STAR CONFIGURATION POINTS AND GENERIC PLANE CURVES ENRICO CARLINI AND ADAM VAN TUYL 0 Abstract. Let ℓ1,...ℓl be l lines in P2 such that no three lines meet in a point. Let 1 X(l)be the setofpoints {ℓ ∩ℓ | 1≤i<j ≤l}⊆P2. We call X(l)astarconfiguration. i j 0 We describe all pairs (d,l) such that the generic degree d curve in P2 contains a X(l). 2 n a J 1. Introduction 5 2 The problem of studying subvarieties of algebraic varieties is a crucial one in algebraic G] geometry, e.g., the case of divisors. The study of subvarieties of hypersurfaces in Pn has a particularly rich history. For example, one can look for the existence of m dimensional A linear spaces on generic hypersurfaces of degree d leading to the theory of Fano varieties . h and to the well known formula relating n,m and d (e.g., see [10, Theorem 12.8]). t a m Because linear spaces are complete intersections, it is natural to look for the existence of complete intersection subvarieties on a generic hypersurface. The case of codimension [ two complete intersections was first studied by by Severi [15], and later generalized and 1 v extended to higher codimensions by Noether, Lefschetz [11] and Groethendieck [9]. Re- 4 cently, in [3], secant vavieties were used to give a complete solution for the existence of 0 complete intersections of codimension r on generic hypersurfaces in Pn when 2r ≤ n+2. 5 4 Fewer results are known if the codimension of the complete intersection is large, i.e., the 1. codimension is close to the dimension of the ambient space. In [16] the case of complete 0 intersection curves is studied and completely solved. The case of complete intersection 0 points on generic surfaces in P3 is considered in [4] where some asymptotic results are 1 : presented. The case of complete intersection points in P2 is a special case of [3], and it is v i completely solved. X Taking our inspiration from [3, 4], we examine the problem of determining when special r a configurations of points lie on a generic degree d plane curve in P2. We shall focus on star configurations of points. Consider l lines in the plane ℓ ,...,ℓ ⊂ P2 such that 1 l ℓ ∩ℓ ∩ℓ = ∅. The set of points X(l) consisting of the l pairwise intersections of the i j k 2 lines ℓi is called a star configuration. In this paper, we ad(cid:0)d(cid:1)ress the following question Question 1.1. For what pairs (d,l) does the generic degree d plane curve contain a star configuration X(l) ⊆ P2? The name star configuration was suggested by A.V. Geramita since X(5) is the ten points of intersections when drawing a star using five lines (see Figure 6.2.4 in [6]). The configurations X(l) appeared in the work of Geramita, Migliore, and Sabourin [8] as the 2000 Mathematics Subject Classification. 14M05,14H50. Key words and phrases. star configurations, generic plane curves. 