# the monster graph and McKay’s observation

While the verdict on a neolithic Scottish icosahedron is still open, let us recall Kostant’s group-theoretic construction of the icosahedron from its rotation-symmetry group $A_5$.

The alternating group $A_5$ has two conjugacy classes of order 5 elements, both consisting of exactly 12 elements. Fix one of these conjugacy classes, say $C$ and construct a graph with vertices the 12 elements of $C$ and an edge between two $u,v \in C$ if and only if the group-product $u.v \in C$ still belongs to the same conjugacy class.

Observe that this relation is symmetric as from $u.v = w \in C$ it follows that $v.u=u^{-1}.u.v.u = u^{-1}.w.u \in C$. The graph obtained is the icosahedron, depicted on the right with vertices written as words in two adjacent elements u and v from $C$, as indicated.

Kostant writes : “Normally it is not a common practice in group theory to consider whether or not the product of two elements in a conjugacy class is again an element in that conjugacy class. However such a consideration here turns out to be quite productive.”

Still, similar constructions have been used in other groups as well, in particular in the study of the largest sporadic group, the monster group $\mathbb{M}$.

There is one important catch. Whereas it is quite trivial to multiply two permutations and verify whether the result is among 12 given ones, for most of us mortals it is impossible to do actual calculations in the monster. So, we’d better have an alternative way to get at the icosahedral graph using only $A_5$-data that is also available for the monster group, such as its character table.

Let $G$ be any finite group and consider three of its conjugacy classes $C(i),C(j)$ and $C(k)$. For any element $w \in C(k)$ we can compute from the character table of $G$ the number of different products $u.v = w$ such that $u \in C(i)$ and $v \in C(j)$. This number is given by the formula

$\frac{|G|}{|C_G(g_i)||C_G(g_j)|} \sum_{\chi} \frac{\chi(g_i) \chi(g_j) \overline{\chi(g_k)}}{\chi(1)}$

where the sum is taken over all irreducible characters $\chi$ and where $g_i \in C(i),g_j \in C(j)$ and $g_k \in C(k)$. Note also that $|C_G(g)|$ is the number of $G$-elements commuting with $g$ and that this number is the order of $G$ divided by the number of elements in the conjugacy class of $g$.

The character table of $A_5$ is given on the left : the five columns correspond to the different conjugacy classes of elements of order resp. 1,2,3,5 and 5 and the rows are the character functions of the 5 irreducible representations of dimensions 1,3,3,4 and 5.

Let us fix the 4th conjugacy class, that is 5a, as our class $C$. By the general formula, for a fixed $w \in C$ the number of different products $u.v=w$ with $u,v \in C$ is equal to

$\frac{60}{25}(\frac{1}{1} + \frac{(\frac{1+\sqrt{5}}{2})^3}{3} + \frac{(\frac{1-\sqrt{5}}{2})^3}{3} – \frac{1}{4} + \frac{0}{5}) = \frac{60}{25}(1 + \frac{4}{3} – \frac{1}{4}) = 5$

Because for each $x \in C$ also its inverse $x^{-1} \in C$, this can be rephrased by saying that there are exactly 5 different products $w^{-1}.u \in C$, or equivalently, that the valency of every vertex $w^{-1} \in C$ in the graph is exactly 5.

That is, our graph has 12 vertices, each with exactly 5 neighbors, and with a bit of extra work one can show it to be the icosahedral graph.

For the monster group, the Atlas tells us that it has exactly 194 irreducible representations (and hence also 194 conjugacy classes). Of these conjugacy classes, the involutions (that is the elements of order 2) are of particular importance.

There are exactly 2 conjugacy classes of involutions, usually denoted 2A and 2B. Involutions in class 2A are called “Fischer-involutions”, after Bernd Fischer, because their centralizer subgroup is an extension of Fischer’s baby Monster sporadic group.

Likewise, involutions in class 2B are usually called “Conway-involutions” because their centralizer subgroup is an extension of the largest Conway sporadic group.

Let us define the monster graph to be the graph having as its vertices the Fischer-involutions and with an edge between two of them $u,v \in 2A$ if and only if their product $u.v$ is again a Fischer-involution.

