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Tag: Galois

Weil descent

Published January 9, 2008 by lievenlb

A classic Andre Weil-tale is his narrow escape from being shot as a Russian spy

The war was a disaster for Weil who was a conscientious objector and so wished to avoid military service. He fled to Finland, to visit Rolf Nevanlinna, as soon as war was declared. This was an attempt to avoid being forced into the army, but it was not a simple matter to escape from the war in Europe at this time. Weil was arrested in Finland and when letters in Russian were found in his room (they were actually from Pontryagin describing mathematical research) things looked pretty black. One day Nevanlinna was told that they were about to execute Weil as a spy, and he was able to persuade the authorities to deport Weil instead.

However, Weil’s wikipedia entry calls this a story too good to be true, and continues

In 1992, the Finnish mathematician Osmo Pekonen went to the archives to check the facts. Based on the documents, he established that Weil was not really going to be shot, even if he was under arrest, and that Nevanlinna probably didn’t do – and didn’t need to do – anything to save him. Pekonen published a paper on this with an afterword by Andre Weil himself. Nevanlinna’s motivation for concocting such a story of himself as the rescuer of a famous Jewish mathematician probably was the fact that he had been a Nazi sympathizer during the war. The story also appears in Nevanlinna’s autobiography, published in Finnish, but the dates don’t match with real events at all. It is true, however, that Nevanlinna housed Weil in the summer of 1939 at his summer residence Korkee at Lohja in Finland – and offered Hitler’s Mein Kampf as bedside reading.

This old spy-story gets a recent twist now that it turns out that Weil’s descent theory of tori has applications to cryptography. So far, I haven’t really defined what tori are, so let us start with some basics.

The simplest (and archetypical) example of an algebraic torus is the multiplicative group(scheme) $\mathbb{G}_m $ over a finite field $\mathbb{F}_q $ which is the affine variety

$\mathbb{V}(xy-1) \subset \mathbb{A}^2_{\mathbb{F}_q} $. that is, the $\mathbb{F}_q $ points of $\mathbb{G}_m $ are precisely the couples ${ (x,\frac{1}{x})~:~x \in \mathbb{F}_q^* } $ and so are in one-to-one correspondence with the non-zero elements of $\mathbb{F}_q $. The coordinate ring of this variety is the ring of Laurant polynomials $\mathbb{F}_q[x,x^{-1}] $ and the fact that multiplication induces a group-structure on the points of the variety can be rephrased by saying that this coordinate ring is a Hopf algebra which is just the Hopf structure on the group-algebra $\mathbb{F}_q[\mathbb{Z}] = \mathbb{F}_q[x,x^{-1}] $. This is the first indication of a connection between tori defined over $\mathbb{F}_q $ and lattices (that is free $\mathbb{Z} $-modules with an action of the Galois group $Gal(\overline{F}_q/F_q) $. In this correspondence, the multiplicative group scheme $\mathbb{G}_m $ corresponds to $\mathbb{Z} $ with the trivial action.

Now take a field extension $\mathbb{F}_q \subset \mathbb{F}_{q^n} $, is there an affine variety, defined over $\mathbb{F}_q $ whose $\mathbb{F}_q $-points are precisely the invertible elements $\mathbb{F}_{q^n}^* $? Sure! Just take the multiplicative group over $\mathbb{F}_{q^n} $ and write the elements x and y as $x = x_1 + x_2 a_2 + \ldots + x_n a_n $ (and a similar expression for y with ${ 1,a_2,\ldots,a_n }$ being a basis of $\mathbb{F}_{q^n}/\mathbb{F}_q $ and write the defning equation $xy-1 $ out, also with respect to this basis and this will then give you the equations of the desired variety, which is usually denoted by $R^1_{\mathbb{F}_{q^n}/\mathbb{F}_q} \mathbb{G}_m $ and called the Weil restriction of scalars torus.

