Computing a Databaseof Bely̆ı Maps

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Michael Musty (Dartmouth College)GSCAGT at Temple University

June 4, 2017

Acknowledgements

This is joint work with:

I Mike KlugI Sam SchiavoneI Jeroen SijslingI John Voight

Example

Find ϕ ∈ Q[x ] of degree 3 such that ϕ(x) and ϕ(x)− 1 have adouble root.

ϕ(x) = ax2(x − 1), ϕ(x)− 1 = a(x − b)2(x − c)

implies

ax3 − ax2 − 1 = ax3 − a(2b + c)x2 + ab(b + 2c)x − ab2c

which yields the solution a, b, c = −27/4, 2/3,−1/3.After a linear change of variables (x ← −2x/3) we get

ϕ(x) = 2x3 + 3x2 = x2(2x + 3)ϕ(x)− 1 = 2x3 + 3x2 − 1 = (2x − 1)(x + 1)2.

We can view ϕ as a branched (ramified) covering map of Riemannsurfaces. . .

Example

Find ϕ ∈ Q[x ] of degree 3 such that ϕ(x) and ϕ(x)− 1 have adouble root.

ϕ(x) = ax2(x − 1), ϕ(x)− 1 = a(x − b)2(x − c)

implies

ax3 − ax2 − 1 = ax3 − a(2b + c)x2 + ab(b + 2c)x − ab2c

which yields the solution a, b, c = −27/4, 2/3,−1/3.After a linear change of variables (x ← −2x/3) we get

ϕ(x) = 2x3 + 3x2 = x2(2x + 3)ϕ(x)− 1 = 2x3 + 3x2 − 1 = (2x − 1)(x + 1)2.

We can view ϕ as a branched (ramified) covering map of Riemannsurfaces. . .

Example

Find ϕ ∈ Q[x ] of degree 3 such that ϕ(x) and ϕ(x)− 1 have adouble root.

ϕ(x) = ax2(x − 1), ϕ(x)− 1 = a(x − b)2(x − c)

implies

ax3 − ax2 − 1 = ax3 − a(2b + c)x2 + ab(b + 2c)x − ab2c

which yields the solution a, b, c = −27/4, 2/3,−1/3.

After a linear change of variables (x ← −2x/3) we get

ϕ(x) = 2x3 + 3x2 = x2(2x + 3)ϕ(x)− 1 = 2x3 + 3x2 − 1 = (2x − 1)(x + 1)2.

We can view ϕ as a branched (ramified) covering map of Riemannsurfaces. . .

Example

Find ϕ ∈ Q[x ] of degree 3 such that ϕ(x) and ϕ(x)− 1 have adouble root.

ϕ(x) = ax2(x − 1), ϕ(x)− 1 = a(x − b)2(x − c)

implies

ax3 − ax2 − 1 = ax3 − a(2b + c)x2 + ab(b + 2c)x − ab2c

which yields the solution a, b, c = −27/4, 2/3,−1/3.After a linear change of variables (x ← −2x/3) we get

ϕ(x) = 2x3 + 3x2 = x2(2x + 3)ϕ(x)− 1 = 2x3 + 3x2 − 1 = (2x − 1)(x + 1)2.

We can view ϕ as a branched (ramified) covering map of Riemannsurfaces. . .

Example

Find ϕ ∈ Q[x ] of degree 3 such that ϕ(x) and ϕ(x)− 1 have adouble root.

ϕ(x) = ax2(x − 1), ϕ(x)− 1 = a(x − b)2(x − c)

implies

ax3 − ax2 − 1 = ax3 − a(2b + c)x2 + ab(b + 2c)x − ab2c

which yields the solution a, b, c = −27/4, 2/3,−1/3.After a linear change of variables (x ← −2x/3) we get

ϕ(x) = 2x3 + 3x2 = x2(2x + 3)ϕ(x)− 1 = 2x3 + 3x2 − 1 = (2x − 1)(x + 1)2.

We can view ϕ as a branched (ramified) covering map of Riemannsurfaces. . .

Example

0 1

x = 1/2x = 0

x = −3/2

x = −1

∞

1

2

3

X

P1

ϕ(x) = 2x3 + 3x2︸ ︷︷ ︸z

Riemann’s Existence Theorem

Let X be a compact connected Riemann surface.

