications.

As an example: Selberg conjectured that if $X$ is any quotient of the upper half-plane by a congruence subgroup of $SL_2(\mathbf Z)$ (i.e. by the kernel of the map $SL_2(\mathbf Z) \rightarrow SL_2(\mathbf Z/N)$, for some integer $N$), then every eigenvalue of the Laplacian on $L^2(X)$ is at least $1/4$.

This is obviously a conjecture in harmonic analysis. One of Langlands observations is that it would follow from his functoriality conjecture. (Indeed, he observed that it is the archimedean analogue of the Ramanujan–Petersson conjecture, which — Langlands showed — also follows from functoriality, by the same argument that gives Selberg’s conjecture.)

Posted by: Matthew Emerton on August 25, 2010 7:07 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Contained in Matt’s excellent answer is very polite criticism of my exclusive focus on the reciprocity conjecture! Of course I can’t really disagree. However, it still might be fun for the non-experts if I do. That is, one could dogmatically maintain that the composed arrow

Diophantine $L$-function $\rightarrow$ automorphic $L$-function $\rightarrow$ standard $L$-function

is the main motivation for the second arrow. And then, even further insist that this second arrow, a special case of functoriality, is the main motivation for general functoriality.

Allow me also to correct a small misprint (this has now been corrected —- TL), since all these things can be frustrating to non-experts (like me). In the first paragraph on functoriality, it should say that a map $H^{\vee} \rightarrow G^{\vee}$ allows us to transfer automorphic forms on $H$ over to forms on $G$.

Posted by: Minhyong Kim on August 25, 2010 11:56 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Dear Minhyong,

Thanks for the correction!

Regarding reciprocity vs. functoriality, one reason I brought up the distinction is because when I first tried to learn these ideas, my conception of the Langlands program, such as it was, focussed more on reciprocity than on functoriality (which was only natural, since I was coming from a background in number theory), and I was surprised, when I started reading Langlands, to discover that a lot of his focus was on the concept of functoriality, something that I had been previously unaware of. It took me some time to figure out what was going on, and to realize that there were really two related, but different, problems (i.e. both reciprocity and functoriality) that were part of what was commonly called “the Langlands program”.

One last remark: something which added to my initial confusion between reciprocity and functoriality is that, just as they are more closely related in the function field case than in the number field case, they are also very closely related when one looks at the local rather than the global situation.

More precisely, the local Langlands conjecture, for a non-archimedean local field $K$, is a local reciprocity statement that (conjecturally) describes representations of reductive groups over $K$ in terms of Galois-theoretic (which one can more or less think of as motivic) data. If one had this statement for all reductive groups, then the local analogue of functoriality would follow directly. (The fact that the local case and the function field case share this feature in common, that reciprocity and functoriality are more closely related than in the number field case, will not be completely surprising to those who know class field theory, because in local class field theory, to pass from $K^{\times}$ to the abelianized Galois group one need only replace a $\mathbf Z$ by its pro-finite completion, a process formally identical to the one that takes place in class field theory for function fields. In summary, both local class field theory and function field class field theory fall into what Tate calls “the

Posted by: Matthew Emerton on August 25, 2010 4:00 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

An itex error seems to have cut off the end of my previous post: it should have read “… what Tate calls “the $\mathbf Z$ case”)”.

Also, “reductive groups over $K$” is technically not quite the correct thing to have written. What I mean is “$K$-valued points of reductive groups over $K$”, so local Langlands is a chapter in the representation theory of locally compact groups (albeit locally compact groups of a very particular kind), rather than of representation theory of algebraic groups in the sense of highest weights and so on. (Nevertheless, the concepts of algebraic group theory do play a major role in the study of the local Langlands conjecture, and in the corresponding global problem, which is the study of automorphic representations.)

Posted by: Matthew Emerton on August 25, 2010 4:06 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Dear Matt,

Of course I agree with all that you write, and find it instructive as usual. I guess one of your points is that in the local case, it’s somehow more natural to think of the Weil-Deligne representation as mere ‘parameters’ for the representation of the (points of the) reductive group. And then, some features of this subordination carry over to the global situation as well and, in any case, contribute to a very complex interaction.

I don’t think I ever told you that, curiously enough, my own exposure to the Langlands’ programme went almost in an opposite direction. I only started thinking seriously about arithmetic things in the last year or so of graduate school. Since I was at Yale in the 80’s, that meant essentially all the exposure to Langlands I had was phrased in terms of functoriality! (Other than the bit of reciprocity dealing with Artin representations, which can easily be subsumed by functoriality.) Then I attended a summer school organized by Langlands in 1990, and was startled to hear him motivating people there mostly in terms of reciprocity. (He started out his first lecture with elliptic curves and BSD.) My own impression is that he allows himself to emphasize one aspect or the other depending on his audience. Of course you’ve read Langlands much more extensively and can present a far more knowledgeable view of the whole picture.

Normally, I’m also wary of thinking about any substantial mathematics in terms of a monolithic ‘ultimate goal.’ But I thought it might make for fun and possibly helpful exposition to do so here.

Here is a question: As alluded to above, using the $L$-group rather than the just the complex dual, it’s easy to view reciprocity for Artin representations as a case of functoriality. Now, I always assumed that it would be easy for experts to set up some version of the $L$-group that would turn even general reciprocity for $l$-adic representations into special functoriality. Is this not the case?

Posted by: Minhyong Kim on August 25, 2010 5:41 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

This might be a good time to advertise this Workshop on Non-abelian Class Field Theory. It’s a small event with David Ben-Zvi, John Coates, Matt Emerton, and Bao Chau Ngo as speakers. The plans for this workshop arose from my earlier threat to bombard David with questions.

Availability of funds for participants other than the speakers is somewhat uncertain. But limited support for the local expenses might be possible. Let me know if you’re seriously interested.

Posted by: Minhyong Kim on August 26, 2010 2:04 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

It would be great if lecture notes would become free available on the workshop’s website.

Posted by: Thomas on August 27, 2010 2:42 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

So Tom, the guy who never installed the gothic fonts on his computer because he didn’t want to learn about Lie algebras, suddenly wants to learn about the Langlands program!

