This is a preprint of: “On the configurations of the singular points and their topological indices for the spatial quadratic polynomial differential systems”, Jaume Llibre, Clàudia Valls, J. Differential Equations, vol. 269, 10571–10586, 2020. DOI: [10.1016/j.jde.2020.07.022] ON THE CONFIGURATIONS OF THE SINGULAR POINTS AND THEIR TOPOLOGICAL INDICES FOR THE SPATIAL QUADRATIC POLYNOMIAL DIFFERENTIAL SYSTEMS JAUME LLIBRE1 AND CLAUDIA VALLS2 Abstract. Using the Euler-Jacobi formula there is a relation between the singular points of a polynomial vector field and their topological indices. Using this formula we obtain the configuration of the singular points together with their topological indices for the polynomial differ- ential systems ẋ = P (x, y, z), ẏ = Q(x, y, z), ż = R(x, y, z) with degrees of P , Q and R equal to two when these systems have the maximum num- ber of isolated singular points, i.e., 8 singular points. In other words we extend the well-known Berlinskii’s Theorem for quadratic polynomial differential systems in the plane to the space. 1. Introduction and statement of the main results Consider in R3 the polynomial differential systems (1) ẋ = P (x, y, z), ẏ = Q(x, y, z), ż = R(x, y, z), where P (x, y, z), Q(x, y, z) and R(x, y, z) are real polynomials of degrees 2, called spatial quadratic polynomial differential system, or simply quadratic systems in what follows. The motivation of our paper comes from the fact that for the planar quadratic polynomial differential systems the characterization of all config- urations of the (topological) indices of the singular points of these systems having four singular points is the well-known Berlinskii’s Theorem proved in [2, 7] and reproved in [5] using the Euler-Jacobi formula. More precisely, the Berlinskii’s Theorem can be stated as follows: Assume that a real pla- nar quadratic polynomial differential system has exactly four real singular points. In this case if the quadrilater formed by these points is convex, then two opposite singular points are anti-saddles (i.e. nodes, foci or centers) and the other two are saddles. If this quadrilater is not convex, then either the three exterior vertices are saddles and the interior vertex is an anti-saddle or the exterior vertices are anti-saddles and the interior vertex is a saddle. 1 Declarationsof interest: none. The corresponding author is Jaume Llibre. 2010 Mathematics Subject Classification. Primary 34A05. Secondary 34C05, 37C10. Key words and phrases. Euler-Jacobi formula, singular points, topological index, polyno- mial differential systems, Berlinskii’s Theorem. 1 2 J. LLIBRE AND C. VALLS We recall that four points form a convex quadrilater if and only if none of the four points is contained in the convex hull of the other three. We want to extend the Berlinskii’s Theorem from the plane to the space, i.e., we shall obtain all the configurations of the singular points and their topological indices for the quadratic systems in R3 having 8 singular points. We recall that if the degree of R is one (that is, if R is a polynomial of degree 1) then the following theorem holds. Assume that a real spatial poly- nomial differential system of degrees (2, 2, 1) has exactly four real singular points. In this case the four singular points are in a plane of R3 (not nec- essarily invariant) and on this plane Berlinskii’s Theorem holds, that is, if the quadrilater formed by these points is convex, then two opposite singular points have index 1 and the other two have index −1. If this quadrilater is not convex, then either the three exterior vertices have index −1 and the interior vertex has index 1, or the exterior vertices have index 1 and the in- terior vertex has index −1. The proof is essentially the same as Berlinskii’s theorem. This result was also observed in [5]. So we will consider the case in which the degrees are (2, 2, 2). Assuming that the quadratic system (1) has 8 singular points, then using the Euler-Jacobi formula we shall obtain the configuration of the singular points and their topological indices. If the number of singular points of a spatial quadratic polynomial differen- tial system (1) is finite, then it is at most 8, see for more details the Bézout’s Theorem (for a proof of this theorem see [18]). When all the singular points have multiplicity one we can apply the Euler-Jacobi formula (see [1] for a proof of a such formula). If system (1) has exactly 8 singular points, then the Jacobian determinant ∂P/∂x ∂P/∂y ∂P/∂z J= ∂Q/∂x ∂Q/∂y ∂Q/∂z ∂R/∂x ∂R/∂y ∂R/∂z evaluated at each singular point does not vanish, and for any polynomial S of degree less than or equal to 2 we have X S(a) (2) = 0, J(a) a∈A where A is the set of singular points of system (1). Given a finite subset of points B of R3 , we denote by B̂ its convex hull,by ∂ B̂ the boundary of B̂, and by #B the cardinal of B. Set A0 = A ∩ ∂ A b and for i ≥ 1 Ai = A ∩ ∂(A \ (A0 \ ∪ · · · ∪ Ai−1 )) for i = 1, 2, . . .. Clearly since A is finite there is an integer q such that Aq+1 = ∅ and Aq 6= ∅. THE EULER-JACOBI FORMULA 3 We say that A has the configuration (K0 ; K1 ; K2 ; . . . ; Kq ) where Ki = #Ai . We say that the singular points of system (1) which belong to Ai are on the i-th level. We recall that if we assume that #A = 8 then the Jacobian determinant J is non-zero at any singular point of system (1), and that the topological indices of these singular points are 1 (respectively −1) if J > 0 (respectively J < 0) (for more details see [17, 12]). Our main result is the extension of Berlinskii’s Theorem to the quadratic systems (1) having 8 singular points, see Theorem 5. 