On the Complex and Real Hessian Polynomials

Abstract

We study some realization problems related to the Hessian polynomials. In particular, we solve the Hessian curve realization problem for degrees zero, one, two, and three and the Hessian polynomial realization problem for degrees zero, one, and two.

1. Introduction

A topic which has been of interest since the XIX century is the study of the parabolic curve of smooth surfaces in real three-dimensional space, as shown in the works of Gauss, Darboux, Salmon [1], Kergosien and Thom [2], Arnold [3], among others.

The parabolic curve of the graph of a smooth function, 𝑓∶ℝ²→ℝ, is the set {(𝑝,𝑓(𝑝))∈ℝ²×ℝ: Hess𝑓(𝑝)=0}, where Hess 𝑓∶=𝑓_𝑥𝑥𝑓_𝑦𝑦−𝑓²_𝑥𝑦. In this case, the Hessian curve of 𝑓, Hess 𝑓(𝑥,𝑦)=0, is a plane curve which is the projection of the parabolic curve into the 𝑥𝑦-plane along the 𝑧-axis. When 𝑓 is a polynomial of degree 𝑛 in two variables, Hess 𝑓 is a polynomial of degree at most 2𝑛−4. Therefore, the Hessian curve is an algebraic plane curve. In this setting there are two natural realization problems related to the Hessian of a polynomial. (1)Hessian curve realization problem. Given an algebraic plane curve 𝑔(𝑥,𝑦)=0 in 𝕂², where 𝕂=ℝ or ℂ, we ask: When is 𝑔(𝑥,𝑦)=0 the Hessian curve of a polynomial 𝑓∈𝕂[𝑥,𝑦]? (2)Hessian polynomial realization problem Given 𝑔∈𝕂[𝑥,𝑦] we ask: When does 𝑓∈𝕂[𝑥,𝑦] exist such that Hess 𝑓=𝑔? If such 𝑓 exists and 𝕂=ℂ (𝕂=ℝ), then 𝑔 is called a complex Hessian polynomial (real Hessian polynomial). We remark that Arnold (see [4]) calls Hessian topology problem to the study of the problem 1 in the real case.

Note that problem 2 contains problem 1. Moreover, in the real case, problem 2 is a global realization problem of a smooth function such as the Gaussian curvature function. In [5], Arnold studies this problem locally.

This work is devoted to problems 1 and 2 for complex and real case of degree less or equal to three. It is divided in two parts. In the first we give the results according to the degree of the polynomials 𝑔. In the second part, we give the proofs and some other results such as a geometric interpretation of Corollary 2.8.

2. Notation and Results

For each nonnegative integer number 𝑛, we define 𝔸^𝕂_𝑛∶={𝑓∈𝕂[𝑥,𝑦]|deg(𝑓)≤𝑛}. Let us consider the Hessian map 𝐻^𝕂_𝑛∶𝔸^𝕂_𝑛→𝔸^𝕂_2𝑛−4,givenby𝑓↦Hess𝑓. (2.1)

We will say that the Hessian map 𝐻^𝕂_𝑛 is complex or real if 𝕂=ℂ or 𝕂=ℝ, respectively. We remark that dim(𝔸^𝕂_𝑛)≥ dim(𝔸^𝕂_2𝑛−4) if 2≤𝑛≤4 and dim(𝔸^𝕂_𝑛)< dim(𝔸^𝕂_2𝑛−4) if 𝑛≥5. In particular, for 𝑛≥4, the image, 𝐻^𝕂_𝑛(𝔸^𝕂_𝑛), is a connected subset of codimension at least three in 𝔸^𝕂_2𝑛−4. This means that, in “general”, a polynomial 𝑔∈𝔸^𝕂_2𝑛−4 is not a Hessian polynomial if 𝑛≥4 under 𝐻^𝕂_𝑛.

In diagram form we have 𝔸^𝕂₂⊊𝔸^𝕂₃⊊𝔸^𝕂₄⊊⋯⊊𝔸^𝕂_𝑛⋯↓_𝐻^𝕂₂↓_𝐻^𝕂₃↓_𝐻^𝕂₄⋯↓_𝐻^𝕂_𝑛⋯𝔸^𝕂₀⊊𝔸^𝕂₂⊊𝔸^𝕂₄⊊⋯⊊𝔸^𝕂_2𝑛−4⋯ (2.2)

In virtue of the previous remarks, we introduce the fiber of 𝑔∈𝔸^𝕂_2𝑛−4 under 𝐻^𝕂_𝑛 as the set (𝐻^𝕂_𝑛)⁻¹(𝑔)∶={𝑓∈𝔸^𝕂_𝑛∣𝐻^𝕂_𝑛(𝑓)=𝑔}. (2.3) Even when we consider the fibers, (𝐻^𝕂_𝑛)⁻¹(𝑔), for different values of 𝑛, we are interested in knowing if the fiber (𝐻^𝕂_𝑛)⁻¹(𝑔) is not empty when 𝑛 satisfies 0≤2𝑛−4−deg(𝑔)≤1. If the fiber (𝐻^𝕂_𝑛)⁻¹(𝑔) is empty, the next problem is to see if the fibers (𝐻^𝕂_𝑠)⁻¹(𝑔) are not empty for 𝑠≥𝑛+1 (this problem will not be studied in this work). Another way to study the Hessian polynomial realization problem is by describing the set of all polynomials 𝑓∈𝔸^𝕂_𝑟, with 𝑟≥𝑛=(𝑚+4)/2, such that 𝐻^𝕂_𝑛(𝑓)=𝑔 whenever 𝑔∈𝔸^𝕂_𝑚.

Remark 2.1. Let 𝑔(𝑥,𝑦)=∑^2𝑛−4_𝑖+𝑗=0𝑏_𝑖𝑗𝑥^𝑖𝑦^𝑗 be a polynomial in 𝔸^𝕂_2𝑛−4. There exists a polynomial 𝑓(𝑥,𝑦)=∑^𝑛_𝑖+𝑗=0𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗 in (𝐻^𝕂_𝑛)⁻¹(𝑔) if and only if 𝑓 satisfies the following system of (2𝑛−3)(𝑛−1) equations: ℎ_𝑖𝑗(𝑎_𝑟𝑠)=𝑏_𝑖𝑗,0≤𝑖+𝑗≤2𝑛−4,2≤𝑟+𝑠≤𝑛, (2.4) where ℎ_𝑖𝑗(𝑎_𝑟𝑠) are the coefficients of the Hessian polynomial𝐻^𝕂_𝑛(𝑓). They are also quadratic polynomials in the variables 𝑎_𝑟𝑠.

It is important to note that the computations for solving the system (2.4) are generally very complicated.

In the following results we describe the sets of polynomials of a given degree which are Hessian and those which are not Hessian under a specific Hessian map.

2.1. Degree of 𝑔 Equal to Zero

Proposition 2.2. For each 𝑔∈𝔸^ℂ₀, the set (𝐻^ℂ₂)⁻¹(𝑔) is a quadric in 𝔸^ℂ₂ given by 𝑎²₁₁=4𝑎₂₀𝑎₀₂−𝑔. Moreover, this quadric is singular if and only if 𝑔=0.

Corollary 2.3. In the complex case every element in 𝐴^ℂ₀=ℂ is a complex Hessian polynomial under𝐻^ℂ₂. And, in the real case every element in𝐴^ℝ₀=ℝis a real Hessian polynomial under𝐻^ℝ₂.

Proposition 2.4. For each 𝑔∈𝔸^ℂ₀, the set  (𝐻^ℂ₃)⁻¹(𝑔)  is an analytic subvariety in 𝔸^ℂ₃ which is given by the union of connected analytic subvarieties which are parametrized by the following. (1)𝐹^±₁∶ℂ×ℂ^∗×ℂ^∗×ℂ³→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅,𝑡₆)↦(𝑎₃₀=𝑡²₂/3𝑡₃, 𝑎₂₁=𝑡₂, 𝑎₁₂=𝑡₃,𝑎₀₃=𝑡²₃/3𝑡₂,𝑎₂₀=𝑡₁, 𝑎₁₁=(2𝑡₃𝑡₁±𝑡₂√−𝑔)/𝑡₂, 𝑎₀₂=𝑡₃(𝑡₃𝑡₁±𝑡₂√−𝑔)/𝑡²₂, 𝑎₁₀=𝑡₄,𝑎₀₁=𝑡₅,𝑎₀₀=𝑡₆).(2)𝐹₂∶ℂ×ℂ^∗×ℂ³→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅)↦(𝑎₃₀=0,𝑎₂₁=0,𝑎₁₂=0,𝑎₀₃=0,𝑎₂₀=(𝑡²₁+𝑔)/4𝑡₂, 𝑎₁₁=𝑡₁,𝑎₀₂=𝑡₂, 𝑎₁₀=𝑡₃,𝑎₀₁=𝑡₄,𝑎₀₀=𝑡₅). (3)𝐹^±₃∶ℂ⁵→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅)↦(𝑎₃₀=𝑡₁,𝑎₂₁=0, 𝑎₁₂=0,𝑎₀₃=0,𝑎₂₀=𝑡₂, 𝑎₁₁=±√−𝑔,𝑎₀₂=0, 𝑎₁₀=𝑡₃,𝑎₀₁=𝑡₄,𝑎₀₀=𝑡₅).(4)𝐹^±₄∶ℂ⁵→𝔸^ℂ₃;  (𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅)↦(𝑎₃₀=0, 𝑎₂₁=0,𝑎₁₂=0,𝑎₀₃=𝑡₁, 𝑎₂₀=0,𝑎₁₁=±√−𝑔, 𝑎₀₂=𝑡₂,𝑎₁₀=𝑡₃, 𝑎₀₁=𝑡₄,𝑎₀₀=𝑡₅).

