## The $\mathbb{F}_1$ World Seminar

For some time I knew it was in the making, now they are ready to launch it:

The $\mathbb{F}_1$ World Seminar, an online seminar dedicated to the “field with one element”, and its many connections to areas in mathematics such as arithmetic, geometry, representation theory and combinatorics. The organisers are Jaiung Jun, Oliver Lorscheid, Yuri Manin, Matt Szczesny, Koen Thas and Matt Young.

From the announcement:

“While the origins of the “$\mathbb{F}_1$-story” go back to attempts to transfer Weil’s proof of the Riemann Hypothesis from the function field case to that of number fields on one hand, and Tits’s Dream of realizing Weyl groups as the $\mathbb{F}_1$ points of algebraic groups on the other, the “$\mathbb{F}_1$” moniker has come to encompass a wide variety of phenomena and analogies spanning algebraic geometry, algebraic topology, arithmetic, combinatorics, representation theory, non-commutative geometry etc. It is therefore impossible to compile an exhaustive list of topics that might be discussed. The following is but a small sample of topics that may be covered:

Algebraic geometry in non-additive contexts – monoid schemes, lambda-schemes, blue schemes, semiring and hyperfield schemes, etc.
Arithmetic – connections with motives, non-archimedean and analytic geometry
Tropical geometry and geometric matroid theory
Algebraic topology – K-theory of monoid and other “non-additive” schemes/categories, higher Segal spaces
Representation theory – Hall algebras, degenerations of quantum groups, quivers
Combinatorics – finite field and incidence geometry, and various generalizations”

The seminar takes place on alternating Wednesdays from 15:00 PM – 16:00 PM European Standard Time (=GMT+1). There will be room for mathematical discussion after each lecture.

The first meeting takes place Wednesday, January 19th 2022. If you want to receive abstracts of the talks and their Zoom-links, you should sign up for the mailing list.

Perhaps I’ll start posting about $\mathbb{F}_1$ again, either here, or on the dormant $\mathbb{F}_1$ mathematics blog. (see this post for its history).

## Do we need the sphere spectrum?

Last time I mentioned the talk “From noncommutative geometry to the tropical geometry of the scaling site” by Alain Connes, culminating in the canonical isomorphism (last slide of the talk)

Or rather, what is actually proved in his paper with Caterina Consani BC-system, absolute cyclotomy and the quantized calculus (and which they conjectured previously to be the case in Segal’s Gamma rings and universal arithmetic), is a canonical isomorphism between the $\lambda$-rings
$\mathbb{Z}[\mathbb{Q}/\mathbb{Z}] \simeq \mathbb{W}_0(\overline{\mathbb{S}})$
The left hand side is the integral groupring of the additive quotient-group $\mathbb{Q}/\mathbb{Z}$, or if you prefer, $\mathbb{Z}[\mathbf{\mu}_{\infty}]$ the integral groupring of the multiplicative group of all roots of unity $\mathbf{\mu}_{\infty}$.

The power maps on $\mathbf{\mu}_{\infty}$ equip $\mathbb{Z}[\mathbf{\mu}_{\infty}]$ with a $\lambda$-ring structure, that is, a family of commuting endomorphisms $\sigma_n$ with $\sigma_n(\zeta) = \zeta^n$ for all $\zeta \in \mathbf{\mu}_{\infty}$, and a family of linear maps $\rho_n$ induced by requiring for all $\zeta \in \mathbf{\mu}_{\infty}$ that
$\rho_n(\zeta) = \sum_{\mu^n=\zeta} \mu$
The maps $\sigma_n$ and $\rho_n$ are used to construct an integral version of the Bost-Connes algebra describing the Bost-Connes sytem, a quantum statistical dynamical system.

On the right hand side, $\mathbb{S}$ is the sphere spectrum (an object from stable homotopy theory) and $\overline{\mathbb{S}}$ its ‘algebraic closure’, that is, adding all abstract roots of unity.

The ring $\mathbb{W}_0(\overline{\mathbb{S}})$ is a generalisation to the world of spectra of the Almkvist-ring $\mathbb{W}_0(R)$ defined for any commutative ring $R$, constructed from pairs $(E,f)$ where $E$ is a projective $R$-module of finite rank and $f$ an $R$-endomorphism on it. Addition and multiplication are coming from direct sums and tensor products of such pairs, with zero element the pair $(0,0)$ and unit element the pair $(R,1_R)$. The ring $\mathbb{W}_0(R)$ is then the quotient-ring obtained by dividing out the ideal consisting of all zero-pairs $(E,0)$.

