Variations on a Theme: Fields of Definition, Fields of Moduli, Automorphisms, and Twists

Transcription

1 Variations on a Theme: Fields of Definition, Fields of Moduli, Automorphisms, and Twists Michelle Manes (mmanes@math.hawaii.edu) ICERM Workshop Moduli Spaces Associated to Dynamical Systems 17 April, 2012

2 Definitions Definition Let φ Rat N d. A field K /K is a field of definition for φ if φ f Rat N d (K ) for some f PGL N+1.

3 Definitions Definition Let φ Rat N d. A field K /K is a field of definition for φ if φ f Rat N d (K ) for some f PGL N+1. Definition Let φ Rat N d, and define G φ = {σ G K φ σ is K equivalent to φ}. The field of moduli of φ is the fixed field K G φ.

4 Definitions The field of moduli of φ is the smallest field L with the property that for every σ Gal(K /L) there is some f σ PGL N+1 such that φ σ = φ fσ.

5 Definitions The field of moduli of φ is the smallest field L with the property that for every σ Gal(K /L) there is some f σ PGL N+1 such that φ σ = φ fσ. The field of moduli for φ is contained in every field of definition.

6 Definitions The field of moduli of φ is the smallest field L with the property that for every σ Gal(K /L) there is some f σ PGL N+1 such that φ σ = φ fσ. The field of moduli for φ is contained in every field of definition. Equality???

7 FOD = FOM criterion Proposition (Hutz, M.) Let ξ M N d (K ) be a dynamical system with Aut φ = {id}, and let D = N j=0 d N. If gcd(d, N + 1) = 1, then K is a field of definition of ξ.

8 FOD = FOM criterion Idea: If [φ] Md N (K ), then you get a cohomology class f : Gal( K /K ) PGL N+1 σ f σ

9 FOD = FOM criterion Idea: If [φ] Md N (K ), then you get a cohomology class f : Gal( K /K ) PGL N+1 σ f σ Twists of P N are in 1-1 correspondence with cocyles: i : P N X σ i 1 i σ [φ] M N d (K ) cocycle c φ X cφ K is FOD for φ c φ trivial X cφ /K When gcd(d, N + 1) = 1, we can find a K -rational zero-cycle on X cφ.

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.

11 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 P 1 doesn t require Aut(φ) = id. Proof requires knowledge of the possible automorphism groups and cohomology lifting.

12 Example (Silverman) φ(z) = i ( z 1 z+1) 3. So Q(i) is a field of definition for φ. Let σ represent complex conjugation, then φ σ = φ f for f = 1 z. Hence, Q is the field of moduli for φ. K is a field of definition for φ iff 1 N K (i)/k (K (i) ).

13 Normal Form for M 2 Lemma (Milnor) Let φ Rat 2 have multipliers λ 1, λ 2, λ 3. 1 If not all three multipliers are 1, φ is conjugate to a map of the form: z 2 + λ 1 z λ 2 z If all three multipliers are 1, φ is conjugate to: z + 1 z.

14 Normal Form for M 2 Lemma (Milnor) Let φ Rat 2 have multipliers λ 1, λ 2, λ 3. 1 If not all three multipliers are 1, φ is conjugate to a map of the form: z 2 + λ 1 z λ 2 z If all three multipliers are 1, φ is conjugate to: z + 1 z. Possible that φ K (z) but the conjugate map is not.

15 Arithmetic Normal Form for M 2 Theorem (M., Yasufuku) Let φ Rat 2 (K ) have multipliers λ 1, λ 2, λ 3. 1 If the multipliers are distinct or if exactly two multipliers are 1, then φ(z) is conjugate over K to 2z 2 + (2 σ 1 )z + (2 σ 1 ) z 2 + (2 + σ 1 )z + 2 σ 1 σ 2 K (z), where σ 1 and σ 2 are the first two symmetric functions of the multipliers. Furthermore, no two distinct maps of this form are conjugate to each other over K.

16 Arithmetic Normal Form for M 2 Theorem (M., Yasufuku) 2 If λ 1 = λ 2 1 and λ 3 λ 1 or if λ 1 = λ 2 = λ 3 = 1, then ψ is conjugate over K to a map of the form φ k,b (z) = kz + b z with k = λ 1+1 2, and b K. Furthermore, two such maps φ k,b and φ k,b are conjugate over K if and only if k = k ; they are conjugate over K if in addition b/b (K ) 2.