1 2 E. CARLINIAND A.VAN TUYL support of a set of double points whose Hilbert function exhibited an extremal behaviour. More recently, Cooper, Harbourne, and Teitler [6] computed the Hilbert function of any homogeneous set of fatpoints supported on X(l). Bocci and Harbourne [1] used star configurations (and their generalizations) to compare the symbolic and regular powers of an ideal. Further properties of star configurations continue to be uncovered; e.g., ongoing work of Geramita, Harbourne, Migliore [7]. We can answer Question 1.1 because we can exploit the rich algebraic structure of star configurations. In particular, we will require the fact that one can easily write down a list of generators for IX(l), the defining ideal of X(l), as well as the fact that the Hilbert function of X(l) is the same as the Hilbert function of l generic points in P2. Using 2 these properties, among others, we give the following solu(cid:0)t(cid:1)ion to Question 1.1: Theorem. 6.3 Let l ≥ 2. Then the generic degree d plane curve contains a star configu- ration X(l) if and only if (i) l = 2 and d ≥ 1, or (ii) l = 3 and d ≥ 2, or (iii) l = 4 and d ≥ 3, or (iv) l = 5 and d ≥ 5. Our proof is broken down into a number of cases. It will follow directly from the generators of I that Question 1.1 can have no solution for d < l −1. Using a simple X(l) dimension counting argument, Theorem 3.1 shows that for l ≥ 6, there is no solution to Question 1.1. The cases l = 2 and l = 3 are trivial cases, so the bulk of the paper will be devoted to the cases that l = 4 and l = 5. To prove these cases, we rephrase Question 1.1 into a purely ideal theoretic question (see Lemma 4.3). Precisely, we construct a new ideal I fromthe generatorsof I andthe X(l) linear forms defining the lines ℓ ,...,ℓ . We then show that Question 1.1 is equivalent to 1 l determining whether I = (C[x,y,z]) . We then answer this new algebraic reformulation. d d The proofs for the cases (d,l) = (3,4) and (4,5) are, we believe, especially interesting. To prove that Question 1.1 is true for (3,4), we exploit the natural group structure of the plane cubic curve to find a X(4) on the curve. To show the non-existence of a solution for (4,5), we require the classical theory of Lu¨roth quartics. Lu¨roth quartics are the plane quartics that pass through a X(5). Lu¨roth quartics have the property of forming a hypersurface in the space of plane quartics. The existence of this hypersurface is the obstruction for the existence of a solution when (d,l) = (4,5). Our paper is structured as follows. In Section 2, we describe the needed algebraic propertiesofstar configurations. InSection3, wegive someasymptotic results. InSection 4, in preparation for the last two sections, we rephrase Question 1.1 into an equivalent algebraic problem. Sections 5 and 6 deal with the cases l = 4 and l = 5, respectively. Acknowledgements This paper began when the first author visited the second at Lake- head University. Theorem 6.3 was inspired by computer experiments using CoCoA [5]. The first author was partially supported by the Giovani Ricercatori grant 2008 of the Politecnico di Torino. Both authors acknowledge the financial support of NSERC. STAR CONFIGURATION POINTS AND GENERIC PLANE CURVES 3 2. Star Configurations Throughout this paper, we set S = C[x,y,z] and we denote its dth homogeneous piece with S . Moreover, we fix standard monomial bases in each S in such a way that PS will d d d be identified with PNd where N = d+2 −1. We recall the relevant definitions concerning d 2 star configurations of points in P2.