2.2. Degree of 𝑔 Equal to One

Proposition 2.5. For each 𝑔(𝑥,𝑦)=𝑏₁₀𝑥+𝑏₀₁𝑦+𝑏₀₀∈𝔸^ℂ₂ of degree one, the set (𝐻^ℂ₃)⁻¹(𝑔) is an analytic subvariety in 𝔸^ℂ₃ which is given by the union of analytic subvarieties parametrized by the following. (1)For 𝑏₁₀𝑏₀₁=0,  𝐹₁∶ℂ^∗×ℂ×ℂ³⧵{4𝑏²₀₁𝑡²₁−4𝑏₁₀𝑏₀₁𝑡₁𝑡₂+𝑏²₁₀𝑡²₂+𝑏₀₀𝑏²₁₀=0}→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅)↦(𝑎₃₀ = (1/3)(𝑏³₁₀𝑡₁/(4𝑏²₀₁𝑡²₁−4𝑏₁₀𝑏₀₁𝑡₁𝑡₂+𝑏²₁₀𝑡²₂+𝑏₀₀𝑏²₁₀)), 𝑎₂₁=𝑏₀₁𝑏²₁₀𝑡₁/(4𝑏²₀₁𝑡²₁−4𝑏₁₀𝑏₀₁𝑡₁𝑡₂+𝑏²₁₀𝑡²₂+𝑏₀₀𝑏²₁₀), 𝑎₁₂=𝑏²₀₁𝑏₁₀𝑡₁/(4𝑏²₀₁𝑡²₁−4𝑏₁₀𝑏₀₁𝑡₁𝑡₂+𝑏²₁₀𝑡²₂+𝑏₀₀𝑏²₁₀), 𝑎₀₃=𝑏³₀₁𝑡₁/3(4𝑏²₀₁𝑡²₁−4𝑏₁₀𝑏₀₁𝑡₁𝑡₂+𝑏²₁₀𝑡²₂+𝑏₀₀𝑏²₁₀), 𝑎₂₀=𝑡₁,𝑎₁₁=𝑡₂,𝑎₀₂=(𝑡²₂+𝑏₀₀)/4𝑡₁, 𝑎₁₀=𝑡₃,𝑎₀₁=𝑡₄,𝑎₀₀=𝑡₅).(2)For 𝑏₁₀𝑏₀₁≠0, 𝐹^±₂∶ℂ^∗×ℂ³→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄)↦(𝑎₃₀=(1/3)(𝑡₁𝑏₁₀/𝑏₀₁), 𝑎₂₁=𝑡₁,𝑎₁₂=𝑏₀₁𝑡₁/𝑏₁₀, 𝑎₀₃=(1/3)(𝑏²₀₁𝑡₁/𝑏²₁₀), 𝑎₂₀=0,𝑎₁₁=±−𝑏₀₀, 𝑎₀₂=(1/4)(𝑏₀₁(±4𝑡₁−𝑏₀₀+𝑏₁₀)/𝑏₁₀𝑡₁), 𝑎₁₀=𝑡₂,𝑎₀₁=𝑡₃,𝑎₀₀=𝑡₄).

From this proposition we have the following.

Corollary 2.6. Every complex polynomial of degree one is a complex Hessian polynomial under 𝐻^ℂ₃. And every real polynomial of degree one is a real Hessian polynomial under 𝐻^ℝ₃.

2.3. Degree of 𝑔 Equal to Two

Let 𝑔₁,𝑔₂∈𝕂[𝑥,𝑦]. We say that 𝑔₁ is in the orbit of 𝑔₂ if they are equivalent by an affine transformation of the plane 𝕂².

Theorem 2.7. Complex Case
(1)The complex polynomials of degree two 𝑦²𝑥−𝑦²𝑦²−𝑥²−𝑘,𝑘∈ℂ (2.5) are complex Hessian polynomials under 𝐻^ℂ₃. Moreover, a polynomial 𝑔∈𝔸^ℂ₂ of degree two is a complex Hessian polynomial under 𝐻^ℂ₃ if and only if it belongs to the orbit of one of those polynomials. (2)The complex polynomials, 𝑦²+𝑟∈𝔸^ℂ₂, where 𝑟∈ℂ^∗, are not complex Hessian polynomials under 𝐻^ℂ₃. Moreover, a polynomial 𝑔∈𝔸^ℂ₂ of degree two is not a complex Hessian polynomial under 𝐻^ℂ₃ if and only if it belongs to the orbit of one of those polynomials.
Real case
(1)The real polynomials of degree two −𝑦²𝑥−𝑦²𝑦²−𝑥²−𝑟₁,𝑟₁>0−𝑦²−𝑥²+𝑟₂,𝑟₂>0 (2.6) are real Hessian polynomials under  𝐻^ℝ₃. Moreover, a polynomial 𝑔∈𝔸^ℝ₂ of degree two is a real Hessian polynomial under  𝐻^ℝ₃ if and only if it belongs to the orbit of one of those polynomials. (2)The real polynomials of degree two 𝑦²𝑦²−𝑥²−𝑟₃,𝑟₃≤0𝑦²+𝑥²−𝑟₄,𝑟₄∈ℝ−𝑦²−𝑥²+𝑟₅,𝑟₅𝑦≤0₂+𝑥𝑦²−𝑟₆,𝑟₆∈ℝ^∗−𝑦²−𝑟₇,𝑟₇∈ℝ^∗ (2.7) are not real Hessian polynomials under 𝐻^ℝ₃. Moreover, a polynomial 𝑔∈𝔸^ℝ₂ of degree two is not a real Hessian polynomial under 𝐻^ℝ₃ if and only if it belongs to the orbit of one of those polynomials.

It is well known that the complex affine classification of conics is given by the normal forms: 𝑦²=𝑥 (parabola), 𝑦²=𝑥²+1 (general conic), 𝑦²=𝑥² (line pair), 𝑦²=1 (parallel lines), and finally 𝑦²=0 (double line). Therefore, we have the following corollary.

Corollary 2.8. All the complex affine conics, except the parallel lines, are complex Hessian curves of polynomials in 𝔸^ℂ₃.

In the next section we will give a geometric proof of this corollary.

Proposition 2.9. Let 𝑔(𝑥,𝑦)=𝑏₂₀𝑥²+𝑏₁₁𝑥𝑦+𝑏₀₂𝑦²+𝑏₀₀ be a polynomial in 𝔸^ℂ₂ with 𝑏₂₀(4𝑏₂₀𝑏₀₂−𝑏²₁₁)≠0. Then the set (𝐻^ℂ₃)⁻¹(𝑔) is an analytic subvariety on 𝔸^ℂ₃ which is the union of analytic subvarieties parametrized by the following. (1)𝐹^±₁∶ℂ⁴→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄)↦(𝑎₃₀=𝑡₁,𝑎₂₁=(1/2)(3𝑏₁₁𝑡₁±9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀/𝑏₂₀),𝑎₁₂ = (−6𝑡₁𝑏₀₂𝑏₂₀+3𝑏₁₁𝑡₁±9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀)/2𝑏²₂₀, 𝑎₀₃ = (−1/6𝑏²₂₀) (±𝑏₀₂9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀+6𝑏₁₁𝑡₁𝑏₀₂−(3𝑏₁₁𝑡₁±9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀)𝑏²₁₁/𝑏₂₀+3𝑏₁₁𝑏₀₂𝑡₁), 𝑎₂₀=𝑏₂₀𝑏₀₀/4𝑏₂₀𝑏₀₂−𝑏²₁₁, 𝑎₁₁=𝑏₁₁𝑏₀₀/4𝑏₂₀𝑏₀₂−𝑏²₁₁, 𝑎₀₂=𝑏₀₂𝑏₀₀/(4𝑏₂₀𝑏₀₂−𝑏²₁₁), 𝑎₁₀=𝑡₂, 𝑎₀₁=𝑡₃,𝑎₀₀=𝑡₄).(2)𝐹^±₂: ℂ⁴→𝔸^ℂ₃; (𝑡₁,𝑡₂,𝑡₃,𝑡₄)↦(𝑎₃₀=𝑡₁,𝑎₂₁=(3𝑏₁₁𝑡₁±9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀)/2𝑏₂₀, 𝑎₁₂ = (3𝑏₁₁𝑡₁−6𝑡₁𝑏₀₂𝑏₂₀±9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀)/2𝑏²₂₀, 𝑎₀₃ = (−1/6𝑏²₂₀) (±𝑏₀₂9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀+3𝑏₀₂𝑏₁₁𝑡₁+6𝑏₁₁𝑡₁𝑏₀₂−(3𝑏³₁₁𝑡₁±𝑏²₁₁9𝑡²₁𝑏²₁₁−36𝑡²₁𝑏₀₂𝑏₂₀−𝑏³₂₀)/𝑏₂₀), 𝑎₂₀ = −𝑏₂₀𝑏₀₀/4𝑏₂₀𝑏₀₂−𝑏²₁₁, 𝑎₁₁ = −𝑏₁₁𝑏₀₀/4𝑏₂₀𝑏₀₂−𝑏²₁₁, 𝑎₀₂ = −𝑏₀₂𝑏₀₀/4𝑏₂₀𝑏₀₂−𝑏²₁₁, 𝑎₁₀ = 𝑡₂,𝑎₀₁ = 𝑡₃,𝑎₀₀ = 𝑡₄).

From the study of the previous fibers, a natural question arises. What is the relation between the set of critical points of  𝐻^ℂ_𝑛  and the set of singular Hessian curves defined by polynomials in 𝔸^ℂ_2𝑛−4?

We define 𝔸^ℂ_𝑛𝑅∶={𝑓∈ℂ[𝑥,𝑦]∣2≤deg(𝑓)≤𝑛}. Let us describe the relation between the set of critical points of 𝐻^ℂ_3𝑅∶𝔸^ℂ_3𝑅→𝔸^ℂ₂ and the set of polynomials in 𝔸^ℂ₂ such that they define singular Hessian curves.

Proposition 2.10. If 𝑓∈𝔸^ℂ_3𝑅 is a critical point of the map 𝐻^ℂ_3𝑅∶𝔸^ℂ_3𝑅→𝔸^ℂ₂, then the Hessian curve 𝐻^ℂ_3𝑅(𝑓)(𝑥,𝑦)=0 is singular or it has degree one.

For the general case, we have the following conjecture.

