Secant varieties. Marin Petkovic. November 23, 2015

Transcription

1 Secant varieties Marin Petkovic November 23, 2015 Abstract The goal of this talk is to introduce secant varieies and show connections of secant varieties of Veronese variety to the Waring problem. 1 Secant varieties Let X P n be a variety over k, where k is algebraically closed field of characteristic 0 (or just assume k = C). Definition 1.1. For s 1 the s-th higher secant variety of X is σ s (X) = P 1,...,P s X P 1,..., P s. Remark 1.2. Although in some cases closure is not necessary, for twisted cubic in P 3 the union of secants is not closed. Lemma 1.3. If σ i (X) = σ i+1 (X), then σ i (X) is a linear subspace of P N. Proof. Let z P 1,...,P i X P 1,..., P i be nonsingular point on σ i (X). Then z, x σ i+1 for each x X. Therefore z, x T z (σ i+1 ) = T z (σ i ). Then σ i T z (σ i ). Since they have the same dimension, σ i = T z (σ i ), i.e. σ i is linear. Proposition 1.4. If X is nondegenerate (not contained in a hyperplane), then there exists r 1 such that X = σ 1 (X) σ 2 (X)... σ r (X) = P N. Now we can ask two natural questions about higher secant varieties. First, what is the minimal r such that σ r (X) = P N? More generally, what is the dimension dim(σ i (X))? Proposition 1.5. dim(σ i (X)) min{in + i 1, N}. 1

2 We will call the number min{in + i 1, N} the expected dimension of σ i (X) and denote it with expdim(σ i (X)). If dim(σ i (X)) expdim(σ i (X)), we say that X is i-defective. Remark 1.6. If X is i-defective and σ i (X) P N, then X is j-defective, for all j i, because of dim(σ i+1 (X)) dim(σ i (X)) + n + 1. Example 1.7 (Veronese surface). Let X be the Veronese surface, that is the image of the map v 2 : P 2 P 5 (x : y : z) (x 2 : xy : xz : y 2 : yz : z 2 ). The expected dimension of σ 2 (X) is 5, but we will show that σ 2 (X) P 5, so X is 2-defective. To prove this, we identify elements of P 5 with 3 3 symmetric matrices. Notice that the elements of X are symmetric matrices of rank 1. Thus σ 2 (X) consists of all linear combinations of 2 rank 1 symmetric matrices, thus of rank 2. Hence σ 2 (X) P 5, i.e. X is 2-defective. Lemma 1.8 (Terracini s lemma). Let P 1,..., P s X be general points and P P 1,..., P s σ s (X) be a general point. Then the tangent space to σ s (X) in P is T P (σ s (X)) = T P1 (X),..., T Ps (X). Example 1.9 (Twisted cubic curve). Let X P 3 be a twisted cubic curve, that is the image of the map v 3 : P 1 P 3 defined with v 3 (x : y) = (x 3 : x 2 y : xy 2 : y 3 ). Then dim(σ 2 (X)) = dim(t P (σ 2 (X))) for generic point P. lemma, we have T P (σ 2 (X)) = T P1 (X), T P2 (X) By Terracini s for some points P 1, P 2 X. Since in general case T P1 (X), T P2 (X) do not intersect, they span 3-dimensional space, that is P 3. 2 Veronese varieties and the Waring problem Definition 2.1. The d-th Veronese map v d : P n P N, where N = ( ) n+d d 1, is defined with v d (x 0 :... : x n ) = (x i0 0 xi1 1 xin n i i n = d) The image of v d is called the Veronese variety. Remark 2.2. Veronese map can also be defined as a map taking linear form L to L d. Taking appropriate basis and identifying P 1 with the space of linear forms this map corresponds to v d in the above definition. We will denote the point corresponding to F with [F ]. Notice that [F ] P N is in X if and only if F = L d for some linear form L. 2

