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Twists of the Klein curve and of the Fricke-


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Twists of the Klein curve and of the Fricke-

Macbeath curve

Master’s Thesis Mathematics August 2015

Student: C.A. Verschoor

First Supervisor: Prof.dr. J.Top

Second Supervisor: dr. A.V. Kiselev



1 The Klein Curve 4

1.1 The Fermat Curve . . . 4

1.2 Twists of curves . . . 6

1.3 Twists of the Klein Curve . . . 6

2 The Fricke Macbeath Curve 9 2.1 A model over Q . . . 9

2.2 Counting points on M . . . 11

2.3 Twists of M . . . 18

2.3.1 Case q = pn ≡ ±1 (mod 7) . . . 18

2.3.2 Case q = pn 6≡ ±1 (mod 7) . . . 20

3 Conclusions 22 Appendices 23 Appendix A Algorithms 23 A.1 Klein Curve . . . 23

A.2 Fricke Macbeath Curve . . . 25

A.2.1 Model . . . 25

A.2.2 Splitting of M . . . 26

A.2.3 Twists . . . 32



In this thesis we study properties of the Klein curve and the Fricke Macbeath curve. These curves of genus 3 respectively 7, that are the unique curves of genus g ≤ 7 satisfying the property that the automorphism group has cardinality 84(g − 1). Of course over a finite field, this bound is not necessarily maintained.

The first such encounter is when studying the Klein Curve over a finite field of characteristic 3. In this special case we will therefore try to construct twists of the Klein curve and ultimately try to count points on them. Continuing in this fashion we are given a model for the Fricke Macbeath curve over the rationals.

Using this we can investigate how we can count points on the Fricke Macbeath curve over finite fields. And finally we try to construct models of twists for the Fricke Macbeath curve over these fields and count points on those. The reason we want to count points on the Fricke Macbeath curve and its twists in particular is due to the amount of automorphisms on these curves. This extends work of Iwan Duursma [3] and of Top and Meagher [7] (all only discussing the Klein Curve). While writing this thesis, Hidalgo [5] published an article where he writes that Bradley Brock obtained a singular model for the Fricke Macbeath curve given by a single equation.


1 The Klein Curve

The Klein Curve K is the unique curve of genus 3 that is a Hurwitz curve, which means that its automorphism group contains 84 · (g − 1) = 168 elements over Q. The best known model of the Klein curve is given by¯

K = Z X3Y + Y3Z + Z3X From [4] we know that AutQ¯(K) ∼= PSL(2, 7).

Now instead of looking over characteristics 0, we are more interested in finite fields of characteristic 3, since then the cardinality of the automorphism group of K is greater than 168. Our job is to count the points on K and its twists over finite fields of characteristic 3. So in this section we denote K = ¯F3.

Over this characteristic we see that [4] and [9] show that the Klein curve and Fermat curve are isomorphic over F32, hence they are twists. So constructing twists of the Klein curve, is equivalent to constructing twists of the Fermat curve.

1.1 The Fermat Curve

A model for the Fermat curve, that we will use in this case is:

F4= Z X4+ Y4+ Z4

This curve has special properties over K, which can be seen from the fact that for any a ∈ F9 we have

a4= a3a = NormF9/F3(a)

Using this observation and noting we have 3 variables X, Y and Z, we define a bilinear form h • , • i : ¯F33× ¯F33→ ¯F3 given by


 a1




 b1



 +






Denote v ·w..= hv, wi for any v, w ∈ ¯F33. This dot product shares properties with the ordinary dot product. It is not hard to check that it is symmetric, bilinear and non-degenerate. Now let ~fr : ¯F33→ ¯F33 be the morphism on the vector space K3induced by the Frobenius endomorphism fr : x 7→ x3. It acts on a vector by applying fr to each coefficient, so


 a1




 a31 a32 a33

Then we can write the defining polynomial of F4 in the following way:

X4+ Y4+ Z4=

 X Y Z

· ~fr

 X Y Z

Now to find the automorphism group of F4 over K, we highlight that an automorphism A ∈ AutK(F4) can be represented by a matrix A ∈ PGL3(k) ∼=


PSL3(k), for k a finite extension of F3, such that when we apply A to any point on our curve then it remains on the curve itself. In other words there exists λ ∈ K such that


 X Y Z

· ~frA

 X Y Z

= λ

 X Y Z

· ~fr

 X Y Z

Letting v =

 X Y Z

, then by the properties of ~fr we have ~frAv =frA~frv, where we denote frA as applying Frobenius to all coefficients of the matrix A. Now using the linearity of the dot product (frA)|Av · ~frv = λv · ~frv. But this means that we want to have (frA)|A = λI where I ∈ PSL3(K) the 3×3 identity matrix.

Or written differently (frA)| = λA−1. Taking determinants on both sides the LHS gives det((frA)|) = (det A)3= 1 because of application of Frobenius. On the RHS we get det(λA−1) = λ3(det A)−1 = λ3. So combining them we see 1 = λ3 and hence we must have λ = 1 since λ ∈ ¯F3.