(cid:0) (cid:1) Let l ≥ 2 be an integer. The scheme X(l) ⊂ P2 is said to be a star configuration if it consists of l distinct points which are the pairwise intersections of l distinct lines, say 2 ℓ1,...,ℓl, w(cid:0)he(cid:1)re no three lines pass through the same point. We will also call X(l) a star configuration set of points. Note that when l = 2, then X(2) consists of a single point. When l = 3, then X(3) is any set of three points in P2, provided the three points do not lie on the same line. It is clear that any degree d ≥ 1 plane curve contains a point. Furthermore, when d ≥ 2, the generic degree d plane curve will contain three points not lying on a line. These remarks take care of the trivial cases of Theorem 6.3, which we summarize as a lemma: Lemma 2.1. The generic degree d plane curve contains a star configuration X(l) with l = 2, respectively l = 3, if and only if d ≥ 1, respectively d ≥ 2. Given a X(l), for each i = 1,...,l, we let L denote a linear form in S defining the line i 1 ℓ . The defining ideal of I is then given by i X(l) I = (L ,L ). X(l) i j \i6=j We will sometimes write a point of X(l) as p , where p is the point defined by the ideal i,j i,j (L ,L ), i.e., p is the point of intersection of the lines ℓ and ℓ . The following lemma i j i,j i j describes the generators of I ; we will exploit this fact throughout the paper. X(l) Lemma 2.2. Let l ≥ 2, and let X(l) denote the star configuration constructed from the lines ℓ ,...,ℓ . If L is a linear form defining ℓ for i = 1,...,l, then 1 l i i I = (Lˆ ,...,Lˆ ) where Lˆ = L . X(l) 1 l i j Yj6=i For a proof of this fact, see [2, Claim in Proposition 3.4]. From this description of the generators, we see that I is generated in degree l − 1. Thus, for d < l −1, there are X(l) no plane curves of degree d that contain a X(l) since (IX(l))d = (0). Hence, as a corollary, we get some partial information about Question 1.1. Corollary 2.3. If d < l −1, then the generic degree d curve does not contain a X(l). It is useful to recall that star configuration points have the same Hilbert function as generic points. More precisely, by [8, Lemma 7.8], we have Lemma 2.4. Let HF(X(l),t) = dimC(S/IX(l))t denote the Hilbert function of S/IX(l). Then t+2 l HF(X(l),t) = min , for all t ≥ 0. (cid:26)(cid:18) 2 (cid:19) (cid:18)2(cid:19)(cid:27) 4 E. CARLINIAND A.VAN TUYL 3. An asymptotic result In [3] it is shown that the generic degree d plane curve contains a 0-dimensional com- plete intersection scheme of type (a,b) whenever a,b ≤ d (actually the result holds for any degree d plane curve). Thus, arbitrarily large complete intersections can be found on generic plane curves of degree high enough. But the same does not hold for star configurations as shown by Theorem 3.1. UsingthepreviousdescriptionofI weintroduceaquasi-projectivevarietyparametriz- X(l) ing star configurations. Namely, we consider D ⊂ Pˇ2 ×···×Pˇ2 l l | {z } such that (ℓ ,...,ℓ ) ∈ D if and only if no three of the lines ℓ are passing through the 1 l l i same point; here, Pˇ2 denotes the dual projective space. Theorem 3.1. If l > 5 is an integer, then the generic degree d plane curve does not contain any star configuration X(l). Proof. It is enough to consider the case d ≥ l−1 (see Corollary 2.3). Let PS be the space d parametrizing degree d planes