Special cubic fourfolds


 Gabriel Gray
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1 Special cubic fourfolds 1 Hodge diamonds Let X be a cubic fourfold, h H 2 (X, Z) be the (Poincaré dual to the) hyperplane class. We have h 4 = deg(x) = 3. By the Lefschetz hyperplane theorem, one knows that the natural restriction H n (P 5, Z) H n (X, Z) is an isomorphism provided that n < dim(x) 1 = 3, so that in particular H 2 (X, Z) = Zh. By Poincaré duality, we have that H 6 (X, Z) is free on one generator, and this generator g satisfies h, g = 1. Since g h 3, one checks that g = h3. Hence so far the Hodge diamond 3 of X looks like ????? How do we compute the missing row? One starts with Euler s exact seqeunce, Teorema 1.1. There is an exact sequence Some standard algebra gives Proposition 1.2. Let 0 Ω 1 P n O( 1)n+1 P n O P n 0. 0 A B C 0 be a short exact sequence of locally free sheaves. If C has rank 1, then 0 Λ p A Λ p B Λ p 1 A C 0 is exact for any p 1. Likewise, if A has rank 1, then is exact. 0 A Λ p 1 C Λ p B Λ p C 0 Taking the pth exterior power then gives 1
2 Teorema 1.3. There is an exact sequence Taking cohomology of this we obtain H q 1 (P n, O( p)) (n+1 p 0 Ω p P O( p) (n+1 p ) n P Ω p 1 n P 0. n ) H q 1 ( P n, Ω p 1 P n ) H q (P n, Ω p P n ) H q (P n, O( p)) (n+1 p ), and since (most of the time) H q 1 (P n, O( p)) = H q (P n, O( p)) = (0) we obtain H q (P n, Ω p P n ) = C δp,q. Now let X be a hypersurface. We want a similar SES for the sheafs of differentials on X, and we want to reduce to the case of P n. Everything is taken care of by the following SES s: Teorema 1.4. There are short exact sequences 1. 0 Ω p P ( d) Ωp P Ωp P X O X ( d) Ω 1 P X Ω 1 X O X ( d) Ω p 1 X Ωp P X Ω p X 0 Proof. 1. Follows from 0 O P ( d) O P O X 0 upon tensoring with Ω p P 2. Direct computation 3. Take exterior powers in (2). Now we have (essentially) everything we need, but writing down explicit formulas is still cumbersome. Things get marginally better by looking at Euler characteristics: one gets h p,n 1 p (X) = ( 1) n 1 p χ(ω p X ) + ( 1)n + 1 and from here everything is just a boring exercise. I can t resist mentioning Teorema 1.5. (Hirzebruch) Let Then H(d) = p,q (h pq (d) δ pq )x p y q. H(d) = (1 + y)d 1 (1 + x) d 1 (1 + x) d y (1 + y) d x. From here, a direct (but cumbersome) computation gives the missing row of the Hodge diamond of X,
3 2 The Fano variety of lines is a deformation of S [2] Let me sketch a proof of the following statement: Teorema 2.1. F is a deformation of S [2], where S is a degree 14 K3 surface in P 8. This relies on two results I will just admit. First, the deformation space of S [2] is a smooth projective variety of dimension 20. On the other hand, cubic hypersurfaces depend on 20 moduli: the space of monomials of degree 3 in 6 variables has dimension ( ) = 56. The projectivization has dimension 55. There is an action of SL 6 which is transitive on isomorphic threefolds. The dimension of SL 6 is = 35. Hence the dimension of the moduli space is = 20. Secondly, the association X F (X) is injective on isomorphism classes: this is because one can reconstruct F (X) from its Hodge structure (Torelli), and we shall show below that H 4 (X, Z) 0 = H 2 (F, Z) 0 ( 1) as polarized Hodge structures. So distinct deformations of X should go to distinct deformations of F (X), and therefore to show that the space of deformations of S [2] and of F (X) are the same it suffices to show that they have a point in common. In particular, it suffices to construct a cubic fourfold whose Fano variety of lines is S [2] for S a K3 as in the statement. I will just indicate the construction: the cubic we construct is