MATH 250B: COMPLEX ALGEBRAIC GEOMETRY

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1 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Contents 1. Complex manifolds vs almost complex manifolds Definition Complex linear structure Almost complex manifolds Distributions and Frobenius Theorem Newlander-Nirenberg Theorem 8 2. Complex differential forms Definitions Coordinate invariant definitions Local exactness of the complex One complex variable Back to differential forms Dolbeault complex Kähler manifolds Kähler forms Kähler metrics Kähler manifolds Connections Examples Harmonic forms L 2 forms and the Hodge star operator Adjoint operators Laplacian Differential operators Symbols and elliptic operators Fundamental Theorem Serre duality Lefschetz operator and applications Hodge decomposition Lefschetz decomposition Lefschetz decomposition in cohomology Representations of sl 2 (C) Polarization Outline of Kodaira s theorem Construction of L Hypercohomology and spectral sequences Definition of hypercohomology 59 1

2 2 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Introduction 62 References 62

3 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 3 1. Complex manifolds vs almost complex manifolds The first step is to understand the difference between complex manifolds and real manifolds Definition. Definition A (topological) n-manifold is a Hausdorff space together with a collection of pairs called charts (U, ϕ) where U are open subsets of which cover and ϕ : U R n is a homeomorphism of U onto an open subset V of R n. The structure of a manifold is given by the transition mappings. These are ϕ ij = ϕ j ϕ 1 i : ϕ i (U i U j ) V j where (U i, ϕ i : U i = Vi ) and (U j, ϕ j : U j = Vj ) are two charts which intersect. If these transition maps are C k (k times continuously differentiable) then M is a C k -manifold. If they are analytic (given locally by converging power series) then M is an analytic manifold. Given a chart (U, ϕ) we get local coordinates which are function x 1,, x n : U R given by composition: x i = p i ϕ : U R n R Definition A complex n-manifold is a real 2n manifold with charts (U, ϕ : U C n ) so that the transition maps are holomorphic (complex differentiable and thus complex analytic). This formal definition does not help understand what we are talking about. We will later go through the proof that complex differentiable functions are analytic. But these are more formalities which do not help to visualize it. There is some beautiful mathematics here that I want to show you Complex linear structure. The first point to understand is that complex structure is locally defined. A complex manifold locally looks like C n. For a coordinate chart (U, ϕ) we get n complex coordinate functions z i : U C given by 2n real coordinate functions x i, y i : U R given by z i = x i + iy i. The tangent plane at one point T,x is a complex vector space of dimension n: T,x = C n Recall the definition of the tangent plane T,x. Definition A tangent vector to at x 0 is an R-linear function: χ : C k () R satisfying the Leibnitz rule: χ(fg) = g(x 0 )χ(f) + f(x 0 )χ(g). Tangent vectors are point derivations on C k (), the ring of C k functions R. Since is a real 2n-manifold, we learned in elementary school that the tangent vectors form a 2n-dimensional real vector space spanned by the 2n tangent vectors x0, x0, x0,, x0 x 1 x n y 1 y n

4 4 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Call this vector space T,x0,R = R 2n. When is a complex manifold, this vector space has a complex structure given by x0 = i x0 y j x j Definition Take the complexified tangent space T,x0,R C = C 2n This has a complex basis of the 2n vectors = 1 ( i ) z j 2 x j y j = 1 ( + i ) z j 2 x j y j Definition A (real-) differentiable (C 1 ) function f :, x C (notation means f is only defined in a nbh U of x) is holomorphic if f z j = 0 for j = 1,, n. This is equivalent to saying that the derivative of f is a complex linear map: f : T C at each point in the domain of f. Example: f(z) = az1 m = a(x 1 + iy 1 ) m. Then, by the chain rule, ( ) f = a(x 1 + iy 1 ) m = am(x 1 + iy 1 ) m 1 x 1 x 1 ( ) f = a(x 1 + iy 1 ) m = ami(x 1 + iy 1 ) m 1 y 1 y 1 So, ( f + i ) = amz1 m 1 amz1 m 1 = 0 = 2 f y 1 y 1 z 1 More generally, if f is C-linear then ( ) ( f = f i ) ( ) = if y j x j x j ( f i ) ( = f i 2 ) ( ) = f y j x j x j ( (1.1) f C + i ) = 0 x j y j Note that x j + i y j is a vector field over an open subset U on the complexified tangent bundle T R C which is a complex 2n dimensional vector bundle over. Here f C : T,x C C

