This is a preprint of: “Twin polynomial vector fields of arbitrary degree”, Jaume Llibre, Clàudia Valls, Bull. Braz. Math. Soc. (N.S.), vol. 53, 295–306, 2022. DOI: [10.1007/s00574-021-00259-4] TWIN POLYNOMIAL VECTOR FIELDS OF ARBITRARY DEGREE JAUME LLIBRE1 AND CLAUDIA VALLS2 Abstract. In this paper we study polynomial vector fields on C2 of degree larger than 2 with n2 isolated singularities. More precisely, we show that if two poly- nomial vector fields share n2 − 1 singularities with the same spectra (trace and determinant) and from these singularities n2 − 2 have the same positions, then both vector fields have identical position and spectra at all the singularities. More- over we also show that if two polynomial vector fields share n2 − 1 singularities with the same positions and from these singularities n2 − 2 have the same spectra, then both vector fields have identical position and spectra at all the singularities. Moreover we also prove that generic vector fields of degree n > 2 have no twins and that for any n > 2 there exist two uniparametric families of twin vector fields, i.e. two different families of vector fields having exactly the same singular points and for each singular point both vector fields have the same spectrum. 1. Introduction and statement of the main results Consider the polynomial vector fields on the affine plane C2 and denote by Pn the space of all polynomial vector fields ∂ ∂ χ = P (x, y) + Q(x, y) , ∂x ∂y such that P and Q are polynomials and have degree at most n. By Bezout’s theorem, a generic element of Pn has exactly n2 isolated singularities. We denote by P bn the space of vector fields Pn that have n isolated singularities. Since χ has the maximum 2 number of singularities the determinant of the linear part of χ at each singular point is nonzero. So the eigenvalues at any singular point are nonzero, i.e. all singular points are non degenerate (see for more details [7]). The space P bn is endowed with a structure of a complex affine space identifying all the (n + 1)(n + 2) coefficients of the polynomials P and Q with a point of C(n+1)(n+2) . This topology in the set of the polynomial vector fields of degree n is called the topology of the coefficients and bn is an open subset of Pn . P We will say that a property is a generic property for the class of vector fields Pbn if the set of vector fields having this property contains a (non-empty) Zariski open and dense subset of P bn . We say that two vector fields χ1 and χ2 are affine equivalent if there exists an affine map T so that χ2 (x, y) = DT · χ1 (T −1 (x, y)). 2010 Mathematics Subject Classification. Primary 34A05. Secondary 34C05, 37C10. Key words and phrases. Euler-Jacobi formula, singular points, topological index, polynomial dif- ferential systems, Berlinskii’s Theorem. 