groups

Seating the first few thousand Knights

Feb 3rd

The Knight-seating problems asks for a consistent placing of n-th Knight at an odd root of unity, compatible with the two different realizations of the algebraic closure of the field with two elements. The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers.

The odd Knights of the round table-problem asks for a specific one-to-one correspondence between two realizations of ‘the’ algebraic closure $\overline{\mathbb{F}_2}$ of the field of two elements.

The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The addition on $\overline{\mathbb{F}_2}$ is then recovered by inducing an involution on the odd roots, pairing the one corresponding to x to the one corresponding to x+1.

The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers. Conway proves in ONAG that this becomes an algebraically closed field of characteristic two and that $\overline{\mathbb{F}_2}$ is the subfield of all ordinals smaller than $\omega^{\omega^{\omega}}$ . The finite ordinals (the natural numbers) form the quadratic closure of $\mathbb{F}_2$ .

On the natural numbers the Conway-addition is binary addition without carrying and Conway-multiplication is defined by the properties that two different Fermat-powers $N=2^{2^i}$ multiply as they do in the natural numbers, and, Fermat-powers square to its sesquimultiple, that is $N^2=\frac{3}{2}N$ . Moreover, all natural numbers smaller than $N=2^{2^{i}}$ form a finite field $\mathbb{F}_{2^{2^i}}$ . Using distributivity, one can write down a multiplication table for all 2-powers.

The Knight-seating problems asks for a consistent placing of n-th Knight K_n at an odd root of unity, compatible with the two different realizations of $\overline{\mathbb{F}_2}$ . Last time, we were able to place the first 15 Knights as below, and asked where you would seat $K_{16}$

K_4 was placed at $e^{2\pi i/15}$ as 4 was the smallest number generating the ‘Fermat’-field $\mathbb{F}_{2^{2^2}}$ (with multiplicative group of order 15) subject to the compatibility relation with the generator 2 of the smaller Fermat-field $\mathbb{F}_2$ (with group of order 15) that 4^5=2 .

To include the next Fermat-field $\mathbb{F}_{2^{2^3}}$ (with multiplicative group of order 255) consistently, we need to find the smallest number n generating the multiplicative group and satisfying the compatibility condition $n^{17}=4$ . Let’s first concentrate on finding the smallest generator : as 2 is a generator for 1st Fermat-field $\mathbb{F}_{2^{2^1}}$ and 4 a generator for the 2-nd Fermat-field $\mathbb{F}_{2^{2^2}}$ a natural conjecture might be that 16 is a generator for the 3-rd Fermat-field $\mathbb{F}_{2^{2^3}}$ and, more generally, that $2^{2^i}$ would be a generator for the next field $\mathbb{F}_{2^{2^{i+1}}}$ .

However, an “exercise” in the 1978-paper by Hendrik Lenstra Nim multiplication asks : “Prove that $2^{2^i}$ is a primitive root in the field $\mathbb{F}_{2^{2^{i+1}}}$ if and only if i=0 or 1.”

I’ve struggled with several of the ‘exercises’ in Lenstra’s paper to the extend I feared Alzheimer was setting in, only to find out, after taking pen and paper and spending a considerable amount of time calculating, that they are indeed merely exercises, when looked at properly… (Spoiler-warning : stop reading now if you want to go through this exercise yourself).

In the picture above I’ve added in red the number x(x+1)=x^2+1 to each of the involutions. Clearly, for each pair these numbers are all distinct and we see that for the indicated pairing they make up all numbers strictly less than 8.

