# On the Complex and Real Hessian Polynomials

Authors: Emigdio Martínez-Ojeda; Adriana Ortiz-Rodríguez
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## 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.


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## Body

## 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:=fxxfyy-fxy2. In this case, the Hessian curve off, 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 polynomialsg. 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 numbern, we define 𝔸n𝕂:={f∈𝕂[x,y]|deg(f)≤n}. Let us consider the Hessian map(2.1)Hn𝕂:𝔸n𝕂→𝔸2n-4𝕂,givenbyf↦Hessf.We will say that the Hessian mapHn𝕂 is complex or real if 𝕂=ℂ or 𝕂=ℝ, respectively. We remark that dim(𝔸n𝕂)≥ dim(𝔸2n-4𝕂) if 2≤n≤4 and dim(𝔸n𝕂)< dim(𝔸2n-4𝕂) if n≥5. In particular, for n≥4, the image, Hn𝕂(𝔸n𝕂), is a connected subset of codimension at least three in 𝔸2n-4𝕂. This means that, in “general”, a polynomial g∈𝔸2n-4𝕂 is not a Hessian polynomial if n≥4 under  Hn𝕂.In diagram form we have(2.2)𝔸2𝕂⊊𝔸3𝕂⊊𝔸4𝕂⊊⋯⊊𝔸n𝕂⋯↓H2𝕂↓H3𝕂↓H4𝕂⋯↓Hn𝕂⋯𝔸0𝕂⊊𝔸2𝕂⊊𝔸4𝕂⊊⋯⊊𝔸2n-4𝕂⋯In virtue of the previous remarks, we introducethe fiber ofg∈𝔸2n-4𝕂underHn𝕂  as the set (2.3)(Hn𝕂)-1(g):={f∈𝔸n𝕂∣Hn𝕂(f)=g}.
Even when we consider the fibers, (Hn𝕂)-1(g), for different values of n, we are interested in knowing if the fiber (Hn𝕂)-1(g) is not empty when n satisfies 0≤2n-4-deg(g)≤1. If the fiber (Hn𝕂)-1(g) is empty, the next problem is to see if the fibers (Hs𝕂)-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∈𝔸r𝕂, with r≥n=(m+4)/2, such that Hn𝕂(f)=g whenever g∈𝔸m𝕂.Remark 2.1.
Letg(x,y)=∑i+j=02n-4bijxiyj be a polynomial in 𝔸2n-4𝕂. There exists a polynomial f(x,y)=∑i+j=0naijxiyj in (Hn𝕂)-1(g) if and only if f satisfies the following system of (2n-3)(n-1) equations:
(2.4)hij(ars)=bij,0≤i+j≤2n-4,2≤r+s≤n,
where hij(ars) are the coefficients of the Hessian polynomialHn𝕂(f). They are also quadratic polynomials in the variables ars.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 ofg Equal to Zero

Proposition 2.2.
For eachg∈𝔸0ℂ, the set (H2ℂ)-1(g) is a quadric in 𝔸2ℂ given by a112=4a20a02-g. Moreover, this quadric is singular if and only if g=0.Corollary 2.3.
In the complex case every element inA0ℂ=ℂ is a complex Hessian polynomial underH2ℂ. And, in the real case every element inA0ℝ=ℝis a real Hessian polynomial underH2ℝ.Proposition 2.4.
For eachg∈𝔸0ℂ, the set  (H3ℂ)-1(g)  is an analytic subvariety in 𝔸3ℂ which is given by the union of connected analytic subvarieties which are parametrized by the following.  (1)
F1±:ℂ×ℂ*×ℂ*×ℂ3→𝔸3ℂ; (t1,t2,t3,t4,t5,t6)↦(a30=t22/3t3, a21=t2, a12=t3,a03=t32/3t2,a20=t1, a11=(2t3t1±t2-g)/t2, a02=t3(t3t1±t2-g)/t22, a10=t4,a01=t5,a00=t6).(2)
F2:ℂ×ℂ*×ℂ3→𝔸3ℂ;    (t1,t2,t3,t4,t5)↦(a30=0,a21=0,a12=0,a03=0,a20=(t12+g)/4t2, a11=t1,a02=t2,  a10=t3,a01=t4,a00=t5).(3)
F3±:ℂ5→𝔸3ℂ; (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)
F4±:ℂ5→𝔸3ℂ;  (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 ofg Equal to One

Proposition 2.5.
For eachg(x,y)=b10x+b01y+b00∈𝔸2ℂ of degree one, the set (H3ℂ)-1(g) is an analytic subvariety in 𝔸3ℂ which is given by the union of analytic subvarieties parametrized by the following.  (1)
Forb10b01=0,  F1:ℂ*×ℂ×ℂ3∖{4b012t12-4b10b01t1t2+b102t22+b00b102=0}→𝔸3ℂ; (t1,t2,t3,t4,t5)↦(a30 = (1/3)(b103t1/(4b012t12-4b10b01t1t2+b102t22+b00b102)), a21=b01b102t1/(4b012t12-4b10b01t1t2+b102t22+b00b102),  a12=b012b10t1/(4b012t12-4b10b01t1t2+b102t22+b00b102),  a03=b013t1/3(4b012t12-4b10b01t1t2+b102t22+b00b102), a20=t1,a11=t2,a02=(t22+b00)/4t1, a10=t3,a01=t4,a00=t5).(2)
Forb10b01≠0, F2±:ℂ*×ℂ3→𝔸3ℂ; (t1,t2,t3,t4)↦(a30=(1/3)(t1b10/b01), a21=t1,a12=b01t1/b10, a03=(1/3)(b012t1/b102), 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 underH3ℂ. And every real polynomial of degree one is a real Hessian polynomial under H3ℝ.


