Formality of Kähler manifolds

Transcription

1 Formality of Kähler manifolds Aron Heleodoro February 24, 2015 In this talk of the seminar we like to understand the proof of Deligne, Griffiths, Morgan and Sullivan [DGMS75] of the formality of Kähler manifolds. Given a compact Kähler manifold, the statement of formality is Theorem 1. The commutative differential graded algebra (Ω (, R), d dr ) of de Rham forms is quasi-isomorphic as algebras to (H (, R), 0). In particular, since the later has zero differential it is actually just a graded commutative algebra, which implies that all higher products of the former 1 vanishes. The proof of the theorem above is very straightforward once we have the crucial technical piece, which is the dd c -lemma. To put it in context we will first recall some concepts about Kähler geometry and the Hodge theory of compact Kähler manifolds. 1 Kähler manifolds Let be a complex Riemannian manifold. If one denotes its metric by g and the almost complex structure by J, we assume the metric is hermitian, i.e. g(u, v) = g(ju, Jv). Then one has that the imaginary part of the metric is proportional to ω(v, u) = g(ju, v). Definition 1. is said to be Kähler if one of the following equivalent conditions hold: (i) ω is closed, i.e. dω = 0; 1 For instance, the Massey products we learned about in Eric s talk. 1

2 (ii) the almost complex structure gives a flat section with respect to the canonical connection 2 on T, i.e. for any v Γ(, T ), v J = 0; (iii) locally on there are holomorphic coordinates which express the metric without linear terms, i.e. for all x there exist (z 1,..., z n ) a coordinate system around x, such that g ij (z, z) = δ ij + O( z 2 ). Remark. It is an exercise to check that these definitions agree. Morally the first condition can be seem as the first indication of the topological nature of the Kähler condition, and the third is interesting because it justifies the following principle. Any identity involving g and only its first derivatives on is true if such identity holds over C n for the flat metric. This can be used to great extent to proof the so-called Kähler identities, about which we will not say anymore here. Example. (i) let D n C n be the unit disk, with Kähler form given by ω(z, z) = i 2 ( 1 z 2). (ii) consider P n with the metric 3 given on each U i (the affine patch where z i 0) by ( n ω(z, z) = i ) 2π z j. (iii) since any complex submanifold of a Kähler manifold has a Kähler structure by simply restricting the metric to it, as can be checked from the first definition above. One obtains that any complex projective algebraic variety has at least one Kähler structure, or more informally is a Kähler manifold. (iv) consider C 2 /(Z Z) with the metric induced by (v, u) = 1 2 vt w, by Riemann s criterion V/Γ for V a complex vector space with an herminitian inner product (, ) and Γ a lattice in V is embeddable in projective space if and only if (, ) Λ has integer-valued imaginary part. j=0 2 The Levi-Civita connection associated to the metric g. 3 This is called the Fubini-Study metric. z i 2

3 (v) by [Siu83], any K3-surface is Kähler. Remark. Notice that example (iv) above is a compact Kähler manifold which is not algebraic. Actually the nice work [Kod54] shows that a Kähler manifold (, ω) can be embedded in P N for some N > 0 if and only if [ω] H 2 (, Z) 4. 2 Hodge theory One can define abstractly what a Hodge structure of weight k is on an abelian group (of finite type) H Z. It is given by a decomposition of H Z Z C = p+q=k H p,q, such that H p,q = H q,p. The main result is the following Theorem 2. For a compact Kähler manifold of dimension n, H k (; Z) has a Hodge structure of weight k, for all 0 k n. We can be more explicitly about how to construct the direct summands of this decomposition. Let Ω p,q be the vector bundle Λp (Ω 1,0 ) Λq (Ω 0,1 ), where Ω 1,0 (resp. Ω0,1 ) is the eigenspace of Ω1 where J 5 has eigenvalue +i (resp. i). Then one obtains that H p,q () is the subset of classes of H p+q (; C) which can be represented by a closed form in Γ(, Ω p,q ). Remark. The above theorem implies that the odd cohomology groups have even dimension. So if one consider C 2 modulo the action of Z given by (z 1, z 2 ) (λ 1 z 1, λ 2 z 2 ), with λ 1, λ 2 of modulus less than 1. Then the manifold obtained 6 has b 1 = 1, hence can not possibly be Kähler. Remark. A conjecture of Kodaira was that the pairity of H 1 was the only obstruction to a complex surface being Kähler. This is actually a theorem and the last case to be verified apparently was that of K3 surfaces, which is example (v) of last section. 4 Notice this gives directly Riemann s criterion mentioned in the example above. 5 We denote by J both the operator on vector fields and on one-forms. 6 These are known as Hopf surfaces. 3