By Conway’s simplicity rules (or by checking) the pair (16,17) gives the number 8. In other words, the equation x^2+x+8 is an irreducible polynomial over $\mathbb{F}_{16}$ having as its roots in $\mathbb{F}_{256}$ the numbers 16 and 17. But then, 16 and 17 are conjugated under the Galois-involution (the Frobenius $y \mapsto y^{16}$ ). That is, we have $16^{16}=17$ and $17^{16}=16$ and hence $16^{17}=8$ . Now, use the multiplication table in $\mathbb{F}_{16}$ given in the previous post (or compute!) to see that 8 is of order 5 (and NOT a generator). As a consequence, the multiplicative order of 16 is 5×17=85 and so 16 cannot be a generator in $\mathbb{F}_{256}$ . For general i one uses the fact that $2^{2^i}$ and $2^{2^i}+1$ are the roots of the polynomial $x^2+x+\prod_{j<i} 2^{2^j}$ over $\mathbb{F}_{2^{2^i}}$ and argues as before.

Right, but then what is the minimal generator satisfying $n^{17}=4$ ? By computing we see that the pairings of all numbers in the range 16…31 give us all numbers in the range 8…15 and by the above argument this implies that the 17-th powers of all numbers smaller than 32 must be different from 4. But then, the smallest candidate is 32 and one verifies that indeed $32^{17}=4$ (use the multiplication table given before).

Hence, we must place Knight $K_{32}$ at root $e^{2 \pi i/255}$ and place the other Knights prior to the 256-th at the corresponding power of 32. I forgot the argument I used to find-by-hand the requested place for Knight 16, but one can verify that $32^{171}=16$ so we seat $K_{16}$ at root $e^{342 \pi i/255}$ .

But what about Knight $K_{256}$ ? Well, by this time I was quite good at squaring and binary representations of integers, but also rather tired, and decided to leave that task to the computer.

If we denote Nim-addition and multiplication by $\oplus$ and $\otimes$ , then Conway’s simplicity results in ONAG establish a field-isomorphism between $~(\mathbb{N},\oplus,\otimes)$ and the field $\mathbb{F}_2(x_0,x_1,x_2,\hdots )$ where the x_i satisfy the Artin-Schreier equations

$x_i^2+x_i+\prod_{j < i} x_j = 0$

and the i-th Fermat-field $\mathbb{F}_{2^{2^i}}$ corresponds to $\mathbb{F}_2(x_0,x_1,\hdots,x_{i-1})$ . The correspondence between numbers and elements from these fields is given by taking $x_i \mapsto 2^{2^i}$ . But then, wecan write every 2-power as a product of the x_i and use the binary representation of numbers to perform all Nim-calculations with numbers in these fields.

Therefore, a quick and dirty way (and by no means the most efficient) to do Nim-calculations in the next Fermat-field consisting of all numbers smaller than 65536, is to use sage and set up the field $\mathbb{F}_2(x_0,x_1,x_2,x_3)$ by

R.< x,y,z,t > =GF(2)[]
S.< a,b,c,d >=R.quotient((x^2+x+1,y^2+y+x,z^2+z+x*y,t^2+t+x*y*z))

To find the smallest number generating the multiplicative group and satisfying the additional compatibility condition $n^{257}=32$ we have to find the smallest binary number $i_1i_2 \hdots i_{16}$ (larger than 255) satisfying

(i1*a*b*c*t+i2*b*c*t+i3*a*c*t+i4*c*t+i5*a*b*t+i6*b*t+
i7*a*t+i8*t+i9*a*b*c+i10*b*c+i11*a*c+i12*c+i13*a*b+
i14*b+i15*a+i16)^257=a*c

It takes a 2.4GHz 2Gb-RAM MacBook not that long to decide that the requested generator is 1051 (killing another optimistic conjecture that these generators might be 2-powers). So, we seat Knight $K_{1051}$ at root $e^{2 \pi i/65535}$ and can then arrange seatings for all Knight queued up until we reach the 65536-th! In particular, the first Knight we couldn’t place before, that is Knight $K_{256}$ , will be seated at root $e^{65826 \pi i/65535}$ .

If you’re lucky enough to own a computer with more RAM, or have the patience to make the search more efficient and get the seating arrangement for the next Fermat-field, please drop a comment.

