 # A proof of Hilbert's theorem on ternary quartic forms with the ladder technique

This paper proposes a totally constructive approach for the proof of Hilbert's theorem on ternary quartic forms. The main contribution is the ladder technique, with which the Hilbert's theorem is proved vividly.

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## 1 Introduction

In 1888, David Hilbert published a paper (Hilbert, 1888) on the problem whether a positive semi-definite real polynomial is inevitably a sum of squares of other real polynomials. This work is influential and inspires a lot of great works in researchers even today. In the exceptional case, Hilbert proved that a positive semi-definite real ternary quartic form can be written as a sum of squares of quadratic forms. Hilbert’s proof is very brief but rather hard, because some complicated mathematical tools are used (see comments in Albrecht and Claus (2012) and Choi and Lam (1977)). Furthermore, Hilbert’s method did not lend itself to a really practical construction. In 1977, Choi and Lam (Choi and Lam, 1977) showed a graceful elementary proof, in which only the rudiments of real analysis and the representation theorem of convex set are utilized. However,this method is not constructive either. Compared with Choi and Lam (1977), though there are some similarities in the rudiments of real analysis, the proof of the present paper has the following two conspicuous differences. First, we take the ladder technique instead of the representation theorem of convex set. Furthermore, the method of our proof is constructive whereas the one in Choi and Lam (1977) is not, since the representation theorem of convex set is an existence theorem. Thus, for a positive semi-definite ternary quartic form, its explicit representation as a sum of squares can be constructed step by step according to the method addressed in this paper. From this perspective, we emphasize that this is not a completely new proof but, rather, a completely constructive account of the proof in Choi and Lam (1977).

The rest of this paper is organized as follows. In section 2, four lemmas are presented. Each of them is like a step of the ladder, and the highest one (Lemma 4) achieves the goal. With the ladder technique, the Hilbert’s theorem is proved easily. In section 3, the computation of real zero points is discussed, which provides the concrete details for the construction of sum of squares.

## 2 Proof of Hilbert’s theorem

Firstly, the following are required. A form is said to be positive semi-definite (psd) if

 f(P)≥0, ∀P∈R3,

and represents the set of all positive semi-definite ternary quartic forms. SOS stands for the set of sum of squares of polynomials.

Given satisfying , we call and are same if there is a non-zero number such that . That is, we discuss the zeros of in the real projective space . The set of zeros in is denoted by , and stands for the number of elements of . The unit sphere is written as .

Next, we will present four lemmas.

###### Lemma 1.

If the psd quartic form has no zero in (i.e. ), then there is a quadratic form such that is positive semi-definite and has at least one zero in (i.e. ).

###### Proof.

Since is continuous on the unit sphere , inevitably attains a minimum . It is obvious that and

 f−λ(x2+y2+z2)2≥0.

Let , and then is positive semi-definite and . ∎

###### Lemma 2.

If the psd quartic form has only one zero in (i.e. ), then there are quadratic forms and non-negative real numbers and with such that is positive semi-definite and .

###### Proof.

By coordinates transformation , we can assume that , and can be written as

 f=x2p(y,z)+2xq(y,z)+r(y,z), (1)

where , and . There are three cases of to be discussed.

Case (i): ( 0 form). It is clearly that  is necessarily a 0 form. Then

 f=r(y,z).

Since  has only one zero , it holds that for all Furthermore,  is continuous on the unit circle , so that  has the minimum  on . It is obviously that  and

 f−λ(y2+z2)2≥0.

Let . Thus is positive semi-definite and .

Case (ii): The rank of is . By coordinate transformation, assume that . The condition implies that is divided by (the coefficient of in  is 0). That is, there is satisfying . Substituting it into (1) one has

 f=x2y2+2xyq1+r=(xy+q1)2+r−q21.

Notice that is not a 0 form, otherwise  would have infinite zeros on . There are two situations to be considered.

(a)   has a zero   on  , then it has at least 2 zeros  and  on  . Hence is positive semi-definite and  .

(b)   has no zero on , then   yields case (i), so that there is a quadratic form  such that  is positive semi-definite and .

