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Introduction to Hodge Theory and K3 Surfaces Contents Introduction to Hodge Theory and K3 surfaces Contents I Hodge Theory (Pierre Py) 2 1 Kähler manifold and Hodge decomposition 2 1.1 Introduction . 2 1.2 Hermitian and Kähler metric on Complex Manifolds . 4 1.3 Characterisations of Kähler metrics . 5 1.4 The Hodge decomposition . 5 2 Ricci Curvature and Yau's Theorem 7 3 Hodge Structure 9 II Introduction to Complex Surfaces and K3 Surfaces (Gianluca Pacienza) 10 4 Introduction to Surfaces 10 4.1 Surfaces . 10 4.2 Forms on Surfaces . 10 4.3 Divisors . 11 4.4 The canonical class . 12 4.5 Numerical Invariants . 13 4.6 Intersection Number . 13 4.7 Classical (and useful) results . 13 5 Introduction to K3 surfaces 14 1 Part I Hodge Theory (Pierre Py) Reference: Claire Voisin: Hodge Theory and Complex Algebraic Geometry 1 Kähler manifold and Hodge decomposition 1.1 Introduction Denition 1.1. Let V be a complex vector space of nite dimension, h is a hermitian form on V . If h : V × V ! C such that 1. It is bilinear over R 2. C-linear with respect to the rst argument 3. Anti-C-linear with respect to the second argument i.e., h(λu, v) = λh(u; v) and h(u; λv) = λh(u; v) 4. h(u; v) = h(v; u) 5. h(u; u) > 0 if u 6= 0 Decompose h into real and imaginary parts, h(u; v) = hu; vi − i!(u; v) (where hu; vi is the real part and ! is the imaginary part) Lemma 1.2. h ; i is a scalar product on V , and ! is a simpletic form, i.e., skew-symmetric. Note. − h ; i determines ! and conversely Proof. By the property 4. h ; i is real symmetric and ! is skew-symmetric. hu; ui = h(u; u) > 0 so h ; i is scalar product Let u0 2 V such that !(u0; v) = 08v 2 V . This equates to h(u0; v) is real for all v. Also h(u0; iv) is real for all v, but h(u0; iv) = −ih(u0; v) 2 = therefore, h(u0; v) = 0. Hence u0 = 0, as h is non-degenerate. So ! is non-degenerate. Now we show − h ; i determines !: !(u; v) = −=h(u; v) = =(i2h(u; v)) = =(ih(iu; v)) = hiu; vi, so !(u; v) = hiu; vi. Lemma 1.3. !(u; iu) > 0 for all u 6= 0. Proof. Plug in v = iu in the last part of the previous lemma. Denition 1.4. We say that a skew-symmetric form on a complex vector space is positive if it has the above property (of lemma 1.3) ( !(iu; iv) = !(u; v) If h(iu; iv) = h(u; v) then (∗) hiu; ivi = hu; vi Exercise. Prove that a 2-form on ! on V satisfy (∗) if and only if it is of type (1; 1) 2 Let be a -vector space of . Let be coordinates on and be V C dimC V = n = 2k z1; : : : ; zn V e1; : : : ; en a basis such that P for . Dene P and P for . v = i ziei v 2 V dzj : ziei 7! zj dzj : ziei 7! zj 1 ≤ j ≤ n Then . We have . If dz1; : : : ; dzn; dz1; : : : ; dzn 2 HomR(V; C) dimR HomR(V; C) = 2 · 2n = 4n λ 2 C and , dene . So can be viewed as a -vector space, then φ 2 HomR(V; C) λφ : v 7! λφ(v) HomR(V; C) C . dimC HomR(V; C) = 2n Exercise. is a basis for as a -vector space dz1; : : : ; dzn; dz1; : : : ; dzn HomR(V; C) C Exercise. 1. An element is -linear if and only if can be written as Pn φ 2 HomR(V; C) C φ φ = i=1 αidzi where αi 2 C. 2. is anti -linear map if and only if can be written as Pn where . φ C φ φ = i=1 βidzi βi 2 C Let be the set of . is a -linear alternating form to . I fi1 < ··· < ikg dzI = dzi1 ^ dzi2 ^ · · · ^ dzik k V C is a -linear alternating form to dzI = dzi1 ^ dzi2 ^ · · · ^ dzik k V C Denition 1.5. A -form on (with values in ), is of type , with if P k α V C (p; q) p+q = k α = jIj=p;jJj=q λI;J dzI ^ dzJ for λI;J 2 C. Any -form is a sum of forms of type for and . Then P p;q k α (p; q) 0 ≤ p; q ≤ k p + q = k α = p+q=k α Example. Let V be of dimension 2 k = 1 (1; 0)-forms are C-linear maps from V ! C (0; 1)-forms are anti-C-linear maps from V ! C k = 2 (2; 0)-forms are dz1 ^ dz2 (1; 1) forms are spanned by dzi ^ dzj for i; j 2 f1; 2g (0; 2)-forms are dz1 ^ dz2 Exercise. The type of a form does not depend on the choice of basis. n Example. Let V = C , zi = xi + iyi then dz + dz dx ^ dz = 1 1 ^ dz 1 2 2 2 dz ^ dz dz ^ dz = 1 2 + 1 2 2 2 | {z } | {z } (2;0)−form (1;1)−form Example. If X is a complex surface, z1; z2 are local coordinate on X, then a 2-form is a combination • dz1 ^ dz2 a (2; 0)-form 8 >dz1 ^ dz2 > <>dz ^ dz • 1 1 are (1; 1)-forms dz ^ dz > 2 2 > :dz2 ^ dz1 • dz1 ^ dz2 a (0; 2)-form Summary: If h is a hermitian form, ! = −=h is a (1; 1)-form and is positive (i.e, !