This gives me newfound courage. I’ve always been afraid of heights. I think tomorrow I’ll climb Mount Everest.

Posted by: John Baez on August 27, 2010 6:57 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Less jokingly, Tom: did you ever study Galois theory? That might be an easier angle. You could learn the basic idea of class field theory pretty easily: that’s about abelian Galois groups. The Langlands program can be seen as a generalization to nonabelian Galois groups. So then you could say: “I don’t understand this stuff, but I know it’s a generalization of something cool.”

Posted by: John Baez on August 27, 2010 7:01 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

People, watch out! There’s a snark in the waters!

Less jokingly, I’m a little surprised that no one else has mentioned class field theory up until now (relating abelian Galois extensions to how primes split), although quadratic reciprocity, which started it all, has gotten a mention or two.

Posted by: Todd Trimble on August 27, 2010 7:55 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I deliberately avoided the non-abelian class field theory angle because it’s a somewhat misleading and limiting view, at least the way it’s usually presented. It says that the $L$-function associated to a a finite-image representation of the Galois group of a number field is automorphic. In the language we’ve been using here already, it’s the reciprocity conjecture for motives of dimension zero.

Posted by: Minhyong Kim on August 27, 2010 11:47 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

In that case, could one still say something like: for example, class field theory, one of the crown jewels of algebraic number theory, is in retrospect just a very special aspect of the overall grand program? (I don’t know the right words to use, because I don’t work in this area.)

I have a book, An Introduction to the Langlands Program, written by various experts and presumably intended to be useful to non-experts. One of the chapters, by Stephen Kudla, is Tate’s Thesis which is claimed to be a fundamental starting point for the modern study of automorphic representations. My understanding is that Tate’s thesis is considered a landmark in the modern formulations of class field theory, so this is where I got the idea.

But again, I write this from a standpoint of ignorance, as a mathematician who would like to be better educated about current events.

Posted by: Todd Trimble on August 27, 2010 1:51 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

‘In that case, could one still say something like: for example, class field theory, one of the crown jewels of algebraic number theory, is in retrospect just a very special aspect of the overall grand program?’

That’s right! Now if we take the general reciprocity conjecture I mentioned:

all motivic $L$-functions are automorphic $L$-functions

it is indeed the main goal of the Langlands program (or one of two main goals, in deference to Matt’s objection), of which ‘usual’ non-abelian class field theory is one tiny portion. However, this general form is so grand it’s hopeless. This makes many experts understandably uncomfortable even to state it clearly. So in expositions, they prefer to stick to some version relatively close to known results, and portions of the programme endowed with enough concrete examples to give a definite feel for the subject. At some level, I think they also have in mind young people who might actually end up working on the subject, who then might be better served by having a doable portion explained to them in some detail. I only looked briefly into that book you mention, but perhaps Artin $L$-functions are the only arithmetic geometric $L$-functions they bring up at all. This is quite a common style that undoubtedly serves an important function. (Artin $L$-functions hardly feel ‘arithmetic-geometric’ at all, exactly because they come from geometry of dimension zero: $Spec(K)\rightarrow Spec(\mathbb{Q})$ for an algebraic number field $K$.)

When I was thinking about Tom’s original question, what came to mind was a colloquium-style answer with at least some attempt to intimate grandeur, addressed less at the student and more at mathematicians in other areas. In particular, I didn’t expect that most readers would even know the point of class field theory. (Even now, I suspect many good mathematicians reading this will be asking ‘so what *is* the point of class field theory?’)

With such considerations in mind, ideas surrounding Tate’s thesis, splitting laws and all that seemed to carry exactly that kind of esoteric beauty difficult to hear without frustration in a one-hour lecture. So I thought to go right for the form of the conjecture that I myself found most inspiring, which then could be understood over time by anyone who pays occasional attention with more than casual interest. (This isn’t to say it will be *proved* over time by such an observer!)

My feeling is there are some ideas whose grand scope can be appreciated even with just a vague feeling for the words that express them. But then, we can’t be too bashful in presenting such ideas.

Posted by: Minhyong Kim on August 27, 2010 2:40 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks for re-emphasizing this, Minhyong. In the spirit of Tom’s post, let me try to ask some naive questions about the main reciprocity slogan which you spoke on back here:

• Every motivic L-function is an automorphic L-function.

First, when you describe a motivic L-function as encoding information on the “arithmetic of varieties” (via special values of the L-function), the picture I have is that we are secretly thinking of varieties as schemes which are certain nice functors

$X: Alg_R \to Set$

from commutative $R$-algebras over a commutative ring $R$ to sets. If for example the variety is defined over $\mathbb{Z}$, then there is an associated scheme where $R = \mathbb{Z}$, and by “arithmetic of the variety” I guess you mean to study the structure of $X(A)$ where $A$ is something like the ring of integers in a number field, or perhaps something more local like a finite field $A/p$, or a localization $A_p$, or something like that. Is that far from the truth?

Then you say

The special values in question often lie outside the domains of convergence of the initial definitions as Euler products, and hence, call for analytic continuation

which immediately puts me in mind of the classical Riemann zeta function and its analytic continuation, or the zeta function of a number field. So then one is led to ask: what is the variety in question here, whose L-function is the zeta function? (I know full well this must be baby stuff for the experts.)

Okay, here I’ll see if I can back into an answer based on snippets I’ve seen here and there. Let me start with the classical zeta function. We have the Euler product

$\zeta(s) = \sum_{n \geq 1} \frac1{n^s} = \prod_{primes p} (1 - \frac1{p^s})^{-1}$

and each of the “local factors” is supposed to do something like encode the number of points of the variety when we specialize to $A = \mathbb{Z}/p$. Hm, is that right?

