This is a preprint of: “On the centers of cubic polynomial differential systems with four invariant straight lines”, Jaume Llibre, Topol. Methods Nonlinear Anal., vol. 55(2), 387–402, 2020. DOI: [10.12775/TMNA.2020.004] ON THE CENTERS OF CUBIC POLYNOMIAL DIFFERENTIAL SYSTEMS WITH FOUR INVARIANT STRAIGHT LINES JAUME LLIBRE Abstract. Assume that a cubic polynomial differential system in the plane has four invariant straight lines in generic position, i.e. they are not paral- lel and no more than two straight lines intersect in a point. Then such a differential system only can have 0, 1 or 3 centers. 1. Introduction and statement of the main results A center of a differential system in R2 is an equilibrium point p for which there exists a neighborhood U of p such that U \ {p} is filled by periodic orbits. The equilibrium point p is a focus if there exists a neighborhood U of p where all the orbits in U \ {p} spiral tending to p either in backward, or in forward time. These definitions of focus and center goes back to Dulac [10] and Poincaré [23]. The problem of distinguish between a focus or a center (known as the center- focus problem) is a classical problem in the qualitative theory of planar polynomial differential systems, which is related to the Hilbert 16th problem, see Hilbert [14], Ilyashenko [15], Li [19]. For the polynomial differential systems of degree 2 the center–focus problem was solved by Bautin [3], Kapteyn [16, 17], Schlomiuk [24], Vulpe [25], Żola̧dek [28]. However, this problem remains unsolved for polynomial differential systems of degree 3, simply called from now on cubic systems. The centers of many different subclasses of cubic differential systems have been studied. Thus reversible cubic systems with a center has been classified by Żola̧dek [30, 31], and Buzzi et al [4]. the centers of cubic systems without quadratic terms also were classified by Malkin [21], Vulpe and Sibirskii [26], Żola̧dek [29], ... The classification of the Hamiltonian linear type centers and the nilpotent ones for cubic systems without quadratic terms has been recently classified by Colak et al [5, 6, 7, 8]. The centers of the Kolmogorov cubic systems have been analyzed and classified in [20]. In this work we study the centers for the cubic systems having four invariant straight lines in generic position, i.e. they are not parallel and no more than two straight lines intersect in a point. Our main result is the following. 2010 Mathematics Subject Classification. 37K10, 37C27, 37K05. Key words and phrases. cubic system, cubic polynomial differential systems, centers, invariant straight line. 1 http://www.gsd.uab.cat 2 J. LLIBRE Theorem 1. Cubic systems having four invariant straight lines in generic position only can have 0, 1 or 3 centers. Theorem 1 is proved in section 3. In section 2 we summarize some well known results that we shall use for proving Theorem 1. 2. Preliminary results 2.1. Equilibrium points. We say that an equilibrium point p of a differential system in R2 is elementary if the determinant of the linear part of the system at p is non–zero. An elementary equilibrium p is hyperbolic if the two eigenvalues of the linear part of the system at p have non-zero real part. The elementary equilibria only can be saddles with topological index −1 and the corresponding determinant neg- ative; nodes or foci, these last two with topological index 1 and the corresponding determinant positive. We recall that the trace of the matrix of the linear part of a planar differential system at an elementary node is not zero. For a proof and definitions of saddle, node and focus and their properties here mentioned see for instance Chapter 2 of [11]. It is known that an elementary equilibrium point of an analytic differential sys- tem in R2 which is not hyperbolic, has eigenvalues purely imaginary, and only can be a focus or a center, for more details see [1]. 