Plane Polynomials and Hamiltonian Vector Fields Determined by Their Singular Points

Abstract

Let $\Sigma (f)$ be the singular points of a polynomial $f \in \mathbb {K}[x,y]$ in the plane $\mathbb {K}^2$, where $\mathbb {K}$ is $\mathbb {R}$ or $\mathbb {C}$. Our goal is to study the singular point map $\mathfrak {S}_d$, it sends polynomials f of degree d to their singular points $\Sigma (f)$. Very roughly speaking, a polynomial f is essentially determined when any other g sharing the singular points of f satisfies that $f = \lambda g$; here both are polynomials of degree d, $\lambda \in \mathbb {K}^*$. In order to describe the degree d essentially determined polynomials, a computation of the required number of isolated singular points $\delta (d)$ is provided. A dichotomy appears for the values of $\delta (d)$; depending on a certain parity, the space of essentially determined polynomials is an open or closed Zariski set. We compute the map $\mathfrak {S}_{3}$, describing under what conditions a configuration of 4 points leads to a degree 3 essentially determined polynomial. Furthermore, we describe explicitly configurations supporting degree 3 non essential determined polynomials. The quotient space of essentially determined polynomials of degree 3 up to the action of the affine group $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$ determines a singular $\mathbb {K}$-analytic surface.

1 Introduction

Very roughly speaking, the singular point map sends polynomials $f \in \mathbb {K}[x,y]$, of degree d, to their singular points

(1)

where $\Sigma (f) \doteq I(f_x, f_y)$ is the affine algebraic variety (not necessarily reduced) generated by the ideal of partial derivatives of f, see Definition 3. Under what conditions is a degree d polynomial $f \in \mathbb {K}[x,y]$ essentially determined by its singular points $\Sigma (f) \subset \mathbb {K}^2$? Our approximation route uses a finite dimensional framework. Let $\mathbb {K}[x,y]^0_{\le d}$ be the $\mathbb {K}$-vector space of polynomials having at most degree d ($\ge 3$) and a zero independent term, and let $\mathcal {P}=\{ (x_\iota , y_\iota ) \}$ be a configuration of n different points in the plane. The linear projective subspace of the polynomials with singular points at least in $\mathcal {P}$, denoted as

(2)

is well defined. We say that a polynomial f is essentially determined by $\mathcal {P}$ when $\mathcal {L}_d(\mathcal {P})$ is a projective point $\{ \lambda f \ \vert \ \lambda \in \mathbb {K}^*\}$, see Definition 4. All this leads us to the following.

Interpolation problem for singular points. Let $\mathcal {P} \subset \mathbb {K}^2$ be a configuration of n different points, we try to determine the projective subspace $\mathcal {L}_{d}(\mathcal {P})$ of polynomials of at most degree d with singular points at least in $\mathcal {P}$.

This problem has several novel features. The singular values $\{ c_\iota \} \subset \mathbb {K}$ of f can appear in different level curves $\{ f(x,y)-c_\iota =0 \}$; it is natural in Hamiltonian vector field theory and moduli spaces of polynomials, see Wightwick [18] and Fernández de Bobadilla [11]. This is the main difference from the widely considered problem of linear systems of curves in $\mathbb{C}\mathbb{P}^2$, e.g., Miranda [15] and Ciliberto [8].

Very roughly speaking, for degree $d \ge 3$ the relevant data are the cardinality and position of the configuration $\mathcal {P}$, as a candidate to be a singular point configuration $\Sigma (f)$. For degree 3, the prescription of 4 singular points is suitable. For degree $d \ge 4$, however, the generic configuration $\mathcal {P}$ with $(d-1)^2$ points is too restrictive. Thus, the fiber $\mathfrak {S}_d^{-1}(\mathcal {P})$ will be generically empty. It follows that the position of the configurations $\mathcal {P}$ coming from polynomials is the hardest part to characterize. At this first stage, we consider mainly $\mathcal {P}$ as isolated points of multiplicity one, Remark 1 provides an explanation. Our first result describes the role of cardinality $\delta (d)$ of $\mathcal {P}$ in Eq. (2), see Proposition 1.

Dichotomy of the required number of singular points. If the dimension of $\mathbb {K}[x,y]^0_{\le d}$ is odd (resp. even), then the configurations $\{\mathcal {P} \}$ with $\delta (d)$ points and $dim_\mathbb {K}(\mathcal {L}_d(\mathcal {P})) \ge 0$ determine an open (resp. closed) Zariski set in the space of configurations with $\delta (d)$ points, denoted as ${ Conf}(\mathbb {K}^2, \delta (d))$.

We compute the singular point map $\mathfrak {S}_{3}$. Thus, a description for the 4 singular point configurations $\{ \mathcal {P}\}$ with essentially determined polynomials is provided. Recall that the affine group $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$ acts on the space of polynomials, see Eq. (20). This action is rich enough and yet treatable for degree 3. Let

be an arrangement of six lines from two nested triangles, where of them is $\triangle =\{ (0,0), (1, 0), (0,1) \}$. See Fig. 1a. We prove the following result.

Theorem 1

Let f be a degree 3 polynomial having at least 4 singular points $\Sigma (f)$.

• (1) f is essentially determined if and only if up to affine transformation the four singular points are

  

• (2) f is not essentially determined if and only if up to affine transformation the four singular points are

  

  Moreover, in this case $\Sigma (f)$ can be four isolated points or two parallel lines.

In simple words, the 4-th point $(x_4, y_4)$ generically determines the polynomial f. We compute the fundamental domain for this $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$-action and obtain a tessellation of $\mathbb {K}^2= \{(x_4, y_4)\}$ with 24 tiles, as seen in Fig. 3. As expected, some interesting phenomena occur for configurations with nontrivial isotropy groups in $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$, Fig. 4 illustrates this. For degree $d \ge 3$, a particular family of configurations is the grid of $(d-1)^2$ points, from the intersection of two families of d parallel lines in $\mathbb {K}^2$, see Definition 8. They provide examples of nonessential determined polynomials with $(d-1)^2$ Morse singular points. A remaining open question is are these grids of $(d-1)^2$ points the unique mechanism in order to produce non essential determined Morse polynomials?

From the point of view of vector fields; under what conditions the singular points (i.e., zeros) of a Hamiltonian vector field determine it in a unique way? This is a very general and interesting issue in real and complex foliation theory, studied by Gómez-Mont and Kempf [13], Artes et al. [4], Campillo and Olivares [6] and Ramírez [17]. See Corollary 6. These related results are described in Sect. 7.

The content of this work is as follows. In Sects. 2 and 3, we study the problem of the dimension of linear systems for polynomials with singular points, using the degree as a parameter. In Sect. 4, we characterize polynomials essentially determined by their configurations of singular points; this proves Theorem 1. In Sect. 5, we focus on the degree 4 case. For each configuration of 6 points, we obtain a plane curve of degree 6 by parametrizing the essentially determined polynomials, see Proposition 2. Section 6 explores the behavior of pencils of Hamiltonian vector fields with common simple singularities.

2 Linear Systems $\mathcal {L}_{d} (\mathcal {P})$

Let $\mathbb {K}[x,y]^0_{\le d}$ (resp. $\mathbb {K}[x,y]^0_{= d}$) be the $\mathbb {K}$-vector space of polynomials with at most degree $d \ge 3$ (resp. the set for degree $=d$) and a zero independent term. Consider

(3)

from which the $\mathbb {K}$-dimension of $\mathbb {K}[x,y]_{\le d}^0$ is $\frac{1}{2}(d^2 + 3d)$ and its projectivization is

(4)

where $[\ \ ]$ denotes a projective class. Recall that

$$\begin{aligned} { Conf}(\mathbb {K}^2,n)= & {} \big \{ \, \mathcal {P}= \{ (x_1, y_1), \ldots , (x_n, y_n )\} \ \vert \nonumber \\{} & {} (x_\iota , y_\iota ) \ne (x_j, y_j )\quad \hbox {for}\quad \iota \ne j \big \} / Sym (n) \end{aligned}$$

(5)

is the space of unordered configurations of n points in $\mathbb {K}^2$, where the symmetric group Sym(n) in n elements acts by exchanging the points. The configuration space ${ Conf}(\mathbb {K}^2, n)$ is a $\mathbb {K}$-analytic manifold.

Definition 1

Given a configuration $\mathcal {P} \in { Conf}(\mathbb {K}^2,n ),$ the linear system of polynomials of at most degree d with singular points at least in $\mathcal {P}$ is the projective subspace

$$\begin{aligned} \mathcal {L}_{d}(\mathcal {P}) = \big \{ [f] \ \vert \ \mathcal {P} \subseteq \{f_x(x,y) =0\} \cap \{ f_y(x,y) =0 \} \big \} \subset Proj\big ( \mathbb {K}[x,y]^0_{\le d} \big ).\nonumber \\ \end{aligned}$$

(6)

In algebraic geometry language, $\{ f_x (x,y) = 0\}$ and $\{f_y (x,y) =0\}$ belong to the linear system of algebraic curves

$$\begin{aligned} \mathcal {L}_{d-1} \big ( - \Sigma _{\alpha = 1}^n (x_\iota , y_\iota ) \big ). \end{aligned}$$

See [8, 15]. In several places, however we consider $f_x$, $f_y$ as functions and not just as algebraic curves.

The polynomials of at most degree d, the Hamiltonian polynomial vector fields and the polynomial vector fields, of at most degree $d-1$, are related by linear maps

$$\begin{aligned} \begin{array}{ccccc} \mathbb {K}[x,y]^0_{\le d} &{} {\mathop {\longleftrightarrow }\limits ^{\cong }} &{} Ham(\mathbb {K}^2)_{\le d-1} &{} \longrightarrow &{} \mathfrak {X} (\mathbb {K}^2)_{\le d-1}\\ &{}&{}&{}&{} \\ f &{} \longleftrightarrow &{} X_f = -f_y \frac{\partial }{\partial x} + f_x \frac{\partial }{\partial y} &{} \longrightarrow &{} X_f. \end{array} \end{aligned}$$

In the space of Hamiltonian vector fields, $\mathcal {L}_{d} (\mathcal {P})$ determines a linear subspace

$$\begin{aligned} \{ \lambda X_f \ \vert \ \mathcal {P} \subseteq \mathcal {Z}(\lambda X_f),\ \lambda \in \mathbb {K}^* \} \subset Ham(\mathbb {K}^2)_{\le d-1}. \end{aligned}$$

Set theoretically, the zeros $\mathcal {Z}(\lambda X_f)$ of the vector field $X_f$ coincide with $\{f_x(x,y)=0 \} \cap \{f_y(x,y)= 0\}$.

Definition 2

Let $f \in \mathbb {K}[x,y]$ be a nonconstant polynomial. Over $\mathbb {K}=\mathbb {C}$, the Milnor number of $X_f$ at a zero point $(x_\iota , y_\iota ) \in \mathcal {Z}(X)$ is

$$\begin{aligned} \mu _{(x_\iota , y_\iota )}(X) = \dim _\mathbb {C}\frac{\mathcal {O}_{\mathbb {C}^2, (x_\iota , y_\iota )} }{ <-f_{y}, f_x >}, \end{aligned}$$

where $\mathcal {O}_{\mathbb {C}^2, (x_\iota , y_\iota )}$ is the local ring of holomorphic functions at the point $(x_\iota , y_\iota )$ and $<-f_{y}, f_x>$ is the ring generated by the partial derivatives.

