Variations on a Theme: Fields of Definition, Fields of Moduli

Fields of Definition / Fields of Moduli Automorphism Groups Twists

Variations on a Theme:Fields of Definition, Fields of Moduli,

Automorphisms, and Twists

Michelle Manes ([email protected])

ICERM WorkshopModuli Spaces Associated to Dynamical Systems

17 April, 2012

1


Definitions

Definition

Let φ ∈ RatNd . A field K ′/K is a field of definition for φ ifφf ∈ RatNd (K ′) for some f ∈ PGLN+1.

Definition

Let φ ∈ RatNd , and define

Gφ = {σ ∈ GK | φσ is K equivalent to φ}.

The field of moduli of φ is the fixed field KGφ.

2


Definitions

Definition

Let φ ∈ RatNd . A field K ′/K is a field of definition for φ ifφf ∈ RatNd (K ′) for some f ∈ PGLN+1.

Definition

Let φ ∈ RatNd , and define

Gφ = {σ ∈ GK | φσ is K equivalent to φ}.

The field of moduli of φ is the fixed field KGφ.

3


Definitions

The field of moduli of φ is the smallest field L with theproperty that for every σ ∈ Gal(K/L) there is somefσ ∈ PGLN+1 such that φσ = φfσ .

The field of moduli for φ is contained in every field ofdefinition.

Equality???

4


Definitions



Equality???

5


Definitions



Equality???

6


FOD = FOM criterion

Proposition (Hutz, M.)

Let ξ ∈ MNd (K ) be a dynamical system with Autφ = {id},

and let D =∑N

j=0 dN .

If gcd(D,N + 1) = 1, then K is a field of definition of ξ.

7


FOD = FOM criterion

Idea: If [φ] ∈ MNd (K ), then you get a cohomology class

f : Gal(K̄/K )→ PGLN+1

σ 7→ fσ

Twists of PN are in 1-1 correspondence with cocyles:

i : PN → X

σ 7→ i−1iσ

[φ] ∈ MNd (K ) ; cocycle cφ ; Xcφ

K is FOD for φ⇐⇒ cφ trivial ⇐⇒ Xcφ/K

When gcd(D,N + 1) = 1, we can find a K -rationalzero-cycle on Xcφ

.

8


FOD = FOM criterion

Idea: If [φ] ∈ MNd (K ), then you get a cohomology class

f : Gal(K̄/K )→ PGLN+1

σ 7→ fσ

Twists of PN are in 1-1 correspondence with cocyles:

i : PN → X

σ 7→ i−1iσ

[φ] ∈ MNd (K ) ; cocycle cφ ; Xcφ

K is FOD for φ⇐⇒ cφ trivial ⇐⇒ Xcφ/K

When gcd(D,N + 1) = 1, we can find a K -rationalzero-cycle on Xcφ

.9


FOD = FOM criterion

If N = 1, then D = d + 1, and the test is on gcd(d + 1,2).

Corollary (Silverman)If d is even, then the field of moduli is a field of definition.

Result in P1 doesn’t require Aut(φ) = id.

Proof requires knowledge of the possible automorphismgroups and “cohomology lifting.”

10


FOD = FOM criterion

If N = 1, then D = d + 1, and the test is on gcd(d + 1,2).

Corollary (Silverman)If d is even, then the field of moduli is a field of definition.

Result in P1 doesn’t require Aut(φ) = id.

Proof requires knowledge of the possible automorphismgroups and “cohomology lifting.”

11


Example (Silverman)

φ(z) = i(

z−1z+1

)3. So Q(i) is a field of definition for φ.

Let σ represent complex conjugation, then

φσ = φf for f = −1z.

Hence, Q is the field of moduli for φ.

K is a field of definition for φ iff −1 ∈ NK (i)/K (K (i)∗).

12


Normal Form for M2

Lemma (Milnor)Let φ ∈ Rat2 have multipliers λ1, λ2, λ3.

1 If not all three multipliers are 1, φ is conjugate to amap of the form:

z2 + λ1zλ2z + 1

.

2 If all three multipliers are 1, φ is conjugate to:

z +1z.

Possible that φ ∈ K (z) but the conjugate map is not.

13


Normal Form for M2

Lemma (Milnor)Let φ ∈ Rat2 have multipliers λ1, λ2, λ3.

1 If not all three multipliers are 1, φ is conjugate to amap of the form:

z2 + λ1zλ2z + 1

.

