# Bring’s curve: Old and New

H. W. Braden and Linden Disney-Hogg  School of Mathematics and Maxwell
Institute for Mathematical Sciences
The University of Edinburgh
Edinburgh EH9 3FD, Scotland, U.K<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Bring’s curve, the unique Riemann surface of genus-4 with automorphism group
$S_{5}$, has many exceptional properties. We review, give new proofs of, and
extend a number of these including giving the complete realisation of the
automorphism group for a plane curve model, identifying a new elliptic
quotient of the curve and the modular curve $X_{0}(50)$, providing a complete
description of the orbit decomposition of the theta characteristics, and
identifying the unique invariant characteristic with the divisor of the Szëgo
kernel. In achieving this we have used modern computational tools in Sagemath,
Macaulay2, and Maple, for which notebooks demonstrating calculations are
provided.

###### 2020 Mathematics Subject Classification:

14E07, 14H55, 14Q05, 14K02, 14K25

##  1\. Introduction

Bring’s curve, a genus-4 Riemann surface first introduced by Erland Bring in
1786 in relation to solutions of the quintic equation, is the unique genus-4
curve with automorphism group $S_{5}$, the largest possible group for curves
of this genus. This automorphism group acts transitively on the Weierstrass
points of the curve, which are all weight one, a very rare property of curves.
The curve can moreover be seen to be an example of an elliptic modular
surface, and has a unique (even) theta characteristic invariant under the
action of the whole automorphism group. It shares all these properties with
Klein’s curve, a genus-3 curve which has been studied extensively [Lev01].

We aim to give a unified review of many of these aspects of Bring’s curve,
extend these results and provide connections between them. This will include
providing an explicit presentation of the $S_{5}$ group in terms of birational
automorphisms of a plane-curve model, using this to completely describe the
quotient structure of the curve, identifying fixed points of these quotients
with geometrically significant points on the curve including the Weierstrass
points, and utilising these geometric points to describe the orbit
decomposition of odd theta characteristics on the curve.

Moreover, we will construct an explicit birational map between the canonical
embedding of the curve and a plane-curve model. This will allow us to use
modern tools in SageMath, Maple, and Macaulay2 (all interfaced through Sage)
to clarify and fix errors in the existing literature as well as finding the
orbit decomposition of even theta characteristics of the curve. We shall also
calculate the unique invariant theta characteristic on the curve as a vector
in the Jacobian, making explicit previous work of others. These calculations
are presented in Jupyter notebooks, available at
https://github.com/DisneyHogg/Brings_Curve.

The structure of the papers is as follows, in §2 we recall basic properties
from the literature, provide the birational map between models of the curve
and describe the period matrix and automorphism group of the curve. In §3 we
identify geometrically relevant orbits of points under the action of the
automorphism group, notably the Weierstrass points $W$, for which we construct
the associated holomorphic differentials with vanishing of order $4$ at $W$
and the meromorphic functions with poles only at $W$; indeed we shall describe
the divisors of these completely. (Recall, for comparison, that while the pole
of the Weierstrass $\wp$-function is $z=0$, describing its zeros is
nontrivial.) In §4 we use the action of the automorphism group to explicitly
describe the quotient structure of the curve. Using representation theory we
elucidate much of the structure of the Jacobian of Bring’s curve, giving new
derivations of previously known results, and identifying new quotients of the
curve. In doing so, we also discover isomorphisms (as opposed to isogenies) of
quotients not explained by group theory alone, and these deserve further
study. In §5 we will develop the understanding of the theta characteristics on
the curve, including explicitly the unique invariant one. Bring’s curve has
arisen in many disparate mathematical areas and we will use this paper to also
make some of these results and interconnections more widely known; we will
however not focus on either the modular aspects of Bring’s curve or extension
to positive characteristic here. Because of the wide-ranging nature of this
paper we will conclude in §6 by summarising our new results having by then
placed these in context. We draw attention here to two areas we feel merit
further explanation.

This work answers questions first raised in [BN12], and we will leave many
expository results to that paper. For a source on background classical
material, see [FK92]. Unless otherwise said explicitly, we shall assume the
Riemann surfaces we discuss are all smooth.

Acknowledgements. We are grateful to Vanya Cheltsov for discussions. The
research of LDH is supported by a UK Engineering and Physical Sciences
Research Council (EPSRC) studentship.

##  2\. Defining Equations and Elementary Properties

In this section we shall introduce Bring’s curve describing its basic
properties, its period matrix and its automorphisms. We begin with two
descriptions: one as a model in $\mathbb{P}^{4}$ and the second as a plane
curve. In particular we shall construct the birational map between these
models, which we believe new. Such a map enables us to concretely translate
calculations from one model to another and will be helpful later in §4 when we
look at quotients of the curve.

######  Definition 2.1.

Bring’s Curve $\mathcal{B}$ is defined in $\mathbb{P}^{4}$ by the
(homogeneous) equations111Note here we use the notation $H_{k}$ of [Bur83] for
the symmetric sum, whereas [Edg78] uses $S_{k}$.

(1) $H_{k}:=\sum_{i=1}^{5}x_{i}^{k}=0,\quad k=1,2,3,$

where we have taken the coordinates
$[x_{1}:x_{2}:x_{3}:x_{4}:x_{5}]\in\mathbb{P}^{4}$.

We will call (1) the $\mathbb{P}^{4}$-model of Bring’s curve. Historically,
this curve was attributed to Bring by Klein in his lectures on the icosahedron
[Kle88]. Bring’s improvement on the Tschirnhaus transformation enabled the
general quintic to be brought to the Bring-Jerrard form $x^{5}+ex+f=0$
[VSA03]: if $x_{i}$, $i=1,\dots,5,$ are the roots of this equation then the
symmetric sums $\sum_{i}x_{i}^{k}=0$ for $k=1,2,3$, and so the above curve
emerges. Green [Gre78] exploits this relation to Bring’s curve to (implicitly)
solve the quintic using automorphic forms. At this stage we may observe that
$S_{5}\leq\operatorname{Aut}(\mathcal{B})$.

######  Remark 2.2.

In the above definition we have implicitly taken
$\mathbb{P}^{4}=\mathbb{CP}^{4}$, and indeed this will be our definition
throughout. At various junctures we will however highlight the number fields
within $\mathbb{C}$ that are relevant fields of definition for morphisms of
the curve, or geometrically important points on the curve.

For many calculations a plane model of a curve in $\mathbb{P}^{2}$ is useful.
One of the main tools we will utilise in this paper is the Riemann surfaces
module of SageMath [The21] which requires such a model. As such we introduce
the following.

######  Definition 2.3.

The Hulek-Craig (HC) model of Bring’s curve is the (singular) plane curve in
$\mathbb{P}^{2}$ given by

(2) $F(X,Y,Z):=X(Y^{5}+Z^{5})+(XYZ)^{2}-X^{4}YZ-2(YZ)^{3}=0,$

taking homogeneous coordinates $[X:Y:Z]\in\mathbb{P}^{2}$.

This model was used in [Hul87, Cra02] where they studied the curves modular
properties. Indeed Craig gives a parameterization of the curve in terms of
modular functions following work of Ramanujan. Craig does not identify the
curve as Bring’s, whereas Hulek finds the equation as a hyperplane section of
a surface that parameterises elliptic normal curves of degree 5, and then
refers to previous results to connect it with Bring. This connection was first
made in [Nar78], where the curve is attributed to Klein. Indeed, because of
this interpretation, one can write down Bring’s curve as

$\overline{\mathcal{H}/\Gamma_{0}(2,5)}\quad\text{where}\quad\Gamma_{0}(2,5)=\Gamma_{0}(2)\cap\Gamma(5)=\left\\{\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in SL_{2}(\mathbb{Z})\,\big{|}\,\begin{array}[]{l}a\equiv
d\equiv 1\pmod{5}\\\ \phantom{d\equiv}\;b\equiv 0\pmod{5}\\\
\phantom{d\equiv}\;c\equiv 0\pmod{10}\end{array}\right\\}.$

To our knowledge, [Hul87] is the first occurrence of this plane model of the
curve; Klein will write it down in [Kle88, p. 165, p. 242] but the relation to
a plane model of Bring’s curve is not made explicit. We will call (2) the HC-
model of Bring’s curve. In the following subsection we shall show the
birational equivalence of (2) and (1).

Another plane model of Bring’s curve is given by
$y^{5}-(x+1)x^{2}(x-1)^{-1}=0$, introduced in [Web05, Proposition 3.1] by
considering the curve as a cyclic cover of $\mathbb{P}^{1}$, see §4.

###  2.1. Basic Properties

Towards proving the birational equivalence of the two models given for Bring’s
curve we begin with some basic properties of the curve. We introduce affine
coordinates $(x,y)=(X/Z,Y/Z)$ such that the HC-model is $0=f(x,y)=F(x,y,1)$.
Then
$\operatorname{Res}_{y}(f(x,y),\partial_{y}f(x,y))=x^{4}(x^{5}-1)^{2}(256x^{10}-837x^{5}+3456)$.
Investigation of the vanishing of this resultant leads to the following.

######  Lemma 2.4 ([BN12]).

The only singular (double) points in the HC-model of the curve are
$V_{k}=[\zeta^{k}:\zeta^{2k}:1]$ for $k=0,\dots,4$, where $\zeta=\exp(2\pi
i/5)$, and $V_{5}=[1:0:0]$.

We shall return these points in due course; each of these singular points
desingularizes to two points. We also find

######  Lemma 2.5.

In the HC-model we have local expansions about the following points on the
curve

$\displaystyle a$ $\displaystyle:=[0:0:1]\simeq[2t^{3}:t:1],$ $\displaystyle
b$ $\displaystyle:=[0:1:0]\simeq[2t^{2}:1/t:1],$ $\displaystyle c$
$\displaystyle:=[1:0:0]_{2}\simeq[1:t:t^{4}],$ $\displaystyle d$
$\displaystyle:=[1:0:0]_{1}\simeq[1:t^{4}:t].$

We see from this that

######  Corollary 2.6.

$\operatorname{div}(x)=3a+2b-4c-d,\qquad\operatorname{div}(y)=a-b-3c+3d,\qquad
R_{F}=2a+b+3c+\sum_{i}r_{i},$

where $R_{F}$ is the ramification divisor corresponding to the map
$x:\mathcal{B}\to\mathbb{P}^{1}$ and the $r_{i}$ are the roots of the
polynomial $256x^{10}-837x^{5}+3456$ appearing in the resultant.

######  Lemma 2.7 ([BN12]).

The genus of Bring’s curve is $g=4$ and we have the ordered basis of
(unnormalised) differentials

$\displaystyle v_{1}$
$\displaystyle=\frac{(y^{3}-x)\mathrm{d}x}{\partial_{y}f(x,y)},$
$\displaystyle v_{2}$
$\displaystyle=\frac{(y^{2}x-1)\mathrm{d}x}{\partial_{y}f(x,y)},$
$\displaystyle v_{3}$
$\displaystyle=\frac{(y-x^{2})\mathrm{d}x}{\partial_{y}f(x,y)},$
$\displaystyle v_{4}$
$\displaystyle=\frac{y(x^{2}-y)\mathrm{d}x}{\partial_{y}f(x,y)}.$

These differentials, while independent as holomorphic differentials, do
satisfy an algebraic relation.

######  Lemma 2.8.

The canonical embedding of Bring’s curve in
$\mathbb{P}^{3}=\left\\{[v_{1}:v_{2}:v_{3}:v_{4}]\right\\}$ lies on the
quadric

(3) $\mathcal{Q}:v_{1}v_{2}+v_{3}v_{4}=0.$

While all full-rank quadrics in $\mathbb{P}^{3}$ are isomorphic to
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ [Vak10], the standard proof of which goes
via showing every such quadric can be written in some coordinate system as
$\mathcal{Q}$, we will subsequently want an explicit form of the isomorphism.

######  Proposition 2.9.

$\mathcal{Q}\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$.

###### Proof.

Consider the map $\mathcal{Q}\to\mathbb{P}^{1}\times\mathbb{P}^{1}$, a
modification of the Segre embedding, given by

$\varphi:\mathbb{P}^{1}\times\mathbb{P}^{1}\rightarrow\mathbb{P}^{3},\qquad\varphi([u:v],[z:w])=[uz:vw:vz:-uw]:=[v_{1}:v_{2}:v_{3}:v_{4}].$

Then $\varphi([u:v],[z:w])\in\mathcal{Q}$.

The map $\varphi$ is injective for suppose
$\varphi([u:v],[z:w])=\varphi([u^{\prime}:v^{\prime}],[z^{\prime}:w^{\prime}])$.
Consider first that $v_{4}\neq 0$. Then

$-\frac{v_{1}}{v_{4}}=\frac{z}{w}=\frac{z^{\prime}}{w^{\prime}},\qquad-\frac{v_{2}}{v_{4}}=\frac{v}{u}=\frac{v^{\prime}}{u^{\prime}},$

and so correspond to the same point in $\mathbb{P}^{1}\times\mathbb{P}^{1}$.
Next suppose that $0=-v_{4}=uw=u^{\prime}w^{\prime}$ and without loss of
generality that $u=0$. If $u^{\prime}=0$ we have injectivity, so suppose that
$u^{\prime}\neq 0$. Then $w^{\prime}=0$ and
$[0:vw:vz:0]=[u^{\prime}z^{\prime}:0:v^{\prime}z^{\prime}:0]$ requiring
$z^{\prime}=0$ and a contradiction. Thus we have that $\varphi$ is injective.

To show the surjectivity of $\varphi$ we may assume without loss of generality
that $v_{4}\neq 0$. Then set

$-\frac{v_{1}}{v_{4}}=\frac{z}{w},\qquad-\frac{v_{2}}{v_{4}}=\frac{v}{u}.$

Indeed we can spot the inverse map
$\varphi^{-1}:\mathcal{Q}\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ to be given by

$\varphi^{-1}([v_{1}:v_{2}:v_{3}:v_{4}])=\left\\{\begin{array}[]{cc}([v_{1}:v_{3}],[v_{3}:v_{2}])&\left\\{v_{1},v_{3}\right\\}\neq\left\\{0\right\\}\neq\left\\{v_{2},v_{3}\right\\},\\\
([-v_{4}:v_{2}],[v_{3}:v_{2}])&\left\\{v_{1},v_{3}\right\\}=\left\\{0\right\\}\neq\left\\{v_{2},v_{3}\right\\},\\\
([v_{1}:v_{3}],[v_{1}:-v_{4}])&\left\\{v_{1},v_{3}\right\\}\neq\left\\{0\right\\}=\left\\{v_{2},v_{3}\right\\},\\\
([-v_{4}:v_{2}],[v_{1}:-v_{4}])&\left\\{v_{1},v_{3}\right\\}=\left\\{0\right\\}=\left\\{v_{2},v_{3}\right\\}.\end{array}\right.$

∎

######  Proposition 2.10 ([BN12]).

The canonical divisor class on Bring’s curve is
$[\mathcal{K}_{\mathcal{B}}]=[a+2b+3c]$.

###### Proof.

This may be shown analytically or (as given in the accompanying notebook)
entirely using computer algebra. To show this analytically we use
$\operatorname{div}(dx)=-2(4c+d)+R_{F}$. ∎

We now establish the main result of this subsection.

######  Proposition 2.11.

The HC-model (2) is a model of Bring’s curve (1).

###### Proof.

We will prove this by explicitly constructing the birational transform. To do
so we make use of the proof of Theorem 3 in [Dye95], where the author
considers a particular Clebsch hexagon222For the purposes of this definition a
hexagon is a set of 6 points in $\mathbb{P}^{2}$, no three of which are
collinear, called the vertices. A Brianchon point of a hexagon is a non-vertex
point through which 3 edges (the lines joining two distinct vertices) pass. A
Clebsch hexagon is a hexagon with 10 Brianchon points [Dye91]., constructs a
pencil of plane sextics from this, and finds Bring’s curve as the canonical
model of a distinguished point in this pencil. By assuming that the HC-model
is already the distinguished curve in a pencil, we can construct the
birational map. Note this is a fundamentally different approach to the that
originally taken with the HC-model, which was derived from considerations of
the modular theory of the curve.

Explicitly, Dye introduces $j$ as a solution to $j^{2}-j-1=0$ and then defines
the pencil of curves333We are using the coordinates $[x,y,z]$ here, distinct
from $[X,Y,Z]$, to highlight that these are not those of the HC-model.
$S_{\lambda}=S+\lambda\left\lvert\mathcal{C}\right\rvert^{3}$ where

(4)
$\begin{split}S(x,y,z)&=(x+jy)^{6}+(x-jy)^{6}+(y+jz)^{6}+(y-jz)^{6}+(z+jx)^{6}+(z-jx)^{6},\\\
\mathcal{C}(x,y,z)&=x^{2}+y^{2}+z^{2}.\end{split}$

Next Dye considers the Clebsch hexagon $H$ with vertices

$(1,\pm j,0),\quad(0,1,\pm j),\quad(\pm j,0,1),$

for which the corresponding 10 Brianchon points are at

$(\pm j^{2},1,0),\quad(0,\pm j^{2},1),\quad(1,0,\pm j^{2}),\quad(1,\pm 1,\pm
1).$

Dye shows that there is a unique member of the pencil, which he calls
$\Gamma$, that contains the vertices. Moreover, $\Gamma$ has the vertices and
only the vertices as double points.

