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ceci n’est pas un corps

To Gavin Wraiht a mathematical phantom is a “nonexistent entity which ought to be there but apparently is not; but nevertheless obtrudes its effects so convincingly that one is forced to concede a broader notion of existence”. Mathematics’ history is filled with phantoms getting the kiss of life.

Nobody will deny the ancient Greek were pretty good at maths, but still they were extremely unsure about the status of zero as a number. They asked themselves, “How can nothing be something?”, and, paradoxes such as of Zeno’s depend in large part on that uncertain interpretation of zero. It lasted until the 9th century before Indian scholars were comfortable enough to treat 0 just as any other number.

Italian gamblers/equation-solvers of the early 16th century were baffled by the fact that the number of solutions to quartic equations could vary, seemingly arbitrary, from zero to four until Cardano invented ‘imaginary numbers’ and showed that there were invariably four solutions provided one allows these imaginary or ‘phantom’ numbers.

Similar paradigm shifts occurred in mathematics much more recently, for example the discovery of the quaternions by William Hamilton. This object had all the telltale signs of a field-extension of the complex numbers, apart from the fact that the multiplication of two of its numbers a.b did not necessarely give you the same result as multiplying the other way around b.a.

Hamilton was so shaken by this discovery (which he made while walking along the Royal canal in Dublin with his wife on october 16th 1843) that he carved the equations using his penknife into the side of the nearby Broom Bridge (which Hamilton called Brougham Bridge), for fear he would forget it. Today, no trace of the carving remains, though a stone plaque does commemorate the discovery.
It reads :

Here as he walked by
on the 16th of October 1843
Sir William Rowan Hamilton
in a flash of genius discovered
the fundamental formula for
quaternion multiplication
$i^2 = j^2 = k^2 = i j k = −1 $
& cut it on a stone of this bridge

The fact that this seems to be the least visited tourist attraction in Dublin tells a lot about the standing of mathematics in society. Fortunately, some of us go to extreme lengths making a pilgrimage to Hamilton’s bridge…

In short, the discovery of mathematical objects such as 0, the square root of -1, quaternions or octonions, often allow us to make great progress in mathematics at the price of having to bend the existing rules slightly.

But, to suggest seriously that an unobserved object should exist when even the most basic arguments rule against its existence is a different matter entirely.

Probably, you have to be brought up in the surrealistic tradition of artists such as Renee Magritte, a guy who added below a drawing of a pipe a sentence saying “This is not a pipe” (Ceci n’est pas une pipe).

In short, you have to be Belgian…

Jacques Tits is a Belgian (today he is also a citizen of a far less surrealistic country : France). He is the ‘man from Uccle’ (in Mark Ronan’s bestselling Symmetry and the Monster), the guy making finite size replicas of infinite Lie groups. But also the guy who didn’t want to stop there.

He managed to replace the field of complex numbers $\mathbb{C} $ by a finite field $\mathbb{F}_q $, consisting of precisely $q=p^n $ a prime-power elements, but wondered what this group might become if $q $ were to go down to size $1 $, even though everyone knew that there couldn’t be a field $\mathbb{F}_1 $ having just one element as $0 \not= 1 $ and these two numbers have to be in any fields DNA.

Tits convinced himself that this elusive field had to exists because his limit-groups had all the characteristics of a finite group co-existing with a Lie group, its companion the Weyl group. Moreover, he was dead sure that the finite geometry associated to his versions of Lie groups would also survive the limit process and give an entirely new combinatorial geometry, featuring objects called ‘buildings’ containing ‘appartments’ glued along ‘walls’ and more terms a real-estate agent might use, but surely not a mathematician…

At the time he was a researcher with the Belgian national science foundation and, having served that agency twenty years myself, I know he had to tread carefully not to infuriate the more traditional committee-members that have to decide on your grant-application every other year. So, when he put his thoughts in writing



he added a footnote saying : “$K_1 $ isn’t generally considered a field”. I’m certain he was doing a Magritte :

$\mathbb{F}_1 $ (as we call today his elusive field $~K_1~ $)

ceci n’est pas un corps

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the buckyball curve

We are after the geometric trinity corresponding to the trinity of exceptional Galois groups

The surfaces on the right have the corresponding group on the left as their group of automorphisms. But, there is a lot more group-theoretic info hidden in the geometry. Before we sketch the $L_2(11) $ case, let us recall the simpler situation of $L_2(7) $.

There are some excellent web-page on the Klein quartic and it would be too hard to try to improve on them, so we refer to John Baez’ page and Greg Egan’s page for more details.

The Klein quartic is the degree 4 projective plane curve defined by the equation $x^3y+y^3z+z^3x=0 $. It can be tiled with a set of 24 regular heptagons, or alternatively with a set of 56 equilateral triangles and these two tilings are dual to each other




In the triangular tiling, there are 56 triangles, 84 edges and 24 vertices. The 56 triangles come in 7 bunches of 8 each and we give the 7 bunches of triangles each a different color as in the pictures below made by Greg Egan. Observe that in the hyperbolic tiling all triangles look alike, but in the picture on the left most of them get warped as we try to embed the quartic in 3-space (which is impossible to do properly). The non-warped triangles (the red ones) come into pairs, the top and bottom triangles of a triangular prism, one prism at each of the four ‘vertices’ of a tetrahedron.

The automorphism group $L_2(7) $ acts on these triangles as $S_4 $ acts on the triangles in a truncated cube.




The buckyball construction from a conjugacy class of order 11 elements from $L_2(11) $ recalled last time, has an analogon $L_2(7) $, leading to the truncated cube.

