# Tag: Klein

The usual argument to show that the group of all orientation-preserving symmetries of the Klein quartic is the simple group $L_2(7)$ of order $168$ goes like this:

There are two families of $7$ truncated cubes on the Klein quartic. The triangles of one of the seven truncated cubes in the first family have as center the dots, all having the same colour. The triangles of one of the truncated cubes in the second family correspond to the squares all having the same colour.

If you compare the two colour schemes, you’ll see that every truncated cube in the first family is disjoint from precisely $3$ truncated cubes in the second family.

That is, we can identify the truncated cubes of the first family with the points in the Fano plane $\mathbb{P}^2(\mathbb{F}_2)$, and those of the second family with the lines in that plane.

The Klein quartic consists of $24$ regular heptagons, so its rotation symmetry group consists of $24 \times 7 = 168$ rotations,each preserving the two families of truncated cubes. This is exactly the same number as there are isomorphisms of the Fano plane, $PGL_3(\mathbb{F}_2) = L_2(7)$. Done!

For more details, check John Baez’ excellent page on the Klein quartic, or the Buckyball curve post.

Here’s another ‘look-and-see’ proof, starting from Klein’s own description of his quartic.

Look at the rotation $g$, counter-clockwise with angle $2\pi / 7$ fixing the center of the central blue heptagon, and a similar rotation $h$ fixing the center of one of the neighbouring red heptagons.

The two vertices of the edge shared by the blue and red heptagon are fixed by $g.h$ and $h.g$, respectively, so these rotations must have order three (there are $3$ heptagons meeting in the vertex).

That is, the rotation symmetry group $G$ of the Klein quartic has order $168$, and contains two elements $g$ and $h$ of order $7$, such that the subgroup generated by them contains elements of order $3$.

This is enough to prove that the $G$ must be simple and therefore isomorphic to $L_2(7)$!

The following elegant proof is often attributed to Igor Dolgachev.

If $G$ isn’t simple there is a maximal normal subgroup $N$ with $G/N$ simple .

The only non-cyclic simple group having less elements that $168$ is $A_5$ but this cannot be $G/N$ as $60$ does not divide $168$.

So, $G/N$ must be cyclic of order $2,3$ or $7$ (the only prime divisors of $168=2^3.3.7$).

Order $2$ is not possible as any group $N$ of order $84=2^2.3.7$ can just have one Sylow $7$-subgroup. Remember that the number of $7$-Sylows of $N$ must divide $2^2.3=12$ and must be equal to $1$ modulo $7$. And $G$ (and therefore $N$) has at least two different cyclic subgroups of order $7$.

Order $3$ is impossible as this would imply that the normal subgroup $N$ of order $2^3.7=56$ must contain all $7$-Sylows of $G$, and thus also an element of order $3$. But, $3$ does not divide $56$.

Order $7$ is a bit more difficult to exclude. This would mean that there is a normal subgroup $N$ of order $2^3.3=24$.

$N$ (being normal) must contain all Sylow $2$-subgroups of $G$ of which there are either $1$ or $3$ (the order of $N$ is $2^3.3=24$).

If there is just one $S$ it should be a normal subgroup with $G/S$ (of order $21$) having a (normal) unique Sylow $7$-subgroup, but then $G$ would have a normal subgroup of index $3$, which we have excluded.

The three $2$-Sylows are conjugated and so the conjugation morphism
$G \rightarrow S_3$
is non-trivial and must have image strictly larger than $C_3$ (otherwise, $G$ would have a normal subgroup of index $3$), so must be surjective.

But, $S_3$ has a normal subgroup of index $2$ and pulling this back, $G$ must also have a normal subgroup of index two, which we have excluded. Done!

Sunday january 2nd around 18hr NeB-stats went crazy.

Referrals clarified that the post ‘What is the knot associated to a prime?’ was picked up at Reddit/math and remained nr.1 for about a day.

Now, the dust has settled, so let’s learn from the experience.

A Reddit-mention is to a blog what doping is to a sporter.

You get an immediate boost in the most competitive of all blog-stats, the number of unique vistors (blue graph), but is doesn’t result in a long-term effect, and, it may even be harmful to more essential blog-stats, such as the average time visitors spend on your site (yellow graph).

For NeB the unique vistors/day fluctuate normally around 300, but peaked to 1295 and 1733 on the ‘Reddit-days’. In contrast, the avg. time on site is normally around 3 minutes, but dropped the same days to 44 and 30 seconds!

