And this means that under the rational map given by L, C maps to a line in X. This gives a map S[2] → F1(X). Furthermore, since F1(X) and S[2] are both 4-dimensional, the map is rational if it is generally injective. So assume thatp0+q0 gives the same line asp+q. This means that the surface P1×P1⊂P2×P2determined byp0, q0 intersects that given byp, qalong a curve.
But if we consider the Chow ring ofP2×P2, we find that this is impossible.
We have
CH•(P2×P2=Z[a, b]/(a3, b3) [3264, p. 51],
witha, bthe pullbacks of the hyperplane classes. So the class of ourP1×P1s isab, and two such surfaces will intersect in a single point. Hence the map is generally injective.
However, we cannot hope to have an isomorphismS[2]−∼→F(X) for a K3 surface S constructed in this fashion. We sketch the argument of Hassett:
[Bea+16, p. 41]
IfSis a smooth intersection of hypersurfaces of types (1,2,(2,1)) inP4, then the classes (1,0),(0,1) correspond to linear systems of curves onS. LetC1and C2be curves representing these. We have (by adjunction) that C12=C22= 2, C1C2= 5. ThenC1+C2 represents (1,1) and so gives a degree 14 polarization f ofS[2].
However, if S[2]'F(X), then the Plücker polarization g equals 2f−5δ ([BD85]), where 2δis the image of the exceptional divisor in S[2].Then we have
C1.g=C1(2f)−C1(5δ) = 14−15 =−1 which is ridiculous.
7.2 The Fermat cubic
Let
Y =Z(x30+x31+· · ·+x35)
7.2. The Fermat cubic be the Fermat cubic fourfold. An explicit but boring computation shows thatY has exactly 45 Eckardt points, where all but two coordinates are zero. Through every Eckardt point, there are 27 planes. It is not difficult to see that each plane inF passing through one Eckardt point must pass through three others. Hence there are at least 45· 273 = 405 planes in F. It is also possible to show that these 405 planes, together with the Chern classes of the tautological subbundle ofG(2,6) restricted toS, generateH2,2(X,Z), by seeing that their intersection matrix has rank 232 (Proposition 3.2.1).
We will see that the Hodge Conjecture holds for the Fano scheme of the Fermat cubic.
Lemma 7.2.1. Pick any two disjoint planes inY, and let the induced rational map (as in Proposition 7.1.1) beφ. The K3 surfaceS which is the indeterminacy locus of φ, is isomorphic to the surface
x2u+y2v+z2w=xu2+yv2+zw2= 0 inP2×P2.
Proof. This is true if we can write the equation of the Fermat cubic as Z(x2u+y2v+z2w−xu2+yv2+zw2)⊂P5
. To see this, note that the expression is a sum of three terms each depending only on two variables, so this cubic must be isomorphic toY. (In fact, letη be a nonreal cube root of 1. Then settingx=−η2α−ηβ, u=ηα+η2β leads to
xu2−x2u=α3+β3.)
Lemma 7.2.2. The Picard numberρ(S)of S is 20.
Proof. We note that S contains at least twenty-four lines. These fall in two groups, which we will denote lx, ly, . . . , lw (the first, six of these) and lx,ω, lx,−ω, lx,−1, . . . , lw,ω, lw,−ω, lw,−1(the second type, eighteen of these). Here ω is a nonreal cube root of −1. We assume that x, y, z are the first three coordinates onP2×P2,u, v, w the last three. Then the lines are
• lx={(1 : 0 : 0)×(0 :v:w)} ...
• lw={(x:y: 0)×(0 : 0 : 1)},
• lx,ω={(0 : 1 :ω)×(u:v:ωv)} ...
• lw,−1={(x:−x:z)×(1 :−1 : 0)}
wherex, y, z, u, v, ware free variables. That is, of the first type, the subscript indicates the nonzero coordinate among either the first (x, y, z) or second (u, v, w) triplet, and for the second type, the subscript indicates the zero coordinate and the quotient of the two other coordinates in one triplet.
It is now easily checked that we have the following intersection pairings for the linelx:
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7. Cubic fourfolds and special K3 surfaces
• lx.lv=lx.lw= 1,
• lx.lu,ω=lx.lu,−ω=lx.lu,−1= 1
• lx.lx=−2.
• lx.any other line = 0 and for the linelx,ω:
• lx,ω.lu= 1
• lx,ω.lu,ω = 1
• lx,ω.lx,ω =−2.
• lx,ω.any other line = 0
The intersection numbers for the other lines are completely analogous.
That a line on a K3 surface S has self-intersection 2 follows from the adjunction formula 2g−2 =C(C+KS) – sinceS is K3, its canonical divisor is trivial.
These numbers give a size 24×24 intersection matrix, which has rank 20.
Then, we apply the following result from [Mor84, p. 15]:
Proposition 7.2.3. Let S be a K3 surface where ρ(S) ≥ 19. Then S×S satisfies the Hodge conjecture.
Now note thatS[2]is formed fromS2by blowing up the diagonal and taking the quotient by Z/2, as detailed in Proposition 3.2.1. It follows that if the Hodge conjecture holds forS2, it will also hold forS[2].
So for the Fermat cubic, S has a Picard group of rank 20. It follows that the Hodge conjecture is satisfied forS[2]
We conclude by applying to the following lemma:
Lemma 7.2.4. Suppose thath:X 99K∼ Y is a birational equivalence of varieties of dimension≤4. ThenX satisfies the (Integral) Hodge conjecture if and only if Y does it.
