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Tag: Grothendieck

Klein’s dessins d’enfant and the buckyball

Published June 30, 2008 by lievenlb

We saw that the icosahedron can be constructed from the alternating group $A_5 $ by considering the elements of a conjugacy class of order 5 elements as the vertices and edges between two vertices if their product is still in the conjugacy class.

This description is so nice that one would like to have a similar construction for the buckyball. But, the buckyball has 60 vertices, so they surely cannot correspond to the elements of a conjugacy class of $A_5 $. But, perhaps there is a larger group, somewhat naturally containing $A_5 $, having a conjugacy class of 60 elements?

This is precisely the statement contained in Galois’ last letter. He showed that 11 is the largest prime p such that the group $L_2(p)=PSL_2(\mathbb{F}_p) $ has a (transitive) permutation presentation on p elements. For, p=11 the group $L_2(11) $ is of order 660, so it permuting 11 elements means that this set must be of the form $X=L_2(11)/A $ with $A \subset L_2(11) $ a subgroup of 60 elements… and it turns out that $A \simeq A_5 $…

Actually there are TWO conjugacy classes of subgroups isomorphic to $A_5 $ in $L_2(11) $ and we have already seen one description of these using the biplane geometry (one class is the stabilizer subgroup of a ‘line’, the other the stabilizer subgroup of a point).

Here, we will give yet another description of these two classes of $A_5 $ in $L_2(11) $, showing among other things that the theory of dessins d’enfant predates Grothendieck by 100 years.

In the very same paper containing the first depiction of the Dedekind tessellation, Klein found that there should be a degree 11 cover $\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}} $ with monodromy group $L_2(11) $, ramified only in the three points ${ 0,1,\infty } $ such that there is just one point lying over $\infty $, seven over 1 of which four points where two sheets come together and finally 5 points lying over 0 of which three where three sheets come together. In 1879 he wanted to determine this cover explicitly in the paper “Ueber die Transformationen elfter Ordnung der elliptischen Funktionen” (Math. Annalen) by describing all Riemann surfaces with this ramification data and pick out those with the correct monodromy group.




He manages to do so by associating to all these covers their ‘dessins d’enfants’ (which he calls Linienzuges), that is the pre-image of the interval [0,1] in which he marks the preimages of 0 by a bullet and those of 1 by a +, such as in the innermost darker graph on the right above. He even has these two wonderful pictures explaining how the dessin determines how the 11 sheets fit together. (More examples of dessins and the correspondences of sheets were drawn in the 1878 paper.)

The ramification data translates to the following statements about the Linienzuge : (a) it must be a tree ($\infty $ has one preimage), (b) there are exactly 11 (half)edges (the degree of the cover),
(c) there are 7 +-vertices and 5 o-vertices (preimages of 0 and 1) and (d) there are 3 trivalent o-vertices and 4 bivalent +-vertices (the sheet-information).

Klein finds that there are exactly 10 such dessins and lists them in his Fig. 2 (left). Then, he claims that one the two dessins of type I give the correct monodromy group. Recall that the monodromy group is found by giving each of the half-edges a number from 1 to 11 and looking at the permutation $\tau $ of order two pairing the half-edges adjacent to a +-vertex and the order three permutation $\sigma $ listing the half-edges by cycling counter-clockwise around a o-vertex. The monodromy group is the group generated by these two elements.

Fpr example, if we label the type V-dessin by the numbers of the white regions bordering the half-edges (as in the picture Fig. 3 on the right above) we get
$\sigma = (7,10,9)(5,11,6)(1,4,2) $ and $\tau=(8,9)(7,11)(1,5)(3,4) $.

Nowadays, it is a matter of a few seconds to determine the monodromy group using GAP and we verify that this group is $A_{11} $.

Of course, Klein didn’t have GAP at his disposal, so he had to rule out all these cases by hand.

gap> g:=Group((7,10,9)(5,11,6)(1,4,2),(8,9)(7,11)(1,5)(3,4));
Group([ (1,4,2)(5,11,6)(7,10,9), (1,5)(3,4)(7,11)(8,9) ])
gap> Size(g);
19958400
gap> IsSimpleGroup(g);
true

Klein used the fact that $L_2(11) $ only has elements of orders 1,2,3,5,6 and 11. So, in each of the remaining cases he had to find an element of a different order. For example, in type V he verified that the element $\tau.(\sigma.\tau)^3 $ is equal to the permutation (1,8)(2,10,11,9,6,4,5)(3,7) and consequently is of order 14.

Perhaps Klein knew this but GAP tells us that the monodromy group of all the remaining 8 cases is isomorphic to the alternating group $A_{11} $ and in the two type I cases is indeed $L_2(11) $. Anyway, the two dessins of type I correspond to the two conjugacy classes of subgroups $A_5 $ in the group $L_2(11) $.

But, back to the buckyball! The upshot of all this is that we have the group $L_2(11) $ containing two classes of subgroups isomorphic to $A_5 $ and the larger group $L_2(11) $ does indeed have two conjugacy classes of order 11 elements containing exactly 60 elements (compare this to the two conjugacy classes of order 5 elements in $A_5 $ in the icosahedral construction). Can we construct the buckyball out of such a conjugacy class?