Now before continuing we highlight the fact that over PSL3(K) we only have matrices of determinant 1 and so taking the inverse is nothing more than taking the adjugate of the matrix and this adjugate commutes with taking Frobenius. This fact follows from the definition of the adjugate matrix and using the commutativity the diagram:

Mat2(¯F3)det //




Mat2(¯F3)det // ¯F3

The adjugate is the matrix that contains determinants of minors of your starting matrix and on some position’s there are minus signs. But this means that taking inverse and taking Frobenius are commuting operations. Now using the relationship (frA)|= A−1 we obtain that:

fr2A =fr((A−1)|)

= ((frA)−1)|

= (((A−1)|)−1)|

= A

This means fr2 should act trivially on the components of A, whence it means A ∈ PSL3(F9) and fr can now be seen as conjugation and hence we define the Hermitian of A to be A† ..= (frA)|. So combining this we get A ∈ {M ∈ PSL(F9) : MM = I} ∼= PSU3(3) = U3(3). Here U3(3) is the group of all 3 × 3 matrices over F32 with unit determinant such that AA = I. Note that this definition differs from the definition given in software such as Sage, Magma, or gap, see [8] for more information. The two definitions give different matri- ces, although there is a correspondence between them. To still work with our definition of U3(3) we use generators described in [1].

Hence we get AutK(F4) ∼= PSU3(3), using this automorphism group we can calculate the twists of F4 and therefore also the twists of K indirectly.


Furthermore over F9, F4 and K are isomorphic, so AutK(K) ∼= PSU3(3), which is a group of size 6048.

1.2 Twists of curves

To use these automorphisms to get twists of curves, we will need to introduce some theory. By [7] we have that for a smooth projective curve C over a field k we have that

Theorem 1.1. there exists a bijection

θ : Twist(C) → H1(Gk, Autk¯(C)) where we denote Gk = Gal(¯k/k).

In our case we have k = Fq and K = ¯k where q a power of 3. The bijection θ takes a twist ψ : C ⊗ ¯k → C0⊗ ¯k and sends it to the cocycle fψ : σ 7→ ψ−1σψ.

The map that goes back (see [7] for more details) is formed on the function field of the curve. Given a cocycle f ∈ H1(Gk, Aut¯k(C)) it defines an action on k(C) by the rule¯

x 7→ σ ◦ x ◦ (f (σ−1) ⊗ σ−1)

for all σ ∈ Gk. The invariants of ¯k(C) under this action give a new function field k(C0).

To calculate twists of C is therefore in essence equal to calculating the set H1(Gk, Autk¯(C)), but this process is still a bit heavy. To calculate it quickly we need Frobenius conjugation classes.

Definition 1.1. Two elements g, h ∈ Autk¯(C) are called Frobenius conjugate if there is an element x ∈ Aut¯k(C) such that xg = h(frx), where fr ∈ Gk denotes the Frobenius endomorphism.

Frobenius conjugacy defines an equivalence relation and therefore we call equivalence classes of this equivalence relation a Frobenius conjugacy classes.

A theorem proven in [7], helps us to calculate twists more easily Theorem 1.2. There is a bijection

H1(Gk, Aut¯k(C)) → {Frobenius Conjugation Classes of Aut¯k(C)}

The bijection is given by the map sending a Frobenius conjugacy class [α]

to f := fr 7→ α, and the other way around we have that a cocyle f maps to to a Frobenius conjugacy class [f (fr)]. It is easy to check by the rules of cocycles that this indeed gives a bijection.

1.3 Twists of the Klein Curve

In this section we calculate twists of the Klein curve over finite fields of charac- teristic 3. More specifically we want to count points on these twists. So consider K and F4over a field k = F3m, for some m > 0.

A first objective is to calculate the Frobenius conjugacy classes of Aut¯F3m(F4) for all m. To do this we first note that Aut¯F3m(F4) = U3(3) which consists of ma- trices with coefficients in F9. Hence the Frobenius endomorphism fr : x 7→ x3m


acts trivially on U3(3) when m ≡ 0 (mod 2) and it is an involution when m ≡ 1 (mod 2). So we only need to calculate Frobenius conjugacy classes for m = 1 and m = 2.

The case m = 2 is the case when the Frobenius conjugacy classes are equal to the conjugacy classes, because Frobenius acts trivial in this case.

The case m = 1 can be calculated naively. That is, we pick an element g ∈ U3(3) and for all elements x ∈ U3(3) we calculate xg ·frx−1, this yields one Frobenius conjugation class. Then make a loop where we keep picking an ele- ment g ∈ U3(3) that is not in any previously found Frobenius conjugacy class, and do the same thing. The loop halts when each element is in a Frobenius conjugacy class.

Given a Frobenius conjugacy class represented by the automorphism A we want to find the corresponding twist. A very naive way which works for twists over small extensions is finding the twist ψ that corresponds to A = ψ ◦frψ, by letting ψ be a 3 × 3 matrix containing 9 variables ψij, where we let the determinant of ψ be 1. These conditions can be written as an ideal I in ¯F311, . . . , ψ33]. We know over which extension this twist is isomorphic to the original curve, namely take the least n > 0 such that A ◦frA ◦fr2A ◦ . . . ◦frnA = id. Then ψ should become an isomorphism over F3nm, hence to find a twist we may assume I is an ideal in F3mn11, . . . , ψ33].

Now to find the twist map ψ : F4 → C defined over F3m, we need to find one solution in Z(I) and embed ψ inside PSL3(F3mn) such that it becomes an isomorphism over F4⊗ F3n. Finding this solution in Z(I) can be a pain when n is really large, therefore in the section about the Fricke Macbeath curve we will give a way of calculating twists using Linear algebra, that can be applied to the Klein curve as well.

When a twist C is found we are interested in counting the points on the twist. This can also be done naively, but in general there are better ways to do this. Let I(C) be the ideal corresponding to C, then we could calculate it’s Gr¨obner basis and then perform triangularization to find the points on the curve.

Since we are only interested in points on the curve over F3m and because using naive methods can be very time consuming, we will give an alternative to counting points on C.

