Chapter 17 Computation of Some Hodge Numbers

Transcription

1 Chapter 17 Computation of Some Hodge Numbers The Hodge numbers of a smooth projective algebraic variety are very useful invariants. By Hodge theory, these determine the Betti numbers. In this chapter, we turn to the practical matter of actually computing these for a number of examples such as projective spaces, hypersurfaces, and double covers. The GAGA theorem, Theorem , allows us to do this by working in the algebraic setting, where we may employ some of the tools developed in the earlier chapters Hodge Numbers of n Let S = k[x 0,...,x n ] and = n k sheaf of differentials. for some field k. We first need to determine the roposition There is an exact sequence 0 Ω 1 O ( 1) n+1 O 0. roof. Let Ω S = Sdx i = S n+1 be the module of Kähler differentials of S. Construct the graded S-module M = Γ (Ω 1 )=Γ (An+1 {0},π Ω 1 ), where π : A n+1 {0} is the projection. This can be realized as the submodule Ω S consisting of those forms that annhilate the tangent spaces of the fibers of π. The tangent space of the fiber over [x 0,...,x n ] is generated by the Euler vector field x i xi. Thus a 1-form f i dx i lies in M if and only if f i x i = 0. Next, we have to check the gradings. Ω S has a grading such that the dx i lie in degree 0. Under the natural grading of M = Γ (Ω 1), sections of Ω 1(U i) that are generated by ( ) x j d = x idx j x j dx i x i x 2 i 279

2 Computation of Some Hodge Numbers should have degree 0. Thus the gradings on Ω S and M are off by a shift of one. To conclude, we have an exact sequence of graded modules 0 M Ω S ( 1) m 0, (17.1.1) where m =(x 0,...,x n ) and the first map sends dx i to x i. Since m S, (17.1.1) implies the result. Dualizing yields the sequence of Example roposition Suppose we are given an exact sequence of locally free sheaves If A has rank one, then 0 A B C 0. 0 A p 1 C p B p C 0 is exact for any p 1. If C has rank one, then is exact for any p 1. 0 p A p B p 1 A C 0 roof. We prove the first statement, where rank(a )=1, by induction, leaving the second as an exercise. When p = 1 we have the original sequence. In general, the maps in the putative exact sequence need to be explained. The last map λ : p B p C is the natural one. The multiplication B p 1 B p B restricts to give µ : A p 1 B p B. We claim that µ factors through a map α : A p 1 C p B. For this we can, by induction, appeal to the exactness of Since A has rank one, 0 A p 2 C p 1 B p 1 C 0. µ A (A p 2 C ) = 0, so µ factors as claimed. Therefore all the maps in the sequence are defined. Exactness can be checked on stalks. For this the sheaves can be replaced by free modules. Let {b 0,b 1,...} be a basis for B with b 0 spanning A. Then the images b 1, b 2,... give a basis for C. Then the above maps are given by α : b 0 b i1 b ip 1 b 0 b i1 b ip 1, λ : b i1 b ip b i1 b ip, where > i 2 > i 1 > 0. The exactness is immediate.

3 17.1 Hodge Numbers of n 281 Corollary There is an exact sequence 0 Ω p O ( p) (n+1 p ) Ω p 1 0 and in particular, Ω n = O ( n 1). roof. This follows from the above proposition and roposition , together with the isomorphism p [O ( 1) n+1 ] = O ( p) (n+1 p ). This corollary can be understood from another point of view. Using the notation introduced in the proof of roposition , we can extend the map Ω S ( 1) m to an exact sequence where 0 [ n+1 Ω S ]( n 1) δ [ 2 Ω S ]( 2) δ Ω S ( 1) δ m 0, (17.1.2) δ(dx i1 dx ip )= ( 1) p x i j dx i1 ˆ dx i j dx ip is contraction with the Euler vector field. The sequence is called the Koszul complex, and it is one of the basic workhorses of homological algebra [33, Chapter 17]. The associated sequence of sheaves is 0 [ n+1 O n+1 ]( n 1) [ 2 O n+1 ]( 2) [O n+1 ]( 1) O 0. If we break this up into short exact sequences, then we obtain exactly the sequences in Corollary roposition H q (,Ω p )= { k if p = q n, 0 otherwise. roof. When p = 0, this follows from Theorem In general, the same theorem together with Corollary implies H q (Ω p ) = H q 1 (Ω p 1 ). Therefore, we get the result by induction. When k = C, this gives a new proof of the formula for Betti numbers of n given in Section 7.2. By a somewhat more involved induction we can obtain the following theorem of Raoul Bott:

