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Tag: hyperbolic

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…

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the buckyball curve

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

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

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

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




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

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




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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

References :

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

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

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Arnold’s trinities

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

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

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

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

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

1 : “The first trinity everyone knows is”

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

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

with corresponding Dynkin-Coxeter diagrams

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

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

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

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



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



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

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

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



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

7 : The seventh trinity involves complex polynomials in one variable

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

8 : The eight one is another beauty

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

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

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

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

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

14 : The exceptional Galois groups

explained last time.

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

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

16 : The three generations of sporadic groups

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

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