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International Journal of Mathematics and Mathematical Sciences Volume 2010, Article ID 962719,22pages

doi:10.1155/2010/962719

Research Article

On the Complex and Real Hessian Polynomials

Emigdio Mart´ınez-Ojeda

1

and Adriana Ortiz-Rodr´ıguez

2

1Universidad Aut´onoma de la Ciudad de M´exico (UACM), Calzada Ermita Iztapalapa 4163, Col. Lomas de Zaragoza C.P., Delegaci´on Iztapalapa, 09620 M´exico, DF, Mexico

2Instituto de Matem´aticas (IMATE-D.F.), Universidad Nacional Aut´onoma de M´exico (UNAM), Area de la Inv. Cient. Circuito Exterior, C.U., C.P, 04510 M´exico, DF, Mexico

Correspondence should be addressed to Adriana Ortiz-Rodr´ıguez,aortiz@matem.unam.mx

Received 25 November 2009; Accepted 8 March 2010

Academic Editor: Mihai Putinar

Copyrightq2010 E. Mart´ınez-Ojeda and A. Ortiz-Rodr´ıguez. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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, Salmon1, Kergosien and Thom2, Arnold3, among others.

The parabolic curve of the graph of a smooth function, f : R2 R, is the set {p, fp ∈ R2×R: Hessfp 0}, where Hessf : f

xxfyyfxy2 . In this case, the Hessian

curve of f, Hessfx, y 0, is a plane curve which is the projection of the parabolic curve into thexy-plane along thez-axis. Whenfis a polynomial of degreenin two variables, Hess

fis 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.

1Hessian curve realization problem. Given an algebraic plane curvegx, y 0 inK2, whereKR or C, we ask: When isgx, y 0 the Hessian curve of a polynomial

f ∈Kx, y?

2Hessian polynomial realization problem. Given g ∈ Kx, ywe ask: When does f

Kx, yexist such that Hessf g? If such f exists andK C K R, then

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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 Corollary2.8.

2. Notation and Results

For each nonnegative integer numbern,we define AKn :{f ∈Kx, y| degfn}. Let us consider the Hessian map

HnK:AKn −→ AK2n4, given byf−→Hessf. 2.1

We will say that the Hessian map HnK is complex or real if K C or K R, respectively. We remark that dimAKn≥dimAK2n4if 2 ≤n≤ 4 and dimAKn<dimAK2n4 ifn≥5. In particular, forn≥4, the image, HnKAKn,is a connected subset of codimension at least three inAK2n4. This means that, in “general”, a polynomial g ∈AK2n4is not a Hessian polynomial if n≥4 underHnK.

In diagram form we have

AK

2 AK3 AK4 · · · AKn· · · ↓H2K↓H3K↓H4K· · · ↓HnK· · ·

AK

0 AK2 AK4 · · · AK2n−4· · ·

2.2

In virtue of the previous remarks, we introduce the fiber of g ∈AK2n4underHnKas the set

HnK−1g:f ∈AKn |HnKfg. 2.3

Even when we consider the fibers, HnK−1g, for different values of n, we are interested in knowing if the fiber HnK−1g is not empty whennsatisfies 0 ≤ 2n−4−degg≤ 1.

If the fiber HnK−1g is empty, the next problem is to see if the fibers HsK−1g are not empty for sn1 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 polynomialsf ∈AKr, with rn m4/2,such that HnKf gwhenever g∈AKm.

Remark 2.1. Letgx, y 2inj40bijxiyj be a polynomial in AK2n4. There exists a polynomial

fx, y nij0aijxiyj in HnK−1g if and only iffsatisfies the following system of2n

3n−1equations:

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wherehijarsare the coefficients of the Hessian polynomial HnKf. They are also quadratic

polynomials in the variables ars.

It is important to note that the computations for solving the system2.4are 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 eachg∈AC0, the setH2C−1gis a quadric inAC2 given by a2

114a20a02−g. Moreover, this quadric is singular if and only ifg0.

Corollary 2.3. In the complex case every element in AC0 C is a complex Hessian polynomial underH2C. And, in the real case every element inAR0 Ris a real Hessian polynomial underH2R.

Proposition 2.4. For eachg∈AC0 , the setH3C−1gis an analytic subvariety inAC3 which is given by the union of connected analytic subvarieties which are parametrized by the following.

1F1± :C×C∗×C∗×C3 AC

3;t1, t2, t3, t4, t5, t6→a30t22/3t3, a21t2, a12t3, a03

t2

3/3t2, a20 t1, a11 2t3tt2√−g/t2, a02 t3t3tt2√−g/t22, a10 t4, a01

t5, a00t6.

2F2 : C×C∗×C3 → AC3;t1, t2, t3, t4, t5 → a30 0, a21 0, a12 0, a03 0, a20

t2

1g/4t2, a11 t1, a02t2, a10t3, a01 t4, a00t5.