2. Preliminary results First of all we observe that if a configuration exists for a polynomial vector field X with degrees (2, 2, 2) and #A = 8, then it is possible to construct the same configuration but changing the sign of all the indices of the singular points, i.e. the points with index +1 become with index −1 and vice versa. For doing that it is enough to take the vector field Y = (−P, Q, R) instead of the vector field X = (P, Q, R). In the proof of Theorem 5 we will denote by {p1 , . . . , p8 } the set of points − of A. If the index of pk is +1 we shall denote it by p+ k , and similarly for pk . Also we will denote by Πijk the plane Πijk (x, y, z) = 0 through the three points pi , pj and pk where i, j, k ∈ {1, . . . , 8}. We note that three singular points of a polynomial differential system (1) of degree (2, 2, 2) having 8 singular points cannot be collinear, otherwise the straight line containing them is a factor of the polynomials P , Q and R, in contradiction that system (1) has exactly 8 singular points. Proposition 1. Let X = (P, Q, R) be a polynomial differential system in R3 with finitely many singular points and with max(deg P, deg Q, deg R) = n. Then, any plane cannot contain more than n2 singular points. Proof. Note that the polynomial system P = Q = R = 0 restricted to a plane is a polynomial system in R2 with degree n, and so by Bézout’s Theorem it cannot contain more than n2 singular points in this plane.  In our case since n = 2 we cannot have more than four singular points in any plane of a quadratic system (1). Proposition 2. Let X = (P, Q, R) be the vector field associated to the qua- dratic system (1) with 8 singular points. If four of these singular points are contained in a plane, then the other four singular points are also contained in a plane. Proof. By Proposition 1 we know that at most four singular points of the vector field X are contained in a plane. Assume that four singular points 4 J. LLIBRE AND C. VALLS p1 , . . . , p4 are in a plane and the remaining singular points p5 , . . . , p8 are not contained in a plane. Without loss of generality we can assume that p8 is not contained in the plane generated by {p5 , p6 , p7 }. Then applying the Euler-Jacobi formula (2) with the polynomial S = Π123 Π567 we reach to a contradiction. Indeed, we have 8 X Π123 (pi )Π567 (pi ) Π123 (p8 )Π567 (p8 ) = = 0, J(pi ) J(p8 ) i=1 so Π123 (p8 )Π123 (p8 ) = 0. Since Π123 (p8 ) 6= 0, otherwise five singular points will be contained in a plane, we have that Π567 (p8 ) = 0 in contradiction that the singular point p8 was assumed that it is contained in the plane Π567 (x, y, z) = 0 generated by the three singular points {p5 , p6 , p7 }.  From Proposition 2 it follows immediately the next corollary. Corollary 3. If a quadratic system (1) has 8 singular points, then they must satisfy one of the following two conditions. (i) Four singular points are contained in a plane and the other four singular points are also contained in another plane. (ii) In any plane there are at most three singular points of the eight possible. Then all the faces of the polyhedron of the convex hull of the eight singular points are triangles. The next result follows from the theory of the topological index, see [3]. Proposition 4. The topological index of an isolated singular point of system (1) remains constant under sufficiently small continuous perturbations of the coefficients of system (1). 2.1. Polyhedra with 4, 5, 6, 7 and 8 vertices. A graph is a pair (V, E) where V is a set whose elements are called vertices and E is a set of two-sets vertices, whose elements are called edges. A planar graph is a graph that can be embedded in a plane, i.e. it can be drawn on the plane in such a way that its edges intersect only at the endpoints, which are vertices. A 3-connected planar graph is a connected graph having more than 3 vertices whenever fewer than 3 vertices are removed. An n-polyhedral graph is a 3-connected planar graph on n vertices. Every convex polyhedron can be represented in the plane by a 3-connected planar graph. This can be done enlarging one of the faces of the convex polyhedron and projecting on it the other faces. Conversely, by a theorem of Steinitz (see [19, 20]) as restated by Grünbaum (see p. 235 of [13]), every 3-connected planar graph can be realized as a convex polyhedron (see [10]). THE EULER-JACOBI FORMULA 5 The number of distinct polyhedral graphs having 4, 5, 6, 7 and 8 vertices are 1, 2, 7, 34 and 257 (see p. 424 of [13] and [8, 9, 10]). In Figure 1 we provide the n-polyhedral graphs for n = 4, 5, 6, 7 which provide all the polyedra with vertices 4, 5, 6, 7. We do not provide the 257 8-polyhedral graphs, but for instance they are given in [13] or in http://mathworld.wolfram.com/OctahedralGraph.html. We note that polyhedral graphs are sometimes simply known as polyhe- dra, as we do in Figure 1 and when it would be convenient. 2.2. Possible polyhedra for the convex hull of the 8 singular points of a quadratic system. Note that any configuration of the form (3; ∗) cannot occur because three points determine a plane and then its convex hull is a triangle so the other 5 points must be in the interior of the triangle, and therefore the 8 points must be on a plane contradicting Proposition 1. Clearly, the configurations (2; ∗) and (1; ∗) also cannot occur. For the configuration of the form (4; 3; 1) proceeding as we did for the case (3; ∗) we have that the four points in the first and second levels must be in a plane, but then in view of Proposition 2 the four points in the 0 level must also be in the same plane contradicting Proposition 1. So this case is also not possible. The configuration (4; 2; 2) is not possible because the four singular points of the first and second level are collinear and we have seen that this is not possible. The configuration (5; 2; 1) is also non possible because the three singular points in the first and second levels are collinear. In short the possible configurations are (8), (7; 1), (6; 2), (5; 3) and (4; 4). Moreover, in view of Proposition 1 the previous configurations of the form (K; ∗) cannot have the K points in a plane. We numerate the 34 convex polyhedra with seven vertices of Figure 1(d) from 1 to 34 starting from the first polyhedron on the left hand side of the first row of Figure 1(d). We also numerate the 7 convex polyhedra with six vertices of Figure 1(c) from 1 to 7 starting from the first polyhedron on the left hand side of the first row of Figure 1(c). In a similar way we numerate the convex polyhedra with five vertices. Using Corollary 3 we have the following possible polyhedra for the convex hull of the 8 singular points of a quadratic system (1) when these 8 points are in the boundary of the convex hull, i.e. for the configuraion (8): (8)(i) If there is a plane containing four singular points we have the follow- ing four possibilities for the polyhedron defined by the convex hull of the eight vertices. (8)(i.1) A quadrilateral-quadrilateral prism is a prism with two convex quadrilater bases which do not share any edge, the lateral faces can be triangles or quadrilaters, these polyhedra have eight ver- tices. 