Computing the Field of Moduli of the KFT family

The computation of the field of moduli of a given closed Riemann surface is in general a very difficult task. In this note we consider the family of closed Riemann surfaces of genus three admitting the symmetric group in four letters as a group of conformal automorphisms and we provide the computations of the corresponding field of moduli. Subjclass [2010] : 30F10, 14H37, 14H45, 14Q05. ∗Partially supported by project Fondecyt 1110001 and UTFSM 12.13.01


Introduction
There is a natural one to one correspondence between birational isomorphism classes of non-singular irreducible projective complex algebraic curves and conformal classes of closed Riemann surfaces.If C 1 and C 2 are nonsingular complex irreducible projective algebraic curves, we denote by the symbol C 1 ∼ = C 2 to indicate that they are birationally equivalent (that is, the corresponding closed Riemann surfaces are conformally equivalent).We denote by Gal(C/Q) the group of field automorphisms of C and, if K is a subfield of C, then we denote by Gal(C/K) the subgroup of Gal(C/Q) formed by those elements acting as the identity on K.It is known that the fixed subfield of Gal(C/K) still K (since C is algebraically closed of characteristic zero).If P ∈ C[x 0 , ..., x n ] is any polynomial and if σ ∈ Gal(C/Q), then P σ ∈ C[x 0 , ..., x n ] will denote the polynomial obtained by applying σ to each of the coefficients of P .
Let C ⊂ P n be a non-singular irreducible projective complex algebraic curve, say defined by the homogeneous polynomials P 1 , ..., P r ∈ C[x 0 , ..., x n ].If σ ∈ Gal(C/Q), then the polynomials P σ 1 , ..., P σ r define a new non-singular irreducible projective complex algebraic curve C σ .The field of moduli of C, denoted by M(C), is the fixed subfield of the group G C = {σ ∈ Gal(C/Q) : In this way, if S is a closed Riemann surface and C is any non-singular irreducible projective algebraic curve defining S, then we may set G S = G C and define the field of moduli of S as M(S) = M(C).
A field of definition of a closed Riemann surface S is a subfield K of C for which it is possible to find a non-singular irreducible projective complex algebraic curve C, whose Riemann surface structure is conformally equivalent to S, defined by homogeneous polynomials with coefficients in K; it is said that S is definable over K.If the closed Riemann surface S is definable over K, say by the algebraic curve C, and σ ∈ Gal(C/K), then C σ = C; so it follows that M(S) < K, that is, every field of definition of S contains its field of moduli.By results of Koizumi [13] it is also known that M(S) is equal to the intersection of all the fields of definition of S and, by results of Hammer-Herrlich [8], S is always definable over a finite extension of its field of moduli.
If the genus of S is zero, then S is conformally equivalent to the Riemann sphere b C; so it can be defined over its field of moduli Q.If S has genus one, then it can be described by a curve of the form (Legendre normal form) E λ : y 2 = x(x−1)(x−λ), where λ ∈ C−{0, 1}.A direct consequence of the fact that any two conformal automorphisms of order two with fixed points (necessarily four fixed points) of E λ are conjugate in the group of conformal automorphisms, is that the field of moduli of E λ is Q(j(λ)), where j is the classical j-function.It is also known that E λ can be defined over Q(j(λ)) [17,Chapter III,Prop. 1.4].
Let us assume, from now on, that S has genus at least two and let Aut(S) be its full group of conformal automorphisms.It is a well known fact that |Aut(S)| ≤ 84(g − 1) (Hurwitz's bound) [11].In this case, M(S) is not in general a field of definition of S, as it is shown by explicit examples provided by Earle [4] and Shimura [16] in the case of hyperelliptic case and by the author [9] in the non-hyperelliptic case.Sufficient conditions for S to be definable over its field of moduli are given by Weil's Galois descent theorem [19].If Aut(S) is trivial, then (as a direct consequence of Weil's Galois descent theorem) S can be defined over its field of moduli.Unfortunately, Weil's conditions are in general very difficult to check if Aut(S) is non-trivial.But, in the particular case that S/Aut(S) has signature of the form (0; a, b, c) (one says that S is quasiplatonic), Wolfart [21] proved that S can be defined over its field of moduli (which is a number field by Belyi's theorem [1]).The computation of the field of moduli of S is in general a difficult task.Moreover, if we have computed explicitly the field of moduli, to determine if S can be defined over it is also a difficult problem (except for some simple cases).Even, if we already have explicitly the field of moduli and we know that the surface can be defined over it, it is a very hard problem to compute an algebraic curve defined over it that represents the surface.
In this paper we work out the family of closed Riemann surfaces of genus three admitting the symmetric group in four letters S 4 as a group of conformal automorphisms.It is well known that, up to conformal equivalence, there is only one such hyperelliptic surface; which is described by the hyperelliptic curve C : y 2 = x 8 + 14x 4 + 1; so it is already defined over its field of moduli Q.In the non-hyperelliptic case, there are exactly two conformal classes for which the full group of conformal automorphisms is bigger than S 4 : Fermat's curve F : x 4 + y 4 + z 4 = 0 and Klein's curve K : x 3 y + y 3 z + z 3 x = 0.It is well known that |Aut(F )| = 96 and that |Aut(K)| = 168 and that F/Aut(F ) has signature (0; 2, 4, 8) and that K/Aut(K) has signature (0; 2, 3, 7).Again, these quasiplatonic surfaces are defined over their field of moduli Q.In [15] there is provided a pencil of non-singular quartic curves C λ , where λ ∈ P = C − {−2, −1, 2}, called the KFT family.Each non-hyperelliptic Riemann surface of genus three admitting S 4 as a group of conformal automorphisms is represented by a quartic in the KFT family and, conversely, every member of the KFT family is of such type of Riemann surfaces.For three especial values of λ ∈ P the quartics correspond to Klein's quartic and Fermat's quartic.For the rest of the parameters, they correspond to closed Riemann surfaces of genus three with full group of automorphisms isomorphic to S 4 ; the quotient orbifold of signature (0; 2, 2, 2, 3).In Section 2 we will compute the field of moduli of each memeber of the KFT family and will notice that the quartics provided in the KFT family, with the exception of Klein's quartic, are already defined over them.
Recently, we have noticed the paper [6] on which the KFT family (and also other families of genus three curves) has been considered and their results may also be used to compute the corresponding fields of moduli.