Theorem (Riemann Existence)

There exists a non-constant meromorphic function on X thatrepresents X as a branched cover of P1 (with branch locus a finitesubset of P1) as in the previous slide.

A consequence of this theorem is the equivalence between Riemannsurfaces and algebraic curves.

Riemann’s Existence Theorem

Let X be a compact connected Riemann surface.

Theorem (Riemann Existence)

There exists a non-constant meromorphic function on X thatrepresents X as a branched cover of P1 (with branch locus a finitesubset of P1) as in the previous slide.

A consequence of this theorem is the equivalence between Riemannsurfaces and algebraic curves.

Riemann’s Existence Theorem

Let X be a compact connected Riemann surface.

Theorem (Riemann Existence)

There exists a non-constant meromorphic function on X thatrepresents X as a branched cover of P1 (with branch locus a finitesubset of P1) as in the previous slide.

A consequence of this theorem is the equivalence between Riemannsurfaces and algebraic curves.

Bely̆ı’s Theorem

We say X is defined over a subfield L of C if there existsf (z ,w) ∈ L[z ,w ] such that the field of meromorphic functions onX is isomorphic to

C(z)[w ](f (z ,w)) .

For example, P1 is defined over Q:Let f (z ,w) = w ∈ Q[z ,w ]. Then

C(z)[w ](w) = C(z)

which is the meromorphic functions on P1.

Question: How do we know when X is defined over Q?

Bely̆ı’s Theorem

We say X is defined over a subfield L of C if there existsf (z ,w) ∈ L[z ,w ] such that the field of meromorphic functions onX is isomorphic to

C(z)[w ](f (z ,w)) .

For example, P1 is defined over Q:

Let f (z ,w) = w ∈ Q[z ,w ]. Then

C(z)[w ](w) = C(z)

which is the meromorphic functions on P1.

Question: How do we know when X is defined over Q?

Bely̆ı’s Theorem

We say X is defined over a subfield L of C if there existsf (z ,w) ∈ L[z ,w ] such that the field of meromorphic functions onX is isomorphic to

C(z)[w ](f (z ,w)) .

For example, P1 is defined over Q:Let f (z ,w) = w ∈ Q[z ,w ]. Then

C(z)[w ](w) = C(z)

which is the meromorphic functions on P1.

Question: How do we know when X is defined over Q?

Bely̆ı’s Theorem

We say X is defined over a subfield L of C if there existsf (z ,w) ∈ L[z ,w ] such that the field of meromorphic functions onX is isomorphic to

C(z)[w ](f (z ,w)) .

For example, P1 is defined over Q:Let f (z ,w) = w ∈ Q[z ,w ]. Then

C(z)[w ](w) = C(z)

which is the meromorphic functions on P1.

Question: How do we know when X is defined over Q?

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A curve X over C can be defined over Q if and only if there existsa branched covering of compact connected Riemann surfacesϕ : X → P1 unramified (unbranched) above P1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map. The degree of a Bely̆ı map isthe number of sheets in the covering.

In the 1980s, Grothendieck spoke of Bely̆ı’s result saying: Never,without a doubt, was such a deep and disconcerting result provedin so few lines!

There is a zoo of objects in bijection with the set of Bely̆ı maps.

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A curve X over C can be defined over Q if and only if there existsa branched covering of compact connected Riemann surfacesϕ : X → P1 unramified (unbranched) above P1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map.

The degree of a Bely̆ı map isthe number of sheets in the covering.

In the 1980s, Grothendieck spoke of Bely̆ı’s result saying: Never,without a doubt, was such a deep and disconcerting result provedin so few lines!

There is a zoo of objects in bijection with the set of Bely̆ı maps.

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A curve X over C can be defined over Q if and only if there existsa branched covering of compact connected Riemann surfacesϕ : X → P1 unramified (unbranched) above P1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map. The degree of a Bely̆ı map isthe number of sheets in the covering.

In the 1980s, Grothendieck spoke of Bely̆ı’s result saying: Never,without a doubt, was such a deep and disconcerting result provedin so few lines!

There is a zoo of objects in bijection with the set of Bely̆ı maps.

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A curve X over C can be defined over Q if and only if there existsa branched covering of compact connected Riemann surfacesϕ : X → P1 unramified (unbranched) above P1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map. The degree of a Bely̆ı map isthe number of sheets in the covering.