I’ll leave you with another Lenstra-exercise which shouldn’t be too difficult for you to solve now : “Prove that $x^3=2^{2^i}$ has three solutions in $\mathbb{N}$ for each $i \geq 2$ .”

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Conway, fun, Galois, games, ONAG, ordinals

The odd knights of the round table

Jan 28th

Posted by lievenlb in games

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Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights $\{ K_1,K_2,K_3,\hdots \}$ , waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights K_a and K_b a place at an odd root of unity, say $\omega_a$ and $\omega_b$ , such that the seat at the odd root of unity $\omega_a \times \omega_b$ must be given to the Knight $K_{a \otimes b}$ , where $a \otimes b$ is the Nim-multiplication of and . Which place would you offer to Knight $K_{16}$ , or Knight K_n , or, if you’re into ordinals, Knight $K_{\omega}$ ?”

What does this have to do with finite fields? Well, consider the simplest of all finite field $\mathbb{F}_2 = \{ 0,1 \}$ and consider its algebraic closure $\overline{\mathbb{F}_2}$ . Last year, we’ve run a series starting here, identifying the field $\overline{\mathbb{F}_2}$ , following John H. Conway in ONAG, with the set of all ordinals smaller than $\omega^{\omega^{\omega}}$ , given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers $n \oplus m$ can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, $9 \oplus 1 = 1001+1 = 1000 = 8$ . Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers $F_n = 2^{2^n}$ . We then demand that $F_n \otimes m = F_n \times m$ whenever m < F_n and $F_n \otimes F_n = \frac{3}{2}F_n$ . Distributivity wrt. $\oplus$ can then be used to calculate arbitrary Nim-products. For example, $8 \otimes 3 = (4 \otimes 2) \otimes (2 \oplus 1) = (4 \otimes 3) \oplus (4 \otimes 2) = 12 \oplus 8 = 4$ . Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure $\overline{\mathbb{F}_2}$ is identified with the subfield of all ordinals smaller than $\omega^{\omega^{\omega}}$ . For those of you who don’t feel like going transfinite, the subfield $~(\mathbb{N},\oplus,\otimes)$ is identified with the quadratic closure of $\mathbb{F}_2$ .

The connection between $\overline{\mathbb{F}_2}$ and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video

At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of $\overline{\mathbb{F}_2}$ can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form $\mathbb{F}_{2^n}$ for some n, and, as the group of units of any finite field is cyclic, x is an element of order 2^n-1 . Hence, $\mathbb{F}_{2^n}- \{ 0 \}$ can be identified with the set of 2^n-1 -roots of unity, with $e^{2 \pi i/n}$ corresponding to a generator of the unit-group. So, all elements of $\overline{\mathbb{F}_2}$ correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds¹.

Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of $\overline{\mathbb{F}_2}$ and the odd roots of unity $\mu_{odd}$ respecting multiplication, how can we recover the addition on $\overline{\mathbb{F}_2}$ ? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as $x+y=x \times s(\frac{y}{x}) = x \times (\frac{y}{x}+1)$ . For example, all information about the finite field $\mathbb{F}_{2^4}$ is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as

Okay, we now have two identifications of the algebraic closure $\overline{\mathbb{F}_2}$ : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.

To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!

Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above

To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are

It is known that the finite subfields of $~(\mathbb{N},\oplus,\otimes)$ are precisely the sets of numbers smaller than the Fermat-powers F_n . So, the first one is all numbers smaller than F_1=4 (check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root $e^{2 \pi i/3}$ . The next subfield are all numbers smaller than F_2 = 16 and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that $k^{\otimes 5}=2$ . Verify that this number is 4 and that this forces the identification and coupling given above.

The next finite subfield would consist of all natural numbers smaller than F_3=256 . Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that $k^{\otimes 17}=4$ which would fix an identification at that level. Then, the next level would be all numbers smaller than F_4=65536 and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.

Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).

