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Category: groups

214066877211724763979841536000000000000

If you Googled this number a week ago, all you’d get were links to the paper by Melanie Wood Belyi-extending maps and the Galois action on dessins d’enfants.

In this paper she says she can separate two dessins d’enfants (which couldn’t be separated by other Galois invariants) via the order of the monodromy group of the inflated dessins by a certain degree six Belyi-extender.

She gets for the inflated $\Delta$ the order 19752284160000 and for inflated $\Omega$ the order 214066877211724763979841536000000000000 (see also this post).

After that post I redid the computations a number of times (as well as for other Belyi-extenders) and always find that these orders are the same for both dessins.

And, surprisingly, each time the same numbers keep popping up.

For example, if you take the Belyi-extender $t^6$ (power-map) then it is pretty easy to work out the generators of the monodromy group of the extended dessin.

For example, there is a cycle $(1,2)$ in $x_{\Omega}$ and you have to replace it by
\[
(11,12,13,14,15,16,21,22,23,24,25,26) \]
and similarly for other cycles, always replace number $k$ by $k1,k2,k3,k4,k5,k6$ (these are the labels of the edges in the extended dessin corresponding to edge $k$ in the original dessin, starting to count from the the ‘spoke’ of the $6$-star of $t^6$ corresponding to the interval $(0,e^{\frac{4 \pi i}{3}})$, going counterclockwise). So the edge $(0,1)$ corresponds to $k3$, and for $y$ you take the same cycles as in $y_{\Omega}$ replacing number $k$ by $k3$.

Here again, you get for both extended diagrams the same order of the monodromy group, and surprise, surprise: it is 214066877211724763979841536000000000000.

Based on these limited calculations, it seems to be that the order of the monodromy group of the extended dessin only depends on the degree of the extender, and not on its precise form.

I’d hazard a (probably far too optimistic) conjecture that the order of the monodromy groups of a dessin $\Gamma$ and the extended dessin $\gamma(\Gamma)$ for a Belyi-extender $\gamma$ of degree $d$ are related via
\[
\# M(\gamma(\Gamma)) = d \times (\# M(\Gamma))^d \]
(or twice that number), except for trivial settings such as power-maps extending stars.

Edit (august 19): In the comments Dominic shows that in “most” cases the monodromy group of $\gamma(\Gamma)$ should be the wreath product on the monodromy groups of $\gamma$ and $\Gamma$ which has order
\[
\# M(\Gamma)^d \times \# M(\gamma) \]
which fits in with the few calculations i did.

We knew already that the order of the monodromy groups op $\Delta$ and $\Omega$ is $1814400$, and sure enough
\[
6 \times 1814400^6 = 214066877211724763979841536000000000000. \]

If you extend $\Delta$ and $\Omega$ by the power map $t^3$, you get the orders
\[
17919272189952000000 = 3 \times 1814400^3 \]
and if you extend them with the degree 3 extender mentioned in the dessinflateurs-post you get 35838544379904000000, which is twice that number. (Edit : the order of the monodromy group of the extender is $6$, see also above)

As much as i like the Belyi-extender idea to construct new Galois invariants, i fear it’s a dead end. (Always glad to be proven wrong!)

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the mystery Manin-Marcolli monoid

A Belyi-extender (or dessinflateur) $\beta$ of degree $d$ is a quotient of two polynomials with rational coefficients
\[
\beta(t) = \frac{f(t)}{g(t)} \]
with the special properties that for each complex number $c$ the polynomial equation of degree $d$ in $t$
\[
f(t)-c g(t)=0 \]
has $d$ distinct solutions, except perhaps for $c=0$ or $c=1$, and, in addition, we have that
\[
\beta(0),\beta(1),\beta(\infty) \in \{ 0,1,\infty \} \]

Let’s take for instance the power maps $\beta_n(t)=t^n$.

For every $c$ the degree $n$ polynomial $t^n – c = 0$ has exactly $n$ distinct solutions, except for $c=0$, when there is just one. And, clearly we have that $0^n=0$, $1^n=1$ and $\infty^n=\infty$. So, $\beta_n$ is a Belyi-extender of degree $n$.

A cute observation being that if $\beta$ is a Belyi-extender of degree $d$, and $\beta’$ is an extender of degree $d’$, then $\beta \circ \beta’$ is again a Belyi-extender, this time of degree $d.d’$.

That is, Belyi-extenders form a monoid under composition!