3F3± : C5 AC

3;t1, t2, t3, t4, t5 → a30 t1, a21 0, a12 0, a03 0, a20 t2, a11 ±√−g, a02 0, a10 t3, a01t4, a00t5.

4F4± : C5 AC

3;t1, t2, t3, t4, t5 → a30 0, a21 0, a12 0, a03 t1, a20 0, a11 ±√−g, a02 t2, a10t3, a01 t4, a00 t5.

2.2. Degree of

g

Equal to One

Proposition 2.5. For eachgx, y b10xb01yb00 ∈AC2 of degree one, the set H3C−1g is an analytic subvariety inAC3 which is given by the union of analytic subvarieties parametrized by the following.

1For b10b01 0, F1 : C∗×C×C3 \ {4b201t12 −4b10b01t1t2 b210t22 b00b210 0} →

AC

3;t1, t2, t3, t4, t5→a30 1/3b310t1/4b201t21−4b10b01t1t2b210t22b00b210, a21

b01b210t1/4b201t21−4b10b01t1t2b210t22b00b210, a12 b201b10t1/4b201t21−4b10b01t1t2

b2

10t22b00b210, a03b013 t1/34b201t21−4b10b01t1t2b210t22b00b210, a20t1, a11t2, a02

t22b00/4t1, a10t3, a01 t4, a00t5.

2For b10b01/0, F2± : C∗ ×C3 → AC3;t1, t2, t3, t4 → a30 1/3t1b10/b01, a21

t1, a12 b01t1/b10, a03 1/3b201t1/b102 , a20 0, a11 ± −b00, a02

1/4b01±4t1 −b00b10/b10t1, a10 t2, a01t3, a00t4.

From this proposition we have the following.

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2.3. Degree of

g

Equal to Two

Let g1, g2 ∈ Kx, y. We say that g1 is in the orbit of g2 if they are equivalent by an affine transformation of the planeK2.

Theorem 2.7.

Complex Case

1The complex polynomials of degree two

y2 xy2 y2x2k, kC 2.5

are complex Hessian polynomials under H3C. Moreover, a polynomialg∈AC2 of degree two is a complex Hessian polynomial under H3Cif and only if it belongs to the orbit of one of those polynomials.

2The complex polynomials, y2 r ∈ AC2, where r ∈ C∗, are not complex Hessian polynomials under H3C. Moreover, a polynomialg ∈ AC2 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

1The real polynomials of degree two

y2 xy2 y2x2r

1, r1>0 −y2−x2r2, r2>0 2.6

are real Hessian polynomials underH3R. Moreover, a polynomialg ∈ AR2 of degree two is a real Hessian polynomial underH3Rif and only if it belongs to the orbit of one of those polynomials.

2The real polynomials of degree two

y2 y2x2r

3, r3≤0 y2x2−r4, r4∈R −y2−x2r5, r5≤0

y2x y2r

6, r6 ∈R∗ −y2−r7, r7∈R∗

2.7

are not real Hessian polynomials under H3R. Moreover, a polynomialg∈AR2 of degree two is not a real Hessian polynomial under H3Rif 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 xparabola,y2 x21general conic,y2 x2 line pair,y2 1 parallel lines, and finallyy20double 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 inAC3.

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Proposition 2.9. Let gx, y b20x2 b11xy b02y2 b00 be a polynomial in AC2 with

b204b20b02 −b211/0. Then the setH3C−1gis an analytic subvariety onAC3 which is the union of analytic subvarieties parametrized by the following.

1F1± : C4 AC

3; t1, t2, t3, t4 → a30 t1, a21 1/23b11t

9t21b112 −36t21b02b20−b320/b20, a12=−6t1b02b203b11t

9t21b211−36t21b02b20−b320

/2b2

20, a03 = −1/6b220±b02

9t2

1b211−36t21b02b20−b203 6b11t1b02 − 3b11t1 ±

9t2

1b112 −36t21b02b20−b320b211/b20 3b11b02t1, a20 b20

b00/4b20b02−b211, a11

b11

b00/4b20b02−b211,a02 b02

b00/4b20b02−b211,a10t2,a01 t3, a00t4.

2F2±: C4 AC

3; t1, t2, t3, t4 → a30 t1, a21 3b11t1 ±

9t21b112 −36t12b02b20−b320/2b20, a12 = 3b11t1 − 6t1b02b20 ±

9t2

1b112 −36t21b02b20−b320/2b220, a03 = −1/6b202 ±b02

9t2

1b112 −36t21b02b20−b320 3b02b11t1 6b11t1b02 − 3b113t1 ± b211

9t2

1b112 −36t21b02b20−b320/b20, a20 =b20

b00/4b20b02−b211,a11 =b11

b00/4b20b02−b112 ,a02 =b02

b00/4b20b02−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 ofHnCand the set of singular Hessian curves defined by polynomials inAC2n4?