### 2.3. Degree ofg Equal to Two

Letg1,g2∈𝕂[x,y]. We say that g1is in the orbit ofg2 if they are equivalent by an affine transformation of the plane 𝕂2.Theorem 2.7.
Complex Case
(1)
The complex polynomials of degree two(2.5)y2x-y2y2-x2-k,k∈ℂ
are complex Hessian polynomials under H3ℂ. Moreover, a polynomial g∈𝔸2ℂ of degree two is a complex Hessian polynomial under H3ℂ if and only if it belongs to the orbit of one of those polynomials.(2)
The complex polynomials,y2+r∈𝔸2ℂ, where r∈ℂ*, are not complex Hessian polynomials under H3ℂ. Moreover, a polynomial g∈𝔸2ℂ of degree two is not a complex Hessian polynomial under H3ℂ if and only if it belongs to the orbit of one of those polynomials.
Real case
(1)
The real polynomials of degree two(2.6)-y2x-y2y2-x2-r1,r1>0-y2-x2+r2,r2>0
are real Hessian polynomials under  H3ℝ. Moreover, a polynomial g∈𝔸2ℝ of degree two is a real Hessian polynomial under  H3ℝ if and only if it belongs to the orbit of one of those polynomials.(2)
The real polynomials of degree two(2.7)y2y2-x2-r3,r3≤0y2+x2-r4,r4∈ℝ-y2-x2+r5,r5≤0y2+xy2-r6,r6∈ℝ*-y2-r7,r7∈ℝ*
are not real Hessian polynomials under H3ℝ. Moreover, a polynomial g∈𝔸2ℝ of degree two is not a real Hessian polynomial under H3ℝ 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 in𝔸3ℂ.In the next section we will give a geometric proof of this corollary.Proposition 2.9.
Letg(x,y)=b20x2+b11xy+b02y2+b00 be a polynomial in 𝔸2ℂ with b20(4b20b02-b112)≠0. Then the set (H3ℂ)-1(g) is an analytic subvariety on 𝔸3ℂ which is the union of analytic subvarieties parametrized by the following.  (1)
F1±:ℂ4→𝔸3ℂ; (t1,t2,t3,t4)↦(a30=t1,a21=(1/2)(3b11t1±9t12b112-36t12b02b20-b203/b20),a12 = (-6t1b02b20+3b11t1±9t12b112-36t12b02b20-b203)/2b202, a03 = (-1/6b202) (±b029t12b112-36t12b02b20-b203+6b11t1b02-(3b11t1±9t12b112-36t12b02b20-b203)b112/b20+3b11b02t1),  a20=b20b00/4b20b02-b112, a11=b11b00/4b20b02-b112,  a02=b02b00/(4b20b02-b112),  a10=t2, a01=t3,a00=t4).(2)
F2±: ℂ4→𝔸3ℂ;   (t1,t2,t3,t4)↦(a30=t1,a21=(3b11t1±9t12b112-36t12b02b20-b203)/2b20, a12 = (3b11t1-6t1b02b20±9t12b112-36t12b02b20-b203)/2b202, a03 = (-1/6b202)  (±b029t12b112-36t12b02b20-b203+3b02b11t1+6b11t1b02-(3b113t1±b1129t12b112-36t12b02b20-b203)/b20),  a20 = -b20b00/4b20b02-b112, a11 = -b11b00/4b20b02-b112,  a02 = -b02b00/4b20b02-b112, 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 ofHnℂ  and the set of singular Hessian curves defined by polynomials in 𝔸2n-4ℂ?We define𝔸nRℂ≔{f∈ℂ[x,y]∣2≤deg(f)≤n}. Let us describe the relation between the set of critical points of H3Rℂ:𝔸3Rℂ→𝔸2ℂ and the set of polynomials in 𝔸2ℂ such that they define singular Hessian curves.Proposition 2.10.
Iff∈𝔸3Rℂ is a critical point of the map H3Rℂ:𝔸3Rℂ→𝔸2ℂ, then the Hessian curve H3Rℂ(f)(x,y)=0 is singular or it has degree one.For the general case, we have the following conjecture.Conjecture 2.11.
Iff∈𝔸nR is a critical point of the map  HnRℂ:𝔸nRℂ→𝔸2n-4ℂ, then its Hessian curve,  Hessf(x,y)=0, is singular or has degree less than  2n-4.


### 2.4. Degree ofg 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 Table1 of complex cubic curves (see [6]) are complex Hessian curves under H4ℂ.(2)
The curves defined by Table2 of complex cubic curves are not complex Hessian curves under H4ℂ.Real Case
(1)
The normal form of curves (see [7]) defined in Table 3 is a real Hessian curve under H4ℝ. Moreover, a curve g∈𝔸3ℝ of degree three is a real Hessian curve under H4ℝ 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 Table4 is not a real Hessian polynomial under  H4ℝ. Moreover, a curve g∈𝔸3ℝ of degree three is not a real Hessian curve under  H4ℝ if and only if its polynomial belongs to the orbit of one of those polynomials.Table 1
Complex Hessian cubic curves.
xy2=bx2+cx+d,b,c,d∈ℂxy2+ey=bx2+cx+d,e∈ℂ*,b,c,d∈ℂy2=ax3+bx2+cx+d,a∈ℂ*,b,c,d∈ℂTable 2
Complex cubic curves which are not complex Hessian curves.
xy2+ey=ax3+bx2+cx+d,a∈ℂ*,b,c,d,e∈ℂxy=ax3+bx2+cx+d,a∈ℂ*,b,c,d∈ℂy=ax3+bx2+cx+d,a∈ℂ*,b,c,d∈ℂx3+bx2+cx+d=0,b,c,d∈ℂTable 3
Real Hessian polynomials.
x2y+y2+x+a1y+a2=0,a1,a2∈ℝx2y+y2+y+a3=0,a3∈(-∞,0)∪(0,1/4)x2y+y2-y+a4=0,a4<-3/4x2y+y2-1=0x2y+3y=0x2y-3y=0x2y=0x3+a5x-y2-1=0,a5∈ℝ,x3-y2+a6x+1=0,a6∈ℝx3-y2+x=0x3-y2-x=0x3-y2=0Table 4
Real polynomials which are not Hessian polynomials.
xy2-x(x-3)2+b1x+b2y-b3=0,b1,b2,b3∈ℝxy2-x3+b4x+2y-b5=0,b5∈ℝxy2-x3+b6x-1=0,b6∈ℝxy2-x3+x=0xy2-x3-x=0xy2-x3=0x3+xy2-6y2+b7x+b8y+b9=0,b7,b8,b9∈ℝx3+xy2+b10x+3y+b11=0,b10,b11∈ℝx3+xy2+b12x+1=0,b12∈ℝx3+xy2+3x=0x3+xy2-3x=0x3+xy2=0x2y+y2+y+c=0,c∈{0}∪[1/4,∞)x2y+y2-y+d=0,d≥-3/4x2y+y2+1=0x2y+y2=0x2y+3x+3y+b13=0,b13∈ℝx2y+3y+1=0x2y+3x-3y+b14=0,b14∈ℝx2y-3y+1=0x2y+3x+1=0x2y+3x=0x2y+1=0x3-xy+1=0x3-xy=0x3-y=0x3+b15x+1=0,b15∈ℝx3-3x=0x3+3x=0x3=0