(u; iu) > 0). Conversely if a (1; 1) form is positive it arises as ! = −=h for some hermitian form h. 3 1.2 Hermitian and Kähler metric on Complex Manifolds Let M be a complex manifold. Convention: Each tangent space of M, TxM is a complex vector space and write J (or Jx) for the 2 endomorphism Jx : TxM ! TxM dened by v 7! iv.(J = − id) Denition 1.6. A hermitian metric on M is the following. For each x 2 M, hx is a hermitian metric on 1 TxM and hx is C on M. So as before we can write h = h ; i − i!. The h ; i is a Riemannian metric on M and ! is a (1; 1) form on M Denition 1.7. We say h is Kähler if ! is closed, i.e., d! = 0. Example. If , that is is a Riemann surface, then any hermitian metric is Kähler.: • dimC M = 1 M Why? d! by denition is a 3-form on a 2-dimension R-manifold, so it must be zero. • @; @ operators: If f : M ! C is a function. df is a 1-form and dfx : TxM ! C. We can decompose as k k+1 dfx = @fx + @fx . So df = @f + @f. @ and @ extend to operators from Ω ! Ω (where |{z} |{z} C−linear C−antilinear k Ω is the C-value k-forms) dened by @(α ^ β) = @α ^ β + (−1)jαjα ^ @β @(α ^ β) = @α ^ β + (−1)jαjα ^ @β Exercise. If α is a (p; q)-form (it is of type (p; q) at each point), then dα is the sum of a (p + 1; q) form and a (p; q + 1) form: @ is the (p + 1; q) form and @ is the (p; q + 1)piece • M = n : We dene a close positive (that is positive on each point on M) (1; 1)-form (it must be PC the imaginary part of a hermitian metric) is dened by 1 2 . Check that it is ![z] = 2πi @@ log(jjzjj ) ∗ well dened, (does not dene on the ane piece): hint: if f : U ! C is holomorphic, check that @@ log jfj2 = 0. n n • If T = C =Λ, Λ a lattice of C , then any constant coecient metric is Kähler. • If (M; h) is Kähler, and Σ ⊂ M is a C-submanifold. Then (Σ; hjΣ) is Kähler. As d(!jΣ) = d!jΣ ) !jΣ is closed. Lemma 1.8. Let M be a complex manifold of C-dimension n with hermitian metric h. The Riemannian volume form of is equal to !n . h ; i n! (If V is C-vector space with C-basis e1; : : : ; en then e1; Je1; e2; Je2; : : : ; en; Jen is a positive real basis. That is it has a canonical basis) Corollary 1.9. Let M be a closed complex manifold, i.e., compact with no boundary. Then 8k 2 f1; : : : ng, !k = ! ^ · · · ^ ! is closed and non-zero in cohomology, i.e., ! is not exact. | {z } k times Proof. If !k = dα for some α, then !n = !k ^!n−k = dα^!n−k = d(α^!n−k): Hence by Stoke's theorem n , but !n Vol , hence contradiction. M ! = 0 M n! = (M) > 0 ´ ´ 4 So 2n HDR(M; R) 6= 0 p n Corollary 1.10. If M is compact, Kähler and Σ ⊂ M closed C-submanifold, then the homology class [Σ] 2 H2p(M), the fundamental class of Σ, is non-zero. Proof. !p Vol , then is not homologeous to . 0 < Σ p! = hΣ Σ 0 ´ Exercise. If is a compact manifold and and a Kähler metric, and ∗ . Prove X dimC X ≥ 2 h φ : X ! R+ that φh is Kähler if and only if φ is constant. 1.3 Characterisations of Kähler metrics Let (M; h) be a complex manifold with hermitian metric. Recall that r is the Levi-Civita connection of <(h) = h ; i which is a Riemannian metric. Theorem 1.11. The following are equivalent: 1. h is Kähler 2. For any vector eld X on U ⊂ M (open set) then r(JX) = J(rX) Proof.
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