5 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 5 is the C-linear extension of f : T,x C considered as an R-linear map Almost complex manifolds. Definition An almost complex manifold is a C k real 2n-manifold together with a C k endomorphism I of its tangent bundle so that I 2 = id. I x : T,x T,x This is a local structure. A neighborhood of each point is (equivalent to) an open subset U R 2n with a complex structure on the tangent plane (also R 2n ) at each point. An almost complex structure is equivalent to an honest complex manifold structure on if, for each such nbh U, there is a differentiable embedding ϕ : U C n so that, at each point x U, the derivative ϕ : T U,x C n is complex linear. I.e., (1.2) ϕ (Iv) = iϕ (v) (and a linear isomorphism). We will rephrase this condition to resemble (1.1). Since U is a real manifold, the derivative ϕ : T U,x C n is apriori only linear over R. Tensoring this R linear map with C (and composing with the C-linear map C n R C C n ) will always give a C-linear map: Condition (1.2) is equivalent to: ϕ C : T U,x R C C n (1.3) ϕ C (v + iiv) = 0 Proof: Since ϕ C is C-linear, ϕ C (v + iiv) = ϕ C (v) + iϕ C (Iv) = ϕ C (v) + i 2 ϕ C (v) = 0 The condition that ϕ is a linear isomorphism translates into ϕ C being an epimorphism. We will rephrase Equation (1.3) in a fancy language and derive necessary and sufficient conditions for an almost complex manifold to be equivalent to a complex manifold. Remark For any C-vector space V and any C-linear map J : V V with J 2 = id V, V will decompose as V = V 1,0 V 0,1 where V 1,0 is the i-eigenspace of J and V 0,1 is the i-eigenspace of J. For each v V, the components are: v 1,0 = 1 (v ijv), 2 v0,1 = 1 (v + ijv) 2 Definition Let be an almost complex manifold (this includes complex manifolds). Define T 1,0, T 0,1 to be the subbundles of the complexified tangent bundle T,C := T R C which are the i and i eigenspaces of the C-linear endomorphism I C.

6 6 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Equation (1.3) says that ϕ C : T C C n is a complex linear epimorphism (one each fiber) with kernel equal to T 0,1. We need conditions under which such a map exists Distributions and Frobenius Theorem. For simplicity we assume all structures are C. Definition A k-distribution on a real n-manifold is defined to be a k-dimensional subbundle F T of the tangent bundle of. The distribution is called integrable if, in a nbh U of each point x, there is a submersion So that the kernel of ϕ U is F U. ϕ : U V R n k Theorem (Frobenius). A distribution F on is integrable if and only if [F, F ] F I.e., for any two C 1 vector fields σ, τ on with values in F, [σ, τ] is in F. Hiding a ton of technicalities under the rug this will give: Theorem (Newlander-Nirenberg). An almost complex manifold is equivalent to a complex manifold if and only if the distribution T 0,1 T C is integrable. I.e., iff [T 0,1, T 0,1 ] T 0, Review of bracket [σ, τ]. A C k -vector field is defined to be a derivation where l > k. In other words, σ : C l () C k () σ(fg) = σ(f)g + fσ(g). It is a standard fact that all C k -derivations are given locally by σ = σ i x i where σ i is a C k function on the coordinate chart U. Conversely, any such σ gives a derivation C k+1 () C k (). Thus, even if σ is only defined on C l () C k+1 (), it will extend uniquely to the larger domain. The bracket [σ, τ] is defined to be the C k 1 derivation given by [σ, τ] = στ τσ : C k+1 () k 1 () Proposition If σ = σ i x i τ = τ j x j Proof. [σ, τ] = ( σ i τ j x i τ i σj x i then ) x j στ = σ i τ j + σ i τ j x i x j x i x j