called a Pfaffian cubic. Proof. Let V be a 6dimensional complex vector space. We let G be the subvariety of P(Λ 2 V ) given by tensors of rank 2 (up to scalar). Let be the subvariety of tensors of rank at most 4, and define G P(Λ 2 V ) in the same way. One sees that G is the Grassmannian G(2, V ); moreover, is a cubic hypersurface in P(Λ 2 V ) (=it is the subset of degenerate bilinear alternating forms) while G is of dimension 8 and degree 14. Take a generic linear subspace L P(Λ 2 V ) of dimension 8 and set S = G L, X = L. Then X is a cubic hypersurface of dimension 4 and S is a K3. Let s show that S [2] is the variety of lines of X (and by the way, X is rational). Let s first construct the map S [2] F (X). Let P, Q S G, suppose they are distinct. They are 2planes in V such that P +Q is a 4plane. Consider the linear forms in L that restrict to 0 on P +Q. These are clearly all in, hence in X, and they form a linear subspace. This subspace is at least a line since we are imposing 4 conditions on a 5dimensional subspace, so it is a line in X. Conversely, start with a line in X, that is, a pencil ϕ t of degenerate alternating forms on V. One can show that there is a unique 4plane that is isotropic for all the ϕ t. We would like this plane to be P + Q with P, Q S G = G(2, V ). Let M be the 5plane generated by G(2, K). Then M L is a line (because M L is the pencil we started with), and this line intersect the quadric G(2, K) in 2 points: this gives a map F (X) S [2]. Corollary 2.2. The cohomology group H 2 (F (X), Z) is H 2 (F, Z) = H 2 (S, Z) Zδ, where the sum is orthogonal, δ 2 = 2, and 2δ is the exceptional divisor of S [2]. 3
4 3 Lattices So far, so good: we know that L := H 4 (X, Z) is a 23dimensional lattice. Its signature is (21, 2) by the RiemannHodge bilinear relations, namely the following statement. Consider the bilinear form h H 4 (X, Z) H 4 (X, Z) Z = H 8 (X, Z) (α, β) X α β; then different pieces of the Hodge decomposition are orthogonal with respect to h, and moreover h is ( 1) p definite on H p,q. We want to know more, namely, we want to know the structure of L = (H 4 (X, Z), h) as a lattice with a bilinear form. This is actually not very hard, due to the following result: Teorema 3.1. Let Λ be an indefinite, unimodular lattice of signature (n +, n ). Then 1. if Λ is even of index τ := n + n, then τ 0 (mod 8) and according to the sign of τ; 2. if Λ is odd, then Λ = (1) n + ( 1) n. Λ = E τ/8 8 U n or Λ = E 8 ( 1) τ/8 U n + Now simply observe that h 2 h 2 = h 4 = 3 is odd, so L is odd and therefore it is equal to (1) 21 ( 1) 2. The tricky part comes when we want to understand the primitive cohomology L 0 := h. Let me just remind myself of what primitive cohomology is: the primitive part P n k of the cohomology is the kernel of the multiplication by h k, which is a map H n k H n+k+2. We also have a Lefschetz decomposition theorem, which tells us that primitive classes really are primitive in some sense: H n (X, C) = k h k P n 2k (X, C). The problem with L 0 is that it is not unimodular. To study it, we relate it to the cohomology of the Fano variety of lines of X. Definition 3.2. The Fano variety of lines of X is the subvariety of the Grassmannian G(1, 5) parametrizing lines contained in X. It is a smooth fourfold. Let Z F X be the universal line, i.e. the variety of pairs (l, x) with x l. Letting p, q denote the projections, we define the AbelJacobi map α as α = p q : H 4 (X, Z) H 2 (F, Z). Let M := H 2 (F, Z), and let g be the hyperplane class on F (induced from the Grassmannian). Now α(h 2 ) is the (cohomology class of the) lines meeting a codimension2 subspace of P 5. This is a hyperplane, so α(h 2 ) = g. Let M 0 be the primitive cohomology of F. One can define a bilinear form on M (the Beauville canonical form) as follows: 4