A concrete example? Take $\mathbb{F}_9 = \mathbb{F}_3(\sqrt{-1}) $ and write $x=x_1+x_2 \sqrt{-1} $ and $y=y_1+y_2 \sqrt{-1} $, then the defining equation $xy-1 $ becomes

$~(x_1y_1-x_2y_2) + (x_1y_2-x_2y_1) \sqrt{-1} = 1 $

whence $R^1_{\mathbb{F}_9/\mathbb{F}_3} = \mathbb{V}(x_1y_1-x_2y_2-1,x_1y_2-x_2y_1) \subset \mathbb{A}^4_{\mathbb{F}_3} $, the intersection of two quadratic hypersurfaces in 4-dimensional space.

Why do we call $R^1 \mathbb{G}_m $ a _torus_? Well, as with any variety defined over $\mathbb{F}_q $ we can also look at its points over a field-extension, for example over the algebraic closure $\overline{\mathbb{F}}_q $ and then it is easy to see that

$R^1_{\mathbb{F}_{q^n}/\mathbb{F}_q} \mathbb{G}_m (\overline{\mathbb{F}}_q) = \overline{\mathbb{F}}_q^* \times \ldots \times \overline{\mathbb{F}}_q^* $ (n copies)

and such algebraic groups are called tori. (To understand terminology, the compact group corresponding to $\mathbb{C}^* \times \mathbb{C}^* $ is $U_1 \times U_1 = S^1 \times S^1 $, so a torus).

In fact, it is already the case that the $\mathbb{F}_{q^n} $ points of the restriction of scalar torus are $\mathbb{F}_{q^n}^* \times \ldots \times \mathbb{F}_{q^n}^* $ and therefore we call this field a splitting field of the torus.

This is the general definition of an algebraic torus : a torus T over $\mathbb{F}_q $ is an affine group scheme over $\mathbb{F}_q $ such that, if we extend scalars to the algebraic closure (and then it already holds for a finite extension) we get an isomorphism of affine group schemes

$T \times_{\mathbb{F}_q} \overline{\mathbb{F}}_q = \overline{\mathbb{F}}_q^* \times \ldots \times \overline{\mathbb{F}}_q^* = (\overline{\mathbb{F}}_q^*)^{n} $

in which case we call T a torus of dimension n. Clearly, the Galois group $Gal(\overline{\mathbb{F}}_q^*/\mathbb{F}_q) $ acts on the left hand side in such a way that we recover $T $ as the orbit space for this action.

Hence, anther way to phrase this is to say that an algebraic torus is the Weil descent of an action of the Galois group on the algebraic group $\overline{\mathbb{F}}_q^* \times \ldots \times \overline{\mathbb{F}}_q^* $.

Of course we can also rephrase this is more algebraic terms by looking at the coordinate rings. The coordinate ring of the algebraic group $~(\overline{\mathbb{F}}_q^*)^n $ is the group-algebra of the rank n lattice $\mathbb{Z}^n = \mathbb{Z} \oplus \ldots \oplus \mathbb{Z} $ (the free Abelian group of rank n), that is,
$\overline{\mathbb{F}}_q [ \mathbb{Z}^n ] $. Now the Galois group acts both on the field $\overline{\mathbb{F}}_q $ as on the lattice $\mathbb{Z}^n $ coming from the action of the Galois group on the extended torus $T \times_{\mathbb{F}_q} \overline{\mathbb{F}}_q $. In fact, it is best to denote this specific action on $\mathbb{Z}^n $ by $T^* $ and call $T^* $ the character group of $T $. Now, we recover the coordinate ring of the $\mathbb{F}_q $-torus $T $ as the ring of invariants

$\mathbb{F}_q[T] = \overline{\mathbb{F}}_q [T^*]^{Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)} $

