I am waiting impatiently for a more refined approach that will include meta-statistics:

Did you know? France has never lost a game where a single US-based newspaper presented more than three human-interest stories concerning the opposing team players in the three days before the game.

Did you know? It is the first time since the invention of the personal computer that the BBC has listed more than 243 statistical facts about a game.

Did you know? Three out of four statistical facts about the Italy-Switzerland game have involved numbers larger than thirteen.

These will be the days…

]]>Of course, even without being present, the written text of the seminar is always available later to learn what the talks were about. But, especially when the subject is not close to something I know, it’s often much better to have seen first a one-hour presentation which distills the most important information, which may well be a bit hidden in the written text for non-specialists. So it is rather wonderful to see that, since last March, the lectures of the Bourbaki seminar are recorded and made available on Youtube on the channel of the Institut Henri Poincaré.

Here is the playlist for the March seminars, with lectures by Golse (on Bolztmann’s equation), Bolthausen (spin glasses), Benoist (curves on K3 surfaces) and myself (guess…), and here that of last Saturday, with lectures by A. Valette (the Kadison-Singer problem), Smulevici (general relativity), Hales (formal proofs) and Coquand (dependent type theory and univalent foundations; with Voevodsky, who was present, explaining at the end his choice of the word “univalent”).

The links to the videos (as well as links to the texts) can also be found on the web page of Bourbaki.

(There was also mention that the seminar was streamed as it happened, but I don’t know if that’s really the case; if yes, will fashionable bars in Manhattan and Princeton start opening at 4 A.M. on selected Saturdays to offer croissants, coffee, cognac, and the Bourbaki talks?)

]]>But no! The actual question is to compute times ! We must correct this! But it’s just as easy without starting from scratch: we turn the “plus” cross a quarter turn on the left-hand side:

and then switch the digits on the right-hand side:

This is a fun little random fact about integers and decimal expansions, certainly.

But there’s a bit more to it than that: it is in fact independent of the choice of base , in the sense that if we pick any other integer , and consider base expansions, then we also have

as well as

(where we underline individual digits in base expansion.)

At this point it is natural to ask if there are any other Léo-pairs to base , i.e., pairs of digits in base such that the base expansions of the sum and the product of and are related by switching the two digits (where we always get two digits in the result by viewing a one-digit result as ).

It turns out that, whatever the base , the only such pairs are and the “degenerate” case .

To see this, there are two cases: either the addition leads to a carry, or not.

If it does, this means that where . The sum is then

So this is a Léo-pair if and only if

This equation, in terms of and , becomes

which holds if and only if . Since the factors are integers and non-negative, this is only possible if , which means , the solution found by Léo.

Now suppose there is no carry. This means that we have and . Then

and we have a Léo-pair if and only if

i.e., if and only if .

This is not an uninteresting little equation! For a fixed (which could now be any non-zero rational), this defines a simple quadratic curve. Without the restrictions on the size of the solution , there is always a point on this curve, namely

This does not fit our conditions, of course. But we can use it to find all other integral solutions, as usual for quadratic curves. First, any line through intersects the curve in a a second point, which has rational coordinates if the line is also defined by rational coefficients, and conversely.

Doing this, some re-arranging and checking leads to the parameterization

of the rational solutions to , where is an arbitrary non-zero rational number. In this case, this can also be found more easily by simply writing the equation in the form

Now assume that is an integer, and we want to be integers. This holds if and only if is an integer such that .

Such solutions certainly exist, but do they satisfy the digit condition? The answer is yes if and only if , which means , giving the expected degenerate pair. Indeed, to have , the parameter must be a negative divisor of . We write with positive. Then to have non-negative digits, we must have

the first one of these inequalities means , while the second means that …

where the Royal Society has a conference center. This was quite a fun occasion, and not only because the curtains had one the most interesting decorative pattern I have seen during years of extensive study:

The talks were only 30 minutes long; because of this and the absence of blackboards, they were all beamer talks, and I’ve uploaded my slides here. They might be useful even for participants in the meeting, since I only had time to cover the first 60% or so of what I had planned, which was a survey of my most recent work with É. Fouvry and Ph. Michel (available here, and which I hope to discuss in more detail in a later post)…

]]>Can anybody make a guess of what is the first line? It is

What is the **** historical perspective?

but I can’t read the missing fourth word!

And what is the Kierkergaardian idea, really?

]]>However, we can also think of automorphic representations as points in the unitary dual of the relevant adélic group , and this has a natural topology (the Fell topology), for which it then makes sense to ask whether the set of all cuspidal automorphic representations is discrete or not.

The two notions might well be unrelated: for instance, if we take the group and the direct sum over of the principal series with parameter , we obtain a representation containing only “discrete” spectrum, but parameterized by a “non-discrete” subset of the unitary dual.

It is nevertheless true that the discrete spectrum is discrete in the unitary dual, in the automorphic case. I am sure this is well-known. (Unless the topology on the unitary dual is much weirder than I expect; my expertise in this respect is quite limited, but I remember asking A. Venkatesh who told me that the pathologies of the unitary dual topology are basically irrelevant as far as the automorphic case is concerned). I think this must be well-known, but I don’t remember seeing it mentioned anywhere (hence this post…)

Here is the argument, at least over number fields, and for . Let be a given cuspidal representation. The unitary dual topology on the adélic group is, if I understand right, the restricted direct product topology with respect to the unramified spectrum. So a neighborhood of is determined by a finite set of places and corresponding neighborhoods of for . We want to find such a neighborhood in which is the unique automorphic cuspidal representation.