Whereas some of the Reddits spend enough time to read the post and comment on it, the vast majority zap from one link to the next. Having monitored the Reddit/math page for two weeks, I’m convinced that post only made it because it was visually pretty good. The average Reddit/math-er is a viewer more than a reader…

So, should I go for shorter, snappier, more visual posts?

Let’s compare Reddits to those coming from the three sites giving NeB most referrals : Google search, MathOverflow and Wikipedia.

This is the traffic coming from Reddit/math, as always the blue graph are the unique visitors, the yellow graph their average time on site, blue-scales to the left, yellow-scales to the right.

Here’s the same graph for Google search. The unique visitors/day fluctuate around 50 and their average time on site about 2 minutes.

The math-related search terms most used were this month : ‘functor of point approach’, ‘profinite integers’ and ‘bost-connes sytem’.

More rewarding to me are referrals from MathOverflow.

The number of visitors depends on whether the MathO-questions made it to the front-page (for example, the 80 visits on december 15, came from the What are dessins d’enfants?-topic getting an extra comment that very day, and having two references to NeB-posts : The best rejected proposal ever and Klein’s dessins d’enfant and the buckyball), but even older MathO-topics give a few referrals a day, and these people sure take their time reading the posts (+ 5 minutes).

Other MathO-topics giving referrals this month were Most intricate and most beautiful structures in mathematics (linking to Looking for F-un), What should be learned in a first serious schemes course? (linking to Mumford’s treasure map (btw. one of the most visited NeB-posts ever)), How much of scheme theory can you visualize? (linking again to Mumford’s treasure map) and Approaches to Riemann hypothesis using methods outside number theory (linking to the Bost-Connes series).

Finally, there’s Wikipedia

giving 5 to 10 referrals a day, with a pretty good time-on-site average (around 4 minutes, peaking to 12 minutes). It is rewarding to see NeB-posts referred to in as diverse Wikipedia-topics as ‘Fifteen puzzle’, ‘Field with one element’, ‘Evariste Galois’, ‘ADE classification’, ‘Monster group’, ‘Arithmetic topology’, ‘Dessin d’enfant’, ‘Groupoid’, ‘Belyi’s theorem’, ‘Modular group’, ‘Cubic surface’, ‘Esquisse d’un programme’, ‘N-puzzle’, ‘Shabat polynomial’ and ‘Mathieu group’.

What lesson should be learned from all this data? Should I go for shorter, snappier and more visual posts, or should I focus on the small group of visitors taking their time reading through a longer post, and don’t care about the appallingly high bounce rate the others cause?

The Monster is the largest of the 26 sporadic simple groups and has order

808 017 424 794 512 875 886 459 904 961 710 757 005 754 368 000 000 000

= 2^46 3^20 5^9 7^6 11^2 13^3 17 19 23 29 31 41 47 59 71.

It is not so much the size of its order that makes it hard to do actual calculations in the monster, but rather the dimensions of its smallest non-trivial irreducible representations (196 883 for the smallest, 21 296 876 for the next one, and so on).

In characteristic two there is an irreducible representation of one dimension less (196 882) which appears to be of great use to obtain information. For example, Robert Wilson used it to prove that The Monster is a Hurwitz group. This means that the Monster is generated by two elements g and h satisfying the relations

$g^2 = h^3 = (gh)^7 = 1$

Geometrically, this implies that the Monster is the automorphism group of a Riemann surface of genus g satisfying the Hurwitz bound 84(g-1)=#Monster. That is,

g=9619255057077534236743570297163223297687552000000001=42151199 * 293998543 * 776222682603828537142813968452830193

Or, in analogy with the Klein quartic which can be constructed from 24 heptagons in the tiling of the hyperbolic plane, there is a finite region of the hyperbolic plane, tiled with heptagons, from which we can construct this monster curve by gluing the boundary is a specific way so that we get a Riemann surface with exactly 9619255057077534236743570297163223297687552000000001 holes. This finite part of the hyperbolic tiling (consisting of #Monster/7 heptagons) we’ll call the empire of the monster and we’d love to describe it in more detail.

Look at the half-edges of all the heptagons in the empire (the picture above learns that every edge in cut in two by a blue geodesic). They are exactly #Monster such half-edges and they form a dessin d’enfant for the monster-curve.

If we label these half-edges by the elements of the Monster, then multiplication by g in the monster interchanges the two half-edges making up a heptagonal edge in the empire and multiplication by h in the monster takes a half-edge to the one encountered first by going counter-clockwise in the vertex of the heptagonal tiling. Because g and h generated the Monster, the dessin of the empire is just a concrete realization of the monster.

