Category: groups

the “uninteresting” case p=5

Published July 4, 2008 by lieven

I was hoping you would write a post on the ‘uninteresting case’ of p=5 in this context. Note that the truncated tetrahedron has (V,E,F)=(12,18,8) which is a triple that appears in the ternary (cyclic) geometry for the cube. This triple can be 4 hexagons and 4 triangles (the truncated tetrahedron) OR 4 pentagons and 4 squares!

Kea commented and I didnt know the answer to the ‘obvious’ question :

how can one get the truncated tetrahedron from either of the two conjugacy classes of order 5 elements in $L_{2} (5) = A_{5}$ , each consisting of 12 elements.

Fortunately the groups involved are small enough to enable hand-calculations. Probably there is a more elegant way to do this, but I was already happy to find this construction…

This time, there is just one conjugacy class of subgroups isomorphic to $A_{4}$ (the symmetry group of the (truncated) tetrahedron) in $L_{2} (5) = A_{5}$ . Take one of the two conjugacy classes C of 5-cycles in $A_{5}$ and use the following notation for its 12 elements :

A=(1,2,3,4,5), B=(1,2,4,5,3), C=(1,2,5,3,4), D=(1,3,5,4,2), E=(1,3,2,5,4), F=(1,3,4,2,5), G=(1,5,4,3,2), H=(1,5,3,2,4), I=(1,5,2,4,3), J=(1,4,2,3,5), K=(1,4,5,2,3), L=(1,4,3,5,2)

We’d like to view these elements as the vertices of a truncated tetrahedron, so we need to find the 4 triangles and the 6 connecting edges between them. The first task calls for order 3 elements, the second one for order two elements.

Take a conjugacy class of order 3 elements in $A_{4}$ say $T = (2, 4, 3), (1, 2, 3), (1, 3, 4), (1, 4, 2)$ and observe that when one computes the products of T with a fixed 5-cycle in the conjugacy class C there is a unique element among the four obtained that belongs to the conjugacy class C. This gives a cyclic action on C with orbits of length 3 (the triangles). Here they are :

A–> J –> F –> A, B–>C–>H–>B, D–>G –> E–>D, I–>L–>K–>I

For the edges, take the conjugacy class $S = (1, 2) (3, 4), (1, 3) (2, 4), (1, 4) (2, 3)$ of order two elements in $A_{4}$ and compute for any 5-cycle c in C the products c_S_c and observe that among the elements obtained there is again one element belonging to C. This gives the following pairing

A<-->C, B<-->I, D<-->F, E<-->H, G<-->L and J<-->K and a bit of puzzling shows that all this can indeed be realized within a truncated tetrahedron (on the right). As to her other request

… and how about a post on how 1 + 4 + 9 + … + 24^2 = 70^2 is REALLY a statement about unifying cusps and holes (genus) as degrees of freedom in quantum geometry.

The scarecrow will need to take some time to think before giving his answer…

the buckyball curve

Published July 2, 2008 by lieven

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 $x^{3} y + y^{3} z + z^{3} x = 0$ . 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 $L_{2} (7) ≃ P G L_{3} (F_{2})$ is the isomorphism group of the projective plane over the finite field $F_{2}$ 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 $Γ (7)$ . Recall that a modular interpretation of the Klein quartic is as $H / Γ (7)$ where $H$ is the upper half-plane on which the modular group $Γ = P S L_{2} (Z)$ acts via Moebius transformations, that is, to a 2×2 matrix corresponds the transformation

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

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 $Γ = C_{2} * C_{3} \to S_{4}$ .

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

$H^{5} ≃ C_{2} * C_{5} = ⟨ z \mapsto - \frac{1}{z}, z \mapsto z + ϕ ⟩$

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

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

Hence, Martin and Singerman define the buckyball curve as the modular quotient $X = H / Λ$ 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).

Klein’s dessins d’enfant and the buckyball

Published June 30, 2008 by lieven

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 $L_{2} (p) = P S L_{2} (F_{p})$ 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 $A ≃ A_{5}$ …

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 $P_{C}^{1} \to P_{C}^{1}$ 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 $τ$ of order two pairing the half-edges adjacent to a +-vertex and the order three permutation $σ$ 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
$σ = (7, 10, 9) (5, 11, 6) (1, 4, 2)$ and $τ = (8, 9) (7, 11) (1, 5) (3, 4)$ .

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

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

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

Klein used the fact that $L_{2} (11)$ only has elements of orders 1,2,3,5,6 and 11. So, in each of the remaining cases he had to find an element of a different order. For example, in type V he verified that the element $τ . (σ . τ)^{3}$ 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 $b = [x, a] = x^{- 1} a^{- 1} x a$ belongs again to D. The unique hexagonal side having vertex x connects it to the element $b . x$ which belongs again to C as $b . x = (a x)^{- 1} . x . (a x)$ .

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 $τ x = b . x$ .