ICM2014 — Jim Arthur plenary lecture

The main other thing I did on day two of the congress was go to a reception in the evening hosted by the French Embassy. It was less formal than that makes it sound, and as I circulated I met a number of people I hadn’t seen for quite a while, as well as others I had got to know at the congress itself. The French ambassador, who was disconcertingly young looking, gave a speech, as did Artur Avila (as you know, Avila, like Ngo four years ago, is French), and one other person, whose name I’ve annoyingly forgotten. One interesting nugget of information to come out of those speeches was that Paris is planning to bid for the 2022 ICM. If that bid is successful, then Avila will have two successive ICMs in his home country. There was plenty of food at the reception, so, as I had hoped, I didn’t need to think about finding supper. When we arrived, we were asked for our business cards. In common with approximately 99.9% of mathematicians, I don’t have a business card, but for cardless people it was sufficient to write our names on little bits of paper. This, it turned out, was to be entered in a draw for a bottle of wine. When the time came, it was Avila who drew out the pieces of paper. Apparently, this is a Korean custom. There were in fact two bottles going, so two chances to win, which sort of became three chances when the person who should have won the first bottle wasn’t there to claim it. And so, for the first time in my life … not really. I have never won anything in a raffle and that puzzling sequence continued.

The next morning kicked off (after breakfast at the place on the corner opposite my hotel, which served decent espressos) with Jim Arthur, who gave a talk about the Langlands programme and his role in it. He told us at the beginning that he was under strict instructions to make his talk comprehensible — which is what you are supposed to do as a plenary lecturer, but this time it was taken more seriously, which resulted in a higher than average standard. Ingrid Daubechies deserves a lot of credit for that. He explained that in response to that instruction, he was going to spend about two thirds of his lecture giving a gentle introduction to the Langlands programme and about one third talking about his own work. In the event he messed up the timing and left only about five minutes for his contribution, but for everybody except him that was just fine: we all knew he was there because he had done wonderful work, and most of us stood to learn a lot more from hearing about the background than from hearing about the work itself.

I’ve made a few attempts to understand the Langlands programme — not by actually studying it, you understand, but by attending general-audience talks or reading general-audience articles. It’s a bit of a two-steps-forward (during the talk) and one-step-back (during the weeks and months after the talk) process, but this was a very good lecture and I really felt I learned things from it. Some of them I immediately forgot, but have in my notes, and perhaps I’ll fix them slightly better in my brain by writing about them here.

For example, if you had asked me what the central problem in algebraic number theory is, I would never have thought of saying this. Given a fixed polynomial P and a prime p, we can factorize P into irreducibles over the field \mathbb{F}_p. It turns out to be inconvenient if any of these irreducible factors occurs with multiplicity greater than 1, so an initial assumption is that P has distinct roots over \mathbb{C} (or at least I think that’s the assumption). [Insert: looking at my notes, I realize that a better thing to write is \mathbb{E}, the splitting field of P, rather than \mathbb{C}, though I presume that that gives the same answer.] But even then, it may be that over some primes p there are repeated irreducible factors. The word “ramified”, which I had always been slightly scared of, comes in here. I can’t remember what ramifies over what, or which way round is ramified and which unramified, so let me quickly look that up. Hmm, that was harder than I expected, because the proper definition is to do with rings, extension fields and the like. But it appears that “ramified” refers to the case where you have multiplicity greater than 1 somewhere. For the purposes of this post, let’s say that a prime p is ramified (I’ll take the polynomial P as given) if P has an irreducible factor over \mathbb{F}_p with multiplicity greater than 1. The main point to remember is that the set S of ramified primes is small. I think Arthur said that it was always finite.

So what is the fundamental problem of algebraic number theory? Well, when you decompose a polynomial into irreducible factors, those factors have degrees. If the degree of P is d, then the degrees of the irreducible factors form a partition of d: that is, a collection of positive integers that add up to d. The question is this: which (unramified) primes give rise to which partitions of d?

How on earth is that the fundamental problem of algebraic number theory? What’s interesting about it? Aren’t number theorists supposed (after a long and circuitous route) to be solving Diophantine equations and things like that?

Arthur gave us a pretty convincing partial answer to these questions by discussing the example P(x)=x^2+1. The splitting field is \mathbb{Q}(\sqrt{-1}) — that is, rational linear combinations of 1 and i — and the only ramified prime is 2. (The reason 2 is ramified is that over \mathbb{F}_2 we have x^2+1=(x+1)^2.)

