neverendingbooks

sporadic simple games

Published July 13, 2008 by lievenlb

About a year ago I did a series of posts on games associated to the Mathieu sporadic group $M_{12} $, starting with a post on Conway’s puzzle M(13), and, continuing with a discussion of mathematical blackjack. The idea at the time was to write a book for a general audience, as discussed at the start of the M(13)-post, ending with a series of new challenging mathematical games. I asked : “What kind of puzzles should we promote for mathematical thinking to have a fighting chance to survive in the near future?”

Now, Scientific American has (no doubt independently) taken up this lead. Their July 2008 issue features the article Rubik’s Cube Inspired Puzzles Demonstrate Math’s “Simple Groups” written by Igor Kriz and Paul Siegel.

By far the nicest thing about this article is that it comes with three online games based on the sporadic simple groups, the Mathieu groups $M_{12} $, $M_{24} $ and the Conway group $.0 $.

the M(12) game

Scrambles to an arbitrary permutation in $M_{12} $ and need to use the two generators $INVERT=(1,12)(2,11)(3,10)(4,9)(5,8)(6,7) $ and $MERGE=(2,12,7,4,11,6,10,8,9,5,3) $ to return to starting position.

Here is the help-screen :

They promise the solution by july 27th, but a few-line GAP-program cracks the puzzle instantly.

the M(24) game

Similar in nature, again using two generators of $M_{24} $. GAP-solution as before.

This time, they offer this help-screen :

the .0 game

Their most original game is based on Conway’s $.0 $ (dotto) group. Unfortunately, they offer only a Windows-executable version, so I had to install Bootcamp and struggle a bit with taking screenshots on a MacBook to show you the game’s starting position :

Dotto:

Dotto, our final puzzle, represents the Conway group Co0, published in 1968 by mathematician John H. Conway of Princeton University. Co0 contains the sporadic simple group Co1 and has exactly twice as many members as Co1. Conway is too modest to name Co0 after himself, so he denotes the group “.0” (hence the pronunciation “dotto”).

In Dotto, there are four moves. This puzzle includes the M24 puzzle. Look at the yellow/blue row in the bottom. This is, in fact, M24, but the numbers are arranged in a row instead of a circle. The R move is the “circle rotation to the right”: the column above the number 0 stays put, but the column above the number 1 moves to the column over the number 2 etc. up to the column over the number 23, which moves to the column over the number 1. You may also click on a column number and then on another column number in the bottom row, and the “circle rotation” moving the first column to the second occurs. The M move is the switch, in each group of 4 columns separated by vertical lines (called tetrads) the “yellow” columns switch and the “blue” columns switch. The sign change move (S) changes signs of the first 8 columns (first two tetrads). The tetrad move (T) is the most complicated: Subtract in each row from each tetrad 1/2 times the sum of the numbers in that tetrad. Then in addition to that, reverse the signs of the columns in the first tetrad.

Strategy hints: Notice that the sum of squares of the numbers in each row doesn’t change. (This sum of squares is 64 in the first row, 32 in every other row.) If you manage to get an “8”in the first row, you have almost reduced the game to M24 except those signs. To have the original position, signs of all numbers on the diagonal must be +. Hint on signs: if the only thing wrong are signs on the diagonal, and only 8 signs are wrong, those 8 columns can be moved to the first 8 columns by using only the M24 moves (M,R).

GAMAP 2008

Published July 7, 2008 by lievenlb

Next week, our annual summer school Geometric and Algebraic Methods with Applications in Physics will start, once again (ive lost count which edition it is).

Because Isar is awol to la douce France, I’ll be responsible (once again) for the web-related stuff of the meeting. So, here a couple of requests to participants/lecturers :

if you are giving a mini-course and would like to have your material online, please contact me and i’ll make you an author of the Arts blog.
if you are a student attending the summerschool and would love to do some Liveblogging about the meeting, please do the same.

I’ll try to do some cross-posting here when it comes to my own lectures (and, perhaps, a few others). For now, I settled on ‘What is noncommutative geometry?’ as a preliminary title, but then, I’m in the position to change the program with a few keystrokes, so I’ll probably change it by then (or remove myself from it altogether…).