Theorem 1.3. Let C be a curve defined over a finite field k with Frobenius fr and let A ∈ Aut¯k(C) and C0 be the twist over k corresponding to the Frobenius conjugacy class [A]. Then

#C0 = #{Q ∈ C(¯k) :frQ = A−1Q}

Proof. Let P ∈ C0 then frP = P , because fr works trivial over k. Let ψ : C ⊗ ¯k → C0⊗ ¯k be the twist map corresponding to A. Then

ψ−1(frP ) = ψ−1(P )


Now A = ψ−1frψ so

ψ−1(P ) = ψ−1(frP )

= (A ◦ (frψ)−1)(frP )

= (A ◦fr−1))(frP )

= A(fr−1(P )))

So take Q = ψ−1(P ) then we have that A−1Q =frQ like we wanted.

Instead of having 9 variables as we did when using the naive method of finding twists, in this case we don’t need the twist and the only variables come from Q. The conditions Q ∈ F4(¯k) with frQ = A−1Q can be written as an ideal I in ¯F3[Q1, Q2, Q3]. Now noting that the twist lives over a finite extension of F3m, we can project this ideal I into F3nm[Q1, Q2, Q3] and calculate #Z(I).

Note that F4 is projective and I is affine, so we don’t want to count the point (0, 0, 0), and we want F3nm-multiples of any affine point to be equivalent. Hence

#C0 = (#Z(I) − 1)/3nm, as just a naive way to use the theorem.

As a test for this theorem, we counted the number of points over fields F3and F32. First we count points on the twists corresponding to Frobenius conjugacy classes C defined over F3, where ω is a zero of x2−x−1. A table is shown below, a similar table can be generated by the implementation found in appendix A.1, where generators α and β are given in [1]. The representative shown has the smallest factorization into α and β of the class.

C [I] [α] [β] β−1



# 4 4 13 1 4 4

For F32 = F3[ω] satisfying ω2− ω − 1 = 0 we made a table of numbers of points on the twist of F4 corresponding to a conjugacy class C

C [I] [α] [β] αβ−1

[βα] αβ3


# 28 4 13 7 7 16 1

C β−2α







# 13 13 10 10 10 4 4

A faster method to count points on curves is needed in the case of the Fricke Macbeath curve, this will be done by using linear algebra. This method can also be applied to find numbers of points on the Klein curve K and the Fermat curve F4, which is left as an exercise to the reader.


2 The Fricke Macbeath Curve

The Fricke Macbeath curve is the only Hurwitz Curve of genus g = 7, e.g. the unique curve M of genus 7 whose automorphism group satisfies the Hurwitz bound, i.e. # AutQ¯(M) = 84 · (g − 1) = 504. More specifically AutQ¯(M) ∼= PSL(2, 8). A model for this curve was found in [6] and we use a transformation of this model to analyze the Fricke Macbeath curve in more detail.

2.1 A model over Q

In [6] a model for the Fricke Macbeath curve was constructed over Q(ζ) where ζ is a seventh root of unity. It has the property that all of its automorphisms are defined over Q. The projective model was generated by ten homogeneous equations over Q(ζ)[y1, . . . , y7] and where we take αj:= ζj+ ζ−j:

Xy2i =X

ζiy2i =X

ζ−iyi2= 0 α4y1y7+ α2y2y5+ α1y4y6= 0 α4y2y1+ α2y3y6+ α1y5y7= 0 α4y3y2+ α2y4y7+ α1y6y1= 0 α4y4y3+ α2y5y1+ α1y7y2= 0 α4y5y4+ α2y6y2+ α1y1y3= 0 α4y6y5+ α2y7y3+ α1y2y4= 0 α4y7y6+ α2y1y4+ α1y3y5= 0

The automorphism group of this model can be generated by the two matrices

T = 1 2

−1 0 0 1 1 −1 0

0 0 −1 −1 1 0 −1

0 1 −1 1 0 1 0

−1 −1 −1 0 −1 0 0

−1 1 0 −1 0 0 1

1 0 −1 0 0 −1 1

0 1 0 0 −1 −1 −1

, W =

0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0

Note that T = ((V U )−1W )2W , where U, V, W are the generator matrices from [6]. Looking at these matrices U, V, W we see that also U|, V|, W|are in the automorphism group. Hence in this case taking transposed W or T doesn’t matter yet. This will however matter when we convert it to the new model.

This new model over Q was given to us by Maxim Hendriks and can also be found at [5]. It is defined by the following homogeneous equations over


Z[x1, . . . , x7]:

− x1x2+ x3x4− x3x5+ x2x6− x4x6− x5x6+ x1x7− x3x7,

− x22+ x1x3− x23+ 2x2x5+ x4x5− x25+ x1x6+ x2x7− x4x7, x21+ x22− x1x3− x23− x2x4− x2x5+ x3x6− x2x7+ 2x5x7− x27, x1x4− x3x4− 2x1x5− x3x5− x2x6+ x5x6+ 2x1x7+ x6x7,

x21− x22− 2x1x3+ x23− x2x4− x2x5+ x4x5+ x25+ x3x6− x26+ 2x2x7− x5x7, x1x2+ 2x2x3− x1x5− x5x6− 2x1x7− x3x7+ 2x6x7,

− 2x1x2+ 2x2x3− x1x4− x1x5+ 2x5x6+ 2x1x7− 2x3x7− x6x7, 2x21+ x1x3+ x4x5− x25− x1x6+ x26+ 3x2x7− x4x7− x27,

2x21+ x22− x1x3+ x23+ x4x5+ x25+ x1x6− 2x3x6+ x26+ x2x7− x4x7− 2x5x7+ x27, x21+ x1x3+ x24+ 2x2x5+ x4x5− x1x6+ 2x3x6+ x26− 3x2x7− x4x7+ 3x27

Its automorphism group is generated by R and S defined over PSL7(Q(α)) where α = ζ + ζ−1.

