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Category: noncommutative

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 H1 properly, and how this led them to ‘absolute’ algebraic geometry, meaning over the sphere spectrum 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…

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the mystery Manin-Marcolli monoid

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

Let’s take for instance the power maps βn(t)=tn.

For every c the degree n polynomial tnc=0 has exactly n distinct solutions, except for c=0, when there is just one. And, clearly we have that 0n=0, 1n=1 and n=. So, βn is a Belyi-extender of degree n.

A cute observation being that if β is a Belyi-extender of degree d, and β is an extender of degree d, then ββ is again a Belyi-extender, this time of degree d.d.

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

In our example, βnβm=βn.m. So, the power-maps are a sub-monoid of the Belyi-extenders, isomorphic to the multiplicative monoid N× 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 β(t)=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 if β()=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 if β()=1).

Now comes the fun part.

Because β has exactly d pre-images for all real numbers λ in the open interval (0,1) (and β 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 βn(t)=tn, we have just one black dot at 0 (being the only solution of tn=0), and n white dots at the n-th roots of unity (the solutions of xn1=0). Any λ(0,1) has as its n pre-images the numbers ζi.λn with ζi 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 , so projecting from gives a homeomorphism between the 2-sphere and C{}.

To get all information of the dessin (including possible dots at infinity) it is best to slice the sphere open along the real segments (,0) and (1,) 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 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 PC1{0,1,}=S2{0,1,} permuting the set {0,1,}.

They form the symmetric group S3 on 3-letters and correspond to the Belyi-extenders
t, 1t, 1t, 11t, t1t, tt1
You can compose these units with an extender to get anther extender of the same degree where the roles of 0,1 and are changed.

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

Manin and Marcolli use this and claim that you can transform any extender η to an extender γ by composing with a unit, such that γ(0)=0,γ(1)=1 and γ()=.

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

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
β(t)=t2(t1)(t43)3
with β(0)=0=β(1) and β()=1.

So, a first property of the mystery Manin-Marcolli monoid EMMM must surely be that all its elements γ(t) map {0,1,} onto {0,1,}, for they use this property a number of times, for instance to construct a monoid map
EMMMM2(Z)+γ[dm101]
where d is the degree of γ 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 ζ(t)=27t2(t1)24(t2t+1)3 with dessin



But then, you may argue, this extender sends all of 0,1 and to 0, so it cannot belong to EMMM.

Here’s a trick to construct Belyi-extenders from Belyi-maps β:P1P1, defined over Q and having the property that there are rational points in the fibers over 0,1 and .

Let’s take an example, the ‘monstrous dessin’ corresponding to the congruence subgroup Γ0(2)



with map β(t)=(t+256)31728t2.

As it stands, β is not a Belyi-extender because it does not map 1 into {0,1,}. But we have that
256β1(0), β1(), and 512,64β1(1)
(the last one follows from (t+256)21728t3=(t512)2(t+64)).

We can now pre-compose β with the automorphism (defined over Q) sending 0 to 256, 1 to 64 and fixing to get a Belyi-extender
γ(t)=(192t)31728(192t256)2
which maps γ(0)=0, γ(1)=1 and γ()= (so belongs to EMMM) with the same dessin, which is not a tree,

That is, EMMM can at best consist only of those Belyi-extenders γ(t) that map {0,1,} onto {0,1,} 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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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).

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