Because the centralizer subgroup is $2.\mathbb{B}$, the number of vertices is equal to $97239461142009186000 = 2^4 * 3^7 * 5^3 * 7^4 * 11 * 13^2 * 29 * 41 * 59 * 71$.

From the general result recalled before we have that the valency in all vertices is equal and to determine it we have to use the character table of the monster and the formula. Fortunately GAP provides the function ClassMultiplicationCoefficient to do this without making errors.

 gap> table:=CharacterTable("M"); CharacterTable( "M" ) gap> ClassMultiplicationCoefficient(table,2,2,2); 27143910000 

Perhaps noticeable is the fact that the prime decomposition of the valency $27143910000 = 2^4 * 3^4 * 5^4 * 23 * 31 * 47$ is symmetric in the three smallest and three largest prime factors of the baby monster order.

Robert Griess proved that one can recover the monster group $\mathbb{M}$ from the monster graph as its automorphism group!

As in the case of the icosahedral graph, the number of vertices and their common valency does not determine the monster graph uniquely. To gain more insight, we would like to know more about the sizes of minimal circuits in the graph, the number of such minimal circuits going through a fixed vertex, and so on.

Such an investigation quickly leads to a careful analysis which other elements can be obtained from products $u.v$ of two Fischer involutions $u,v \in 2A$. We are in for a major surprise, first observed by John McKay:

Printing out the number of products of two Fischer-involutions giving an element in the i-th conjugacy class of the monster,
where i runs over all 194 possible classes, we get the following string of numbers :

 97239461142009186000, 27143910000, 196560, 920808, 0, 3, 1104, 4, 0, 0, 5, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 

That is, the elements of only 9 conjugacy classes can be written as products of two Fischer-involutions! These classes are :

• 1A = { 1 } written in 97239461142009186000 different ways (after all involutions have order two)
• 2A, each element of which can be written in exactly 27143910000 different ways (the valency)
• 2B, each element of which can be written in exactly 196560 different ways. Observe that this is the kissing number of the Leech lattice leading to a permutation representation of $2.Co_1$.
• 3A, each element of which can be written in exactly 920808 ways. Note that this number gives a permutation representation of the maximal monster subgroup $3.Fi_{24}’$.
• 3C, each element of which can be written in exactly 3 ways.
• 4A, each element of which can be written in exactly 1104 ways.
• 4B, each element of which can be written in exactly 4 ways.
• 5A, each element of which can be written in exactly 5 ways.
• 6A, each element of which can be written in exactly 6 ways.

Let us forget about the actual numbers for the moment and concentrate on the orders of these 9 conjugacy classes : 1,2,2,3,3,4,4,5,6. These are precisely the components of the fundamental root of the extended Dynkin diagram $\tilde{E_8}$!

This is the content of John McKay’s E(8)-observation : there should be a precise relation between the nodes of the extended Dynkin diagram and these 9 conjugacy classes in such a way that the order of the class corresponds to the component of the fundamental root. More precisely, one conjectures the following correspondence

This is similar to the classical McKay correspondence between finite subgroups of $SU(2)$ and extended Dynkin diagrams (the binary icosahedral group corresponding to extended E(8)). In that correspondence, the nodes of the Dynkin diagram correspond to irreducible representations of the group and the edges are determined by the decompositions of tensor-products with the fundamental 2-dimensional representation.

Here, however, the nodes have to correspond to conjugacy classes (rather than representations) and we have to look for another procedure to arrive at the required edges! An exciting proposal has been put forward recently by John Duncan in his paper Arithmetic groups and the affine E8 Dynkin diagram.

It will take us a couple of posts to get there, but for now, let’s give the gist of it : monstrous moonshine gives a correspondence between conjugacy classes of the monster and certain arithmetic subgroups of $PSL_2(\mathbb{R})$ commensurable with the modular group $\Gamma = PSL_2(\mathbb{Z})$. The edges of the extended Dynkin E(8) diagram are then given by the configuration of the arithmetic groups corresponding to the indicated 9 conjugacy classes! (to be continued…)

# noncommutative F_un geometry (2)

Last time we tried to generalize the Connes-Consani approach to commutative algebraic geometry over the field with one element $\mathbb{F}_1$ to the noncommutative world by considering covariant functors

$N~:~\mathbf{groups} \rightarrow \mathbf{sets}$

which over $\mathbb{C}$ resp. $\mathbb{Z}$ become visible by a complex (resp. integral) algebra having suitable universal properties.