R = 1 14

2α2 + 3α − 7 3α2 + 4α + 1 4α2 + 2α − 6 −α2 + α + 2 −4α2 − 3α + 1 −7α2 − 5α + 10 3α2 − 3α − 6 4α2 + 8α − 4 4α2 + α − 11 −9α2 − 3α + 14 2α2 + α − 3 −5α2 − 5α + 2 −2α2 − α + 3 α2 + 6α + 5 2α2 + 4α − 2 6α2 + 3α − 9 α2 + α − 6 −2α2 − α + 3 −α2 + 3α − 2 −2α2 − 7α + 1 −α2 − 4α + 5

14α2 + 7α − 21 7α + 7 −7α2 + 7α + 14 0 7α2 − 7 0 −7α2 − 7α + 14

6α2 + 9α − 7 5α2 − α − 4 −11α2 − 7α + 16 −α2 − 2α + 1 −3α2 − 4α − 1 α2 + 2α − 1 5α − 3 6α2 − 10 3α2 − 5 4α2 + 8α − 4 −α2 − 3 −4α2 + 2 −3α2 − 4α − 1 3α2 − 5 4α2 + 11α − 3 5α2 − 13 −8α2 − 4α + 12 −α2 + α + 2 −6α2 − 5α + 13 α2 − α − 2 −α2 + 3α − 2

S =1 7

−α2 + 4 −α − 5 2α2 + 2α − 5 0 −α + 2 −2α2 − α + 3 −α + 2

3α + 1 −α2 − 2α + 1 −2α2 − 3α α − 2 −3α2 − α + 7 α2 + α + 1 3α2 + 3α − 4

2α + 3 α2 − 4 2α2 + α − 3 0 α2 + 3 −α2 − 2α + 1 α2 − 4

−7α2 + 14 0 −7α 0 0 −7 0

−α2 + 2α −3α − 1 −2α2 − α + 3 α2 − 4 −α2 + α + 2 −α + 2 2α2 + 3α 2α2 + 3α 5α2 + 2α − 10 −α2 + α + 2 0 −2α2 − 5α + 4 −α2 − α − 1 −2α2 + 2α + 4 α2 + 2α − 1 α2 − 2α −3α2 − 3α + 4 −α2 − α − 1 −3α2 + 5 −α2 + 4 2α2 + α − 3

Note that these matrices are transposed, such that the implementation in computer algebra packages is better. Using these packages it is easy to check that these are automorphisms of the model for the Fricke-Macbeath curve given, and that together they generate the group PSL(2, 8).

Now one could be interested in the isomorphism that transforms the model Mold as given in [6] to a model Mnew as given by Maxim Hendriks. Let φ : Mold → Mnew be such a map, then it induces an isomorphism Cφ : Aut(Mold) → Aut(Mnew) where A 7→ φAφ−1. So we can try to map the generators T and W to the generators R and S in this way. By checking orders we see that under Cφ we need to map T 7→ R and W 7→ S. Hence we need to find φ such that R = φT φ−1 and S = φW φ−1. Then we can rewrite this to the two Sylvester equations Rφ = φT and Sφ = φW . The general technique to solve these equations is through vectorization of φ.

Definition 2.1. Let

M =

a11 a12 · · · a1n a21 a22 · · · a2n ... ... . .. ... am1 am2 · · · amn

 be a m × n matrix, then the vectorization of M is defined as

Vec(M ) := (a11, a21, . . . , am1, a12, a22, . . . , am2, . . . , a1n, a2n, . . . , amn)|


When M is seen as a k-vector for some field k this means that basically M ∈ km⊗ kn. Now Vec gives us the isomorphism km⊗ kn∼= kmn. The inverse of Vec is denoted by Matm×n and turns the vector back into its original form.

When there is no confusion on what m and n are, we just write Mat := Matm,n. In the case of finding φ we take m = n = 7.

Now for solving an equation of the form Aφ = φB for some fixed 7 × 7 matrices A and B. We note that A works on the codomain of φ and B works on the domain of φ. Hence if we vectorize φ, the equation in its vectorized form becomes (id ⊗A) Vec(φ) = (B|⊗ id) Vec(φ). Hence Vec(φ) ∈ ker(id ⊗A − B|⊗ id).

In our case where Rφ = φT and Sφ = φW , we have

Vec(φ) ∈ ker(id ⊗R − T|⊗ id) ∩ ker(id ⊗S − W|⊗ id)

Then picking a non-zero Vec(φ) such that Mat(Vec(φ)) = φ has non-zero determinant from this should give us the required transformation φ. In our case by intersecting the kernels we find a vector space of dimension 1, which gives a possible φ as:

φ = 1 7

0 0 α2− α − 2 −α2− α − 1 0 2α2− 1 0

α2− 2α 0 0 0 3α + 1 0 −α2− 2α + 1

0 0 −α2− α − 1 −2α2+ 1 0 α2− α − 2 0

0 7α 0 0 0 0 0

−α2− 2α + 1 0 0 0 −α2+ 2α 0 −3α − 1

0 0 −3α − 1 −α2+ 2α 0 α2+ 2α − 1 0

−3α − 1 0 0 0 α2+ 2α − 1 0 α2− 2α

 Something interesting that we can say about this φ, is that after permutation

of some variables this becomes a block matrix φ0 looking like

φ0= 1 7

7α 0 0 0 0 0 0

0 α2− α − 2 −α2− α − 1 2α2− 1 0 0 0

0 −α2− α − 1 −2α2+ 1 α2− α − 2 0 0 0

0 −3α − 1 −α2+ 2α α2+ 2α − 1 0 0 0

0 0 0 0 α2− 2α 3α + 1 −α2− 2α + 1

0 0 0 0 −α2− 2α + 1 −α2+ 2α −3α − 1

0 0 0 0 −3α − 1 α2+ 2α − 1 α2− 2α

 So this shows that the transformation was constructed by taking one variable

a multiple of α. Then two sets of 3 variables are each independently transformed into each other by a linear transformation. From hereon we use the new model