7 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 7 Switch σ, τ: τσ = τ i σj + τ i σ j x i x j x i x j Subtract to get the Proposition. Example If σ is the vertical vector field σ = x n then [ ] [σ, τ] =, τ = τ j x n x j n x j This is the vertical component of the derivative of the section τ of T in the direction of the vector σ in. (The horizontal component is just σ.) The proof of Frobenius Theorem uses the following basic fact. Theorem (Constant Rank Theorem). Let f : M N be a C 1 map between C 1 manifolds. Suppose that the derivative f : T M T N has rank k at every point in M. Then the image f(m) of f is (locally) a k-submanifold of N. Locally refers to the case when f is an immersion which has self-crossings. Proof of Frobenius Theorem. The bracket condition is necessary: Given U open and ϕ : U V R n k a submersion, we can use ϕ to give the last n k local coordinates x k+1,, x n by x k+j = p j ϕ : U V p j R. Then a section of F is σ = k i=1 σi x i and k ( ) [σ, τ] = σ i τx j i τ i σx j i x j j=1 also lies in F where we use the notation f x = f. This trivial proof hides an important x concept that we need to come back to later. Sufficiency is by induction on k. For k = 1, choose a C 1 vector field σ tangent to F. Then σ generates a flow ψ : V ( ɛ, ɛ) U so that ψ(x, t) = σ(ψ(x, t)) t where V U is an n 1 dimensional set transverse to the vector field σ. By making V, U smaller we may assume that ψ : V ( ɛ, ɛ) = U and that V is an open subset of R n 1. Taking the projection to V gives the integration of the distribution F spanned by σ: ϕ : U = V ( ɛ, ɛ) V R n 1 For k 2, first choose a nonzero vector field σ in F defined in a nbd of any chosen point x. Then we get a smaller open nbd U = V ( ɛ, ɛ) as above with last coordinate tangent to σ: σ = x n

8 8 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Let F be the k 1 distribution on V given by F y = F y T V is spanned by F y and σ and σ is in the kernel of ϕ, for all y V. Since F y ϕ (F y ) = F y Claim 1 For any (y, t) ϕ 1 (y), ϕ (F (y,t) ) = F y. This implies ϕ 1 (F ) = F. Pf: Let K be the restriction of F to ϕ 1 (y) = y ( ɛ, ɛ). (So, K = R k ( ɛ, ɛ) has dimension k + 1.) Then the condition [σ, τ] F for all sections τ of F implies that, for all (y, t) ϕ 1 (y) and all w = (v, t) K over (y, t), T K,w = R k+1 we have: ϕ θ (T K,w ) = ϕ (F (y,t) ) = ϕ (F (y,t) ) where F (y,t) = F (y,t) T V t and θ : y ( ɛ, ɛ) U since T K,w is spanned by F (y,t) and the vector τ (σ) and ( ) θ τ (σ) = (τ), σ = ([σ, τ], 1) F (y,t) R x n Thus ϕ has constant rank k 1 on K. Therefore, by the Constant Rank Theorem, ϕ (K) is a k 1 dimensional manifold (possibly by making ɛ smaller). But ϕ (K) contains ϕ (F y ) = F y = R k 1. So, ϕ (K) = F y. Claim 2 F satisfies the bracket condition [F, F ] F. Pf: Since F and T V satisfy the bracket condition, so does their intersection. By induction on k, the k 1 distribution F on V can be integrated giving a map ρ : V, x R n k with ker ρ = F. So ρ ϕ : U, x R n k is the integral of F Newlander-Nirenberg Theorem. We need the following complex version of Frobenius Theorem. See [Voisin] for explanation. Theorem Suppose that is a complex m manifold and E is a k dimensional holomorphic subbundle of the tangent bundle of. Then, E can be integrated, i.e., near each x, E = ker ϕ for some locally defined holomorphic, x C m k if and only if [E, E] E. This is used for m = 2n and k = n. Comments on proof of Newlander-Nirenberg Theorem. The important part of the proof for us is the trivial part which states that, if is a complex manifold, then [ T 0,1, T ] 0,1 T 0,1 It is also important to note that this is equivalent to the condition [ T 1,0, T ] 1,0 T 1,0 (See Remark )

9 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 9 2. Complex differential forms Some approaches to this topic obscure the difficulties using deceptively simple notation. [ I will explain why the equation ( + ) 2 = 0 is equivalent to the bracket conditions T 0,1, T ] 0,1 T 0,1 and [ T 1,0, T ] 1,0 T 1, Definitions. We assume that (, I) is an almost complex manifold which satisfies the bracket conditions. Thus is a complex manifold. But it is easier to keep track of what we are doing if we use the notation of an almost complex manifolds and we have actions of I = I 1 and i = 1 i on T C = T 1,0 T 0,1. Definition Let Ω,C := Hom R (T, C) = Hom C (T C, C) This is a C 2n bundle over. On a coordinate chart U, let dz j = dx j + idy j This is the homomorphism T U C given by: and dz j sends the other basis elements dz j : dz j : x j 1 y j i x k, y k, k j to 0. Similarly, dz j = dx j idy j Note that dz j is the composition of dz j with complex conjugation c which is an R-linear automorphism of C: dz j = c dz j As in the case of T C, let Ω,C = Ω 1,0 Ω0,1 be the decomposition of Ω,C into the i and i eigenspaces of I. Then dz j Ω 1,0 I x j = y j and the dz j form a basic for Ω 1,0 written as α = n α j dz j j=1 in the sense that every α Ω1,0 since can be where α j are C maps C (differentiable over R). Similarly, dz j forms a basis of Ω 0,1. Proposition If f : C is a (real) differentiable function (f = g + ih, g, h : R) then df = dg + idh Ω,C is given in local coordinates by df = f z j dz j + f z j dz j