At this point I’m going to cheat and dip inside a book I happen to have close to hand (Ireland and Rosen, A Classical Introduction to Modern Number Theory) where I see something like this definition of a zeta function attached to a variety over $\mathbb{Z}/p$:

$Z_X(u) = \exp(\sum_{k \geq 1} \frac{N_k u^k}{k})$

Here $N_k$ is the number of points in $X(\mathbb{F}_{p^k})$ where the argument of $X$ is a finite field (unique up to isomorphism) with $p^k$ elements. Okay, this definition doesn’t look extremely motivated at first blush, but at least one sees that it is connected somehow with the “arithmetic of the variety”. Reading further, setting $u = p^{-s}$, this gives

$Z_X(p^{-s}) = \exp(\sum_{k \geq 1} \frac{N_k p^{-s k}}{k})$

and if all the $N_k$ were (seemingly boringly) equal to 1, we’d arrive at

$\sum_{k \geq 1} \frac{p^{-s k}}{k} = -\log(1 - p^{-s})$

whose exponential is the Euler factor $(1 - \frac1{p^s})^{-1}$ in the zeta function. Now $N_k = 1$ counts the number of points in

$Spec(\mathbb{Z})(\mathbb{F}_{p^k}) = \hom(\mathbb{Z}, \mathbb{F}_{p^k})$

At this point, I’m going to go out on a limb and guess that the classical zeta function is going to be the motivic L-function attached to the “variety” (scheme) $Spec(\mathbb{Z})$, and that the zeta function of a number field $k$ is the motivic zeta function attached to $Spec(\mathcal{O}(k))$, where $\mathcal{O}(k)$ refers to the ring of algebraic integers in $k$.

Okay, so here I’m guessing the general idea is that the motivic L-function of a variety is some vast generalization of the classical zeta function which is invariably built on local Euler factors, each having to do with arithmetic attached to a prime divisor of the variety.

Pressing on: to get at the really juicy arithmetic information that such motivic L-functions can be expected to encode, you said one generally has to pass to values of the analytic continuation (outside the region of convergence of the Euler product) of the L-function.

Again, one can’t help thinking of the classical zeta function, which is famously analytically continued according to a functional equation which looks something like this:

$Z(s) = Z(1-s)$

where $Z(s)$ is a product of the zeta function with an extra Gamma function factor thrown in. (And here I seem to have heard that the extra Gamma factor is secretly an Euler factor attached to the “prime at infinity”. Here we really mean that an ordinary prime is connected with a nonarchimedean valuation on $\mathbb{Q}$, but there is also the usual archimedean valuation which is dubbed “the prime at infinity”. This extra prime is important to take into account to get the full picture.)

Next, you remark

All known cases of this analytic continuation rely on the identification of the arithmetic-geometric L-function with an automorphic L-function

Hm, okay. I get the idea that for the classical zeta function, the automorphic function (or whatever you call it) in question is supposed to be a theta function. In another comment, you jotted down

$L(f, s) = \frac1{(2\pi)^s \Gamma(s)} \int_0^\infty f(i y) y^s \frac{d y}{y}$

and somewhere in my past I do seem to remember seeing things like

$Z(s) = \zeta(s) (2\pi)^s \Gamma(s) = \int_0^\infty \theta(i y) y^s \frac{d y}{y}$

Moreover, I am given to understand that in essence, Riemann wrote down this formula. Moreover, he must have known that the theta function $\theta(z)$ transforms nicely under the substitutions $z \mapsto z + 1$ and $z \mapsto -1/z$ (these transformations generate $SL_2(\mathbb{Z})$, which is a certain modular group), and these nice transformation properties lead him to the functional equation

$Z(s) = Z(1-s)$

which forms the basis of the analytic continuation.

“Automorphic representation” would seem to refer to an especially clean way of expressing the transformation rules for the theta function (or automorphic function) $f$, so that some expression or some differential form associated with $f$ is invariant under the action of the modular group. So I’m guessing “automorphic representation” refers to a space of differential forms on which some modular group acts, in which this pretty expression lives.

So in broad brushstrokes, I’m guessing that the koan

• Every motivic L-function is an automorphic L-function.

is a concentrated expression for a program which goes something like this: for every “motivic L-function” attached to a variety, there is some “reductive group” $G$ (like $SL_2$) and a representation of $G$ in which there lives some invariant theta-function-like object $f$ so that the motivic L-function can be expressed as some integral involving $f$, and such that the invariance of the automorphic form (or whatever it is) attached to $f$ leads directly to a functional equation for the motivic L-function which allows the desired analytic continuation.

I’m probably way, way, way off, but these are the guesses I’ve extracted thus far.

Posted by: Todd Trimble on August 27, 2010 6:21 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Almost everything you write is remarkably accurate!

Here are a few remarks: arithmetic geometers are themselves a bit confused about the distinction between zeta functions and $L$-functions. But roughly, a zeta function refers to the $L$-function of a whole variety, rather than just ‘piece’ of it (a motive, or an $H^i$). And then, when speaking about zeta functions, they’re also somewhat confused about whether to focus mostly on varieties over $\Spec(\mathbb{Q})$ or ‘integral models’ over $\Spec(\mathbb{Z})$. (Integral models are what you get by clearing denominators from the equations. But there are better and worse ways to do this depending on your choice of equations.) But if we stick to the former, the zeta function of a variety is actually that of a sufficiently nice integral model. (This notion might not make sense in general unless you remove finitely many Euler factors.) The zeta function of a variety $X$ is built by choosing a model, reducing modulo $p$, and counting points exactly as you’ve done. So for a number field $F$, the integral model is $\Spec(O_F)$. And yes, the Riemann zeta function is the zeta function for $\Spec(Q)$ (or its integral model $\Spec(Z)$). It’s useful to know that the exponential formula you’ve written down for a variety over a finite field can also be written as a product $Z_X(p^{-s})=\prod_x(1-N(x)^{-s})^{-1}.$ Here, $x$ runs over the ‘closed points of $X$’ (in the sense of scheme theory) and $N(x)$ is the number of points in the residue field at $x$, which is a finite field. If these words don’t make sense, consider the case where $X=\Spec(A)$. Then the $x$ are maximal ideals $m$, and the residue fields are just $A/m$.

Your last paragraph on the ‘reductive group attached to a variety’ is also very accurate, except it’s here that it’s best to go to this strange notion of a motive. Attaching a reductive group, even conjecturally, seems to work best if we break the variety up into its ‘irreducible components’ and attach a reductive group to each component separately. You could say such a procedure is the main motivation for the theory of motives.