Because g is of order two and h is of order three, the two permutations they determine on the dessin, gives a group epimorphism $C_2 \ast C_3 = PSL_2(\mathbb{Z}) \rightarrow \mathbb{M}$ from the modular group $PSL_2(\mathbb{Z})$ onto the Monster-group.

In noncommutative geometry, the group-algebra of the modular group $\mathbb{C} PSL_2$ can be interpreted as the coordinate ring of a noncommutative manifold (because it is formally smooth in the sense of Kontsevich-Rosenberg or Cuntz-Quillen) and the group-algebra of the Monster $\mathbb{C} \mathbb{M}$ itself corresponds in this picture to a finite collection of ‘points’ on the manifold. Using this geometric viewpoint we can now ask the question What does the Monster see of the modular group?

To make sense of this question, let us first consider the commutative equivalent : what does a point P see of a commutative variety X?

Evaluation of polynomial functions in P gives us an algebra epimorphism $\mathbb{C}[X] \rightarrow \mathbb{C}$ from the coordinate ring of the variety $\mathbb{C}[X]$ onto $\mathbb{C}$ and the kernel of this map is the maximal ideal $\mathfrak{m}_P$ of
$\mathbb{C}[X]$ consisting of all functions vanishing in P.

Equivalently, we can view the point $P= \mathbf{spec}~\mathbb{C}[X]/\mathfrak{m}_P$ as the scheme corresponding to the quotient $\mathbb{C}[X]/\mathfrak{m}_P$. Call this the 0-th formal neighborhood of the point P.

This sounds pretty useless, but let us now consider higher-order formal neighborhoods. Call the affine scheme $\mathbf{spec}~\mathbb{C}[X]/\mathfrak{m}_P^{n+1}$ the n-th forml neighborhood of P, then the first neighborhood, that is with coordinate ring $\mathbb{C}[X]/\mathfrak{m}_P^2$ gives us tangent-information. Alternatively, it gives the best linear approximation of functions near P.
The second neighborhood $\mathbb{C}[X]/\mathfrak{m}_P^3$ gives us the best quadratic approximation of function near P, etc. etc.

These successive quotients by powers of the maximal ideal $\mathfrak{m}_P$ form a system of algebra epimorphisms

$\ldots \frac{\mathbb{C}[X]}{\mathfrak{m}_P^{n+1}} \rightarrow \frac{\mathbb{C}[X]}{\mathfrak{m}_P^{n}} \rightarrow \ldots \ldots \rightarrow \frac{\mathbb{C}[X]}{\mathfrak{m}_P^{2}} \rightarrow \frac{\mathbb{C}[X]}{\mathfrak{m}_P} = \mathbb{C}$

and its inverse limit $\underset{\leftarrow}{lim}~\frac{\mathbb{C}[X]}{\mathfrak{m}_P^{n}} = \hat{\mathcal{O}}_{X,P}$ is the completion of the local ring in P and contains all the infinitesimal information (to any order) of the variety X in a neighborhood of P. That is, this completion $\hat{\mathcal{O}}_{X,P}$ contains all information that P can see of the variety X.

In case P is a smooth point of X, then X is a manifold in a neighborhood of P and then this completion
$\hat{\mathcal{O}}_{X,P}$ is isomorphic to the algebra of formal power series $\mathbb{C}[[ x_1,x_2,\ldots,x_d ]]$ where the $x_i$ form a local system of coordinates for the manifold X near P.

Right, after this lengthy recollection, back to our question what does the monster see of the modular group? Well, we have an algebra epimorphism

$\pi~:~\mathbb{C} PSL_2(\mathbb{Z}) \rightarrow \mathbb{C} \mathbb{M}$

and in analogy with the commutative case, all information the Monster can gain from the modular group is contained in the $\mathfrak{m}$-adic completion

$\widehat{\mathbb{C} PSL_2(\mathbb{Z})}_{\mathfrak{m}} = \underset{\leftarrow}{lim}~\frac{\mathbb{C} PSL_2(\mathbb{Z})}{\mathfrak{m}^n}$

where $\mathfrak{m}$ is the kernel of the epimorphism $\pi$ sending the two free generators of the modular group $PSL_2(\mathbb{Z}) = C_2 \ast C_3$ to the permutations g and h determined by the dessin of the pentagonal tiling of the Monster’s empire.