3 Proposition 2.3. The tangent space to the Veronese variety at the point [L d ] is [L d 1 M] ; M linear form. Example 2.4. If n = 1, we call the Veronese varieties rational normal curves. The curve v 2 (P 1 ) P 2 is just the irreducible conic, so σ 2 (v 2 (P 1 ) = P 2. The curve v 3 (P 1 ) P 3 is the twisted cubic curve. The higher secant varieties of the Veronese variety are connected with the Waring problem for forms, which we will introduce now. Definition 2.5. Let F be a degree s homogeneous form. The Waring rank of F is the minimum s such that F = L d L d s for some linear forms L i. We denote the Waring rank of F with rk(f ). The big Waring problem: Find the minimal integer g(n, d) such that rk(f ) g(n, d), for a generic homogeneous form F of degree d in n + 1 variables. The little Waring problem: Find the minimal integer G(n, d) such that rk(f ) G(n, d), for all homogeneous forms F of degree d in n + 1 variables. Remark 2.6. It is clear that solving the big Waring problem is equivalent to finding minimal s such that σ s (X) = P N for Veronese variety X = v d (P n ) P N. This is not enough to solve the little Waring problem, because of the closure in the definition of secant varieties. Example 2.7. Let X = v 2 (P 1 ) P 2 be an irreducible conic. Then σ 2 (X) = P 2, so g(1, 2) = 2. Moreover, since in this case the closure is not necessary in the definition of σ 2 (X), we also get G(1, 2) = 2. Example 2.8. Let X = v 3 (P 1 ) P 3 be the twisted cubic curve. Then σ 2 (X) = P 3 so g(1, 3) = 2. But in this case, there are forms with rank 3. We can explain this geometrically. Let [F ] be a point in P 3 corresponding to F. Let π be the projection of X from the point [F ]. Then the image is the plane cubic curve which has to be singular. If the singularity is a node, then [F ] lies in on a secant of X, so F = L L 3 2. But if singularity is a cusp, then [F ] lies on a tangent line of X, and not on a secant, and rk(f ) = 3. Let P 1,..., P s X = P n be distinct points. Then each hyperplane H containing all tangent spaces T vd (P i)(x) corresponds to a degree d hypersurface in P n that is singular in points P 1,..., P s. Such hypersurfaces are defined by a homogeneous degree d form contained in p p 2 s, where p i is the ideal of the point P i. Thus if (p p 2 s) d is not empty, σ s (X) P N. Moreover, since two different hypersurfaces give two different hyperplanes, we have the following result: 3

4 Lemma 2.9. Let P 1,..., P s P n be generic points. Then dim s (σ(x)) = N dim(p p 2 s) d. From the expected value for dim(σ s (X)) we can easily calculate the expected value for g(n, d) to be ) ( n+d d n + 1 Theorem 2.10 (Alexander and Hirtschowitz). Let F be a generic degree d form in n + 1 variables. Then ) except in the following cases: d = 2, where rk(f ) = n + 1 rk(f ) = d = 4, n = 2, where rk(f ) = 6 d = 4, n = 3, where rk(f ) = 10 d = 3, n = 4, where rk(f ) = 8 d = 4, n = 4, where rk(f ) = 15.. ( n+d d n + 1 It is easy to explain the case d = 2 using the symmetric matrices. Indeed, a general n + 1 n + 1 matrix is regular, that is of rank n + 1. Hence it is a sum of n + 1 matrices of rank 1. Therefore the rank of associated form is n + 1. Example 2.11 (case (d = 4, n = 2)). Let X = v 4 (P 2 ) P 14. We prove that σ 5 (X) P 14. By the lemma and the discussion before, it is enough to show that there exists a degree 4 hypersurface through 5 generic points that is singular in those 5 points. Example 2.12 (case d = 4, n = 3, 4). To show that g(4, 3) > 9 and g(4, 4) > 14 it is sufficient to find a quadric hypersurface in P 3 through 9 given points, and a quadric hypersurface in P 4 through 14 points. 3 Apolarity Let S = k[x 1,..., x n ] and T = k[y 1,..., y n ]. We define the action of T on S with y i.x j = x j x i and extend it to T. For a graded ring R, we denote with R d the d-th graded subring. 4