2.2 Counting points on M

A naive way to count points is by brute forcing, or using techniques with cal- culating groebner bases and triangularizations. But in this chapter we try to accelerate this by splitting an abelian variety isogenous to Jac(M) defined over a finite field K, and then counting points on M using the trace of frobenius calculated from the factors of Jac(M). We can test how M behaves over certain finite fields, and we immediately see that over characteristics 2 and 7 the curve becomes singular. So we want to avoid those cases, we will later see that these are the cases where Jac(M) has bad reduction.


To split Jac(M), we find quotient curves of M. Before we move to finite fields we first take K = Q. The easiest ways to make quotient curves is to look for a low order automorphism inside AutK¯(M) defined over the ground field K.

The one we consider is the automorphism ∆ = Diag(1, −1, 1, −1, −1, 1, −1) ∈ AutK¯(M) which is in diagonal form of order 2 and is defined over K. A quotient curve of M with respect to ∆ is a curve X such that K(M)h∆i= K(X ).

To know the genus of this new curve X , we can use Riemann-Hurwitz formula applied to the theory of algebraic curves.

Theorem 2.1 (Riemann-Hurwitz). Let π : S0 → S be a surjective covering map of Riemann surfaces of degree N then

χ(S0) = N χ(S) − X

P ∈S0

(eP − 1)

where eP is the ramification index at P and χ = 2−2g is the Euler characteristic.

In our case π will be the map induced from π : K(M)h∆i → K(M) and since ∆ has order two, this map will have degree N = 2. And lastly eP = 2 whenever P is fixed under ∆, and eP = 1 otherwise. Now by a quick calculation shown in A.2.2 one sees that P

P ∈S0(eP − 1) = 4. Hence (2 − 2g(M)) = 2(2 − 2g(X )) − 4 gives us that g(X ) = 3.

Due to the simplicity of our ∆ we see that K(M)h∆i is generated by (x1, x3, x6, x22, x2x4, x2x5, x2x7, x24, x4x5, x4x7, x25, x5x7, x27)

so we could take these as our new variables ξ1, . . . , ξ13 and then construct our genus 3 curve X from that with the corresponding relations between the ξi and the equations we have for M.

But the simplicity of ∆ actually makes us suggest to try projecting from P6 → P2 by (x1 : . . . : x7) 7→ (x1 : x3 : x6). This is just a gamble, but one checks that by Riemann-Hurwitz that this gives a curve of genus 3 and the function field of the resulting curve is a subfield of K(M)h∆i. By a calculation as shown in appendix A.2.2, one obtains a non-singular curve of genus 3 that therefore should be isomorphic to X .

A model for X over P2 is given by the variety:

X = Z(5y14+ 12y31y2+ 6y12y22− 4y1y23+ 4y24− 28y13y3 + 16y12y2y3− 24y1y22y3+ 16y13y3+ 24y12y23− 10y22y32

− 12y1y33+ 8y2y33+ 3y34)

Now using this X we prove the following statement:

Proposition 2.1. For some elliptic curve E we have Jac(M) ∼ Jac(X )2× E

Proof. Let τ := SR be an automorphism of M of order 2 and defined over the ground field and let

π : M −→ X

(x1: . . . : x7) 7−→ (x1: x3: x6)


as defined above. Let ω1, ω2, ω3be a basis for the space of holomorphic differ- entials H0(X , Ω1). (Note that the dimension of H0(X , Ω1) is equal to the genus of X ). Now this gives us 3 linearly independent differentials πω1, πω2, πω3∈ H0(M, Ω1). Now we could look at the map π ◦ τ and pullback the ωi through this. This again gives 3 linearly independent differentials τπω1, τπω2, τπω3∈ H0(M, Ω1).

To prove the proposition we need to show that the subspace of H0(M, Ω1) generated by πω1, πω2, πω3, τπω1, τπω2, τπω3has dimension 6. Hence these differentials all need to be linearly independent. This will be computed in appendix A.2.2.

Since the space generated by πω1, πω2, πω3, τπω1, τπω2, τπω3 has dimension 6, we see that we still miss one differential that generates H0(M, Ω1), hence this differental must belong to an elliptic curve E. So Jac(M) splits, upto isogeny, into two parts Jac(X ) and one part E.

Now experimentally we can try to find what E should be. We know that over a finite field Fq, counting points on M will now be equal to counting points twice on X and once on E in the sense that #M = q + 1 − tM, #X = q + 1 − tX and #E = q + 1 − tE and now tM = 2tX+ tE. Hence by counting points on X and counting points on M we can count points on E and compare that with elliptic curves in a database. This can be done manually, or programmatically knowing that the conductor of E is of the form 2a7bc for some a, b, c ∈ Z≥1

The elliptic curve E by experimentation when c = 1 should then be isoge- nous to the curve given by the equation y2 = x3+ x2− 114x − 127 which has conductor 22· 72= 196. To prove that this is indeed the right curve we take the automorphism ρ := (SR−1S)3 also defined over the ground field, but of order 3. First we show that M/hρi = E. And then by a calculation we show that E is isogenous to the elliptic curve y2= x3+ x2− 114x − 127

To make it ourselves a bit easier using ρ we will perform a change of coor- dinates, such that ρ 7→ ˜ρ where ˜ρ is of the form

 1



where A, B, C are 2 × 2 matrices defined over Q with eigenvalues γ and γ2, where γ is a third root of unity.