In $L_2(7) $ there are two conjugacy classes of subgroups isomorphic to $S_4 $ (the rotation-symmetry group of the cube) as well as two conjugacy classes of order 7 elements, each consisting of precisely 24 elements, say C and D. The normalizer subgroup of C has order 21, so there is a cyclic group of order 3 acting non-trivially on the conjugacy class C with 8 orbits consisting of three elements each. These are the eight triangles of the truncated cube identified above as the red triangles.

Shifting perspective, we can repeat this for each of the seven different colors. That is, we have seven truncated cubes in the Klein quartic. On each of them a copy of $S_4 $ acts and these subgroups form one of the two conjugacy classes of $S_4 $ in the group $L_2(7) $. The colors of the triangles of these seven truncated cubes are indicated by bullets in the picture above on the right. The other conjugacy class of $S_4 $’s act on ‘truncated anti-cubes’ which also come in seven bunches of which the color is indicated by a square in that picture.

If you spend enough time on it you will see that each (truncated) cube is completely disjoint from precisely 3 (truncated) anti-cubes. This reminds us of the Fano-plane (picture on the left) : it has 7 points (our seven truncated cubes), 7 lines (the truncated anti-cubes) and the incidence relation of points and lines corresponds to the disjointness of (truncated) cubes and anti-cubes! This is the geometric interpretation of the group-theoretic realization that $L_2(7) \simeq PGL_3(\mathbb{F}_2) $ is the isomorphism group of the projective plane over the finite field $\mathbb{F}_2 $ on two elements, that is, the Fano plane. The colors of the picture on the left indicate the colors of cubes (points) and anti-cubes (lines) consistent with Egan’s picture above.

Further, the 24 vertices correspond to the 24 cusps of the modular group $\Gamma(7) $. Recall that a modular interpretation of the Klein quartic is as $\mathbb{H}/\Gamma(7) $ where $\mathbb{H} $ is the upper half-plane on which the modular group $\Gamma = PSL_2(\mathbb{Z}) $ acts via Moebius transformations, that is, to a 2×2 matrix corresponds the transformation

[tex]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[/tex] <----> $ z \mapsto \frac{az+b}{cz+d} $

Okay, now let’s briefly sketch the exciting results found by Pablo Martin and David Singerman in the paper From biplanes to the Klein quartic and the buckyball, extending the above to the group $L_2(11) $.

There is one important modification to be made. Recall that the Cayley-graph to get the truncated cube comes from taking as generators of the group $S_4 $ the set ${ (3,4),(1,2,3) } $, that is, an order two and an order three element, defining an epimorphism from the modular group $\Gamma= C_2 \ast C_3 \rightarrow S_4 $.

We have also seen that in order to get the buckyball as a Cayley-graph for $A_5 $ we need to take the generating set ${ (2,3)(4,5),(1,2,3,4,5) } $, so a degree two and a degree five element.

Hence, if we want to have a corresponding Riemann surface we’d better not start from the action of the modular group on the upper half-plane, but rather the action via Moebius transformations of the
Hecke group

$H^5 \simeq C_2 \ast C_5 = \langle z \mapsto -\frac{1}{z}, z \mapsto z+ \phi \rangle $

where $\phi = \frac{1 + \sqrt{5}}{2} $ is the golden ratio.

But then, there is an epimorphism $H^5 \rightarrow L_2(11) $ (as this group is generated by one element of degree 2 and one of degree 5) and let $\Lambda $ denote its kernel. Observe that $\Lambda $ is the analogon of the modular subgroup $\Gamma(7) $ used above to define the Klein quartic.

Hence, Martin and Singerman define the buckyball curve as the modular quotient $X=\mathbb{H}/\Lambda $ which is a Riemann surface of genus 70.

The terminlogy is motivated by the fact that, precisely as we got 7 truncated cubes in the Klein quartic, we now get 11 truncated icosahedra (that is, buckyballs) in $X $. The 11 coming, analogous to the Klein case, from thefact that there are precisely two conjugacy classes of subgroups of $L_2(11) $ isomorphic to $A_5 $, each class containing precisely eleven elements!
The 60 vertices of the buckyball again correspond to the fact that there are 60 cusps in this case.

So, what is the analogon of the Fano plane in this case? Well, observe that the Fano-plane is a biplane of order two. That is, if we take as ‘points’ the points of the Fano plane and as ‘lines’ the complements of lines in the Fano plane then this defines a biplane structure. This means that any two distinct ‘points’ are contained in two distinct ‘lines’ and that two distinct ‘lines’ intersect in two distinct ‘points’. A biplane is said to be of order k is each ‘line’ consist of k-2 ‘points’. As the complement of a line in the Fano plane consists of 4 points, the Fano plane is therefore a biplane of order 2. The intersection pattern of cubes and anti-cubes in the Klein quartic is this biplane structure on the Fano plane.

In a similar way, Martin and Singerman show that the two conjugacy classes of subgroups isomorphic to $A_5 $ in $L_2(11) $, each containing exactly 11 elements, correspond to 11 embedded buckyballs (and 11 anti-buckyballs) in the buckyball-curve $X $ and that the intersection relations among them describe the combinatorial structure of a biplane of order three if we view the 11 buckys as ‘points’ and the anti-buckys as ‘lines’.

That is, the buckyball curve is a perfect geometric counterpart of the Klein quartic for the two trinities

At the Arcadian Functor, Kea also has a post on this in which she conjectures that the Kac-Moody algebra of E11 may be related to the buckyball curve.

References :

David Singerman, “Klein’s Riemann surface of genus 3 and regular embeddings of finite projective planes” Bull. London Math. Soc. 18 (1986) 364-370.

Pablo Martin and David Singerman, “From biplanes to the Klein quartic and the Buckyball” (note that this is a preliminary version, please contact David Singerman for the latest version).

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the buckyball symmetries

The buckyball is without doubt the hottest mahematical object at the moment (at least in Europe). Recall that the buckyball (middle) is a mixed form of two Platonic solids



the Icosahedron on the left and the Dodecahedron on the right.