Remark 1

1. Over $\mathbb {K}=\mathbb {C}$, if $(x_\iota , y_\iota )$ is an isolated singular point of f, then the notions of multiplicity for the intersection of the curves $\{f_x(x,y)=0\} \cap \{f_y(x,y)=0\}$ and the Milnor number for $X_f$ coincide; see [14, p. 174].

2. A priori, we consider each point $(x_\iota , y_\iota ) \in \mathcal {P}$ in (6) with multiplicity of intersection 1 for the algebraic curves $\{ f_x(x,y)= 0\}$ and $\{ f_y(x,y)=0\}$.

3. By Bézout’s theorem, the maximal number of isolated singularities of $X_f$ on $\mathbb {C}^2$ is $(d-1)^2$. In this case, all the affine singularities are of multiplicity 1.

4. Moreover, the maximal number of isolated singularities of $X_f$ extended to $\mathbb{C}\mathbb{P}^2$ is

$$\begin{aligned} (d-1)^2 + d. \end{aligned}$$

Here the upper bound d comes from the intersection of a generic projectivized level curve $\{ f=c\}$ with the line at infinity; see [6, 13] for the case of rational vector fields, which are not necessarily Hamiltonian.

Let $\mathbb {A}^2_\mathbb {K}= \hbox {Spec}\, \mathbb {K}[x,y]$ be the affine scheme of the affine plane $\mathbb {K}^2$, see [10, pp. 48–49].

Definition 3

The singular point map of degree d is

$$\begin{aligned} \begin{array}{rcl} \mathfrak {S}_d : \mathbb {K}[x,y]_{= d} &{}\longrightarrow &{} \hbox {Spec}\, \mathbb {K}[x,y]\\ &{}&{} \\ f &{}\longmapsto &{}\Sigma (f)= I(f_x, f_y), \end{array} \end{aligned}$$

(7)

sending a polynomial of degree d to its singular points $\Sigma (f)$ as an affine algebraic variety (not necessarily reduced) generated by the ideal of partial derivatives of f.

In fact, $\Sigma (f)$ can be understood as a subscheme, with support at the points $\{f_x(x,y)=0\}\cap \{f_y(x,y)= 0\}$, where the sheaf of ideals is defined by the germs of $I(f_x, f_y)$; compare with [6, 10, p. 100]. In a set theoretical language, $\Sigma (f)$ determines points and even algebraic curves. In the study of rational vector fields on $\mathbb{C}\mathbb{P}^2$ however, the case of foliations with singularities along curves is removed, see [6, 13].

Remark 2

The simplest case of the interpolation problem for singular points occurs when $\Sigma (f)$ is a finite set of points of multiplicity 1, i.e., $\{ f_x(x,y)=0\}$ and $\{f_y(x,y) = 0\}$ have transversal intersections. The $\Sigma (f)$ is a configuration in ${ Conf}(\mathbb {K}^2, n)$, for $0 \le n \le (d-1)^2$.

Our former task is as follows: Given a configuration $\mathcal {P}$, which is $dim_\mathbb {K}(\mathcal {L}_{d}(\mathcal {P}))$?

To be clear, three relevant data must be considered the degree d of the polynomials $\{ f \}$, the cardinality n and the position of the configuration $\mathcal {P}$. The following diagram explains:

The natural concepts are as follows.

Definition 4

Let $f \in \mathbb {K}[x,y]_{\le d}^0$ be a polynomial and let $\mathcal {P}$ be a configuration of n points in $\mathbb {K}^2$.

• (1) A polynomial f is essentially determined by $\mathcal {P}$ when $[f]=\mathcal {L}_{d}(\mathcal {P})$.

• (2) A polynomial f is nonessentially determined by $\mathcal {P}$ when $[f]\in \mathcal {L}_{d}(\mathcal {P})$ and $dim_\mathbb {K}( \mathcal {L}_{d}(\mathcal {P})) \ge 1$.

• (3) $\mathcal {P}$ is a forbidden configuration (for polynomials of at most degree d) when $\mathcal {L}_{d}(\mathcal {P})= \emptyset$.

• (4) The set of degree d essentially determined polynomials is

  $$\begin{aligned} \mathcal {E}_{d} \doteq \bigcup _{ \tiny {\mathcal {P} } } \mathcal {L}_{d} (\mathcal {P}) \subset Proj \big (\mathbb {K}[x,y]^0_{\le d} \big ), \end{aligned}$$

  (9)

  where the union is over all configurations $\{ \mathcal {P} \}$ such that $dim_\mathbb {K}(\mathcal {L}_{d} (\mathcal {P}))= 0$.

Remark 3

• (1) The strict set theoretical inclusion $\mathcal {P} \varsubsetneq \Sigma (f)$ can be satisfied for essentially determined polynomials f. For example, in the case of a product of three lines, one possesses a multiplicity 1, say $f=L_1^2 L_2$.

• (2) The set of degree 3 essentially determined polynomials $\mathcal {E}_{3}$ is a union of projective spaces; however, it is not a projective space, as Proposition 1 will show.

• (3) As expected, many of the projective classes in $\mathcal {E}_{d}$ arise from Morse polynomials. The converse is not true, as seen in Corollary 7.

Table 1 Dimensions and values for the interpolation problem

3 On the Number of Required Singular Points

A novel aspect of the interpolation problem for singular points is its cardinality; the configurations having a certain number $\delta (d)$ of points determine open or closed Zariski sets in $\mathbb {K}[x,y]_{\le d}^0$. As a key point, the dimension $\frac{1}{2} (d^3+ 3d)$ of $\mathbb {K}[x,y]_{\le d}^0$ can be even or odd. Starting with degree $d=4$, the pattern of these dimensions is 4-periodic; even, even, odd odd, $\ldots$. See the third column in Table 1.

Proposition 1

(A dichotomy of the number $\delta (d)$ of required singular points). Let $\mathbb {K}[x,y]^0_{\le d}$ be the set of polynomials having at most degree $d \ge 3$, and let

$$\begin{aligned} \delta (d) \doteq \left\{ \begin{array}{ll} \frac{1}{4}(d^2+3d-2)&{} \quad \hbox {when}\quad \frac{1}{2}(d^2+3d)~\hbox {is odd}, \\ \frac{1}{4}(d^2+3d) &{}\quad \hbox {when}\quad \frac{1}{2}(d^2+3d)~\hbox {is even}. \end{array}\right. \end{aligned}$$

(10)

1. If the dimension of $\mathbb {K}[x,y]^0_{\le d}$ is odd, then the configurations $\{\mathcal {P} \}$ with $\delta (d)$ points and $dim_\mathbb {K}(\mathcal {L}_d(\mathcal {P})) \ge 0$ determine an open Zariski set in ${ Conf}(\mathbb {K}^2, \delta (d))$.

2. If the dimension of $\mathbb {K}[x,y]^0_{\le d}$ is even, then the configurations $\{\mathcal {P} \}$ with $\delta (d)$ points and $dim_\mathbb {K}(\mathcal {L}_d(\mathcal {P})) \ge 0$ determine a closed Zariski set in ${ Conf}(\mathbb {K}^2, \delta (d))$.

Proof

Let $f(x,y) \in \mathbb {K}[x,y]^0_{\le d}$ be a polynomial as in (3). Assume that $\mathcal {P}=\{ (x_\iota , y_\iota ) \ | \ \iota =1,\ldots , n\}$ is set theoretically contained in $\Sigma (f)$. A priori, each point $(x_\iota , y_\iota ) \in \mathcal {P}$ will drop the dimension of the vector space $\mathbb {K}[x,y]^0_{\le d}$ by 2. In the linear framework, this leads to a linear system of 2n equations:

$$\begin{aligned} f_x(x_\iota ,y_\iota ) = f_y(x_\iota ,y_\iota ) =0, \quad \iota =1,\ldots , n, \end{aligned}$$

(11)

with $\{a_{\iota j}\}$ as variables. Following Bézout’s theorem for a moment, let us consider a configuration with $n=(d-1)^2$ points. We have a linear map

$$\begin{aligned}{} & {} \phi : \mathbb {K}[x,y]^0_{ \le d} \cong \mathbb {K}^{\frac{1}{2}(d^2 + 3d)} \longrightarrow \mathbb {K}^{2(d-1)^2} \nonumber \\{} & {} \quad f \longmapsto \big (f_x(x_1, y_1), \ldots , f_x(x_{(d-1)^2}, y_{(d-1)^2}), \nonumber \\{} & {} \qquad \qquad f_y(x_1, y_1), \ldots , f_y(x_{(d-1)^2}, y_{(d-1)^2})\big ). \end{aligned}$$

(12)

The interpolation matrix $\phi$ depends on $\mathcal {P}$, and for notational simplicity we omit this dependence. The matrix $\phi$ has $\frac{1}{2}(d^2 + 3d)$ columns, $2 (d-1)^2$ rows and a very particular shape because of the partial derivatives involved in it, see Eqs. (17), (33) for explicit examples with $d= 3$, 4.

For degree $d = 3$ and a configuration $\mathcal {P}$ of 4 points; however, then the rank of the matrix $\phi$ associated with $\mathcal {P}$ is 8 if and only if $dim_\mathbb {K}(\mathcal {L}_{3} (\mathcal {P}) )= 0$. If we consider degree $d \ge 4$, then the number of rows of $\phi$ is bigger than the number of columns. We must reduce the number n of required points in the configurations $\mathcal {P}$, this $n < (d-1)^2$. The number $\delta (d)$ in (10) determines two possibilities.

Case 1 in (10). For $\mathcal {P}$ with $\delta (d)=\frac{1}{4}(d^2+3d-2)$ points, the interpolation matrix $\phi$ has $\frac{1}{2}(d^2+3d)$ odd columns and $\frac{1}{2}(d^2+3d-2)$ even rows, for example for $(d+1)=3, 6, 7$. Moreover,

$$\begin{aligned} \hbox {(number of columns of }\phi )-1 = \hbox {(number of rows of }\phi ). \end{aligned}$$

The dimension of the kernel of $\phi$ is at least one, thus $dim_K(\mathcal {L}_d(\mathcal {P})) \ge 0$. There are $\frac{1}{2}(d^2+3d)$ minors $A_j$ from the matrix $\phi (x_1, y_1, \ldots , x_{\delta (d)}, y_{\delta (d)})$. The complement of the algebraic equations

$$\begin{aligned} \{ \Pi _j det(A_j(x_1, y_1, \ldots , x_{\delta (d)}, y_{\delta (d)})) =0 \} \subset { Conf}(\mathbb {K}, \delta (d)) \end{aligned}$$

describes the set of configurations having $dim_K (\mathcal {L}_d(\mathcal {P}))=0$, corresponding to the essentially determined polynomials. These configurations of $\delta (d)$ points in ${ Conf}_{\delta (d)} (\mathbb {K}^2)$ determine an open Zariski and dense set, which is the second part of assertion (1).

Case 2 in (10). The dimension of $\mathbb {K}[x,y]_{\le n}^0$ is even and we assume $\frac{1}{4}(d^2+3d) \in \mathbb {N}$ points in $\mathcal {P}$. The interpolation matrix $\phi$ is square of even size, and there are $\frac{1}{2}(d^2+3d)$ columns and rows; for example when $d= 4, 5$.