As it is a hopeless task to determine the Monster-empire explicitly, it seems even more hopeless to determine the kernel $\mathfrak{m}$ let alone the completed algebra… But, (surprise) we can compute $\widehat{\mathbb{C} PSL_2(\mathbb{Z})}_{\mathfrak{m}}$ as explicitly as in the commutative case we have $\hat{\mathcal{O}}_{X,P} \simeq \mathbb{C}[[ x_1,x_2,\ldots,x_d ]]$ for a point P on a manifold X.

Here the details : the quotient $\mathfrak{m}/\mathfrak{m}^2$ has a natural structure of $\mathbb{C} \mathbb{M}$-bimodule. The group-algebra of the monster is a semi-simple algebra, that is, a direct sum of full matrix-algebras of sizes corresponding to the dimensions of the irreducible monster-representations. That is,

$\mathbb{C} \mathbb{M} \simeq \mathbb{C} \oplus M_{196883}(\mathbb{C}) \oplus M_{21296876}(\mathbb{C}) \oplus \ldots \ldots \oplus M_{258823477531055064045234375}(\mathbb{C})$

with exactly 194 components (the number of irreducible Monster-representations). For any $\mathbb{C} \mathbb{M}$-bimodule $M$ one can form the tensor-algebra

$T_{\mathbb{C} \mathbb{M}}(M) = \mathbb{C} \mathbb{M} \oplus M \oplus (M \otimes_{\mathbb{C} \mathbb{M}} M) \oplus (M \otimes_{\mathbb{C} \mathbb{M}} M \otimes_{\mathbb{C} \mathbb{M}} M) \oplus \ldots \ldots$

and applying the formal neighborhood theorem for formally smooth algebras (such as $\mathbb{C} PSL_2(\mathbb{Z})$) due to Joachim Cuntz (left) and Daniel Quillen (right) we have an isomorphism of algebras

$\widehat{\mathbb{C} PSL_2(\mathbb{Z})}_{\mathfrak{m}} \simeq \widehat{T_{\mathbb{C} \mathbb{M}}(\mathfrak{m}/\mathfrak{m}^2)}$

where the right-hand side is the completion of the tensor-algebra (at the unique graded maximal ideal) of the $\mathbb{C} \mathbb{M}$-bimodule $\mathfrak{m}/\mathfrak{m}^2$, so we’d better describe this bimodule explicitly.

Okay, so what’s a bimodule over a semisimple algebra of the form $S=M_{n_1}(\mathbb{C}) \oplus \ldots \oplus M_{n_k}(\mathbb{C})$? Well, a simple S-bimodule must be either (1) a factor $M_{n_i}(\mathbb{C})$ with all other factors acting trivially or (2) the full space of rectangular matrices $M_{n_i \times n_j}(\mathbb{C})$ with the factor $M_{n_i}(\mathbb{C})$ acting on the left, $M_{n_j}(\mathbb{C})$ acting on the right and all other factors acting trivially.

That is, any S-bimodule can be represented by a quiver (that is a directed graph) on k vertices (the number of matrix components) with a loop in vertex i corresponding to each simple factor of type (1) and a directed arrow from i to j corresponding to every simple factor of type (2).

That is, for the Monster, the bimodule $\mathfrak{m}/\mathfrak{m}^2$ is represented by a quiver on 194 vertices and now we only have to determine how many loops and arrows there are at or between vertices.

Using Morita equivalences and standard representation theory of quivers it isn’t exactly rocket science to determine that the number of arrows between the vertices corresponding to the irreducible Monster-representations $S_i$ and $S_j$ is equal to

$dim_{\mathbb{C}}~Ext^1_{\mathbb{C} PSL_2(\mathbb{Z})}(S_i,S_j)-\delta_{ij}$

Now, I’ve been wasting a lot of time already here explaining what representations of the modular group have to do with quivers (see for example here or some other posts in the same series) and for quiver-representations we all know how to compute Ext-dimensions in terms of the Euler-form applied to the dimension vectors.

