2-solvable Bely̆ı maps

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Michael Musty, Dartmouth College (grad)math.dartmouth.edu/˜mjmusty

Maine Québec Number Theory ConferenceOctober 14, 2017

1 / 15

math.dartmouth.edu/~mjmusty

Outline

1. What is a 2-solvable Bely̆ı map?2. Motivation3. Algorithm to compute explicitly

3.1 Find permutation triples3.2 Compute equations

4. Explicit examples

2 / 15

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A smooth projective curve X over C can be defined over Q if andonly if there exists a branched covering of compact connectedRiemann surfaces ϕ : X → P1 unramified (unbranched) aboveP1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map.

In the 1980s, Grothendieck described a bijection between Bely̆ımaps and dessins d’enfants. Gal(Q/Q) acts on these sets.

3 / 15

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A smooth projective curve X over C can be defined over Q if andonly if there exists a branched covering of compact connectedRiemann surfaces ϕ : X → P1 unramified (unbranched) aboveP1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map.

In the 1980s, Grothendieck described a bijection between Bely̆ımaps and dessins d’enfants. Gal(Q/Q) acts on these sets.

3 / 15

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A smooth projective curve X over C can be defined over Q if andonly if there exists a branched covering of compact connectedRiemann surfaces ϕ : X → P1 unramified (unbranched) aboveP1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map.

In the 1980s, Grothendieck described a bijection between Bely̆ımaps and dessins d’enfants. Gal(Q/Q) acts on these sets.

3 / 15

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A smooth projective curve X over C can be defined over Q if andonly if there exists a branched covering of compact connectedRiemann surfaces ϕ : X → P1 unramified (unbranched) aboveP1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map.

In the 1980s, Grothendieck described a bijection between Bely̆ımaps and dessins d’enfants.

Gal(Q/Q) acts on these sets.

3 / 15

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A smooth projective curve X over C can be defined over Q if andonly if there exists a branched covering of compact connectedRiemann surfaces ϕ : X → P1 unramified (unbranched) aboveP1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map.

In the 1980s, Grothendieck described a bijection between Bely̆ımaps and dessins d’enfants. Gal(Q/Q) acts on these sets.

3 / 15

A Zoo of Bijections

{dessins withd edges

}

Transitive permutationTriples (σ0, σ1, σ∞) ∈S3d with σ∞σ1σ0 = 1

{Index d triangle sub-

groups Γ ≤ ∆(a, b, c)

}

{Bely̆ı maps of degree d }

Degree d function fieldextensions K ⊇ C(z)with discriminant sup-ported above 0, 1,∞

∼

∼

∼

∼

All up to the appropriate version of equivalence in each category.

4 / 15

A Zoo of Bijections

{dessins withd edges

}

Transitive permutationTriples (σ0, σ1, σ∞) ∈S3d with σ∞σ1σ0 = 1

{Index d triangle sub-

groups Γ ≤ ∆(a, b, c)

}

{Bely̆ı maps of degree d }

Degree d function fieldextensions K ⊇ C(z)with discriminant sup-ported above 0, 1,∞

∼

∼

∼

∼

All up to the appropriate version of equivalence in each category.

4 / 15

A Zoo of Bijections

{dessins withd edges

}

Transitive permutationTriples (σ0, σ1, σ∞) ∈S3d with σ∞σ1σ0 = 1

{Index d triangle sub-

groups Γ ≤ ∆(a, b, c)

}

{Bely̆ı maps of degree d }

Degree d function fieldextensions K ⊇ C(z)with discriminant sup-ported above 0, 1,∞

∼

∼

∼

∼

All up to the appropriate version of equivalence in each category.4 / 15

Example K

1 0

x1 = 1

x1 = −1

x1 = 0

∞

1

2

X1

X0 = P1

x0 = ϕ(x1) = x21= (x1 + 1)(x1 − 1) + 1

5 / 15

Example K

21

((1 2), (1)(2), (1 2)

)[∆(2,∞, 2) : Γ] = 2

x0 = ϕ(x1) = x21C(x21)⊆ C(x1)

∼

∼

∼

∼

6 / 15

2-solvable (Galois) Bely̆ı maps

X0 = P1

X1

X2

...

Xr

2

2

2

2

K (X0) = K (P1)

K (X1)

K (X2)

...

K (Xr )

2

2

2

2

7 / 15

2-solvable (Galois) Bely̆ı maps

X0 = P1

X1

X2

...

