Dessins d'enfants, Beijing, 04-2013

Grothendieck’s dessins d’enfants Children’s drawings of Grothendieck

谭蕾

, Université d’Angers

2013

, Beijing

Two key results

Definition. A complex polynomialP ∈C[X]is calledBelyi if it has only two critical values, at 0 and 1 (can be chosen to be at other points).

Two such polynomialsf,g arecovering equivalent if there is an affine mapM(z) =az+b,a6=0, such that

C →^M C g & .f

C

. Theorem(Grothendieck). In each equivalence class there is a representative polynomialP withalgebraic numbers as coefficients.

Theorem(Belyi) (1) Given any finite setA of algebraic numbers there is a polynomialP with rational coefficients such that P(A)∪ {critical values of P} ⊂ {0,1}.

(2) Given any rational mapf with algebraic numbers as critical values (for examplef ∈Q[X]), there is a polynomialP with rationalcoefficients such thatcr. val(P◦f)⊂ {0,1,∞}.

Basic facts

Ifcv(f)⊂A, thencv(P◦f)⊂P(A)∪cv(P).

Proof. Pullback byP then byf a small disc outside P(A)∪cv(P).

Q={rational numbers},

Q={algebraic numbers}={a∈C|P(a) =0 for someP ∈Q[X]}.

Qis countable, dense inC, a field, algebraically closed. Moreover, the roots of a polynomial system inQ[X₁,· · ·,X_n]form either an unbounded set, or a finite set with algebraic coordinates.

Proof of Grothendieck’s thm : ∃P ◦ M monic and fixing 0

AnyP monic,P(0) =0 and cv(P)⊂ {0,1} must have algebraic coefficients.

Proof. LetP be monic. Let {a_i,i ∈I}=P⁻¹(0), with multiplicity s_i for a_i, and {b_j,j ∈J}=P⁻¹(1) with multiplicity t_j for b_j, d =P

s_i =P

t_j. Then Y

I

(z−a_i)^si =P(z) =Y

J

(z−b_j)^tj+1. Equalizing the coefficientson both sides above we get that a={a_i}andb={b_j} must satisfy a system of polynomial equationsF(a,b) =0, withF=Fs_i,t_j ∈(Q[a,b])^d. It is known

{zeros ofF}=

eitheran unbounded algebraic variety or a finite set with algebraic coordinates . Choose now thesi andtj’s so that the critical values ofP are in {0,1}, in other words, P

(s_i−1) +P

(t_j −1) =d −1 (s_i andt_j’s are called thepassport of P ).

Geometric proof of the boundedness of F(a, b) = 0

Let(a,b) ={a_i,bj} be a solution ofF(a,b) =0.

LetP(z) =Y

I

(z−a_i)^si =Y

J

(z −b_j)^tj +1.Thus the critical values ofP are in {0,1} ⊂[0,1].

C r D C rP⁻¹([0,1])

z^d ↓ ↓P

C r D ^Riemann−→

1 4^(z+

1 z)+1

2

C r[0,1]

Now usingP(0) =0 and theKoebe 1/4-theorem, we have {a_i,b_j |i ∈I,j ∈J}=P⁻¹({0,1})⊂P⁻¹([0,1])⊂D(0, 4

4^1d).

(Bezout=⇒the number of solutions = the number of normalized Belyi polynomials with a given passport =d^d−2 counted with multiplicity.)

Geometric proof of the boundedness of F(a, b) = 0

Let(a,b) ={a_i,bj} be a solution ofF(a,b) =0.

LetP(z) =Y

I

(z−a_i)^si =Y

J

(z −b_j)^tj +1.Thus the critical values ofP are in {0,1} ⊂[0,1].

C r D

∃ 1

4^1/d−→^z+O(1)

biholomorphic C rP⁻¹([0,1])

z^d ↓ ↓P

C r D ^Riemann−→

1 4^(z+

1 z)+1

2

C r[0,1]

Now usingP(0) =0 and theKoebe 1/4-theorem, we have {a_i,b_j |i ∈I,j ∈J}=P⁻¹({0,1})⊂P⁻¹([0,1])⊂D(0, 4

4^1d ).

(Bezout=⇒the number of solutions = the number of normalized Belyi polynomials with a given passport =d^d−2 counted with multiplicity.)

Action of the absolute Galois group Γ = Aut Q ( Q )

Known fact :Fix(Γ) =Q. LetP(X) =P

a_iXⁱ with a_i ∈Q. Letσ ∈Aut Q(Q).

a_i P ^derivative−→ P^{0 take zeros}−→ cpts −→^P cvs

σ↓ σ↓ σ ↓ ↓σ ↓σ

σ(a_i) Q ^derivative−→ Q^{0 take zeros}−→ cpts −→^Q cvs

alg. conjugacy

Q3z −→^P P(z)

σ ↓ ↓σ

σ(z) −→^Q Q(z)

σ(P ◦L) =σ(P)◦σ(L).

Proof of Belyi’s theorem

Given any finite setAof algebraic numbers there is a polynomial P ∈Q[X]such that P(A)∪ {critical values of P} ⊂ {0,1}.

Step 1. Reducing the number of points outsideQ: B₀ =Ar Q

P0−→∈Q^[X]

say d^o=d 0

↓

B₁={0} ∪cv(P₀)r Q

P1∈−→Q^[X]

d^o≤d−1 0

↓

{0} ∪cv(P₁)r Q · · · withP₁(z) = Y

δ∈B1

(z−δ). Anyσ∈Γmust fixeP₀ thus permutes B₁ and fixes the coefficients ofP₁. So these coefficients are all inQ. So∃P ∈Q[X]such that Y :=P(A)∪ {critical values ofP} ⊂Q.