One final remark on your passage

‘by “arithmetic of the variety” I guess you mean to study the structure of $X(A)$ where A is something like the ring of integers in a number field.’

This brings up a very important point. Unfortunately, $L$-functions seem to get at the structure of points in number fields only in very special cases, namely, when $X$ is an abelian variety. For other varieties, the arithmetic contained in the $L$-function consists of various ‘abelianized’ bits of information, like groups of algebraic cycles or $K$-theory. The whole motivic philosophy tends to focus on such invariants, which capture sets of points only when the variety itself is abelian. The situation is essentially by design: You may have heard that in the theory of motives, one replaces maps between varieties by correspondences. So it’s not surprising that information about points, that is, certain sets of actual maps, gets lost in the process. Nevertheless, to my mind, this is a dissatisfactory state of affairs.

Posted by: Minhyong Kim on August 28, 2010 2:17 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Some responses to Minhyong and Matthew:

Minhyong: Thanks for your patient explanations! This is actually very helpful (like Tom, I don’t plan on retraining as a mathematician to understand all of this, but it’s very pleasant to be able to ask an expert some basic questions and get such nice replies).

It may be time for me to look at that lecture by Langlands you pointed to over here. But here are some responses.

The zeta function of a variety $X$ is built by choosing a model, reducing modulo $p$, and counting points exactly as you’ve done.

I guess I don’t really understand how one chooses an integral model, but I have to assume the zeta function of $X$ is well-defined, i.e., doesn’t depend on the choice.

It’s useful to know that the exponential formula you’ve written down for a variety over a finite field can also be written as a product

$Z_X(p^{−s})= \prod_x (1 − N(x)^{−s})^{−1}.$

Here, $x$ runs over the ‘closed points of $X$’ (in the sense of scheme theory) and $N(x)$ is the number of points in the residue field at $x$, which is a finite field. If these words don’t make sense

They make a lot of sense; thanks! This is very helpful. I guess the $p$ refers to the finite field $\mathbb{Z}_p$, but all of this generalizes to any finite field with $q = p^k$ elements; just replace the argument $p^{-s}$ by $q^{-s}$.

except it’s here that it’s best to go to this strange notion of a motive.

Of course you’re right that I blithely elided over any mention of motives. The reason being that I know basically nothing about them: ‘motive’ is one of those words which for me is charged with mystery.

When I saw it in your comment, I ran to wikipedia where I got some enlightenment, as least as far as ‘pure motives’ are concerned, because they resonate a lot with various constructions in the calculus of relations which I know a bit about. In the categorical language I am familiar with, a pure motive appears to be an object in the Cauchy or idempotent-splitting completion of the (locally posetal) bicategory of algebraic varieties and Zariski-closed relations. (There’s probably a slew of things here that I’m happily skipping over that are needed to make this really precise, like Chow rings and Chow’s moving lemma and whatnot. Apparently one should also think of the data of pure motives as including a grade $m$, and I can vaguely see why this might be important, but for the time being I’m skipping over that as well. If I shouldn’t, no doubt you’d be able to say why.) This type of thing is important in categorical logic, so its presence here is inviting and intriguing to me.

But apparently there are these mysterious mixed motives as well. The wikipedia article is vague on these critters. I’m hoping the motives you’re referring to are pure motives, not mixed motives.

$L$-functions seem to get at the structure of points in number fields only in very special cases, namely, when $X$ is an abelian variety. For other varieties, the arithmetic contained in the $L$-function consists of various ‘abelianized’ bits of information, like groups of algebraic cycles or $K$-theory. The whole motivic philosophy tends to focus on such invariants

Ah, this sounds like a really vital piece of wisdom! Thanks for sharing it.

In another comment, Matthew Emerton helpfully wrote:

The one that Minhyong explained, to the effect that motivic $L$-functions should also be automorphic $L$-functions, is known as reciprocity… A simple illustration of this kind of phenomenon is given by the famous result, due essentially to Gauss, that the Galois group of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$ (here $\zeta_n$ is a primitive $n$th root of 1), is canonically isomorphic to $(\mathbb{Z}/n)^\times$, where an element $a$ in this group of units acts on $\zeta_n$ just by raising to the $a$th power. The group $(\mathbb{Z}/n)^\times$ is a very simple example of an automorphic object (the underlying reductive group is just $GL_1$), and it is eminently understandable. With this isomorphism in hand, it is straightforward to solve the Diophantine problem of describing all subfields of $\mathbb{Q}(\zeta_n)$. For example, taking $n = p$, an odd prime, it is easy to deduce that $\mathbb{Q}(\zeta_p)$ has a unique quadratic subfield, equal to $\mathbb{Q}(\sqrt{\pm p})$ (the sign being chosen so that $\pm p \equiv 1 \mod 4$), and using the description of $Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q}$) afforded in this way, one can easily deduce (for example) the law of quadratic reciprocity.

This puts quadratic reciprocity in a light I didn’t know about. So we have $(\mathbb{Z}_n)^\times = GL_1(\mathbb{Z}_n)$, and I guess attached to this reductive group there’s going to be an automorphic representation in which a suitable “theta function” or automorphic function (better known as a quadratic Gauss sum) lives [obviously I’m trying to fit in your comment with my ramblings here]. This Gauss sum lives at the heart of the class field theory approach to quadratic reciprocity, if I understand things correctly.

This provides me with some interesting food for thought…

Posted by: Todd Trimble on August 28, 2010 6:32 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

In the categorical language I am familiar with, a pure motive appears to be an object in the Cauchy or idempotent-splitting completion of the (locally posetal) bicategory of algebraic varieties and Zariski-closed relations.

That sounds worth expanding on. I like it when categorical logic type constructions appear in mainstream maths. There is a page (which you mostly wrote) on Cauchy complete categories. Does the example above count as an example of completion of a Poset-enriched category?

Posted by: David Corfield on September 6, 2010 5:11 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I guess so! Although I think there are a few wrinkles that need ironing out (see below). And while I guess a note to that effect could go into the nLab, I am not even close to conversant with motives, so it’s hard for me to say how useful this general point of view is.