## 2.1. Degree ofg Equal to Zero

Proposition 2.2.
For eachg∈𝔸0ℂ, the set (H2ℂ)-1(g) is a quadric in 𝔸2ℂ given by a112=4a20a02-g. Moreover, this quadric is singular if and only if g=0.Corollary 2.3.
In the complex case every element inA0ℂ=ℂ is a complex Hessian polynomial underH2ℂ. And, in the real case every element inA0ℝ=ℝis a real Hessian polynomial underH2ℝ.Proposition 2.4.
For eachg∈𝔸0ℂ, the set  (H3ℂ)-1(g)  is an analytic subvariety in 𝔸3ℂ which is given by the union of connected analytic subvarieties which are parametrized by the following.  (1)
F1±:ℂ×ℂ*×ℂ*×ℂ3→𝔸3ℂ; (t1,t2,t3,t4,t5,t6)↦(a30=t22/3t3, a21=t2, a12=t3,a03=t32/3t2,a20=t1, a11=(2t3t1±t2-g)/t2, a02=t3(t3t1±t2-g)/t22, a10=t4,a01=t5,a00=t6).(2)
F2:ℂ×ℂ*×ℂ3→𝔸3ℂ;    (t1,t2,t3,t4,t5)↦(a30=0,a21=0,a12=0,a03=0,a20=(t12+g)/4t2, a11=t1,a02=t2,  a10=t3,a01=t4,a00=t5).(3)
F3±:ℂ5→𝔸3ℂ; (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)
F4±:ℂ5→𝔸3ℂ;  (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 ofg Equal to One

Proposition 2.5.
For eachg(x,y)=b10x+b01y+b00∈𝔸2ℂ of degree one, the set (H3ℂ)-1(g) is an analytic subvariety in 𝔸3ℂ which is given by the union of analytic subvarieties parametrized by the following.  (1)
Forb10b01=0,  F1:ℂ*×ℂ×ℂ3∖{4b012t12-4b10b01t1t2+b102t22+b00b102=0}→𝔸3ℂ; (t1,t2,t3,t4,t5)↦(a30 = (1/3)(b103t1/(4b012t12-4b10b01t1t2+b102t22+b00b102)), a21=b01b102t1/(4b012t12-4b10b01t1t2+b102t22+b00b102),  a12=b012b10t1/(4b012t12-4b10b01t1t2+b102t22+b00b102),  a03=b013t1/3(4b012t12-4b10b01t1t2+b102t22+b00b102), a20=t1,a11=t2,a02=(t22+b00)/4t1, a10=t3,a01=t4,a00=t5).(2)
Forb10b01≠0, F2±:ℂ*×ℂ3→𝔸3ℂ; (t1,t2,t3,t4)↦(a30=(1/3)(t1b10/b01), a21=t1,a12=b01t1/b10, a03=(1/3)(b012t1/b102), 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 underH3ℂ. And every real polynomial of degree one is a real Hessian polynomial under H3ℝ.


