#### 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, , is the set : , where Hess . In this case, * the Hessian curve of *, Hess , is a plane curve which is the projection of the parabolic curve into the -plane along the -axis. When is a polynomial of degree in two variables, Hess is a polynomial of degree at most . 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 in , where or , we ask: When is the Hessian curve of a polynomial ? (2)*Hessian polynomial realization problem* Given we ask: When does exist such that Hess ? If such exists and (), then 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 . 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 we define . Let us consider the * Hessian map *

We will say that the Hessian map is * complex* or * real* if or , respectively. We remark that dim dim if and dim dim if . In particular, for , the image, is a connected subset of codimension at least three in . This means that, in “general”, a polynomial is not a Hessian polynomial if under .

In diagram form we have

In virtue of the previous remarks, we introduce * the fiber of ** under * as the set
Even when we consider the fibers, , for different values of , we are interested in knowing if the fiber is not empty when satisfies If the fiber is empty, the next problem is to see if the fibers are not empty for (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 , with such that whenever .

*Remark 2.1. *Let be a polynomial in . There exists a polynomial in if and only if satisfies the following system of equations:
where are the coefficients of the Hessian polynomial. They are also quadratic polynomials in the variables .

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 Equal to Zero

Proposition 2.2. *For each , the set is a quadric in given by Moreover, this quadric is singular if and only if .*

Corollary 2.3. *In the complex case every element in is a complex Hessian polynomial under. And, in the real case every element inis a real Hessian polynomial under.*

Proposition 2.4. *For each , the set is an analytic subvariety in which is given by the union of connected analytic subvarieties which are parametrized by the following. *(1)*; , , , , , *(2)*; ,,, , . *(3)*; , , , .*(4)*; , , , , . *

##### 2.2. Degree of Equal to One

Proposition 2.5. *For each of degree one, the set is an analytic subvariety in which is given by the union of analytic subvarieties parametrized by the following. *(1)*For , ; = , , , , , .*(2)*For , ; , , , , , . *

From this proposition we have the following.

Corollary 2.6. *Every complex polynomial of degree one is a complex Hessian polynomial under . And every real polynomial of degree one is a real Hessian polynomial under .*

##### 2.3. Degree of Equal to Two

Let . We say that * is in the orbit of * if they are equivalent by an affine transformation of the plane .

Theorem 2.7. *Complex Case*

()*The complex polynomials of degree two
are complex Hessian polynomials under . Moreover, a polynomial of degree two is a complex Hessian polynomial under if and only if it belongs to the orbit of one of those polynomials. *()*The complex polynomials, where are not complex Hessian polynomials under . Moreover, a polynomial of degree two is not a complex Hessian polynomial under if and only if it belongs to the orbit of one of those polynomials.**Real case*

(1)*The real polynomials of degree two
are real Hessian polynomials under . Moreover, a polynomial of degree two is a real Hessian polynomial under if and only if it belongs to the orbit of one of those polynomials. *(2)*The real polynomials of degree two
are not real Hessian polynomials under . Moreover, a polynomial of degree two is not a real Hessian polynomial under 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: (parabola), (general conic), (line pair), (parallel lines), and finally (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 .*

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

Proposition 2.9. *Let be a polynomial in with . Then the set is an analytic subvariety on which is the union of analytic subvarieties parametrized by the following. *(1)*; = , = () , , , , , *(2)*: ; , = , = () , = , = , = , = = = . *

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

We define . Let us describe the relation between the set of critical points of and the set of polynomials in such that they define singular Hessian curves.

Proposition 2.10. *If is a critical point of the map , then the Hessian curve is singular or it has degree one.*

For the general case, we have the following conjecture.

Conjecture 2.11. *If is a critical point of the map , then its Hessian curve, Hess is singular or has degree less than .*

##### 2.4. Degree of 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 .*(2)*The curves defined by Table 2 of complex cubic curves are not complex Hessian curves under .**Real Case*

(1)*The normal form of curves (see [7]) defined in Table 3 is a real Hessian curve under . Moreover, a curve of degree three is a real Hessian curve under 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 . Moreover, a curve of degree three is not a real Hessian curve under if and only if its polynomial belongs to the orbit of one of those polynomials.*

#### 3. Proofs

Let us consider the complex Hessian map and recall that the fiber of under is the set

*Proof. *Let . A direct calculus shows that , and . Therefore,
For each , consider . To show that it is enough to show because a direct substitution shows . Therefore, let us consider , that is, . Hence and the first part of the claim is done. Finally, the derivative of is given by
Therefore, we conclude the proof of the result.

*Proof. *The polynomial satisfies that in complex or real case.

Lemma 3.1. *If , then the map is
**
where the coefficients satisfy the following system of quadratic equations:
*

*Proof. *For we have that, by Lemma 3.1, is equivalent to the following system of equations:
Let be the set obtained by the union of the image of parametrizations . To prove that , it is enough to show that because a direct substitution shows that . Now, to prove we will consider two cases: Case 1 is when and Case 2 is when .*Case 1. *In this case, from a direct substitution we obtain , , , and (3.8), (3.9), and (3.10). To solve these equations we will assume the following.*Subcase 1.1. *. In this case we obtain ; . All this values together are contained in the set whose parametrization is .*Subcase 1.2. *. This case will be subdivided in two subcases.(). First, we obtain from (3.10). Later, from a substitution of together with the value of in (3.8) we get . This equation implies . All these values together are contained in the set whose parametrization is .(). In this case, from (3.8) we obtain
A substitution of in (3.10) gives us the quadratic equation in the variable:
Solving this quadratic equation we get . Finally, from a substitution of in (3.11) we get . From these values we obtain the parametrization .*Case 2. *From a direct calculus we obtain and the equations:
To solve these four equations we will consider two cases.*Subcase 2.1. *. From a direct substitution we get and the two equations:
To solve these two equations we will consider two subcases.(). We obtain and . From all of these values we get the parametrization .(). We obtain . From all these values we get the parametrization when .*Subcase 2.2. *. We obtain and the three equations:
To solve these three equations we will consider two subcases.(). We obtain and then the parametrization .(). We obtain , , as well as . All these values together are included in the parametrization when . Therefore, we have obtained all parametrizations in the proposition and the proof is done.