17 Arithmetic Normal Form for M 2 Theorem (M., Yasufuku) 3 If λ 1 = λ 2 = λ 3 = 2, then φ is conjugate over K to θ d,k (z) = kz2 2dz + dk z 2 2kz + d, with k K, d K, and k 2 d. All such maps are conjugate over K. Furthermore, θ d,k (z) and θ d,k (z) are conjugate over K if and only if ugly, but easily testable condition

18 φ Hom 1 d Aut(φ) is conjugate to one of the following: 1 Cyclic group of order n: C n = ζ n z. 2 Dihedral group of order 2n: D n = 3 Tetrahedral group: A 4 = 4 Octahedral group: S 4 = 5 Icosahedral group: A 5 = ζ n z, 1 z z, 1 ( z + 1 z, i z 1 iz, 1 z, i ( z + 1 z 1 ζ 5 z, 1 ( z, ζ5 + ζ 1 5 z ( ζ 5 + ζ 1 5. ). ). ) z + 1 ).

19 φ Hom 2 d 1 Diagonal Abelian Groups (Cyclic Group of order n): H = ζa n ζn b 0, gcd(a, n) = 1 or gcd(b, n) = Proposition Let r be the number of solutions to x 2 1 mod n. There are n + r/2 ϕ(n)/2 representations of C n of the form ζ n ζn a

20 φ Hom 2 d 2 Subgroups of the form ζ p a i b i, 0 c i d i where the lower right 2 2 matrices come from embedding the PGL 2 automorphism groups.

21 φ Hom 2 d 3 Subgroups that don t come from embedding PGL 2. (Lots of them.) , , ,

22 Higher Dimensions

23 Higher Dimensions This slide intentionally left blank.

24 Computing the Absolute Automorphism Group Algorithm (Faber, M., Viray) Input: a nonconstant rational function φ K (z), an Aut φ ( K )-invariant subset T = {τ 1,..., τ n } P 1 (E) with n 3. Output: the set Aut φ ( K )

25 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 PGL 2 ( K ) by solving the linear system s(τ 1 ) = τ i, s(τ 2 ) = τ j, s(τ 3 ) = τ k. if s φ = φ s: append s to L. return L.

26 Computing the Automorphism Group for a Given Map Proposition (Faber, M., Viray) Let K be a number field and let φ K (z) a rational function of degree d 2. Define S 0 to be the set of rational primes given by { S 0 = {2} p odd : p 1 2 } [K : Q] and p d(d 2 1), and let S be the (finite) set of places of K of bad reduction for φ along with the places that divide a prime in S 0. Then red v : Aut φ (K ) Aut φ (F v ) is a well-defined injective homomorphism for all places v outside S.

27 Realizing Maps with a Given Automorphism Group Given a finite subgroup Γ PGL 2, Doyle & McMullen give a way to construct all rational maps φ Rat d with Γ Aut(φ). 2 d n inv. hom. one-form 1 1 inv. hom. rational map Fdx + Gdy 1 1 φ = G F

28 Realizing Maps with a Given Automorphism Group It is enough to find all (relative) invariant homogeneous polynomials, i.e. for each γ Γ there is a character χ: γ F = F(γ x) = χ(γ)f( x).

29 Realizing Maps with a Given Automorphism Group It is enough to find all (relative) invariant homogeneous polynomials, i.e. for each γ Γ there is a character χ: γ F = F(γ x) = χ(γ)f( x). 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.

30 Two useful gadgets Molien Series: Given a finite group Γ (and character χ), outputs the power series dim ( K [ x] k) Γ t k. k=0 Reynolds Operator: Given a finite group Γ, (character χ), and all homogeneous monomials of a given degree, outputs all (relative) Γ-invariants of that degree.

31 Example Γ = C 4 = ( ) i 0 0 1

32 Example Molien Series: Γ = C 4 = ( ) i t 2 + t 4 + t 6 + 3t 8 + 3t t t t 16 + O(t 18 )

33 Example Molien Series: Γ = C 4 = ( ) i t 2 + t 4 + t 6 + 3t 8 + 3t t t t 16 + O(t 18 ) Invariants of degree 8: xy x 2 y 2 x 3 y 3 x 4 y 4 x 8 y 8

34 Example Molien Series: Γ = C 4 = ( ) i t 2 + t 4 + t 6 + 3t 8 + 3t t t t 16 + O(t 18 ) Invariants of degree 8: xy x 2 y 2 x 3 y 3 x 4 y 4 x 8 y 8 Some maps with Γ Aut(φ): φ 1 (z) = z z 3 φ 2 (z) = z9 + 9z z 8 1