curves and define the following incidence correspondence Σ = {(C,X(l)) : C ⊃ X(l)} ⊂ PS ×D . d,l d l We also consider the natural projection maps ψ : Σ −→ D and φ : Σ −→ PS . d,l d,l l d,l d,l d Clearly we have that φ is dominant if and only if Question 1.1 has anaffirmative answer. d,l Using a standard fiber dimension argument, we see that dimΣ ≤ dimD +dimψ−1(X(l)) d,l l d,l for a generic star configuration X(l), where dimD = 2l and by Lemma 2.4 l d+2 l dimψd−,l1(X(l)) = dimC IX(l) d −1 = (cid:18) 2 (cid:19)−(cid:18)2(cid:19)−1. (cid:0) (cid:1) Hence the map φ is dominant only if dimΣ −dimPS ≥ 0 and this is equivalent to d,l d,l d l l(5−l) 2l− = ≥ 0. (cid:18)2(cid:19) 2 (cid:3) Thus the result is proved. By Lemma 2.1 and the above result, we only need to treat the cases l = 4 and 5. We postpone these cases to first rephrase Question 1.1 into an equivalent algebraic question. STAR CONFIGURATION POINTS AND GENERIC PLANE CURVES 5 4. Restatement of Question 1.1 We derive some technical results, moving our Question 1.1 back and forth between questions in algebra and questions in geometry. First, we notice the following trivial fact: Lemma 4.1. Let {F = 0} be an equation of the degree d curve C ⊂ P2. Then C contains a star configuration X(l) only if l ˆ F = M L i i Xi=1 where the forms M have degree d −l + 1 and the forms Lˆ are defined as Lˆ = L i i i j6=i i for some linear forms L1,...,Ll. Q Hence, it is natural to perform the following geometric construction. We define a map of affine varieties Φ : S ×···×S ×S ×···×S −→ S d,l 1 1 d−l+1 d−l+1 d l l such that | {z } | {z } l Φ (L ,...,L ,M ,...,M ) = M Lˆ d,l 1 l 1 l i i Xi=1 We then rephrase our problem in terms of the map Φ : d,l Lemma 4.2. Let d,l be nonnegative integers with d ≥ l − 1. Then the following are equivalent: (i) Question 1.1 has an affirmative answer for d and l; (ii) the map Φ is a dominant map. d,l Proof. Lemma 4.1 proves that (i) implies (ii). To prove the other direction, it is enough to show that for a generic formF, the fiber Φ−1(F) contains a set of l linear forms defining d,l a star configuration. More precisely, define ∆ ⊂ S ×···×S ×S ×···×S as 1 1 d−l+1 d−l+1 follows: there exists a 6= b 6= c such that ∆ = (L ,...,L ,M ,...,M ) . (cid:26) 1 l 1 l (cid:12) La,Lb,Lc are linearly dependent (cid:27) (cid:12) (cid:12) Then we want to show that Φ−1(F) 6⊂ ∆.(cid:12) d,l We proceed by contradiction, assuming that the generic fiber of Φ is contained in ∆. d,l Then ∆ would be a component of the domain of Φ . Thus a contradiction as the latter d,l (cid:3) is an irreducible variety being the product of irreducible varieties. Using the map Φ we can now translate Question 1.1 into an ideal theoretic question. d,l Lemma 4.3. Let d,l be non-negative integers. Consider generic forms L ,...,L ∈ S 1 l 1 and M ,...,M ∈ S . Set 1 l d−l+1 Lˆ = L and Lˆ = L , for i 6= j. i j i,j h Yj6=i h6=Yi,h6=j 6 E. CARLINIAND A.VAN TUYL Also set Q = M Lˆ +···+M Lˆ = M Lˆ , 1 2 1,2 l 1,l i 1,i Xi6=1 . . . Q = M Lˆ +···+M Lˆ = M Lˆ . l 1 l,1 l−1 l,l−1 i l,i Xi6=l With this notation, form the ideal ˆ ˆ I = (L ,...,L ,Q ,...,Q ) ⊂ S. 1 l 1 l Then the following are equivalent: (i) Question 1.1 has an affirmative answer for d and l; (ii) I = S . d d Proof. Using Lemma 4.2 we just have to show that Φ is a dominant map. In order to do d,l this we will determine the tangent space to the image of Φ in a generic point q = Φ (p), d,l d,l where p = (L ,...,L ,M ,...,M ) and we denote with T this affine tangent space. 