Let . We say that * is in the orbit of * (or * is in the orbit of *) if they are equivalent by an affine transformation of the plane (where or ).

*Remark 3.2. *If is in the orbit of , then is a Hessian polynomial if and only if is a Hessian polynomial. This remark is due to the equality where and .

*Proof of Corollary 2.6. *Note that the polynomial is a complex Hessian polynomial under because On the other hand, every complex polynomial of degree one is in the orbit of . 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 be a polynomial of degree one in with constant term. Then the expression is equivalent, by Lemma 3.1, to the system of equations:

Denote by the union of the images of and . We shall prove that After some calculus it is proved that Now, we shall prove that

Suppose that . Multiply (3.19) by and (3.20) by and subtract the two obtained equations. It gives

From (3.16) and (3.18) we obtain and respectively, which we insert in (3.22) to obtain
From (3.23) we obtain if and we insert it in (3.16) to obtain
Analogous, we insert it in (3.18) to obtain
When we put (3.23), (3.24), and (3.25) in (3.17), it fulfills identically.*Case 1. *Suppose that . From (3.21) it has
From (3.19)
From (3.23) we obtain and we replace it in (3.27). We replace also from (3.24) and to obtain
We associate the terms containing
We replace this last expression of in (3.23) and we obtain
We replace the expression of in (3.24) and (3.25) and we have
Note that those last four expressions are in the image of .*Case 2. *Suppose that . From (3.21) we obtain
From (3.19)
We replace the value of (obtained from (3.24)) and the expression in this last equation to obtain
The expression of is obtained from (3.23). The expressions of this case are in the image of .

*Proof of Theorem 2.7. **Complex Case*

()Let be a quadratic polynomial. By an affine transformation of the complex plane the polynomial is equivalent to the prenormal form where .If , then is in the orbit of where . If and then is in the orbit of . Finally, if then is in the orbit of where .Note that the polynomial satisfies The polynomial verifies For each , the polynomial fulfills ()To verify that the polynomials where 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 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 is in the orbit of one of the normal forms: where

*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 be a polynomial with real coefficients. Then is a real Hessian polynomial if and only if there exists a polynomial totally imaginary on the set .*

*Proof of Lemma 3.4. *) By hypothesis there exists such that . Therefore, consider the totally imaginary polynomial , which satisfies .

) By hypothesis there exists totally imaginary such that . Therefore, is a real polynomial such that Hence, is a real Hessian polynomial.

*Proof of Corollary 2.8. * Let us consider the map given by
If and , then
For each fixed let us consider given by .

Now, we are interested to describe the conditions in under which the image under of , is not a plane.

Lemma 3.5. *The set is not a plane is given by the union of the following sets:
*

*Proof. *The Jacobian matrix of is
is not a plane if and only if has not maximal rank, which is equivalent to solve the system of equations given by the three minor equal to zero. That is,
The sets and are obtained by solving the last system of equations.

Lemma 3.6. *Let be a complex polynomial such that . **Case 1. *Suppose . (1)*If then is a parallel line to the -axis in . *(2)*If then is the point .**Case 2. *If then is the line

*Proof. * Let In virtue of Lemma 3.5 we have the following cases*Case 1. *If then, by (3.36), (1)If then (2)If then is the point .*Case 2. *If then, by (3.36),
Let be a nonzero vector (by hypothesis) and . Note that . Therefore, the set is the same set of .

Let us consider the cone We shall describe the set when .

Lemma 3.7. *Let . *(1)*If is a parallel line to the -axis in , then (a) is a parallel line to the -axis in whenever ; (b) is the point whenever . *(2)

*If is the point and , then is the point .*(3)

*If is the line , then (a)*

*is the point if where ;*(b)*is the line whenever .**Proof. * In virtue of Lemma 3.6 we have the following cases. (1)If is a parallel line to the -axis, then (a)If , then is a parallel line to the -axis. (b)If , then and is a point. (2)If is the point , then if . (3)If is the line , then . It means that
Therefore, if , then and we obtain a line on . If , then is a point.

With this lemma we finish the proof of corollary.

*Proof of Proposition 2.9. *Let . Then the expression is equivalent, by Lemma 3.1, to the system of equations:
Let be the union of the images of and . A direct substitution shows that . Therefore, to finish the proof it is enough to show that . 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 where is not a real Hessian curve (the other cases are analogous). To do that, we will prove that there are no real cuartic polynomial in that is, there are no real polynomial satisfying the system of equations:

A Groebner bases for (3.44)–(3.48) with the graded reverse lexicographic monomial order is

The set of common zeroes of these last 7 polynomials is the union of the sets
On the one hand, replacing the solution (3.60) in (3.49)–(3.58), particularly in (3.56) we obtain . On the other hand, replacing the solution (3.61) in (3.49)–(3.58), we obtain the system
From the last equation we obtain that and from the sixth equation,

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):
From (3.65) we have that or . If then (3.64) becomes