Tag: Conway

Monstrous dessins 2

Published July 24, 2019 by lievenlb

Let’s try to identify the $\Psi(n) = n \prod_{p|n}(1+\frac{1}{p})$ points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ with the lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ from the standard lattice $L_1$ in Conway’s big picture.

Here are all $24=\Psi(12)$ lattices at hyperdistance $12$ from $L_1$ (the boundary lattices):

You can also see the $4 = \Psi(3)$ lattices at hyperdistance $3$ (those connected to $1$ with a red arrow) as well as the intermediate $12 = \Psi(6)$ lattices at hyperdistance $6$.

The vertices of Conway’s Big Picture are the projective classes of integral sublattices of the standard lattice $\mathbb{Z}^2=\mathbb{Z} e_1 \oplus \mathbb{Z} e_2$.

Let’s say our sublattice is generated by the integral vectors $v=(v_1,v_2)$ and $w=(w_1.w_2)$. How do we determine its class $L_{M,\frac{g}{h}}$ where $M \in \mathbb{Q}_+$ is a strictly positive rational number and $0 \leq \frac{g}{h} < 1$?

Here’s an example: the sublattice (the thick dots) is spanned by the vectors $v=(2,1)$ and $w=(1,4)$

Well, we try to find a basechange matrix in $SL_2(\mathbb{Z})$ such that the new 2nd base vector is of the form $(0,z)$. To do this take coprime $(c,d) \in \mathbb{Z}^2$ such that $cv_1+dw_1=0$ and complete with $(a,b)$ satisfying $ad-bc=1$ via Bezout to a matrix in $SL_2(\mathbb{Z})$ such that
\[
\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} v_1 & v_2 \\ w_1 & w_2 \end{bmatrix} = \begin{bmatrix} x & y \\ 0 & z \end{bmatrix} \]
then the sublattice is of class $L_{\frac{x}{z},\frac{y}{z}~mod~1}$.

In the example, we have
\[
\begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 7 \end{bmatrix} \]
so this sublattice is of class $L_{\frac{1}{7},\frac{4}{7}}$.

Starting from a class $L_{M,\frac{g}{h}}$ it is easy to work out its hyperdistance from $L_1$: let $d$ be the smallest natural number making the corresponding matrix integral
\[
d. \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} u & v \\ 0 & w \end{bmatrix} \in M_2(\mathbb{Z}) \]
then $L_{M,\frac{g}{h}}$ is at hyperdistance $u . w$ from $L_1$.

Now that we know how to find the lattice class of any sublattice of $\mathbb{Z}^2$, let us assign a class to any point $[c:d]$ of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.

As $gcd(c,d)=1$, by Bezout we can find a integral matrix with determinant $1$
\[
S_{[c:d]} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]
But then the matrix
\[
\begin{bmatrix} a.n & b.n \\ c & d \end{bmatrix} \]
has determinant $n$.

Working backwards we see that the class $L_{[c:d]}$ of the sublattice of $\mathbb{Z}^2$ spanned by the vectors $(a.n,b.n)$ and $(c,d)$ is of hyperdistance $n$ from $L_1$.

This is how the correspondence between points of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$ and classes in Conway’s big picture at hyperdistance $n$ from $L_1$ works.

Let’s do an example. Take the point $[7:3] \in \mathbb{P}^1(\mathbb{Z}/12\mathbb{Z})$ (see last time), then
\[
\begin{bmatrix} -2 & -1 \\ 7 & 3 \end{bmatrix} \in SL_2(\mathbb{Z}) \]
so we have to determine the class of the sublattice spanned by $(-24,-12)$ and $(7,3)$. As before we have to compute
\[
\begin{bmatrix} -2 & -7 \\ 7 & 24 \end{bmatrix} \begin{bmatrix} -24 & -12 \\ 7 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 0 & -12 \end{bmatrix} \]
giving us that the class $L_{[7:3]} = L_{\frac{1}{12}\frac{3}{4}}$ (remember that the second term must be taken $mod~1$).

If you do this for all points in $\mathbb{P}^1(\mathbb{Z}/12\mathbb{Z})$ (and $\mathbb{P}^1(\mathbb{Z}/6\mathbb{Z})$ and $\mathbb{P}^1(\mathbb{Z}/3 \mathbb{Z})$) you get this version of the picture we started with

You’ll spot that the preimages of a canonical coordinate of $\mathbb{P}^1(\mathbb{Z}/m\mathbb{Z})$ for $m | n$ are the very same coordinate together with ‘new’ canonical coordinates in $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.

To see that this correspondence is one-to-one and that the index of the congruence subgroup
\[
\Gamma_0(n) = \{ \begin{bmatrix} p & q \\ r & s \end{bmatrix}~|~n|r~\text{and}~ps-qr=1 \} \]
in the full modular group $\Gamma = PSL_2(\mathbb{Z})$ is equal to $\Psi(n)$ it is useful to consider the action of $PGL_2(\mathbb{Q})^+$ on the right on the classes of lattices.

