faculteit Wiskunde en Natuurwetenschappen

## 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

### Contents

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 = p^{n} ≡ ±1 (mod 7) . . . 18

2.3.2 Case q = p^{n} 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

### Introduction

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 X^{3}Y + Y^{3}Z + Z^{3}X
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 F^{3}^{2}, 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 X^{4}+ Y^{4}+ Z^{4}

This curve has special properties over K, which can be seen from the fact
that for any a ∈ F^{9} we have

a^{4}= a^{3}a = Norm_{F}_{9}_{/F}_{3}(a)

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

*

a1

a2

a3

,

b1

b2

b3

+

=

3

X

i=1

aibi

Denote v ·w^{..}= hv, wi for any v, w ∈ ¯F^{3}3. 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 : ¯F^{3}3→ ¯F^{3}3 be the morphism on the vector space
K^{3}induced by the Frobenius endomorphism fr : x 7→ x^{3}. It acts on a vector by
applying fr to each coefficient, so

~fr

a1

a2

a3

=

a^{3}_{1}
a^{3}_{2}
a^{3}_{3}

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

X^{4}+ Y^{4}+ Z^{4}=

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) ∼=

PSL_{3}(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

A

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 =^{fr}A~frv, where
we denote ^{fr}A as applying Frobenius to all coefficients of the matrix A. Now
using the linearity of the dot product (^{fr}A)^{|}Av · ~frv = λv · ~frv. But this means
that we want to have (^{fr}A)^{|}A = λI where I ∈ PSL_{3}(K) the 3×3 identity matrix.

Or written differently (^{fr}A)^{|} = λA^{−1}. Taking determinants on both sides the
LHS gives det((^{fr}A)^{|}) = (det A)^{3}= 1 because of application of Frobenius. On
the RHS we get det(λA^{−1}) = λ^{3}(det A)^{−1} = λ^{3}. So combining them we see
1 = λ^{3} 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} //

fr•

¯F3

fr

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 (^{fr}A)^{|}= A^{−1} we obtain that:

fr^{2}A =^{fr}((A^{−1})^{|})

= ((^{fr}A)^{−1})^{|}

= (((A^{−1})^{|})^{−1})^{|}

= A

This means fr^{2} should act trivially on the components of A, whence it means
A ∈ PSL3(F^{9}) and fr can now be seen as conjugation and hence we define the
Hermitian of A to be A^{† ..}= (^{fr}A)^{|}. So combining this we get A ∈ {M ∈
PSL(F9) : M^{†}M = I} ∼= PSU3(3) = U_{3}(3). Here U_{3}(3) is the group of all
3 × 3 matrices over F3^{2} with unit determinant such that A^{†}A = 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, F_{4} and K are isomorphic, so Aut_{K}(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) → H^{1}(G_{k}, Autk¯(C))
where we denote Gk = Gal(¯k/k).

In our case we have k = F^{q} and K = ¯k where q a power of 3. The bijection θ
takes a twist ψ : C ⊗ ¯k → C^{0}⊗ ¯k and sends it to the cocycle fψ : σ 7→ ψ^{−1}◦^{σ}ψ.

The map that goes back (see [7] for more details) is formed on the function
field of the curve. Given a cocycle f ∈ H^{1}(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 ∈ G_{k}. The invariants of ¯k(C) under this action give a new function
field k(C^{0}).

To calculate twists of C is therefore in essence equal to calculating the set
H^{1}(G_{k}, 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(^{fr}x), 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

H^{1}(G_{k}, 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 = F^{3}^{m}, 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 F^{9}. Hence the Frobenius endomorphism fr : x 7→ x^{3}^{m}

acts trivially on U_{3}(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 ·^{fr}x^{−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 ◦^{fr}ψ, by letting
ψ be a 3 × 3 matrix containing 9 variables ψ_{ij}, where we let the determinant of
ψ be 1. These conditions can be written as an ideal I in ¯F3[ψ_{11}, . . . , ψ_{33}]. We
know over which extension this twist is isomorphic to the original curve, namely
take the least n > 0 such that A ◦^{fr}A ◦^{fr}^{2}A ◦ . . . ◦^{fr}^{n}A = id. Then ψ should
become an isomorphism over F^{3}^{nm}, hence to find a twist we may assume I is an
ideal in F^{3}^{mn}[ψ11, . . . , ψ33].