However, we didn’t specify what we meant by a complex noncommutative variety (resp. an integral noncommutative scheme). In particular, we claimed that the $\mathbb{F}_1$-‘points’ associated to the functor

$D~:~\mathbf{groups} \rightarrow \mathbf{sets} \qquad G \mapsto G_2 \times G_3$ (here $G_n$ denotes all elements of order $n$ of $G$)

were precisely the modular dessins d’enfants of Grothendieck, but didn’t give details. We’ll try to do this now.

For algebras over a field we follow the definition, due to Kontsevich and Soibelman, of so called “noncommutative thin schemes”. Actually, the thinness-condition is implicit in both Soule’s-approach as that of Connes and Consani : we do not consider R-points in general, but only those of rings R which are finite and flat over our basering (or field).

So, what is a noncommutative thin scheme anyway? Well, its a covariant functor (commuting with finite projective limits)

$\mathbb{X}~:~\mathbf{Alg}^{fd}_k \rightarrow \mathbf{sets}$

from finite-dimensional (possibly noncommutative) $k$-algebras to sets. Now, the usual dual-space operator gives an anti-equivalence of categories

$\mathbf{Alg}^{fd}_k \leftrightarrow \mathbf{Coalg}^{fd}_k \qquad A=C^* \leftrightarrow C=A^*$

so a thin scheme can also be viewed as a contra-variant functor (commuting with finite direct limits)