4 To state the lemma we need to prove formality we will reformulate the Hodge decomposition in a more algebraic language. Let s denote by A p,q = Γ(, Ω p,q ). This forms a double complex, whose total complex is the de Rham complex of the manifold, i.e. A = Γ(, Ω ). This implies one has two filtration on A, i.e. F p (A k ) = i p A i,k i, F q (A k ) = i q A k i,i. This naturally induce two spectral sequences for calculating H k (; C), namely E p,q 1 = H q (, Ω p,0 )7 and E p,q 1 = H p (, Ω 0,q ). The statement of Theorem 2 is equivalent to Theorem 3. For a compact Kähler manifold: (1) the above spectral sequences degenerate at E 1, and (2) F and F are complementary filtrations, i.e. A k F p (A k ) F q (A k ) for p + q + 1 = k for all k. Remark. The condition (2) is very important, for example all complex surfaces satisfy condition (1), though the Hopf surface mentioned above can not have a Hodge structure because its first Betti number is 1. Remark. Normally the proofs of the Hodge decomposition in either way stated above involve some analysis of the harmonic forms on a manifold. Essentially one can use the fact that cohomology classes on a Riemannian manifold can be represented by harmonic forms, and then using the Kähler form one obtains that harmonic forms with respect to the de Rham operator are also harmonic with respect to the Laplacian associated to the and operator. There is, however, also a more algebraic way of concluding at least condition (1) of the above theorem, for the class of projective algebraic varieties by using the work [DI87] 8. 3 dd c -lemma Here is a purely algebraic proposition which will be useful to deduce the lemma we need. 7 This is sometimes called the Frölicher spectral seuquence. 8 Though, I do not know if one can obtain the whole Hodge decomposition from it, since as explained in the previous remark one also needs condition (2). 4

5 Proposition ( 1. Let (K,, d, d ) be a double complex, whose total complex we denote by K, d), with d = d + d. Then the following are equivalent: (i) for all n, one has that in H n ( K ). Ker(d ) Ker(d ) Im(d ) = Im(d d ), (1) (ii) the two spectral sequences associated to the filtrations F and F of K degenerate at E 1 and the filtrations are complementary. Proof. The direction (i) (ii) follows from the following. Consider an element x in ( E p,q 1 = H p+q ) F p Kp+q / p 1 F Kp+q, we need to check that d (x) vanishes. Now d (x) belongs to the lefthand side of (1), thus there is y E p 1,q 1 1, such that x = d d (y), so d (x) is zero in E p+1,q 1 because it is d -exact. The argument is completely symmetric, so the same proves that it degenerates for the F filtration. The complementarity follows because i+j=k E i,j 1 = H k ( K ), and for each p, F p H k ( K ) = i+j=k,i p E i,j 1 and F q H k ( K ) = i+j=k,j q E i,j 1. Conversely, consider x K p,q such that d (x) = d (x) = 0 and that there exist y K p 1,q with d y = x. On the (p + q = k)th cohomology of K the class of x is zero, since it is cohomologous to d y, which has degree (p + 1, q 1) which is complementary to (p, q) which is where we assumed x to belong. Hence one can write x = da = db for either a F p (K, ) or b F q (K, ), which gives that in the (k 1)th cohomology, because the filtrations are complementary one can represent the class [a b] as a 1 + b 1 + dc, where a 1 F p ( K k 1 ) and b 1 F q ( K k 1 ), with da 1 = db 1 = 0 and c K k 2. Now apply d to a: d (a) = d b + d a 1 + d b 1 + d dc = x. Firstly, d(a) = x, however since d (x) lives in degree (p + 1, q 1) it vanishes, so actually d a = x. Now d b lives in degree (p 1, q + 1) so has to vanish, the same applies to d b 1, finally da 1 = 0, however since d a 1 is the only term with degree (p, q) it has to vanishes as well. This implies that x = d d c and we are done. 5

6 We now apply the above proposition to the double complex given by A p,q. This implies that on H n (; C) one has that Ker( ) Ker( ) Im( ) = Im( ). Consider d = + and d c = i ( ). Let s rephrase it as follows Lemma 1. Let x A k such that d(x) = dc (x) = 0 and there exist y A k 1 with x = dy. Then there is z A k 2 such that x = dd c (z). 9 4 Formality We are finally in condition to prove the main theorem of this talk. Let s consider the following diagram of cdga s (A, d) i ( A,d c, d) j (H d c(), d H). 10 Here ( A,d c, d) is defined to be the subalgebra of (A, d) formed by the d c -closed forms. I claim that the formality result follows from the claims 1. H d c() H (; C); 2. i and j are isomorphisms 11 ; 3. d H = 0. Item 1. is standard and is just the fact that Ω,0 constant sheaf C. Item 2. has four parts: is a resolution of the a) i is surjective - let x A k with dx = 0, then dc x satisfy the conditions of the lemma, so there exist z A k 1 such that d c d(z) = d c (x). Now [x] = [x dz] in H k (A ), with dc (x dz) = 0. Hence there exist γ H k (A,d c) such that i (γ) = [x dz] = [x]. 9 This lemma is symmetric in changing d by d c. 10 Note that these are maps of commutative differential graded algebras. Indeed, d c (a b) = d c (a) b + a d c (b), so i is a map of algebras, and a d c (b) = d c (a b) d c (a) b, so Im(d c ) is an ideal. 11 These are just the maps induced in cohomology by the natural inclusion i and quotient j. 6