Right, so for every Monster-irreducible $S_i$ we have to determine the corresponding dimension-vector $~(a_1,a_2;b_1,b_2,b_3)$ for the quiver

$\xymatrix{ & & & & \vtx{b_1} \\ \vtx{a_1} \ar[rrrru]^(.3){B_{11}} \ar[rrrrd]^(.3){B_{21}} \ar[rrrrddd]_(.2){B_{31}} & & & & \\ & & & & \vtx{b_2} \\ \vtx{a_2} \ar[rrrruuu]_(.7){B_{12}} \ar[rrrru]_(.7){B_{22}} \ar[rrrrd]_(.7){B_{23}} & & & & \\ & & & & \vtx{b_3}}$

Now the dimensions $a_i$ are the dimensions of the +/-1 eigenspaces for the order 2 element g in the representation and the $b_i$ are the dimensions of the eigenspaces for the order 3 element h. So, we have to determine to which conjugacy classes g and h belong, and from Wilson’s paper mentioned above these are classes 2B and 3B in standard Atlas notation.

So, for each of the 194 irreducible Monster-representations we look up the character values at 2B and 3B (see below for the first batch of those) and these together with the dimensions determine the dimension vector $~(a_1,a_2;b_1,b_2,b_3)$.

For example take the 196883-dimensional irreducible. Its 2B-character is 275 and the 3B-character is 53. So we are looking for a dimension vector such that $a_1+a_2=196883, a_1-275=a_2$ and $b_1+b_2+b_3=196883, b_1-53=b_2=b_3$ giving us for that representation the dimension vector of the quiver above $~(98579,98304,65663,65610,65610)$.

Okay, so for each of the 194 irreducibles $S_i$ we have determined a dimension vector $~(a_1(i),a_2(i);b_1(i),b_2(i),b_3(i))$, then standard quiver-representation theory asserts that the number of loops in the vertex corresponding to $S_i$ is equal to

$dim(S_i)^2 + 1 – a_1(i)^2-a_2(i)^2-b_1(i)^2-b_2(i)^2-b_3(i)^2$

and that the number of arrows from vertex $S_i$ to vertex $S_j$ is equal to

$dim(S_i)dim(S_j) – a_1(i)a_1(j)-a_2(i)a_2(j)-b_1(i)b_1(j)-b_2(i)b_2(j)-b_3(i)b_3(j)$

This data then determines completely the $\mathbb{C} \mathbb{M}$-bimodule $\mathfrak{m}/\mathfrak{m}^2$ and hence the structure of the completion $\widehat{\mathbb{C} PSL_2}_{\mathfrak{m}}$ containing all information the Monster can gain from the modular group.

But then, one doesn’t have to go for the full regular representation of the Monster. Any faithful permutation representation will do, so we might as well go for the one of minimal dimension.

That one is known to correspond to the largest maximal subgroup of the Monster which is known to be a two-fold extension $2.\mathbb{B}$ of the Baby-Monster. The corresponding permutation representation is of dimension 97239461142009186000 and decomposes into Monster-irreducibles

$S_1 \oplus S_2 \oplus S_4 \oplus S_5 \oplus S_9 \oplus S_{14} \oplus S_{21} \oplus S_{34} \oplus S_{35}$

(in standard Atlas-ordering) and hence repeating the arguments above we get a quiver on just 9 vertices! The actual numbers of loops and arrows (I forgot to mention this, but the quivers obtained are actually symmetric) obtained were found after laborious computations mentioned in this post and the details I’ll make avalable here.

Anyone who can spot a relation between the numbers obtained and any other part of mathematics will obtain quantities of genuine (ie. non-Inbev) Belgian beer…

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

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

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).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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?

“Ne pleure pas, Alfred ! J’ai besoin de tout mon courage pour mourir à vingt ans!”

We all remember the last words of Evariste Galois to his brother Alfred. Lesser known are the mathematical results contained in his last letter, written to his friend Auguste Chevalier, on the eve of his fatal duel. Here the final sentences :

Tu prieras publiquement Jacobi ou Gauss de donner leur avis non sur la verite, mais sur l’importance des theoremes.
Apres cela il se trouvera, j’espere, des gens qui trouvent leur profis a dechiffrer tout ce gachis.
Je t’embrasse avec effusion.
E. Galois, le 29 Mai 1832

A major result contained in this letter concerns the groups $L_2(p)=PSL_2(\mathbb{F}_p)$, that is the group of $2 \times 2$ matrices with determinant equal to one over the finite field $\mathbb{F}_p$ modulo its center. $L_2(p)$ is known to be simple whenever $p \geq 5$. Galois writes that $L_2(p)$ cannot have a non-trivial permutation representation on fewer than $p+1$ symbols whenever $p > 11$ and indicates the transitive permutation representation on exactly $p$ symbols in the three ‘exceptional’ cases $p=5,7,11$.