6 J. LLIBRE AND C. VALLS (a) The polyhedron (b) The polyhedra with 5 vertices. with 4 vertices. (c) The polyhedra with 6 vertices. (d) The polyhedra with 7 vertices. Figure 1. All convex polyedra with 4, 5, 6 and 7 vertices. (8)(i.2) A quadrilateral-triangular prism is a prism with a convex quadri- later and a triangle as bases (the triangle of the bases has a THE EULER-JACOBI FORMULA 7 singular point in its interior), the lateral faces are triangles or quadrilaters, these polyhedra have seven vertices. Then, from Figure 1(d) these polyhedra are one of the following 8, 9, 10, 14, 17, 19, 20, 30 and 32 of Figure 1(d). (8)(i.3) A triangular-triangular prism is a prism with two triangular bases containing each base a singular point in its interior, the lateral faces are triangles or quadrilaters, these polyhedra have six vertices, see the polyhedra 5 and 6 of Figure 1(c). (8)(i.4) A tetrahedron with a point in the interior of the four triangles of its faces, of course this tetrahedron has four vertices. (8)(ii) If there is not a plane containing four singular points, then it is a polyhedron with twelve triangular faces, eighteen edges and eight vertices. Applying the Euler-Jacobi formula (2) with the polynomial S equal to the product of a plane containing four vertices of a face of the polyhedron with 7 vertices, with the plane determined by the remainder 3 vertices, we obtain that the polyhedra 1, 2, 3, 4, 6, 7, 8, 9, 10, 11 ,12 ,13 ,14 ,15 ,17 , 19, 20, 21, 23, 24, 25, 26, 27, 29, 30 and 32 of Figure 1(d) cannot be a convex hull of type (7; 1). Since the polyhedra 5, 22 and 34 of Figure 1(d) have a face with more than 4 vertices by Proposition 1 they cannot be the convex hull of type (7; 1). Therefore the unique possible polyhedra of type (7;1) are (7;1) the polyhedra 16, 18, 28, 31 and 33 of Figure 1(d), which have ten triangular faces, fifteen edges and seven vertices. The singular point which is not a vertex of this polyhedra is contained in the interior of the polyhedra. We analyze the convex hulls of type (6;2). (6;2)(i) If there is a plane containing four singular points, then we have four possibilities. (6;2)(i.1) A quadrilateral bipyramid is formed by two pyramids glued by the same quadrilateral base in such a way that the two vertices of the two pyramids and the two points in the 1st level are in the same plane, again by Proposition 2. See the polyhedron 5 of Figure 1(c). Note that in the polyhedron 5 we have projected the quadrilateral bipyramid in a lateral triangular face. (6;2)(i.2) A quadrilateral-edge pyramid is a pyramid with a convex quadri- lateral as base and instead of a vertex the pyramid has an edge E, see the polyhedra 2 and 3 of Figure 1(c). The two singular points in the interior of these polyhedra and the edge E are contained in a plane due to Proposition 2. (6;2)(i.3) A triangular-edge pyramid is a pyramid with a triangular base having in the interior an additional singular point and instead of a vertex the pyramid has an edge E, see the polyhedron 1 8 J. LLIBRE AND C. VALLS of Figure 1(c). The two singular points in the interior of this polyhedron and the edge E are contained in a plane due to Proposition 2. (6;2)(i.4) A truncated triangular pyramid it has two triangular bases with one singular point in the interior of each triangle, see the convex polyhedron 6 of Figure 1(c). The two singular points in the interior of this polyhedron must be in the same plane than each one of the lateral edges of this polyhedron due to Proposition 2. (6;2)(ii) If there is no a plane containing four singular points, then it must be a polyhedron with eight triangular faces, twelve edges and six vertices, see the polyhedra 4 and 5 of Figure 1(c). We note that the polyhedra 7 cannot be a convex hull of type (6;2) due to the existence of a pentagon. Now we study the polyhedra which are convex hull of type (5;3). (5;3)(i) If there is a plane containing four singular points, then a quadrilateral pyramid is a pyramid with a quadrilateral base in such a way that the vertex of the pyramid which is not contained in the base is in the plane generated by the three points in the 1st level, see the polyhedron 2 of Figure 1(b). (5;3)(ii) If there is not a plane containing four singular, then it must be a triangular bipyramid which is formed by two pyramids glued by the same triangular base in such a way that the two vertices of the two pyramids and the three points in the 1st level never are more than three in the same plane. This polyhedron has six triangular faces, nine edges and five vertices. See the polyhedron 1 of Figure 1(b). Finally the polyhedron which is a convex hull of type (4;4) is: (4;4) A tetrahedron such that the four points in the 1st level are not con- tained in a plane, and never four of the eight singular points are contained in a plane. If one of the fourteen previous described possible convex hulls of the 8 singular points of a quadratic system (1) is realizable, in what follows we will only say that the corresponding polyhedron is realizable. 