4 Computation of Some Hodge Numbers Theorem (Bott). H q ( n,ω p n (r)) = 0 unless (a) p = q, r = 0, (b) q = 0,r> p, (c) or q = n, r < n + p. roof. A complete proof will be left for the exercises. We give the proof for p 1. For p = 0, this is a consequence of Theorem We now turn to p = 1. Corollary implies that H q 1 (O(r)) H q (Ω 1 (r)) H q (O(r 1)) e (17.1.3) is exact. We use constants e,e,... for exponents whose exact values are immaterial for the argument. The sequence (17.1.3), along with Theorem , forces H q (Ω 1 (r)) = 0 in the following four cases: q = 0,r < 1; q = 1,r < 0; 1 < q < n; q = n,r n + 1. The remaining cases are q = 0,r = 1 and q = 1,r > 0. The trick is to apply Corollary with other values of p. This yields exact sequences H q (O(r 2)) e H q (Ω 1 (r)) H q+1 (Ω 2 (r)) H q+1 (O(r 2)) e, H q+1 (O(r 3)) e H q+1 (Ω 2 (r)) H q+2 (Ω 3 (r)) H q+2 (O(r 3)) e,... leading to isomorphisms H 0 (Ω 1 (1)) = H 1 (Ω 2 (1)) = = H n 1 (Ω n (1)) = H n 1 (O( n)) = 0. Likewise, for r > 0, H 1 (Ω 1 (r)) = H 2 (Ω 2 (r)) = = H n (Ω n (r)) = H n (O( n 1 + r)) = 0. Exercises Finish the proof of roposition Given an exact sequence 0 A B C 0 of locally free sheaves, prove that the top exterior power detb is isomorphic to (deta ) (detc ). Use this to rederive the formula for Ω n n Finish the proof of Theorem Hodge Numbers of a Hypersurface We now let X = n k be a nonsingular hypersurface defined by a degree-d polynomial.

5 17.2 Hodge Numbers of a Hypersurface 283 roposition The restriction map is an isomorphism when p + q < n 1. H q ( n,ω p ) Hq (X,Ω p X ) We give two proofs, one now, over C, and another later for general k. roof. Let k = C. The weak Lefschetz theorem, Theorem , implies that the restriction map H i (X an,c) H i (Y an,c) is an isomorphism for i < n 1. The proposition is a consequence of this together with the the canonical Hodge decomposition (Theorem ) and GAGA (Theorem ). As a corollary, we can calculate many of the Hodge numbers of X. Corollary The Hodge numbers h pq (X) equal δ pq, where δ pq is the Kronecker symbol, when n 1 p + q < 2n 2. roof. We give a proof when k = C. For p + q < n 1, this follows from the above proposition and roposition For p + q > n 1, this follows from GAGA and Corollary We prepare for the second proof by establishing a few key lemmas. Lemma There is an exact sequence 0 Ω p ( d) Ω p Ω p X 0. (Recall that Ω p X is shorthand for i i Ω p, where i : X is the inclusion.) roof. Tensor with Ω p to get 0 O ( d) O O X 0 0 Ω p O ( d) Ω p O Ω p O X 0 = 0 Ω p ( d) Ω p = = Ω p X 0 For the last isomorphism, it is simply a matter of expanding the notation. Observe that if ι : X is the inclusion, then Ω p X = ι ι Ω p = ι (ι 1 Ω p O X) = Ω p O X. Lemma O X ( d) Ω 1 X ΩX 1 0 is exact.