By looking at the matrix ρ closely and making an Eigen decomposition ρ = V ΛV−1 we see that by using

D = Diag(1, 1/2, 3/2, 1/2, −1/2, 1/2, 3/2),

W =

1 0 0 0 0 0 0

0 1 0 0 1 0 0

0 −23γ −13 0 0 23γ +13 0 0

0 0 1 0 0 1 0

0 0 −23γ −13 0 0 23γ +13 0

0 0 0 1 0 0 1

0 0 0 −23γ −13 0 0 23γ +13


We can make this Eigen decomposition into a decomposition ρ = QSQ−1 where Q = DW V and

S =

1 0 0 0 0 0 0

0 −1/2 −1/2 0 0 0 0

0 3/2 −1/2 0 0 0 0

0 0 0 −1/2 3/2 0 0

0 0 0 −1/2 −1/2 0 0

0 0 0 0 0 −1/2 −1/2

0 0 0 0 0 3/2 −1/2

 such that S is in the form we wanted.

Using Q as a coordinate transformation we can construct a new model for M where S is an automorphism. We call the induced coordinate transformation from the old model to the new model κ. Then over this new model H0(M, Ω1) will be generated by ei := (κ−1)πωi and ei+3 := (κ−1)τπωi for i = 1, 2, 3 and an extra differential e7. We take e7 to be a differential that is invariant under S. We know that such e7should exist, cause it is exactly the differential pulled back from a generator of H0(M/hρi, Ω1).

Using this fact we now try to see if e7is linearly independent with respect to the other ei. This can be done by taking (id +S + S2) which maps e77→ 3e7. And it maps all differentials orthogonal to e7, to 0. This is because such differen- tials are linear combinations of elements in the γ Eigenspace or γ2Eigenspace of S. When β is in the γ Eigenspace then (id +S +S2)β = β +γβ +(−1−γ)β = 0, and similarly for γ2 = −1 − γ. The calculation in appendix A.2.2 shows that indeed (id +S + S2)ei= 0 for i = 1, 2, 3.

For i = 4, 5, 6 we find the relation Sei= ei−3, see appendix A.2.2. Hence (id +S + S2)ei= (id +S + S2)Sei= (id +S + S2)ei−3= 0 So e7 is indeed linearly independent of e1, . . . , e6 and hence M/hρi = E.

To show that E is isogenous to the curve given by the equation y2 = x3+ x2− 114x − 127 is a bit more involved. For this I refer to appendix A.2.2 for an implementation. The basic idea is as follows.

1. Using the new model make a projection to P2 on the first 3 coordinates.

We obtain a singular curve G of genus 7. Now over this new curve, S can be projected to an automorphism S0= π ◦ S.

2. Now we construct G/hS0i by calculating generators for K(G)hS0iand then expressing the equations for G in terms of these invariants, and adding extra relations between the generators.

3. Now again we project G/hS0i to the first 3 coordinates, and we have obtained a non-singular curve H of genus 1 and having one defining ho- mogeneous polynomial h ∈ Z[x, y, z].

4. We know that the affine patch of h where z = 1 has even degree for all y coordinates, hence we replace y2 by ˜y in this affine patch, and then take the projective closure using coordinates (X : Y : Z).

5. Now the defining polynomial of h will be parameterized over variables (t : t0) in P1 such that the affine part t0 = 1 of ˜y = Y /Z equals f (t) for some f ∈ Q(t).


6. Now we write y2= ˜y = Y /Z = f (t) and simplify this. Then what we get is y2= −7t4− 28t3− 56t2− 28

7. Then we perform a quadratic twist y 7→ √

−7 · y such that we obtain a curve with minimal model y2= t3− t2− 2t + 1

8. Then we perform the same quadratic twist again and calculate a minimal model. We obtain our wanted y2= t3+ t2− 114t − 127

Tracing back the steps one notices that E is isogenous to the curve given by y2= t3+ t2− 114t − 127.

Now let K = Q(α) where α a zero of x3+x2−2x−1, then all automorphisms are defined over K. And we can also see how Jac(M) splits over this field.

For this we need to check if Jac(X ) can split further over this field. To do this we first calculate the automorphism group of X = Z(f ) in a very na¨ıve way, namely by taking a 3 × 3 matrix M filled with 9 variables defined over Q and then solving f ◦ M = λf for some variable λ ∈¯ ¯

Q. And including the condition det(M ) = 1 and for simplicity we just take the subgroup of AutK¯(X ) of all M such that the corresponding λ = 1. Since AutK¯(X ) ⊂ PSL(3, ¯Q) all possible λ are third roots of unity. Suppose that we take λ a third root of unity, then the corresponding automorphism will not be defined over Q(α), so we are not interested in those. The calculation for finding the subgroup of AutK¯(X ) corresponding to all automorphisms where λ = 1 is done in appendix A.2.2. We find that this subgroup is generated by the involution

A = 1 7

−2µ − 7 µ2+ 5µ −1/2µ2− µ + 7/2 1/2µ2+ µ − 7/2 µ 1/2µ2+ 4µ + 7/2

−1µ2− 5µ 3/2µ2+ 9µ + 7/2 µ

and its conjugates. Here µ is a root of x3+ 7x2+ 7x − 7. This equation splits over K(x) as (x + 2α + 3)(x − 2α2− 2α + 5)(x − 1 + 2α2), so we may pick µ = −2α − 3.