To start, we can identify the 12 pentagons of the buckyball from a conjugacy class C of order 11 elements. If $x \in C $, then so do $x^3,x^4,x^5 $ and $x^9 $, whereas the powers ${ x^2,x^6,x^7,x^8,x^{10} } $ belong to the other conjugacy class. Hence, we can divide our 60 elements in 12 subsets of 5 elements and taking an element x in each of these, the vertices of a pentagon correspond (in order) to $~(x,x^3,x^9,x^5,x^4) $.

Group-theoretically this follows from the fact that the factorgroup of the normalizer of x modulo the centralizer of x is cyclic of order 5 and this group acts naturally on the conjugacy class of x with orbits of size 5.

Finding out how these pentagons fit together using hexagons is a lot subtler… and in The graph of the truncated icosahedron and the last letter of Galois Bertram Kostant shows how to do this.



Fix a subgroup isomorphic to $A_5 $ and let D be the set of all its order 2 elements (recall that they form a full conjugacy class in this $A_5 $ and that there are precisely 15 of them). Now, the startling observation made by Kostant is that for our order 11 element $x $ in C there is a unique element $a \in D $ such that the commutator$~b=[x,a]=x^{-1}a^{-1}xa $ belongs again to D. The unique hexagonal side having vertex x connects it to the element $b.x $which belongs again to C as $b.x=(ax)^{-1}.x.(ax) $.

Concluding, if C is a conjugacy class of order 11 elements in $L_2(11) $, then its 60 elements can be viewed as corresponding to the vertices of the buckyball. Any element $x \in C $ is connected by two pentagonal sides to the elements $x^{3} $ and $x^4 $ and one hexagonal side connecting it to $\tau x = b.x $.

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The F_un folklore

Published June 7, 2008 by lievenlb

All esoteric subjects have their own secret (sacred) texts. If you opened the Da Vinci Code (or even better, the original The Holy blood and the Holy grail) you will known about a mysterious collection of documents, known as the “Dossiers secrets“, deposited in the Bibliothèque nationale de France on 27 April 1967, which is rumoured to contain the mysteries of the Priory of Sion, a secret society founded in the middle ages and still active today…

The followers of F-un, for $\mathbb{F}_1 $ the field of one element, have their own collection of semi-secret texts, surrounded by whispers, of which they try to decode every single line in search of enlightenment. Fortunately, you do not have to search the shelves of the Bibliotheque National in Paris, but the depths of the internet to find them as huge, bandwidth-unfriendly, scanned documents.

The first are the lecture notes “Lectures on zeta functions and motives” by Yuri I. Manin of a course given in 1991.

One can download a scanned version of the paper from the homepage of Katia Consani as a huge 23.1 Mb file. Of F-un relevance is the first section “Absolute Motives?” in which

“…we describe a highly speculative picture of analogies between arithmetics over $\mathbb{F}_q $ and over $\mathbb{Z} $, cast in the language reminiscent of Grothendieck’s motives. We postulate the existence of a category with tensor product $\times $ whose objects correspond not only to the divisors of the Hasse-Weil zeta functions of schemes over $\mathbb{Z} $, but also to Kurokawa’s tensor divisors. This neatly leads to teh introduction of an “absolute Tate motive” $\mathbb{T} $, whose zeta function is $\frac{s-1}{2\pi} $, and whose zeroth power is “the absolute point” which is teh base for Kurokawa’s direct products. We add some speculations about the role of $\mathbb{T} $ in the “algebraic geometry over a one-element field”, and in clarifying the structure of the gamma factors at infinity.” (loc.cit. p 1-2)

I’d welcome links to material explaining this section to people knowing no motives.

The second one is the unpublished paper “Cohomology determinants and reciprocity laws : number field case” by Mikhail Kapranov and A. Smirnov.

This paper features in blog-posts at the Arcadian Functor, in John Baez’ Weekly Finds and in yesterday’s post at Noncommutative Geometry.

You can download every single page (of 15) as a separate file from here. But, in order to help spreading the Fun-gospel, I’ve made these scans into a single PDF-file which you can download as a 2.6 Mb PDF. In the introduction they say :

“First of all, it is an old idea to interpret combinatorics of finite sets as the $q \rightarrow 1 $ limit of linear algebra over the finite field $\mathbb{F}_q $. This had lead to frequent consideration of the folklore object $\mathbb{F}_1 $, the “field with one element”, whose vector spaces are just sets. One can postulate, of course, that $\mathbf{spec}(\mathbb{F}_1) $ is the absolute point, but the real problem is to develop non-trivial consequences of this point of view.”

They manage to deduce higher reciprocity laws in class field theory within the theory of $\mathbb{F}_1 $ and its field extensions $\mathbb{F}_{1^n} $. But first, let us explain how they define linear algebra over these absolute fields.