Hence, the restriction of scalars torus $R^1_{\mathbb{F}_{q^n}/\mathbb{F}_q} \mathbb{G}_m $ is an n-dimensional torus over $\mathbb{F}_q $ and its corresponding character group is the free Abelian group of rank n which can be written as $\mathbb{Z}[x]/(x^n-1) = \mathbb{Z}1 \oplus \mathbb{Z}x \oplus \ldots \oplus \mathbb{Z}x^{n-1} $ and where the action of the cyclic Galois group $Gal(\mathbb{F}_{q^n}/\mathbb{F}_q) = C_n = \langle \sigma \rangle $ s such that the generator $\sigma $ as as multiplication by $x $. That is, in this case the character group is a permutation lattice meaning that the $\mathbb{Z} $-module has a basis which is permuted under the action of the Galois group. Next time we will encounter more difficult tori sich as the crypto-torus $T_n $.

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ECSTR aka XTR

Published January 5, 2008 by lievenlb

The one thing that makes it hard for an outsider to get through a crypto-paper is their shared passion for using nonsensical abbreviations. ECSTR stands for “Efficient Compact Subgroup Trace Representation” and we are fortunate that Arjen Lenstra and Eric Verheul shortened it in their paper The XTR public key system to just XTR. As both of them speak Dutch, they will know why Ive chosen a magpie-picture on the left… Btw. there is a nice MSRI-talk by Lenstra, starting off with a couple of jokes on what ECSTR is NOT meant to abbreviate (one of them being ‘Elliptic Curve Systems Too Risky’… (( I may even start to share their passion… )) ).

The XTR-system uses safety of $\mathbb{F}_{p^6} $ in the Diffie-Hellman key exchange while transmitting only $2=\phi(6) $ pits. The first question one asks is : why the jump from $N=2 $ from last time to $N=6 $? Well, remember that (conjecturally) we want to use safety of $\mathbb{F}_q $ for $q=p^N $ while using only $\phi(N) $ pits. That is, we want to have $N log(p) $ large (for safety) while at the same time $\phi(N) log(p) $ small (for efficiency). Thus, the most useful N’s to consider are those in the sequence

$N=1,~2,~6=2.3,~30=2.3.5,~210=2,3,5,7,~\ldots $

that is, the products of the first so many prime numbers. The number of elements of the cyclic group $\mathbb{F}_q^* $ is equal to

$p^6-1 = (p-1)(p+1)(p^2+p+1)(p^2-p+1) $

and that the subgroup of order $p-1 $ can be embedded in $\mathbb{F}_p^* $, that of order $p+1 $ can be embedded in $\mathbb{F}_{p^2}^* $, that of order $p^2+p+1 $ can be embedded in $\mathbb{F}_{p^3}^* $, BUT that the subgroup of order $\Phi_6(p)=p^2-p+1 $ CANNOT be embedded in any $\mathbb{F}_{p^i}^* $ for $i = 1,2,3 $, or in other words, the $p^2-p+1 $ subgroup is as hard as $\mathbb{F}_{p^6}^* $. So, let us take a generator $g $ of the subgroup $T_6 $ of order $p^2-p+1 $ and do the Diffie-Hellman trick with it in a modified manner.

Galois groups of finite fields are cyclic and generated by the Frobenius $x \mapsto x^p $. In particular, the Galois group $Gal(\mathbb{F}_{p^6}/\mathbb{F}_{p^2}) = C_3 $ is cyclic of order three and consists of the auromorphisms ${ 1=id, \sigma = (x \mapsto x^{p^2}), \sigma^2 = (x \mapsto x^{p^4}) } $, so the corresponding trace map is given by

$Tr~:~\mathbb{F}_{p^6} \rightarrow \mathbb{F}_{p^2} \qquad Tr(x) = x + x^{p^2} + x^{p^4} $

So, how do we perform our key-exchange using my secret number $a $ and yours $b $? Well, I’ll send you $Tr(g^a) $ and as this is an element of the quadratic extension $\mathbb{F}_{p^2} $ I’ll need just 2 pits instead of 6 and you will send me likewise $Tr(g^b) $. I claim that the common key we (and only we) can compute is $Tr(g^{ab}) $. How does this work?

