A Computational Algebra approach to Intersection Theory and Enumerative Geometry

Transcription

1 A Computational Algebra approach to Intersection Theory and Enumerative Geometry TU Kaiserslautern Summer School in Algorithmic Mathematics Munich, August 2012

2 Outline Linear supspaces on hypersurfaces 1 Linear supspaces on hypersurfaces 2 3

3 Grassmannians and Fano schemes Algorithm and implementation Linear supspaces on hypersurfaces Question Given a general hypersurface X of degree d in P n. How many linear subspaces of dimension k in P n does X contain? In most of the cases, the answers are infinite. However, if d, k, n N satisfy ( ) d + k = (k + 1)(n k), k then the answers should be finite.

4 Example Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation Figure: There are 27 lines on a general cubic hypersurface in P 3.

5 Grassmannians and Fano schemes Algorithm and implementation Grassmannians and Fano schemes Definition For each k = 1, 2,..., n 1, the Grassmannian G(k, n) is the set of all vector subspaces V C n of dimension k. G(1, n + 1) = P n. G(2, n + 1) = {lines in P n }. G(k + 1, n + 1) = {k-planes in P n }. Definition Let X P n be a hypersurface of degree d. Let F k (X ) be the set of linear subspaces of dimension k on X. Then F k (X ) G(k + 1, n + 1) is called the k-th Fano scheme of X.

6 Expected dimension of F k (X ) Grassmannians and Fano schemes Algorithm and implementation We denote φ(k, d, n) = (k + 1)(n k) ( ) d+k. 1 F k (X ) is empty, if φ(k, d, n) < O. 2 F k (X ) is smooth of expected dimension φ(k, d, n), otherwise. Therefore, if φ(k, d, n) = 0, then F k (X ) is zero-dimensional and the number of linear subspaces of dimension k on X is equal to the degree of F k (X ). d

7 Degree of F k (X ) Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation Theorem Let Sym d S be the d-th symmetric power of the dual of tautological subbundle S on G(k + 1, n + 1). Then [F k (X )] = c ( d+k d ) (Symd S ). Here we denoted c i (E) the i-th Chern class of vector bundle E. This implies that the degree of F k (X ) is equal to the degree of the top Chern class of Sym d S.

8 Algorithm Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation Input: d, k, n N. Output: The number of k-planes on a general hypersurface of degree d in P n. 1 Construct G(k + 1, n + 1) with tautological subbundle S. 2 Compute Sym d S and its top Chern class. 3 Compute the integral of top Chern class.

9 Implementation Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation Old: New: Maple - Schubert : S. Katz and S. A. Stromme. Macaulay2 - Schubert2: D. R. Grayson, M. E. Stillman, S. A. Stromme, D. Eisenbud and C. Crissman. Sage. Singular. Our package is Schubert3 which developed on Sage 5.0 and written by Python programming language. This package supports computation in Intersection Theory on smooth varieties. It deals with abstract varieties, vector bundles on abstract varieties, and morphisms between abstract varieties.

10 Implementation Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation A variety will be given by dimenion, variables, degrees, and relations such that we can compute its Chow ring. Sometime its relations are not known, but we must know its monomial values. This helps us return the degree of cycle classes (integration). A vector bundle on a variety is given by its rank and total Chern class or its Chern character. A morphism between varieties is given by its pullback which is a homomorphism of the Chow rings corresponding to these varieties.

11 Implementation Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation class Variety(SageObject): def variables(self): def chow_ring(self): def monomial_values(): def integral(self): class Grassmannian(Variety): def tautological_subbundle(self): class VectorBundle(SageObject): def dual(self): def symmetric_power(self): def top_chern_class(self):

12 Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation How many lines on a general cubic hypersurfaces in P 3? 27 How many lines on a general quintic hypersurfaces in P 4? 2875 More generally, how many lines on a general degree d = 2 n 3 hypersurfaces in P n?