Here is a first principle : in doing linear algebra over these fields, there is no additive structure but only scalar multiplication by field elements. So, what are vector spaces over the field with one element? Well, as scalar multiplication with 1 is just the identity map, we have that a vector space is just a set. Linear maps are just set-maps and in particular, a linear isomorphism of a vector space onto itself is a permutation of the set. That is, linear algebra over $\mathbb{F}_1 $ is the same as combinatorics of (finite) sets.

A vector space over $\mathbb{F}_1 $ is just a set; the dimension of such a vector space is the cardinality of the set. The general linear group $GL_n(\mathbb{F}_1) $ is the symmetric group $S_n $, the identification via permutation matrices (having exactly one 1 in every row and column)

Some people prefer to view an $\mathbb{F}_1 $ vector space as a pointed set, the special element being the ‘origin’ $0 $ but as $\mathbb{F}_1 $ doesnt have a zero, there is also no zero-vector. Still, in later applications (such as defining exact sequences and quotient spaces) it is helpful to have an origin. So, let us denote for any set $S $ by $S^{\bullet} = S \cup { 0 } $. Clearly, linear maps between such ‘extended’ spaces must be maps of pointed sets, that is, sending $0 \rightarrow 0 $.

The field with one element $\mathbb{F}_1 $ has a field extension of degree n for any natural number n which we denote by $\mathbb{F}_{1^n} $ and using the above notation we will define this field as :

$\mathbb{F}_{1^n} = \mu_n^{\bullet} $ with $\mu_n $ the group of all n-th roots of unity. Note that if we choose a primitive n-th root $\epsilon_n $, then $\mu_n \simeq C_n $ is the cyclic group of order n.

Now what is a vector space over $\mathbb{F}_{1^n} $? Recall that we only demand units of the field to act by scalar multiplication, so each ‘vector’ $\vec{v} $ determines an n-set of linear dependent vectors $\epsilon_n^i \vec{v} $. In other words, any $\mathbb{F}_{1^n} $-vector space is of the form $V^{\bullet} $ with $V $ a set of which the group $\mu_n $ acts freely. Hence, $V $ has $N=d.n $ elements and there are exactly $d $ orbits for the action of $\mu_n $ by scalar multiplication. We call $d $ the dimension of the vectorspace and a basis consists in choosing one representant for every orbits. That is, $~B = { b_1,\ldots,b_d } $ is a basis if (and only if) $V = { \epsilon_n^j b_i~:~1 \leq i \leq d, 1 \leq j \leq n } $.

So, vectorspaces are free $\mu_n $-sets and hence linear maps $V^{\bullet} \rightarrow W^{\bullet} $ is a $\mu_n $-map $V \rightarrow W $. In particular, a linear isomorphism of $V $, that is an element of $GL_d(\mathbb{F}_{1^n}) $ is a $\mu_n $ bijection sending any basis element $b_i \rightarrow \epsilon_n^{j(i)} b_{\sigma(i)} $ for a permutation $\sigma \in S_d $.

An $\mathbb{F}_{1^n} $-vectorspace $V^{\bullet} $ is a free $\mu_n $-set $V $ of $N=n.d $ elements. The dimension $dim_{\mathbb{F}_{1^n}}(V^{\bullet}) = d $ and the general linear group $GL_d(\mathbb{F}_{1^n}) $ is the wreath product of $S_d $ with $\mu_n^{\times d} $, the identification as matrices with exactly one non-zero entry (being an n-th root of unity) in every row and every column.

This may appear as a rather sterile theory, so let us give an extremely important example, which will lead us to our second principle for developing absolute linear algebra.

Let $q=p^k $ be a prime power and let $\mathbb{F}_q $ be the finite field with $q $ elements. Assume that $q \cong 1~mod(n) $. It is well known that the group of units $\mathbb{F}_q^{\ast} $ is cyclic of order $q-1 $ so by the assumption we can identify $\mu_n $ with a subgroup of $\mathbb{F}_q^{\ast} $.

Then, $\mathbb{F}_q = (\mathbb{F}_q^{\ast})^{\bullet} $ is an $\mathbb{F}_{1^n} $-vectorspace of dimension $d=\frac{q-1}{n} $. In other words, $\mathbb{F}_q $ is an $\mathbb{F}_{1^n} $-algebra. But then, any ordinary $\mathbb{F}_q $-vectorspace of dimension $e $ becomes (via restriction of scalars) an $\mathbb{F}_{1^n} $-vector space of dimension $\frac{e(q-1)}{n} $.

Next time we will introduce more linear algebra definitions (including determinants, exact sequences, direct sums and tensor products) in the realm the absolute fields $\mathbb{F}_{1^n} $ and remarkt that we have to alter the known definitions as we can only use the scalar-multiplication. To guide us, we have the second principle : all traditional results of linear algebra over $\mathbb{F}_q $ must be recovered from the new definitions under the vector-space identification $\mathbb{F}_q = (\mathbb{F}_q^{\ast})^{\bullet} = \mathbb{F}_{1^n} $ when $n=q-1 $. (to be continued)

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un-doing the Grothendieck?