Given any element $x \in T_6 \subset \mathbb{F}_{p^6}^* $ we can compute the 3-element set $C_x = { x,\sigma(x),\sigma(x^2) } $ and hence the characteristic polynomial
$~(t-x)(t-\sigma(x))(t-\sigma^2(x)) $

$ = t^3 – (x+\sigma(x)+\sigma^2(x))t^2 + (x \sigma(x)+ x\sigma^2(x)+\sigma(x)\sigma^2(x))t – x \sigma(x)\sigma^2(x) $

The first coefficient $x+\sigma(x)+\sigma^2(x) $ is the trace $Tr(x) $ and the second and third coefficients are respectively $Tr(x \sigma(x)) $ and the norm $N(x) $. Now, if $x \in T_6 $ one can show that

$Tr(x \sigma(x)) = Tr(x)^p $ and $N(x)=1 $

That is, from knowing only $Tr(x) $ we can compute the characteristic polynomial and hence recover the 3-element set ${ h,\sigma(h),\sigma^2(h) } $!

If I give you $Tr(g^a) $ you can compute from it the 3-set ${ g^a,\sigma(g^a),\sigma^2(g^a) } $ and raise them all the the b-th power (b being your secret number) to obtain

${ g^{ab},\sigma(g^a)^b,\sigma^2(g^a)^b } = { g^{ab},\sigma(g^{ab}),\sigma^2(g^{ab}) } $

but then you also know our shared key $Tr(g^{ab}) = g^{ab}+\sigma(g^{ab})+\sigma^2(g^{ab}) $… Done!

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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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Vacation reading

Published July 26, 2007 by lievenlb

Im in the process of writing/revising/extending the course notes for next year and will therefore pack more math-books than normal.

These are for a 3rd year Bachelor course on Algebraic Geometry and a 1st year Master course on Algebraic and Differential Geometry. The bachelor course was based this year partly on Miles Reid’s Undergraduate Algebraic Geometry and partly on David Mumford’s Red Book, but this turned out to be too heavy going. Next year I’ll be happy if they know enough on algebraic curves. The backbone of these two courses will be Fulton’s old but excellent Algebraic curves. It’s self contained (unlike Hartshorne’s book that assumes a prior course on commutative algebra), contains a lot of exercises and goes on to the Brill-Noether proof of Riemann-Roch. Still, Id like to extend it with the introductory chapter and the chapters on Riemann surfaces from Complex Algebraic Curves by Frances Kirwan, a bit on elliptic and modular functions from Elliptic curves by Henry McKean and Victor Moll and the adelic proof of Riemann-Roch and applications of it to the construction of algebraic codes from Algebraic curves over finite fields by Carlos Moreno. If time allows Id love to include also the chapter on zeta functions but I fear this will be difficult.

These are to spice up a 2nd year Bachelor course on Representations of Finite Groups with a tiny bit of Galois representations, both as motivation and to wet their appetite for elliptic curves and algebraic geometry. Ive received Fearless Symmetry by Avner Ash and Robert Gross only yesterday and find it hard to stop reading. It attempts to explain Galois representations and generalized reciprocity laws to a general audience and from what I read so far, they really do a terrific job. Another excellent elementary introduction to elliptic curves and Galois representations is in Invitation to the Mathematics of Fermat-Wiles by Yves Hellegouarch. On a gossipy note, the appendix “The origin of the elliptic approach to Fermat’s last theorem” is fun reading. Finally, Ill also take Introduction to Fermat’s Last Theorem by Alf van der Poorten along simply because I love his writing style.

These are included just for fun. The Poincare Conjecture by Donal O’Shea because I know far too little about it, Letters to a Young Mathematician by Ian Stewart because I like the concept of the book and finally The sensual (quadratic) form by John Conway because I need to have at all times at least one Conway-book nearby.

