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Category: number theory

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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Dessinflateurs

Published August 10, 2019 by lievenlb

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

Published August 3, 2019 by lievenlb

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
\[
\Gamma = C_2 \ast C_3 = \langle s,u~|~s^2=1=u^3 \rangle \]
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

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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the Riemann hypothesis and Psi

Published July 14, 2019 by lievenlb

Last time we revisited Robin’s theorem saying that 5040 being the largest counterexample to the bound
\[
\frac{\sigma(n)}{n~log(log(n))} < e^{\gamma} = 1.78107... \] is equivalent to the Riemann hypothesis.

There’s an industry of similar results using other arithmetic functions. Today, we’ll focus on Dedekind’s Psi function
\[
\Psi(n) = n \prod_{p | n}(1 + \frac{1}{p}) \]
where $p$ runs over the prime divisors of $n$. It is series A001615 in the online encyclopedia of integer sequences and it starts off with

1, 3, 4, 6, 6, 12, 8, 12, 12, 18, 12, 24, 14, 24, 24, 24, 18, 36, 20, 36, 32, 36, 24, 48, 30, 42, 36, 48, 30, 72, 32, 48, 48, 54, 48, …

and here’s a plot of its first 1000 values



To understand this behaviour it is best to focus on the ‘slopes’ $\frac{\Psi(n)}{n}=\prod_{p|n}(1+\frac{1}{p})$.

So, the red dots of minimal ‘slope’ $\approx 1$ correspond to the prime numbers, and the ‘outliers’ have a maximal number of distinct small prime divisors. Look at $210 = 2 \times 3 \times 5 \times 7$ and its multiples $420,630$ and $840$ in the picture.

For this reason the primorial numbers, which are the products of the fist $k$ prime numbers, play a special role. This is series A002110 starting off with

1, 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870,…

In Patrick Solé and Michel Planat Extreme values of the Dedekind $\Psi$ function, it is shown that the primorials play a similar role for Dedekind’s Psi as the superabundant numbers play for the sum-of-divisors function $\sigma(n)$.

That is, if $N_k$ is the $k$-th primorial, then for all $n < N_k$ we have that the 'slope' at $n$ is strictly below that of $N_k$ \[ \frac{\Psi(n)}{n} < \frac{\Psi(N_k)}{N_k} \] which follows immediately from the fact that any $n < N_k$ can have at most $k-1$ distinct prime factors and $p \mapsto 1 + \frac{1}{p}$ is a strictly decreasing function.

Another easy, but nice, observation is that for all $n$ we have the inequalities
\[
n^2 > \phi(n) \times \psi(n) > \frac{n^2}{\zeta(2)} \]
where $\phi(n)$ is Euler’s totient function
\[
\phi(n) = n \prod_{p | n}(1 – \frac{1}{p}) \]
This follows as once from the definitions of $\phi(n)$ and $\Psi(n)$
\[
\phi(n) \times \Psi(n) = n^2 \prod_{p|n}(1 – \frac{1}{p^2}) < n^2 \prod_{p~\text{prime}} (1 - \frac{1}{p^2}) = \frac{n^2}{\zeta(2)} \] But now it starts getting interesting.

In the proof of his theorem, Guy Robin used a result of his Ph.D. advisor Jean-Louis Nicolas



known as Nicolas’ criterion for the Riemann hypothesis: RH is true if and only if for all $k$ we have the inequality for the $k$-th primorial number $N_k$
\[
\frac{N_k}{\phi(N_k)~log(log(N_k))} > e^{\gamma} \]
From the above lower bound on $\phi(n) \times \Psi(n)$ we have for $n=N_k$ that
\[
\frac{\Psi(N_k)}{N_k} > \frac{N_k}{\phi(N_k) \zeta(2)} \]
and combining this with Nicolas’ criterion we get
\[
\frac{\Psi(N_k)}{N_k~log(log(N_k))} > \frac{N_k}{\phi(N_k)~log(log(N_k)) \zeta(2)} > \frac{e^{\gamma}}{\zeta(2)} \approx 1.08… \]
In fact, Patrick Solé and Michel Planat prove in their paper Extreme values of the Dedekind $\Psi$ function that RH is equivalent to the lower bound
\[
\frac{\Psi(N_k)}{N_k~log(log(N_k))} > \frac{e^{\gamma}}{\zeta(2)} \]
holding for all $k \geq 3$.

Dedekind’s Psi function pops up in lots of interesting mathematics.

In the theory of modular forms, Dedekind himself used it to describe the index of the congruence subgroup $\Gamma_0(n)$ in the full modular group $\Gamma$.

In other words, it gives us the number of tiles needed in the Dedekind tessellation to describe the fundamental domain of the action of $\Gamma_0(n)$ on the upper half-plane by Moebius transformations.

When $n=6$ we have $\Psi(6)=12$ and we can view its fundamental domain via these Sage commands:


G=Gamma0(6)
FareySymbol(G).fundamental_domain()

giving us the 24 back or white tiles (note that these tiles are each fundamental domains of the extended modular group, so we have twice as many of them as for subgroups of the modular group)



But, there are plenty of other, seemingly unrelated, topics where $\Psi(n)$ appears. To name just a few:

  • The number of points on the projective line $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.
  • The number of lattices at hyperdistance $n$ in Conway’s big picture.
  • The number of admissible maximal commuting sets of operators in the Pauli group for the $n$ qudit.

and there are explicit natural one-to-one correspondences between all these manifestations of $\Psi(n)$, tbc.