To get a canonical model for $\Gamma$, which Dye show’s has genus 4, we now
need a little theory from [SR49, pp. 122-124], namely that a generic cubic
surface in $\mathbb{P}^{3}$ is birational to a system of plane cubics through
6 base points. We apply this taking these six points to be the vertices of the
hexagon $H$. Such a generic cubic surface $F\subset\mathbb{P}^{3}$ can be
written as the vanishing of a determinant

$\begin{vmatrix}u_{1}&v_{1}&w_{1}\\\ u_{2}&v_{2}&w_{2}\\\
u_{3}&v_{3}&w_{3}\end{vmatrix}=0,$

where $u_{1},\dots,w_{3}$ are linear homogeneous functions of the
$\mathbb{P}^{3}$ coordinates, that is to say $P=[L_{1}:L_{2}:L_{3}:L_{4}]\in
F\subset\mathbb{P}^{3}$ if and only if there exists
$P^{\prime}=[X:Y:Z]\in\mathbb{P}^{2}$ such that

$\displaystyle Xu_{i}(P)+Yv_{i}(P)+Zw_{i}(P)=0.$

Thinking of $P^{\prime}$ as a point in a plane $\Pi$ we get a birational
transformation $\Pi\leftrightarrow F$, $P^{\prime}\leftrightarrow P$. The map
$\Psi:\Pi\to F$ will have the $L_{a}$ as homogeneous cubics in $X,Y,Z$. To see
this rewrite the determinant equation as (for $i=1,2,3$)

$\sum_{a=1}^{4}a_{ia}(P^{\prime})L_{a}=0$

for some $a_{ia}$ linear homogeneous in the $X,Y,Z$. On each affine patch
$L_{a}\neq 0$ solving this involves inverting a $3\times 3$ matrix whose
entries are linear homogeneous polynomials in $X,Y,Z$. Likewise, given the
$L_{a}$, we have a 3-parameter family of cubics given by

$aL_{1}+bL_{2}+cL_{3}+dL_{4}=0\quad\text{for}\quad[a:b:c:d]\in\mathbb{P}^{3}.$

A cubic in $\mathbb{P}^{2}$ has 10 projective coefficients, and so a
3-parameter family is defined by 6 constraints. Generically we can take those
constraints to come in the form of intersection with 6 generic points
$O_{i}\in\Pi$.

For our purposes, the six points we want to intersect with are the vertices of
the Clebsch hexagon, which are the double points of the exceptional curve
$\Gamma$. Assuming the the HC-model gives such an exceptional curve, we take
the points $V_{k}$ identified in Lemma 2.4. Hence, if we write a generic cubic
in $X,Y,Z$ as

$a_{0}X^{3}+a_{1}X^{2}Y+a_{2}X^{2}Z+a_{3}XY^{2}+a_{4}XYZ+a_{5}XZ^{2}+a_{6}Y^{3}+a_{7}Y^{2}Z+a_{8}YZ^{2}+a_{9}Z^{3}$

the equations on the coefficients we get (coming from intersecting with
$V_{k}$ ($k=0,\dots,4$) and $V_{5}$ respectively) are

$\displaystyle
a_{0}\zeta^{3k}+a_{1}\zeta^{4k}+a_{2}\zeta^{2k}+a_{3}+a_{4}\zeta^{3k}+a_{5}\zeta^{k}+a_{6}\zeta^{k}+a_{7}\zeta^{4k}+a_{8}\zeta^{2k}+a_{9}$
$\displaystyle=0,$ $\displaystyle a_{0}$ $\displaystyle=0.$

Setting the coefficients of $\zeta^{nk}$ to be zero gives us the 3-parameter
family of cubics

$\displaystyle 0$
$\displaystyle=aX^{2}Y+bX^{2}Z+cXY^{2}+dXZ^{2}-dY^{3}-aY^{2}Z-bYZ^{2}-cZ^{3},$
$\displaystyle:=aL_{1}+bL_{2}+cL_{3}+dL_{4}.$

Comparing the coefficients we see that our map into $\mathbb{P}^{3}$ is
(essentially) the canonical embedding. as

$[v_{1}:v_{2}:v_{3}:v_{4}]=[-L_{4}:L_{3}:-L_{2}:L_{1}].$

One can check, using for example Gröbner bases, that the $L_{a}$ satisfy the
equation

$L_{2}L_{4}^{2}-L_{1}^{2}L_{4}-L_{1}L_{3}^{2}+L_{2}^{2}L_{3}=0.$

This is the cubic we call $F$. One can use the package Cremona [Sta18] in
Macaulay2 to check that the map $\Psi$ is birational. Note one needs to make
sure that the range is chosen such that the map is explicitly surjective, not
just use the implicit knowledge that the map is surjective onto its image.
Doing so and asking for the inverse map gives

$\displaystyle[X:Y:Z]=[L_{2}^{2}-L_{1}L_{3}:L_{1}L_{4}:L_{2}L_{4}].$

For example, we can see

$\displaystyle\frac{L_{2}^{2}-L_{1}L_{3}}{L_{1}L_{4}}$
$\displaystyle=\frac{\left(X^{2}Z-YZ^{2}\right)^{2}-\left(X^{2}Y-Y^{2}Z\right)\left(XY^{2}-Z^{3}\right)}{\left(X^{2}Y-Y^{2}Z\right)\left(XZ^{2}-Y^{3}\right)},$
$\displaystyle=\frac{X^{4}Z^{2}-X^{2}YZ^{3}-X^{3}Y^{3}+XY^{4}Z}{X^{3}YZ^{2}-XY^{2}Z^{3}-X^{2}Y^{4}+Y^{5}Z}=\frac{X}{Y},$

and

$\frac{L_{1}L_{4}}{L_{2}L_{4}}=\frac{L_{1}}{L_{2}}=\frac{X^{2}Y-Y^{2}Z}{X^{2}Z-YZ^{2}}=\frac{(X^{2}Z-YZ^{2})(Y/Z)}{X^{2}Z-YZ^{2}}=\frac{Y}{Z}.$

We now have a cubic surface in $\mathbb{P}^{3}$ corresponding to the system of
curves intersecting the $V_{i}$. From [Hir86, pp. 198-201] we know there
exists a coordinate system in which this cubic can be written as the subset of
$\mathbb{P}^{4}$

$H_{3}=0=H_{1}$

(this is called ‘the’ Clebsch surface). To look for such a coordinate system,
we write this surface in $\mathbb{P}^{3}$ as

$\displaystyle 0$
$\displaystyle=x_{1}^{2}x_{2}+x_{1}x_{2}^{2}+x_{1}^{2}x_{3}+x_{2}^{2}x_{3}+x_{1}x_{3}^{2}+x_{2}x_{3}^{2}+x_{1}^{2}x_{4}+x_{2}^{2}x_{4}+x_{3}^{2}x_{4}+x_{1}x_{4}^{2}+x_{2}x_{4}^{2}+x_{3}x_{4}^{2}$
$\displaystyle\phantom{=}+2x_{1}x_{2}x_{3}+2x_{1}x_{2}x_{4}+2x_{1}x_{3}x_{4}+2x_{2}x_{3}x_{4}.$

From here, one can use the fact that the 10 Eckardt points444An Eckardt point
of a surface is a point where three lines contained within the surface
intersect [Hir86]. of the cubic in $L_{a}$ form the 10 vertices of a
pentahedron [Hir86], and that in the coordinate system of the Clebsch surface
in $\mathbb{P}^{4}$ these lie at the permutations of $[1:-1:0:0:0]$ [Edg78].
This gives us a possible way a spotting the transform if we can calculate the
Eckardt points of $F$. To do this, we use the identification from [Dye95],
that the 3 lines in $F$ intersecting to give an Eckardt point come from the 3
edges of $H$ intersecting at a Brianchon point. To this end we find the images
of the $V_{i}V_{j}$ for which we give a generating set of the ideal
corresponding to the line, for example

  1. (1)

$\Psi(V_{0}V_{5}):\left\langle L_{4}-L_{3},L_{2}-L_{1}\right\rangle$,

  2. (2)

$\Psi(V_{1}V_{2}):\left\langle
L_{4}+(\zeta^{2}+\zeta+1)L_{2}+(\zeta^{4}-\zeta^{2}-\zeta)L_{1},L_{3}+(\zeta^{3}+2\zeta^{2}+\zeta)L_{2}+(\zeta^{3}+\zeta^{2}+\zeta)L_{1}\right\rangle$,

  3. (3)

$\Psi(V_{3}V_{4}):\left\langle
L_{4}+(-\zeta^{2}-\zeta)L_{2}+(\zeta^{3}+2\zeta^{2}+\zeta)L_{1},L_{3}+(\zeta^{3}-\zeta-1)L_{2}+(\zeta^{2}+\zeta+1)L_{1}\right\rangle$.

Then

$\Psi(V_{0}V_{5})\cap\Psi(V_{1}V_{2})\cap\Psi(V_{3}V_{4})=[-\zeta^{3}-\zeta^{2}-1:-\zeta^{3}-\zeta^{2}-1:1:1]=\Psi(V_{0}V_{5}\cap
V_{1}V_{2}\cap V_{3}V_{4}).$

One can do likewise to find the other Eckardt points.

Armed now with the knowledge of the Eckardt points, we can find appropriate
projective transforms $A$ that biject the sets of Eckardt points by acting
$(x_{1},x_{2},x_{3},x_{4})^{T}=A(L_{1},L_{2},L_{3},L_{4})^{T}$, for example

$A=\begin{pmatrix}\zeta^{3}&-1&-\zeta^{2}&\zeta\\\
1&-\zeta^{3}&-\zeta&\zeta^{2}\\\ \zeta^{2}&-\zeta&-1&\zeta^{3}\\\
\zeta&-\zeta^{2}&-\zeta^{3}&1\end{pmatrix}.$

We can then use again Macaulay2 to check that this transform then gives us the
correct cubic in $\mathbb{P}^{3}$. Moreover, if we now restrict to the HC-
model in the system of curves that intersect the $V_{i}$, that is we impose
the condition that $F(X,Y,Z)=0$, we get the final degree-2 polynomial on the
$x_{i}$ ($H_{2}=0$). This quadric is in fact the Schur quadric corresponding
to a distinguished double-six555A double six is a collection of 12 lines
$a_{1},\dots,a_{6},b_{1},\dots,b_{6}$ in $\mathbb{P}^{3}$, arranged as
$\begin{array}[]{cccccc}a_{1}&a_{2}&a_{3}&a_{4}&a_{5}&a_{6}\\\
b_{1}&b_{2}&b_{3}&b_{4}&b_{5}&b_{6}\end{array}$ so that each line is disjoint
from those in the same row and column, but intersects the other 5 lines. of
lines that necessarily exist on a cubic surface constructed as above [Dye95].
An equivalent rational map is given in [Dol12, p. 557], but the inverse is not
provided.

To see this working, let’s consider $c$ and $d$, and see that these are
desingualrised on the smooth canonical embedding. Taking $[X:Y:Z]=[1:t:t^{4}]$
we get

$[L_{1}:L_{2}:L_{3}:L_{4}]=[1:t^{3}:t+t^{6}:-t^{2}],$

and so taking the limit we get $[1:0:0]_{2}\mapsto[1:0:0:0]$ in $L$
coordinates. Acting with $A$, we get the point in $\mathbb{P}^{4}$ given by

$[x_{1}:x_{2}:x_{3}:x_{4}:x_{5}]=[\zeta^{3}:1:\zeta^{2}:\zeta:\zeta^{4}].$

Repeating the process taking $[X:Y:Z]=[1:t^{4}:t]$ gives
$[1:0:0]_{1}\mapsto[0:1:0:0]$ in $L$ coordinates, which is equivalently

$[x_{1}:x_{2}:x_{3}:x_{4}:x_{5}]=[1:\zeta^{3}:\zeta:\zeta^{2}:\zeta^{4}].$

This makes sense, as the change $Y\leftrightarrow Z$ corresponds to
$L_{1}\leftrightarrow L_{2}$, $L_{3}\leftrightarrow L_{4}$.

Moreover, to clarify our last point on the imposition of $H_{2}=0$, we
consider the point $[X:Y:Z]=[0:1:1]$ which does not lie on the HC-model of the
curve. Under our birational map this corresponds to the point

$\displaystyle[L_{1}:L_{2}:L_{3}:L_{4}]$
$\displaystyle=[1:1:1:1]\Rightarrow[x_{1}:x_{2}:x_{3}:x_{4}:x_{5}]$
$\displaystyle=[2-\sqrt{5}:-2+\sqrt{5}:-1:1:0].$

This point does lie on the Clebsch diagonal surface given by $H_{1}=0=H_{3}$,
but does not satisfy $H_{2}=0$. ∎

Note that in the above proof of the equivalence of the Riemann surfaces, the
birational map we constructed was defined over $\mathbb{Q}[\zeta]$. As such,
to equate the two algebraic curves we need to be working over a field
containing $\mathbb{Q}[\zeta]$. We will later see when looking at quotients of
Bring’s curve that it is insufficient to work over $\mathbb{Q}$. Indeed, note
that over $\mathbb{Q}$ there are no solutions to the equations defining
Bring’s curve in the $\mathbb{P}^{4}$-model.

###  2.2. The Period Matrix

We record now a result from the literature.

######  Theorem 2.12 ([RR92], [GAR00]).

Define the matrices $M$, $M_{S}$ by

$M=\begin{pmatrix}4&1&-1&1\\\ 1&4&1&-1\\\ -1&1&4&1\\\
1&-1&1&4\end{pmatrix},\quad M_{S}=\begin{pmatrix}4&-1&-1&-1\\\ -1&4&-1&-1\\\
-1&-1&4&-1\\\ -1&-1&-1&4\end{pmatrix}.$

A Riemann matrix for Bring’s curve is given by $\tau_{\mathcal{B}}=\tau_{0}M$,
where $\tau_{0}=-0.5+0.186676i\,(6.d.p)$ is given by the conditions

$j(\tau_{0})=-\frac{29^{3}\times 5}{2^{5}},\quad j(5\tau_{0})=-\frac{25}{2}.$

Further, there exists a symplectic transformation such that
$\tau_{\mathcal{B}}=\tau_{0}M_{S}$.

###### Proof.

This was first shown in [RR92], but the equations for $j(\tau_{0})$,
$j(5\tau_{0})$ were incorrectly swapped as first noted666Note the
approximation of $\tau_{0}$ in [BN12] contains a typographic error. in [BN12],
where the above period matrix was also calculated via the HC-model.

Moreover, in [Web05], a period matrix is constructed with respect to a
homology basis with intersection matrix

$I_{W}=\begin{pmatrix}0&1&1&-1&-1&0&0&1\\\ -1&0&1&1&0&1&0&-1\\\
-1&-1&0&1&0&-1&0&0\\\ 1&-1&-1&0&0&0&1&0\\\ 1&0&0&0&0&1&1&-1\\\
0&-1&1&0&-1&0&1&1\\\ 0&0&0&-1&-1&-1&0&1\\\ -1&1&0&0&1&-1&-1&0\end{pmatrix},$

that is given as $\Omega=(A,B)$ where the columns $A_{k},B_{k}$ are

$A_{k}=\begin{pmatrix}\zeta^{k}(1-\zeta^{2})\\\
\zeta^{2k+3/2}(1-\zeta^{4})(l\Phi-1)\\\ \zeta^{4k+3}(1-\zeta^{3})\Phi(1-l)\\\
\zeta^{3k+2}(1-\zeta)l\end{pmatrix},\quad
B_{k}=\begin{pmatrix}\zeta^{2k+3/2}(1-\zeta^{4})(l\Phi-1)\\\
\zeta^{4k+3}(1-\zeta^{3})\Phi(1-l)\\\ \zeta^{3k+2}(1-\zeta)l\\\
\zeta^{k}(1-\zeta^{2})\end{pmatrix},$

for $k=0,\dots,3$, $\Phi=\frac{1}{2}(1+\sqrt{5})$, and
$l=\left\lvert{\frac{I(-1,0)}{I(-\infty,-1)}}\right\rvert\approx 0.848641$
where

$I(a,b)=\int_{a}^{b}(t-1)^{-1/5}t^{-3/5}(t+1)^{-4/5}dt.$

Note [Web05, Lemma 5.1] uses $\zeta=\exp(2\pi i/10)$, whereas we take
$\zeta=\exp(2\pi i/5)$. As such, the columns of $\Omega$ look different to
those of Weber in terms of the exponent of $\zeta$, but they do indeed agree.

One can find (using Sage) the matrix

$C=\begin{pmatrix}0&1&-2&1&1&1&0&-1\\\ 0&1&-1&1&0&0&-1&0\\\
0&0&-1&0&0&1&1&0\\\ 1&1&-2&2&0&1&-1&-1\\\ 0&0&0&0&0&0&1&1\\\
0&0&1&0&0&0&0&0\\\ 0&0&0&0&0&0&0&1\\\ 0&0&0&1&0&0&0&0\end{pmatrix}$

such that $C^{T}I_{W}C=J_{g}=\begin{pmatrix}0&\operatorname{Id}_{g}\\\
-\operatorname{Id}_{g}&0\end{pmatrix}$, i.e. $C$ ‘canonicalises’ the homology
basis. This means we get a Riemann matrix
$\tau_{W}=(AC)^{-1}(BC)=C^{-1}A^{-1}BC$, and one can numerically find that the
matrix

$R=\begin{pmatrix}\delta&\beta\\\
\gamma&\alpha\end{pmatrix}=\begin{pmatrix}0&0&0&1&0&0&1&2\\\
0&-1&1&0&-1&-2&1&1\\\ 1&0&1&0&2&1&1&1\\\ 0&1&0&0&1&2&0&0\\\ 1&0&0&0&2&1&0&1\\\
1&0&1&-1&1&1&1&-1\\\ -1&0&0&1&-1&-1&1&2\\\ 1&0&1&-1&1&2&1&-1\end{pmatrix}$

relates $\tau_{W}$ to $\tau_{\mathcal{B}}$ by

(5) $\tau_{\mathcal{B}}=(\delta+\tau_{W}\gamma)^{-1}(\beta+\tau_{W}\alpha).$

Here $R$ is a symplectic transform with respect the standard symplectic form
$J_{g}$.