35 Example Γ = ( ) (also cyclic of order 4).

36 Example Γ = ( ) Invariants of degree 8: (also cyclic of order 4). x 2 + y 2 x x 6 y 2 20x 4 y x 2 y 6 + y 8 x 4 + 2x 2 y 2 + y 4 x 8 4x 6 y x 4 y 4 4x 2 y 6 + y 8 x 6 + 3x 4 y 2 + 3x 2 y 4 + y 6 x 7 y 7x 5 y 3 + 7x 3 y 5 xy 7 Some maps with Γ Aut(φ): φ 1 (z) = z7 + 24z 6 + 3z 5 40z 4 + 3z z 2 + z + 8 8z 7 z z 5 3z 4 40z 3 3z z 1 φ 2 (z) = z ( 3z 6 39z z 2 13 ) 13z 6 73z z 2 3

37 Exact Automorphism Groups? Proposition (Hutz, M.) Let A 4 = z, 1 ( ) z + 1 z, i z 1 and S 4 = iz, 1 ( ) z + 1 z, i. z 1 If φ Q(z) satisfies A 4 Aut(φ), then in fact Aut(φ) = S 4.

38 Exact Automorphism Groups? Proposition (Hutz, M.) Let A 4 = z, 1 ( ) z + 1 z, i z 1 and S 4 = iz, 1 ( ) z + 1 z, i. z 1 If φ Q(z) satisfies A 4 Aut(φ), then in fact Aut(φ) = S 4. Question How to construct maps φ K (z) with Γ = Aut(φ) (or decide there are none)? How to construct maps φ K (z) with a subgroup of Aut(φ) conjugate to (or equal to) Γ?

39 Automorphisms and Twists K -equivalence classes Twist(φ/K ) = of maps ψ Hom N d (K ). such that ψ is K -equivalent to φ Twists give automorphisms of the map φ: f φf 1 = (f φf 1 ) σ = f σ φ(f 1 ) σ. φ = f 1 f σ φ(f σ ) 1 f f 1 f σ Aut(φ).

40 Uniform Bounds on Preperiodic Points for Twists Proposition (Levy, M., Thompson) Let φ Hom N d (K ). There is a uniform bound B φ such that for all ψ Twist(φ/K ), # PrePer(ψ, P N K ) B φ. Idea: The degree of the field of definition of the twisting map f is bounded by # Aut(φ). Apply Northcott.

41 Cohomology and Twists For an object X, twists give automorphisms: g σ : X σ(i 1 ) Y A twist gives a one-cocyle: i X. g : Gal( K /K ) Aut(X) σ i σ(i 1 )

42 Cohomology and Twists For an object X, twists give automorphisms: g σ : X σ(i 1 ) Y A twist gives a one-cocyle: i X. g : Gal( K /K ) Aut(X) σ i σ(i 1 ) Does every one-cocyle come from a twist?

43 Cohomology and Twists For an object X, twists give automorphisms: g σ : X σ(i 1 ) Y A twist gives a one-cocyle: i X. g : Gal( K /K ) Aut(X) σ i σ(i 1 ) Does every one-cocyle come from a twist? For algebraic varieties, yes. For morphisms, sometimes. { ξ H ( Twist(φ/K ) = 1 Gal( K /K ), Aut(φ) ) } : ξ becomes trivial in H ( 1 Gal( K /K )., PGL N+1

44 Describing Twists Question Given φ Rat d, can we write an explicit formula for all twists of φ?

45 Describing Twists Question Given φ Rat d, can we write an explicit formula for all twists of φ? Done for Rat 2 by Arithmetic Normal Form Theorem.

46 Describing Twists Question Given φ Rat d, can we write an explicit formula for all twists of φ? Done for Rat 2 by Arithmetic Normal Form Theorem. If Aut(φ) = {ζ n z}, then we have an isomorphism K /K n Twist(φ/K ) ( b φ z n ) b n. b

47 Describing Twists Question Given φ Rat d, can we write an explicit formula for all twists of φ? Done for Rat 2 by Arithmetic Normal Form Theorem. If Aut(φ) = {ζ n z}, then we have an isomorphism K /K n Twist(φ/K ) ( b φ z n ) b n. b More general presentation of C n? Other automorphism groups? Higher dimensions?