$\mathbb{X}~:~\mathbf{Coalg}^{fd}_k \rightarrow \mathbf{Sets}$

In particular, we are interested to associated to any {tex]k $-algebra$A $its representation functor :$\mathbf{rep}(A)~:~\mathbf{Coalg}^{fd}_k \rightarrow \mathbf{Sets} \qquad C \mapsto Alg_k(A,C^*) $This may look strange at first sight, but$C^* $is a finite dimensional algebra and any$n $-dimensional representation of$A $is an algebra map$A \rightarrow M_n(k) $and we take$C $to be the dual coalgebra of this image. Kontsevich and Soibelman proved that every noncommutative thin scheme$\mathbb{X} $is representable by a$k $-coalgebra. That is, there exists a unique coalgebra$C_{\mathbb{X}} $(which they call the coalgebra of ‘distributions’ of$\mathbb{X} $) such that for every finite dimensional$k $-algebra$B $we have$\mathbb{X}(B) = Coalg_k(B^*,C_{\mathbb{X}}) $In the case of interest to us, that is for the functor$\mathbf{rep}(A) $the coalgebra of distributions is Kostant’s dual coalgebra$A^o $. This is the not the full linear dual of$A $but contains only those linear functionals on$A $which factor through a finite dimensional quotient. So? You’ve exchanged an algebra$A $for some coalgebra$A^o $, but where’s the geometry in all this? Well, let’s look at the commutative case. Suppose$A= \mathbb{C}[X] $is the coordinate ring of a smooth affine variety$X $, then its dual coalgebra looks like$\mathbb{C}[X]^o = \oplus_{x \in X} U(T_x(X)) $the direct sum of all universal (co)algebras of tangent spaces at points$x \in X $. But how do we get the variety out of this? Well, any coalgebra has a coradical (being the sun of all simple subcoalgebras) and in the case just mentioned we have$corad(\mathbb{C}[X]^o) = \oplus_{x \in X} \mathbb{C} e_x $so every point corresponds to a unique simple component of the coradical. In the general case, the coradical of the dual coalgebra$A^o $is the direct sum of all simple finite dimensional representations of$A $. That is, the direct summands of the coalgebra give us a noncommutative variety whose points are the simple representations, and the remainder of the coalgebra of distributions accounts for infinitesimal information on these points (as do the tangent spaces in the commutative case). In fact, it was a surprise to me that one can describe the dual coalgebra quite explicitly, and that$A_{\infty} $-structures make their appearance quite naturally. See this paper if you’re in for the details on this. That settles the problem of what we mean by the noncommutative variety associated to a complex algebra. But what about the integral case? In the above, we used extensively the theory of Kostant-duality which works only for algebras over fields… Well, not quite. In the case of$\mathbb{Z} $(or more general, of Dedekind domains) one can repeat Kostant’s proof word for word provided one takes as the definition of the dual$\mathbb{Z} $-coalgebra of an algebra (which is$\mathbb{Z} $-torsion free)$A^o = { f~:~A \rightarrow \mathbb{Z}~:~A/Ker(f)~\text{is finitely generated and torsion free}~} $(over general rings there may be also variants of this duality, as in Street’s book an Quantum groups). Probably lots of people have come up with this, but the only explicit reference I have is to the first paper I’ve ever written. So, also for algebras over$\mathbb{Z} $we can define a suitable noncommutative integral scheme (the coradical approach accounts only for the maximal ideals rather than all primes, but somehow this is implicit in all approaches as we consider only thin schemes). Fine! So, we can make sense of the noncommutative geometrical objects corresponding to the group-algebras$\mathbb{C} \Gamma $and$\mathbb{Z} \Gamma $where$\Gamma = PSL_2(\mathbb{Z}) $is the modular group (the algebras corresponding to the$G \mapsto G_2 \times G_3 $-functor). But, what might be the points of the noncommutative scheme corresponding to$\mathbb{F}_1 \Gamma $??? Well, let’s continue the path cut out before. “Points” should correspond to finite dimensional “simple representations”. Hence, what are the finite dimensional simple$\mathbb{F}_1 $-representations of$\Gamma $? (Or, for that matter, of any group$G $) Here we come back to Javier’s post on this : a finite dimensional$\mathbb{F}_1 $-vectorspace is a finite set. A$\Gamma $-representation on this set (of n-elements) is a group-morphism$\Gamma \rightarrow GL_n(\mathbb{F}_1) = S_n $hence it gives a permutation representation of$\Gamma $on this set. But then, if finite dimensional$\mathbb{F}_1 $-representations of$\Gamma $are the finite permutation representations, then the simple ones are the transitive permutation representations. That is, the points of the noncommutative scheme corresponding to$\mathbb{F}_1 \Gamma $are the conjugacy classes of subgroups$H \subset \Gamma $such that$\Gamma/H $is finite. But these are exactly the modular dessins d’enfants introduced by Grothendieck as I explained a while back elsewhere (see for example this post and others in the same series). # noncommutative F_un geometry (1) It is perhaps surprising that Alain Connes and Katia