Conjecture 2.11. If 𝑓∈𝔸_𝑛𝑅 is a critical point of the map 𝐻^ℂ_𝑛𝑅∶𝔸^ℂ_𝑛𝑅→𝔸^ℂ_2𝑛−4, then its Hessian curve, Hess𝑓(𝑥,𝑦)=0, is singular or has degree less than 2𝑛−4.

2.4. Degree of 𝑔 Equal to Three

The following theorem is one of the most important of this paper.

Theorem 2.12. Complex Case
(1)The curves defined by Table 1 of complex cubic curves (see [6]) are complex Hessian curves under 𝐻^ℂ₄.(2)The curves defined by Table 2 of complex cubic curves are not complex Hessian curves under 𝐻^ℂ₄.Real Case
(1)The normal form of curves (see [7]) defined in Table 3 is a real Hessian curve under 𝐻^ℝ₄. Moreover, a curve 𝑔∈𝔸^ℝ₃ of degree three is a real Hessian curve under 𝐻^ℝ₄ if and only if its polynomial belongs to the orbit of one of those polynomials. (2)The normal form of curves defined by the polynomials in Table 4 is not a real Hessian polynomial under  𝐻^ℝ₄. Moreover, a curve 𝑔∈𝔸^ℝ₃ of degree three is not a real Hessian curve under  𝐻^ℝ₄ if and only if its polynomial belongs to the orbit of one of those polynomials.

Table 1 

Complex Hessian cubic curves.

Table 2 

Complex cubic curves which are not complex Hessian curves.

Table 3 

Real Hessian polynomials.

Table 4 

Real polynomials which are not Hessian polynomials.

3. Proofs

Let us consider the complex Hessian map 𝐻^ℂ_𝑛∶𝔸^ℂ_𝑛→𝔸^ℂ_2𝑛−4𝑓↦Hess𝑓 and recall that the fiber of 𝑔∈𝔸_2𝑛−4 under 𝐻^ℂ_𝑛 is the set (𝐻^ℂ_𝑛)⁻¹(𝑔)∶={𝑓∈𝔸^ℂ_𝑛∣𝐻^ℂ_𝑛(𝑓)=𝑔}.

Proof. Let 𝑓(𝑥,𝑦)=∑²_𝑖+𝑗=0𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗∈𝔸^ℂ₂. A direct calculus shows that 𝑓_𝑥𝑥=2𝑎₂₀, 𝑓_𝑥𝑦=𝑎₁₁, and 𝑓_𝑦𝑦=2𝑎₀₂. Therefore, 𝐻^ℂ₂(𝑓)=4𝑎₂₀𝑎₀₂−𝑎²₁₁. (3.1) For each 𝑔∈𝔸^ℂ₀, consider 𝑆(𝑔)={(𝑎₀₀,𝑎₁₀,𝑎₀₁,𝑎₂₀,𝑎₁₁,𝑎₀₂)∶𝑎²₁₁=4𝑎₂₀𝑎₀₂−𝑔}. To show that (𝐻^ℂ₂)⁻¹(𝑔)=𝑆(𝑔) it is enough to show (𝐻^ℂ₂)⁻¹(𝑔)⊂𝑆 because a direct substitution shows 𝑆⊂(𝐻^ℂ₂)⁻¹(𝑔). Therefore, let us consider 𝑓∈(𝐻^ℂ₂)⁻¹(𝑔), that is, 𝐻^ℂ₂(𝑓)=4𝑎₂₀𝑎₀₂−𝑎²₁₁=𝑔. Hence 𝑎²₁₁=4𝑎₂₀𝑎₀₂−𝑔 and the first part of the claim is done. Finally, the derivative of 𝐻^ℂ₂ is given by 𝐷𝐻^ℂ₂(𝑓)=(0,0,0,4𝑎₀₂,−2𝑎₁₁,4𝑎₂₀). (3.2) Therefore, we conclude the proof of the result.

Proof. The polynomial 𝑓(𝑥,𝑦)=𝑥²+(𝑔/4)𝑦² satisfies that 𝐻^𝕂₂(𝑓)=𝑔 in complex or real case.

Lemma 3.1. If 𝑓(𝑥,𝑦)=∑³_𝑖+𝑗=0𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗∈𝔸^ℂ₃, then the map 𝐻^ℂ₃∶𝐴^ℂ₃→𝐴^ℂ₂ is 𝐻^ℂ₃(𝑓)=𝑏₂₀𝑥²+𝑏₁₁𝑥𝑦+𝑏₀₂𝑦²+𝑏₁₀𝑥+𝑏₀₁𝑦+𝑏₀₀, (3.3) where the 𝑏_𝑟𝑠 coefficients satisfy the following system of quadratic equations: 𝑏₂₀=−4𝑎²₂₁+12𝑎₃₀𝑎₁₂,𝑏₁₁=−4𝑎₂₁𝑎₁₂+36𝑎₃₀𝑎₀₃,𝑏₀₂=−4𝑎²₁₂+12𝑎₂₁𝑎₀₃,𝑏₁₀=12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁,𝑏₀₁=12𝑎₂₀𝑎₀₃+4𝑎₂₁𝑎₀₂−4𝑎₁₂𝑎₁₁,𝑏₀₀=−𝑎²₁₁+4𝑎₂₀𝑎₀₂. (3.4)

Proof. For 𝑔∈𝔸^ℂ₀ we have that, by Lemma 3.1, 𝐻^ℂ₃(𝑓)=𝑔 is equivalent to the following system of equations: 0=−4𝑎²₂₁+12𝑎₃₀𝑎₁₂,0=−4𝑎₂₁𝑎₁₂+36𝑎₃₀𝑎₀₃,0=−4𝑎²₁₂+12𝑎₂₁𝑎₀₃,0=12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁,0=12𝑎₂₀𝑎₀₃+4𝑎₂₁𝑎₀₂−4𝑎₁₂𝑎₁₁,𝑔=−𝑎²₁₁+4𝑎₂₀𝑎₀₂. (3.5)(3.6)(3.7)(3.8)(3.9)(3.10) Let 𝑆_𝑔 be the set obtained by the union of the image of parametrizations 𝐹₁,…,𝐹₄. To prove that (𝐻^ℂ₃)⁻¹(𝑔)=𝑆_𝑔, it is enough to show that (𝐻^ℂ₃)⁻¹(𝑔)⊂𝑆_𝑔 because a direct substitution shows that 𝑆_𝑔⊂(𝐻^ℂ₃)⁻¹(𝑔). Now, to prove (𝐻^ℂ₃)⁻¹(𝑔)⊂𝑆_𝑔 we will consider two cases: Case 1 is when 𝑎₁₂≠0 and Case 2 is when 𝑎₁₂=0.Case 1. In this case, from a direct substitution we obtain 𝑎₃₀=𝑎²₂₁/3𝑎₁₂, 𝑎₀₃=𝑎²₁₂/3𝑎₂₁, 𝑎₁₂∈ℂ^∗, and (3.8), (3.9), and (3.10). To solve these equations we will assume the following.
Subcase 1.1. 𝑎₂₀=0. In this case we obtain 𝑎₁₁=±√−𝑔; 𝑎₀₂=±𝑎₁₂𝑎₂₁√−𝑔/𝑎²₂₁. All this values together are contained in the set whose parametrization is 𝐹₁.
Subcase 1.2. 𝑎₂₀≠0. This case will be subdivided in two subcases.(1)𝑎₀₂=0. First, we obtain 𝑎₁₁ from (3.10). Later, from a substitution of 𝑎₁₁ together with the value of 𝑎₃₀ in (3.8) we get 𝑎₂₀𝑎₁₂±𝑎₂₁√−𝑔=0. This equation implies 𝑎₁₁=2𝑎₂₀𝑎₁₂±𝑎₂₁√−𝑔/𝑎₂₁. All these values together are contained in the set whose parametrization is 𝐹₁.(2)𝑎₀₂≠0. In this case, from (3.8) we obtain𝑎₁₁=1𝑎₂₁𝑎₁₂(𝑎²₂₁𝑎₀₂+𝑎₂₀𝑎²₁₂). (3.11) A substitution of 𝑎₁₁ in (3.10) gives us the quadratic equation in the 𝑎₂₀ variable: 𝑎⁴₂₁𝑎²₀₂−2𝑎²₂₁𝑎²₁₂𝑎₀₂+(𝑎²₂₀𝑎⁴₁₂+𝑔𝑎²₂₁𝑎²₁₂)=0. (3.12) Solving this quadratic equation we get 𝑎₀₂=𝑎₁₂(𝑎₂₀𝑎₁₂±𝑎₂₁√−𝑔)/𝑎²₂₁. Finally, from a substitution of 𝑎₀₂ in (3.11) we get 𝑎₁₁=(2𝑎₂₀𝑎₁₂±𝑎₂₁√−𝑔)/𝑎₂₁. From these values we obtain the parametrization 𝐹₁.Case 2. From a direct calculus we obtain 𝑎₂₁=0 and the equations: 0=𝑎₃₀𝑎₀₃,0=𝑎₃₀𝑎₀₂,0=𝑎₂₀𝑎₀₃,𝑔=−𝑎²₁₁+4𝑎₂₀𝑎₀₂. (3.13) To solve these four equations we will consider two cases.
Subcase 2.1. 𝑎₀₂=0. From a direct substitution we get 𝑎₁₁=±√−𝑔 and the two equations: 0=𝑎₃₀𝑎₀₃,0=𝑎₃₀𝑎₀₂. (3.14) To solve these two equations we will consider two subcases.(1)𝑎₀₃=0. We obtain 𝑎₃₀∈ℂ and 𝑎₂₀∈ℂ. From all of these values we get the parametrization 𝐹₃.(2)𝑎₀₃≠0. We obtain 𝑎₃₀=𝑎₂₀=0. From all these values we get the parametrization 𝐹₄ when 𝑎₀₂=0.
Subcase 2.2. 𝑎₀₂≠0. We obtain 𝑎₂₀=(𝑎²₁₁+𝑔)/4𝑎₀₂ and the three equations: 0=𝑎₃₀𝑎₀₃,0=𝑎₃₀𝑎₀₂,0=𝑎₂₀𝑎₀₃. (3.15) To solve these three equations we will consider two subcases.(1)𝑎₀₃=0. We obtain 𝑎₃₀=0 and then the parametrization 𝐹₂.(2)𝑎₀₃≠0. We obtain 𝑎₃₀=0, 𝑎₂₀=0, as well as 𝑎₁₁=±√−𝑔. All these values together are included in the parametrization 𝐹₄ when 𝑎₀₂∈ℂ^∗. Therefore, we have obtained all parametrizations in the proposition and the proof is done.