Case (iii): The rank of is . By coordinate transformation, may take the form . Consequently, we claim that on . Otherwise if for , then is a zero of , conflicting with the fact that only has the zero . Thus  attains the minimum on . As a result,

 (pr−q2)−λ(y2+z2)3≥0,

with equality at . Hence

 f−λ(y2+z2)2

is still positive semi-definite by discriminant, and has at least two zeros: and .

It follows from case (i) to case (iii) that Lemma 2 holds. ∎

###### Lemma 3.

If the psd quartic form has only two zeros in (i.e. ), then there are quadratic forms and non-negative real numbers with such that is positive semi-definite and .

###### Proof.

Change coordinates so that . Write

 f=x2p(y,z)+2xzq(y,z)+z2r(y,z), (2)

where and are quadratic forms with   and . Next the proof will be split into two cases.

Case (i): If at least one of and has a zero on , then let be the zero of without loss of generality. Substituting it into (2), one yields

 f(x,y0,z0)=x2p(y0,z0)+2xz0q(y0,z0).

Then  is also a zero of , hence . Substituting them into quadratic form , it is reduced to the form

 r(y,z)=tz2, t>0.

From , must be divided by . Thus , where is a linear form. Therefore,

 pr−q2=tz2(p−q21)≥0⟹p−q21≥0,

and

 f=x2p+2xz2√tq1+z2(tz2)=(√tz2+xq1)2+x2(p−q21).

Then  is positive semi-definite and has infinite zeros . Hence

 |Z(f−(√tz2+xq1)2)|>3.

Case (ii): If neither nor is supposed to have a zero on  (both and are strictly positive on ), then discuss two possibilities of the discriminant .

(a) The discriminant  has a zero , and let . Then one has

 f1(x,y,z)=f(x+λz,y,z)=x2p+2xz(q+λp)+z2(r+2λq+λ2p).

Since  is a zero of , is reduced to the case (i).

(b) The discriminant  has no zero on . Since  and are strictly positive on , the function

 pr−q2p(y2+z2)

is also strictly positive on , and its minimum . Thus and  is positive semi-definite. If  has zeros, then  is reduced to the case (i). In contrast, if  has no zero on , then is reduced to the case (ii) (a).

Because we have considered all two cases, we can conclude that Lemma 3 holds. ∎

###### Lemma 4.

If the number of zeros of the psd quartic form in is more than 3 (i.e. ), then is a sum of squares of quadratic forms.

###### Proof.

The proof will be split into two cases.

Case(i): If there are three zeros not on the same line, then by arranging coordinates, we may assume that

 f(1,0,0)=f(0,1,0)=f(0,0,1)=0.

This implies that the degree of each variable is less than or equal to . Write

 f=ax2y2+by2z2+cz2x2+a1x2yz+b1y2xz+c1z2xy,a,b,c,a1,b1,c1∈R.

Do substitution , and can be transformed into a positive semi-definite form in . It is obvious that is a sum of squares of quadratic forms.

Case (ii): If all of the zeros are on the same line, then it can be split into two subcases.

(a) has infinite zeros. Then has linear factors by Bzout’s theorem. Since is positive semi-definite, the degree of the linear factors is even. Hence they are

 l2h2, l4,

where is linear form and is positive semi-definite quadratic form (maybe degenerate). It is obvious that all of them are sum of squares of quadratic forms.

(b) has finite zeros. we claim that this case will never happen. The reason is as follows.

Arrange coordinates so that with . Write

 f=x2p(y,z)+2xzq(y,z)+z2r(y,z), (3)

where and are quadratic forms with . Therefore,

 f(x0,y0,0)=x20p(y0,0)=0.

Furthermore, yields , where is a quadratic form only in and . Let

 p=ay2+2byz+cz2,a,b,c∈R.

Consequently , and by . Thus divides  by (3), and all of the points on the line are zeros of . This contradicts the premise that has finite zeros on the line . ∎

With the above 4 lemmas, Hilbert’s theorem will be proved with a chart.

###### Theorem 1 (Hilbert).

A positive semi-definite ternary quartic form over the reals can be written as a sum of squares of quadratic forms.