We defineACnR:{f ∈Cx, y|2≤degfn}. Let us describe the relation between the set of critical points ofH3CR :AC3R → AC2 and the set of polynomials inAC2 such that they define singular Hessian curves.

Proposition 2.10. If f ∈ AC3R is a critical point of the mapH3CR : AC3R → AC2, then the Hessian curveH3CRfx, y 0 is singular or it has degree one.

For the general case, we have the following conjecture.

Conjecture 2.11. If f ∈ AnRis a critical point of the mapHnRC : ACnR → AC2n4, then its Hessian

curve, Hessfx, y 0,is singular or has degree less than 2n4.

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

1The curves defined by Table1of complex cubic curves (see [6]) are complex Hessian curves under H4C.

2The curves defined by Table2of complex cubic curves are not complex Hessian curves under

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Table 1: Complex Hessian cubic curves.

xy2bx2cxd, b, c, dC xy2eybx2cxd, eC, b, c, dC y2ax3bx2cxd, aC, b, c, dC

Table 2: Complex cubic curves which are not complex Hessian curves.

xy2eyax3bx2cxd, aC, b, c, d, eC xyax3bx2cxd, aC, b, c, dC yax3bx2cxd, aC, b, c, dC x3bx2cxd0, b, c, dC

Real Case

1The normal form of curves (see [7]) defined in Table3is a real Hessian curve under H4R. Moreover, a curveg ∈AR3 of degree three is a real Hessian curve under H4Rif and only if its polynomial belongs to the orbit of one of those polynomials.

2The normal form of curves defined by the polynomials in Table 4 is not a real Hessian polynomial under H4R. Moreover, a curveg∈AR3 of degree three is not a real Hessian curve underH4Rif and only if its polynomial belongs to the orbit of one of those polynomials.

3. Proofs

Let us consider the complex Hessian mapHnC:ACn → AC2n4 f→Hessf and recall that the fiber ofg∈A2n−4underHnCis the setHnC−1g:{f ∈AnC |HnCf g}.

Proof of the Proposition2.2. Let fx, y 2ij0aijxiyj ∈ AC2. A direct calculus shows that

fxx2a20,fxya11,andfyy2a02. Therefore,

H2Cf4a20a02−a211. 3.1

For eachg∈AC0, considerSg {a00, a10, a01, a20, a11, a02:a2114a20a02−g}. To show that

H2C−1g Sgit is enough to showH2C−1gSbecause a direct substitution shows

SH2C−1g. Therefore, let us considerfH2C−1g, that is,H2Cf 4a20a02−a211 g. Hencea211 4a20a02−gand the first part of the claim is done. Finally, the derivative ofH2Cis given by

DH2Cf 0,0,0,4a02,−2a11,4a20. 3.2

Therefore, we conclude the proof of the result.

Proof of the Corollary2.3. The polynomialfx, y x2 g/4y2 satisfies thatHK

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[image:7.600.107.505.232.418.2]

Table 3: Real Hessian polynomials.

x2yy2xa

1ya20, a1, a2∈R x2yy2ya30, a3∈−∞,0∪0,1/4 x2yy2ya

40, a4<−3/4 x2yy2−10

x2y3y0 x2y3y0

x2y0 x3a

5xy2−10, a5∈R, x3y2a

6x10, a6∈R x3−y2x0

x3y2x0 x3y20

Table 4: Real polynomials which are not Hessian polynomials.

xy2xx32b

1xb2yb30, b1, b2, b3∈R xy2−x3b4x2yb50, b5∈R xy2x3b

6x−10, b6∈R xy2−x3x0

xy2x3x0 xy2x30

x3xy26y2b

7xb8yb90, b7, b8, b9∈R x3xy2b10x3yb110, b10, b11∈R x3xy2b

12x10, b12∈R x3xy23x0

x3xy23x0 x3xy20

x2yy2yc0, c∈ {0} ∪1/4, x2yy2yd0, d≥ −3/4

x2yy210 x2yy20

x2y3x3yb

130, b13∈R x2y3y10

x2y3x3yb

140, b14∈R x2y−3y10

x2y3x10 x2y3x0

x2y10 x3xy10

x3xy0 x3y0

x3b

15x10, b15∈R x3−3x0

x33x0 x30

Lemma 3.1. Iffx, y 3ij0aijxiyj ∈AC3, then the mapH3C:AC3AC2 is

H3Cfb20x2b11xyb02y2b10xb01yb00, 3.3

where thebrscoefficients satisfy the following system of quadratic equations:

b20−4a22112a30a12,

b11−4a21a1236a30a03,

b02−4a21212a21a03,

b1012a30a024a20a12−4a21a11,

b0112a20a034a21a02−4a12a11,

b00−a2114a20a02.