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the Riemann hypothesis and 5040

Published July 7, 2019 by lievenlb

Yesterday, there was an interesting post by John Baez at the n-category cafe: The Riemann Hypothesis Says 5040 is the Last.

The 5040 in the title refers to the largest known counterexample to a bound for the sum-of-divisors function
\[
\sigma(n) = \sum_{d | n} d = n \sum_{d | n} \frac{1}{d} \]

In 1983, the french mathematician Guy Robin proved that the Riemann hypothesis is equivalent to
\[
\frac{\sigma(n)}{n~log(log(n))} < e^{\gamma} = 1.78107... \] when $n > 5040$.

The other known counterexamples to this bound are the numbers 3,4,5,6,8,9,10,12,16,18,20,24,30,36,48,60,72,84,120,180,240,360,720,840,2520.




In Baez’ post there is a nice graph of this function made by Nicolas Tessore, with 5040 indicated with a grey line towards the right and the other counterexamples jumping over the bound 1.78107…



Robin’s theorem has a remarkable history, starting in 1915 with good old Ramanujan writing a part of this thesis on “highly composite numbers” (numbers divisible by high powers of primes).

His PhD. adviser Hardy liked his result but called them “in the backwaters of mathematics” and most of it was not published at the time of Ramanujan’s degree ceremony in 1916, due to paper shortage in WW1.



When Ramanujan’s paper “Highly Composite Numbers” was first published in 1988 in ‘The lost notebook and other unpublished papers’ it became clear that Ramanujan had already part of Robin’s theorem.

Ramanujan states that if the Riemann hypothesis is true, then for $n_0$ large enough we must have for all $n > n_0$ that
\[
\frac{\sigma(n)}{n~log(log(n))} < e^{\gamma} = 1.78107... \] When Jean-Louis Nicolas, Robin's PhD. adviser, read Ramanujan's lost notes he noticed that there was a sign error in Ramanujan's formula which prevented him from seeing Robin's theorem.

Nicolas: “Soon after discovering the hidden part, I read it and saw the difference between Ramanujan’s result and Robin’s one. Of course, I would have bet that the error was in Robin’s paper, but after recalculating it several times and asking Robin to check, it turned out that there was an error of sign in what Ramanujan had written.”

If you are interested in the full story, read the paper by Jean-Louis Nicolas and Jonathan Sondow: Ramanujan, Robin, Highly Composite Numbers, and the Riemann Hypothesis.

What’s the latest on Robin’s inequality? An arXiv-search for Robin’s inequality shows a flurry of activity.

For starters, it has been verified for all numbers smaller that $10^{10^{13}}$…

It has been verified, unconditionally, for certain classes of numbers:

  • all odd integers $> 9$
  • all numbers not divisible by a 25-th power of a prime

Rings a bell? Here’s another hint:

According to Xiaolong Wu in A better method than t-free for Robin’s hypothesis one can replace the condition of ‘not divisible by an N-th power of a prime’ by ‘not divisible by an N-th power of 2’.

Further, he claims to have an (as yet unpublished) argument that Robin’s inequality holds for all numbers not divisible by $2^{42}$.

So, where should we look for counterexamples to the Riemann hypothesis?

What about the orders of huge simple groups?

The order of the Monster group is too small to be a counterexample (yet, it is divisible by $2^{46}$).

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G+ recovery 5 : #IUTeich (3)

Published February 5, 2019 by lievenlb

The final part, starting with a reaction of Mochizuki himself on my previous IUTeich-posts.

July 7th, 2013

Mochizuki update

While i’ve been away from G+, someone emailed Mochizuki my last post on the problem i have with Frobenioids1. He was kind enough to forward  me M’s reply:

“There is absolutely nothing difficult or subtle going on here (e.g., by comparison to the portion of the theory cited in the discussion preceding the statement of this “problem”!).  The nontrivial result is the fact that the degree is rational, i.e., the initial portion of [FrdI], Theorem 6.4, (iii), which is a consequence of a highly nontrivial result in transcendence theory due to Lang (i.e., [FrdI], Lemma 6.5, (ii)).  Once one knows this rationality, the conclusion that distinct prime numbers are not confused with one another is a formal consequence (i.e., no complicated subtle arguments!) of the fact that the ratio of the natural logarithm of any two distinct prime numbers is never rational.
Sincerely, Shinichi Mochizuki”

Being back, i’ll give it another go.

Having read the shared article below (2/5/2019 ‘the paradox of the proof’), it’s comforting to know that other people, including  +Aise Johan de Jong and +Cathy O’Neil, are also frustrated by M’s opaque latest writings.

July 11th, 2013

Mochizuki’s Frobenioid reconstruction: the final bit

In “The geometry of Frobenioids 1” Mochizuki ‘dismantles’ arithmetic schemes and replaces them by huge categories called Frobenioids. Clearly, one then wants to reconstruct the schemes from these categories and we almost understood how he manages to to this, modulo the ‘problem’ that there might be auto-equivalences of $Frob(\mathbb{Z})$, the Frobenioid corresponding to $\mathbf{Spec}(\mathbb{Z})$ i.e.the collection of all prime numbers, reshuffling distinct prime numbers.