## 2.3. Degree ofg Equal to Two

Letg1,g2∈𝕂[x,y]. We say that g1is in the orbit ofg2 if they are equivalent by an affine transformation of the plane 𝕂2.Theorem 2.7.
Complex Case
(1)
The complex polynomials of degree two(2.5)y2x-y2y2-x2-k,k∈ℂ
are complex Hessian polynomials under H3ℂ. Moreover, a polynomial g∈𝔸2ℂ of degree two is a complex Hessian polynomial under H3ℂ if and only if it belongs to the orbit of one of those polynomials.(2)
The complex polynomials,y2+r∈𝔸2ℂ, where r∈ℂ*, are not complex Hessian polynomials under H3ℂ. Moreover, a polynomial g∈𝔸2ℂ of degree two is not a complex Hessian polynomial under H3ℂ if and only if it belongs to the orbit of one of those polynomials.
Real case
(1)
The real polynomials of degree two(2.6)-y2x-y2y2-x2-r1,r1>0-y2-x2+r2,r2>0
are real Hessian polynomials under  H3ℝ. Moreover, a polynomial g∈𝔸2ℝ of degree two is a real Hessian polynomial under  H3ℝ if and only if it belongs to the orbit of one of those polynomials.(2)
The real polynomials of degree two(2.7)y2y2-x2-r3,r3≤0y2+x2-r4,r4∈ℝ-y2-x2+r5,r5≤0y2+xy2-r6,r6∈ℝ*-y2-r7,r7∈ℝ*
are not real Hessian polynomials under H3ℝ. Moreover, a polynomial g∈𝔸2ℝ of degree two is not a real Hessian polynomial under H3ℝ 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 in𝔸3ℂ.In the next section we will give a geometric proof of this corollary.Proposition 2.9.
Letg(x,y)=b20x2+b11xy+b02y2+b00 be a polynomial in 𝔸2ℂ with b20(4b20b02-b112)≠0. Then the set (H3ℂ)-1(g) is an analytic subvariety on 𝔸3ℂ which is the union of analytic subvarieties parametrized by the following.  (1)
F1±:ℂ4→𝔸3ℂ; (t1,t2,t3,t4)↦(a30=t1,a21=(1/2)(3b11t1±9t12b112-36t12b02b20-b203/b20),a12 = (-6t1b02b20+3b11t1±9t12b112-36t12b02b20-b203)/2b202, a03 = (-1/6b202) (±b029t12b112-36t12b02b20-b203+6b11t1b02-(3b11t1±9t12b112-36t12b02b20-b203)b112/b20+3b11b02t1),  a20=b20b00/4b20b02-b112, a11=b11b00/4b20b02-b112,  a02=b02b00/(4b20b02-b112),  a10=t2, a01=t3,a00=t4).(2)
F2±: ℂ4→𝔸3ℂ;   (t1,t2,t3,t4)↦(a30=t1,a21=(3b11t1±9t12b112-36t12b02b20-b203)/2b20, a12 = (3b11t1-6t1b02b20±9t12b112-36t12b02b20-b203)/2b202, a03 = (-1/6b202)  (±b029t12b112-36t12b02b20-b203+3b02b11t1+6b11t1b02-(3b113t1±b1129t12b112-36t12b02b20-b203)/b20),  a20 = -b20b00/4b20b02-b112, a11 = -b11b00/4b20b02-b112,  a02 = -b02b00/4b20b02-b112, 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 ofHnℂ  and the set of singular Hessian curves defined by polynomials in 𝔸2n-4ℂ?We define𝔸nRℂ≔{f∈ℂ[x,y]∣2≤deg(f)≤n}. Let us describe the relation between the set of critical points of H3Rℂ:𝔸3Rℂ→𝔸2ℂ and the set of polynomials in 𝔸2ℂ such that they define singular Hessian curves.Proposition 2.10.
Iff∈𝔸3Rℂ is a critical point of the map H3Rℂ:𝔸3Rℂ→𝔸2ℂ, then the Hessian curve H3Rℂ(f)(x,y)=0 is singular or it has degree one.For the general case, we have the following conjecture.Conjecture 2.11.
Iff∈𝔸nR is a critical point of the map  HnRℂ:𝔸nRℂ→𝔸2n-4ℂ, then its Hessian curve,  Hessf(x,y)=0, is singular or has degree less than  2n-4.


## 2.4. Degree ofg 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 Table1 of complex cubic curves (see [6]) are complex Hessian curves under H4ℂ.(2)
The curves defined by Table2 of complex cubic curves are not complex Hessian curves under H4ℂ.Real Case
(1)
The normal form of curves (see [7]) defined in Table 3 is a real Hessian curve under H4ℝ. Moreover, a curve g∈𝔸3ℝ of degree three is a real Hessian curve under H4ℝ 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 Table4 is not a real Hessian polynomial under  H4ℝ. Moreover, a curve g∈𝔸3ℝ of degree three is not a real Hessian curve under  H4ℝ if and only if its polynomial belongs to the orbit of one of those polynomials.Table 1
Complex Hessian cubic curves.
xy2=bx2+cx+d,b,c,d∈ℂxy2+ey=bx2+cx+d,e∈ℂ*,b,c,d∈ℂy2=ax3+bx2+cx+d,a∈ℂ*,b,c,d∈ℂTable 2
Complex cubic curves which are not complex Hessian curves.
xy2+ey=ax3+bx2+cx+d,a∈ℂ*,b,c,d,e∈ℂxy=ax3+bx2+cx+d,a∈ℂ*,b,c,d∈ℂy=ax3+bx2+cx+d,a∈ℂ*,b,c,d∈ℂx3+bx2+cx+d=0,b,c,d∈ℂTable 3
Real Hessian polynomials.
x2y+y2+x+a1y+a2=0,a1,a2∈ℝx2y+y2+y+a3=0,a3∈(-∞,0)∪(0,1/4)x2y+y2-y+a4=0,a4<-3/4x2y+y2-1=0x2y+3y=0x2y-3y=0x2y=0x3+a5x-y2-1=0,a5∈ℝ,x3-y2+a6x+1=0,a6∈ℝx3-y2+x=0x3-y2-x=0x3-y2=0Table 4
Real polynomials which are not Hessian polynomials.
xy2-x(x-3)2+b1x+b2y-b3=0,b1,b2,b3∈ℝxy2-x3+b4x+2y-b5=0,b5∈ℝxy2-x3+b6x-1=0,b6∈ℝxy2-x3+x=0xy2-x3-x=0xy2-x3=0x3+xy2-6y2+b7x+b8y+b9=0,b7,b8,b9∈ℝx3+xy2+b10x+3y+b11=0,b10,b11∈ℝx3+xy2+b12x+1=0,b12∈ℝx3+xy2+3x=0x3+xy2-3x=0x3+xy2=0x2y+y2+y+c=0,c∈{0}∪[1/4,∞)x2y+y2-y+d=0,d≥-3/4x2y+y2+1=0x2y+y2=0x2y+3x+3y+b13=0,b13∈ℝx2y+3y+1=0x2y+3x-3y+b14=0,b14∈ℝx2y-3y+1=0x2y+3x+1=0x2y+3x=0x2y+1=0x3-xy+1=0x3-xy=0x3-y=0x3+b15x+1=0,b15∈ℝx3-3x=0x3+3x=0x3=0