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Proof of the Proposition2.4. Forg ∈AC0 we have that, by Lemma3.1, H3Cf gis equivalent to the following system of equations:

0−4a22112a30a12, 3.5

0−4a21a1236a30a03, 3.6

0−4a21212a21a03, 3.7

012a30a024a20a12−4a21a11, 3.8

012a20a034a21a02−4a12a11, 3.9

ga2114a20a02. 3.10

LetSg be the set obtained by the union of the image of parametrizationsF1, . . . , F4. To prove thatH3C−1g Sg, it is enough to show thatH3C−1gSg because a direct substitution

shows thatSgH3C−

1g. Now, to proveHC 3−

1gS

gwe will consider two cases: Case

1is whena12/0 and Case2is whena120.

Case 1. In this case, from a direct substitution we obtaina30a221/3a12,a03 a212/3a21,a12 ∈

C∗, and3.8,3.9, and3.10. To solve these equations we will assume the following.

Subcase 1.1. a20 0. In this case we obtaina11±√−g;a02±a12a21√−g/a221. All this values together are contained in the set whose parametrization isF1.

Subcase 1.2. a20/0. This case will be subdivided in two subcases.

1a02 0. First, we obtaina11from3.10. Later, from a substitution ofa11 together with the value ofa30 in 3.8 we geta20a12 ±a21√−g 0. This equation implies

a112a20a12±a21√−g/a21. All these values together are contained in the set whose parametrization isF1.

2a02/0. In this case, from3.8we obtain

a11 1

a21a12

a221a02a20a212

. 3.11

A substitution ofa11in3.10gives us the quadratic equation in thea20variable:

a421a202−2a221a212a02

a220a412ga221a2120. 3.12

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Case 2. From a direct calculus we obtaina210 and the equations:

0a30a03,

0a30a02,

0a20a03,

ga2114a20a02.

3.13

To solve these four equations we will consider two cases.

Subcase 2.1. a02 0. From a direct substitution we geta11 ±√−gand the two equations:

0a30a03,

0a30a02.

3.14

To solve these two equations we will consider two subcases.

1a03 0. We obtain a30 ∈ C and a20 ∈ C. From all of these values we get the parametrizationF3.

2a03/0. We obtaina30a200. From all these values we get the parametrizationF4 whena020.

Subcase 2.2. a02/0. We obtaina20 a211g/4a02and the three equations:

0a30a03,

0a30a02,

0a20a03.

3.15

To solve these three equations we will consider two subcases.

1a030. We obtaina300 and then the parametrizationF2.

2a03/0. We obtaina30 0,a20 0, as well asa11 ±√−g. All these values together are included in the parametrizationF4whena02 ∈C∗. Therefore, we have obtained all parametrizations in the proposition and the proof is done.

Letg1, g2∈Kx, y. We say thatg1is in the orbit of g2org2is in the orbit of g1if they are equivalent by an affine transformation of the planeK2 where KRorC.

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 Hess1/detTfT HessfT,

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Proof of Corollary2.6. Note that the polynomial gx, y x ∈ AC2 is a complex Hessian polynomial under H3C because fx, y 1/12x3y2 HC

3−

1g. On the other hand,

every complex polynomial of degree one is in the orbit ofg. By Remark3.2we have that every complex polynomial of degree one is a Hessian polynomial. The real case is analogous.

Proof of Proposition2.5. Letgx, y b10xb01yb00be a polynomial of degree one inAC2 with constant term. Then the expression H3Cf gis equivalent, by Lemma3.1, to the system of equations:

0−a2213a30a12, 3.16

0−a21a129a30a03, 3.17

0−a2123a21a03, 3.18

b1012a30a024a20a12−4a21a11, 3.19

b0112a20a034a21a02−4a12a11, 3.20

b00−a2114a20a02. 3.21

Denote bySgthe union of the images ofF1andF2. We shall prove that H3C−1g

Sg. After some calculus it is proved that SgH3C−1g. Now, we shall prove that

H3C−1gSg.

Suppose thata21, a12/0. Multiply3.19by a21and3.20by a12and subtract the two obtained equations. It gives

4a20a212−4a02a22112a30a02a12−12a03a20a21 b10a12−b01a21. 3.22

From3.16and3.18we obtain a2213a30a12 and a2129a30a03,respectively, which we insert in3.22to obtain

a12b10b01a21. 3.23

From3.23we obtain a12 b01a21/b10 ifb10/0 and we insert it in3.16to obtain

a30

b10a21 3b01

. 3.24

Analogous, we insert it in3.18to obtain

a03

b2 01a21 3b2

10

. 3.25

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Case 1. Suppose thata20/0. From3.21it has

a02

b00a211 4a20 .