2 If all three multipliers are 1, φ is conjugate to:

z +1z.

Possible that φ ∈ K (z) but the conjugate map is not.

14


Arithmetic Normal Form for M2

Theorem (M., Yasufuku)Let φ ∈ Rat2(K ) have multipliers λ1, λ2, λ3.

1 If the multipliers are distinct or if exactly twomultipliers are 1, then φ(z) is conjugate over K to

2z2 + (2− σ1)z + (2− σ1)

−z2 + (2 + σ1)z + 2− σ1 − σ2∈ K (z),

where σ1 and σ2 are the first two symmetric functionsof the multipliers.

Furthermore, no two distinct maps of this form areconjugate to each other over K .

15



Theorem (M., Yasufuku)

2 If λ1 = λ2 6= 1 and λ3 6= λ1 or if λ1 = λ2 = λ3 = 1, thenψ is conjugate over K to a map of the form

φk ,b(z) = kz +bz

with k = λ1+12 , and b ∈ K ∗.

Furthermore, two such maps φk ,b and φk ′,b′ areconjugate over K if and only if k = k ′; they areconjugate over K if in addition b/b′ ∈ (K ∗)2.

16



Theorem (M., Yasufuku)

3 If λ1 = λ2 = λ3 = −2, then φ is conjugate over K to

θd ,k (z) =kz2 − 2dz + dkz2 − 2kz + d

,

with k ∈ K ,d ∈ K ∗, and k2 6= d .

All such maps are conjugate over K . Furthermore,θd ,k (z) and θd ′,k ′(z) are conjugate over K if and only if

ugly, but easily testable condition

17


φ ∈ Hom1d

Aut(φ) is conjugate to one of the following:

1 Cyclic group of order n: Cn = 〈ζnz〉 .

2 Dihedral group of order 2n: Dn =

⟨ζnz,

1z

⟩.

3 Tetrahedral group: A4 =

⟨−z,

1z, i(

z + 1z − 1

)⟩.

4 Octahedral group: S4 =

⟨iz,

1z, i(

z + 1z − 1

)⟩.

5 Icosahedral group:

A5 =

⟨ζ5z,−1

z,

(ζ5 + ζ−1

5

)z + 1

z −(ζ5 + ζ−1

5

) ⟩ .18


φ ∈ Hom2d

1 Diagonal Abelian Groups (Cyclic Group of order n):

H =

ζan 0 00 ζb

n 00 0 1

, gcd(a,n) = 1 or gcd(b,n) = 1.

Proposition

Let r be the number of solutions to x2 ≡ 1 mod n. Thereare n + r/2− ϕ(n)/2 representations of Cn of the formζn 0 0

0 ζan 0

0 0 1

.

19


φ ∈ Hom2d

2 Subgroups of the form⟨ζp 0 00 ai bi0 ci di

⟩ ,where the lower right 2× 2 matrices come fromembedding the PGL2 automorphism groups.

20


φ ∈ Hom2d

3 Subgroups that don’t come from embedding PGL2.(Lots of them.)

⟨0 1 01 0 00 0 1

,

0 0 11 0 00 1 0,

−1 0 0

0 1 00 0 −1

,

1 0 00 −1 00 0 −1

⟩

21


Higher Dimensions

This slide intentionally left blank.

22


Higher Dimensions

This slide intentionally left blank.

23


Computing the Absolute Automorphism Group

Algorithm (Faber, M., Viray)Input:

a nonconstant rational function φ ∈ K (z),an Autφ(K̄ )-invariant subset T = {τ1, . . . , τn} ⊂ P1(E)with n ≥ 3.

Output: the set Autφ(K̄ )

24


Computing the Absolute Automorphism Group

Algorithm (Faber, M., Viray)create an empty list L.

for each triple of distinct integers i , j , k ∈ {1, . . . ,n}:compute s ∈ PGL2(K̄ ) by solving the linear system

s(τ1) = τi , s(τ2) = τj , s(τ3) = τk .

if s ◦ φ = φ ◦ s: append s to L.

return L.

25


Computing the Automorphism Group for a Given Map

Proposition (Faber, M., Viray)Let K be a number field and let φ ∈ K (z) a rationalfunction of degree d ≥ 2. Define S0 to be the set ofrational primes given by

S0 = {2} ∪{

p odd :p − 1

2

∣∣∣[K : Q] and p | d(d2 − 1)

},

and let S be the (finite) set of places of K of bad reductionfor φ along with the places that divide a prime in S0. Thenredv : Autφ(K )→ Autφ(Fv ) is a well-defined injectivehomomorphism for all places v outside S.