In the 1980s, Grothendieck spoke of Bely̆ı’s result saying: Never,without a doubt, was such a deep and disconcerting result provedin so few lines!

There is a zoo of objects in bijection with the set of Bely̆ı maps.

Bely̆ı’s Theorem

Theorem (G.V. Bely̆ı 1979)

A curve X over C can be defined over Q if and only if there existsa branched covering of compact connected Riemann surfacesϕ : X → P1 unramified (unbranched) above P1 \ {0, 1,∞}.

Such a map is called a Bely̆ı map. The degree of a Bely̆ı map isthe number of sheets in the covering.

In the 1980s, Grothendieck spoke of Bely̆ı’s result saying: Never,without a doubt, was such a deep and disconcerting result provedin so few lines!

There is a zoo of objects in bijection with the set of Bely̆ı maps.

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,∞

∼

∼

∼

∼

Example

0 1

x = 1/2x = 0

x = −3/2

x = −1

∞

1

2

3

X

P1

ϕ(x) = 2x3 + 3x2︸ ︷︷ ︸z

Example

3 2 1

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

)[∆(2, 2, 3) : Γ] = 3

ϕ(x) = 2x3 + 3x2︸ ︷︷ ︸z

C

2x3 + 3x2︸ ︷︷ ︸z

⊆ C(x)

∼

∼

∼

∼

Motivation

In his 1984 work Esquisse d’un Programme, Grothendieck describesan action of Gal(Q/Q) on the set of dessins d’enfants.

About this discovery, Grothendieck said: I do not believe that amathematical fact has ever struck me quite so strongly as this one.

This yields a Galois action on everything in the zoo.

Motivation

In his 1984 work Esquisse d’un Programme, Grothendieck describesan action of Gal(Q/Q) on the set of dessins d’enfants.

About this discovery, Grothendieck said: I do not believe that amathematical fact has ever struck me quite so strongly as this one.

This yields a Galois action on everything in the zoo.

Motivation

In his 1984 work Esquisse d’un Programme, Grothendieck describesan action of Gal(Q/Q) on the set of dessins d’enfants.

About this discovery, Grothendieck said: I do not believe that amathematical fact has ever struck me quite so strongly as this one.

This yields a Galois action on everything in the zoo.

Motivation

In his 1984 work Esquisse d’un Programme, Grothendieck describesan action of Gal(Q/Q) on the set of dessins d’enfants.

About this discovery, Grothendieck said: I do not believe that amathematical fact has ever struck me quite so strongly as this one.

This yields a Galois action on everything in the zoo.

Database

Also in Grothendieck’s Esquisse d’un Programme, he writes aboutthe computation of specific examples of Bely̆ı maps:Exactly which are the conjugates of a given oriented map? (Visibly,there is only a finite number of these.) I considered some concretecases (for coverings of low degree) by various methods, J. Malgoireconsidered some others–I doubt there is a uniform method forsolving the problem by computer.

In 2014, KMSV provide a general purpose numerical method forcomputing Bely̆ı maps using power series expansions of modularforms.

Database

Also in Grothendieck’s Esquisse d’un Programme, he writes aboutthe computation of specific examples of Bely̆ı maps:Exactly which are the conjugates of a given oriented map? (Visibly,there is only a finite number of these.) I considered some concretecases (for coverings of low degree) by various methods, J. Malgoireconsidered some others–I doubt there is a uniform method forsolving the problem by computer.

In 2014, KMSV provide a general purpose numerical method forcomputing Bely̆ı maps using power series expansions of modularforms.

Database

Also in Grothendieck’s Esquisse d’un Programme, he writes aboutthe computation of specific examples of Bely̆ı maps:Exactly which are the conjugates of a given oriented map? (Visibly,there is only a finite number of these.) I considered some concretecases (for coverings of low degree) by various methods, J. Malgoireconsidered some others–I doubt there is a uniform method forsolving the problem by computer.

In 2014, KMSV provide a general purpose numerical method forcomputing Bely̆ı maps using power series expansions of modularforms.

DatabaseWe are currently tabulating a database of all Bely̆ı maps in lowdegree to be included in the LMFDB www.lmfdb.org.