1 2 J. LLIBRE AND C. VALLS We denote by Sing (χ) the set of singular points of the vector field χ. If p ∈ Sing (χ), we define the spectrum of χ at p as the two eigenvalues of the linearization matrix   Px Qy , Qx Qy (x,y)=p that is, the spectrum is the unordered set of eigenvalues of Dχ(p). Note that if M ∈ GL2 (C) then the spectrum of M carries exactly the same information as the pair Spec (M ) = (trM, det M ) being trM the trace of the matrix M and det M the determinant of the matrix M . In all the paper the spectrum of a vector field χ will be thought as the pair of the trace and the determinant. If X is a topological space and m ≥ 1 let Sm denote the symmetric group of m elements. Then X m /Sm is the quotient of the usual action of Sm on X m permuting its components. The set of spectra of singularities of a generic polynomial vector 2 field χ of degree n belongs to the space Sn = (C2 )n /Sn2 which is an irreducible bn → Sn . affine algebraic variety. We have a well defined map Specn : P We say that two vector fields from the class P bn have the same spectra of singu- larities if they have the same image under the map Specn . Note that the above definition takes into account the spectra of singularities and does not take into ac- count the position of them. Our main aim is to understand the pair of vector fields that share both position and spectra of singularities, that is, our main aim is to provide results in the following spirit: consider two polynomial vector fields χ and b having each of them n2 isolated singularities and label these points as p1 , . . . , pn2 χ and pb1 , . . . , pbn2 , respectively. Assume that (1) pi = pbi for i = 1, . . . , M and Dχ(pi ) ∼ Dχ b(b pi ) for i = 1, . . . , N where A ∼ B denotes that the matrices A and B are similar that is, they have the same spectrum. Then for certain values of M and N we want to see when pi = pbi and Dχ(pi ) ∼ Dχ pi ) for all i = 1, . . . , n2 . b(b Once it has been established that two vector fields agree on position of their sin- gularities and their corresponding spectra, it is natural to ask whether these vector fields are identical or not. This question gives rise to the concept of twin vector fields: two different vector fields are said to be twins if they agree on position and spectra at all their singular points. This corresponds to M = N = n2 in (1). The corresponding question for quadratic polynomial vector fields, that is, when n = 2 was studied in [8]. There the author proved that if two quadratic vector fields have the same spectra (namely M = n2 = 4 in equation (1)), then after an affine transformation we can achieve that all points share the same position and spectra (that is M = N = 4 in (1)). Furthermore it is proved that a generic quadratic vector field indeed admits a unique twin vector field. The results of [8] are inspired in the papers [5, 6]. Similar results for polynomial vector fields of degree greater than n = 2 are not currently available in the literature and this is the main aim of this paper. The fact that n > 2 makes the analysis much more intricate in particular TWIN POLYNOMIAL VECTOR FIELDS 3 because even when n = 3 after an affine transformation we can achieve that three points share the same position but there are still six free points. The following are our results. We always assume that we are under the assump- tions of equation (1). Theorem 1. The following statements hold for n ≥ 2: (i) If two vector fields have the same spectra at N = n2 − 1 singular points, then they have the same spectra at all n2 singular points; (ii) If in addition they have the same position at the original M = n2 − 1 points, then they also have the same position for the final singular point. Note that it follows from Theorem 1 that