6 Computation of Some Hodge Numbers roof. We have a natural epimorphism Ω 1 X ΩX 1 corresponding to restriction of 1-forms. We just have to determine the kernel. Let f be a defining polynomial of X, and let M = Γ (Ω 1) and M = Γ (ΩX 1). We embed M as a submodule of Ω S( 1) as in the proof of roposition In particular, the symbols dx i have degree 1. Then ker[m/ fm M] is a free S/( f )-module generated by df = i f dx i. x i Thus it is isomorphic to S/( f )( d). Corollary Ω p 1 X ( d) Ω p X Ω p X 0. roof. Apply roposition to the lemma. For the second proof of roposition , it is convenient to prove something stronger. roposition If p + q < n 1, then { H q (X,Ω p X ( r)) = H q ( n,ω p ) if r = 0, 0 if r > 0. roof. We prove this by induction on p. For p = 0, this follows from the long exact sequences associated to 0 O ( d r) O ( r) O X ( r) 0 and Theorem In general, by induction and Corollary we deduce H q (X,Ω p X ( r)) = Hq (Ω p ( r) X) for r 0 and p + q < n 1. Lemma and Theorem give { H q (Ω p ( r) H q ( n,ω p X )= ) if r = 0, 0 if r > 0. Exercises Using Exercise , deduce a version of the adjunction formula ΩX n 1 = O X (d n 1) with X as above Compute H q (X,Ω n 1 X ) for all q.

7 17.3 Hodge Numbers of a Hypersurface II Hodge Numbers of a Hypersurface II As in the previous section, X n is a nonsingular degree-d hypersurface. By Corollary , the Hodge numbers h pq (X) equal δ pq when n 1 p + q < 2(n 1). So the only thing left to compute are the Hodge numbers in the middle. The formulas simplify a bit by setting h pq 0 (X)=hpq (X) δ pq. These can be expressed by the Euler characteristics: Lemma h p,n 1 p 0 (X)=( 1) n 1 p χ(ω p X ) + ( 1)n. We can calculate these Hodge numbers by hand using the following recurrence formulas. roposition (a) (b) χ(ω p p ( n + 1 (i)) = ( 1) j j=0 p j )( i p + j + n ( ) ( ) i + n i + n d χ(o X (i)) =. n n (c) χ(ω p X (i)) = χ(ω p (i)) χ(ω p p 1 (i d)) χ(ωx (i d)). roof. Corollary yields the recurrence ( ) χ(ω p n + 1 (i)) = χ(o (i p)) χ(ω p 1 p (i)). Therefore (a) follows by induction on p. The base case was obtained previously in (16.2.4). Lemma and Corollary imply χ(ω p X (i)) = χ(ω p (i) X ) χ(ω p 1 X (i d)) = χ(ω p (i)) χ(ω p n ). p 1 (i d)) χ(ωx (i d)). When p = 0, the right side can be evaluated explicitly to obtain (b). Corollary The Hodge numbers of X depend only on d and n, and are given by polynomials in these variables. In principle, formulas for all the Hodge numbers can be calculated using the above recurrence formulas. For example, ( ) ( ) n d d 1 h 0,n 1 (X) = ( 1) n =, (17.3.1) n n ( ) ( ) ( )] h 1,n 2 n 1 n d 1 n 2d 0 (X) = ( 1) [(n n + 1) (n + 1) +. n n n

8 Computation of Some Hodge Numbers But this gets quite messy as p increases. So, instead, we give a closed form for the generating function below. Let h pq (d) denote the pqth Hodge number of a smooth hypersurface of degree d in p+q+1. Define the formal power series in x and y. Theorem (Hirzebruch). H(d)= (h pq (d) δ pq )x p y q pq H(d)= (1 + y)d 1 (1 + x) d 1 (1 + x) d y (1 + y) d x. Corollary Hodge symmetry h pq = h qp holds for smooth hypersurfaces in projective space over arbitrary fields. Remark Hodge symmetry can fail for arbitrary smooth projective varieties in positive characteristic [90]. Corollary If X n+1 C b n (X)=h n/2,n/2 (X)=2. roof. has degree 2, then b n (X)=0 if n is odd; otherwise, H(2)= 1 1 xy. By expanding the series H(3) for a few terms, we obtain the following corollary: Corollary If X n+1 has degree 3, the middle hodge numbers are for n 5. 1,1 0,7,0 0,5,5,0 0,1,21,1,0 0,0,21,21,0,0 Although this result can be deduced from the previous formulas in principle, Hirzebruch [63, ] obtained this from his general Riemann Roch theorem. His original formula gave a generating function for χ(ω p X ); the above form can be obtained by a change of variables, cf. [30, Example XI Corollary 2.4]. Similar formulas are available for complete intersections. We will be content to work out the case y = 0. On one side we have the generating function h n0 (d)x n. By (17.3.1), this equals ( ) d 1 x n = (1 + x)d 1 1, n + 1 x which is what one gets by substituting y = 0 into Hirzebruch s formula.

