Kowalski – Page 79 – E. Kowalski's blog

a long the… riverrun

If you had to write a circular novel, that can be started at any point of the narrative and that would circle back to itself, how long would it be? Probably, most mathematicians would say that either 63, 628, or 6283, pages would be the most appropriate.

And so it’s probably not surprising that the one circular novel I know, Joyce’s “Finnegans wake“, is indeed (in all editions I’ve seen) 628 pages long. Amusingly, this is not mentioned in the introduction to the one I have (which is, also, about the only thing I’ve read of the book, with the exception of the first few and last paragraphs, and isolated snatches here and there).

The reason for this post is really that today is Bloomsday, the day when, fictionally, the action of the earlier “Ulysses” evolves, and that Zürich is well-connected with Joyce. Not only is he buried there, very close to the zoo, so I can use one appropriate quotation

As the lion in our teargarten remembers the nenuphars of his Nile…

from “Finnegans wake” (one of those few snatches I have actually read), but he also wrote parts of “Ulysses” in a house on Universitätstrasse, quite close to the main building of ETH Zürich where I work, as memorialized by the plaque below:

(I’ve heard that there are a few other such places in Zürich, the reason being that, unable to pay the rent, Joyce had to move frequently…)

Tidbits of terminology and other folklore

Two small and independent interrogative remarks:

(1) Nowadays, an extension of a group G by a group K is a group H fitting in a short exact sequence

$1\rightarrow K\rightarrow H\rightarrow G\rightarrow 1$

in other words, and rather counterintuitively, the group G is a quotient of the group which is the extension. When did this terminology originate? A paper of Alan Turing (entitled, rather directly, “The extensions of a group”) defines “extension”, in the very first paragraph, exactly in the opposite (naïve) way, quoting Schreier and Baer who, presumably, had the same convention.

(2) There’s a whole lot of discussion here and there about the mystical “field with one element”; usually, papers of Tits from around 1954 are mentioned as being the source of the whole “idea”; however, the following earlier quote from a 1951 paper of R. Steinberg (“A geometric approach to the representations of the full linear group over a Galois field”, 1951, p. 279, TAMS 71, 274–282) seems to also contain a germ of the often mentioned analogy between the formulas for the order of the Weyl groups and those of groups of Lie type over a field with q elements, the former being obtained by specializing the latter for q=1:

In closing this section, a remark on the analogy between G and H seems to be in order. Instead of considering G as a group of linear transformations of a vector space, we could consider G as a collineation group of a finite (n-1)-dimensional geometry. If q=1, the vector space fails to exist but the finite geometry does exist and, in fact, reduces to the n vertices of a simplex with a collineation group isomorphic to H. “

In this citation, G is GL(n,F_q), and H is the symmetric group on n letters.

(3) Here’s a third question: when did the terminology “Galois field” become more or less obsolete within the pure mathematics community?

Yet another property of quadratic fields with extra units

Here is an amusing exercise that is suitable for a course on basic algebraic number theory: let p be a prime number. Consider integral solutions (a,f) to

$4p=a^2+3f^2$

with a and f positive. The claim is that, if p is congruent to one mod 3, there are three distinct solutions (a,f), (b,g), (c,h), and if they are ordered

$1\leq a<b<c$

then we have

$c=a+b$

For example, in the case p=541, we find a=17, b=29, c=46:

$4p=2164=17^2+3\times 5^5=29^2+3^3\times 7^2=46^2+3\times 2^4$

The pedagogic value of the exercise is that while it looks like something that one could prove by a simple brute force computation, this is not so easy to do, while it becomes elementary knowing the basic facts about factorizations of primes in quadratic fields (and units of imaginary quadratic fields, in this case Q(√-3).