Published January 23, 2008 by lievenlb

(via the Arcadian Functor) At the time of the doing the Perelman-post someone rightfully commented that “making a voluntary retreat from the math circuit to preserve one’s own well-being (either mental, physical, scientific …)” should rather be called doing the Grothendieck as he was the first to pull this stunt.

On Facebook a couple of people have created the group The Petition for Alexander Grothendieck to Return from Exile. As you need to sign-up to Facebook to use this link and some of you may not be willing to do so, let me copy the description.

Alexander Grothendieck was born in Berlin, Germany on March 28, 1928. He was one of the most important and enigmatic mathematicians of the 20th century. After a lengthy and very productive career, highlighted by the awarding of the Fields Medal and the Crafoord Prize (the latter of which he declined), Grothendieck disappeared into the French countryside and ceased all mathematical activity. Grothendieck has lived in self-imposed exile since 1991.

We recently spotted Grothendieck in the “Gentleman’s Choice” bar in Montreal, Quebec. He was actually a really cool guy, and we spoke with him for quite some time. After a couple of rounds (on us) we were able to convince him to return from exile, under one stipulation – we created a facebook petition with 1729 mathematician members!

If 1729 mathematicians join this group, then Alexander Grothendieck will return from exile!!

1729 being of course the taxicab-curve number. The group posts convincing photographic evidence (see above) for their claim, has already 201 members (the last one being me) and has this breaking news-flash

Last week Grothendieck, or “the ‘Dieck” as we affectionately refer to him, returned to Montreal for a short visit to explain some of the theories he has been working on over the past decade. In particular, he explained how he has generalised the theory of schemes even further, to the extent that the Riemann Hypothesis and a Unified Field Theory are both trivial consequences of his work.

You know what to do!

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recycled : dessins

Published December 27, 2007 by lievenlb

In a couple of days I’ll be blogging for 4 years… and I’m in the process of resurrecting about 300 posts from a database-dump made in june. For example here’s my first post ever which is rather naive. This conversion program may last for a couple of weeks and I apologize for all unwanted pingbacks it will produce.

I’ll try to convert chunks of related posts in one go, so that I can at least give them correct self-references. Today’s work consisted in rewriting the posts of my virtual course, in march of this year, on dessins d’enfants and its connection to noncommutative geometry (a precursor of what Ive been blogging about recently). These posts were available through the PDF-archive but are from now on open to the internal search-function. Here are the internal links and a short description of their contents

Besides, I’ve added a few scattered old posts, many more to follow…

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profinite groups survival guide

Published December 14, 2007 by lievenlb

Even if you don’t know the formal definition of a profinte group, you know at least one example which explains the concept : the Galois group of the algebraic numbers $Gal = Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ aka the absolute Galois group. By definition it is the group of all $\mathbb{Q} $-isomorphisms of the algebraic closure $\overline{\mathbb{Q}} $. Clearly, it is an object of fundamental importance for mathematics but in spite of this very little is known about it. For example, it obviously is an infinite group but, apart from the complex conjugation, try to give one (1!) other nontrivial element… On the other hand we know lots of finite quotients of $Gal $. For, take any finite Galois extension $\mathbb{Q} \subset K $, then its Galois group $G_K = Gal(K/\mathbb{Q}) $ is a finite group and there is a natural onto morphism $\pi_K~:~Gal \rightarrow G_K $ obtained by dividing out all $K $-automorphisms of $\overline{\mathbb{Q}} $. Moreover, all these projections fit together nicely. If we take a larger Galois extension $K \subset L $ then classical Galois theory tells us that there is a projection $\pi_{LK}~:~G_L \rightarrow G_K $ by dividing out the normal subgroup of all $K $-automorphisms of $L $ and these finite maps are compatible with those from the absolute Galois group, that is, for all such finite Galois extensions, the diagram below is commutative

[tex]\xymatrix{Gal \ar[rr]^{\pi_L} \ar[rd]_{\pi_K} & & G_L \ar[ld]^{\pi_{LK}} \\
& G_K &}[/tex]

By going to larger and larger finite Galois extensions $L $ we get closer and closer to the algebraic closure $\overline{Q} $ and hence a better and better finite approximation $G_L $ of the absolute Galois group $Gal $. Still with me? Congratulations, you just rediscovered the notion of a profinite group! Indeed, the Galois group is the projective limit

$Gal = \underset{\leftarrow}{lim}~G_L $

over all finite Galois extensions $L/\mathbb{Q} $. If the term ‘projective limit’ scares you off, it just means that all the projections $\pi_{KL} $ coming from finite Galois theory are compatible with those coming from the big Galois group as before. That’s it : profinite groups are just projective limits of finite groups.