Step 2. Reducing the number of points inside Q

Given any finite setY ⊂Q, we will show that there isL∈Q[X] such thatL(Y)∪ {critical values ofL} ⊂ {0,1}.

For any increasing triplea,b,c ∈Q, a<b<c ^αz+β−→

α,β∈Q

0< p

p+q <1 ^z

p(1−z)^q

magic map−→ 0<s, s ∈Q.

Dessin d’enfants

For any meromorphic mapF :S →CwithS a compact Riemann surface such thatcv(F)⊂ {0,1,∞}. The set D_F :=F⁻¹([0,1])is called the dessin ofF.

In the case of a Belyi polynomialP,D_P is an embedded tree in C whose vertices are bi-colored.

The set of covering equivalence classesBelyi(C)/∼=Belyi(Q)/∼ is in bijection with the set of embedded bi-colored trees (up to homeo⁺ preserving the coloring).

Any bounded arc linking 0 and 1 will lift to a homeomorphic dessin.

Action of the absolute Galois group Γ = Aut Q ( Q )

LetP(X) =P

a_iXⁱ with a_i ∈Q. Letσ ∈Aut Q(Q).

a_i P ^derivative−→ P^{0 take zeros}−→ cpts −→^P cvs

σ↓ σ↓ ↓σ ↓σ ↓σ

σ(a_i) Q ^derivative−→ Q^{0 take zeros}−→ cpts −→^Q cvs

alg. conjugacy

Q3z −→^P P(z)

σ ↓ ↓σ

σ(z) −→^Q Q(z)

σ(P ◦L) =σ(P)◦σ(L).

Theorem. The groupΓ acts faithfully : 1. onBelyi(Q)and on Belyi(Q)/∼ (Lenstra-Schneps) ; 2. on the set of dynamical Belyi polynomials andcommutes with iterations (Pilgrim). The pull-back dessins Hausdorff-converge to the Julia set, which contains a dense set of algebraic numbers (Pilgrim).

Open problem. See the action on trees.

Proof of faithfulness

Letid 6=γ ∈Γ. Then∃α∈Q r Q such that γ(α)6=α. Setf such thatf⁰(z) =z(z−1)²(z−α)³. By Belyi∃g ∈Q[X]such that g◦f ∈Bel(Q). Thenγ.(g ◦f) can not be equivalent tog ◦f. Assume the contrary,γ.(g◦f) =g ◦f ◦M, withM affine. So g◦γ.f =γ.(g◦f) =g◦f ◦M. By thedecomposition theorem, there is affineB such that γ.f =B◦f ◦M. This implies thatM transports the critical points off to that ofγ.f. But due to the pairwise distinct powers of the critical points,M must fixe 0 and 1 and so much be identity. ButM(α) =γ.α6=α. A contradiction.

Decomposition theorem. If g₁◦f₁ =g₂◦f₂ with the same degree decomposition, thenf₂=affine(f₁).

Proof. Look at the preimage of a line segment linking the critical values.

Dessin as trajectories of polynomial vector fields

A Belyi passport(si,tj) is calledclean iftj ≡2. Any Belyi polynomialQ can be turned into a cleanB by post composing z 7→4z(1−z). In terms of the dessin, one just needs to add a vertex in each edge to be the new colored 1 vertices and color the old vertices by 0. So d old vertices =⇒d −1 new vertices . Theorem (Pilgrim). Any clean Belyi polynomial is isomorphic to a B(z)of the form −Q

I,|I|=d(z−a_i)^si, with a_i pairwise distinct, P

Ia_i =0 andP

Is_i =deg(B) =Pd−1

j=1 t_j =2(d−1). AnyB induces a monic centered polynomialP_B(z) =Q

I(z−a_i) and

˙

z =P_B(z) ←−^B∗

pullsbackz˙² =4(z²−z³) with separatricesB⁻¹([−∞,0]). The edges of the dessin B⁻¹([0,1]) are trajectories. Eacha_i receives s_i separatrices.

The vector fieldz˙ =P_B(z) is generic, monic centered with constant analytic invariant(= πi

d−1).

Bijection : from vector fields to Belyi polynomials

LetP(z) =Q

I,|I|=d(z−ai) be a monic centered polynomial such that the vector fieldP(z) ∂

∂z is generic : all roots are simple and no homoclinic connections of the 2(d −1)separatrix germs at ∞(in particular eacha_i is either a source or a sink)

Eacha_i receives s_i >0 separatrices (asa_i is not a center), and P

Is_i =2(d −1) (every separatrix germ lands at somea_i).

So(s₁,· · · ,s_d)is an integer partition of 2(d−1).

Within the same landing pattern there is a unique choice of a= (a_i)_I such that the analytic invariant ofP is

{ πi

d −1,· · ·, πi

d −1}. For thisP, setB_P(z) =−Q

I(z−a_i)^si. Then (by Pilgrim)B_P is a clean Belyi polynomial. This is obtained by uniformizing and then squaring the zones ofz˙ =P(z) to get a global holomorphic mapB which turns out to beB_P.

Counting resultThe number of normalized clean Belyi polynomial of given degree is the catalan number.

(Tomasini) There is a closed form for the number of equivalence classes of clean Belyi polynomials as well as the number of combinatorial invariants of generic polynomial vector fields.

QuestionGiven(s₁,s₂,· · · ,s_d) a partition of 2(d −1), how many landing patterns are there ? A landing pattern is a non-crossing partition of{0,· · · ,2d −3} into d sets of cardinality s1,· · ·,s_d so that each set contains only numbers of the same parity (either all are odd or all are even).