The ring $\mathbb{W}_0(R)$ becomes a $\lambda$-ring via the Frobenius endomorphisms $F_n$ sending a pair $(E,f)$ to the pair $(E,f^n)$, and we also have a collection of linear maps on $\mathbb{W}_0(R)$, the ‘Verschiebung’-maps which send a pair $(E,f)$ to the pair $(E^{\oplus n},F)$ with
$F = \begin{bmatrix} 0 & 0 & 0 & \cdots & f \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$
Connes and Consani define a notion of modules and their endomorphisms for $\mathbb{S}$ and $\overline{\mathbb{S}}$, allowing them to define in a similar way the rings $\mathbb{W}_0(\mathbb{S})$ and $\mathbb{W}_0(\overline{\mathbb{S}})$, with corresponding maps $F_n$ and $V_n$. They then establish an isomorphism with $\mathbb{Z}[\mathbb{Q}/\mathbb{Z}]$ such that the maps $(F_n,V_n)$ correspond to $(\sigma_n,\rho_n)$.

But, do we really have the go to spectra to achieve this?

All this reminds me of an old idea of Yuri Manin mentioned in the introduction of his paper Cyclotomy and analytic geometry over $\mathbb{F}_1$, and later elaborated in section two of his paper with Matilde Marcolli Homotopy types and geometries below $\mathbf{Spec}(\mathbb{Z})$.

Take a manifold $M$ with a diffeomorphism $f$ and consider the corresponding discrete dynamical system by iterating the diffeomorphism. In such situations it is important to investigate the periodic orbits, or the fix-points $Fix(M,f^n)$ for all $n$. If we are in a situation that the number of fixed points is finite we can package these numbers in the Artin-Mazur zeta function
$\zeta_{AM}(M,f) = exp(\sum_{n=1}^{\infty} \frac{\# Fix(M,f^n)}{n}t^n)$
and investigate the properties of this function.

To connect this type of problem to Almkvist-like rings, Manin considers the Morse-Smale dynamical systems, a structural stable diffeomorphism $f$, having a finite number of non-wandering points on a compact manifold $M$.

From Topological classification of Morse-Smale diffeomorphisms on 3-manifolds

In such a situation $f_{\ast}$ acts on homology $H_k(M,\mathbb{Z})$, which are free $\mathbb{Z}$-modules of finite rank, as a matrix $M_f$ having only roots of unity as its eigenvalues.

Manin argues that this action is similar to the action of the Frobenius on etale cohomology groups, in which case the eigenvalues are Weil numbers. That is, one might view roots of unity as Weil numbers in characteristic one.

Clearly, all relevant data $(H_k(M,\mathbb{Z}),f_{\ast})$ belongs to the $\lambda$-subring of $\mathbb{W}_0(\mathbb{Z})$ generated by all pairs $(E,f)$ such that $M_f$ is diagonalisable and all its eigenvalues are either $0$ or roots of unity.

If we denote for any ring $R$ by $\mathbb{W}_1(R)$ this $\lambda$-subring of $\mathbb{W}_0(R)$, probably one would obtain canonical isomorphisms

– between $\mathbb{W}_1(\mathbb{Z})$ and the invariant part of the integral groupring $\mathbb{Z}[\mathbb{Q}/\mathbb{Z}]$ for the action of the group $Aut(\mathbb{Q}/\mathbb{Z}) = \widehat{\mathbb{Z}}^*$, and

– between $\mathbb{Z}[\mathbb{Q}/\mathbb{Z}]$ and $\mathbb{W}_1(\mathbb{Z}(\mathbf{\mu}_{\infty}))$ where $\mathbb{Z}(\mathbf{\mu}_{\infty})$ is the ring obtained by adjoining to $\mathbb{Z}$ all roots of unity.

## Alain Connes on his RH-project

In recent months, my primary focus was on teaching and family matters, so I make advantage of this Christmas break to catch up with some of the things I’ve missed.

Peter Woit’s blog alerted me to the existence of the (virtual) Lake Como-conference, end of september: Unifying themes in Geometry.

In Corona times, virtual conferences seem to sprout up out of nowhere, everywhere (zero costs), giving us an inflation of YouTubeD talks. I’m always grateful to the organisers of such events to provide the slides of the talks separately, as the generic YouTubeD-talk consists merely in reading off the slides.

Allow me to point you to one of the rare exceptions to this rule.

When I downloaded the slides of Alain Connes’ talk at the conference From noncommutative geometry to the tropical geometry of the scaling site I just saw a collage of graphics from his endless stream of papers with Katia Consani, and slides I’d seen before watching several of his YouTubeD-talks in recent years.

Boy, am I glad I gave Alain 5 minutes to convince me this talk was different.

For the better part of his talk, Alain didn’t just read off the slides, but rather tried to explain the thought processes that led him and Katia to move on from the results on this slide to those on the next one.

If you’re pressed for time, perhaps you might join in at 49.34 into the talk, when he acknowledges the previous (tropical) approach ran out of steam as they were unable to define any $H^1$ properly, and how this led them to ‘absolute’ algebraic geometry, meaning over the sphere spectrum $\mathbb{S}$.

Sadly, for some reason Alain didn’t manage to get his final two slides on screen. So, in this case, the slides actually add value to the talk…

## the mystery Manin-Marcolli monoid

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

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

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

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

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

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

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

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

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

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

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

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

Now comes the fun part.

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

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

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

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

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

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

What are its invertible elements?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Monstrous dessins 1

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 $\Gamma_0(n)$.