Continue readingVacation reading
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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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permutation representations of monodromy groups

Published March 20, 2007 by lievenlb

Today we will explain how curves defined over
$\overline{\mathbb{Q}} $ determine permutation representations
of the carthographic groups. We have seen that any smooth projective
curve $C $ (a Riemann surface) defined over the algebraic
closure $\overline{\mathbb{Q}} $ of the rationals, defines a
_Belyi map_ $\xymatrix{C \ar[rr]^{\pi} & & \mathbb{P}^1} $ which is only ramified over the three points
$\\{ 0,1,\infty \\} $. By this we mean that there are
exactly $d $ points of $C $ lying over any other point
of $\mathbb{P}^1 $ (we call $d $ the degree of
$\pi $) and that the number of points over $~0,1~ $ and
$~\infty $ is smaller than $~d $. To such a map we
associate a _dessin d\’enfant_, a drawing on $C $ linking the
pre-images of $~0 $ and $~1 $ with exactly $d $
edges (the preimages of the open unit-interval). Next, we look at
the preimages of $~0 $ and associate a permutation
$\tau_0 $ of $~d $ letters to it by cycling
counter-clockwise around these preimages and recording the edges we
meet. We repeat this procedure for the preimages of $~1 $ and
get another permutation $~\tau_1 $. That is, we obtain a
subgroup of the symmetric group $ \langle \tau_0,\tau_1
\rangle \subset S_d $ which is called the monodromy
group of the covering $\pi $.

For example, the
dessin on the right is
associated to a degree $8 $ map $\mathbb{P}^1 \rightarrow
\mathbb{P}^1 $ and if we let the black (resp. starred) vertices be
the preimages of $~0 $ (respectively of $~1 $), then the
corresponding partitions are $\tau_0 = (2,3)(1,4,5,6) $
and $\tau_1 = (1,2,3)(5,7,8) $ and the monodromy group
is the alternating group $A_8 $ (use
GAP ).

But wait! The map is also
ramified in $\infty $ so why don\’t we record also a
permutation $\tau_{\infty} $ and are able to compute it from
the dessin? (Note that all three partitions are needed if we want to
reconstruct $C $ from the $~d $ sheets as they encode in
which order the sheets fit together around the preimages). Well,
the monodromy group of a $\mathbb{P}^1 $ covering ramified only
in three points is an epimorphic image of the fundamental
group of the sphere
minus three points $\pi_1(\mathbb{P}^1 – { 0,1,\infty
}) $ That is, the group of all loops beginning and
ending in a basepoint upto homotopy (that is, two such loops are the
same if they can be transformed into each other in a continuous way
while avoiding the three points).

This group is generated by loops
$\sigma_i $ running from the basepoint to nearby the i-th
point, doing a counter-clockwise walk around it and going back to be
basepoint $Q_0 $ and the epimorphism to the monodromy group is given by sending

$\sigma_1 \mapsto \tau_0~\quad~\sigma_2 \mapsto
\tau_1~\quad~\sigma_3 \mapsto \tau_{\infty} $

Now,
these three generators are not independent. In fact, this fundamental
group is

$\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =
\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2
\sigma_3 = 1 \rangle $

To understand this, let us begin
with an easier case, that of the sphere minus one point. The fundamental group of the plane minus one point is
$~\mathbb{Z} $ as it encodes how many times we walk around the
point. However, on the sphere the situation is different as we can make
our walk around the point longer and longer until the whole walk is done
at the backside of the sphere and then we can just contract our walk to
the basepoint. So, there is just one type of walk on a sphere minus one
point (upto homotopy) whence this fundamental group is trivial. Next,
let us consider the sphere minus two points