1 l 1 l q The elements of the tangent space T are obtained as q d Φ (L +tL′,...,L +tL′,M +tM′,...,M +tM′) dt(cid:12) d,l 1 1 l l 1 1 l l (cid:12)t=0 when we vary the(cid:12)forms L′ ∈ S and M′ ∈ S . By a direct computation we see that (cid:12) i 1 i d−l+1 the elements of T have the form q M′Lˆ +···+M′Lˆ +L′(M Lˆ +···+M Lˆ )+···+ 1 1 l l 1 2 1,2 l 1,l +L′(M Lˆ +···+M Lˆ )+···+L′(M Lˆ +···+M Lˆ ), j 1 1,j l j,l l 1 1,2 l−1 1,l−1 where Lˆ = L and Lˆ = L , for i 6= j. i j6=i i i,j h6=i,h6=j h Since theQL′ ∈ S and M′ ∈ SQ can be chosen freely, we obtain that I = T . (cid:3) i 1 i d−l+1 d q Remark 4.4. Lemma 4.3 is an effective tool to give a positive answer to each special issue of Question 1.1. Given d and l we construct the ideal I as described by choosing forms Li and Mi. Then we compute dimCId using a computer algebra system, e.g., CoCoA. If dimCId = dimCSd, by upper semi-continuity of the dimension, we have proved that our question has an affirmative answer for these given d and l. In fact, the dimension can decrease only on a proper closed subset. On the contrary, if dimCId < dimCSd we do not have any proof. Question 1.1 can both have a negative (for any choice of forms the inequality holds) or an affirmative answer (our choice of forms is too special and another choice will give an equality). But, if we choose our forms generic enough, we do have a strong indication that the answer to Question 1.1 should be negative. Example 4.5. We use CoCoA [5] to consider the example (d,l) = (4,4). --we define the ring we want to use Use S::=QQ[x,y,z]; --we choose our forms L1:=x; STAR CONFIGURATION POINTS AND GENERIC PLANE CURVES 7 L2:=y; L3:=z; L4:=x+y+z; M1:=x+y-z; M2:=-x+2y+2z; M3:=2x-y-z; M4:=x+y+2z; --we build the forms Qi Q1:=M2*L3*L4+M3*L2*L4+M4*L2*L3; Q2:=M1*L3*L4+M3*L1*L4+M4*L1*L3; Q3:=M1*L3*L4+M2*L1*L4+M4*L1*L2; Q4:=M1*L2*L3+M2*L1*L3+M3*L1*L2; --we define the ideal I I:=Ideal(L2*L3*L4,L1*L3*L4,L1*L2*L4,L1*L2*L3,Q1,Q2,Q3,Q4); --we compute the Hilbert function of S/I in degree 4 Hilbert(S/I,4); Evaluating the code above we get Hilbert(S/I,4); 0 ------------------------------- and this means that dimCS4 − dimCI4 = 0. Hence we showed that the generic plane quartic contains a X(4). 5. The l = 4 case In this section we consider Question 1.1 when l = 4. We begin with a special instance of Question 1.1, that is, when d = 3, since it has a nice geometric proof which takes advantage of the group structure on the curve. Lemma 5.1. Let C ⊂ P2 be a smooth cubic curve. Then there exists a star configuration X(4) ⊂ C. Proof. We will use the group law on C and hence we fix a point p ∈ C serving as identity. 0 Then we choose a point p ∈ C such that 2p = p and a generic point p ∈ C. Consider 2 2 0 1 the line joining p and p and let p be the third intersection point with C. Joining p 1 2 3 3 and p and taking the third intersection we get the point p +p . Now join p +p and 0 1 2 1 2 p and let p be the third intersection point. Then joining p and p we get p +2p = p . 