# How to Reconstruct a Planar Map From its Branching Curve

Eriola Hoxhaj and Josef Schicho
RISC, Johannes Kepler University of Linz, Austria

###### Abstract

We present an algorithm for constructing a map
$\mathbb{P}^{2}\to\mathbb{P}^{2}$ with a given branching curve. The stepping
stone is the ramification curve, which is obtained as the linear normalization
of the branching curve.

## Introduction

We provide an algorithm that deals with the following reconstruction problem:
given a plane curve $B\subset\mathbb{P}^{2}$, compute a map
$f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ such that $B$ is the branching curve of
this map, assuming that the map is generic. By [CF11], genericity implies that
the ramification curve is smooth, the branching curve has only nodes and
cusps, and the map from the ramification curve to the branching curve is
birational.

The problem is motivated by potential applications in computer vision: a
picture of a parametrised surface can be mathematically described by the
composition of the parameterization followed by a central projection to the
image plane, and the apparent contour of the parameterised surface is the
branching curve of the composed map. A solution to the reconstruction problem,
if it is unique, should allow to obtain information about the parameterised
surface.

The uniqueness of the solution, up to unavoidable projective transformations,
is guaranteed by a much more general result in [Kul99]: any generic map
$f:S_{1}\to S_{2}$ of degree at least $5$ is determined up to right
composition with isomorphism to $S_{1}$ by its branching curve in $S_{2}$. In
our case, the map $f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ is defined by
polynomials of degree $d$ that do not have a common zero, and this implies
that the degree of the map is $d^{2}$. For $d\geq 3$, uniqueness is therefore
guaranteed; our constructive algorithm could be considered as another proof
for this special case. The case $d=2$ is classical. Here the branching curve
is a sextic with 9 cusps, dual to a smooth cubic. Chisini [Chi44] showed that
there are four maps $S_{1}\to\mathbb{P}^{2}$, up to right equivalence. In
three of these four solutions, $S_{1}$ is isomorphic to $\mathbb{P}^{2}$. The
three solutions turned out to be left-right-equivalent, i.e., there are
isomorphic up to composing with an isomorphism on the right and with an
automorphism on the left that preserves the branching curve.

Our algorithm computes first the linear normalization $R$ of the given
branching curve (see [GKZ94] for a discussion of the linear normalization of a
projective variety). We show that in our case – assuming that the
reconstruction problem has a solution – the linear normalization is the
ramification curve of a generic projection from the Veronese surface
$V_{d}\in\mathbb{P}^{\frac{(d(d+3)}{2}}$ to $\mathbb{P}^{2}$. The computation
of the linear normalization can in theory be achieved by computing the
normalization of the cone over the branching curve. However, the curve $B$ has
degree $3d(d-1)$ and $\frac{9(d^{2}-2)(d-1)^{2}}{2}$ singularities (nodes and
cusps), so the algorithms available in special purpose computer algebra
systems would not be fast enough even for the case $d=3$. Hence we provide a
new algorithm for the linear normalization, adapted to the situation of curves
that have only nodes and cusps. It works well for $d=3$ and $d=4$.

Once we have the ramification curve $R\in\mathbb{P}^{\frac{d(d+3)}{2}}$ (in
case $d\geq 3$), we show that there is a unique Veronese surface $V_{d}$
containing $R$, and we compute the equations of this Veronese surface. This
step uses syzygies in a way similar to [vB07], where a curve is given too, and
certain specific varieties containing the curve need to be constructed. The
classical case $d=2$ is treated by a different approach, using the Hessian
pencil of the dual curve (see [AD09]).

The Veronese surface $V_{d}$ is isomorphic to $\mathbb{P}^{2}$. The last step
of our algorithm is to compute an explicit isomorphism $\mathbb{P}^{2}\to
V_{d}$, i.e., a parametrisation of $V_{d}$. This is even more expensive than
the computation of the linear normalization, see for instance [dGHPS06] for
the case $d=3$. The hardest part is a Diophantine subproblem: one needs to
find a point in $V_{d}$ first. Unfortunately, the available algorithms for
solving this Diophantine problems over $\mathbb{Q}$ are too slow to handle
random examples with $d=3$. Hence this substep of the algorithm is just
theoretical. Here we assume that we can find two distinct rational points and
proceed to computing a parametrization. We illustrate this step by examples in
positive characteristics: here it is easy to find rational points
probabilistically, by intersecting with random linear subspaces of codimension
two.

Here is an outline of the paper. Section 1 gives a precise specification of
the problem, including the equivalence relation we need to take into account
in order to speak of uniqueness of the solution. Section 2 is dedicated to the
case $d=2$. In Section 3, we give our new algorithm for linear normalization
and use it to construct the ramification curve $R$. Section 4 is dedicated to
the construction of the unique Veronese surface containing $R$. In Section 5,
we give a parametrization algorithm and put the other parts together.

##  1 Generic Planar Maps and their Branching Curves

Let $k$ be a computable field. A regular map
$f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ is a called a planar map. Every planar map
is defined by three homogeneous polynomials $F_{0},F_{1},F_{2}\in k[s,t,u]$ of
the same degree $d$ without a common zero; we say that $d$ is the polynomial
degree of the map $f$. This number is not equal to the degree of $f$, i.e.,
the cardinality of the preimage of a generic point. Specifically, the degree
of the map is $d^{2}$, by Bezout’s theorem.

Let $f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ be a planar map of polynomial degree
$d\geq 2$. The ramification curve $R$ of $f$ is the scheme defined by the
homogeneous Jacobi matrix
$\det\left(\frac{\partial{(F_{0},F_{1},F_{2})}}{\partial{(s,t,u)}}\right)$.
This is a curve of degree $3(d-1)$ (maybe reducible, maybe with multiple
components). The branching curve $B$ is defined as the image of the
ramification curve. We say that the planar map $f$ is generic if it satisfies
the following conditions:

  * •

the ramification curve $R$ is nonsingular (in particular irreducible);

  * •

the restriction of $f$ to $R$ is birational to $B$;

  * •

the branching curve $B$ has only nodes and cusps.