Xr

2

2

2

2

K (X0) = K (P1)

K (X1)

K (X2)

...

K (Xr )

2

2

2

2

7 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.Suppose p does not divide #G. Then there exists a number fieldM such that p is unramified in M and ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot: Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.

Suppose p does not divide #G. Then there exists a number fieldM such that p is unramified in M and ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot: Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.Suppose p does not divide #G.

Then there exists a number fieldM such that p is unramified in M and ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot: Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.Suppose p does not divide #G. Then there exists a number fieldM such that p is unramified in M and

ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot: Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.Suppose p does not divide #G. Then there exists a number fieldM such that p is unramified in M and ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot: Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.Suppose p does not divide #G. Then there exists a number fieldM such that p is unramified in M and ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot:

Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Beckmann’s Theorem

Theorem (Beckmann-Kazez 1989)

Let ϕ : X → P1 be a Bely̆ı map with monodromy group G.Suppose p does not divide #G. Then there exists a number fieldM such that p is unramified in M and ϕ is defined over M withgood reduction at all primes p of M above p.

Upshot: Every 2-solvable Bely̆ı curve we write down has goodreduction away from p = 2.

8 / 15

Lifting example K

X̃

X

P1

ϕ̃

ψ

ϕ

K (X̃ )

K (X )

K (P1)

2

2

{1}

H

G̃

2

2

G = Gal(K (X )/K (P1)) G ∼=〈(

(1 2), (1)(2), (1 2))〉≤ S2

G̃ = Gal(K (X̃ )/K (P1)) G̃ ∼= 〈σ̃〉 ≤ S4H = Gal(K (X̃ )/K (X )) H ∼= 〈(1 3)(2 4)〉 ≤ S4

1 H G̃ G 1

σ̃ σ

ι f

?

9 / 15

Lifting example K

X̃

X

P1

ϕ̃

ψ

ϕ

K (X̃ )

K (X )

K (P1)

2

2

{1}

H

G̃

2

2

G = Gal(K (X )/K (P1)) G ∼=〈(

(1 2), (1)(2), (1 2))〉≤ S2

G̃ = Gal(K (X̃ )/K (P1)) G̃ ∼= 〈σ̃〉 ≤ S4H = Gal(K (X̃ )/K (X )) H ∼= 〈(1 3)(2 4)〉 ≤ S4

1 H G̃ G 1

σ̃ σ

ι f

?

9 / 15

Lifting example K

X̃

X

P1

ϕ̃

ψ

ϕ

K (X̃ )

K (X )

K (P1)

2

2

{1}

H

G̃

2

2

G = Gal(K (X )/K (P1)) G ∼=〈(

(1 2), (1)(2), (1 2))〉≤ S2

G̃ = Gal(K (X̃ )/K (P1)) G̃ ∼= 〈σ̃〉 ≤ S4H = Gal(K (X̃ )/K (X )) H ∼= 〈(1 3)(2 4)〉 ≤ S4

1 H G̃ G 1

σ̃ σ

ι f

?

9 / 15

Lifting example K

X̃

X

P1

ϕ̃

ψ

ϕ

K (X̃ )

K (X )

K (P1)

2

2

{1}

H

G̃

2

2

G = Gal(K (X )/K (P1)) G ∼=〈(

(1 2), (1)(2), (1 2))〉≤ S2

G̃ = Gal(K (X̃ )/K (P1)) G̃ ∼= 〈σ̃〉 ≤ S4H = Gal(K (X̃ )/K (X )) H ∼= 〈(1 3)(2 4)〉 ≤ S4

1 H G̃ G 1

σ̃ σ

ι f

?9 / 15

Lifting example K

σ = (σ0, σ1, σ∞) =(

(1 2), (1)(2), (1 2))∈ S32

τ = (1 3)(2 4) ∈ S4G̃ = 〈σ̃〉 ≤ S4

1 〈τ〉 G̃ 〈σ〉 1

σ̃ σ

ι f

?

f −1(σ0) f −1(σ1) f −1(σ∞)

(1 2)(3 4) (1)(2)(3)(4) (1 2)(3 4)(1 4)(2 3) (1 3)(2 4) (1 4)(2 3)(1 4 3 2) (1 4 3 2)(1 2 3 4) (1 2 3 4)