Proof. The Weak Factorization Theorem ([Abr+99]) tells us that h can be decomposed into a diagram
Z
π1
~~
π2
X h //Y
whereπ1, π2are compositions of blow-ups along smooth curves or surfaces. But on these the Hodge conjecture is known. So consider the decompositions of Chow groups and Hodge structures of blow-ups. The class map is compatible with these decompositions. Hence we have: IfY is the blow-up ofX along the smooth centreS such that the Hodge conjecture holds onS, then the Hodge conjecture holds forY if and only if it holds forX. The conclusion follows.
7.2. The Fermat cubic
Corollary 7.2.5. The Fano scheme of the Fermat cubic fulfills the Hodge conjecture.
Remark 7.2.6. In fact, we have shown that: IfX is a cubic fourfold such that F1(X) is birationally equivalent to the Hilbert square of a K3 surface of Picard rank 19 or 20, the Hodge conjecture holds forF1(X).
From this point, most of the ideas are due to R. Laterveer (personal com-munication).
How do we find such cubic fourfolds? One idea is the following lemma.
Lemma 7.2.7. Suppose thatX is a cubic fourfold with associated K3 surface S, in the sense thatF(X)−∼→S[2]. Then the Picard number ρ(S)of S equals dimQH2,2(X,Q)−1.
Proof. This follows from the construction Proposition 3.2.1 and the properties of the Abel-Jacobi map. Let Htr4(X,Q)denote the transcendental part of H4(X,Q), i.e., the subspace ofH4(X,Q) orthogonal to every algebraic class . (That the Hodge conjecture holds for cubic fourfolds means that we can say
Htr4(X,Q) =H2,2(X,Q)⊥.) This is mapped onto Htr2,2(F(X),Q).
But since H2,2(F(X),Q) 'H1,1(S, Q)⊕QE where E is the exceptional divisor of the blowup, it follows that
ρ(S) = rkH2,2(S,Q) = rkH2,2(X,Q).
Let us conclude by discussing a possible generalization. We suspect that it is possible to adapt this argument to prove that the Hodge conjecture holds for every power ofF1(Y). It is at least true that:
Lemma 7.2.8. For everym, iand every cubic hypersurfaceX, there is a split injection
CHi−2m(F1(X)m)Q,−→CHi(Xm)Q.
Proof. This follows from [Lat16]. He shows, using the Grothendieck ring construction from [GS14], that there is an isomorphism
CHj−2(F1(X)×M)Q⊕CHj(W ×M)Q
−∼→CHj(X[2]×M)Q
whereW is the projectivization ofTPn|X. Then we can apply this isomorphism iteratively withM =F1(X)m, X[2]×F1(X)m−1, . . . , X[2]m−1×F1(X).
This result is probably valid for cohomology as well. It is known by work of Shioda ([Shi79]), that the (generalized) Hodge conjecture holds for everyYm. So if we can apply the above lemma in both Chow-theoretic and cohomological settings, we have a way of moving cycles back and forth betweenYmandF1m, and this should be enough to prove:
Conjecture 7.2.9. LetY be the Fermat cubic of dimension n. Then, for every m, the Hodge conjecture holds forF1(Y)m.
In fact, it should be possible to show
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7. Cubic fourfolds and special K3 surfaces
Conjecture 7.2.10. Let X be a cubic fourfold,S a K3 surface. Assume that F(X)99K∼ S[2], and thatS has Picard number ≥19. Then the Fano variety F1(X)mfulfills the Hodge Conjecture for anym≥1.
CHAPTER 8
Cones on special cubic fourfolds
In this chapter, we describe the cones of nef and effective cycles on cubic fourfolds and related varieties.
We start by giving a very explicit computation of the cones on cubic fourfolds containing a plane. We set up machinery for translating results about cubic fourfolds into results for the Fano schemes of lines on the fourfolds using the Beauville-Bogomolov form and the Abel-Jacobi map. Then we discuss special cubic fourfolds –in the sense of Hassett– for small discriminants, while discussing some conjectures of Hassett and Tschinkel. The techniques used for a fourfold containing a plane admit a wide generalization to the case of an even-dimensional hypersurface containing a complete intersection of half the dimension, which we will see in Chapter 9.
8.1 Nef cycles
We will especially need to checknefness of various cycles in the cubic hypersur-faces - and later on in other varieties. We have already mentioned it, but let us define:
Definition 8.1.1. Let X be a projective variety of dimension n. A k-cycle S∈CHk(X) isnef if, for any effectiven−kcycleT,
S.T ≥0.
This generalizes the usual definition of nefness for divisors.
We know that nef divisors are limits of ample divisors, but this geometric intuition is no longer true when speaking of higher codimension cycles. It is, for instance, true that a cycle can be nef even without being the limit of effective cycles ([Deb+10]).
So determining whether a higher-codimension cycle is nef can be a subtle problem. For smooth subvarieties, however, there is a simple criterion:
Theorem 8.1.2([FL82]).LetX be a projective variety (not necessarily smooth), Y ⊂X a smooth local complete intersection of pure codimensionk, and NY /X
the normal sheaf ofY in X. IfSymmNY /X is globally generated for somem, thenY is nef.
This is not too surprising if we consider the definition of intersection numbers through the "deformation to the normal cone" as in [Ful98]. Essentially, the
8. Cones on special cubic fourfolds
intersection products of an effective subvariety is determined by the positivity properties of its normal bundle. The condition in the theorem is exactly that the normal bundle is a nef vector bundle.