Let 𝑔₁,𝑔₂∈𝕂[𝑥,𝑦]. We say that 𝑔₁ is in the orbit of 𝑔₂ (or 𝑔₂ is in the orbit of 𝑔₁) if they are equivalent by an affine transformation of the plane 𝕂² (where 𝕂=ℝ or ℂ).

Remark 3.2. If 𝑔₁ is in the orbit of 𝑔, then 𝑔₁ is a Hessian polynomial if and only if 𝑔 is a Hessian polynomial. This remark is due to the equality Hess((1/det𝑇)(𝑓∘𝑇))=(Hess𝑓)∘𝑇, where 𝑇∈Aff(2,𝕂), and Hess𝑓=𝑔.

Proof of Corollary 2.6. Note that the polynomial 𝑔(𝑥,𝑦)=𝑥∈𝔸^ℂ₂ is a complex Hessian polynomial under 𝐻^ℂ₃ because 𝑓(𝑥,𝑦)=(1/12)𝑥³+𝑦²∈(𝐻^ℂ₃)⁻¹(𝑔). On the other hand, every complex polynomial of degree one is in the orbit of 𝑔. By Remark 3.2 we have that every complex polynomial of degree one is a Hessian polynomial. The real case is analogous.

Proof of Proposition 2.5. Let 𝑔(𝑥,𝑦)=𝑏₁₀𝑥+𝑏₀₁𝑦+𝑏₀₀ be a polynomial of degree one in 𝔸^ℂ₂ with constant term. Then the expression 𝐻^ℂ₃(𝑓)=𝑔 is equivalent, by Lemma 3.1, to the system of equations: 0=−𝑎²₂₁+3𝑎₃₀𝑎₁₂,0=−𝑎₂₁𝑎₁₂+9𝑎₃₀𝑎₀₃,0=−𝑎²₁₂+3𝑎₂₁𝑎₀₃𝑏₁₀,=12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁𝑏₀₁,=12𝑎₂₀𝑎₀₃+4𝑎₂₁𝑎₀₂−4𝑎₁₂𝑎₁₁𝑏₀₀,=−𝑎²₁₁+4𝑎₂₀𝑎₀₂. (3.16)(3.17)(3.18)(3.19)(3.20)(3.21)
Denote by 𝑆_𝑔 the union of the images of 𝐹₁ and 𝐹₂. We shall prove that (𝐻^ℂ₃)⁻¹(𝑔)=𝑆_𝑔. After some calculus it is proved that 𝑆_𝑔⊂(𝐻^ℂ₃)⁻¹(𝑔). Now, we shall prove that (𝐻^ℂ₃)⁻¹(𝑔)⊂𝑆_𝑔.
Suppose that 𝑎₂₁,𝑎₁₂≠0. Multiply (3.19) by 𝑎₂₁ and (3.20) by 𝑎₁₂ and subtract the two obtained equations. It gives 4𝑎₂₀𝑎²₁₂−4𝑎₀₂𝑎²₂₁+12𝑎₃₀𝑎₀₂𝑎₁₂−12𝑎₀₃𝑎₂₀𝑎₂₁=𝑏₁₀𝑎₁₂−𝑏₀₁𝑎₂₁. (3.22)
From (3.16) and (3.18) we obtain 𝑎²₂₁=3𝑎₃₀𝑎₁₂ and 𝑎²₁₂=9𝑎₃₀𝑎₀₃, respectively, which we insert in (3.22) to obtain 𝑎₁₂𝑏₁₀=𝑏₀₁𝑎₂₁. (3.23) From (3.23) we obtain 𝑎₁₂=𝑏₀₁𝑎₂₁/𝑏₁₀ if 𝑏₁₀≠0 and we insert it in (3.16) to obtain 𝑎₃₀=𝑏₁₀𝑎₂₁3𝑏₀₁. (3.24) Analogous, we insert it in (3.18) to obtain 𝑎₀₃=𝑏²₀₁𝑎₂₁3𝑏²₁₀. (3.25) When we put (3.23), (3.24), and (3.25) in (3.17), it fulfills identically.
Case 1. Suppose that 𝑎₂₀≠0. From (3.21) it has 𝑎₀₂=𝑏₀₀+𝑎²₁₁4𝑎₂₀. (3.26) From (3.19) 𝑎₁₂=4𝑎₂₁𝑎₁₁−12𝑎₃₀𝑎₀₂+𝑏₁₀4𝑎₂₀. (3.27) From (3.23) we obtain 𝑎₂₁ and we replace it in (3.27). We replace also 𝑎₃₀ from (3.24) and 𝑎₀₂ to obtain 12𝑎²₂₀𝑏²₀₁𝑎₁₂=12𝑎₁₁𝑎₂₀𝑎₁₂𝑏₁₀𝑏₀₁−3𝑎₁₂𝑏²₁₀(𝑏₀₀+𝑎²₁₁)+3𝑎₂₀𝑏₁₀𝑏²₀₁. (3.28) We associate the terms containing 𝑎₁₂𝑎₁₂=𝑎₂₀𝑏₁₀𝑏²₀₁4𝑎²₂₀𝑏²₀₁−4𝑎₁₁𝑎₂₀𝑏₁₀𝑏₀₁+𝑏²₁₀𝑏₀₀+𝑏²₁₀𝑎²₁₁. (3.29) We replace this last expression of 𝑎₁₂ in (3.23) and we obtain 𝑎₂₁=𝑎₂₀𝑏²₁₀𝑏₀₁4𝑎²₂₀𝑏²₀₁−4𝑎₁₁𝑎₂₀𝑏₁₀𝑏₀₁+𝑏²₁₀𝑏₀₀+𝑏²₁₀𝑎²₁₁. (3.30) We replace the expression of 𝑎₂₁ in (3.24) and (3.25) and we have 𝑎₃₀=𝑎₂₀𝑏³₁₀3(4𝑎²₂₀𝑏²₀₁−4𝑎₁₁𝑎₂₀𝑏₁₀𝑏₀₁+𝑏²₁₀𝑏₀₀+𝑏²₁₀𝑎²₁₁),𝑎₀₃=𝑎₂₀𝑏³₀₁3(4𝑎²₂₀𝑏²₀₁−4𝑎₁₁𝑎₂₀𝑏₁₀𝑏₀₁+𝑏²₁₀𝑏₀₀+𝑏²₁₀𝑎²₁₁). (3.31) Note that those last four expressions are in the image of 𝐹₁.
Case 2. Suppose that 𝑎₂₀=0. From (3.21) we obtain 𝑎²₁₁=−𝑏₀₀. (3.32) From (3.19) 𝑎₀₂=4𝑎₂₁𝑎₁₁+𝑏₁₀12𝑎₃₀. (3.33) We replace the value of 𝑎₃₀ (obtained from (3.24)) and the expression 𝑎₁₁ in this last equation to obtain 𝑎₀₂=𝑏₀₁(±4𝑎₂₁−𝑏₀₀+𝑏₁₀)4𝑎₂₁𝑏₁₀. (3.34) The expression of 𝑎₁₂ is obtained from (3.23). The expressions of this case are in the image of 𝐹₂.

Proof of Theorem 2.7. Complex Case
(1)Let 𝑔(𝑥,𝑦)=𝑎𝑥²+𝑏𝑥𝑦+𝑐𝑦²+𝑑𝑥+𝑒𝑦+ℎ∈𝔸^ℂ₂ be a quadratic polynomial. By an affine transformation of the complex plane the polynomial 𝑔 is equivalent to the prenormal form 𝑃(𝑥,𝑦)=𝑦²−𝐴𝑥²−𝐵𝑥−𝐶, where 𝐴,𝐵,𝐶∈ℂ.If 𝐴≠0, then 𝑃 is in the orbit of 𝑦²−𝑥²+𝑘, where 𝑘∈ℂ. If 𝐴=0 and 𝐵≠0, then 𝑃 is in the orbit of 𝑦²+𝑥. Finally, if 𝐴=0,𝐵=0,𝐶≠0, then 𝑃 is in the orbit of 𝑦²−𝑚, where 𝑚∈ℂ.Note that the polynomial 𝑓(𝑥,𝑦)=(√−1/2)𝑥𝑦²∈𝔸^ℂ₃ satisfies 𝐻^ℂ₃(𝑓)=𝑦². The polynomial 𝑓(𝑥,𝑦)=𝑥𝑦²/2+𝑥²/2∈𝔸^ℂ₃ verifies 𝐻^ℂ₃(𝑓)=𝑥−𝑦². For each 𝑘∈ℂ, the polynomial 𝑓_𝑘(𝑥,𝑦)=𝑦³/6+𝑥²𝑦/2−(√−𝑘/2)𝑥²+(√−𝑘/2)𝑦²∈𝔸^ℂ₃ fulfills 𝐻^ℂ₃(𝑓_𝑘)=𝑦²−𝑥²−𝑘.(2)To verify that the polynomials 𝑦²+𝑟, where 𝑟∈ℂ^∗, are not complex Hessian polynomials, we used the computer algebra system Maple 9.5. In particular, the Groebner package with the graded reverse lexicographic monomial order. We obtained a reduced Groebner basis for the system 𝐻^ℂ₃(𝑓)=𝑦²+𝑟. The Groebner basis obtained was 1. So, by the Weak Nullstellensatz Theorem there is no solution to this system of equations.
Real Case
Analogous to the complex case, after a composition with an affine transformation of the real plane, the real quadratic polynomial 𝑔(𝑥,𝑦)=𝑎𝑥²+𝑏𝑥𝑦+𝑐𝑦²+𝑑𝑥+𝑒𝑦+ℎ is in the orbit of one of the normal forms: 𝑦²−𝑥²−𝑞₁,𝑦²+𝑥²−𝑞₂,−𝑦²−𝑥²−𝑞₃,𝑥−𝑦²,𝑦²+𝑥,𝑦²−𝑞₄,−𝑦²−𝑞₅, where 𝑞₁,…,𝑞₅∈ℝ.