3.26

From3.19

a12

4a21a11−12a30a02b10 4a20

. 3.27

From3.23we obtain a21and we replace it in3.27. We replace also a30 from3.24and

a02to obtain

12a220b201a1212a11a20a12b10b01−3a12b210

b00a211

3a20b10b201. 3.28

We associate the terms containing a12

a12

a20b10b201 4a2

20b201−4a11a20b10b01b210b00b102 a211

. 3.29

We replace this last expression ofa12in3.23and we obtain

a21

a20b210b01 4a2

20b201−4a11a20b10b01b210b00b102 a211

. 3.30

We replace the expression ofa21in3.24and3.25and we have

a30

a20b310

34a220b201−4a11a20b10b01b210b00b210a211 ,

a03

a20b301 34a2

20b201−4a11a20b10b01b210b00b210a211 .

3.31

Note that those last four expressions are in the image of F1.

Case 2. Suppose thata20 0. From3.21we obtain

a211b00. 3.32

From3.19

a02 4a21a11b10 12a30

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We replace the value of a30obtained from3.24and the expression a11in this last equation to obtain

a02

b01

±4a21 −b00b10

4a21b10 . 3.34

The expression of a12 is obtained from3.23. The expressions of this case are in the image of F2.

Proof of Theorem2.7.

Complex Case

1Letgx, y ax2 bxycy2dxeyh ∈ AC2 be a quadratic polynomial. By an affine transformation of the complex plane the polynomialgis equivalent to the prenormal form Px, y y2Ax2BxC,where A, B, CC.

IfA /0, thenPis in the orbit of y2x2k,wherekC. If A0 and B /0,then

P is in the orbit of y2x. Finally, if A 0, B 0, C /0,thenP is in the orbit of

y2−m,where m∈C.

Note that the polynomialfx, y √−1/2xy2 AC

3 satisfies H3Cf y2. The polynomial fx, y xy2/2x2/2 ∈ AC3 verifies H3Cf xy2. For each

k ∈C, the polynomialfkx, y y3/6x2y/2− √

k/2x2 √k/2y2 AC 3 fulfills H3Cfk y2−x2−k.

2To 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 H3Cf y2r .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 gx, y ax2bxycy2dxeyh is in the orbit of one of the normal forms: y2x2q

1, y2x2− q2,y2−x2−q3, xy2, y2x, 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. Letg∈AR2n4be a polynomial with real coefficients. Thengis a real Hessian polynomial if and only if there exists a polynomialftotally imaginary on the set HnC−1−g.

1By Lemma3.4and Proposition2.9we have that the polynomials y2x2r 1,y2−

x2r

2,withr1, r2>0 are real Hessian polynomials. To finish this part we note that the polynomialfx, y xy2/2x2/2 AR

3 verifies H3Rf xy2 and that

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2By Lemma3.4and Proposition2.9we have that the polynomials −x2y2, x2

y2, y2x2r

3, y2x2−r4,y2−x2r5,withr3, r5<0, r4∈R,are not real Hessian polynomials. On the other hand, the polynomials y2r

6,y2−r7,where r6, r7∈R∗, are not real Hessian polynomials because the complex polynomial y2r, rCis

not complex Hessian polynomial. To show that the polynomials y2and y2xare not real Hessian polynomials, we use the same method of Groebner basis realized in the complex case.

Proof of Lemma3.4.⇒By hypothesis there existsfx, y nrs0arsxrys ∈ ARn such that

HnRf g. Therefore, consider the totally imaginary polynomialifx, y rns0 iars xrys,

which satisfiesHnCif i2HC

nfg.

⇐By hypothesis there existsftotally imaginary such thatHnCfg. Therefore,if

is a real polynomial such thatHnRif i2HC

nf g.Hence,gis a real Hessian polynomial.

Proof of Corollary2.8. Let us consider the map ψ:AC3 ×C2 C3{w

1, w2, w3} given by

f, p−→fxx

p, fxy

p, fyy

p. 3.35

Iffx, y 3ij0aijxiyj∈AC3 andp x, y, then

ψf, p6a30x2a21y2a20,2a21x2a12y2a11,2a12x6a03y2a02

. 3.36

For each fixed f∈AC3 let us consider ψf :C2 → C3 given by ψfp:ψf, p.

Now, we are interested to describe the conditions inf under which the image under

ψf of C2, ψfC2, is not a plane.

Lemma 3.5. The set :{f∈AC3 |ψfC2 is not a plane} is given by the union of the following

sets:

CP1:

f∈AC3 |a30 a21a120

,

CP2:

f∈AC3 |a30∈C∗, a12

a2 21 3a30, a03

a321

27a230

.

3.37

Proof. The Jacobian matrix ofψf is

⎝3aa2130 aa2112

a12 3a03 ⎞

. 3.38

ψfC2is 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,

3a12a30−a2210, 3a21a03−a122 0, 9a03a30−a21a12 0. 3.39

(14)

Lemma 3.6. Letfx, y 3ij0aijxiyj be a complex polynomial such that fSψ.

Case 1. Suppose fCP1.

1Ifa03/0,then ψfC2is a parallel line to thew3-axis inC3.

2Ifa030,then ψfC2is the point2a20, a11,2a02.