10 10 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Proof. Expand it out: 1 2 ( f x j i f y j Definition Ω p,q which are given locally by α = I,J ) (dx j + idy j ) + 1 ( f + i f ) (dx j idy j ) 2 x j y j = ( 1 f dx j i f ) idy j + same 2 x j y j Ωp+q,C = f x j dx j + f y j dy j = df is the space of p + q-forms on with coefficients in C α I,J dz i1 dz i2 dz ip dz j1 dz j2 dz jq = }{{}}{{} dz I dz J α I,J dz I dz J where the sum is over all pairs of multi-indices I = (i 1, i 2,, i p ) and J = (j 1, j 2,, j q ) and α I,J C (, C). We need to show that this is independent of the choice of local coordinates. Given that, the following are obvious. I,J Proposition d = + where dω p,q Ω k,c = p+q=k Ωp+1,q : Ω p,q : Ω p,q Ω p,q Ω p,q+1 Ωp+1,q Ωp,q+1 2 = 0, + = 0, 2 = 0 Proof. The key part is the second line. Since dα I,J = α I,J z j dz j + αi,j z j dz j, dα = α I,J z j dz j dz I dz J + αi,j dz j dz I dz J z j Ω p+1,q Ω p,q+1 The important properties of the boundary operators, have been trivialized! The key step is independence of coordinates which I need to explain.

11 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Coordinate invariant definitions. Suppose that is a real n-manifold and Ω k is the space of k-forms on. Elements are alternating C ()-multilinear maps ω : ΓT ΓT C () Alternating means ω(σ 1,, σ k ) = 0 if the σ i are not distinct (equivalently, the sign changes if we switch two of them). C ()-multilinear means ω(f 1 σ 1,, f n σ k ) = f 1 f 2 f n ω(σ 1,, σ k ) for f i : R in C (). This second condition is equivalent to saying that ω is determined by its value at each point. I.e., it is a section of the k-th tensor power of the cotangent bundle. Given k vector fields σ 1,, σ k on we have ω(σ 1,, σ k ) : R. In local coordinates we have ω = ω I dx I I where the sum is over all multi-indices I = (i 1, i 2,, i k ) and dx I = dx i1 dx i2 dx ik. The differential d : Ω k Ωk+1 in the derham complex of has the following coordinate free definition. k dω(σ 0,, σ k ) = ( 1) i σ i (ω(σ 0,, σ i,, σ k )) + 0 i<j k i=0 ( 1) i+j ω([σ i, σ j ], σ 0,, σ i,, σ j,, σ k ) For example, when k = 1 and α Ω 1 we have dα(σ, τ) = σ(α(τ)) τ(α(σ)) α([σ, τ]) Proposition Suppose that ω = I ωi dx I Ω k. Then dω = ω I dx i dx I x I i i Proof. dω = c J dx J. To find c J for J = (j 0,, j k ): ( c j = dω(σ 0,, σ k ) = dω,, x j0 where σ i = x ji. But, [σ i, σ j ] = 0 for these vector fields (because their coefficients are constant functions). So, we only get the first sum in the definition of dω. But ω(σ 0,, σ i,, σ k ) = ω I x jk where I = (j 0,, ĵ i,, j n ). Call this I = δ i J. So, dω = c J dx J = ( 1) i ωδ ij dx J x J J i ji which is the same as the other equation since ( 1) i dx J = dx ji dx δi J Ω p,q, already defined, has the following coordinate free characterization. )