For a fourth proof, we may consider the Riemann matrix numerically calculated
by SageMath, and find a transform to $\tau_{\mathcal{B}}$, as is done in the
corresponding notebook. ∎

Riera and Rodriguez (hereafter abbreviated to R&R) constructed the constraints
on $\tau_{0}$ via $j$-invariants by considering the quotients by group actions
of the curve to elliptic curves [RR92]. They show that these constraints give
a unique value of $\tau_{0}$ modulo $\Gamma_{0}(5)$, or equivalently in the
language of [GAR00] that $\tau_{0}$ gives a distinguished point in the modular
curve $X_{0}(5)$. As we will see later in §4.4, there are additional quotients
to elliptic curves not considered by R&R, which have $j$-invariants

(6) $\displaystyle j(15\tau_{0})$ $\displaystyle=-\frac{5^{2}\times
241^{3}}{2^{3}},$ (7) $\displaystyle j(3\tau_{0})$
$\displaystyle=\frac{5\times 211^{3}}{2^{15}}.$

The latter is identified in relation to Bring’s curve in [Ser08, Exercise
8.3.2c]. Serre says that this curve (50H) and 50E (using the naming convention
of [BK75]) with $j(5\tau_{0})$ are 15-isogenous over $\mathbb{Q}$. This
isogeny is not too mysterious when we think on the level of the corresponding
elliptic curves over $\mathbb{C}$ as $\mathbb{C}/\left\langle
1,3\tau_{0}\right\rangle$ and $\mathbb{C}/\left\langle
1,5\tau_{0}\right\rangle$, wherein the isogeny $\mathbb{C}/\left\langle
1,3\tau_{0}\right\rangle\to\mathbb{C}/\left\langle 1,5\tau_{0}\right\rangle$
is the composition of the quotients by the maps $z\mapsto z+\tau_{0}$ and
$z\mapsto z+1/5$ respectively. There is a complete $\mathbb{Q}$-isogeny class
of elliptic curves of order four with periods $\tau_{0},3\tau_{0},5\tau_{0}$
and $15\tau_{0}$ [BK75].

It follows from the above that $\tau_{0}$ is transcendental. By a theorem of
Schneider [BW08], if $j(z)$ is rational, $z$ is either transcendental or an
element of a quadratic imaginary field with class number 1. However, the
$j$-invariants of elements of these fields are known to be integers, which
doesn’t occur in the case of $\tau_{0}$. In Weber’s form of the period matrix,
the transcendentality comes about because of the constant $l$, which is a
ratio of Schwarz-Christoffel integrals which arise from the map of a Euclidean
quadrilateral to a hyperbolic quadrilateral.

###  2.3. The Automorphism Group

As previously noted, the $\mathbb{P}^{4}$-model of Bring’s curve shows that
the automorphism group $\operatorname{Aut}(\mathcal{B})$ contains $S_{5}$.
Hurwitz’s theorem on the automorphism group
$G=\operatorname{Aut}(\mathcal{C})$ of a curve of genus $g\geq 2$ yields that
$|G|\in\\{84(g-1),48(g-1),40(g-1),\ldots\\}$ and that if $p$ is a prime where
$p\big{|}|G|$, then $p\in\\{2,\ldots,g,g+1,2g+1\\}$. For genus $4$ this means
$p\in\\{2,3,5\\}$ and so the maximal automorphism group of a genus $4$ curve
has order strictly less than $84(g-1)$. In 1895 Wiman showed777An English
translation of Wiman’s paper has been produced [DHBD22].

######  Proposition 2.13 ([Wim95]).

$\operatorname{Aut}(\mathcal{B})=S_{5}$. This is the maximal possible
automorphism group for a genus-4 surface, and Bring’s curve is the only curve
to achieve it.

Wiman’s proof was enumerative and produced equations for the curves with a
given automorphism group. He had previously dealt with the hyperelliptic
curves and their automorphism groups and for non-hyperelliptic curves
$\mathcal{C}$ of genus 4 he began with the canonical embedding of the curve in
$\mathbb{P}^{3}$ which meant that their automorphisms could be taken to be
collineations, and so forming a subgroup of $PGL_{4}(\mathbb{C})$. Wiman’s
approach then used the fact (see [Har77, IV.5.2.2]) that the canonical
embedding of a non-hyperelliptic curve of genus 4 is the complete intersection
of an irreducible cubic surface and a unique quadric surface (which is either
irreducible, or a cone). The uniqueness of the quadric $Q$ means that
automorphisms of the curve become automorphisms of $Q$. In the case where $Q$
is of full rank we have that it is isomorphic to
$\mathbb{P}^{1}\times\mathbb{P}^{1}$, hence we know
$\operatorname{Aut}(\mathcal{C})$ is isomorphic to a finite subgroup of
$\operatorname{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})=C_{2}\ltimes(PGL_{2}(\mathbb{C})\times
PGL_{2}(\mathbb{C}))$ and that using the coordinates of
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ Wiman could express the curve
$\mathcal{C}$ as an equation of bidegree $(3,3)$ in terms of these. In the
singular case $Q$ is a quadric cone and by projecting from a point of the cone
meant that $\mathcal{C}$ could be expressed as a plane sextic. In both cases
Wiman could write equations for possible genus $4$ curves and his strategy was
to look at the restrictions on these imposed by symmetries of orders $2$, $3$
and $5$. For curves without an order $5$ element the maximal order of symmetry
group was $72$ for $Q$ either a smooth quadric or cone. Including an order $5$
element yielded a maximal symmetry group of order $120$ only in the case of
the smooth quadric with the resulting curve being Bring’s. In the case of
Bring’s curve the quadric $Q$ is the quadric $\mathcal{Q}$ we determined
earlier. Different proofs of Wiman’s result may be found in [KK90, MSSV02].

It is clear from the $\mathbb{P}^{4}$-model that
$\operatorname{Aut}(\mathcal{B})$ may be realised as projective transforms via
the permutation representation of $S_{5}$ acting on the subspace
$\sum_{i}x_{i}=0$; it is also clear that it may be be realised as a subgroup
of $PGL_{4}(\mathbb{C})$ via the induced action on the differentials or
equivalently the $L_{a}$’s. What is non-trivial is the following fact.

######  Theorem 2.14 ([Dye95]).

The $A_{5}$ subgroup of $\operatorname{Aut}(\mathcal{B})$ can be realised as a
group of collineations in the HC-model, that is, can be realised as a subgroup
of $PGL_{3}(\mathbb{C})$.

###### Proof.

We will be explicit about the construction here as this will be profitable
later; [BN12] explains how this representation follows from [Dye95]. The group
$A_{5}$ has two inequivalent irreducible three-dimensional representations,
one of which is given by $\left\langle R,S\right\rangle$ where

$R:=\frac{1}{\sqrt{5}}\begin{pmatrix}1&2&2\\\
1&\zeta^{2}+\zeta^{-2}&\zeta+\zeta^{-1}\\\
1&\zeta+\zeta^{-1}&\zeta^{2}+\zeta^{-2}\end{pmatrix},\,{R}^{2}=I,\,S:=\begin{pmatrix}1&&\\\
&\zeta&\\\ &&\zeta^{-1}\end{pmatrix},\ {S}^{5}=I,\,(RS)^{3}=I.$

Note the other inequivalent irreducibel 3-dimensional representation comes
from replacing $\zeta$ in the above with $\zeta^{2}$ or $\zeta^{3}$. The
invariants of the representation $\left\langle R,S\right\rangle$ (when acting
on $(X,Y,Z)^{T}$ via left multiplication) are

$\displaystyle i_{2}$
$\displaystyle=\left(\frac{X}{2}\right)^{2}+\sum_{k=0}^{4}\left(\frac{\frac{X}{2}+Y\zeta^{k}+Z\zeta^{-k}}{\sqrt{5}}\right)^{2}=\frac{1}{2}X^{2}+2YZ,$
$\displaystyle i_{6}$
$\displaystyle=\left(\frac{X}{2}\right)^{6}+\sum_{k=0}^{4}\left(\frac{\frac{X}{2}+Y\zeta^{k}+Z\zeta^{-k}}{\sqrt{5}}\right)^{6},$
$\displaystyle i_{10}$
$\displaystyle=\left(\frac{X}{2}\right)^{10}+\sum_{k=0}^{4}\left(\frac{\frac{X}{2}+Y\zeta^{k}+Z\zeta^{-k}}{\sqrt{5}}\right)^{10},$
$\displaystyle i_{15}$
$\displaystyle=\left\lvert\frac{\partial\\{i_{2},i_{6},i_{10}\\}}{\partial\\{X,Y,Z\\}}\right\rvert.$

There is a polynomial relation between $i_{15}^{2}$ and $i_{2}$, $i_{6}$,
$i_{10}$. In particular the vanishing of $i_{6}-\lambda i_{2}^{3}$ gives us
Dye’s one-parameter family of $A_{5}$-invariant sextics in $\mathbb{P}^{2}$;
this pencil appears to have first been studied by Winger [Win25]. This pencil
yields genus-10 curves for generic $\lambda$ [Dye95].888A variation on this
pencil has been used to explain why Bring’s curve is uniquely defined as an
$A_{5}$-invariant curve of genus 4, but there is a 1-parameter family of
dimension-4 $A_{5}$-invariant principally polarised abelian varieties,
deforming $\operatorname{Jac}\mathcal{B}$ [Zi21, LZ21]. The paper [Mel03]
gives further interesting representation theoretic perspectives on Bring’s
curve. The special value of $\lambda=13/100$999For $\lambda=17/180$ the curve
is of genus 0; for $\lambda=1/10$ the curve is reducible. yields a genus-4
curve, namely Bring’s curve, as we have that

$\frac{1}{12}\left(100i_{6}-13i_{2}^{3}\right)=X(Y^{5}+Z^{5})+(XYZ)^{2}-X^{4}YZ-2(YZ)^{3}=F(X,Y,Z).$

∎

To complete our picture, we use the following result.

######  Proposition 2.15.

The map

$U:\left(x,y\right)\mapsto\left(-\frac{y^{5}+x^{3}y-3xy^{2}+1}{(y-x^{2})(y^{3}-x)},-\frac{y^{2}x-1}{y^{3}-x}\right)$

is an automorphism of the HC-model, and together with $R$ and $S$ generates
the entire automorphism group $S_{5}$.

###### Proof.

The proof that $U$ is an automorphism is simple algebraic verification. In
order to find this map, we adapted the methods of [BSZ19]. To see that
$\left\langle R,S,U\right\rangle\cong S_{5}$, note that $U$ is of order $4$,
for example by checking that it has the orbit $a\mapsto c\mapsto b\mapsto
d\mapsto a$. As such, $U$ corresponds to an odd permutation under the
isomorphism $\operatorname{Aut}(\mathcal{B})\cong S_{5}$, and a single odd
permutation and all of $A_{5}$ together generate $S_{5}$. ∎

By fixing a map from the HC-model to the $\mathbb{P}^{4}$-model as we did in
the proof of Proposition 2.11, we have fixed an isomorphism from the
automorphism group of the curve (in the HC-model) to $S_{5}$, which we shall
denote $\psi:\left\langle R,S,U\right\rangle\to S_{5}$. For example, it is
simple to verify that we have $U^{2}([X:Y:Z])=[X:Z:Y]$. We see
$U^{2}([L_{1}:L_{2}:L_{3}:L_{4}])=[L_{2}:L_{1}:L_{4}:L_{3}]$, and so
$U^{2}([x_{1}:x_{2}:x_{3}:x_{4}:x_{5}])=[x_{2}:x_{1}:x_{4}:x_{3}:x_{5}]$, that
is $(12)(34)$. Through similar calculation we can find

$\displaystyle\psi(R)$
$\displaystyle=(13)(24),\quad\psi(S)=(13425),\quad\psi(U)=(1324).$

There are myriad choices that can be made in constructing the birational
transformation (such as the labelling of the coordinates in $\mathbb{P}^{3}$
and the ordering of the rows of $A$, or indeed composing with any automorphism
of the $\mathbb{P}^{4}$-model), and changing these would give different
isomorphisms to $S_{5}$.

As we have previously described, the uniqueness of the quadric $Q$ whose
intersection with a cubic yields the canonical model of the curve leads to an
isomorphism of the automorphism group of the curve and a subgroup of
$\operatorname{Aut}(Q)$. For Bring’s curve
$\mathcal{Q}\cong\mathbb{P}^{1}\times\mathbb{P}^{1}$ and we obtain an
isomorphism from $S_{5}$ to a subgroup of
$\operatorname{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})=\mathbb{Z}_{2}\ltimes(PGL_{2}(\mathbb{C})\times
PGL_{2}(\mathbb{C}))$ which we now write down. Let $([u:v],[z:w])$ be the
coordinates on $\mathbb{P}^{1}\times\mathbb{P}^{1}$, then using the birational
map constructed in Proposition 2.11 one can conjugate the standard irreducible
4-dimensional representation of $S_{5}$ on $[x_{1}:x_{2}:x_{3}:x_{4}]$ to an
action on $[v_{1}:v_{2}:v_{3}:v_{4}]$. The resulting action of $(12)$ is
(projectively)

(8) $\displaystyle\begin{pmatrix}v\\\ u\end{pmatrix}$
$\displaystyle\mapsto\begin{pmatrix}j&-1\\\
-1&-j\end{pmatrix}\begin{pmatrix}w\\\ z\end{pmatrix}:=A\begin{pmatrix}w\\\
z\end{pmatrix},\quad\begin{pmatrix}w\\\
z\end{pmatrix}\mapsto\begin{pmatrix}-1&j-1\\\
j-1&1\end{pmatrix}\begin{pmatrix}v\\\ u\end{pmatrix}:=B\begin{pmatrix}v\\\
u\end{pmatrix},$

where $j=-\zeta^{3}-\zeta^{2}$ satisfies $j^{2}-j-1=0$; that is it is the $j$
defined by Dye. One can show that $(34)$ has the same action, where the other
root $j^{\prime}=\zeta^{3}+\zeta^{2}+1$ is taken. One can check that $AB=1\in
PGL_{2}(\mathbb{C})$, consistent with the fact that $(12)^{2}=1\in S_{5}$.
Note the transposition interchanges the two copies of $\mathbb{P}^{1}$, which
is the action of the semi-direct product with $\mathbb{Z}_{2}$. Combining the
two transforms one gets that $(12)(34)$ acts as

$[u:v]\mapsto[-v:u]\,,\quad[z:w]\mapsto[-w:z].$

This fixes each copy of $\mathbb{P}^{1}$ and is the antipodal map on each.
Moreover, we can calculate the action of $(145)$ on the copies to be

(9) $\displaystyle\begin{pmatrix}v\\\ u\end{pmatrix}$
$\displaystyle\mapsto\begin{pmatrix}\zeta&-\zeta^{3}-\zeta\\\
-\zeta^{2}-1&-\zeta^{2}\end{pmatrix}\begin{pmatrix}v\\\
u\end{pmatrix},\quad\begin{pmatrix}w\\\
z\end{pmatrix}\mapsto\begin{pmatrix}\zeta^{3}+1&\zeta^{2}\\\
\zeta^{4}&-\zeta^{3}-\zeta\end{pmatrix}\begin{pmatrix}w\\\ z\end{pmatrix}.$

As $\left\langle(12)(34),(145)\right\rangle\cong A_{5}$, we discover that the
action of $A_{5}$ does not interchange the two copies of $\mathbb{P}^{1}$, but
odd-parity elements in $S_{5}$ do. Here $A_{5}$ is given by the diagonal
embedding in $PGL_{2}(\mathbb{C})\times PGL_{2}(\mathbb{C})$.

##  3\. Geometric Points

We now have a good understanding of how the automorphism group acts on the
curve, and so before looking at quotient Riemann surfaces in §4 we want to
first consider orbits of points that have geometric significance on the curve.
These points will have important connections to the function theory of the
curve; they are also related to physical aspects of Euclidean realisations
(i.e. can be immersed in Euclidean 3-space) of the curve. Such orbits are
characterised by the following result from Wiman.

######  Proposition 3.1 ([Wim95]).

There are only 3 orbits orbits of points of size less than $120$ on
$\mathcal{B}$ and these have sizes 24, 30, and 60 respectively.

###### Proof.

By [FK92, III.7.7] we know the stabilizer of a point must be a cyclic group.
The cyclic subgroups of $S_{5}$ are $C_{2}$, $C_{3}$, $C_{4}$, $C_{5}$,
$C_{6}$; the corresponding orbits would thus be of (respective) sizes $60$,
$40$, $30$, $24$ or $20$. To obtain Wiman’s result we must show that $C_{3}$
does not fix a point on Bring’s curve. Using Riemann-Hurwitz we have

$3=120(g-1)+30a_{60}+40a_{40}+45a_{30}+48a_{24}+50a_{20}$

where $a_{k}\geq 0$ are the number of $S_{5}$ orbits of size $k$. There are no
solutions to this for $g\geq 1$. For $g=0$ we have the unique solution
$1=a_{60}=a_{30}=a_{24}$. This shows there can be no points with $C_{3}$
stabilizer. ∎

These points and corresponding geometric structures are important when
relating Clebsch’s diagonal surface to Hilbert modular surfaces [Hir76,
Bel04]. We identify these orbits as the geometric points on the curve defined
in [SW97].101010Being a geometric point on a curve is a priori not an
interesting statement unless we know the corresponding map is regular, as we
have in this case. Explicitly they are the vertices, face-centres, and edge-
centres of the the universal map $\left\\{5,4\right\\}_{6}$ \- the Petrie
polygon (as defined in [CM80, §8.6]) of degree 6 coming from the tiling of the
hyperbolic disk by pentagons, where 4 meet at a vertex [Sin88]. It is noted in
[Web05] that this tessellation has a Euclidean realisation as a
dodecadodecahedron (Figure 1(a)). This has 30 vertices, 60 edges, and 24
faces, giving genus

$g=1-\frac{V-E+F}{2}=4,$

as we expect. In a recent paper, this connection to the dodecadodecahedron was
used to identify Bring’s curve as the moduli space of equilateral plane
pentagons up to the action of the conformal group [Ram22]. The (small)
stellated dodecahedron (Figure 1(b)) also has genus 4 (having $V=12=F$,
$E=30$), coming from the tessellation
$\left\\{5/2,5\right\\}\cong\left\\{5,5|3\right\\}$, which can be interpreted
as adding 3 ‘holes’ to the $\left\\{5,5\right\\}$ tessellation [CM80, §8.5].
This $\left\\{5,5|3\right\\}$ tessellation has automorphism group $C_{2}\times
A_{5}$, which is an index-2 subgroup of $C_{2}\times S_{5}$, the automorphism
group of $\left\\{5,4\right\\}_{6}$. This is due to the map $D_{1}$ defined in
[Hen13, §3.1], which maps the dodecadodecahedron to the small sellated
dodecahedron. We include both these tessellations in Figure 2 below. Klein
connects the small stellated dodecahedron to Bring’s curve through a degree-3
covering $\mathcal{B}\to\mathbb{P}^{1}$ constructed from the hyperbolic
triangles giving the tessellation [Web05], and we will see this map later in
§5 in a different context.