First, we can fix a neighborhood of the archimedean (Langlands) parameters that is relatively compact. Next, we note (or claim…) that for a finite place , the exponent of the conductor is locally constant, so we get neighborhoods of , with the unramified spectrum when is unramified at , such that all automorphic representations in

have arithmetic conductor equal to that of . With the archimedean condition, it follows that the Iwaniec-Sarnak analytic conductor of cuspidal representations in is *bounded*.

However, a basic “height-like” property is that there are only finitely many cuspidal representations with bounded analytic conductor (this is proved by Michel-Venkatesh in Section 2.6.5 of this paper, and a different proof based on spherical codes, as suggested by Venkatesh, is due to Brumley). Thus almost isolates , in the sense that only finitely many other cuspidal representations can lie in . Denote by this set of representations.

Now, the unitary dual is (or should be…) at least minimally separated so that, for any two cuspidal representations and , there is an open set which contains and not , say . Then

is an open neighborhood of which only contains the cuspidal representation …

I had in mind a case where the groups involved are finite, or close enough, so that the reverse inclusion is indeed obviously true. But it is amusing to see that in fact what I wrote is correct in all cases: if is a subgroup of an arbitrary group and satisfies , then in fact : taking the inverse of the inclusion gives

I find this interesting because, when it comes to the normalizer, the analogue of this fact is not true: the condition is not, in general, equivalent with .

(I found this explained in an exercise in Bourbaki, Algèbre, Chapitre I, p. 134, Exercice 27, or page 146 of the English edition; here is a simple case of the counterexample from that exercise: consider the group of permutations of ; consider the subgroup which is the pointwise stabilizer of , and the element which is just

Then we have , because the left-hand side is the pointwise stabilizer of which is a subset of . But is not equal to , because is the pointwise stabilizer of , and there are elements in which fix but not .)

It seems natural to ask: what exactly is the set of all such that is equivalent to ? This set contains the line for , and also the coordinates axes and (since we then deal with cosets of ).

In fact, one can determine : it is the complement of the line , for (which corresponds to conjugation). Indeed, suppose for instance that . Let and be such that .

We can write

and

Since , *and since* is the common exponent of and at the two extremities of the right-hand side, it follows that this right-hand side is contained in .

A similar argument works for , using

where again it is crucial that the coefficient appears on both sides, and that it is .

Since we already know that , and in fact that for (the same setting as the counterexample above works, because the we used is an -th power for every ), we have therefore determined the set …

]]>This certainly can happen: the two obvious cases are when , or when is an involution that happens to be in the normalizer of .

In fact the general case is just a tweak of this last case: we have if and only if and , or in other words, if belongs to the normalizer, and is an involution modulo .

This is of course easy to check. I then asked myself: what about the possibility that

,

where and are arbitrary integers? Can one classify when this happens? The answer is another simple exercise (that I will probably use when I teach Algeba I next semester): this is the case if and only if and , where is the gcd of and . In particular, for all pairs where and are coprime, the condition above implies that belongs to the normalizer of .

Here is the brief argument: having fixed , let

This set is easily seen to be a subgroup of . Furthermore, note that implies that , which in turns means that and .

Hence if , we get

,

so that

contains , which is just

where . But means exactly that .

Thus we have got the first implication. Conversely, the conclusion means exactly that

But then

shows that .

To finish, how did I get to this situation? This can arise quite naturally as follows: one has a collection of representations of a fixed group , and an action of a group latex on these representations (action up to isomorphism really).

For a given representation , we can then define a group

and also a subset

where denotes the contragredient representation.

It can be that is empty, but let us assume it is not. Then has two properties: (1) it is a coset of (because acts on simply transitively); (2) we have for all (because the contragredient of the contragredient is the representation itself).

This means that, for some , we have , and furthermore

since for all .

By the previous discussion, we therefore get a good handle on the structure of : either it is empty, or it is of the form for some such that and normalizes in . In particular, if is trivial (which happens often), either is empty, or it consists of a single element which is an involution of .

]]>Il a publié il y a deux ans (…) un ouvrage relatif au sentiment de l’Infini sur la rive occidentale du lac Victoria-Nyanza et cette année un opuscule moins important, mais conduit d’une plume alerte, parfois même acérée, sur le fusil à répétition dans l’armée bulgare, qui l’ont mis tout à fait hors de pair.

or, in translation:

He has published two years ago (…) a book concerning the feeling of Infinity on the occidental shore of the Victoria-Nyanza lake, and this year another booklet, less important but written with a lively, and even piercing, pen, on the repeating rifle in the bulgarian army, which have made him rather peerless.

This reminds of the curiously little-known hilarious “Antrobus stories” of diplomatic mishaps:

It was during one of those long unaccountable huffs between ourselves and the Italians. You know the obscure vendettas which break out between Missions? Often they linger on long after the people who threw the first knife have been posted away. I have no idea how this huff arose. I simply inherited it from bygone dips whose bones were now dust. It was in full swing when I arrived — everyone applying freezing-mixture to the Italians and getting the Retort Direct in exchange. (…) So while bows were still exchanged for protocol reasons they were only, so to speak, from above the waist. A mere contortion of the dickey, if you take me, as a tribute to manners. A slight Inclination, accompanied by a

moue. Savage work, old lad, savage work!

(from “The game’s the thing”, where a soccer game between the English and Italian embassies rather degenerates.)

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