The field of moduli of the KFT Family
In this section we consider the family of closed Riemann surfaces of genus three admitting the symmetric group S 4 as a group of conformal automorphisms and we provide the field of moduli of these surfaces and explicit equations in these fields.

The hyperelliptic case
As the hyperelliptic involution is in the center of the group of conformal automorphisms of a hyperelliptic Riemann surface, it is not difficult to see that there is exactly one, up to biholomorphisms, hyperelliptic Riemann surface S 0 of genus 3 with a group of conformal automorphisms isomorphic to S 4 .If we quotient S 0 by the hyperelliptic involution, then we obtain that the 8 cone points of order 2 should be invariant under the action of a group of Möbius transformations isomorphic to S 4 .This permits to see that Aut(S 0 ) = S 4 ⊕ Z/2Z and that S 0 /Aut(S 0 ) has signature (0; 2, 4, 6).Using the above information, one can see that S 0 can be represented by the algebraic curve C : y 2 = x 8 + 14x 4 + 1, that is, S 0 can be defined over Q.

The non-hyperelliptic case
A well known fact is the topological rigidity property on the action of the group S 4 as group of conformal automorphisms of closed non-hyperelliptic Riemann surfaces of genus g = 3.
Theorem 1 (Broughton 2).If (S, H) and (R, K) are so that S and R are non-hyperelliptic Riemann surfaces of genus 3 and H ∼ = K ∼ = S 4 are respective group of conformal automorphisms, then there is an orientation preserving homeomorphism between the surfaces conjugating the groups.
In Section 3 we provide a simple proof, based on Fuchsian groups, of Theorem 1 as a matter of completeness.
Remark 2. The hyperelliptic Riemann surface C : y 2 = x 8 + 14x 4 + 1 admits two different subgroups H 1 and H 2 inside Aut(C) with H j ∼ = S 4 so that C/H 1 has signature (0; 2, 4, 6) and C/H 2 has signature (0; 2, 2, 2, 3).If (S, H) is so that S is non-hyperelliptic Riemann surface and S 4 ∼ = H < Aut(S), then there is a an orientation preserving homeomorphism f : S → C so that H 2 = fHf −1 .A description of these Riemann surfaces, from the point of view of Schottky uniformizations, can be found in [10].
Let S be a non-hyperelliptic closed Riemann surface of genus g = 3 and let S 4 ∼ = H < Aut(S).As a consequence of the Riemann-Hurwitz formula [5], the orbifold S/H has signature (0; 2, 2, 2, 3).It follows that the locus, in the moduli space of genus three Riemann surfaces, of the classes of Riemann surfaces admitting the non-hyperelliptic action of S 4 is one-complex dimensional.
As a consequence of Singerman's list [18] of maximal Fuchsian groups, one has that either Aut(S) = H or S/Aut(S) has signature of the form (0; a, b, c).
If S/Aut(S) has signature of the form (0; a, b, c), then S is quasiplatonic and so it can be defined over its field of moduli [21].The set of these quasiplatonic surfaces form a finite subset up to conformal equivalence.Apart from the hyperelliptic case, there are only other two such Riemann surfaces; Fermat's curve F : x 4 + y 4 + z 4 = 0 and Klein's curve K : x 3 y + y 3 z + z 3 x = 0.It is well known that |Aut(F )| = 96 and that |Aut(K)| = 168 and that F/Aut(F ) has signature (0; 2, 4, 8) and that K/Aut(K) has signature (0; 2, 3, 7).All of these quasiplatonic surfaces are defined over their field of moduli, that is, over Q.If Aut(S) = H, the generic situation, then we will see that S can be defined over its field of moduli and we will in fact compute it.