If we assume $\mathcal {P}$ such that $\{ det(\phi (x_1, y_1, \ldots , x_{\delta (d)},y_{\delta (d)}))\ne 0 \}$, then the only vector in the $\{ a_{\iota j} \}$ variables solving the linear system (11) is zero. The set of desired polynomials is empty.

The configuration with nonempty polynomials

$$\begin{aligned} \left\{ \mathcal {P} \ \vert \ det(\phi (x_1, y_1, \ldots , x_{\delta (d)},y_{\delta (d)}))\ne 0 \right\} \subset { Conf}(\mathbb {K}, \delta (d)) \end{aligned}$$

determines an algebraic set. $\square$

Recalling (4), the expected projective dimension of $\mathcal {L}_{d}(\mathcal {P})$, which is the linear system of polynomials of at most degree d with singular points at least in $\mathcal {P} \in { Conf}(\mathbb {K}^2, n)$, is

$$\begin{aligned} max \left\{ \frac{1}{2}(d^2+3d-2)-2n, \ -1 \right\} . \end{aligned}$$

In Sect. 5, we provide an alternative for studying the even dimension case in Proposition 1.

4 Essentially Determined Polynomials of Degree 3

4.1 A Linear System

In order to apply elementary methods, we introduce a very simple configuration of 4 points, depending essentially on the fourth one $(x_4, y_4)$. Secondly, we must find a polynomial $f(x_4, y_4, x,y)$ with a singular point set containing the above simple configuration. Let

$$\begin{aligned} \mathscr {A} \doteq \big \{ x y (x+y-1) (x+y) (x-1) (y-1) = 0\big \} \end{aligned}$$

(13)

be an arrangement of six $\mathbb {K}$-lines; it is illustrated in Fig. 1a.

Fig. 1

a The line arrangement $\mathscr {A}$ (of double lines) and the triangle $\triangle = \{V_1, \, V_2, \, V_3\}$. b The analogous objects under the linear map R, sending $\mathscr {A}$ to ${\texttt {A}}$ and $\triangle$ to $\Delta$

Lemma 1

Let

$$\begin{aligned} \mathcal {P}= & {} \{V_1=(0,0), \, V_2=(1,0), \, V_3 =(0,1), \, (x_4, y_4) \}\\ {}{} & {} \in { Conf}(\mathbb {K}^2, 4), \quad (x_4, y_4) \notin \mathscr {A}, \end{aligned}$$

be a fourt point configuration. The polynomial

$$\begin{aligned} f(x_4, y_4, x, y)&=( y_4^2(y_4-1)(-1+2 x_4 +y_4)(2x^3 -3x^2)\nonumber \\&\quad +x_4^2(x_4-1) ( -1 + x_4+2 y_4)(2y^3 - 3y^2)\nonumber \\&\quad -6x_4y_4(x_4-1)(y_4-1) (x^2y +xy^2-xy) ) a_6 \nonumber \\&\quad \in \mathbb {K}[x,y]_{=3}, \end{aligned}$$

(14)

for $a_6 \in \mathbb {K}^*$ is well defined and $\mathcal {P} = \Sigma \big ( f(x_4, y_4, x,y) \big )$.

It will be convenient to write Eq. (14) as a map to the space of polynomials

$$\begin{aligned} f(x_4, y_4, \ \ , \ \ ): \mathbb {K}^2 \backslash \mathscr {A} \longrightarrow \mathbb {K}[x,y]_{=3}, \quad (x_4, y_4) \longmapsto f(x_4, y_4, x,y). \end{aligned}$$

(15)

Proof

Let the following be a polynomial

$$\begin{aligned} f(x,y)= & {} a_1 x^3+a_2 x^2 y+a_3 x y^2+a_4 y^3+a_5 x^2 +a_6 x y+a_7 y^2+a_8 x+a_9 y \nonumber \\{} & {} \in \mathbb {K}[x,y]^0_{\le 3}. \end{aligned}$$

(16)

For notational simplicity, only one subindex $a_{\iota }$ is considered. Let $\{ (x_\iota , y_\iota ) \ \vert \ \iota = 1,\ldots , 4\}$ be an arbitrary configuration, and we require $(a_1, \ldots , a_9 )$ to be solutions of the linear system

$$\begin{aligned} \left( \begin{array}{ccccccccc} &{}&{}&{} \vdots &{}&{}&{}&{}&{} \\ 3x_\iota ^2 &{} 2x_\iota y_\iota &{} y_\iota ^2 &{} 0 &{} 2x_\iota &{} y_\iota &{} 0 &{} 1 &{} 0 \\ 0 &{} x_\iota ^2 &{} 2x_\iota y_\iota &{} 3y_\iota ^2 &{} 0 &{} x_\iota &{} 2y_\iota &{} 0 &{} 1 \\ &{}&{}&{} \vdots &{}&{}&{}&{}&{} \end{array} \right) \left( \begin{array}{c} a_1 \\ \vdots \\ a_9 \end{array} \right) = \left( \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} \right) . \end{aligned}$$

(17)

The interpolation matrix $\phi$ in (17) has 9 columns and 8 rows. The choice $\mathcal {P} = \{ (0,0), (1,0), (0,1), (x_4, y_4)\}$ determines the linear system with only two equations

$$\begin{aligned} \begin{array}{c} f_x (x,y) = 3a_1 x^2 + 2a_2 xy + a_3 y^2 + 2a_5 x + a_6 y + a_8 =0,\\ f_y (x,y) = a_2 x^2 + 2a_3 xy + 3a_4 y^2 + a_6 x + 2a_7 y + a_9=0. \end{array} \end{aligned}$$

Obviously, $(0,0) \in \mathcal {P}$ implies the vanishing of the linear part $f_x (0, 0) = a_8 = 0 = a_9 =f_y(0,0)$. The linear conditions imposed by (1, 0) and (0, 1) are

$$\begin{aligned}{} & {} \left\{ \begin{array}{lccl} f_x (1, 0) = 3a_1 + 2a_5 =0 &{}&{}&{} a_1 = - \frac{2}{3} a_5, \\ f_y (1, 0) = a_2 + a_6 =0 &{}&{}&{} a_6 = - a_2, \end{array} \right. \\{} & {} \left\{ \begin{array}{lccl} f_x (0, 1) = a_3 + a_6 = 0 &{} &{} &{} a_6= -a_3, \\ f_y (0, 1) = 3a_4 + 2a_7 = 0 &{} &{} &{} a_4 = \frac{-2}{3} a_7. \end{array} \right. \end{aligned}$$

The solution of this system

$$\begin{aligned} \begin{array}{ll} f(x_4, y_4, x, y)&{}= a_6 \left( \dfrac{ y_4 (-1+2 x_4 +y_4)}{3x_4 (x_4-1) }x^3 - x^2y - xy^2 + \dfrac{x_4( -1 + x_4+2 y_4)}{3 y_4(y_4-1) }y^3 \right. \\ &{} \left. \quad + \dfrac{y_4(1-2 x_4 -y_4)}{2 x_4 (x_4-1) }x^2 + xy + \dfrac{x_4(1-x_4 -2 y_4)}{2 y_4 (y_4-1) }y^2\right) \\ &{}\in \mathbb {K}[x,y]_{=3} \end{array} \end{aligned}$$

(18)

has rational coefficients. If we normalize, we get Eq. (14). $\square$

Corollary 1

Let

$$\begin{aligned} \mathcal {P}_1= \{(0,0), (1,0), (0,1), R_1 \doteq (1, 1) \} \in { Conf}(\mathbb {K}^2, 4) \end{aligned}$$

be a four point configuration, and then $dim_\mathbb {K}(Proj(\mathcal {L}_3(\mathcal {P}_1 )))=1$.

We say that, $R_1=(1,1)$ is a rhombus point; see Fig. 1.

Proof

By replacing in $\phi$ the points in $\mathcal {P}_1$, a direct calculation shows that the equivalent $9 \times 8$ matrix has a rank 7, where the null space of $\phi$ is given by the vectors $(0, 0, 0, -2/3, 0, 0, 1, 0, 0)$ and $(-2/3, 0, 0, 0, 1, 0, 0, 0, 0)$. The linear combination of the corresponding polynomials leads to

$$\begin{aligned} f( {\texttt {a,d,}}x,y)= a \left( 2x^3 - 3x^2 \right) + d \left( 2y^3 - 3y^2 \right) ,\quad [a,\,d] \in \mathbb {K}\mathbb {P}^1. \end{aligned}$$

(19)

$\square$

Remark 4

Behavior of the linear system at $\mathscr {A}$. Let $\mathcal {P}=\{ (0,0), (1,0), (0,1), (x_4, y_4) \}$ be a configuration.

1. If $(x_4, y_4)$ tends to be in a line

$$\begin{aligned} L_\alpha \subset \mathscr {A} \big \{ \mathscr {A}\big \} \backslash \{ R_1=(1,1), R_2 =(-1,1), R_3= (1,-1) \}, \end{aligned}$$

then the polynomial $f(x_4, y_4,x,y)$ in (17) has two lines of singular points in the respective pair of parallel $\mathbb {K}$-lines $L_\alpha$, $L_\beta$, in the arrangement $\{\mathscr {A}(x,y) = 0\}$. Figure 4 provides a sketch up to affine transformations.

2. If $(x_4, y_4)$ tends to be the vertex $(0,0) \in \triangle$, then the polynomial $f(x_4, y_4,x,y)$ in (16) becomes

$$\begin{aligned} f(0,0, x,y) = \frac{1}{3}(x^3 + y^3) - (x^2 y + xy^2) - \frac{1}{2} (x^2 + y^2) + xy. \end{aligned}$$

As is expected, the curve $\{f(0,0, x, y)= 0 \}$ has a cusp of multiplicity 2 at (0,0), see Fig. 4. The same is valid if $(x_4, y_4)$ tends to be any other vertex (1, 0), (0, 1) of $\triangle$. Figure 4 shows f(1, 0, x, y), corresponding to $V_2=(0,1)$ denoted as ${\texttt {V}}_2$ in the figure.

Remark 5

Let $\mathcal {P}$ be any configuration of four points. Thus $\mathcal {L}_3(\mathcal {P}) \ne \emptyset$: there exists a nonconstant degree 3 polynomial with singular points at least in $\mathcal {P}$.

4.2 Affine Classification of Quadrilateral Configurations

We now study the independence of the previous results §4.1, with respect to the coordinate system.

A valuable tool in the study of polynomials of degree 3 is the action of the group of affine automorphisms of $\mathbb {K}^2$, say $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$. It is a six $\mathbb {K}$-dimensional Lie group. Let $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$ acts on the space of polynomials of degree d as

$$\begin{aligned} \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2) \times \mathbb {K}[x,y]_{=d} \longrightarrow \mathbb {K}[x,y]_{=d}, \ \ \ (T,f)\longmapsto f\circ T. \end{aligned}$$

(20)

This action is rich enough and yet treatable. The affine group acts on configurations such as

$$\begin{aligned} \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2) \times { Conf}(\mathbb {K}^2, n) \longrightarrow { Conf}(\mathbb {K}^2, n), \ \ \ (T,\mathcal {P}) \longmapsto T^{-1}(\mathcal {P}) . \end{aligned}$$

(21)

Thus, if $f \in \mathbb {K}[x,y]_{=d}$ has n isolated singular points, say $\mathcal {P} \in { Conf}(\mathbb {K}^2, n)$, then $f \circ T$ has singular points at $T^{-1}(\mathcal {P})$. Hence, a useful associated object is the quotient space of quadrilateral configurations up to affine transformations.