2.2. Poincaré compactification. Let S2 = {y ∈ R3 : ||y|| = 1} be the 2– dimensional sphere, here || · || denotes the Euclidean norm in R3 . The Poincaré compactification of a polynomial vector field X , denoted by p(X ), is an induced vec- tor field on the sphere S2 defined as follows. For additional details on the Poincaré compactification see Chapter 5 of [11]. Let Ty S2 be the tangent space to S2 at the point y. Consider X defined in the plane T(0,0,1) S2 identified with R2 . The central projection f : T(0,0,1) S2 → S2 sends every point y of the plane T(0,0,1) S2 to two points on S2 , the ones which are in the intersection of S2 with the straight line passing through the point y and the center of the sphere S2 , i.e. the origin of coordinates. This map defines two copies of X , one in the open northern hemisphere H+ and other in the open southern hemisphere H− . Denote by X 0 the vector field Df ◦ X defined on S2 except on its equator S1 = {y = (y1 , y2 , y3 ) ∈ S2 : y3 = 0}. Clearly the infinity of R2 ≡ T(0,0,1) S2 is identified with the equator S1 of S2 . The polynomial vector field X 0 of degree n can be extended to an analytic vector field p(X ) on S2 taking p(X ) = y3n−1 X 0 . On S2 \ S1 = H+ ∪ H− there are two copies of X , one in H+ and the other in H− , and knowing the behaviour of p(X ) around S1 , we know the behaviour of X at infinity. The Poincaré compactification has the property that S1 is invariant under the flow of p(X ). The finite equilibrium points of X or of p(X ) are the equilibrium points of X , while the equilibrium points of p(X ) contained in S1 , i.e. at infinity, are called the CENTERS FOR CUBIC POLYNOMIAL DIFFERENTIAL SYSTEMS 3 infinite equilibrium points of X or of p(X ). We note that the infinity equilibrium points appear in pairs diametrically opposed. Let R[x, y] be the ring of real polynomials in the variables x and y. A planar polynomial differential system of degree m is a differential system (1) ẋ = P (x, y), ẏ = Q(x, y), with P, Q ∈ R[x, y] satisfying max{deg P, deg Q} = m. As usual the dot denotes derivative with respect to an independent variable t, usually called the time. We denote by ∂ ∂ X =P +Q ∂x ∂y the polynomial vector field associated to system (1). For working with the vector field p(X ) on the sphere S2 we use the six local charts given by Uk = {y ∈ S2 : yk > 0}, Vk = {y ∈ S2 : yk < 0} for k = 1, 2, 3. The corresponding local maps φk : Uk → R2 and ψk : Vk → R2 are defined as φk (y) = −ψk (y) = (ym /yk , yn /yk ) for m < n and m, n 6= k. We denote by z = (u, v) the value of φk (y) or ψk (y) for any k, so the coordinates (u, v) will play different roles depending on the local chart we are considering. The points in the infinity S1 in any chart have the coordinate v = 0. The expression for p(X ) in local chart (U1 , φ1 ) is given by      m 1 u 1 u u̇ = v −uP , +Q , , v v v v (2)   1 u v̇ = −v m+1 P , . v v The expression for (U2 , φ2 ) is      u 1 u 1 u̇ = v m P , − uQ , , v v v v (3)   u 1 v̇ = −v m+1 Q , , v v and for (U3 , φ3 ) is u̇ = P (u, v), v̇ = Q(u, v). The expression for p(X ) in the charts (Vk , ψk ) is the same as for (Uk , φk ) multiplied by (−1)d−1 , for k = 1, 2, 3. 2.3. Poincaré–Hopf Theorem. The next result is the Poincaré–Hopf Theorem for a vector field defined on the 2-dimensional sphere S2 , see an elementary proof in Chapter 6 of [11]. Theorem 2. For every tangent vector field on the sphere S2 with a finite number of equilibria, the sum of their topological indices is 2. 