If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group $~(\mathbb{Z}/m\mathbb{Z})^*$ whence , so the m-roots of unity lie within those of order [↩]

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arxiv, Connes, Conway, fun, Galois, games, geometry, ONAG, ordinals, simples

Olivier Messiaen & Mathieu 12

Dec 31st

Posted by lievenlb in Bourbaki

5 comments

To mark the end of 2009 and 6 years of blogging, two musical compositions with a mathematical touch to them. I wish you all a better 2010!

Remember from last time that we identified Olivier Messiaen as the ‘Monsieur Modulo’ playing the musical organ at the Bourbaki wedding. This was based on the fact that his “modes à transposition limitée” are really about epimorphisms between modulo rings Z/12Z→Z/3Z and Z/12Z→Z/4Z.

However, Messiaen had more serious mathematical tricks up his sleeve. In two of his compositions he did discover (or at least used) one of the smaller sporadic groups, the Mathieu group $M_{12}$ of order 95040 on which we have based a whole series of Mathieu games two and a half years ago.

Messiaen’s ‘Ile de fey 2′ composition for piano (part of Quatre études de rythme (“Four studies in rhythm”), piano (1949–50)) is based on two concurrent permutations. The first is shown below, with the underlying motive rotational permutation shown.

This gives the permutation (1,7,10,2,6,4,5,9,11,12)(3,8). A second concurrent permutation is based on the permutation (1,6,9,2,7,3,5,4,8,10,11) and both of them generate the Mathieu group $M_{12}$ . This can be seen by realizing the two permutations as the rotational permutations

and identifying them with the Mongean shuffles generating $M_{12}$ . See for example, Dave Benson’s book “Music: A Mathematical Offering”, freely available online.

Clearly, Messiaen doesn’t use all of its 95040 permutations in his piece! Here’s how it sounds. The piece starts 2 minutes into the clip.

The second piece is “Les Yeux dans les Roues” (The Eyes in the Wheels), sixth piece from the “Livre d’Orgue” (1950/51).

According to Hauptwerk, the piece consists of a melody/theme in the pedal, accompanied by two fast-paced homorhythmic lines in the manuals. The pedal presents a sons-durées theme which is repeated six times, in different permutations. Initially it is presented in its natural form. Afterwards, it is presented alternatively picking notes from each end of the original form. Similar transformations are applied each time until the sixth, which is the retrograde of the first. The entire twelve-tone analysis (pitch only, not rhythm) of the pedal is shown below:

That is we get the following five permutations which again generate Mathieu 12 :

a=(2,3,5,9,8,10,6,11,4,7,12)
b=(1,2,4,8,9,7,11,3,6,12)(5,10)=ea

c=(1,12,11,9,5,4,6,2,10,7)(3,8)=ed
d=(1,11,10,8,4,5,3,7,2,9,6)
e=(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)

Here’s the piece performed on organ :

Considering the permutations $X=d.a^{-1}$ and $Y=(a.d^2.a.d^3)^{-1}$ one obtains canonical generators of $M_{12}$ , that is, generators satisfying the defining equations of this sporadic group

$X^2=Y^3=(XY)^{11}=[X,Y]^6=(XYXYXY^{-1})^6=1$

I leave you to work out the corresponding dessin d’enfant tonight after a couple of glasses of champagne! It sure has a nice form. Once again, a better 2010!

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blogging, bourbaki, Conway, games, groups, Mathieu, puzzle, simple groups, sporadic groups

E(8) from moonshine groups

May 15th

Posted by lievenlb in groups

1 comment

Moonshine and E(8)

the monster graph and McKay’s observation
Conway’s big picture
looking for the moonshine picture
E(8) from moonshine groups

Are the valencies of the 171 moonshine groups are compatible, that is, can one construct a (disconnected) graph on the 171 vertices such that in every vertex (determined by a moonshine group G) the vertex-valency coincides with the valency of the corresponding group? Duncan describes a subset of 9 moonshine groups for which the valencies are compatible. These 9 groups are characterized as those moonshine groups G having width 1 at the cusp and such that their intersection with the modular group is big.