5 (g, M 0 ) = 0; (g, g) = 6; (x, y) = 1 6 g2 xy for x, y M 0. Proposition 3.3. (Nikulin, integral symmetric bilinear forms and some of their applications) Let Λ be a unimodular lattice, let K be a saturated (nondegenerate) sublattice of K, and let K be its orthogonal complement. Then the discriminants of K and K are equal up to sign. In particular: Corollary 3.4. The discriminant of the primitive cohomology L 0 = H 4 (X, Z) 0 is ±3. Teorema 3.5. The AbelJacobi map α induces an isomorphism between L 0 and M 0. Moreover, we have ( ) (α(x), α(y)) = x, y for all x, y L 0. Proof. Recall: Z q (X, h) and p (F, g) α = p q. Sooo... many many things to do. First of all, it is clear that α sends H 4 (X, Z) H 2 (F, Z)( 1), because integration along the fibers kill one holomorphic and one antiholomorphic differential. Injectivity follows from the formula (α(x), α(y)) = x, y and the nondegeneracy of,. Surjectivity then follows because both lattices have the same rank and the same (absolute value of the) discriminant (indeed one can explicitly compute that the discriminant of M 0 is 3, and the discriminant of L 0 is ±3). Hence in the end it suffices to establish ( ). To do this, start with any class x H 4 (X, Z). Since Z F is a P 1 bundle, its Chow ring is p CH(F ) p CH(F )ζ, where ζ is the class of the bundle itself. Since q (h) is a birational section of p (the general line meets the general hyperplane in 1 point), we can take ζ = q (h). In particular we can write where x i H 2i (F, Z). Applying p we get q (x) = p (x 1 )q (h) p (x 2 ) (1) α(x) = x 1 p q (h) + 0 x 1 = α(u). 5
6 In particular, if x = h 2 is the square of the hyperplane class on X, we have q (h 2 ) = p (g 1 )q (h) p (g 2 ), g 1 = α(h 2 ) = g. Now suppose that u is in the primitive cohomology of X, i.e. it is orthogonal to h. Then one can multiply (1) by q (h) to obtain 0 = p (x 1 )q (h 2 ) p (x 2 )q (h) = p (x 1 ) (p (g)q (h) p (g 2 )) p (x 2 )q (h) = (p (x 1 g) p (x 2 )) q (h) p (x 1 g 2 ) whence x 1 g 2 = 0 and x 2 = x 1 g. Now let s square (1): q (x 2 ) = p (x 2 1)q (h 2 ) + p (x 2 2) 2p (x 2 )p (x 1 )q (h) = p (x 2 1) (p (g 1 )q (h) p (g 2 )) + p (x 2 2) 2p (x 2 )p (x 1 )q (h) = p (x 2 1)p (g 1 )q (h) + p (x 2 2) 2p (x 2 )p (x 1 )q (h) = p (gx 2 1)q (h) + p (x 2 2) 2p (gx 1 )p (x 1 )q (h) = p (g 2 x 2 1) p (gx 2 1)q (h) = p (g 2 α(x) 2 ) p (gα(x) 2 )q (h) Now multiply by p (g). The first term on the RHS disappears, and we are down to p (g)q (x 2 ) = p (g 2 α(x) 2 )q (h). We re almost done: clearly q (x 2 ) = x 2 q ([X]) and p (g 2 α(x) 2 ) = g 2 α(x) 2 p ([F ]), so we just need to compute the intersection numbers p ([F ]).q (h), q ([X]).p (g). Since these are numbers, we can push them and pull them all we like; in particular, to compute the fist one we push down to F, so that it becomes [F ]p q (h), i.e. the degree of the map q (h) F. Given a generic line l F, this amounts to asking: how many points are there on X that lie both on l and on X? Since the generic line meets the generic hyperplane in 1 point, the answer is 1. For the second, we push down to X, so that the question becomes: what is degree of p (g) X? Now we know that p (g) is the set of lines that meet a generic codimension 2 linear subspace P, so the question becomes: given a generic x X, how many lines are there in X that pass through x and meet P? This is the same as asking for how many lines there are on X that pass through x and are contained in P, x = P 4, that is, in a generic hyperplane. Now a section of X by a generic hyperplane is a smooth cubic threefold, so we already know the answer: through the general point on a smooth cubic threefold there are 6 lines. Conclusion: 6x 2 = g 2 α(x) 2, i.e. x, x = (x, x). 6