2 4 4 0 1 2 1 Hence, the points p ,p ,p ,p +p ,p ,p ∈ C are a star configuration. (cid:3) 0 1 2 1 2 3 4 We now consider the general situation: Theorem 5.2. Let C ⊂ P2 be a generic degree d ≥ 3 plane curve. Then there exists a star configuration X(4) ⊂ C. Proof. The case d = 3 is treated in Lemma 5.1, while for d > 3 we will use Lemma 4.3 and the notation introduced therein. For d = 4, we produced an explicit example in Example 4.5 where I = S , and this is enough to conclude by semi-continuity. 4 4 8 E. CARLINIAND A.VAN TUYL To deal with the general case d ≥ 5, we use the structure of the coordinate ring of a star configuration. Given four generic linear forms L ,L ,L ,L we consider the star 1 2 3 4 configuration they define, say X = {p : 1 ≤ i < j ≤ 4}. The coordinate ring of X is i,j S A = . Lˆ ,...,Lˆ 1 4 (cid:16) (cid:17) When d ≥ 2, dimCAd = 6. To prove that Id = Sd, we want to find linear forms N ,...,N such that the six forms N Q are linearly independent in A . To check whether 1 6 i i 4 elements in A are linearly independent it is enough to consider their evaluations at the points p . Consider the following evaluation matrix i,j Q Q Q Q 1 2 3 4 p M L L M L L 0 0 1,2 2 3 4 1 3 4 p M L L 0 M L L 0 (5.1) 1,3 3 2 4 1 2 4 p M L L 0 0 M L L 1,4 4 2 3 1 2 3 p 0 M L L M L L 0 2,3 3 1 4 2 1 4 p 0 M L L 0 M L L 2,4 4 1 3 2 1 3 p 0 0 M L L M L L 3,4 4 1 2 3 1 2 obtained by evaluating each Q at the points of the configuration where we denote, by i abuse of notation, M L L (p ) with M L L . For example, since Q = M L L + h i j m,n h i j 2 1 3 4 M L L +M L L , Q (p ) = M L L (p ) because L vanishes at p , while Q (p ) = 3 1 4 4 1 3 2 2,3 3 1 4 2,3 3 2,3 2 1,4 0 since L and L vanish at p . Now choose the forms M with degM = d−3 ≥ 2 so 4 1 1,4 i i that M (p ) = M (p ) = M (P ) = 0 2 2,3 3 3,4 4 2,4 and no other vanishing occurs at the points of the star configuration. Notice that this is possible as dimCAd = 6 for d ≥ 2. The matrix (5.1) can be represented as Q Q Q Q 1 2 3 4 p ∗ ∗ 0 0 1,2 p ∗ 0 ∗ 0 (5.2) 1,3 p ∗ 0 0 ∗ 1,4 p 0 ∗ 0 0 2,3 p 0 0 0 ∗ 2,4 p 0 0 ∗ 0 3,4 where ∗ denotes a non-zero scalar. As this matrix has rank four, then the forms Q are i linearly independent in A. To represent N Q in A it is enough to multiply the j-th element of the i-th column of i i (5.1) by the evaluation of N at p . Hence, if we choose the forms N such that they do i j,i i not vanish at any point of X, the evaluation matrix of the N Q has the same non-zero i i pattern as (5.2). Thus the forms N Q ,...,N Q are linearly independent in A. 1 1 4 4 STAR CONFIGURATION POINTS AND GENERIC PLANE CURVES 9 To complete the proof we need two more forms, and we choose L Q and L Q whose 1 3 1 2 evaluation matrix at the points p is i,j p p p p p p 1,2 1,3 1,4 2,3 2,4 3,4 (5.3) L Q 0 0 0 0 0 ∗ 1 3 L Q 0 0 0 ∗ 0 0 1 2 (cid:3) These rows are linearly independent with the columns of (5.2), completing the proof. 