Repeat the foregoing to the walk $\sigma_2 $, that
is, strech the upper part of the circular tour all over the backside of
the sphere and then we see that we can move it to fit with the walk
$\sigma_1$ BUT for the orientation of the walk! That is, if we do this
modified walk $\sigma_1 \sigma_2^{\’} $ we just made the
trivial walk. So, this fundamental group is $\langle
\sigma_1,\sigma_2~\mid~\sigma_1 \sigma_2 = 1 \rangle =
\mathbb{Z} $ This is also the proof of the above claim. For,
we can modify the third walk $\sigma_3 $ continuously so that
it becomes the walk $\sigma_1 \sigma_2 $ but
with the reversed orientation ! As $\sigma_3 =
(\sigma_1 \sigma_2)^{-1} $ this allows us to compute the
\’missing\’ permutation $\tau_{\infty} = (\tau_0
\tau_1)^{-1} $ In the example above, we obtain
$\tau_{\infty}= (1,2,6,5,8,7,4)(3) $ so it has two cycles
corresponding to the fact that the dessin has two regions (remember we
should draw ths on the sphere) : the head and the outer-region. Hence,
the pre-images of $\infty$ correspond to the different regions of the
dessin on the curve $C $. For another example,
consider the degree 168 map

$K \rightarrow \mathbb{P}^1 $

which is the modified orbit map for the action of
$PSL_2(\mathbb{F}_7) $ on the Klein quartic.
The corresponding dessin is the heptagonal construction of the Klein
quartic

Here, the pre-images of 1 correspond to the midpoints of the
84 edges of the polytope whereas the pre-images of 0 correspond to the
56 vertices. We can label the 168 half-edges by numbers such that
$\tau_0 $ and $\tau_1 $ are the standard generators b
resp. a of the 168-dimensional regular representation (see the atlas
page ).
Calculating with GAP the element $\tau_{\infty} = (\tau_0
\tau_1)^{-1} = (ba)^{-1} $ one finds that this permutation
consists of 24 cycles of length 7, so again, the pre-images of
$\infty $ lie one in each of the 24 heptagonal regions of the
Klein quartic. Now, we are in a position to relate curves defined
over $\overline{Q} $ via their Belyi-maps and corresponding
dessins to Grothendiecks carthographic groups $\Gamma(2) $,
$\Gamma_0(2) $ and $SL_2(\mathbb{Z}) $. The
dessin gives a permutation representation of the monodromy group and
because the fundamental group of the sphere minus three
points $\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =
\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2
\sigma_3 = 1 \rangle = \langle \sigma_1,\sigma_2
\rangle $ is the free group op two generators, we see that
any dessin determines a permutation representation of the congruence
subgroup $\Gamma(2) $ (see this
post where we proved that this
group is free). A clean dessin is one for which one type of
vertex has all its valancies (the number of edges in the dessin meeting
the vertex) equal to one or two. (for example, the pre-images of 1 in
the Klein quartic-dessin or the pre-images of 1 in the monsieur Mathieu
example ) The corresponding
permutation $\tau_1 $ then consists of 2-cycles and hence the
monodromy group gives a permutation representation of the free
product $C_{\infty} \ast C_2 =
\Gamma_0(2) $ Finally, a clean dessin is said to be a
quilt dessin if also the other type of vertex has all its valancies
equal to one or three (as in the Klein quartic or Mathieu examples).
Then, the corresponding permutation has order 3 and for these
quilt-dessins the monodromy group gives a permutation representation of
the free product $C_2 \ast C_3 =
PSL_2(\mathbb{Z}) $ Next time we will see how this lead
Grothendieck to his anabelian geometric approach to the absolute Galois
group.

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The best rejected proposal ever

Published March 7, 2007 by lievenlb


The Oscar in
the category The Best Rejected Research Proposal in Mathematics
(ever) goes to … Alexander Grothendieck
for his proposal Esquisse d’un Programme, Grothendieck\’s research program from 1983, written as
part of his application for a position at the CNRS, the French
equivalent of the NSF. An English translation is
available.

Here is one of the problems discussed :
Give TWO non-trivial elements of
$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ the _absolute_
Galois group of the algebraic closure of the rational numbers
$\overline{\mathbb{Q}} $, that is the group of all
$\mathbb{Q} $-automorphisms of $\overline{\mathbb{Q}} $. One element
most of us can give (complex-conjugation) but to find any other
element turns out to be an extremely difficult task.