4 J. LLIBRE 2.4. Arrangement of four straight lines. We consider an arbitrary arrangement of four straight lines in generic position. Then doing an affine change of variables we can assume that these four straight lines are x = 0, y = 0, x + ay + b = 0 and x + cy + d = 0, satisfying that (4) abcd(a − c)(b − d)(bc − ad) 6= 0. For more information on arrangements of straight lines see for instance [22]. 2.5. Normal forms of polynomial differential systems having a set of given generic invariant algebraic curves. The algebraic curve C(x, y) = 0 of R2 is called an invariant algebraic curve of system (1) if P Cx + QCy = KC, for some real polynomial K(x, y). With this definition it is easy to check that the algebraic curve C = 0 is formed by orbits of system (1). Theorem 3. Let Ci = 0 for i = 1, · · · , p be irreducible invariant algebraic curves in p P R2 , and let r = degree(Ci ). We assume that all Ci satisfy the following generic i=1 conditions: (i) There are no points at which Ci and its first derivatives are all vanish. (ii) The highest order terms of Ci have no repeated factors. (iii) If two curves intersect at a point in the finite plane, they are transversal at this point. (iv) There are no more than two curves Ci = 0 meeting at any point in the finite plane. (v) There are no two curves having a common factor in the highest order terms. If r = m + 1 then any polynomial vector field X of degree m tangent to all Ci = 0 is of the form   p X  Y p    (5) X = αi  Cj  XCi , i=1   j=1 j 6= i with αi ∈ R. Theorem 2 is proved in [9], where also it is explained the history of this result. Let U be an open subset of R2 . A non–locally constant function H : U → R is a first integral of system (1) in U if it is constant on all the orbits of system (1) contained in U , i.e. ∂H ∂H Ḣ = X H = P +Q = 0, ∂x ∂y in all the points of U . A function R : U → R not identically zero is an integrating factor of system (1) if it satisfies ∂(RP ) ∂(RQ) + = 0. ∂x ∂y CENTERS FOR CUBIC POLYNOMIAL DIFFERENTIAL SYSTEMS 5 A first integral H associated to this integrating factor R is Z H(x, y) = − R(x, y)P (x, y)dy + f (x), where H(x, y) must satisfy ∂H/∂x = RQ. Theorem 4. Under the assumptions of Theorem 3 the polynomial vector field (5) has the integrating factor 1/(C1 · · · Cp ). For a proof see the paper of Christopher and Kooij [18]. 2.6. Bezout Theorem. For a proof of the next result known as the Bezout The- orem see for instance [12]. Theorem 5. Let R(x, y) and S(x, y) be two real polynomials. If both polynomials do not share a non-trivial common factor, then the algebraic system of equations R(x, y) = S(x, y) = 0 has at most degree(R)degree(S) solutions in R2 . We recall that a simple equilibrium (x0 , y0 ) of system (1) is an equilibrium sat- isfying   ∂(P, Q) det (x0 , y0 ) 6= 0. ∂(x, y) If a polynomial differential system (1) of degree m has its maximum number of equilibria m2 , then all its equilibria are simple, and consequently elementary. See for a proof [2] or [13]. 3. Proof of Theorem 1 We consider the four straight lines in generic position described in subsection 2.4. Applying Theorem 3 we obtain all cubic systems having the mentioned four straight lines as invariant algebraic curves, which are 2 ẋ = −x bdβ + (b + d)βx + (ad(β + γ) + bc(β + δ))y  + βx 2 +(a(β + γ) + c(β + δ))xy + ac(β + γ + δ)y , (6) 2 ẏ = y bdα + (b(α + δ) + d(α + γ))x + (ad  + bc)αy + (α + γ + δ)x 2 +(a(α + δ) + c(α + γ))xy + aαcy . By Theorem 4 we know that the function 1 R(x, y) = xy(x + ay + b)(x + cy + d) is an integrating factor of system (6), and from subsection 2.5 the associated first integral to this integrating factor is (7) H(x, y) = xα y β (x + ay + b)γ (x + cy + d)δ . Claim 1: If some of parameters α, β, γ and δ is zero, then the cubic system (6) has at most one center. 