26


Realizing Maps with a Given Automorphism Group

Given a finite subgroup Γ ∈ PGL2, Doyle & McMullen givea way to construct all rational maps

φ ∈⋃

2≤d≤n

Ratd with Γ ⊆ Aut(φ).

inv. hom. one-form 1−1←→ inv. hom. rational map

Fdx + Gdy 1−1←→ φ = −GF

27



It is enough to find all (relative) invariant homogeneouspolynomials, i.e. for each γ ∈ Γ there is a character χ:

γ∗F = F (γx̄) = χ(γ)F (x̄).

λ = (xdy − ydx)/2. Every invariant one-form has the form:

Fλ + dG,

where deg F + 2 = deg G,F and G invariant with the same character.

28



It is enough to find all (relative) invariant homogeneouspolynomials, i.e. for each γ ∈ Γ there is a character χ:

γ∗F = F (γx̄) = χ(γ)F (x̄).

λ = (xdy − ydx)/2. Every invariant one-form has the form:

Fλ + dG,

where deg F + 2 = deg G,F and G invariant with the same character.

29


Two useful gadgets

Molien Series: Given a finite group Γ (and character χ),outputs the power series

∞∑k=0

dim(K [x̄ ]Γ

k

)tk .

Reynolds Operator: Given a finite group Γ, (characterχ), and all homogeneous monomials of agiven degree, outputs all (relative) Γ-invariantsof that degree.

30


Example

Γ = C4 =

⟨(i 00 1

)⟩

Molien Series:

1 + t2 + t4 + t6 + 3t8 + 3t10 + 3t12 + 3t14 + 5t16 + O(t18)

Invariants of degree ≤ 8:

xy x2y2 x3y3

x4y4 x8 y8

Some maps with Γ ⊆ Aut(φ):

φ1(z) =z4 + 16

z3 φ2(z) =z9 + 9zz8 − 1

31


Example

Γ = C4 =

⟨(i 00 1

)⟩Molien Series:

1 + t2 + t4 + t6 + 3t8 + 3t10 + 3t12 + 3t14 + 5t16 + O(t18)


xy x2y2 x3y3

x4y4 x8 y8


φ1(z) =z4 + 16

z3 φ2(z) =z9 + 9zz8 − 1

32


Example

Γ = C4 =

⟨(i 00 1

)⟩Molien Series:

1 + t2 + t4 + t6 + 3t8 + 3t10 + 3t12 + 3t14 + 5t16 + O(t18)


xy x2y2 x3y3

x4y4 x8 y8


φ1(z) =z4 + 16

z3 φ2(z) =z9 + 9zz8 − 1

33


Example

Γ = C4 =

⟨(i 00 1

)⟩Molien Series:

1 + t2 + t4 + t6 + 3t8 + 3t10 + 3t12 + 3t14 + 5t16 + O(t18)


xy x2y2 x3y3

x4y4 x8 y8


φ1(z) =z4 + 16

z3 φ2(z) =z9 + 9zz8 − 1

34


Example

Γ =

⟨(1 −11 1

)⟩(also cyclic of order 4).


x2 + y2 x8 + 12x6y2 − 20x4y4 + 12x2y6 + y8

x4 + 2x2y2 + y4 x8 − 4x6y2 + 22x4y4 − 4x2y6 + y8

x6 + 3x4y2 + 3x2y4 + y6 x7y − 7x5y3 + 7x3y5 − xy7


φ1(z) = − z7 + 24z6 + 3z5 − 40z4 + 3z3 + 72z2 + z + 88z7 − z6 + 72z5 − 3z4 − 40z3 − 3z2 + 24z − 1

φ2(z) = −z(3z6 − 39z4 + 73z2 − 13

)13z6 − 73z4 + 39z2 − 3

35


Example

Γ =

⟨(1 −11 1

)⟩(also cyclic of order 4).


x2 + y2 x8 + 12x6y2 − 20x4y4 + 12x2y6 + y8

x4 + 2x2y2 + y4 x8 − 4x6y2 + 22x4y4 − 4x2y6 + y8

x6 + 3x4y2 + 3x2y4 + y6 x7y − 7x5y3 + 7x3y5 − xy7


φ1(z) = − z7 + 24z6 + 3z5 − 40z4 + 3z3 + 72z2 + z + 88z7 − z6 + 72z5 − 3z4 − 40z3 − 3z2 + 24z − 1

φ2(z) = −z(3z6 − 39z4 + 73z2 − 13

)13z6 − 73z4 + 39z2 − 3

36


Exact Automorphism Groups?