12 12 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Lemma A k-form on with coefficients in C: ω Ω k,c p + q = k if and only if it has the following property. Suppose that σ 1,, σ k are sections of either T 1,0 0,1 or T. Then ω(σ 1,, σ k ) = 0 lies in Ωp,q where unless exactly p of the σ i lie in T 1,0 and q of them lie in T 0,1. Note that this characterization makes sense for any almost complex manifold. In other words, Ω p,q is well-defined for any almost complex. The following observation from [Voisin] is the crucial concept underlying the construction of the double complex which was hidden in the other approach (which is also correct). Ω, Theorem The differential d : Ω k Ωk+1 sends Ωp,q only if the bracket conditions holds: [T 0,1, T 0,1 ] T 0,1 into Ωp+1,q and similarly for T 1,0 Ω p,q+1. if and Proof. Look at the definition of dω. Assume the bracket conditions. In the first sum, one of the σ i is deleted. So, in order to have p holomorphic and q anti-holomorphic sections we needed to start with p + 1, q or p, q + 1. In the second sum, if σ i, σ j are in T 1,0 then so is [σ i, σ j ]. So, the number of terms in then so is [σ i, σ j ]. 0,1 and the other in T then [σ i, σ j ] is in T,C = T 1,0 decreases by one in that case. Similarly, if σ i, σ j are in T 0,1 Finally, if one of σ i, σ j is in T 1,0 T 1,0 T 0,1. Conversely, suppose the bracket condition does not hold. Then, e.g, there might be σ i, σ j in T 1,0 so that [σ i, σ j ] has a T 0,1 component. Such a component will be linearly independent from all other terms, so there is a ω Ω p,q+1 so that dω is nonzero on that term and zero on all other terms giving dω a Ω p+2,q component. Remark (1) The equation d 2 = 0 which implies that 2 = 0, 2 = 0, + = 0 follows from the definition since any differential form on with coefficients in C is the sum of two real forms ω = α + iβ and dω = dα + idβ. So, d 2 ω = d 2 α + id 2 β = 0. (2) The bracket condition holds for T 1,0 0,1 if and only if it holds for T because these are related by complex conjugation (which acts on the second factor of T C) and [σ, τ] = [σ, τ] So, conjugating both sides of [T 0,1, T 0,1 ] T 0,1 gives, T 0,1 0,1 ] = [T, T 0,1 1,0 ] = [T, T 1,0 ] T 0,1 = T 1,0 [T 0, Local exactness of the complex. Review: We have a complex manifold. This is an almost complex manifold (, I) satisfying the bracket condition [T 0,1, T 0,1 ] T 0,1. Complexified differential forms are: Ω k,c = Ω k C = p+q=k Ω p,q

13 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 13 In local coordinates, elements have the form α = I,J α I,J dz I dz J where I = (i 1,, i p ), J = (j 1,, j q ). We proved that the standard differential d : Ω k Ωk+1 only has terms in degree (1, 0) and (0, 1): where α = I,J α = I,J It follows immediately that d = + : Ω p,q Ωp+1,q α I,J dz I dz J = I,J,i α I,J dz I dz J = I,J,i Ω p,q+1 α I,J dz i dz i dz I dz J α I,J dz i dz i dz I dz J = 0, + = 0, = 0 Today, we will prove the local exactness of : Theorem Suppose α Ω p,q with q > 0 so that α = 0 in a nbh of some point x. Then there is a β Ω p,q 1 so that β = α in a nbh of x. To prove this we have to go back to Chapter I One complex variable. We start with Stokes Theorem: dα = α M n M if M is a (real) n-manifold α Ω n 1 M and either (1) M is compact or (2) α has compact support We apply this to the case n = 2 and M C. (1) M is compact in C. Suppose α = f(z)dz where f is complex differentiable ( f = 0). Then z ( ) f f dα = dz + z z dz dz = 0 since dz dz = 0 and f = 0. So, z 0 = f(z)dz = 0 M This implies: Theorem (Cauchy s formula). If f is C 1 and f = 0 then, for all z < 1, z f(z) = 1 f(ζ) 2πi ζ z dζ D 1

14 14 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Proof. By Stokes thm this integral is equal to: 1 f(ζ) (2.1) 2πi ζ z dζ D ε where D ε is the ε-disk around z. But (as we will see in a minute) this integral gives the average value of f(ζ) on D ε. So, as ε 0, this converges to f(z). Since it is independent of ε, it must be equal to f(z) for all ε. This formula shows that f(z) is analytic since the integrand is given by a converging (geometric) series which can be differentiated term by term by the dominated convergence theorem. (2) 2nd example. Suppose f is not holomorphic but has compact support. Then (2.1) is not equal to f(z), but it is still equal to the average value of f(ζ) as we now verify: Let ζ = ζ z. Then ζ = ε. So, ζ = εe iθ, dζ = iζ dθ and ( ) = 1 2πi 2π 0 f(z + ζ ) ζ iζ dθ = 1 2π 2π 0 f(z + ζ )dθ = ave f(ζ) This converges to f(z) as ε 0. By Stokes thm, ( ) = 1 ( ) f(z + ζ ) 2πi C D ε(z) ζ dζ dζ f(z) ζ f(z) = 1 1 f(z + ζ ) 2πi C ζ ζ dζ dζ f(z) = 1 1 f(z + ζ ) dζ dζ 2πi C ζ z f(z) = 1 1 z 2πi C ζ f(z + ζ ) dζ dζ }{{} So, f(z) = g where z g(z) = 1 f(ζ) dζ dζ 2πi C ζ z If f is C k then this integral can be (partially) differentiated k times under the integral sign (integral over C D ε ). So, g is C k. Theorem Any C k function f : C C is locally equal to g z for some Ck function g : C C. We need the multi-variable version of this: Lemma Suppose that U, V are open subsets of C p, C q and f : U V C is holomorphic in the first p variables. I.e., f z i = 0 for i p. Then, locally, f = g z p+1 where g : U V C is also holomorphic in the first p variables. =g(z)