At times, I feel it would be more fun to do a few talks on Math-blogging. An entertaining hour could be spend on the forensic investigation of the recent Riemann-Hypothesis-hype in (a good part of) the math-blogosphere…

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the “uninteresting” case p=5

Published July 4, 2008 by lievenlb

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…

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the buckyball curve

Published July 2, 2008 by lievenlb

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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Klein’s dessins d’enfant and the buckyball

Published June 30, 2008 by lievenlb

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 $.

the buckyball symmetries

Published June 27, 2008 by lievenlb

The buckyball is without doubt the hottest mahematical object at the moment (at least in Europe). Recall that the buckyball (middle) is a mixed form of two Platonic solids

the Icosahedron on the left and the Dodecahedron on the right.

For those of you who don’t know anything about football, it is that other ball-game, best described via a quote from the English player Gary Lineker

“Football is a game for 22 people that run around, play the ball, and one referee who makes a slew of mistakes, and in the end Germany always wins.”

We still have a few days left hoping for a better ending… Let’s do some bucky-maths : what is the rotation symmetry group of the buckyball?

For starters, dodeca- and icosahedron are dual solids, meaning that if you take the center of every face of a dodecahedron and connect these points by edges when the corresponding faces share an edge, you’ll end up with the icosahedron (and conversely). Therefore, both solids (as well as their mixture, the buckyball) will have the same group of rotational symmetries. Can we at least determine the number of these symmetries?

Take the dodecahedron and fix a face. It is easy to find a rotation taking this face to anyone of its five adjacent faces. In group-slang : the rotation automorphism group acts transitively on the 12 faces of the dodecohedron. Now, how many of them fix a given face? These can only be rotations with axis through the center of the face and there are exactly 5 of them preserving the pentagonal face. So, in all we have $12 \times 5 = 60 $ rotations preserving any of the three solids above. By composing two of its elements, we get another rotational symmetry, so they form a group and we would like to determine what that group is.

There is one group that springs to mind $A_5 $, the subgroup of all even permutations on 5 elements. In general, the alternating group has half as many elements as the full permutation group $S_n $, that is $\frac{1}{2} n! $ (for multiplying with the involution (1,2) gives a bijection between even and odd permutations). So, for $A_5 $ we get 60 elements and we can list them :

the trivial permutation$~() $, being the identity.
permutations of order two with cycle-decompostion $~(i_1,i_2)(i_3,i_4) $, and there are exactly 15 of them around when all numbers are between 1 and 5.
permutations of order three with cycle-form $~(i_1,i_2,i_3) $ of which there are exactly 20.
permutations of order 5 which have to form one full cycle $~(i_1,i_2,i_3,i_4,i_5) $. There are 24 of those.

Can we at least view these sets of elements as rotations of the buckyball? Well, a dodecahedron has 12 pentagobal faces. So there are 4 nontrivial rotations of order 5 for every 2 opposite faces and hence the dodecaheder (and therefore also the buckyball) has indeed 6×4=24 order 5 rotational symmetries.

The icosahedron has twenty triangles as faces, so any of the 10 pairs of opposite faces is responsible for two non-trivial rotations of order three, giving us 10×2=20 order 3 rotational symmetries of the buckyball.

The order two elements are slightly harder to see. The icosahedron has 30 edges and there is a plane going through each of the 15 pairs of opposite edges splitting the icosahedron in two. Hence rotating to interchange these two edges gives one rotational symmetry of order 2 for each of the 15 pairs.

And as 24+20+15+1(identity) = 60 we have found all the rotational symmetries and we see that they pair up nicely with the elements of $A_5 $. But do they form isomorphic groups? In other words, can the buckyball see the 5 in the group $A_5 $.

In a previous post I’ve shown that one way to see this 5 is as the number of inscribed cubes in the dodecahedron. But, there is another way to see the five based on the order 2 elements described before.

If you look at pairs of opposite edges of the icosahedron you will find that they really come in triples such that the planes determined by each pair are mutually orthogonal (it is best to feel this on ac actual icosahedron). Hence there are 15/3 = 5 such triples of mutually orthogonal symmetry planes of the icosahedron and of course any rotation permutes these triples. It takes a bit of more work to really check that this action is indeed the natural permutation action of $A_5 $ on 5 elements.

Having convinced ourselves that the group of rotations of the buckyball is indeed the alternating group $A_5 $, we can reverse the problem : can the alternating group $A_5 $ see the buckyball???

Well, for starters, it can ‘see’ the icosahedron in a truly amazing way. Look at the conjugacy classes of $A_5 $. We all know that in the full symmetric group $S_n $ elements belong to the same conjugacy class if and only if they have the same cycle decomposition and this is proved using the fact that the conjugation f a cycle $~(i_1,i_2,\ldots,i_k) $ under a permutation $\sigma \in S_n $ is equal to the cycle $~(\sigma(i_1),\sigma(i_2),\ldots,\sigma(i_k)) $ (and this gives us also the candidate needed to conjugate two partitions into each other).