3. The main result We denote by i(a) the (topological) index of a singular point a ∈ A of the quadratic P system (1). It wasPproved in [15] that for a quadratic system (1) either a∈A iX (a) = 0, or | a∈A iX (a)| = 2. Our main result is the following. THE EULER-JACOBI FORMULA 9 Theorem 5. For quadratic systems (1) having 8 singular points the follow- ing polyhedra are realizable. (8)(i.1) If a quadrilateral-quadrilateral prism is realizable, then the indices in the two quadrilaters of the bases are 1, −1, 1, −1 in counterclockwise sense. (8)(i.3) If a triangular-triangular prism is realizable, then the indices of the vertices of the two triangular bases are 1, 1, 1 and −1, −1, −1, and the indices of the interior singular point at every one of these triangles is different from the ones of the vertices of the triangle. (8)(ii) The configuration of the indices of the vertices of a polyhedron with twelve triangular faces, eighteen edges and eight vertices realizable by a quadratic system can be obtained as follows. Take two triangu- lar faces without intersection, then the two remaining vertices have different index. Repeating this for every pair of triangular faces with- out intersection the configuration of the indices of this polyhedron is obtained. (6;2)(i.1) If a quadrilateral bipyramid is realizable, then the indices of the quadrilater of the basis are 1, 1, −1 − 1 in counterclockwise sense, the indices of the two vertices outside the quadrilateral are different, and the indices of the two interior singular points are different. (6;2)(i.2) If a quadrilateral-edge pyramid is realizable, then the indices of the quadrilater of the basis are 1, 1, −1, −1 in counterclockwise sense, the indices of the two vertices of the edge E are different and the indices of the two interior points are also different. (6;2)(i.3) If a truncated triangular pyramid is realizable, then the indices of the vertices of the two triangular bases are 1, 1, 1 and −1, −1, −1, and the indices of the two interior points of the pyramid are different. (6;2)(ii) If the polyhedra 4 and 5 of Figure 1(c) are realizable, then the indices of two triangular faces without intersection are 1, 1, 1 and −1, −1, −1, and the indices of the two interior singular points of the polyhedra are different. (5;3)(i) If a quadrilateral pyramid is realizable then it has the following two possible configurations of indices. First the indices of the vertices of the quadrilateral basis are 1, 1, 1, −1, the vertex of the pyramid which is not contained in the basis has index −1, and the three interior singular points have indices 1, −1, −1; of course we can reverse all the signs of these indices. Second the indices of the vertices of the quadrilateral basis are 1, 1, −1, −1 in counterclockwise, the vertex of the pyramid which is not contained in the basis has index 1, and the three interior singular points have indices 1, −1, −1; of course we can reverse all the signs of these indices. (5;3)(ii) If a triangular bipyramid is realizable, then it has the following three possible configurations of indices. First the indices of the three ver- tices of the triangular base (where both pyramids are glued) are 1, 1, 10 J. LLIBRE AND C. VALLS 1, the indices of the two vertices of the bipyramid which are not in the triangular base are −1, −1, and the three interior singular points have indices 1, −1, −1; of course we can reverse all the signs of these indices. Second the indices of the three vertices of the triangular base (where both pyramids are glued) are 1, 1, −1, the indices of the two vertices of the bipyramid which are not in the triangular base are 1, −1, and the three interior singular points have indices 1, −1, −1; of course we can reverse all the signs of these indices. Third the indices of the triangular base are 1, −1, −1, the indices of the two vertices are 1, 1 and the three interior singular points have indices 1, −1, −1; of course we can reverse all the signs of these indices. (a) There are quadratic systems (1) realizing the configurations of the indices of all the previous statements. (b) The polyhedra of (8)(i.2), (8)(i.4), (7; 1), (6; 2)(i.3) and (4; 4) are not realizable. Proof of statement (8)(i.1). Let Q1 and Q2 be the two quadrilaters of the bases of the quadrilateral-quadrilateral prism. Let Π1 be a plane containing Q1 . Let Π2 be a plane through an edge of Q2 containing only this edge of the quadrilateral-quadrilateral prism. Applying the Euler-Jacobi formula (2) with S = Π1 Π2 we get that the two remaing vertices of Q2 must have dif- ferent index. Repeating this process for every edge of Q1 and Q2 statement (8)(i.1) follows.  Take P = x2 − x, Q = y 2 − y and R = z 2 − z. Then the quadratic system (1) has the singular points (0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1) and (0, 1, 1), and their indices are −1, 1, −1, 1, 1, −1, 1 and −1, respectively. So the configuration (8)(i.1) is realizable. Proof of statement (8)(i.3). Let Q1 and Q2 be the two triangles of the bases of the triangular-triangular prism, and let p1 and p2 be the singular points at the interior of the triangles Q1 and Q2 respectively. Let Π1 be the plane containing Q1 . Let Π2 be a plane defined by an edge of Q2 containing only this edge of the triangular-triangular prism. Applying the Euler-Jacobi formula (2) with S = Π1 Π2 we get that the remaining vertex of the triangle Q2 and p2 must have different index. Repeating this process for every edge of Q2 we obtain that the three vertices of Q2 have the same index different from the index of p2 . Doing a similar process for Q1 we obtain that the three vertices of Q1 have the same index different from the index of p1 . Taking into account that the absolute value of the sum of all the indices of the singular points is 0 or 2, the indices of the vertices of Q1 are different from the indices of the vertices of Q2 .  Take P = −2x − 2xy + xz, Q = −4x2 − 2y + 2y 2 + yz and R = −z + z 2 . Then the quadratic system (1) has the singular points (−1, −1, 0), (1, −1, 0), THE EULER-JACOBI FORMULA 11 (0, 1, 0), (−1/2, −1/2, 1), (1/2, −1/2, 1), (0, 1/2, 1), (0, 0, 0) and (0, 0, 1), and their indices are 1, 1, 1, −1, −1, −1, −1 and 1, respectively. So the configu- ration (8)(i.3) is realizable. Proof of statement (8)(ii). Let Π1 and Π2 the two planes defined by two triangular faces without intersection, then applying the Euler-Jacobi formula with S = Π1 Π2 we get that the two remaining vertices of the polyhedron have different index. Repeating this for every pair of triangular faces without intersection the configuration of the indices of this polyhedron is obtained. Note that in this way we can describe the configuration of the indices of the vertices of all the polyhedra with twelve triangular faces, eighteen edges and eight vertices realizable by the quadratic systems. In what follows we provide an example.  √ 2 − 6y 2 − z 2 , Q = y − 2y 2 / 3 − 2yz and R = √ Take P√ = 6 + z − 6x √ 3 3x2 − 3y 2 + x(6y + 3 3(−1 + 2z)). √ Then the quadratic √ system (1) has the singular √ points (1, √ 0, 0), (1/2, 3/2, 0), (−1/2, 3/2, 0), (−1, 0, 1), (−1/2, − 3/2, 1), (1/2, − 3/2, 1), (0, 0, 3) and (0, 0, −2), and their indices are −1, 1, −1, 1, −1, 1, −1 and 1, respectively. So this quadratic system re- alize a configuration (8)(ii). Proof of statement (6; 2)(i.1). Let p7 and p8 be two interior points of a quadrilateral bipyramid, and pk for k = 1, 2, 3, 4 the vertices of the quadri- lateral base of the bipyramid ordered in counterclockwise sense. The two vertices of the bipyramid which are not in the quadrilateral base are denoted by p5 and p6 . Applying the Euler-Jacobi formula to Π145 Π236 we get that p7 and p8 must have opposite index. Let L be the straight line defined by the points p7 and p8 . By Proposition 4 we can assume that L intersects the interior of the triangular face with vertices pi pj pk being i, j, k distinct and {i, j, k} ⊂ {1, 2, 3, 4, 5, 6}. In this case the indices of pi , pj and pk are equal. Indeed, we can choose the vertex pi such that the plane Π78i separates the points pj and pk and then applying the Euler-Jacobi formula to C = Πrst Π78i with r, s, t distinct, {r, s, t} ⊂ {1, 2, 3, 4, 5, 6} and {r, s, t} ∩ {i, j, k} = ∅, we get that pj and pk have the same indices. In a similar way pi and pk , and pj and pk have the same indices. Again by Proposition 4 we can assume that L intersects the interior of the triangular faces T1 and T2 . First assume that the two triangular faces T1 and T2 intersected by L have no edges in common. Since the absolute value of the sum of the indices of all singular points must be either zero or two, the indices of the vertices of T1 are different from the indices of the vertices of T2 , and this configurations of indices is realizable (see below). 12 J. LLIBRE AND C. VALLS Suppose now that the two triangular faces T1 and T2 intersected by L have an edge in common. Then the indices of the vertices of T1 ∪ T2 are equal. Since the indices of p7 and p8 are different, and again since the absolute value of the sum of the indices of all singular points must be either zero or two, the indices of the remaining singular points must be different from the ones in T1 ∪ T2 . So we have three points with an index and five with the opposite index. Two of the singular points of the three having the same index are vertices of the quadrilateral base and the other is in the interior of the bipyramid. Let Π0 be the plane defined by the three points with the same index. This plane separates the other five singular points with the same index either in four points in one side and the other in the other side, or three points in one side and two in the other side. In the first case the unique point p in a side of the plane Π0 must be a vertex of the quadrilateral bipyramid. Let Π1 be a plane through p whose intersection with the bypiramid is only p. Applying the Euler-Jacobi formula to S = Π0 Π1 we reach a contradiction, because all the singular points which are not in S = 0 are in the same connected component of R3 \ S, and all of them have the same index. In the second case let Π2 be the parallel plane to Π0 containing the two points in a side of the plane Π0 . Then applying the Euler-jacobi formula to S = Π0 Π2 we obtain again a contradiction.  Take P = yz, Q = x2 −y 2 +5xz and R = −4+4y 2 −399xz/5+z 2 . Then the quadratic system (1) has the singular points (1, −1, 0), (1, 1, 0), (−1, 1, 0), (−1, −1, 0), (0, 0, 2), (0, 0, −2), (1/2, 0, −1/10) and (−1/2, 0, 1/10), and their indices are 1, 1, −1, −1, −1, 1, −1 and 1, respectively. So the configuration (6;2)(i.1) described in Theorem 5 is realizable. Proof of statement (6; 2)(i.2). Let Q1 be the quadrilateral basis of the quadri- lateral-edge pyramid, and let p1 and p2 be the singular points at the interior of pyramid. Let Π1 be the plane containing Q1 , and let Π2 be a plane which intersection with the pyramid be only the edge E. Applying the Euler-Jacobi formula with S = Π1 Π2 we get that p1 and p2 have different index. Let Π3 be the plane containing the edge E and p1 and p2 . The plane Π3 intersects either two opposite edges of Q1 or two consecutive edges of Q1 . If Π3 intersects two opposite (respectively consecutive) edges of Q1 let E4 and E5 be the two edges of Q1 which are not intersected by Π3 . If Πk is a plane whose intersection with the pyramid only contains Ek for k = 4, 5, then applying the Euler-Jacobi formula with S = Π3 Πk we get that the two vertices of Ej with j ∈ {4, 5} and j 6= k have different index (respectively the same index) for k = 4, 5. Let E6 and E7 be the two edges of Q1 which are intersected by Π3 . If Πk is a plane whose intersection with the pyramid only contains Ek for k = 6, 7, then applying the Euler-Jacobi formula with S = Π3 Πk we get that the two vertices of Ej with j ∈ {6, 7} and j 6= k have the same index for k = 6, 7. Let Π8 be a plane containing the points p1 THE EULER-JACOBI FORMULA 13 and p2 without intersection with the edge E, then applying the Euler-Jacobi formula with S = Π1 Π8 we get that the two vertices of the edge E have different index. In the case that Π3 intersects two consecutive edges of Q1 the indices of the quadrilater of the basis are 1, 1, 1 − 1, the indices of the edge E are different and the indices of the interior points are also different. Let Π9 be the plane defined by the three points with the same index. This plane separates the other five points with the same index either in four points in one side and the other point in the other side, or three points in one side and two points in the other side. Now we arrive to a contradiction using the same arguments as in the last part o the proof of the case (6.2)(i.1).  