These groups come equipped with a natural topology : the Krull topology. Again, this notion is best clarified by considering the absolute Galois group. Now that we have $Gal $ we would like to extend the classical Galois correspondence between subgroups and subfields $\mathbb{Q} \subset K \subset \overline{\mathbb{Q}} $ and between normal subgroups and Galois subfields. For each finite Galois extension $K/\mathbb{Q} $ we have a normal subgroup of finite index, the kernel $U_K=Ker(\pi_K) $ of the projection map above. Let us take the set of all $U_K $ as a fundamental system of neighborhoods of the identity element in $Gal $. This defines a topology on $Gal $ and this is the Krull topology. As every open subgroup has finite index it is clear that this turns $Gal $ into a compact topological group. Its purpose is that we can now extend the finite Galois correspondence to Krull’s Galois theorem :

There is a bijective lattice inverting Galois correspondence between the set of all closed subgroups of $Gal $ and the set of all subfields $\mathbb{Q} \subset F \subset \overline{\mathbb{Q}} $. Finite field extensions correspond in this bijection to open subgroups and the usual normal subgroup and factor group correspondences hold!

So far we had a mysterious group such as $Gal $ and reconstructed it from all its finite quotients as a projective limit. Now we can reverse the situation : suppose we have a wellknown group such as the modular group $\Gamma = PSL_2(\mathbb{Z}) $, then we can look at the set of all its normal subgroups $U $ of finite index. For each of those we have a quotient map to a finite group $\pi_U~:~\Gamma \rightarrow G_U $ and clearly if $U \subset V $ we have a quotient map of finite groups $\pi_{UV}~:~G_U \rightarrow G_V $ compatible with the quotient maps from $\Gamma $

[tex]\xymatrix{\Gamma \ar[rr]^{\pi_U} \ar[rd]_{\pi_V} & & G_U \ar[ld]^{\pi_{UV}} \\
& G_V &}[/tex]

For the family of finite groups $G_U $ and groupmorphisms $\pi_{UV} $ we can ask for the ‘best’ group mapping to each of the $G_U $ compatible with the groupmaps $G_{UV} $. By ‘best’ we mean that any other group with this property will have a morphism to the best-one such that all quotient maps are compatible. This ‘best-one’ is called the projective limit

$\hat{\Gamma} = \underset{\leftarrow}{lim}~G_U $

and as a profinite group it has again a Krull topology making it into a compact group. Because the modular group $\Gamma $ had quotient maps to all the $G_U $ we know that there must be a groupmorphism to the best-one
$\phi~:~\Gamma \rightarrow \hat{\Gamma} $ and therefore we call $\hat{\Gamma} $ the profinite compactification (or profinite completion) of the modular group.

A final remark about finite dimensional representations. Every continuous complex representation of a profinite group like the absolute Galois group $Gal \rightarrow GL_n(\mathbb{C}) $ has finite image and this is why they are of little use for people studying the Galois group as it conjecturally reduces the study of these representations to ‘just’ all representations of all finite groups. Instead they consider representations to other topological fields such as p-adic numbers $Gal \rightarrow GL_n(\mathbb{Q}_p) $ and call these Galois representations.

For people interested in Grothendieck’s dessins d’enfants, however, continuous complex representations of the profinite compactification $\hat{\Gamma} $ is exactly their object of study and via the universal map $\phi~:~\Gamma \rightarrow \hat{\Gamma} $ above we have an embedding

$\mathbf{rep}_c~\hat{\Gamma} \rightarrow \mathbf{rep}~\Gamma $

of them in all finite dimensional representations of the modular group (
and we have a similar map restricted to simple representations). I hope this clarifies a bit obscure terms in the previous post. If not, drop a comment.

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Anabelian vs. Noncommutative Geometry

Published December 12, 2007 by lievenlb

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 finite type of the prime field) on (profinite) 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 profinite compactification 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!

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group think 2

Published May 23, 2007 by lievenlb

Someone from down under commented on the group think post yesterday :

Nice post, but I might humbly suggest that there’s not much in it that anyone would disagree with. I’d be interested in your thoughts on the following:
1. While many doomed research programs have the seven symptoms you mention, so do some very promising research programs. For instance, you could argue that Grothendieck’s school did. While it did eventually explode, it remains one of the high points in the history of mathematics. But at the time, many people (Mordell, Siegel) thought it was all garbage. Indeed there was even doubt into the late eighties. Is there anything close to a necessary and sufficient condition that an outsider can use to get some idea of whether a research group is doing work that will last?
2. Pretty much everyone thinks they’re underappreciated. It’s easy to advise them to pull a Perelman because it costs you nothing. But most unappreciated researchers are unappreciated for a good reason. How can unappreciated researchers decide whether their ideas really are good or not before spending ten years of their lives finding out?

First the easy bit : the ‘do a Perelman’-sentence seems to have been misread by several people (probably due to my inadequate English). I never suggested ‘unappreciated researchers’ to pull a Perelman but rather the key figures in seemingly successful groups making outrageous claims for power-reasons. Here is what I actually wrote