Since the degree of x^2+1 is 2, the two partitions of the degree are (2) and (1,1). The first occurs if and only if x^2+1 cannot be factorized over \mathbb{F}_p, which is the same as saying that -1 is not a quadratic residue. So in this case, the question becomes, “For which odd primes p is -1 a quadratic residue?” to which the answer is, famously, all primes congruent to 1 mod 4. So Arthur’s big grown-up question is a generalization of a familiar classical result of number theory.

To answer the question for quadratic polynomials, Gauss’s law of quadratic reciprocity is a massive help. I think it is correct to say that the Langlands programme is all about trying to find vast generalizations of quadratic reciprocity that will address the far more general question about the degrees of irreducible factors of arbitrary polynomials. But perhaps it is more general still — at the time of writing I’m not quite sure.

Actually, I think I am sure. One thing Arthur described was Artin L-functions, which are a way of packaging up the data I’ve just described. Here is the definition he gave. You start with a representation r:\mathrm{Gal}(\mathbb{E}/\mathbb{Q})\to\mathrm{GL}(N,\mathbb{C}) of the Galois group of P. For simplicity he assumed that the Galois group was actually S_n (where n is the degree of P). Then for each unramified prime p the partition of n you get can be thought of as the cycle type of a permutation and thus as a conjugacy class in S_n. The image of this conjugacy class under r is a conjugacy class in \mathrm{GL}(N,\mathbb{C}), which is denoted by c_p(r). The Artin L-function is then defined to be

\displaystyle s\mapsto\prod_{p\notin S}\mathrm{det}(1-c_p(r)p^{-s})^{-1}.

It is easy to see that the determinant is well-defined — it follows from the fact that conjugate linear maps have the same determinant.

If you expand out this product, you get a Dirichlet series, of which this is the Euler product. And Dirichlet series that have Euler products are basically L-functions. Just as the Riemann zeta function packages up lots of important information about the primes, so the Artin L-functions package up lots of important information about the fundamental problem of algebraic number theory discussed earlier.

One interesting thing that Arthur told us was that in order to do research in this area, you have to use results from many different areas. This makes it difficult to get started, so most young researchers start by scouring the textbooks for the key theorems and using them as black boxes, understanding them fully only much later.

For example, certain Riemannian manifolds are particularly important, because automorphic forms come from solutions to differential equations (based on the Laplacian) on those manifolds. Arthur didn’t tell us exactly what these “special Riemannian manifolds” were, but he did say that they corresponded to reductive algebraic groups. (An algebraic group is roughly speaking a group defined using polynomials. For example, \mathrm{SO}(n) is algebraic, because the condition of having determinant 1 is expressible as a polynomial in the entries of a matrix, and the group operation, matrix multiplication, is also a polynomial operation. What “reductive” means I don’t know.) He then said that many beginners memorize ten key theorems about reductive algebraic groups and don’t bother themselves with the proofs.

Where does Langlands come into all this? He defined some L-functions that have a formula very similar to the formula for Artin L-functions: in fact, all you have to do is replace the r in that formula with a \pi. So a lot depends on what \pi is. Apparently it’s an automorphic representation. I’m not sure what those are.

A big conjecture is that every arithmetic L-function is an automorphic L-function. This would give us a non-Abelian class field theory. (Classical class field theory studies Abelian field extensions, and can tell you things like which numbers are cubic residues mod p.)

This conjecture is a special case of Langlands’s famous principle of functoriality, which Artin described as the fundamental problem. (OK, I’ve already described something else as the fundamental problem, but this is somehow the real fundamental problem.) I can’t resist stating the problem, because it looks as though it ought to be easy. I can imagine getting hooked on it in a parallel life, because it screams out, “Think about me in the right way and I’ll drop out.” Of course, that’s a very superficial impression, and probably once one actually does think about it, one quickly loses any feeling that it should at some sufficiently deep level be easy.

The principle says this.

Conjecture. Given two groups G and G', an automorphic representation \pi' of G' and an analytic homomorphism between their dual groups

\displaystyle \rho:\hat{G'}\to\hat G

there is an automorphic representation \pi of G such that c(\pi)=\rho(c(\pi')); that is,

c_p(\pi) = \rho(c_p(\pi'))

as conjugacy classes in \hat{G}.

To me it looks like the kind of trivial-but-not-trivially-trivial statement one proves in a basic algebra course, but obviously it is far more than that.