Proposition (Hutz, M.)Let

A4 =

⟨−z,

1z, i(

z + 1z − 1

)⟩and S4 =

⟨iz,

1z, i(

z + 1z − 1

)⟩.

If φ ∈ Q(z) satisfies A4 ⊆ Aut(φ), then in fact Aut(φ) = S4.

QuestionHow to construct maps φ ∈ K (z) with Γ = Aut(φ) (ordecide there are none)?

How to construct maps φ ∈ K (z) with a subgroup ofAut(φ) conjugate to (or equal to) Γ?

37


Exact Automorphism Groups?

Proposition (Hutz, M.)Let

A4 =

⟨−z,

1z, i(

z + 1z − 1

)⟩and S4 =

⟨iz,

1z, i(

z + 1z − 1

)⟩.

If φ ∈ Q(z) satisfies A4 ⊆ Aut(φ), then in fact Aut(φ) = S4.

QuestionHow to construct maps φ ∈ K (z) with Γ = Aut(φ) (ordecide there are none)?

How to construct maps φ ∈ K (z) with a subgroup ofAut(φ) conjugate to (or equal to) Γ?

38


Automorphisms and Twists

Twist(φ/K ) =

K -equivalence classesof maps ψ ∈ HomN

d (K )

such that ψ is K -equivalent to φ

.

Twists give automorphisms of the map φ:

fφf−1 = (fφf−1)σ

= f σφ(f−1)σ.

φ = f−1f σφ(f σ)−1f

f−1f σ ∈ Aut(φ).

39


Uniform Bounds on Preperiodic Points for Twists

Proposition (Levy, M., Thompson)

Let φ ∈ HomNd (K ). There is a uniform bound Bφ such that

for all ψ ∈ Twist(φ/K ),

# PrePer(ψ,PNK ) ≤ Bφ.

Idea: The degree of the field of definition of the twistingmap f is bounded by # Aut(φ). Apply Northcott.

40


Cohomology and Twists

For an object X , twists give automorphisms:

gσ : Xσ(i−1)−−−−→ Y i−→ X .

A twist gives a one-cocyle:

g : Gal(K̄/K )→ Aut(X )

σ 7→ i ◦ σ(i−1)

Does every one-cocyle come from a twist?

For algebraic varieties, yes. For morphisms, sometimes.

Twist(φ/K ) =

{ξ ∈ H1

(Gal(K̄/K ),Aut(φ)

):

ξ becomes trivial in H1(Gal(K̄/K

),PGLN+1

}.

41




gσ : Xσ(i−1)−−−−→ Y i−→ X .



σ 7→ i ◦ σ(i−1)



Twist(φ/K ) =

{ξ ∈ H1


):


),PGLN+1

}.

42




gσ : Xσ(i−1)−−−−→ Y i−→ X .



σ 7→ i ◦ σ(i−1)



Twist(φ/K ) =

{ξ ∈ H1


):


),PGLN+1

}.

43


Describing Twists

QuestionGiven φ ∈ Ratd , can we write an explicit formula for alltwists of φ?

Done for Rat2 by Arithmetic Normal Form Theorem.If Aut(φ) = {ζnz}, then we have an isomorphism

K ∗/K ∗n → Twist(φ/K )

b 7→

φ(

z n√

b)

n√

b

.More general presentation of Cn? Otherautomorphism groups? Higher dimensions?

44


Describing Twists


Done for Rat2 by Arithmetic Normal Form Theorem.

If Aut(φ) = {ζnz}, then we have an isomorphism


b 7→

φ(

z n√

b)

n√

b


45


Describing Twists




b 7→

φ(

z n√

b)

n√

b

.

More general presentation of Cn? Otherautomorphism groups? Higher dimensions?

46


Describing Twists




b 7→

φ(

z n√

b)

n√

b


47

Variations on a Theme: Fields of Definition, Fields of Moduli

Documents

Transcript of Variations on a Theme: Fields of Definition, Fields of Moduli