Category: groups

Sylvester’s synthemes

Published March 19, 2020 by lievenlb

I was running a bachelor course on representations of finite groups and a master course on simple (mainly sporadic) groups until Corona closed us down. Perhaps these blog-posts can be useful to some.

A curious fact, with ripple effect on Mathieu sporadic groups, is that the symmetric group $S_6$ has an automorphism $\phi$, different from an automorphism by conjugation.

In the course notes the standard approach was given, based on the $5$-Sylow subgroups of $S_5$.

Here’s the idea. Let $S_6$ act by permuting $6$ elements and consider the subgroup $S_5$ fixing say $6$. If such an odd automorphism $\phi$ would exist, then the subgroup $\phi(S_5)$ cannot fix one of the six elements (for then it would be conjugated to $S_5$), so it must act transitively on the six elements.

The alternating group $A_5$ is the rotation symmetry group of the icosahedron

Any $5$-Sylow subgroup of $A_5$ is the cyclic group $C_5$ generated by a rotation among one of the six body-diagonals of the icosahedron. As $A_5$ is normal in $S_5$, also $S_5$ has six $5$-Sylows.

More lowbrow, such a subgroup is generated by a permutation of the form $(1,2,a,b,c)$, of which there are six. Good old Sylow tells us that these $5$-Sylow subgroups are conjugated, giving a monomorphism
\[
S_5 \rightarrow Sym(\{ 5-Sylows \})\simeq S_6 \]
and its image $H$ is a subgroup of $S_6$ of index $6$ (and isomorphic to $S_5$) which acts transitively on six elements.

Left multiplication gives an action of $S_6$ on the six cosets $S_6/H =\{ \sigma H~:~\sigma \in S_6 \}$, that is a groupmorphism
\[
\phi : S_6 \rightarrow Sym(\{ \sigma H \}) = S_6 \]
which is our odd automorphism (actually it is even, of order two). A calculation shows that $\phi$ sends permutations of cycle shape $2.1^4$ to shape $2^3$, so can’t be given by conjugation (which preserves cycle shapes).

An alternative approach is given by Noah Snyder in an old post at the Secret Blogging Seminar.

Here, we like to identify the six points $\{ a,b,c,d,e,f \}$ with the six points $\{ 0,1,2,3,4,\infty \}$ of the projective line $\mathbb{P}^1(\mathbb{F}_5)$ over the finite field $\mathbb{F}_5$.

There are $6!$ different ways to do this set-theoretically, but lots of them are the same up to an automorphism of $\mathbb{P}^1(\mathbb{F}_5)$, that is an element of $PGL_2(\mathbb{F}_5)$ acting via Mobius transformations on $\mathbb{P}^1(\mathbb{F}_5)$.

$PGL_2(\mathbb{F}_5)$ acts $3$-transitively on $\mathbb{P}^1(\mathbb{F}_5)$ so we can fix three elements in each class, say $a=0,b=1$ and $f=\infty$, leaving six different ways to label the points of the projective line
\[
\begin{array}{c|cccccc}
& a & b & c & d & e & f \\
\hline
1 & 0 & 1 & 2 & 3 & 4 & \infty \\
2 & 0 & 1 & 2 & 4 & 3 & \infty \\
3 & 0 & 1 & 3 & 2 & 4 & \infty \\
4 & 0 & 1 & 3 & 4 & 2 & \infty \\
5 & 0 & 1 & 4 & 2 & 3 & \infty \\
6 & 0 & 1 & 4 & 3 & 2 & \infty
\end{array}
\]
A permutation of the six elements $\{ a,b,c,d,e,f \}$ will result in a permutation of the six classes of $\mathbb{P}^1(\mathbb{F}_5)$-labelings giving the odd automorphism
\[
\phi : S_6 = Sym(\{ a,b,c,d,e,f \}) \rightarrow Sym(\{ 1,2,3,4,5,6 \}) = S_6 \]
An example: the involution $(a,b)$ swaps the points $0$ and $1$ in $\mathbb{P}^1(\mathbb{F}_5)$, which can be corrected via the Mobius-automorphism $t \mapsto 1-t$. But this automorphism has an effect on the remaining points
\[
2 \leftrightarrow 4 \qquad 3 \leftrightarrow 3 \qquad \infty \leftrightarrow \infty \]
So the six different $\mathbb{P}^1(\mathbb{F}_5)$ labelings are permuted as
\[
\phi((a,b))=(1,6)(2,5)(3,4) \]
showing (again) that $\phi$ is not a conjugation-automorphism.

Yet another, and in fact the original, approach by James Sylvester uses the strange terminology of duads, synthemes and synthematic totals.

A duad is a $2$-element subset of $\{ 1,2,3,4,5,6 \}$ (there are $15$ of them).
A syntheme is a partition of $\{ 1,2,3,4,5,6 \}$ into three duads (there are $15$ of them).
A (synthematic) total is a partition of the $15$ duads into $5$ synthemes, and they are harder to count.

There’s a nice blog-post by Peter Cameron on this, as well as his paper From $M_{12}$ to $M_{24}$ (after Graham Higman). As my master-students have to work their own way through this paper I will not spoil their fun in trying to deduce that

Two totals have exactly one syntheme in common, so synthemes are ‘duads of totals’.
Three synthemes lying in disjoint pairs of totals must consist of synthemes containing a fixed duad, so duads are ‘synthemes of totals’.
Duads come from disjoint synthemes of totals in this way if and only if they share a point, so points are ‘totals of totals’

My hint to the students was “Google for John Baez+six”, hoping they’ll discover Baez’ marvellous post Some thoughts on the number $6$, and in particular, the image (due to Greg Egan) in that post

which makes everything visually clear.

The duads are the $15$ red vertices, the synthemes the $15$ blue vertices, connected by edges when a duad is contained in a syntheme. One obtains the Tutte-Coxeter graph.

The $6$ concentric rings around the picture are the $6$ synthematic totals. A band of color appears in one of these rings near some syntheme if that syntheme is part of that synthematic total.

If $\{ t_1,t_2,t_3,t_4,t_5,t_6 \}$ are the six totals, then any permutation $\sigma$ of $\{ 1,2,3,4,5,6 \}$ induces a permutation $\phi(\sigma)$ of the totals, giving the odd automorphism
\[
\phi : S_6 = Sym(\{ 1,2,3,4,5,6 \}) \rightarrow Sym(\{ t_1,t_2,t_3,t_4,t_5,t_6 \}) = S_6 \]

214066877211724763979841536000000000000

Published August 18, 2019 by lievenlb

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

Published August 14, 2019 by lievenlb

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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