15 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 15 Proof. Replace V with C q and multiply f with C function on V with compact support. Then let g(u, z p+1,, z n ) = 1 f(u, ζ, z p+2,, z n ) dζ dζ 2πi C ζ z p Back to differential forms. We are proving Theorem Suppose that α Ω p,q with q > 0 so that α = 0. If α = α I,J dz I dz J I,J then α = α I,J dz k dz I dz J z k I,J,k For this to be zero, the sum of all terms with the same dz I must be zero. So, we may assume p = 0 and α = α J dz J Ω 0,q J Let t be the smallest index which occurs in any J. Then Every α J is holomorphic in every z i where i < t since the term α J z i dz i dz J must be zero (it cannot be cancelled with another term in α = 0.) α = dz t β + α where the smallest index that occurs in either α or β is t > t and β = K β K dz K where each β K is holomorphic in z 1,, z t 1. By Lemma 2.4.3, there are g K holomorphic in z 1,, z t 1 so that g K = β K z t Then, g K dz K = β K dz t dz K + terms with smallest index > t K K So, α K gk dz K has smallest index > t. By downward induction on t, α = γ for some γ Ω 0,q. For p > 0 we have α = α I dz I and α I = 0 for each I. So, α I = γ I for some γ I Ω 0,q 1 and α = γ I dz I I

16 16 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 2.6. Dolbeault complex. Although Ω p,q is a bicomplex with vertical and horizontal boundary maps,, we will see that, for a holomorphic bundle, only gives a welldefined boundary map. Let E be a holomorphic bundle over a complex manifold. By definition this is locally trivial, we have local trivalizations ϕ i : E Ui U i C k whose transition maps ϕ ij = ϕ 1 j ϕ i give a k k matrix of holomorphic maps U i U j M k (C) = C k2 let A 0,q (E) = Ω 0,q E be the space of (0, q)-forms on with coefficients in E. In local coordinates an element α A 0,q (E) is given by where each α i Ω 0,q U. Let α be given by α = (α 1,, α k ) α = (α 1,, α k ) Given two coordinate charts (U i, ϕ i ), (U j, ϕ j ), we need to know that α defined by this formula agree on U i U j. Then we can conclude that is a well defined operator : A 0,q (E) A 0,q+1 (E) Proposition The boundary α = (α 1,, α k ) is well defined. I.e., if α is changed to α = (α 1,, α k ) by holomorphic change of coordinates α t = (ϕ ij ) st α s then α is related to α by the same change in coordinates. Proof. This follows from the Leibnitz rule and the fact that ϕ ij is holomorphic. So, α t = (ϕ ij ) st α s = (ϕ ij ) st α s Definition The Dolbeault complex of a homomorphic bundle E is A 0,q 1 (E) A 0,q (E) A 0,q+1 (E) I forgot to mention the key point: Proposition The kernel of : A 0,0 (E) = Ω (E) A 0,1 (E) is the space of holomorphic sections of E.