Still, I sometimes wonder… when I look at the Wikipedia page on motives and see formulae like

$\alpha \circ \beta \coloneqq \pi_{X Z *}(\pi_{X Y}^*(\alpha) \cdot \pi_{Y Z}^*(\beta)),$

that formally looks exactly like the kinds of formulae one meets in relational calculus: if $R: X \to Y$ and $S: Y \to Z$ are relations or spans, one has

$R \circ S = \pi_{1 3 *}(\pi_{1 2}^{-1}(R) \cap \pi_{2 3}^{-1}(S))$

where $\pi_{i j}$ denotes projection of $X \times Y \times Z$ onto the $i j$-factors, $\pi_*$ indicates direct image, and $\pi^{-1}$ indicates inverse image. (In words: pull back predicates $R(x, y)$ and $S(y, z)$ so that they are defined on a common domain $X \times Y \times Z$, then intersect, then existentially quantify by taking a direct image which bounds the variable $y$. The result is $\exists_y R(x, y) \wedge S(y, z)$, as in relational composition.) Clearly, this analogy is by intentional design, but for some reason one rarely hears people say it like that or make the connection explicit. At least I don’t.

Maybe that’s because, if I understand matters correctly, there are some wrinkles here. The $\cdot$ in the algebraic geometry formula is the intersection multiplication in the Chow ring, but it’s not a naive intersection; for example, when you intersect a variety with itself, you don’t perform the usual set-theoretic intersection. Rather, you perturb things a bit to get a transverse intersection. So really what you do is consider equivalence classes of subvarieties (rational equivalence classes I think they’re called), and then allow yourself to move around in the equivalence classes until you find good representatives so that their naive intersection is a good transverse intersection. There are some technical results like “Chow’s moving lemma” that say you can get away with this and get a well-defined product in this way, but I don’t know the details of this.

Still, it is very tempting to hope that there is some nice “spanny” or cartesian bicategory niche into which all of this fits.

Posted by: Todd Trimble on September 6, 2010 7:41 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Todd wrote:

I guess I don’t really understand how one chooses an integral model, but I have to assume the zeta function of X is well-defined, i.e., doesn’t depend on the choice.

You’d probably be interested in Matthew Emerton’s remark on precisely this question here. I’d been wondering about the same thing.

Personally I find it much less stressful to think of the zeta function as being defined for an integral scheme — or more precisely, a ‘finite type’ integral scheme (this condition guarantees that your counts of points come out finite). Then I can leave it to the experts to worry about how to define zeta functions for rational algebraic varieties — which gets us into the question you’re asking about.

Anyway, this the approach that Jim and I took when we categorified zeta functions using the theory of species. (Note that the ‘Dirichlet product’ used here is one of those five monoidal structures on species that I kept harping about over on the n-Forum.)

Posted by: John Baez on August 31, 2010 8:59 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Matthew Emerton’s remark

I remember reading that, but it might have been during a massive binge of skimming posts and comments after returning from internetless vacation. So I hadn’t internalized that discussion. Now I agree with you, it’s really cool!

(I’ve been paying at least a little attention to what you and Jim have been cooking up, and that also is very interesting…)

Posted by: Todd Trimble on September 1, 2010 12:07 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

John wrote:

So Tom, the guy who never installed the gothic fonts on his computer because he didn’t want to learn about Lie algebras, suddenly wants to learn about the Langlands program!

I don’t think I was aware that the Langlands programme had anything to do with Lie algebras. Though given its reputation for having something to do with just about everything, it didn’t surprise me that it does.

For those who weren’t there, John’s crack at me is probably based on what I wrote here:

A confession: for me, some terminology and notation inspires fear and insecurity. I can barely look at a gothic or fraktur letter — I mean the ones that Lie algebras are done in — without feeling narrow and ignorant.

[…]

Fortunately, my browser isn’t set up to show gothic letters on the $n$-Café.

Unfortunately this got misinterpreted by some (John and Urs, stand up!) as “I don’t like Lie algebras because I don’t like gothic letters.”

(In case it’s not clear, the last line of that quotation was a joke. I don’t really find ‘Let [unreadable character] be a Lie algebra’ more welcoming than ‘Let [gothic $g$] be a Lie algebra’.)

Posted by: Tom Leinster on August 27, 2010 12:58 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Tom wrote:

I don’t think I was aware that the Langlands programme had anything to do with Lie algebras.

Not mainly. I was mainly just eager to tease you about your distaste for gothic letters (sorry). But Lie algebras and their corresponding algebraic groups are closely akin, so when Grojnowski says:

The Langlands program describes the representation theory of reductive groups.

he’s hoping that you instantly envision the nicest of the reductive algebraic groups, namely the simple ones, which mainly means our friends $SL(n)$ and $SO(n)$ and $Sp(n)$, together with a few ‘exceptional’ weirdos like $E_8$

… but a key part of befriending these objects is learning to love their Lie algebras… and here I will try to make your browser show you undreadable characters: $\mathfrak{sl}(n)$ and $\mathfrak{so}(n)$ and $\mathfrak{sp}(n)$.

The point is this: an algebraic group is a group that’s also an algebraic variety, or scheme if you prefer. Like any variety it has a tangent space, but because groups ‘look the same everywhere’ you know almost everything there is to know about an algebraic group once you deeply understand its tangent space at the identity element — and that’s a Lie algebra.

The way I sometimes put it is that a Lie algebra is like a slide rule for studying a group: it lets us turn problems involving multiplication into problems involving addition. An algebraic group has a multiplication, but its Lie algebra is a vector space, where we can add, and things are easier there.

This is all just preliminary stuff to soften up your resistance, of course… the real meat of a Lie algebra comes from the nonabelian nature of the corresponding group, which is something you didn’t need to worry about when you were studying logarithms and exponentials and slide rules back in school.

But: while I know algebraic groups and their Lie algebras pretty well by now, the Langlands program is still pretty darn scary. Sort of like how I enjoy climbing the hills behind my house in Riverside, but am far from ready to tackle Mount Everest.

The really scary part of the Langlands program, to me, is that nobody seems able to give a hand-wavy argument as to why it should be true. There’s a delightfully terse statement of the conjectured “geometrical Langlands correspondence” as an equivalence of categories here, but I’ve never seen a clear explanation of why we should expect such an equivalence of categories.