Time to wrap up this series on John Duncan‘s paper Arithmetic groups and the affine E8 Dynkin diagram in which he gives a realization of the extended E(8)-Dynkin diagram (together with its isotropic root vector) from the moonshine groups, compatible with McKay’s E(8)-observation.

In the previous post we have described all 171 moonshine groups using Conway’s big picture. This description will allow us to associate two numbers to a moonshine group $G \subset PSL_2(\mathbb{R})$ . Recall that for any such group we have a positive integer such that

$\Gamma_0(N) \subset G \subset \Gamma_0(h,\frac{N}{h})+$

where is the largest divisor of 24 such that h^2 | N . Let us call $n_G=\frac{N}{h}$ the dimension of (Duncan calls this number the ‘normalized level’) as it will give us the dimension component at the vertex determined by .

We have also seen last time that any moonshine group is of the form $G = \Gamma_0(n_G || h)+e,f,g$ , that is, $G/\Gamma_0(n_G ||h)$ is an elementary abelian group $~(\Z/2\Z)^m$ generated by Atkin-Lehner involutions. Let’s call v_G=m+1 the valency of the group as it will give s the valency of the vertex determined by .

It would be nice to know whether the valencies of the 171 moonshine groups are compatible, that is, whether one can construct a (disconnected) graph on the 171 vertices such that in each vertex (determined by a moonshine group ) the vertex-valency coincides with the valency of the corresponding group.

Duncan describes a subset of 9 moonshine groups for which the valencies are compatible. These 9 groups are characterized as those moonshine groups having width 1 at the cusp and such that their intersection with the modular group $\Gamma = PSL_2(\Z)$ is big, more precisely the index $[\Gamma : \Gamma \cap G] \leq 12$ and $[\Gamma : \Gamma \cap G]/[G : \Gamma \cap G] \leq 3$ .

They can be described using the mini-moonshine picture on the right. They are :

The modular group itself $1=\Gamma$ , being the stabilizer of the lattice 1. This group has clearly dimension and valency equal to one.

The modular subgroup $2=\Gamma_0(2)$ being the point-wise stabilizer of the lattices 1 and 2 (so it has valency one and dimension two, and, its normalizer $2+ =\Gamma_0(2)+$ which is the set-wise stabilizer of the lattices 1 and 2 and the one Atkin-Lehner involution interchanges both. So, this group has valency two (as we added one involution) as well as dimension two.

Likewise, the groups $3+=\Gamma_0(3)+$ and $5+=\Gamma_0(5)+$ are the stabilzer subgroups of the red 1-cell (1,3) resp. the green 1-cell (1,5) and hence have valency two (as we add one involution) and dimensions 3 resp. 5.

The group $4+=\Gamma_0(4)+$ stabilizes the (1|4)-thread and as we add one involution must have valency 2 and dimension 4.

On the other hand, the group $6+=\Gamma_0(6)+$ stabilizes the unique 2-cell in the picture (having lattices 1,2,3,6) so this time we will add three involutions (horizontal and vertical switches and their product the antipodal involution). Hence, for this group the valency is three and its dimension is equal to six.

Remain the two groups connected to the mini-snakes in the picture. The red mini-snake (top left hand) is the ball with center 3 and hyperdistance 3 and determines the group $3||3=\Gamma_0(3||3)$ which has valency one (we add no involutions) and dimension 3. The blue mini-snake (the extended D(5)-Dynkin in the lower right corner) determines the group $4||2+=\Gamma(4||2)+$ which has valency two and dimension 4.

The valencies of these 9 moonshine groups are compatible and they can be arranged in the extended E(8) diagram depicted below

Moreover, the dimensions of the groups give the exact dimension-components of the isotropic root of the extended E(8)-diagram. Further, the dimension of the group is equal to the order of the elements making up the conjugacy class of the monster to which exactly the given groups correspond via monstrous moonshine and hence compatible with John McKay‘s original E(8)-observation!