The stabilizer of $L_1$ is the full modular group $\Gamma$ and the stabilizer of any class is a suitable conjugate of $\Gamma$. For example, for the class $L_n$ (that is, of the sublattice spanned by $(n,0)$ and $(0,1)$, which is of hyperdistance $n$ from $L_1$) this stabilizer is
\[
Stab(L_n) = \{ \begin{bmatrix} a & \frac{b}{n} \\ c.n & d \end{bmatrix}~|~ad-bc = 1 \} \]
and a very useful observation is that
\[
Stab(L_1) \cap Stab(L_n) = \Gamma_0(n) \]
This is the way Conway likes us to think about the congruence subgroup $\Gamma_0(n)$: it is the joint stabilizer of the classes $L_1$ and $L_n$ (as well as all classes in the ‘thread’ $L_m$ with $m | n$).

On the other hand, $\Gamma$ acts by rotations on the big picture: it only fixes $L_1$ and maps a class to another one of the same hyperdistance from $L_1$.The index of $\Gamma_0(n)$ in $\Gamma$ is then the number of classes at hyperdistance $n$.

To see that this number is $\Psi(n)$, first check that the classes at hyperdistance $p^k$ for $p$ a prime number and for all $k$ for the $p+1$ free valent tree with root $L_1$, so there are exactly $p^{k-1}(p+1)$ classes as hyperdistance $p^k$.

To get from this that the number of hyperdistance $n$ classes is indeed $\Psi(n) = \prod_{p|n}p^{v_p(n)-1}(p+1)$ we have to use the prime- factorisation of the hyperdistance (see this post).

The fundamental domain for the action of $\Gamma_0(12)$ by Moebius tranfos on the upper half plane must then consist of $48=2 \Psi(12)$ black or white hyperbolic triangles

Next time we’ll see how to deduce the ‘monstrous’ Grothendieck dessin d’enfant for $\Gamma_0(12)$ from it

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Monstrous dessins 1

Published July 22, 2019 by lievenlb

Dedekind’s Psi-function $\Psi(n)= n \prod_{p |n}(1 + \frac{1}{p})$ pops up in a number of topics:

$\Psi(n)$ is the index of the congruence subgroup $\Gamma_0(n)$ in the modular group $\Gamma=PSL_2(\mathbb{Z})$,
$\Psi(n)$ is the number of points in the projective line $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$,
$\Psi(n)$ is the number of classes of $2$-dimensional lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ in Conway’s big picture from the standard lattice $L_1$,
$\Psi(n)$ is the number of admissible maximal commuting sets of operators in the Pauli group of a single qudit.

The first and third interpretation have obvious connections with Monstrous Moonshine.

Conway’s big picture originated from the desire to better understand the Moonshine groups, and Ogg’s Jack Daniels problem
asks for a conceptual interpretation of the fact that the prime numbers such that $\Gamma_0(p)^+$ is a genus zero group are exactly the prime divisors of the order of the Monster simple group.

Here’s a nice talk by Ken Ono : Can’t you just feel the Moonshine?

For this reason it might be worthwhile to make the connection between these two concepts and the number of points of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$ as explicit as possible.

Surely all of this is classical, but it is nicely summarised in the paper by Tatitscheff, He and McKay “Cusps, congruence groups and monstrous dessins”.

The ‘monstrous dessins’ from their title refers to the fact that the lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ from $L_1$ are permuted by the action of the modular groups and so determine a Grothendieck’s dessin d’enfant. In this paper they describe the dessins corresponding to the $15$ genus zero congruence subgroups $\Gamma_0(n)$, that is when $n=1,2,3,4,5,6,7,8,9,10,12,13,16,18$ or $25$.

Here’s the ‘monstrous dessin’ for $\Gamma_0(6)$

But, one can compute these dessins for arbitrary $n$, describing the ripples in Conway’s big picture, and try to figure out whether they are consistent with the Riemann hypothesis.

We will get there eventually, but let’s start at an easy pace and try to describe the points of the projective line $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$.