(a) Dodecadodecahedron.

(b) Small Stellated Dodecahedron.

Figure 1. Geometric realisations.

(a) $\left\\{5,4\right\\}$ tessellation.

(b) $\left\\{5,5\right\\}$ tessellation.

Figure 2. Hyperbolic tilings.

###  3.1. Weierstrass Points

We recall the definition of an important class of points.

######  Definition 3.2 ([McM14], Proposition 12.6, Theorem 12.7).

A Weierstrass Point on a curve $X$ of genus $g$ with canonical embedding
$X_{c}\subseteq\mathbb{P}^{g-1}$ is a point $P$ satisfying one of the
following equivalent conditions:

  1. (1)

there exists a hyperplane $H\subset\mathbb{P}^{g-1}$such that $H\cap X_{c}$
has multiplicity at least $g$ at $P$,

  2. (2)

there exists a holomorphic differential on $X$ vanishing at $P$ with order
$\geq g$,

  3. (3)

there exists a meromorphic function on $X$ with poles just at $P$ of order
$\leq g$, and

  4. (4)

the Wronskian determinant given by
$\operatorname{Wr}(P)=\det\left(\frac{d^{i}\omega_{j}}{dz^{i}}\right)_{i,j=0,\dots,g-1}$,
where $\\{\omega_{j}\\}$ is a basis of holomorphic differentials and $z$ is a
local coordinate around $P$, vanishes.

In [SW97] the Weierstrass points are shown (implicitly) to correspond to the
edge-centres of the universal map. This shows that the order 2 rotation that
permutes the vertices and face-centres adjacent to an edge-centre will
preserve the corresponding Weierstrass point. They can also be interpreted
geometrically as pairwise-symmetric distributed along the edges of the small
stellated dodecahedron. As Weber identifies the order-3 symmetry as the
rotation about the axis through opposite vertices of the unstellated
dodecahedron, one might wonder from this picture whether Weierstrass points
are fixed points of the order-3 permutations in the group. This turns out not
be the case and a counting argument helps elucidate: the small stellated
dodecahedron has 12 faces, which in turn means we want to have
$\frac{60}{12}=5$ Weierstrass points per face. Hence where three faces overlap
there must be three Weierstrass points ‘stacked’ there, which are invisibly
permuted by the action.

Having now identified the Weierstrass points as some of the geometric points,
we give a concrete result about what the Weierstrass points are.

######  Proposition 3.3 ([Edg78]).

Bring’s curve has 60 Weierstrass points, on which
$\operatorname{Aut}(\mathcal{B})$ acts transitively. Letting
$\\{\alpha,\beta,\gamma\\}$ be the roots of the cubic $x^{3}+2x^{2}+3x+4$,
these are given in the $\mathbb{P}^{4}$-model by $W_{ijk}$ where, for example,

$W_{345}=[1:1:\alpha:\beta:\gamma].$

###### Proof.

Edge, working in the $\mathbb{P}^{4}$-model, identifies the Weierstrass points
with the 60 intersections of the curve with the 10 planes
$\Pi_{ij}=\left\\{x_{i}=x_{j}\right\\}$. To do this, Edge quotes [Wim95] to
show that these intersection points are stalls111111A stall of a plane curve
is a point where the osculating (hyper)plane has a 4-point intersection
[Edg71]., i.e. inflection points of certain linear series, and for the
canonical embedding these are exactly the Weierstrass points [ACGH85, p. 37].
Simple algebra then gives the exact expression we write down. This viewpoint
makes it clear that the Weierstrass points at the intersection with $\Pi_{ij}$
are preserved by the transposition121212In Theorem 4 of [Dye95], these
transpositions are associated with the Brianchon points of the Clebsch hexagon
$H$ from the proof of Proposition 2.11. $(ij)$ only, so have orbits of size
$\frac{120}{2}=60$, and as the automorphism restricts to a permutation of the
Weierstrass points, the action is then transitive. ∎

With our naming convention, note $W_{ijk}$ is defined by $x_{i}=\alpha$,
$x_{j}=\beta$, $x_{k}=\gamma$. If we choose a different labelling of the roots
of the cubic, this would give a different labelling of the Weierstrass points.
The Weierstrass points split as $60=6\times 10$, with 6 Weierstrass points
being fixed by each of the 10 involutions in $S_{5}$.

The property that the automorphism group acts transitively on the Weierstrass
points is very rare, as characterised by the following result.

######  Theorem 3.4 ([LS12], Theorem 15).

If $X$ is a Riemann surface of genus $g>2$ with $g^{3}-g$ Weierstrass points
on which $\operatorname{Aut}X$ acts transitively then either

  * •

$g=4$ and $X$ is Bring’s curve,

  * •

$g=3$ and $X$ is Klein’s curve, or

  * •

$g=3$ and $\operatorname{Aut}X\cong S_{4}$.

######  Remark 3.5.

As we have an explicit birational map from our plane model to our canonical
embedding of the curve, we can get the explicit forms of the Weierstrass
points in the HC-model using Edge’s identification of the Weierstrass points
in the $\mathbb{P}^{4}$-model. If one does not have this information, it is
still possible to calculate the Weierstrass points using Sage and some
educated guesswork. Using computer algebra, and the characterisation of
Weierstrass points as zeros of the Wronskian determinant, one can check that
in the HC-model the Weierstrass points have base coordinates at the 60 roots
of the polynomial equations

$\displaystyle 0$
$\displaystyle=x^{12}-32x^{11}-114x^{10}-200x^{9}+100x^{8}+48x^{7}-936x^{6}+1728x^{5}-2000x^{4}+3200x^{3}$
$\displaystyle\phantom{=}-2624x^{2}+768x-64,$ $\displaystyle 0$
$\displaystyle=x^{24}-24x^{23}+1306x^{22}-2864x^{21}+10096x^{20}-32704x^{19}-5704x^{18}-41824x^{17}$
$\displaystyle\phantom{=}+43056x^{16}+831616x^{15}+837856x^{14}+992256x^{13}+2603136x^{12}+1238016x^{11}$
$\displaystyle\phantom{=}+1560576x^{10}+5584896x^{9}+3357696x^{8}+3838976x^{7}+5856256x^{6}+2543616x^{5}$
$\displaystyle\phantom{=}+2200576x^{4}+1355776x^{3}+454656x^{2}+65536x+4096,$
$\displaystyle 0$
$\displaystyle=x^{24}+56x^{23}+1176x^{22}-1784x^{21}-3904x^{20}+36096x^{19}+12776x^{18}-211904x^{17}$
$\displaystyle\phantom{=}+304736x^{16}+431616x^{15}+339456x^{14}-1985664x^{13}-625344x^{12}+1034496x^{11}$
$\displaystyle\phantom{=}+3512576x^{10}-584704x^{9}-3572224x^{8}-2018304x^{7}+3303936x^{6}+3055616x^{5}$
$\displaystyle\phantom{=}+1099776x^{4}+45056x^{3}+229376x^{2}-16384x+4096.$

One can check that the Galois group of each polynomial is $C_{4}\times S_{3}$.
We have seen that in the HC-model the automorphisms have coefficients in
$\mathbb{Z}[\zeta]$, and so we know the splitting field must be an extension
of $\mathbb{Q}[\zeta]$, which accounts for the $C_{4}$ factor in the Galois
group. $S_{3}$ has a subgroup of order $2$, corresponding to an extension of
degree-2, and a brute force calculation shows that we also wish to adjoin
$i\sqrt{2}$. This has already nearly reduced the problem, and then one needs a
small moment of inspiration to find the last thing to adjoin. Looking at
[RR92] then may inspire one to adjoin the real root of the polynomial
$x^{3}+7x^{2}+8x+4$, say $\xi$, and this gives the full splitting field. We
can solve the cubic explicitly using Cardano’s formula to find

$\xi=-\frac{1}{3}\left\\{7+\left[\sqrt[3]{145+30\sqrt{6}}+\sqrt[3]{145-30\sqrt{6}}\right]\right\\},$

and as such we could also take our splitting field to be

$\mathbb{Q}\left[\zeta,i\sqrt{2},\sqrt[3]{145+30\sqrt{6}}\right].$

We observe by Cardano’s formula that
$\alpha,\beta,\gamma\in\mathbb{Q}[i\sqrt{2},\sqrt[3]{30+15\sqrt{6}}]$, and
this latter field is isomorphic to
$\mathbb{Q}[i\sqrt{2},\sqrt[3]{145+30\sqrt{6}}]$. With these expression for
the Weierstrass points, and the explicit knowledge of the automorphism group
as an action on affine coordinates, we can find explicitly the transposition
that preserves a given Weierstrass point.

Note that an alternative method for computing Weierstrass points
algorithmically has been implemented in Magma [Hes02b].

We now have seen how our Weierstrass points satisfy characterisations 1 and 4
of Definition 3.2, and we complete the picture with the following result which
we believe new.

######  Proposition 3.6.

Define the points $P_{345}$, $P^{\prime}_{345}$ corresponding to the
Weierstrass point $W_{345}$ (and similarly for the other Weierstrass points)
by

$\displaystyle P_{345}$
$\displaystyle=[\delta^{\prime}:\delta:(-43\alpha^{2}-113\alpha-92)\beta/112$
$\displaystyle\phantom{=}-(13\alpha^{2}-27\alpha-20)/28:(43\alpha^{2}+25\alpha-4)\beta/112+(13\alpha^{2}+3\alpha-24)/28:1]$
$\displaystyle P^{\prime}_{345}$
$\displaystyle=[\delta:\delta^{\prime}:(-43\alpha^{2}-113\alpha-92)\beta/112$
$\displaystyle\phantom{=}-(13\alpha^{2}-27\alpha-20)/28:(43\alpha^{2}+25\alpha-4)\beta/112+(13\alpha^{2}+3\alpha-24)/28:1].$

Here $\alpha$, $\beta$ were defined in Proposition 3.3 and $\delta$,
$\delta^{\prime}$ are the roots of

$x^{2}-\left[\frac{(11\alpha+12)\beta+4(3\alpha+2)}{14}\right]x+\left[\frac{23(155\alpha+388)\beta+92(97\alpha+172)}{6272}\right].$

Then there is a holomorphic differential $\nu_{345}$ on Bring’s curve with
divisor

$4W_{345}+P_{345}+P_{345}^{\prime},$

and a meromorphic function on Bring’s curve with divisor

$P_{145}+P_{145}^{\prime}+P_{245}+P_{245}^{\prime}-4W_{345}.$

###### Proof.

Note that algorithmically it is possible to construct these through the work
in [Hes02a], the tools for which are partially implemented in Sage but not
with enough generality for us to use out-the-box. As such we need a different
approach.

We know that for any Weierstrass point $W$ there is a hyperplane
$H\subset\mathbb{P}^{3}$ intersecting the canonical embedding with
multiplicity $g=4$ at $W$ (that is $H\cap\mathcal{B}=4W+P+P^{\prime}$ for some
$P,P^{\prime}\in\mathcal{B}$). Such a hyperplane gives a holomorphic
differential $\nu_{ijk}$ with $(\nu_{ijk})=4W_{ijk}+P_{ijk}+P^{\prime}_{ijk}$.
From [Edg78] we know that for Bring’s curve the osculating plane131313The
osculating plane at a point $P$ on a curve is the limiting plane through $P$,
$P^{\prime}$, $P^{\prime\prime}$ as $P^{\prime},P^{\prime\prime}\to P$ on the
curve [Sal62]. at $W$ intersects 4 times, and so this is the plane we are
looking for. As Edge gives a formula for the osculating plane (attributed to
Hesse), we can explicitly calculate the remaining 2 intersections with the
curve in terms of a polynomial roots. This gives us the first result for the
divisor of the meromorphic differential.

Furthermore, from [Edg81] we have a tritangent plane which has intersection
with Bring’s curve

$2(W_{145}+W_{245}+W_{345}).$

We will discuss this plane (and others like it) more in §5.2, but for now all
we need is that this means there is a holomorphic differential
$\omega_{45\alpha}^{(1)}$ on $\mathcal{B}$ with
$(\omega_{45\alpha}^{(1)})=2(W_{145}+W_{245}+W_{345})$. As such we get the
divisor of the meromorphic function

$\left(\frac{\nu_{145}\nu_{245}}{\left(\omega_{45\alpha}^{(1)}\right)^{2}}\right)=P_{145}+P_{145}^{\prime}+P_{245}+P_{245}^{\prime}-4W_{345}.$

As we can calculate the formula for all these planes explicitly if we wish, we
could (in principle) construct the corresponding function and differential. ∎

######  Remark 3.7.

We are able to verify the results in Proposition 3.6 using the Abel-Jacobi map
implemented in the Sage [DH21]; see the notebooks.

###  3.2. Vertices and Face-Centres

Using Figure 2 of [BN12] we are able to link the Petrie polygon to the R&R
model of the curve, and this gives us a concrete expression for the remaining
geometric points.

######  Proposition 3.8.

The face-centres are exactly the points fixed by an order-5 automorphism of
the curve. They are given in the $\mathbb{P}^{4}$-model by the permutations of
$[1:\zeta:\zeta^{2}:\zeta^{3}:\zeta^{4}]$. The corresponding points in the HC-
model are the desingularizations of $V_{k}$, together with $a$, $b$, $c$, $d$,
and

  1. (1)

$[-2\zeta^{3}-2\zeta^{2}:\zeta^{3}+\zeta^{2}-1:1]=[\sqrt{5}+1:-\sqrt{5}/2-3/2:1]=[2j:j^{\prime}-2:1]$,

  2. (2)

$[-2\zeta^{3}-2:-2\zeta^{3}-\zeta-1:1]$,

  3. (3)

$[-2\zeta^{2}-2\zeta:-\zeta^{3}+\zeta+1:1]$,

  4. (4)

$[2\zeta^{3}+2\zeta^{2}+2\zeta:\zeta^{3}+2\zeta^{2}+2\zeta+1:1]$,

  5. (5)

$[-2\zeta^{2}-2:\zeta^{3}-\zeta^{2}+\zeta:1]$,

  6. (6)

$[2\zeta^{3}+2\zeta+2:\zeta^{2}-\zeta+1:1]$,

  7. (7)

$[-2\zeta^{3}-2\zeta:2\zeta^{3}+\zeta^{2}+\zeta+2:1]$

  8. (8)

$[2\zeta^{3}+2\zeta^{2}+2:-\zeta^{3}-\zeta^{2}-2:1]=[-\sqrt{5}+1:\sqrt{5}/2-3/2:1]=[2j^{\prime}:j-2:1]$,

  9. (9)

$[2\zeta^{2}+2\zeta+2:-\zeta^{3}-2\zeta^{2}-\zeta:1]$,

  10. (10)

$[-2\zeta-2:-\zeta^{2}-2\zeta-1:1]$,

where $j,j^{\prime}$ are the roots identified by Dye mentioned in §2.3.

###### Proof.

Certainly the order-5 rotation about a face-centre fixes that centre, and so a
Riemann-Hurwitz counting argument gives us that the $24$ face-centres are the
fixed points of order-5 automorphisms. It is a simple matter of computation to
verify the given expressions are fixed; this may be done in Sage. ∎

The Galois group $\operatorname{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})\cong C_{4}$
acts element-wise on the face-centres and the orbits partition the set of
face-centres into 6 sets of 4. Each set of 4 face-centres form the vertices of
a quadrilateral whose edges lie in $\left\\{H_{2}=0\right\\}$ [Edg78].
Moreover, the faces of the dodecadodecahedron corresponding to the face-
centres in each quadrilateral are parallel.

We have a similar result for the vertices.

######  Proposition 3.9.

The vertices are exactly the points fixed by an order-4 automorphism of the
curve. They are given in the $\mathbb{P}^{4}$-model by the permutations of
$[1:i:-1:-i:0]$.

We could use our birational map as above to give the vertices in HC
coordinates, but these are not very illuminating. For example, that vertex
$[1:i:-1:-i:0]$ maps to

$[X:Y:Z]=[(3\zeta^{3}-\zeta^{2}+2\zeta+1)i+3\zeta^{3}+3\zeta^{2}+5:(2\zeta^{3}+2\zeta+1)i+2\zeta^{3}+2\zeta^{2}+2:1]$

though $[1:-1:i:-i:0]$ maps to $[1:(-1+i)/2:(-1+i)/2]$.