For those of you who don’t know anything about football, it is that other ball-game, best described via a quote from the English player Gary Lineker

“Football is a game for 22 people that run around, play the ball, and one referee who makes a slew of mistakes, and in the end Germany always wins.”

We still have a few days left hoping for a better ending… Let’s do some bucky-maths : what is the rotation symmetry group of the buckyball?

For starters, dodeca- and icosahedron are dual solids, meaning that if you take the center of every face of a dodecahedron and connect these points by edges when the corresponding faces share an edge, you’ll end up with the icosahedron (and conversely). Therefore, both solids (as well as their mixture, the buckyball) will have the same group of rotational symmetries. Can we at least determine the number of these symmetries?

Take the dodecahedron and fix a face. It is easy to find a rotation taking this face to anyone of its five adjacent faces. In group-slang : the rotation automorphism group acts transitively on the 12 faces of the dodecohedron. Now, how many of them fix a given face? These can only be rotations with axis through the center of the face and there are exactly 5 of them preserving the pentagonal face. So, in all we have $12 \times 5 = 60 $ rotations preserving any of the three solids above. By composing two of its elements, we get another rotational symmetry, so they form a group and we would like to determine what that group is.

There is one group that springs to mind $A_5 $, the subgroup of all even permutations on 5 elements. In general, the alternating group has half as many elements as the full permutation group $S_n $, that is $\frac{1}{2} n! $ (for multiplying with the involution (1,2) gives a bijection between even and odd permutations). So, for $A_5 $ we get 60 elements and we can list them :

  • the trivial permutation$~() $, being the identity.
  • permutations of order two with cycle-decompostion $~(i_1,i_2)(i_3,i_4) $, and there are exactly 15 of them around when all numbers are between 1 and 5.
  • permutations of order three with cycle-form $~(i_1,i_2,i_3) $ of which there are exactly 20.
  • permutations of order 5 which have to form one full cycle $~(i_1,i_2,i_3,i_4,i_5) $. There are 24 of those.

Can we at least view these sets of elements as rotations of the buckyball? Well, a dodecahedron has 12 pentagobal faces. So there are 4 nontrivial rotations of order 5 for every 2 opposite faces and hence the dodecaheder (and therefore also the buckyball) has indeed 6×4=24 order 5 rotational symmetries.

The icosahedron has twenty triangles as faces, so any of the 10 pairs of opposite faces is responsible for two non-trivial rotations of order three, giving us 10×2=20 order 3 rotational symmetries of the buckyball.

The order two elements are slightly harder to see. The icosahedron has 30 edges and there is a plane going through each of the 15 pairs of opposite edges splitting the icosahedron in two. Hence rotating to interchange these two edges gives one rotational symmetry of order 2 for each of the 15 pairs.

And as 24+20+15+1(identity) = 60 we have found all the rotational symmetries and we see that they pair up nicely with the elements of $A_5 $. But do they form isomorphic groups? In other words, can the buckyball see the 5 in the group $A_5 $.

In a previous post I’ve shown that one way to see this 5 is as the number of inscribed cubes in the dodecahedron. But, there is another way to see the five based on the order 2 elements described before.

If you look at pairs of opposite edges of the icosahedron you will find that they really come in triples such that the planes determined by each pair are mutually orthogonal (it is best to feel this on ac actual icosahedron). Hence there are 15/3 = 5 such triples of mutually orthogonal symmetry planes of the icosahedron and of course any rotation permutes these triples. It takes a bit of more work to really check that this action is indeed the natural permutation action of $A_5 $ on 5 elements.

Having convinced ourselves that the group of rotations of the buckyball is indeed the alternating group $A_5 $, we can reverse the problem : can the alternating group $A_5 $ see the buckyball???

Well, for starters, it can ‘see’ the icosahedron in a truly amazing way. Look at the conjugacy classes of $A_5 $. We all know that in the full symmetric group $S_n $ elements belong to the same conjugacy class if and only if they have the same cycle decomposition and this is proved using the fact that the conjugation f a cycle $~(i_1,i_2,\ldots,i_k) $ under a permutation $\sigma \in S_n $ is equal to the cycle $~(\sigma(i_1),\sigma(i_2),\ldots,\sigma(i_k)) $ (and this gives us also the candidate needed to conjugate two partitions into each other).

Using this trick it is easy to see that all the 15 order 2 elements of $A_5 $ form one conjugacy class, as do the 20 order 3 elements. However, the 24 order 5 elements split up in two conjugacy classes of 12 elements as the permutation needed to conjugate $~(1,2,3,4,5) $ to $~(1,2,3,5,4) $ is $~(4,5) $ but this is not an element of $A_5 $.

Okay, now take one of these two conjugacy classes of order 5 elements, say that of $~(1,2,3,4,5) $. It consists of 12 elements, 12 being also the number of vertices of the icosahedron. So, is there a way to identify the elements in the conjugacy class to the vertices in such a way that we can describe the edges also in terms of group-computations in $A_5 $?

Surprisingly, this is indeed the case as is demonstrated in a marvelous paper by Kostant “The graph of the truncated icosahedron and the last letter of Galois”.

Two elements $a,b $ in the conjugacy class C share an edge if and only if their product $a.b \in A_5 $ still belongs to the conjugacy class C!

So, for example $~(1,2,3,4,5).(2,1,4,3,5) = (2,5,4) $ so there is no edge between these elements, but on the other hand $~(1,2,3,4,5).(5,3,4,1,2)=(1,5,2,4,3) $ so there is an edge between these! It is no coincidence that $~(5,3,4,1,2)=(2,1,4,3,5)^{-1} $ as inverse elements correspond in the bijection to opposite vertices and for any pair of non-opposite vertices of an icosahedron it is true that either they are neighbors or any one of them is the neighbor of the opposite vertex of the other element.