Now to find the twist map ψ : F4 → C defined over F3^{m}, we need to find
one solution in Z(I) and embed ψ inside PSL3(F^{3}^{mn}) such that it becomes an
isomorphism over F4⊗ F3^{n}. 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 F^{3}^{m} 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 C^{0} be the twist over k corresponding to the Frobenius
conjugacy class [A]. Then

#C^{0} = #{Q ∈ C(¯k) :^{fr}Q = A^{−1}Q}

Proof. Let P ∈ C^{0} then ^{fr}P = P , because fr works trivial over k. Let ψ :
C ⊗ ¯k → C^{0}⊗ ¯k be the twist map corresponding to A. Then

ψ^{−1}(^{fr}P ) = ψ^{−1}(P )

Now A = ψ^{−1}◦^{fr}ψ so

ψ^{−1}(P ) = ψ^{−1}(^{fr}P )

= (A ◦ (^{fr}ψ)^{−1})(^{fr}P )

= (A ◦^{fr}(ψ^{−1}))(^{fr}P )

= A(^{fr}(ψ^{−1}(P )))

So take Q = ψ^{−1}(P ) then we have that A^{−1}Q =^{fr}Q 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 ^{fr}Q = A^{−1}Q can be written as an
ideal I in ¯F3[Q1, Q2, Q3]. Now noting that the twist lives over a finite extension
of F^{3}^{m}, we can project this ideal I into F^{3}^{nm}[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 F^{3}^{nm}-multiples of any affine point to be equivalent. Hence

#C^{0} = (#Z(I) − 1)/3^{nm}, as just a naive way to use the theorem.

As a test for this theorem, we counted the number of points over fields F3and
F3^{2}. First we count points on the twists corresponding to Frobenius conjugacy
classes C defined over F3, where ω is a zero of x^{2}−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}

β^{2}

αβαβ^{−1}

# 4 4 13 1 4 4

For F^{3}^{2} = F^{3}[ω] satisfying ω^{2}− 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] αβ^{3}

β^{2}

# 28 4 13 7 7 16 1

C β^{−2}α

β^{2}α

βαβ^{−1}α

(βα)^{2}β^{3}

(βα)^{3}β^{−2}

(β^{−2}α)^{3}

(β^{2}α)^{3}

# 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)[y^{1}, . . . , y7] and where we take αj:= ζ^{j}+ ζ^{−j}:

Xy^{2}_{i} =X

ζ^{i}y^{2}_{i} =X

ζ^{−i}y_{i}^{2}= 0
α4y1y7+ α2y2y5+ α1y4y6= 0
α4y2y1+ α2y3y6+ α1y5y7= 0
α4y3y2+ α2y4y7+ α1y6y1= 0
α_{4}y_{4}y_{3}+ α_{2}y_{5}y_{1}+ α_{1}y_{7}y_{2}= 0
α4y5y4+ α2y6y2+ α1y1y3= 0
α_{4}y_{6}y_{5}+ α_{2}y_{7}y_{3}+ α_{1}y_{2}y_{4}= 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 )^{−1}W )^{2}W , 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, . . . , x_{7}]:

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

− x^{2}_{2}+ x1x3− x^{2}_{3}+ 2x2x5+ x4x5− x^{2}_{5}+ x1x6+ x2x7− x4x7,
x^{2}_{1}+ x^{2}_{2}− x_{1}x_{3}− x^{2}_{3}− x_{2}x_{4}− x_{2}x_{5}+ x_{3}x_{6}− x_{2}x_{7}+ 2x_{5}x_{7}− x^{2}_{7},
x1x4− x3x4− 2x1x5− x3x5− x2x6+ x5x6+ 2x1x7+ x6x7,

x^{2}_{1}− x^{2}_{2}− 2x1x3+ x^{2}_{3}− x2x4− x2x5+ x4x5+ x^{2}_{5}+ x3x6− x^{2}_{6}+ 2x2x7− x5x7,
x_{1}x_{2}+ 2x_{2}x_{3}− x1x_{5}− x5x_{6}− 2x1x_{7}− x3x_{7}+ 2x_{6}x_{7},

− 2x1x2+ 2x2x3− x1x4− x1x5+ 2x5x6+ 2x1x7− 2x3x7− x6x7,
2x^{2}_{1}+ x1x3+ x4x5− x^{2}_{5}− x1x6+ x^{2}_{6}+ 3x2x7− x4x7− x^{2}_{7},

2x^{2}_{1}+ x^{2}_{2}− x_{1}x_{3}+ x^{2}_{3}+ x_{4}x_{5}+ x^{2}_{5}+ x_{1}x_{6}− 2x_{3}x_{6}+ x^{2}_{6}+ x_{2}x_{7}− x_{4}x_{7}− 2x_{5}x_{7}+ x^{2}_{7},
x^{2}_{1}+ x1x3+ x^{2}_{4}+ 2x2x5+ x4x5− x1x6+ 2x3x6+ x^{2}_{6}− 3x2x7− x4x7+ 3x^{2}_{7}

Its automorphism group is generated by R and S defined over PSL_{7}(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 =

a_{11} a_{12} · · · a_{1n}
a_{21} a_{22} · · · a_{2n}
... ... . .. ...
am1 am2 · · · amn

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

Vec(M ) := (a_{11}, a_{21}, . . . , a_{m1}, a_{12}, a_{22}, . . . , a_{m2}, . . . , a_{1n}, a_{2n}, . . . , a_{mn})^{|}

When M is seen as a k-vector for some field k this means that basically
M ∈ k^{m}⊗ k^{n}. Now Vec gives us the isomorphism k^{m}⊗ k^{n}∼= k^{mn}. 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 α^{2}− α − 2 −α^{2}− α − 1 0 2α^{2}− 1 0

α^{2}− 2α 0 0 0 3α + 1 0 −α^{2}− 2α + 1

0 0 −α^{2}− α − 1 −2α^{2}+ 1 0 α^{2}− α − 2 0

0 7α 0 0 0 0 0

−α^{2}− 2α + 1 0 0 0 −α^{2}+ 2α 0 −3α − 1

0 0 −3α − 1 −α^{2}+ 2α 0 α^{2}+ 2α − 1 0

−3α − 1 0 0 0 α^{2}+ 2α − 1 0 α^{2}− 2α

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

of some variables this becomes a block matrix φ^{0} looking like

φ^{0}= 1
7

7α 0 0 0 0 0 0

0 α^{2}− α − 2 −α^{2}− α − 1 2α^{2}− 1 0 0 0

0 −α^{2}− α − 1 −2α^{2}+ 1 α^{2}− α − 2 0 0 0

0 −3α − 1 −α^{2}+ 2α α^{2}+ 2α − 1 0 0 0

0 0 0 0 α^{2}− 2α 3α + 1 −α^{2}− 2α + 1

0 0 0 0 −α^{2}− 2α + 1 −α^{2}+ 2α −3α − 1

0 0 0 0 −3α − 1 α^{2}+ 2α − 1 α^{2}− 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 : S^{0} → S be a surjective covering
map of Riemann surfaces of degree N then

χ(S^{0}) = N χ(S) − X

P ∈S^{0}

(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 e_{P} = 2
whenever P is fixed under ∆, and e_{P} = 1 otherwise. Now by a quick calculation
shown in A.2.2 one sees that P

P ∈S^{0}(e_{P} − 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, x^{2}_{2}, x2x4, x2x5, x2x7, x^{2}_{4}, x4x5, x4x7, x^{2}_{5}, x5x7, x^{2}_{7})

