# Recovery of Plane Curves from Branch Points

Daniele Agostini  Hannah Markwig  Clemens Nollau
Victoria Schleis  Javier Sendra–Arranz  and Bernd Sturmfels

###### Abstract

We recover plane curves from their branch points under projection onto a line.
Our focus lies on cubics and quartics. These have $6$ and $12$ branch points
respectively. The plane Hurwitz numbers $40$ and $120$ count the orbits of
solutions. We determine the numbers of real solutions, and we present exact
algorithms for recovery. Our approach relies on $150$ years of beautiful
algebraic geometry, from Clebsch to Vakil and beyond.

##  1 Introduction

Arthur Cayley in 1879 was the first to use “algorithm” to title a discrete
geometry paper. In [7] he identifies the finite vector space
$(\mathbb{F}^{2})^{6}$ with the $64$ theta characteristics of a plane quartic
curve, i.e. the $28$ bitangents and the $36$ symmetric determinantal
representations. The present paper can be viewed as a sequel. Our Table
LABEL:table:40covers is very much in the spirit of [7].

One century after Cayley, algorithms in discrete geometry became a field in
its own right, in large part thanks to Eli Goodman. We are proud to dedicate
this article to Eli’s memory. Eli obtained his PhD in 1967 with Heisuke
Hironaka. He had important publications in algebraic geometry (e.g. [14])
before embarking on his distinguished career on the discrete side.

Consider the map $\pi:\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{1}$ that takes
a point $(x:y:z)$ in the projective plane to the point $(x:y)$ on the
projective line. Geometrically, this is the projection with center
$p=(0:0:1)$. We restrict $\pi$ to the curve $V(A)$ defined by a general
ternary form of degree $d$,

$A(x,y,z)\,\,\,=\,\sum_{i+j+k=d}\\!\alpha_{ijk}\,x^{i}y^{j}z^{k}.$ (1)

The resulting $d:1$ cover $V(A)\rightarrow\mathbb{P}^{1}$ has $d(d-1)$ branch
points, represented by a binary form

$B(x,y)\,\,\,=\sum_{i+j=d(d-1)}\\!\\!\beta_{ij}\,x^{i}y^{j}.$ (2)

Passing from the curve to its branch points defines a rational map from the
space $\mathbb{P}^{\binom{d+2}{2}-1}$ with coordinates $\alpha$ to the space
$\mathbb{P}^{d(d-1)}$ with coordinates $\beta$. Algebraically, this is the map

$\mathbb{P}^{\binom{d+2}{2}-1}\,\dashrightarrow\,\,\mathbb{P}^{d(d-1)}\,,\,\,A\,\mapsto\,{\rm
discr}_{z}(A).$ (3)

This is the discriminant of $A$ with respect to the last variable. That
discriminant is a binary form $B$ of degree $d(d-1)$ in $x,y$ whose
coefficients are polynomials of degree $2d-2$ in $\alpha$.

We here study the Inverse Problem, namely recovery of the curve from its
branch points. Given the binary form $B$, our task is to compute all ternary
forms $\hat{A}$ such that ${\rm discr}_{z}(\hat{A})=B$. This is a system of
$d(d-1)+1$ polynomial equations of degree $2d-2$ in the $\binom{d+2}{2}$
unknowns $\alpha$. Solving this system means computing a fiber of the map (3)
over $B$. Recovery is not unique because ${\rm discr}_{z}(A)$ is invariant
under the action of the subgroup $\mathcal{G}$ of ${\rm PGL}(3)$ given by

$\qquad g\,\,:\,\,x\mapsto g_{0}x\,,\,\,y\mapsto g_{0}y\,,\,\,z\mapsto
g_{1}x+g_{2}y+g_{3}z\qquad\hbox{with $\,g_{0}g_{3}\not=0$.}$ (4)

By [23, Corollary 5.2.1], the fiber over $B$ is a finite union of
$\mathcal{G}$-orbits. Their number $\mathfrak{h}_{d}$ is the plane Hurwitz
number of degree $d$. Our task is to compute representatives for all
$\mathfrak{h}_{d}$ orbits.

######  Example 1.1 ($d=2$).

For conics we have $\mathfrak{h}_{2}=1$ and recovery is easy. Our polynomials
are

$\begin{matrix}A&=&\alpha_{200}x^{2}+\alpha_{110}xy+\alpha_{101}xz+\alpha_{020}y^{2}+\alpha_{011}yz+\alpha_{002}z^{2},\\\
{\rm
discr}_{z}(A)&=&(4\alpha_{002}\alpha_{200}-\alpha_{101}^{2})x^{2}\,+\,(4\alpha_{002}\alpha_{110}-2\alpha_{011}\alpha_{101})xy\,+\,(4\alpha_{002}\alpha_{020}-\alpha_{011}^{2})y^{2},\\\
B&=&\beta_{20}x^{2}+\beta_{11}xy+\beta_{02}y^{2}.\end{matrix}$

The equations ${\rm discr}_{z}(\hat{A})=B$ describe precisely one
$\mathcal{G}$-orbit in $\mathbb{P}^{5}$. A point in that orbit is

$\hat{A}\,\,=\,\,\frac{1}{4}\beta_{20}x^{2}+\frac{1}{4}\beta_{11}xy-\beta_{02}yz+\beta_{02}z^{2}.$

Up to the $\mathcal{G}$-action, this is the unique solution to our recovery
problem for plane conics. $\diamond$

Plane Hurwitz numbers $\mathfrak{h}_{d}$ were studied in Ongaro’s 2014 PhD
thesis and in his work with Shapiro [23, 25]. These served as the inspiration
for our project. Presently, the only known nontrivial values are
$\mathfrak{h}_{3}=40$ and $\mathfrak{h}_{4}=120$. The former value is due to
Clebsch [9, 10]. We first learned it from [23, Proposition 5.5.2]. The latter
value was computed by Vakil in [29]. The plane Hurwitz number
$\mathfrak{h}_{4}=120$ was presented with the extra factor $(3^{10}-1)/2$ in
[23, eqn. (5.14)] and in [25, p. 608]. However, that factor is not needed; see
Remark 2.8.

The parameter count above implies that the closure of the image of (3) is a
variety $\mathcal{V}_{d}$ of dimension $\binom{d+2}{2}-4$ in an ambient space
of dimension $d(d-1)$. For $d=2,3$, the two dimensions agree, so recovery is
possible for generic $B$. For $d\geq 4$, the constraint $B\in\mathcal{V}_{d}$
is nontrivial. For instance, $\mathcal{V}_{4}$ is a hypersurface of degree
$3762$ in $\mathbb{P}^{12}$, as shown by Vakil [29].

This article is organized as follows. In Section 2 we approach our problem
from the perspective of computer algebra. We establish a normal form with
respect to the $\mathcal{G}$-action, and we identify the base locus of the map
(3). This allows to state the recovery problem as a polynomial system with
finitely many solutions over the complex numbers $\mathbb{C}$. The number of
solutions is $\mathfrak{h}_{3}=40$ for cubics, and it is
$\mathfrak{h}_{4}=120$, provided $B$ lies on the hypersurface
$\mathcal{V}_{4}$.

In Section 3 we establish the relationship to Hurwitz numbers that count
abstract coverings of $\mathbb{P}^{1}$. We encode such coverings by monodromy
graphs, and we determine the real Hurwitz numbers for our setting. A highlight
is Table LABEL:table:40covers, which matches the $40$ monodromy
representations for $d=3$ with combinatorial labels taken from Clebsch [10]
and Elkies [11].

In Section 4 we exhibit the Galois group for the $40$ solutions when $d=3$,
and we discuss different realizations of this group. Theorem 4.1 implies that
it agrees with the Galois group for the $27$ lines on the cubic surface.
Following classical work of Clebsch [9, 10], we show that the recovery of the
$39$ other cubics from the given cubic $A$ can be solved in radicals.

Section 5 builds on work of Vakil [29]. It relates the recovery of quartic
curves to tritangents of sextic space curves and to del Pezzo surfaces of
degree one. Theorem 5.1 determines the possible number of real solutions.
Instances with $120$ rational solutions can be constructed by blowing up the
plane $\mathbb{P}^{2}$ at $8$ rational points. We conclude with Theorem 5.6
which connects the real structure of $8$ points in $\mathbb{P}^{2}$ with that
of the $12$ branch points in $\mathbb{P}^{1}$.

This article revolves around explicit computations, summarized in Algorithms
2.10, 3.9, 4.5, 5.4, 5.5. Our software and other supplementary material is
available at the repository website MathRepo [12] of MPI-MiS via the link
https://mathrepo.mis.mpg.de/BranchPoints .

##  2 Normal Forms and Polynomial Systems

We identify $\mathbb{P}^{\binom{d+2}{2}-1}$ with the space of plane curves (1)
of degree $d$ and use as homogeneous coordinates the $\alpha_{ijk}$. The
following subspace of that projective space has codimension three:

$L_{d}\,\,=\,\,V(\,\alpha_{10\,d-1}\,,\,\alpha_{d-1\,10}\,,\,\alpha_{00d}-\alpha_{01\,d-1}\,).$
(5)

We now show that this linear space serves as normal form with respect to the
group action on fibers of (3). The group that acts is the three-dimensional
group $\mathcal{G}\subset{\rm PGL}(3)$ given in (4).

######  Theorem 2.1.

Let $A$ be a ternary form of degree $d\geq 3$ such that

$\displaystyle\alpha_{00d}\left(\,\sum_{k=0}^{d-1}\frac{(k+1)(-1)^{k}}{d^{k}}\alpha_{10\,d-1}^{k}\alpha_{00d}^{d-k-1}\alpha_{d-k-1\,0\,k+1}\right)\,\,\neq\,\,0.$
(6)

The orbit of $\,A$ under the $\mathcal{G}$-action on
$\,\mathbb{P}^{\binom{d+2}{2}-1}$ intersects the linear space $L_{d}$ in one
point.