Indeed, the equation means that p is the norm of the integral ideal generated by

$\frac{a}{2}+ \frac{f\sqrt{-3}}{2}$

It is known that only primes congruent to 1 modulo 3 are norms in this field, hence the first condition on p. Then, it is known that the ideal above is unique up to conjugation. So the only possible extra solutions, given one of them, are obtained by multiplying by a unit of the field, and isolating the “coefficient of 1”, or in other words taking the trace to Q. Since the units are

$\pm 1,\, \pm j=\frac{\mp 1\pm \sqrt{-3}}{2},\, \pm j^2$

simply multiplying shows there are three positive solutions:

$a,\,\frac{a+3f}{2},\,\frac{|a-3f|}{2}$

Depending on the sign of a-3f, the conclusion follows by considering two cases.

[This minor property of quadratic fields was motivated by the question of finding interesting examples of relations between the zeros of zeta functions of algebraic curves over finite fields; for quite a bit more about this – both results of independence and examples of relations -, see my preprint on the subject, in particular Section 6 for the examples.]

Équidistribution, or équirépartition ?

For years, I have been convinced that the proper French translation of “equidistribution” was not the faux ami (false friend) “équidistribution”, but rather the word “équirépartition”. The latter is for instance used by Serre (and Bourbaki).

But then I realized recently that Deligne uses “équidistribution” in his great paper containing his second proof of the Riemann Hypothesis over finite fields, which contains in particular his famous equidistribution theorem (see Section 3.5, entitled “Application: théorèmes d’équidistribution”).

Since, in fact, neither word appears in the French dictionaries I have available (unsurprisingly: “equidistribution” is not in the OED), and since moreover “distribution” and “répartition” do appear and are identified as synonyms, it seems now that in fact both words should be acceptable…

There is no hyperbolic Minkowski theorem

Minkowski’s classic theorem of “geometry of numbers” states that any convex subset of Rⁿ which is symmetric (with respect to the origin) and of volume (with respect to Lebesgue measure) larger than 2ⁿ contains a non-zero integral point.

This theorem is used, in particular, in the classical treatment of Dirichlet’s Unit Theorem in algebraic number theory. While teaching this topic last year, I wondered whether there was an hyperbolic analogue, in the following sense, where H is the hyperbolic plane in the Poincaré model:

does there exist a constant C such that any geodesically convex subset B of the hyperbolic plane H with hyperbolic area at least C which is geodesically symmetric with respect to the point i contains at least one point z of the form g.i, where g is an element of SL(2,Z) and g.i refers to the usual action by fractional linear transformations, with z not equal to i.

Here, the subset B is geodesically convex if it contains the geodesic segment between any two points, and symmetric if, whenever x is in B, the point on the geodesic from i to x which is at distance d(i,x) from i, but in the opposite direction, is also in B.

It turns out that the answer is “No”. Indeed, C. Bavard gave the following example:

let B be a euclidean half-cone with base vertex at 0, axis the vertical axis, and angle at the origin small enough, then B does not contain any “integral” point except i, but has infinite hyperbolic area. Moreover, it is easily seen that B is convex and symmetric in the hyperbolic sense, since hyperbolic geodesics are vertical half-lines and half-circles meeting the real line at right angles.

To check the claim, it is enough to show that for any integral point z=g.i distinct from i, the ratio |x|/y has a positive lower bound, where z=x+iy (this will show that the angle from the vertical axis is bounded from below, so the point is not in a cone like the one above with sufficiently small angle). But z is given by (ai+b)/(ci+d) with a, b, c, d integers and ad-bc=1, and this ratio is simply |ac+bd|. Being an integer, either it is 0, or it is at least 1. Manipulating things, one checks that the first case only occurs for matrices in SL(2,Z) which are orthogonal matrices, and those fix i, so the point is then z=i. Hence, except for this case, the ratio is at least 1 and this concludes the argument.

It is interesting to see what breaks down in the (very simple) proofs of Minkowski’s theorem in the plane. In the first proof found on page 33 of the 5th edition of Hardy and Wright’s “An introduction to the theory of numbers” (visible here), the problem is that there is no way to dilate the convex region B in a homogeneous way compatible with the SL(2,Z) action. In other words, SL(2,Z) is essentially a maximal discrete subgroup of SL(2,R) (maybe it is maximal? I can’t find a reference).