Skip to content →

Monstrous dessins 1

Dedekind’s Psi-function Ψ(n)=np|n(1+1p) pops up in a number of topics:

  • Ψ(n) is the index of the congruence subgroup Γ0(n) in the modular group Γ=PSL2(Z),
  • Ψ(n) is the number of points in the projective line P1(Z/nZ),
  • Ψ(n) is the number of classes of 2-dimensional lattices LMgh at hyperdistance n in Conway’s big picture from the standard lattice L1,
  • Ψ(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 Γ0(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 P1(Z/nZ) 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 LMgh at hyperdistance n from L1 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 Γ0(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 Γ0(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 P1(Z/nZ).

Over a field k the points of P1(k) correspond to the lines through the origin in the affine plane A2(k) and they can represented by projective coordinates [a:b] which are equivalence classes of couples (a,b)k2{(0,0)} under scalar multiplication with non-zero elements in k, so with points [a:1] for all ak together with the point at infinity [1:0]. When n=p is a prime number we have #P1(Z/pZ)=p+1. Here are the 8 lines through the origin in A2(Z/7Z)



Over an arbitrary (commutative) ring R the points of P1(R) again represent equivalence classes, this time of pairs
(a,b)R2 : aR+bR=R
with respect to scalar multiplication by units in R, that is
(a,b)(c,d)  iff λR : a=λc,b=λd
For P1(Z/nZ) we have to find all pairs of integers (a,b)Z2 with 0a,b<n with gcd(a,b)=1 and use Cremona’s trick to test for equivalence:
(a,b)=(c,d)P1(Z/nZ) iff adbc0 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 §1.3 in the Tatitscheff, He and McKay paper suffices:

  • Consider the action of (Z/nZ) on {0,1,,n1}=Z/nZ and let D be the set of the smallest elements in each orbit,
  • For each dD compute the stabilizer subgroup Gd for this action and let Cd be the set of smallest elements in each Gd-orbit on the set of all elements in Z/nZ coprime with d,
  • Then P1(Z/nZ)={[c:d] | dD,cCd}.

Let’s work this out for n=12 which will be our running example (the smallest non-squarefree non-primepower):

  • (Z/12Z)={1,5,7,11}C2×C2,
  • 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},
  • G0=C2×C2, G1={1}, G2={1,7}, G3={1,5}, G4={1,7} and G6=C2×C2,
  • 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 C2×C2 giving us just one additional point [1:6].

This gives us all 24=Ψ(12) points of P1(Z/12Z) (strangely, op page 43 of the T-H-M paper they use different representants).

One way to see that #P1(Z/nZ)=Ψ(n) comes from a consequence of the Chinese Remainder Theorem that for the prime factorization n=p1e1pkek we have
P1(Z/nZ)=P1(Z/p1e1Z)××P1(Z/pkekZ)
and for a prime power pk we have canonical representants for P1(Z/pkZ)
[a:1] for a=0,1,,pk1 and[1:b] for b=0,p,2p,3p,,pkp
which shows that #P1(Z/pkZ)=(p+1)pk1=Ψ(pk).

Next time, we’ll connect P1(Z/nZ) to Conway’s big picture and the congruence subgroup Γ0(n).

Published in geometry groups math noncommutative number theory