Of course, that could be part of why people consider it nontrivial.

Posted by: John Baez on August 28, 2010 10:35 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

There’s a delightfully terse statement of the conjectured “geometrical Langlands correspondence” as an equivalence of categories

namely:

$\mathcal{O}$-modules on $Loc_{{}^L G}$ are equivalent to $\mathcal{D}$-modules on $Bun_G$.

(I originally learned this from a useful introduction by Pantev back then.)

but I’ve never seen a clear explanation of why we should expect such an equivalence of categories.

We can restate the above equivalence in vague broad-stroke but more suggestive words:

$\mathcal{O}$-modules $\sim$ vector bundles

$\mathcal{D}$-modules $\sim$ flat vector bundles

$Bun_G$ $\sim$ space of $G$-bundles

$Loc_{{}^L G}$ $\sim$ space of flat ${}^L G$-bundles

With that dictionary, the geometric Langlands duality seems to read like the following slogan:

All vector bundles on the space of flat ${}^L G$-bundles
are equivalent to
flat vector bundles on the space of all $G$-bundles.

That might begin to sound like something that resonates.

Posted by: Urs Schreiber on August 28, 2010 11:48 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

That’s starting to sound good, Urs.

I think someone out there should take some idea like this:

All vector bundles on the space of flat ${}^L G$-bundles
are equivalent to
flat vector bundles on the space of all $G$-bundles.

and turn it into a theorem.

If someone can find a nice conceptual proof of a ‘toy model’ of the Langlands conjectures in some very simplified context, then maybe someone else can gradually crank up the level of complexity until they get a nice proof of the real thing.

Posted by: John Baez on September 1, 2010 7:55 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

The correspondence isn’t something you expect to hold with the simplifications

“complex of quasi-coherent sheaves” -> “vector bundle”

“complex of algebraic D-modules” -> “flat vector bundle”.

There is a baby case of the geometric Langlands correspondence which has been worked out (by Laumon): the case where G is a torus (say, G = GL_1).

In this case, there’s a simpler version of the equivalence available: there is an equivalence
between quasi-coherent complexes on
Bun_G and quasi-coherent complexes on Bun_G^.
(i.e., you get an equivalence if you ignore the flat connections on both sides, and the actual result is a jazzed up version of this).

When G = GL_1 (so that G^ = G), Bun_G is just the moduli stack of line bundles on your algebraic curve X. This is in turn just a fancy version of the Jacobian of X (over the complex numbers, this is given concretely as the complex torus H^1(X;R)/H^1(X;Z) ).
There are two complications: Bun_G is a stack (because line bundles have automorphisms), and Bun_G has several components (because line bundles can have have different degrees). But these complications are “Langlands dual” to one another, so it’s safe to ignore them as long as you ignore both simultaneously.

So in this special case G = GL_1, the Langlands correspondence is a jazzed up version of the statement that there is an equivalence of “quasi-coherent complexes on J(X)” with itself. This equivalence is not the identity functor: it’s given by the Fourier-Mukai transform. The existence of this equivalence is an avatar of the algebro-geometric fact that J(X) is self-dual as an abelian variety.
(In terms of the description J(X) = H^1(X;R) / H^1(X;Z), it comes from the fact that H^1(X) is self-dual, since X satisfies Poincare duality).

In particular, the geometric Langlands correspondence doesn’t preserve properties like “being a vector bundle” in general. For example, the Fourier-Mukai transform of a trivial vector bundle is a skyscraper sheaf. Analogous phenomenon in a simpler setting: one of the simplest functions (on R) you can think of is the constant function f(t) = 1. The Fourier transform of f is Dirac’s delta function, which exists only as a distribution.

Posted by: Jacob Lurie on September 1, 2010 9:05 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Thanks.

One comment on this here, then I have a question:

The correspondence isn’t something you expect to hold with the simplifications

I wasn’t suggesting that one can actually replace complexes of QC-sheaves by vector bundles. What I tried to suggest in reaction to John’s query as to why there is Langlands duality was that when formulated in these “vague broad-stroke but more suggestive words” it becomes more manifest that the statement of the correspondence involves a curious symmetry where we have bundle-like structures over spaces of bundles, and in one case there is a flat connection here, in the other case it is there. I should have said it the other way round and said that both are stack-like structures. But I tried to produce a simple-sounding slogan.

Anyway. Here is the question:

starting from the simple case of Fourier-Mukai-transforms, how much of geometric Langlands duality in more generality can be understood in terms of the Ben-Zvi/Nadler-type derived integral transforms?

(I gather part of the answer is in the followup, but I haven’t absorbed that yet.)

Posted by: Urs Schreiber on September 1, 2010 9:29 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

There is a generalization of the geometric Langlands story which makes the symmetry between the two sides more apparent. On the moduli stack Bun_G, there is a canonical line bundle Det (the determinant bundle).
For any scalar c, it makes sense to consider
D-modules on Bun_G twisted by Det^c.
The “quantum version” of geometric Langlands
posits roughly that there should be an equivalence between Det^c-twisted D-modules on Bun_G and
Det^c’-twisted D-modules on Bun_G’, where G’ is the Langlands dual of G and c’ = -1/c. The usual formulation can be regarded as a limiting case of this, where c approaches zero.

essentially any functor can be realized as an integral transform (there should be some technical caveats here having to do with the issue that Bun_G is a very large stack, so you need to be careful how all of the categories are defined). So if the geometric Langlands equivalence exists, it will be given by an integral kernel.

Posted by: Jacob Lurie on September 2, 2010 12:54 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

There is a generalization of the geometric Langlands story which makes the symmetry between the two sides more apparent.

Ah, thanks, the quantum geometric Langlands correspondence.

Can you give me a conceptual idea of the relevance and origin of those sheaves of twisted differential operators $\mathcal{D}^{k,\lambda}$? I follow the construction in Frenkel’s notes, but the twisting looks somewhat ad hoc . I bet it’s not, but I don’t quite see why it is these twists that give a nice symmetric correspondence.