An aspect of these groupthinking science-groups that worries me most of all is their making of exagerated claims to potential applications, not supported (yet) by solid proof. Short-time effect may be to attract more people to the subject and to keep doubting followers on board, but in the long term (at least if the claimed results remain out of reach) this will destroy the subject itself (and, sadly enough, also closeby subjects making no outrageous claims!). My advice to people making such claims is : do a Perelman! Rather than doing a PR-job, devote yourself for as long as it takes to prove your hopes, somewhere in splendid isolation and come back victoriously. I have a spare set of keys if you are in search for the perfect location!
Before I will try to answer both questions let me stress that this is just my personal opinion to which I attach no particular value. Sure, I will forget things and will over-stress others. You can always leave a comment if you think I did, but I will not enter a discussion. I think it is important that a person develops his or her own scientific ethic and tries to live by it. 1. Is there anything close to a necessary and sufficient condition that an outsider can use to get some idea of whether a research group is doing work that will last? Clearly, the short answer to this is “no”. Still, there are some signs an outsider might pick up to form an opinion. – What is the average age of the leading people in the group? (the lower, the better) – The percentage of talks given by young people at a typical conference of the group (the higher, the better) – The part of a typical talk in the subject spend setting up notation, referring to previous results and namedropping (the lower, the better) – The number of group-outsiders invited to speak at a typical conference (the higher, the better) – The number of self-references in a typical paper (the lower, the better) – The number of publications by the group in non-group controlled journals (the higher, the better) – The number of group-controlled journals (the lower, the better) – The readablity of survey papers and textbooks on the subject (the higher, the better) – The complexity of motivating examples not covered by competing theories (the lower, the better) – The number of subject-gurus (the higher, the better) – The number of phd-students per guru (the lower, the better) – The number of main open problems (the higher, the better) – The Erdoes-like number of a typical group-member wrt. John Conway (the lower, the better) Okay, Im starting to drift but I hope you get the point. It is not that difficult to set up your own tools to measure the amount to which a scientific group suffers from group think. Whether the group will make a long-lasting contribution is another matter which is much harder to predict. Here, I would go for questions like : – Does the theory offer a new insight into classical & central mathematical objects such as groups, curves, modular forms, Dynkin diagrams etc. ? – Does the theory offer tools to reduce the complexity of a problem or does is instead add a layer of technical complexity? That is, are they practicing mathematics or obscurification? 2. How can unappreciated researchers decide whether their ideas really are good or not before spending ten years of their lives finding out? Here is my twofold advice to all the ‘unappreciated’ : (1) be at least as critical to your own work as you are to that of others (it is likely you will find out that you are rightfully under-appreciated compared to others) and (2) enjoy the tiny tokens of appreciation because they are likely all that you will ever get. Speaking for myself, I do not feel unappreciated compared to what I did. I did prove a couple of good results to which adequate reference is given and I had a couple of crazy ideas which were ridiculed by some at the time. A silly sense of satisfaction comes from watching the very same people years later fall over each other trying to reclaim some of the credit for these ideas. Okay, it may not have the same status of recognition as a Fields medal or a plenary talk at the ICM but it is enough to put a smile on my face from time to time and to continue stubbornly with my own ideas.

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recap and outlook

Published May 7, 2007 by lievenlb

After a lengthy spring-break, let us continue with our course on noncommutative geometry and $SL_2(\mathbb{Z}) $-representations. Last time, we have explained Grothendiecks mantra that all algebraic curves defined over number fields are contained in the profinite compactification
$\widehat{SL_2(\mathbb{Z})} = \underset{\leftarrow}{lim}~SL_2(\mathbb{Z})/N $ of the modular group $SL_2(\mathbb{Z}) $ and in the knowledge of a certain subgroup G of its group of outer automorphisms. In particular we have seen that many curves defined over the algebraic numbers $\overline{\mathbb{Q}} $ correspond to permutation representations of $SL_2(\mathbb{Z}) $. The profinite compactification $\widehat{SL_2}=\widehat{SL_2(\mathbb{Z})} $ is a continuous group, so it makes sense to consider its continuous n-dimensional representations $\mathbf{rep}_n^c~\widehat{SL_2} $ Such representations are known to have a finite image in $GL_n(\mathbb{C}) $ and therefore we get an embedding $\mathbf{rep}_n^c~\widehat{SL_2} \hookrightarrow \mathbf{rep}_n^{ss}~SL_n(\mathbb{C}) $ into all n-dimensional (semi-simple) representations of $SL_2(\mathbb{Z}) $. We consider such semi-simple points as classical objects as they are determined by – curves defined over $\overline{Q} $ – representations of (sporadic) finite groups – modlart data of fusion rings in RCTF – etc… To get a feel for the distinction between these continuous representations of the cofinite completion and all representations, consider the case of $\hat{\mathbb{Z}} = \underset{\leftarrow}{lim}~\mathbb{Z}/n \mathbb{Z} $. Its one-dimensional continuous representations are determined by roots of unity, whereas all one-dimensional (necessarily simple) representations of $\mathbb{Z}=C_{\infty} $ are determined by all elements of $\mathbb{C} $. Hence, the image of $\mathbf{rep}_1^c~\hat{\mathbb{Z}} \hookrightarrow \mathbf{rep}_1~C_{\infty} $ is contained in the unit circle

and though these points are very special there are enough of them (technically, they form a Zariski dense subset of all representations). Our aim will be twofold : (1) when viewing a classical object as a representation of $SL_2(\mathbb{Z}) $ we can define its modular content (which will be the noncommutative tangent space in this classical point to the noncommutative manifold of $SL_2(\mathbb{Z}) $). In this way we will associate noncommutative gadgets to our classical object (such as orders in central simple algebras, infinite dimensional Lie algebras, noncommutative potentials etc. etc.) which give us new tools to study these objects. (2) conversely, as we control the tangentspaces in these special points, they will allow us to determine other $SL_2(\mathbb{Z}) $-representations and as we vary over all classical objects, we hope to get ALL finite dimensional modular representations. I agree this may all sound rather vague, so let me give one example we will work out in full detail later on. Remember that one can reconstruct the sporadic simple Mathieu group $M_{24} $ from the dessin d’enfant