##  4\. Quotients by Subgroups

Both [RR92] and [Web05] consider the quotient of $\mathcal{B}$ by the action
of subgroups of $S_{5}$. In this section we shall study the various quotients
of Bring’s curve of nonzero genus and the relations between them, both
clarifying and extending previous work. In §4.1 we will use the Riemann-
Hurwitz theorem to describe possible quotients. In §4.2 we shall note the
various relationships we expect between the quotients just on group theoretic
grounds, while in §4.3-§4.5 we turn to their explicit construction. In so
doing we discover a number of curious isomorphisms beyond those expected. In
§4.6 we summarise our calculations and relate them to known isogeny results.
Throughout we will use the following: for any subgroup $H$ of the
automorphisms $\operatorname{Aut}(\mathcal{C})$ of a curve $\mathcal{C}$,
$H\leq\operatorname{Aut}(\mathcal{C})$, then the normaliser
$N_{\operatorname{Aut}(\mathcal{C})}(H)$ acts on the $H$-orbits and
$N_{\operatorname{Aut}(\mathcal{C})}(H)/H\leq\operatorname{Aut}(\mathcal{C}/H)$;
if $g\in N_{\operatorname{Aut}(\mathcal{C})}(H)$ we will denoted by
$\overline{g}$ the $H$-coset of $N_{\operatorname{Aut}(\mathcal{C})}(H)$
containing this. The quotient curves of this section are summarised in Figures
3 and 5.

###  4.1. Genera of Quotients

Our first step is to know the topology of the quotients we are going to find.
To this end, we give the following result.

######  Proposition 4.1.

The data of the quotients of $\mathcal{B}$ by subgroups
$\left\langle\sigma\right\rangle\leq S_{5}$ is summarised by the following
table:

$\sigma$ | Example | $\left\lvert{cl(\sigma)}\right\rvert$ | $\left\lvert{\operatorname{Fix(\sigma)}}\right\rvert$ | Fixed Points | $N_{S_{5}}(\left\langle\sigma\right\rangle)/\left\langle\sigma\right\rangle$ | $g\left(\mathcal{B}/\left\langle\sigma\right\rangle\right)$
---|---|---|---|---|---|---
$(12)$ | - | 10 | 6 | Edges | $S_{3}$ | 1
$(123)$ | $RS$ | 20 | 0 | - | $V_{4}$ | 2
$(12)(34)$ | $R$ | 15 | 2 | Vertices | $V_{4}$ | 2
$(1234)$ | $U$ | 30 | 2 | Vertices | $C_{2}$ | 1
$(123)(45)$ | - | 20 | 0 | - | $C_{2}$ | 0
$(12345)$ | $S$ | 24 | 4 | Face-centres | $C_{4}$ | 0

Here $cl(\sigma)$ is the conjugacy class of $\sigma$,
$\operatorname{Fix}(\sigma)=\\{P\in\mathcal{B}\,|\,\sigma(P)=P\\}$.

###### Proof.

As conjugate elements yield isomorphic quotients we need only to give one
$\sigma$ per conjugacy class. The first three columns follow from the group
theory we have previously shown in §2.3, the fourth and fifth follow from §3
and the sixth is elementary group theory. The final column remaining is then a
Riemann-Hurwitz argument for genus, which will we demonstrate for the quotient
by $(2345)$ as in [RR92].

For general $\sigma$, denoting by $\pi$ the projection
$\mathcal{B}\to{\mathcal{B}}/{\left\langle\sigma\right\rangle}:=H$, Riemann-
Hurwitz says

$\displaystyle g_{\mathcal{B}}-1$
$\displaystyle=(\deg\pi)(g_{H}-1)+\frac{1}{2}B,$

where $B$ is the degree of ramification of $\pi$, which in the case of a
quotient by a group action corresponds to the fixed point structure of
$\sigma$.

Consider $(24)(35)$. A fixed point of $\mathcal{B}$ under this, given by
projective coordinates $x_{i},i=1,\dots,5$, must have

$(x_{1},x_{4},x_{5},x_{2},x_{3})=(\lambda x_{1},\lambda x_{2},\lambda
x_{3},\lambda x_{4},\lambda x_{5}).$

From this we can see $\lambda=\pm 1$. Taking $\lambda=1$ gives no solutions,
but taking $\lambda=-1$ one finds the equations

$x_{1}=0,\quad x_{1}^{2}+2x_{2}^{2}+2x_{3}^{2}=0,\quad x_{1}^{3}=0,$

which gives the 2 fixed points $[0:1:\pm i:-1:\mp i]$. Hence we have

$\displaystyle 4-1=2(g_{H}-1)+\frac{1}{2}(1+1)\Rightarrow g_{H}=2.$

Moving now to $(2345)$, thinking about the possible branching structure we get
from Riemann-Hurwitz

$3=4(g_{H}-1)+\frac{1}{2}\left[\sum_{P\in\operatorname{Fix}(\sigma)}3+\sum_{P\in\operatorname{Fix}(\sigma^{2})\setminus\operatorname{Fix}(\sigma)}1\right].$

Fixed points of $(2345)$ will correspond to points such that

$(x_{1},x_{3},x_{4},x_{5},x_{2})=(\lambda x_{1},\lambda x_{2},\lambda
x_{3},\lambda x_{4},\lambda x_{5}),$

and we get the constraint $\lambda^{4}=1$. The equations of the curve become

$\displaystyle x_{1}+x_{2}(1+\lambda+\lambda^{2}+\lambda^{3})$
$\displaystyle=0,$ $\displaystyle
x_{1}^{2}+x_{2}^{2}(1+\lambda^{2}+1+\lambda^{2})$ $\displaystyle=0,$
$\displaystyle x_{1}^{3}+x_{2}^{3}(1+\lambda^{3}+\lambda^{2}+\lambda)$
$\displaystyle=0,$

and checking the possible cases one finds that the only fixed points of
$(2345)$ are $[0:1:\pm i:-1:\mp i]$. These are also the only fixed points of
$(24)(35)=(2345)^{2}$ and so the second sum vanishes, hence we find $g_{H}=1$.
∎

R&R [RR92] provides a nice visual interpretation of the quotient by a 4-cycle,
namely think of a sphere with 4 handles attached around an equator, then
$(2345)$ is the cycle rotating these handles to each other by a quarter turn
about the axis through the center of the equator. The fixed points are then
where this axis intersects the sphere. Edge [Edg78], citing Wiman, says that
Bring’s curve ‘‘is, in ten different ways, in $(2,1)$ correspondence with a
plane curve of genus 1". The 10 $(2,1)$ correspondences noted by Edge are
exactly the 10 quotients by a transposition giving a $2:1$ map
$\mathcal{B}\to\mathcal{E}$, where $\mathcal{E}$ is an elliptic curve. Such a
map is called a bielliptic structure, and Bring’s curve is the unique genus-4
curve to have 10 such structures [CDC05]. This table also lets us reconstruct
the results about the gonality of Bring’s curve from [GWW10].

###  4.2. Relations between Quotients

We now consider the various relationships we might expect between the quotient
curves of Bring’s curve. Recall (again from Riemann-Hurwitz) that if
$\mathcal{C}$ is a curve of genus $g\geq 2$ and
$H\leq\operatorname{Aut}(\mathcal{C})$ is such that $|H|>4(g-1)$, a so called
‘large automorphism group’, then the genus of the quotient curve
$g_{\mathcal{C}/H}=0$. For Brings’s curve we are therefore interested in
subgroups of $S_{5}$ of order less than or equal to $12$, and as we have seen
the quotient by a 5-cycle leads to genus $0$ we may exclude subgroups
containing such. The relevant conjugacy classes of subgroups are
then141414Note that here we are using the GAP convention for the dihedral
group that $|{D_{n}}|=n$. This is the opposite convention to Sage which uses
$|{D_{n}}|=2n$.

  1. (a)

$\left\langle(12)(34)\right\rangle\cong C_{2}$,
$\left\langle(12),(34)\right\rangle\cong V_{4}$,
$\left\langle(1324)\right\rangle\cong C_{4}$,
$\left\langle(1324),(12)\right\rangle\cong D_{8}$. Each of these groups $H$
have the same normaliser:
$N_{S_{5}}(H)=\left\langle(1324),(12)\right\rangle\cong D_{8}$.

  2. (b)

$\left\langle(12)\right\rangle\cong C_{2}$,
$\left\langle(345)\right\rangle\cong C_{3}$,
$\left\langle(345),(12)\right\rangle\cong C_{6}$,
$\left\langle(345),(34)\right\rangle\cong S_{3}$,
$\left\langle(345),(12)(34)\right\rangle\cong S_{3}^{\prime}$,
$\left\langle(345),(12),(34)\right\rangle\cong D_{12}\cong S_{3}\times C_{2}$.
Each of these groups $H$ have the same normaliser:
$N_{S_{5}}(H)=\left\langle(345),(12),(34)\right\rangle\cong D_{12}$.

  3. (c)

$\left\langle(12)(34),(13)(24)\right\rangle\cong V_{4}$,
$\left\langle(234),(12)(34)\right\rangle\cong A_{4}$. Each of these groups $H$
have the same normaliser:
$N_{S_{5}}(\left\langle(234),(12)(34)\right\rangle)=\left\langle(234),(12),(34)\right\rangle\cong
S_{4}$.

The groupings here are such that if $H$, $H^{\prime}$ are from the same item
then they share the same normaliser $N=N_{S_{5}}(H)=N_{S_{5}}(H^{\prime})$. In
particular each of $H$, $H^{\prime}$ and $HH^{\prime}=H^{\prime}H$ are normal
in $N$ and we have commutativity of the following diagram of quotients

${\mathcal{B}}$${\mathcal{B}/H}$${\mathcal{B}/H^{\prime}}$${\mathcal{B}/HH^{\prime}}$$\scriptstyle{H}$$\scriptstyle{H^{\prime}}$$\scriptstyle{HH^{\prime}}$$\scriptstyle{HH^{\prime}/H}$$\scriptstyle{HH^{\prime}/H^{\prime}}$

where the arrows are labelled by the symmetry being quotiented; by an
isomorphism theorem we have of course that $HH^{\prime}/H\cong
H^{\prime}/(H\cap H^{\prime})$ and $HH^{\prime}/H^{\prime}\cong H/(H\cap
H^{\prime})$. If, say $H\leq H^{\prime}$, then $HH^{\prime}=H^{\prime}$ and
one side of this diagram collapses.

Now a Riemann-Hurwitz calculation shows that the each of the quotients by
$\left\langle(12),(34)\right\rangle$, $\left\langle(1324),(12)\right\rangle$,
$\left\langle(345),(34)\right\rangle$ and
$\left\langle(345),(12),(34)\right\rangle$ is of genus $0$ and so not being
considered. Thus from the preceding discussion we have the following relations
amongst quotients for the subgroups of (a), (b) and (c):

${\mathcal{B}}$${\mathcal{C}}$${\mathcal{E}}$${(a)}$$\scriptstyle{(12)(34)}$$\scriptstyle{(1324)}$${\mathcal{B}}$${\mathcal{E}^{\prime}}$${\mathcal{E}^{\prime\prime}}$${\mathcal{C}^{\prime}}$${\mathcal{E}^{\prime\prime\prime}}$${(b)}$$\scriptstyle{(12)}$$\scriptstyle{(345)}$$\scriptstyle{S_{3}^{\prime}}$$\scriptstyle{C_{6}}$$\scriptstyle{(345)}$$\scriptstyle{(12)}$$\scriptstyle{\overline{(12)(34)}}$${\mathcal{B}}$${\mathcal{C}}$${\mathcal{E}^{(iv)}}$${\mathcal{E}^{(v)}}$${(c)}$$\scriptstyle{V_{4}}$$\scriptstyle{(12)(34)}$$\scriptstyle{A_{4}}$$\scriptstyle{\overline{(234)}}$

Here $\mathcal{C},\mathcal{C}^{\prime}$ are genus 2 curves and
$\mathcal{E}$,…,$\mathcal{E}^{(v)}$ elliptic curves that we cannot yet specify
purely on group theoretic grounds. We turn now to their specification and
indeed we shall find some interesting identities between them, which will then
be summarised in §4.6.

###  4.3. Quotients by a 4- and 2,2-cycle

Armed with the knowledge of the genus of the quotients we expect we shall now
write them explicitly. We will begin with the 2,2-cycle corresponding to
$U^{2}$ with the 4-cycle $U$ arising in the discussion. Recall that we have

$U^{2}:[X,Y,Z]\rightarrow[X,Z,Y],\quad\psi(U^{2})=(12)(34).$

The normaliser of $\left\langle(12)(34)\right\rangle$ in $S_{5}$ is
$N_{S_{5}}(\left\langle(12)(34)\right\rangle)=\left\langle(12),(1324)\right\rangle\cong
D_{8}$. As we remarked earlier,
$N_{S_{5}}(\left\langle(12)(34)\right\rangle)/\left\langle(12)(34)\right\rangle\cong
V_{4}$ are symmetries which remain when we go to the quotient
$\mathcal{B}/\left\langle U^{2}\right\rangle$ and we expect the quotient
genus-2 curve to have (at least) this $V_{4}$ symmetry group. One of these
involutions, which we will later see to be
$\overline{(12)}=\overline{(34)}$,151515The
$\left\langle(12)(34)\right\rangle$ cosets of the normaliser are
$\left\\{e,(12)(34)\right\\}$, $\left\\{(12),(34)\right\\}$,
$\left\\{(13)(24),(14)(23))\right\\}$, and $\left\\{(1324),(1423)\right\\}$.
is the hyperelliptic involution of the quotient curve and so further
quotienting by this symmetry gives $\mathbb{P}^{1}$. Quotienting by the other
two involutions will yield elliptic curves, and we will complete this
construction now, both from the perspective of the HC-model, and the
$\mathbb{P}^{4}$-model.

Starting in the HC-model, in order to get the first quotient
$\mathcal{B}/\left\langle U^{2}\right\rangle$ we express our curve in terms of
the invariants of $U^{2}$: $X$, $T:=Y+Z$, and $V:=YZ$. Then

$\displaystyle 0$ $\displaystyle=X(Y^{5}+Z^{5})+(XYZ)^{2}-X^{4}YZ-2(YZ)^{3},$
$\displaystyle=X(T^{5}-5T^{3}V+5TV^{2})+X^{2}V^{2}-X^{4}V-2V^{3}.$

In $\mathbb{P}^{1,1,2}$ this is our genus-$4$ curve. Setting161616Equally one
may set $X=1$ but we note that setting $V=1$ yields a genus 4 curve as here
$V$ is not of weight $1$.  $T=1$ and viewing

$0=X(1-5V+5V^{2})+X^{2}V^{2}-X^{4}V-2V^{3}$

as the affine part of a projective curve we have (after a not-very-
illuminating transformation, for which Maple was used) the hyperelliptic curve

(10) $\mathcal{C}_{1}:\ B^{2}=A^{6}+4A^{5}+10A^{3}+4A+1.$

This is the genus-2 curve of [RR92] with automorphism group $V_{4}$.
Calculating the Igusa-Clebsch invariants (which give the
$\overline{\mathbb{Q}}$-isomorphism class of a curve) and searching the LMFDB
[LMF22] shows that this is a model for the modular curve $X_{0}(50)$, which we
verifying by directly calculating a model for $X_{0}(50)$ over $\mathbb{Q}$
using [Shi95]. The substitutions $A=(2+A^{\prime})/(2-A^{\prime})$,
$B=4B^{\prime}/(2-A^{\prime})^{3}$ make the $V_{4}$ symmetry clearer, giving

$(B^{\prime})^{2}+\left[(A^{\prime})^{6}-5(A^{\prime})^{4}-40(A^{\prime})^{2}-80\right]=0,$

where we have the hyperelliptic involution $B^{\prime}\rightarrow-B^{\prime}$
and the map $A^{\prime}\rightarrow-A^{\prime}$ generating $V_{4}$. We shall
now be explicit about how these work.

As previously mentioned, quotienting $\mathcal{C}_{1}$ by either of the two
non-hyperelliptic involutions yields an elliptic curve, with each involution
having two fixed points from Riemann-Hurwitz. Quotienting by
$(B^{\prime},A^{\prime})\rightarrow(B^{\prime},-A^{\prime})$ by introducing
$A^{\prime\prime}=-(A^{\prime})^{2}$ yields

(11)
$\mathcal{E}_{1}:(B^{\prime})^{2}=(A^{\prime\prime})^{3}-5(A^{\prime\prime})^{2}-40(A^{\prime\prime})-80\
\text{ with $j$-invariant }\ j_{\mathcal{E}_{1}}=-\frac{5\times
29^{3}}{2^{5}}=j(\tau_{0}).$

The fixed points are $(A^{\prime},B^{\prime})=(0,\pm\sqrt{-80})$,
corresponding to the two points $(X,V)=([1\pm\sqrt{5}]/2,-1\pm\sqrt{5}/2)$,
which are the images of the four vertices $[1:\pm i:\mp i:-1:0]$ and $[1:\pm
i:-1\mp i:0]$ respectively, depending on sign. We recognise $\mathcal{E}_{1}$
to be the elliptic curve $E_{1}$ in [RR92].

We may also quotient $\mathcal{C}_{1}$ by
$(B^{\prime},A^{\prime})\rightarrow(-B^{\prime},-A^{\prime})$. To do this
write $\mathcal{C}_{1}$ as

$(B^{\prime})^{2}C^{4}+\left[(A^{\prime})^{6}-5(A^{\prime})^{4}C^{2}-40(A^{\prime})^{2}C^{4}-80C^{6}\right]=0,$

from where we see the same automorphism acts as $C\to-C$. Quotienting by this
action by introducing $C^{\prime}=-C^{2}$ yields

$(B^{\prime})^{2}(C^{\prime})^{2}=(A^{\prime})^{6}-5(A^{\prime})^{4}(C^{\prime})-40(A^{\prime})^{2}(C^{\prime})^{2}-80(C^{\prime})^{3}.$

Setting $A^{\prime}=1$ and taking $B^{\prime\prime}=B^{\prime}C^{\prime}$
gives the standard elliptic form

(12)
$\mathcal{E}_{2}:(B^{\prime\prime})^{2}=1-5(C^{\prime})-40(C^{\prime})^{2}-80(C^{\prime})^{3},\quad
j_{\mathcal{E}_{2}}=-\frac{25}{2}=j(5\tau_{0}).$

The fixed point of this involution is $[B^{\prime}:A^{\prime}:C]=[1:0:0]$;
this is a singular point where the desingularization corresponds to the 2
points at infinity, or correspondingly $(X,V)=(\mp i/2-1/2,1/4)$, which are
the images of the two vertices $[1:-1:\pm i:\mp i:0]$. We recognise
$\mathcal{E}_{2}$ to be $E_{2}$ in [RR92].171717There is a typo in the
constant term of R&R’s $E_{2}$ if it is to have the stated $j$-invariant.