Once again, I would love to hear when someone has more information on the cell-decomposition of the moonshine picture or if someone can extend the moonshine E(8)-graph, possibly to include all 171 moonshine groups.

Previous in series

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arxiv, Conway, groups, modular, monster, moonshine

looking for the moonshine picture

May 11th

Posted by lievenlb in groups

4 comments

Moonshine and E(8)

the monster graph and McKay’s observation
Conway’s big picture
looking for the moonshine picture
E(8) from moonshine groups

We have seen that Conway’s big picture helps us to determine all arithmetic subgroups of $PSL_2(\mathbb{R})$ commensurable with the modular group $PSL_2(\Z)$ , including all groups of monstrous moonshine.

As there are exactly 171 such moonshine groups, they are determined by a finite subgraph of Conway’s picture and we call the minimal such subgraph the moonshine picture. Clearly, we would like to determine its structure.

On the left a depiction of a very small part of it. It is the minimal subgraph of Conway’s picture needed to describe the 9 moonshine groups appearing in Duncan’s realization of McKay’s E(8)-observation. Here, only three primes are relevant : 2 (blue lines), 3 (reds) and 5 (green). All lattices are number-like (recall that $M \frac{g}{h}$ stands for the lattice $\langle M e_1 + \frac{g}{h} e_2, e_2 \rangle$ ).

We observe that a large part of this mini-moonshine picture consists of the three p-tree subgraphs (the blue, red and green tree starting at the 1-lattice $1 = \langle e_1,e_2 \rangle$ . Whereas Conway’s big picture is the product over all p-trees with p running over all prime numbers, we observe that the mini-moonshine picture is a very small subgraph of the product of these three subtrees. In fact, there is just one 2-cell (the square 1,2,6,3).

Hence, it seems like a good idea to start our investigation of the full moonshine picture with the determination of the p-subtrees contained in it, and subsequently, worry about higher dimensional cells constructed from them. Surely it will be no major surprise that the prime numbers p that appear in the moonshine picture are exactly the prime divisors of the order of the monster group, that is p=2,3,5,7,11,13,17,19,23,29,31,41,47,59 or 71. Before we can try to determine these 15 p-trees, we need to know more about the 171 moonshine groups.

Recall that the proper way to view the modular subgroup $\Gamma_0(N)$ is as the subgroup fixing the two lattices L_1 and L_N , whence we will write $\Gamma_0(N)=\Gamma_0(N|1)$ , and, by extension we will denote with $\Gamma_0(X|Y)$ the subgroup fixing the two lattices L_X and L_Y .

As $\Gamma_0(N)$ fixes L_1 and L_N it also fixes all lattices in the (N|1)-thread, that is all lattices occurring in a shortest path from L_1 to L_N (on the left a picture of the (200|1)-thread).

If $N=p_1^{a_1} p_2^{a_2} \hdots p_k^{a_k}$ , then the (N|1)-thread has 2^k involutions as symmetries, called the Atkin-Lehner involutions. For every exact divisor e || N (that is, e|N and $gcd(e,\frac{N}{e})=1$ we have an involution W_e which acts by sending each point in the thread-cell corresponding to the prime divisors of to its antipodal cell-point and acts as the identity on the other prime-axes. For example, in the (200|1)-thread on the left, W_8 is the left-right reflexion, $W_{25}$ the top-bottom reflexion and $W_{200}$ the antipodal reflexion. The set of all exact divisors of N becomes the group $~(\Z/2\Z)^k$ under the operation $e \ast f = \frac{e \times f}{gcd(e,f)^2}$ .

Most of the moonshine groups are of the form $\Gamma_0(n|h)+e,f,g,...$ for some N=h.n such that h | 24 and h^2 | N . The group $\Gamma_0(n|h)$ is then conjugate to the modular subgroup $\Gamma_0(\frac{n}{h})$ by the element $\begin{bmatrix} h & 0 \\ 0 & 1 \end{bmatrix}$ . With $\Gamma_0(n|h)+e,f,g,...$ we mean that the group $\Gamma_0(n|h)$ is extended with the involutions W_e,W_f,W_g,... . If we simply add all Atkin-Lehner involutions we write $\Gamma_0(n|h)+$ for the resulting group.