Let $\alpha = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and consider for $p=5,7,11$ the involutions on $\mathbb{P}^1_{\mathbb{F}_p} = \mathbb{F}_p \cup { \infty }$ (on which $L_2(p)$ acts via Moebius transformations)

$\pi_5 = (0,\infty)(1,4)(2,3) \quad \pi_7=(0,\infty)(1,3)(2,6)(4,5) \quad \pi_{11}=(0,\infty)(1,6)(3,7)(9,10)(5,8)(4,2)$

(in fact, Galois uses the involution $~(0,\infty)(1,2)(3,6)(4,8)(5,10)(9,7)$ for $p=11$), then $L_2(p)$ leaves invariant the set consisting of the $p$ involutions $\Pi = { \alpha^{-i} \pi_p \alpha^i~:~1 \leq i \leq p }$. After mentioning these involutions Galois merely writes :

Ainsi pour le cas de $p=5,7,11$, l’equation modulaire s’abaisse au degre p.
En toute rigueur, cette reduction n’est pas possible dans les cas plus eleves.

Alternatively, one can deduce these permutation representation representations from group isomorphisms. As $L_2(5) \simeq A_5$, the alternating group on 5 symbols, $L_2(5)$ clearly acts transitively on 5 symbols.

Similarly, for $p=7$ we have $L_2(7) \simeq L_3(2)$ and so the group acts as automorphisms on the projective plane over the field on two elements $\mathbb{P}^2_{\mathbb{F}_2}$ aka the Fano plane, as depicted on the left.

This finite projective plane has 7 points and 7 lines and $L_3(2)$ acts transitively on them.

For $p=11$ the geometrical object is a bit more involved. The set of non-squares in $\mathbb{F}_{11}$ is

${ 1,3,4,5,9 }$

and if we translate this set using the additive structure in $\mathbb{F}_{11}$ one obtains the following 11 five-element sets

${ 1,3,4,5,9 }, { 2,4,5,6,10 }, { 3,5,6,7,11 }, { 1,4,6,7,8 }, { 2,5,7,8,9 }, { 3,6,8,9,10 },$

${ 4,7,9,10,11 }, { 1,5,8,10,11 }, { 1,2,6,9,11 }, { 1,2,3,7,10 }, { 2,3,4,8,11 }$

and if we regard these sets as ‘lines’ we see that two distinct lines intersect in exactly 2 points and that any two distinct points lie on exactly two ‘lines’. That is, intersection sets up a bijection between the 55-element set of all pairs of distinct points and the 55-element set of all pairs of distinct ‘lines’. This is called the biplane geometry.

The subgroup of $S_{11}$ (acting on the eleven elements of $\mathbb{F}_{11}$) stabilizing this set of 11 5-element sets is precisely the group $L_2(11)$ giving the permutation representation on 11 objects.

An alternative statement of Galois’ result is that for $p > 11$ there is no subgroup of $L_2(p)$ complementary to the cyclic subgroup

$C_p = { \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}~:~x \in \mathbb{F}_p }$

That is, there is no subgroup such that set-theoretically $L_2(p) = F \times C_p$ (note this is of courese not a group-product, all it says is that any element can be written as $g=f.c$ with $f \in F, c \in C_p$.

However, in the three exceptional cases we do have complementary subgroups. In fact, set-theoretically we have

$L_2(5) = A_4 \times C_5 \qquad L_2(7) = S_4 \times C_7 \qquad L_2(11) = A_5 \times C_{11}$

and it is a truly amazing fact that the three groups appearing are precisely the three Platonic groups!

Recall that here are 5 Platonic (or Scottish) solids coming in three sorts when it comes to rotation-automorphism groups : the tetrahedron (group $A_4$), the cube and octahedron (group $S_4$) and the dodecahedron and icosahedron (group $A_5$). The “4” in the cube are the four body diagonals and the “5” in the dodecahedron are the five inscribed cubes.

That is, our three ‘exceptional’ Galois-groups correspond to the three Platonic groups, which in turn correspond to the three exceptional Lie algebras $E_6,E_7,E_8$ via McKay correspondence (wrt. their 2-fold covers). Maybe I’ll detail this latter connection another time. It sure seems that surprises often come in triples…

Finally, it is well known that $L_2(5) \simeq A_5$ is the automorphism group of the icosahedron (or dodecahedron) and that $L_2(7)$ is the automorphism group of the Klein quartic.

So, one might ask : is there also a nice curve connected with the third group $L_2(11)$? Rumour has it that this is indeed the case and that the curve in question has genus 70… (to be continued).

Reference