To get a handle on
this problem, Grothendieck introduced his _’Dessins d’enfants’_
(Children’s drawings). Recall from last session the pictures of the
left and right handed Monsieur Mathieu

The left hand side drawing was associated to a map
$\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}} $ which was
defined over the field $\mathbb{Q} \sqrt{-11} $ whereas the right side
drawing was associated to the map given when one applies to all
coefficients the unique non-trivial automorphism in the Galois group
$Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) $ (which is
complex-conjugation). Hence, the Galois group
$Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) $ acts _faithfully_ on the
drawings associated to maps $\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} \rightarrow
\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} $ which are ramified only over
the points ${ 0,1,\infty } $.

Grothendieck’s idea was to
extend this to more general maps. Assume that a projective smooth curve
(a Riemann surface) X is defined over the algebraic numbers
$\overline{\mathbb{Q}} $ and assume that there is a map $X
\rightarrow \mathbb{P}^1_{\mathbb{C}} $ ramified only over the points
${ 0,1,\infty } $, then we can repeat the procedure of last time and
draw a picture on X consisting of d edges (where d is the degree
of the map, that is the number of points lying over another point of
$\mathbb{P}^1_{\mathbb{C}} $) between white resp. black points (the
points of X lying over 1 (resp. over 0)).

Call such a drawing a
‘dessin d\’enfant’ and look at the collection of ALL dessins
d’enfants associated to ALL such maps where X runs over ALL curves
defined over $\overline{\mathbb{Q}} $. On this set, there is an action
of the absolute Galois group
$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ and if this action would be
faithful, then this would give us insight into this
group. However, at that time even the existence of a map $X \rightarrow
\mathbb{P}^1 $ ramified in the three points ${ 0,1,\infty } $
seemed troublesome to prove, as Grothendieck recalls in his proposal

In more erudite terms, could it be true that
every projective non-singular algebraic curve defined over a number
field occurs as a possible ‚ modular curve‚ parametrising
elliptic curves equipped with a suitable rigidification? Such a
supposition seemed so crazy that I was almost embarrassed to submit
it to the competent people in the domain. Deligne when I consulted
him found it crazy indeed, but didn’t have any counterexample up
his sleeve. Less than a year later, at the International Congress in
Helsinki, the Soviet mathematician Bielyi announced exactly that result,
with a proof of disconcerting simplicity which fit into two little
pages of a letter of Deligne ‚ never, without a doubt, was such a
deep and disconcerting result proved in so few lines!

In
the form in which Bielyi states it, his result essentially says that
every algebraic curve defined over a number field can be obtained as
a covering of the projective line ramified only over the points 0,
1 and infinity. This result seems to have remained more or less
unobserved. Yet, it appears to me to have considerable importance. To
me, its essential message is that there is a profound identity
between the combinatorics of finite maps on the one hand, and the
geometry of algebraic curves defined over number fields on the
other. This deep result, together with the algebraic- geometric
interpretation of maps, opens the door onto a new, unexplored world within reach of all, who pass by without seeing it.

Belyi’s proof is indeed relatively easy
(full details can be found in the paper Dessins d’enfants on the
Riemann sphere by Leila
Schneps). Roughly it goes as follows : as both X and the map are
defined over $\overline{\mathbb{Q}} $ the map is only ramified over
(finitely many) $\overline{\mathbb{Q}} $-points. Let S be the finite
set of all Galois-conjugates of these points and consider the polynomial

$f_0(z_0) = \prod_{s \in S} (z_0 -s) \in
\mathbb{Q}[z_0] $

Now, do a
resultant trick. Consider the
polynomial $f_1(z_1) = Res_{z_0}(\frac{d f_0}{d
z_0},f_0(z_0)-z_1) $ then the roots of $f_1(z_1) $ are exactly the
finite critical values of $f_0 $, $f_1 $ is again defined over
$\mathbb{Q} $ and has lower degree (in $z_1 $) than $f_0 $ in $z_1 $.
Continue this trick a finite number of times untill you have constructed
a polynomial $f_n(z_n) \in \mathbb{Q}[z_n] $ of degree zero.