13 Linear supspaces on hypersurfaces Grassmannians and Fano schemes Algorithm and implementation sage: def line(n):...: d = 2*n-3...: G = Grassmannian(2,n+1)...: S = G.tautological_subbundle().dual()...: B = S.symmetric_power(d)...: return G.integral(B.top_chern_class())...: sage: line(3) 27 sage: line(4) 2875

14 Grassmannians and Fano schemes Algorithm and implementation Linear subspaces on hypersurfaces More generally, if k, d, n N satisfy ( ) d+k k = (k + 1)(n k), then the number of the linear subspaces of dimension k on a general hypersurface of degree d in P n is finite. sage: def linear_subspace(k,d,n):...: G = Grassmannian(k+1,n+1)...: S = G.tautological_subbundle().dual()...: B = S.symmetric_power(d)...: return G.integral(B.top_chern_class())...: sage: linear_subspace(3,3,8)

15 Conics on a quintic threefold Conics intersecting 8 lines Conics on a quintic threefold Question Let X be a general quintic threefold in P 4. How many rational curves of degree d does X contain? In case d = 2, there are conics on X. This number was computed by S. Katz in Here is how the computation can be made with Sage. class ProjectiveBundle(Variety): def OO(self): class VectorBundle(SageObject): def tensor(self, arg): def minus(self): def plus(self, arg):

16 Conics on a quintic threefold Conics on a quintic threefold Conics intersecting 8 lines Let G(3, 5) be the Grassmannian of planes in P 4, with tautological subbundle S. The space of conics in P 4 is the projective bundle P((Sym 2 S) ) of lines in the dual of vector bundle Sym 2 S on G(3, 5). sage: G = Grassmannian(3,5) sage: S = G.tautological_subbundle() sage: B = S.symmetric_power(2).dual() sage: P = ProjectiveBundle(B)

17 Conics on a quintic threefold Conics on a quintic threefold Conics intersecting 8 lines The class of conics contained in a general quintic threefold is the top Chern class of vector bundle A = Sym 5 S Sym 3 S O( 1). sage: V = P.OO().tensor(S.symmetric_power(3)) sage: A = V.minus().plus(S.symmetric_power(5)) The number of conics contained in a general quintic threefold is the degree of top Chern class of A. sage: P.integral(A.top_chern_class())

18 Conics on a quintic threefold Conics intersecting 8 lines Conics intersecting 8 lines Question Given 8 lines L 1,..., L 8 P 3 in general position. How many plane conics in P 3 meeting all eight? Let G(3, 4) be the Grassmannian of 2-planes in P 3, with tautological subbundle S. The sapce of conics in P 3 is projective bundle P((Sym 2 S) ) of lines in the dual of vector bundle Sym 2 S on G(3, 4).

19 Conics intersecting 8 lines Conics on a quintic threefold Conics intersecting 8 lines The class of conics meeting a line L in P 3 is D L = 2ω + ζ, where ω denotes the hyperplane class on G and ζ = c 1 (O P((Sym 2 S) )(1)). The number of conics intersecting 8 lines in general position is the intersection number DL. 8 P((Sym 2 S) )

20 Conics intersecting 8 lines Conics on a quintic threefold Conics intersecting 8 lines sage: G = Grassmannian(3,4) sage: S = G.tautological_subbundle() sage: B = S.symmetric_power(2).dual() sage: P = ProjectiveBundle(B) sage: v = P.variables() sage: c = 2*v[6]+v[0]; c S1 + 2*q1 sage: P.integral(c^8) 92

21 Excess intersection Algorithm Implementation C 1 = Z(x 0 x 1 ), C 2 = Z(x 0 x 2 ). C 1 C 2 = {p = (1, 0, 0)} {L = Z(x 0 )}. Since the point p is a transverse intersection so P 2 [p] = 1. Must find somehow to define the intersection number of C 1 and C 2 along L such that P 2 [L] = 3.

22 Excess intersection Algorithm Implementation The excess intersection formula gives us an effective way to compute the degree of connected components where the intersection is not transverse. Let X be a smooth variety. We denote TX is the tangent bundle on X. If Z X be a smooth subvariety, we denote TX Z be the restriction of TX to Z. We claim that TZ is a subbundle of TX Z. Definition We define the normal bundle of Z in X as the quotient N Z/X = (TX Z )/TZ.