Over a field $k$ the points of $\mathbb{P}^1(k)$ correspond to the lines through the origin in the affine plane $\mathbb{A}^2(k)$ and they can represented by projective coordinates $[a:b]$ which are equivalence classes of couples $(a,b) \in k^2- \{ (0,0) \}$ under scalar multiplication with non-zero elements in $k$, so with points $[a:1]$ for all $a \in k$ together with the point at infinity $[1:0]$. When $n=p$ is a prime number we have $\# \mathbb{P}^1(\mathbb{Z}/p\mathbb{Z}) = p+1$. Here are the $8$ lines through the origin in $\mathbb{A}^2(\mathbb{Z}/7\mathbb{Z})$

Over an arbitrary (commutative) ring $R$ the points of $\mathbb{P}^1(R)$ again represent equivalence classes, this time of pairs
\[
(a,b) \in R^2~:~aR+bR=R \]
with respect to scalar multiplication by units in $R$, that is
\[
(a,b) \sim (c,d)~\quad~\text{iff}~\qquad \exists \lambda \in R^*~:~a=\lambda c, b = \lambda d \]
For $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ we have to find all pairs of integers $(a,b) \in \mathbb{Z}^2$ with $0 \leq a,b < n$ with $gcd(a,b)=1$ and use Cremona’s trick to test for equivalence:
\[
(a,b) = (c,d) \in \mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})~\quad \text{iff}~\quad ad-bc \equiv 0~mod~n \]
The problem is to find a canonical representative in each class in an efficient way because this is used a huge number of times in working with modular symbols.

Perhaps the best algorithm, for large $n$, is sketched in pages 145-146 of Bill Stein’s Modular forms: a computational approach.

For small $n$ the algorithm in $\S 1.3$ in the Tatitscheff, He and McKay paper suffices:

Consider the action of $(\mathbb{Z}/n\mathbb{Z})^*$ on $\{ 0,1,…,n-1 \}=\mathbb{Z}/n\mathbb{Z}$ and let $D$ be the set of the smallest elements in each orbit,
For each $d \in D$ compute the stabilizer subgroup $G_d$ for this action and let $C_d$ be the set of smallest elements in each $G_d$-orbit on the set of all elements in $\mathbb{Z}/n \mathbb{Z}$ coprime with $d$,
Then $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})= \{ [c:d]~|~d \in D, c \in C_d \}$.

Let’s work this out for $n=12$ which will be our running example (the smallest non-squarefree non-primepower):

$(\mathbb{Z}/12\mathbb{Z})^* = \{ 1,5,7,11 \} \simeq C_2 \times C_2$,
The orbits on $\{ 0,1,…,11 \}$ are
\[
\{ 0 \}, \{ 1,5,7,11 \}, \{ 2,10 \}, \{ 3,9 \}, \{ 4,8 \}, \{ 6 \} \]
and $D=\{ 0,1,2,3,4,6 \}$,
$G_0 = C_2 \times C_2$, $G_1 = \{ 1 \}$, $G_2 = \{ 1,7 \}$, $G_3 = \{ 1,5 \}$, $G_4=\{ 1,7 \}$ and $G_6=C_2 \times C_2$,
$1$ is the only number coprime with $0$, giving us $[1:0]$,
$\{ 0,1,…,11 \}$ are all coprime with $1$, and we have trivial stabilizer, giving us the points $[0:1],[1:1],…,[11:1]$,
$\{ 1,3,5,7,9,11 \}$ are coprime with $2$ and under the action of $\{ 1,7 \}$ they split into the orbits
\[
\{ 1,7 \},~\{ 3,9 \},~\{ 5,11 \} \]
giving us the points $[1:2],[3:2]$ and $[5:2]$,
$\{ 1,2,4,5,7,8,10,11 \}$ are coprime with $3$, the action of $\{ 1,5 \}$ gives us the orbits
\[
\{ 1,5 \},~\{ 2,10 \},~\{ 4,8 \},~\{ 7,11 \} \]
and additional points $[1:3],[2:3],[4:3]$ and $[7:3]$,
$\{ 1,3,5,7,9,11 \}$ are coprime with $4$ and under the action of $\{ 1,7 \}$ we get orbits
\[
\{ 1,7 \},~\{ 3,9 \},~\{ 5,11 \} \]
and points $[1:4],[3:4]$ and $[5,4]$,
Finally, $\{ 1,5,7,11 \}$ are the only coprimes with $6$ and they form a single orbit under $C_2 \times C_2$ giving us just one additional point $[1:6]$.

This gives us all $24= \Psi(12)$ points of $\mathbb{P}^1(\mathbb{Z}/12 \mathbb{Z})$ (strangely, op page 43 of the T-H-M paper they use different representants).

One way to see that $\# \mathbb{P}^1(\mathbb{Z}/n \mathbb{Z}) = \Psi(n)$ comes from a consequence of the Chinese Remainder Theorem that for the prime factorization $n = p_1^{e_1} … p_k^{e_k}$ we have
\[
\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z}) = \mathbb{P}^1(\mathbb{Z}/p_1^{e_1} \mathbb{Z}) \times … \times \mathbb{P}^1(\mathbb{Z}/p_k^{e_k} \mathbb{Z}) \]
and for a prime power $p^k$ we have canonical representants for $\mathbb{P}^1(\mathbb{Z}/p^k \mathbb{Z})$
\[
[a:1]~\text{for}~a=0,1,…,p^k-1~\quad \text{and} \quad [1:b]~\text{for}~b=0,p,2p,3p,…,p^k-p \]
which shows that $\# \mathbb{P}^1(\mathbb{Z}/p^k \mathbb{Z}) = (p+1)p^{k-1}= \Psi(p^k)$.