One quite nice thing that Arthur did was to draw an extended analogy with a situation that held in physics a century or so ago. It was observed that the absorption spectra of starlight had black lines where certain frequencies were absent, and these corresponded to the wavelengths emitted by familiar elements. This suggested that the chemistry of stars was similar to the chemistry on earth. Furthermore, because these absorption spectra were red-shifted to various extents, it also suggested that the stars were moving away from us, and ultimately suggested the Big Bang theory. However, exactly why these black lines appeared was a mystery, which was not solved until the formulation of quantum mechanics.

Something like this is how Arthur sees number theory today. Automorphic forms tell us about other number-theoretic worlds. Spectra come from differential equations that are quite similar to the Schrödinger equation — in particular, they are based on Laplacians — that come from the geometry of the special Riemannian manifolds I mentioned above. But exactly how the connection between the number theory and the spectral theory works is still a mystery.

To end on a rather different note, the one other thing I got out of this excellent talk was to see Gerhard “I was at ICM2014” Paseman, of Mathoverflow fame. Later I even got to meet him, and he gave me a Mathoverflow teeshirt. I became aware of him because there were some small technical problems during the talk, and GP offered advice from the audience.

21 Responses to “ICM2014 — Jim Arthur plenary lecture”

  1. Mike Shulman Says:

    To seize on the most trivial and inconsequential aspect of this post: is it really true that 99.9% of mathematicians don’t have a business card? One of the first things I had to do upon starting my tenure-track job was to confirm all my contact info with the EA so she could order my business cards.

  2. Matheus Says:

    Actually, Avila is Brazilian.

  3. Anonymous Says:

    Wikipedia and Youtube seems to think that the name should be James Arthur.

  4. grpaseman Says:

    (In case the first take, uh, didn’t take.) Thanks for the nod to me and MathOverflow. Mariano Suarez-Alvarez and I (and doubtless many others) are grateful for your efforts here and on mathoverflow.net. (If for some reason you need another shirt, let Mariano or I know, and we’ll hook you up.) I look forward to your next MathOverflow contribution.

    Gerhard “Yes, I Was At ICM2014” Paseman, 2014.08.27

  5. David Roberts Says:

    A reductive group is essentially the algebraic group version of a compact Lie group (the nicest sort of Lie group). Over the complex numbers, they are precisely the complexifications of compact Lie groups over the reals, for instance GL(n,C) is the complexification of U(n). In the algebraic case it’s a bit more complicated.

  6. Anonymous Says:

    On a side note…why are the fields medalists lectures not available on https://www.youtube.com/user/ICM2014VOD ?

  7. Vincent Says:

    A very nice read! Just one request: could you re-order the sentence that introduces $c_p(r)$ in such a way that it is clear that $c_p(r)$ is not a determinant (i.e. a number) but rather a conjugacy class? Not to nitpick, but it took me a while to figure out and it is really needed to make sense of the rest of the post.

  8. Anonymous Says:

    These are just video of the people giving the laudation talks. I was asking why there were no videos of the medalists giving lectures. I know there was a lecture given by Bhargava, but no video on the ICM channel. Not finding any lectures of the other 3 either..

    • gowers Says:

      The videos of the lectures by the medallists have been spliced on to the laudation talks. So if you want to see the medallists, go to the laudation talks, observe that the videos last about 90 minutes, and start watching at about 30 minutes.

  9. Anonymous Says:

    Thank You! So Mirzakhani didn’t give a lecture, did she?

    • gowers Says:

      I don’t know the answer to that. Her plenary lecture was cancelled on the grounds that she would be giving a Fields medallist’s lecture, but that was scheduled for after I left and I can’t now find it on Youtube.

    • grpaseman Says:

      Her Fields medal lecture was also cancelled, with no reason announced. (I posit that it was either illness or reluctance; it was a letdown regardless of the reason.)

      Gerhard “Or Too Much Gender Publicity” Paseman, 2014.09.04

    • BQ Says:

      The word “letdown” is not appropriate here, since there must be a reason. Reluctance is not a possible reason; she is not that type of person.

  10. ashwingokhale95 Says:

    Not related to this post I assume,what makes Riemann Hypothesis one of the hardest problems mathematics?

  11. ICM 2018 at Rio, ICM Trivia Questions, the Sao Paulo General Assembly Meeting, and Other Excitements | Combinatorics and more Says:

    […] blog from Rio 2018” as he did in ICM 2010 and ICM 2014? Here is, for example,  Tim’s beautiful post on Arthur plenary lecture.  This post mentioned France’s bid for ICM2022, and indeed the choice was between two great […]

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