If we take $u=(1,2,3,4,5) $ and $v=(5,3,4,1,2) $ (or any two elements of the conjugacy class such that u.v is again in the conjugacy class), then one can describe all the vertices of the icosahedron group-theoretically as follows



Isn’t that nice? Well yes, you may say, but that is just the icosahedron. Can the group $A_5 $ also see the buckyball?

Well, let’s try a similar strategy : the buckyball has 60 vertices, exactly as many as there are elements in the group $A_5 $. Is there a way to connect certain elements in a group according to fixed rules? Yes, there is such a way and it is called the Cayley Graph of a group. It goes like this : take a set of generators ${ g_1,\ldots,g_k } $ of a group G, then connect two group element $a,b \in G $ with an edge if and only if $a = g_i.b $ or $b = g_i.a $ for some of the generators.

Back to the alternating group $A_5 $. There are several sets of generators, one of them being the elements ${ (1,2,3,4,5),(2,3)(4,5) } $. In the paper mentioned before, Kostant gives an impressive group-theoretic proof of the fact that the Cayley-graph of $A_5 $ with respect to these two generators is indeed the buckyball!

Let us allow to be lazy for once and let SAGE do the hard work for us, and let us just watch the outcome. Here’s how that’s done

A=PermutationGroup([‘(1,2,3,4,5)’,'(2,3)(4,5)’])
B=A.cayley_graph()
B.show3d()

The outcone is a nice 3-dimensional picture of the buckyball. Below you can see a still, and, if you click on it you will get a 3-dimensional model of it (first click the ‘here’ link in the new window and then you’d better control-click and set the zoom to 200% before you rotate it)





Hence, viewing this Cayley graph from different points we have convinced ourselves that it is indeed the buckyball. In fact, most (truncated) Platonic solids appear as Cayley graphs of groups with respect to specific sets of generators. For later use here is a (partial) survey (taken from Jaap’s puzzle page)



Tetrahedron : $C_2 \times C_2,[(12)(34),(13)(24),(14)(23)] $
Cube : $D_4,[(1234),(13)] $
Octahedron : $S_3,[(123),(12),(23)] $
Dodecahedron : IMPOSSIBLE
Icosahedron : $A_4,[(123),(234),(13)(24)] $



Truncated tetrahedron : $A_4,[(123),(12)(34)] $
Cuboctahedron : $A_4,[(123),(234)] $
Truncated cube : $S_4,[(123),(34)] $
Truncated octahedron : $S_4,[(1234),(12)] $
Rhombicubotahedron : $S_4,[(1234),(123)] $
Rhombitruncated cuboctahedron : IMPOSSIBLE
Snub cuboctahedron : $S_4,[(1234),(123),(34)] $



Icosidodecahedron : IMPOSSIBLE
Truncated dodecahedron : $A_5,[(124),(23)(45)] $
Truncated icosahedron : $A_5,[(12345),(23)(45)] $
Rhombicosidodecahedron : $A_5,[(12345),(124)] $
Rhombitruncated icosidodecahedron : IMPOSSIBLE
Snub Icosidodecahedron : $A_5,[(12345),(124),(23)(45)] $

Again, all these statements can be easily verified using SAGE via the method described before. Next time we will go further into the Kostant’s group-theoretic proof that the buckyball is the Cayley graph of $A_5 $ with respect to (2,5)-generators as this calculation will be crucial in the description of the buckyball curve, the genus 70 Riemann surface discovered by David Singerman and
Pablo Martin which completes the trinity corresponding to the Galois trinity

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Arnold’s trinities version 2.0

Arnold has written a follow-up to the paper mentioned last time called “Polymathematics : is mathematics a single science or a set of arts?” (or here for a (huge) PDF-conversion).

On page 8 of that paper is a nice summary of his 25 trinities :



I learned of this newer paper from a comment by Frederic Chapoton who maintains a nice webpage dedicated to trinities.

In his list there is one trinity on sporadic groups :

where $F_{24} $ is the Fischer simple group of order $2^{21}.3^{16}.5^2.7^3.11.13.17.23.29 = 1255205709190661721292800 $, which is the third largest sporadic group (the two larger ones being the Baby Monster and the Monster itself).

I don’t know what the rationale is behind this trinity. But I’d like to recall the (Baby)Monster history as a warning against the trinity-reflex. Sometimes, there is just no way to extend a would be trinity.

The story comes from Mark Ronan’s book Symmetry and the Monster on page 178.

Let’s remind ourselves how we got here. A few years earlier, Fischer has created his ‘transposition’ groups Fi22, Fi23, and Fi24. He had called them M(22), M(23), and M(24), because they were related to Mathieu’s groups M22,M23, and M24, and since he used Fi22 to create his new group of mirror symmetries, he tentatively called it $M^{22} $.
It seemed to appear as a cross-section in something even bigger, and as this larger group was clearly associated with Fi24, he labeled it $M^{24} $. Was there something in between that could be called $M^{23} $?
Fischer visited Cambridge to talk on his new work, and Conway named these three potential groups the Baby Monster, the Middle Monster, and the Super Monster. When it became clear that the Middle Monster didn’t exist, Conway settled on the names Baby Monster and Monster, and this became the standard terminology.

Marcus du Sautoy’s account in Finding Moonshine is slightly different. He tells on page 322 that the Super Monster didn’t exist. Anyone knowing the factual story?

Some mathematical trickery later revealed that the Super Monster was going to be impossible to build: there were certain features that contradicted each other. It was just a mirage, which vanished under closer scrutiny. But the other two were still looking robust. The Middle Monster was rechristened simply the Monster.

And, the inclusion diagram of the sporadic simples tells yet another story.