In previous posts i’ve simplified things a lot, leaving out the ‘Arakelov’ information contained in the infinite primes, and feared that this lost info might be crucial to understand the final bit. Mochizuki’s email also points in that direction.

Here’s what i hope to have learned this week:

the full Frobenioid $Frob(\mathbb{Z})$

objects of $Frob(\mathbb{Z})$ consist of pairs $(q,r)$ where $q$ is a strictly positive rational number and $r$ is a real number.

morphisms are of the form $f=(n,a,(z,s)) : (q,r) \rightarrow (q’,r’)$ (where $n$ and $z$ are strictly positive integers, $a$ a strictly positive rational number and $s$ a positive real number) subject to the relations that

\[
q^n.z = q’.a  \quad \text{and} \quad n.r+s = r’+log(a) \]

$n$ is called the Frobenious degree of $f$ and $(z,s)$ the divisor of $f$.

All this may look horribly complicated until you realise that the isomorphism classes in $Frob(\mathbb{Z})$ are exactly the ‘curves’ $C(\alpha)$ consisting of all pairs $(q,r)$ such that $r-log(q)=\alpha$, and that morphisms with the same parameters as $f$ send points in $C(\alpha)$ to points in $C(\beta)$ where

\[
\beta = n.\alpha + s – log(a) \]

So, the isomorphism classes can be identified with the real numbers $\mathbb{R}$ and special linear morphisms of type $(1,a,(1,s))$ and their compositions are compatible with the order-structure and addition on $\mathbb{R}$.

Crucial is Mochizuki’s observation that $C(0)$ are precisely the ‘Frobenious-trivial’ objects in $Frob(\mathbb{Z})$ (i’ll spare you the details but it is a property on having sufficiently many nice endomorphisms).

Now, consider an auto-equivalence $E$ of $Frob(\mathbb{Z})$. It will induce a map between the isoclasses $E : \mathbb{R} \rightarrow \mathbb{R}$ which is additive and as Frob-trivs are mapped under $E$ to Frob-trivs this will map $0$ to $0$, so $E$ will be an additive group-endomorphism on $\mathbb{R}$ hence of the form $x \rightarrow r.x$ for some fixed real number $r$, and we want to show that $r=1$.

Linear irreducible morphisms are of type $(1,a,(p,0))$ where $p$ is a prime number and they map $C(\alpha)$ to $C(\alpha-log(p))$. As irreducibles are preserved under equivalence this means that for each prime $p$ there must exist a prime $q$ such that $r log(p) = log(q)$.

Now if $r$ is an irrational number, there must be at least three triples $(p_1,q_1),(p_2,q_2)$ and $(p_3,q_3)$ satisfying $r log(p_i) = log(q_i)$ but this contradicts a fairly hard result, due to Lang, that for 6 distinct primes l_1,…,l_6 there do not exist positive rational numbers $a,b$ such that

\[
log(l_1)/log(l_2) = a log(l_3)/log(l_4) = b log(l_5)/log(l_6) \]

So, $r$ must be rational and of the form $n/m$, but then $r=1$ (if not the correspondence $r.log(p) = log(q)$ gives $p^n=q^m$ contradicting unique factorisation). This then shows that under the auto-equivalence each prime $p$ (corresponding to a linear irreducible map) is send to itself.

This was the remaining bit left to show that the Frobenioid corresponding to any Galois extension of the rationals contains enough information to reconstruct from it the schemes of all rings of integers in intermediate fields.

August 15th, 2013

Szpiro’s Marabout-Flash on number theory

For travellers into Mochizuki-territory the indispensable rough guide is Lucien Szpiro’s ‘Marabout-Flash de théorie des nombres algébriques’, aka section I.1.3 in  ‘Séminaire sur les pinceaux arithmétiques: la conjecture de Mordell’.

Marabout Flash was a Franco-Belge series of do-it-yourself booklets, quite popular in the 60ties and 70ties, on almost every aspect of everyday’s life.

In just a couple of pages Szpiro describes how one can extend the structure sheaf of a fractional ideal of a ring of integers to a ‘metrized’ (or Arakelov) line-bundle on the completed prime spectrum (including the infinite places). These bundles then satisfy properties similar to those of line-bundles on smooth projective curves, including a version of the Riemann-Roch theorem and a criterium to have non-zero global sections.

These (fairly simple) results then quickly lead to proofs of the first major results in number theory such as the Hermite-Minkowski theorem and Dirichlet’s unit theorem.

December 29th, 2014

Mochizuki in denial

From M’s 2014 IUTeich-Progress-Report (17 pages, the 2013-edition was only 7 pages long):

“Activities surrounding IUTeich appears to be in a stage of transition from a focus on verification to a focus on dissemination.”

If only…

He further lists hypotheses as to why nobody (apart from his 3 disciples Yamashita,Saidi and Hoshi) makes a serious effort to “study the theory carefully and systematically from the beginning”:

1. it is too long (1500-2500 pages)
2. there’s lack of textbooks on anabelian geometry
3. we are obsessed with the Langlands program
4. there’s little room for generalisations
5. it may not be directly useful for our own research

But then, why should anyone make such an effort, as:

“With the exception of the handful of researchers already involved in the verification activities concerning IUTeich, every researcher in arithmetic geometry throughout the world is a complete novice with respect to the mathematics surrounding IUTeich, and hence, in particular, is simply not qualified to issue a definite judgment concerning the validity of IUTeich on the basis of a ‘deep understanding’ arising from his/her previous research achievements.”

