# E. Kowalski's blog

Comments on mathematics, mostly.

26.06.2014

### Bourbaki seminar online

Filed under: France,Mathematics @ 11:42

Whenever I can, I like to attend the Bourbaki Seminar in Paris, but that’s not always feasible.

June 2014

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?)

11.06.2014

### Leo’s first theorem

Filed under: Exercise,Mathematics @ 11:46

I learnt the following from my son Léo: the teacher asks to compute $9+9$; that’s easy
$9\ +\ 9\ =\ 18.$
But no! The actual question is to compute $9$ times $9$! 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:
$9\ \times\ 9$
and then switch the digits on the right-hand side:
$9\ \times\ 9\ =\ 81.$

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 $10$, in the sense that if we pick any other integer $b\geq 2$, and consider base $b$ expansions, then we also have

$(b-1)\ +\ (b-1)\ =\ 2b-2\ =\ b+(b-2)= \underline{1}\,\underline{b-2}$

as well as

$(b-1)\ \times\ (b-1)\ =\ (b-1)^2=b(b-2)+1= \underline{b-2}\,\underline{1},$

(where we underline individual digits in base $b$ expansion.)

At this point it is natural to ask if there are any other Léo-pairs $(x,y)$ to base $b$, i.e., pairs of digits in base $b$ such that the base $b$ expansions of the sum and the product of $x$ and $y$ are related by switching the two digits (where we always get two digits in the result by viewing a one-digit result $z$ as $\underline{0}\, \underline{z}$).

It turns out that, whatever the base $b$, the only such pairs are $(b-1,b-1)$ and the “degenerate” case $(0,0)$.

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

If it does, this means that $y=b-z$ where $x>z$. The sum is then

$x+y=b+(x-z)=\underline{1}\,\underline{x-z}.$

So this is a Léo-pair if and only if

$xy=\underline{x-z}\,\underline{1}.$

This equation, in terms of $x$ and $z$, becomes

$x(b-z)=b(x-z)+1,$

which holds if and only if $z(b-x)=1$. Since the factors are integers and non-negative, this is only possible if $z=b-x=1$, which means $x=y=b-1$, the solution found by Léo.

Now suppose there is no carry. This means that we have $0\leq x,y\leq b-1$ and $x+y\leq b-1$. Then
$x+y=\underline{0}\,\underline{x+y},$
and we have a Léo-pair if and only if
$xy=\underline{x+y}\,\underline{0},$
i.e., if and only if $xy=b(x+y)$.

This is not an uninteresting little equation! For a fixed $b$ (which could now be any non-zero rational), this defines a simple quadratic curve. Without the restrictions on the size of the solution $(x,y)$, there is always a point on this curve, namely
$(x_0,y_0)=(2b,2b).$
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 $(x_0,y_0)$ 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

$\begin{cases} x=b+k\\ y=b+\frac{b^2}{k}\end{cases}$

of the rational solutions to $xy=b(x+y)$, where $k$ 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
$0=xy-bx-by=(x-b)(y-b)-b^2\ldots$

Now assume that $b\geq 2$ is an integer, and we want $(x,y)$ to be integers. This holds if and only if $k$ is an integer such that $k\mid b^2$.

Such solutions certainly exist, but do they satisfy the digit condition? The answer is yes if and only if $k=-b$, which means $x=y=0$, giving the expected degenerate pair. Indeed, to have $x, the parameter $k$ must be a negative divisor of $k^2$. We write $k=-d$ with $d\mid k^2$ positive. Then to have non-negative digits, we must have
$\begin{cases} x=b-d\geq 0\\ y=b-\frac{b^2}{d}\geq 0\end{cases},$
the first one of these inequalities means $b\geq d$, while the second means that $b\leq d$

06.06.2014

### Chicheley Hall meeting

Filed under: Mathematics,Travel @ 20:40

Just after the end of the semester last week, I went to a short meeting organized by J. Keating, Z. Rudnick and T. Wooley, in the stately English house of Chicheley Hall,

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)…

18.05.2014

### Is the Kierkegaardian idea true? and other queries

Filed under: Language,Literature,Mathematics @ 20:22

In February, I was invited to give talks in Bristol and Oxford, and I spent the night after the second talk in a guest room of Worcester College. While looking at the brochure explaining the history of this college, I noticed that a previous guest had left a cryptic inscription, which I took a photograph of:

Kierkegaard

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?

11.05.2014

### The discrete spectrum is discrete

Filed under: Exercise,Mathematics @ 19:06

No, this post is not an exercise in tautological reasoning: the point is that the word “discrete” is relatively overloaded. In the theory of automorphic forms, “discrete spectrum” (or “spectre discret”) is the same as “cuspidal spectrum”, and refers to those automorphic representations (of a given group $G$ over a given global field $F$) which are realized as closed subspaces of the relevant representation space on $L^2(G(\mathbf{Q})\backslash G(\mathbf{A}_F))$, by opposition with the “continuous spectrum”, whose components may fail to be actual subrepresentations.

However, we can also think of automorphic representations as points in the unitary dual of the relevant adélic group $G(\mathbf{A}_F)$, 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 $\mathrm{SL}_2(\mathbf{R})$ and the direct sum over $t\in\mathbf{Q}$ of the principal series with parameter $t$, 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 $\mathrm{GL}_n$. Let $\pi$ 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 $\pi$ is determined by a finite set $S$ of places and corresponding neighborhoods $U_s$ of $\pi_s$ for $s\in S$. We want to find such a neighborhood in which $\pi$ is the unique automorphic cuspidal representation.

First, we can fix a neighborhood $U_{\infty}$ of the archimedean (Langlands) parameters that is relatively compact. Next, we note (or claim…) that for a finite place $v$, the exponent of the conductor is locally constant, so we get neighborhoods $U_v$ of $\pi_v$, with $U_v$ the unramified spectrum when $\pi$ is unramified at $v$, such that all automorphic representations in
$U=U_{\infty}\times \prod_{v} U_v$
have arithmetic conductor equal to that of $\pi$. With the archimedean condition, it follows that the Iwaniec-Sarnak analytic conductor of cuspidal representations in $U$ 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 $U$ almost isolates $\pi$, in the sense that only finitely many other cuspidal representations can lie in $U$. Denote by $X$ this set of representations.

Now, the unitary dual is (or should be…) at least minimally separated so that, for any two cuspidal representations $\pi_1$ and $\pi_2$, there is an open set which contains $\pi_1$ and not $\pi_2$, say $V_{\pi_1,\pi_2}$. Then
$U'=U\cap \bigcap_{\rho\in X-\{\pi\}} V_{\pi,\rho}$
is an open neighborhood of $\pi$ which only contains the cuspidal representation $\pi$

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