Anyhow, this inclusion diagram is helpful in seeing the three generations of the Happy Family (as well as the Pariahs) of the sporadic groups, terminology invented by Robert Griess in his 100+p Inventiones paper on the construction of the Monster (which he liked to call, for obvious reasons, the Friendly Giant denoted by FG).
The happy family appears in Table 1.1. of the introduction.




It was this picture that made me propose the trinity on the left below in the previous post. I now like to add another trinity on the right, and, the connection between the two is clear.

Here $Golay $ denotes the extended binary Golay code of which the Mathieu group $M_{24} $ is the automorphism group. $Leech $ is of course the 24-dimensional Leech lattice of which the automorphism group is a double cover of the Conway group $Co_1 $. $Griess $ is the Griess algebra which is a nonassociative 196884-dimensional algebra of which the automorphism group is the Monster.

I am aware of a construction of the Leech lattice involving the quaternions (the icosian construction of chapter 8, section 2.2 of SPLAG). Does anyone know of a construction of the Griess algebra involving octonions???

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Arnold’s trinities

Referring to the triple of exceptional Galois groups $L_2(5),L_2(7),L_2(11) $ and its connection to the Platonic solids I wrote : “It sure seems that surprises often come in triples…”. Briefly I considered replacing triples by trinities, but then, I didnt want to sound too mystic…

David Corfield of the n-category cafe and a dialogue on infinity (and perhaps other blogs I’m unaware of) pointed me to the paper Symplectization, complexification and mathematical trinities by Vladimir I. Arnold. (Update : here is a PDF-conversion of the paper)

The paper is a write-up of the second in a series of three lectures Arnold gave in june 1997 at the meeting in the Fields Institute dedicated to his 60th birthday. The goal of that lecture was to explain some mathematical dreams he had.

The next dream I want to present is an even more fantastic set of theorems and conjectures. Here I also have no theory and actually the ideas form a kind of religion rather than mathematics.
The key observation is that in mathematics one encounters many trinities. I shall present a list of examples. The main dream (or conjecture) is that all these trinities are united by some rectangular “commutative diagrams”.
I mean the existence of some “functorial” constructions connecting different trinities. The knowledge of the existence of these diagrams provides some new conjectures which might turn to be true theorems.

Follows a list of 12 trinities, many taken from Arnold’s field of expertise being differential geometry. I’ll restrict to the more algebraically inclined ones.

1 : “The first trinity everyone knows is”

where $\mathbb{H} $ are the Hamiltonian quaternions. The trinity on the left may be natural to differential geometers who see real and complex and hyper-Kaehler manifolds as distinct but related beasts, but I’m willing to bet that most algebraists would settle for the trinity on the right where $\mathbb{O} $ are the octonions.

2 : The next trinity is that of the exceptional Lie algebras E6, E7 and E8.

with corresponding Dynkin-Coxeter diagrams

Arnold has this to say about the apparent ubiquity of Dynkin diagrams in mathematics.

Manin told me once that the reason why we always encounter this list in many different mathematical classifications is its presence in the hardware of our brain (which is thus unable to discover a more complicated scheme).
I still hope there exists a better reason that once should be discovered.

Amen to that. I’m quite hopeful human evolution will overcome the limitations of Manin’s brain…

3 : Next comes the Platonic trinity of the tetrahedron, cube and dodecahedron



Clearly one can argue against this trinity as follows : a tetrahedron is a bunch of triangles such that there are exactly 3 of them meeting in each vertex, a cube is a bunch of squares, again 3 meeting in every vertex, a dodecahedron is a bunch of pentagons 3 meeting in every vertex… and we can continue the pattern. What should be a bunch a hexagons such that in each vertex exactly 3 of them meet? Well, only one possibility : it must be the hexagonal tiling (on the left below). And in normal Euclidian space we cannot have a bunch of septagons such that three of them meet in every vertex, but in hyperbolic geometry this is still possible and leads to the Klein quartic (on the right). Check out this wonderful post by John Baez for more on this.



4 : The trinity of the rotation symmetry groups of the three Platonics

where $A_n $ is the alternating group on n letters and $S_n $ is the symmetric group.

Clearly, any rotation of a Platonic solid takes vertices to vertices, edges to edges and faces to faces. For the tetrahedron we can easily see the 4 of the group $A_4 $, say the 4 vertices. But what is the 4 of $S_4 $ in the case of a cube? Well, a cube has 4 body-diagonals and they are permuted under the rotational symmetries. The most difficult case is to see the $5 $ of $A_5 $ in the dodecahedron. Well, here’s the solution to this riddle



there are exactly 5 inscribed cubes in a dodecahedron and they are permuted by the rotations in the same way as $A_5 $.

7 : The seventh trinity involves complex polynomials in one variable

the Laurant polynomials and the modular polynomials (that is, rational functions with three poles at 0,1 and $\infty $.

8 : The eight one is another beauty

Here ‘numbers’ are the ordinary complex numbers $\mathbb{C} $, the ‘trigonometric numbers’ are the quantum version of those (aka q-numbers) which is a one-parameter deformation and finally, the ‘elliptic numbers’ are a two-dimensional deformation. If you ever encountered a Sklyanin algebra this will sound familiar.

This trinity is based on a paper of Turaev and Frenkel and I must come back to it some time…

The paper has some other nice trinities (such as those among Whitney, Chern and Pontryagin classes) but as I cannot add anything sensible to it, let us include a few more algebraic trinities. The first one attributed by Arnold to John McKay

13 : A trinity parallel to the exceptional Lie algebra one is

between the 27 straight lines on a cubic surface, the 28 bitangents on a quartic plane curve and the 120 tritangent planes of a canonic sextic curve of genus 4.

14 : The exceptional Galois groups

explained last time.

15 : The associated curves with these groups as symmetry groups (as in the previous post)

where the ? refers to the mysterious genus 70 curve. I’ll check with one of the authors whether there is still an embargo on the content of this paper and if not come back to it in full detail.