7 b) i is injective - let x A,d k such that c [i (x)] = 0, i.e. x = dy for some y A k 1. Then since d(x) = dc (x) = 0, we have that x = d c d(z) for some z A k 2, i.e. [x] = 0 in Hk 1 (A,d c). c) j is surjective - let α Hd k c(), choose a lift x A,d k c, such that d c (x) = 0. Then dx satisfies the lemma, so there exists a y A k 1 such that dx = dd c (y), this implies that [x d c y] surjects onto α. d) j is injective - let x A,d k c, represent [x] Hk (A,d k c) such that [j (x)] = 0, that is there exist y A k 1,d such that x = d c (y). Since dx = d c x = 0, c there exists z A k 2,d such that x = dd c z, that is [x] = 0 in H k (A c,d c). Item 3. consider α Hd k c(), and let x A,d k be a lift, i.e. c dc x = 0. Then d H α = [dx]. Now by the lemma again any lift of dx is d c -exact hence [dx] = 0 in H k+1 d (). c This concludes the proof of formality for Kähler manifolds. In the last section we will give an application of formality, for that we need a functorial form of the result. Theorem 4. Let f : Y be a holomorphic map between compact Kähler manifolds, then f : AY A is determined (up to homotopy) by f : H (Y ; C) H (; C). The proof is straightforward from the above theorem, we leave it to the reader. 5 Application Let = D 1 D 2 D 3, where each divisor D i (i = 1, 2, 3) is Kähler and the intersections D 1 D 2, D 2 D 3 and D 3 D 1 are transverse and themselves Kähler submanifolds of each divisor and similarly for D 1 D 2 D 3. Recall we denote by A k = Γ(, Ωk ). Consider the complex E k = {(x 1, x 2, x 3 ) x i A kdi, x i Di Dj = x j Di Dj } where i, j = 1, 2, 3. This forms a complex whose differential is just the restriction of the differential on A. Proposition 2. H n (, E ) Hn (; C) 7

8 Proof. One just notes that for all k, E k A k because the data (x 1, x 2, x 3 ) such that x 1 D1 D 2 = x 2 D1 D 2, and so forth, define an element x Ω k by the definition of sheaf. Then one uses the standard fact that Ω C12 are quasi-isomorphic. Now we define the following complex: B = A D1 δ A D2 δ A D3 0. Here one puts D 1 = D 1 D 2 D 3, D2 = D 1 D 2 D 2 D 3 D 3 D 1 and D 3 = D 1 D 2 D 3 ; and the differential δ is defined as δ(x) = ( 1) deg(x) x D1 D 2, for x A D 1, and similarly for the corresponding restrictions on the other terms. One remarks that A D 2 D 1 is defined to be A D 1 D 2 with a minus sign. This is done so that δ δ = 0, and d δ = δ d. Note that since 0 E B 0 is exact, H n (; C) H n (B ). Now we consider the filtration of B by the number of divisors, namely F 1 (B ) = A D1, F 2 (B ) = A D1 The associated spectral sequence has δ A D2,... E p,q 1 = H p (gr q B ) = H p (A Dq) = H p ( I =q A D I ) = I =q H p (A D I ), here D {1,2} = D 1 D 2 and similarly for other I. The final claim is Theorem 5. The above spectral sequence degenerate at E 1. Proof. Consider the above construction with A,d and c H ( ; C) for each respective space. Since they are all levelwise quasi-isomorphic it is equivalent to consider the spectral sequence with E p,q 1 = I =q H p (H d c(di )), since H d c(di ) have zero differentials the spectral sequence collapses. 12 The underline means we are considering the associated constant sheaf on. 8

9 References [DGMS75] Pierre Deligne, Phillip Griffiths, John Morgan, and Dennis Sullivan. Real homotopy theory of kähler manifolds. Inventiones mathematicae, 29(3): , [DI87] Pierre Deligne and Luc Illusie. Relèvements modulop 2 et décomposition du complexe de de rham. Inventiones Mathematicae, 89(2): , [Kod54] [Siu83] Kunihiko Kodaira. On kahler varieties of restricted type an intrinsic characterization of algebraic varieties). Annals of Mathematics, pages 28 48, Y-T Siu. Every k3 surface is kähler. Inventiones mathematicae, 73(1): ,