Definition 5

The space of generic quadrilateral configurations is

$$\begin{aligned} \mathcal {Q}= & {} \left\{ \mathcal {P}_0= \{(x_{1\, 0}, y_{1 \, 0}), \ldots , (x_{4 \, 0}, y_{4\, 0})\} \ \Big \vert \ \begin{array}{l} \hbox {quadrilateral configurations} \\ \hbox {having no three collinear vertices } \\ \hbox {or determining two parallel lines} \end{array} \right\} \nonumber \\{} & {} \subsetneq { Conf}(\mathbb {K}^2, 4). \end{aligned}$$

(22)

Note that a quadrilateral configuration $\mathcal {P}_0$ does not have order. It determines several quadrilaterals, i.e., with a cyclic order in its vertices. Let

$$\begin{aligned} \triangle= & {} \{V_1=(0, 0), V_2 = (1,0), V_3=(0,1) \}, \\ \Delta= & {} \{ {\texttt {V}}_1=(0, 0), {\texttt {V}}_2 = (1,0), \texttt {V}_3=(1/2, \sqrt{3}/2) \} \end{aligned}$$

be two triangles. Consider a linear transformation $R \in GL(2, \mathbb {K})$ such that $R(\triangle ) =\Delta$, $R(V_2) = \texttt {V}_2$ and $R(V_3) = \texttt {V}_3$, see Fig. 1. The affine symmetries of $\Delta$,

$$\begin{aligned} Sym(3)= \{ \sigma _\alpha \in \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2 ) \ \vert \ \sigma _\alpha (\Delta ) = \Delta , \ \alpha \in 1, \ldots , 6 \}, \end{aligned}$$

(23)

are isomorphic to the symmetric group of order 3: with three reflections $\sigma _2, \sigma _4, \sigma _6$ (with the axis in the lines $\texttt {N}_1, \, \texttt {N}_2, \, \texttt {N}_3$) and their products $\sigma _1=id, \sigma _3, \sigma _5$; see Fig. 1b. By abusing the notation, Sym(3) also denotes the affine symmetries of $\triangle$.

Thus, we use three coordinate systems as follows. Let $\mathcal {P}_0= \{ (x_{1\,0}, y_{1\,0}), \ldots , (x_{4\,0}, y_{4\,0})\}$ as in (22). By using the affine action, we reduce $\mathcal {P}_0$ to $\{(x_4, y_4) \}$ or $\{ (\texttt {x}_4, \texttt {y}_4) \}$. There are affine maps $T_j \in \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$ as follows

By notational simplicity, we also denote by $\mathcal {P}$ the configuration on the right side.

A key point is the number of affine maps $\{ T_j \}$, depending on $\mathcal {P}_0$ to be computed in Corollary 2.

In accordance with Figs. 1 and 3, the triangles $\triangle$, $\Delta$ determine the points, line arrangements and regions below.

${\varvec{\cdot }}$ Three rhombus points $R_1, \, R_2, \, R_3$ (resp. $\texttt {R}_1, \ \texttt {R}_2, \, \texttt {R}_3$).

${\varvec{\cdot }}$ Four center points $C_1, \, C_2, \, C_3, \, C_4$ (resp. $\texttt {C}_1, \, \texttt {C}_2, \, \texttt {C}_3, \, \texttt{C}_4$).

${\varvec{\cdot }}$ A six line arrangement $\mathscr {A}=L_1\cup \cdots \cup L_6$ (resp. $\texttt {A}=\texttt {L}_1 \cup \cdots \cup \texttt{L}_6$) sketched as six double lines. $\mathscr {A}$ was already described in the introduction and in (13).

${\varvec{\cdot }}$ A six line arrangement $\mathscr {B}=N_1\cup \cdots \cup N_6$ (resp. $\texttt {B}=\texttt {N}_1 \cup \cdots \cup \texttt{N}_6$) sketched as six blue lines, where $N_1, \, N_2, \, N_3$ are the axis of symmetry of $\triangle$. The lines $N_1, \, N_2, \, N_3$ are fixed under $\sigma _1, \, \sigma _2, \, \sigma _3$ in $\hbox { Aff}\hspace{1.42271pt}(\mathbb {R}^2)$ leaving invariant $\triangle$. The lines $N_4, N_5, N_6$ determine the triangle $C_1, C_2, C_3$.

Naturally, these points and arrangements correspond to under the map R in (24).

${\varvec{\cdot }}$ In case $\mathbb {K}= \mathbb {R}$, we have two open connected regions in $\mathbb {R}^2$; convex quadrilateral configurations when $(x_4, y_4) \in Q_1$ (aquamarine) and nonconvex for $Q_2$ (magenta).

Analogously, we have $\texttt {Q}_1=R(Q_1)$ and $\texttt {Q}_2= R(Q_2)$. Moreover, the boundary of $Q_1$, $Q_2$ shall be described by using the isotropy of the respective configurations.

Lemma 2

Let $\mathcal {P} \in \mathcal {Q}$ be a generic quadrilateral configuration in $\mathbb {K}^2$ as in (22). If the affine isotropy group of $\mathcal {P}$

$$\begin{aligned} \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{\mathcal {P}} \doteq \{ T \in \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2) \ \vert \ T^{-1}(\mathcal {P}) = \mathcal {P} \} \end{aligned}$$

is nontrivial, then it is isomorphic to one of the subgroups below.

Case 1. $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{\mathcal {P}} \cong Sym(3)$ if and only if up to affine transformation $\mathcal {P}$ has vertices in an equilateral triangle and its center.

Case 2. $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{\mathcal {P}} \cong \mathbb {Z}_2 \times \mathbb {Z}_4$ if and only if up to affine transformation $\mathcal {P}$ is a rhombus; its vertices determine a pair of two parallel lines.

Case 3. $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{\mathcal {P}} \cong \mathbb {Z}_2$ if and only if up to affine transformation

(i) $\mathcal {P} = \{(0,0), (1,0), (1/2, \sqrt{3}/2), (\texttt{x}_4,\texttt {y}_4) \}$ where $(\texttt {x}_4, \texttt {y}_4)$ is a fixed point under the reflection $\sigma ^{\prime }_2$ with axis $\texttt {N}_2$ in the isotropy of the triangle $\Delta$ and it is different of the center of $\Delta$,

(ii) Conversely, $\mathcal {P}$ is a trapezoid and its vertices determine two parallel lines, different from a rhombus. $\square$

Corollary 2

Let $\mathcal {P}_0$ be a generic quadrilateral configuration. The following assertions are equivalent.

• (1) $\mathcal {P}_0$ has a trivial isotropy group $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{\mathcal {P}_0} =id$.

• (2) There are 24 affine transformations $R \circ T_j$ in (24), sending $\mathcal {P}_0$ to $\{(0,0), (1,0)$, $(1/2, \sqrt{3}/2), (\texttt {x}_4, \texttt {y}_4) \}$. $\square$

Now we compute the orbit $\{ R \circ T_j (\mathcal {P}_0) \}_{j=1}^{24}$ in terms of the fourth point in $\{(\texttt {x}_4, \texttt{y}_4) \} \in \mathbb {R}^2$. Certainly, the orbit has obvious elements given by the affine symmetries of $\Delta$. The nonintuitive transformations between quadrilateral configurations $R \circ T_j (\mathcal {P}_0)$ are computed in the following result.

Lemma 3

Let

$$\begin{aligned} \{ \underbrace{(0,0), (1,0), (1/2, \sqrt{3}/2),}_{\Delta } \texttt {V}_4 =(\texttt {x}_4, \texttt {y}_4) \} \end{aligned}$$

be a generic quadrilateral configuration and consider a vertex $\texttt {V}_j \in \Delta$. There exist three $\mathbb {K}$-rational diffeomorphisms (different from the identity)

$$\begin{aligned} \texttt {g} (\texttt {V}_{j}, \ \ ): \mathbb {K}^2 \backslash \texttt {A} \longrightarrow \mathbb {K}^2 \backslash \texttt {A} , \quad \texttt {V}_4 \longmapsto \texttt {g}(\texttt{V}_j, \texttt {V}_4) , \quad j \in 1, 2, 3, \end{aligned}$$

(25)

such that the quadrilateral configurations

$$\begin{aligned} \{(0,0), (1,0), (1/2, \sqrt{3}/2), \texttt {V}_4 \} \ \ \hbox { and }\ \ \{(0,0), (1,0), (1/2, \sqrt{3}/2), \texttt {g}(\texttt {V}_j, \texttt {V}_4) \} \end{aligned}$$

are $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$-equivalent.

We note that $\texttt {g}(\texttt {V}_j, \ \ )$ are nonaffine maps.

Proof

The choice of one vertex $\texttt {V}_j \in \Delta$, determines an opposite side $\Delta$. Without loss of generality, we consider the vertex $\texttt {V}_2= (1,0) \in \Delta$ and $\texttt {L}_1 =\{\texttt {y}- \sqrt{3}{} \texttt {x}= 0\} \subset \texttt {A}$ is the opposite side; see Fig. 2.

For fixed $j=2$, we consider $\texttt {V}_4$. Let $\texttt {L}$ be the line by $\texttt {V}_4$ and $\texttt {V}_2$; $\texttt {L}$ is the red line in Fig. 2. We assume that $\texttt {L}_1$ and $\texttt {L}$ are nonparallel.

There exists a unique $\mathbb {K}$-affine embedding

$$\begin{aligned} \mathfrak {j}: \mathbb {K}\longrightarrow \mathbb {K}^2, \quad \hbox {with}\quad \mathfrak {j}(\mathbb {K})= \texttt {L}, \quad \mathfrak {j}(1) = \texttt {V}_2, \quad \mathfrak {j}(0) = \texttt {L}_1 \cap \texttt {L} \doteq \texttt {0}. \end{aligned}$$

The definition of the map in $\texttt {L}$ is

$$\begin{aligned} \texttt {g}(\texttt {V}_2, \ \ ) : \texttt {L}\backslash \mathfrak {j}(0) \longrightarrow \texttt {L} \backslash \mathfrak {j}(0), \quad \texttt {V}_4 \longmapsto \mathfrak {j} \left( \frac{1}{ \mathfrak {j}^{-1} (\texttt{x}_4, \texttt {y}_4)} \right) . \end{aligned}$$

(26)

Secondly, we shall extend this definition for $\texttt {V}_4 \in \mathbb {K}^2 \backslash \texttt {L}_1$. In order to avoid cumbersome computations, the coordinates $\{ (x,y) \}$ in (24) are more suitable. Assume $\mathcal {P}= \{(0,0), (1,0), (0,1), (x_4, y_4) \}$, the vertex is $V_2=(1,0) \in \triangle$ and $L_1 = \{x_4=0 \}$ is the opposite side. The analogous definition provides the rational map

$$\begin{aligned} \begin{array}{rcl} g(V_2, \ \ ): \mathbb {K}^2 \backslash \{x_4 (x_4-1)= 0 \} &{} \longrightarrow &{} \mathbb {K}^2 \backslash \{x_4 (x_4-1)= 0 \},\\ V_4 = (x_4, y_4) &{} \longmapsto &{} \Big ( \dfrac{1}{x_4}, \dfrac{-y_4 +y_4 x_4}{x_4 -1}\Big ). \end{array} \end{aligned}$$

(27)

It enjoys the properties described below.