The KFT family
It is well known that the canonical embedding of a non-hyperelliptic Riemann surface of genus 3 is a non-singular projective algebraic curves of degree 4 (a quartic) in the complex projective plane P 2 .A description of such quartics for the family of non-hyperelliptic Riemann surfaces of genus 3 admitting S 4 as group of conformal automorphisms has been done in [15]; called the KFT family.A study of such a family from the point of view of idempotents has been done in [7].This family has been studied in [3,14,20].The quartics in the KFT family are of the form [15] where λ ∈ P = C − {−2, −1, 2}.The curves C λ , where λ ∈ {±2, −1} are singular quartics.
The group H ∼ = S 4 , for each member of of the KFT family, is generated by the transformations As a consequence of Theorem 1, every non-hyperelliptic Riemann surface of genus 3 admitting a group of conformal automorphisms isomorphic to S 4 is represented by one of the curves in the KFT family.Conversely, every curve C λ , with λ ∈ P, is a closed Riemann surface of genus 3 admitting the group H as a group of conformal automorphisms.As for each σ ∈ Gal(C/Q) one has that C σ λ = C σ(λ) , it follows that the orbits under the action of Gal(C/Q) of such a family are given as: 1. the orbit of C π (containing exactly all curves of the form C λ , where λ is transcendental); and 2. the orbits of the curves C λ 1 ,. . ., C λ n ,. . ., where λ 1 , . . ., λ n , . . .∈ Q is a maximal collection of algebraic numbers non-equivalent under the absolute Galois group Gal(Q/Q)

Equivalence of curves
In order to find explicitly the field of moduli M(C λ ), we first need to provide conditions on λ 1 and λ 2 for C λ 1 and C λ 2 to be conformally equivalent.
As λ ∈ P, we have that λ 2 − λ − 2 6 = 0.In this way, for each simplyconnected subset D of P we may choose one of the branchs of √ λ 2 − λ − 2 to get an analytic map f (λ) = √ λ 2 − λ − 2 defined over D. Let us fix λ ∈ P and let us consider the map As the polynomials x 2 y 2 z 2 and x 2 + y 2 + z 2 are invariant under A and B, we obtain that C is a regular branched cover with H as Deck group of cover transformations.
If we fix one of the two values of √ λ 2 − λ − 2, then the branch values of Q are given by the points ∞ (of order 3), 0 (of order 2) and the following two points (each of order 2)

´3
Notice that the branch value 0 is the projection under Q of the fixed points of the conjugates of A 2 and that the branch values l 1 (λ) and l 2 (λ) are the projections of the fixed points of those elements of order two conjugate to B.
The map π λ = T λ • Q defines a regular branched cover with H as Deck group of cover transformations whose branch values of order 2 are ∞, 0 and 1 and the one of order 3 is Remark 4. If we change to the other value of √ λ 2 − λ − 2 the roles of l 1 (λ) and l 2 (λ) get interchanged.In this case, the Möbius transformation that fixes 0 and sends l 1 (λ) to ∞ and l 2 (λ)) to 1 is given by b and the branch values of the cover map b C are ∞, 0 and 1 (of order 2) and the one of order 3 is η(µ(λ)) = µ(λ)/(µ(λ) − 1).In this way, to each λ ∈ P we have associated the two values µ(λ) and η(µ(λ)) depending on the choice of √ λ 2 − λ − 2. Let us also notice that is well defined over all P.
Let P 0 be the subset of P consisting of those values for which Aut(C λ ) 6 = H.Then P 0 is a set of isolated points.Notice that if 1) and ( 3) of the next result was also obtained, by a different method, in [6].
are equivalent to Klein's curve and the curve C 0 is Fermat's curve.

