# Tag: Marcolli

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!

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!

The Bulletin of the AMS just made this paper by Julia Mueller available online: “On the genesis of Robert P. Langlands’ conjectures and his letter to Andre Weil” (hat tip +ChandanDalawat and +DavidRoberts on Google+).

It recounts the story of the early years of Langlands and the first years of his mathematical career (1960-1966)leading up to his letter to Andre Weil in which he outlines his conjectures, which would become known as the Langlands program.

Langlands letter to Weil is available from the IAS.

The Langlands program is a vast net of conjectures. For example, it conjectures that there is a correspondence between

– $n$-dimensional representations of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, and

– specific data coming from an adelic quotient-space $GL_n(\mathbb{A}_{\mathbb{Q}})/GL_n(\mathbb{Q})$.

For $n=1$ this is essentially class field theory with the correspondence given by Artin’s reciprocity law.

Here we have on the one hand the characters of the abelianised absolute Galois group

$Gal(\overline{\mathbb{Q}}/\mathbb{Q})^{ab} \simeq Gal(\mathbb{Q}(\pmb{\mu}_{\infty})/\mathbb{Q}) \simeq \widehat{\mathbb{Z}}^{\ast}$

and on the other hand the connected components of the idele class space

$GL_1(\mathbb{A}_{\mathbb{Q}})/GL_1(\mathbb{Q}) = \mathbb{A}_{\mathbb{Q}}^{\ast} / \mathbb{Q}^{\ast} = \mathbb{R}_+^{\ast} \times \widehat{\mathbb{Z}}^{\ast}$

For $n=2$ it involves the study of Galois representations coming from elliptic curves. A gentle introduction to the general case is Mark Kisin’s paper What is … a Galois representation?.

One way to look at some of the quantum statistical systems studied via non-commutative geometry is that they try to understand the “bad” boundary of the Langlands space $GL_n(\mathbb{A}_{\mathbb{Q}})/GL_n(\mathbb{Q})$.

Here, the Bost-Connes system corresponds to the $n=1$ case, the Connes-Marcolli system to the $n=2$ case.

If $\mathbb{A}’_{\mathbb{Q}}$ is the subset of all adeles having almost all of its terms in $\widehat{\mathbb{Z}}_p^{\ast}$, then there is a well-defined map

$\pi~:~\mathbb{A}’_{\mathbb{Q}}/\mathbb{Q}^{\ast} \rightarrow \mathbb{R}_+ \qquad (x_{\infty},x_2,x_2,\dots) \mapsto | x_{\infty} | \prod_p | x_p |_p$

The inverse image of $\pi$ over $\mathbb{R}_+^{\ast}$ are exactly the idele classes $\mathbb{A}_{\mathbb{Q}}^{\ast}/\mathbb{Q}^{\ast}$, so we can view them as the nice locus of the horrible complicated quotient of adele-classes $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}^*$. And we can view the adele-classes as a ‘closure’ of the idele classes.

But, the fiber $\pi^{-1}(0)$ has horrible topological properties because $\mathbb{Q}^*$ acts ergodically on it due to the fact that $log(p)/log(q)$ is irrational for distinct primes $p$ and $q$.

This is why it is better to view the adele-classes not as an ordinary space (one with bad topological properties), but rather as a ‘non-commutative’ space because it is controlled by a non-commutative algebra, the Bost-Connes algebra.

For $n=2$ there’s a similar story with a ‘bad’ quotient $M_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$, being the closure of an ‘open’ nice piece which is the Langlands quotient space $GL_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$.

A well-known link between Conway’s Big Picture and non-commutative geometry is given by the Bost-Connes system.

This quantum statistical mechanical system encodes the arithmetic properties of cyclotomic extensions of $\mathbb{Q}$.

The corresponding Bost-Connes algebra encodes the action by the power-maps on the roots of unity.

It has generators $e_n$ and $e_n^*$ for every natural number $n$ and additional generators $e(\frac{g}{h})$ for every element in the additive group $\mathbb{Q}/\mathbb{Z}$ (which is of course isomorphic to the multiplicative group of roots of unity).

The defining equations are
$\begin{cases} e_n.e(\frac{g}{h}).e_n^* = \rho_n(e(\frac{g}{h})) \\ e_n^*.e(\frac{g}{h}) = \Psi^n(e(\frac{g}{h}).e_n^* \\ e(\frac{g}{h}).e_n = e_n.\Psi^n(e(\frac{g}{h})) \\ e_n.e_m=e_{nm} \\ e_n^*.e_m^* = e_{nm}^* \\ e_n.e_m^* = e_m^*.e_n~\quad~\text{if (m,n)=1} \end{cases}$

Here $\Psi^n$ are the power-maps, that is $\Psi^n(e(\frac{g}{h})) = e(\frac{ng}{h}~mod~1)$, and the maps $\rho_n$ are given by
$\rho_n(e(\frac{g}{h})) = \sum e(\frac{i}{j})$
where the sum is taken over all $\frac{i}{j} \in \mathbb{Q}/\mathbb{Z}$ such that $n.\frac{i}{j}=\frac{g}{h}$.

