Tag: moonshine

Monstrous dessins 1

Published July 22, 2019 by lieven

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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RH and the Ishango bone

Published December 8, 2018 by lieven

“She simply walked into the pond in Kensington Gardens Sunday morning and drowned herself in three feet of water.”

This is the opening sentence of The Ishango Bone, a novel by Paul Hastings Wilson. It (re)tells the story of a young mathematician at Cambridge, Amiele, who (dis)proves the Riemann Hypothesis at the age of 26, is denied the Fields medal, and commits suicide.

In his review of the novel on MathFiction, Alex Kasman casts he story in the 1970ties, based on the admission of the first female students to Trinity.

More likely, the correct time frame is in the first decade of this century. On page 121 Amiele meets Alain Connes, said to be a “past winner of the Crafoord Prize”, which Alain obtained in 2001. In fact, noncommutative geometry and its interaction with quantum physics plays a crucial role in her ‘proof’.

The Ishango artefact only appears in the Coda to the book. There are a number of theories on the nature and grouping of the scorings on the bone. In one column some people recognise the numbers 11, 13, 17 and 19 (the primes between 10 and 20).

In the book, Amiele remarks that the total number of lines scored on the bone (168) “happened to be the exact total of all the primes between 1 and 1000” and “if she multiplied 60, the total number of lines in one side column, by 168, the grand total of lines, she’d get 10080,…,not such a far guess from 9592, the actual total of primes between 1 and 100000.” (page 139-140)

The bone is believed to be more than 20000 years old, prime numbers were probably not understood until about 500 BC…

More interesting than these speculations on the nature of the Ishango bone is the description of the tools Amiele thinks to need to tackle the Riemann Hypothesis:

“These included algebraic geometry (which combines commutative algebra with the language and problems of geometry); noncommutative geometry (concerned with the geometric approach to associative algebras, in which multiplication is not commutative, that is, for which $x$ times $y$ does not always equal $y$ times $x$); quantum field theory on noncommutative spacetime, and mathematical aspects of quantum models of consciousness, to name a few.” (page 115)

The breakthrough came two years later when Amiele was giving a lecture on Grothendieck’s dessins d’enfant.

“Dessin d’enfant, or ‘child’s drawing’, which Amiele had discovered in Grothendieck’s work, is a type of graph drawing that seemed technically simple, but had a very strong impression on her, partly due to the familiar nature of the objects considered. (…) Amiele found subtle arithmetic invariants associated with these dessins, which were completely transformed, again, as soon as another stroke was added.” (page 116)

Amiele’s ‘disproof’ of RH is outlined on pages 122-124 of “The Ishango Bone” and is a mixture of recognisable concepts and ill-defined terms.

“Her final result proved that Riemann’s Hypothesis was false, a zero must fall to the east of Riemann’s critical line whenever the zeta function of point $q$ with momentum $p$ approached the aelotropic state-vector (this is a simplification, of course).” (page 123)

More details are given in a footnote:

“(…) a zero must fall to the east of Riemann’s critical line whenever:

\[
\zeta(q_p) = \frac{( | \uparrow \rangle + \Psi) + \frac{1}{2}(1+cos(\Theta))\frac{\hbar}{\pi}}{\int(\Delta_p)} \]

(…) The intrepid are invited to try the equation for themselves.” (page 124)

Wilson’s “The Ishango Bone” was published in 2012. A fair number of topics covered (the Ishango bone, dessin d’enfant, Riemann hypothesis, quantum theory) also play a prominent role in the 2015 paper/story by Michel Planat “A moonshine dialogue in mathematical physics”, but this time with additional story-line: monstrous moonshine…

Such a paper surely deserves a separate post.

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the monster dictates her picture

Published February 3, 2018 by lieven

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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