6 J. LLIBRE Proof of Claim 1. We assume that α = 0, the proof if β, γ or δ is zero is exactly the same. Then the equilibria of system (6) are all the points of the straight line x = 0 and the points   bc − ad d − b (−b, 0), (−d, 0), , , a−c a−c   b(c(β + δ) − aδ) − d(a(β + γ) − cγ) β(d − b) , . . (a − c)(β + γ + δ) (a − c)(β + γ + δ) Since the unique equilibrium point which is not on an invariant straight line is the last one if exists, i.e. if β + γ + δ 6= 0, it is clear that when α = 0 at most we can have one center.  From Claim 1 in what follows we can assume that (8) αβγδ 6= 0. Since the four invariant straight lines are in generic position they intersect in six equilibrium points, namely       b d bc − ad d − b (9) (0, 0), (−b, 0), (−d, 0), 0, − , 0, − , , . a c a−c a−c Note that from (4) all these equilibria exist. The determinant of the linear part of system (6) in these equilibria are b2 (bc − ad)2 −b2 d2 αβ, −b2 (b − d)2 βγ, −(b − d)2 d2 βδ, − αγ, a2 (10) d2 (bc − ad)2 (b − d)2 (bc − ad)2 − αδ, − γδ, c2 (a − c)2 respectively. From (4) and (8) all these determinants never vanish, consequently all the equilibria (9) are elementary. Therefore, by subsection 2.1 they only can be saddles, nodes, focus or centers, but since they are on invariant straight lines they are either saddles or nodes. Saddles when their corresponding determinant is negative, and nodes when they are positive. On the other hand by Bezout Theorem (see Theorem 5) the cubic system (6) has at most nine equilibria, since six of them are on invariant straight lines it follows that system (6) has at most three centers. Claim 2: The trace of the matrix of the linear part of the cubic system (6) at an equilibrium point which is not on the invariant straight lines is zero, or system (6) has at most one center. Proof of Claim 2. The equilibria which are not on the invariant straight lines are solution of the system ẋ/x = ẏ/y = 0, with the exception of the equilibrium point ((bc − ad)/(a − c), (d − b)/(a − c)). We isolate x2 from the equation ẋ/x = 0 and substitute it into the equation ẏ/y = 0. Now from this last equation we isolate x CENTERS FOR CUBIC POLYNOMIAL DIFFERENTIAL SYSTEMS 7 and we obtain x = −p1 (y)/p2 (y) assuming that p2 (y) 6= 0, where p1 (y) = bc(αδ + (β + δ)(γ + δ))) + (ad(αγ + (β + γ)(γ + δ))y +ac(γ + δ)(α + β + γ + δ) + bβd(γ + δ)y 2 , p2 (y) = β(bγ + dδ) + (α + β + γ + δ)(aγ + cδ)y. Later on we shall study the case p2 (y) = 0. Substituting x = −p1 (y)/p2 (y) into the equation ẋ/x = 0 we get the following equation only in the variable y: βγδ(b − d + (a − c)y)p3 (y) = 0, p2 (y)2 where  p3 (y) = bβ 2 d(b − d) + β bd(a(α + 2(β + γ) + δ) − c(α + 2β + γ + 2δ))  2 2 −ad (α + β + γ) + b c(α + β + δ) y + (α + β + γ + δ)   a d(β + γ) + ac(b(α + 2β + δ) − d(α + 2β + γ)) − bc (β + δ) y 2 2 2 +ac(a − c)(α + β + γ + δ)2 y 3 . The real roots of the polynomial p3 (y) provide the coordinates y of the equilibria which are not on the invariant straight lines, and substituting these roots into x = −p1 (y)/p2 (y) we obtain the coordinate x of these equilibria. Substituting x = −p1 (y)/p2 (y) into the divergence of the cubic system (6), i.e. into ∂ ẋ/∂x + ∂ ẏ/∂y we obtain p3 (y)p4 (y) (11) divergence = , p2 (y)2 where p4 (y) = bγ(β(γ  + 2δ) − δ(α + γ + δ)) + dδ(γ(α − 2β + γ) + δ(γ − β)) + aγ(α(γ − 2δ) + β(γ + 2δ) + (γ + δ)(γ − 2δ)) − cδ δ(α + β − γ)  2  −2γ(α − β + γ) + δ y. Since the real roots of the polynomial p3 (y) provide the y coordinate of the equilibria which are not on the invariant straight lines, and this polynomial is a factor of the divergence of the system, it follows that the linear part of system (6) at all the equilibria which are not on the invariant straight lines is zero when the polynomial p2 (y) is not the zero polynomial. Now assume that p2 (y) = 0, looking at the two coefficients of this polynomial we consider four cases. Case 1: β = 0 and δ = −α − γ. In this case it is easy to check that all the equilibrium points of system (6) are on the invariant straight lines, consequently this system has no centers. 