23 Excess intersection Algorithm Implementation Let X 1, X 2 be smooth subvarieties of a smooth variety X, Z be a connected component of X 1 X 2 and set m = codim X 1 + codim X 2 codim Z. We write (X 1 X 2 ) Z for the part of X 1 X 2 supported on Z and call it the equivalence class of Z for the intersection X 1 X 2. Let N i be the restriction of N Xi /X to Z. Then (X 1 X 2 ) Z = [ ] m c(n1 )c(n 2 ). c(n Z/X )

24 Algorithm Linear supspaces on hypersurfaces Excess intersection Algorithm Implementation Input: X 1, X 2 be smooth hypersurfaces in P n and Z be a connected component of X 1 X 2. Output: The degree of equivalence class of Z for the intersection X 1 X 2 in A(Z). 1 Construct morphism i : Z P n and its pullback i : A(P n ) A(Z). 2 Compute the normal bundles N 1, N 2 and N Z/P n. 3 Compute the excess bundle (N 1 N 2 )/N Z/P n. 4 Compute the top Chern class of excess bundle in A(Z). 5 Compute degree of this top Chern class.

25 Excess intersection Algorithm Implementation Determine the normal bundles Let Y P n be a smooth hypersurface of degree d, and let N be the normal bundle of Y in P n. Then N O P n(d) Y. This alows us determine the normal bundles N 1, N 2 as follows: N i O P n(d i ) Z.

26 Implementation Linear supspaces on hypersurfaces Excess intersection Algorithm Implementation class ProjectiveSpace(Variety): def hyperplane(self): class Morphism(SageObject): def upperstar(self, c): def normal_bundle(self): def excess_bundle(self, VectorBundle):

27 Example 1 Linear supspaces on hypersurfaces Excess intersection Algorithm Implementation sage: P1 = ProjectiveSpace(1) sage: P2 = ProjectiveSpace(2) sage: h = P1.hyperplane() sage: H = P2.hyperplane() sage: f = Morphism(P1,P2,[h]) sage: B = VectorBundle(P2,2,chern_class=(1+2*H)^2) sage: E = f.excess_bundle(b) sage: P1.integral(E.top_chern_class()) 3

28 Excess intersection Algorithm Implementation Example in Fulton s book Let Q 1, Q 2, Q 3 be 3 quadrics in P 3 which contain a line L but are otherwise general, Q 1 Q 2 Q 3 = L {p 1,..., p k }, where p i be the points. Determine k? sage: P1 = ProjectiveSpace(1); P3 = ProjectiveSpace(3) sage: h = P1.hyperplane(); H = P3.hyperplane() sage: f = Morphism(P1,P3,[h]) sage: B = VectorBundle(P3,3,chern_class=(1+2*H)^3) sage: E = f.excess_bundle(b) sage: P1.integral(E.top_chern_class()) 4 = k = = 4.

29 Excess intersection Algorithm Implementation Example in Fulton s book Let Q 1, Q 2, Q 3, Q 4 be 4 quadrics in P 4 such that Q 1 Q 2 Q 3 Q 4 = C {p 1,..., p k }, where C be a smooth rational quartic curve and p i be the points. Determine k? sage: P1 = ProjectiveSpace(1); P4 = ProjectiveSpace(4) sage: h = P1.hyperplane(); H = P4.hyperplane() sage: f = Morphism(P1,P4,[4*h]) sage: B = VectorBundle(P4,4,chern_class=(1+2*H)^4) sage: E = f.excess_bundle(b) sage: P1.integral(E.top_chern_class()) 14 = k = = 2.

30 Excess intersection Algorithm Implementation Example in Fulton s book Question How many smooth conics in P 2 tangent to each of five general lines? The conics tangent to a given line form a hypersurface in P 5 of degree 2. Thus the cycle of conics tangent to 5 general lines is the intersection of 5 quadrics Q 1,..., Q 5 in P 5. However, the Veronese surface Z of double lines (Z = P 2 ) is a connected component of Q 1 Q 5. The equivalence class of Z for Q 1 Q 5 has degree (Q 1 Q 5 ) Z = 31. Z

31 Excess intersection Algorithm Implementation Example in Fulton s book sage: P2 = ProjectiveSpace(2) sage: P5 = ProjectiveSpace(5) sage: h = P2.hyperplane(); H = P5.hyperplane() sage: f = Morphism(P2,P5,[2*h]) sage: B = VectorBundle(P5,5,chern_class=(1+2*H)^5) sage: E = f.excess_bundle(b) sage: P2.integral(E.top_chern_class()) 31 This implies that there is = 1 smooth conic P 2 tangent to each of five general lines.

32 Excess intersection Algorithm Implementation Thank you for attention!