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Category: groups

the “uninteresting” case p=5

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

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

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 $.

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the buckyball symmetries

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

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

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

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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F_un and braid groups

Recall that an n-braid consists of n strictly descending elastic strings connecting n inputs at the top (named 1,2,…,n) to n outputs at the bottom (labeled 1,2,…,n) upto isotopy (meaning that we may pull and rearrange the strings in any way possible within 3-dimensional space). We can always change the braid slightly such that we can divide the interval between in- and output in a number of subintervals such that in each of those there is at most one crossing.

n-braids can be multiplied by putting them on top of each other and connecting the outputs of the first braid trivially to the inputs of the second. For example the 5-braid on the left can be written as $B=B_1.B_2 $ with $B_1 $ the braid on the top 3 subintervals and $B_2 $ the braid on the lower 5 subintervals.

In this way (and using our claim that there can be at most 1 crossing in each subinterval) we can write any n-braid as a word in the generators $\sigma_i $ (with $1 \leq i < n $) being the overcrossing between inputs i and i+1. Observe that the undercrossing is then the inverse $\sigma_i^{-1} $. For example, the braid on the left corresponds to the word

$\sigma_1^{-1}.\sigma_2^{-1}.\sigma_1^{-1}.\sigma_2.\sigma_3^{-1}.\sigma_4^{-1}.\sigma_3^{-1}.\sigma_4 $

Clearly there are relations among words in the generators. The easiest one we have already used implicitly namely that $\sigma_i.\sigma_i^{-1} $ is the trivial braid. Emil Artin proved in the 1930-ies that all such relations are consequences of two sets of ‘obvious’ relations. The first being commutation relations between crossings when the strings are far enough from each other. That is we have

$\sigma_i . \sigma_j = \sigma_j . \sigma_i $ whenever $|i-j| \geq 2 $


=

The second basic set of relations involves crossings using a common string

$\sigma_i.\sigma_{i+1}.\sigma_i = \sigma_{i+1}.\sigma_i.\sigma_{i+1} $


=

Starting with the 5-braid at the top, we can use these relations to reduce it to a simpler form. At each step we have outlined to region where the relations are applied


=
=
=

These beautiful braid-pictures were produced using the braid-metapost program written by Stijn Symens.

Tracing a string from an input to an output assigns to an n-braid a permutation on n letters. In the above example, the permutation is $~(1,2,4,5,3) $. As this permutation doesn’t change under applying basic reduction, this gives a group-morphism

$\mathbb{B}_n \rightarrow S_n $

from the braid group on n strings $\mathbb{B}_n $ to the symmetric group. We have seen before that the symmetric group $S_n $ has a F-un interpretation as the linear group $GL_n(\mathbb{F}_1) $ over the field with one element. Hence, we can ask whether there is also a F-un interpretation of the n-string braid group and of the above group-morphism.

Kapranov and Smirnov suggest in their paper that the n-string braid group $\mathbb{B}_n \simeq GL_n(\mathbb{F}_1[t]) $ is the general linear group over the polynomial ring $\mathbb{F}_1[t] $ over the field with one element and that the evaluation morphism (setting t=0)

$GL_n(\mathbb{F}_1[t]) \rightarrow GL_n(\mathbb{F}1) $ gives the groupmorphism $\mathbb{B}_n \rightarrow S_n $

The rationale behind this analogy is a theorem of Drinfeld‘s saying that over a finite field $\mathbb{F}_q $, the profinite completion of $GL_n(\mathbb{F}_q[t]) $ is embedded in the fundamental group of the space of q-polynomials of degree n in much the same way as the n-string braid group $\mathbb{B}_n $ is the fundamental group of the space of complex polynomials of degree n without multiple roots.

And, now that we know the basics of absolute linear algebra, we can give an absolute braid-group representation

$\mathbb{B}_n = GL_n(\mathbb{F}_1[t]) \rightarrow GL_n(\mathbb{F}_{1^n}) $

obtained by sending each generator $\sigma_i $ to the matrix over $\mathbb{F}_{1^n} $ (remember that $\mathbb{F}_{1^n} = (\mu_n)^{\bullet} $ where $\mu_n = \langle \epsilon_n \rangle $ are the n-th roots of unity)

$\sigma_i \mapsto \begin{bmatrix}
1_{i-1} & & & \\
& 0 & \epsilon_n & \\
& \epsilon_n^{-1} & 0 & \\
& & & 1_{n-1-i} \end{bmatrix} $

and it is easy to see that these matrices do indeed satisfy Artin’s defining relations for $\mathbb{B}_n $.

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Galois’ last letter

“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”

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Dedekind or Klein ?

The black&white psychedelic picture on the left of a tessellation of the hyperbolic upper-halfplane, was called the Dedekind tessellation in this post, following the reference given by John Stillwell in his excellent paper Modular Miracles, The American Mathematical Monthly, 108 (2001) 70-76.

But is this correct terminology? Nobody else uses it apparently. So, let’s try to track down the earliest depiction of this tessellation in the literature…

Stillwell refers to Richard Dedekind‘s 1877 paper “Schreiben an Herrn Borchard uber die Theorie der elliptische Modulfunktionen”, which appeared beginning of september 1877 in Crelle’s journal (Journal fur die reine und angewandte Mathematik, Bd. 83, 265-292).