No Mochizuki, the next phase will not be dissemination, it will be denial.

Other remarkable sentences are:

“IUTeich is ‘the correct theory’ in the sense that it leads one to doubt the existence of any sort of ‘alternative proof’, i.e. via essentially different techniques, of the ABC Conjecture.”

and:

“the status of IUTeich in the field of arithmetic geometry constitutes a sort of faithful miniature model of the status of pure mathematics in human society.”

Already looking forward to the 2015 ‘progress’ report…

October 8th, 2015

Proud to be working at a well-known university

First time I’m mentioned in “Nature”, they issue this correction:

Corrected: An earlier version of this story incorrectly located the University of Antwerp in the Netherlands. It is in Belgium. The text has been updated.

Not particularly proud of the quote they took from my blog though:

“Is it just me, or is Mochizuki really sticking up his middle finger to the mathematical community”.

December 17th, 2017

We’re heading for a bad ending

Yesterday, my feeds became congested by (Japanese) news saying that Mochizuki’s (claimed) proof of the abc-conjecture had been vetted and considered fit to be published in a respectable journal.

Today, long term supporters of M’s case began their Echternachian-retreat after finding out that Mochizuki himself is the editor in charge of that respectable journal.

Attached is Ed Frenkel’s retraction of his previous tweet. Also, Taylor Dupuy deemed the latest action a bridge too far, see here.

Peter Woit did a great job in his recent post
pinpointing a critical argument in the 500-page long papers having as its “proof” that it followed trivially from the definitions…

I’d expect any board of editors to resign in such a case.
I fear this can only end badly.

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G+ recovery 4 : #IUTeich (2)

Published February 5, 2019 by lievenlb

Here’s batch 2 of my old google+ posts on ‘Inter Universal Teichmuller theory’, or rather on the number theoretic examples of Frobenioids.

June 5th, 2013

Mochizuki’s categorical prime number sieve

And now for the interesting part of Frobenioids1: after replacing a bunch of arithmetic schemes and maps between them by a huge category, we will reconstruct this classical picture by purely categorical means.

Let’s start with the simplest case, that of the ‘baby arithmetic Frobenioid’ dismantling $\mathbf{Spec}(\mathbb{Z}) (that is, the collection of all prime numbers) and replacing it by the category having as its objects

all $(a)$ where $a$ is a strictly positive rational number

and morphisms labeled by triples $(n,r,z)$ where $n$ and $z$ are strictly positive integers and $r$ is a strictly positive rational number and connecting two objects

\[
(n,r,z) : (a) \rightarrow (b) \quad \text{ if and only if } \quad a^n.z=b.r \]

Composition of morphisms is well-defined and looks like $(m,s,v) \circ (n,r,u) = (m.n,r^m.s,u^m.v)$ as one quickly checks.

The challenge is to recover all prime numbers back from this ‘Frobenioid’. We would like to take an object $(a)$ and consider the maps $(1,1,p)$ from it for all prime numbers $p$, but cannot do this as categorically we have to drop all labels of objects and arrows. That is, we have to recognize the map $(1,1,p)$ among all maps starting from a given object.

We can identify all isomorphisms in the category and check that they are precisely the morphisms labeled $(1,r,1)$. In particular, this implies that all objects are isomorphic and that there is a natural correspondence between arrows leaving $(a)$ and arrows leaving $(b)$ by composing them with the iso $(1,b/a,1) : (b) \rightarrow (a)$.

Another class of arrows we can spot categorically are the ‘irreducibles’, which are maps $f$ which are not isos but have the property that in any factorization $f=g \circ h$ either $g$ or $h$ must be an iso. One easily verifies from the composition rule that these come in two flavours:

– those of Frobenius type : $(p,r,1)$ for any prime number $p$
– those of Order type : $(1,r,p)$ for any prime number $p$

We would like to color the froBs Blue and the oRders Red, but there seems to be no way to differentiate between the two classes by purely categorical means, until you spot Mochizuki’s clever little trick.

start with a Red say $(1,r,q)$ for a prime number $q$ and compose it with the Blue $(p,1,1)$, then you get the morphism $(p,r^p,q^p)$ which you can factor as a composition of $p+1$ irreducibles

\[
(p,r^p,q^p) = (1,r,q) \circ (1,r,q) \circ …. \circ (1,r,q)o(p,1,1) \]

and if $p$ grows, so will the number of factors in this composition.

On the other hand, if you start with a Blue and compose it with either a Red or a Blue irreducible, the obtained map cannot be factored in more irreducibles.

Thus, we can identify the Order-type morphisms as those irreducibles $f$ for which there exists an irreducible $g$ such that the composition $g \circ f$ can be factored as the composition in at least $n$ irreducibles, where we can take n arbitrarily large.

Finally we say that two Reds out of $(a)$ are equivalent iff one is obtained from the other by composing with an isomorphism and it is clear that the equivalence classes are exactly the arrows labeled $(1,r,p)$ for fixed prime number $p$.

So we do indeed recover all prime numbers from the category.

Similarly, we can see that equivalence classes of Frobs from $(a)$ are of the form $(p,r,1)$ for fixed prime $p$. An amusing fact is that we can recover the prime $p$ for a Frob by purely categorical ways using the above long factorization of a composition with a Red.

There seems to be no categorical way to determine the prime number associated to an equivalence class of Order-morphisms though… Or, am i missing something trivial?