5 Definition 3.1. For F S homogenous we define the anihilator with F = { T F = 0}. Definition 3.2. Let V, W be finite dimensional k-vector spaces and B : V W k a bilinear map. We say that B is a perfect pairing if the induced maps V W and W V are isomorphisms. Proposition 3.3. The bilinear map S d T d k defined with (F, ) F is a perfect pairing. Definition 3.4. For an ideal I T we define the Hilbert function of T/I with for t 0. HF (T/I, t) = dim(t/i) t, In the case I = F we will use the notation H F (t) instead HF (T/F, t). Let m be the maximal ideal (y 1,..., y n ) T. We define the sockle of the ring (T/F ) with Soc(T/F ) = {x T/F xm = 0}. It is an ideal in T/F and it is easy to check { dim Soc(T/F 0 if i d ) i = 1 if i = d Proposition 3.5. Hilbert function of F for F of degree d vanishes for t > d, H F (0) = H F (d) = 1, and is symmetric with respect to d+1 2. Proof. The first part is obvious since F = 0 for each of degree > d. From the perfect pairing property it follows that not all T d vanish in F, so H F (d) = 1. It is obvious that H F (0) = 1. Let R = T/F ). To prove that H F is symmetric, it is enough to prove that the multiplication R i R d i R d = k is perfect pairing. Let x R i such that xy = 0 for each y R d i. We will prove that x is in Soc(R). If y R l for 0 < l < d i, using downward induction we have xym = 0, that is xy Soc(R). Since deg(xy) < d this is a contradiction. Thus y Soc(R), so y = 0. Example 3.6. Let F k[x 0, x 1 ] be homogenous cubic. We will compute all the possible Waring ranks of F. The Hilbert function of F is determined by the value for t = 1, so there is only two possibilities: H F (1) = 1 or H F (1) = 2. Notice that since (F ) 1 = 0 rank of F cannot be 1. Assume the first case. Then (F ) 1 = 1. The space of all linear forms L such that 1 L = 0 is not S 1 so it is one dimensional, that is equal to L 1 for some L 1. Thus we can take L 0 such that 1 L 0 = 1, and S 1 = L 0, L 1. Then there exists a polynomial G such that G(L 0, L 1 ) = al bl 2 0L 1 + cl 0 L dl 3 1 = F (x 0, x 1 ). 5

6 Since 1 (L 0 ) = 1 and 1 (L 1 ) = 0 we get 0 = 1 G(L 0, L 1 ) = 3aL bL 0 L 1 + cl 2 1, so F (x 0, x 1 ) = dl 3 1, and rk(f ) = 1. In the second case, we have (F ) 2 = Q. Then Q =, where, T 1. Suppose that, are linearly independent. We can find a basis L, L for S 1 such that L = L = 1 and L = L = 0. Then we take polynomial G as before and Show that G(L, L ) al 3 dl 3 is a zero polynomial ( G = 0 for all T 3 ). Thus rk(f ) = 2. Suppose now that Q = 2 and that F = N 3 + M 3, that is rk(f ) = 2. Then we can take N, M such that M M = N N = 1 and M N = N M = 0. Then M N F and that is a contradiction since (F ) 2 = (Q). Notice that in the previous example for the forms of rank 1, F ( ), and ( ) is an ideal of a point in P 1. Similar, in the case of rank 2, F ( ) which is an ideal of two points in P 1. Since (F ) 1 = 0 there is no ideals of one point in F. Also, in the case of rank 3, there are no ideals of one or two points in F, but there are ideals of three points. Lemma 3.7 (Apolarity Lemma). F = L d L d s F contains an ideal of s distinct points. Remark 3.8. If I F, then obviously HF (T/I, t) H F (t). Example 3.9. For F = x 0 x 2 1 we see that F = ( 2 0, 3 1), so it does not contain an ideal of one or two points. Therefore rk(f ) = 3 (( ) is an ideal of 3 points). More generally, we can conclude that rk(x 0 x d 1) = d + 1. Example 3.10 (defective cases for Alexander Hirowitz theorem). Let F be a generic form of degree 4, n = 2. The Hilbert function is given with H F (1) = 3, H F (2) = 6, that is, no first or second order differentials vanish in F. If I is an ideal of five points, we have HF (T/I, 2) = 5 (i.e. I contains a quadric), so F contains no ideals of five points. Similarly, in the case n = 3 we have H F (2) = ( ) = 10 > 9, and HF (T/I, 2) = 9 for and ideal I of nine points. Thus I F. In the case n = 4 we have H F (2) = ( ) = 15 > 14 so F does not contain an ideal of 14 points. References [1] Enrico Carlini, Nathan Grieve, and Luke Oeding, Four Lectures on Secant Varieties, Volume 76 of the series Springer Proceedings in Mathematics & Statistics pp ,