Now diagonalize A = B−1DB where D = Diag(1, −1, −1) and then change coordinates of X by B. We obtain a new model:

X ∼= Z(x4+ (2α2− 4α + 2)x2y2+ (−4α2+ 9α − 5)y4+ (10α2− 8α − 6)x2z2 + (−34α2+ 28α + 18)y2z2+ (81α2− 65α − 45)z4)

With respect to this new model we can take the quotient under x 7→ −x over the affine part z = 1 and write the equation in the form η2= ay4+ by2+ c, for some a, b, c ∈ K. Taking ˜y = yη and ˜x = y2 we obtain a map of degree 2 to a curve of the form ˜y2= a03+ b02+ c0x for some a˜ 0, b0, c0∈ K. The constructed elliptic curve will be called Ex.

Similarly define elliptic curves Eyand Ezby taking automorphisms y 7→ −y and z 7→ −z respectively. Now from Ex we use an isogeny to Ex/hP i over the group generated by the point P = (0, 0) of order 2. Using V´elu’s formulae [10]

we find the curve Ex,2 = Ex/hP i. We do exactly the same for Ey and Ez to obtain Ey,2 = Ey/hP i and Ez,2 = Ez/hP i. A further calculation shows that all these quotients Ex,2, Ey,2 and Ez,2 are isomorphic to each other, but more interestingly they are isomorphic to E/K. So we get the following diagram:









= // Ey,2






These Ex, Ey, Ezare therefore all isogenous to E/K, however Ex, Ey, Ezare not isomorphic to each other, since they have different j-invariants. So this give us the following corollary:

Corollary 2.1. Over K = Q(α) we have Jac(M) ∼ E7

Note that this cannot happen over Q as X has no non-trivial automorphisms defined over Q.

Let K be a finite field Fq of characteristics unequal to 2 or 7 and let U ⊆ Aut¯Fq(M) be the group of automorphisms induced by the automor- phisms from AutQ¯(M). This can be done by noting that all automorphisms A ∈ AutQ¯(M) have denominators dividing 14. As 14 is invertible in Fq, we can use the reduction rp : Q(α) → Fp(α) composed with the inclusion ιn : Fp(α) → Fpn(α), where pn= q. Applying ιn◦ rp to coefficients of all auto- morphisms in AutQ¯(M), results in the subgroup U we want. Now we check for which q, we have that U is defined over the ground field.

The Galois group of the splitting field of x3+x2−2x−1 over Q is isomorphic to the cyclic group of order 3, because α = ζ + ζ−1 where ζ a seventh root of unity. Hence x3+ x2 − 2x − 1 ∈ Fq[x] either splits completely or remains inert when char(K) 6= 7. We know that when q ≡ 1 (mod 7) then Fq contains ζ and hence α. And when q ≡ −1 (mod 7) then Fq2 contains ζ, but then trF

q2/Fq(ζ) = ζ + ζq = ζ + ζ−1= α ∈ Fq. For all other q note that ζ + ζ−1= α, so ζ2+ 1 − αζ = 0. Hence the inclusion Fq(α) ⊂ Fq(ζ) has degree at most 2.

But Fq ⊂ Fq(α) has degree dividing 3.

For q ≡ 2, 4 (mod 7) we have q3≡ 1 (mod 7), hence ζ ∈ Fq3 and hence the inclusion Fq(α) ⊂ Fq must have degree 1. So Fq(α) ∼= Fq3 in this case.

For q ≡ 3, 5 (mod 7) we have q6≡ 1 (mod 7), hence ζ ∈ Fq6 and hence the inclusion Fq(α) ⊂ Fq must have degree 2. So Fq(α) ∼= Fq3 in this case. We obtain the following proposition.

Proposition 2.2. Let M be defined over K = Fq. For q 6= ±1 (mod 7) we have Jac(M) ∼ Jac(X )2× E. And for q ≡ ±1 (mod 7) we then have Jac(M) ∼ E7 Corollary 2.2. Jac(M) has bad reduction if and only if the finite field K has characteristic 2 or 7.

Proof. This follows from the fact that E has conductor 196 = 22· 72, and hence has bad reduction over 2 and 7. Also X has bad reduction over 2 and 7, this is because we can take the defining ideal I of Sing(X ) and calculate I ∩ Z in magma, see appendix A.2.2. The generator of the ideal I ∩ Z is 38416 = 24· 74, hence X has bad reduction over primes 2 and 7. So Jac(M) has bad reduction over primes 2 and 7, while any other prime has good reduction.


Now when q 6≡ ±1 (mod 7) another thing happens, Ex, Ey and Ezas shown above are now defined over the extension Fq[α] by the same reduction process.

It turns out Ex, Ey and Ez are conjugates of each other, as seen in appendix A.2.2 over Q(α). So Frobenius acts on the points of Ex, Ey and Ezas follows:

Ex Ey


(e1, e2) (e3, e4)

(e5, e6)




Here (e1, e2), (e3, e4) and (e5, e6) denote generators of the points of order `n on the curves Ex, Ey and Ez respectively, for any odd prime ` 6= char(Fq), with e3, e4 the images of e1, e2under fr and e5, e6the images of e3, e4 under fr. And e1, e2 can now be obtained from linear combinations of images of e5, e6 under fr.

Hence fr as a matrix acting on the Tate module of Jac(X ) looks like

0 0 0 0 ∗ ∗

0 0 0 0 ∗ ∗

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

and hence has trace equal to 0.