Next time, we’ll connect $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ to Conway’s big picture and the congruence subgroup $\Gamma_0(n)$.

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Rarer books: Singmaster’s notes

Published February 16, 2018 by lievenlb

David Singmaster‘s “Notes on Rubik’s magic cube” are a collectors item, but it is still possible to buy a copy. I own a fifth edition (august 1980).

These notes capture the Rubik craze of those years really well.

Here’s a Conway story, from Siobhan Roberts’ excellent biography Genius at Play.

The ICM in Helsinki in 1978 was Conway’s last shot to get the Fields medal, but this was the last thing on his mind. He just wanted a Rubik cube (then, iron-curtain times, only sold in Hungary), so he kept chasing Hungarians at the meeting, hoping to obtain one. Siobhan writes (p. 239):

“The Fields Medals went to Pierre Deligne, Charles Fefferman, Grigory Margulis, and Daniel Quillen. The Rubik’s cube went to Conway.”

After his Notes, David Singmaster produced a follow-up newsletter “The Cubic Circular”. Only 5 magazines were published, of which 3 were double issues, between the Autumn of 1981 and the summer of 1985.

These magazines were reproduced on this page.

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taking stock

Published February 4, 2018 by lievenlb

The one thing harder than to start blogging after a long period of silence is to stop when you think you’re still in the flow.

(image credit Putnam Consulting)

The Januari 1st post a math(arty) 2018 was an accident. I only wanted to share this picture, of a garage-door with an uncommon definition of prime numbers, i saw the night before.

I had been working on a better understanding of Conway’s Big Picture so I had material for a few follow-up posts.

It was never my intention to start blogging on a daily basis.

I had other writing plans for 2018.

For years I’m trying to write a math-book for a larger audience, or at least to give it an honest try.

My pet peeve with such books is that most of them are either devoid of proper mathematical content, or focus too much on the personal lives of the mathematicians involved.

An inspiring counter-example is ‘Closing the gap’ by Vicky Neal.

From the excellent review by Colin Beveridge on the Aperiodical Blog:

“Here’s a clever way to structure a maths book (I have taken copious notes): follow the development of a difficult idea or discovery chronologically, but intersperse the action with background that puts the discovery in context. That’s not a new structure – but it’s tricky to pull off: you have to keep the difficult idea from getting too difficult, and keep the background at a level where an interested reader can follow along and either say “yes, that’s plausible” or better “wait, let me get a pen!”. This is where Closing The Gap excels.”

So it is possible to publish a math-book worth writing. Or at least, some people can pull it off.

Problem was I needed to kick myself into writing mode. Feeling forced to post something daily wouldn’t hurt.

Anyway, I was sure this would have to stop soon. I had plans to disappear for 10 days into the French mountains. Our place there suffers from frequent power- and cellphone-cuts, which can last for days.

Thank you Orange.fr for upgrading your network to the remotest of places. At times, it felt like I was working from home.

I kept on blogging.

Even now, there’s material lying around.

I’d love to understand the claim that non-commutative geometry may offer some help in explaining moonshine. There was an interesting question on an older post on nimber-arithmetic I feel I should be following up. I’ve given a couple of talks recently on $\mathbb{F}_1$-material, parts of which may be postable. And so on.

Problem is, I would stick to the same (rather dense) writing style.

Perhaps it would make more sense to aim for a weekly (or even monthly) post over at Medium.

Medium offers no MathJax support forcing me to write differently about maths, and for a broader potential audience.

I may continue to blog here (or not), stick to the current style (or try something differently). I have not the foggiest idea right now.

If you have suggestions or advice, please leave a comment or email me.

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the monster dictates her picture

Published February 3, 2018 by lievenlb

The monstrous moonshine picture is a sub-graph of Conway’s Big Picture on 218 vertices. These vertices are the classes of lattices needed in the construction of the 171 moonshine groups. That is, moonshine gives us the shape of the picture.

(image credit Friendly Monsters)

But we can ask to reverse this process. Is the shape of the picture dictated by group-theoretic properties of the monster?

That is, can we reconstruct the 218 lattices and their edges starting from say the conjugacy classes of the monster and some simple rules?

Look at the the power maps for the monster. That is, the operation on conjugacy classes sending the class of $g$ to that of $g^k$ for all divisors $k$ of the order of $g$. Or, if you prefer, the $\lambda$-ring structure on the representation ring.

Rejoice die-hard believers in $\mathbb{F}_1$-theory, rejoice!

Here’s the game to play.

Let $g$ be a monster element of order $n$ and take $d=gcd(n,24)$.

(1) : If $d=8$ and a power map of $g$ gives class $8C$ add $(n|4)$ to your list.