Using this trick it is easy to see that all the 15 order 2 elements of $A_5 $ form one conjugacy class, as do the 20 order 3 elements. However, the 24 order 5 elements split up in two conjugacy classes of 12 elements as the permutation needed to conjugate $~(1,2,3,4,5) $ to $~(1,2,3,5,4) $ is $~(4,5) $ but this is not an element of $A_5 $.

Okay, now take one of these two conjugacy classes of order 5 elements, say that of $~(1,2,3,4,5) $. It consists of 12 elements, 12 being also the number of vertices of the icosahedron. So, is there a way to identify the elements in the conjugacy class to the vertices in such a way that we can describe the edges also in terms of group-computations in $A_5 $?

Surprisingly, this is indeed the case as is demonstrated in a marvelous paper by Kostant “The graph of the truncated icosahedron and the last letter of Galois”.

Two elements $a,b $ in the conjugacy class C share an edge if and only if their product $a.b \in A_5 $ still belongs to the conjugacy class C!

So, for example $~(1,2,3,4,5).(2,1,4,3,5) = (2,5,4) $ so there is no edge between these elements, but on the other hand $~(1,2,3,4,5).(5,3,4,1,2)=(1,5,2,4,3) $ so there is an edge between these! It is no coincidence that $~(5,3,4,1,2)=(2,1,4,3,5)^{-1} $ as inverse elements correspond in the bijection to opposite vertices and for any pair of non-opposite vertices of an icosahedron it is true that either they are neighbors or any one of them is the neighbor of the opposite vertex of the other element.

If we take $u=(1,2,3,4,5) $ and $v=(5,3,4,1,2) $ (or any two elements of the conjugacy class such that u.v is again in the conjugacy class), then one can describe all the vertices of the icosahedron group-theoretically as follows

Isn’t that nice? Well yes, you may say, but that is just the icosahedron. Can the group $A_5 $ also see the buckyball?

Well, let’s try a similar strategy : the buckyball has 60 vertices, exactly as many as there are elements in the group $A_5 $. Is there a way to connect certain elements in a group according to fixed rules? Yes, there is such a way and it is called the Cayley Graph of a group. It goes like this : take a set of generators ${ g_1,\ldots,g_k } $ of a group G, then connect two group element $a,b \in G $ with an edge if and only if $a = g_i.b $ or $b = g_i.a $ for some of the generators.

Back to the alternating group $A_5 $. There are several sets of generators, one of them being the elements ${ (1,2,3,4,5),(2,3)(4,5) } $. In the paper mentioned before, Kostant gives an impressive group-theoretic proof of the fact that the Cayley-graph of $A_5 $ with respect to these two generators is indeed the buckyball!

Let us allow to be lazy for once and let SAGE do the hard work for us, and let us just watch the outcome. Here’s how that’s done

A=PermutationGroup([‘(1,2,3,4,5)’,'(2,3)(4,5)’])
B=A.cayley_graph()
B.show3d()

The outcone is a nice 3-dimensional picture of the buckyball. Below you can see a still, and, if you click on it you will get a 3-dimensional model of it (first click the ‘here’ link in the new window and then you’d better control-click and set the zoom to 200% before you rotate it)

Hence, viewing this Cayley graph from different points we have convinced ourselves that it is indeed the buckyball. In fact, most (truncated) Platonic solids appear as Cayley graphs of groups with respect to specific sets of generators. For later use here is a (partial) survey (taken from Jaap’s puzzle page)

Tetrahedron : $C_2 \times C_2,[(12)(34),(13)(24),(14)(23)] $
Cube : $D_4,[(1234),(13)] $
Octahedron : $S_3,[(123),(12),(23)] $
Dodecahedron : IMPOSSIBLE
Icosahedron : $A_4,[(123),(234),(13)(24)] $

Truncated tetrahedron : $A_4,[(123),(12)(34)] $
Cuboctahedron : $A_4,[(123),(234)] $
Truncated cube : $S_4,[(123),(34)] $
Truncated octahedron : $S_4,[(1234),(12)] $
Rhombicubotahedron : $S_4,[(1234),(123)] $
Rhombitruncated cuboctahedron : IMPOSSIBLE
Snub cuboctahedron : $S_4,[(1234),(123),(34)] $