${\varvec{\cdot }}$ $g(V_2, \ \ )$ is a birational map of $\mathbb {K}^2$.

${\varvec{\cdot }}$ $g^{-1}(V_2, \ \ ) = g(V_2, \ \ )$, it is an involution.

Fig. 2

The point $\texttt {g} ( \texttt {V}_2, \texttt {V}_4 )$ determines an affine map T between generic quadrilateral configurations

${\varvec{\cdot }}$ The point $V_2$ and the line $\{ x=-1\}$ are fixed under $g(V_2, \ \ )$.

${\varvec{\cdot }}$ The poles of the map $g(V_2, \ \ )$ are localized at $\{ x=0\}$ and $\{ x-1=0\} \backslash \{(0,1) \}$. Thus, strictly speaking the map is a $\mathbb {K}$-analytic diffeomorphism on $\mathbb {K}^2 \backslash \{ x(x-1)=0\}$. In the synthetic definition (26), $\texttt {L}_1$ and $\texttt {L}$ are nonparallel. This construction originates the pole of $g(V_2, \ \ )$ at $\{x-1=0\}$.

${\varvec{\cdot }}$ A straightforward computations shows that the line arrangements $\mathscr {A}$ and $\mathscr {B}$ (double and blue lines in Fig. 3) are poles or remain invariants under $g_2(V_2, \ \ )$.

In summary, we define (26) as

$$\begin{aligned} \texttt {g}(\texttt {V}_2, \ \ ) \doteq R \circ g(V_2, \ \ ) \circ R^{-1}. \end{aligned}$$

Finally, given $\texttt {V}_4$ and $\texttt {g}(\texttt {V}_2, \texttt {V}_4)$, there exists a unique transformation $T \in \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$, which leaves the line $\texttt {L}_1$ fixed so that $T(\texttt {V}_4) = \texttt{g}(\texttt {V}_2, \texttt {V}_4)$; see Fig. 3. Under T, the quadrilateral configurations

$$\begin{aligned} \{ (0,0), (1,0), (1/2, \sqrt{3}/2), \texttt {V}_4 \}\quad \hbox {and}\quad \{ (0,0), (1,0), (1/2, \sqrt{3}/2), T (\texttt {V}_4)\} \end{aligned}$$

are affine equivalent.

The other vertices of the triangle $\Delta$ determine rational maps $\texttt {g}(\texttt {V}_1,), \, \texttt {g}(\texttt {V}_3,)$, both enjoy analogous properties. $\square$

Remark 6

Three blue lines in Fig. 3 correspond to the fixed points under the reflection symmetries Sym(3) of $\Delta$. By using (26), the complete configuration of six blue lines $N_1, \ldots , N_6$ is invariant under the three transformations $\texttt {g}(\texttt {V}_j, \ \ )$. We leave this assertion for the reader.

Fig. 3

The plane $\mathbb {R}^2 \backslash \mathtt A$ with coordinates $\{\texttt {x}_4, \texttt {y}_4 \}$ parametrizes the quadrilateral configurations $\{ \texttt {V}_2, \texttt {V}_2,\texttt {V}_3, \texttt {V}_4=(\texttt{x}_4, \texttt {y}_4) \}$. The pair tile $\mathtt Q= \texttt {Q}_1 \cup \texttt{Q}_2$ is a fundamental domain for the moduli space of quadrilateral configurations, up to $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$-equivalence. There are 24 copies of the fundamental region $\texttt {Q}$. We colored $\texttt {Q}_2$ and its copies with pink or blue (resp. $\texttt {Q}_1$ and its copies aquamarine or magenta) tiles for strictly convex (resp. non convex) quadrilateral configurations

Lemma 4

• 1. The quotient space of generic quadrilateral configurations up to affine transformations, given by

  $$\begin{aligned} \pi : \mathcal {Q} \longrightarrow \mathcal {Q} / \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2),\quad \{(x_{1\,0}, y_{1\, 0}), \ldots , (x_{4\, 0}, y_{4\, 0} ) \} \longmapsto [ (\texttt {x}_4, \texttt {y}_4 ) ], \end{aligned}$$

  (28)

  is a $\mathbb {K}$-analytic surface $\mathtt Q$.

• 2. For $\mathbb {K}= \mathbb {C}$, the quotient $\mathtt Q$ is a connected complex surface.

• 3. For $\mathbb {K}= \mathbb {R}$, the quotient has two connected components $\mathtt Q = \texttt {Q}_1 \cup \texttt {Q}_2$ and singular points with local models $\mathbb {K}^2 / \mathbb {Z}_2$ or $\mathbb {K}^2 /Sym(3)$.

Some comments are in order. Figure 3 illustrates the fundamental domains for $\pi$ over $\mathbb {K}=\mathbb {R}$. The double lines $\texttt{A}= \texttt {L}_1 \cup \cdots \cup \texttt {L}_6$ in Figs. 1, 2, 3 and 4 correspond to forbidden positions for $(\texttt {x}_4,\texttt {y}_4)$. Moreover, $(\texttt {x}_4,\texttt{y}_4) \in \texttt {Q}_1$ determines a nonconvex quadrilateral configuration; $(\texttt {x}_4,\texttt {y}_4) \in \texttt {Q}_2$ determines a strictly convex quadrilateral configuration.

Proof

The set theoretical construction of the quotient is simple, and we describe its projection $\pi$ in (28). Given $\mathcal {P}_0 \in \mathcal {Q}$, we apply an affine transformation $R \circ T_j$ in (24) sending it to

$$\begin{aligned} R \circ T_j(\mathcal {P})= \{(0, 0), (1, 0), (1/2, \sqrt{3}/ 2), \texttt {V}_4=(\texttt {x}_4, \texttt {y}_4) \}. \end{aligned}$$

Case 1. The isotropy is trivial $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{\mathcal {P}} =id$. There are exactly 24 different choices for $R\circ T_j$, as in Lemma 2; we have that $\pi$ has as a target $\mathbb {K}^2=\{(\texttt {x}_4, \texttt {y}_4)\}$.

In order to describe its analytic properties, recall that the Klein four-group K is isomorphic to $\mathbb {Z}_2 \times \mathbb {Z}_2$. It is such that each element is self-inverse (composing it with itself produces the identity) and composing any two of the three nonidentity elements produces the third one; see [2, p. 87]. Moreover, the group Sym(4) is of order 24, having a Klein four-group K as a proper normal subgroup; thus $Sym(3) = Sym(4)/K$. We recognize

$$\begin{aligned} K =\{ id, \, \texttt {g}(\texttt {V}_j, \ \ ) \ \vert \ j \in 1,2,3 \} \end{aligned}$$

as the group in Lemma 3. Recall (23) and consider the homomorphism given by

$$\begin{aligned} \varphi : Sym(3) \longrightarrow Aut(K), \ \ \ \sigma \longmapsto \sigma ^{-1}_\alpha \circ \texttt {g}(\texttt {V}_j, \ \ ) \circ \sigma _\alpha (\texttt {x}_4, \texttt {y}_4). \end{aligned}$$

The semidirect product of K and Sym(3) determined by $\varphi$ is $Sym(4) = K \rtimes _\varphi Sym(3),$ see [2, p. 133]. Hence, we have a representation of Sym(4) in the birational transformations of $\mathbb {K}^2\backslash \texttt{A}$ and

$$\begin{aligned} \mathtt Q = \frac{\mathcal {Q}}{\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)} = \frac{\mathbb {K}^2 \backslash \texttt{A} }{Sym(4)} \end{aligned}$$

(29)

is the quotient space. See [16] for a general theory of the quotients of complex manifolds under a discontinuous group of automorphisms. Assertion (1) is done.

For assertion (2), we assume $\mathbb {K}=\mathbb {C}$; note that $\mathbb {K}^2 \backslash \texttt {A}$ is a connected complex manifold. The local behavior of this complex quotient at the points with nontrivial isotropy $\mathbb {Z}_2$ at the lines $\texttt {N}_1, \, \texttt {N}_2, \, \texttt {N}_3$ is known to be nonsingular (because of Chevalley [7], see also [12]). For $\texttt {C}$ the isotropy is Sym(3) and the same references describe the local structure of the quotient.

For assertion (3), we assume $\mathbb {K}=\mathbb {R}$, clearly the convexity or non convexity of a quadrilateral configurations are affine invariants, from where there are two connected components. At the points $\texttt{C}, \ldots ,\texttt {C}_4$ and lines $\texttt {N}_1, \, \texttt {N}_2, \, \texttt{N}_3$ where the isotropy of the quadrilateral configurations is nontrivial, the quotient (29) has singularities; it is an orbifold. $\square$

As final step in the proof of Theorem 1, we consider the action on projective classes

$$\begin{aligned} \mathcal {A}:\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2) \times Proj(\mathbb {K}[x,y]_{=3}) \longrightarrow Proj(\mathbb {K}[x,y]_{=3}), \ \ \ (T, [f]) \longmapsto [f \circ T].\nonumber \\ \end{aligned}$$

(30)

This action provides an $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$-bundle structure on $\mathbb {K}[x,y]_{=3}$. Denote the stabilizer or isotropy group of $[f] \in Proj(\mathbb {K}[x,y]_{=3})$ by

$$\begin{aligned} \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{[f]} \doteq \{T\in \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2) \ \vert \ f \circ T = \lambda f, \ \lambda \in \mathbb {K}^* \}. \end{aligned}$$

Equations (15) and (24) provide bijective correspondence between the generic quadrilateral configuration in $(\texttt {x}_4, \texttt {y}_4) \in \mathbb {K}^2 \backslash \texttt {A}$ and projective classes of polynomials $[f( R^{-1}(\texttt {x}_4,\texttt {y}_4),x,y)]$. If $\mathcal {P} \in \texttt{Q}$, then we verify that the isotropy of the quadrilateral configuration $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_\mathcal {P}$ is isomorphic to $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{[f]}$. Thus, we have a section

$$\begin{aligned} f \circ R^{-1}: \mathbb {K}^2\backslash \{ \texttt {A} \} \longrightarrow Proj(\mathbb {K}[x, y]_{=3}), \ \ \ (\texttt {x}_4, \texttt {y}_4) \longmapsto [f( R^{-1}(\texttt {x}_4,\texttt {y}_4),x,y)] \end{aligned}$$

and a diagram

where $\pi$ is the projection of classes from the action (30). The $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K})$-orbit of a projective class $[f] \in \mathbb {K}[x,y]_{=3}$ is homeomorphic to $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2) / \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{[f]}$. Obviously, $\mathbb {K}[x,y]_{=3, id}$ is open and dense in $\mathbb {K}[x,y]_{=3}$.

The proof of assertion 1, Theorem 1 is done.

Remark 7

It is well known (as a seen for instance in [9, p. 53]) that if we consider

$$\begin{aligned} \mathbb {K}[x,y]_{=3,id} \doteq \{ f \in \mathbb {K}[x,y]_{=3} \ \vert \ \hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)_{f} ={id}\}, \end{aligned}$$

then the restricted action in $\mathbb {K}[x,y]_{id}$, determines a principal fiber $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$-bundle structure. In particular, the quotient $\mathbb {K}[x,y]_{=3, id}/\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$ is a two dimensional $\mathbb {K}$-analytic manifold.