17 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Kähler manifolds Having a Hermitian metric on a complex manifold puts strong restrictions on its cohomology Kähler forms. We started with the basic concept of a Kähler form. Suppose that V is a vector space over C: V = C n and W R = Hom(V, R), W C = W R C = W 1,0 W 0,1 W 1,1 = W 1,0 W 0,1. We want to look at W 1,1 R = W 1,1 2 W R. Lemma ω W 1,1 R form so that if and only if ω : V V R is a skew symmetric R-bilinear (3.1) ω(iu, Iv) = ω(u, v) for all u, v V. Proof. First note that the condition ω(iu, Iv) = ω(u, v) is equivalent to the condition (3.2) ω(u, Iv) + ω(iu, v) = 0 since I 2 = 1. By definition, W 1,1 R is the set of skew-symmetric forms ω on V so that ω C = ω C lies in W 1,1. This is equivalent to the condition that ω C vanishes on pairs of vectors from V 1,0 or from V 0,1. But V 1,0 is the set of all vectors of the form where u V and ũ = u iiu ω(ũ, ṽ) = ω(u iiu, v iiv) = ω(u, v) ω(iu, Iv) i(ω(u, Iv) + ω(iu, v)) which is zero if and only if (3.1) and (3.2) hold. Note that (3.2) implies that g(u, v) := ω(u, Iv) is a symmetric bilinear pairing g : V V R since g(v, u) = ω(v, Iu) = ω(iu, v) = ω(u, Iv) = g(u, v) Definition A hermitian form on a complex vector space V is defined to be an map h : V V C so that (1) h(u, v) is C-linear in u (2) h(u, v) is C-anti-linear in v (3) h(v, u) = h(u, v) Note that (3) implies that h(v, v) R. The form h is said to be positive definite if h(v, v) > 0 for all v 0. A positive definite hermitian form on V is called a Hermitian metric on V.

18 18 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Proposition There is a 1 1 correspondence between hermitian forms h on V and ω W 1,1 R given by ω = Ih and h(u, v) = g(u, v) iω(u, v) where g : V V R is given by g(u, v) = ω(u, Iv). Exercise Show that there is a 1-1 correspondence between hermitian forms h and symmetric forms g satisfying g(iu, Iv) = g(u, v) Kähler metrics. Suppose that (M, I) is a complex manifold. Then a Hermitian metric h on M is a Hermitian metric h x on the tangent space T M,x at each point which varies smoothly with x M. Associated to h we have: ω = Ih which is a 2-form on M which is also in Ω 1,1 M ω(iu, Iv) = ω(u, v) for any two vector vectors u, v T M,x at each point x M. which is equivalent to the equation Proposition The real part of a Hermitian metric h is a Riemannian metric g on M which is also invariant under I: g(u, v) = ω(u, Iv) = g(iu, Iv) Proof. We know that g is a symmetric real form. If v 0 T M,x is a nonzero vector, h(v, v) = h(v, v) is a positive real number. So, So, g is a Riemannian metric on M. g(v, v) = h(v, v) > 0 Note that M is oriented since any complex vector space has a natural real orientation. Theorem The volume form on M associated to g = Rh is equal to ω n /n!. To prove this, we need the matrices for h, ω in local coordinates. Take a coordinate chart U with holomorphic coordinates centered at x 0 U (z(x 0 ) = (0, 0,, 0)) z = (z 1,, z n ) : U C n and z = (z 1,, z n ). Then dz j, dz j form bases for A 1,0, A 0,1. These are the bundles with fibers W 1,0, W 0,1 for V = T M,x. So h A 1,1 is given by h = α ij dz i dz j where α ij : M C (notation: α ij Ω M ). These are given by: ( ) α ij = h, x i x j Since h(u, v) = h(v, u), α ji = α ji.

19 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 19 Since the negative imaginary part of a complex number z is given by i (z z), we 2 have: ω = i (αij dz i dz j α ji dz i dz j ) = i αij dz i dz j 2 2 Proof of Theorem We show equality at the point x 0 using complex ortho-normal coordinates at that point. By a C-linear change of the coordinates z = (z 1,, z n ), we can arrange for to be ortho-normal at the point x 0. In other words, Then, at x 0, x i α ij (x 0 ) = h x0 ( x i, ) = δ ij x j So, ω x0 = i 2 dzj dz j = dx j dy j ω n = π S n dx π(1) dy π(1) dx π(2) dy π(2) dx π(n) dy π(n) which is n! times the volume form dx 1 dy 1 dx n dy n at the point x 0. Since this hold for every point, ω n /n! is the volume form at each point Kähler manifolds. Definition A Kähler manifold is a complex manifold with a Hermitian metric h so that ω is closed (dω = 0). Corollary On a compact Kähler manifold M, for every 1 k n, the form ω k is closed but not exact. I.e., [ω k ] 0 H 2k (M). Proof. ω k is clearly closed: If ω k = dα then dω k = kω k 1 dω = 0 d(α ω n k ) = dα ω n k = ω n which is impossible since the volume form is nonzero in H 2n (M) when M is an oriented compact manifold. Corollary A compact complex k-submanifold N of a Kähler manifold M cannot be the boundary of a (real) submanifold of M. This follows easily from Stokes Theorem if we take the Kähler form on N to be the pull-back of the Kähler form on M. See [Voisin].