######  Remark 2.2.

This statement is false for $d=2$. The $\mathcal{G}$-orbit of $A$ consists of
the conics

$\displaystyle
gA\,=\,(\alpha_{002}g_{1}^{2}+\alpha_{101}g_{0}g_{1}+\alpha_{200}g_{0}^{2})x^{2}+(2\alpha_{002}g_{1}g_{2}+\alpha_{011}g_{0}g_{1}\,+\,\alpha_{101}g_{0}g_{2}+\alpha_{110}g_{0}^{2})xy\,\,+$
$\displaystyle(2\alpha_{002}g_{1}g_{3}{+}\alpha_{101}g_{0}g_{3})xz+(\alpha_{002}g_{2}^{2}{+}\alpha_{011}g_{0}g_{2}{+}\alpha_{020}g_{0}^{2})y^{2}+(2\alpha_{002}g_{2}g_{3}{+}\alpha_{011}g_{0}g_{3})yz\\!+\\!\alpha_{002}g_{3}^{2}z^{2}.$

For generic $\alpha$, no choice of $g\in\mathcal{G}$ makes both the
$xy$-coefficient and the $xz$-coefficient zero. Note that the parenthesized
sum in (6) is the zero polynomial for $d=2$, but not for $d\geq 3$.

###### Proof of Theorem 2.1.

The unique point in $\,L_{d}\,\cap\,\mathcal{G}A\,$ is found by computation.
Without loss of generality, we set $g_{0}=1$. Next we set
$g_{1}=-\frac{1}{d}\alpha_{10\,d-1}/\alpha_{00d}$ because the coefficient of
$xz^{d-1}$ in $gA$ equals $(d\alpha_{00d}g_{1}+\alpha_{10\,d-1})g_{3}^{d-1}$.
The polynomial $gA$ arises from $A$ by the coordinate change $z\mapsto
g_{1}x+g_{2}y+g_{3}z$. Thus, a monomial $x^{i}y^{j}z^{d-i-j}$ contributes the
expression $x^{i}y^{j}(g_{1}x+g_{2}y+g_{3}z)^{d-i-j}$ to $gA$. This
contributes to the monomials
$x^{i^{\prime}}y^{j^{\prime}}z^{d-i^{\prime}-j^{\prime}}$ with $i^{\prime}\geq
i$ and $j^{\prime}\geq j$. The coefficient of $x^{d-1}y$ in $gA$ arises from
the following subsum of $A$:

$\sum_{i=0}^{d-1}\alpha_{i0\,d-i}\,x^{i}z^{d-i}\,+\,\sum_{i=0}^{d-1}\alpha_{i1\,d-i-1}\,x^{i}yz^{d-i-1},$

after inserting the coordinate change. Thus the coefficient of $x^{d-1}y$ in
$gA$ equals

$\sum_{i=0}^{d-1}\alpha_{i0\,d-i}(d-i)\,g_{1}^{d-i-1}g_{2}\,+\,\sum_{i=0}^{d-1}\alpha_{i1\,d-i-1}\,g_{1}^{d-i-1}.$

Inserting our above result for $g_{1}$, and setting the coefficient of
$x^{d-1}y$ to zero, we can solve this affine-linear equation for $g_{2}$,
obtaining a rational function in the $\alpha_{ijk}$ as solution for $g_{2}$.

Next, we equate the coefficients of $yz^{d-1}$ and $z^{d}$. The first can be
computed from the subsum $\,\alpha_{00d}z^{d}\,+\,\alpha_{01\,d-1}yz^{d-1}$
and equals
$\,\alpha_{00d}\,d\,g_{2}g_{3}^{d-1}\,+\,\alpha_{01\,d-1}\,g_{3}^{d-1}$. The
second is computed from the $z^{d}$ coefficient of $A$ only, and we find it to
be $\alpha_{00d}\cdot g_{3}^{d}$. Setting these two equal and solving for
$g_{3}$, we obtain
$\,g_{3}=\frac{1}{\alpha_{00d}}\,(\alpha_{00d}\,d\,g_{2}+\alpha_{01\,d-1})$.
Inserting our result for $g_{2}$, we obtain a rational function in the
$\alpha_{ijk}$ as solution for $g_{3}$. ∎

######  Example 2.3.

To be explicit, we display the solution in the two cases of primary interest.
For cubics $(d=3)$, the unique point $gA$ in $\,L_{3}\,\cap\,\mathcal{G}A\,$
is given by the group element $g$ with

$g_{0}=1,\,\,g_{1}\,=\,-\frac{\alpha_{102}}{3\alpha_{003}},\,\,g_{2}\,=\,\frac{9\alpha_{003}^{2}\alpha_{210}-3\alpha_{003}\alpha_{102}\alpha_{111}+\alpha_{012}\alpha_{102}^{2}}{3\alpha_{003}(3\alpha_{003}\alpha_{201}-\alpha_{102}^{2})},$
$g_{3}\,\,=\,\,\frac{9\alpha_{003}^{3}\alpha_{210}+3\alpha_{003}\alpha_{012}\alpha_{201}-3\alpha_{003}^{2}\alpha_{102}\alpha_{111}+\alpha_{003}\alpha_{012}\alpha_{102}^{2}-\alpha_{102}^{2}\alpha_{012}}{\alpha_{003}(3\alpha_{003}\alpha_{201}-\alpha_{102}^{2})}.$

For quartics $(d=4)$, the unique point $gA$ in $\,L_{4}\,\cap\,\mathcal{G}A\,$
is given by $g\in\mathcal{G}$, where

$g_{0}=1,\,\,g_{1}\,=\,-\frac{\alpha_{103}}{4\alpha_{004}},\,\,g_{2}\,=\,\frac{64\alpha_{004}^{3}\alpha_{310}-16\alpha_{004}^{2}\alpha_{103}\alpha_{211}+4\alpha_{004}\alpha_{103}^{2}\alpha_{112}-\alpha_{013}\alpha_{103}^{3})}{8\alpha_{004}(8\alpha_{004}^{2}\alpha_{301}-4\alpha_{004}\alpha_{103}\alpha_{202}+\alpha_{103}^{3})},\,\,$

and $g_{3}$ is the quotient of
$64\alpha_{004}^{4}\alpha_{310}+16\alpha_{004}^{2}\alpha_{013}\alpha_{301}-16\alpha_{004}^{3}\alpha_{103}\alpha_{211}-8\alpha_{004}\alpha_{013}\alpha_{103}\alpha_{202}$
$+\,4\alpha_{004}^{2}\alpha_{103}^{2}\alpha_{112}+2\alpha_{103}^{3}\alpha_{013}-\alpha_{004}\alpha_{013}\alpha_{103}^{3}$
and
$2\alpha_{004}\cdot(8\alpha_{004}^{2}\alpha_{301}-4\alpha_{004}\alpha_{103}\alpha_{202}+\alpha_{103}^{3})$.

One can derive similar formulas for the transformation to normal form when
$d\geq 5$. The denominator in the expressions for $g$ is the polynomial of
degree $d$ in $\alpha$ shown in (6). $\diamond$

Our task is to solve ${\rm discr}_{z}(\hat{A})=B$, for a fixed binary form
$B$. This equation is understood projectively, meaning that we seek $\hat{A}$
in $\mathbb{P}^{\binom{d+2}{2}-1}$ such that ${\rm discr}_{z}(\hat{A})$
vanishes at all zeros of $B$ in $\mathbb{P}^{1}$. By Theorem 2.1, we may
assume that $\hat{A}$ lies in the subspace $L_{d}$. Our system has extraneous
solutions, namely ternary forms $\hat{A}$ whose discriminant vanishes
identically. They must be removed when solving our recovery problem. We now
identify them geometrically.

######  Proposition 2.4.

The base locus of the discriminant map (3) has two irreducible components.
These have codimension $3$ and $2d-1$ respectively in
$\,\mathbb{P}^{\binom{d+2}{2}-1}$. The former consists of all curves that are
singular at $\,p=(0:0:1)$, and the latter is the locus of non-reduced curves.

###### Proof.

The binary form ${\rm discr}_{z}(A)$ vanishes identically if and only if the
univariate polynomial function $z\mapsto A(u,v,z)$ has a double zero $\hat{z}$
for all $u,v\in\mathbb{C}$. If $p$ is a singular point of the curve $V(A)$
then $\hat{z}=0$ is always such a double zero. If $A$ has a factor of
multiplicity $\geq 2$ then so does the univariate polynomial $z\mapsto
A(u,v,z)$, and the discriminant vanishes. Up to closure, we may assume that
this factor is a linear form, so there are $\binom{d}{2}-1+2$ degrees of
freedom. This shows that the family of nonreduced curves $A$ has codimension
$2d-1=(\binom{d+2}{2}-1)-(\binom{d}{2}+1)$. The two scenarios define two
distinct irreducible subvarieties of $\mathbb{P}^{\binom{d+2}{2}-1}$. For $A$
outside their union, the binary form ${\rm discr}_{z}(A)$ is not identically
zero. ∎

We now present our solution to the recovery problem for cubic curves. Let $B$
be a binary sextic with six distinct zeros in $\mathbb{P}^{1}$. We are looking
for a ternary cubic in the normal form

$A\,\,=\,\,\alpha_{300}x^{3}+\alpha_{201}x^{2}z+\alpha_{111}xyz+\alpha_{102}xz^{2}+\alpha_{030}y^{3}+\alpha_{021}y^{2}z+yz^{2}+z^{3}.$

Here we assume $p=(0:0:1)\not\in V(A)$, so that $\alpha_{012}=\alpha_{003}=1$.
We saw this in Theorem 2.1. The remaining six coefficients $\alpha_{ijk}$ are
unknowns. The discriminant has degree three in these:

$\\!{\rm
discr}_{z}(A)\\!=\\!(4\alpha_{201}^{3}+27\alpha_{300}^{2})x^{6}+(12\alpha_{111}\alpha_{201}^{2}-18\alpha_{201}\alpha_{300})x^{5}y+\cdots+(4\alpha_{021}^{3}-\alpha_{021}^{2}-\cdots+4\alpha_{030})y^{6}.$

This expression is supposed to vanish at each of the six zeros of $B$. This
gives a system of six inhomogeneous cubic equations in the six unknowns
$\alpha_{ijk}$. In order to remove the extraneous solutions described in
Proposition 2.4, we further require that the leading coefficient of the
discriminant is nonzero. We can write our system of cubic constraints in the
$\alpha_{ijk}$ as follows:

$\begin{matrix}\quad{\rm
rank}\begin{bmatrix}4\alpha_{201}^{3}{+}27\alpha_{300}^{2}&12\alpha_{111}\alpha_{201}^{2}{-}18\alpha_{201}\alpha_{300}&\cdots&4\alpha_{021}^{3}{-}\alpha_{021}^{2}-\cdots+4\alpha_{030}\\\
\beta_{60}&\beta_{51}&\cdots&\beta_{06}\end{bmatrix}\,\leq\,1\vskip 3.0pt plus
1.0pt minus 1.0pt\\\ {\rm and}\quad
4\alpha_{201}^{3}+27\alpha_{300}^{2}\not=0.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{matrix}$
(7)

This polynomial system exactly encodes the recovery of plane cubics from six
branch points.