Next let us quotient the $\mathbb{P}^{4}$-model directly by the action of
$(12)(34)$ and compare with these quotients just obtained from the HC-model.
To this end we introduce semi-invariants of the action of $(1324)=\psi(U)$,
defined by

$(s_{1},s_{2},s_{3},s_{4})^{T}=\begin{pmatrix}1&1&1&1\\\ 1&-1&i&-i\\\
1&1&-1&-1\\\ 1&-1&-i&i\end{pmatrix}(x_{1},x_{2},x_{3},x_{4})^{T}.$

These are constructed such that $U\cdot s_{j}=i^{j-1}s_{j}$, and so we have
$U^{2}$-invariants
$[s:t:u:v]:=[s_{1}:s_{2}s_{4}:s_{3}:s_{4}^{2}]\in\mathbb{P}^{1,2,1,2}$.181818Note
one could have instead taken $v=s_{2}^{2}$ for the last invariant, but the
choice above happens to be better around the vertices that are the fixed
points of the 4-cycle. Moreover, recognise that the invariants taken are
defined over the field extension $\mathbb{Q}[i]$ and not $\mathbb{Q}$. In
terms of these invariants we have

$\displaystyle H_{2}$ $\displaystyle=\frac{1}{4}\left(5s^{2}+u^{2}+2t\right),$
$\displaystyle H_{3}$
$\displaystyle=\frac{3}{16}\left(-5s^{3}+su^{2}+\frac{t^{2}u}{v}+2st+uv\right).$

Eliminating $t$ from these equations, and setting $s=1$, we can use Maple to
get the genus-2 curve in Weierstrass form

(13) $\mathcal{C}_{1}:\ y^{2}=100-25x^{2}-10x^{4}-x^{6}$

where $x=u$ and $y=-10+2uv$. The roots of the sextic here are a Möbius
transform of the curve $\mathcal{C}_{1}$ previously given (namely
$x\mapsto\sqrt{20}/x$), so these curves are the same over
$\mathbb{Q}[\sqrt{5}]$. Note the reason we see $\mathbb{Q}[\sqrt{5}]$ here is
that it is the degree-2 subfield of $\mathbb{Q}[\zeta]$, the field required to
have equivalence of the Hulek-Craig and $\mathbb{P}^{4}$-models of Bring’s
curve. We now aim to identify the $V_{4}$ of this curve described earlier with
the quotient of the normaliser
$\left\langle(12),(1324)\right\rangle/\left\langle(12)(34)\right\rangle$. One
can check that $(12):[s:t:u:v]\to[s:t:u:t^{2}/v]$. This fixes $x$, and so must
be the automorphism $y\to-y$; that is $\overline{(12)}=\overline{(34)}$ is the
hyperelliptic involution of $\mathcal{C}_{1}$. Indeed one can check

$\displaystyle y=-10+2uv$ $\displaystyle\mapsto-10+2\frac{t^{2}u}{v},$
$\displaystyle=-10-2\left(-5+u^{2}+2t+uv\right),$
$\displaystyle=-10-2\left(-10+uv+(5+u^{2}+2t)\right),$
$\displaystyle=10-2uv=-y.$

Likewise, $(1324):[s:t:u:v]\to[s:t:-u:-v]$, and so leads to the automorphism
$(x,y)\to(-x,y)$. Now if $\mathsf{x}=x^{2}$, the elliptic curve
$y^{2}=100-25\mathsf{x}-10\mathsf{x}^{2}-\mathsf{x}^{3}$ has $j$-invariant
$-25/2$ and so we may identify
$\mathcal{E}_{2}\cong\mathcal{B}/\left\langle(12)(34),(1324)\right\rangle=\mathcal{B}/\left\langle(1324)\right\rangle$.
Similarly $(13)(24):[s:t:u:v]\to[s:t:-u:-t^{2}/v]$ leads to $(x,y)\to(-x,-y)$
and we may identify
$\mathcal{E}_{1}\cong\mathcal{B}/\left\langle(12)(34),(13)(24)\right\rangle$.
Note when we compare with [RR92], we differ from R&R in the identification of
which quotient is being taken and their ascribing of $\tau_{0}$ and
$5\tau_{0}$. Our results agree with those of [Web05], wherein the author
describes an order-4 rotation (which he calls $\phi$, but we shall call
$\tilde{U}$), and calculates the quotient by its action
$T=\mathcal{B}/\left\langle\tilde{U}\right\rangle$ to be

$T:\ y^{2}=4x^{3}-75x-1475,\quad j_{T}=-\frac{25}{2}=j(5\tau_{0}).$

Note that our strategy of semi-invariants can be used directly to calculate
the quotient by the 4-cycle $(1324)$, something we were unable to do in the
HC-model because of the non-linearity of the automorphism $U$. To do so
introduce new variables invariant under $(1324)$ (and so necessarily under
$(12)(34)$) given by $u^{\prime}=uv$, $v^{\prime}=v^{2}$. These let us rewrite
the defining equations of Bring’s curve as

$\displaystyle H_{2}$
$\displaystyle=\frac{1}{4}\left(5s^{2}+\frac{(u^{\prime})^{2}}{v^{\prime}}+2t\right),$
$\displaystyle H_{3}$
$\displaystyle=\frac{3}{16}\left(-5s^{3}+\frac{s(u^{\prime})^{2}}{v^{\prime}}+\frac{t^{2}u^{\prime}}{v^{\prime}}+2st+u^{\prime}\right),$

and we can apply Maple to find a Weierstrass form of the resulting elliptic
curve. This is exactly the process of quotienting by $(x,y)\mapsto(-x,-y)$ as
above, but in a different language.

Furthermore, from our previous investigation using group theory, we know that
the quotient $\mathcal{B}/V_{4}$, where this $V_{4}$ is the one containing
only 2,2-cycles, can further be quotiented by $(234)$ to give
$\mathcal{B}/A_{4}$. Rather than attempt to construct this quotient using the
invariants previously calculated on the 2,2-quotient, we step back and recall
that the invariant ring $\mathbb{Q}[x_{1},\dots,x_{n}]^{A_{n}}$ is generated
by the symmetric polynomials $s_{k}=\sum_{i=1}^{n}x_{i}^{k}$ for $k=1,\dots,n$
and the Vandermonde polynomial $V=\prod_{1\leq i<j\leq n}(x_{j}-x_{i})$. As
generators of the invariant algebra we know that there must be a relation
between $V^{2}$ and the other generators as $V^{2}$ is an $S_{n}$-invariant,
and the $S_{n}$-invariant algebra is generated by the $s_{k}$. Taking $n=4$ in
the case of Bring’s curve, the relations imposed from the curve are
$s_{2}+s_{1}^{2}=0=s_{3}-s_{1}^{3}$, and this gives for the additional
relation,

(14)
$4s_{4}^{3}-\frac{373}{16}s_{1}^{4}s_{4}^{2}+\frac{431}{8}s_{1}^{8}s_{4}-\frac{701}{16}s_{1}^{12}+V^{2}=0.$

Setting $s_{1}=1$ (as we may do at all points on the quotient except those
points coming from the vertices on Bring’s curve) we see this is clearly an
elliptic curve $\mathcal{E}_{3}$ with hyperelliptic involution $V\to-V$, and
for which we can calculate the $j$-invariant to be $\frac{211^{3}\times
5}{2^{15}}=j(3\tau_{0})$.191919This is the curve 15-isogenous to
$\mathcal{E}_{2}$ noted by Serre, see §2.2. Further quotienting by the
hyperelliptic involution then corresponds to the quotient $\mathcal{B}/S_{4}$,
which is $\mathbb{P}^{1}$ as expected, being the quotient by a large
automorphism group.

###  4.4. Quotients by a 3-cycle

We may utilise the same methods illustrated above for the 2,2-cycles to
calculate the quotient of Bring’s curve by a 3-cycle. We will work with the
3-cycle $(345)$ for purely aesthetic resaons. Now the normaliser of
$\left\langle(345)\right\rangle$ in $S_{5}$ is $N_{S_{5}}(\left\langle
345\right\rangle))=\left\langle(12),(34),(345)\right\rangle\cong D_{12}$ and
we expect the quotient genus-2 curve
$\mathcal{B}/\left\langle(345)\right\rangle$ to have $D_{12}/C_{3}\cong V_{4}$
symmetry group. Again one of these involutions, which we will later see to be
$\overline{(34)}$,202020The cosets of $\left\langle(345)\right\rangle$ in the
normaliser are now $\left\\{e,(345),(354)\right\\}$,
$\left\\{(12),(12)(345),(12)(354)\right\\}$,
$\left\\{(34),(45),(35))\right\\}$, and
$\left\\{(12)(34),(12)(45),(12)(35))\right\\}$. is the hyperelliptic
involution on the curve and so further quotienting by this symmetry gives
$\mathbb{P}^{1}$. Quotienting by the other two involutions again yields
elliptic curves we shall now describe. As the quotients of both the HC-model
and $\mathbb{P}^{4}$-model via semi-invariants proceed analogously we shall
present here only the $\mathbb{P}^{4}$-model calculations.

Letting $\rho$ be a primitive cube-root, we take semi-invariants of the action
of $(345)$

$(s_{1},s_{2},s_{3},s_{4})^{T}=\begin{pmatrix}1&0&0&0\\\ 0&1&1&1\\\
0&1&\rho&\rho^{2}\\\
0&1&\rho^{2}&\rho\end{pmatrix}(x_{2},x_{3},x_{4},x_{5})^{T}.$

We correspondingly take invariants
$[s:t:u:v]=[s_{1}:s_{2}:s_{3}s_{4}:s_{3}^{3}]\in\mathbb{P}^{1,1,2,3}$.212121Note
in this case that the invariant $v$ is not defined over $\mathbb{Q}$, but over
the cyclotomic extension $\mathbb{Q}[\rho]$. In terms of these variables we
have

$\displaystyle H_{2}$
$\displaystyle=\frac{2}{3}\left(3s^{2}+3st+2t^{2}+u\right),$ $\displaystyle
H_{3}$
$\displaystyle=\frac{1}{9}\left(-27s^{2}t-27st^{2}-8t^{3}+v+6tu+\frac{u^{3}}{v}\right).$

Eliminating $u$ from these and setting $s=1$, we can use Maple to get the
genus-2 curve in Weierstrass form

(15)
$\mathcal{C}_{2}:y^{2}=108\left(4+12x+95x^{2}+170x^{3}+155x^{4}+72x^{5}+16x^{6}\right)$

where $x=t$ and $y=-90t-90t^{2}-40t^{3}+4v$. Examining the roots of the sextic
confirms that $\mathcal{C}_{2}$ is a genuinely distinct genus-2 curve from
$\mathcal{C}_{1}$. Indeed, this curve does not currently exist in the
LMFDB222222This is to be expected, as it does not satisfy the conditions to be
included in curves investigated in [BSS+16], but one can check that over
$\mathbb{Q}[\sqrt{5}]$ it is isomorphic to the curve 2500.a.400000.1 given by
$y^{2}=-7x^{6}-8x^{5}+10x^{3}-8x-7$. Moreover, introducing
$x^{\prime},y^{\prime}$ by $x={-2}/{(1+x^{\prime})}$,
$y={y^{\prime}}/{(1+x^{\prime})^{3}}$, yields

$(y^{\prime})^{2}=432\left[(x^{\prime})^{6}+80(x^{\prime})^{4}+125(x^{\prime})^{2}+50\right].$

This makes the $V_{4}$ symmetry evident, being generated by
$x^{\prime}\to-x^{\prime}$ and $y^{\prime}\to-y^{\prime}$. The elliptic curve
obtained from quotienting by the action $x^{\prime}\to-x^{\prime}$ is

(16)
$\mathcal{E}_{4}:(y^{\prime})^{2}=432\left[(x^{\prime\prime})^{3}+80(x^{\prime\prime})^{2}+125x^{\prime\prime}+50\right],\quad
j_{\mathcal{E}_{4}}=-\frac{5^{2}\times 241^{3}}{2^{3}}=j(15\tau_{0}).$

To our knowledge this elliptic curve has not been previously noted in
discussions of Bring’s curve. The elliptic curve obtained from quotienting by
the action $(x^{\prime},y^{\prime})\to(-x^{\prime},-y^{\prime})$ is the
previously seen

$\mathcal{E}_{1}:(y^{\prime\prime})^{2}=432\left[1+80z^{\prime}+125(z^{\prime})^{2}+50(z^{\prime})^{3}\right],\quad
j_{\mathcal{E}_{1}}=-\frac{5\times 29^{3}}{2^{5}}=j(\tau_{0}).$

We now wish to identify the quotient
$\left\langle(12),(34),(345)\right\rangle/\left\langle(345)\right\rangle\cong
V_{4}$ with the $V_{4}$ just described. It requires a little effort to see
$(12):[s:t:u:v]\to[-s-t:t:u:v]$, which corresponds to
$x^{\prime}\to-x^{\prime}$, and that $(34):[s:t:u:v]\to[s:t:u:u^{3}/v]$, which
fixes $x^{\prime}$ and so must be the map $y^{\prime}\to-y^{\prime}$. This
means that the remaining involution
$(x^{\prime},y^{\prime})\to(-x^{\prime},-y^{\prime})$ comes from the group
element $(12)(34)$. As such we identify
$\mathcal{E}_{4}\cong\mathcal{B}/\left\langle(12),(345)\right\rangle$, and we
have a curious isomorphism, namely

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As before we can also consider distinguished points on the quotients coming
from fixed points. Whereas $(345)$ has no fixed points when acting on
$\mathcal{B}$ each involution on $\mathcal{C}_{2}$ that gives a quotient to an
elliptic curve has 2 fixed points. The fixed points of $\overline{(12)}$ are
the orbits under $(345)$ of the six fixed points of $(12)$ that are
Weierstrass points. The fixed points of $\overline{(12)(34)}$ are the orbits
under $(345)$ of the $3\times 2$ fixed points of $(12)(34)$, $(12)(45)$, and
$(12)(35)$ that are vertices.

###  4.5. Quotients by a transposition

The strategy of quotienting from the $\mathbb{P}^{4}$-model using semi-
invariants from the previous sections can also be used to calculate the
quotient of Bring’s curve by a transposition. We take semi-invariants

$(s_{1},s_{2},s_{3},s_{4})^{T}=\begin{pmatrix}1&1&0&0\\\ 1&-1&0&0\\\
0&0&1&1\\\ 0&0&1&-1\end{pmatrix}(x_{1},x_{2},x_{3},x_{4})^{T},$

and in terms of these variables the defining equations of Bring’s curve become

$\displaystyle H_{2}$
$\displaystyle=\frac{1}{2}(s_{1}^{2}+s_{2}^{2})+s_{3}^{2}+s_{4}^{2}+(s_{1}+s_{3}+s_{4})^{2},$
$\displaystyle H_{3}$
$\displaystyle=\frac{1}{4}(s_{1}^{3}+3s_{1}s_{2}^{2})+s_{3}^{3}+s_{4}^{3}-(s_{1}+s_{3}+s_{4})^{3}.$

Taking invariants $[s:t:u:v]=[s_{1}:s_{2}^{2}:s_{3}:s_{4}]$ and eliminating
$t$ yields the elliptic curve

(17) $\mathcal{E}_{2}:\ 4s^{3}+8s^{2}u+7su^{2}+u^{3}+sv^{2}-uv^{2}=0,\quad
j_{\mathcal{E}_{2}}=-\frac{25}{2}=j(5\tau_{0}).$

We observe that (also curiously) the quotient of $\mathcal{B}$ by a
transposition is isomorphic to the quotient by a 4-cycle. It is not clear that
there is any apriori reason for this to be the case. We are however able to
relate the elliptic curves $\mathcal{E}_{2}$ and $\mathcal{E}_{4}$ as follows.