if n ≥ 2 and equation (1) is satisfied with M = N = n2 −1, then both vector fields χ and χ b have the same singular points and the same spectra. This corresponds to M = N = n2 in equation (1), with the additional constraint that χ 6= χ b. The proof of Theorem 1 is given in section 2. The particular case for n = 2 was first treated in [8] and the proof for any n follows the same lines but we include it in the paper for completeness. Theorem 2. Consider equation (1) with M = n2 − 2 and N = n2 − 1 with n > 2. b have the same singular points and the same spectra Then both vector fields χ and χ 2 and so M = N = n in equation (1). The proof of Theorem 2 is given in section 3. Theorem 3. Consider equation (1) with M = n2 − 1 and N = n2 − 2 with n > 2. b have the same singular points and the same spectra Then both vector fields χ and χ 2 and so M = N = n in equation (1). The proof of Theorem 3 is given in section 4. We recall here that as mentioned above, the space P bn has dimension (n+1)(n+2). bn . Requiring that another vector field Let χ be a given vector field from the class P b∈P χ bN agrees on position on M points imposes 2M conditions on the parameters of χ b). Similarly, requiring that Dχ(pi ) ∼ Dχ b (that is on the polynomials defining χ b(b pi ) at N singular points imposes 2N conditions on the parameters. Thus we impose 2M + 2N conditions on (n + 1)(n + 2) parameters. For the values of M and N in Theorems 2 and 3, we have 2M + 2N = 4n2 − 6 > (n + 1)(n + 2). Even on the cubic case the gap is quite big. The problem of finding the smallest value of 2M + 2N that guarantees that if equation (1) is satisfied then all singular points agree on position and spectra, although is an interesting question, it is a very hard problem and is out of the scope of this paper. In Theorems 2 and 3 we provide conditions and study the situation of having two vector fields agree on position and spectra at all their singularities. Now we want to establish whether the two vector fields are identical or not. We will see that when n > 2 vector fields that are twin vector fields are rather uncommon. The proof of this claim is the content of the following result. We set Rx = ∂R/∂x and Ry = ∂R/∂y for any polynomial R. 4 J. LLIBRE AND C. VALLS ∂ Theorem 4. Let χ = P (x, y) ∂x ∂ + Q(x, y) ∂y bn . The following statements hold: ∈P (a) If this vector field has a twin vector field then the four polynomials (2) Px , Qx , Py , Qy are linearly dependent over C. (b) If the polynomials in (2) are linearly dependent over C and admit a non- trivial linear combination (3) (a − 1)Px + bQx + cPy + (d − 1)Qy ≡ 0 satisfying ad − bc = 1, then χ admits a twin vector field. The proof of Theorem 4 is given in section 5. We have the following corollary from Theorem 4. Corollary 5. A generic vector field of degree n > 2 has no twins. Note that the polynomials Px , Qx , Py and Qy will generically be of degree n − 1. The space of polynomials of two variables of degree n−1 has dimension n(n+1)/2 > 4, whenever n > 2. That is the reason why twin vector fields are uncommon in higher degrees while in the quadratic case Px , Qx , Py and Qy are always linearly dependent and it is not surprising that in [8] the author proves that generic quadratic vector fields admit always a twin vector field. Taking