(2) : Otherwise, look at the smallest power of $g$ such that the class is one of $12J,8F,6F,4D, 3C,2B$ or $1A$ and add $(n|e)$ where $e$ is the order of that class, or, if $n > 24$ and $e$ is even add $(n | \frac{e}{2})$.

A few examples:

For class 20E, $d=4$ and the power maps give classes 4D and 2B, so we add $(20|2)$.

For class 32B, $d=8$ but the power map gives 8E so we resort to rule (2). Here the power maps give 8E, 4C and 2B. So, the best class is 4C but as $32 > 24$ we add $(32|2)$.

For class 93A, $d=3$ and the power map gives 3C and even though $93 > 24$ we add $(93|3)$.

This gives us a list of instances $(n|e)$ with $n$ the order of a monster element. For $N=n \times e$ look at all divisors $h$ of $24$ such that $h^2$ divides $N$ and add to your list of lattices those of the form $M \frac{g}{h}$ with $g$ strictly smaller than $h$ and $(g,h)=1$ and $M$ a divisor of $\frac{N}{h^2}$.

This gives us a list of lattices $M \frac{g}{h}$, which is an $h$-th root of unity centered as $L=M \times h$ (see this post). If we do this for all lattices in the list we can partition the $L$’s in families according to which roots of unity are centered at $L$.

This gives us the moonshine picture. (modulo mistakes I made)

The operations we have to do after we have our list of instances $(n|e)$ is pretty straightforward from the rules we used to determine the lattices needed to describe a moonshine group.

Perhaps the oddest part in the construction are the rules (1) and (2) and the prescribed conjugacy classes used in them.

One way to look at this is that the classes $8C$ and $12J$ (or $24J$) are special. The other classes are just the power-maps of $12J$.

Another ‘rationale’ behind these classes may come from the notion of harmonics (see the original Monstrous moonshine paper page 312) of the identity element and the two classes of involutions, 2A (the Fischer involutions) and 2B (the Conway involutions).

For 1A these are : 1A,3C

For 2A these are : 2A,4B,8C

For 2B these are : 2B,4D,6F,8F,12J,24J

These are exactly the classes that we used in (1) and (2), if we add the power-classes of 8C.

Perhaps I should take some time to write all this down more formally.

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A forgotten type and roots of unity (again)

Published January 29, 2018 by lievenlb

The monstrous moonshine picture is the finite piece of Conway’s Big Picture needed to understand the 171 moonshine groups associated to conjugacy classes of the monster.

Last time I claimed that there were exactly 7 types of local behaviour, but I missed one. The forgotten type is centered at the number lattice $84$.

Locally around it the moonshine picture looks like this
\[
\xymatrix{42 \ar@{-}[dr] & 28 \frac{1}{3} \ar@[red]@{-}[d] & 41 \frac{1}{2} \ar@{-}[ld] \\ 28 \ar@[red]@{-}[r] & \color{grey}{84} \ar@[red]@{-}[r] \ar@[red]@{-}[d] \ar@{-}[rd] & 28 \frac{2}{3} \\ & 252 & 168} \]

and it involves all square roots of unity ($42$, $42 \frac{1}{2}$ and $168$) and $3$-rd roots of unity ($28$, $28 \frac{1}{3}$, $28 \frac{2}{3}$ and $252$) centered at $84$.

No, I’m not hallucinating, there are indeed $3$ square roots of unity and $4$ third roots of unity as they come in two families, depending on which of the two canonical forms to express a lattice is chosen.

In the ‘normal’ expression $M \frac{g}{h}$ the two square roots are $42$ and $42 \frac{1}{2}$ and the three third roots are $28, 28 \frac{1}{3}$ and $28 \frac{2}{3}$. But in the ‘other’ expression
\[
M \frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M}) \]
(with $g.g’ \equiv 1~mod~h$) the families of $2$-nd and $3$-rd roots of unity are
\[
\{ 42 \frac{1}{2} = (\frac{1}{2},\frac{1}{168}), 168 = (0,\frac{1}{168}) \} \]
and
\[
\{ 28 \frac{1}{3} = (\frac{1}{3},\frac{1}{252}), 28 \frac{2}{3} = (\frac{2}{3},\frac{1}{252}), 252 = (0 , \frac{1}{252}) \} \]
As in the tetrahedral snake post, it is best to view the four $3$-rd roots of unity centered at $84$ as the vertices of a tetrahedron with center of gravity at $84$. Power maps in the first family correspond to rotations along the axis through $252$ and power maps in the second family are rotations along the axis through $28$.

In the ‘normal’ expression of lattices there’s then a total of 8 different local types, but two of them consist of just one number lattice: in $8$ the local picture contains all square, $4$-th and $8$-th roots of unity centered at $8$, and in $84$ the square and $3$-rd roots.