10 / 15

Lifting example K

σ = (σ0, σ1, σ∞) =(

(1 2), (1)(2), (1 2))∈ S32

τ = (1 3)(2 4) ∈ S4G̃ = 〈σ̃〉 ≤ S4

1 〈τ〉 G̃ 〈σ〉 1

σ̃ σ

ι f

?

f −1(σ0) f −1(σ1) f −1(σ∞)

(1 2)(3 4) (1)(2)(3)(4) (1 2)(3 4)(1 4)(2 3) (1 3)(2 4) (1 4)(2 3)(1 4 3 2) (1 4 3 2)(1 2 3 4) (1 2 3 4)

10 / 15

Lifting example K

σ = (σ0, σ1, σ∞) =(

(1 2), (1)(2), (1 2))∈ S32

τ = (1 3)(2 4) ∈ S4G̃ = 〈σ̃〉 ≤ S4

1 〈τ〉 G̃ 〈σ〉 1

σ̃ σ

ι f

?

f −1(σ0) f −1(σ1) f −1(σ∞)

(1 2)(3 4) (1)(2)(3)(4) (1 2)(3 4)(1 4)(2 3) (1 3)(2 4) (1 4)(2 3)(1 4 3 2) (1 4 3 2)(1 2 3 4) (1 2 3 4)

10 / 15

Lifting example K

σ = (σ0, σ1, σ∞) =(

(1 2), (1)(2), (1 2))∈ S32

τ = (1 3)(2 4) ∈ S4G̃ = 〈σ̃〉 ≤ S4

1 〈τ〉 G̃ 〈σ〉 1

σ̃ σ

ι f

?

f −1(σ0) f −1(σ1) f −1(σ∞)

(1 2)(3 4) (1)(2)(3)(4) (1 2)(3 4)(1 4)(2 3) (1 3)(2 4) (1 4)(2 3)(1 4 3 2) (1 4 3 2)(1 2 3 4) (1 2 3 4)

10 / 15

Lifting example K

There are 32 possible σ from these lists.

Only 6 such triples correspond to Bely̆ı maps.

3 of these triples corresponding to distinct Bely̆ı maps (up toisomorphism).

G̃ ∼= Z/4Z((1 4 3 2), (1)(2)(3)(4), (1 2 3 4)

)(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

G̃ ∼= Z/2Z× Z/2Z((1 2)(3 4), (1 4)(2 3), (1 3)(2 4)

)

11 / 15

Lifting example KThere are 32 possible σ from these lists.

Only 6 such triples correspond to Bely̆ı maps.

3 of these triples corresponding to distinct Bely̆ı maps (up toisomorphism).

G̃ ∼= Z/4Z((1 4 3 2), (1)(2)(3)(4), (1 2 3 4)

)(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

G̃ ∼= Z/2Z× Z/2Z((1 2)(3 4), (1 4)(2 3), (1 3)(2 4)

)

11 / 15

Lifting example KThere are 32 possible σ from these lists.

Only 6 such triples correspond to Bely̆ı maps.

3 of these triples corresponding to distinct Bely̆ı maps (up toisomorphism).

G̃ ∼= Z/4Z((1 4 3 2), (1)(2)(3)(4), (1 2 3 4)

)(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

G̃ ∼= Z/2Z× Z/2Z((1 2)(3 4), (1 4)(2 3), (1 3)(2 4)

)

11 / 15

Lifting example KThere are 32 possible σ from these lists.

Only 6 such triples correspond to Bely̆ı maps.

3 of these triples corresponding to distinct Bely̆ı maps (up toisomorphism).

G̃ ∼= Z/4Z((1 4 3 2), (1)(2)(3)(4), (1 2 3 4)

)(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

G̃ ∼= Z/2Z× Z/2Z((1 2)(3 4), (1 4)(2 3), (1 3)(2 4)

)

11 / 15

Lifting example KThere are 32 possible σ from these lists.

Only 6 such triples correspond to Bely̆ı maps.

3 of these triples corresponding to distinct Bely̆ı maps (up toisomorphism).