I mean, to come back to John’s question: what’s going on here, why would be expect something like quantum geometric Langlands to hold?

(I guess one answer is: because it’s part of the physics of the B-model, as shown by Kapustin, and maybe that’s the best explanation there is.)

essentially any functor can be realized as an integral transform […] So if the geometric Langlands equivalence exists, it will be given by an integral kernel.

All right, but is there a natural candidate for a correspondence-span stack? For quantum geometric Langlands? (I gather there is not.)

Posted by: Urs Schreiber on September 2, 2010 9:16 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

For the question of “why should one expect a geometric Langlands correspondence”, I can think of a few answers, but none of them are great.

1) By analogy with the Langlands program in number theory. (Which raises the question of why you should believe in the usual Langlands correspondence, which I find much more mysterious.)

2) The work of Kapustin-Witten, showing that it’s predicted by dualities in physics. This I have basically no understanding of, so I can’t comment.

3) It is related to more concrete duality phenomena, like the assertion that the generic fibers of the Hitchin fibration for G are abelian varieties which are dual to the generic fibers of the Hitchin fibration for G^.

4) If you want to construct the Langlands dual group G^ without referencing roots and weights, you probably want to be consider something like the geometric Satake isomorphism, which identifies its representation category with the category of equivariant D-modules on a space Gr_G. You can think of Gr_G as a space of G-bundles on a curve X which are trivialized away from a point x, so there is an obvious relationship between Gr_G and Bun_G. So from this point of view, G^ is defined in a way that is closely related to D-modules on Bun_G. (Of course, this raises the question of why you should expect the geometric Satake isomorphism, which is really a special case of the general program.)

5) Because it is known to be true for a torus for a very conceptual reason. (But then you might ask: why shouldn’t we believe in a correspondence between quasi-coherent sheaves on Bun_G and quasi-coherent sheaves on Bun_G^?)

For your question about the sheaves of twisted differential operators: I also don’t have a very good answer for you, except that there aren’t many other choices. The Kapustin-Witten suggestion that geometric Langlands is a special case of an equivalence of topological field theories suggests that we should be able to obtain both sides of the correspondence by assembling local information,
as in the cobordism hypothesis. This doesn’t quite match the mathematician’s story (we’re doing algebraic geometry rather than topology, and the category of D-modules on Bun_G(X) is not a topological invariant of X), but it does suggest an approach: namely, establish some equivalence of categories locally (some version of the geometric Satake isomorphism) and try to “integrate” to get the global correspondence. One can show that considering twisted D-modules is essentially the only way to deform the locally defined categories. So if you wanted to consider a generalization of geometric Langlands which you could attack using this strategy, you don’t have many options.

As for writing down the integral kernel: the idea of obtaining the global geometric Langlands correspondence by “integrating” the local correspondence suggests some candidates for what the integral kernel might be (but nothing as simple as a line bundle on the product, as in the torus case).
It may be possible to make some educated guesses.
(Proving that one of these guesses gives an equivalence of categories is another matter.)

Posted by: Jacob Lurie on September 2, 2010 12:40 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Regarding Kapustin-Witten, Langlands becomes a manifestation of 4D S-duality. In the abelian case, this can be understood as a sort of Fourier transform (ie, abelian duality), and this is a nonabelian generalization of that. I’ve been told that this is rather analogous to the math where one can think of class field theory (which I don’t really understand) as the abelian case.

For Urs, it might be helpful to remember that, at least from the physics point of view, all these D-modules (twisted and not) are really A-branes in disguise. In particular, it’s not the B-model; it’s the A model in disguise for the quantum case (if my memory is right).

Posted by: Aaron Bergman on September 2, 2010 1:51 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Langlands becomes a manifestation of 4D S-duality

Yes, but is that an explanation or a reformulation? I don’t know. But I was hoping that if we see a nice formal mathematical “reason” for geometric Langlands duality, this would conversely also shed light on S-duality and in fact on the definition of the field theories involved.

I am thinking of something that Ben-Zvi and Nadler sort of seem to be hinting at in the intro of their The character theory of a complex group, where it kind of looks as if there were a way to take two fully dualizable objects built from $D$-module reps, defining two extended TFTs and then establishing some relation between these. So the physics picture in precise math terms

it might be helpful to remember that, at least from the physics point of view, all these D-modules (twisted and not) are really A-branes in disguise. In particular, it’s not the B-model

I was just glancing at literature on quantum Langlands, and noticed that here Kapustin seems to be saying he explains that with B-branes. But haven’t yet had a chance to look at any details.

Posted by: Urs Schreiber on September 2, 2010 2:01 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I wrote in reply to Aaron:

Langlands becomes a manifestation of 4D S-duality

Yes, but is that an explanation or a reformulation?

I’ll correct myself: actually, $S$-duality is supposed to have a genuine “explanation”, namely as the remnant conformal invariance of the theory on a stack of 5-branes after compactification on a torus. Maybe there is no simpler “explanation” for geometric Langlands than this (which is quite good as an explanation, but also quite out of reach as far as making it precise goes).

Posted by: Urs Schreiber on September 2, 2010 3:23 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

and turn it into a theorem.

Right. I have been thinking just a little bit about it. What I can prove is this:

There is a canonical natural morphism from
flat $\infty$-bundles on the space all $G$-bundles
to
all $\infty$-bundles on the space of flat $G$-bundles.

(Even for $G$ an $\infty$-group.)

I made that remark in a bit more detail on the $n$Forum here. It is a formal consequence of the definition of differential cohomology in an $(\infty,1)$-topos that I am thinking about and so holds in quite a bit of generality. An obvious guess would be that geometric Langlands might be about extra conditions (e.g. holomorphicity) that make this morphism an equivalence. But I don’t know.

Posted by: Urs Schreiber on September 1, 2010 1:21 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

all $\infty$-bundles on the space of flat $G$-bundles.

Where does $G$ become ${}^L G$? Will this approach only work for self-dual $G$?

Posted by: David Corfield on September 1, 2010 1:33 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Where does $G$ become ${}^L G$?