Perhaps surprisingly, if we redo everything in the ‘other’ expression (and use the other families of roots of unity), then the moonshine picture has only 7 types of local behaviour. The forgotten type $84$ appears to split into two occurrences of other types (one with only square roots of unity, and one with only $3$-rd roots).

I wonder what all this has to do with the action of the Bost-Connes algebra on the big picture or with Plazas’ approach to moonshine via non-commutative geometry.

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What we (don’t) know

Published January 26, 2018 by lievenlb

Do we know why the monster exists and why there’s moonshine around it?

The answer depends on whether or not you believe that vertex operator algebras are natural, elegant and inescapable objects.

the monster

Simple groups often arise from symmetries of exceptionally nice mathematical objects.

The smallest of them all, $A_5$, gives us the rotation symmetries of the icosahedron. The next one, Klein’s simple group $L_2(7)$, comes from the Klein quartic.

The smallest sporadic groups, the Mathieu groups, come from Steiner systems, and the Conway groups from the 24-dimensional Leech lattice.

What about the largest sporadic simple, the monster $\mathbb{M}$?

In his paper What is … the monster? Richard Borcherds writes (among other characterisations of $\mathbb{M}$):

“3. It is the automorphism group of the monster vertex algebra. (This is probably the best answer.)”

But, even Borcherds adds:

“Unfortunately none of these definitions is completely satisfactory. At the moment all constructions of the algebraic structures above seem artificial; they are constructed as sums of two or more apparently unrelated spaces, and it takes a lot of effort to define the algebraic structure on the sum of these spaces and to check that the monster acts on the resulting structure.
It is still an open problem to find a really simple and natural construction of the monster vertex algebra.”

Here’s 2 minutes of John Conway on the “one thing” he really wants to know before he dies: why the monster group exists.

moonshine

Moonshine started off with McKay’s observation that 196884 (the first coefficient in the normalized j-function) is the sum 1+196883 of the dimensions of the two smallest simple representations of $\mathbb{M}$.

Soon it was realised that every conjugacy class of the monster has a genus zero group (or ‘moonshine group’) associated to it.

Borcherds proved the ‘monstrous moonshine conjectures’ asserting that the associated main modular function of such a group is the character series of the action of the element on the monster vertex algebra.

Here’s Borcherds’ ICM talk in Berlin on this: What is … Moonshine?.

Once again, the monster vertex algebra appears to be the final answer.

However, in characterising the 171 moonshine groups among all possible genus zero groups one has proved that they are all of the form:

(ii) : $(n|h)+e,g,\dots$

In his book Moonshine beyond the Monster, Terry Gannon writes:

“We now understand the significance, in the VOA or CFT framework, of transformations in $SL_2(\mathbb{Z})$, but (ii) emphasises that many modular transformations relevant to Moonshine are more general (called the Atkin-Lehner involutions).
Monstrous moonshine will remain mysterious until we can understand its Atkin-Lehner symmetries.”

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the moonshine picture – at last

Published January 23, 2018 by lievenlb

The monstrous moonshine picture is the subgraph of Conway’s big picture consisting of all lattices needed to describe the 171 moonshine groups.

It consists of:

– exactly 218 vertices (that is, lattices), out of which

– 97 are number-lattices (that is of the form $M$ with $M$ a positive integer), and

– 121 are proper number-like lattices (that is of the form $M \frac{g}{h}$ with $M$ a positive integer, $h$ a divisor of $24$ and $1 \leq g \leq h$ with $(g,h)=1$).

The $97$ number lattices are closed under taking divisors, and the corresponding Hasse diagram has the following shape

Here, number-lattices have the same colour if they have the same local structure in the moonshine picture (that is, have a similar neighbourhood of proper number-like lattices).

There are 7 different types of local behaviour:

The white numbered lattices have no proper number-like neighbours in the picture.

The yellow number lattices (2,10,14,18,22,26,32,34,40,68,80,88,90,112,126,144,180,208 = 2M) have local structure

\[
\xymatrix{M \ar@{-}[r] & \color{yellow}{2M} \ar@{-}[r] & M \frac{1}{2}} \]

which involves all $2$-nd (square) roots of unity centered at the lattice.

The green number lattices (3,15,21,39,57,93,96,120 = 3M) have local structure

\[
\xymatrix{& M \ar@[red]@{-}[d] & \\ M \frac{1}{3} \ar@[red]@{-}[r] & \color{green}{3M} \ar@[red]@{-}[r] & M \frac{2}{3}} \]

which involve all $3$-rd roots of unity centered at the lattice.

The blue number lattices (4,16,20,28,36,44,52,56,72,104 = 4M) have as local structure

\[
\xymatrix{M \frac{1}{2} \ar@{-}[d] & & M \frac{1}{4} \ar@{-}[d] \\
2M \ar@{-}[r] & \color{blue}{4M} \ar@{-}[r] & 2M \frac{1}{2} \ar@{-}[d] \\
M \ar@{-}[u] & & M \frac{3}{4}} \]

and involve the $2$-nd and $4$-th root of unity centered at the lattice.