It is well-known that the set of coefficients vectors of planar maps that
satisfy the conditions above is a Zariski-open and dense subset in the set of
all coefficient vectors (see [CF11]). From now on, assume that $f$ is a
generic planar map. Then $\deg(B)=d\deg(R)=3d(d-1)$. Since $R$ is smooth, its
geometric genus is equal to the arithmetic genus
$\frac{(3(d-1)-1)(3(d-1)-2)}{2}$. The geometric genus is a birationally
invariant, thus the geometric genus of $B$ is the same as the geometric genus
of $R$. From the genus formula for planar curves, we can find the number of
singularities (in our case nodes and cusps)

$\frac{(3d(d-1)-1)(3d(d-1)-2)}{2}-\frac{(3(d-1)-1)(3(d-2)-2)}{2}=\frac{9(d^{2}-2)(d-1)^{2}}{2}.$

We say that two planar maps $f_{1},f_{2}:\mathbb{P}^{2}\to\mathbb{P}^{2}$ are
right equivalent if there is a projective isomorphism
$t:\mathbb{P}^{2}\to\mathbb{P}^{2}$ such that $f_{2}=f_{1}\circ t$. Similarly,
we say that $f_{1},f_{2}:\mathbb{P}^{2}\to\mathbb{P}^{2}$ are left equivalent
if there is a projective isomorphism $t:\mathbb{P}^{2}\to\mathbb{P}^{2}$ such
that $f_{2}=t\circ f_{1}$. Left equivalent maps have the same ramification
curve and projectively equivalent branching curves. Right equivalent maps have
the same branching curve and projectively equivalent ramification curves.

The goal of this paper is to provide an algorithm for the following problem:

Problem DiscToMap

1:an irreducible curve $B$ of degree $3d(d-1)$ in $\mathbb{P}^{2}$ that has
only nodal and cuspidal singularities.

2:a generic planar map $f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ that has $B$ as a
branching curve, if such a map exists.

If the no generic planar map exists such that the branching curve is the given
curve, then the algorithm detects this. We also prove that for $d\geq 3$, the
solution is unique up to right equivalence; but, as mentioned in the
introduction, the uniqueness is already a consequence of Kulikov’s result
[Kul99].

The first step in our algorithm will be to compute the linear normalization of
$B$, which we now define, following [GKZ94]. For two integers $n>m>2$, a
rational map $\mathbb{P}^{n}\dashrightarrow\mathbb{P}^{m}$ given by an
$(m+1)\times(n+1)$ matrix of full rank is called a projection; its center is
the projectivised nullset of the defining matrix. Let $X\subset\mathbb{P}^{n}$
be an irreducible variety that is not contained in a hyperplane. The
projection $p:\mathbb{P}^{n}\dashrightarrow\mathbb{P}^{m}$ is called a regular
projection if the center of $p$ does not intersect $X$ and $p$ maps $X$
birationally to its image $p(X)$. Regular projections preserve the degree,
i.e., $\deg(X)=\deg(p(X))$. The linear normalization of $X$ is an irreducible
variety $\tilde{X}$ in some $\mathbb{P}^{N}$ together with a regular
projection $p:\mathbb{P}^{N}\dashrightarrow\mathbb{P}^{n}$ that maps
$\tilde{X}$ to $X$ and which is universal in the following sense: for any
irreducible variety $X^{\prime}$ in some $\mathbb{P}^{M}$ together with a
regular projection $p^{\prime}:\mathbb{P}^{M}\dashrightarrow\mathbb{P}^{n}$
that maps $X^{\prime}$ to $X$, there is a unique regular projection
$q:\mathbb{P}^{N}\dashrightarrow\mathbb{P}^{M}$ that maps $\tilde{X}$ to
$X^{\prime}$ and that satisfies $p=p^{\prime}\circ q$. It is well-known that
the linear normalizations always exists and is unique up to right equivalence.
If the linear normalization is a projective isomorphism, then we say that $X$
us linearly normal. A normal variety $X\subset\mathbb{P}^{n}$ is linearly
normal if and only if the restriction map of linear forms
$\Gamma(\mathbb{P}^{n},{\cal O}_{\mathbb{P}^{n}}(1))\to\Gamma(X,{\cal
O}_{X}(1))$ is surjective.

######  Example 1.1.

The linear normalization of the nodal cubic in $\mathbb{P}^{2}$ with equation
$y^{2}z-x^{2}z-x^{3}=0$ is the twisted cubic in $\mathbb{P}^{3}$ with
equations $zy-wx=w^{2}-z^{2}-xz=wy-xz-x^{2}=0$. The regular projection is
$(x:y:z:w)\mapsto(x:y:z)$ with center $(0:0:0:1)$.

The linear normalization is useful because it is projectively isomorphic to
the ramification curve; the precise statement is below. We write
$\nu_{d}:\mathbb{P}^{2}\to\mathbb{P}^{\frac{d(d+3)}{2}}$ for a Veronese map
defined by a basis of forms if degree $d$, and $V_{d}$ for the Veronese
surface $\nu_{d}(\mathbb{P}^{2})$. The map $\nu_{d}$ is unique up to left
equivalence.

######  Proposition 1.2.

Assume that $B$ is the branching curve of a generic planar map
$f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ of polynomial degree $d\geq 2$. Let
$\tilde{B}\subset\mathbb{P}^{N}$,
$p:\mathbb{P}^{N}\dashrightarrow\mathbb{P}^{2}$ be the linear normalization of
$B$. Then $N=\frac{d(d+3)}{2}$, and there is a Veronese embedding
$\nu_{d}:\mathbb{P}^{2}\to V_{d}\subset\mathbb{P}^{N}$ such that the diagram

${\tilde{B}\subset
V_{d}\subset\mathbb{P}^{N}}$${R\subset\mathbb{P}^{2}}$${B\subset\mathbb{P}^{2}}$$\scriptstyle{p}$$\scriptstyle{\nu_{d}}$$\scriptstyle{f}$

commutes. The Veronese embedding $\nu_{d}$ maps the ramification curve $R$ of
$f$ isomorphically to $\tilde{B}$. In particular, $\tilde{B}$ is contained in
$V_{d}$.

###### Proof.

Let $B^{\prime}\subset\mathbb{P}^{\frac{d(d+3)}{2}}$ be the image of $R$ under
the Veronese map. The map $f$ is defined by three forms of degree $d$, hence
there is a projection map
$p^{\prime}:\mathbb{P}^{\frac{d(d+3)}{2}}\to\mathbb{P}^{2}$ such that
$f=p^{\prime}\circ\nu_{d}$. The center of the projection $p^{\prime}$ does not
intersect $V_{d}$, because the intersection is the image of the base locus of
$f$ and this base locus is empty by assumption. Hence $p^{\prime}$ is a
regular projection from $B^{\prime}$ to $B$. By the universality property of
the linear normalization, there is a regular projection
$q:\mathbb{P}^{N}\to\mathbb{P}^{\frac{d(d+3)}{2}}$ from $\tilde{B}$ to
$B^{\prime}$.