######  Corollary 2.5.

For general $\beta_{ij}$, the system (7) has $\mathfrak{h}_{3}=40$ distinct
solutions $\alpha\in\mathbb{C}^{6}$.

###### Proof.

The study of cubic curves tangent to a pencil of six lines goes back to Cayley
[6]. The formula $\mathfrak{h}_{3}=40$ was found by Clebsch [9, 10]. We shall
discuss his remarkable work in Section 4. A modern proof for
$\mathfrak{h}_{3}=40$ was given by Kleiman and Speiser in [21, Corollary 8.5].

We here present the argument given in Ongaro’s thesis [23]. By [23,
Proposition 5.5.2], every covering of $\mathbb{P}^{1}$ by a plane cubic curve
is a shift in the group law of that elliptic curve followed by a linear
projection from a point in $\mathbb{P}^{2}$. This implies that the classical
Hurwitz number, which counts such coverings, coincides with the plane Hurwitz
number $\mathfrak{h}_{3}$. The former is the number of six-tuples
$\tau=(\tau_{1},\tau_{2},\tau_{3},\tau_{4},\tau_{5},\tau_{6})$ of permutations
of $\\{1,2,3\\}$, not all equal, whose product is the identity, up to
conjugation. We can choose $\tau_{1},\ldots,\tau_{5}$ in $3^{5}=243$ distinct
ways. Three of these are disallowed, so there are $240$ choices. The symmetric
group $\mathbb{S}_{3}$ acts by conjugation on the tuples $\tau$, and all
orbits have size six. The number of classes of allowed six-tuples is thus
$240/6=40$. This is our Hurwitz number $\mathfrak{h}_{3}$. Now, the assertion
follows from Theorem 2.1, which ensures that the solutions of (7) are
representatives. ∎

We next turn to another normal form, shown in (8), which has desirable
geometric properties. Let $A$ be a ternary form (1) with $a_{00\,d}\not=0$. We
define a group element $g\in\mathcal{G}$ by

$g_{0}=1\,,\,\,g_{1}=-\frac{a_{10\,d-1}}{d\cdot
a_{00d}}\,,\,\,g_{2}=-\frac{a_{01\,d-1}}{d\cdot a_{00d}}\,,\,\,g_{3}=1.$

The coefficients of $xz^{d-1}$ and $yz^{d-1}$ in $gA$ are zero. Thus, after
this transformation, we have

$A\,\,=\,\,z^{d}\,+\,A_{2}(x,y)\cdot z^{d-2}\,+\,A_{3}(x,y)\cdot
z^{d-3}\,+\,\cdots\,+\,A_{d-1}(x,y)\cdot z\,+\,A_{d}(x,y).$ (8)

Here $A_{i}(x,y)$ is an arbitrary binary form of degree $i$. Its $i+1$
coefficients are unknowns. The group $\mathcal{G}$ still acts by rescaling
$x,y$ simultaneously with arbitrary non-zero scalars
$\lambda\in\mathbb{C}^{*}$.

We next illustrate the utility of (8) by computing the planar Hurwitz number
for $d{=}4$. Consider a general ternary quartic $A$. We record its $12$ branch
points by fixing the discriminant $B={\rm discr}_{z}(A)$. Let $\hat{A}\in
L_{4}$ be an unknown quartic in the normal form specified in Theorem 2.1, so
$\hat{A}$ has $13$ terms, $11$ of the form $\alpha_{ijk}x^{i}y^{j}z^{k}$ plus
$yz^{3}$ and $z^{4}$. Our task is to solve the following system of $12$
polynomial equations of degree five in the $11$ unknowns $\alpha_{ijk}$:

Find all quartics $\hat{A}$ such that ${\rm discr}_{z}(\hat{A})$ is a non-zero
multiple of the binary form $B$.  (9)

The number of solutions of this system was found by Vakil [29] with geometric
methods.

######  Theorem 2.6.

Let $B=\sum_{i+j=12}\beta_{ij}x^{i}y^{j}$ be the discriminant with respect to
$z$ of a general ternary quartic $A$. Then the polynomial system (9) has
$\mathfrak{h}_{4}=120$ distinct solutions $\alpha\in\mathbb{C}^{11}$.

The hypothesis ensures that $B$ is a point on Vakil’s degree $3762$
hypersurface $\mathcal{V}_{4}$ in $\mathbb{P}^{12}$. This is a necessary and
sufficient condition for the system (9) to have any solution at all.

######  Corollary 2.7.

If we prescribe $11$ general branch points on the line $\mathbb{P}^{1}$ then
the number of complex quartics $A$ such that ${\rm discr}_{z}(A)$ vanishes at
these points is equal to $120\cdot 3762=451440$.

###### Proof.

Consider the space $\mathbb{P}^{12}$ of binary forms of degree $12$. Vanishing
at $11$ general points defines a line in $\mathbb{P}^{12}$. That line meets
the hypersurface $\mathcal{V}_{4}$ in $3762$ points. By Theorem 2.6, each of
these points in $\mathcal{V}_{4}\subset\mathbb{P}^{12}$ has precisely $120$
preimages $A$ in $\mathbb{P}^{14}$ under the map (3). ∎

######  Remark 2.8.

It was claimed in [23, equation (5.14)] and [25, page 608] that
$\mathfrak{h}_{3}$ is equal to $120\cdot(3^{10}-1)/2=3542880$. That claim is
not correct. The factor $(3^{10}-1)/2$ is not needed.

###### Proof of Theorem 2.6.

We work with the normal form (8). Up to the $\mathcal{G}$-action, the triples
$(A_{2},A_{3},A_{4})$ are parametrized by the $11$-dimensional weighted
projective space $\mathbb{P}(2^{3},3^{4},4^{5})$. Following Vakil [29], we
consider a second weighted projective space of dimension $11$, namely
$\,\mathbb{P}(3^{5},2^{7})$. The weighted projective space
$\mathbb{P}(3^{5},2^{7})$ parametrizes pairs $(U_{2},U_{3})$ where
$U_{i}=U_{i}(x,y)$ is a binary form of degree $2i$, up to a common rescaling
of $x,y$ by some $\lambda\in\mathbb{C}^{*}$.

We define a rational map between our two weighted projective spaces as
follows:

$\begin{matrix}\nu\,:\,\mathbb{P}(2^{3},3^{4},4^{5})\,\dashrightarrow\,\mathbb{P}(3^{5},2^{7})\,,\,\,(A_{2},A_{3},A_{4})\,\mapsto\,(U_{2},U_{3}),\qquad\qquad\vskip
3.0pt plus 1.0pt minus 1.0pt\\\ \qquad{\rm where}\quad
U_{2}\,=\,-4A_{4}-\frac{1}{3}A_{2}^{2}\quad{\rm and}\quad
U_{3}\,=\,A_{3}^{2}-\frac{8}{3}A_{2}A_{4}+\frac{2}{27}A_{2}^{3}.\end{matrix}$
(10)

We compose this with the following map into the space $\mathbb{P}^{12}$ of
binary forms of degree $12$:

$\mu\,:\,\mathbb{P}(3^{5},2^{7})\,\dashrightarrow\,\mathbb{P}^{12}\,,\,\,(U_{2},U_{3})\,\mapsto\,4\cdot
U_{2}^{3}+27\cdot U_{3}^{2}.$ (11)

The raison d’être for the maps (10) and (11) is that they represent the
formula of the discriminant ${\rm discr}_{z}(A)$ of the special quartic in
(8). Thus, modulo the action of $\mathcal{G}$, we have

$\pi\,\,=\,\,\mu\,\circ\,\nu,$

where $\pi:\mathbb{P}^{14}\rightarrow\mathbb{P}^{12}$ is the branch locus map
in (3). One checks this by a direct computation.

Vakil proves in [29, Proposition 3.1] that the map $\nu$ is dominant and its
degree equals $120$. We also verified this statement independently via a
numerical calculation in affine coordinates using HomotopyContinuation.jl [2],
and we certified its correctness using the method in [1]. This implies that
the image of the map $\mu$ equals the hypersurface $\mathcal{V}_{4}$. In
particular, $\mathcal{V}_{4}$ is the locus of all binary forms of degree $12$
that are sums of the cube of a quartic and the square of a sextic. Vakil
proves in [29, Theorem 6.1] that the map $\mu$ is birational onto its image
$\mathcal{V}_{4}$. We verified this statement by a Gröbner basis calculation.
This result implies that both $\nu$ and $\pi$ are maps of degree $120$, as
desired. ∎

######  Remark 2.9.

We also verified that $\mathcal{V}_{4}$ has degree $3762$, namely by solving
$12$ random affine-linear equations on the parametrization (11). The common
Newton polytope of the resulting polynomials has normalized volume $31104$.
This is the number of paths tracked by the polyhedral homotopy in
HomotopyContinuation.jl. We found $22572=3762\times 6$ complex solutions. The
factor $6$ arises because $U_{2}$ and $U_{3}$ can be multiplied by roots of
unity.

######  Algorithm 2.10.

We implemented a numerical recovery method based on the argument used to prove
Theorem 2.6. The input is a pair $(U_{2},U_{3})$ as above. The output consists
of the $120$ solutions in the subspace $L_{4}\simeq\mathbb{P}^{11}$ seen in
(5). We find these by solving the equations

$A_{1}A_{3}-4A_{0}A_{4}-\frac{1}{3}A_{2}^{2}\,=\,U_{2}\quad{\rm and}\quad
A_{1}^{2}A_{4}+A_{0}A_{3}^{2}-\frac{8}{3}A_{0}A_{2}A_{4}-\frac{1}{3}A_{1}A_{2}A_{3}+\frac{2}{27}A_{2}^{3}\,=\,U_{3}.$
(12)

By [29, Equation (5)], these represent the discriminant for quartics
$A=\sum_{i=0}^{4}A_{i}z^{4-i}$. To be precise, (12) is a system of $12=5+7$
equations in the $12$ unknown coefficients of $A\in L_{4}$. These have $120$
complex solutions, found easily with HomotopyContinuation.jl [2].