Take P = −1/3+z −2x2 /3+y 2 , Q = yz and R = −x2 +y 2 +z 2 . Then the quadratic system (1) has the singular points (1, −1, 0), (1, 1, 0), (−1, 1, 0), (−1, −1, 0), (1, 0, 1), (−1, 0, 1), (1/2, 0, 1/2) and (−1/2, 0, 1/2), and their indices are −1, −1, 1, 1, −1, 1, 1 and −1, respectively. So the configuration (6;2)(i.2) is realizable. Proof of statement (6; 2)(i.3). Let Q1 and Q2 be the two triangles of the bases of the truncated triangular pyramid, and let p1 and p2 be the singular points at the interior of pyramid. Let Πk be the plane containing Qk for k = 1, 2. Applying the Euler-Jacobi formula with S = Π1 Π2 we get that p1 and p2 have different index. Let E be a lateral edge of the pyramid. Let Π3 the plane containing the edge E and p1 and p2 . Let Ei∗ be the edge of the triangle Qi of the truncated pyramid which has no intersection with E. Let Π∗i be a plane such that its intersection with the truncated pyramid is Ei∗ . Applying the Euler-Jacobi formula with S = Π3 Π∗i we get that the two vertices of Qj with j 6= i which are not contained in E must have the same index. Repeating this process for every lateral edge of the truncated pyramid we obtain that the three vertices of Qk have the same index and since the sum of the absolute value of the indices is zero or two, the vertices of Q1 and Q2 must have different index.  Take P = 4 − 4y − 13z − 8x2 + 2yz + 10z 2 , Q = 5x(2 + 2y − z) and R = −4x2 +y(−2+2y+z). Then the quadratic system (1) has the singular points (−1, −1, 0), (1, −1, 0), (0, 1, 0), (−1/2, −1/2, 1), (1/2, −1/2, 1), (0, 1/2, 1), (0, 0, 1/2) and (0, 0, 4/5), and their indices are −1, −1, −1, 1, 1, 1, 1 and −1, respectively. So the configuration (6;2)(i.3) is realizable. Proof of statement (6; 2)(ii). The proof of this statement follows using the same arguments of the proof of statement (6;2)(i.1).  14 J. LLIBRE AND C. VALLS Take q √  P = 180x(y − z + 1) − 5 3 1041 − 88 (z − 1)z,  √  √  Q = 9000 2x2 + y − 1 + −13500y − 3 1041 + 25213 z + 3 1041 − 17338 z 2 , R = y 2 + 14 yz − 1 18 (z − 1)(25z − 18). Then the quadratic system (1) has the singular points (−1, −1, 0), (1, −1, 0), (0, −1/4, 1), (−1/4, 0, 1), (1/4, 0, 1), (0, 1, 0), r  ! 1 1 √  1  √  3 3023 + 137 1041 , − 9 + 1041 , 100 30 120 5 and r  ! 1 1 √  1 1 − 3 1041 − 88 , , , 120 5 3 2 and their indices are −1, −1, −1, 1, 1, 1, −1 and 1, respectively. So the con- figuration (6;2)(ii) is realizable. Proof of statement (5; 3)(i). Let Q1 be the quadrilateral basis of the quadri- lateral pyramid, q the vertex of the pyramid that is not contained in the base, and let p1 , p2 and p3 be the singular points in the interior of the pyramid. Let Π1 be the plane containing Q1 , and let Π2 be a plane whose intersection with the pyramid is only the vertex q. Applying the Euler-Jacobi formula to S = Π1 Π2 we get that the three singular points p1 , p2 , p3 cannot have the same index. Let Π3 be the plane containing p1 , p2 , p3 (which also contains q). Note that the plane Π3 separates the points in the quadrilateral of the basis as follows: (I) one point p4 in one side of the plane Π3 and the remaining three points on the other side of this plane; (II) two points in one side of the plane Π3 and the remaining two points on the other side of this plane. In case (I) let Π4 be a plane defined by an edge of Q1 (which does not contain p4 ) containing only this edge of the quadrilateral pyramid. Applying the Euler-Jacobi formula to S = Π3 Π4 we obtain that the two remaining vertices of the quadrilateral must have the same index. Take now Π∗ the plane containing an edge of Q1 having as vertex p4 . Then the Euler-Jacobi formula applied to S = Π3 Π∗ implies that the index of p5 is different from the index of p4 . In short we get that three vertices of the quadrilateral have the same index and the other that we denote by p5 has different index, and p5 6= p4 . Now we claim that the index of q must be the same than the index of p5 . Indeed, assume that it is not the case. Using that the absolute value of the sum of the indices is either 0 or 2 there are two points in {p1 , p2 , p3 } having the same index than p5 denote them by pi1 and pi2 . Now let Π5 be the plane determined by the points p5 , pi1 and pi2 . The plane Π5 separates THE EULER-JACOBI FORMULA 15 the other five singular points with the same index either in four points in one side and one point r in the other side, or three points in one side and two points in the other side. In the first case the point r is one of the vertices of the pyramid. Let Π6 be a plane through r whose intersection with the pyramid is only the point r. Applying the Euler-Jacobi formula to S = Π5 Π6 we reach a contradiction, because all the singular points which are not in S = 0 are in the same connected component of R3 \ S and all of them have the same index. In the second case the two unique points in a side of the plane Π5 are r and the remaining interior point of the pyramid. Let Π7 be a plane containing the two points on one side of the plane Π5 and such that the other three points separated by Π5 remain in the same side of Π7 . Applying the Euler-Jacobi formula to S = Π5 Π7 we obtain again a contradiction. Hence the claim is proved. If only one pi1 of the interior points of the polyhedron has the same index than p5 and q, taking the plane passing through p5 , q, pi1 and using similar arguments to the ones used for proving the claim we reach a contradiction. Hence, the indices of the interior singular points are two equal to the index of p5 and the other different. This completes the characterization of the indices in case (I). In case (II) let Π4 be a plane defined by an edge E of Q1 containing only this edge of the quadrilateral pyramid and such that E has no intersection with the plane Π3 . Applying the Euler-Jacobi formula to S = Π3 Π4 we ob- tain that the two remaining vertices of the quadrilateral must have different index. Repeating these arguments for the other three edges of Q1 we get that indices of the vertices of the quadrilateral are 1, 1, −1, −1 in counter- clokwise. Now similar arguments to the case (I) show that if the index of q is 1, then the indices of the three interior singular points are 1, −1, −1, and if the index of q is −1, then the indices of the three interior singular points must be 1, 1, −1. This completes the configurations of the indices in case (II).  