June 7th, 2013

Mochizuki’s Frobenioids for the Working Category Theorist

Many of you, including +David Roberts  +Charles Wells  +John Baez  (and possibly others, i didn’t look at all comments left on all reshares of the past 4 posts in this MinuteMochizuki project) hoped that there might be a more elegant category theoretic description of Frobenioids, the buzz-word apparently being ‘Grothendieck fibration’ …

Hence this attempt to deconstruct Frobenioids. Two caveats though:

– i am not a category theorist (the few who know me IRL are by now ROFL)
– these categories are meant to include all arithmetic information of number fields, which is a messy business, so one should only expect clear cut fibrations in easy situation such as principal ideal domains (think of the integers $\mathbb{Z}$).

Okay, we will try to construct the Frobenioid associated to a number field $K$ (that is, a finite dimensional extension of the rationals $\mathbb{Q}$) with ring of integers $R$ (the integral closure of $\mathbb{Z}$ in $K$). For a concrete situation, look at the quadratic case.

The objects will be fractional ideals of K which are just the R-submodules $I$ of $K$ such that there in an $r$ in $R$ such that $I.r$ is a proper ideal of $R$. Dedekind showed that any such thing can be written uniquely as a product

\[
I = P_1^{a_1} … P_k^{a_k} \]

where the $P_i$ are prime ideals of $R$ and the $a_i$ are integers (if they are all natural numbers, I will be a proper ideal of $R$). Clearly, if one multiplies two fractional ideals $I$ and $J$, the result $I.J$ is again a fractional ideal, so they form a group and by Dedekind’s trick this group is the free Abelian group on all prime ideals of $R$.

Next, we define an equivalence relation on this set, calling two fractional ideals $I$ and $J$ equivalent if there is a $k$ in $K$ such that $I=J.k$ (or if you prefer, if they are isomorphic as $R$-modules).

We have a set with an equivalence relation and hence a groupoid where these is a unique isomorphism between any two equivalent objects. This groupoid is precisely the groupoid of isomorphisms of the Frobenioid we’re after.

The number of equivalence classes is finite and these classes correspond to the element of a finite group $Cl(R)$ called the class group of $R$ which is the quotient group of ideals modulo principal ideals (so if your $R$ is a principal ideal domain there is just one component). The ‘groups’ corresponding to each connected component of the groupoid are all isomorphic to the quotient group of the units in $K$ by the units in $R$.

Next, we will add the other morphisms. By definition they are all compositions of irreducibles which come in 2 flavours:

– the order-morphisms $P$ for any prime ideal $P$ of $R$ sending $I$ to $I.P$. Typically, these maps switch between different equivalence classes (unless $P$ itself is principal). We can even explicitly compute small norm prime ideals which will generate all elements in the class group $Cl(R)$.
– the power-maps $[p]$ for any prime number $p$ which sends $I$ to $I^p$. The nature of these maps really depend on the order of the component in the finite group $Cl(R)$.

Well, that’s it basically for the layer of the Frobenioid corresponding to the number field $K$. (You have to repeat all this for any subfield between $\mathbb{Q}$ and $K$). A cute fact is that all endomorphism-monoids of objects in the layer of K are all isomorphic as abstract monoid to the skew-monoid

$\mathbb{N}^x_{>0} x Prin(R)$ of the multiplicative group of all strictly positive integers $n$ with the monoid of all principal ideals in R with multiplication defined by

\[
(n,Ra).(m,Rb)=(nm,Rab^n) \]

The only extra-type morphisms we still have to include are those between the different layers of the Frobenioid, the green ones which M calls the pull-back morphisms.
They are of the following form: if $R_1$ and $R_2$ are rings of integers in the fields $K_1$ contained in $K_2$, then for any ringmorphism $\sigma : R_1 \rightarrow R_2$ one can extend a fractional ideal $I$ of $R_1$ to $K_2$ by considering $R_2.\sigma(I)$. These then give the morphism $r_2\sigma(I) \rightarrow I$ and as we will see in a next instalment, they encode the splitting behaviour of prime ideals.

a question for category people

Take the simplest situation, that of the integers $\mathbb{Z}$. So, we have just a groupoid with extra morphisms generated by the order-maps $o_p$ and the power maps $f_p$. The endo-ring of any object is then isomorphic top the abstract group generated by the $f_p$ and $o_p$ and satisfying following relations

\[
o_p.o_q=o_q.o_p \]
\[
f_p.f_q=f_q.f_p \]
\[
f_p.o_q=o_q^p.f_p \]

My question now is: if for two different primes $p$ and $q$ i switch their role in the endo-ring of 1 object and propagate this via all isos to all morphisms, do i get a category equivalence? (or am i missing something?).  (tbc)

June 11th, 2013

my problem with Mochizuki’s Frobenioid1

Let us see how much arithmetic information can be reconstructed from an arithmetic Frobenioids. Recall that for a fixed finite Galois extension $L$ of $\mathbb{Q}$ this is a category with objects all fractional ideals in subfields of $L$, and maps generated by multiplication-maps with ideals in rings of integers, power-maps and Galois-extension maps.

When all objects and morphisms are labeled it is quite easy to reconstruct the Galois field $L$ from it as well as all maps between prime spectra of rings of integers in intermediate fields, which after all was the intended use of Frobenioids, to ‘dismantle’ these arithmetic schemes and endow them with extra structure given by the power-maps.