Composing
the original map with the maps $f_j $ in succession yields that all
ramified points of this composition are
$\mathbb{Q} $-points! Now, we only have to limit the number of
these ramified $\mathbb{Q} $-points (let us call this set T) to three.

Take any three elements of T, then there always exist integers $m,n
\in \mathbb{Z} $ such that the three points go under a linear
fractional transformation (a Moebius-function associated to a matrix in
$PGL_2(\mathbb{Q}) $) to ${ 0,\frac{m}{m+n},1 } $. Under the
transformation $z \rightarrow \frac{(m+n)^{m+n}}{m^m
n^n}z^m(1-z)^n $ the points 0 and 1 go to 0 and
$\frac{m}{m+n} $ goes to 1 whence the ramified points of the
composition are one less in number than T. Continuing in this way we
can get the set of ramified $\mathbb{Q} $-points of a composition at
most having three elements and then a final Moebius transformation gets
them to ${ 0,1,\infty } $, done!

As a tribute for this clever
argument, maps $X \rightarrow \mathbb{P}^1 $ ramified only in 0,1 and
$\infty $ are now called Belyi morphisms. Here is an example of
a Belyi-morphism (and the corresponding dessin d’enfants) associated to
one of the most famous higher genus curves around : the Klein
quartic (if you haven’t done
so yet, take your time to go through this marvelous pre-blog post by
John Baez).

One can define the Klein quartic as the plane projective
curve K with defining equation in
$\mathbb{P}^2_{\\mathbb{C}} $ given by $X^3Y+Y^3Z+Z^3X = 0 $ K has
a large group of automorphism, namely the simple group of order
168 $G = PSL_2(\mathbb{F}_7) =
SL_3(\mathbb{F}_2) $ It is a classical fact (see for example
the excellent paper by Noam Elkies The Klein quartic in number theory) that the quotient map $K \rightarrow K/G =
\mathbb{P}^1_{\mathbb{C}} $ is ramified only in the points
0,1728 and $\infty $ and the number of points of K lying over them
are resp. 56, 84 and 24. Now, compose this map with the Moebius
transormation taking ${ 0,1728,\infty } \rightarrow { 0,1,\infty } $
then the resulting map is a Belyi-map for the Klein quartic. A
topological construction of the Klein quartic is fitting 24 heptagons
together so that three meet in each vertex, see below for the gluing
data-picture in the hyperbolic plane : the different heptagons are given
a number but they appear several times telling how they must fit
together)

The resulting figure has exactly $\frac{7 \times 24}{2} =
84 $ edges and the 84 points of K lying over 1 (the white points in
the dessin) correspond to the midpoints of the edges. There are exactly
$\frac{7 \times 24}{3}=56 $ vertices corresponding to the 56 points
lying over 0 (the black points in the dessin). Hence, the dessin
d\’enfant associated to the Klein quartic is the figure traced out by
the edges on K. Giving each of the 168 half-edges a
different number one assigns to the white points a permutation of order
two and to the three-valent black-points a permutation of order three,
whence to the Belyi map of the Klein quartic corresponds a
168-dimensional permutation representation of $SL_2(\mathbb{Z}) $,
which is not so surprising as the group of automorphisms is
$PSL_2(\mathbb{F}_7) $ and the permutation representation is just the
regular representation of this group.

Next time we will see how
one can always associate to a curve defined over
$\overline{\mathbb{Q}} $ a permutation representation (via the Belyi
map and its dessin) of one of the congruence subgroups $\Gamma(2) $ or
$\Gamma_0(2) $ or of $SL_2(\mathbb{Z}) $ itself.

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