Remark 8

For $\mathbb {K}= \mathbb {R}$, the fundamental domain $\texttt {Q}_1 \cup \texttt {Q}_2$ determines the bifurcation diagram of the respective Hamiltonian vector fields, see Fig. 4. By construction, $\texttt {Q}_1$ has two boundaries and one vertex $\texttt{C}$ and $\texttt {Q}_2$ has one boundary (without extreme points).

Fig. 4

Bifurcation diagram of the real Hamiltonian vector fields $X_{f \circ R^{-1}}$ according to the position of four singular points in the fundamental region $\mathtt Q$. At the rhombus point $\mathtt R_1$, the configuration of four points $\mathcal {P}=\{(0,0), (1,0), (0,1), \texttt {R}_1=(1,1) \} \subset \Sigma (f_\theta )$ is common; see Example 6. The upper row illustrates the topology of $\{ f_\theta (x,y) \ \vert \ \theta \in [0,\pi /2 ]\}$. A saddle connection bifurcation occurs for $\theta = \pi / 4$. See https://github.com/alexander-arredondo/Mathematica-code-for-Essentially-determined-polynomials-of-degree-3/commit/e6a08f9a20da7b23d7a72beff8290af3a23260dc for a code animation in Mathematica of this situation

We summarize the results in Table 2.

Table 2 Dimension, generators and isotropy for $\mathcal {L}_3(\mathcal {P})$, where $\mathcal {P}$ is a configuration with 4 points (3 simple points and a double one in the last row)

Example 1

Relation to the classification of cubic plane curves. The Hesse pencil of cubic curves is

$$\begin{aligned} \{ z^3 + x^3 + y^3 - 3 \mu zxy = 0\},\quad \hbox {resp.}~\{ x^3 + y^3 - 3 \mu xy +1 = 0\},\quad \mu \in \mathbb {C}^*, \end{aligned}$$

in the projective plane $\mathbb {C}\mathbb {P}^2= \{ [z,x, y] \}$, resp. the affine plane; see [3]. The key property is that any nonsingular cubic plane is projectively equivalent to a member of the Hesse pencil. The singular points of the affine Hesse polynomial

$$\begin{aligned} f(\mu , x, y )= x^3 + y^3 - 3 \mu xy +1 \end{aligned}$$

determine a generic quadrilateral configuration

$$\begin{aligned} \left\{ (0,0), (\mu ,\mu ), \big (-\zeta _1 \mu ,\, \zeta _2 \mu \big ), \big (\zeta _2 \mu , \, -\zeta _1 \mu \big ) \right\} \subset \mathbb {C}^2 \backslash \mathbb {R}^2, \end{aligned}$$

where $\{ 1, \zeta _2, \zeta _3 \}$ are the cube roots of unity. In order to translate it to our language, up to the linear transformation $M_\mu : \mathbb {C}^2 \longrightarrow \mathbb {C}^2, \ (x,y) \longmapsto \big (\mu x - \zeta _2 \mu y, \mu x + \zeta _3 \mu y \big )$. The quadrilateral configuration changes to

$$\begin{aligned} \mathcal {P}= \{ (0,0), (1,0), (0,1), ( 2 \zeta _1 \mu ^2, \big (1 + \zeta _2 \big )\mu ^2) \}. \end{aligned}$$

By Theorem 1, the affine Hesse polynomial

$$\begin{aligned} f(\mu , \ \, \ \ ) \circ M (x,y) = \mu ^3 \left( 2 x^3 - 3 x (-1 + y) y - 3 x^2 (1 + y) + y^2 (-3 + 2 y) \right) +1 \end{aligned}$$

is essentially determined. Since these quadrilateral configurations are nonreal, they are different from those given in Fig. 4.

4.3 Nonessential Determined Polynomials of Degree 3

By completeness, we describe the polynomials arising from the configurations

$$\begin{aligned} \mathcal {P}= \{(0,0), (1,0), (0,1), (x_4, y_4) \} \in { Conf}(\mathbb {K}^2, 4), \ (x_4, y_4) \in \mathscr {A}. \end{aligned}$$

Lemma 5

• 1. Let $\mathcal {P}= \{ (0, 0), (1, 0), (x_3, 0), (x_4, y_4) \}$, with $x_3 \ne 0, 1$ and $y_4\ne 0$, then $dim_\mathbb {K}(Proj(\mathcal {L}_3(\mathcal {P})))= 0$.

• 2. Let $\mathcal {P}= \{ (0, 0), (1, 0), (x_3, 0), (x_4, 0) \}$ be a configuration, then $dim_\mathbb {K} (Proj(\mathcal {L}_3(\mathcal {P})))= 2$.

Proof

In assertion (1), up to an affine transformation we can assume $y_4=1$. The corresponding cubic polynomial takes the form $f(x,y)= a_4 \left( 2y^3-3y^2 \right)$, where $a_4 \in \mathbb {K}^*$.

For assertion (2), we search for polynomials $f(x,y) \in \mathbb {K}[x,y]^0_{\le 3}$ with at least 4 affine collinear singular points. The matrix of Eq. (17) results in the cubic polynomials

$$\begin{aligned} f(x,y)= a_3 x y^2+a_4y^3+a_7y^2 = y^2(a_3 x+a_4 y + a_7), \quad [a_3, a_4, a_7] \in \mathbb {K}\mathbb {P}^2, \end{aligned}$$

with a line of singular points in $\{ y=0\}$. $\square$

Example 2

The elementary methods provide an insight in the case of a double point in $\Sigma (f)$. Let $\mathcal {P}_2=\{ (0,0),(1,0),(0,1), (0,0) \}$ be such a configuration. A basis for $\mathcal {L}_3 (\mathcal {P}_2)$ is

$$\begin{aligned} x^3 - 3 x^2, \quad y^3 - 3 y^2, \quad x^2y+xy^2-xy. \end{aligned}$$

The first and second polynomials have lines of singularities, while the third one has four isolated critical points. The family of polynomials is

$$\begin{aligned} f(a_1,a_2, a_4, x,y )= & {} a_1 (x^3 - 3 x^2)+ a_2 ( x^2y+xy^2-xy )+ a_4(y^3 - 3 y^2), \\{} & {} {[}a_1, a_2, a_4] \in \mathbb {K}\mathbb {P}^2. \end{aligned}$$

As is expected, for values $\left\{ (a_1, a_2, a_4=a_2^2 /9 a_1 \right\}$ the 2-dimensional family $f(a_1,a_2, a_4, x,y)$ determines polynomials with three isolated singular points, one of them of multiplicity 2, see Fig. 4.

5 Degree 4 Polynomials

Let

$$\begin{aligned} f(x,y)=a_1 x^4+a_2 x^3 y+ \cdots + a_{13} x+a_{14} y \, \in \mathbb {K}[x,y]^0_{\le 4} \end{aligned}$$

(32)

be a polynomial as in (3). Here by notational simplicity, we have avoided the double subindex, and let $\mathcal {P} = \{ (x_\iota , y_\iota ) \ \vert \ \iota \in 1, \ldots , 7 \}$ be a configuration of seven points. The associated linear system for Eq. (32) is

$$\begin{aligned}{} & {} \left( \begin{array}{cccccccccccccc} &{} &{} &{} &{} &{}\vdots &{} &{} &{} &{} &{} &{} &{} &{} \\ 4x_\iota ^3 &{} 3x_\iota ^2y_\iota ^2 &{} 2x_\iota y_\iota ^2 &{} y_\iota ^3 &{} 0 &{} 3x_\iota ^3 &{} 2x_\iota y_\iota &{} y_\iota ^2 &{} 0 &{} 2x_\iota &{} y_\iota &{} 0 &{} 1 &{} 0 \\ 0 &{} x_\iota ^3 &{} 2x_\iota ^2y_\iota &{} 3x_\iota y_\iota ^2 &{} 4y_\iota ^3 &{} 0 &{} x_\iota ^2 &{} 2x_\iota y_\iota &{} 3y_\iota ^2 &{} 0 &{} x_\iota &{} 2y_\iota &{} 0 &{} 1 \\ &{}&{}&{}&{}&{}\vdots &{}&{}&{}&{}&{}&{}&{}&{} \end{array}\right) \nonumber \\{} & {} \quad \left( \begin{array}{c} a_1 \\ \vdots \\ a_{14} \end{array} \right) = \overline{0},\quad \iota = 1, \ldots , 7. \end{aligned}$$

(33)

The interpolation matrix $\phi$, Eq. (33), is square. Hence, for an open and dense set of configurations $\{ \mathcal {P}\} \subset { Conf}(\mathbb {K}^2, 7)$ such that $\{ det(\phi )=0 \}$, the resulting space of polynomials of degree 4 with having these $\mathcal {P}$ as critical points is empty. In order to overcome this situation, we introduce the following concept.

Definition 6

Assume $\mathbb {K}[x,y]^0_{\le d}$ with an even dimension and $\delta (d) = \frac{1}{4} \left( d^2 + 3d \right)$ as in (10). Given a configuration $\mathcal {P}_0 \in { Conf}(\mathbb {K}^2, \delta (d)-1)$, consider a point $(x, y) \in \mathbb {K}^2$ and

$$\begin{aligned} \mathcal {P}_1= \Big \{ \underbrace{(x_1, y_1), \ldots , (x_{\delta (d)-1}, y_{\delta (d)-1}) }_{ \mathcal {P}_0 }, (x, y) \Big \}\, \in { Conf}(\mathbb {K}^2, \delta (d)). \end{aligned}$$

The interpolation algebraic curve of $\mathcal {P}_0$ is

$$\begin{aligned} \mathcal {I} = \left\{ det \big ( {\phi } ( x_1, y_1, \ldots , x_{\delta (d)-1}, y_{\delta (d)-1}, x, y) \big )= 0 \right\} \,\hbox {in}\,\mathbb {K}^2. \end{aligned}$$

Obviously, $\mathcal {I}$ depends on $\mathcal {P}_0$, by notational simplicity we omit this dependence. Thus, we have a map

$$\begin{aligned} \mathcal {P}_0 = \{ (x_1, y_1), \ldots , (x_{\delta (d)-1}, y_{\delta (d)-1} ) \} \longmapsto \mathcal {I}. \end{aligned}$$

Proposition 2

Assume $\mathbb {K}[x,y]^0_{\le d}$ with even dimension.

• 1. The interpolation curve $\mathcal {I}$ of $\mathcal {P}_0$ describes the position of the $\delta (d)$-th point such that $dim_\mathbb {K}(\mathcal {L}_{d} (\mathcal {P}_1)) \ge 0$.

• 2. There exists a Zariski open set $\{ \mathcal {P}_0 \} \subset { Conf}(\mathbb {K}^2, \delta (d)-1)$ such that the associated $\{ \mathcal {I} \}$ are algebraic curves of degree $2d-2$ in $\mathbb {K}^2$.