Conway’s Big Picture has as its vertices the (equivalence classes of) lattices $M,\frac{g}{h}$ with $M \in \mathbb{Q}_+$ and $\frac{g}{h} \in \mathbb{Q}/\mathbb{Z}$.

The Bost-Connes algebra acts on the vector-space with basis the vertices of the Big Picture. The action is given by:
$\begin{cases} e_n \ast \frac{c}{d},\frac{g}{h} = \frac{nc}{d},\rho^m(\frac{g}{h})~\quad~\text{with m=(n,d)} \\ e_n^* \ast \frac{c}{d},\frac{g}{h} = (n,c) \times \frac{c}{nd},\Psi^{\frac{n}{m}}(\frac{g}{h})~\quad~\text{with m=(n,c)} \\ e(\frac{a}{b}) \ast \frac{c}{d},\frac{g}{h} = \frac{c}{d},\Psi^c(\frac{a}{b}) \frac{g}{h} \end{cases}$

This connection makes one wonder whether non-commutative geometry can shed a new light on monstrous moonshine?

This question is taken up by Jorge Plazas in his paper Non-commutative geometry of groups like $\Gamma_0(N)$

Plazas shows that the bigger Connes-Marcolli $GL_2$-system also acts on the Big Picture. An intriguing quote:

“Our interest in the $GL_2$-system comes from the fact that its thermodynamic properties encode the arithmetic theory of modular functions to an extend which makes it possible for us to capture aspects of moonshine theory.”

Looks like the right kind of paper to take along when I disappear next week for some time in the French mountains…

Last fall, Matilde Marcolli gave a course at CalTech entitled Oral Presentation: The (Martial) Art of Giving Talks. The purpose of this course was to teach students “how to effectively communicate their work in seminars and conferences and how to defend it from criticism from the audience”.

The lecture notes contain basic information on the different types of talks and how to prepare them. But they really shine when it comes to spotting the badasses in the public and how to respond to their interference. She identifies 5 badsass-types : the empreror and the hierophant (see below), the chariot (the one with a literal mind, asking continuously for details), the fool (the one who happens to sit in the talk but doesn’t belong there) and the magician (the quick smartass).

I’ll just quote here the description of, and most effective strategy against, the first two badass-types. Please have a look at the whole paper, it is a good read!

“The Emperor is the typical figure of power and authority in a given field. It refers to those people who have a tendency to think that the whole field is their own private property, and in particular that only what they do in the field is important, that the work of all others is derivative and that in any case they are not being quoted enough. These are typically pathological narcissists, so one needs to take this into account in interacting with them.
The trouble of having The Emperor in your audience is that he (it is rarely she) can very easily disrupt your presentation completely, by continuous interruptions, by running his own commentary while you are trying to stay focused on delivering your talk and by distracting the rest of the audience.
The Emperor is by far one of the most dangerous encounters you can make in the wilderness of the conference rooms.”

Counter-measure : “Keep in mind that the Emperor is a pathological narcissist: part of the reason why he keeps interrupting your talk is because he cannot stand the fact that, during those fifty minutes, the attention of the audience is focused on you and not on him. His continuous interruptions and complaints are a way to try to divert the attention of the audience back to him and away from you. That your talk gets disrupted in the process, he could not care the less.
A good way to try to avoid the worst case scenario is to make sure (if you know in advance you may be having the Emperor in the audience) that you arrange in your talk to make frequent references to him and his work. In this way, he will hopefully feel that his need to be at the center of attention is sufficiently satisfied that he can let you continue with your talk. Effectiveness: high.”

“The Hierophant represents a priestly figure. What this refers to here is the type of character who feels entitled to represent (and defend) a certain “orthodoxy”, a certain school of thought, or a certain group of people within the field.
Typically the hierophants are the minions and lackeys of the Emperor, his entourage and fan club, those who think that the Emperor represents the only and true orthodoxy in the field and that anything that is done in a different way should be opposed and suppressed.
These characters are generally less disruptive than the Emperor himself, as they are really only fighting you on someone else’s behalf. Nonetheless, they can sometime manage to seriously disrupt your presentation.”

Counter-measure : “This is essentially the same advise as in the case of the Emperor. To an objection that substantially is of the form: “This is not the right way to do things because this is not what what we do (= what the Emperor does)”, which is what you expect to hear from the Hierophant, you can reply along lines such as: “There is also another approach to this problem, developed by the Emperor and his school, which is a very interesting approach that gave nice and important results. However, this is not what I am talking about today: I am talking here about a different approach, and I will be focusing only on the specific features of this other approach…”
Something along these lines would recognize “their” work without having to make any concession on their approach being the only game in town.
Effectiveness: high (unless the Emperor is also present and is delegating to his hierophants the task of attacking you: in that case they won’t give up so easily and the effectiveness of this line of defense becomes medium/low).”