The purple number lattices (6,30,42,48,60 = 6M) have local structure

\[
\xymatrix{& M \frac{1}{3} \ar@[red]@{-}[d] & 2M \frac{1}{3} & M \frac{1}{6} \ar@[red]@{-}[d] & \\
M \ar@[red]@{-}[r] & 3M \ar@{-}[r] \ar@[red]@{-}[d] & \color{purple}{6M} \ar@{-}[r] \ar@[red]@{-}[u] \ar@[red]@{-}[d] & 3M \frac{1}{2} \ar@[red]@{-}[r] \ar@[red]@{-}[d] & M \frac{5}{6} \\
& M \frac{2}{3} & 2M \frac{2}{3} & M \frac{1}{2} & } \]

and involve all $2$-nd, $3$-rd and $6$-th roots of unity centered at the lattice.

The unique brown number lattice 8 has local structure

\[
\xymatrix{& & 1 \frac{1}{4} \ar@{-}[d] & & 1 \frac{1}{8} \ar@{-}[d] & \\
& 1 \frac{1}{2} \ar@{-}[d] & 2 \frac{1}{2} \ar@{-}[r] \ar@{-}[d] & 1 \frac{3}{4} & 2 \frac{1}{4} \ar@{-}[r] & 1 \frac{5}{8} \\
1 \ar@{-}[r] & 2 \ar@{-}[r] & 4 \ar@{-}[r] & \color{brown}{8} \ar@{-}[r] & 4 \frac{1}{2} \ar@{-}[d] \ar@{-}[u] & \\
& & & 1 \frac{7}{8} \ar@{-}[r] & 2 \frac{3}{4} \ar@{-}[r] & 1 \frac{3}{8}} \]

which involves all $2$-nd, $4$-th and $8$-th roots of unity centered at $8$.

Finally, the local structure for the central red lattices $12,24 = 12M$ is

\[
\xymatrix{
M \frac{1}{12} \ar@[red]@{-}[dr] & M \frac{5}{12} \ar@[red]@{-}[d] & M \frac{3}{4} \ar@[red]@{-}[dl] & & M \frac{1}{6} \ar@[red]@{-}[dr] & M \frac{1}{2} \ar@[red]@{-}[d] & M \frac{5}{6} \ar@[red]@{-}[dl] \\
& 3M \frac{1}{4} \ar@{-}[dr] & 2M \frac{1}{6} \ar@[red]@{-}[d] & 4M \frac{1}{3} \ar@[red]@{-}[d] & 2M \frac{1}{3} \ar@[red]@{-}[d] & 3M \frac{1}{2} \ar@{-}[dl] & \\
& 2M \frac{1}{2} \ar@[red]@{-}[r] & 6M \frac{1}{2} \ar@{-}[dl] \ar@[red]@{-}[d] \ar@{-}[r] & \color{red}{12M} \ar@[red]@{-}[d] \ar@{-}[r] & 6M \ar@[red]@{-}[d] \ar@{-}[dr] \ar@[red]@{-}[r] & 2M & \\
& 3M \frac{3}{4} \ar@[red]@{-}[dl] \ar@[red]@{-}[d] \ar@[red]@{-}[dr] & 2M \frac{5}{6} & 4M \frac{2}{3} & 2M \frac{2}{3} & 3M \ar@[red]@{-}[dl] \ar@[red]@{-}[d] \ar@[red]@{-}[dr] & \\
M \frac{1}{4} & M \frac{7}{12} & M \frac{11}{12} & & M \frac{1}{3} & M \frac{2}{3} & M}
\]

It involves all $2$-nd, $3$-rd, $4$-th, $6$-th and $12$-th roots of unity with center $12M$.

No doubt this will be relevant in connecting moonshine with non-commutative geometry and issues of replicability as in Plazas’ paper Noncommutative Geometry of Groups like $\Gamma_0(N)$.

Another of my pet follow-up projects is to determine whether or not the monster group $\mathbb{M}$ dictates the shape of the moonshine picture.

That is, can one recover the 97 number lattices and their partition in 7 families starting from the set of element orders of $\mathbb{M}$, applying some set of simple rules?

One of these rules will follow from the two equivalent notations for lattices, and the two different sets of roots of unities centered at a given lattice. This will imply that if a number lattice belongs to a given family, certain divisors and multiples of it must belong to related families.

If this works out, it may be a first step towards a possibly new understanding of moonshine.

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a SNORTgo endgame

Published January 21, 2018 by lievenlb

SNORT, invented by Simon NORTon is a map-coloring game, similar to COL. Only, this time, neighbours may not be coloured differently.

SNORTgo, similar to COLgo, is SNORT played with go-stones on a go-board. That is, adjacent stones must have the same colour.