In the first realization of the index configurations for the quadrilateral pyramids the√ two planes containing four singular points are the planes z = 0√and x − 3y = 0. Consider the quadratic √ system (1) given by P = ( 3y − x)z, Q = −6 − x + x2 + 13z/2 + 3yz − 3z 2 /2, R = −2 + y 2 + y(−1 + z) + 13z/6 − z 2 /2. This quadratic system has the √ singular points (−2, −1, 0), (3, √ −1, 0), (3, 2, 0), (−2, 2, 0), (0, 0, 4/3), (1, 1/ 3, 1), (0, 0, 3) and (−1, −1/ 3, 1), these singular points have indices 1, 1, 1, −1, 1, −1, −1 and −1, respectively. Now we realize the second index configuration for the quadrilateral pyra- mids of type (5;3)(i).The two planes containing four singular points are the planes z = 0 and y = 0. Consider the quadratic system (1) given by P = yz, Q = x2 − y 2 − xz/10, R = 3 − 3y 2 − 4z + 5xz/2 + z 2 . This quadratic system has the singular points (−1, 1, 0), (−1, −1, 0), (1, −1, 0), 16 J. LLIBRE AND C. VALLS (1, 1, 0), (0, 0, 1), (1/5, 0, 2), (0, 0, 3) and (3/25, 0, 6/5), these singular points have indices 1, 1, −1, −1, −1, −1, 1 and 1, respectively. Proof of statement (5; 3)(ii). Let T1 be the base of the triangular bipyramid, and p1 , p2 and p3 be the singular points in the interior of the bipyramid. Let Π1 be the plane defined by a face of the bipyramid and Π2 a plane defined by another face of the bipyramid such that these two faces only share one point. Applying the Euler-Jacobi formula to S = Π1 Π2 we get that the three singular points p1 , p2 , p3 cannot have the same index. We claim that the sum of the indices of the singular points is zero. Assume that the sum of the indices of the singular points is 2 (we can assume that it is positive). Then there are three points with negative index that we denote by q1 , q2 , q3 . Note that at least one of them is in the 1st level. Now let Π5 be the plane determined by the points q1 , q2 , q3 . The plane Π5 separates the other five singular points with the same index either in four points in one side and one point r in the other side, or three points in one side and two points in the other side. In the first case the point r is one of the vertices. Let Π6 be a plane through r whose intersection with the bipyramid is only the point r. Applying the Euler-Jacobi formula to S = Π5 Π6 we reach a contradiction, because all the singular points which are not in S = 0 are in the same connected component of R3 \ S and all of them have the same index. In the second case the two unique points in a side of the plane Π5 are denoted by r and s. Let Π7 be a plane containing the two points r and s and such that leave the other three points separated by the plane Π5 in one side of Π7 . Applying the Euler-Jacobi formula to S = Π5 Π7 we obtain again a contradiction. Hence the claim is proved. In view of the claim, taking into account that p1 , p2 , p3 cannot have the same index, we get the following three configurations: (I) The indices of the three vertices of the triangular base are 1, 1, 1; the indices of the two vertices of the bipyramid which are not in the triangular base are −1, −1 and the three interior points must be of the form 1, −1, −1. (II) The indices of the three vertices of the triangular base are 1, 1, −1; the indices of the two vertices of the bipyramid which are not in the triangular base are 1, −1 and the three interior points must be of the form 1, −1, −1. (III) The indices of the three vertices of the triangular base are 1, −1, −1; the indices of the two vertices of the bipyramid which are not in the triangular base are 1, 1 and the three interior points must be of the form 1, −1, −1.  THE EULER-JACOBI FORMULA 17 Now we realize the three index configurations for the quadrilateral pyra- mids of type (5;3)(ii). For the first realization take P = (42 + 91x + 252y − 14z + 35x2 + 126xy + 157xz + 244yz)/42, Q = (−56x − 168y − 21z − 28x2 − 84xy − 101xz − 146yz + 7z 2 )/7 and R = (−56y − 14x2 − 14xy − 31xz + 28y 2 − 6yz)/28. Then the quadratic system (1) has the singular points (−2, −1, 0), (−2, 2, 0), (3, −1, 0) in the vertices of the triangular bases of the bipyramide, (0, 0, 3) and (3, 2, −2) in the two vertices of the bipyramide which are not in the triangular basis, and (1, −1/2, 1), (−1, 1/2, 1) and (1/4, −1/4, −1/4) in the interior of the bipyramide, and their indices are 1, 1, 1, −1, −1, 1, −1 and −1, respectively. So the first realization of the configuration (5;3)(ii) is done. For the second realization take P = (2016 + 3500x + 28000y − 672z + 406x2 + 3500xy + 12751xz + 4502yz)/2016, Q = (−112x − 896y − 63z − 14x2 − 386xz − 100yz + 21z 2 )/21 and R = (196x + 1232y + 14x2 + 196xy + 647xz +336y 2 +454yz)/336. Then the quadratic system (1) has the singular points (−8, −1, 0), (8, −1, 0), (−8, 2, 0) in the vertices of the triangular bases of the bipyramide, (0, 0, 3) and (−125455790514, 1122399166369, −1469697563719)/172072592267 in the two vertices of the bipyramide which are not in the triangular basis, and (1, −1/2, 1), (−1, 1/2, 1) and (3, 2, −2) in the interior of the bipyramide, and their indices are 1, 1, −1, −1, 1, −1, 1 and −1, respectively. So the second realization of the configuration (5;3)(ii) is done. For the fifth realization take P = x + xy + 4xz + yz, Q = 4x − 2x2 − y + 4xy + y 2 + 15xz and R = −9 − 36x + 18x2 + 9y − 36xy − 269xz + z 2 . Then the quadratic system (1) has the singular points (0, 1, 0), (−1, −1, 0), (1, −1, 0), (0, 0, 3), (0, 0, −3), (1/4, −1/4, −1/4), (1/4, 1/4, −1/4) and (9/53, −19/53, −18/53), and their indices are 1, 1, −1, −1, 1, 1, −1 and −1, respectively. So the third realization of the configuration (5;3)(ii) done. Proof that the polyhedra (8)(i.2) are not realizable. Let Q1 and Q2 be the quadrilateral and the triangle of the bases of the quadrilateral-triangular prism respectively, and let p be the singular point at the interior of triangle of the bases. Let Π1 be a plane containing Q1 . Let Π2 be a plane defined by an edge of Q2 containing only this edge of the quadrilateral-triangular prism. Applying the Euler-Jacobi formula (2) with S = Π1 Π2 we get that the remaining vertex of Q2 and p must have different index. Repeating this process for every edge of Q2 we obtain that the three vertices of Q2 have the same index different from the index of p. Let Π3 be a plane containing one edge E of Q1 and no other points of the polyhedron, and let Π4 be the plane defined by Q2 . Applying the Euler-Jacobi formula (2) with S = Π3 Π4 we get that the two vertices of Q1 not contained in E must have different index. Repeating these arguments we obtain that the index configuration on Q1 is 1, −1, 1, −1 in counterclockwise sense. 