Consani, two icons of noncommutative geometry, restrict themselves to define commutative algebraic geometry over$\mathbb{F}_1 $, the field with one element. My guess of why they stop there is as good as anyone’s. Perhaps they felt that there is already enough noncommutativity in Soule’s gadget-approach (the algebra$\mathcal{A}_X $as in this post may very well be noncommutative). Perhaps they were only interested in the Bost-Connes system which can be entirely encoded in their commutative$\mathbb{F}_1 $-geometry. Perhaps they felt unsure as to what the noncommutative scheme of an affine noncommutative algebra might be. Perhaps … Remains the fact that their approach screams for a noncommutative extension. Their basic object is a covariant functor$N~:~\mathbf{abelian} \rightarrow \mathbf{sets} \qquad A \mapsto N(A) $from finite abelian groups to sets, together with additional data to the effect that there is a unique minimal integral scheme associated to$N $. In a series of posts on the Connes-Consani paper (starting here) I took some care of getting rid of all scheme-lingo and rephrasing everything entirely into algebras. But then, this set-up can be extended verbatim to noncommuative$\mathbb{F}_1 $-geometry, which should start from a covariant functor$N~:~\mathbf{groups} \rightarrow \mathbf{sets} $from all finite groups to sets. Let’s recall quickly what the additional info should be making this functor a noncommutative (affine) F_un scheme : There should be a finitely generated$\mathbb{C} $-algebra$R $together with a natural transformation (the ‘evaluation’)$e~:~N \rightarrow \mathbf{maxi}(R) \qquad N(G) \mapsto Hom_{\mathbb{C}-alg}(R, \mathbb{C} G) $(both$R $and the group-algebra$\mathbb{C} G $may be noncommutative). The pair$(N, \mathbf{maxi}(R)) $is then called a gadget and there is an obvious notion of ‘morphism’ between gadgets. The crucial extra ingredient is an affine$\mathbb{Z} $-algebra (possibly noncommutative)$S $such that$N $is a subfunctor of$\mathbf{mini}(S)~:~G \mapsto Hom_{\mathbb{Z}-alg}(S,\mathbb{Z} G) $together with the following universal property : any affine$\mathbb{Z} $-algebra$T $having a gadget-morphism$~(N,\mathbf{maxi}(R)) \rightarrow (\mathbf{mini}(T),\mathbf{maxi}(T \otimes_{\mathbb{Z}} \mathbb{C})) $comes from a$\mathbb{Z} $-algebra morphism$T \rightarrow S $. (If this sounds too cryptic for you, please read the series on C-C mentioned before). So, there is no problem in defining noncommutative affine F_un-schemes. However, as with any generalization, this only makes sense provided (a) we get something new and (b) we have interesting examples, not covered by the restricted theory. At first sight we do not get something new as in the only example we did in the C-C-series (the forgetful functor) it is easy to prove (using the same proof as given in this post) that the forgetful-functor$\mathbf{groups} \rightarrow \mathbf{sets} $still has as its integral form the integral torus$\mathbb{Z}[x,x^{-1}] $. However, both theories quickly diverge beyond this example. For example, consider the functor$\mathbf{groups} \rightarrow \mathbf{sets} \qquad G \mapsto G \times G $Then, if we restrict to abelian finite groups$\mathbf{abelian} $it is easy to see (again by a similar argument) that the two-dimensional integer torus$\mathbb{Z}[x,y,x^{-1},y^{-1}] $is the correct integral form. However, this algebra cannot be the correct form for the functor on the category of all finite groups as any$\mathbb{Z} $-algebra map$\phi~:~\mathbb{Z}[x,y,x^{-1},y^{-1}] \rightarrow \mathbb{Z} G $determines (and is determined by) a pair of commuting units in$\mathbb{Z} G $, so the above functor can not be a subfunctor if we allow non-Abelian groups. But then, perhaps there isn’t a minimal integral$\mathbb{Z} $-form for this functor? Well, yes there is. Take the free group in two letters (that is, all words in noncommuting$x,y,x^{-1} $and$y^{-1} $satisfying only the trivial cancellation laws between a letter and its inverse), then the corresponding integral group-algebra$\mathbb{Z} \mathcal{F}_2 $does the trick. Again, the proof-strategy is the same. Given a gadget-morphism we have an algebra map$f~:~T \mapsto \mathbb{C} \mathcal{F}_2 $and we have to show, using the universal property that the image of$T $is contained in the integral group-algebra$\mathbb{Z} \mathcal{F}_2 $. Take a generator$z $of$T $then the degree of the image$f(z) $is bounded say by$d $and we can always find a subgroup$H \subset \mathcal{F}_2 $such that$\mathcal{F}_2/H $is a fnite group and the quotient map$\mathbb{C} \mathcal{F}_2 \rightarrow \mathbb{C} \mathcal{F}_2/H $is injective on the subspace spanned by all words of degree strictly less than$d+1 $. Then, the usual diagram-chase finishes the proof. What makes this work is that the free group$\mathcal{F}_2 $has ‘enough’ subgroups of finite index, a property it shares with many interesting discrete groups. Whence the blurb-message : if the integers$\mathbb{Z} $see a discrete group$\Gamma $, then the field$\mathbb{F}_1 $sees its profinite