16 : The three generations of sporadic groups

Do you have other trinities you’d like to worship?

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Quiver-superpotentials

It’s been a while, so let’s include a recap : a (transitive) permutation representation of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is determined by the conjugacy class of a cofinite subgroup $\Lambda \subset \Gamma $, or equivalently, to a dessin d’enfant. We have introduced a quiver (aka an oriented graph) which comes from a triangulation of the compactification of $\mathbb{H} / \Lambda $ where $\mathbb{H} $ is the hyperbolic upper half-plane. This quiver is independent of the chosen embedding of the dessin in the Dedeking tessellation. (For more on these terms and constructions, please consult the series Modular subgroups and Dessins d’enfants).

Why are quivers useful? To start, any quiver $Q $ defines a noncommutative algebra, the path algebra $\mathbb{C} Q $, which has as a $\mathbb{C} $-basis all oriented paths in the quiver and multiplication is induced by concatenation of paths (when possible, or zero otherwise). Usually, it is quite hard to make actual computations in noncommutative algebras, but in the case of path algebras you can just see what happens.

Moreover, we can also see the finite dimensional representations of this algebra $\mathbb{C} Q $. Up to isomorphism they are all of the following form : at each vertex $v_i $ of the quiver one places a finite dimensional vectorspace $\mathbb{C}^{d_i} $ and any arrow in the quiver
[tex]\xymatrix{\vtx{v_i} \ar[r]^a & \vtx{v_j}}[/tex] determines a linear map between these vertex spaces, that is, to $a $ corresponds a matrix in $M_{d_j \times d_i}(\mathbb{C}) $. These matrices determine how the paths of length one act on the representation, longer paths act via multiplcation of matrices along the oriented path.

A necklace in the quiver is a closed oriented path in the quiver up to cyclic permutation of the arrows making up the cycle. That is, we are free to choose the start (and end) point of the cycle. For example, in the one-cycle quiver

[tex]\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar[ld]^b \\ & \vtx{} \ar[lu]^c &}[/tex]

the basic necklace can be represented as $abc $ or $bca $ or $cab $. How does a necklace act on a representation? Well, the matrix-multiplication of the matrices corresponding to the arrows gives a square matrix in each of the vertices in the cycle. Though the dimensions of this matrix may vary from vertex to vertex, what does not change (and hence is a property of the necklace rather than of the particular choice of cycle) is the trace of this matrix. That is, necklaces give complex-valued functions on representations of $\mathbb{C} Q $ and by a result of Artin and Procesi there are enough of them to distinguish isoclasses of (semi)simple representations! That is, linear combinations a necklaces (aka super-potentials) can be viewed, after taking traces, as complex-valued functions on all representations (similar to character-functions).

In physics, one views these functions as potentials and it then interested in the points (representations) where this function is extremal (minimal) : the vacua. Clearly, this does not make much sense in the complex-case but is relevant when we look at the real-case (where we look at skew-Hermitian matrices rather than all matrices). A motivating example (the Yang-Mills potential) is given in Example 2.3.2 of Victor Ginzburg’s paper Calabi-Yau algebras.

Let $\Phi $ be a super-potential (again, a linear combination of necklaces) then our commutative intuition tells us that extrema correspond to zeroes of all partial differentials $\frac{\partial \Phi}{\partial a} $ where $a $ runs over all coordinates (in our case, the arrows of the quiver). One can make sense of differentials of necklaces (and super-potentials) as follows : the partial differential with respect to an arrow $a $ occurring in a term of $\Phi $ is defined to be the path in the quiver one obtains by removing all 1-occurrences of $a $ in the necklaces (defining $\Phi $) and rearranging terms to get a maximal broken necklace (using the cyclic property of necklaces). An example, for the cyclic quiver above let us take as super-potential $abcabc $ (2 cyclic turns), then for example

$\frac{\partial \Phi}{\partial b} = cabca+cabca = 2 cabca $

(the first term corresponds to the first occurrence of $b $, the second to the second). Okay, but then the vacua-representations will be the representations of the quotient-algebra (which I like to call the vacualgebra)

$\mathcal{U}(Q,\Phi) = \frac{\mathbb{C} Q}{(\partial \Phi/\partial a, \forall a)} $

which in ‘physical relevant settings’ (whatever that means…) turn out to be Calabi-Yau algebras.

But, let us return to the case of subgroups of the modular group and their quivers. Do we have a natural super-potential in this case? Well yes, the quiver encoded a triangulation of the compactification of $\mathbb{H}/\Lambda $ and if we choose an orientation it turns out that all ‘black’ triangles (with respect to the Dedekind tessellation) have their arrow-sides defining a necklace, whereas for the ‘white’ triangles the reverse orientation makes the arrow-sides into a necklace. Hence, it makes sense to look at the cubic superpotential $\Phi $ being the sum over all triangle-sides-necklaces with a +1-coefficient for the black triangles and a -1-coefficient for the white ones. Let’s consider an index three example from a previous post


[tex]\xymatrix{& & \rho \ar[lld]_d \ar[ld]^f \ar[rd]^e & \\
i \ar[rrd]_a & i+1 \ar[rd]^b & & \omega \ar[ld]^c \\
& & 0 \ar[uu]^h \ar@/^/[uu]^g \ar@/_/[uu]_i &}[/tex]

In this case the super-potential coming from the triangulation is

$\Phi = -aid+agd-cge+che-bhf+bif $

and therefore we have a noncommutative algebra $\mathcal{U}(Q,\Phi) $ associated to this index 3 subgroup. Contrary to what I believed at the start of this series, the algebras one obtains in this way from dessins d’enfants are far from being Calabi-Yau (in whatever definition). For example, using a GAP-program written by Raf Bocklandt Ive checked that the growth rate of the above algebra is similar to that of $\mathbb{C}[x] $, so in this case $\mathcal{U}(Q,\Phi) $ can be viewed as a noncommutative curve (with singularities).