Icosidodecahedron : IMPOSSIBLE
Truncated dodecahedron : $A_5,[(124),(23)(45)] $
Truncated icosahedron : $A_5,[(12345),(23)(45)] $
Rhombicosidodecahedron : $A_5,[(12345),(124)] $
Rhombitruncated icosidodecahedron : IMPOSSIBLE
Snub Icosidodecahedron : $A_5,[(12345),(124),(23)(45)] $

Again, all these statements can be easily verified using SAGE via the method described before. Next time we will go further into the Kostant’s group-theoretic proof that the buckyball is the Cayley graph of $A_5 $ with respect to (2,5)-generators as this calculation will be crucial in the description of the buckyball curve, the genus 70 Riemann surface discovered by David Singerman and
Pablo Martin which completes the trinity corresponding to the Galois trinity

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Arnold’s trinities version 2.0

Published June 20, 2008 by lievenlb

Arnold has written a follow-up to the paper mentioned last time called “Polymathematics : is mathematics a single science or a set of arts?” (or here for a (huge) PDF-conversion).

On page 8 of that paper is a nice summary of his 25 trinities :

I learned of this newer paper from a comment by Frederic Chapoton who maintains a nice webpage dedicated to trinities.

In his list there is one trinity on sporadic groups :

where $F_{24} $ is the Fischer simple group of order $2^{21}.3^{16}.5^2.7^3.11.13.17.23.29 = 1255205709190661721292800 $, which is the third largest sporadic group (the two larger ones being the Baby Monster and the Monster itself).

I don’t know what the rationale is behind this trinity. But I’d like to recall the (Baby)Monster history as a warning against the trinity-reflex. Sometimes, there is just no way to extend a would be trinity.

The story comes from Mark Ronan’s book Symmetry and the Monster on page 178.

Let’s remind ourselves how we got here. A few years earlier, Fischer has created his ‘transposition’ groups Fi22, Fi23, and Fi24. He had called them M(22), M(23), and M(24), because they were related to Mathieu’s groups M22,M23, and M24, and since he used Fi22 to create his new group of mirror symmetries, he tentatively called it $M^{22} $.
It seemed to appear as a cross-section in something even bigger, and as this larger group was clearly associated with Fi24, he labeled it $M^{24} $. Was there something in between that could be called $M^{23} $?
Fischer visited Cambridge to talk on his new work, and Conway named these three potential groups the Baby Monster, the Middle Monster, and the Super Monster. When it became clear that the Middle Monster didn’t exist, Conway settled on the names Baby Monster and Monster, and this became the standard terminology.

Marcus du Sautoy’s account in Finding Moonshine is slightly different. He tells on page 322 that the Super Monster didn’t exist. Anyone knowing the factual story?

Some mathematical trickery later revealed that the Super Monster was going to be impossible to build: there were certain features that contradicted each other. It was just a mirage, which vanished under closer scrutiny. But the other two were still looking robust. The Middle Monster was rechristened simply the Monster.

And, the inclusion diagram of the sporadic simples tells yet another story.

Anyhow, this inclusion diagram is helpful in seeing the three generations of the Happy Family (as well as the Pariahs) of the sporadic groups, terminology invented by Robert Griess in his 100+p Inventiones paper on the construction of the Monster (which he liked to call, for obvious reasons, the Friendly Giant denoted by FG).
The happy family appears in Table 1.1. of the introduction.

It was this picture that made me propose the trinity on the left below in the previous post. I now like to add another trinity on the right, and, the connection between the two is clear.

Here $Golay $ denotes the extended binary Golay code of which the Mathieu group $M_{24} $ is the automorphism group. $Leech $ is of course the 24-dimensional Leech lattice of which the automorphism group is a double cover of the Conway group $Co_1 $. $Griess $ is the Griess algebra which is a nonassociative 196884-dimensional algebra of which the automorphism group is the Monster.

I am aware of a construction of the Leech lattice involving the quaternions (the icosian construction of chapter 8, section 2.2 of SPLAG). Does anyone know of a construction of the Griess algebra involving octonions???

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Monstrous frustrations

Published June 19, 2008 by lievenlb

Thanks for clicking through… I guess.

If nothing else, it shows that just as much as the stock market is fueled by greed, mathematical reasearch is driven by frustration (or the pleasure gained from knowing others to be frustrated).

I did spend the better part of the day doing a lengthy, if not laborious, calculation, I’ve been postponing for several years now. Partly, because I didn’t know how to start performing it (though the basic strategy was clear), partly, because I knew beforehand the final answer would probably offer me no further insight.