In our example, $\beta_n \circ \beta_m = \beta_{n.m}$. So, the power-maps are a sub-monoid of the Belyi-extenders, isomorphic to the multiplicative monoid $\mathbb{N}_{\times}$ of strictly positive natural numbers.



In their paper Quantum statistical mechanics of the absolute Galois group, Yuri I. Manin and Matilde Marcolli say they use the full monoid of Belyi-extenders to act on all Grothendieck’s dessins d’enfant.

But, they attach properties to these Belyi-extenders which they don’t have, in general. That’s fine, as they foresee in Remark 2.21 of their paper that the construction works equally well for any suitable sub-monoid, as long as this sub-monoid contains all power-map exenders.

I’m trying to figure out what the maximal mystery sub-monoid of extenders is satisfying all the properties they need for their proofs.

But first, let us see what Belyi-extenders have to do with dessins d’enfant.



In his user-friendlier period, Grothendieck told us how to draw a picture, which he called a dessin d’enfant, of an extender $\beta(t) = \frac{f(t)}{g(t)}$ of degree $d$:

Look at all complex solutions of $f(t)=0$ and label them with a black dot (and add a black dot at $\infty$ if $\beta(\infty)=0$). Now, look at all complex solutions of $f(t)-g(t)=0$ and label them with a white dot (and add a white dot at $\infty$ if $\beta(\infty)=1$).

Now comes the fun part.

Because $\beta$ has exactly $d$ pre-images for all real numbers $\lambda$ in the open interval $(0,1)$ (and $\beta$ is continuous), we can connect the black dots with the white dots by $d$ edges (the pre-images of the open interval $(0,1)$), giving us a $2$-coloured graph.

For the power-maps $\beta_n(t)=t^n$, we have just one black dot at $0$ (being the only solution of $t^n=0$), and $n$ white dots at the $n$-th roots of unity (the solutions of $x^n-1=0$). Any $\lambda \in (0,1)$ has as its $n$ pre-images the numbers $\zeta_i.\sqrt[n]{\lambda}$ with $\zeta_i$ an $n$-th root of unity, so we get here as picture an $n$-star. Here for $n=5$:

This dessin should be viewed on the 2-sphere, with the antipodal point of $0$ being $\infty$, so projecting from $\infty$ gives a homeomorphism between the 2-sphere and $\mathbb{C} \cup \{ \infty \}$.

To get all information of the dessin (including possible dots at infinity) it is best to slice the sphere open along the real segments $(\infty,0)$ and $(1,\infty)$ and flatten it to form a ‘diamond’ with the upper triangle corresponding to the closed upper semisphere and the lower triangle to the open lower semisphere.

In the picture above, the right hand side is the dessin drawn in the diamond, and this representation will be important when we come to the action of extenders on more general Grothendieck dessins d’enfant.

Okay, let’s try to get some information about the monoid $\mathcal{E}$ of all Belyi-extenders.

What are its invertible elements?

Well, we’ve seen that the degree of a composition of two extenders is the product of their degrees, so invertible elements must have degree $1$, so are automorphisms of $\mathbb{P}^1_{\mathbb{C}} – \{ 0,1,\infty \} = S^2-\{ 0,1,\infty \}$ permuting the set $\{ 0,1,\infty \}$.

They form the symmetric group $S_3$ on $3$-letters and correspond to the Belyi-extenders
\[
t,~1-t,~\frac{1}{t},~\frac{1}{1-t},~\frac{t-1}{t},~\frac{t}{t-1} \]
You can compose these units with an extender to get anther extender of the same degree where the roles of $0,1$ and $\infty$ are changed.

For example, if you want to colour all your white dots black and the black dots white, you compose with the unit $1-t$.

Manin and Marcolli use this and claim that you can transform any extender $\eta$ to an extender $\gamma$ by composing with a unit, such that $\gamma(0)=0, \gamma(1)=1$ and $\gamma(\infty)=\infty$.

That’s fine as long as your original extender $\eta$ maps $\{ 0,1,\infty \}$ onto $\{ 0,1,\infty \}$, but usually a Belyi-extender only maps into $\{ 0,1,\infty \}$.

Here are some extenders of degree three (taken from Melanie Wood’s paper Belyi-extending maps and the Galois action on dessins d’enfants):



with dessin $5$ corresponding to the Belyi-extender
\[
\beta(t) = \frac{t^2(t-1)}{(t-\frac{4}{3})^3} \]
with $\beta(0)=0=\beta(1)$ and $\beta(\infty) = 1$.