However, this is not the case for all such algebras. For example, the vacualgebra associated to the second index three subgroup (whose fundamental domain and quiver were depicted at the end of this post) has growth rate similar to that of $\mathbb{C} \langle x,y \rangle $…

I have an outlandish conjecture about the growth-behavior of all algebras $\mathcal{U}(Q,\Phi) $ coming from dessins d’enfants : the algebra sees what the monodromy representation of the dessin sees of the modular group (or of the third braid group).
I can make this more precise, but perhaps it is wiser to calculate one or two further examples…

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ECSTR aka XTR

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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quivers versus quilts

We have associated to a subgroup of the modular group $PSL_2(\mathbb{Z}) $ a quiver (that is, an oriented graph). For example, one verifies that the fundamental domain of the subgroup $\Gamma_0(2) $ (an index 3 subgroup) is depicted on the right by the region between the thick lines with the identification of edges as indicated. The associated quiver is then

\[
\xymatrix{i \ar[rr]^a \ar[dd]^b & & 1 \ar@/^/[ld]^h \ar@/_/[ld]_i \\
& \rho \ar@/^/[lu]^d \ar@/_/[lu]_e \ar[rd]^f & \\
0 \ar[ru]^g & & i+1 \ar[uu]^c}
\]

The corresponding “dessin d’enfant” are the green edges in the picture. But, the red dot on the left boundary is identied with the red dot on the lower circular boundary, so the dessin of the modular subgroup $\Gamma_0(2) $ is

\[
\xymatrix{| \ar@{-}[r] & \bullet \ar@{-}@/^8ex/[r] \ar@{-}@/_8ex/[r] & -}
\]

Here, the three red dots (all of them even points in the Dedekind tessellation) give (after the identification) the two points indicated by a $\mid $ whereas the blue dot (an odd point in the tessellation) is depicted by a $\bullet $. There is another ‘quiver-like’ picture associated to this dessin, a quilt of the modular subgroup $\Gamma_0(2) $ as studied by John Conway and Tim Hsu.

On the left, a quilt-diagram copied from Hsu’s book Quilts : central extensions, braid actions, and finite groups, exercise 3.3.9. This ‘quiver’ has also 5 vertices and 7 arrows as our quiver above, so is there a connection?

A quilt is a gadget to study transitive permutation representations of the braid group $B_3 $ (rather than its quotient, the modular group $PSL_2(\mathbb{Z}) = B_3/\langle Z \rangle $ where $\langle Z \rangle $ is the cyclic center of $B_3 $. The $Z $-stabilizer subgroup of all elements in a transitive permutation representation of $B_3 $ is the same and hence of the form $\langle Z^M \rangle $ where M is called the modulus of the representation. The arrow-data of a quilt, that is the direction of certain edges and their labeling with numbers from $\mathbb{Z}/M \mathbb{Z} $ (which have to satisfy some requirements, the flow rules, but more about that another time) encode the Z-action on the permutation representation. The dimension of the representation is $M \times k $ where $k $ is the number of half-edges in the dessin. In the above example, the modulus is 5 and the dessin has 3 (half)edges, so it depicts a 15-dimensional permutation representation of $B_3 $.

If we forget the Z-action (that is, the arrow information), we get a permutation representation of the modular group (that is a dessin). So, if we delete the labels and directions on the edges we get what Hsu calls a modular quilt, that is, a picture consisting of thick edges (the dessin) together with dotted edges which are called the seams of the modular quilt. The modular quilt is merely another way to depict a fundamental domain of the corresponding subgroup of the modular group. For the above example, we have the indicated correspondences between the fundamental domain of $\Gamma_0(2) $ in the upper half-plane (on the left) and as a modular quilt (on the right)

That is, we can also get our quiver (or its opposite quiver) from the modular quilt by fixing the orientation of one 2-cell. For example, if we fix the orientation of the 2-cell $\vec{fch} $ we get our quiver back from the modular quilt


\[
\xymatrix{i \ar[rr]^a \ar[dd]^b & & 1 \ar@/^/[ld]^h \ar@/_/[ld]_i \\
& \rho \ar@/^/[lu]^d \ar@/_/[lu]_e \ar[rd]^f & \\
0 \ar[ru]^g & & i+1 \ar[uu]^c}
\]

This shows that the quiver (or its opposite) associated to a (conjugacy class of a) subgroup of $PSL_2(\mathbb{Z}) $ does not depend on the choice of embedding of the dessin (or associated cuboid tree diagram) in the upper half-plane. For, one can get the modular quilt from the dessin by adding one extra vertex for every connected component of the complement of the dessin (in the example, the two vertices corresponding to 0 and 1) and drawing a triangulation from them (the dotted lines or ‘seams’).

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the modular group and superpotentials (1)

Here I will go over the last post at a more leisurely pace, focussing on a couple of far more trivial examples. Here’s the goal : we want to assign a quiver-superpotential to any subgroup of finite index of the modular group. So fix such a subgroup $\Gamma’ $ of the modular group $\Gamma=PSL_2(\mathbb{Z}) $ and consider the associated permutation representation of $\Gamma $ on the left-cosets $\Gamma/\Gamma’ $. As $\Gamma \simeq C_2 \ast C_3 $ this representation is determined by the action of the order 2 and order 3 generators of the modular group. There are a number of combinatorial gadgets to control the subgroup $\Gamma’ $ and the associated permutation representation : (generalized) Farey symbols and dessins d’enfants.