However, in this reconstruction process we are only allowed t use the category structure, so all objects and morphisms are unlabelled (the situation top left) and we want to reconstruct from it the different layers of the Frobenioid (corresponding to the different subfields) and divide all arrows according to their type (situation bottom left).

First we can look at all isomorphisms. They will divide the category in the dashed regions, some of them will be an entire layer (for example for $\mathbb{Q}$) but in general a finite number of these regions will make up the full layer of a subfield (the regions labeled by the elements of the ideal class group).

Another categorical notion we can use are ‘irreducible morphisms’, that is a morphism $f$ which is not an iso but having the property that in each factorisation $f = g \circ h$ either $g$ or $h$ must be an iso. If we remember the different types of morphisms in our Frobenioid we see that the irreducibles come in 3 flavours:

– oRder-maps (Red) : multiplication by a prime ideal $P$ of the ring of integers of the subfield
– froBenius or power-maps (Blue) sendingg a fractional ideal $I$ to $I^p$ for a prime number $p$
– Galois-maps (Green) extending ideals for a subfield $K$ to $K’$ having no intermediate field.

We would like to determine the colour of these irreducibles purely categorical. The idea is that reds have the property that they can be composed with another irreducible (in fact, of power type) such that the composition can again be decomposed in irreducibles and that there is no a priori bound on the number of these terms (this uses the fact that $[p] \circ Q = Q \circ Q \circ … Q \circ [p]$ and that there are infinitely many prime numbers $p$). One checks that compositions of order or Galois maps with irreducibles have factorisation with a bounded number of irreducibles.

The most interesting case is the composition of a Galois map with an order map $P$, this can be decomposed alternatively as order maps in the bigger field followed by a Galois map, the required order maps are the bigger primes $Q_i$ occurring in the decomposition of the extended ideal

\[
S.P = Q_1.Q_2….Q_k \]

but the number $k$ of this decomposition is bounded by the dimension of the bigger field over the smaller one. Summarizing:

– we can determine all the red maps, which will then give us also the different layers
– we can determine the green maps as they move between different layers
– to the remaining blue ones we can even associate their label $[p]$ by the observed property of composition with order maps.

Taking an object in a layer, we get the set of prime ideals of the ring on integers in that field as the set of all red arrows leaving that object unto equivalence (by composing with an isomorphism), so we get the prime spectra $\mathbf{Spec}(S)$.

For a ring-extension  $R \rightarrow S$ we also can recover the cover map $\mathbf{Spec}(S) \rightarrow \mathbf{Spec}(R)$

Indeed, composing the composition of the Galois map with the order-map $P$ and decomposing it alternatively will give us the finite number of prime ideals $Q_i$ of $S$ lying over $P$. That is, we get all splitting behaviour of prime ideals in intermediate field-extensions.

Let $K$ be a Galois subfield of $L$ then we have a way to see how a prime ideals in $\mathbf{Spec}(\mathbb{Z})$ splits, ramifies or remains inert in $K$ and so by Chebotarev density this gives us the dimension of $K$ as well as the Galois group. And, if we could label the prime ideal by a prime number $p$, we could even reconstruct $K$ itself as $K$ is determined once we know all prime numbers which completely split.

The problem i have is that i do not see a categorical way to label the red arrows in $Frob(\mathbb{Z})$ by prime numbers. Mochizuki says we can do this in the proof of Thm 6.4(iii) by using the fact that the $log(p)$ are linearly independent over $\mathbb{Q}$.

This suggests that one might use the ‘Arakelov information’ (that is the archimidean valuations) to do this (the bit i left out so far), but i do not see this in the case of $\mathbb{Q}$ as there is just 1 extra (real) valuation determined by the values of the nonarchimidean valuations.

Probably i am missing something so all sorts of enlightenment re welcome!

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G+ recovery 3 : #IUTeich (1)

Published February 4, 2019 by lievenlb

Perhaps the ‘hottest’ topic I got involved with on Google+ was Mochizuki’s (claimed) proof of the ABC-conjecture.

Rereading these posts in chronological order shows my changing attitude to this topic, from early skepticism, over attempts to understand at least one pre-IUTeich paper (Frobenioids 1) to a level of belief, to … resignation.

Here’s the first batch of #IUTeich posts (here IUTeich stands for Mochizuki’s ‘Inter Universal Teichmuller theory).

June 12th, 2012

Linked post: ABC conjecture rumor at the Secret Blogging Seminar.

My understanding is that Google+ is for spreading (or debunking) such rumours. Sure would love to have more details!

May 26th, 2013

The Mochizuki craze continues…
Now Ted Nelson claims Mochizuki must be none other than the mysterious Satoshi Nakamoto, inventor of BitCoin.
I’ve seen BitCoin explained in 60 second on YouTube. It would be great if someone did something similar and try to explain inter-universal Teichmuller theory in 60 seconds…

May 27th, 2013

in IUTeich the theta function corresponds to the gaze of the little girl into the “small house”

Yesterday i was hoping for a 60 second introduction to inter-universal Teichmuller theory. Today i learned that Mochizuki himself provided such a thing.

Well, sort of…

In a one-page post he gives an ‘explanation’ of IUTeich via Sokkuri-Animation (Sokkuri being the Japanese for mirror-image). 