The normaliser of $\left\langle(12)\right\rangle$ in $S_{5}$ is
$N_{S_{5}}(\left\langle(12)\right\rangle)=\left\langle(12),(34),(345)\right\rangle\cong
D_{12}$ where now
$N_{S_{5}}(\left\langle(12)\right\rangle)/\left\langle(12)\right\rangle\cong
S_{3}$.232323The cosets of $\left\langle(12)\right\rangle$ in the normaliser
are $\left\\{e,(12)\right\\}$, $\left\\{(345),(12)(345)\right\\}$,
$\left\\{(354),(12)(354)\right\\}$, $\left\\{(34),(12)(34)\right\\}$,
$\left\\{(35),(12)(35)\right\\}$, and $\left\\{(45),(12)(45)\right\\}$.  We
may immediately identify the action of $\overline{(34)}$ as the hyperelliptic
involution on $\mathcal{E}_{2}$ as it acts as $v\to-v$, but we also retain
another group action, that of $\overline{(345)}$. To understand this action
recall that there are six fixed points of the action of $(12)$ on
$\mathcal{B}$, which are all Weierstrass points. This gives six more
distinguished points on the curve $\mathcal{E}_{2}$ in addition to the two
images of the vertices we have seen previously. The six Weierstrass points
fixed by $(12)$ break into two orbits of three under $(345)$ which we denote
by $\\{H_{i}\\}$, $\\{H_{i}^{\prime}\\}$ ($i=1,2,3$). Just as $(345)$ gives
Bring’s curve as an unramified cover of the genus $2$ curve $\mathcal{C}_{2}$,
quotienting each by a transposition has $\overline{(345)}$ yielding an
unramified automorphism of the quotient curves. Hence there exists a quotient
from $\mathcal{E}_{2}=\mathcal{B}/\langle(12)\rangle$ to another elliptic
curve $\mathcal{E}_{4}=\mathcal{B}/\left\langle(12),(345)\right\rangle$ such
that the following diagram commutes,

${\mathcal{B}}$${\mathcal{C}_{2}}$${\mathcal{E}_{2}}$${\mathcal{E}_{4}}$$\scriptstyle{(345)}$$\scriptstyle{(12)}$$\scriptstyle{\overline{(12)}}$$\scriptstyle{\overline{(345)}}$

To view this action on the elliptic curve $\mathcal{E}_{2}$, we calculate that
$(345):[s:t:u:v]\to[s:t:-s-u/2-v/2:s+3u/2-v/2]$. In the $s=1$ affine chart
$4+8u+7u^{2}+u^{3}+(1-u)v^{2}=0$ we can take as a cohomology basis

${\eta:=\frac{du}{2v(1-u)}},$

and it is a simple algebraic calculation to see that this differential is
invariant under the action of $\overline{(345)}$. If $\\{\overline{H}_{i}\\}$,
$\\{\overline{H^{\prime}}_{i}\\}$ are the images of the Weierstrass points on
$\mathcal{E}_{2}$ such that ${(345)}({H}_{i})={H}_{i+1}$,
$\overline{(345)}(\overline{H}_{i})=\overline{H}_{i+1}$ and so forth, the
invariance of $\eta$ tells us that

$\int_{\overline{H}_{1}}^{\overline{H}_{2}}\eta=\int_{\overline{H}_{2}}^{\overline{H}_{3}}\eta=\ldots,\quad\int_{\overline{H^{\prime}}_{1}}^{\overline{H^{\prime}}_{2}}\eta=\int_{\overline{H^{\prime}}_{2}}^{\overline{H^{\prime}}_{3}}\eta=\ldots,\quad\int_{\overline{H}_{1}}^{\overline{H^{\prime}}_{1}}\eta=\int_{\overline{H}_{2}}^{\overline{H^{\prime}}_{2}}\eta=\ldots$

This means that if $\Lambda=\left\langle 1,5\tau_{0}\right\rangle$ is the
period lattice of the elliptic curve $\mathcal{E}_{2}$ then
$\int_{\overline{H}_{i}}^{\overline{H}_{i+1}}\eta$ and
$\int_{\overline{H^{\prime}}_{i}}^{\overline{H^{\prime}}_{i+1}}\eta$ are the
same fixed element of $\Lambda/3$. By choosing an appropriate basis we may
take $\overline{(345)}:z\to z+1/3$ for $z\in\mathbb{C}/\left\langle
1,5\tau_{0}\right\rangle$. Then the quotient map will be a 3:1 isogeny of
elliptic curves and we have the period of $\mathcal{E}_{4}$ to be $15\tau_{0}$
and $j_{\mathcal{E}_{4}}=j(15\tau_{0})$. Under this isogeny the six images on
$\mathcal{E}_{2}$ of the Weierstrass points $[1:1:\alpha:\beta:\gamma]$ are
mapped to two on $\mathcal{E}_{4}$. Indeed, one may construct the isogeny
explicitly using the Weierstrass-$\wp$ function and the Abel-Jacobi map via
the diagram

${\mathcal{E}_{2}}$${\mathcal{E}_{4}}$${J(\mathcal{E}_{2})}$${J(\mathcal{E}_{4})}$$\scriptstyle{P\mapsto\int_{\infty}^{P}\eta}$$\scriptstyle{z\mapsto
3z}$$\scriptstyle{z\mapsto(\wp(z){,}\wp^{\prime}(z))}$

where $J(\mathcal{E})$ represents the Jacobian of the elliptic curve viewed at
$\mathbb{C}/\left\langle 1,\tau\right\rangle$.

######  Remark 4.2.

This calculation may be verified numerically, as shown in the notebooks.

######  Remark 4.3.

Note that when we call $\overline{(345)}$ an automorphism of the elliptic
curve, we have to think about it only in the category of projective varieties
with morphisms given by rational maps. To consider the group structure on an
elliptic curve $E$ we need to also give a base point $O\in E$ which acts as
the additive identity. A morphisms from pair $(E,O)\to(E^{\prime},O^{\prime})$
must map $O\to O^{\prime}$. If we denote the group of automorphisms of $E$ as
a projective variety as $\operatorname{Aut}(E)$, and the group of $O$-fixing
automorphisms as $\operatorname{Aut}_{O}(E)$, then we have short exact
sequence

$0\to T_{E}\to\operatorname{Aut}(E)\to\operatorname{Aut}_{O}(E)\to 0,$

where $T_{E}$ is the group of translations of $E$. It is a classical theorem
that $\operatorname{Aut}_{O}(E)\in\\{C_{2},C_{4},C_{6}\\}$ (here we are over
$\mathbb{C}$).

Here $\overline{(345)}$ is indeed a translation. To see this take variables
$x,y$ defined by

$\displaystyle u$ $\displaystyle=\frac{-70s^{3}+6sx}{2(25s^{2}+3x)},\quad
v=\frac{-3y}{25s^{2}+3x},$

which in the affine patch where $s=1$ give $\mathcal{E}_{2}$ as the curve
$x^{3}-25/3x+2950/27+y^{2}=0$. In these coordinates $(345)$ acts as

$\displaystyle x$
$\displaystyle\mapsto\frac{5(275-3x+15y)}{3(65+15x-3y)},\quad
y\mapsto\frac{20(55-15x-3y)}{(65+15x-3y)}.$

Indeed for the projective coordinates $[X:Y:Z]$ of our curve
$X^{3}-25/3XZ^{2}+2950/27Z^{3}+Y^{2}Z=0$ we have under $\overline{(345)}$ that

$\displaystyle[X:Y:Z]$
$\displaystyle\mapsto[{5(275Z/3-X+5Y)}:20(55Z-15X-3Y):65Z+15X-3Y]$
$\displaystyle\mapsto[275Z/3-X-5Y:-220Z+60X-12Y:13Z+3X+(3Y)/5]\mapsto[X:Y:Z].$

When working with a Weierstrass model of an elliptic curve it is standard to
take the distinguished point to be $\infty=[0:1:0]$ and we see in terms of
$(x,y)$ that $\infty\mapsto(-25/3,20)\mapsto(-25/3,-20)\mapsto\infty$.

###  4.6. Summarising

We can collect the information of the quotients we have seen into the
following diagram. Solid arrows represent a covering map coming from a
quotient, whereas ‘squiggly’ arrows indicate isomorphisms that are unexplained
by group theory alone.

${\mathcal{B}}$${\mathcal{E}_{5\tau_{0}}}$${\mathcal{C}_{2}}$${\mathcal{C}_{1}}$${\mathcal{E}_{15\tau_{0}}}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$${\mathcal{E}_{\tau_{0}}}$${\mathcal{E}_{\tau_{0}}}$${\mathcal{E}_{5\tau_{0}}}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$${\mathcal{E}_{3\tau_{0}}}$${\mathbb{P}^{1}}$
Figure 3. Quotient structure of Bring’s curve.

This corresponds to the quotients by the following groups, where an arrow now
indicates that a source group is normal in the target, whereas a dashed line
just indicates that a group is a subgroup. The label used corresponds to the
list in §4.2, except where that label is ambiguous and the exact group must be
specified.

${\left\\{e\right\\}}$${\left\langle(12)\right\rangle}$${C_{3}}$${\left\langle(12)(34)\right\rangle}$${C_{6}}$${S_{3}}$${\left\langle(12),(34)\right\rangle}$${S_{3}^{\prime}}$${\left\langle(12)(34),(13)(24)\right\rangle}$${C_{4}}$${D_{12}}$${D_{8}}$${A_{4}}$${S_{4}}$
Figure 4. Subgroup structre of $S_{5}$ corresponding to the quotients of
Bring’s curve.

Finally, we recall from a Riemann-Hurwitz argument that each elliptic curves
that comes a quotient from a genus-2 curve with $V_{4}$ symmetry has 2 marked
points which are the branch points of the covering map, equivalently the
images of the fixed points of the involution being quotiented by. We have
shown that the preimages of these on Bring’s curve are geometric points. In
Figure 5 we illustrate the quotient structure highlighting these points. (We
do not decorate the curve $\mathcal{B}/A_{4}\cong\mathcal{E}_{3\tau_{0}}$, but
include it so as to show every quotient with genus greater than 0.) Note the
nodes of this diagram do not correspond directly to the placements of those
above.

${6{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times},4{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\otimes},2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\ocircle},4{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{\square}}{\includegraphics[height=19.91814pt]{g4.png}}}$${6{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}{\includegraphics[height=19.91814pt]{g1.png}}}$${2{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times},2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\ocircle}\includegraphics[height=19.91814pt]{g2.png}}$${2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\ocircle},2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{\square}}{\includegraphics[height=19.91814pt]{g2.png}}}$${2{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}{\includegraphics[height=19.91814pt]{g1.png}}}$${2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\ocircle}{\includegraphics[height=19.91814pt]{g1.png}}}$${2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\square}{\includegraphics[height=19.91814pt]{g1.png}}}$${2{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\ocircle}{\includegraphics[height=19.91814pt]{g1.png}}}$
Figure 5. Quotient structure including marked points. The points
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}$
correspond to the Weierstrass points $[1:1:\alpha:\beta:\gamma]$ (and
permutations of $\alpha,\beta,\gamma$), the points
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\otimes}$
to the vertices $[1:-1:0:\pm i:\mp i]$ and $[1:-1:\pm i:0:\mp i]$, the points
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\ocircle}$
to the vertices $[1:-1:\pm i:\mp i:0]$, and the points
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{\square}}$
correspond to the vertices $[1:\pm i:\mp i:-1:0]$ and $[1:\pm i:-1:\mp:0]$.

With the information of the quotients, we can now discuss the isogeny class
and isomorphism class of the Jacobian of Bring’s curve with the following
results.

######  Proposition 4.4 ([RR92], §4, [Web05], Corollary 5.5).

The $\mathbb{C}$-isogeny class of the Jacobian of Bring’s curve is
$\mathcal{E}_{3}^{4}$.

###### Proof.

As we will later want to strengthen this result, we will use methods which
relate subvarieties of the Jacobian to idempotents, and thus to subgroups of
the automorphism group. Namely we will use [LR04, §4.2], which gives the
isogeny decomposition in terms of Prym varieties of the Jacobian of a curve
with $A_{5}$ action. Note, because of comments made before, Bring’s curve is
the unique genus-4 curve for which we could apply this argument. The result
follows as, using the notation of Lange, $X_{A_{4}}=\mathcal{E}_{3}$,
$X_{D_{5}}=X_{Z_{5}}=Y=\mathbb{P}^{1}$.

This proof also follows from [LR04, Proposition 5.1], which uses only the
action of $S_{4}$ on the curve. Moreover, one could use [KR89, Theorem C]
taking the subgroups to be
$\left\langle(12)\right\rangle,\left\langle(34)\right\rangle$, $C_{4}$ and
$A_{4}$. Alternatively, using the isomorphism $A_{5}\cong
PSL_{2}(\mathbb{F}_{5})$ and [KR89, Example 2] on can find
$\operatorname{Jac}\mathcal{B}\sim\operatorname{Jac}\mathcal{C}_{1}\times\operatorname{Jac}\mathcal{C}_{2}$
and proceed from there. These proof strategies all follow the same approach of
looking for idempotents.

One may also obtain the result following the same method as R&R. We use that
fact that isogenies act on the period matrix by right multiplication by
matrices $R\in M_{4}(\mathbb{Z})$, and so we have the required isogeny by
taking $\lambda=1$ in the identity

$\begin{pmatrix}\lambda^{-1}\operatorname{Id}\end{pmatrix}\begin{pmatrix}1&\tau
M\end{pmatrix}\begin{pmatrix}\lambda\operatorname{Id}&0\\\
0&M^{\prime}\end{pmatrix}=\begin{pmatrix}\operatorname{Id}&\dfrac{5}{\lambda}\tau\operatorname{Id}\end{pmatrix},$

where

$M^{\prime}:=5M^{-1}=\begin{pmatrix}2&-1&1&-1\\\ -1&2&-1&1\\\ 1&-1&2&-1\\\
-1&1&-1&2\end{pmatrix}.$

∎

Note that we are able to construct the quotient $\mathcal{B}/A_{4}$ directly
from the $\mathbb{P}^{4}$-model using transformations with coefficients in
$\mathbb{Q}$, and the resulting curve is defined over $\mathbb{Q}$, so by
[KR89, Remark 6] the above proposition can be strengthened to a statement
about the $\mathbb{Q}$-isogeny class of the Jacobian of Bring’s curve.
Calculating using Sage we find that the $\mathbb{Q}$-isogeny class of
$\mathcal{B}/A_{4}$ is 50a using the Cremona labels for elliptic curve
$\mathbb{Q}$-isogeny classes. Hence the following result holds.

######  Proposition 4.5 ([Ser08], Exercise 8.3.2(b)).

The $\mathbb{Q}$-isogeny class of the Jacobian of the $\mathbb{P}^{4}$-model
of Bring’s curve is $(50a)^{4}$.

Note in the above proposition we had to be careful to specify the Jacobian of
the $\mathbb{P}^{4}$-model of Bring’s curve, as we have seen that HC-model is
not birationally equivalent over a field that doesn’t contain
$\mathbb{Q}[\zeta]$. The $\mathbb{Q}$-isogeny class of the elliptic curve
$\mathcal{B}/\left\langle(12)(34),(13)(24)\right\rangle$ calculated via the
HC-model is 50b. Similarly, the $\mathbb{Q}$-isogeny class of the two elliptic
curves covered by $\mathcal{C}_{2}$ is 450b, and the computation of the
quotient required the coefficient field to be $\mathbb{Q}[\rho]$. In order to
not have a contradiction with Proposition 4.5, we must have that the
$\mathbb{Q}$-isogeny classes 50b and 50a merge over $\mathbb{Q}[\sqrt{5}]$,
and that the isogeny classes 450b and 50a merge over $\mathbb{Q}[\rho]$, which
is indeed the case.242424The isogeny classes 450b and 50b merge over
$\mathbb{Q}[\sqrt{-15}]$.

One could use computational tools such as those in [BSS+16, Lom19]252525To
make the reconstruction process of Lombardo more accessible, we recreated in
Sage some of the functions implemented by Lombardo in Magma. to numerically
find the $\mathbb{Q}$-isogeny class of the Jacobian of the HC-model of Bring’s
curve, as we did in the notebooks. Such computational results using
idempotents can be helpful for developing our understanding, for example one
can use computer algebra to search for relations between the characters
$\operatorname{Ind}_{H}^{S_{5}}(1_{H})$, that is the characters of $S_{5}$
induced from the trivial representation of $H\leq S_{5}$. Doing so gives
relations between subvarieties of the Jacobian of Bring’s curve following
[KR89, Theorem 3], for example

(18) $\displaystyle J_{\left\langle(12)\right\rangle}\times
J_{\left\langle(12)(34),(13)(24)\right\rangle}$ $\displaystyle\sim
J_{\left\langle(12)(34)\right\rangle}\times
J_{\left\langle(12),(34)\right\rangle},$ $\displaystyle
J_{\left\langle(12)(34),(13)(24)\right\rangle}\times J_{S_{3}}$
$\displaystyle\sim J_{\left\langle(12),(34)\right\rangle}\times
J_{S_{3}^{\prime}}.$

Using Riemann-Hurwitz arguments these would let us say that
$\mathcal{B}/\left\langle(12)\right\rangle\sim\mathcal{B}/C_{4}$ and
$\mathcal{B}/\left\langle(12)(34),(13)(24)\right\rangle\sim\mathcal{B}/S_{3}^{\prime}$
without having to do any calculation, entirely from group theory. The reason
why these isogenies are actually isomorphisms is not clear.

Proposition 4.4 can also be strengthened in a different direction.

######  Proposition 4.6 ([GAR00]).

The Jacobian of Bring’s curve is isomorphic as a complex torus to
$\mathcal{E}_{2}^{3}\times\mathcal{E}_{1}$.

###### Proof.

The proof in [GAR00] is very general, considering Jacobians whose period
matrix is invariant under $S_{n}$ for any $n$. The isomorphism from the period
matrix given in [RR92] is

$C\begin{pmatrix}1&\tau M\end{pmatrix}\begin{pmatrix}D&0\\\
0&E\end{pmatrix}=\begin{pmatrix}1,\tau\operatorname{diag}(5,5,5,1)\end{pmatrix}$

where

$C=\begin{pmatrix}-1&1&-1&2\\\ 1&0&0&1\\\ 0&-1&0&1\\\
0&0&0&1\end{pmatrix},D=\begin{pmatrix}0&1&0&-1\\\ 0&0&-1&1\\\ -1&-1&-1&4\\\
0&0&0&1\end{pmatrix},E=\begin{pmatrix}-1&1&0&0\\\ 1&0&-1&0\\\ -2&-1&-1&1\\\
1&0&0&0\end{pmatrix}.$

∎

Note it would not be possible that the Jacobian is isomorphic to the product
of the elliptic curves as a principally polarised abelian variety, as it is
well know that the Jacobian of any smooth compact Riemann surface with
canonical principal polarisation is irreducible. In particular this means the
matrix $\begin{pmatrix}D&0\\\ 0&E\end{pmatrix}$ of the proposition is not
symplectic.

##  5\. Theta Characteristics

We now investigate the theta characteristics of Bring’s curve, which in
[Ati71] were shown to be equivalent to spin structures, and as such we will
use the terms interchangeably. These are objects which have been studied
greatly in a variety of contexts in algebraic geometry, number theory, and
even string theory, though in this paper we will not discuss these
applications and focus on calculation. Among other results we shall identify
the unique invariant spin structure of Bring’s curve.