Finally, whenever $h \not= 1$ there is a subgroup $\Gamma_0(n||h)+e,f,g,...$ which is the kernel of a character $\lambda$ being trivial on $\Gamma_0(N)$ and on all involutions W_e for which every prime dividing also divides $\frac{n}{h}$ , evaluating to $e^{\frac{2\pi i}{h}}$ on all cosets containing $\begin{bmatrix} 1 & \frac{1}{h} \\ 0 & 1 \end{bmatrix}$ and to $e^{\pm \frac{2 \pi i }{h}}$ for cosets containing $\begin{bmatrix} 1 & 0 \\ n & 0 \end{bmatrix}$ (with a + sign if $\begin{bmatrix} 0 & -1 \\ N & 0 \end{bmatrix}$ is present and a – sign otherwise). Btw. it is not evident at all that this is a character, but hard work shows it is!

Clearly there are heavy restrictions on the numbers that actually occur in moonshine. In the paper On the discrete groups of moonshine, John Conway, John McKay and Abdellah Sebbar characterized the 171 arithmetic subgroups of $PSL_2(\mathbb{R})$ occuring in monstrous moonshine as those of the form $G = \Gamma_0(n || h)+e,f,g,...$ which are

(a) of genus zero, meaning that the quotient of the upper-half plane by the action of $G \subset PSL_2(\mathbb{R})$ by Moebius-transformations gives a Riemann surface of genus zero,
(b) the quotient group $G/\Gamma_0(nh)$ is a group of exponent 2 (generated by some Atkin-Lehner involutions), and
(c) every cusp can be mapped to $\infty$ by an element of $PSL_2(\mathbb{R})$ which conjugates the group to one containing $\Gamma_0(nh)$ .

Now, if $\Gamma_0(n || h)+e,f,g,...$ is of genus zero, so is the larger group $\Gamma_0(n | h)+e,f,g,...$ , which in turn, is conjugated to the group $\Gamma_0(\frac{n}{h})+e,f,g,...$ . Therefore, we need a list of all groups of the form $\Gamma_0(\frac{n}{h})+e,f,g,...$ which are of genus zero. There are exactly 123 of them, listed on the right.

How does this help to determine the structure of the p-subtree of the moonshine picture for the fifteen monster-primes p? Look for the largest p-power p^k such that p^k+e,f,g... appears in the list. That is for p=2,3,5,7,11,13,17,19,23,29,31,41,47,59,71 these powers are resp. 5,3,2,2,1,1,1,1,1,1,1,1,1,1,1. Next, look for the largest p-power p^l dividing 24 (that is, 3 for p=2, 1 for p=3 and 0 for all other primes). Then, these relevant moonshine groups contain the modular subgroup $\Gamma_0(p^{k+2l})$ and are contained in its normalizer in $PSL_2(\mathbb{R})$ which by the Atkin-Lehner theorem is precisely the group $\Gamma_0(p^{k+l}|p^l)+$ .

Right, now the lattices fixed by $\Gamma_0(p^{k+2l})$ (and permuted by its normalizer), that is the lattices in our p-subtree, are those that form the $~(p^{k+2l}|1)$ -snake in Conway-speak. That is, the lattices whose hyper-distance to the $~(p^{k+l}|p^l)$ -thread divides 24. So for all primes larger than 2 or 3, the p-tree is just the ~(p^l|1) -thread.

For p=3 the 3-tree is the (243|1)-snake having the (81|3)-thread as its spine. It contains the following lattices, all of which are number-like.

Depicting the 2-tree, which is the (2048|1)-snake may take a bit longer… Perhaps someone should spend some time figuring out which cells of the product of these fifteen trees make up the moonshine picture!

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Conway, groups, modular, monster, moonshine, Riemann

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