9 17.4 Double Covers 287 Exercises rove Lemma Calculate the Hodge numbers of a degree-d surface in 3 (a) using the recurrence formulas, (b) using the generating function. Compare the expressions rove that for every fixed d and p, there exists q 0 such h pq (d) =0 for q q Double Covers Our goal is to compute the Hodge numbers for another natural class of examples that generalize hyperelliptic curves. Let f (x 0,...,x n ) C[x 0,...,x n ] be a homogeneous polynomial of degree 2d such that the hypersurface D n = defined by f = 0 is nonsingular. Let π : X be the double cover branched along D (Example 3.4.9). By construction, this is gotten by gluing the affine varieties defined by y 2 i = f (x 0,...,1,...,x n ) over U i. It follows that X is nonsingular. Using these coordinates, it is also clear that the local coordinate ring O X (π 1 U i ) is a free O(U i )-module generated by 1 and y i. Globally, we have π O X = O L, where L is the line bundle locally generated by y i. The ratios y i /y j give a cocycle for L, from which it easily follows that L = O(±d). To get the correct sign, we need to observe that L is a nontrivial ideal in π O X, so it has no nonzero global sections. Therefore we obtain the following: Lemma π O X = O O ( d). It is worth observing that the summands O and O ( d) are exactly the invariant and anti-invariant parts under the action of the Galois group, which is generated by the involution σ : y i y i. A more abstract, but less ad hoc, argument involves observing that O O ( d) is a sheaf of algebras, and defining X as its relative spectrum [35]. Then the lemma becomes a tautology. We need an extension of the previous lemma to forms: Lemma (Esnault Viehweg). There are isomorphisms for every p. π Ω p X = Ω p (Ω p (logd) O( d)) roof. We sketch the proof. See [35, pp. 6 7] for further details. The Galois group acts on π Ω p X. We check that the invariant and anti-invariant parts correspond to

10 Computation of Some Hodge Numbers Ω p and Ω p (logd) O( d) respectively. It is enough to do this for the associated analytic sheaves. By the implicit function theorem, we can choose new analytic local coordinates such that X is given locally by y 2 = x 1. Then y,x 2,...,x n are coordinates on X, so that their derivatives locally span ΩX 1. It follows that a local basis for π ΩX 1 is given by ydy = 1 2 dx 1,dx 2,...,dx n, dy = y dx 1,ydx 2,...,ydx n. }{{} 2x } 1 {{} invariant anti-invariant The forms in the first group are invariant and give a local basis for Ω 1. The remainder are anti-invariant and form a local basis for Ω 1 (logd) O( d). By taking wedge products, we get a similar decomposition for p-forms. Corollary H q (X,Ω p X ) = H q (,Ω p ) Hq (,Ω p (logd) O( d)). roof. Let {U i } be the standard affine cover of n. Then Ũ i = π 1 U i gives an affine cover of X. We can compute H q (X,Ω p X ) using the Čech complex Č({Ũ i },Ω p X )=Č({U i },Ω p ) Č({U i },Ω p (logd)), which decomposes into a sum. This decomposition passes to cohomology. Corollary We have h pq (X) =δ pq + dimcoker[h q 1 (Ω p 1 D ( d)) Hq (Ω p ( d))] +dimker[h q (Ω p D ( d)) Hq+1 (Ω p ( d))] in general, and h pq (X)=δ pq if p + q < n. roof. This follows from (12.6.4) from 12.6 together with Bott s vanishing theorem, Theorem We can obtain more explicit formulas by combining this with earlier results. Exercises When n = 2, check that h 02 (X)= (d 1)(d 2) 2 and h 11 (X)=3d 2 3d Verify that h pq (X)=δ pq also holds when p + q > n.