completion$\hat{\Gamma} = \underset{\leftarrow}{lim}~\Gamma/ H $So, yes, we get something new by extending the Connes-Consani approach to the noncommutative world, but do we have interesting examples? As “interesting” is a subjective qualification, we’d better invoke the authority-argument. Alexander Grothendieck (sitting on the right, manifestly not disputing a vacant chair with Jean-Pierre Serre, drinking on the left (a marvelous picture taken by F. Hirzebruch in 1958)) was pushing the idea that profinite completions of arithmetical groups were useful in the study of the absolute Galois group$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $, via his theory of dessins d’enfants (children;s drawings). In a previous life, I’ve written a series of posts on dessins d’enfants, so I’ll restrict here to the basics. A smooth projective$\overline{\mathbb{Q}} $-curve$X $has a Belyi-map$X \rightarrow \mathbb{P}^1_{\overline{\mathbb{Q}}} $ramified only in three points${ 0,1,\infty } $. The “drawing” corresponding to$X $is a bipartite graph, drawn on the Riemann surface$X_{\mathbb{C}} $obtained by lifting the unit interval$[0,1] $to$X $. As the absolute Galois group acts on all such curves (and hence on their corresponding drawings), the action of it on these dessins d’enfants may give us a way into the multiple mysteries of the absolute Galois group. In his “Esquisse d’un programme” (Sketch of a program if you prefer to read it in English) he writes : “C’est ainsi que mon attention s’est portée vers ce que j’ai appelé depuis la “géométrie algêbrique anabélienne”, dont le point de départ est justement une étude (pour le moment limitée à la caractéristique zéro) de l’action de groupe de Galois “absolus” (notamment les groupes$Gal(\overline{K}/K) $, ou$K $est une extension de type fini du corps premier) sur des groupes fondamentaux géométriques (profinis) de variétés algébriques (définies sur$K $), et plus particulièrement (rompant avec une tradition bien enracinée) des groupes fondamentaux qui sont trés éloignés des groupes abéliens (et que pour cette raison je nomme “anabéliens”). Parmi ces groupes, et trés proche du groupe$\hat{\pi}_{0,3} $, il y a le compactifié profini du groupe modulaire$SL_2(\mathbb{Z}) $, dont le quotient par le centre$\pm 1 $contient le précédent comme sous-groupe de congruence mod 2, et peut s’interpréter d’ailleurs comme groupe “cartographique” orienté, savoir celui qui classifie les cartes orientées triangulées (i.e. celles dont les faces des triangles ou des monogones).” and a bit further, he writes : “L’élément de structure de$SL_2(\mathbb{Z}) $qui me fascine avant tout, est bien sur l’action extérieure du groupe de Galois$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $sur le compactifié profini. Par le théorème de Bielyi, prenant les compactifiés profinis de sous-groupes d’indice fini de$SL_2(\mathbb{Z}) $, et l’action extérieure induite (quitte à passer également à un sous-groupe overt de$Gal(\overline{\mathbb{Q}},\mathbb{Q}) $), on trouve essentiellement les groupes fondamentaux de toutes les courbes algébriques définis sur des corps de nombres$K $, et l’action extérieure de$Gal(\overline{K}/K) $dessus.” So, is there a noncommutative affine variety over$\mathbb{F}_1 $of which the unique minimal integral model is the integral group algebra of the modular group$\mathbb{Z} \Gamma $(with$\Gamma = PSL_2(\mathbb{Z}) $? Yes, here it is$N_{\Gamma}~:~\mathbf{groups} \rightarrow \mathbf{sets} \qquad G \mapsto G_2 \times G_3 $where$G_n $is the set of all elements of order$n $in$G $. The reason behind this is that the modular group is the free group product$C_2 \ast C_3 $. Fine, you may say, but all this is just algebra. Where is the noncommutative complex variety or the noncommutative integral scheme in all this? Well, we can introduce them too but as this post is already 1300 words long, I’ll better leave this for another time. In case you cannot stop thinking about it, here’s the short answer. The complex noncommutative variety has as its ‘points’ all finite dimensional simple complex representations of the modular group, and the ‘points’ of the noncommutative$\mathbb{F}_1 $-scheme are exactly the (modular) dessins d’enfants… # what does the monster see? The Monster is the largest of the 26 sporadic simple groups and has order 808 017 424 794 512 875 886 459 904 961 710 757 005 754 368 000 000 000 = 2^46 3^20 5^9 7^6 11^2 13^3 17 19 23 29 31 41 47 59 71. It is not so much the size of its order that makes it hard to do actual calculations in the monster, but rather the dimensions of its smallest non-trivial irreducible representations (196 883 for the smallest, 21 296 876 for the next one, and so on). In characteristic two there is an irreducible representation of one dimension less (196 882) which appears to be of great use to obtain information. For example, Robert Wilson used it to prove that The Monster is a Hurwitz group. This means that the Monster is generated by two elements g and h satisfying the relations$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}}