Corollary 2.3. Let q 6≡ ±1 (mod 7) then #X (Fq) = q + 1 and #M(Fq) =


And from proposition 2.2 we find

Corollary 2.4. Let q ≡ ±1 (mod 7) and take t = q + 1 − #E(Fq) then

#X (Fq) = q + 1 − 3t and #M(Fq) = q + 1 − 7t

From Corollary 2.3 and Corollary 2.4 we can now make a table of how many points lie on M defined over a certain finite field and compare that to records found at [2] for genus 7 curves. When a number in the table is a record we mark it with green. When it is a questionable record (since it is a record according to the table and is within 10% of the maximum, but the trace of Frobenius is

”small”) we mark it with orange. The table shows #M(Fpn).


p\n 1 2 3 4 5

3 3 15 84 75 213

5 3 27 252 675 3183

11 15 135 828 14715 161625

13 0 324 2688 27540 362880

17 15 315 5796 83475 1417575 19 21 399 6468 129675 2477811

p 23 29 31 37 41 43 47 53 59

#M(Fpn) 21 72 39 39 0 72 57 51 51

p 61 67 71 73 79 83 89 97

#M(Fpn) 63 75 72 75 93 0 75 168

So we have 5 records, namely over finite fields F33, F53, F132, F133, F173, as is backed up from the theory one would expect records to occur at q ≡ ±1 (mod 7). And this is exactly what we see happening for these values. Now one could wonder if any twist of M can improve records on finding points on finite fields, so this is what we investigate in the next chapter.

2.3 Twists of M

To calculate twists of M we use the theory as described in section 1.2 and in [7].

From this we know that obtaining twists is equivalent to obtaining Frobenius conjugacy classes of AutK¯(M). In our case K is finite and we take the subgroup U ⊆ AutK¯(M) induced from AutQ¯M ∼= PSL(2, 8). We see that basically when q ≡ ±1 (mod 7) the Frobenius conjugacy classes of U are nothing more than the conjugacy classes. And when q 6≡ ±1 (mod 7) then we have to calculate them ourselves.

2.3.1 Case q = pn≡ ±1 (mod 7)

Let C = [C|] be a conjugacy class of U with representative C|. In this case we can make a cocycle in H1(Gk, Aut¯k(M)), namely σ : fr 7→ C. Then we know there exists a twist-map ψ : C ⊗ ¯k → C0⊗ ¯k such that ψ−1frψ = C. Note that we need to be cautious since we defined our U by transposed matrices.

A few methods to find ψ are the naive methods explained in the part about the Klein curve. But these methods are too slow for this case. Instead, we solve ψ−1frψ = C by linear algebra. Over a finite field extension of Fq, the frobenius fr becomes a linear automorphism. We also have the property that frψ = ψC or when we take ψ0= ψ|thenfrψ0= C|ψ0. Now we know ψ lives over a finite field extension of degree m and C just lives over the ground field, we will soon see what m we should take in this calculation. Then we can write this equation over the vector space (L7

i=1Fq· xi) ⊗ Fmq as

(id7⊗ Fr)Ψ0 = (C|⊗ idm0

where Fr ∈ SL(m, k) denotes the matrix belonging to fr as linear map over the vector space Fmq and where Ψ0 is the inflation of ψ0 seen as linear map over (L7

i=1Fq·xi)⊗Fmq . Hence if we can find a linear independent basis v1, . . . , v7for (a subspace of) ker(id7⊗ Fr −C|⊗idm) then we can reconstruct Ψ0= (v1 . . . v7)


as a 7 × 7m matrix, which will deflate to a 7 × 7 matrix corresponding to ψ0 under the map Fmq → Fqm.

The problem now is that for this method to work, this ker(id7⊗ Fr −C|⊗ idm) needs to be at least 7-dimensional. So m needs to be at least so high that σm = id as cocycle and the eigenvalues of (C|)−1 must get canceled out by eigenvalues of Fr. This is because

ker(id7⊗ Fr −C|⊗ idm) = ker((C|)−1⊗ Fr − id7m) and hence we look for the +1-eigenspace of (C|)−1⊗ Fr.

We are in the case q ≡ ±1 (mod 7) so we are lucky, because here C is defined over the ground field. So let c be the multiplicative order of C then σc= id and Fr contains all c-th roots of unity as eigenvalue. So

dim(ker(id7⊗ Fr −C|⊗ id7)) ≥ 7, hence we can take m = c.

The last problem, which didn’t occur in the Klein Curve since it had only one defining polynomial, is finding a model of the twist curve over Fq. We can do this by applying trace maps trµl to the coefficients of the defining polynomials, and then combining all resulting polynomials. Here µ is a generator of Fqm= Fq(µ), we take 0 ≤ l ≤ m − 1 and for a ∈ Fqm we have

tra: Fqm −→ Fq

b 7−→ tr(ab)

These maps are all linear and because of this we have tra+ trb = tra+b, hence the trµl form a basis for a Fq-vector space that give us qmdistinct linear maps on Fqm → Fq. Note that as a Fq-vector space





Fq· µi.

Hence the vector space htr1, trµ, . . . , trµm−1i ∼= Fmq .

By applying trµl to each coefficient of each defining polynomial of the twist model for each l, we then get a model over Fq by merging all resulting polyno- mials.

It is known that PSL(2, 8) only contains elements of order 2, 3, 7 and 9. So in our case the kernel calculation is done using matrix calculations over Fq of size at most 63 × 63. So the time complexity for this algorithm is mostly based on how well a computer algebra package can do arithmetic over Fq.

Now PSL(2, 8) has 9 conjugacy classes and this gives us 9 twists, of which one is the trivial twist. And we can then calculate points on these twists for q ≡ ±1 (mod 7), as is done in the following table, which we can generate by using implementations found in appendix A.2.3. Note that q = 135, 175and 195 took too long to finish completely, that is why the row q = 135only contains two numbers and rows q = 175 and q = 195 are omitted from the table. Coloring is similar to the table we saw before, and it shows the number of points on the twist of M(Fq) corresponding to the conjugacy class C.



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