Definition 3.3. We say that a complex polynomial is totally imaginary if the real part of all its coefficients is zero.

Lemma 3.4. Let 𝑔∈𝔸^ℝ_2𝑛−4 be a polynomial with real coefficients. Then 𝑔 is a real Hessian polynomial if and only if there exists a polynomial 𝑓 totally imaginary on the set (𝐻^ℂ_𝑛)⁻¹(−𝑔).

(1)By Lemma 3.4 and Proposition 2.9 we have that the polynomials 𝑦²−𝑥²−𝑟₁,−𝑦²−𝑥²−𝑟₂, with 𝑟₁,𝑟₂>0 are real Hessian polynomials. To finish this part we note that the polynomial 𝑓(𝑥,𝑦)=𝑥𝑦²/2+𝑥²/2∈𝔸^ℝ₃ verifies 𝐻^ℝ₃(𝑓)=𝑥−𝑦² and that 𝐻^ℝ₃(𝑥𝑦²/2)=−𝑦².(2)By Lemma 3.4 and Proposition 2.9 we have that the polynomials −𝑥²+𝑦²,−𝑥²−𝑦²,𝑦²−𝑥²−𝑟₃,𝑦²+𝑥²−𝑟₄,−𝑦²−𝑥²+𝑟₅, with 𝑟₃,𝑟₅<0,𝑟₄∈ℝ, are not real Hessian polynomials. On the other hand, the polynomials 𝑦²−𝑟₆,−𝑦²−𝑟₇, where 𝑟₆,𝑟₇∈ℝ^∗, are not real Hessian polynomials because the complex polynomial 𝑦²+𝑟,𝑟∈ℂ^∗ is not complex Hessian polynomial. To show that the polynomials 𝑦² and 𝑦²+𝑥 are not real Hessian polynomials, we use the same method of Groebner basis realized in the complex case.

Proof of Lemma 3.4. ⇒) By hypothesis there exists 𝑓(𝑥,𝑦)=∑^𝑛_𝑟+𝑠=0𝑎_𝑟𝑠𝑥^𝑟𝑦^𝑠∈𝔸^ℝ_𝑛 such that 𝐻^ℝ_𝑛(𝑓)=𝑔. Therefore, consider the totally imaginary polynomial 𝑖𝑓(𝑥,𝑦)=∑^𝑛_𝑟+𝑠=0𝑖𝑎_𝑟𝑠𝑥^𝑟𝑦^𝑠, which satisfies 𝐻^ℂ_𝑛(𝑖𝑓)=𝑖²𝐻^ℂ_𝑛(𝑓)=−𝑔.
⇐) By hypothesis there exists 𝑓 totally imaginary such that 𝐻^ℂ_𝑛(𝑓)=−𝑔. Therefore, 𝑖𝑓 is a real polynomial such that 𝐻^ℝ_𝑛(𝑖𝑓)=𝑖²𝐻^ℂ_𝑛(𝑓)=𝑔. Hence, 𝑔 is a real Hessian polynomial.

Proof of Corollary 2.8. Let us consider the map 𝜓∶𝔸^ℂ₃×ℂ²→ℂ³={(𝑤₁,𝑤₂,𝑤₃)} given by (𝑓,𝑝)↦(𝑓_𝑥𝑥(𝑝),𝑓_𝑥𝑦(𝑝),𝑓_𝑦𝑦(𝑝)). (3.35) If 𝑓(𝑥,𝑦)=∑³_𝑖+𝑗=0𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗∈𝔸^ℂ₃ and 𝑝=(𝑥,𝑦), then 𝜓(𝑓,𝑝)=(6𝑎₃₀𝑥+2𝑎₂₁𝑦+2𝑎₂₀,2𝑎₂₁𝑥+2𝑎₁₂𝑦+2𝑎₁₁,2𝑎₁₂𝑥+6𝑎₀₃𝑦+2𝑎₀₂). (3.36) For each fixed 𝑓∈𝔸^ℂ₃ let us consider 𝜓_𝑓∶ℂ²→ℂ₃ given by 𝜓_𝑓(𝑝)∶=𝜓(𝑓,𝑝).

Now, we are interested to describe the conditions in 𝑓 under which the image under 𝜓_𝑓 of ℂ²,𝜓_𝑓(ℂ²), is not a plane.

Lemma 3.5. The set 𝑆𝜓∶={𝑓∈𝔸^ℂ₃|𝜓_𝑓(ℂ²) is not a plane} is given by the union of the following sets: 𝐶𝑃₁∶={𝑓∈𝔸^ℂ₃∣𝑎₃₀=𝑎₂₁=𝑎₁₂=0},𝐶𝑃₂∶={𝑓∈𝔸^ℂ₃∣𝑎₃₀∈ℂ^∗,𝑎₁₂=𝑎²₂₁3𝑎₃₀,𝑎₀₃=𝑎³₂₁27𝑎²₃₀}. (3.37)

Proof. The Jacobian matrix of 𝜓_𝑓 is 3𝑎₃₀𝑎₂₁𝑎₂₁𝑎₁₂𝑎₁₂3𝑎₀₃. (3.38)𝜓_𝑓(ℂ²) is not a plane if and only if 𝐽𝜓_𝑓 has not maximal rank, which is equivalent to solve the system of equations given by the three minor equal to zero. That is, 3𝑎₁₂𝑎₃₀−𝑎²₂₁=0,3𝑎₂₁𝑎₀₃−𝑎²₁₂=0,9𝑎₀₃𝑎₃₀−𝑎₂₁𝑎₁₂=0. (3.39) The sets 𝐶𝑃₁ and 𝐶𝑃₂ are obtained by solving the last system of equations.

Lemma 3.6. Let 𝑓(𝑥,𝑦)=∑³_𝑖+𝑗=0𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗 be a complex polynomial such that 𝑓∈𝑆𝜓. Case 1. Suppose 𝑓∈𝐶𝑃₁. (1)If 𝑎₀₃≠0, then 𝜓_𝑓(ℂ²) is a parallel line to the 𝑤₃-axis in ℂ³. (2)If 𝑎₀₃=0, then 𝜓_𝑓(ℂ²) is the point (2𝑎₂₀,𝑎₁₁,2𝑎₀₂).Case 2. If 𝑓∈𝐶𝑃₁, then 𝜓_𝑓(ℂ²) is the line 𝑙_𝑓={(6𝑎₃₀𝑥+2𝑎₂₀,2𝑎₂₁𝑥+𝑎₁₁,2𝑎²₂₁3𝑎₃₀𝑥+2𝑎₀₂)∣𝑥∈ℂ}. (3.40)

Proof. Let 𝑓(𝑥,𝑦)=∑³_𝑖+𝑗=0𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗∈𝔸^ℂ₃. In virtue of Lemma 3.5 we have the following cases
Case 1. If 𝑓∈𝐶𝑃₁, then, by (3.36), 𝜓_𝑓(ℂ²)={(2𝑎₂₀,𝑎₁₁,6𝑎₃₀𝑦+2𝑎₀₂)∈ℂ³∣𝑦∈ℂ}.(1)If 𝑎₀₃≠0, then 𝜓_𝑓(ℂ²)={(2𝑎₂₀,𝑎₁₁,6𝑎₃₀𝑦+2𝑎₀₂)∈ℂ³∣𝑦∈ℂ}.(2)If 𝑎₀₃=0, then 𝜓_𝑓(ℂ²) is the point (2𝑎₂₀,𝑎₁₁,2𝑎₀₂).
Case 2. If 𝑓∈𝐶𝑃₂, then, by (3.36), 𝜓_𝑓(ℂ²)=(6𝑎₃₀𝑥+2𝑎₂₁𝑦+2𝑎₂₀,2𝑎₂₁𝑥+2𝑎²₂₁3𝑎₃₀𝑦+𝑎₁₁,2𝑎³₂₁9𝑎²₃₀𝑦+2𝑎²₂₁3𝑎₃₀𝑥+2𝑎₀₂). (3.41) Let 𝑣₁=(6𝑎₃₀,2𝑎₂₁,2𝑎²₂₁/3𝑎₃₀) be a nonzero vector (by hypothesis) and 𝑣₂=(2𝑎₂₁,2𝑎₂₁/3𝑎₃₀,2𝑎³₂₁/9𝑎²₃₀). Note that 𝑣₂=(𝑎₂₁/3𝑎₃₀)𝑣₁. Therefore, the set 𝑙_𝑓 is the same set of 𝜓_𝑓(ℂ²).

Let us consider the cone 𝐶={(𝑤₁,𝑤₂,𝑤₃)∈ℂ³∣𝑤₁𝑤₃−𝑤²₂=0}. We shall describe the set 𝜓_𝑓(ℂ²)∩𝐶 when 𝑓∈𝑆𝜓.

