Let’s try to identify the Ψ(n)=n∏p|n(1+1p)Ψ(n)=n∏p|n(1+1p) points of P1(Z/nZ) with the lattices LMgh at hyperdistance n from the standard lattice L1 in Conway’s big picture.
Here are all 24=Ψ(12) lattices at hyperdistance 12 from L1 (the boundary lattices):
You can also see the 4=Ψ(3) lattices at hyperdistance 3 (those connected to 1 with a red arrow) as well as the intermediate 12=Ψ(6) lattices at hyperdistance 6.
The vertices of Conway’s Big Picture are the projective classes of integral sublattices of the standard lattice Z2=Ze1⊕Ze2.
Let’s say our sublattice is generated by the integral vectors v=(v1,v2) and w=(w1.w2). How do we determine its class LM,gh where M∈Q+ is a strictly positive rational number and 0≤gh<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 SL2(Z) such that the new 2nd base vector is of the form (0,z). To do this take coprime (c,d)∈Z2 such that cv1+dw1=0 and complete with (a,b) satisfying ad−bc=1 via Bezout to a matrix in SL2(Z) such that
[abcd][v1v2w1w2]=[xy0z]
then the sublattice is of class Lxz,yz mod 1.
In the example, we have
[01−12][2114]=[1407]
so this sublattice is of class L17,47.
Starting from a class LM,gh it is easy to work out its hyperdistance from L1: let d be the smallest natural number making the corresponding matrix integral
d.[Mgh01]=[uv0w]∈M2(Z)
then LM,gh is at hyperdistance u.w from L1.
Now that we know how to find the lattice class of any sublattice of Z2, let us assign a class to any point [c:d] of P1(Z/nZ).
As gcd(c,d)=1, by Bezout we can find a integral matrix with determinant 1
S[c:d]=[abcd]
But then the matrix
[a.nb.ncd]
has determinant n.
Working backwards we see that the class L[c:d] of the sublattice of Z2 spanned by the vectors (a.n,b.n) and (c,d) is of hyperdistance n from L1.
This is how the correspondence between points of P1(Z/nZ) and classes in Conway’s big picture at hyperdistance n from L1 works.
Let’s do an example. Take the point [7:3]∈P1(Z/12Z) (see last time), then
[−2−173]∈SL2(Z)
so we have to determine the class of the sublattice spanned by (−24,−12) and (7,3). As before we have to compute
[−2−7724][−24−1273]=[−130−12]
giving us that the class L[7:3]=L11234 (remember that the second term must be taken mod 1).
If you do this for all points in P1(Z/12Z) (and P1(Z/6Z) and P1(Z/3Z)) you get this version of the picture we started with
You’ll spot that the preimages of a canonical coordinate of P1(Z/mZ) for m|n are the very same coordinate together with ‘new’ canonical coordinates in P1(Z/nZ).
To see that this correspondence is one-to-one and that the index of the congruence subgroup
Γ0(n)={[pqrs] | n|r and ps−qr=1}
in the full modular group Γ=PSL2(Z) is equal to Ψ(n) it is useful to consider the action of PGL2(Q)+ on the right on the classes of lattices.
The stabilizer of L1 is the full modular group Γ and the stabilizer of any class is a suitable conjugate of Γ. For example, for the class Ln (that is, of the sublattice spanned by (n,0) and (0,1), which is of hyperdistance n from L1) this stabilizer is
Stab(Ln)={[abnc.nd] | ad−bc=1}
and a very useful observation is that
Stab(L1)∩Stab(Ln)=Γ0(n)
This is the way Conway likes us to think about the congruence subgroup Γ0(n): it is the joint stabilizer of the classes L1 and Ln (as well as all classes in the ‘thread’ Lm with m|n).
On the other hand, Γ acts by rotations on the big picture: it only fixes L1 and maps a class to another one of the same hyperdistance from L1.The index of Γ0(n) in Γ is then the number of classes at hyperdistance n.
To see that this number is Ψ(n), first check that the classes at hyperdistance pk for p a prime number and for all k for the p+1 free valent tree with root L1, so there are exactly pk−1(p+1) classes as hyperdistance pk.
To get from this that the number of hyperdistance n classes is indeed Ψ(n)=∏p|npvp(n)−1(p+1) we have to use the prime- factorisation of the hyperdistance (see this post).
The fundamental domain for the action of Γ0(12) by Moebius tranfos on the upper half plane must then consist of 48=2Ψ(12) black or white hyperbolic triangles
Next time we’ll see how to deduce the ‘monstrous’ Grothendieck dessin d’enfant for Γ0(12) from it