Proof.
Given any two points λ 1 , λ 2 ∈ P, we may consider a simply connected domain D ⊂ P containing the points λ 1 and λ 2 .Once this is done, we make a choice for a analytic branch of √ λ 2 − λ − 2 in D. Using such a choice, we have fixed the choices of π λ (z) and of µ(λ) for λ ∈ D (both are analytic on the parameter λ ∈ D).
In this way, we obtain that We assume now on that λ 1 6 = −5/2.
Let us consider the Riemann orbifolds O j = C λ j /hABi which has signature (1; 3, 3).It was obtained in [15] that the j-invariant of the underlying Riemann surface structure T j of O j is Similarly, we may consider the orbifolds (all of them of genus one) obtained by quotient C λ j by the cyclic groups hAi, hBi and hA 2 i.The corresponding j-invariantes are Next we make a comparison of j 2 , j 3 , j 4 , j 2,2 for the three above possible values for λ 2 and those for λ 1 (this can be done with any computational software) and we obtain that the only possibility is λ 2 = λ 1 (as λ 1 6 = −5/2) Proof.If λ ∈ P 0 , then C λ is either Klein's curve or Fermat's curve, both of them can be defined over Q.

Fuchsian uniformization of the KFT family
For each λ 1 ∈ P there are three values for λ ∈ P so that G(λ) = G(λ 1 ); one of them being clearly λ 1 and the others two are given by We have that in P there is no solution for λ 1 = λ 3 or for λ 2 = λ 3 and there is exactly one solution for λ 1 = λ 2 , this being for λ 1 = −5/2.Notice that in this case λ 3 = 2 / ∈ P corresponds to the hyperelliptic curve admitting S 4 as a group of conformal automorphisms.
The curves C λ 1 , C λ 2 and C λ 3 can also be seen as follows from a Fuchsian uniformization's point of view.Let us consider the orbifold of signature (0; 2, 2, 2, 3) whose cone points of order 2 are given by ∞, 0, 1 and the cone point of order 3 is µ(λ 1 ) = µ 1 (once we have fixed a value for √ λ 2 − λ − 2).Let us consider a Fuchsian group acting on the unit disc D and a universal branched cover P : D → b C with Γ as Deck group of covering transformations so that the fixed point of x 3 projects by P to 0, the fixed point of x 1 projects to ∞ and the fixed point of x 2 projects to 1.The fixed point of x 1 x 2 x 3 projects to µ 1 .
As a consequence of results due to L. Keen [12] there is a fundamental domain for Γ given by a suitable hyperbolic triangle ∆ 1 , say with sides s 11 , s 12 and s 13 counted in counterclockwise order, so that x j is an involution with fixed point at the middle side of s 1j .
In order to find the torsion free normal subgroups F of Γ so that Γ/F ∼ = S 4 , up to inner conjugation in Γ, we only need to find all possible different surjective homomorphisms Θ : Γ → S 4 with torsion free kernel up to postcomposition with automorphisms of S 4 .
All of the above also provides a proof of Theorem 1.
Remark 8. Let us denote the internal angles of the triangle ∆ 1 by θ 1 , θ 2 and θ 3 , so that θ 1 is the angle between the sides s 11 and s 12 , θ 2 is the angle between s 12 and s 13 and θ 3 is the angle between s 13 and s 11 .Clearly, θ 1 + θ 2 + θ 3 = 2π/3.In the particular case when θ 1 /2 = θ 2 = θ 3 = π/6, there is a conformal automorphism U : D → D of order 4 with the same fixed points as for x 3 (so As a consequence, this is the case corresponding to G(λ) = 4.The curve C −5/2 is uniformized by any of the two Fuchsian groups appearing in (2) and ( 3) and a hyperelliptic one is uniformized by the one appearing in (1).