G̃ ∼= Z/4Z((1 4 3 2), (1)(2)(3)(4), (1 2 3 4)

)(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

G̃ ∼= Z/2Z× Z/2Z((1 2)(3 4), (1 4)(2 3), (1 3)(2 4)

)11 / 15

4T1-[4,2,4]-4-22-4-g1

(σ0, σ1, σ∞) =(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

X0

X1 : x21 = x0

X2 :x21 =x0

x22 =x31−x1

K (X0) = K (x0)

K (X1) = K(x0)[x1](x21−x0)

K (X2) = K(X1)[x2](x22−(x31−x1))

0 1 ∞

(0, 0) (1, 1) (1,−1) ∞

(0, 0, 0) (1, 1, 0) (1,−1, 0) ∞

12 / 15

4T1-[4,2,4]-4-22-4-g1

(σ0, σ1, σ∞) =(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

X0

X1 : x21 = x0

X2 :x21 =x0

x22 =x31−x1

K (X0) = K (x0)

K (X1) = K(x0)[x1](x21−x0)

K (X2) = K(X1)[x2](x22−(x31−x1))

0 1 ∞

(0, 0) (1, 1) (1,−1) ∞

(0, 0, 0) (1, 1, 0) (1,−1, 0) ∞

12 / 15

4T1-[4,2,4]-4-22-4-g1

(σ0, σ1, σ∞) =(

(1 4 3 2), (1 3)(2 4), (1 4 3 2))

X0

X1 : x21 = x0

X2 :x21 =x0

x22 =x31−x1

K (X0) = K (x0)

K (X1) = K(x0)[x1](x21−x0)

K (X2) = K(X1)[x2](x22−(x31−x1))

0 1 ∞

(0, 0) (1, 1) (1,−1) ∞

(0, 0, 0) (1, 1, 0) (1,−1, 0) ∞

12 / 15

16T6-[8,8,4]-88-88-4444-g5

We now exhibit a genus 5 Bely̆ı map ϕ : X → P1 defined byx0 ∈ K (X ) with monodromy group C8 : C2.

X is cut out by the following equations in A5:

x21 = x0x3x24 = x1x2 + x1 + x23x3x24 = x1 + x1x23x3x44 = x33 + 2x1x24

2x23 x44 = x22 + 2x33 x24 + 2x23 − 2x3x24 + 2x44 + 1x33 = x2x3 + x23 x24 − x24

x23 x24 = x2x24 + x33 + x3x23 x44 = x43 + x23 + x44

13 / 15

16T6-[8,8,4]-88-88-4444-g5

We now exhibit a genus 5 Bely̆ı map ϕ : X → P1 defined byx0 ∈ K (X ) with monodromy group C8 : C2.

X is cut out by the following equations in A5:

x21 = x0x3x24 = x1x2 + x1 + x23x3x24 = x1 + x1x23x3x44 = x33 + 2x1x24

2x23 x44 = x22 + 2x33 x24 + 2x23 − 2x3x24 + 2x44 + 1x33 = x2x3 + x23 x24 − x24

x23 x24 = x2x24 + x33 + x3x23 x44 = x43 + x23 + x44

13 / 15

16T6-[8,8,4]-88-88-4444-g5

We now exhibit a genus 5 Bely̆ı map ϕ : X → P1 defined byx0 ∈ K (X ) with monodromy group C8 : C2.

X is cut out by the following equations in A5:

x21 = x0x3x24 = x1x2 + x1 + x23x3x24 = x1 + x1x23x3x44 = x33 + 2x1x24

2x23 x44 = x22 + 2x33 x24 + 2x23 − 2x3x24 + 2x44 + 1x33 = x2x3 + x23 x24 − x24

x23 x24 = x2x24 + x33 + x3x23 x44 = x43 + x23 + x44

13 / 15

16T6-[8,8,4]-88-88-4444-g5

We now exhibit a genus 5 Bely̆ı map ϕ : X → P1 defined byx0 ∈ K (X ) with monodromy group C8 : C2.

X is cut out by the following equations in A5:

x21 = x0x3x24 = x1x2 + x1 + x23x3x24 = x1 + x1x23x3x44 = x33 + 2x1x24

2x23 x44 = x22 + 2x33 x24 + 2x23 − 2x3x24 + 2x44 + 1x33 = x2x3 + x23 x24 − x24

x23 x24 = x2x24 + x33 + x3x23 x44 = x43 + x23 + x44

13 / 15

Acknowledgements

Thanks to the following for helpful discussions:

I Sam SchiavoneI Jeroen SijslingI John Voight

14 / 15

Thanks for listening!

https://math.dartmouth.edu/˜mjmusty/32.html

15 / 15

https://math.dartmouth.edu/~mjmusty/32.html