Tag: modular

Anabelian vs. Noncommutative Geometry

Published December 12, 2007 by lieven

This is how my attention was drawn to what I have since termed
anabelian algebraic geometry, whose starting point was exactly a study
(limited for the moment to characteristic zero) of the action of absolute
Galois groups (particularly the groups $Gal(\overline{K}/K) $, where K is an extension of finite type of the prime field) on (profinite) geometric fundamental
groups of algebraic varieties (defined over K), and more particularly (breaking with a well-established tradition) fundamental groups which are very far
from abelian groups (and which for this reason I call anabelian). Among
these groups, and very close to the group $\hat{\pi}_{0,3} $ , there is the profinite compactification of the modular group $SL_2(\mathbb{Z}) $, whose quotient by its centre
$\{ \pm 1 \} $ contains the former as congruence subgroup mod 2, and can also be
interpreted as an oriented cartographic group, namely the one classifying triangulated oriented maps (i.e. those whose faces are all triangles or
monogons).

The above text is taken from Alexander Grothendieck‘s visionary text Sketch of a Programme. He was interested in the permutation representations of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ as they correspond via Belyi-maps and his own notion of dessins d’enfants to smooth projective curves defined over $\overline{\mathbb{Q}} $. One can now study the action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ on these curves and their associated dessins. Because every permutation representation of $\Gamma $ factors over a finite quotient this gives an action of the absolute Galois group as automorphisms on the profinite compactification

$\hat{\Gamma} = \underset{\leftarrow}{lim}~\Gamma/N $

where the limit is taken over all finite index normal subgroups $N \triangleleft PSL_2(\mathbb{Z}) $. In this way one realizes the absolute Galois group as a subgroup of the outer automorphism group of the profinite group $\hat{\Gamma} $. As a profinite group is a compact topological group one should study its continuous finite dimensional representations which are precisely those factoring through a finite quotient. In the case of $\hat{\Gamma} $ the simple continuous representations $\mathbf{simp}_c~\hat{\Gamma} $ are precisely the components of the permutation representations of the modular group. So in a sense, anabelian geometry is the study of these continuous simples together wirth the action of the absolute Galois group on it.

In noncommutative geometry we are interested in a related representation theoretic problem. We would love to know the simple finite dimensional representations $\mathbf{simp}~\Gamma $ of the modular group as this would give us all simples of the three string braid group $B_3 $. So a natural question presents itself : how are these two ‘geometrical’ objects $\mathbf{simp}_c~\hat{\Gamma} $ (anabelian) and $\mathbf{simp}~\Gamma $ (noncommutative) related and can we use one to get information about the other?

This is all rather vague so far, so let us work out a trivial case to get some intuition. Consider the profinite completion of the infinite Abelian group

$\hat{\mathbb{Z}} = \underset{\leftarrow}{lim}~\mathbb{Z}/n\mathbb{Z} = \prod_p \hat{\mathbb{Z}}_p $

As all simple representations of an Abelian group are one-dimensional and because all continuous ones factor through a finite quotient $\mathbb{Z}/n\mathbb{Z} $ we see that in this case

$\mathbf{simp}_c~\hat{\mathbb{Z}} = \mu_{\infty} $

is the set of all roots of unity. On the other hand, the simple representations of $\mathbb{Z} $ are also one-dimensional and are determined by the image of the generator so

$\mathbf{simp}~\mathbb{Z} = \mathbb{C} – { 0 } = \mathbb{C}^* $

Clearly we have an embedding $\mu_{\infty} \subset \mathbb{C}^* $ and the roots of unity are even dense in the Zariski topology. This might look a bit strange at first because clearly all roots of unity lie on the unit circle which ‘should be’ their closure in the complex plane, but that’s because we have a real-analytic intuition. Remember that the Zariski topology of $\mathbb{C}^*$ is just the cofinite topology, so any closed set containing the infinitely many roots of unity should be the whole space!

Let me give a pedantic alternative proof of this (but one which makes it almost trivial that a similar result should be true for most profinite completions…). If $c $ is the generator of $\mathbb{Z} $ then the different conjugacy classes are precisely the singletons $c^n $. Now suppose that there is a polynomial $a_0+a_1x+\ldots+a_mx^m $ vanishing on all the continuous simples of $\hat{\mathbb{Z}} $ then this means that the dimensions of the character-spaces of all finite quotients $\mathbb{Z}/n\mathbb{Z} $ should be bounded by $m $ (for consider $x $ as the character of $c $), which is clearly absurd.