##  3 Hurwitz Combinatorics

The enumeration of Riemann surfaces satisfying fixed ramification was
initiated by Hurwitz in his 1891 article [19]. Hurwitz numbers are a widely
studied subject, seen as central to combinatorial algebraic geometry. For
basics see [4, 5, 15, 20, 23] and the references therein.

This paper concerns a general projection $V(A)\rightarrow\mathbb{P}^{1}$ of a
smooth plane curve of degree $d$ and genus $g=\binom{d-1}{2}$. In Section 2 we
studied the inverse problem of recovering $A$ from the $d(d-1)$ simple branch
points. We now relate the plane Hurwitz numbers $\mathfrak{h}_{d}$ to the
Hurwitz numbers $H_{d}$ that count abstract covers. To be precise, $H_{d}$ is
the number of degree $d$ covers $f$ of $\mathbb{P}^{1}$ by a genus
$\binom{d-1}{2}$ curve $C$ having $d(d-1)$ fixed simple branch points. Each
cover $f:C\rightarrow\mathbb{P}^{1}$ is weighted by
$\frac{1}{|\mathrm{Aut}(f)|}$. Following [5], the number $H_{d}$ can be found
by counting monodromy representations, i.e. homomorphisms from the fundamental
group of the target minus the branch points to the symmetric group over the
fiber of the base point.

######  Lemma 3.1 (Hurwitz [19]).

The Hurwitz number $H_{d}$ equals $1/d!$ times the number of tuples of
transpositions $\tau=(\tau_{1},\tau_{2},\ldots,\tau_{d\cdot(d-1)})$ in the
symmetric group $\mathbb{S}_{d}$ satisfying

$\tau_{d\cdot(d-1)}\circ\dots\circ\tau_{2}\circ\tau_{1}=\mathrm{id},$

where the subgroup generated by the $\tau_{i}$ acts transitively on the set
$\\{1,2,\dots,d\\}$.

######  Proposition 3.2.

For $d\geq 3$, the plane Hurwitz number is less than or equal to the classical
Hurwitz number that counts abstract covers. In symbols, we have
$\,\mathfrak{h}_{d}\,\leq\,H_{d}$.

The restriction $d\geq 3$ is needed because of the weighted count, with
automorphisms. For $d=2$, we have $H_{2}=1/2$ because of the existence of a
non-trivial automorphism for maps $\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$.
For higher $d$, the covers coming from projections of plane curves do not have
automorphisms, so we can count them without this weight. This establishes
Proposition 3.2.

The two cases of primary interest in this paper are $d=3$ and $d=4$. From the
proofs of Corollary 2.5 and Theorem 2.6, we infer that the two cases exhibit
rather different behaviors.

######  Corollary 3.3.

For linear projections of cubic curves and quartic curves in $\mathbb{P}^{2}$,
we have

$\qquad\qquad\mathfrak{h}_{3}\,=\,H_{3}\,=\,40\qquad{\rm
and}\qquad\mathfrak{h}_{4}\,=\,120\,\,<\,\,H_{4}\,=\,7528620.$

The count in Lemma 3.1 can be realized by combinatorial objects known as
monodromy graphs. These occur in different guises in the literature. We here
use the version that is defined formally in [15, Definition 3.1]. These
represent abstract covers in the tropical setting of balanced metric graphs.
We next list all monodromy graphs for $d=3$.

######  Example 3.4 (Forty monodromy graphs).

For $d=3$, Lemma 3.1 yields $H_{3}=40$ six-tuples
$\tau=(\tau_{1},\tau_{2},\ldots,\tau_{6})$ of permutations of $\\{1,2,3\\}$,
up to the conjugation action by $\mathbb{S}_{3}$. In Table
LABEL:table:40covers we list representatives for these $40$ orbits (see also
[24, Table 1]). Each tuple $\tau$ determines a monodromy graph as in [4, Lemma
4.2] and [15, Section 3.3]. Reading from the left to right, the diagram
represents the cycle decompositions of the permutations
$\tau_{i}\circ\cdots\circ\tau_{1}$ for $i=1,\ldots,6$. For instance, for the
first type $\mathcal{A}_{1}$, we start at ${\rm id}=(1)(2)(3)$, then pass to
$(12)(3)$, next to $(123)$, then to $(12)(3)$, etc. On the right end, we are
back at ${\rm id}=(1)(2)(3)$.

Table 1: The monodromy graphs for the $H_{3}=40$ coverings of $\mathbb{P}^{1}$ by a genus one curve. Eight of the $40$ coverings are real, and the certifying edge coloring is shown in the graph. The two rightmost columns, labeled Clebsch and $\,\mathbb{P}^{3}(\mathbb{F}_{3})$, will be explained in Section 4.  Type | Real? |  Six-Tuple $\tau$ | Monodromy Graph | Clebsch | $\\!\mathbb{P}^{3}(\mathbb{F}_{3})\\!$
---|---|---|---|---|---
$\mathcal{A}_{1}$ $\mathcal{A}_{2}$ |  ✓$(12)$ ✓$(12)$ |  $(12)(13)(13)(13)(13)(12)$ $(12)(13)(13)(23)(23)(12)$ |  |  $123$ $1a$ |  $0010$ $0100$
$\mathcal{A}_{3}$ $\mathcal{A}_{4}$ $\mathcal{A}_{11}$ $\mathcal{A}_{12}$ |  ✗ ✗ ✗ ✗ |  $(12)(13)(13)(13)(23)(13)$ $(12)(13)(13)(13)(12)(23)$ $(12)(13)(13)(23)(12)(13)$ $(12)(13)(13)(23)(13)(23)$ |  |  $348$ $357$ $7b$ $4c$ |  $1022$ $1012$ $1102$ $1201$
$\mathcal{A}_{5}$ $\mathcal{A}_{6}$ $\mathcal{A}_{7}$ $\mathcal{A}_{13}$ $\mathcal{A}_{14}$ $\mathcal{A}_{15}$ |  ✗ ✗ ✗ ✗ ✗ ✗ |  $(12)(13)(23)(23)(13)(12)$ $(12)(13)(23)(23)(23)(13)$ $(12)(13)(23)(23)(12)(23)$ $(12)(13)(23)(12)(23)(12)$ $(12)(13)(23)(12)(12)(13)$ $(12)(13)(23)(12)(13)(23)$ |  |  $456$ $267$ $168$ $1b$ $7c$ $4a$ |  $1020$ $1011$ $0012$ $1100$ $1201$ $0101$
$\mathcal{A}_{8}$ $\mathcal{A}_{9}$ $\mathcal{A}_{10}$ $\mathcal{A}_{16}$ $\mathcal{A}_{17}$ $\mathcal{A}_{18}$ |  ✗ ✗ ✗ ✗ ✗ ✗ |  $(12)(13)(12)(12)(13)(12)$ $(12)(13)(12)(12)(23)(13)$ $(12)(13)(12)(12)(12)(23)$ $(12)(13)(12)(13)(23)(12)$ $(12)(13)(12)(13)(12)(13)$ $(12)(13)(12)(13)(13)(23)$ |  |  $789$ $159$ $249$ $1c$ $7a$ $4b$ |  $1010$ $0010$ $1021$ $1200$ $0102$ $1101$
$\mathcal{B}_{1}$ $\mathcal{B}_{2}$ |  ✓(id) ✓(id) |  $(12)(12)(13)(13)(12)(12)$ $(12)(12)(13)(13)(23)(23)$ |  |  base $147$ |  $1000$ $0001$
$\mathcal{C}^{\ell}_{1}$ $\mathcal{C}^{\ell}_{2}$ $\mathcal{C}^{\ell}_{3}$ |  ✓$(12)$ ✗ ✗ |  $(12)(12)(12)(13)(13)(12)$ $(12)(12)(12)(13)(23)(13)$ $(12)(12)(12)(13)(12)(23)$ |  |  $2a$ $8b$ $5c$ |  $0110$ $1112$ $1222$
$\mathcal{C}^{r}_{1}$ $\mathcal{C}^{r}_{2}$ $\mathcal{C}^{r}_{3}$ |  ✓$(12)$ ✗ ✗ |  $(12)(13)(13)(12)(12)(12)$ $(12)(13)(23)(13)(13)(13)$ $(12)(13)(12)(23)(23)(23)$ |  |  $3a$ $6b$ $9c$ |  $0120$ $1121$ $1211$
$\mathcal{D}^{\ell}_{1}$ |  ✓(id) | $(12)(12)(12)(12)(13)(13)$ |  | $369$ | $1002$
$\mathcal{D}^{r}_{1}$ |  ✓(id) | $(12)(12)(13)(13)(13)(13)$ |  | $258$ | $1001$
$\mathcal{E}^{\ell}_{1}$ $\mathcal{E}^{\ell}_{3}$ $\mathcal{E}^{\ell}_{5}$ |  ✗ ✗ ✗ |  $(12)(12)(13)(23)(13)(12)$ $(12)(12)(13)(23)(23)(13)$ $(12)(12)(13)(23)(12)(23)$ |  |  $2b$ $8c$ $5a$ |  $1110$ $1221$ $0111$
$\mathcal{E}^{\ell}_{2}$ $\mathcal{E}^{\ell}_{4}$ $\mathcal{E}^{\ell}_{6}$ |  ✗ ✗ ✗ |  $(12)(12)(13)(12)(23)(12)$ $(12)(12)(13)(12)(12)(13)$ $(12)(12)(13)(12)(13)(23)$ |  |  $2c$ $5b$ $8a$ |  $1220$ $1111$ $0112$
$\mathcal{E}^{r}_{1}$ $\mathcal{E}^{r}_{3}$ $\mathcal{E}^{r}_{5}$ |  ✗ ✗ ✗ |  $(12)(13)(23)(13)(12)(12)$ $(12)(13)(13)(12)(13)(13)$ $(12)(13)(13)(12)(23)(23)$ |  |  $3c$ $6c$ $9b$ |  $1210$ $1212$ $1122$
$\mathcal{E}^{r}_{2}$ $\mathcal{E}^{r}_{4}$ $\mathcal{E}^{r}_{6}$ |  ✗ ✗ ✗ |  $(12)(13)(12)(23)(12)(12)$ $(12)(13)(12)(23)(13)(13)$ $(12)(13)(23)(13)(23)(23)$ |  |  $3b$ $6a$ $9a$ |  $1120$ $0121$ $0122$

To identify real monodromy representations (see Lemma 3.5), we give a coloring
as in [15, Definition 3.5]. Using [15, Lemma 3.5] we find eight real covers
among the $40$ complex covers. We use [15, Lemma 2.3] to associate the real
covers to their monodromy representations.