This
dessin determines a 24-dimensional permutation representation (of
$M_{24} $ as well of $SL_2(\mathbb{Z}) $) which
decomposes as the direct sum of the trivial representation and a simple
23-dimensional representation. We will see that the noncommutative
tangent space in a semi-simple representation of
$SL_2(\mathbb{Z}) $ is determined by a quiver (that is, an
oriented graph) on as many vertices as there are non-isomorphic simple
components. In this special case we get the quiver on two points
$\xymatrix{\vtx{} \ar@/^2ex/[rr] & & \vtx{} \ar@/^2ex/[ll]
\ar@{=>}@(ur,dr)^{96} } $ with just one arrow in each direction
between the vertices and 96 loops in the second vertex. To the
experienced tangent space-reader this picture (and in particular that
there is a unique cycle between the two vertices) tells the remarkable
fact that there is **a distinguished one-parameter family of
24-dimensional simple modular representations degenerating to the
permutation representation of the largest Mathieu-group**. Phrased
differently, there is a specific noncommutative modular Riemann surface
associated to $M_{24} $, which is a new object (at least as far
as I’m aware) associated to this most remarkable of sporadic groups.
Conversely, from the matrix-representation of the 24-dimensional
permutation representation of $M_{24} $ we obtain representants
of all of this one-parameter family of simple
$SL_2(\mathbb{Z}) $-representations to which we can then perform
noncommutative flow-tricks to get a Zariski dense set of all
24-dimensional simples lying in the same component. (Btw. there are
also such noncommutative Riemann surfaces associated to the other
sporadic Mathieu groups, though not to the other sporadics…) So this
is what we will be doing in the upcoming posts (10) : explain what a
noncommutative tangent space is and what it has to do with quivers (11)
what is the noncommutative manifold of $SL_2(\mathbb{Z}) $? and what is its connection with the Kontsevich-Soibelman coalgebra? (12)
is there a noncommutative compactification of $SL_2(\mathbb{Z}) $? (and other arithmetical groups) (13) : how does one calculate the noncommutative curves associated to the Mathieu groups? (14) : whatever comes next… (if anything).
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anabelian geometry

Published March 22, 2007 by lievenlb

Last time we saw
that a curve defined over $\overline{\mathbb{Q}} $ gives rise
to a permutation representation of $PSL_2(\mathbb{Z}) $ or one
of its subgroups $\Gamma_0(2) $ (of index 2) or
$\Gamma(2) $ (of index 6). As the corresponding
monodromy group is finite, this representation factors through a normal
subgroup of finite index, so it makes sense to look at the profinite
completion of $SL_2(\mathbb{Z}) $, which is the inverse limit
of finite
groups $\underset{\leftarrow}{lim}~SL_2(\mathbb{Z})/N $
where N ranges over all normalsubgroups of finite index. These
profinte completions are horrible beasts even for easy groups such as
$\mathbb{Z} $. Its profinite completion
is

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

where the right hand side
product of p-adic integers ranges over all prime numbers! The
_absolute Galois group_
$G=Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ acts on all curves
defined over $\overline{\mathbb{Q}} $ and hence (via the Belyi
maps ans the corresponding monodromy permutation representation) there
is an action of $G $ on the profinite completions of the
carthographic groups.

This is what Grothendieck calls anabelian
algebraic geometry

Returning to the general
case, since finite maps can be interpreted as coverings over
$\overline{\mathbb{Q}} $ of an algebraic curve defined over
the prime field $~\mathbb{Q} $ itself, it follows that the
Galois group $G $ of $\overline{\mathbb{Q}} $ over
$~\mathbb{Q} $ acts on the category of these maps in a
natural way.
For instance, the operation of an automorphism
$~\gamma \in G $ on a spherical map given by the rational
function above is obtained by applying $~\gamma $ to the
coefficients of the polynomials P , Q. Here, then, is that
mysterious group $G $ intervening as a transforming agent on
topologico- combinatorial forms of the most elementary possible
nature, leading us to ask questions like: are such and such oriented
maps ‚conjugate or: exactly which are the conjugates of a given
oriented map? (Visibly, there is only a finite number of these).
I considered some concrete cases (for coverings of low degree) by
various methods, J. Malgoire considered some others ‚ I doubt that
there is a uniform method for solving the problem by computer. My
reflection quickly took a more conceptual path, attempting to
apprehend the nature of this action of G.
One sees immediately
that roughly speaking, this action is expressed by a certain
outer action of G on the profinite com- pactification of the
oriented cartographic group $C_+^2 = \Gamma_0(2) $ , and this
action in its turn is deduced by passage to the quotient of the
canonical outer action of G on the profinite fundamental group
$\hat{\pi}_{0,3} $ of
$(U_{0,3})_{\overline{\mathbb{Q}}} $ where
$U_{0,3} $ denotes the typical curve of genus 0 over the
prime field Q, with three points re- moved.
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(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 (break- ing 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 ±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).