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
P^{6} → P^{2} 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 P^{2} is given by the variety:

X = Z(5y_{1}^{4}+ 12y^{3}_{1}y_{2}+ 6y_{1}^{2}y_{2}^{2}− 4y_{1}y_{2}^{3}+ 4y_{2}^{4}− 28y_{1}^{3}y_{3}
+ 16y_{1}^{2}y_{2}y_{3}− 24y1y^{2}_{2}y_{3}+ 16y_{1}^{3}y_{3}+ 24y_{1}^{2}y^{2}_{3}− 10y^{2}_{2}y_{3}^{2}

− 12y1y^{3}_{3}+ 8y2y_{3}^{3}+ 3y_{3}^{4})

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 ω_{1}, ω_{2}, ω_{3}be a basis for the space of holomorphic differ-
entials H^{0}(X , Ω^{1}). (Note that the dimension of H^{0}(X , Ω^{1}) is equal to the genus
of X ). Now this gives us 3 linearly independent differentials π^{∗}ω1, π^{∗}ω2, π^{∗}ω3∈
H^{0}(M, Ω^{1}). Now we could look at the map π ◦ τ and pullback the ωi through
this. This again gives 3 linearly independent differentials τ^{∗}π^{∗}ω1, τ^{∗}π^{∗}ω2, τ^{∗}π^{∗}ω3∈
H^{0}(M, Ω^{1}).

To prove the proposition we need to show that the subspace of H^{0}(M, Ω^{1})
generated by π^{∗}ω1, π^{∗}ω2, π^{∗}ω3, τ^{∗}π^{∗}ω1, τ^{∗}π^{∗}ω2, τ^{∗}π^{∗}ω3has 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 π^{∗}ω1, π^{∗}ω2, π^{∗}ω3, τ^{∗}π^{∗}ω1, τ^{∗}π^{∗}ω2, τ^{∗}π^{∗}ω3 has
dimension 6, we see that we still miss one differential that generates H^{0}(M, Ω^{1}),
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 − t_{M}, #X = q + 1 − t_{X}
and #E = q + 1 − tE and now t_{M} = 2t_{X}+ 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 2^{a}7^{b}c 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 y^{2} = x^{3}+ x^{2}− 114x − 127 which has
conductor 2^{2}· 7^{2}= 196. To prove that this is indeed the right curve we take the
automorphism ρ := (SR^{−1}S)^{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 y^{2}= x^{3}+ x^{2}− 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

A B

C

where A, B, C are 2 × 2 matrices defined over Q with eigenvalues γ and γ^{2},
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 −^{2}_{3}γ −^{1}_{3} 0 0 ^{2}_{3}γ +^{1}_{3} 0 0

0 0 1 0 0 1 0

0 0 −^{2}_{3}γ −^{1}_{3} 0 0 ^{2}_{3}γ +^{1}_{3} 0

0 0 0 1 0 0 1

0 0 0 −^{2}_{3}γ −^{1}_{3} 0 0 ^{2}_{3}γ +^{1}_{3}

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 H^{0}(M, Ω^{1})
will be generated by ei := (κ^{−1})^{∗}π^{∗}ωi and ei+3 := (κ^{−1})^{∗}τ^{∗}π^{∗}ωi 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 H^{0}(M/hρi, Ω^{1}).

Using this fact we now try to see if e7is linearly independent with respect to
the other e_{i}. This can be done by taking (id +S + S^{2})^{∗} which maps e_{7}7→ 3e7.
And it maps all differentials orthogonal to e_{7}, to 0. This is because such differen-
tials are linear combinations of elements in the γ Eigenspace or γ^{2}Eigenspace of
S^{∗}. When β is in the γ Eigenspace then (id +S +S^{2})^{∗}β = β +γβ +(−1−γ)β = 0,
and similarly for γ^{2} = −1 − γ. The calculation in appendix A.2.2 shows that
indeed (id +S + S^{2})^{∗}ei= 0 for i = 1, 2, 3.