We divide the $40$ classes into five types, $\mathcal{A}$ to $\mathcal{E}$,
depending on the combinatorial type of the graph. Types $\mathcal{A}$ and
$\mathcal{B}$ are symmetric under reflection of the ends, $\mathcal{C}$,
$\mathcal{D}$ and $\mathcal{E}$ are not. An upper index $\ell$ indicates that
the cycle of the graph is on the left side of the graph, while $r$ indicates
that it is on the right side. The number of classes of each type is the
multiplicity in [4, Lemma 4.2] and [24, Table 1]. Each class starts with the
real types, if there are any, and proceeds lexicographically in $\tau$. In the
table, the edges of the monodromy graphs are labeled by the cycle they
represent. If the edge is unlabeled, then the corresponding cycle is either
clear from context or varies through all possible cycles in $\mathbb{S}_{3}$
of appropriate length. $\diamond$

We now turn to branched covers that are real. In the abstract setting of
Hurwitz numbers $H_{d}$, this has been studied in [3, 15, 20]. A cover
$f:C\rightarrow\mathbb{P}^{1}$ is called real if the Riemann surface $C$ has
an involution which is compatible with complex conjugation on the Riemann
sphere $\mathbb{P}^{1}$. The branch points in $\mathbb{P}^{1}$ can be real or
pairs of complex conjugate points. We let $H_{d}(r)$ be the weighted count of
degree $d$ real covers $f$ of $\mathbb{P}^{1}$ by a genus $\binom{d-1}{2}$
curve $C$ having $d(d-1)$ fixed simple branch points, of which $r$ are real.
As before, each cover $f:C\rightarrow\mathbb{P}^{1}$ is weighted by
$\frac{1}{|\mathrm{Aut}(f)|}$. The following result appears in [3, Section
3.3].

######  Lemma 3.5.

The real Hurwitz number $H^{_{d}}(r)$ equals $1/d!$ times the number of tuples
$\tau$ as in Lemma 3.1 for which there exists an involution
$\sigma\in\mathbb{S}_{3}$ such that

$\sigma\circ\tau_{i}\circ\dots\circ\tau_{1}\circ\sigma=(\tau_{1}\circ\dots\circ\tau_{i})^{-1}$

for $i=1,\dots,r-1$ and
$\sigma\circ\tau_{r+i}\circ\sigma=\tau_{r^{\prime}+1-i}$ for
$i=1,\dots,r^{\prime}$, where $r$ is the number of real branch points and
$r^{\prime}$ the number of pairs of complex conjugate branch points.

Geometrically, this means that, for a pair of complex conjugate points
$q_{1},q_{2}$, under complex conjugation the arc $\gamma_{1}$ around $q_{1}$
maps to $-\gamma_{2}$, where $\gamma_{2}$ is the arc around $q_{2}$. Our next
result says that the real Hurwitz number for $d=3$ does not depend on $r$ and
$r^{\prime}=6-2r$.

######  Proposition 3.6.

We have $H_{3}(r)=8$ for $r=6,4,2,0$.

###### Proof.

We prove this by investigating all monodromy representations in Table
LABEL:table:40covers. Using explicit computations, we identify all six-tuples
$\tau$ that satisfy the conditions in Lemma 3.5. For a cover with $6$ real
branch points, we obtain $8$ real monodromy representations, of types
$\mathcal{A}_{1},\mathcal{A}_{2},\mathcal{B}_{1},\mathcal{B}_{2},\mathcal{C}^{l}_{1},\mathcal{C}^{r}_{1},\mathcal{D}^{l}_{1}$
and $\mathcal{D}^{r}_{1}$, listed in Table LABEL:table:40covers with coloring.
For a cover with $4$ real branch points and a pair of complex conjugate branch
points, we again obtain $8$ real monodromy representations. These are the
types
$\mathcal{A}_{3},\mathcal{A}_{12},\mathcal{B}_{1},\mathcal{B}_{2},\mathcal{C}^{l}_{2},\mathcal{C}^{r}_{1},\mathcal{D}^{l}_{1}$
and $\mathcal{D}^{r}_{1}$. For two real branch points and two complex
conjugate pairs, we again obtain $8$ real monodromy representations, namely of
types
$\mathcal{A}_{9},\mathcal{A}_{12},\mathcal{B}_{1},\mathcal{B}_{2},\mathcal{D}^{l}_{1},\mathcal{D}^{r}_{1},\mathcal{E}^{\ell}_{3}$
and $\mathcal{E}^{r}_{1}$. Finally, for three pairs of complex conjugate
branch points, we find the $8$ types
$\mathcal{A}_{5},\mathcal{A}_{17},\mathcal{B}_{1},\mathcal{B}_{2},\mathcal{D}^{l}_{1},\mathcal{D}^{r}_{1},\mathcal{E}^{\ell}_{3}$
and $\mathcal{E}^{r}_{5}$. ∎

The situation is more interesting for $d=4$, where we obtained the following
result:

######  Theorem 3.7.

The real Hurwitz numbers for degree $4$ coverings of $\mathbb{P}^{1}$ by genus
$3$ curves are

$\begin{matrix}H_{4}(12)=20590,&H_{4}(10)=15630,&H_{4}(8)=11110,&H_{4}(6)=7814,\\\
&H_{4}(4)=5654,&H_{4}(2)=4070,\,&H_{4}(0)=4350.\end{matrix}$

###### Proof.

This is found by a direct computation using Oscar [26]. We start by
constructing a list of all monodromy representations of degree $4$ and genus
$3$. As monodromy representations occur in equivalence classes, we construct
only one canonical representative for each class. This is the element of the
equivalence class that is minimal with respect to the lexicographic ordering.
The resulting list of $7528620$ monodromy representations was computed in
about $6.5$ hours. In other words, we embarked on a table just like Table
LABEL:table:40covers, but its number of rows is now $7528620$ instead of $40$.
Those are the two numbers seen in Corollary 3.3.

We next applied Cadoret’s criterion in [3, Section 3.3, formula $(\star)$] to
our big table. This criterion was stated in Lemma 3.5. We start with our
$7528620$ tuples $\tau$, computed as just described, and mentioned in Lemma
3.1. According to Cadoret’s criterion, we must check for each $12$-tuple
$\tau$ whether there exists an involution $\sigma$ that satisfies certain
equations in the symmetric group $\mathbb{S}_{4}$. These depend on the number
$r$ of real branch points. Note that $r=\\{0,2,4,\ldots,12\\}$. For
$r=2,4,\ldots,12$, the only possible involutions $\sigma$ are $id$, $(12)$,
$(34)$ and $(12)(34)$, by the structure of the canonical representative
computed for the list. For $r=0$, all involutions in $\mathbb{S}_{4}$ can
appear. For each involution $\sigma$ and each value of $r$, it took between
$5$ and $30$ minutes to scan our big table, and to determine how many
$12$-tuples $\tau$ satisfy Cadoret’s criterion for the pair $(r,\sigma)$. For
each $r$, we collected the number of tuples $\tau$ for which the answer was
affirmative. This gave the numbers stated in Theorem 3.7. ∎

We next relate this Hurwitz combinatorics to the polynomial systems in Section
2. Recall that we seek orbits of the group $\mathcal{G}$ acting on
$\mathbb{P}^{\binom{d+2}{2}-1}$. An orbit is called real if it has the form
$\mathcal{G}A$ where $A$ is a ternary form with real coefficients. Since
$\mathcal{G}$ is defined over $\mathbb{R}$, an orbit is real if and only if
its unique intersection point with the linear space $L_{d}$ in Theorem 2.1 is
real. Thus, identifying the real orbits among those with prescribed branch
points is equivalent to deciding how many of the $\mathfrak{h}_{d}$ complex
solutions in our exact formulations (7) and (9) are real.

Suppose that the given binary form $B\in\mathcal{V}_{d}$ has real
coefficients, and let $r$ denote the number of real zeros of $B$. In addition,
there are $d(d-1)-2r$ pairs of complex conjugate zeros. It turns out that for
$d=3$ the number of real solutions is independent of the number $r$. Our
census of real plane Hurwitz numbers for quartics will be presented in Section
5.

######  Corollary 3.8.

The real plane Hurwitz number for cubics equals eight. To be precise, the
system (7) always has $8$ real solutions, provided the given parameters
$\beta_{ij}$ are real and generic.

###### Proof.

This is derived from Corollary 3.3 and Proposition 3.6. Namely, we use the
fact that plane covers are in bijection with abstract covers. Let
$C\rightarrow\mathbb{P}^{1}$ be a real cover by an elliptic curve $C$. The
involution of $C$ that is referred to in the proof of Corollary 2.5 is real as
well. Another proof, following Clebsch [9, 10], appears in Section 4. ∎

######  Algorithm 3.9.

We implemented numerical recovery for cubics that matches Table
LABEL:table:40covers. The input is a binary sextic $B$ with real coefficients.
The output consists of $40$ cubics $A$ in $L_{3}$ along with their labeling by
$\mathcal{A}_{1},\mathcal{A}_{2},\ldots,\mathcal{E}_{6}^{r}$. The cubics are
found with HomotopyContinuation.jl by solving (7). We next fix loops
$\gamma_{1},\gamma_{2},\ldots,\gamma_{6}$ around the six roots of $B$ that are
compatible with complex conjugation on the Riemann sphere $\mathbb{P}^{1}$. If
all six roots are real then we use [15, Construction 2.4]. For each cubic $A$,
we track the three roots $z$ of $A(x,y,z)=0$ as $(x:y)$ cycles along
$\gamma_{i}$. The resulting permutation of the three roots is the
transposition $\tau_{i}$. This process maps $A$ to a tuple $\tau$ in Table
LABEL:table:40covers. This is unique up to conjugacy by $\mathbb{S}_{3}$. The
$8$ real cubics $A$ are mapped to the $8$ real monodromy representations, in
the proof of Proposition 3.6.