Still, it gives the final answer to a problem that may be of interest to anyone vaguely interested in Moonshine :

What does the Monster see of the modular group?

I know at least two of you, occasionally reading this blog, understand what I was trying to do and may now wonder how to repeat the straightforward calculation. Well the simple answer is : Google for the number 97239461142009186000 and, no doubt, you will be able to do the computation overnight.

One word of advice : don’t! Get some sleep instead, or make love to your partner, because all you’ll get is a quiver on nine vertices (which is pretty good for the Monster) but having an horrible amount of loops and arrows…

If someone wants the details on all of this, just ask. But, if you really want to get me exited : find a moonshine reason for one of the following two numbers :

$791616381395932409265430144165764500492= 2^2 * 11 * 293 * 61403690769153925633371869699485301 $

(the dimension of the monster-singularity upto smooth equivalence), or,

$1575918800531316887592467826675348205163= 523 * 1655089391 * 15982020053213 * 113914503502907 $

(the dimension of the moduli space).

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Arnold’s trinities

Published June 17, 2008 by lievenlb

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?

bloomsday 2 : BistroMath

Published June 16, 2008 by lievenlb

Exactly one year ago this blog was briefly renamed MoonshineMath. The concept being that it would focus on the mathematics surrounding the monster group & moonshine. Well, I got as far as the Mathieu groups…

After a couple of months, I changed the name back to neverendingbooks because I needed the freedom to post on any topic I wanted. I know some people preferred the name MoonshineMath, but so be it, anyone’s free to borrow that name for his/her own blog.

Today it’s bloomsday again, and, as I’m a cyclical guy, I have another idea for a conceptual blog : the bistromath chronicles (or something along this line).

Here’s the relevant section from the Hitchhikers guide

Bistromathics itself is simply a revolutionary new way of understanding the behavior of numbers. …
Numbers written on restaurant checks within the confines of restaurants do not follow the same mathematical laws as numbers written on any other pieces of paper in any other parts of the Universe.
This single statement took the scientific world by storm. It completely revolutionized it.So many mathematical conferences got hold in such good restaurants that many of the finest minds of a generation died of obesity and heart failure and the science of math was put back by years.

Right, so what’s the idea? Well, on numerous occasions Ive stated that any math-blog can only survive as a group-blog. I did approach a lot of people directly, but, as you have noticed, without too much success… Most of them couldnt see themselves contributing to a blog for one of these reasons : it costs too much energy and/or it’s way too inefficient. They say : career-wise there are far cleverer ways to spend my energy than to write a blog. And… there’s no way I can argue against this.

Whence plan B : set up a group-blog for a fixed amount of time (say one year), expect contributors to write one or two series of about 4 posts on their chosen topic, re-edit the better series afterwards and turn them into a book.

But, in order to make a coherent book proposal out of blog-post-series, they’d better center around a common theme, whence the BistroMath ploy. Imagine that some of these forgotten “restaurant-check-notes” are discovered, decoded and explained. Apart from the mathematics, one is free to invent new recepies or add descriptions of restaurants with some mathematical history, etc. etc.

One possible scenario (but I’m sure you will have much better ideas) : part of the knotation is found on a restaurant-check of some Italian restaurant. This allow to explain Conway’s theory of rational tangles, give the perfect way to cook spaghetti to experiment with tangles and tell the history of Manin’s Italian restaurant in Bonn where (it is rumoured) the 1998 Fields medals were decided…

But then, there is no limit to your imagination as long as it somewhat fits within the framework. For example, I’d love to read the transcripts of a chat-session in SecondLife between Dedekind and Conway on the construction of real numbers… I hope you get the drift.

I’m not going to rename neverendingbooks again, but am willing to set up the BistroMath blog provided

Five to ten people are interested to participate
At least one book-editor shows an interest
update : (16/06) contacted by first publisher

You can leave a comment or, if you prefer, contact me via email (if you’re human you will have no problem getting my address…).

Clearly, people already blogging are invited and are allowed to cross-post (in fact, that’s what I will do if it ever gets so far). Finally, if you are not willing to contribute blog-posts but like the idea and are willing to contribute to it in any other way, we are still auditioning for chanting monks

The small group of monks who had taken up hanging around the major research institutes singing strange chants to the effect that the Universe was only a figment of its own imagination were eventually given a street theater grant and went away.

And, if you do not like this idea, there will be another bloomsday-idea next year…

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