Py = 0 or Qx = 0 in Theorem 4 we have the family of twin vector fields provided in the following theorem. Theorem 6. Let P (x) and Q(y) be polynomials each having n distinct roots. Let ∂ ∂ χ = P (x) + Q(y) ∂x ∂y and consider the following one-parameter family of vector fields   ∂ ∂ (P (x) + bQ(y)) + Q(y) , b ∈ C , ∂x ∂y   ∂ P (x) + (cP (x) + Q(y))d ∂y , b ∈ C . ∂ ∂x Then each vector field in each family is a twin vector field of χ (except for b = 0 or c = 0) when the vector fields are identical. Our final comment is the following: whenever N = n2 − 1 in system (1) the Euler-Jacobi formula (see section 2 for its explicit statement) easily implies that the two vector fields have the same spectra (see the first part of the proof of Theorem 1). Moreover, when M = n2 − 1 in system (1), the Cayley-Bacharach theorem (see section 4 for its explicit statement) easily implies that the two vector fields have the same singular set (see the first part in the proof of Theorem 3). Therefore, the first non trivial cases are M = n2 − 1 with N = n2 − 2 and M = n2 − 2 with N = n2 − 1 which are precisely the two cases treated in Theorems 2 and 3, respectively. TWIN POLYNOMIAL VECTOR FIELDS 5 2. Proof of Theorem 1 The main result that we use for proving Theorems 2, 3 and 6 is the Euler-Jacobi formula which can be stated as follows. Theorem 7 (Euler-Jacobi formula). If P and Q are polynomials in C[x, y] of degree n whose divisors intersect transversally in n2 different points p1 , . . . , pn2 ∈ C2 and g(x, y) is a polynomial of degree at most 2n − 3, then n2 X g(pk ) = 0, k=1 J(pk ) where J(x, y) is the Jacobian determinant of P and Q, that is,   ∂(P, Q) Px Qy  J(x, y) = det = det = det Dχ(x, y) . ∂(x, y) Qx Qy For a proof of Theorem 7 see [1]. Proof of Theorem 1. Assume that two vector fields χ and χ b have the same spectra 2 at N = n − 1 singular points. Denote by an2 and b an2 the determinant of Dχ and Dχ 2 2 b b at pn , and by bn and bn the traces of Dχ and Dχ 2 b at pn2 , respectively. Note that from the Euler-Jacobi formula with g1 (x, y) = 1 and g2 (x, y) = tr(Dχ(x, y)) we get 1 1 = which yields an2 = b an2 , an2 ban2 and bn2 bbn2 = which yields bn2 = bbn2 . an2 b an2 Therefore, both vector fields χ and χb have the same spectra at all n2 singular points. This completes the proof of statement (i). Now assume that in addition we have M = n2 −1 and we denote by pn2 = (xn2 , yn2 ) and pbn2 = (b xn2 , ybn2 ) the positions of pn2 of χ and of pbn2 of χ b, respectively. Applying the Euler-Jacobi formula with g1 (x, y) = x and g2 (x, y) = y we get xn2 bn2 x yn2 ybn2 = and = an2 b an2 an2 b an2 which yields xn2 = xbn2 and yn2 = ybn2 . Therefore, both vector fields χ and χ b have the same singular locus and the proof of statement (ii) is complete.  3. Proof of Theorem 2 We continue to denote by an2 and b an2 the determinant of Dχ(pn2 ) and Dχ b(pn2 ), respectively, and by bn2 and bbn2 the traces of Dχ(pn2 ) and Dχ b(pn2 ), respectively. It follows from the proof of Theorem 1 (i) that an2 = b an2 and bn2 = bbn2 . Therefore, the spectrum of pn2 and pbn2 is the same and so M = n2 . Now we also continue to denote by pn2 = (xn2 , yn2 ) and pn2 −1 = (xn2 −1 , yn2 −1 ) the positions of the points for the vector field χ, and by pbn2 = (b xn2 , ybn2 ) and pbn2 −1 = xn2 −1 , ybn2 −1 ) the positions of the points for the vector field χ (b b. Applying the Euler 6 J. LLIBRE AND C. VALLS Jacobi formula with g1 (x, y) = x, g2 (x, y) = y, g3 (x, y) = tr(Dχ(p))x, g4 (x, y) = tr(Dχ(p))y and taking into account that ai = bai and bi = bbi for i = 1, . . . , n2 , we get xn2 −1 xn2 bn2 −1 x x bn2 + = + , an2 −1 an2 an2 −1 an2 yn2 −1 yn2 ybn2 −1 ybn2 + = + , an2 −1 an2 an2 −1 an2 (4) bn2 −1 xn2 −1 bn2 xn2 bn2 −1 xbn2 −1 bn2 x bn2 + = + , an2 −1 an2 an2 −1 an2 bn2 −1 yn2 −1 bn2 yn2 bn2 −1 ybn2 −1 bn2 ybn2 + = + . an2 −1 an2 an2 −1 an2 Multiplying by bn2 −1 the first equality in (4) and subtracting the third from it, and doing the same with the second and the fourth equalities we get (bn2 −1 − bn2 )xn2 (bn2 −1 − bn2 )b xn2 (bn2 −1 − bn2 )yn2 (bn2 −1 − bn2 )b yn2 = , = . a2n a2n a2n a2n We have two cases: either bn2 −1 − bn2 6= 0 or bn2 −1 − bn2 = 0. If bn2 −1 − bn2 6= 0, then xn2 = x bn2 and yn2 = ybn2 . Then the third and second equalities in (4) imply that xn2 −1 = x bn2 −1 as well as yn2 −1 = ybn2 −1 . In short pn2 −1 = pbn2 −1 , pn2 = pbn2 and so N = n2 . This completes the proof of the theorem in this case. If bn2 −1 = bn2 then we apply the Euler-Jacobi formula with g1 (x, y) = x2 , g2 (x, y) = x3 , g3 (x, y) = y 2 and g4 (x, y) = y 3 and we get x2n2 −1 x2n2 b2n2 −1 x b2n2 x x2n2 −1 − x b2n2 −1 b2 2 − x2n2 x + = + , i.e = n , an2 −1 an 2 an2 −1 an 2 an2 −1 an2 x3n2 −1 x3 2 b3n2 −1 x b3 2 x x3n2 −1 − x b3n2 −1 b3 2 − x3n2 x + n = + n , i.e = n , an2 −1 an 2 an2 −1 an 2 an2 −1 an2 (5) yn2 2 −1 y22 yb2 2 yb2 2 yn2 2 −1 − ybn2 2 −1 yb2 2 − yn2 2 + n = n −1 + n , i.e = n , an2 −1 an2 an2 −1 an2 an2 −1 an2 yn3 2 −1 y32 ybn3 2 −1 yb3 2 yn3 2 −1 − ybn3 2 −1 yb3 2 − x3n2 + n = + n , i.e = n . an2 −1 an2 an2 −1 an2 an2 −1 an 2 Note that the first and second relations in (4) can be written as xn2 −1 − xbn2 −1 bn2 − xn2 x yn2 −1 − ybn2 −1 ybn2 − yn2 (6) = , = . an2 −1 an2 an2 −1 an2 It follows from the first identity in (6) that either x bn2 −1 = xn2 −1 and so x bn2 = xn2 , bn2 −1 6= xn2 −1 and x or x bn2 6= xn2 . In this last case from the first and second relations in (5) we get bn2 −1 + xn2 −1 = x x bn2 + xn2 , (7) b2n2 −1 + x x bn2 −1 xn2 −1 + x2n2 −1 = x bn2 xn2 + x2n2 b2n2 + x So taking the square of the first relation in (7) and substracting it in the second one we obtain bn2 xn2 = x x bn2 −1 xn2 −1 . TWIN POLYNOMIAL VECTOR FIELDS 7 Therefore, using (6) and (7) we have 1 1 2 xn2 − xn2 )2 = 2 (b (b xn2 −1 − xn2 −1 )2 an2 an2 −1 1 = 2 (b x2n2 −1 − 2b xn2 −1 xn2 −1 + x2n2 −1 ) an2 −1 1 1 = 2 (b x2n2 − 2bxn2 xn2 + x2n2 ) = 2 (b xn2 − xn2 )2 an2 −1 an2 −1 and so an2 −1 = ±an2 . Now proceeding exactly as we did with x but now with y using (6) and (7) we get that either ybn2 −1 = yn2 −1 and so ybn2 = yn2 , or ybn2 −1 6= yn2 −1 and ybn2 6= yn2 and again an2 −1 = ±an2 . In short we have four possibilities that we will study separately. As we will see, only Case 1 may occur. Case 1: x bn2 −1 = xn2 −1 , x bn2 = xn2 , ybn2 −1 = yn2 −1 and ybn2 = yn2 . In this case, pbn2 −1 = pn2 −1 , pbn2 = pn2 , and so N = n2 , as claimed in the statement of the theorem. bn2 = xn2 , ybn2 −1 6= yn2 −1 , ybn2 6= yn2 and an2 −1 = ±an2 . Note bn2 −1 = xn2 −1 , x Case 2: x that from (7) we have x bn2 −1 = xn2 −1 = xn2 = x bn2 . Moreover, if an2 −1 = an2 then from (6) and (7) we obtain ybn2 −1 − yn2 −1 = ybn2 − yn2 and ybn2 −1 + yn2 −1 = ybn2 + yn2 and so ybn2 −1 = ybn2 and yn2 −1 = yn2 , but then pbn2 −1 = pbn2 and pn2 −1 = pn2 which is not possible. On the other hand, if an2 −1 = −an2 then ybn2 −1 − yn2 −1 = −b yn2 + yn2 and ybn2 −1 + yn2 −1 = ybn2 + yn2 which yields ybn2 −1 = yn2 and yn2 −1 = ybn2 , but then pbn2 −1 = pn2 and pn2 −1 = pbn2 and in this case both vector fields χ and χ b have the same singular points and the determinants of pn2 −1 and pn2 have different signs. Now it follows from the second and fourth relations in (5) with an2 −1 = −an2 and x bn2 −1 = xn2 , and xn2 −1 = x bn2 , ybn2 −1 = yn2 and yn2 −1 = ybn2 that b3n2 − x x b3n2 −1 b3n2 − x x b3n2 −1 − = a2n a2n and ybn3 2 − ybn3 2 −1 ybn3 2 − ybn3 2 −1 − = a2n a2n which yields xb3n2 = xb3n2 −1 and ybn3 2 = ybn3 2 −1 and so x bn2 = x bn2 −1 and ybn2 = ybn2 −1 , but then pbn2 −1 = pbn2 and pn2 −1 = pn2 which is not possible. Case 3: xbn2 −1 6= xn2 −1 , x bn2 6= xn2 , ybn2 −1 = yn2 −1 , ybn2 = yn2 and an2 −1 = ±an2 . In this case interchanging the roles of x and y in Case 2 we arrive to the same conclusions as in Case 2, that is, this case is not possible. 8 J. LLIBRE AND C. VALLS Case 4: x bn2 −1 6= xn2 −1 , x bn2 6= xn2 , ybn2 −1 6= yn2 −1 , ybn2 6= yn2 and an2 −1 = ±an2 . If an2 −1 = an2 then from (6) and (7) we get bn2 −1 − xn2 −1 = x x bn2 − xn2 bn2 −1 + xn2 −1 = x and x bn2 + xn2 , bn2 −1 = x which yields x bn2 and xn2 −1 = xn2 and proceeding analogously as in Case 2 we get that ybn −1 = ybn2 and yn2 −1 = yn2 . But then pbn2 −1 = pbn2 and pn2 −1 = pn2 2 which is not possible. If an2 −1 = −an2 then bn2 −1 − xn2 −1 = −b x xn2 + xn2 bn2 −1 + xn2 −1 = x and x bn2 + xn2 which yields xbn2 −1 = xn2 and xn2 −1 = xbn2 , and proceeding analogously as in Case 2 we get that ybn2 −1 = yn2 and yn2 −1 = ybn2 . But then pbn2 −1 = pn2 and pn2 −1 = pbn2 and in this case both vector fields χ and χ b have the same singular points and the determinants of pn2 −1 and pn2 have different signs. Proceeding as in Case 2 we reach a contradiction. This concludes the proof of the theorem. 4. Proof of Theorem 3 To prove Theorem 3 we state the well-known Cayley-Bacharach theorem for poly- nomials. See [3, Theorem CB5] for a proof. Theorem 8 (Cayley-Bacharach Theorem). Let X1 , X2 ∈ P2 be plane curves of de- grees d1 and d2 respectively, intersecting in d1 d2 points Γ = X1 ∩X2 = {p1 , . . . , pd1 d2 }, and assume that Γ is the disjoint union of subsets Γ0 and Γ00 . Set s = d1 + d2 − 3. If k ≤ s is a non-negative integer, then the dimension of the vector space of polyno- mials of degree k vanishing on Γ0 modulo those containing Γ is equal to the failure of Γ00 to impose independent conditions on the polynomials of degree s − k. We recall that the failure of Γ00 to impose independent conditions on polynomials of degree s − k is the difference between the cardinality of Γ00 and the rank of the linear conditions on the polynomials of degree s − k imposed by Γ00 . Proof of Theorem 3. Take χ = P (x, y) ∂x∂ ∂ + Q(x, y) ∂y bn . Taking into account ∈ P that since n > 2 the curves of degree n passing through the n2 points of P ∩ Q provide an overdetermined linear system, it follows from Theorem 8 (with X1 = P , X2 = Q, d1 = d2 = k = n, Γ0 being