To show that $q$ is an isomorphism, we show that $R^{\prime}$ is linearly
normal. The restriction map $\Gamma(\mathbb{P}^{\frac{d(d+3)}{2}},{\cal
O}_{\mathbb{P}^{\frac{d(d+3)}{2}}}(1))\to\Gamma(B^{\prime},{\cal
O}_{B^{\prime}}(1))$ of linear forms corresponds to the restriction map
$\Gamma(\mathbb{P}^{2},{\cal O}_{\mathbb{P}^{2}}(1))\to\Gamma(B,{\cal
O}_{B}(d))$ of forms of degree $d$. This map is surjective because $R$ is a
hypersurface and therefore arithmetically Cohen-Macaulay. ∎

Proposition 1.2 suggests the following strategy for solving our problem:

Strategy for Algorithm DiscToMap

1:an irreducible curve $B$ of degree $3d(d-1)$ in $\mathbb{P}^{2}$.

2:a generic planar map $f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ that has $B$ as a
branching curve.

3:

4:Compute the linear normalization $p:\tilde{B}\to B$ (see section 3).

5:Compute a Veronese surface $V_{d}$ containing $\tilde{B}$ (see section 4).

6:Compute a Veronese map $\nu_{d}:\mathbb{P}^{2}\to V_{d}$ (see section 5).

7:Return $p\circ\nu_{d}:\mathbb{P}^{2}\to\mathbb{P}^{2}$.

We will follow this strategy for $d\geq 3$. For $d=2$, we could not find a way
to do step 3. Fortunately, there is an alternative, to be described in the
next section.

##  2 Maps of Polynomial Degree Two

In this section, we assume that $k$ has characteristic zero and is
algebraically closed.

The branching curve $B$ of a generic planar map of polynomial degree two is a
sextic of genus one with nine cusps. By the Plücker formula for plane curves,
the dual curve $\check{B}$ is a nonsingular cubic. On the other hand, the
ramification curve $R$ is also a smooth cubic. Both $R$ and $\check{B}$ are
birationally equivalent to $B$, hence they are birationally equivalent to each
other. Two cubic curves are birationally equivalent if and only if they are
projectively equivalent; so, we already know $R$ up to projective equivalence.

######  Lemma 2.1.

Every nonsingular cubic is the ramification curve of some generic planar map
of polynomial degree two.

###### Proof.

Let $F\in k[s,t,u]$ be a cubic form defining a nonsingular cubic curve. The
gradient $\nabla(F)$ defines a planar map of polynomial degree two. Its
ramification curve coincides with the Hessian $H(F)$. Hence the statement is a
consequent of the next lemma. ∎

######  Lemma 2.2.

Every nonsingular cubic is the Hessian of exactly three other nonsingular
cubics.

###### Proof.

We are going to construct a nonsingular cubic form $G$ whose Hessian is a
given cubic form $F$ defining a nonsingular cubic curve. We claim that $G$ can
be chosen of the form $aF+bH(F)$ with suitable constants $a,b\in k$. In order
to prove the claim, we assume that $F$ is in Hessian normal form

$F_{\lambda}=s^{3}+t^{3}+u^{3}+\lambda stu,\ \lambda^{3}+27\neq 0.$

This is without loss of generality: any nonsingular cubic can be transformed
to Hessian normal form by some projective transformation (see [AD09]). The
condition $\lambda^{3}+27\neq 0$ is equivalent to the statement that
$\nabla(F)$ does not have zeroes, or in other words that $F$ defines a
nonsingular curve.

The Hessian of a Hessian normal form is also a Hessian normal form:
specifically,

$H(F_{\mu})=F_{-\frac{\mu^{3}+108}{3\mu^{2}}}.$

So we just need to find $\mu$ such that
$-\frac{\mu^{3}+108}{3\mu^{2}}=\lambda$, and this is always possible. The fact
that the discriminant of $\mu^{3}+3\lambda\mu^{2}+108$ with respect to $\mu$
is equal to $\lambda^{3}+27$ (up to some multiplicative constant) shows that
there are exactly three solutions. The fact that the resultant of
$\mu^{3}+3\lambda\mu^{2}+108$ and $\mu^{3}+27$ with respect to $\mu$ is equal
to $\lambda^{3}+27$ shows that all three solutions are equations of
nonsingular cubic curves.

Assume, indirectly, that $G^{\prime}$ is the equation of a nonsingular cubic
which is not in Hessian normal form such that $H(G^{\prime})=F$. Let $\Gamma$
be the pencil of cubics generated by $G^{\prime}$ and $H(G^{\prime})=F$. The
Hessian of any cubic in $\Gamma$ is again in $\Gamma$. In particular, $H(F)$
is in $\Gamma$. Since $F$ and $H(F)$ are distinct, they generate $\Gamma$, and
this shows that $\Gamma$ is the pencil of Hessian normal forms. But this
contradicts the assumption that $G^{\prime}$ is not in Hessian normal form. ∎

The three solutions are classical [Chi44] – see also [Cat86] for a very
explicit discussion including also the fourth map from a surface – not
isomorphic to $\mathbb{P}^{2}$ – to $\mathbb{P}^{2}$ with the same branching
curve.

Here is the algorithm for the reconstruction of a planar map of polynomial
degree 2 that corresponds to the constructive proof above.

Algorithm DiscToMapCase2

1:an irreducible sextic $B$ with nine cusps.

2:a generic planar map $f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ that has $B$ as a
branching curve.

3:

4:Compute the dual $\check{B}$ of $B$ (a nonsingular cubic).

5:Compute the Hessian $H(\check{B})$.

6:Compute $\mu\in k$ such that the Hessian of $H(\check{B})+\mu\check{B}$ is
equal to $\check{B}$ (three solutions).

7:Return $\nabla(H(\check{B})+\mu\check{B})$.