Lemma 3.7. Let 𝑓∈𝑆𝜓. (1)If 𝜓_𝑓(ℂ²) is a parallel line to the 𝑤₃-axis in ℂ³, then (a)𝜓_𝑓(ℂ²)∩𝐶 is a parallel line to the 𝑤₃-axis in ℂ³ whenever 𝑎₂₀=0; (b)𝜓_𝑓(ℂ²)∩𝐶 is the point (2𝑎₂₀,𝑎₁₁,(𝑎²₁₁−4𝑎₂₀𝑎₀₂)/2𝑎₂₀+2𝑎₂₀) whenever 𝑎₂₀≠0. (2)If 𝜓_𝑓(ℂ²) is the point (2𝑎₂₀,𝑎₁₁,2𝑎₀₂) and 4𝑎₂₀𝑎₀₂−𝑎²₁₁=0, then 𝜓_𝑓(ℂ²)∩𝐶 is the point (2𝑎₂₀,𝑎₁₁,2𝑎₀₂). (3)If 𝜓_𝑓(ℂ²) is the line 𝑙_𝑓, then (a)𝜓_𝑓(ℂ²)∩𝐶 is the point (6𝑎₃₀𝑥₀+2𝑎₂₀,2𝑎₂₁𝑥₀+𝑎₁₁,(2𝑎²₂₁/3𝑎₃₀)𝑥₀+2𝑎₀₂) if 𝛼=(1/3𝑎₃₀)(36𝑎²₃₀𝑎₀₂+4𝑎₂₀𝑎²₂₁−12𝑎₂₁𝑎₁₁𝑎₃₀)≠0, where 𝑥₀=(1/𝛼)(𝑎²₁₁−4𝑎₂₀𝑎₀₂); (b)𝜓_𝑓(ℂ²)∩𝐶 is the line 𝑙_𝑓 whenever 𝛼=0.

Proof. In virtue of Lemma 3.6 we have the following cases. (1)If 𝜓_𝑓(ℂ²)∩𝐶 is a parallel line to the 𝑤₃-axis, then 𝜓_𝑓(ℂ²)∩𝐶={(2𝑎₂₀,𝑎₁₁,6𝑎₀₃𝑦+2𝑎₀₂)∈ℂ³∣𝑦∈ℂ,2𝑎₂₀(6𝑎₀₃𝑦+2𝑎₀₂)−𝑎²₁₁=0}.(a)If 𝑎₂₀=0, then 𝜓_𝑓(ℂ²)∩𝐶 is a parallel line to the 𝑤₃-axis. (b)If 𝑎₂₀≠0, then 𝑦=(𝑎²₁₁−4𝑎₂₀𝑎₀₂)/12𝑎₂₀6𝑎₀₃ and 𝜓_𝑓(ℂ²)∩𝐶 is a point. (2)If 𝜓_𝑓(ℂ²)∩𝐶 is the point (2𝑎₂₀,𝑎₁₁,2𝑎₀₂), then 𝜓_𝑓(ℂ²)∩𝐶=(2𝑎₂₀,𝑎₁₁,2𝑎₀₂) if 4𝑎₂₀𝑎₀₂−𝑎²₁₁=0. (3)If 𝜓_𝑓(ℂ²)∩𝐶 is the line 𝑙_𝑓, then 𝜓_𝑓(ℂ²)∩𝐶=𝑙_𝑓∩{(6𝑎₃₀𝑥+2𝑎₂₀)((2𝑎²₂₁/3𝑎₃₀)𝑥+2𝑎₂₀)−(2𝑎₂₁𝑥+𝑎₁₁)²=0}. It means that 𝛼𝑥+4𝑎₂₀𝑎₀₂−𝑎²₁₁=0,where𝛼=36𝑎²₃₀𝑎₀₂+4𝑎₂₀𝑎²₂₁−12𝑎₂₁𝑎₁₁𝑎₃₀3𝑎₃₀. (3.42) Therefore, if 𝛼=0, then 4𝑎₂₀𝑎₀₂−𝑎²₁₁=0 and we obtain a line on 𝐶. If 𝛼≠0, then 𝜓_𝑓(ℂ²)∩𝐶 is a point.

With this lemma we finish the proof of corollary.

Proof of Proposition 2.9. Let 𝑔(𝑥,𝑦)=𝑏₂₀𝑥²+𝑏₁₁𝑥𝑦+𝑏₀₂𝑦²+𝑏₀₀∈𝔸^ℂ₂. Then the expression 𝐻^ℂ₃(𝑓)=𝑔 is equivalent, by Lemma 3.1, to the system of equations: 𝑏₀₂=−4𝑎²₂₁+12𝑎₃₀𝑎₁₂,𝑏₁₁=−4𝑎₂₁𝑎₁₂+36𝑎₃₀𝑎₀₃,𝑏₀₂=−4𝑎²₁₂+12𝑎₂₁𝑎₀₃,0=12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁,0=12𝑎₂₀𝑎₀₃+4𝑎₂₁𝑎₀₂−4𝑎₁₂𝑎₁₁,𝑏₀₀=−𝑎²₁₁+4𝑎₂₀𝑎₀₂. (3.43) Let 𝑆 be the union of the images of 𝐹₁ and 𝐹₂. A direct substitution shows that 𝑆⊂(𝐻^ℂ₃)⁻¹(𝑔). Therefore, to finish the proof it is enough to show that (𝐻^ℂ₃)⁻¹(𝑔)⊂𝑆. To check this last sentence we have used the computer algebra system Maple 9.5.

Proof. Real Case
We will proof that the plane curve 𝑔(𝑥,𝑦)=0, where 𝑔(𝑥,𝑦)=𝑥𝑦²−𝑥³+𝑥, is not a real Hessian curve (the other cases are analogous). To do that, we will prove that there are no real cuartic polynomial 𝑓 in (𝐻^ℂ₄)⁻¹(𝑔); that is, there are no real polynomial satisfying the system of equations: 24𝑎₄₀𝑎₂₂−9𝑎²₃₁=0,72𝑎₄₀𝑎₁₃−12𝑎₃₁𝑎₂₂=0,144𝑎₄₀𝑎₀₄+18𝑎₃₁𝑎₁₃−12𝑎²₂₂=0,72𝑎₃₁𝑎₀₄−12𝑎₂₂𝑎₁₃=0,24𝑎₀₄𝑎₂₂−9𝑎²₁₃=0,12𝑎₃₀𝑎₂₂+24𝑎₄₀𝑎₁₂−12𝑎₃₁𝑎₂₁=−1,36𝑎₃₀𝑎₁₃+72𝑎₄₀𝑎₀₃−12𝑎₂₁𝑎₂₂=0,36𝑎₀₃𝑎₃₁+72𝑎₀₄𝑎₃₀−12𝑎₁₂𝑎₂₂=1,12𝑎₀₃𝑎₂₂+24𝑎₀₄𝑎₂₁−12𝑎₁₃𝑎₁₂=0,24𝑎₄₀𝑎₀₂+4𝑎₂₀𝑎₂₂+12𝑎₃₀𝑎₁₂−6𝑎₃₁𝑎₁₁−4𝑎²₂₁=0,12𝑎₃₁𝑎₀₂+12𝑎₂₀𝑎₁₃+36𝑎₃₀𝑎₀₃−4𝑎₂₁𝑎₁₂−8𝑎₁₁𝑎₂₂=0,24𝑎₀₄𝑎₂₀+4𝑎₀₂𝑎₂₂+12𝑎₀₃𝑎₂₁−6𝑎₁₃𝑎₁₁−4𝑎²₁₂=0,12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁=1,12𝑎₀₃𝑎₂₀+4𝑎₀₂𝑎₂₁−4𝑎₁₂𝑎₁₁=0,4𝑎₂₀𝑎₀₂−𝑎²₁₁=0. (3.44)(3.45)(3.46)(3.47)(3.48)(3.49)(3.50)(3.51)(3.52)(3.53)(3.54)(3.55)(3.56)(3.57)(3.58)
A Groebner bases for (3.44)–(3.48) with the graded reverse lexicographic monomial order is 3𝑎²₁₃−8𝑎₂₂𝑎₀₄,𝑎₂₂𝑎₁₃−6𝑎₃₁𝑎₀₄,2𝑎²₂₂−3𝑎₃₁𝑎₁₃−24𝑎₄₀𝑎₀₄,𝑎₃₁𝑎₂₂−6𝑎₄₀𝑎₁₃,3𝑎²₃₁−8𝑎₄₀𝑎₂₂,𝑎₁₃𝑎₃₁𝑎₀₄−16𝑎₄₀𝑎²₀₄,𝑎₃₁𝑎₄₀𝑎₁₃−16𝑎²₄₀𝑎₀₄. (3.59)
The set of common zeroes of these last 7 polynomials is the union of the sets {𝑎₄₀=𝑎₄₀,𝑎₃₁=0,𝑎₂₂=0,𝑎₁₃=0,𝑎₀₄=0},{𝑎₄₀=𝑎⁴₁₃256𝑎³₀₄,𝑎₃₁=𝑎³₁₃16𝑎²₀₄,𝑎₂₂=3𝑎²₁₃8𝑎₀₄,𝑎₁₃=𝑎₁₃,𝑎₀₄≠0}. (3.60)(3.61) On the one hand, replacing the solution (3.60) in (3.49)–(3.58), particularly in (3.56) we obtain 0=1. On the other hand, replacing the solution (3.61) in (3.49)–(3.58), we obtain the system 4𝑎₂₀𝑎₀₂−𝑎²₁₁=0,12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁−1=0,−4𝑎₁₂𝑎₁₁+4𝑎₂₁𝑎₀₂+12𝑎₂₀𝑎₀₃=0,12𝑎₃₀𝑎₁₂+3𝑎⁴₁₃𝑎₀₂32𝑎³₀₄−4𝑎²₂₁−3𝑎³₁₃𝑎₁₁8𝑎²₀₄+3𝑎₂₀𝑎²₁₃2𝑎₀₄=0,36𝑎₃₀𝑎₀₃+12𝑎₂₀𝑎₁₃−3𝑎²₁₃𝑎₁₁𝑎₀₄−4𝑎₂₁𝑎₁₂+3𝑎³₁₃𝑎₀₂4𝑎²₀₄=0,−6𝑎₁₃𝑎₁₁+24𝑎₂₀𝑎₀₄+3𝑎²₁₃𝑎₀₂2𝑎₀₄+12𝑎₂₁𝑎₀₃−4𝑎²₁₂=0,−3𝑎³₁₃𝑎₂₁4𝑎²₀₄+9𝑎₃₀𝑎²₁₃2𝑎₀₄+3𝑎⁴₁₃𝑎₁₂32𝑎³₀₄+1=0,9𝑎⁴₁₃𝑎₀₃32𝑎³₀₄+36𝑎₃₀𝑎₁₃−9𝑎₂₁𝑎²₁₃2𝑎₀₄=0,9𝑎³₁₃𝑎₀₃4𝑎²₀₄−9𝑎²₁₃𝑎₁₂2𝑎₀₄+72𝑎₃₀𝑎₀₄−1=0,−12𝑎₁₃𝑎₁₂+9𝑎²₁₃𝑎₀₃2𝑎₀₄+24𝑎₂₁𝑎₀₄=0. (3.62) From the last equation we obtain that 𝑎₂₁=−𝑎₁₃(−8𝑎₁₂𝑎₀₄+3𝑎₁₃𝑎₀₃)/16𝑎²₀₄ and from the sixth equation, 𝑎₂₀=−6𝑎²₁₃𝑎₀₂𝑎₀₄−24𝑎₁₃𝑎₁₁𝑎²₀₄+24𝑎₁₃𝑎₀₃𝑎₁₂𝑎₀₄−9𝑎²₁₃𝑎²₀₃−16𝑎²₁₂𝑎²₀₄96𝑎³₀₄. (3.63)
When we replace these expressions in the last four equations of the last system (we do not write the other equations because we do not need it): −18𝑎⁴₁₃𝑎₁₂𝑎₀₄+9𝑎⁵₁₃𝑎₀₃+288𝑎₃₀𝑎²₁₃𝑎³₀₄+64𝑎⁴₀₄64𝑎⁴₀₄=0,9𝑎₁₃(32𝑎₃₀𝑎³₀₄−2𝑎₁₂𝑎²₁₃𝑎₀₄+𝑎³₁₃𝑎₀₃)8𝑎³₀₄=0,9𝑎³₁₃𝑎₀₃−18𝑎₁₂𝑎²₁₃𝑎₀₄+288𝑎₃₀𝑎³₀₄−4𝑎²₀₄4𝑎²₀₄=0,0=0. (3.64)(3.65)(3.66)(3.67) From (3.65) we have that 𝑎₁₃=0 or 32𝑎₃₀𝑎³₀₄−2𝑎₁₂𝑎²₁₃𝑎₀₄+𝑎³₁₃𝑎₀₃=0. If 𝑎₁₃=0, then (3.64) becomes −1=0.
If 32𝑎₃₀𝑎³₀₄−2𝑎₁₂𝑎²₁₃𝑎₀₄+𝑎³₁₃𝑎₀₃=0, then 𝑎₃₀=(2𝑎₁₂𝑎²₁₃𝑎₀₄−𝑎³₁₃𝑎₀₃)/32𝑎³₀₄. Substituing 𝑎₃₀ in (3.66), we obtain −1=0.