n2 − 1 points and Γ00 only one point) that any polynomial R of degree n which vanishes on n2 − 1 points in P ∩ Q also vanishes on the remaining one. Therefore, since by assumption M = n2 − 1, we immediately conclude that the two vector fields χ and χb have the same singular locus and so M = n2 . We take the same notation as in the proof of Theorem 2. More precisely, we denote by an2 −1 , an2 , b an2 −1 , b an2 the determinants of Dχ(pn2 −1 ), Dχ(pn2 ), Dχ b(pn2 −1 ) and Dχ pn2 ), respectively, and by bn2 −1 , bn2 , bbn2 −1 , bbn2 the traces of Dχ(pn2 −1 ), Dχ(pn2 ),Dχ b(b b(pn2 −1 ) and Dχ b(b pn2 ), respectively. Moreover, the points will be denoted as pk = (xk , yk ) for k = 1, . . . , n2 . Note that from the Euler-Jacobi fromula with g1 (x, y) = 1, g2 (x, y) = tr(Dχ(x, y)), g3 (x, y) = tr(Dχ(p)x, g4 (x, y) = x, g5 (x, y) = tr(Dχ(p)y) and g6 (x, y) = y we also TWIN POLYNOMIAL VECTOR FIELDS 9 get 1 1 1 1 + = + an2 −1 an2 ban2 −1 b an2 bn2 −1 bn2 bbn2 −1 bbn2 + = + , an2 −1 an2 b an2 −1 b an2 bn2 −1 xn2 −1 bn2 xn2 bbn2 −1 xn2 −1 bbn2 x bn2 + = + , an2 −1 an2 b an2 −1 b an2 (8) xn2 −1 xn2 xn2 −1 x bn2 + = + , an2 −1 an2 b an2 −1 b an2 bn2 −1 yn2 −1 bn2 yn2 bbn2 −1 yn2 −1 bbn2 ybn2 + = + , an2 −1 an2 b an2 −1 b an2 yn2 −1 yn2 yn2 −1 ybn2 + = + an2 −1 an2 ban2 −1 b an2 Multiplying the first equation in (8) by xn2 −1 and subtracting it to the fourth iden- tity, and multiplying the first equation in (8) by yn2 −1 and subtracting it to the sixth identity we get xn2 −1 − xn2 xn2 −1 − x bn2 yn2 −1 − yn2 yn2 −1 − ybn2 (9) = and = . an2 b an2 an2 b an2 Since either xn2 −1 6= xn2 or yn2 −1 6= yn2 , we get from (9) that an2 = b an2 and then the first identity in (8) implies that also an2 −1 = b an2 −1 . Now multiplying the second equation in (8) by xn2 −1 and subtracting it to the third identity, and multiplying the second equation in (8) by yn2 −1 and subtracting it to the fifth identity we get bn2 −1 (xn2 −1 − xn2 ) bn2 −1 (xn2 −1 − xbn2 ) = , an2 an2 (10) bn2 −1 (yn2 −1 − yn2 ) bn2 −1 (yn2 −1 − ybn2 ) = . an2 an2 Since either xn2 −1 6= xn2 or yn2 −1 6= yn2 , we get from (10) that bn2 = bbn2 and then the second identity in (8) implies that also bn2 −1 = bbn2 −1 . In short N = n2 and the proof of the theorem is complete.  5. Proof of Theorem 4 To prove Theorem 6 we state the well-known Max Noether’s fundamental theorem and a closely related proposition. For a proof of both of them see [4, Section 5 of Chapter 5]. Let P2 be the projective plane. For F ∈ C[X, Y, Z] we write F∗ = F (X, Y, 1). Moreover, when p = [x : y : 1] then Op (P2 ) is canonically isomorphic to O(x,y) (C2 ) can be regarded as an element of the local ring Op (P2 ). We recall that Op (P2 ) is the ring of rational functions on P2 that are defined at p. Let p ∈ P2 , F, G curves with no common component through p and H be another curve. We say that Noether’s conditions are satisfied at p (with respect to F, G 10 J. LLIBRE AND C. VALLS and H) if H∗ ∈ (F∗ , G∗ ) ⊂ Op (P2 ) that is, there are a, b ∈ Op (P2 ) such that H∗ = aF∗ + bG∗ . Theorem 9 (Max Noether’s fundamental theorem). Let F, G, H be projective plane curves. Assume that F and G have no common components. Then there is an equation H = AF + BG (with A and B forms of degrees deg(H) − deg(F ) and deg(H) − deg(G), respectively) if