## 3. Proofs

Let us consider the complex Hessian mapHnℂ:𝔸nℂ→𝔸2n-4ℂf↦Hessf and recall that the fiber of g∈𝔸2n-4 under Hnℂ is the set  (Hnℂ)-1(g):={f∈𝔸nℂ∣Hnℂ(f)=g}.Proof.
Letf(x,y)=∑i+j=02aijxiyj∈𝔸2ℂ. A direct calculus shows that fxx=2a20, fxy=a11, and fyy=2a02. Therefore,
(3.1)H2ℂ(f)=4a20a02-a112.
For each g∈𝔸0ℂ, consider S(g)={(a00,a10,a01,a20,a11,a02):a112=4a20a02-g}. To show that (H2ℂ)-1(g)=S(g) it is enough to show (H2ℂ)-1(g)⊂S because a direct substitution shows S⊂(H2ℂ)-1(g). Therefore, let us consider f∈(H2ℂ)-1(g), that is, H2ℂ(f)=4a20a02-a112=g. Hence a112=4a20a02-g and the first part of the claim is done. Finally, the derivative of H2ℂ is given by
(3.2)DH2ℂ(f)=(0,0,0,4a02,-2a11,4a20).
Therefore, we conclude the proof of the result.Proof.
The polynomialf(x,y)=x2+(g/4)y2 satisfies that H2𝕂(f)=g in complex or real case.Lemma 3.1.
Iff(x,y)=∑i+j=03aijxiyj∈𝔸3ℂ, then the map H3ℂ:A3ℂ→A2ℂ is
(3.3)H3ℂ(f)=b20x2+b11xy+b02y2+b10x+b01y+b00,
where the brs coefficients satisfy the following system of quadratic equations:
(3.4)b20=-4a212+12a30a12,b11=-4a21a12+36a30a03,b02=-4a122+12a21a03,b10=12a30a02+4a20a12-4a21a11,b01=12a20a03+4a21a02-4a12a11,b00=-a112+4a20a02.Proof.
Forg∈𝔸0ℂ we have that, by Lemma 3.1, H3ℂ(f)=g is equivalent to the following system of equations:
(3.5)0=-4a212+12a30a12,(3.6)0=-4a21a12+36a30a03,(3.7)0=-4a122+12a21a03,(3.8)0=12a30a02+4a20a12-4a21a11,(3.9)0=12a20a03+4a21a02-4a12a11,(3.10)g=-a112+4a20a02.
Let Sg be the set obtained by the union of the image of parametrizations F1,…,F4. To prove that (H3ℂ)-1(g)=Sg, it is enough to show that (H3ℂ)-1(g)⊂Sg because a direct substitution shows that Sg⊂(H3ℂ)-1(g). Now, to prove (H3ℂ)-1(g)⊂Sg 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 obtaina30=a212/3a12, a03=a122/3a21, 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/a212. 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 obtain(3.11)a11=1a21a12(a212a02+a20a122).
A substitution of a11 in (3.10) gives us the quadratic equation in the a20 variable:
(3.12)a214a022-2a212a122a02+(a202a124+ga212a122)=0.
Solving this quadratic equation we get a02=a12(a20a12±a21-g)/a212. 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 obtaina21=0 and the equations:
(3.13)0=a30a03,0=a30a02,0=a20a03,g=-a112+4a20a02.
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:
(3.14)0=a30a03,0=a30a02.
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=(a112+g)/4a02 and the three equations:
(3.15)0=a30a03,0=a30a02,0=a20a03.
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.Letg1,g2∈𝕂[x,y]. We say that g1is in the orbit ofg2 (or g2is in the orbit ofg1) if they are equivalent by an affine transformation of the plane 𝕂2 (where 𝕂=ℝ or ℂ).Remark 3.2.
Ifg1 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 Corollary2.6.
Note that the polynomialg(x,y)=x∈𝔸2ℂ is a complex Hessian polynomial under H3ℂ because f(x,y)=(1/12)x3+y2∈(H3ℂ)-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 Proposition2.5.
Letg(x,y)=b10x+b01y+b00 be a polynomial of degree one in 𝔸2ℂ with constant term. Then the expression H3ℂ(f)=g is equivalent, by Lemma 3.1, to the system of equations:
(3.16)0=-a212+3a30a12,(3.17)0=-a21a12+9a30a03,(3.18)0=-a122+3a21a03,(3.19)b10=12a30a02+4a20a12-4a21a11,(3.20)b01=12a20a03+4a21a02-4a12a11,(3.21)b00=-a112+4a20a02.
Denote bySg the union of the images of F1 and F2. We shall prove that (H3ℂ)-1(g)=Sg. After some calculus it is proved that Sg⊂(H3ℂ)-1(g). Now, we shall prove that (H3ℂ)-1(g)⊂Sg.
Suppose thata21,a12≠0. Multiply (3.19) by a21 and (3.20) by a12 and subtract the two obtained equations. It gives
(3.22)4a20a122-4a02a212+12a30a02a12-12a03a20a21=b10a12-b01a21.
From (3.16) and (3.18) we obtain a212=3a30a12 and a122=9a30a03, respectively, which we insert in (3.22) to obtain
(3.23)a12b10=b01a21.
From (3.23) we obtain a12=b01a21/b10 if b10≠0 and we insert it in (3.16) to obtain
(3.24)a30=b10a213b01.
Analogous, we insert it in (3.18) to obtain
(3.25)a03=b012a213b102.
When we put (3.23), (3.24), and (3.25) in (3.17), it fulfills identically.
Case 1.
Suppose thata20≠0. From (3.21) it has
(3.26)a02=b00+a1124a20.
From (3.19)
(3.27)a12=4a21a11-12a30a02+b104a20.