For completeness we recall the main definition that we shall use in this
paper.

######  Definition 5.1.

A theta characteristic is a half-canonical divisor class, i.e. a divisor class
$[D]$ such that $2[D]=[\mathcal{K}_{\mathcal{B}}]$.

######  Lemma 5.2.

Call a theta characteristic odd/even based on the parity of it’s index of
speciality. Then on a Riemann surface of genus $g$ there are
$2^{g-1}(2^{g}-1)$ odd theta characteristics and $2^{g-1}(2^{g}+1)$ even theta
characteristics.

There are many equivalent definitions of parity for a theta characteristic,
and while the above is not computationally the easiest, it is one of the
easier to state. [Far12] provides a nice overview of these and their
connections.

######  Example 5.3.

It is a computational exercise to verify
$\mathcal{K}_{\mathcal{B}}\sim\left(\frac{-xv_{3}v_{4}}{v_{2}}\right)=2(3a+b-c)$,
and so $\Delta=3a+b-c$ is a theta characteristic on Bring’s curve. A simple
calculation in Sage shows that it is even.

###  5.1. Abel-Jacobi and the Riemann Constant Vector

We will now take a detour to lay out some notation and definitions in
preparation for later.

######  Definition 5.4.

The Abel-Jacobi map on a Riemann surface $\mathcal{C}$ with basis of
differentials $\left\\{\omega_{j}\right\\}$, based at $Q\in\mathcal{C}$, is
given by

$(\mathcal{A}_{Q}(P))_{j}=\int_{Q}^{P}\omega_{j}.$

This is well defined up to periods on the curve. Letting $\tau$ be the Riemann
matrix of $\mathcal{C}$, the Riemann Constant Vector (RCV) based at $Q$ is
given by

$(K_{Q})_{j}=-\frac{1+\tau_{jj}}{2}+\sum_{k\neq
j}^{g}\oint_{a_{k}}\omega_{k}(P)(\mathcal{A}_{Q}(P))_{j}.$

The RCV is related to the canonical divisor by
$2K_{Q}\equiv\mathcal{A}_{Q}(\mathcal{K}_{\mathcal{C}})$. Indeed, [DP08] shows
how to find $K_{Q}$ from this relation given the theta function. We then have
the following theorem about the divisor of the Szëgo kernel $\Delta_{S}$,
defined in [Fay73].

######  Theorem 5.5 ([Fay73]).

$K_{Q}=\mathcal{A}_{Q}(\Delta_{S})$, and as such $\Delta_{S}$ is a theta
characteristic.

The definition of the RCV also requires that the differentials are normalised
with respect to the homology basis taken $\left\\{a_{j},b_{j}\right\\}$ such
that $\oint_{a_{j}}\omega_{i}=\delta_{ij}$. As such hidden in the above result
is the fact that the Szëgo kernel divisor is homology dependent, because the
RCV is.

###  5.2. Tritangent Planes and Odd Characteristics

Recall that when we consider the canonical embedding of a curve in
$\mathbb{P}^{g-1}$, the intersection with a hyperplane gives an effective
element of the canonical divisor class of the curve, as it corresponds to the
zero-locus of a differential on the curve. A tritangent plane is a plane that
intersects the curve tangentially in three points. If a hyperplane is tangent
at each intersection (i.e. the intersection is of order 2) then this naturally
constructs an effective theta divisor on the curve, and moreover every such
theta characteristic $\theta$ gives a hyperplane [CKRSN19]. For any sextic
that is the intersection of a smooth quadric and smooth quartic in
$\mathbb{P}^{3}$, for example Bring’s curve, the 120 tritangent planes are in
1-1 correspondence with the 120 odd theta characteristics [HL17, Theorem 2.2].

On Bring’s curve we have the following result about these planes.

######  Proposition 5.6 ([Edg81]).

The 120 tritangent planes on Bring’s curve split into two classes, 60 in each
class;

  1. (1)

those where all three contact points are Weierstrass points, and

  2. (2)

those where only one contact points is a Weierstrass point.

Planes in the first and second class respectively have equations (recalling
notation from §3.1)

$\displaystyle\Pi_{\alpha jk}^{(1)}$
$\displaystyle:=\left\\{\frac{x_{j}}{\beta}-\frac{x_{k}}{\gamma}=0\right\\},$
$\displaystyle\Pi_{ijk}^{(2)}$
$\displaystyle:=\left\\{(\alpha-1)(\alpha+4)x_{i}+(\beta-1)(\beta+4)x_{j}+(\gamma-1)(\gamma+4)x_{k}=0\right\\},$

where $[x_{i}]\in\mathbb{P}^{4}$ and $i,j,k$ distinct.

To clarify the notation we have used for the planes, recall that the position
of the indices on Weierstrass points $W_{ijk}$ indicates which root was equal
to $x_{i}$. The same principle holds for $\Pi_{\alpha
jk}^{(1)},\Pi^{(1)}_{i\beta k},$ and $\Pi^{(1)}_{ij\gamma}$.

######  Corollary 5.7.

The orbit decomposition of odd theta characteristics on Bring’s curve is

(19) $120=20+20+20+60.$

###### Proof.

The characteristics coming from the triangent planes are

$\displaystyle T_{\alpha jk}^{(1)}$
$\displaystyle=\sum_{\begin{subarray}{c}i=1\\\ i\neq
j,k\end{subarray}}^{5}W_{ijk},$ $\displaystyle T_{ijk}^{(2)}$
$\displaystyle=W_{ijk}+O_{ijk}^{+}+O_{ijk}^{-}.$

Here we define the points $O_{ijk}^{\pm}$ by, for example,

$O_{345}^{\pm}=\left[\frac{1\pm i\sqrt{15}}{2}:\frac{1\mp
i\sqrt{15}}{2}:\alpha^{2}+\alpha+1,\beta^{2}+\beta+1,\gamma^{2}+\gamma+1\right].$

A simple orbit-stabiliser argument then gives the orbit decomposition as, for
example, $T_{\alpha 45}^{(1)}$ is stabilised by the symmetric group
$S_{\\{1,2,3\\}}$ and $T_{345}^{(2)}$ is stabilised by $S_{\\{1,2\\}}$. ∎

######  Remark 5.8.

While we were able to proceed analytically here characterising the orbits of
the odd characteristics using the work of earlier authors who had identified
the tritangent planes of the curve we also may obtain the orbit structure
numerically using Sage. Given a homology representation of the automorphism
group, the algorithm to compute the orbits is exact, but the current
implementation in Sage requires floating point arithmetic to calculate this
representation.

###  5.3. Even Characteristics

In the previous subsection, we were able to fully characterise the odd theta
characteristics on Bring’s curve without too much difficulty through the use
of existing work giving the stalls of the canonical embedding. The story for
even characteristics is different, as the line bundle corresponding to a
generic even theta characteristic has no sections, which is an obstacle to
their study [TZ11]. In [DK93] the authors were able to use the Scorza
correspondence to calculate the orbit decomposition of even spin structures on
Klein’s curve, but at the time of writing the theory of the Scorza map has not
been developed far enough to cover Bring’s curve. The Scorza quartic necessary
to extend Dolgachev and Kanev’s work beyond $g=3$, wherein it becomes a
codimension-1 hypersurface in $\mathbb{P}^{g-1}$, has been proven to exist for
a generic even spin curve (that is a pair consisting of a curve and an even
theta characteristic on it); it has only been explicitly found for trigonal
curves in $g=4$ [TZ11]. In [Bur83], motivated by its realisation via an
elliptic modular surface, Burns identified a theta characteristic on Bring’s
curve invariant under the $A_{5}$ subgroup of the automorphism group, though
he described this characteristic only in terms of two line bundles on the
curve, not directly in terms of points on the curve. We shall now fully
classify the orbits of the even characteristics and give an explicit
description of Burns’ divisor in the process.

We have two distinct methods of probing the orbit decomposition of the theta
characteristics on Bring’s curve.

  1. (1)

Use the method of [KS06], wherein theta characteristics are identified with
vectors in $\mathbb{F}_{2}^{2g}$, and the action of automorphisms given by the
homology representation of the automorphism group of the curve as found using
the methods of [BSZ19].

  2. (2)

Identify theta characteristic with the $2^{2g}$ translates by half-lattice
vectors of a half-canonical vector in the Jacobian of the curve, done using
the implementation of the Abel-Jacobi map developed by Disney-Hogg [DH21], and
the action of automorphisms given by the cohomology representation of the
automorphism group of the curve as found using the methods of [BSZ19].

The two methods have different strengths, namely that the first uses exact
computations rather than approximations with multi-precision arithmetic, but
the second can concretely relate theta characteristics from their
representation to an actual divisor (class). Both methods not only verify
Corollary 5.7 but also gives the following result.

######  Theorem 5.9.

The orbit decomposition of even theta characteristics on Bring’s curve is

(20) $136=1+5+5+5+10+10+10+30+30+30.$

######  Corollary 5.10.

Bring’s curve has a unique theta characteristic invariant under the action of
the automorphism group, which is also the theta characteristic invariant under
the $A_{5}$ subgroup found in [Bur83].

The existence of the unique invariant theta characteristic was known in
[BN12], but it had not been identified. This we rectify with the following
result.

######  Theorem 5.11.

The theta characteristic $\Delta$ (defined in Example 5.3) is the unique
invariant theta characteristic in on Bring’s curve.

###### Proof.

We first consider the action of $S$ on $a$, $b$, $c$, $d$. We have

$\displaystyle a:=[0:0:1]$
$\displaystyle\simeq[2t^{3}:t:1]\xrightarrow{S}[2t^{3}:\zeta
t:\zeta^{-1}]=[2\zeta t^{3}:\zeta^{2}t:1]=[2(\zeta^{2}t)^{3}:\zeta^{2}t:1]$
$\displaystyle=[2\epsilon^{3}:\epsilon:1],$ $\displaystyle b:=[0:1:0]$
$\displaystyle\simeq[2t^{2}:1/t:1]\xrightarrow{S}[2t^{2}:\zeta/t:\zeta^{-1}]=[2\zeta
t^{2}:\zeta^{2}/t:1]=[2(t/{\zeta^{2}})^{2}:\zeta^{2}/t:1]$ $\displaystyle
c:=[1:0:0]_{2}$ $\displaystyle\simeq[1:t:t^{4}]\xrightarrow{S}[1:\zeta
t:\zeta^{-1}t^{4}]=[1:\zeta t:(\zeta t)^{4}],$ $\displaystyle d:=[1:0:0]_{1}$
$\displaystyle\simeq[1:t^{4}:t]\xrightarrow{S}[1:\zeta
t^{4}:\zeta^{-1}t]=[1:(\zeta^{-1}t)^{4}:\zeta^{-1}t].$

Thus $a$, $b$, $c$ and $d$ are invariant under the symmetry $S$ and
consequently $\Delta=3a+b-c$ is also invariant.

Similarly, as mentioned in Proposition 2.15, the action of $U$ on $a,b,c,d$
can be calculated as

$a\mapsto c\mapsto b\mapsto d\mapsto a,$

and so

$\Delta=3a+b-c\mapsto 3c+d-b=3a+b-c-(3a+2b-4c-d)=\Delta-(x)\sim\Delta.$

We have thus shown that $\Delta$ is invariant under $\left\langle
S,U\right\rangle$. To complete the proof that $\Delta$ is the invariant theta
characteristic, one could attempt to show that $\Delta$ is invariant under the
action of $R$ by direct computation, but this proves to be difficult. It is
instead better to check in Sage that the unique spin structure invariant under
the whole automorphism group is actually also the unique spin structure
invariant under the subgroup generated by $S$ and $U$. In fact, by Theorem 1.2
of [KS06] and our work in §4, we know there is a unique theta characteristic
invariant under $\left\langle S\right\rangle$262626Under the action of $S$,
the orbit decompositions are $120=24\times 5$ and $136=1+27\times 5$. , and
this completes the proof. ∎

This proof strategy is similar to the identification of the unique invariant
theta characteristic on Klein’s curve in [KS06].272727The final part of the
proof in that paper, showing invariance under the order-2 generator of
$PSL_{2}(\mathbb{F}_{7})$, is in our opinion incomplete; Kallel (private
correspondence) believes our reasoning correct. The theorem remains correct
nevertheless, as one can check using their methodology that invariance under
the order-7 generator is enough to specify the unique invariant spin
structure, as the decompositions are $28=4\times 7$ and $36=1+5\times 7$. It
is a curious coincidence that in both cases it was a generator of order 2 for
which the direct calculation was difficult.

We now want to make the connection to [Bur83]. Recall Lemma 3 and Proposition
2.9. This gives us two degree-3 maps $f_{i}:\mathcal{B}\to\mathbb{P}^{1}$,
namely $f_{i}=\left.\pi_{i}\circ\varphi^{-1}\right\rvert_{\mathcal{B}}$,
$i=1,2$, where $\varphi$ was the isomorphism
$\mathbb{P}^{1}\times\mathbb{P}^{1}\to\mathcal{Q}$, and $\pi_{i}$ the
projection to the two factors of $\mathbb{P}^{1}\times\mathbb{P}^{1}$. What
are the corresponding divisors? Working in the $L_{a}$ coordinates, we can use
Sage to find that $f_{1}^{-1}([1:0])=2[0:0:0:1]+[1:0:0:0]=2b+c:=L^{\prime}$,
while $f_{2}^{-1}([1:0])=2[0:1:0:0]+[0:0:0:1]=2d+b:=L$.

######  Proposition 5.12.

The divisors $L$, $L^{\prime}$ satisfy the properties described in [Bur83],
namely

$\displaystyle\Delta$ $\displaystyle\sim
3(L^{\prime}-L)+L,\quad\mathcal{K}_{\mathcal{B}}\sim L+L^{\prime},\quad 0\sim
5(L^{\prime}-L).$

###### Proof.

This is straightforward verification from the definitions. ∎

Note we can connect this back to the degree-3 map given by Klein. We know from
[Web05] that we expect this map to be branched at face-centres of the
$\left\\{5,5|3\right\\}$ tessellation, and these come from face-centres of the
$\left\\{5,4\right\\}_{6}$ tessellation, and indeed recall we saw that
$a,b,c,d$ were face-centres.

An additional characterisation of the invariant theta characteristic is given
by the following result.

######  Proposition 5.13.

In the homology basis of [RR92], the RCV satisfies

$K_{a}=\frac{1}{10}(3,2,-2,-3)+\Im(\tau_{0})(1,-2,-2,1)i=\mathcal{A}_{a}(\Delta).$

As such, in the R&R homology basis, the unique invariant theta characteristic
is the divisor (class) of the Szëgo kernel, i.e. $\Delta=\Delta_{S}$.

###### Proof.

The first equality is shown analytically in [BN12]. The second is shown
numerically in the corresponding notebooks, using the Abel-Jacobi map
developed by Disney-Hogg [DH21]. To verify the RCV, we implemented the
methodology of [DP08] using the theta function in Sage developed by Bruin and
Ganjian [Bru21]. While these calculations are numerical in nature, the
calculations can be done with arbitrary binary precision. We were satisfied by
calculating with 400 binary digits of precision, giving an absolute error of
less than $10^{-118}$ for the first equality, and of less than $10^{-23}$ for
the second equality. ∎

##  6\. Conclusion

We have now seen that Bring’s curve has many interesting properties and
appears in a number of quite different settings; with this background now
established we can summarise the main new results of the paper. We have:

  1. (1)

Constructed an explicit birational map between the HC- and
$\mathbb{P}^{4}$-model of the curve, Proposition 2.11.

  2. (2)

Constructed an explicit transform (5) between the period matrix of Weber and
that of R&R.

  3. (3)

Provided a complete realisation of the automorphism group of the curve in the
HC-model (Propositions 2.14 and 2.15) and the canonical model (8,9).

  4. (4)

Unified the functional, algebraic and geometric pictures of the distinguished
orbits on the curve (those of size 24, 30, and 60) by explicitly identifying
these points (Propositions 3.3,3.8,3.9) and identifying the divisors of the
holomoprhic differentials and meromorphic functions associated to the
Weierstrass points (Proposition 3.6).

  5. (5)

Completed the classification of, and relations between, the quotients of
Bring’s curve (Figures 3 and 4), finding a previously unidentified elliptic
curve in the process (14). We identify one of the known quotient curves with
the modular curve $X_{0}(50)$.

  6. (6)

Proven in a new way the $\mathbb{Q}$-isogeny class of the Jacobian
(Proposition 4.5) and established a number of isogenies between (products of)
the various Prym varieties associated with the quotients (18).

  7. (7)

Computed the orbit decomposition of both the odd and even theta
characteristics on the curve (19,20) and determined the unique invariant theta
characteristic (Proposition 5.11) identifying this with the Szëgo divisor
(Proposition 5.13) and a divisor identified by Burns (Proposition 5.12).

At various points we have corrected the literature.

In addition to our written account there are the accompanying notebooks.
Throughout we have provided examples of how modern computational tools can
simplify calculation and provide insight to aid discovery. The notebooks
utilise SageMath, Maple, and Macaulay2 (all interfaced through Sage) and the
code may be easily modified for other curves.

Finally we would draw attention to two areas that (for us) merit attention.
First are the isomorphisms in Figure 3 we have been unable to explain within
the quotient structure of the curve. Perhaps this is something that the
modular theory of the Bring’s curve can shed light on. The second area deals
with the orbit decompositions of the theta characteristics. In the case of
even characteristics our construction was purely numeric and not analytic: a
solution to this likely involves extending the theory of the Scorza quartic.
Further, for both even and odd characteristics, there is a ‘‘threeness ’’we
observe that we have found no satisfying reason for. We look forward to the
clarification of these.

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  * [Ati71] Michael F. Atiyah, _Riemann surfaces and spin structures_ , Annales scientifiques de l’École Normale Supérieure, Series 4. 4 (1971), no. 1, 47--62.
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