and a bit further, on page
250

I would like to conclude this rapid outline
with a few words of commentary on the truly unimaginable richness
of a typical anabelian group such as $SL_2(\mathbb{Z}) $
doubtless the most remarkable discrete infinite group ever
encountered, which appears in a multiplicity of avatars (of which
certain have been briefly touched on in the present report), and which
from the point of view of Galois-Teichmuller theory can be
considered as the fundamental ‚building block‚ of the
Teichmuller tower
The element of the structure of
$Sl_2(\mathbb{Z}) $ which fascinates me above all is of course
the outer action of G on its profinite compactification. By
Bielyi’s theorem, taking the profinite compactifications of subgroups
of finite index of $Sl_2(\mathbb{Z}) $, and the induced
outer action (up to also passing to an open subgroup of G), we
essentially find the fundamental groups of all algebraic curves (not
necessarily compact) defined over number fields K, and the outer
action of $Gal(\overline{K}/K) $ on them at least it is
true that every such fundamental group appears as a quotient of one
of the first groups.
Taking the anabelian yoga
(which remains conjectural) into account, which says that an anabelian
algebraic curve over a number field K (finite extension of Q) is
known up to isomorphism when we know its mixed fundamental group (or
what comes to the same thing, the outer action of
$Gal(\overline{K}/K) $ on its profinite geometric
fundamental group), we can thus say that
all algebraic
curves defined over number fields are contained in the profinite
compactification $\widehat{SL_2(\mathbb{Z})} $ and in the
knowledge of a certain subgroup G of its group of outer
automorphisms!

To study the absolute
Galois group $Gal(\overline{\mathbb{\mathbb{Q}}}/\mathbb{Q}) $ one
investigates its action on dessins denfants. Each dessin will be part of
a finite family of dessins which form one orbit under the Galois action
and one needs to find invarians to see whether two dessins might belong
to the same orbit. Such invariants are called _Galois invariants_ and
quite a few of them are known.

Among these the easiest to compute
are

  • the valency list of a dessin : that is the valencies of all
    vertices of the same type in a dessin
  • the monodromy group of a dessin : the subgroup of the symmetric group $S_d $ where d is
    the number of edges in the dessin generated by the partitions $\tau_0 $
    and $\tau_1 $ For example, we have seen
    before     that the two
    Mathieu-dessins

form a Galois orbit. As graphs (remeber we have to devide each
of the edges into two and the midpoints of these halfedges form one type
of vertex, the other type are the black vertices in the graphs) these
are isomorphic, but NOT as dessins as we have to take the embedding of
them on the curve into account. However, for both dessins the valency
lists are (white) : (2,2,2,2,2,2) and (black) :
(3,3,3,1,1,1) and one verifies that both monodromy groups are
isomorphic to the Mathieu simple group $M_{12} $ though they are
not conjugated as subgroups of $S_{12} $.

Recently, new
Galois invariants were obtained from physics. In Children’s drawings
from Seiberg-Witten curves
the authors argue that there is a close connection between Grothendiecks
programme of classifying dessins into Galois orbits and the physics
problem of classifying phases of N=1 gauge theories…

Apart
from curves defined over $\overline{\mathbb{Q}} $ there are
other sources of semi-simple $SL_2(\mathbb{Z}) $
representations. We will just mention two of them and may return to them
in more detail later in the course.

Sporadic simple groups and
their representations There are 26 exceptional finite simple groups
and as all of them are generated by two elements, there are epimorphisms
$\Gamma(2) \rightarrow S $ and hence all their representations
are also semi-simple $\Gamma(2) $-representations. In fact,
looking at the list of ‘standard generators’ of the sporadic
simples

(here the conjugacy classes of the generators follow the
notation of the Atlas project) we see that all but
possibly one are epimorphic images of $\Gamma_0(2) = C_2 \ast
C_{\infty} $ and that at least 12 of then are epimorphic images
of $PSL_2(\mathbb{Z}) = C_2 \ast
C_3 $.

Rational conformal field theories Another
source of $SL_2(\mathbb{Z}) $ representations is given by the
modular data associated to rational conformal field theories.

These
representations also factor through a quotient by a finite index normal
subgroup and are therefore again semi-simple
$SL_2(\mathbb{Z}) $-representations. For a readable
introduction to all of this see chapter 6 \”Modular group
representations throughout the realm\” of the
book Moonshine beyond the monster the bridge connecting algebra, modular forms and physics by Terry
Gannon. In fact, the whole book
is a good read. It introduces a completely new type of scientific text,
that of a neverending survey paper…

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