7 We have thus proved that L 0 = M 0. The cohomology lattice of a K3 is wellknown, and is equal to Λ := H 3 ( E 8 ) 2 ; we deduce that the cohomology M = H 2 (F, Z) is Λ ( 2), and by taking an explicit orthogonal we find ( ) 2 1 M 0 = H 2 ( E ) 2. Remark 3.6. To compute the orthogonal one needs to know the class of a polarization. It turns out that this is 2l 5δ, where l is a hyperplane class in H 2 (S, Z) and 2δ is the class of the exceptional divisor in S [2]. One also needs to know δ 2 = 2. 4 The definition of special cubic fourfolds and related topics Teorema 4.1. For a very general cubic fourfold X one has H 2,2 (X, Z) = Zh 2. Here is a very vague sketch of proof of a stronger statement. To begin with, a complete marking is an isomorphism H 4 (X, Z) L that carries the square of the hyperplane section to a fixed section h 2 with selfintersection 3. Given a cubic fourfold with a complete marking, one can identify the primitive cohomology of X with the orthogonal complement of h 2 L. There is a distinguished subspace of H 4 (X, Z) 0 given by which enjoys the following properties: F (X) = H 3,1 (X, C) 1. F (X) is isotropic with respect to the intersection form; 2. The Hermitian form (α, β) α, β is positive on F (X). Let Q be the quadric in P(L 0,C ) given by the quadratic form corresponding to the intersection form, and let U be the open subset of it on which the positivity condition holds. Notice that L 0 has dimension 22, so U has dimension 20. U has two connected components, which we denote by D and D (the difference is in whether the negativedefinite part is sent to itself or minus itself). There is also an action of the group Γ of automorphisms of L that preserve both the intersection form and the distinguished class. Now Γ contains an index2 subgroup Γ that stabilizes D ; the quotient Γ\D =: D is called the global period domain, and exists as an analytic space. Voisin has proved that the map τ : C D, where C is the coarse moduli space of smooth cubic fourfolds, is an open immersion. One can also prove that it is algebraic. 7
8 5 Some special intersection lattices Suppose that X contains a surface S not homologous to a complete intersection. intersection form on the lattice generated by h 2 and S looks like The h 2 S h 2 3 deg(s) S deg(s) S S The mysterious entry S S is the selfintersection of S in X, which as is wellknown is the Euler class of the normal bundle of S in X. Hence to compute this selfintersection we study the normal bundle. Proposition 5.1. For any bundle V one has c 1 (V ) = c 1 (det V ). Proof. By the splitting principle we may assume that the vector bundle is a direct sum of k line bundles. The first Chern class of direct sum of line bundles is the sum of the first Chern classes of the line bundles, which in turn is the Chern class of the tensor product of those bundles, which is isomorphic the top wedge power of the original vector bundle. Proposition 5.2. Let X be a complex manifold. One has c 1 (X) = c 1 (K X ). Proof. By Serre duality, K X = Λ n (TX ), hence c 1 (K X ) = c 1 (Λ n T X) = c 1 (T X) = c 1 (T X ). The last statement follows from the fact that T X TX Chern class, and that c 1 (V W ) = c 1 (V ) + c 1 (W ). is a trivial bundle, hence has trivial Proposition 5.3. (Chern class computations) Let S be a surface in X. One has or (better) c 2 (N X/S ) = c 2 (T S ) + c 1 (T S ) 2 c 1 (T X )c 1 (T S ) + c 2 (T X ) c 2 (N X/S ) = χ(s) + K 2 S + 3K S H + 6H 2, where H is the restriction to S of the hyperplane section on X. Proof. Consider the short exact sequence defining the normal bundle, and take Chern classes: we obtain 0 T S T X S N X/S 0 c(n X/S ) = c(t X S ) 1 + c 1 (T S ) + c 2 (T S ) 8