6. The l = 5 case It remains to consider the case that l = 5. The case l = 5 and d = 4 was classically studied. A quartic containing a star configuration of ten points, that is, a X(5). For more details, see [12, 13], and for a modern treatment we refer to [14]. In particular, we require the following property of Lu¨roth quartics: Theorem (Theorem 11.4 of [14]). Lu¨roth quartics form a hypersurface of degree 54 in the space of plane quartics. Corollary 6.1. Question 1.1 has a negative answer for (d,l) = (4,5). Proof. We notice that the previous theorem is enough to give a negative answer to our question for l = 5 and d = 4. In fact, a generic plane quartic is not a Lu¨roth quartic, and hence, no star configuration X(5) can be found on it. (cid:3) For the remaining values d > 4, we give an affirmative answer to Question 1.1: Theorem 6.2. The generic degree d > 4 plane curve contains a star configuration X(5). Proof. For the case d = 5 we produce an explicit example and then we conclude by semi-continuity. For the general case, we produce a proof similar to Theorem 5.2. For the d = 5 case, let our linear forms be given by L = x, L = y,L = z, L = x+y +z, and L = 2x−3y +5z. 1 2 3 4 5 To construct the polynomials Q ,...,Q , we make use of the following forms: 1 5 M = x+y−z, M = −x+2y+2z, M = 2x−y−z, M = 3x+y−z, andM = 4x−4y+3z. 1 2 3 4 5 Form the ideal I as in Lemma 4.3. Using CoCoA [5] to compute I , we find that I = S . 5 5 5 For the d ≥ 6 case, we use the notation of Lemma 4.3, and fix linear forms L ,...,L , 1 5 and use p to denote the ten points of X(5) defined by the linear forms L . We construct i,j i the evaluation table 10 E. CARLINIAND A.VAN TUYL Q Q Q Q Q 1 2 3 4 5 p M L L L M L L L 0 0 0 1,2 2 3 4 5 1 3 4 5 p M L L L 0 M L L L 0 0 1,3 3 2 4 5 1 2 4 5 p M L L L 0 0 M L L L 0 1,4 4 2 3 5 1 2 3 5 p M L L L 0 0 0 M L L L (6.1) 1,5 5 2 3 4 1 2 3 4 p 0 M L L L M L L L 0 0 2,3 3 1 4 5 2 1 4 5 p 0 M L L L 0 M L L L 0 2,4 4 1 3 5 2 1 3 5 p 0 M L L L 0 0 M L L L 2,5 5 1 3 4 2 1 3 4 p 0 0 M L L L M L L L 0 3,4 4 1 2 5 3 1 2 5 p 0 0 M L L L 0 M L L L 3,5 5 1 2 4 3 1 2 4 p 0 0 0 M L L L M L L L 4,5 5 1 2 3 4 1 2 3 obtained by evaluating each Q at the points of the configuration where we denote, by i abuse of notation, M L L L (p ) with M L L L . h i j r m,n h i j r We work in the coordinate ring of the star configuration A = S . Now consider (Lˆ1,...,Lˆ5) (6.2) L Q ,L Q ,L Q ,L Q ,L Q ,L Q ,L Q ,L Q ,L Q ,L Q 5 1 2 1 1 2 3 2 2 3 4 3 3 4 5 4 4 5 1 5 and we want to show that they are linearly independent in A for a generic choice of the forms M , degM = d−4. Again, it is enough to show this for a special choice of forms. i i We choose the forms M in such a way that i M (p ) = M (p ) = M (p ) = 0, 1 1,5 4 3,4 5 4,5 M (p ) = M (p ) = M (p ) = M (p ) = 0, 2 1,2 2 2,5 3 1,3 3 2,3 and the following are non-zero M (p ),M (p ),M (p ), M (p ),M (p ), M (p ),M (p ), 1 1,2 1 1,3 1 1,4 2 2,3 2 2,4 3 3,4 1 3,5 M (p ),M (p ),M (p ), M (p ),M (p ),M (p ). 4 1,4 4 2,4 4 4,5 5 1,5 5 2,5 5 3,5 NoticethattheseconditionscanbesatisfiedbecausedegMi ≥ 2anddimCA2 = 6,dimCAe = 10 for e ≥ 3. Then, evaluating the forms in (6.2) at the points p , we obtain the matrix i,j L Q L Q L Q L Q L Q L Q L Q L Q L Q L Q 5 1 2 1 1 2 3 2 2 3 4 3 3 4 5 4 4 5 1 5 p 0 0 0 ∗ 0 0 0 0 0 0 1,2 p 0 0 0 0 ∗ ∗ 0 0 0 0 1,3 p ∗ ∗ 0 0 0 0 ∗ ∗ 0 0 1,4 p 0 ∗ 0 0 0 0 0 0 0 0 (6.3) 1,5 p 0 0 0 0 0 ∗ 0 0 0 0 2,3 p 0 0 ∗ ∗ 0 0 ∗ ∗ 0 0 2,4 p 0 0 ∗ ∗ 0 0 0 0 0 0 2,5 p 0 0 0 0 0 0 0 ∗ 0 0 3,4 p 0 0 0 0 ∗ ∗ 0 0 ∗ ∗ 3,5 p 0 0 0 0 0 0 0 0 0 ∗ 4,5