Lemma 3.8. The set, 𝐶𝑣(3), of critical points of the map 𝐻^ℂ_3𝑅∶𝐴^ℂ_3𝑅→𝐴^ℂ₂ is the union of the following six sets: 𝑆₁={𝑎₀₂=9𝑎₀₃𝑎₁₁𝑎₃₀−6𝑎₀₃𝑎₂₁𝑎₂₀−𝑎₁₂𝑎₁₁𝑎₂₁+2𝑎²₁₂𝑎₂₀2(3𝑎₁₂𝑎₃₀−𝑎²₂₁),3𝑎₁₂𝑎₃₀−𝑎²₂₁𝑆≠0},₂={𝑎₁₁=0,𝑎₂₁=0,𝑎₁₂𝑆=0},₃={𝑎₁₁=2𝑎₁₂𝑎₂₀𝑎₂₁,𝑎₃₀=𝑎²₂₁3𝑎₁₂,𝑎₂₁𝑎₁₂𝑆≠0},₄={𝑎₃₀=0,𝑎₂₁=0,𝑎₁₂𝑆=0},₅={𝑎₂₀=0,𝑎₃₀=0,𝑎₂₁𝑆=0},₆={𝑎₁₂=𝑎²₂₁3𝑎₃₀,𝑎₀₃=𝑎³₂₁27𝑎²₃₀,𝑎₃₀≠0}. (3.68)

Proof. Let 𝑓(𝑥,𝑦)=∑³_𝑖+𝑗=2𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗∈𝔸^ℂ_3𝑅. The Jacobian matrix of 𝐻^ℂ_3𝑅 is, 𝐽(𝐻^ℂ_3𝑅)=2𝑎₀₂−𝑎₁₁2𝑎₂₀00006𝑎₀₃−2𝑎₁₂2𝑎₂₁02𝑎₀₂−2𝑎₁₁6𝑎₂₀2𝑎₁₂−2𝑎₂₁6𝑎₃₀6𝑎₀₂−2𝑎₁₁2𝑎₂₀000006𝑎₀₃−4𝑎₁₂6𝑎₂₁00018𝑎₀₃−2𝑎₁₂−2𝑎₂₁18𝑎₃₀0006𝑎₁₂−4𝑎₂₁6𝑎₃₀0. (3.69) Let us denote by 𝑀_𝑘,𝑘=1,…,7, the 6×6 matrix obtained from 𝐽(𝐻^ℂ_3𝑅) by deleting the column 7−𝑘+1. On the other hand, 𝑓 is a critical point of 𝐻^ℂ_3𝑅 if and only if det(𝑀_𝑘)=0 for all 𝑘=1,…,7, that is, if and only if the following system of seven equations is satisfied: (−9𝑎₂₁𝑎₁₂𝑎₀₃+27𝑎₃₀𝑎²₀₃+2𝑎³₁₂)(𝐹)=0,(𝑎²₁₂𝑎₂₁−6𝑎²₂₁𝑎₀₃+9𝑎₁₂𝑎₃₀𝑎₀₃)(𝐹)=0,(𝑎²₂₁𝑎₁₂−6𝑎²₁₂𝑎₃₀+9𝑎₂₁𝑎₃₀𝑎₀₃)(𝐹)=0,(2𝑎³₂₁−9𝑎₁₂𝑎₂₁𝑎₃₀+27𝑎²₃₀𝑎₀₃)(𝐹)=0,(−3𝑎₀₃𝑎₂₁𝑎₁₁+𝑎²₁₂𝑎₁₁−𝑎₂₁𝑎₁₂𝑎₀₂+9𝑎₀₃𝑎₃₀𝑎₀₂)(𝐹)=0,(3𝑎₀₂𝑎₁₂𝑎₃₀+𝑎²₁₂𝑎₂₀−3𝑎₀₃𝑎₂₁𝑎₂₀−𝑎²₂₁𝑎₀₂)(𝐹)=0,(𝑎₁₁𝑎²₂₁−𝑎₁₂𝑎₂₁𝑎₂₀+9𝑎₂₀𝑎₃₀𝑎₀₃−3𝑎₁₂𝑎₁₁𝑎₃₀)(𝐹)=0, (3.70) where 𝐹=−9𝑎₀₃𝑎₁₁𝑎₃₀+6𝑎₀₃𝑎₂₁𝑎₂₀+6𝑎₀₂𝑎₁₂𝑎₃₀−2𝑎²₂₁𝑎₀₂+𝑎₁₂𝑎₁₁𝑎₂₁−2𝑎²₁₂𝑎₂₀. (3.71) Therefore 𝑓(𝑥,𝑦)=∑³_𝑖+𝑗=2𝑎_𝑖𝑗𝑥^𝑖𝑦^𝑗 satisfies the previous system of seven equations if and only if 𝑓∈𝑆=⋃⁶_𝑘=1𝑆_𝑘, where 𝑆 is the union of the six solutions 𝑆_𝑗. Therefore, the proof is done.

Lemma 3.9. Consider 𝑓∈𝔸^ℂ₃ and 𝑔=𝐻₃(𝑓). The curve 𝑔=0 is singular if and only if −2𝑎²₁₂𝑎₂₀+6𝑎₁₂𝑎₀₂𝑎₃₀+6𝑎₂₁𝑎₀₃𝑎₂₀+𝑎₂₁𝑎₁₂𝑎₁₁−2𝑎²₂₁𝑎₀₂−9𝑎₀₃𝑎₁₁𝑎₃₀=0,−𝑎²₂₁𝑎²₁₂−18𝑎₂₁𝑎₁₂𝑎₃₀𝑎₀₃+27𝑎²₃₀𝑎²₀₃+4𝑎³₁₂𝑎₃₀+4𝑎³₂₁𝑎₀₃≠0, (3.72) or, −9𝑎₁₂𝑎₂₁𝑎₀₃𝑎₂₀−𝑎₂₁𝑎₁₁𝑎²₁₂−𝑎₁₂𝑎²₂₁𝑎₀₂+27𝑎₃₀𝑎²₀₃𝑎₂₀+2𝑎₂₀𝑎³₁₂+1.8+6𝑎²₂₁𝑎₀₃𝑎₁₁−9𝑎₃₀𝑎₂₁𝑎₀₃𝑎₀₂+6𝑎₃₀𝑎₀₂𝑎²₁₂−9𝑎₃₀𝑎₀₃𝑎₁₁𝑎₁₂=01,−𝑎²₂₁𝑎²₁₂−18𝑎₂₁𝑎₁₂𝑎₃₀𝑎₀₃+27𝑎²₃₀𝑎²₀₃+4𝑎³₁₂𝑎₃₀+4𝑎³₂₁𝑎₀₃=0.2.1, (3.73)

Proof. Suppose 𝑓(𝑥,𝑦)=∑³_𝑟+𝑠=0𝑎_𝑟𝑠𝑥^𝑟𝑦^𝑠; then 𝐻₃(𝑓)=−𝑎²₁₁+4𝑎₂₀𝑎₀₂+(12𝑎₃₀𝑎₀₂+4𝑎₂₀𝑎₁₂−4𝑎₂₁𝑎₁₁)𝑥+(12𝑎₂₀𝑎₀₃−4𝑎₁₂𝑎₁₁+4𝑎₂₁𝑎₀₂)𝑦+(−4𝑎²₂₁+12𝑎₃₀𝑎₁₂)𝑥²+(−4𝑎₂₁𝑎₁₂+36𝑎₃₀𝑎₀₃)𝑦𝑥+(−4𝑎²₁₂+12𝑎₂₁𝑎₀₃)𝑦². (3.74) The curve 𝑔=𝐻^ℂ₃(𝑓) is singular if and only if the system 𝑔_𝑥(𝑝)=𝑔_𝑦(𝑝)=𝑔(𝑝)=0 has a solution for some 𝑝 in ℂ². The system formed by 𝑔_𝑥=0,𝑔_𝑦=0 is (6𝑎₃₀𝑎₁₂−2𝑎²₂₁)𝑥+(9𝑎₃₀𝑎₀₃−𝑎₂₁𝑎₁₂)𝑦+3𝑎₃₀𝑎₀₂+𝑎₂₀𝑎₁₂−𝑎₂₁𝑎₁₁=0,(9𝑎₃₀𝑎₀₃−𝑎₂₁𝑎₁₂)𝑥+(6𝑎₂₁𝑎₀₃−2𝑎²₁₂)𝑦+3𝑎₂₀𝑎₀₃−𝑎₁₂𝑎₁₁+𝑎₂₁𝑎₀₂=0. (3.75) The proof concludes by analyzing the system when its determinant is distinct from zero and when it is zero.