Below is a table detailing how many Bely̆ı maps (up toisomorphism) there are of given degree and genus.

d g = 0 g = 1 g = 2 g = 3 g > 3 total

2 1 0 0 0 0 13 2 1 0 0 0 34 6 2 0 0 0 85 12 6 2 0 0 206 38 29 7 0 0 747 89 50 13 3 0 1558 261 217 84 11 0 5739 583 427 163 28 6 1207

www.lmfdb.org

DatabaseWe are currently tabulating a database of all Bely̆ı maps in lowdegree to be included in the LMFDB www.lmfdb.org.

Below is a table detailing how many Bely̆ı maps (up toisomorphism) there are of given degree and genus.

d g = 0 g = 1 g = 2 g = 3 g > 3 total

2 1 0 0 0 0 13 2 1 0 0 0 34 6 2 0 0 0 85 12 6 2 0 0 206 38 29 7 0 0 747 89 50 13 3 0 1558 261 217 84 11 0 5739 583 427 163 28 6 1207

www.lmfdb.org

Galois action for example7T5-[3,4,4]-331-421-421-g0

Let τ ∈ Gal(Q/Q) be the element interchanging ±√

7...

λ =( 1

3087(

173√

7 + 343))

ϕ = λ ·x3(

x − 1729(68√

7 + 236))1 (

x − 19(20− 4

√7))3

(x − 421

(√7 + 3

))2 (x − 421

(√7 + 1

))4

ϕ− 1 = λ ·

(x− 1189

(44√

7+140))4(

x− 17(

12√

7−28))2(

x− 114(

3√

7+7))1(

x− 421(√

7+3))2(

x− 421(√

7+1))4

Galois action for example7T5-[3,4,4]-331-421-421-g0

Let τ ∈ Gal(Q/Q) be the element interchanging ±√

7...

λ =( 1

3087(

173√

7 + 343))

ϕ = λ ·x3(

x − 1729(68√

7 + 236))1 (

x − 19(20− 4

√7))3

(x − 421

(√7 + 3

))2 (x − 421

(√7 + 1

))4

ϕ− 1 = λ ·

(x− 1189

(44√

7+140))4(

x− 17(

12√

7−28))2(

x− 114(

3√

7+7))1(

x− 421(√

7+3))2(

x− 421(√

7+1))4

Galois action for example7T5-[3,4,4]-331-421-421-g0

Let τ ∈ Gal(Q/Q) be the element interchanging ±√

7...

λ =( 1

3087(

173√

7 + 343))

ϕ = λ ·x3(

x − 1729(68√

7 + 236))1 (

x − 19(20− 4

√7))3

(x − 421

(√7 + 3

))2 (x − 421

(√7 + 1

))4

ϕ− 1 = λ ·

(x− 1189

(44√

7+140))4(

x− 17(

12√

7−28))2(

x− 114(

3√

7+7))1(

x− 421(√

7+3))2(

x− 421(√

7+1))4

Galois action for example7T5-[3,4,4]-331-421-421-g0

Let τ ∈ Gal(Q/Q) be the element interchanging ±√

7...

λ =( 1

3087(

173√

7 + 343))

ϕ = λ ·x3(

x − 1729(68√

7 + 236))1 (

x − 19(20− 4

√7))3

(x − 421

(√7 + 3

))2 (x − 421

(√7 + 1

))4

ϕ− 1 = λ ·

(x− 1189

(44√

7+140))4(

x− 17(

12√

7−28))2(

x− 114(

3√

7+7))1(

x− 421(√

7+3))2(

x− 421(√

7+1))4

Galois action for example7T5-[3,4,4]-331-421-421-g0

σ0 = (1 3 7)(2)(4 5 6)σ1 = (1 4 5 3)(2 7)(6)σ∞ = (1 2 7 5)(3)(4 6)

σ0 = (1 3 7)(2)(4 5 6)σ1 = (1 6 3 2)(4 5)(7)σ∞ = (1 7 6 4)(2 3)(5)

τ

Galois action for example7T5-[3,4,4]-331-421-421-g0

2 7

3

1

5

4

6

7

3

1

2 6

5

4

τ

Galois action for example7T5-[3,4,4]-331-421-421-g0

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×

bc

b ×

bc

b

×

bc b

×

bc

b

×

bc

b

× 1

23

4

5

67

τ

Thanks for listening!