######  Example 2.3.

Let $B$ be the curve with equation

$32x^{6}+196x^{4}y^{2}+72x^{3}y^{3}+436x^{2}y^{4}-280xy^{5}+144y^{6}$
$+16x^{5}z+96x^{4}yz+352x^{3}y^{2}z+140x^{2}y^{3}z+700xy^{4}z-272y^{5}z$
$-32x^{4}z^{2}+44x^{3}yz^{2}+52x^{2}y^{2}z^{2}+184xy^{3}z^{2}+209y^{4}z^{2}-16x^{3}z^{3}$
$-80x^{2}yz^{3}-266xy^{2}z^{3}+128y^{3}z^{3}+x^{2}z^{4}-36xyz^{4}-132y^{2}z^{4}+2yz^{5}=0.$

This is a sextic with nine cusps, three of them being real (see Figure 1). Its
dual is the cubic curve $\check{B}:t^{2}u-(s^{2}-u^{2})(s-2u)+ts^{2}/2=0$. The
three numbers such that such that the hessian of $H(\check{B})+\mu\check{B}$
is equal to the $\check{B}$ are
$\mu_{1}=-31.37404937,\mu_{2}=-14.20865677,\mu_{3}=45.58270614$. We obtain
three planar map of polynomials of degree 2 with the sextic $B$ as its
branching curve. The picture Figure 1 shows the number of real preimages for
each connected component of the complement of $B$.

Figure 1: The branching curve $B$ separates the plane into three regions. The
points in the beige region have no real preimage points for the two first maps
and four real preimages for the third map. The points in the blue sky region
have two real preimage points and the points in the yellow region have four
real preimage points, for all three maps.

##  3 Linear Normalization

As mentioned in Section 1, a normal variety $X\subset\mathbb{P}^{n}$ is
linearly normal if and only if the restriction map of linear forms
$\Gamma(\mathbb{P}^{n},{\cal O}_{\mathbb{P}^{n}}(1))\to\Gamma(X,{\cal
O}_{X}(1))$ is surjective. This suggests an algorithm for the computation of
the linear normalization of $X$: first, compute its projective normalization
$N(X)$. Second, take a basis for the linear space of degree one elements of
the graded coordinate ring of $N(X)$, and use this basis to define a map into
projective space. The image of this map is the linear normalization.

This algorithm would be correct, but unfortunately it is too costly for the
computation of the linear normalization of branching curves. Already for
$d=3$, the branching curve is a plane curve of degree 18 with 126 singular
points with coordinates in a large field extension, and this is not feasible
for the available programs for the computation of the normalization in
Singular or Macaulay2. Here we use a more efficient method that takes
advantage of the special situation that computes the degree 1 elements without
computing the whole integral closure. Also, we make use of the assumption that
the given plane curve has only nodes and cusps as its singular points.

Let $k$ be a field and $B\in\mathbb{P}_{k}^{2}$ be an irreducible plane curve
defined by a homogeneous polynomial $F\in k[x,y,z]$. The graded coordinate
ring of $C$ is denoted by $S:=k[x,y,z]/\langle F\rangle$. Its integral closure
in its quotient field $\mathrm{QF}(S)$ is denoted by $\tilde{S}$. The ring
$\tilde{S}$ is also the graded coordinate ring of a projective curve $N(B)$,
the projective normalization of $B$. The ring inclusion
$S\hookrightarrow\tilde{S}$ induces a projection map $N(B)\to B$.

######  Remark 3.1.

It may happen that the graded ring $\tilde{S}$ is not generated by linear
elements. Then $N(B)$ is naturally embedded in a weighted projective space,
where the weights are the degrees of ring generators. This paper only contains
examples where $\tilde{S}$ is generated in degree 1.

The conductor ideal $C$ is defined as the conductor of $S$ in $\tilde{S}$,
i.e., the set of all elements $s\in S$ such that $s\tilde{S}\subset S$. It is
an ideal in both rings $S$ and $\tilde{S}$. The preimage of $C$ under the
quotient map $k[x,y,z]\to S$ is also known as the Gorenstein adjoint ideal. It
is equal to the ideal of all polynomials that vanish with order at least $r-1$
at all $r$-fold singular points of $B$, including infinitely near
singularities (see [BDLP17]).

######  Remark 3.2.

In our situation, $B$ has only nodes and cups. Hence all multiplicities of
points of singular points are equal to $2$, and there are no infinitely near
singularities. It follows that the Gorenstein adjoint ideal is equal to the
radical of the singular locus of $B$.

######  Lemma 3.3.

Let $a\in\mathrm{QF}(S)^{\ast}$, and assume that $A\subset\mathrm{QF}(S)$ is a
finitely generated $S$-submodule of the quotient field. If $aA\subseteq A$,
then $a$ is integral over $S$.

###### Proof.

Because $A$ is finitely generated, there exists a positive integer $r$ and a
surjective map $\pi:S^{r}\to A$. Let $\mu:A\to A$ be the multiplication map
$m\mapsto am$. Because $\mu$ is $S$-linear and $S^{r}$ is a projective module,
$\mu$ can be lifted to obtain a commutative diagram

$\begin{matrix}S^{r}&\xrightarrow{\quad\psi\quad}&S^{r}\\\
\bigg{\downarrow}{\pi}&&\bigg{\downarrow}{\pi}\\\
A&\xrightarrow{\quad\mu\quad}&A\end{matrix}$

with some map $\psi:S^{r}\to S^{r}$ represented by some matrix $M\in
S^{r\times r}$. Let $P\in S[T]$ be the characteristic polynomial of $M$. By
the Cayley-Hamilton theorem, we have $P(\psi):S^{r}\to S^{r}$ is the zero map.
The diagram

$\begin{matrix}S^{r}&\xrightarrow{\quad P(\psi)\quad}&S^{r}\\\
\bigg{\downarrow}{\pi}&&\bigg{\downarrow}{\pi}\\\ A&\xrightarrow{\quad
P(\mu)\quad}&A\end{matrix}$

also commutes. The map $P(\mu):A\to A$ is the multiplication by $P(a)$. Since
the $P(\psi)$ is the zero map, the map $P(\mu):A\to A$ is also zero. Then $P$
is a monic polynomial that annihilates $a$. ∎

######  Lemma 3.4.

Let $g\in(C\setminus\\{0\\}$ be a nonzero element of the conductor. Then we
have the equality $\langle g\rangle:_{S}C=g\tilde{S}$.

###### Proof.

$\langle g\rangle:_{S}C\supseteq g\tilde{S}$: any element in $g\tilde{S}$ can
be written as $g\tilde{s}$ for some $\tilde{s}\in\tilde{S}$. If $a\in C$ is
arbitrary, then $a(g\tilde{s})=(\tilde{s}a)g$ is in $\langle g\rangle$. Hence
$g\tilde{S}$ is in $\langle g\rangle:_{S}C$.