9 and taking the homogeneous degree2 part c 2 (N X/S ) = c 2 (T S ) + c 1 (T S ) 2 c 1 (T X S )c 1 (T S ) + c 2 (T X S ). Since c 1 (T S ) = c 1 (K S ) = K S we then obtain c 2 (N X/S ) = c 2 (T S ) + K 2 S + K S c 1 (T X S ) + c 2 (T X S ); we can denote by H the restriction of the hyperplane section to S, so that the formula finally reads c 2 (N X/S ) = χ(s) + K 2 S + 3K S H + 6H 2. Notice that the previous proposition reduces the selfintersection problem to a problem which only involves the cohomology of S, so we have essentially got rid of the embedding. Now let s see what happens in a few cases. 5.1 Planes A plane S = P 2 is something of degree 1, and its topological Euler characteristic is h 0 + h 2 + h 4 = 3. The canonical class is O( 3) and H is the actual hyperplane section. It follows that c 2 (N X/S ) = = 3. The intersection lattice is K 8 = h 2 S h S 1 3 This is probably a good moment to introduce the discriminant of a labelling of a special cubic fourfold. A labelling is nothing but a saturated, rank2 sublattice K H 2 (X, Z) that contains h 2. The discriminant of the labelling is the discriminant of the intersection form restricted to K. In particular, the previous labelling has discriminant Cubic scrolls Proposition 5.4. The cohomology of the structure sheaf is a birational invariant. In particular, for a rational scroll X (which is birational to P 2 ) one has H 1 (X, O X ) = H 2 (X, O X ) = 0. This implies that Pic(X) = H 2 (X, Z); moreover, scrolls are simply connected, so H 1 (X, Z) = 0, and finally using Poincaré s duality we get χ(x) = = 4. As for the canonical class, one can describe Pic(Σ) quite explicitly. Recall that a cubic scroll is the image of a morphism f : P(E) P 4, where E is the vector bundle O P 1( 1) O P 1( 2) on P 1, the morphism f : P(E) P 4 is such that f O P 4(1) = O E (1), and the pullback map H 0 (P 4, O P 4(1)) H 0 (P(E), O E (1)) is an isomorphism. Now there is a section σ : P 1 P(E) whose image D has selfintersection 1, and f(d) is a line on Σ called the directrix. For any t P 1, f(π 1 (t)) is a line called a line of ruling of Σ; denote by F its divisor class. 9
10 Pic(Σ) is generated by D, F, with the hyperplane class given by D + 2F and the canonical class given by 2D 3F. The intersection form is given by D 2 = 1, DF = 1, F 2 = 0, so that 6H 2 = 6(D + 2F ) 2 = 18 3HK S = 3(D + 2F )( 2D 3F ) = 3(2 7) = 15 KS 2 = ( 2D 3F )2 = = 8 χ(s) = 4 and c 2 (N X/S ) = 7. The labelling is K 12 = h 2 S h S 3 7 With the more natural description S = Bl p (P 2 ), one sees that Pic(S) is generated by the exceptional divisor E and by H. The canonical class is K P 2 + E = 3H + E, and the hyperplane class is 2H E (but I don t know why) Here is a geometric explanation: look at the linear system of conics in P 2 vanishing on a given point p. This has projective dimension 5 1 = 4, so it gives a rational map P 2 P 4 which is defined everywhere but at p. The scroll is the blowup of this map, so the pullback of the hyperplane section is the system of conics (i.e. 2H) that vanish at p (i.e. E). 5.3 Two more cases If X contains a quartic scroll Σ 4 or a quintic del Pezzo surface T, the labelling has discriminant 14 and it is given by K 14 = h 2 Σ 4 h Σ = h 2 T h 2 3 5, T = 3h 2 Σ 4 T 5 13 Finally, when X contains two disjoint planes P 1, P 2, one has h 2 P 1 P 2 h P P and this contains a distinguished rank 2 lattice, namely a copy of K 14, with a Σ 4 given by 2h 2 P 1 P 2. 10
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