Case 2. If fCP1,then ψfC2is the line

lf

6a30x2a20,2a21xa11, 2a2

21

3a30x2a02

|x∈C

. 3.40

Proof. Letfx, y 3ij0aijxiyj∈AC3.In virtue of Lemma3.5we have the following cases

Case 1. If fCP1,then, by3.36, ψfC2 {2a20, a11,6a30y2a02∈C3|y∈C}.

1Ifa03/0,then ψfC2 {2a20, a11,6a30y2a02∈C3|y∈C}.

2Ifa030,then ψfC2is the point2a20, a11,2a02.

Case 2. If fCP2,then, by3.36,

ψf

C2

6a30x2a21y2a20,2a21x 2a2

21 3a30

ya11, 2a3

21 9a2

30

y2a

2 21 3a30

x2a02

. 3.41

Let v1 6a30,2a21,2a212 /3a30 be a nonzero vector by hypothesis and v2

2a21,2a21/3a30,2a321/9a230. Note thatv2 a21/3a30v1. Therefore, the setlf is the same set

of ψfC2.

Let us consider the cone C{w1, w2, w3∈C3 |w1w3−w22 0}.We shall describe the set ψfC2∩C when f.

Lemma 3.7. Let fSψ.

1If ψfC2is a parallel line to thew3-axis inC3, then

aψfC2∩C is a parallel line to thew3-axis inC3whenevera20 0;

bψfC2∩C is the point2a20, a11,a112 −4a20a02/2a202a20 whenevera20/0.

2If ψfC2is the point2a20, a11,2a02 and 4a20a02−a211 0, then ψfC2 ∩ Cis the

point2a20, a11,2a02.

3If ψfC2is the linelf, then

aψfC2∩C is the point6a30x02a20,2a21x0a11,2a221/3a30x02a02ifα

1/3a3036a230a024a20a221−12a21a11a30/0,where x0 1/αa211−4a20a02;

(15)

Proof. In virtue of Lemma3.6we have the following cases.

1If ψfC2∩C is a parallel line to thew3-axis, then ψfC2∩C {2a20, a11,6a03y 2a02∈C3|y∈C,2a206a03y2a02−a211 0}.

aIfa20 0, then ψfC2∩C is a parallel line to thew3-axis.

bIfa20/0, theny a211−4a20a02/12a206a03and ψfC2∩C is a point.

2If ψfC2∩C is the point2a20, a11,2a02, then ψfC2∩C 2a20, a11,2a02if 4a20a02−a211 0.

3If ψfC2∩C is the linelf, then ψfC2∩C lf∩ {6a30x2a202a221/3a30x 2a20−2a21xa1120}. It means that

αx4a20a02−a211 0, whereα

36a230a024a20a221−12a21a11a30

3a30 .

3.42

Therefore, ifα0, then 4a20a02−a2110 and we obtain a line onC. Ifα /0, then ψfC2∩C

is a point.

With this lemma we finish the proof of corollary.

Proof of Proposition2.9. Letgx, y b20x2 b11xyb02y2 b00 ∈ AC2. Then the expression

H3Cf gis equivalent, by Lemma3.1, to the system of equations:

b02−4a22112a30a12,

b11−4a21a1236a30a03,

b02−4a21212a21a03,

012a30a024a20a12−4a21a11,

012a20a034a21a02−4a12a11,

b00−a2114a20a02.

3.43

LetSbe the union of the images ofF1andF2. A direct substitution shows thatSH3C− 1g.

Therefore, to finish the proof it is enough to show that H3C−1gS. To check this last sentence we have used the computer algebra system Maple 9.5.

Proof of the Theorem2.12.

Real Case

(16)

real cuartic polynomialf in H4C−1g; that is, there are no real polynomial satisfying the system of equations:

24a40a22−9a231 0, 3.44

72a40a13−12a31a220, 3.45

144a40a0418a31a13−12a2220, 3.46

72a31a04−12a22a130, 3.47

24a04a22−9a2130, 3.48

12a30a2224a40a12−12a31a21 −1, 3.49

36a30a1372a40a03−12a21a220, 3.50

36a03a3172a04a30−12a12a221, 3.51

12a03a2224a04a21−12a13a120, 3.52

24a40a024a20a2212a30a12−6a31a11−4a2210, 3.53

12a31a0212a20a1336a30a03−4a21a12−8a11a22 0, 3.54

24a04a204a02a2212a03a21−6a13a11−4a2120, 3.55

12a30a024a20a12−4a21a11 1, 3.56

12a03a204a02a21−4a12a11 0, 3.57

4a20a02−a2110. 3.58

A Groebner bases for3.44–3.48with the graded reverse lexicographic monomial order is

3a213−8a22a04, a22a13−6a31a04, 2a222 −3a31a13−24a40a04, a31a22−6a40a13,

3a2

31−8a40a22, a13a31a04−16a40a204, a31a40a13−16a240a04.