So, a first property of the mystery Manin-Marcolli monoid $\mathcal{E}_{MMM}$ must surely be that all its elements $\gamma(t)$ map $\{ 0,1,\infty \}$ onto $\{ 0,1,\infty \}$, for they use this property a number of times, for instance to construct a monoid map
\[
\mathcal{E}_{MMM} \rightarrow M_2(\mathbb{Z})^+ \qquad \gamma \mapsto \begin{bmatrix} d & m-1 \\ 0 & 1 \end{bmatrix} \]
where $d$ is the degree of $\gamma$ and $m$ is the number of black dots in the dessin (or white dots for that matter).

Further, they seem to believe that the dessin of any Belyi-extender must be a 2-coloured tree.

Already last time we’ve encountered a Belyi-extender $\zeta(t) = \frac{27 t^2(t-1)^2}{4(t^2-t+1)^3}$ with dessin



But then, you may argue, this extender sends all of $0,1$ and $\infty$ to $0$, so it cannot belong to $\mathcal{E}_{MMM}$.

Here’s a trick to construct Belyi-extenders from Belyi-maps $\beta : \mathbb{P}^1 \rightarrow \mathbb{P}^1$, defined over $\mathbb{Q}$ and having the property that there are rational points in the fibers over $0,1$ and $\infty$.

Let’s take an example, the ‘monstrous dessin’ corresponding to the congruence subgroup $\Gamma_0(2)$



with map $\beta(t) = \frac{(t+256)^3}{1728 t^2}$.

As it stands, $\beta$ is not a Belyi-extender because it does not map $1$ into $\{ 0,1,\infty \}$. But we have that
\[
-256 \in \beta^{-1}(0),~\infty \in \beta^{-1}(\infty),~\text{and}~512,-64 \in \beta^{-1}(1) \]
(the last one follows from $(t+256)^2-1728 t^3=(t-512)^2(t+64)$).

We can now pre-compose $\beta$ with the automorphism (defined over $\mathbb{Q}$) sending $0$ to $-256$, $1$ to $-64$ and fixing $\infty$ to get a Belyi-extender
\[
\gamma(t) = \frac{(192t)^3}{1728(192t-256)^2} \]
which maps $\gamma(0)=0,~\gamma(1)=1$ and $\gamma(\infty)=\infty$ (so belongs to $\mathcal{E}_{MMM}$) with the same dessin, which is not a tree,

That is, $\mathcal{E}_{MMM}$ can at best consist only of those Belyi-extenders $\gamma(t)$ that map $\{ 0,1,\infty \}$ onto $\{ 0,1,\infty \}$ and such that their dessin is a tree.

Let me stop, for now, by asking for a reference (or counterexample) to perhaps the most startling claim in the Manin-Marcolli paper, namely that any 2-coloured tree can be realised as the dessin of a Belyi-extender!

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Dessinflateurs

I’m trying to get into the latest Manin-Marcolli paper Quantum Statistical Mechanics of the Absolute Galois Group on how to create from Grothendieck’s dessins d’enfant a quantum system, generalising the Bost-Connes system to the non-Abelian part of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$.

In doing so they want to extend the action of the multiplicative monoid $\mathbb{N}_{\times}$ by power maps on the roots of unity to the action of a larger monoid on all dessins d’enfants.

Here they use an idea, originally due to Jordan Ellenberg, worked out by Melanie Wood in her paper Belyi-extending maps and the Galois action on dessins d’enfants.



To grasp this, it’s best to remember what dessins have to do with Belyi maps, which are maps defined over $\overline{\mathbb{Q}}$
\[
\pi : \Sigma \rightarrow \mathbb{P}^1 \]
from a Riemann surface $\Sigma$ to the complex projective line (aka the 2-sphere), ramified only in $0,1$ and $\infty$. The dessin determining $\pi$ is the 2-coloured graph on the surface $\Sigma$ with as black vertices the pre-images of $0$, white vertices the pre-images of $1$ and these vertices are joined by the lifts of the closed interval $[0,1]$, so the number of edges is equal to the degree $d$ of the map.

Wood considers a very special subclass of these maps, which she calls Belyi-extender maps, of the form
\[
\gamma : \mathbb{P}^1 \rightarrow \mathbb{P}^1 \]
defined over $\mathbb{Q}$ with the additional property that $\gamma$ maps $\{ 0,1,\infty \}$ into $\{ 0,1,\infty \}$.