##  4 Cubics: Solutions in Radicals

The theme of this paper is the rational map (3) that takes a ternary form to
its $z$-discriminant. This map is finite-to-one onto its image
$\mathcal{V}_{d}$, which is an irreducible variety of dimension
$\binom{d+2}{2}-4$ in $\mathbb{P}^{d(d-1)}$. The general fiber consists of
$\mathfrak{h}_{d}$ complex points. We are curious about the Galois group ${\rm
Gal}_{d}$ associated with this covering. Here Galois group is defined as in
[17]. Informally, ${\rm Gal}_{d}$ is the subgroup of geometry-preserving
permutations of the $\mathfrak{h}_{d}$ solutions.

######  Theorem 4.1.

The Galois group ${\rm Gal}_{3}$ for cubics is the simple group of order
$25920$, namely

${\rm Gal}_{3}\,\,=\,\,{\rm SU}_{4}(\mathbb{F}_{2})\,\,=\,\,{\rm
PSp}_{4}(\mathbb{F}_{3})\,\,=\,\,W(E_{6})/\\!\pm.$ (13)

This is the Weyl group of type $E_{6}$ modulo its center, here realized as
$4\times 4$ matrix groups over the finite fields $\mathbb{F}_{2}$ and
$\mathbb{F}_{3}$. The action of ${\rm Gal}_{3}$ on the $40$ monodromy graphs
in Table LABEL:table:40covers agrees with that of the symplectic group on the
$40$ points in the projective space $\mathbb{P}^{3}$ over $\mathbb{F}_{3}$.

This is a modern interpretation of Clebsch’s work [9, 10] on binary sextics.
We first learned about the role of the Weyl group in (13) through Elkies’
unpublished manuscript [11].

######  Remark 4.2.

The last two columns of Table LABEL:table:40covers identify the $40$ monodromy
representations with $\mathbb{P}^{3}(\mathbb{F}_{3})$ and with Clebsch’s $40$
cubics in [10]. The bijection we give respects the maps ${\rm
Gal}_{3}\simeq{\rm PSp}_{4}(\mathbb{F}_{3})\hookrightarrow\mathbb{S}_{40}$.
But it is far from unique. The same holds for Cayley’s table in [7].

###### Proof of Theorem 4.1.

We consider cubics $\,A=z^{3}+A_{2}(x,y)z+A_{3}(x,y)$. This is the normal form
in (8). The discriminant equals $\,{\rm discr}_{z}(A)=4A_{2}^{3}+27A_{3}^{2}$.
Thus our task is as follows:
Given a binary sextic $B$, compute all pairs of binary forms $(A_{2},A_{3})$
such that $4A_{2}^{3}+27A_{3}^{2}=B$.

This system has $40$ solutions, modulo the scaling of $A_{2}$ and $A_{3}$ by
roots of unity. Setting $U=\sqrt[3]{4}\cdot A_{2}$ and $V=\sqrt{-27}\cdot
A_{3}$, we must solve the following problem: Given a binary sextic $B$,
compute all decompositions into a binary quadric $U$ and a binary cubic $V$:

$B\,\,=\,\,U^{3}-V^{2}.$ (14)

This is precisely the problem addressed by Clebsch in [9, 10]. By considering
the change of his labeling upon altering the base solution, he implicitly
determined the Galois group as a subgroup of $\mathbb{S}_{40}$. The
identification of this group with $W(E_{6})$ modulo its center appears in a
number of sources, including [18, 28]. These sources show that ${\rm Gal}_{3}$
is also the Galois group of the $27$ lines on the cubic surface. Todd [28]
refers to permutations of the $40$ Jacobian planes, and Hunt [18, Table 4.1]
points to the $40$ triples of trihedral pairs. The connection to cubic
surfaces goes back to Jordan in 1870, and it was known to Clebsch.

As a subgroup of the symmetric group $\mathbb{S}_{40}$, our Galois group is
generated by five permutations $\Gamma_{1},\ldots,\Gamma_{5}$. These
correspond to consecutive transpositions $(\gamma_{i}\gamma_{i+1})$ of the six
loops $\gamma_{1},\gamma_{2},\ldots,\gamma_{6}$ in Algorithm 3.9. Each
generator is a product of nine $3$-cycles in $\mathbb{S}_{40}$.

Here are the formulas for $\Gamma_{1},\ldots,\Gamma_{5}$ as permutations of
the $40$ rows in Table LABEL:table:40covers:

$\\!\Gamma_{1}=(\mathcal{A}_{10}\mathcal{A}_{6}\mathcal{A}_{1}\\!)(\mathcal{A}_{8}\mathcal{A}_{7}\mathcal{A}_{3}\\!)(\mathcal{A}_{9}\mathcal{A}_{5}\mathcal{A}_{4}\\!)(\mathcal{A}_{17}\mathcal{A}_{13}\mathcal{A}_{12}\\!)(\mathcal{A}_{18}\mathcal{A}_{14}\mathcal{A}_{2}\\!)(\mathcal{A}_{16}\mathcal{A}_{15}\mathcal{A}_{11}\\!)(\mathcal{E}^{r}_{2}\mathcal{E}^{r}_{6}\mathcal{E}^{r}_{3})(\mathcal{E}^{r}_{4}\mathcal{E}^{r}_{1}\mathcal{E}^{r}_{5})(\mathcal{C}^{r}_{3}\mathcal{C}^{r}_{2}\mathcal{C}^{r}_{1})$
$\Gamma_{2}\,=(\mathcal{E}^{\ell}_{4}\mathcal{A}_{14}\mathcal{A}_{10})(\mathcal{E}^{\ell}_{6}\mathcal{A}_{15}\mathcal{A}_{9})(\mathcal{E}^{\ell}_{2}\mathcal{A}_{13}\mathcal{A}_{8})(\mathcal{B}_{1}\,\mathcal{E}^{r}_{1}\mathcal{E}^{r}_{2})(\mathcal{D}^{r}_{1}\,\mathcal{C}^{r}_{2}\mathcal{C}^{r}_{3})(\mathcal{B}_{2}\,\mathcal{E}^{r}_{6}\mathcal{E}^{r}_{4})(\mathcal{E}^{\ell}_{5}\mathcal{A}_{7}\mathcal{A}_{17})(\mathcal{E}^{\ell}_{1}\mathcal{A}_{5}\mathcal{A}_{16})(\mathcal{E}^{\ell}_{3}\mathcal{A}_{6}\mathcal{A}_{18})$
$\Gamma_{3}\,=\,(\mathcal{C}^{\ell}_{3}\,\mathcal{E}^{\ell}_{5}\,\mathcal{E}^{\ell}_{4})(\mathcal{C}^{\ell}_{1}\,\mathcal{E}^{\ell}_{1}\,\mathcal{E}^{\ell}_{2})(\mathcal{C}^{\ell}_{2}\,\mathcal{E}^{\ell}_{3}\,\mathcal{E}^{\ell}_{6})(\mathcal{A}_{17}\mathcal{A}_{11}\mathcal{A}_{14})(\mathcal{A}_{18}\mathcal{A}_{12}\mathcal{A}_{15})(\mathcal{A}_{16}\mathcal{A}_{2}\mathcal{A}_{13})(\mathcal{E}^{r}_{2}\mathcal{E}^{r}_{1}\mathcal{C}^{r}_{1})(\mathcal{E}^{r}_{4}\mathcal{C}^{r}_{2}\mathcal{E}^{r}_{3})(\mathcal{C}^{r}_{3}\mathcal{E}^{r}_{6}\mathcal{E}^{r}_{5})$
$\Gamma_{4}=(\mathcal{D}^{\ell}_{1}\,\mathcal{C}^{\ell}_{2}\,\mathcal{C}^{\ell}_{3})(\mathcal{E}^{\ell}_{6}\,\mathcal{B}_{2}\,\mathcal{E}^{\ell}_{5})(\mathcal{E}^{\ell}_{2}\,\mathcal{E}^{\ell}_{1}\,\mathcal{B}_{1})(\mathcal{A}_{8}\mathcal{A}_{16}\mathcal{E}^{r}_{2})(\mathcal{A}_{9}\mathcal{E}^{r}_{4}\mathcal{A}_{17})(\mathcal{E}^{r}_{3}\mathcal{A}_{3}\mathcal{A}_{11})(\mathcal{E}^{r}_{5}\mathcal{A}_{12}\mathcal{A}_{4})(\mathcal{A}_{15}\mathcal{E}^{r}_{6}\mathcal{A}_{7})(\mathcal{A}_{13}\mathcal{A}_{5}\mathcal{E}^{r}_{1})$
$\Gamma_{5}=(\mathcal{C}^{\ell}_{3}\mathcal{C}^{\ell}_{2}\mathcal{C}^{\ell}_{1})(\mathcal{E}^{\ell}_{4}\mathcal{E}^{\ell}_{6}\mathcal{E}^{\ell}_{2})(\mathcal{E}^{\ell}_{5}\mathcal{E}^{\ell}_{3}\mathcal{E}^{\ell}_{1})(\mathcal{A}_{10}\mathcal{A}_{9}\mathcal{A}_{8}\\!)(\mathcal{A}_{17}\mathcal{A}_{18}\mathcal{A}_{16}\\!)(\mathcal{A}_{4}\mathcal{A}_{3}\mathcal{A}_{1}\\!)(\mathcal{A}_{11}\mathcal{A}_{12}\mathcal{A}_{2}\\!)(\mathcal{A}_{14}\mathcal{A}_{15}\mathcal{A}_{13}\\!)(\mathcal{A}_{7}\mathcal{A}_{6}\mathcal{A}_{5})$

A compatible bijection with the labels of Clebsch [10, Section 9] is given in
the second-to-last column in Table LABEL:table:40covers. The last column
indices a bijection with the $40$ points in the projective space
$\mathbb{P}^{3}$ over the three-element field $\mathbb{F}_{3}$. This bijection
is compatible with the action of the matrix group ${\rm
PSp}_{4}(\mathbb{F}_{3})$. Here, the five generators above are mapped to
matrices of order $3$:

