Tag: geometry

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)=PSL_2(\mathbb{F}_p) $ has a (transitive) permutation presentation on p elements. For, p=11 the group $L_2(11) $ is of order 660, so it permuting 11 elements means that this set must be of the form $X=L_2(11)/A $ with $A \subset L_2(11) $ a subgroup of 60 elements… and it turns out that $A \simeq A_5 $…

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Arnold’s trinities

Published June 17, 2008 by lieven

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?

Galois’ last letter

Published June 12, 2008 by lieven

“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

Bertram Kostant, “The graph of the truncated icosahedron and the last letter of Galois”