###### Proof.

The following chart will present the process of proof.

 |Z(f)|≥3 (SOS,Lemma 4)↗⇑|Z(f)|≥2∥ Lemma 3↗⇑↘∥|Z(f)|≥1∥ Lemma 2|Z(f)|=2↗⇑↘∥PSD43∥ Lemma 1|Z(f)|=1↘∥|Z(f)|=0

The set is divided into two disjoint subsets, that is, the subset with and the other one with . According to Lemma 1, the former can be transformed into the latter by minus a square of a quadratic form. Furthermore, the subset with can be dealt with in the same way according to Lemma 2 and so forth. Finally the theorem holds according to Lemma 4. ∎

rmk1Remark Theorem 1 is the weak edition of Hilbert’s theorem. The strong edition is that the positive semi-definite ternary quartic forms are sum of no more than three squares of quadratic forms. The proof of the strong edition can see Albrecht and Claus (2012), Powers et al. (2004) and Scheiderer (2010).

rmk2[rmk1]Remark The real zeros are assumed to be known in the above proof. To achieve the goal of really practical construction , we still need some method in solving the real zeros of ternary quartic forms. This problem will be conquered in Section 3, and we will really obtain the practically constructive proof of Hilbert’s theorem.

Next we will present an example. ex1Example (Cirtoaje, 2006) Compute the sum of squares of the following ternary quartic form.

 f= 4(x4+y4+z4)+21(xy+yz+zx)2−10(x2+y2+z2)(xy+yz+zx) −37xyz(x+y+z).

Solution: Since has 4 zeros in the projective space. There are

 (1,1,1),(3,2,2),(2,3,2),(2,2,3).

Select three of them and construct the following matrix (each column is a zero of )

 A=⎡⎢⎣1 3 21 2 31 2 2⎤⎥⎦.

 ⎡⎢⎣xyz⎤⎥⎦=A⎡⎢⎣¯x¯y¯z⎤⎥⎦.

By computing, we have

 f(A[¯x,¯y,¯z]T) = ¯x2¯y2+49¯y2¯z2+¯z2¯x2−¯x2¯y¯z+7¯y2¯z¯x+7¯z2¯y¯x (4) = 12((¯x¯y−¯z¯x)2+(¯x¯y+7¯y¯z)2+(¯z¯x+7¯y¯z)2).

Compute the inverse matrix of ,

 A−1=⎡⎢⎣−2 −2  5 1  0 −1 0  1 −1⎤⎥⎦.

Let

 ⎡⎢⎣¯x¯y¯z⎤⎥⎦=A−1⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣−2x−2y+5zx−zx−y⎤⎥⎦,

and substitute it into (4), and then the sum of squares of is as follows.

 f=12∑(−2x2+5xz+2y2−5yz)2.

## 3 Computation of zeros

In this section, we will discuss the problem of computing zeros. Lemma 4 presents a key of constructing sum of squares, that is, finding out at least three real zeros of a positive semi-definite ternary quartic form. Next we will prove a proposition (Lemma 5) that can solve the problem of computing the real zeros of positive semi-definite ternary forms. The fundamental idea comes from Yang and Xia (2000); Xia and Yang (2016).

###### Lemma 5.

Given positive semi-definite form , let be the derivative of with respect to , and be the resultant of  and with respect to . Then the equation and the equations are equivalent in the projective space .

###### Proof.

It is obvious that . Next we will prove

It is easy to get that the real zero of positive semi-definite form need to satisfy the following equations for solving stationary points.

 ⎧⎪⎨⎪⎩f′x=0,f′y=0,f′z=0. (5)

This is because of the fact that if at least one of the following values

 f′x|(x0,y0,z0), f′y|(x0,y0,z0), f′z|(x0,y0,z0)

is not zero, then can be approximated by linear functions on a neighborhood of the point . So is not nonnegative on a neighborhood of the point . This contradicts the premise that is positive semi-definite. Thus satisfies the following equations.