18 J. LLIBRE AND C. VALLS From the polyhedra 8, 9, 10, 14, 17, 19, 20, 30 and 32 of Figure 1(d) it follows that there exists an edge E of Q1 such that the plane Π5 determined by E and p leaves an edge E* of Q2 on one side of the plane. Let Π6 be the plane determined by the three vertices of the polyhedron which are neither in the plane Π5 nor in the edge E*. Then applying the Euler-Jacobi formula (2) with S = Π5 Π6 we get that the two vertices of E* have different index, a contradiction.  Proof that the polyhedron (8)(i.4) is not realizable. Let Qk for k = 1, 2, 3, 4 be the triangles of the faces of the tetrahedron, and let pk be the singular point in the interior of Qk . Let Πk be the plane determined by Qk . Then applying the Euler-Jacobi formula (2) with S = Π1 Π2 we get that the sin- gular points p3 and p4 have different index. Repeating this argument we get that the indices of pi a pj are different if i 6= j, and this provides a contradiction.  Proof that the polyhedra (7; 1) are not realizable. Let q be the point of A in the 1st-level. Let Πi1 i2 i3 be the plane determined one face of a polyhedron of type (7;1), and let Πj1 j2 j3 be the plane determined by another face which has any poin in the plane Πi1 i2 i3 . Applying the Euler-Jacobi formula to S = Πi1 i2 i3 Πj1 j2 j3 we get that the vertex of the polyhedron not contained in these two planes must have different index than the index of q. In this way we obtain that all the vertices of the polyhedron have the same index, in contradiction with the fact that the sum of the indices of the eight singular points is zero or two.  Proof that the polyhedra (6; 2)(i.3) are not realizable. Consider the plane Π1 containing the edge E and the two interior points, and the plane Π2 contain- ing the edge of the triangle of the base that contains three singular points. The Euler-Jacobi formula applied with S = Π1 Π2 provides a contradiction, because it remains in the Euler-Jacobi formula a unique term which cannot be zero.  Proof that the tetrahedra (4; 4) are not realizable. The four points in the 0 level are contained in the vertices of a tetrahedron. We denote them by p1 , p2 , p3 , p4 . By Proposition 2 the four points in the 1st level, that we denote them by p5 , p6 , p7 , p8 , cannot be contained in a plane and so they are also contained in the vertices of a tetrahedron. Given l1 , l2 ∈ {5, 6, 7, 8} there exists k0 ∈ {1, 2, 3, 4} so that Πk0 ,l1 ,l2 leaves the rest of the points in the 1-st level in the same side of the plane. We denote by pl3 , pl4 the points in the 1-st level different from pl1 , pl2 and by pk1 , pk2 , pk3 the points in the 0 level different from pk0 . Then applying the Euler-Jacobi formula to Πk0 l1 l2 Πk1 k2 k3 we obtain that pl3 and pl4 have different signs. THE EULER-JACOBI FORMULA 19 We repeat this procedure successively for all pairs of points in the 1-st level and we reach to a contradiction. Indeed, taking {l3 , l4 } = {5, 6}, we get that p7 and p8 have different indices; taking {l3 , l4 } = {6, 7} we get that p8 and p5 have different indices and taking {l3 , l4 } = {7, 8} we get that p5 and p6 have different indices. So, p5 and p7 have the same index and p6 and p8 have the same index. However taking now {l3 , l4 } = {5, 7} we get that p6 and p8 must have different index which is not possible. In short, configuration (4;4) is not possible.  Acknowledgements The first author is supported by the Ministerio de Economı́a, Indus- tria y Competitividad, Agencia Estatal de Investigación grant MTM2016- 77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Re- cerca 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/04459/2013. References [1] V. Arnold, A. Varchenko and S. Goussein-Zude, Singularités des applications différentialbes, Mir, Moscow, 1982. [2] A. N. Berlinskii, On the behavior of the integral curves of a differential equation, Izv. Vyssh. Uchebn. Zaved. Mat. 2 (1960), 3–18. [3] R.F. Brown, The Lefschetz fixed point theorem, Scott, Foresman and Company, Glenview, I.L., 1991. [4] C. Chicone and T. Jinghuang, On general properties of quadratic systems, Amer. Math. Monthly 81 (1982), 167–178. [5] A. Cima, A. Gasull and F. Manõsas, Some applications of the Euler-Jacobi for- mula to differential equations, Proc. Amer. Math. 118 (1993), 151–163. [6] A. Cima and J. Llibre, Configurations of fans and nests of limit cycles for polyno- mial vector fields in the plane, J. Differential Equations 82 (1989), 71–97. [7] W.A. Coppel, A survey of quadratic systems, J. Differential Equations 2 (1966), 293–304. [8] H.T. Croft, K.J. Falconer and R.K. Guy, §B15 in Unsolved problems in Geom- etry, New York, Springer–Verlag, 1991. [9] M.B. Dillencourt, Polyhedra of small orders and their Hamiltonian properties, Tech. Rep. 92–91, Info. and Comput. Sci. Dept. Irvine, CA: Univ. Calif. Irvine, 1992. [10] A.J.W. Duijvestijn and P.J. Federico, The number of polyhedral (3–connected planar) graphs, Math. Comput. 37 (1981), 523–532. [11] W. Fulton, Algebraic curves, Mathematic Lecture Note Series, Benjamin, 1974. [12] A. Gasull and J. Torregrosa, Euler-Jacobi formula for double points and appli- cations to quadratic and cubic systems, Bull. Belg. Math. Soc. 6 (1999), 337–346. [13] B. Grünbaum, Convex Polytopes, 2nd ed., New York, Springer–Verlag, 2003. [14] O. Hermes, Die formen der vielflache, J. Reine Angew. Math. 120 (1989), 27–59. [15] A. G. Khovanskii, Index of a polynomial vector field, Funktsional Anal. i Prilozhen 13 (1979), 49–58. [16] J. Llibre and C. Valls, The Euler-Jacobi formula and the planar quadratic-quartic polynomial differential systems, to appear in Proc. Amer. Math. Soc. 20 J. LLIBRE AND C. VALLS [17] N.G. Lloyd, Degree theory, Cambridge Tracts in Mathematics, No. 73. Cambridge University Press, Cambridge-New York-Melbourne, 1978. [18] I.R. Shafarevich, Basic algebraic geometry. Varieties in projective space, Third edition, Translated from the 2007 third Russian edition. Springer, Heidelberg, 2013. [19] E. Steinitz, Über die Eulersche polyederrelationen, Arch. Math. Phys. 11(3) (1906), 86–88. [20] E. Steinitz and H. Rademacher, Vorlesungen über die theorie polyeder, Berlin, 1934. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain Email address: jllibre@mat.uab.cat 2 Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa, Av. Rovisco Pais 1049–001, Lisboa, Portugal Email address: cvalls@math.ist.utl.pt