IUTeich, he says, should be viewed as a sequence of nested universes, which are represented by ‘houses’ in Sokkuri-Animation.
Galois groups and arithmetic fundamental groups behave as though they are made of a “mysterious substance” and in Sokkuri this “mysterious substance” corresponds to the “mysterious stars” that form the link between the “small” and “large” houses.
And, most importantly, the bridge between these nested universes corresponds to the gaze of the little girl into the “small house”…

Well, that explains everything, i think.

There’s a small image in the post but the link no longer works as the service stopped in january (via Google-translate).

A post at Hacker news identifies the image as coming from the YouTube linked below (2/4/2019 : link lost) and  discovers a connection to the paper titled ‘The ABC Conjecture and the Last Judgment by Giotto’ 

Yesterday i seriously doubted Ted Nelson’s claim that Mochizuki might be the anonymous Hatoshi Nakamoto, designer of BitCoin, having found little or no evidence of CypherPunk-parlance in Mochizuki’s papers…

Today, i’m not that certain any more.

May 28th, 2013

MochizukiDenial

Starting head-on with the 4 papers on inter-universal Teichmuller theory (IUTeich for the fans) is probably not the smartest move to enter Mochizuki-territory.

If you glanced through any of these papers or one of his numerous talks, you’ll know he makes a point of ‘extending’ or even ‘partially dismantling’  scheme theory using the new notion of Frobenioids, which should be some generalisation of Galois categories. 

So if you ever feel like wasting some months trying to figure out what his claimed proof of the ABC-conjecture is all about, a more advisable route might be to start with his two papers on ‘the geometry of Frobenioids’.

Lots of people must have tried that entrance before, and some even started a blog to record their progress as did the person (or persons) behind MochizukiDenial. Here’s the idea:

“Mochizuki deniers by contrast believe that the claim is not serious. They believe that the body of Mochizuki’s work contains neither a proof outline nor ideas powerful enough to resolve the ABC conjecture. We might be wrong. How do we propose to determine whether or not we are. In contrast to Mochizuki boosters on the internet, we will do this by determining what it is that Mochizuki’s papers purport to do. Stay tuned.”

Unfortunately, the project was given up after 2 days and three posts…

Today i tried to acquaint myself with the 126 pages of Frobenioids1 and have a splitting headache because i miss an extra 2Gb RAM to remember the 173 (or more) new concepts he introduces. 

May 29th, 2013

a baby Arithmetic Frobenioid

probably i should start-up a YouTube channel MinuteMochizuki but until i do here’s what i learned today: Alexei Bondal once told me that some Russians start a paper at the end and work their way to the front when needed. Sound advice when approaching M-papers!

So today i did start with the last section ‘Some Motivating Examples’ from Frobenioids1 and worked out what the simplest possible ArithmeticFrobenioid might be, that associated to the integers $\mathbb{Z}$.

Brace yourself here it comes. It is the category $\mathbf{C}$ with objects $(a)$ where $a$ is a strictly positive rational number and morphisms
$(a) \rightarrow (b)$ given by a couple $(n,c)$ where $n$ is a strictly positive integer and $c$ a strictly positive rational number subject to the condition that $a^n$ divides $c.b$ (meaning that the quotient is a strictly positive integer).

Cute (trivial) fact: compositions exist in the sense that
\[
(m,d) \circ (n,c)=(n.m,c^m.d). \]
What makes $\mathbf{C}$ a frobenioid is that it comes with a functor $\mathbf{C} \rightarrow \mathbf{F}$ where $\mathbf{F}$ is the category with one object $*$ and morphisms elements of the noncommutative monoid consisting of all couples $(n,c)$ as before and multiplication as composition above, functor sending all $(a)$ to $*$ and morphisms to corresponding element of the monoid.

Mochizuki proves that we can recover $\mathbf{C}$ from the functor $\mathbf{C} \rightarrow \mathbf{F}$ (look at endomorphism-submonoids of the nc-monoid above) and that $\mathbf{C}$ contains enough info to reconstruct the scheme $\mathbf{Spec}(\mathbb{Z})$ from it (again, use the functor).

Next goal: bring in some Galois categories (tbc)…

May 31st, 2013

MinuteMochizuki 2 : a quadratic arithmetic Frobenioid

Let $m$ be a squarefree number not 1 mod 4 and say you want to recover the classical arithmetic scheme cover $\mathbf{Spec}(\mathbb{Z}[\sqrt{m}])$ onto $\mathbf{Spec}(\mathbb{Z})$ (top left hand corner for m=3).
Prime ideals of $\mathbb{Z}$ may remain prime in $\mathbb{Z}[\sqrt{m}]$ (e.g. (5) and (7)) or split into two prime ideals (e.g. (11)) or ramify (only (2) and (3)), all this governed by the Kronecker symbol.
If it remains prime then the quotient $\mathbb{Z}[\sqrt{m}]/(p)$ is the finite field on $p^2$ elements so has a non-trivial Frobenius morphism, which cannot be lifted to an auto of $\mathbb{Z}[\sqrt{m}]$ to cannot be expressed in scheme language.

For this reason (i think) Mochizuki introduced Frobenioids which are categories allowing one to recover the classical scheme cover but also containing info on the power maps $x \rightarrow x^n$ for all $n$.