Proof of Proposition 2.10. Let us consider the sets 𝑆_𝑘,𝑘=1,…,6, of Lemma 3.8. If 𝑓∈𝑆₁, then 𝐻^ℂ₃1(𝑓)=3𝑎₁₂𝑎₃₀−𝑎²₂₁×[𝑥²(36𝑎²₃₀𝑎²₁₂+4𝑎⁴₂₁−24𝑎₃₀𝑎₁₂𝑎²₂₁)+𝑥𝑦(108𝑎²₃₀𝑎₀₃𝑎₁₂+36𝑎₃₀𝑎₀₃𝑎²₂₁+4𝑎³₂₁𝑎₁₂−12𝑎₂₁𝑎²₁₂𝑎₃₀)+𝑦²(4𝑎²₁₂𝑎²₂₁−12𝑎³₁₂𝑎₃₀+36𝑎₂₁𝑎₀₃𝑎₁₂𝑎₃₀−12𝑎³₂₁𝑎₀₃)+𝑥(24𝑎₃₀𝑎²₁₂𝑎₂₀+54𝑎²₃₀𝑎₀₃𝑎₁₁+4𝑎³₂₁𝑎₁₁−4𝑎₂₀𝑎₁₂𝑎²₂₁−36𝑎₃₀𝑎₀₃𝑎₂₁𝑎₂₀−18𝑎₃₀𝑎₁₂𝑎₁₁𝑎₂₁)+𝑦(−12𝑎²₁₂𝑎₁₁𝑎₃₀+18𝑎₂₁𝑎₀₃𝑎₁₁𝑎₃₀+36𝑎₂₀𝑎₀₃𝑎₁₂𝑎₃₀−24𝑎²₂₁𝑎₀₃𝑎₂₀+2𝑎²₂₁𝑎₁₂𝑎₁₁+4𝑎₂₁𝑎²₁₂𝑎₂₀)−3𝑎²₁₁𝑎₁₂𝑎₃₀−12𝑎₀₃𝑎₂₁𝑎²₂₀+18𝑎₂₀𝑎₀₃𝑎₁₁𝑎₃₀−2𝑎₂₀𝑎₁₂𝑎₁₁𝑎₂₁+4𝑎²₁₂𝑎²₂₀+𝑎²₁₁𝑎²₂₁]. (3.76) Denote by ℎ the polynomial 𝐻^ℂ₃(𝑓). Then, 1ℎ𝑥=3𝑎₁₂𝑎₃₀−𝑎²₂₁×[𝑥(72𝑎²₃₀𝑎²₁₂−48𝑎₃₀𝑎₁₂𝑎²₂₁+8𝑎⁴₂₁)+𝑦(108𝑎²₃₀𝑎₀₃𝑎₁₂+4𝑎³₂₁𝑎₁₂−36𝑎₃₀𝑎₀₃𝑎²₂₁−12𝑎₂₁𝑎²₁₂𝑎₃₀)+4𝑎³₂₁𝑎₁₁+54𝑎²₃₀𝑎₀₃𝑎₁₁−4𝑎₂₀𝑎₁₂𝑎²₂₁+24𝑎₃₀𝑎²₁₂𝑎₂₀−36𝑎₃₀𝑎₀₃𝑎₂₁𝑎₂₀−18𝑎₃₀𝑎₁₂𝑎₁₁𝑎₂₁1],ℎ𝑦=3𝑎₁₂𝑎₃₀−𝑎²₂₁×[𝑥(108𝑎²₃₀𝑎₀₃𝑎₁₂−36𝑎₃₀𝑎₀₃𝑎²₂₁−12𝑎₂₁𝑎²₁₂𝑎₃₀+4𝑎³₂₁𝑎₁₂)+𝑦(8𝑎²₁₂𝑎²₂₁−24𝑎³₁₂𝑎₃₀−24𝑎³₂₁𝑎₀₃+72𝑎₂₁𝑎₀₃𝑎₁₂𝑎₃₀)+4𝑎₂₁𝑎²₁₂𝑎₂₀−24𝑎²₂₁𝑎₀₃𝑎₂₀+2𝑎²₂₁𝑎₁₂𝑎₁₁−12𝑎²₁₂𝑎₁₁𝑎₃₀+18𝑎₂₁𝑎₀₃𝑎₁₁𝑎₃₀+36𝑎₂₀𝑎₀₃𝑎₁₂𝑎₃₀]. (3.77) The point (𝑥,𝑦) where the Hessian curve is singular is 𝑥=−2𝑎₂₀𝑎₁₂+𝑎₂₁𝑎₁₁2(3𝑎₁₂𝑎₃₀−𝑎²₂₁),𝑦=−3𝑎₁₁𝑎₃₀−2𝑎₂₁𝑎₂₀2(3𝑎₁₂𝑎₃₀−𝑎²₂₁). (3.78) If 𝑓∈𝑆₂, then 𝐻^ℂ₃(𝑓)=(6𝑎₃₀𝑥+2𝑎₂₀)(6𝑎₀₃𝑦+2𝑎₀₂),ℎ𝑥=6𝑎₃₀(6𝑎₀₃𝑦+2𝑎₀₂),andℎ𝑦=6(6𝑎₃₀𝑥+2𝑎₂₀)𝑎₀₃.
The point (𝑥,𝑦) where the Hessian curve is singular is𝑥=−𝑎₂₀/3𝑎₃₀,𝑦=−𝑎₀₂/3𝑎₀₃.
If 𝑓∈𝑆₃, then 𝐻^ℂ₃(𝑓)=−4(𝑎²₂₁𝑥+𝑎₂₁𝑦𝑎₁₂+𝑎₂₀𝑎₁₂)(𝑦(−3𝑎²₂₁𝑎₀₃+𝑎₂₁𝑎²₁₂)−𝑎²₂₁𝑎₀₂+𝑎₂₀𝑎²₁₂)𝑎₁₂𝑎²₂₁,ℎ𝑥=−4(𝑦(−3𝑎²₂₁𝑎₀₃+𝑎₂₁𝑎²₁₂)−𝑎²₂₁𝑎₀₂+𝑎₂₀𝑎²₁₂)𝑎₁₂,ℎ𝑦=−4(𝑦(−3𝑎²₂₁𝑎₀₃+𝑎₂₁𝑦𝑎²₁₂)−𝑎²₂₁𝑎₀₂+𝑎₂₀𝑎²₁₂)𝑎₂₁−4(𝑎²₂₁𝑥+𝑎₂₁𝑦𝑎₁₂+𝑎₂₀𝑎₁₂)(−3𝑎²₂₁𝑎₀₃+𝑎₂₁𝑎²₁₂)𝑎₁₂𝑎²₂₁. (3.79) The point (𝑥,𝑦) where the Hessian curve is singular is 𝑎𝑥=¹²(−𝑎₂₁𝑎₀₂+3𝑎₂₀𝑎₀₃)𝑎₂₁(−3𝑎₂₁𝑎₀₃+𝑎²₁₂),𝑦=−−(−𝑎²₂₁𝑎₀₂+𝑎₂₀𝑎²₁₂)𝑎₂₁(−3𝑎₂₁𝑎₀₃+𝑎²₁₂). (3.80) If −3𝑎₂₁𝑎₀₃+𝑎²₁₂=0, then 𝐻^ℂ₃(𝑓) is of degree one.
If 𝑓∈𝑆₄, then 𝐻^ℂ₃(𝑓)=12𝑎₂₀𝑎₀₃𝑦+4𝑎₂₀𝑎₀₂−𝑎²₁₁,ℎ𝑥=0,andℎ𝑦=12𝑎₂₀𝑎₀₃.
The Hessian curve in this case has degree one.
If 𝑓∈𝑆₅, then 𝐻^ℂ₃(𝑓)=−(2𝑎₁₂𝑦+𝑎₁₁)²,ℎ𝑥=0,ℎ𝑦=−4(2𝑎₁₂𝑦+𝑎₁₁)𝑎₁₂.
The point (𝑥,𝑦) where the Hessian curve is singular is 𝑦=−𝑎₁₁/2𝑎₁₂.
If 𝑓∈𝑆₆, then 𝐻^ℂ₃(𝑓)=(108𝑎³₃₀𝑎₀₂+12𝑎₂₀𝑎²₂₁𝑎₃₀−36𝑎₂₁𝑎²₃₀𝑎₁₁9𝑎²₃₀)𝑥+(36𝑎₂₁𝑎₀₂𝑎²₃₀+4𝑎₂₀𝑎³₂₁−12𝑎²₂₁𝑎₁₁𝑎₃₀9𝑎²₃₀)𝑦+36𝑎₂₀𝑎₀₂𝑎²₃₀−9𝑎²₁₁𝑎²₃₀9𝑎²₃₀. (3.81) The Hessian curve in this case has degree one.

Acknowledgments

The authors would like to thank L. Ortiz-Bobadilla, E. Rosales-González, and R. Uribe-Vargas for their interesting comments.