$\langle g\rangle:_{S}C\subseteq g\tilde{S}$: let $s\in(\langle
g\rangle:_{S}C\setminus\\{0\\}$. Let $M:=\frac{s}{g}C$. For any $m\in M$,
there is an $a\in C$ such that $m=a\frac{s}{g}$. For any
$\tilde{s}\in\tilde{S}$, we have

$m\tilde{s}=a\frac{s}{g}\tilde{s}=(a\tilde{s})\frac{s}{g}=\frac{(a\tilde{s})s}{g},$

with $(a\tilde{s})\in C$. Because $s$ is in the quotient ideal of $\langle
s\rangle$ by $C$, the numerator is of the form $gs^{\prime}$ for some
$s^{\prime}\in\tilde{S}$. Hence $m\tilde{s}\in S$. Since $\tilde{s}$ was
chosen arbitrarily in $\tilde{S}$, it follows that $m$ is in the conductor
$C$. So, $\frac{s}{g}C\subseteq C$. By Lemma 3.3, $\frac{s}{g}\in\tilde{S}$.
Therefore $s\in g\tilde{S}$. ∎

######  Theorem 3.5.

Let $g\in C$ of degree $d$, $g\neq 0$. Let $V$ be the vector space of elements
of degree $d+1$ in the quotient ideal $I:=\langle g\rangle:C$. Then the
rational map defined by a basis of $V$ maps $B$ to its linear normalization
$\tilde{B}$.

###### Proof.

By Lemma 3.4, $V$ is also the vector space of elements of degree $d+1$ in the
ideal $\langle g\rangle_{\tilde{S}}$. Assume that $(gf_{1},\dots,gf_{k})$ is a
basis of $V$, for some $f_{1},\dots,f_{k}\in\tilde{S}$. Then
$f_{1},\dots,f_{k}$ are homogeneous of degree 1, and they generate the space
of all elements of $\tilde{S}$ of degree 1. Therefore the map defined by
$(f_{1},\dots,f_{k})$ maps $B$ to $\tilde{B}$. The map defined by
$(gf_{1},\dots,gf_{k})$ is the same map, because $g$ is may be cancelled in
the definition of a map into projective space. ∎

Lemma 3.4 and Remark 3.2 imply that Algorithm LinearNormalization for the
computation of the linear normalization of a plane curve is correct.

Algorithm LinearNormalization

1:an irreducible plane curve $B\subset\mathbb{P}^{2}$ whose singularities are
nodes or cusps.

2:its linear normalization $\tilde{B}\subset\mathbb{P}^{n}$ together with a
projection $p:\mathbb{P}^{n}\dashrightarrow\mathbb{P}^{3}$ mapping $\tilde{B}$
to $B$.

3:

4:Compute the radical $I$ of the singular locus of $B$.

5:Choose a homegeneous element $g\in I$ of degree less than $\deg(B)$.

6:Compute the quotient ideal $J:=\langle g,f\rangle:I$, where $f$ is the
equation of $B$.

7:Extend $(xg,yg,zg)$ to a basis $(xg,yg,zg,h_{1},\dots,h_{k})$ for the vector
space of forms in $J$ of degree $(\deg(g)+1)$.

8:Compute the image $\tilde{B}\subset\mathbb{P}^{k+2}$ of $B$ under the
rational map $\mathbb{P}^{2}\dashrightarrow\mathbb{P}^{k+2}$ defined by this
basis.

9:Return $\tilde{B}$ together with the projection to the first three
coordinates.

######  Example 3.6.

The nodal cubic curve in Example 1.1 with equation $F=y^{2}z-x^{2}z-x^{3}=0$
has only one node at $(0:0:1)$. Hence we get $I=\langle x,y\rangle$. We choose
$g:=x$. The quotient ideal is

$J=\langle F,g\rangle:I=\langle x,y^{2}z\rangle:\langle x,y\rangle=\langle
x,yz\rangle.$

The homogeneous part in degree two is generated by $x^{2},xy,xz,yz$. Hence the
linear normalization is the image under the map
$(x:y:z)\mapsto(x^{2}:xy:xz:yz)$; this is the twisted cubic curve that has
been already shown in Example 1.1.

##  4 Construction and Parametrization of the Veronese surface

###  4.1 Construction

Let $B$ be the branching curve of the planar map
$\mathbb{P}^{2}\to\mathbb{P}^{2}$ defined by polynomials of degree $d\geq 3$.
By Proposition 1.2, there exists a Veronese surface $V_{d}$ containing the
linear normalization $\tilde{B}$. In this subsection, we will prove that such
a Veronese surface is unique, and we give an algorithm for the construction of
its ideal. It is known that this ideal is generated by quadratic forms. Note
that the Veronese surface $V_{d}$ is already unique up to projective
coordinate transformation; but for our purpose it is necessary to compute a
Veronese surface containing $\tilde{B}$.

We need to make a case distinction for $d=3$ and $d>3$ – the latter case is
much easier.

####  4.1.1 Degree Three Planar Map

If $d=3$, then $B$ is a curve of degree 18 with 126 double points. The linear
normalization $\tilde{B}$ lies in $\mathbb{P}^{9}$. Our goal is to find a
Veronese surface $V_{3}\subset\mathbb{P}^{9}$ of degree 9 containing
$\tilde{B}$. Proposition 1.2 tells us a little bit more: the given curve
$\tilde{B}$ is the image of an irreducible planar sextic under some Veronese
map $\nu_{3}:\mathbb{P}^{2}\to\mathbb{P}^{9}$. This map is equal to the
standard Veronese embedding by cubic monomials followed by a projective
transformation of $\mathbb{P}^{9}$. The Veronese map maps curves of degree
$3k$ to intersections with hypersurfaces of degree $k$, for each positive
integer $k$. Hence $\tilde{B}$ is the intersection of $V_{3}$ with a quadric
hypersurface.

By computation of the ideal of the standard Veronese surface embedded by cubic
polynonmials, we see that the ideal of $V_{3}$ is generated by 27 quadratic
forms. The ideal of the intersection with another quadric hypersurface is
generated by 28 quadrics. This is the ideal we have, and what we need is the
ideal generated by the 27 quadratic forms.