Recall that the modular group acts on the upper-halfplane (the ‘hyperbolic plane’) by Moebius transformations, so to any subgroup $\Gamma’ $ we can associate a fundamental domain for its restricted action. The dessins and the Farey symbols give us a particular choice of these fundamental domains. Let us consider the two most trivial subgroups of all : the modular group itself (so $\Gamma/\Gamma $ is just one element and therefore the associated permutation representation is just the trivial representation) and the unique index two subgroup $\Gamma_2 $ (so there are two cosets $\Gamma/\Gamma_2 $ and the order 2 generator interchanges these two while the order 3 generator acts trivially on them). The fundamental domains of $\Gamma $ (left) and $\Gamma_2 $ (right) are depicted below

In both cases the fundamental domain is bounded by the thick black (hyperbolic) edges. The left-domain consists of two hyperbolic triangles (the upper domain has $\infty $ as the third vertex) and the right-domain has 4 triangles. In general, if the subgroup $\Gamma’ $ has index n, then its fundamental domain will consist of $2n $ hyperbolic triangles. Note that these triangles are part of the Dedekind tessellation so really depict the action of $PGL_2(\mathbb{Z} $ and any $\Gamma $-hyperbolic triangle consists of one black and one white triangle in Dedekind’s coloring. We will indicate the color of a triangle by a black circle if the corresponding triangle is black. Of course, the bounding edges of the fundamental domain need to be identified and the Farey symbol is a notation device to clarify this. The Farey symbols of the above domains are
[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] and [tex]\xymatrix{\infty \ar@{-}[r]_{\bullet} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] respectively. In both cases this indicates that the two bounding edges on the left are to be identified as are the two bounding edges on the right (so, in particular, after identification $\infty $ coincides with $0 $). Hence, after identification, the $\Gamma $ domain consists of two triangles on the vertices ${ 0,i,\rho } $ (where $\rho=e^{2 \pi i}{6} $) (the blue dots) sharing all three edges, the $\Gamma_2 $ domain consists of 4 triangles on the 4 vertices ${ 0,i,\rho,\rho^2 } $ (the blue dots). In general we have three types of vertices : cusps (such as 0 or $\infty $), even vertices (such as $i $ where there are 4 hyperbolic edges in the Dedekind tessellation) and odd vertices (such as $\rho $ and $\rho^2 $ where there are 6 hyperbolic edges in the tessellation).

Another combinatorial gadget assigned to the fundamental domain is the cuboid tree diagram or dessin. It consists of all odd and even vertices on the boundary of the domain, together with all odd and even vertices in the interior. These vertices are then connected with the hyperbolic edges connecting them. If we color the even vertices red and the odds blue we have the indicated dessins for our two examples (the green pictures). An half-edge is an edge connecting a red and a blue vertex in the dessin and we number all half-edges. So, the $\Gamma $-dessin has 1 half-edge whereas the $\Gamma_2 $-dessin has two (in general, the number of these half-edges is equal to the index of the subgroup). Observe also that every triangle has exactly one half-edge as one of its three edges. The dessin gives all information to calculate the permutation representation on the coset-set $\Gamma/\Gamma’ $ : the action of the order 2 generator of $\Gamma $ is given by taking for each internal red vertex the two-cycle $~(a,b) $ where a and b are the numbers of the two half-edges connected to the red vertex and the action of the order 3 generator is given by taking for every internal blue vertex the three cycle $~(c,d,e) $ where c, d and e are the numbers of the three half-edges connected to the blue vertex in counter-clockwise ordering. Our two examples above are a bit too simplistic to view this in action. There are no internal blue vertices, so the action of the order 3 generator is trivial in both cases. For $\Gamma $ there is also no red internal vertex, whence this is indeed the trivial representation whereas for $\Gamma_2 $ there is one internal red vertex, so the action of the order 2 generator is given by $~(1,2) $, which is indeed the representation representation on $\Gamma/\Gamma_2 $. In general, if the index of the subgroup $\Gamma’ $ is n, then we call the subgroup of the symmetric group on n letters $S_n $ generated by the action-elements of the order 2 and order 3 generator the monodromy group of the permutation representation (or of the subgroup). In the trivial cases here, the monodromy groups are the trivial group (for $\Gamma $) and $C_2 $ (for $\Gamma_2 $).

As a safety-check let us work out all these concepts in the next simplest examples, those of some subgroups of index 3. Consider the Farey symbols

[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\circ} & 1 \ar@{-}[r]_{\circ} & \infty}[/tex] and
[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{1} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

In these cases the fundamental domain consists of 6 triangles with the indicated vertices (the blue dots). The distinction between the two is that in the first case, one identifies the two edges of the left, resp. bottom, resp. right boundary (so, in particular, 0,1 and $\infty $ are identified) whereas in the second one identifies the two edges of the left boundary and identifies the edges of the bottom with those of the right boundary (here, 0 is identified only with $\infty $ but also $1+i $ is indetified with $\frac{1}{2}+\frac{1}{2}i $).

In both cases the dessin seems to be the same (and given by the picture on the right). However, in the first case all three red vertices are distinct hence there are no internal red vertices in this case whereas in the second case we should identify the bottom and right-hand red vertex which then becomes an internal red vertex of the dessin!

Hence, if we order the three green half-edges 1,2,3 starting with the bottom one and counting counter-clockwise we see that in both cases the action of the order 3-generator of $\Gamma $ is given by the 3-cycle $~(1,2,3) $. The action of the order 2-generator is trivial in the first case, while given by the 2-cycle $~(1,2) $ in the second case. Therefore, the monodromy group is the cylic group $C_3 $ in the first case and is the symmetric group $S_3 $ in the second case.

Next time we will associate a quiver to these vertices and triangles as well as a cubic superpotential which will then allow us to define a noncommutative algebra associated to any subgroup of the modular group. The monodromy group of the situation will then reappear as a group of algebra-automorphisms of this noncommutative algebra!

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profinite groups survival guide

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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