Yeah, i don’t see that. All I am really observing is that there is such a canonical morphism on general abstract grounds. Apart from observing that it goes between the same two kinds of objects I don’t know how it relates to Langlands. It might be that under special conditions (like self-dual $G$ and others) it does relate to geometric Langlands. But maybe it doesn’t. I haven’t checked at all.

Posted by: Urs Schreiber on September 1, 2010 1:45 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Instead of restricting attention to self-dual $G$, perhaps the next way to get a good clue is to think about the meaning of the Langlands dual. What’s the most elegant, fundamental relation between a reductive algebraic group and its Langlands dual? I’ve only seen somewhat ad hoc looking definitions involving roots.

I would be happy enough to restrict attention to the case of reductive algebraic groups over $\mathbb{C}$, for the purposes of this ‘toy model’ project!

Unless, of course, the elegant relation between a group and its Langlands dual only becomes clear in a more general context.

Posted by: John Baez on September 1, 2010 3:13 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

MO’s version of this question is here, though I can’t see that one finds a conceptual account there. Any chance of one of those nice concrete dualities with a good honest dualizing object?

Posted by: David Corfield on September 1, 2010 3:55 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

the next way to get a good clue is to think about the meaning of the Langlands dual

Is there a chance that this is a bit of a red herring that more obfuscates a grand story than revealing it? I don’t know, just wondering. I notice that above Minhyong Kim wrote:

even though it’s important at some point, for the big picture, you lose nothing by assuming $G$ to be $GL_n$. In fact, the so-called functoriality conjecture implies that we can always take $G$ to be $GL_n$.

Posted by: Urs Schreiber on September 1, 2010 5:30 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Don’t you still have to worry about it? Is there a difference between an isomorphism between group and dual, and one between group and double dual, as with finite dimensional vector spaces?

I thought this comment from some notes:

For $G = GL(1) = \mathbb{C}^*$ or more generally for $GL(n, \mathbb{C}$) the Langlands dual group is $G$ itself. But even then, and even in the symmetric version of the correspondence, the predicted duality is far from trivial (a self-duality on a set or a category which is not the identity),

was relevant, but maybe not. In any case, the notes look approachable.

Posted by: David Corfield on September 1, 2010 6:57 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Don’t you still have to worry about it?

What I have is this naive and uneducated (and possibly wrong) mental image:

we might start with a “flat vector bundle” ($\mathcal{D}$-module) on the space of all $GL_n$-bundles that happens to be supported in a region where all these $GL_n$ bundles have reductions of their structure groups to some other group $G$. Then we send this “flat vector bundle” through the correspondence and obtain a vector bundle on the space of flat $GL_n$-bundles, which so happens to be supported on just those $GL_n$-bundles whose structure group reduces to ${}^L G$.

Is there a chance that it could work like this? If this is not too far from the truth, then it would indicate why it is quite okay to understand the duality first just for $GL_n$, because the other cases are the same duality, studied in a bit more detail.

But I don’t know. I haven’t studied my Higgs bundles and not my Hecke eigensheaves. I am busy with something else…

Posted by: Urs Schreiber on September 1, 2010 7:14 PM | Permalink | Reply to this
Read the post Can the Langlands program be described in layman's terms?
Weblog: Quora
Excerpt: Judging from [1][2] a layman's explanation of the Langlands program does not yet exist. Even if 'layman' is extended to those with a bachelor in math. This is not all bad on the internet, as unlike quantum mechanics, the Riemann hypothesis or P=NP, it ...
Tracked: September 22, 2011 2:14 PM

### Re: What is the Langlands Programme?

Being a complete novice at math(basic High school Algebra/Geometry)the Langlands program sounds like a cookbook that uses different math disciplines to approach various types of problems the answers to which themselves become part of the recipe combinations used to solve additional mathematics problems. The recipes/recipe combos themselves can become so complicated that THEY need to be solved before being used for “cooking up the solutions” to additional problems. I hope that my description is somewhat on the mark however as I’ve said my qualifications never went beyond the high school level.

Posted by: Chris Rogers on August 6, 2015 2:21 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

It sounds like you’ve been reading Eugenia Cheng’s book.

Posted by: Tom Leinster on August 6, 2015 2:29 PM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Actually I’ve recently purchased the book but haven’t had a chance to read it yet. If my description isn’t appropriate would you be able to give me a description? Most of the threads that I’ve read describe the Langlands program at a level way above my knowledge.At 56 I’ve come to appreciate the beauty of mathematics and had no idea that there were so many diverse fields of the language.

Posted by: Chris Rogers on August 8, 2015 5:48 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

I’m really the wrong person to ask! It’s because I don’t have a good overview of the Langlands programme that I wrote this post in the first place. I definitely found the comments helpful, though, and I now understand more than I did. Maybe someone more expert than me will chime in to say whether they think your description is accurate.

Glad you’ve got Eugenia’s book. I think you’ll like it!

Posted by: Tom Leinster on August 8, 2015 11:31 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Chris Rogers—not the Chris Rogers who was my student—wrote:

Most of the threads that I’ve read describe the Langlands program at a level way above my knowledge.

Then I strongly recommend this book:

• Edward Frenkel, Love and Math, Basic Books, 2013.

It’s written for anyone who has a solid grasp of high school algebra, and it mainly consists of his autobiography together with an attempt to explain the Langlands program in simple, intuitive terms. As a mathematician I found it quite enjoyable, and nonmathematicians also seem to enjoy it, though in some sense it’s a different book for them, since I know exactly what a ‘sheaf’ and a ‘group representation’ is, while they must rely on Frenkel’s hand-waving descriptions of these concepts.

Posted by: John Baez on August 9, 2015 9:31 AM | Permalink | Reply to this

### Re: What is the Langlands Programme?

Having recently received an email from John, I suddenly remembered something that was on my mind from years ago. That is, I had wanted to write something more coherent in this thread in regard to the question ‘why would anyone expect anything like Langlands functoriality’. Well, I gave a talk on this topic this summer. It’s not very good, but the goal was to answer exactly this question. So I’m posting a link here, for better or worse:

http://people.maths.ox.ac.uk/kimm/lectures/functoriality.pdf

Posted by: Minhyong Kim on October 22, 2016 6:03 PM | Permalink | Reply to this