and only if Noether’s conditions are satsified at every p ∈ F ∩ G. Proposition 10. Let P, Q, R ∈ C[X, Y, Z] and p be a point in P ∩ Q. If P and Q intersect transversally at p and R(p) = 0, then the Noether conditions holds at p. Proof of Theorem 4. Assume that χ has a twin vector field χ b. Using Max Noether’s fundamental theorem since χ b and χ have the same singular locus, there exist complex numbers a, b, c, d such that ∂ ∂ (11) b = (aP + bQ) χ + (cP + dQ) . ∂x ∂y Consider the polynomial b − trDχ = (a − 1)Px + bQx + cPx + (d − 1)Qy . R(x, y) = trDχ b have the same spectra, then If χ and χ R(pi ) = 0 for i = 1, . . . , n2 . Since P and Q intersect transversally at each pi because the maximum number of their intersection points is n2 , it follows from Proposition 10 and Theorem 9 that if R 6≡ 0 taking into account that the degree of R is n − 1, there exist two polynomials A and B with deg(A) = −1 and deg(B) = 0 such that R = AP + BQ, which is not possible. So, R ≡ 0 and condition in (3) means that the polynomials Px , Qx , Py and Qy are C-linearly dependent. This concludes the proof of statement a) of the theorem. Assume now that equation (3) is a non-trivial linear combination and satisfies ad − bc = 1. We claim that the vector field χ b given in (11) is a twin vector field of χ. Note that if equation (3) is a trivial linear combination of Px , Py , Qx , Qy then a = d = 1 and b = c = 0 which thus gives χ = χ b implying that χ b is not a twin of χ. Now we prove the claim. It is clear that if P = Q = 0, then any linear combination of them will also be zero. For the converse, we note that we already obtained   T a b aP + bQ and cP + dQ by multiplying the vector (P, Q) by the matrix c d which is invertible since ad − bc = 1. Hence, clearly aP + bQ = cP + dQ = 0 imply P = Q = 0. Therefore both χ and χ b have the same singular locus {p1 , . . . , pn2 }. Moreover, since equation (3) is a non-trivial linear combination, for i = 1, . . . , n2 we have trDχ(pi ) − trbχ(pi ) = (a − 1)Px (pi ) + bQx (pi ) + cPy (qi ) + (d − 1)Qy (pi ) = R(pi ) = 0. Finally, note that for i = 1, . . . , n2 b(pi ) = (ad − bc) det χ(pi ) = det χ(pi ), det χ TWIN POLYNOMIAL VECTOR FIELDS 11 b given in (11) is a twin vector field of χ. This completes the and so the vector field χ proof of statement (b) and concludes the proof of the theorem.  Acknowledgements The authors thank to the reviewer his useful and nice comments and suggestions that helped them to improve the presentation and some results of this paper. The first author is supported by the Ministerio de Economı́a, Industria y Compet- itividad, Agencia Estatal de Investigación grant PID2019-104658GB-I00 (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. The second author is partially supported by FCT/Portugal through UID/MAT/04459/2019. References [1] V. Arnold, A. Varchenko and S. Goussein-Zude, Singularités des applications différentiels, Mir, Moscow, 1982. [2] A. N. 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Control Syst. 23 (2017), 623–633. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain Email address: jllibre@mat.uab.cat 2 Departamento de Matemática, Instituto Superior Técnico, Universidade de Lis- boa, Av. Rovisco Pais 1049–001, Lisboa, Portugal Email address: cvalls@math.ist.utl.pt