8 J. LLIBRE Case 2: β = 0 and γ = −cδ/a. Now the unique equilibrium point of system (6) which is not on the invariant straight lines is   a(bc − ad)α d−b , . (a − c)(a(α + δ) − cδ) a − c Consequently system (6) has at most one center. Case 3: γ = d(α + β)/(b − d) and δ = b(α + β)/(d − b). As in Case 1 in this case system (6) has no equilibria outside the invariant straight lines, so no centers. Case 4: γ = −cδ/a = −dδ/b. This case is not possible because would imply that ad − bc = 0, in contradiction with (4). This completes the proof of this claim.  In order to complete the proof of Theorem 1 we need to prove the following four claims. Claim 3: Provide an example of cubic system (6) having exactly three center. Claim 4: Provide an example of cubic system (6) having exactly one center. Claim 5: Provide an example of cubic system (6) without centers. Claim 6: Show that the cubic systems (6) cannot have only two centers. Figure 1. Phase portrait in the Poincaré disc of the cubic system (12). Proof of Claim 3. Take a = d = β = γ = δ = 1, b = −1, c = −2 and α = −4, then system (6) becomes ẋ = x(1 − 6y − x2 + 2xy + 6y 2 ), (12) ẏ = y(4 − 12y − 2x2 + 3xy + 8y 2 ). This system has three equilibria which are not on the invariant straight lines, namely (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), where x1 is the root 0.648190032939388.. of the polynomial p(x) = 16 − 24x − 3x2 + 3x3 , and y1 is the root 0.630709938165919.. of the polynomial q(y) = −2−3y +6y 2 +6y 3 ; x2 is the root 3.04974949521986.. of the polynomial p(x), and y2 is the root −1.184541590375086.. of the polynomial q(y); CENTERS FOR CUBIC POLYNOMIAL DIFFERENTIAL SYSTEMS 9 x3 is the root −2.697939528159256.. of the polynomial p(x), and y3 is the root −0.446168347790832.. of the polynomial q(y). We note that there are explicit exact expressions for the roots of a cubic polynomial, but we prefer present these roots in this shorter way. The determinants of the matrices of the linear part of system (12) at the equi- librium points (xk , yk ) for k = 1, 2, 3 are: For k = 1 is 0.8200842208962.., a root of the polynomial r(λ) = −4800 + 6000λ − 180λ2 + λ3 ; For k = 2 is 136.2086577457645.., a root of the polynomial r(λ); For k = 3 is 42.9712580333392.., a root of the polynomial r(λ). Since the three determinants at the equilibrium point (xk , yk ) for k = 1, 2, 3 are positive, by subsection 2.1 these equilibria only can be nodes, foci or centers. They cannot be foci because in a neighborhood of them it is defined the first integral (7), also they cannot be nodes because the the trace of the matrix of the linear part of a differential system at a node is not zero, but we know by the Claim 2 that the trace of every equilibrium point which is not on the invariant straight lines is zero. In short these three equilibria are centers. See the phase portrait in the Poincaré disc of the cubic system (12) in Figure 1.  Figure 2. Phase portrait in the Poincar’e disc of the cubic system (13). Proof of Claim 4. Take a = d = α = β = γ = δ = 1, b = −1 and c = −2, then system (6) becomes ẋ = x(1 − 6y − x2 + 2xy + 6y 2 ), (13) ẏ = y(1 − 3y − 3x2 + 2xy + 2y 2 ). This system has three equilibria which are not on the invariant straight lines, namely (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), where x1 is the root 0.11303685689460.. of the polynomial p(x) = 1 − 9x + 12x3 , and y1 is the root 0.702574789251432.. of the polynomial q(y) = −2 − 9y + 24y 3 ; x2 is the root −0.916993298352020.. of the polynomial p(x), and y2 is the root −0.282066739245770.. of the polynomial q(y); 10 J. LLIBRE