# what does the monster see?

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…

# Klein’s dessins d’enfant and the buckyball

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 $L_2(7)$ 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 $\tau x = b.x$.

# Galois’ last letter

“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

# the McKay-Thompson series

Monstrous moonshine was born (sometime in 1978) the moment John McKay realized that the linear term in the j-function

$j(q) = \frac{1}{q} + 744 + 196884 q + 21493760 q^2 + 864229970 q^3 + \ldots$

is surprisingly close to the dimension of the smallest non-trivial irreducible representation of the monster group, which is 196883. Note that at that time, the Monster hasn’t been constructed yet, and, the only traces of its possible existence were kept as semi-secret information in a huge ledger (costing 80 pounds…) kept in the Atlas-office at Cambridge. Included were 8 huge pages describing the character table of the monster, the top left fragment, describing the lower dimensional irreducibles and their characters at small order elements, reproduced below

If you look at the dimensions of the smallest irreducible representations (the first column) : 196883, 21296876, 842609326, … you will see that the first, second and third of them are extremely close to the linear, quadratic and cubic coefficient of the j-function. In fact, more is true : one can obtain these actual j-coefficients as simple linear combination of the dimensions of the irrducibles :

$\begin{cases} 196884 &= 1 + 196883 \\ 21493760 &= 1 + 196883 + 21296876 \\ 864229970 &= 2 \times 1 + 2 \times 196883 + 21296876 + 842609326 \end{cases}$

Often, only the first relation is attributed to McKay, whereas the second and third were supposedly discovered by John Thompson after MKay showed him the first. Marcus du Sautoy tells a somewhat different sory in Finding Moonshine :

McKay has also gone on to find these extra equations, but is was Thompson who first published them. McKay admits that “I was a bit peeved really, I don’t think Thompson quite knew how much I knew.”

By the work of Richard Borcherds we now know the (partial according to some) explanation behind these numerical facts : there is a graded representation $V = \oplus_i V_i$ of the Monster-group (actually, it has a lot of extra structure such as being a vertex algebra) such that the dimension of the i-th factor $V_i$ equals the coefficient f $q^i$ in the j-function. The homogeneous components $V_i$ being finite dimensional representations of the monster, they decompose into the 194 irreducibles $X_j$. For the first three components we have the decompositions

$\begin{cases} V_1 &= X_1 \oplus X_2 \\ V_2 &= X_1 \oplus X_2 \oplus X_3 \\ V_3 &= X_1^{\oplus 2 } \oplus X_2^{\oplus 2} \oplus X_3 \oplus X_4 \end{cases}$

Calculating the dimensions on both sides give the above equations. However, being isomorphisms of monster-representations we are not restricted to just computing the dimensions. We might as well compute the character of any monster-element on both sides (observe that the dimension is just the character of the identity element). Characters are the traces of the matrices describing the action of a monster-element on the representation and these numbers fill the different columns of the character-table above.

Hence, the same integral combinations of the character values of any monster-element give another q-series and these are called the McKay-Thompson series. John Conway discovered them to be classical modular functions known as Hauptmoduln.

In most papers and online material on this only the first few coefficients of these series are documented, which may be just too little information to make new discoveries!

Fortunately, David Madore has compiled the first 3200 coefficients of all the 172 monster-series which are available in a huge 8Mb file. And, if you really need to have more coefficients, you can always use and modify his moonshine python program.

In order to reduce bandwidth, here a list containing the first 100 coefficients of the j-function

jfunct=[196884, 21493760, 864299970, 20245856256, 333202640600, 4252023300096, 44656994071935, 401490886656000, 3176440229784420, 22567393309593600, 146211911499519294, 874313719685775360, 4872010111798142520, 25497827389410525184, 126142916465781843075, 593121772421445058560, 2662842413150775245160, 11459912788444786513920, 47438786801234168813250, 189449976248893390028800, 731811377318137519245696, 2740630712513624654929920, 9971041659937182693533820, 35307453186561427099877376, 121883284330422510433351500, 410789960190307909157638144, 1353563541518646878675077500, 4365689224858876634610401280, 13798375834642999925542288376, 42780782244213262567058227200, 130233693825770295128044873221, 389608006170995911894300098560, 1146329398900810637779611090240, 3319627709139267167263679606784, 9468166135702260431646263438600, 26614365825753796268872151875584, 73773169969725069760801792854360, 201768789947228738648580043776000, 544763881751616630123165410477688, 1452689254439362169794355429376000, 3827767751739363485065598331130120, 9970416600217443268739409968824320, 25683334706395406994774011866319670, 65452367731499268312170283695144960, 165078821568186174782496283155142200, 412189630805216773489544457234333696, 1019253515891576791938652011091437835, 2496774105950716692603315123199672320, 6060574415413720999542378222812650932, 14581598453215019997540391326153984000, 34782974253512490652111111930326416268, 82282309236048637946346570669250805760, 193075525467822574167329529658775261720, 449497224123337477155078537760754122752, 1038483010587949794068925153685932435825, 2381407585309922413499951812839633584128, 5421449889876564723000378957979772088000, 12255365475040820661535516233050165760000, 27513411092859486460692553086168714659374, 61354289505303613617069338272284858777600, 135925092428365503809701809166616289474168, 299210983800076883665074958854523331870720, 654553043491650303064385476041569995365270, 1423197635972716062310802114654243653681152, 3076095473477196763039615540128479523917200, 6610091773782871627445909215080641586954240, 14123583372861184908287080245891873213544410, 30010041497911129625894110839466234009518080, 63419842535335416307760114920603619461313664, 133312625293210235328551896736236879235481600, 278775024890624328476718493296348769305198947, 579989466306862709777897124287027028934656000, 1200647685924154079965706763561795395948173320, 2473342981183106509136265613239678864092991488, 5070711930898997080570078906280842196519646750, 10346906640850426356226316839259822574115946496, 21015945810275143250691058902482079910086459520, 42493520024686459968969327541404178941239869440, 85539981818424975894053769448098796349808643878, 171444843023856632323050507966626554304633241600, 342155525555189176731983869123583942011978493364, 679986843667214052171954098018582522609944965120, 1345823847068981684952596216882155845897900827370, 2652886321384703560252232129659440092172381585408, 5208621342520253933693153488396012720448385783600, 10186635497140956830216811207229975611480797601792, 19845946857715387241695878080425504863628738882125, 38518943830283497365369391336243138882250145792000, 74484518929289017811719989832768142076931259410120, 143507172467283453885515222342782991192353207603200, 275501042616789153749080617893836796951133929783496, 527036058053281764188089220041629201191975505756160, 1004730453440939042843898965365412981690307145827840, 1908864098321310302488604739098618405938938477379584, 3614432179304462681879676809120464684975130836205250, 6821306832689380776546629825653465084003418476904448, 12831568450930566237049157191017104861217433634289960, 24060143444937604997591586090380473418086401696839680, 44972195698011806740150818275177754986409472910549646, 83798831110707476912751950384757452703801918339072000]


This information will come in handy when we will organize our Monstrous Easter Egg Race, starting tomorrow at 6 am (GMT)…