3.59

The set of common zeroes of these last 7 polynomials is the union of the sets

{a40a40, a310, a220, a13 0, a04 0}, 3.60

a40

a4 13 256 a3

04

, a31

a3 13 16a2

04

, a22 3a2

13 8a04

, a13 a13, a04/0

(17)

On the one hand, replacing the solution3.60in3.49–3.58, particularly in3.56we obtain 01. On the other hand, replacing the solution3.61in3.49–3.58, we obtain the system

4a20a02−a2110,

12a30a024a20a12−4a21a11−10,

−4a12a114a21a0212a20a030,

12a30a12 3a413a02

32a304 −4a

2 21−

3a313a11 8a2

04

3a20a213 2a04

0,

36a30a0312a20a13− 3a2

13a11

a04 −

4a21a12 3a3

13a02 4a2

04

0,

−6a13a1124a20a04 3a213a02

2a04 12a21a03−4a 2 12 0,

−3a313a21 4a204

9a30a213 2a04

3a413a12

32a304 10,

9a4 13a03 32a3

04

36a30a13−

9a21a213 2a04

0,

9a3 13a03 4a2

04

−9a213a12 2a04

72a30a04−10,

−12a13a12

9a213a03 2a04

24a21a040.

3.62

From the last equation we obtain that a21 −a13−8a12a04 3a13a03/16a204 and from the sixth equation,

a20 − 6a2

13a02a04−24a13a11a20424a13a03a12a04−9a213a203−16a212a204

96a304 . 3.63

When we replace these expressions in the last four equations of the last systemwe do not write the other equations because we do not need it:

−18a4

13a12a049a513a03288a30a213a30464a404 64a4

04

0, 3.64

9a13

32a30a304−2a12a213a04a313a03

8a3 04

(18)

9a313a03−18a12a213a04288a30a304−4a204

4a204 0, 3.66

00. 3.67

From3.65we have that a130 or 32a30a304−2a12a132 a04a313a03 0. If a13 0,then3.64 becomes −10.

If 32a30a304−2a12a132 a04a313a030,then a30 2a12a213a04−a313a03/32a304. Substituing

a30in3.66, we obtain−10.

Lemma 3.8. The set,Cv3, of critical points of the map H3CR : AC3RAC2 is the union of the following six sets:

S1

a02

9a03a11a30−6a03a21a20−a12a11a212a212a20 23a12a30−a221

,3a12a30−a221/0

,

S2{a110, a210, a120},

S3

a11

2a12a20

a21

, a30

a221

3a12

, a21a12/0

,

S4{a300, a210, a120},

S5{a200, a300, a210},

S6

a12

a2 21 3a30

, a03

a3 21 27a2 30

, a30/0

.

3.68

Proof. Letfx, y 3ij2aijxiyj∈AC3R. The Jacobian matrix ofH3CRis,

JH3CR

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

2a02 −a11 2a20 0 0 0 0

6a03 −2a12 2a21 0 2a02 −2a11 6a20

2a12 −2a21 6a30 6a02 −2a11 2a20 0

0 0 0 0 6a03 −4a12 6a21

0 0 0 18a03 −2a12 −2a21 18a30

0 0 0 6a12 −4a21 6a30 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

(19)

Let us denote by Mk, k 1, . . . ,7,the 6×6 matrix obtained from JH3CRby deleting the

column 7−k1. On the other hand,fis a critical point ofH3CRif and only if detMk 0 for

allk1, . . . ,7, that is, if and only if the following system of seven equations is satisfied:

−9a21a12a0327a30a2032a312

F 0,

a212a21−6a221a039a12a30a03

F 0,

a221a12−6a212a309a21a30a03

F 0,

2a321−9a12a21a3027a230a03

F 0,

−3a03a21a11a212a11−a21a12a029a03a30a02

F 0,

3a02a12a30a212a20−3a03a21a20−a221a02

F 0,

a11a221a12a21a209a20a30a03−3a12a11a30

F 0,

3.70

where

F−9a03a11a306a03a21a206a02a12a30−2a221a02a12a11a21−2a212a20. 3.71

Thereforefx, y 3ij2aijxiyjsatisfies the previous system of seven equations if and only

iffS6k1Sk, whereSis the union of the six solutionsSj. Therefore, the proof is done.

Lemma 3.9. Considerf ∈AC3 and g H3f. The curve g 0 is singular if and only if

−2a212a206a12a02a306a21a03a20a21a12a11−2a221a02−9a03a11a300,

a221a212−18a21a12a30a0327a302 a2034a312a304a321a03/0,

3.72

or,

−9a12a21a03a20−a21a11a212−a12a212 a0227a30a203a202a20a3121.8

6a221a03a11−9a30a21a03a026a30a02a212−9a30a03a11a12 01,

a221a212−18a21a12a30a0327a230a2034a312a304a321a03 0.2.1,

(20)

Proof. Supposefx, y 3rs0 arsxrys; then

H3

fa2

114a20a02 12a30a024a20a12−4a21a11x

12a20a03−4a12a114a21a02y

−4a22112a30a12

x2

−4a21a1236a30a03yx

−4a21212a21a03

y2.