From (3.23) we obtain a21 and we replace it in (3.27). We replace also a30 from (3.24) and a02 to obtain
(3.28)12a202b012a12=12a11a20a12b10b01-3a12b102(b00+a112)+3a20b10b012.
We associate the terms containing a12(3.29)a12=a20b10b0124a202b012-4a11a20b10b01+b102b00+b102a112.
We replace this last expression of a12 in (3.23) and we obtain
(3.30)a21=a20b102b014a202b012-4a11a20b10b01+b102b00+b102a112.
We replace the expression of a21 in (3.24) and (3.25) and we have
(3.31)a30=a20b1033(4a202b012-4a11a20b10b01+b102b00+b102a112),a03=a20b0133(4a202b012-4a11a20b10b01+b102b00+b102a112).
Note that those last four expressions are in the image of F1.
Case 2.
Suppose thata20=0. From (3.21) we obtain
(3.32)a112=-b00.
From (3.19)
(3.33)a02=4a21a11+b1012a30.
We replace the value of a30 (obtained from (3.24)) and the expression a11 in this last equation to obtain
(3.34)a02=b01(±4a21-b00+b10)4a21b10.
The expression of a12 is obtained from (3.23). The expressions of this case are in the image of F2.Proof of Theorem2.7.
Complex Case
(1)
Letg(x,y)=ax2+bxy+cy2+dx+ey+h∈𝔸2ℂ 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.IfA≠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 polynomialf(x,y)=(-1/2)xy2∈𝔸3ℂ satisfies H3ℂ(f)=y2. The polynomial f(x,y)=xy2/2+x2/2∈𝔸3ℂ verifies H3ℂ(f)=x-y2. For each k∈ℂ, the polynomial fk(x,y)=y3/6+x2y/2-(-k/2)x2+(-k/2)y2∈𝔸3ℂ fulfills H3ℂ(fk)=y2-x2-k.(2)
To verify that the polynomialsy2+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 H3ℂ(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 polynomialg(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 acomplex polynomial is totally imaginary if the real part of all its coefficients is zero.Lemma 3.4.
Letg∈𝔸2n-4ℝ 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 (Hnℂ)-1(-g).(1)
By Lemma3.4 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∈𝔸3ℝ verifies H3ℝ(f)=x-y2 and that H3ℝ(xy2/2)=-y2.(2)
By Lemma3.4 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 Lemma3.4.
⇒) By hypothesis there exists f(x,y)=∑r+s=0narsxrys∈𝔸nℝ such that Hnℝ(f)=g. Therefore, consider the totally imaginary polynomial if(x,y)=∑r+s=0niarsxrys, which satisfies Hnℂ(if)=i2Hnℂ(f)=-g.
⇐) By hypothesis there exists f totally imaginary such that Hnℂ(f)=-g. Therefore, if is a real polynomial such that Hnℝ(if)=i2Hnℂ(f)=g. Hence, g is a real Hessian polynomial.Proof of Corollary2.8.
Let us consider the mapψ:𝔸3ℂ×ℂ2→ℂ3={(w1,w2,w3)} given by
(3.35)(f,p)↦(fxx(p),fxy(p),fyy(p)).
If f(x,y)=∑i+j=03aijxiyj∈𝔸3ℂ and p=(x,y), then
(3.36)ψ(f,p)=(6a30x+2a21y+2a20,2a21x+2a12y+2a11,2a12x+6a03y+2a02).
For each fixed f∈𝔸3ℂ let us consider ψf:ℂ2→ℂ3 given by ψf(p):=ψ(f,p).Now, we are interested to describe the conditions inf under which the image under ψf of ℂ2,ψf(ℂ2), is not a plane.Lemma 3.5.
The setSψ:={f∈𝔸3ℂ|ψf(ℂ2) is not a plane} is given by the union of the following sets:
(3.37)CP1:={f∈𝔸3ℂ∣a30=a21=a12=0},CP2:={f∈𝔸3ℂ∣a30∈ℂ*,a12=a2123a30,a03=a21327a302}.Proof.
The Jacobian matrix ofψf is
(3.38)(3a30a21a21a12a123a03).ψf(ℂ2) is not a plane if and only if Jψf has not maximal rank, which is equivalent to solve the system of equations given by the three minor equal to zero. That is,
(3.39)3a12a30-a212=0,3a21a03-a122=0,9a03a30-a21a12=0.
The sets CP1 and CP2 are obtained by solving the last system of equations.Lemma 3.6.
Letf(x,y)=∑i+j=03aijxiyj be a complex polynomial such that f∈Sψ. Case 1.
Supposef∈CP1.  (1)
Ifa03≠0, then ψf(ℂ2) is a parallel line to the w3-axis in ℂ3.(2)
Ifa03=0, then ψf(ℂ2) is the point (2a20,a11,2a02).Case 2.
Iff∈CP1, then ψf(ℂ2) is the line
(3.40)lf={(6a30x+2a20,2a21x+a11,2a2123a30x+2a02)∣x∈ℂ}.Proof.
Letf(x,y)=∑i+j=03aijxiyj∈𝔸3ℂ. In virtue of Lemma 3.5 we have the following cases
Case 1.
Iff∈CP1, then, by (3.36), ψf(ℂ2)={(2a20,a11,6a30y+2a02)∈ℂ3∣y∈ℂ}.(1)
Ifa03≠0, then ψf(ℂ2)={(2a20,a11,6a30y+2a02)∈ℂ3∣y∈ℂ}.(2)
Ifa03=0, then ψf(ℂ2) is the point (2a20,a11,2a02).
Case 2.
Iff∈CP2, then, by (3.36),
(3.41)ψf(ℂ2)=(6a30x+2a21y+2a20,2a21x+2a2123a30y+a11,2a2139a302y+2a2123a30x+2a02).
Let v1=(6a30,2a21,2a212/3a30) be a nonzero vector (by hypothesis) and v2=(2a21,2a21/3a30,2a213/9a302). Note that v2=(a21/3a30)v1. Therefore, the set lf is the same set of ψf(ℂ2).Let us consider the coneC={(w1,w2,w3)∈ℂ3∣w1w3-w22=0}. We shall describe the set ψf(ℂ2)∩C when f∈Sψ.Lemma 3.7.