$\Gamma_{1}=\begin{bmatrix}1\\!&\\!1\\!&\\!2\\!&\\!0\\\
0\\!&\\!1\\!&\\!0\\!&\\!0\\\ 0\\!&\\!0\\!&\\!1\\!&\\!0\\\
0\\!&\\!1\\!&\\!2\\!&\\!1\end{bmatrix}\\!,\,\Gamma_{2}=\begin{bmatrix}1\\!&\\!0\\!&\\!0\\!&\\!0\\\
2\\!&\\!1\\!&\\!0\\!&\\!2\\\ 1\\!&\\!0\\!&\\!1\\!&\\!1\\\
0\\!&\\!0\\!&\\!0\\!&\\!1\end{bmatrix}\\!,\,\Gamma_{3}=\begin{bmatrix}1\\!&\\!1\\!&\\!0\\!&\\!0\\\
0\\!&\\!1\\!&\\!0\\!&\\!0\\\ 0\\!&\\!0\\!&\\!1\\!&\\!0\\\
0\\!&\\!0\\!&\\!0\\!&\\!1\end{bmatrix}\\!,\,\Gamma_{4}=\begin{bmatrix}1\\!&\\!0\\!&\\!0\\!&\\!0\\\
2\\!&\\!1\\!&\\!0\\!&\\!1\\\ 2\\!&\\!0\\!&\\!1\\!&\\!1\\\
0\\!&\\!0\\!&\\!0\\!&\\!1\end{bmatrix}\\!,\,\Gamma_{5}=\begin{bmatrix}1\\!&\\!1\\!&\\!1\\!&\\!0\\\
0\\!&\\!1\\!&\\!0\\!&\\!0\\\ 0\\!&\\!0\\!&\\!1\\!&\\!0\\\
0\\!&\\!2\\!&\\!2\\!&\\!1\end{bmatrix}\\!.$

These are symplectic matrices with entries in $\mathbb{F}_{3}$, modulo scaling
by $(\mathbb{F}_{3})^{*}=\\{\pm 1\\}=\\{1,2\\}$. A computation using GAP [13]
verifies that these groups are indeed isomorphic. In the notation of the atlas
of simple groups, our Galois group (13) is the group $O_{5}(3)$. ∎

######  Remark 4.3.

The fundamental group of the configuration space of six points in the Riemann
sphere $\mathbb{P}^{1}$ is the braid group $B_{6}$, which therefore maps onto
the finite group ${\rm Gal}_{3}$. One checks that the permutations and
matrices listed above satisfy the braid relations that define $B_{6}$:

$\Gamma_{i}\,\Gamma_{i+1}\,\Gamma_{i}\,=\,\Gamma_{i+1}\,\Gamma_{i}\,\Gamma_{i+1}\quad\hbox{for
$i=1,2,3,4$}\quad{\rm
and}\quad\Gamma_{i}\,\Gamma_{j}=\Gamma_{j}\,\Gamma_{i}\quad\hbox{for
$|i-j|\geq 2$.}$

We conclude from Theorem 4.1 that our recovery problem for cubics is solvable
in radicals.

######  Corollary 4.4.

Starting from a given ternary cubic $A$, the $39$ other solutions $(U,V)$ to
the equation (14) can be written in radicals in the coefficients of the binary
forms $A_{2}$ and $A_{3}$.

###### Proof.

Viewing ${\rm Gal}_{3}$ as a group of permutations on the $40$ solutions, we
consider the stabilizer subgroup of the one given solution. That stabilizer
has order $25920/40=3\cdot 216$, and this is the Galois group of the other
$39$ solutions. It contains the Hesse group ${\rm ASL}_{2}(\mathbb{F}_{3})$ as
a normal subgroup of index $3$. Recall that ${\rm ASL}_{2}(\mathbb{F}_{3})$ is
solvable, and has order $216$. It is the Galois group of the nine inflection
points of a plane cubic [17, Section II.2]. Therefore the stabilizer is
solvable, and hence $(U,V)$ is expressable in radicals over $(A_{2},A_{3})$. ∎

Clebsch explains how to write the $39$ solutions in radicals in the
coefficients of $(A_{2},A_{3})$. We now give a brief description of his
algorithm, which reveals the inflection points of a cubic.

######  Algorithm 4.5.

Our input is a pair $(A_{2},A_{3})$ that represents a cubic $A$ as in (8). We
set $\tilde{U}=\sqrt[3]{4}\cdot A_{2}$ and $\tilde{V}=\sqrt{-27}\cdot A_{3}$.
Our output is the list of all pairs $(U,V)$ that satisfy

$U^{3}-{\tilde{U}}^{3}\,\,=\,\,V^{2}-{\tilde{V}}^{2}.$ (15)

Real solutions of (7) correspond to pairs $(U,V)$ such that $U$ and $iV$ are
real, where $i=\sqrt{-1}$. To solve (15), consider the cubic
$\mathcal{C}=\\{(x:y:z)\in\mathbb{P}^{2}:z^{3}-3\tilde{U}z+2\tilde{V}=0\\}$.
The nine inflection lines of $\mathcal{C}$ are defined by the linear forms
$\xi=\alpha x+\beta y$ such that $\xi^{3}-3\tilde{U}\xi+2\tilde{V}$ is the
cube of a linear form $\eta=\gamma x+\delta y$. We write the coefficients
$\alpha,\beta$ of such $\xi$ in radicals. Here, the Galois group is ${\rm
ASL}_{2}(\mathbb{F}_{3})$. We next compute $(\gamma,\delta)$ from
$\eta^{3}=\xi^{3}-3\tilde{U}\xi+2\tilde{V}$. For each of the pairs
$(\alpha,\beta)$ above, this system has three solutions
$(\gamma,\delta)\in\mathbb{C}^{2}$. This leads to $27$ vectors
$(\alpha,\beta,\gamma,\delta)$, all expressed in radicals. For each of these
we set $U=\tilde{U}-\xi^{2}-\xi\eta-\eta^{2}$. Then

$U^{3}-({\tilde{U}}^{3}-{\tilde{V}}^{2})\,=\,-\frac{3}{4}\bigl{(}\eta^{3}+2\eta^{2}\xi+2\eta\xi^{2}+\xi^{3}-2\eta\tilde{U}-\xi\tilde{U}\bigr{)}^{2}.$

The square root of the right hand side is a binary cubic $V$, and the pair
$(U,V)$ satisfies (15). In this manner we construct $27$ solutions to (7),
three for each inflection point of the curve $\mathcal{C}$. Three of the $9$
inflection points are real, and each of them yields one real solution to (7).

We finally compute the $12=39-27$ remaining solutions, of which four are real.
For this, we label the inflection points of $\mathcal{C}$ by $1,2,\ldots,9$
such that the following triples are collinear:

$\underline{123},456,789,\quad\underline{147},\underline{258},\underline{369},\quad
159,267,348,\quad 168,249,357.$ (16)

If $1,2,3$ are real and $\\{4,7\\}$, $\\{5,8\\}$, $\\{6,9\\}$ are complex
conjugates, then precisely the four underlined lines are real. Our labeling
agrees with that used by Clebsch in [10, §9, page 50].

We now execute the formulas in [10, §11, §12]. For each of the $12$ lines in
(16), we compute two solutions $(U,V)$ of (15). These are expressed rationally
in the data $(\alpha,\beta)$ found above. Each new solution $(U,V)$ arises
twice from each triple of lines in (16), so we get $12$ in total.

##  5 Quartics: Del Pezzo Strikes Again

This section features threads from algebraic geometry that inform our recovery
problem for quartics. We begin with the extension of Corollary 3.8 from $d=3$
to $d=4$. In other words, we determine the plane counterparts to the real
Hurwitz numbers in Theorem 3.7.

######  Theorem 5.1.

Consider the polynomial system (9) for recovering quartics, where the
parameters $\beta_{ij}$ are generic reals. The number of real solutions equals
$\,8$, $16$, $32$, $64$, or $120$.

###### Proof.

We use the connection to del Pezzo surfaces of degree one and tritangents of
space sextic curves, described by Vakil in [29]. Such a surface is the blow-up
of $\mathbb{P}^{2}$ at a configuration
$\mathcal{P}=\\{P_{1},P_{2},\ldots,P_{8}\\}$ of eight points. The del Pezzo
surface is real if and only if $\mathcal{P}$ is real. In this case,
$\mathcal{P}$ has $2r$ real points and $4-r$ complex conjugate pairs. Fix
linearly independent cubics $u$ and $v$ that vanish at $\mathcal{P}$ and let
$s$ be a sextic such that $\\{u^{2},uv,v^{2},s\\}$ is a basis for the sextics
vanishing doubly at $\mathcal{P}$. Consider the $2$-$1$ map
$\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{3}$ given by $(u^{2}:uv:v^{2}:s)$.
Its image is the quadric cone $V(x_{0}x_{2}-x_{1}^{2})$ in $\mathbb{P}^{3}$,
and its branch locus is a sextic curve $\mathcal{C}$ on that cone. This is a
canonical curve of genus four, and it has $120$ complex tritangent planes.

By [22, §2], one computes the $120=8+28+56+28$ tritangents in rational
arithmetic from $\mathcal{P}$ as follows: we get $8$ from the exceptional
fibers of the blow-up, $28$ from lines spanned by pairs $\\{P_{i},P_{j}\\}$,
we get $56$ from conics through five points in $\mathcal{P}$, and we get
$28=56/2$ from cubics vanishing doubly at $P_{i}$, non-vanishing at $P_{j}$,
and vanishing simply at $\mathcal{P}\backslash\\{P_{i},P_{j}\\}$. Hence, the
number of real tritangents equals $\,8$, $16$, $32$, $64$, or $120$, depending
on whether $r$ equals $0,1,2,3$ or $4$. Notably, if all points $P_{j}$ are
real $(r=4)$ then all tritangents are real.

The link to recovery of quartics is the isomorphism
$\mathcal{E}\simeq\mathcal{B}$ in [29, Theorem 2.12]. Here $\mathcal{B}$ is
the open subset of $\mathbb{P}(3^{5},2^{7})$ of pairs $(U_{2},U_{3})$ as in
(10) such that $4\cdot U_{2}^{3}+27\cdot U_{3}^{2}$ has $12$ distinct roots
and $\mathcal{E}$ is the space of genus $4$ curves whose canonical model lies
on a quadric cone in $\mathbb{P}^{3}$. Given $(U_{2},U_{3})$, the space sextic
$\mathcal{C}$ has the equation $s^{3}+U_{2}(u,v)s+U_{3}(u,v)$ in
$V(x_{0}x_{2}-x_{1}^{2})$. Its $120$ tritangents correspond to the $120$
preimages of $(U_{2},U_{3})$ under the map $\nu$ in (10), that is, to the
quartics $A$ with ${\rm discr}_{z}(A)=4\cdot U_{2}^{3}+27\cdot U_{3}^{2}$.