For i = 4, 5, 6 we find the relation S^{∗}ei= ei−3, see appendix A.2.2. Hence
(id +S + S^{2})^{∗}ei= (id +S + S^{2})^{∗}S^{∗}ei= (id +S + S^{2})^{∗}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 y^{2} = x^{3}+
x^{2}− 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 P^{2} 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 S^{0}= π ◦ S.

2. Now we construct G/hS^{0}i by calculating generators for K(G)^{hS}^{0}^{i}and then
expressing the equations for G in terms of these invariants, and adding
extra relations between the generators.

3. Now again we project G/hS^{0}i 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 y^{2} 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 : t^{0}) in P^{1} such that the affine part t^{0} = 1 of ˜y = Y /Z equals f (t) for
some f ∈ Q(t).

6. Now we write y^{2}= ˜y = Y /Z = f (t) and simplify this. Then what we get
is y^{2}= −7t^{4}− 28t^{3}− 56t^{2}− 28

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

−7 · y such that we obtain a
curve with minimal model y^{2}= t^{3}− t^{2}− 2t + 1

8. Then we perform the same quadratic twist again and calculate a minimal
model. We obtain our wanted y^{2}= t^{3}+ t^{2}− 114t − 127

Tracing back the steps one notices that E is isogenous to the curve given by
y^{2}= t^{3}+ t^{2}− 114t − 127.

Now let K = Q(α) where α a zero of x^{3}+x^{2}−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 µ^{2}+ 5µ −1/2µ^{2}− µ + 7/2
1/2µ^{2}+ µ − 7/2 µ 1/2µ^{2}+ 4µ + 7/2

−1µ^{2}− 5µ 3/2µ^{2}+ 9µ + 7/2 µ

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

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

X ∼= Z(x^{4}+ (2α^{2}− 4α + 2)x^{2}y^{2}+ (−4α^{2}+ 9α − 5)y^{4}+ (10α^{2}− 8α − 6)x^{2}z^{2}
+ (−34α^{2}+ 28α + 18)y^{2}z^{2}+ (81α^{2}− 65α − 45)z^{4})

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 η^{2}= ay^{4}+ by^{2}+ c, for
some a, b, c ∈ K. Taking ˜y = yη and ˜x = y^{2} we obtain a map of degree 2 to a
curve of the form ˜y^{2}= a^{0}x˜^{3}+ b^{0}x˜^{2}+ c^{0}x for some a˜ ^{0}, b^{0}, c^{0}∈ 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:

E_{x}

2−isog.

E_{y}

2−isog.

E_{z}

2−isog.

Ex,2

∼= // Ey,2

∼=

Ez,2

∼=

oo

E/K

These E_{x}, E_{y}, E_{z}are therefore all isogenous to E/K, however E_{x}, E_{y}, E_{z}are
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) ∼ E^{7}

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 F^{q},
we can use the reduction rp : Q(α) → F^{p}(α) composed with the inclusion
ιn : F^{p}(α) → F^{p}^{n}(α), where p^{n}= 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 x^{3}+x^{2}−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 x^{3}+ x^{2} − 2x − 1 ∈ Fq[x] either splits completely or remains
inert when char(K) 6= 7. We know that when q ≡ 1 (mod 7) then F^{q} contains
ζ and hence α. And when q ≡ −1 (mod 7) then F^{q}^{2} contains ζ, but then
tr_{F}

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

But F^{q} ⊂ Fq(α) has degree dividing 3.

For q ≡ 2, 4 (mod 7) we have q^{3}≡ 1 (mod 7), hence ζ ∈ Fq^{3} and hence the
inclusion Fq(α) ⊂ Fq must have degree 1. So Fq(α) ∼= Fq^{3} in this case.