The upshot being that post-compositions of Belyi’s with Belyi-extenders $\gamma \circ \pi$ are again Belyi maps, and if two Belyi’s $\pi$ and $\pi’$ lie in the same Galois orbit, then so must all $\gamma \circ \pi$ and $\gamma \circ \pi’$.

The crucial Ellenberg-Wood idea is then to construct “new Galois invariants” of dessins by checking existing and easily computable Galois invariants on the dessins of the Belyi’s $\gamma \circ \pi$.

For this we need to know how to draw the dessin of $\gamma \circ \pi$ on $\Sigma$ if we know the dessins of $\pi$ and of the Belyi-extender $\gamma$. Here’s the procedure



Here, the middle dessin is that of the Belyi-extender $\gamma$ (which in this case is the power map $t \rightarrow t^4$) and the upper graph is the unmarked dessin of $\pi$.

One has to replace each of the black-white edges in the dessin of $\pi$ by the dessin of the expander $\gamma$, but one must be very careful in respecting the orientations on the two dessins. In the upper picture just one edge is replaced and one has to do this for all edges in a compatible manner.

Thus, a Belyi-expander $\gamma$ inflates the dessin $\pi$ with factor the degree of $\gamma$. For this reason i prefer to call them dessinflateurs, a contraction of dessin+inflator.

In her paper, Melanie Wood says she can separate dessins for which all known Galois invariants were the same, such as these two dessins,



by inflating them with a suitable Belyi-extender and computing the monodromy group of the inflated dessin.

This monodromy group is the permutation group generated by two elements, the first one gives the permutation on the edges given by walking counter-clockwise around all black vertices, the second by walking around all white vertices.

For example, by labelling the edges of $\Delta$, its monodromy is generated by the permutations $(2,3,5,4)(1,6)(8,10,9)$ and $(1,3,2)(4,7,5,8)(9,10)$ and GAP tells us that the order of this group is $1814400$. For $\Omega$ the generating permutations are $(1,2)(3,6,4,7)(8,9,10)$ and $(1,2,4,3)(5,6)(7,9,8)$, giving an isomorphic group.

Let’s inflate these dessins using the Belyi-extender $\gamma(t) = -\frac{27}{4}(t^3-t^2)$ with corresponding dessin



It took me a couple of attempts before I got the inflated dessins correct (as i knew from Wood that this simple extender would not separate the dessins). Inflated $\Omega$ on top:



Both dessins give a monodromy group of order $35838544379904000000$.

Now we’re ready to do serious work.

Melanie Wood uses in her paper the extender $\zeta(t)=\frac{27 t^2(t-1)^2}{4(t^2-t+1)^3}$ with associated dessin



and says she can now separate the inflated dessins by the order of their monodromy groups. She gets for the inflated $\Delta$ the order $19752284160000$ and for inflated $\Omega$ the order $214066877211724763979841536000000000000$.

It’s very easy to make mistakes in these computations, so probably I did something horribly wrong but I get for both $\Delta$ and $\Omega$ that the order of the monodromy group of the inflated dessin is $214066877211724763979841536000000000000$.

I’d be very happy when someone would be able to spot the error!

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Monstrous dessins 3

A long while ago I promised to take you from the action by the modular group $\Gamma=PSL_2(\mathbb{Z})$ on the lattices at hyperdistance $n$ from the standard orthogonal laatice $L_1$ to the corresponding ‘monstrous’ Grothendieck dessin d’enfant.

Speaking of dessins d’enfant, let me point you to the latest intriguing paper by Yuri I. Manin and Matilde Marcolli, ArXived a few days ago Quantum Statistical Mechanics of the Absolute Galois Group, on how to build a quantum system for the absolute Galois group from dessins d’enfant (more on this, I promise, later).

Where were we?

We’ve seen natural one-to-one correspondences between (a) points on the projective line over $\mathbb{Z}/n\mathbb{Z}$, (b) lattices at hyperdistance $n$ from $L_1$, and (c) coset classes of the congruence subgroup $\Gamma_0(n)$ in $\Gamma$.

How to get from there to a dessin d’enfant?

The short answer is: it’s all in Ravi S. Kulkarni’s paper, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1135.

It is a complete mystery to me why Tatitscheff, He and McKay don’t mention Kulkarni’s paper in “Cusps, congruence groups and monstrous dessins”. Because all they do (and much more) is in Kulkarni.

I’ve blogged about Kulkarni’s paper years ago:

– In the Dedekind tessalation it was all about assigning special polygons to subgroups of finite index of $\Gamma$.

