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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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s5
s6
s6
bc b
×
bc
b
×
bc
b ×
bc
b
×
bc b
×
bc
b
×
bc
b
× 1
23
4
5
67
τ
Thanks for listening!
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