It is instructive to compare the Betti tables for the linear normalization and
the Veronese surface of a random example of a planar map defined by degree
three polynomials. Here they are:

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8
---|---|---|---|---|---|---|---|---|---
total | 1 | 28 | 132 | 294 | 378 | 294 | 132 | 28 | 1
0: | 1 | . | . | . | . | . | . | . | .
1: | . | 28 | 105 | 189 | 189 | 105 | 27 | . | .
2: | . | . | 27 | 105 | 189 | 189 | 105 | 28 | .
3: | . | . | . | . | . | . | . | . | 1
Table 1: Betti table for the normalization of the curve. | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
total | 1 | 27 | 105 | 189 | 189 | 105 | 27 | 1
0: | 1 | . | . | . | . | . | . | .
1: | . | 27 | 105 | 189 | 189 | 105 | 27 | .
2: | . | . | . | . | . | . | . | 1
Table 2: Betti table for the Veronese surface.

The tables reveal that both ideals have the same linear syzygies (a space of
dimension 105). The additional quadratic form in the ideal of $\tilde{B}$ will
not show up in any linear syzygy, it is only related by quadratic syzygies to
the other 27 quadratic forms. We will make this observation precise and use it
for the uniqueness proof and the algorithm.

Throughout this subsection, we denote by $W_{1}$ the 28-dimensional vector
space of quadratic forms in the ideal of a curve $\tilde{B}\in\mathbb{P}^{9}$,
which is assumed to be the image of an irreducible sextic plane curve under a
Veronese map $v_{3}$, and we denote by $W_{2}$ the 27-dimensional subspace of
quadratic forms in the ideal of the Veronese surface.

######  Lemma 4.1.

Let $(Q_{1}^{\prime},\dots,Q_{27}^{\prime})$ be a basis for $W_{2}$, and let
$Q\in W_{1}\setminus W_{2}$. Let $(L_{1},L_{2},\dots,L_{28})$ be a linear
syzygy between quadratic forms $(Q,Q_{1}^{\prime},\dots,Q_{27}^{\prime})$.
Then $L_{1}=0$.

###### Proof.

Let $f_{0},...,f_{9}$ be the cubic forms defining the Veronese map. We
substitute them into the syzygy relation
$L_{1}Q+L_{2}Q_{1}^{{}^{\prime}}+..+L_{28}Q_{27}^{{}^{\prime}}=0$. Since
$Q_{i}^{{}^{\prime}}(f_{0},..,f_{9})=0$ for all $i=1,..,27$, we obtain
$L_{1}(f_{0},..,f_{9})Q(f_{0},..,f_{9})=0$. The polynomial $Q(f_{0},..,f_{9})$
is the pullback of the quadratic form $Q\in I_{\tilde{B}}$, which is the
equation of the ramification curve. Furthermore, $Q(f_{0},..,f_{9})$ is not
zero and not a multiple of the equation of the ramification curve since it has
degree 6. Therefore, $L_{1}(f_{0},..,f_{9})$ must be zero. As a result,
$L_{1}$ vanishes on $V_{3}$, which implies that $L_{1}=0$. ∎

######  Lemma 4.2.

The Veronese surface containing $\tilde{B}$ is unique.

###### Proof.

Assume, indirectly, that $U_{3}$ is another Veronese surface containing
$\tilde{B}$, $U_{3}$ not equal $V_{3}$. Let $Q^{{}^{\prime\prime}}$ be a
quadratic form in the ideal of $U_{3}$ but not in $W_{2}$. Since $U_{3}$
contains $\tilde{B}$, $Q^{{}^{\prime\prime}}$ is in $W_{1}$. By Lemma 4.1, any
linear syzygy of
$Q^{{}^{\prime\prime}},Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}}$ has first
entry equal to zero $(*)$. The vectorspace $W_{4}$ of quadratic forms
vanishing on both Veronese surfaces $U_{3}$ and $V_{3}$ has dimension 26, by
linear algebra. Let $P_{1},..,P_{26}$ be a basis. Then
$Q^{{}^{\prime\prime}},P_{1},..,P_{26}$ is basis for the vector space of
quadratic forms in $I_{U_{3}}$. By Lemma 4.1, the first column of the linear
syzygy matrix $S^{{}^{\prime}}$ of
$(Q^{{}^{\prime\prime}},P_{1},..,P_{26})^{t}$ is zero. On the other hand, the
$S^{{}^{\prime}}$ is the syzygy matrix of a Veronese surface - it is known
that the syzygy module of a Veronese surface is generated by linear forms. So
the first generator $Q^{{}^{\prime\prime}}$ is linearly independent from
$P_{1},..,P_{26}$, and this absurd. ∎

The following algorithm explains each step of constructing the Veronese
surface.

Veronese Surface of Degree3PlanarMap

1:$\tilde{B}\in\mathbb{P}^{9}$, the image of a plane sextic under a Veronese
map

2:the unique Veronese surface $V_{3}$ containing $\tilde{B}$

3:

4:Let $Q_{1},\dots,Q_{28}$ be a set of generators of the ideal of $\tilde{B}$
(they are quadratic)

5:Compute the syzygy matrix S in $k[x_{0},..,x_{9}]^{105\times 28}$ (its
entries are linear forms).

6:Write  S as $\sum S_{i}x_{i}$, $i=0..9$, for some $S_{i}\in k^{105\times
28}$

7:Let $M$ be the vector space generated by the rows of all $S_{i}$;

8:Compute a basis for $M$ (27 row vectors in $k^{28}$)

9:Multiply each basis vector by column vector $(Q_{1},..,Q_{28})^{t}$ and
return products.