11 17.5 Griffiths Residues* Griffiths Residues* In this section, we describe an alternative method for computing the Hodge numbers of a hypersurface due to Griffiths [48], although the point is really that the method gives more, namely a method for computing the Hodge structure (or more precisely the part one gets by ignoring the lattice). Further details and applications can be found in books of Carlson, eters, Müller-Stach [17, 3.2] and Voisin [116, Chapter 6] in addition to Griffiths paper. We work over C in this section. Suppose that X = n+1 is a smooth hypersurface defined by a polynomial f C[x 0,...x n+1 ] of degree d. Let U = X. The exact sequence (12.6.5) yields H n 1 (X) H n () H n+1 (U) H n (X) H n+2 (). The first map is an isomorphism by weak Lefschetz. Therefore H n+1 (U) maps isomorphically onto the primitive cohomology n (X)=ker[H n (X) H n+2 ()]. This is the same as H n (X) if n is odd, and has dimension one less if n is even. The Hodge filtration on F p H n+1 (U) maps onto the Hodge filtration on X with a shift F p 1 n (X). We refer to Section 12.6 for the definition of this and of the pole filtration ole p H n+1 (U). The key step is to compare these filtrations. Theorem (Griffiths). The Hodge filtration F p n (X) coincides with the shifted pole filtration ole p+1 H n+1 (U). This can, in turn, be identified with the quotient H 0 (Ω n+1 ((n p + 1)X)) dh 0 (Ω n((n p)x)). roof. For p = n, this is immediate because F n n (X)=F n+1 H n+1 (U)=H 0 (Ω n+1 (logx)) = H 0 (Ω n+1 (X)). For p = n 1, we use the exact sequence 0 Ω n,cl (logx) Ω n (X) d Ω n+1 (2X) 0, where Ω,cl (...) is the subsheaf of closed forms. Then H 0 (Ω n (X)) d H 0 (Ω n+1 (2X)) H 1 (Ω,cl n (logx)) H1 (Ω n (X)). On the right, the group H 1 (Ω (X)) is equal to H 1 (Ω n (d)) = 0 by Bott s Theorem Thus F n H n+1 (U)=H 1 (Ω,cl n (logx)) is isomorphic to the cokernel of the first map labeled by d. This proves the theorem for this case. The remaining p s can be handled by a similar argument, which is left for the exercises.

12 Computation of Some Hodge Numbers This can be made explicit using the following lemma: Lemma Let ω = ( 1) i x i dx 0 dx i dx n+1. Then { } gω H 0 (Ω n+1 (kx)) = g homogeneous of deg = kd (n + 2). f k With these identifications, elements of n (X) can be represented by homogeneous rational differential forms gω/ f k modulo exact forms. Set R = C[x 0,...,x n+1 ] ( f / x 1,..., f / x n+1 ). The ring inherits a grading R = R i from the polynomial ring. We define a map n (X) R by sending the class of gω/ f k to the class of g. Theorem (Griffiths). Under this map, the intersection of H n p (X,Ω p X ) with n (X) maps isomorphically to R τ(p), where τ(p)=(n p + 1)d (n + 2). This leads to an alternative method for computing the Hodge numbers of a degree-d hypersurface. Corollary h pq (d) δ pq is the coefficient of t τ(p) in (1 + t + + t d 2 ) n+2. roof. We can assume that f = x d 0 +xd 1 + +xd n+1 is the Fermat equation. It suffices to prove that the oincaré series of R, which is the generating function p(t) = dimr i t i, is given by Note that R = C[x 0,...,x n+1 ] (x d 1 0,x d 1 1,...) p(t) = (1 + t + + t d 2 ) n+2. = C[x] (x d 1 ) C[x] (x d 1 (n + 2 times). ) Since oincaré series for graded rings are multiplicative for tensor products, p(t) is the (n + 2) power of the oincaré series of C[x]/(x d 1 ), and this is given by the above formula.

13 17.5 Griffiths Residues* 291 Exercises Using exact sequences and identifications 0 Ω,cl n i n i n i+1 (( j 1)X) Ω (( j 1)X) Ω ( jx) 0 F p H n+1 (U) = H n+1 p (Ω p,cl (logx)) = H n+1 p (Ω p,cl (X)), finish the proof of Theorem