 {f(x,y0,z0)=0,f′x(x,y0,z0)=0. (6)

Then according to the fundamental property of resultant. That is,

rmk3Remark In Lemma 5 cannot include square factors, otherwise would identically equal to 0. Thus the equations in is zero dimension, and we can solve the above equations with various efficient methods such as method of resultant (Kapur et al., 1994; Canny and Manocha, 1993), rational single variable present method (Rouillier, 1999), etc.

rmk4[rmk3]Remark The general version of Lemma 5 (successive resultant method (Yang and Xia, 2000; Xia and Yang, 2016)) also ensure the minimum of Lemma 1, Lemma 2 and Lemma 3 can be computed accurately, and then ensure the above proof of Hilbert’s theorem is completely constructive. However, the following example (example 3) shows that the practical computation may be very complex and finally comes to nothing because of high time complexity.

Next we will present two computing examples.

ex2[ex1]Example (Vasile Cirtoaje)(Cirtoaje, 2006) Compute real zeros of

 f=(x2+y2+z2)2−3(x3y+y3z+z3x).

Solution: Firstly compute

 f′x=4(x2+y2+z2)x−9x2y−3z3.

Then compute the resultant ,

 res(f,f′x,x)=9(13y4−18y3z−y2z2−6yz3+13z4)(−z+y)2(y3−5y2z+6yz2−z3)2.

By Gram matrix method (Choi et al., 1995), one yields

 h = (13y4−18y3z−y2z2−6yz3+13z4) = 13(y2−913yz−713z2)2+286(213yz−51286z2)2+322z4.

Thus is strictly positive on . Hence

 res(f,f′x,x)=0⟺(−z+y)(y3−5y2z+6yz2−z3)=0.

By using real root isolation algorithm (Xia and Yang, 2016), we know that the equation has four real zeros in the projective space :

 (1,1), (α1,1), (α2,1), (α3,1),

where are three positive real zeros of . Substituting the above four points into the polynomial , and we have four real zeros of in the real projective space . There are

 (1,1,1), (1α2,α1,1), (1α3,α2,1), (1α1,α3,1). (7)

Consequently can be written as a sum of squares according to Lemma 4.

ex3[ex1]Example (C. Scheiderer)(Scheiderer, 2016)

In Scheiderer (2016) this form is proven to have the following characters: cannot be written as sum of squares of quadratic forms with rational coefficients. That is to say the answer of Sturmfels’s question (Scheiderer, 2016) is negative. It means that it is very difficult to write as sum of squares for the tools that depend on numerical computation such as SOSTOOLS (Parrilo, 2003).

Next we compute the zeros of in the real projective space.

Firstly we compute

 g′x=4x3+4xy2−6xyz+y3−4zy2+z3.

Then compute the resultant

 res(g,g′x,x)=229y12−1904y11z+5896y10z2+1376y9z3−12176y8z4+6432y7z5+8630y6z6−9472y5z7+952y4z8+3232y3z9−96y2z10+336yz11+229z12.

By using real root isolation algorithm, we know that only has one trivial zero , but  is not the zero of . Hence  (empty set), and is strictly positive on .

By Lemma 1, one yields that there is a positive constant such that

 gt=g−t(x2+y2+z2)2

is still positive semi-definite and has at least a zero in the real projective space . Furthermore, by using successive resultant algorithm (Yang and Xia, 2000; Xia and Yang, 2016), one can compute accurately. , in the interval and the approximation being , is the real zero of the following equation with degree 12.

 1540909743009169408 x12−13437733654176464896 x11+51805978528683065344 x10−116396366581901484032 x9+168975565335348900096 x8−165910705322168135008 x7+111957978056509355125 x6−51652982930080321180 x5+15876922302830413280 x4−3088008227838928440 x3+347409936566531728 x2−19347901948050048 x+380514157362176=0.

It is very difficult to compute the real zeros of for high time complexity.

In Scheiderer (2016) can be written as

 g=14((2x2+βy2−yz+(2+1β)z2)2−β(2xy−y2β+2xzβ+βyz−z2)2),

where is a negative zero of equation .

## 4 Conclusion

In this paper, we give a proof of Hilbert’s theorem by four lemmas and the ladder technique. According to the proof, catching at least three real zeros of a positive semi-definite ternary quartic form is necessary for constructing its sum of squares. Consequently, we present the method of locating the zeros based on the property that zeros of a positive semi-definite form satisfy the equations for stationary points. However the representation of sum of squares using this method may be complex and how to build a clear representation is a new question.

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