## Meta-statistics

One of the most unfortunate developments of modern football, and a clear symptom of the decline of civilization, is the regrettable irruption in the comments of a deluge of factoids that manage to simultaneously give a bad name to team sports and to statistics (“This is the first time in twenty-one competitive games played in the Southern Hemisphere that a French Number 10 player’s backward pass from the left foot has been intercepted by a Dutch player born in Amsterdam”). Roger Couderc certainly did not need this to make a game come to life (of course, technically, he was advantaged by the fact that he was commenting the much greater game of Rugby).

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…

## Bourbaki seminar online

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?)

## Leo’s first theorem

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 $b\leq d$

## Chicheley Hall meeting

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)…

## Is the Kierkegaardian idea true? and other queries

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?