Hence, whenever we have a finitely generated group $G $ for which there is no bound on the number of irreducibles for finite quotients, then morally the continuous simple space for the profinite completion

$\mathbf{simp}_c~\hat{G} \subset \mathbf{simp}~G $

should be dense in the Zariski topology on the noncommutative space of simple finite dimensional representations of $G $. In particular, this should be the case for the modular group $PSL_2(\mathbb{Z}) $.

There is just one tiny problem : unlike the case of $\mathbb{Z} $ for which this space is an ordinary (ie. commutative) affine variety $\mathbb{C}^* $, what do we mean by the “Zariski topology” on the noncommutative space $\mathbf{simp}~PSL_2(\mathbb{Z}) $ ? Next time we will clarify what this might be and show that indeed in this case the subset

$\mathbf{simp}_c~\hat{\Gamma} \subset \mathbf{simp}~\Gamma $

will be a Zariski closed subset!

more iguanodons via kfarey.sage

Published December 11, 2007 by lieven

For what it is worth, Ive computed some more terms in the iguanodon series. Here they are

$L_2(7),M_{12},A_{16},M_{24},A_{28},A_{40},A_{48},A_{60},A_{68},A_{88},A_{96},A_{120},A_{132},A_{148},A_{164},A_{196},\ldots $

By construction, the n-th iguanodon group $Ig_n $ (corresponding to the n-th Farey sequence) is a subgroup of the alternating group on its (half)legs. Hence to prove that all remaining iguanodons are alternating groups boils down to proving that they are sufficiently transitive, for example, by showing that there are permutations of certain cycle-types in the group. Im sure any grouptheorist can crack this problem over lunch, so if you did please drop a comment.

Clearly, I didnt do the calculations in the archaic way of the previous post (as depicted on the left) which consisted in adding a pair of new legs at the proper place in the spine for every new Farey number, write down the two generating permutations, giving them to GAP and check simplicity and the isomorphism type.

Instead I used a nice SAGE-package to compute with Farey-symbols written by Chris Kurth and available from his website. As this package is a good tool to experiment hunting for other dinosaur-series of simple groups coming from series of Farey-symbols, Ill include the details for $Ig_3 $ (the example used to outline the construction of the Iguanodon-series ).

First we need to have the n-th Farey-sequence $F(n) $. There are several short Python programs around to do this, for example this one from the Python-Cookbook. Save it to your sage-directory and name it fareyseq.py and load it into sage via load fareyseq.py. Then typing farey(3) to the sage-prompt spits back

sage: farey(3)
[(1, 3), (1, 2), (2, 3)]

That is, 0 and 1 are not included and Farey-numbers are represented by numerator-denominator couples. The iguanodon-series uses the Fareys upto 1/2, identifies the edges connecting 0 and 1 to $\infty $ and makes all other intervals odd. That is, the corresponding Farey symbol for F(3) is

[tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{\bullet} & \frac{1}{3} \ar@{-}[r]_{\bullet} & \frac{1}{2} \ar@{-}[r]_{\bullet} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

(to add to the confusion, I denote odd intervals by a black-bullet whereas in Kulkarni’s paper they are white…) Anyway, get Kurth’s kfarey-package and save the folder as kfarey in your sage-folder. Kurth uses the following notation for Farey-symbols

The Farey Symbol is a list [a,b,p] where
a is a list of numerators, b a list of denominators, and p the pairing
information. If x[i]=a[i]/b[i]:
inf x0 x1 x2 ... xn inf
  p0 p1 p2 ... pn pn+1
So p[i] is the pairing of the side between x[i-1] and x[i]. The p[i]’s can be
positive integers, indicating pairing between sides, or -2 or -3, meaning
an even or odd pairing respectively.

The above Farey-symbol is therefore represented as
[[0,1,1,1],[1,3,2,1],[1,-3,-3,-3,1]]. The kfarey-function LRCosetRep(F) returns two permutations L and R giving the permutation action of the two generators of the modular group $PSL_2(\mathbb{Z}) $

$~~~L = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}~\qquad \text{and} \qquad~R = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $

on the half-legs of the inguanodon (the dessin corresponding to the Farey-symbol). Here’s the sage transcript

sage: load kfarey/farey.sage
sage: load kfarey/conggroups.sage
sage: load kfarey/LR.sage
sage: ig3=[[0,1,1,1],[1,3,2,1],[1,-3,-3,-3,1]]
sage: LRCosetRep(ig3)
[(1,2,3,9,10,11,6,7,8,4,5)(12), (1,8,4,2,11,6,3,12,10,7,5)(9)]