Letf∈Sψ.  (1)
Ifψf(ℂ2) is a parallel line to the w3-axis in ℂ3, then
(a)
ψf(ℂ2)∩C is a parallel line to the w3-axis in ℂ3 whenever a20=0;(b)
ψf(ℂ2)∩C is the point (2a20,a11,(a112-4a20a02)/2a20+2a20) whenever a20≠0.(2)
Ifψf(ℂ2) is the point (2a20,a11,2a02) and 4a20a02-a112=0, then ψf(ℂ2)∩C is the point (2a20,a11,2a02).(3)
Ifψf(ℂ2) is the line lf, then
(a)
ψf(ℂ2)∩C is the point (6a30x0+2a20,2a21x0+a11,(2a212/3a30)x0+2a02) if α=(1/3a30)(36a302a02+4a20a212-12a21a11a30)≠0, where x0=(1/α)(a112-4a20a02);(b)
ψf(ℂ2)∩C is the line lf whenever α=0.Proof.
In virtue of Lemma3.6 we have the following cases.  (1)
Ifψf(ℂ2)∩C is a parallel line to the w3-axis, then ψf(ℂ2)∩C={(2a20,a11,6a03y+2a02)∈ℂ3∣y∈ℂ,2a20(6a03y+2a02)-a112=0}.
(a)
Ifa20=0, then ψf(ℂ2)∩C is a parallel line to the w3-axis.(b)
Ifa20≠0, then y=(a112-4a20a02)/12a206a03 and ψf(ℂ2)∩C is a point.(2)
Ifψf(ℂ2)∩C is the point (2a20,a11,2a02), then ψf(ℂ2)∩C=(2a20,a11,2a02) if 4a20a02-a112=0.(3)
Ifψf(ℂ2)∩C is the line lf, then ψf(ℂ2)∩C=lf∩{(6a30x+2a20)((2a212/3a30)x+2a20)-(2a21x+a11)2=0}. It means that
(3.42)αx+4a20a02-a112=0,whereα=36a302a02+4a20a212-12a21a11a303a30. Therefore, if α=0, then 4a20a02-a112=0 and we obtain a line on C. If α≠0, then ψf(ℂ2)∩C is a point.With this lemma we finish the proof of corollary.Proof of Proposition2.9.
Letg(x,y)=b20x2+b11xy+b02y2+b00∈𝔸2ℂ. Then the expression H3ℂ(f)=g is equivalent, by Lemma 3.1, to the system of equations:
(3.43)b02=-4a212+12a30a12,b11=-4a21a12+36a30a03,b02=-4a122+12a21a03,0=12a30a02+4a20a12-4a21a11,0=12a20a03+4a21a02-4a12a11,b00=-a112+4a20a02.
Let S be the union of the images of F1 and F2. A direct substitution shows that S⊂(H3ℂ)-1(g). Therefore, to finish the proof it is enough to show that (H3ℂ)-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 curveg(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 (H4ℂ)-1(g); that is, there are no real polynomial satisfying the system of equations:
(3.44)24a40a22-9a312=0,(3.45)72a40a13-12a31a22=0,(3.46)144a40a04+18a31a13-12a222=0,(3.47)72a31a04-12a22a13=0,(3.48)24a04a22-9a132=0,(3.49)12a30a22+24a40a12-12a31a21=-1,(3.50)36a30a13+72a40a03-12a21a22=0,(3.51)36a03a31+72a04a30-12a12a22=1,(3.52)12a03a22+24a04a21-12a13a12=0,(3.53)24a40a02+4a20a22+12a30a12-6a31a11-4a212=0,(3.54)12a31a02+12a20a13+36a30a03-4a21a12-8a11a22=0,(3.55)24a04a20+4a02a22+12a03a21-6a13a11-4a122=0,(3.56)12a30a02+4a20a12-4a21a11=1,(3.57)12a03a20+4a02a21-4a12a11=0,(3.58)4a20a02-a112=0.
A Groebner bases for (3.44)–(3.48) with the graded reverse lexicographic monomial order is
(3.59)3a132-8a22a04,a22a13-6a31a04,2a222-3a31a13-24a40a04,a31a22-6a40a13,3a312-8a40a22,a13a31a04-16a40a042,a31a40a13-16a402a04.
The set of common zeroes of these last 7 polynomials is the union of the sets(3.60){a40=a40,a31=0,a22=0,a13=0,a04=0},(3.61){a40=a134256a043,a31=a13316a042,a22=3a1328a04,a13=a13,a04≠0}.
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
(3.62)4a20a02-a112=0,12a30a02+4a20a12-4a21a11-1=0,-4a12a11+4a21a02+12a20a03=0,12a30a12+3a134a0232a043-4a212-3a133a118a042+3a20a1322a04=0,36a30a03+12a20a13-3a132a11a04-4a21a12+3a133a024a042=0,-6a13a11+24a20a04+3a132a022a04+12a21a03-4a122=0,-3a133a214a042+9a30a1322a04+3a134a1232a043+1=0,9a134a0332a043+36a30a13-9a21a1322a04=0,9a133a034a042-9a132a122a04+72a30a04-1=0,-12a13a12+9a132a032a04+24a21a04=0.
From the last equation we obtain that a21=-a13(-8a12a04+3a13a03)/16a042 and from the sixth equation,
(3.63)a20=-6a132a02a04-24a13a11a042+24a13a03a12a04-9a132a032-16a122a04296a043.
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):(3.64)-18a134a12a04+9a135a03+288a30a132a043+64a04464a044=0,(3.65)9a13(32a30a043-2a12a132a04+a133a03)8a043=0,(3.66)9a133a03-18a12a132a04+288a30a043-4a0424a042=0,(3.67)0=0.
From (3.65) we have that a13=0 or 32a30a043-2a12a132a04+a133a03=0. If a13=0, then (3.64) becomes -1=0.
If32a30a043-2a12a132a04+a133a03=0, then a30=(2a12a132a04-a133a03)/32a043. Substituing a30 in (3.66), we obtain -1=0.Lemma 3.8.
The set,Cv(3), of critical points of the map H3Rℂ:A3Rℂ→A2ℂ is the union of the following six sets:
(3.68)S1={a02=9a03a11a30-6a03a21a20-a12a11a21+2a122a202(3a12a30-a212),3a12a30-a212≠0},S2={a11=0,a21=0,a12=0},S3={a11=2a12a20a21,a30=a2123a12,a21a12≠0},S4={a30=0,a21=0,a12=0},S5={a20=0,a30=0,a21=0},S6={a12=a2123a30,a03=a21327a302,a30≠0}.Proof.
Letf(x,y)=∑i+j=23aijxiyj∈𝔸3Rℂ. The Jacobian matrix of H3Rℂ is,
(3.69)J(H3Rℂ)=(2a02-a112a2000006a03-2a122a2102a02-2a116a202a12-2a216a306a02-2a112a20000006a03-4a126a2100018a03-2a12-2a2118a300006a12-4a216a300).
Let us denote by Mk,k=1,…,7, the 6×6 matrix obtained from J(H3Rℂ) by deleting the column 7-k+1. On the other hand, f is a critical point of H3Rℂ if and only if det(Mk)=0 for all k=1,…,7, that is, if and only if the following system of seven equations is satisfied:
(3.70)(-9a21a12a03+27a30a032+2a123)(F)=0,(a122a21-6a212a03+9a12a30a03)(F)=0,(a212a12-6a122a30+9a21a30a03)(F)=0,(2a213-9a12a21a30+27a302a03)(F)=0,(-3a03a21a11+a122a11-a21a12a02+9a03a30a02)(F)=0,(3a02a12a30+a122a20-3a03a21a20-a212a02)(F)=0,(a11a212-a12a21a20+9a20a30a03-3a12a11a30)(F)=0,
where
(3.71)F=-9a03a11a30+6a03a21a20+6a02a12a30-2a212a02+a12a11a21-2a122a20.
Therefore f(x,y)=∑i+j=23aijxiyj satisfies the previous system of seven equations if and only if f∈S=⋃k=16Sk, where S is the union of the six solutions Sj. Therefore, the proof is done.Lemma 3.9.
Considerf∈𝔸3ℂ and g=H3(f). The curve g=0 is singular if and only if
(3.72)-2a122a20+6a12a02a30+6a21a03a20+a21a12a11-2a212a02-9a03a11a30=0,-a212a122-18a21a12a30a03+27a302a032+4a123a30+4a213a03≠0,
or,
(3.73)-9a12a21a03a20-a21a11a122-a12a212a02+27a30a032a20+2a20a123+1.8+6a212a03a11-9a30a21a03a02+6a30a02a122-9a30a03a11a12=01,-a212a122-18a21a12a30a03+27a302a032+4a123a30+4a213a03=0.2.1,Proof.
Supposef(x,y)=∑r+s=03arsxrys; then
(3.74)H3(f)=-a112+4a20a02+(12a30a02+4a20a12-4a21a11)x+(12a20a03-4a12a11+4a21a02)y+(-4a212+12a30a12)x2+(-4a21a12+36a30a03)yx+(-4a122+12a21a03)y2.
The curve g=H3ℂ(f) is singular if and only if the system gx(p)=gy(p)=g(p)=0 has a solution for some p in ℂ2. The system formed by gx=0,gy=0 is
(3.75)(6a30a12-2a212)x+(9a30a03-a21a12)y+3a30a02+a20a12-a21a11=0,(9a30a03-a21a12)x+(6a21a03-2a122)y+3a20a03-a12a11+a21a02=0.
The proof concludes by analyzing the system when its determinant is distinct from zero and when it is zero.Proof of Proposition2.10.
Let us consider the setsSk,k=1,…,6, of Lemma 3.8. If f∈S1, then
(3.76)H3ℂ(f)=13a12a30-a212×[x2(36a302a122+4a214-24a30a12a212)+xy(108a302a03a12+36a30a03a212+4a213a12-12a21a122a30)+y2(4a122a212-12a123a30+36a21a03a12a30-12a213a03)+x(24a30a122a20+54a302a03a11+4a213a11-4a20a12a212-36a30a03a21a20-18a30a12a11a21a302)+y(-12a122a11a30+18a21a03a11a30+36a20a03a12a30-24a212a03a20+2a212a12a11+4a21a122a20)-3a112a12a30-12a03a21a202+18a20a03a11a30-2a20a12a11a21+4a122a202+a112a212].
Denote by h the polynomial H3ℂ(f). Then,
(3.77)hx=13a12a30-a212×[x(72a302a122-48a30a12a212+8a214)+y(108a302a03a12+4a213a12-36a30a03a212-12a21a122a30)+4a213a11+54a302a03a11-4a20a12a212+24a30a122a20-36a30a03a21a20-18a30a12a11a21a302],hy=13a12a30-a212×[x(108a302a03a12-36a30a03a212-12a21a122a30+4a213a12)+y(8a122a212-24a123a30-24a213a03+72a21a03a12a30)+4a21a122a20-24a212a03a20+2a212a12a11-12a122a11a30+18a21a03a11a30+36a20a03a12a30a302].
The point (x,y) where the Hessian curve is singular is
(3.78)x=-2a20a12+a21a112(3a12a30-a212),y=-3a11a30-2a21a202(3a12a30-a212).
If f∈S2, then H3ℂ(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.
Iff∈S3, then
(3.79)H3ℂ(f)=-4(a212x+a21ya12+a20a12)(y(-3a212a03+a21a122)-a212a02+a20a122)a12a212,hx=-4(y(-3a212a03+a21a122)-a212a02+a20a122)a12,hy=-4(y(-3a212a03+a21ya122)-a212a02+a20a122)a21-4(a212x+a21ya12+a20a12)(-3a212a03+a21a122)a12a212.
The point (x,y) where the Hessian curve is singular is
(3.80)x=a12(-a21a02+3a20a03)a21(-3a21a03+a122),y=--(-a212a02+a20a122)a21(-3a21a03+a122).
If -3a21a03+a122=0, then H3ℂ(f) is of degree one.
Iff∈S4, then H3ℂ(f)=12a20a03y+4a20a02-a112,hx=0,andhy=12a20a03.
The Hessian curve in this case has degree one.
Iff∈S5, then H3ℂ(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.
Iff∈S6, then
(3.81)H3ℂ(f)=(108a303a02+12a20a212a30-36a21a302a119a302)x+(36a21a02a302+4a20a213-12a212a11a309a302)y+36a20a02a302-9a112a3029a302.
The Hessian curve in this case has degree one.



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*Source: 962719-2010-04-26.xml*