– In Modular quilts and cuboid tree diagram it did go on assigning (multiple) cuboid trees to a (conjugacy class) of such finite index subgroup.

– In Hyperbolic Mathieu polygons the story continued on a finite-to-one connection between special hyperbolic polygons and cuboid trees.

– In Farey codes it was shown how to encode such polygons by a Farey-sequence.

– In Generators of modular subgroups it was shown how to get generators of the finite index subgroups from this Farey sequence.

The modular group is a free product
\[
\Gamma = C_2 \ast C_2 = \langle s,u~|~s^2=1=u^3 \rangle \]
with lifts of $s$ and $u$ to $SL_2(\mathbb{Z})$ given by the matrices
\[
S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},~\qquad U= \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \]

As a result, any permutation representation of $\Gamma$ on a set $E$ can be represented by a $2$-coloured graph (with black and white vertices) and edges corresponding to the elements of the set $E$.

Each white vertex has two (or one) edges connected to it and every black vertex has three (or one). These edges are the elements of $E$ permuted by $s$ (for white vertices) and $u$ (for black ones), the order of the 3-cycle determined by going counterclockwise round the vertex.



Clearly, if there’s just one edge connected to a vertex, it gives a fixed point (or 1-cycle) in the corresponding permutation.

The ‘monstrous dessin’ for the congruence subgroup $\Gamma_0(n)$ is the picture one gets from the permutation $\Gamma$-action on the points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$, or equivalently, on the coset classes or on the lattices at hyperdistance $n$.

Kulkarni’s paper (or the blogposts above) tell you how to get at this picture starting from a fundamental domain of $\Gamma_0(n)$ acting on teh upper half-plane by Moebius transformations.

Sage gives a nice image of this fundamental domain via the command


FareySymbol(Gamma0(n)).fundamental_domain()

Here’s the image for $n=6$:



The boundary points (on the halflines through $0$ and $1$ and the $4$ half-circles need to be identified which is indicaed by matching colours. So the 2 halflines are identified as are the two blue (and green) half-circles (in opposite direction).

To get the dessin from this, let’s first look at the interior points. A white vertex is a point in the interior where two black and two white tiles meet, a black vertex corresponds to an interior points where three black and three white tiles meet.

Points on the boundary where tiles meet are coloured red, and after identification two of these reds give one white or black vertex. Here’s the intermediate picture



The two top red points are identified giving a white vertex as do the two reds on the blue half-circles and the two reds on the green half-circles, because after identification two black and two white tiles meet there.

This then gives us the ‘monstrous’ modular dessin for $n=6$ of the Tatitscheff, He and McKay paper:



Let’s try a more difficult example: $n=12$. Sage gives us as fundamental domain



giving us the intermediate picture



and spotting the correct identifications, this gives us the ‘monstrous’ dessin for $\Gamma_0(12)$ from the THM-paper:

In general there are several of these 2-coloured graphs giving the same permutation representation, so the obtained ‘monstrous dessin’ depends on the choice of fundamental domain.

You’ll have noticed that the domain for $\Gamma_0(6)$ was symmetric, whereas the one Sage provides for $\Gamma_0(12)$ is not.

This is caused by Sage using the Farey-code
\[
\xymatrix{
0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_1 & \frac{1}{5} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & 1}
\]

One of the nice results from Kulkarni’s paper is that for any $n$ there is a symmetric Farey-code, giving a perfectly symmetric fundamental domain for $\Gamma_0(n)$. For $n=12$ this symmetric code is

\[
\xymatrix{
0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & \frac{5}{6} \ar@{-}[r]_1 & 1}
\]

It would be nice to see whether using these symmetric Farey-codes gives other ‘monstrous dessins’ than in the THM-paper.

Remains to identify the edges in the dessin with the lattices at hyperdistance $n$ from $L_1$.

Using the tricks from the previous post it is quite easy to check that for any $n$ the monstrous dessin for $\Gamma_0(n)$ starts off with the lattices $L_{M,\frac{g}{h}} = M,\frac{g}{h}$ as below



Let’s do a sample computation showing that the action of $s$ on $L_n$ gives $L_{\frac{1}{n}}$:

\[
L_n.s = \begin{bmatrix} n & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} \]

and then, as last time, to determine the class of the lattice spanned by the rows of this matrix we have to compute

\[
\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -n \end{bmatrix} \]

which is class $L_{\frac{1}{n}}$. And similarly for the other edges.

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Monstrous dessins 2

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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