Giving these two generators to GAP one verifies that they indeed generate $M_{12} $

gap> ig3:=Group((1,2,3,9,10,11,6,7,8,4,5)(12), (1,8,4,2,11,6,3,12,10,7,5)(9));
Group([ (1,2,3,9,10,11,6,7,8,4,5), (1,8,4,2,11,6,3,12,10,7,5) ])
gap> IsSimpleGroup(ig3);
true
gap> IsomorphismTypeInfoFiniteSimpleGroup(ig3);
rec( series := “Spor”, name := “M(12)” )

kfarey has plenty of other useful functions. One can even create an .eps file of the fundamental domain specified by the subgroup of the modular group encoded by the Farey symbol using MakeEpsFile(F). For the above example it returns the picture on the right. Not quite as nice as the one on the left, but surely a lot easier to create.

problema bovinum

Published December 3, 2007 by lieven

Suppose for a moment that some librarian at the Bodleian Library announces that (s)he discovered an old encrypted book attributed to Isaac Newton. After a few months of failed attempts, the code is finally cracked and turns out to use a Public Key system based on the product of two gigantic prime numbers, $2^{32582657}-1 $ and $2^{30402457}-1 $, which were only discovered to be prime recently. Would one deduce from this that Newton invented public key cryptography and that he used alchemy to factor integers? (( Come to think of it, some probably would ))

The cynic in me would argue that it is a hell of a coincidence for this text to surface exactly at the moment in history when we are able to show these numbers to be prime and understand their cryptographic use, and conclude that the book is likely to be a fabrication. Still, stranger things have happened in the history of mathematics…

In 1773, Gotthold Ephraim Lessing at that time librarian at the Herzog-August-Bibliothek discovered and published a Greek epigram in 22 elegiac couplets. The manuscript describes a problem sent by Archimedes to the mathematicians in Alexandria.

In his beautiful book “Number Theory, an approach through history. From Hammurapi to Legendre” Andre Weil asserts (( Chapter I,IX )):

Many mathematical epigrams are known. Most of them state problems of little depth; not so Lessing’s find; there is indeed every reason to accept the attribution to Archimedes, and none for putting it into doubt.

This Problema Bovidum (the cattle problem) is a surprisingly difficult diophantine problem and the simplest complete solution consists of eigth numbers, each having about 206545 digits. As we will see later the final ingredient in the solution is the solution of Pell’s equation using continued fractions discovered by Lagrange in 1768 and published in 1769 in a long memoir. Lagrange’s solution to the Pell equation was inserted in Euler’s “Algebra” which was composed in 1771 but published only in 1773… the very same year as Lessing’s discovery! (( all dates learned from Weil’s book Chp. III,XII ))

Weil’s book doesn’t include the details of the original epigram. The (lost) archeologist in me wanted to see the original Greek 22 couplets as well as a translation. So here they are : (( thanks to the Cattle problem site ))

A PROBLEM

which Archimedes solved in epigrams, and which he communicated to students of such matters at Alexandria in a letter to Eratosthenes of Cyrene.

If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who once upon a time grazed on the fields of the Thrinacian isle of Sicily, divided into four herds of different colours, one milk white, another a glossy black, a third yellow and the last dappled. In each herd were bulls, mighty in number according to these proportions: Understand, stranger, that the white bulls were equal to a half and a third of the black together with the whole of the yellow, while the black were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow. Observe further that the remaining bulls, the dappled, were equal to a sixth part of the white and a seventh, together with all of the yellow. These were the proportions of the cows: The white were precisely equal to the third part and a fourth of the whole herd of the black; while the black were equal to the fourth part once more of the dappled and with it a fifth part, when all, including the bulls, went to pasture together. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd. Finally the yellow were in number equal to a sixth part and a seventh of the white herd. If thou canst accurately tell, O stranger, the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each colour, thou wouldst not be called unskilled or ignorant of numbers, but not yet shalt thou be numbered among the wise.

But come, understand also all these conditions regarding the cattle of the Sun. When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth, and the plains of Thrinacia, stretching far in all ways, were filled with their multitude. Again, when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure, there being no bulls of other colours in their midst nor none of them lacking. If thou art able, O stranger, to find out all these things and gather them together in your mind, giving all the relations, thou shalt depart crowned with glory and knowing that thou hast been adjudged perfect in this species of wisdom.

The Lessing epigram may very well be an extremely laborious hoax but it is still worth spending a couple of posts on it. It gives us the opportunity to retell the amazing history of Pell’s problem rangingfrom the ancient Greeks and Indians, over Fermat and his correspondents, to Euler and Lagrange (with a couple of recent heroes entering the story). And, on top of this, the modular group is all the time just around the corner…