In the quadratic case, the objects of the Frobenioid for $\mathbb{Z}[\sqrt{m}]$ is the union of all fractional ideals (that is submodules of $\mathbb{Q}(\sqrt{m})$ of the form $I.q$ where $I$ is an ideal and $q$ non-zero in $\mathbb{Q}(\sqrt{m}))$ and all fractional ideals of $\mathbb{Z}$ (similar defined). Dedekind already knew they correspond to elements of the free Abelian groups on the set of prime ideals and hence have a natural poset-structure.

Now, there are monoid actions on these two posets, giving another set of arrows (for fractional ideals of $\mathbb{Z}[\sqrt{m}]$ the black arrows indicate the poset and the red arrow is the action. Now, the morphisms in the Frobenioid $\mathbf{C}(m)$ are all compositions of an action arrow followed by a poset-arrow (the green ones in the $\mathbb{Z}[\sqrt{m}]$ part, the red ones in the $\mathbb{Z}$-part. Then there is a third set of arrows encoding the Galois-covering info (the black arrows between the two parts). Again, one verifies that compositions exist.

A very special case of Mochizuki’s first Frobenioid paper is that $\mathbf{C}(m)$ contains enough info to recover the scheme cover and even the quadratic field $\mathbb{Q}(\sqrt{m})$. That is, if $\mathbf{C}(m)$ is equivalent to another such one $\mathbf{C}(m’)$ then $m=m’$. Also note that $\mathbf{C}(m)$ contains info on all power maps, so we have somehow ‘lifted’ the Frobenius-maps from the quotients to $\mathbb{Z}[\sqrt{m}]$ at the expense of ‘partially dismantling scheme theory’ (M’s words).

Another neat fact about $\mathbf{C}(m)$ is that all arrows are epimorphisms (compare this to groupoids where all morphisms are isomorphisms). Of course, this only becomes important for more complicated Galois-settings, ideally for the algebraic closure of $\mathbb{Q}$.

June 1st, 2013

Should I stay or should I blog now?

+Alex Nelson commented on +John Baez reshare of my last post on Mochizuki’s Frobenioids:
 
“I just wish he’d blog these, to make it a wee bit easier to read…and print out…”

Valid point, so i did spend some time to make blog-versions of my two G+ posts on this, the last one is linked below, the first one is here (2/4/2019 : lost post).

I’m not sure whether i should continue with this cross-posting. I kinda liked the quick-and-dirty approach of instant-uploading snaps-shots of doodles here. Writing a blogpost  takes more time.

Let me know if it does make a difference to you.

June 3rd, 2013

Mochizuki’s menagerie of morphisms

After zillions of definitions, Mochizuki almost shows empathy with the reader (on page 26 of Frobenioids1) and remarks  that it may be useful to draw a “Chart of Types of Morphisms in a Frobenioid”  (and does this on page 124).

Extremely useful indeed. Try to figure out what a morphism of Frobenius-type might be. It starts like this: it is an LB-invertible base-isomorphism. An LB-invertible map itself is a co-angular and isometric map. A co-angular map itself is defined by the property that for any factorization aoboc of it, where a is linear, b is an isometric pre-step and either a or c are base-isomorphisms, it follows that b is an isomorphism. A pre-step itself is …. etc.etc.

Today i tried to work out what all this means in the case of an arithmetic Frobenioid. Recall that this is a category with objects the elements of layers of posets, each layer (meant to represent the ‘dismantled’ prime-spectrum of a ring of integers $R$ in a number field $K$) the set of fractional ideals in the number field. In each layer we have operations of three types

– Frobenius-type : sending a fractional ideal $I$ to an n-th power $I^n$ (the blue morphisms)
– Poset-type : given by multiplying $I$ with an ideal $K$ of $R$ (the red morphisms, the ‘steps’ in M-parlance)
– Isomorphisms (the blacks, ‘isometric pre-steps’ according to M) which are either isos of fractional ideals (given by multiplying with a non-zero element $q$ of $K$) or Galois action maps on fractional ideals.

The relevant operation between different layers is that of extension of fractional ideals. So, let $\sigma : R \rightarrow S$ be a ring-map induced by a field-iso on $L$, the number-field of $S$, then this operation sends a fractional ideal $J$ in $K$ to the extended fractional ideal $S.\sigma(J)$ in $L$.

What Mochizuki shows is that any arrow in the arithmetic Frobenioid has an essentially unique factorisation into these four types of morphisms (essentially meaning unto irrelevant  isos trown in at one place and compensated by the inverse at the next place).

Let’s work this out for the baby arithmetic Frobenioid corresponding to the integers.

Objects correspond to strictly positive rational numbers $a$, and morphisms are of the form $\phi=(n,q,z) : a \rightarrow b$  where $n$ and $z$ are strictly positive integers and $q$ a strictly positive rational number satisfying $a^n.z=b.q$.

This map can then be factored as a Frobenius-type map $(n,1,1)$ sending $a$ to $a^n$, followed by a poset-map $(1,1,z)$ sending $a^n$ to $a^n.z$ followed by an isomorphism $(1,q^{-1},1)$ sending $a^n.z$ to $a^n.z.q^{-1}=b$.

These different types of morphisms will become important when we want to reconstruct the arithmetic schemes (and covers) from the category structure of the Frobenioid. The essential trick will be to classify and distinguish the irreducible morphisms which are non-isomorphisms phi such that in any factorisation $\phi = \alpha \circ \beta$, either $\alpha$ or $\beta$ must be an isomorphism. (tbc)

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