###### Proof of correctness.

: Let $M$ be the vector space generated by the rows of the matrices $S_{i}$ as
defined in the algorithm. Suppose, indirectly, that $m\in M$ is such that the
scalar product $Q=m\cdot(Q_{1},\dots,Q_{28})^{t}$ is not in the ideal of
$V_{3}$. If $(Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}})$ is a set of
generators for the ideal of $V_{3}$, then the quadratic forms
$(Q,Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}})$ form another basis for
$W_{1}$. Let $T$ be the transformation matrix from $(Q_{1},..,Q_{28})$ to
$(Q,Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}})$ \- i.e.,
$T\cdot(Q_{1},..,Q_{28})^{t}=(Q,Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}})^{t}$.
Then the syzygy matrix of $(Q,Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}})$ is
$S\cdot T^{-1}$. Then by Lemma 4.1, the first column of the matrix $S\cdot
T^{-1}$ is zero. In particular, for each $i\in{0,\dots,9}$, the first column
of $S_{i}\cdot T^{-1}$ is zero. So, for each row of $S_{i}\cdot T^{-1}$, the
first entry is 0. Since $m$ is a linear combination of rows of the matrices
$S_{i}$, we have that $m^{\prime}=m\cdot T^{-1}$ is a linear combination of
rows of $S_{i}\cdot T^{-1}$. So, the first entry of $m^{{}^{\prime}}$ is zero.
Therefore, the product $Q=m\cdot(Q_{1},..,Q_{28})^{t}=m\cdot T^{-1}\cdot
T\cdot(Q_{1},..,Q_{28})^{t}=m^{{}^{\prime}}\cdot(Q,Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}})^{t}$
is a linear combination of $Q_{1}^{{}^{\prime}},..,Q_{27}^{{}^{\prime}}$, and
it follows that $Q$ is in $W_{2}$ \- contradiction. ∎

####  4.1.2 Degree Greater or Equal to Four Planar Map

If $d>3$, then $\tilde{B}$ is the image of a plane curve $B$ of degree
$3(d-1)$ under a Veronese map $v_{d}:\mathbb{P}^{2}\to\mathbb{P}^{N}$, where
$N=\frac{d(d+3)}{2}$.

######  Lemma 4.3.

The Veronese surface $V_{d}\in\mathbb{P}^{N}$ containing $\tilde{B}$ is
unique. The ideal of $V_{d}$ is generated by all quadratic forms in in the
ideal of $\tilde{B}$.

###### Proof.

It suffices to show the second statement. Assume, indirectly, that $\tilde{B}$
is contained in the zero set of a quadratic form $Q$ not vanishing in $V_{3}$.
Then the pullback of $Q$ is a plane curve of degree $2d$ (maybe reducible).
This plane curve has to contain the curve $B$ as one of its components. But
$2d<3(d-1)$, a contradiction. ∎

Lemma 4.3 implies that we can compute the ideal of $V_{3}$ by the following
algorithm.

Veronese Surface of DegreeBigger3PlanarMap

1:$\tilde{B}\in\mathbb{P}^{9}$, the image of a plane curve of degree $3(d-1)$
under a Veronese map

2:the unique Veronese surface $V_{d}$ containing $\tilde{B}$

3:

4:Compute a basis for the vector space of quadratic forms in the ideal of
$\tilde{B}$

5:Return this basis

###  4.2 Parametrization of the Veronese surface

This section focuses on a specific subproblem: given the ideal of a Veronese
surface in $V_{d}\subset\mathbb{P}^{N}$, where $N:=\frac{d(d+3)}{2}$, find a
Veronese map $\nu_{d}:\mathbb{P}^{2}\to V_{d}$ that parametrizes it. In our
setting, the Veronese map is defined by an arbitrary basis of the vector space
of forms of degree $d$ which is not known, and this makes the subproblem
difficult – if the Veronese map would be defined by monomials, then the
subproblem would be trivial.

As mentioned in the introduction, the problem is computationally even harder
than linear normalization (see [dGHPS06]). The main difficulty is a
Diophantine subproblem: if we are working over $\mathbb{Q}$, and we want to
find a parametrization with coefficients in $\mathbb{Q}$, then we need to find
a rational point on $V$. In this paper, we do not want to deal with this
difficulty. Instead, we assume that we have a black box that can produce
points on $V$, and we give a probabilistic algorithm to construct points on
$V$ in the case that the ground field is finite.

Any Veronese map $\nu_{d}:\mathbb{P}^{2}\to V_{d}$ induces a one-to-one
correspondence between linear forms in $\mathbb{P}^{n}$ and forms of degree
$d$ in $\mathbb{P}^{2}$. This will allow as to construct subspaces
corresponding to spaces of forms of degree $d$ with useful properties, despite
the fact that we do not yet have a Veronese map.

Let $y\in V$ be a point and $k\geq 1$. We define the $k^{th}$ osculating space
$O_{y,k}$ to $y$ as the set of all linear forms $L$ such that $L|_{V}$
vanishes to order at least $k$ on $V$ at $y$ (see [BCGI07] for a broader
treatment of osculating spaces).

######  Lemma 4.4.

Assume that the Veronese map $\nu_{d}:\mathbb{P}^{2}\to V_{d}$ maps some point
$x\in\mathbb{P}^{2}$ to $y\in V$. Let $k$ be in integer in $\\{1,\dots,d\\}$.
Then $O_{y,k}$ corresponds to the vector space of degree $d$ forms that vanish
at $x$ to order at least $k$.

###### Proof.

The pullback $\nu_{d}^{*}$ induces an isomorohism of local rings ${\cal
O}_{V,y}$ and ${\cal O}_{\mathbb{P}^{2},x}$, and this isomorphism preserves
the vanishing order. ∎

Here is a theorem that allows to construct the inverse of a parametrization of
$V$. In order to compute the a parametrization, we just need to invert the
result of that construction.

######  Theorem 4.5.

Let $y_{1}$ and $y_{2}$ be two distinct points on the Veronese surface. Let

$J:=(O_{y_{1},d}\cap O_{y_{2},d-1})+(O_{y_{1},d-1}\cap O_{y_{2},d}).$

Then $\dim(J)=3$, and the rational map defined by a basis of $J$ defines an
isomorphism $V_{d}\to\mathbb{P}^{2}$.

###### Proof.

Let $\nu_{d}:\mathbb{P}^{2}\to V$ be a Veronese map, and set
$x_{i}:=\nu_{d}^{-1}(y_{i})$ for $i=1,2$. Without loss of generality, we
assume $x_{1}=(0:0:1)$ and $x_{2}=(0:1:0)$. Then the vectorspace corresponding
to $(O_{y_{1},d}\cap O_{y_{2},d-1})$ is the space of linear forms that vanish
at $x_{1}$ to order $d$ and at $x_{2}$ to order $d-1$. This is $\langle
s^{d},s^{d-1}t\rangle$, where $s,t,u$ are the homogeneous coordinates of
$\mathbb{P}^{2}$. Similarily, the vectorspace corresponding to
$(O_{y_{1},d-1}\cap O_{y_{2},d})$ is $\langle s^{d},s^{d-1}u\rangle$.
Therefore, the vectorspace corresponding to $J$ is $\langle
s^{d},s^{d-1}t,s^{d-1}u\rangle$. The map defined by a basis of $J$ is the
composition of $v_{d}^{-1}$ with the map
$\tau:\mathbb{P}^{2}\to\mathbb{P}^{2}$ defined by a basis of $\langle
s^{d},s^{d-1}t,s^{d-1}u\rangle$. After cancelling the common factor $s^{d-1}$,
we get a basis for the vectorspace of linear forms. Hence $\tau$ is an
isomorphism. ∎

Theorem 4.5 shows that Algorithm Parametrization of Veronese Surface is
correct.