We claim that a quartic $A$ is real if and only if the corresponding
tritangent is real. One way to see this is via [29, Proposition 4.1], which
asserts that the quartics are obtained from the tritangents via Recillas’
trigonal construction. It is straightforward to check that this construction
preserves the base field. More concretely, this holds because the quartics can
be obtained as the output of Algorithms 5.4 and 5.5, which preserve the base
field as well. ∎

######  Remark 5.2.

Algorithm 2.10 is consistent with Theorem 5.1. Given binary forms
$U_{2},U_{3}$ with real coefficients, it correctly outputs $\,8$, $16$, $32$,
$64$ or $120$ real quartics in the subspace $L_{4}$.

######  Corollary 5.3.

From any general configuration
$\mathcal{P}=\\{P_{1},P_{2},\ldots,P_{8}\\}\subset\mathbb{P}^{2}$ with
coordinates in $\mathbb{Q}$, we obtain an instance of (9) whose $120$ complex
solutions $A$ all have coefficients in $\mathbb{Q}$.

###### Proof.

We construct the $120$ quartics $A$ with rational arithmetic from the
coordinates in $\mathcal{P}$. Fix $j$ and let $w$ be a cubic that vanishes on
$\mathcal{P}\backslash\\{P_{j}\\}$ but does not vanish at $P_{j}$. Consider
the $2$-$1$ map $\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{2}$ given by
$(u:v:w)$. The branch locus of this map is a quartic curve.

We claim that this is the quartic $Q$ that corresponds to the exceptional
fiber of the blow-up at $P_{j}$ in Vakil’s construction. Indeed, let $X$ be
the blowup of $\mathbb{P}^{2}$ at all the $9$ base points of the pencil
$(u:v)$ and let $C\subseteq X$ be a general curve in the pencil. The
intersection of the quartic $Q$ with $C$ is given by [29, Proposition 3.1] as
the set of points $p\in C$ such that
$\mathcal{O}_{C}(2p)\cong\mathcal{O}_{C}(E_{0}+E_{1})$. Alternatively, these
are exactly the branch points of the map $C\to\mathbb{P}^{1}$ induced by the
linear system of $\mathcal{O}_{C}(E_{0}+E_{1})$. Thus, we are done if we can
prove that the restriction of the map $(u:v:w)\colon X\to\mathbb{P}^{2}$ is
given exactly by the linear system above. However, the map of $(u:v:w)$ on $X$
corresponds to the complete linear system $L=3H-E_{2}-\dots-
E_{8}=-K_{X}+E_{0}+E_{1}$ so that
$\mathcal{O}_{C}(L)\cong\mathcal{O}_{C}(-K_{X}+E_{0}+E_{1})$ and we need to
show that $\mathcal{O}_{C}(-K_{X})\cong\mathcal{O}_{C}$. But this follows from
the adjunction formula, using the fact that $C$ is an elliptic curve moving in
a pencil.

Thus, we can recover the quartic as this branch locus of our $2$-$1$ map
$\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{2}$. The ternary form $A$ that
defines this branch locus can be computed from the ideals of $\mathcal{P}$ and
$P_{j}$ using Gröbner bases, and this uses rational arithmetic over the ground
field. This yields eight of the $120$ desired quartics. The remaining $112$
quartics are found by repeating this process after some Cremona
transformations have been applied to the pair $(\mathbb{P}^{2},\mathcal{P})$.
These transformations use only rational arithmetic and they preserve the del
Pezzo surface, but they change the collection of eight $(-1)$-curves that are
being blown down. ∎

The following two algorithms provide more details on the steps in the proof
above.

######  Algorithm 5.4.

The input is a set $\mathcal{P}$ of eight rational points in $\mathbb{P}^{2}$.
The output is eight quartics $A\in\mathbb{Q}[x,y,z]$ that share the same
$z$-discriminant. These are the quartics corresponding to the eight
exceptional divisors $E_{i}$. Write $(r:s:t)$ for the coordinates of the
$\mathbb{P}^{2}$ that contains $\mathcal{P}$. We start by setting up the
following ideal in $\,\mathbb{Q}[r,s,t,x,y,z]$:

$I\,\,\,=\,\,\,\bigl{\langle}\hbox{$2\times 2$ minors
of}\,\begin{pmatrix}u&v&w\\\
x&y&z\end{pmatrix}\bigr{\rangle}\,\,+\,\,\biggl{\langle}{\rm
det}\\!\begin{pmatrix}\partial u/\partial r&\partial u/\partial s&\partial
u/\partial t\\\ \partial v/\partial r&\partial v/\partial s&\partial
v/\partial t\\\ \partial w/\partial r&\partial w/\partial s&\partial
w/\partial t\\\ \end{pmatrix}\biggr{\rangle}.$ (17)

Saturate $I$ with respect to the ideal of $\mathcal{P}\backslash\\{P_{j}\\}$.
After that step, eliminate the three variables $r,s,t$. The result is a
principal ideal in $\mathbb{Q}[x,y,z]$ whose generator is the quartic $A$. We
perform the above two steps for $j=1,2,\ldots,8$, and we output the resulting
eight quartics.

Our second algorithm concerns the Cremona transformations mentioned in the
proof.

######  Algorithm 5.5.

Our input is a set $\mathcal{P}$ of eight rational points in $\mathbb{P}^{2}$.
The output is the remaining set of $112$ quartics $A\in\mathbb{Q}[x,y,z]$ that
also have same $z$-discriminant. Fix a basis of cubics $u,v$ passing through
$\mathcal{P}$. Choose a triple $\\{P_{i},P_{j},P_{k}\\}$ and set up the
associated quadratic Cremona transformation
$\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{2}$. This transforms $\mathcal{P}$
into a new configuration $\mathcal{P}^{\prime}$ and it corresponds to an
automorphism of the del Pezzo surface. Apply Algorithm 5.4 to
$\mathcal{P}^{\prime}$ and record the resulting eight quartics. In applying
Algorithm 5.4, we need some care in choosing a basis
$\\{u^{\prime},v^{\prime}\\}$ for the cubics vanishing at
$\mathcal{P}^{\prime}$: we should choose them in such a way that the elliptic
pencil
$(u^{\prime}:v^{\prime})\colon\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{1}$ is
the composition of the original pencil
$(u:v)\colon\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{1}$ and the quadratic
Cremona transformation. This ensures that the quartics share the same
$z$-discriminant with the original eight quartics. We now repeat this process
until $120$ inequivalent quartics have been found. This is possible because
the Cremona transformations act transitively on the set of $(-1)$ curves on
the del Pezzo surface.

Using Algorithm 5.4, we can find one quartic $A$ and hence the binary form
$B={\rm discr}_{z}(A)$ using rational arithmetic. However, there is an even
more direct method which also shows that $B$ does not depend on the choice of
the point $P_{j}$. Namely, $B$ encodes the $12$ singular curves in the pencil
of cubics through $\mathcal{P}$. Based on this observation, we derive our
final result:

######  Theorem 5.6.

Let $B$ be the binary form of degree $12$ computed from a real configuration
$\mathcal{P}$. Suppose $2r$ of the $8$ points in $\,\mathcal{P}$ are real and
$2s$ of the $12$ zeros of $B$ are real. Then $r\leq s$. Conversely, for all
valid parameters $r\leq s$, some configuration $\mathcal{P}$ realizes the pair
$(r,s)$.

###### Proof.

The $12$ points in $\mathbb{P}^{1}$ which are zeros of $B$ correspond to the
singular fibers under the map
$\mathbb{P}^{2}\rightarrow\mathbb{P}^{1}:(x:y:z)\mapsto(u:v)$. This is the map
given by two cubics $u$,$v$ through $\mathcal{P}$. Algebraically, we seek
points $(\alpha:\beta)\in\mathbb{P}^{1}$ such that the cubic $\beta\cdot
u-\alpha\cdot v$ is singular. There are $12$ such rational cubics, and the
question is how many of those can be real.

The Welschinger invariant $W_{r}(\mathbb{P}^{2},3)$ is a signed count of real
rational cubics passing through $2r$ real points and $4-r$ pairs of complex
conjugate points. By definition, $W_{r}(\mathbb{P}^{2},3)\leq 2s$, the number
of real solutions. Example 4.3 in [27] provides a tropical computation of the
Welschinger invariants for cubics and yields $W_{r}(\mathbb{P}^{2},3)=2r$. We
conclude $2r\leq 2s$.

The last sentence is proved with a direct calculation. By sampling from
rational configurations $\mathcal{P}=\\{P_{1},\ldots,P_{8}\\}$, we found
instances for all $(r,s)$ with $0\leq r\leq s\leq 6$ and $r\leq 4$. These can
be found at our MathRepo website https://mathrepo.mis.mpg.de/BranchPoints . ∎

### Acknowledgements

We thank Arthur Bik, Lou-Jean Cobigo, Yelena Mandelstham, Jared Ongaro, Boris
Shapiro and Simon Telen for helpful discussions. Hannah Markwig, Victoria
Schleis and Bernd Sturmfels were supported by the Deutsche
Forschungsgemeinschaft (DFG, German Research Foundation), Project-ID
286237555, TRR 195. Javier Sendra–Arranz received the support of a fellowship
from the “la Caixa” Foundation (ID 100010434). The fellowship code is
LCF/BQ/EU21/11890110.

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Authors’ addresses:

Daniele Agostini, MPI-MiS Leipzig and Universität Tübingen
<EMAIL_ADDRESS>
Hannah Markwig, Universität Tübingen<EMAIL_ADDRESS>
Clemens Nollau, Universität Leipzig<EMAIL_ADDRESS>
Victoria Schleis, Universität Tübingen<EMAIL_ADDRESS>tuebingen.de

Javier Sendra–Arranz, MPI-MiS Leipzig<EMAIL_ADDRESS>
Bernd Sturmfels, MPI-MiS Leipzig and UC Berkeley<EMAIL_ADDRESS>