x3 is the root 0.803956441457416.. of the polynomial p(x), and y3 is the root −0.420508050005661.. of the polynomial q(y). The determinants of the matrices of the linear part of system (12) at the equi- librium points (xk , yk ) for k = 1, 2, 3 are: For k = 1 is 0.1615865516314450.., a root of the polynomial r(λ) = −108 + 621λ + 288λ2 + 32λ3 ; For k = 2 is −4.269300326605967.., a root of the polynomial r(λ); For k = 3 is −4.89228622502547.., a root of the polynomial r(λ). As in the proof of Claim 3 the equilibrium point having positive determinant is a center, and from subsection 2.1 the two equilibrium points with negative deter- minant are saddles. See the phase portrait in the Poincaré disc of the cubic system (13) in Figure 2.  Figure 3. Phase portrait in the Poincar’e disc of the cubic system (14). Proof of Claim 5. Take a = d = β = γ = δ = 1, b = −1, c = −2 and α = −1, then system (6) becomes ẋ = x(1 − 6y − x2 + 2xy + 6y 2 ), (14) ẏ = y(1 + x2 − 3y + 2y 2 ). This system has only one equilibrium point which is not on the invariant straight lines, namely (x1 , y1 ) = (−0.2616677544145299.., 0.918121959154146..), where x1 is a of the polynomial 1 + 3x + 12x3 , and y1 is a root of the polynomial −2 + 15x − 36x2 + 24x3 . The determinant of the matrix of the linear part of system (14) at the equilibrium point (x1 , y1 ) is −0.946250815810897.., a root of the polynomial 36+117λ+144λ2 + 64λ3 . Since this determinant is negative the equilibrium point (x1 , y1 ) is a saddle. Hence system (14) has no centers. See the phase portrait in the Poincaré disc of the cubic system (14) in Figure 3.  Now we shall study the equilibrium points of system (6) at infinity using the Poincaré compactification described in subsection 2.2. Thus system (6) in the local CENTERS FOR CUBIC POLYNOMIAL DIFFERENTIAL SYSTEMS 11 chart U1 has the expression u̇ = (α + β + γ + δ)(u − u2 − 2u3 ) + (γ − δ)uv + (3α + 3β + γ + 2δ)u2 v (15) −(α + β)uv 2 , v̇ = βv − (β − γ + 2δ)uv − βv 3 − 2(β + γ + δ)u2 v + (3β + γ + 2δ)uv 2 . In this local chart the system has three equilibrium at infinity (0, 0), (−1/a, 0) and (−1/c, 0). The determinants of the matrices of the linear part of system (15) at these equilibria are (16) β(α+β +γ +δ), ((a−c)2 γ(α+β +γ +δ))/a2 and ((a−c)2 δ(α+β +γ +δ))/c2 , respectively. System (6) in the local chart U2 has the expression u̇ = (α + β + γ + δ)(2u + u2 − u3 ) − (3α + 3β + γ + 2δ)uv +(δ − γ)u2 v + (α + β)uv 2 , (17) v̇ = α(2v − 3v 2 + v 3 ) + (α + 2γ − δ)uv − (α + β + γ + δ)u2 v +(δ − γ)uv 2 . The unique infinite equilibrium point of system (17) which is not already in the chart U1 is the origin, and the determinant of the matrix of the linear part of system (17) at the origin is (18) a2 c2 α(α + β + γ + δ). Proof of Claim 6. We have that αβδγ 6= 0, see (8). Then the six equilibria which are in the intersection of the four invariant straight lines are elementary because the determinants of the matrices of the linear part of system (6) at them given in (10) are non–zero, so their topological indices are −1 or 1. From the paragraph previous to this proof the infinite equilibria are also non–elementary again because the corresponding determinants are non-zero, and again their topological indices are −1 or 1. From the proof of Claim 2 we know that at most we have three additional finite equilibria which are not contained on the invariant straight line. We denote by I the sum of the topological indices of these finite equilibria outside the invariant straight lines, of course we take I = 0 if such equilibria do not exist. Now we shall study the possible values of I. We shall prove that the unique possible values for I are −1, 0 or 3. Note that these values of I prevents the existence of only two centers. Indeed, if I = −1 and