Algorithm Parametrization of Veronese Surface

1:$V\subset\mathbb{P}^{\frac{d(d+3)}{2}}$, a Veronese surface

2:$y_{1},y_{2}\in V$

3:a Veronese map $\nu_{d}:\mathbb{P}^{2}\to V_{d}$

4:

5:Compute the osculating spaces
$O_{y_{1},d-1},O_{y_{1},d},O_{y_{1},d-1},O_{y_{1},d}$

6:Compute $J:=(O_{y_{1},d}\cap O_{y_{2},d-1})+(O_{y_{1},d-1}\cap O_{y_{2},d})$

7:Let $\pi:V\to\mathbb{P}^{2}$ be the map defined by a basis of $J$

8:Return $\pi^{-1}:\mathbb{P}^{2}\to V$

If the ground field is finite, then we can find a point on $V$ by the
Algorithm Points on Surface.

Find Points on Surface

1:$V\subset\mathbb{P}^{N}$, an algebraic surface

2:a point $y\in V$

3:

4:found := False

5:While not found Do

6: Pick two random linear forms $L_{1},L_{2}$.

7: Compute the associated primes of $H:=I(V)+\langle L_{1},L_{2}\rangle$.

8: If $H$ has a prime component of a single point $p$ Then

9: found := True

10: $y$ := such a point

11:Return $y$

Our computational experiments indicate that the probability of a random pair
of linear forms to intersect $V$ in a point defined over the ground field is
$\frac{1}{9}$ for $d=3$ and $\frac{1}{16}$ for $d=4$. Hence the expected
number of loops before finding a point is $9$ for $d=3$ and $16$ for $d=4$.

######  Remark 4.6.

The assumption of the ground field is finite clearly stands against our
motivation for the whole reconstruction problem, namely computer vision. We
did the the computations in finite fields mainly to have explicit
computational examples. For computer vision applications, it would actually be
a good idea to change to symbolic-numeric computation, because images have
limited resolution leading to numerical errors. Numerically, computing points
on $V$ is not a Diophantine problem and is easier than computing points
defined over $\mathbb{Q}$. However, for many other subtasks, such as linear
normalization, symbolic-numeric methods are not yet available, and the
sensitivity to numerical errors is not known.

##  5 The Whole Algorithm

For the convenience of the reader, we give here a single algorithm that refers
to the functions in the previous section.

Algorithm PlanarMap

1:an irreducible plane curve $B\subset\mathbb{P}^{2}$ whose singularities are
nodes and cusps. Its degree can be factorized as $3d(d-1)$.

2:a generic planar map $f:\mathbb{P}^{2}\to\mathbb{P}^{2}$ that has $B$ as a
branching curve, if exists.

3:

4:If $d=2$ Then use Algorithm DiscToMapCase2.

5:If $d\geq 3$ Then

6: Compute the linear normalization $\tilde{B}$ of $B$ by Algorithm
LinearNormalization.

7: Compute the Veronese surface containing $\tilde{B}$ for two cases:

8: If $d=3$ Then use the Algorithm Veronese Surface of Degree3PlanarMap

9: If $d\geq 4$ Then use the Algorithm Veronese Surface of
DegreeBigger3PlanarMap)

10: Find two points in the Veronese surface $V$;

11:if ground field is finite, use Algorithm Points on Surface

12: Compute parametrization by Algorithm Parametrization of Veronese Surface.

13: Apply on Veronese surface V projection in the first three coordinates.

14: Return the composition of parametrization with the projection.

An implementation of degree two can be found in
https://github.com/HoxhajEriola/Planar-Map-of-Polynomial-of-Degree-
Two/tree/main

An implementation over finite fields can be found in
https://github.com/HoxhajEriola/Planar-Map-of-polynomial-
degree-3/blob/main/ALG3.m2 and https://github.com/HoxhajEriola/Planar-Map-of-
Polynomial-of-Degree-Greater-or-Equal-to-4/blob/main/ALG4.m2.

We extensively tested numerous random examples, specifically the field
$\mathbb{Z}/32003$. We implemented the algorithm in Macaulay2 and tested it on
a computer Intel(R) Core(TM) i5-1135G7 @ 2.40GHz. Below is a table presenting
the CPU timings, measured in seconds, for the algorithm steps concerning
degree 3 and 4. Algorithm DiscToMapCase2 was implemented in Maple, it runs in
less than 0.3 seconds.

Steps | deg 3 | deg 4
---|---|---
Adjoint ideal | 7.63589 | 200.621
Linear Normalization | 7.74667 | 222.676
Veronese Surface | 12.927 | 218.882
Points on Veronese | 10.9681 | 225.291
Parametrization | 65.0948 | 1145.51
Planar map | 87.91753 | 1126.1
Table 3: Timing of the algorithm for the cases of degree three and four

The examples 2 and 3 have been computed using the computer algebra system
Maple, starting with the map and then computing the branching curve.

Figure 2: The branching curve of a planar map of degree 3, with three real
nodes and six real cusps. It separates the points on the plane into three
distinct regions: the beige area, where each point has three preimage points;
the yellow region, where each point has five preimage points; and the blue sky
region, where each point has seven preimage points. A point which almost looks
like a singularity that is neither a node nor a cusp is magnified – it is a
cusp and a regular part passing close by. Figure 3: The branching curve of
another map of degree 3, with four real nodes and six real cusps. It separates
the points on the plane into three distinct regions: the blue sky region,
where each point has five preimage points; the yellow region, where each point
has six preimage points; and the beige region, where each point has three
preimage points.

## Acknowledgments

This work is part of GRAPES project that has received funding from the
European Union’s Horizon 2020 research and innovation programme under the
Marie Sklodowska-Curie grant agreement No 860843. Algorithm
LinearNormalization is based on ideas from a discussion with Matteo Gallet
(University of Trieste, Italy). We appreciated the discussions with him and
with Niels Lubbes (RICAM, Austrian Academy of Sciences).

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