For q ≡ 3, 5 (mod 7) we have q^{6}≡ 1 (mod 7), hence ζ ∈ Fq^{6} and hence the
inclusion Fq(α) ⊂ Fq must have degree 2. So Fq(α) ∼= Fq^{3} in this case. We
obtain the following proposition.

Proposition 2.2. Let M be defined over K = F^{q}. 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) ∼ E^{7}
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 = 2^{2}· 7^{2}, 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 = 2^{4}· 7^{4},
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, E_{x}, E_{y} and E_{z}as 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 E^{x}, Ey and Ezas follows:

E_{x}
Ey

Ez

(e_{1}, e_{2})
(e3, e4)

(e5, e6)

∈

∈

∈

fr

fr

fr

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(F^{q}), 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) =

#E(F^{q})

And from proposition 2.2 we find

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

#X (F^{q}) = q + 1 − 3t and #M(F^{q}) = 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(Fp^{n}).

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(F^{p}^{n}) 21 72 39 39 0 72 57 51 51

p 61 67 71 73 79 83 89 97

#M(Fp^{n}) 63 75 72 75 93 0 75 168

So we have 5 records, namely over finite fields F^{3}^{3}, F^{5}^{3}, F^{13}^{2}, F^{13}^{3}, F^{17}^{3}, 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 = p^{n}≡ ±1 (mod 7)

Let C = [C^{|}] be a conjugacy class of U with representative C^{|}. In this case we
can make a cocycle in H^{1}(Gk, Aut¯k(M)), namely σ : fr 7→ C. Then we know
there exists a twist-map ψ : C ⊗ ¯k → C^{0}⊗ ¯k such that ψ^{−1}◦^{fr}ψ = 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^{−1}◦^{fr}ψ = C by linear algebra. Over a finite field extension of Fq, the frobenius
fr becomes a linear automorphism. We also have the property that ^{fr}ψ = ψC
or when we take ψ^{0}= ψ^{|}then^{fr}ψ^{0}= C^{|}ψ^{0}. 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) ⊗ F^{m}q as

(id7⊗ Fr)Ψ^{0} = (C^{|}⊗ idm)Ψ^{0}

where Fr ∈ SL(m, k) denotes the matrix belonging to fr as linear map over
the vector space F^{m}q and where Ψ^{0} is the inflation of ψ^{0} seen as linear map over
(L7

i=1Fq·x_{i})⊗F^{m}q . Hence if we can find a linear independent basis v_{1}, . . . , v_{7}for
(a subspace of) ker(id7⊗ Fr −C^{|}⊗idm) then we can reconstruct Ψ^{0}= (v1 . . . v7)

as a 7 × 7m matrix, which will deflate to a 7 × 7 matrix corresponding to ψ^{0}
under the map F^{m}q → Fq^{m}.

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
σ^{m} = 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 σ^{c}= id and
Fr contains all c-th roots of unity as eigenvalue. So

dim(ker(id_{7}⊗ 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 F^{q}. We can do this
by applying trace maps tr_{µ}l to the coefficients of the defining polynomials, and
then combining all resulting polynomials. Here µ is a generator of F^{q}^{m}= F^{q}(µ),
we take 0 ≤ l ≤ m − 1 and for a ∈ F^{q}^{m} we have

tr_{a}: Fq^{m} −→ Fq

b 7−→ tr(ab)

These maps are all linear and because of this we have tra+ trb = tra+b,
hence the tr_{µ}l form a basis for a F^{q}-vector space that give us q^{m}distinct linear
maps on F^{q}^{m} → Fq. Note that as a F^{q}-vector space

Fq^{m}∼=

m−1

M

i=0

Fq· µ^{i}.

Hence the vector space htr_{1}, tr_{µ}, . . . , tr_{µ}m−1i ∼= F^{m}q .

By applying tr_{µ}l 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 F^{q} 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 F^{q}.

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 = 13^{5}, 17^{5}and 19^{5}
took too long to finish completely, that is why the row q = 13^{5}only contains two
numbers and rows q = 17^{5} and q = 19^{5} 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.