SNORT is a ‘hot’ game, meaning that each player is eager to move as most moves will improve your position. In COL players are reluctant to move, because a move limits your next moves.

For this reason, SNORT positions are much harder to evaluate, and one needs the full force of Conways’s ONAG.

Here’s a SNORTgo endgame. Who has a winning strategy?, and what is the first move in that strategy?

The method to approach such an endgame is similar to that in COLgo. First we remove all dead spots from the board.

What remains, are a 4 spots available only to Right (white) and 5 spots available only to Left (bLack). Further, there a 3 ‘live’ regions: the upper righthand corner and the two lower corners.

The value of these corners must be computed inductively.

Here’s the answer:

For example, Right’s best option in the left-most game (corresponding to the upper righthand corner of the endgame) is to put his stone on N12, resulting in a game in which neither player can move (the zero game).

On the other hand, Left can put a stone at either N11, N12 or N13 leaving a game in which she has two more moves, whereas Right han none (the $2$ game).

The other positions are computed similarly.

To get the value of the endgame we have to sum up all these values.

This can either be done using the addition rule given in ONAG, or by using programs in combinatorial game theory.

There’s Combinatorial Game Suite, developed by Aaron Siegel. But, for some reason I can no longer use it on macOS High Sierra.

Fortunately, the older program David Wolfe’s toolkit is still available, and runs on my MacBook.

The sum game evaluates to $\{ \{3|2 \}|-1 \}$, which is a ‘fuzzy’ game, that is, its value is confused with $0$.

This means that the first player to move has a winning strategy in the endgame.

Can you spot the (unique) winning move for Right (white) and one (of two) winning move for Left (bLack)?

Spoiler alert : solution in the comments.

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a non-commutative Jack Daniels problem

Published January 20, 2018 by lievenlb

At a seminar at the College de France in 1975, Tits wrote down the order of the monster group

\[
\# \mathbb{M} = 2^{46}.3^{20}.5^9.7^6.11^2.13^3.17·19·23·29·31·41·47·59·71 \]

Andrew Ogg, who attended the talk, noticed that the prime divisors are precisely the primes $p$ for which the characteristic $p$ super-singular $j$-invariants are all defined over $\mathbb{F}_p$.

Here’s Ogg’s paper on this: Automorphismes de courbes modulaires, Séminaire Delange-Pisot-Poitou. Théorie des nombres, tome 16, no 1 (1974-1975).

Ogg offered a bottle of Jack Daniels for an explanation of this coincidence.

Even Richard Borcherds didn’t claim the bottle of Jack Daniels, though his proof of the monstrous moonshine conjecture is believed to be the best explanation, at present.

A few years ago, John Duncan and Ken Ono posted a paper “The Jack Daniels Problem”, in which they prove that monstrous moonshine implies that if $p$ is not one of Ogg’s primes it cannot be a divisor of $\# \mathbb{M}$. However, the other implication remains mysterious.

Duncan and Ono say:

“This discussion does not prove that every $p ∈ \text{Ogg}$ divides $\# \mathbb{M}$. It merely explains how the first principles of moonshine suggest this implication. Monstrous moonshine is the proof. Does this then provide a completely satisfactory solution to Ogg’s problem? Maybe or maybe not. Perhaps someone will one day furnish a map from the characteristic $p$ supersingular $j$-invariants to elements of order $p$ where the group structure of $\mathbb{M}$ is apparent.”

I don’t know whether they claimed the bottle, anyway.

But then, what is the non-commutative Jack Daniels Problem?

A footnote on the first page of Conway and Norton’s ‘Monstrous Moonshine’ paper says:

“Very recently, A. Pizer has shown these primes are the only ones that satisfy a certain conjecture of Hecke from 1936 relating modular forms of weight $2$ to quaternion algebra theta-series.”

Pizer’s paper is “A note on a conjecture of Hecke”.

Maybe there’s a connection between monstrous moonshine and the arithmetic of integral quaternion algebras. Some hints:

The commutation relations in the Big Picture are reminiscent of the meta-commutation relations for Hurwitz quaternions, originally due to Conway in his booklet on Quaternions and Octonions.

The fact that the $p$-tree in the Big Picture has valency $p+1$ comes from the fact that the Brauer-Severi of $M_2(\mathbb{F}_p)$ is $\mathbb{P}^1_{\mathbb{F}_p}$. In fact, the Big Picture should be related to the Brauer-Severi scheme of $M_2(\mathbb{Z})$.

Then, there’s Jorge Plazas claiming that Connes-Marcolli’s $GL_2$-system might be related to moonshine.

One of the first things I’ll do when I return is to run to the library and get our copy of Shimura’s ‘Introduction to the arithmetic theory of automorphic functions’.

Btw. the bottle in the title image is not a Jack Daniels but the remains of a bottle of Ricard, because I’m still in the French mountains.

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