Proof

For assertion (2), we consider the degree d polynomial

$$\begin{aligned} f(x,y)=a_1 x^d+a_2 x^{d-1} y+ \cdots + a_{\delta (d)-1} x+a_{\delta (d)} y. \end{aligned}$$

After fixing the configuration $\mathcal {P}_0$, the associated linear system only has free variables x, y, and the linear system is as follows

$$\begin{aligned}{} & {} \left( \begin{array}{llllllllllllll} &{} &{} &{} &{} &{} \vdots &{} &{} &{} &{} &{} &{} &{} &{} \\ (d)x^{d-1} &{} (d-1)x^{d-2}y &{} (d-2)x^{d-3} y^2 &{} \cdots &{} 0 &{} (d-1)x^{d-2} &{} \cdots &{} y^2 &{} 0 &{} 2x &{} y &{} 0 &{} 1 &{} 0 \\ 0 &{} x^{d-1} &{} 2x^{d-2}y &{} \cdots &{} 4y^3 &{} 0 &{} x^2 &{} 2x y &{} 3y^2 &{} 0 &{} x &{} 2y &{} 0 &{} 1 \\ &{}&{}&{}&{}&{}\vdots &{}&{}&{}&{}&{}&{}&{}&{} \end{array} \right) \nonumber \\{} & {} \quad \left( \begin{array}{ll} a_1 \\ \vdots \\ a_{\delta (d)} \end{array} \right) = \overline{0}. \end{aligned}$$

(34)

The determinant of this matrix has $x^{2d-2}$ as a higher degree monomial, and we are done. $\square$

We describe some interpolation curves $\mathcal {I}$.

Example 3

Let $f \in \mathbb {K}[x,y]^0_{\le 4}$ be a polynomial having of degree 4 and let $\mathcal {P}_0 = \{ (x_\iota , y_\iota ) \ \vert \ \iota \in 1, \ldots ,6 \}$ be a fixed configuration of six different singular points of f.

1. If three points of $\mathcal {P}_0$ are in a line $\{ x= 0 \}$ and two points are in $\{x=1 \}$, then the interpolation curve $\mathcal {I}$, of $\mathcal {P}_0$, is given by

$$\begin{aligned} \mathcal {I}(x, y) =\left( -1152 y_4^2 y_5^2 (y_4-1)^2 x_6 (x_6-1)\right) x (x-1) (x - x_6) g(x, y). \end{aligned}$$

(35)

The $\mathcal {I}$ is reducible and singular, it is the product of three parallel lines and a polynomial g(x, y) that pass through the six points in $\mathcal {P}_0$.

2. Let $\mathcal {P}_0 = \{ (x_\iota , y_\iota ) \ \vert \ \iota \in 1, \ldots ,6 \}$ be any configuration of six points in the grid of nine points

$$\begin{aligned} \mathcal {G}=\{x(x-1)(x-c_1)=0\} \cap \{y(y-1)(y-c_2)=0 \},\quad \hbox {where}\quad c_1, \, c_2 \notin \{0,1\}. \end{aligned}$$

Therefore, the interpolation curve $\mathcal {I}$, associated with the seventh point $(x_7, y_7)$, is the product of the six lines defining $\mathcal {G}$.

3. Let $\mathcal {P} = \{ (x_\iota , y_\iota ) \ \vert \ \iota \in 1, \ldots ,6 \}$ be a configuration of six singular points of f. If the six points are distributed in a conic Q, then the interpolation curve $\mathcal {I}$, associated to the seventh point $(x_7, y_7)$, contains the conic, which is $\mathcal {I}=Qg$ for some $g \in \mathbb {K}[x,y]^0_{\le 4}$.

A complete study of the interpolation curves $\mathcal {I}$ arising from configurations of six points is the goal of a future project.

6 Polynomial Vector Fields with $(d-1)^2$ Singularities

Now we will consider some special configurations of $(d-1)^2 \ge 4$ points.

Definition 7

Let $\{ F(x,y)=0 \}$ and $\{ G(x,y)=0 \}$ be two algebraic curves in $\mathbb {K}^2$, both of degree $d-1$ $(\ge 2)$. We assume that they have transversal intersections in exactly $(d-1)^2$ affine points; therefore

$$\begin{aligned} \mathcal {P}_{ci}=\{F(x,y)=0\} \cap \{ G(x,y)=0\} \in { Conf}(\mathbb {K}^2, (d-1)^2) \end{aligned}$$

(36)

is a complete intersection configuration. The associated pencil of curves is

$$\begin{aligned} \left\{ \mu F(x,y) + \nu G(x,y) =0 \ \vert \ [\mu , \nu ] \in \mathbb {K}\mathbb {P}^1 \right\} . \end{aligned}$$

(37)

$\mathcal {P}_{ci}$ is the base locus of the pencil of curves.

Corollary 3

An ordered pair of polynomial functions from (37), not just curves, determines a $SL(2, \mathbb {K})$-pencil of polynomial vector fields

$$\begin{aligned} \mathfrak {F}(\mathcal {P}_{ci})= & {} \left\{ \ X_\texttt {M}= -\big ( \texttt{c} F(x,y) + \texttt {d} G(x,y) \big ) \frac{\partial }{\partial x} + \big ( \texttt {a} F(x,y) + \texttt {b} G(x,y) \big ) \frac{\partial }{\partial y} \ \Big \vert \right. \nonumber \\{} & {} \left. \texttt {M}= \left( \begin{array}{cc} \mathtt {-c} &{}\quad \mathtt { -d} \\ \texttt{a} &{}\quad \texttt{b} \end{array} \right) \in SL(2, \mathbb {K}) \ \right\} \end{aligned}$$

(38)

Each vector field $X_\texttt {M}$ has singularities of multiplicity 1 at $\mathcal {P}_{ci}$. $\square$

Lemma 6

Let $\mathcal {U}_d \subseteq \mathfrak {X}(\mathbb {K}^2)_{\le d-1}$ be the open and dense set of polynomial vector fields of degree $d-1$, with exactly $(d-1)^2$ singular points in $\mathcal {P}_{ci} \subset { Conf} (\mathbb {K}^2, (d-1)^2)$. Assume that $\mathcal {P}_{ci}$ has a trivial isotropy group in $\hbox { Aff}\hspace{1.42271pt}(\mathbb {K}^2)$. In $\mathcal {U}_d$ there exists an analytic $SL(2,\mathbb {K})$-bundle structure as follows

Proof

We want to show that a polynomial vector field $X \in \mathfrak {X}(\mathbb {K}^2)_{\le d-1}$ has $(d-1)^2$ singular points exactly at $\mathcal {P}_{ci}$ as in (36) if and only if it is of the shape $X_\texttt {M}$ in (38).

($\Rightarrow$) Let $X=A(x,y)\frac{\partial }{\partial x} + B(x,y)\frac{\partial }{\partial y}$ be a vector field in $\mathfrak {X}(\mathbb {K}^2)_{\le d-1}$. The curve $\mathcal {C}_A \doteq \{ A(x,y)=0\}$ has at most degree $d-1$ and would contain $\mathcal {P}_{ci}$. An open set there exists of values $\{ [\mu , \nu ] \} \subset \mathbb {K}\mathbb {P}^1$ such that for each value the respective curve $\{\mu F + \nu G =0 \}$ in the pencil (37) intersects in a transversal way $\mathcal {C}_A$ at every point of $\mathcal {P}_{ci}$. By Bézout’s theorem, the degree of $\mathcal {C}$ is exactly $d-1$. For any point $p \in \mathcal {C}_A \backslash \mathcal {P}_{ci} \subset \mathbb {K}^2$, there exists a value, say $[-\texttt {c}, -\texttt {d} ]$ in (37), such that its respective curve satisfies $\mathcal {C}_{-\texttt {c} -\texttt {d} } \cap \mathcal {C}_A \supset \widehat{\mathcal {P}} \cup \{p\}$. Hence (again by Bézout’s theorem), both curves coincide as sets and $A= -\texttt {c}F - \texttt {d} G$ as polynomials.

$\square$

Thus, each configuration $\mathcal {P}_{ci}$ has an associated fiber $\big \{ X_\texttt {M} \ \vert \ \texttt {M} \in SL(2, \mathbb {K}) \big \} \subset \mathcal {U}_d$ in (39), which is a family of not necessarily Hamiltonian vector fields. A further goal is the study of the intersection

$$\begin{aligned} \big \{ X_\texttt {M} \ \vert \ \texttt {M} \in SL(2, \mathbb {K}) \big \} \cap Ham(\mathbb {K}^2)_{\le d}. \end{aligned}$$

Corollary 4

A jump phenomena. Let $\mathcal {P} = \{ (0,0), (1,0), (1/2, \sqrt{3}/2), (x_4, y_4) \}$ be a configuration leading to a family of vector fields $\mathfrak {F} (\mathcal {P}) = \{ X_\texttt {m } \ \vert \ \texttt {m} \in SL(2, \mathbb {K}) \}$ as in (38).

• (1) If $(x_4, y_4) \in \mathbb {K}^2 \backslash \texttt {A}$, then there exists one projective class in $\mathfrak {F}(\mathcal {P}) \cap Ham(\mathbb {K}^2)_{\le 2}$.

• (2) If $(x_4, y_4) = \texttt {R}_1, \, \texttt {R}_2$ or $\texttt {R}_3$, then there exists a $\mathbb{K}\mathbb{P}^1$-family of Hamiltonian vector fields $\mathfrak {F}(\mathcal {P})\cap Ham(\mathbb {K}^2)_{\le 2}$. $\square$

Example 4

A family $\big \{ X_\texttt {M} \ \vert \ \texttt {M} \in SL(2, \mathbb {K}) \big \}$ exists in (39) with $(d-1)^2 \ge 4$ points as a base locus and such that its Hamiltonian vector fields $Ham(\mathbb {K}^2)_{\le d-1}= [f]$ determine one projective class.

Consider two algebraic curves such that

$$\begin{aligned} \mathcal {P}_{ci}= \{ \underbrace{y- \mu \Pi _{\iota =1}^d (x-x_\iota )=0}_{ F(x,y)=0} \} \cap \{ \underbrace{x- \nu \Pi _{j=1}^d (y- y_j )=0}_{G(x,y)=0} \}, \ \ \ d \ge 3 \end{aligned}$$

has exactly $(d-1)^2 \ge 4$ points.

It follows that the associated 1-form $\omega _\texttt {m}$ is exact if and only if $\texttt {m}= \left( \begin{array}{cc} \texttt{a} &{}\quad \texttt{0} \\ \texttt{0} &{}\quad \texttt{a} \end{array} \right)$.

In fact, suppose f(x, y) such that $\omega _\texttt {m} =df$, then

$$\begin{aligned} \texttt {a}F(x,y)+\texttt {b}G(x,y)=f_x \quad \hbox {and}\quad \texttt {c}F(x,y)+\texttt{d}G(x,y)=f_y. \end{aligned}$$

As $f_{xy}=f_{yx}$, then $\texttt {a}- \texttt {b} \frac{\partial }{\partial y} \Pi _{j=1}^d (y- y_j ) =-\texttt{c}\frac{\partial }{\partial x}\Pi _{\iota =1}^d (x-x_\iota )+\texttt {d}$, so $\texttt {a}=\texttt {d}$ and $\texttt {b}=\texttt {c}=0$.