we have only two centers these contribute to I with a 2 (because the topological index of a center is 1). Since at most there is one additional finite equilibria outside the invariant straight lines, and if exists its index is −1 or 1 (because the cubic system reaches the maximal number of finite equilibria, nine, and then all the equilibria are elementary, see subsection 2.6), so adding this 1 or −1 to 2 never we can reach −1. On the other hand if this third equilibrium outside the invariant straight lines, the two centers cannot produce that I = 1. If I = 3 and the system has the two centers outside the invariant straight lines, then must be a third equilibrium outside the invariant straight lines with index 1, and by the Claim 2 this third equilibrium is a center, because its trace is zero and 12 J. LLIBRE since they are elementary (because the cubic system has nine finite equilibria) with index one only can be nodes, foci or centers, but the elementary nodes and foci have trace non–zero. Now we shall prove that the unique possible values for I are −1 or 3. Recall that αβδγ 6= 0. Taking into account only the signs of α, β, γ and δ we need only to consider the following three cases because changing the sign of the independent variable of system (6), i.e. changing the sign of the time, we get all the other possibilities and we do not change the topological indices of the equilibria. Case 1: All the signs of α, β, γ and δ are positive. Then, from (10) the sum of the topological indices of the six equilibria which are in the intersection of the invariant straight lines is −6, while the sum of the eight infinite equilibria (recall that these equilibria appears in pairs diametrally opposite in S1 ) is 8, see (16) and (18). Therefore from the Poincaré–Hopf Theorem and taking into account that on the sphere S2 we have two copies of the finite equilibria we obtain −12 + 2I + 8 = 2. Hence I = 3. Case 2: Three of the signs of α, β, γ and δ are positive, and the other is negative. We distinguish three subcases. Subcase 2.1: α + β + γ + δ > 0. Now from (10) the sum of the topological indices of the six equilibria which are in the intersection of the invariant straight lines is 0, and from (16) and (18) the sum of the eight infinite equilibria is 4. So from the Poincaré–Hopf Theorem we have 0 + 2I + 4 = 2, and consequently I = −1. Subcase 2.2: α + β + γ + δ = 0. In this case the polynomial p3 (y) given in the proof of Claim 2, whose real roots are the y coordinate of the equilibria which are outside the invariant straight lines, at most has one real root and if it exists is bd(d − b)β . ad(b(β + γ) + dδ) − bc(bγ + d(β + δ)) So in this case at most one center. Subcase 2.3: α + β + γ + δ < 0. Again from (10) the sum of the topological indices of the six equilibria which are in the intersection of the invariant straight lines is 0, and from (16) and (18) the sum of the eight infinite equilibria is −4. Therefore by the Poincaré–Hopf Theorem we have 0 + 2I − 4 = 2, and I = 3. Case 3: Two of the signs of α, β, γ and δ are positive, and the other two negative. Here we only consider that α + β + γ + δ 6= 0, because the case α + β + γ + δ = 0 follows as in the subcase 2.2. From (10) the sum of the topological indices of the six equilibria which are in the intersection of the invariant straight lines is 2, and from (16) and (18) the sum of the eight infinite equilibria is 0. Therefore by the Poincaré–Hopf Theorem we have 4 + 2I + 0 = 2, and I = −1. This completes the proof of the claim.  Acknowledgements The first author is partially supported by the MINECO grants MTM2013-40998- P and MTM2016-77278-P (FEDER), the AGAUR grant 2014 SGR568, and the projecte MDM-2014-0445 (BGSMath). 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