3.74

The curve g H3Cfis singular if and only if the system gxp gyp gp 0 has a

solution for somepinC2. The system formed byg

x0, gy0 is

6a30a12−2a221

x 9a30a03−a21a12y3a30a02a20a12−a21a110,

9a30a03−a21a12x

6a21a03−2a212

y3a20a03−a12a11a21a020.

3.75

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, k1, . . . ,6,of Lemma3.8. IffS1, then

H3Cf 1

3a12a30−a221

×x236a230a2124a421−24a30a12a221

xy108a2

30a03a1236a30a03a2214a321a12−12a21a212a30

y24a2

12a221−12a312a3036a21a03a12a30−12a321a03

x24a30a212a2054a230a03a114a321a11−4a20a12a221

−36a30a03a21a20−18a30a12a11a21

y−12a212a11a3018a21a03a11a3036a20a03a12a30

−24a221a03a202a212 a12a114a21a212a20

−3a211a12a30−12a03a21a22018a20a03a11a30

−2a20a12a11a214a212a220a211a221

.

(21)

Denote byhthe polynomialH3Cf. Then,

hx 1

3a12a30−a221

×x72a230a212−48a30a12a2218a421

y108a230a03a124a321a12−36a30a03a221−12a21a212a30

4a321a1154a230a03a11−4a20a12a22124a30a212a20

−36a30a03a21a20−18a30a12a11a21

,

hy 1

3a12a30−a221

×x108a230a03a12−36a30a03a221−12a21a212a304a321a12

y8a212a221−24a312a30−24a321a0372a21a03a12a30

4a21a212a20−24a221a03a202a221a12a11−12a212a11a30

18a21a03a11a3036a20a03a12a30

.

3.77

The pointx, ywhere the Hessian curve is singular is

x −2a20a12a21a11

23a12a30−a221

, y−3a11a30−2a21a20

23a12a30−a221

. 3.78

IffS2, then H3Cf 6a30x2a206a03y2a02, hx 6a306a03y2a02, andhy 66a30x2a20a03.

The pointx, ywhere the Hessian curve is singular is xa20/3a30, ya02/3a03. IffS3, then

H3Cf −4

a221xa21ya12a20a12

y−3a221a03a21a212

a221a02a20a212

a12a221

,

hx −4

y−3a221a03a21a212

a221a02a20a212

a12

,

hy−4

y−3a2

21a03a21ya212

a2

21a02a20a212

a21

− 4

a221xa21ya12a20a12

−3a221a03a21a212

a12a221

.

(22)

The pointx, ywhere the Hessian curve is singular is

x a12−a21a023a20a03 a21

−3a21a03a212

, y−−

a221a02a20a212

a21

−3a21a03a212

. 3.80

If−3a21a03a122 0, thenH3Cfis of degree one.

IffS4, then H3Cf 12a20a03y4a20a02−a211, hx0, andhy12a20a03. The Hessian curve in this case has degree one.

IffS5, then H3Cf −2a12ya112, hx0, hy−42a12ya11a12. The pointx, ywhere the Hessian curve is singular is ya11/2a12. IffS6, then

H3Cf

108a330a0212a20a212 a30−36a21a230a11 9a2

30

x

36a21a02a2304a20a321−12a221a11a30 9a2

30

y36a20a02a

2

30−9a211a230 9a2

30

.

3.81

The Hessian curve in this case has degree one.

Acknowledgments

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

References

1 G. Salmon, A Treatise on the Analytic Geometry of Three Dimensions, vol. 2, Chelsea, New York, NY, USA, 5th edition, 1965.

2 Y. L. Kergosien and R. Thom, “Sur les points paraboliques des surfaces,” Comptes Rendus Hebdomadaires des S´eances de l’Acad´emie des Sciences. S´eries A, vol. 290, no. 15, pp. 705–710, 1980.

3 V. I. Arnold, “Remarks on parabolic curves on surfaces and on higher-dimensional M ¨obius-Sturm theory,” Funktsional’nyi Analiz i ego Prilozheniya, vol. 31, no. 4, pp. 3–18, 1997 Russian, english translation in Functional Analysis and Its Applications, vol. 31, no. 4, pp. 227–239, 1997.

4 V. I. Arnold, Arnold’s Problems, Springer, Berlin, Germany, 2004.

5 V. I. Arnold, “On the problem of realization of a given Gaussian curvature function,” Topological Methods in Nonlinear Analysis, vol. 11, no. 2, pp. 199–206, 1998.

6 E. Brieskorn and H. Kn ¨orrer, Plane Algebraic Curves, Birkh¨auser, Basel, Switzerland, 1986.

Figure

Table 3: Real Hessian polynomials.

References

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