By assuming $\omega _\texttt {m }$ is exact and defining $f_\texttt {m} (x,y) = \int ^{(x,y)} \omega _\texttt {m}$, we conclude that

$$\begin{aligned} \mathfrak {F}(\mathcal {P}_{ci}) \cap Ham(\mathbb {K}^2)_{\le d-1} = \mathcal {L}_{d} (\mathcal {P}_{ci})= [f_\texttt {m}]\quad \hbox {and}\quad dim_\mathbb {K}(\mathcal {L}_{d} (\mathcal {P}_{ci} )) = 0 . \end{aligned}$$

(40)

Example 5

A fiber $\big \{ X_\texttt {M} \ \vert \ \texttt {M} \in SL(2, \mathbb {K}) \big \}$ as in (39), with $(d-1)^2\ge 9$ points as a base locus satisfying that

$$\begin{aligned} \big \{ X_\texttt {M} \ \vert \ \texttt {M} \in SL(2, \mathbb {K}) \big \} \cap Ham(\mathbb {K}^2)_{=d} = \emptyset . \end{aligned}$$

Consider two hyperelliptic curves such that

$$\begin{aligned} \widehat{\mathcal {P}}= \{ F(x,y)=y^2- \mu \Pi _{\iota = 1}^d (x-x_\iota )=0 \} \cap \{ G(x,y)= x^2 - \nu \Pi _{j=1}^d (y-y_j)=0 \} \end{aligned}$$

has exactly $(d-1)^2 \ge 9$ points. It follows that $\omega _\texttt{m}$ is nonexact for all $\texttt {m}= \left( \begin{array}{ll} \texttt{a} &{} \texttt{ b} \\ \texttt{c} &{} \texttt{d} \end{array} \right)$. We conclude that

$$\begin{aligned} \mathcal {L}_{d} (\widehat{\mathcal {P}})= \emptyset \quad \hbox {and}\quad dim_\mathbb {K}(\mathcal {L}_{d} ( \widehat{\mathcal {P}})) = -1 . \end{aligned}$$

(41)

In fact, if we suppose f(x, y) such that $\omega _m=df$, then $2\texttt {a}y- \texttt {b}\frac{\partial }{\partial y} \Pi _{j=1}^d (y- y_j )=-\texttt {c}\frac{\partial }{\partial x} \Pi _{\iota =1}^d (x-x_\iota )+2 \texttt {d} x$, so $\texttt {a}=\texttt {b}=\texttt {c}=\texttt {d}=0$.

Corollary 5

There exists a fiber $\mathfrak {F}$ as in (39) having $d^2$ points as a base locus and

$$\begin{aligned} \mathfrak {F}(\widehat{\mathcal {P}}) \cap Ham(\mathbb {K}^2)_{=d} = \mathbb {K}\mathbb {P}^1. \end{aligned}$$

Moreover, $\mathbb {K}\mathbb {P}^1$ minus a finite set determines Morse polynomials.

The above result uses the following very particular configurations.

Definition 8

A grid of $(d-1)^2$ points $\mathcal {G}$ is determined by two sets of $d-1$ parallel lines where one set is transverse to the other: up to affine transformation

$$\begin{aligned} \mathcal {G}= \{ F(x,y)= \Pi _{j=1}^{d-1} (y- y_j )=0 \} \cap \{ G(x,y)=\Pi _{\iota =1}^{d-1} (x-x_\iota )=0\} \end{aligned}$$

with exactly $(d-1)^2 \ge 4$ points; it is a complete intersection.

Proof of the Corollary

The family $X_\texttt {M}$ with a grid of $(d-1)^2$ points is Hamiltonian if and only if

$$\begin{aligned} \texttt {M} \in \left\{ \left( \begin{array}{cc} 0&{}\quad -\texttt {d} \\ \texttt {a} &{}\quad 0 \end{array} \right) \right\} \cong \mathbb {K}^2 \subset SL(2, \mathbb {K}). \end{aligned}$$

In fact, $\omega _\texttt {m} = (\texttt {a} F(x) + \texttt {b}G(y)) dx + ( \texttt {c}F(x) + \texttt {d}G(y)) dy =0$ is exact if and only if $\texttt{b} G(y)_y = \texttt {c} F(x)_x$. The equality holds only for $\texttt {b}= \texttt {c}=0$.

The respective vector subspace of polynomials

$$\begin{aligned}{} & {} \Bigg \{ f(\texttt {a}, \texttt {d}, x,y) = \texttt {a} \int ^{(x,y)} \Pi _{\iota =1}^d (x-x_\iota ) dx + \texttt {d} \int ^{(x,y)} \Pi _{j=1}^d (y-y_j) dy \ \Big \vert \ (\texttt {a}, \texttt {d} ) \nonumber \\{} & {} \quad \in \mathbb {K}^2 \backslash \{\overline{0} \}\Bigg \} \subset \mathbb {K}[x,y]^0_{\le d} \end{aligned}$$

(42)

shows that

$$\begin{aligned} \mathcal {L}_{d} (\mathcal {P}) \supset \{ [f(\texttt {a} ,\texttt {d}, x,y) ] \}\quad \hbox {and} \quad dim_\mathbb {K}(\mathcal {L}_{d} (\mathcal {P})) = 1 . \end{aligned}$$

(43)

For $(\texttt {a}, \texttt {d}) \ne (\texttt {a}, 0), \, (0, \texttt {d})$, each polynomial $f(\texttt {a},\texttt {d}, x,y) \in \mathbb {K}[x,y]^0_{\le d}$ in (42) has $(d-1)^2$ Morse singular points. In fact, at each point $p\in \mathcal {P}$, a very simple observation with the Taylor series shows that $f({\texttt {a}}, {\texttt {d}}, x,y) = \widetilde{{\texttt {a}}} x^2 + \widetilde{{\texttt {b}}} y^2 + \mathcal {O}_3(x,y)$, where $\widetilde{{\texttt {a}}} \widetilde{{\texttt {b}}} \ne 0$.

On the other hand, for $({{\texttt {a}}}, {{\texttt {d}}}) = ({{\texttt {a}}}, 0), \, (0, {\texttt {d}})$ the polynomial $f({\texttt {a}}, {\texttt {d}}, x, y)$ has lines of singular points in $\{ P(x,y)=0\}$ or $\{ Q(x,y)= 0\}$. $\square$

Example 6

Real rotated Hamiltonian vector fields for the grid of 4 points. Let $\mathcal {G} = \{ (0,0), (1,0), (0, 1), R=(1,1)\}$ be a grid, and its space of polynomials is

$$\begin{aligned} f(\texttt {a}, \texttt {d}, x,y)= \texttt {a} \Big (\dfrac{x^3}{3} - \dfrac{x^2}{ 2} \Big ) + \texttt {d} \Big ( \frac{y^3}{3} - \frac{y^2}{ 2} \Big ). \end{aligned}$$

In particular for $\mathbb {K}=\mathbb {R}$, we consider the family

$$\begin{aligned} R_\theta =\left\{ f_\theta (x,y) = \cos (\theta ) \left( \frac{x^3}{3} - \frac{x^2}{2} \right) + \sin (\theta ) \left( \frac{y^3}{3} - \frac{y^2}{2} \right) \ \Big \vert \ \theta \in [0, 2\pi ]\right\} \end{aligned}$$

of polynomials in (42). They originate from a family of rotated vector fields, see Fig. 4. The algebraic curve $\{ f_\theta (x,y)+c=0 \}$ is reducible for $\theta =\pi /4$ and $c={1/6}$. In this case we obtain

$$\begin{aligned} \{ (x+y-1)(2y^2-2xy+2x^2-y-x-1)=0 \}. \end{aligned}$$

The following family of vector fields is related to the results in Ramírez [17, Sect. 5] (non-generic Hamiltonian vector fields, theorem 5); see Fig. 4, upper row.

Corollary 6

The 1-dimensional holomorphic family of Hamiltonian vector fields of the polynomials

$$\begin{aligned} \left\{ f(\texttt {a}, \texttt {d}, x,y)= \texttt {a} \Big (\dfrac{x^3}{3} - \dfrac{x^2}{ 2} \Big ) + \texttt {d} \Big ( \frac{y^3}{3} - \frac{y^2}{ 2} \Big ) \ \big \vert \ \texttt {ad}=1 \right\} \end{aligned}$$

has singularities at $\mathcal {G} = \{ (0,0), (1,0), (0, 1), R=(1,1)\}$ and spectra of eigenvalues

$$\begin{aligned} \big [ [i, \, -i], \, [1,-1], \, [i, \, -i], \, [1,-1] \big ]. \end{aligned}$$

$\square$

Corollary 7

For $d \ge 3$, there exist Morse polynomials $f \in \mathbb {K}[x,y]_{=d}^0$ with $(d-1)^2$ singular points that are not essentially determined. $\square$

7 Closing Remarks

Let $\mathfrak {X}(\mathbb {K}^2)_{\le d-1}$ be the space of polynomial vector fields $\{ X \}$ of at most degree $d-1$ on $\mathbb {K}^2$. A general and natural question is as follows. Under what conditions is a polynomial vector field X on $\mathbb {K}^2$ essentially determined by its configuration of singular points, i.e., its zeros, $\mathcal {Z}(X)$ in $\mathbb {K}^2$?

In simple words, a vector field X is essentially determined (in $\mathfrak {X}(\mathbb {K}^2)_{\le d-1}$) by its configuration of zeros $\mathcal {Z}(X_f)$;

$$\begin{aligned} \hbox {if for any}~Y \in \mathfrak {X}(\mathbb {K}^2)_{\le d-1}~\hbox {satisfying}~\mathcal {Z}(X) \subset \mathcal {Z}(Y)\subset \mathbb {K}^2,\quad \hbox {then}~X = \lambda Y. \end{aligned}$$

Recalling that for affine degree d the number of isolated singularities of the associated singular holomorphic foliation $\mathcal {F}(\mathcal {X})$ on the whole $\mathbb{C}\mathbb{P}^2$ is $(d-1)^2 + d$, the hypothesis of multiplicity 1 must be understood for all these points. Proposition 1 confirms that in the Hamiltonian case only $\delta (d) \le (d-1)^2$ points are required.

Recall which Gómez-Mont and Kempf [13], established in the complex rational case the following deep result, that also enlightens the real case.

A meromorphic vector field $\mathcal {X}$ on $\mathbb{C}\mathbb{P}^m$, $m \ge 2$, of degree $r\ge 2$, with singular points of multiplicity 1 is completely determined by its singular set.

Moreover, Artes et al. [4, 5] prove the following:

A polynomial vector field $\mathcal {X}$ on $\mathbb {K}^2$ of degree 2 is completely determined by the position of its 7 singular points (including the points at infinity).

As far as we know, over $\mathbb {K}=\mathbb {C}$ the more general result is due to Campillo and Olivares [6]:

A singular holomorphic foliation $\mathcal {X}$ on $\mathbb{C}\mathbb{P}^2$ of degree $r\ge 2$, is completely determined by its singular scheme.

See Alcántara et al. [1] for recent developments regarding foliations with multiple points. We summarize our results as follows.

Corollary 8

A polynomial Hamiltonian vector field $X_f$ on $\mathbb {K}^2$ of degree 2 is completely determined (in the space of polynomial vector fields of degree 2, up to a scalar factor $\lambda \in \mathbb {K}^*$) by its zero points, when there are 4 isolated points different from $\{(0,0), \, (1,0), \, (0,1), \, (1,1) \}$, up to affine transformation.

Our hope is that the explicit results in this paper can illustrate the classification of polynomials $\mathbb {K}[x,y]$ up to algebraic equivalence $Aut(\mathbb {K}^2)$; see [11, 18] for this order of ideas. This potential application is the subject of a future project.