There are a few odd things about this paper. To start, it really is the transcript of a (lengthy) letter to Herrn Borchardt (at first, I misread the recipient as Herrn Borcherds which would be really weird…), written on June 12th 1877, just 2 and a half months before it appeared… Even today in the age of camera-ready-copy it would probably take longer.

There isn’t a single figure in the paper, but, it is almost impossible to follow Dedekind’s arguments without having a mental image of the tessellation. He gives a fundamental domain for the action of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ on the hyperbolic upper-half plane (a fact already known to Gauss) and goes on in section 3 to give a one-to-one mapping between this domain and the complex plane using what he calls the ‘valenz’ function $v $ (which is our modular function $j $, making an appearance in moonshine, and responsible for the black&white tessellation, the two colours corresponding to pre-images of the upper or lower half-planes).

Then there is this remarkable opening sentence.

Sie haben mich aufgefordert, eine etwas ausfuhrlichere Darstellung der Untersuchungen auszuarbeiten, von welchen ich, durch das Erscheinen der Abhandlung von Fuchs veranlasst, mir neulich erlaubt habe Ihnen eine kurze Ubersicht mitzuteilen; indem ich Ihrer Einladung hiermit Folge leiste, beschranke ich mich im wesentlichen auf den Teil dieser Untersuchungen, welcher mit der eben genannten Abhandlung zusammenhangt, und ich bitte Sie auch, die Ubergehung einiger Nebenpunkte entschuldigen zu wollen, da es mir im Augenblick an Zeit fehlt, alle Einzelheiten auszufuhren.

Well, just try to get a paper (let alone a letter) accepted by Crelle’s Journal with an opening line like : “I’ll restrict to just a few of the things I know, and even then, I cannot be bothered to fill in details as I don’t have the time to do so right now!” But somehow, Dedekind got away with it.

So, who was this guy Borchardt? How could this paper be published so swiftly? And, what might explain this extreme ‘je m’en fous’-opening ?

Carl Borchardt was a Berlin mathematician whose main claim to fame seems to be that he succeeded Crelle in 1856 as main editor of the ‘Journal fur reine und…’ until 1880 (so in 1877 he was still in charge, explaining the swift publication). It seems that during this time the ‘Journal’ was often referred to as “Borchardt’s Journal” or in France as “Journal de M Borchardt”. After Borchardt’s death, the Journal für die Reine und Angewandte Mathematik again became known as Crelle’s Journal.

As to the opening sentence, I have a toy-theory of what was going on. In 1877 a bitter dispute was raging between Kronecker (an editor for the Journal and an important one as he was the one succeeding Borchardt when he died in 1880) and Cantor. Cantor had published most of his papers at Crelle and submitted his latest find : there is a one-to-one correspondence between points in the unit interval [0,1] and points of d-dimensional space! Kronecker did everything in his power to stop that paper to the extend that Cantor wanted to retract it and submit it elsewhere. Dedekind supported Cantor and convinced him not to retract the paper and used his influence to have the paper published in Crelle in 1878. Cantor greatly resented Kronecker’s opposition to his work and never submitted any further papers to Crelle’s Journal.

Clearly, Borchardt was involved in the dispute and it is plausible that he ‘invited’ Dedekind to submit a paper on his old results in the process. As a further peace offering, Dedekind included a few ‘nice’ words for Kronecker

Bei meiner Versuchen, tiefer in diese mir unentbehrliche Theorie einzudringen und mir einen einfachen Weg zu den ausgezeichnet schonen Resultaten von Kronecker zu bahnen, die leider noch immer so schwer zuganglich sind, enkannte ich sogleich…

Probably, Dedekind was referring to Kronecker’s relation between class groups of quadratic imaginary fields and the j-function, see the miracle of 163. As an added bonus, Dedekind was elected to the Berlin academy in 1880…

Anyhow, no visible sign of ‘Dedekind’s’ tessellation in the 1877 Dedekind paper, so, we have to look further. I’m fairly certain to have found the earliest depiction of the black&white tessellation (if you have better info, please drop a line). Here it is

It is figure 7 in Felix Klein‘s paper “Uber die Transformation der elliptischen Funktionen und die Auflosung der Gleichungen funften Grades” which appeared in may 1878 in the Mathematische Annalen (Bd. 14 1878/79). He even adds the j-values which make it clear why black triangles should be oriented counter-clockwise and white triangles clockwise. If Klein would still be around today, I’m certain he’d be a metapost-guru.

So, perhaps the tessellation should be called Klein’s tessellation??
Well, not quite. Here’s what Klein writes wrt. figure 7

Diese Figur nun – welche die eigentliche Grundlage fur das Nachfolgende abgibt – ist eben diejenige, von der Dedekind bei seiner Darstellung ausgeht. Er kommt zu ihr durch rein arithmetische Betrachtung.

Case closed : Klein clearly acknowledges that Dedekind did have this picture in mind when writing his 1877 paper!

But then, there are a few odd things about Klein’s paper too, and, I do have a toy-theory about this as well… (tbc)

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