20 20 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 3.4. Connections. Recall that if E is a real C k-dimensional vector bundle on a real manifold, a connection on E is an R-linear map: satisfying the Leibnitz equation Note: if we had another one then : C (E) A 1 (E) = Ω,R E = Hom R (T, E) (fσ) = df σ + f σ ( )(fσ) = f( )σ In other words, the difference ϕ = is a homomorphism of Ω -modules. Such morphisms are given by matrices: Suppose that e 1,, e k is a basis of local sections of E. Then ϕ(e i ) = ϕ ij e j where ϕ ij Ω,R. Any section of E is given by f i e i where f i Ω. Then or: ϕ( f i e i ) = f i ϕ(e i ) = f i ϕ ij e j ϕ(f 1,, f k ) = (f 1,, f k )[ϕ ij ] In local coordinates, d is a connection. So, an arbitrary connection is given by = d+ϕ or: (f 1,, f k ) = (df 1,, df k ) + (f 1,, f k )M where M is a k k matrix with entries in Ω,R. If M has a Riemannian metric g then recall that the Levi-Civita connection is the unique connection on T M having the properties: (1) dg(σ, τ) = g( σ, τ) + g(σ, τ) ( is compatible with g) (2) (τ)σ (σ)τ = [σ, τ] (usually written with the notation σ (τ) = (τ)σ) for any two vector fields σ, τ. Suppose that E is a holomorphic bundle on a complex manifold M. Then any connection on E has two components: : C (E) A 1 (E) = A 1,0 (E) A 0,1 (E) We write 1,0, 0,1 for these two components. Last time we showed that E : C (E) A 0,1 (E) given in local coordinates by U (f 1,, f k ) = (f 1,, f k ) is well defined. Proposition Given a Hermitian metric h on a holomorphic bundle E, there is a unique connection on E so that (1) dh(σ, τ) = h( σ, τ) + h(σ, τ) for all σ, τ C (E) ( is compatible with h) (2) 0,1 = E This unique connection is called the Chern connection on E.

21 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 21 Proof. = 1,0 + 0,1 where 0,1 = E and 1,0 is uniquely determined by: since h( 0,1 σ, τ) = 0 and h(σ, 1,0 τ) = 0. dh(σ, τ) = h( 1,0 σ, τ) + h(σ, τ) T 1,0 it,r ṽ = v iiv R v T,R 1,0 Since iṽ = iv + Iv = Ĩv, R(iṽ) = IR(ṽ) making R : T T,R an isomorphism of complex vector spaces with i acting as i on T 1,0 and as I on T,R. Theorem Let (M, I) be a complex manifold with a Hermitian metric h. Then the following are equivalent. (1) h is a Kähler metric (dω = 0). (2) LC (Iσ) = I LC (σ) for every real vector field σ on. (3) The holomorphic part of the Chern connection is equal to the Levi-Civita connection: R( 1,0 ) = LC (So, the Chern connection is ( LC, ).) Proof. (3) (2). The Chern connection is complex linear and the holomorphic part is the part where i = I: R(iσ) = IR(σ) for σ a section of T 1,0. So, LC (Iσ) = R( 1,0 (Iσ)) = R(i 1,0 (σ)) = IR( 1,0 (σ)) = I LC (σ) (2) (1). Since = LC is compatible with g = R(h), we have d(g(σ, τ)) = g( σ, τ) + g(σ, τ) Since g(σ, τ) Ω (), all three terms are 1-forms on. For example, if σ = ξ i α i where ξ i are vector fields and α i are 1-forms on, then g( σ, τ) = g(ξ i, τ)α i Applying both sides to the vector field φ we get: g( σ, τ)(φ) = g(ξ i, τ)α i (φ) = g(ξ i α i (φ), τ) = g( σ(φ), τ) = g( φ σ, τ)

22 22 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Replacing τ with Iτ we get g(σ, Iτ) = ω(σ, τ) and d(ω(σ, τ)) = ω( σ, τ) + ω(σ, τ) Apply both to the vector field φ (and use the rule df(φ) = φ(f)): Cyclically permute the three vector fields: φ(ω(σ, τ)) = ω( φ σ, τ) + ω(σ, φ τ) σ(ω(τ, φ)) = ω( σ τ, φ) + ω(τ, σ φ) τ(ω(φ, σ)) = ω( τ φ, σ) + ω(φ, τ σ) Now use the coordinate invariant definition of dω: dω(φ, σ, τ) = φ(ω(σ, τ)) σ(ω(φ, τ)) + τ(ω(φ, σ)) ω([φ, σ], τ) + ω([φ, τ], σ) ω([σ, τ], φ) This is zero since [φ, σ] = φ σ σ φ. (1) (3). The proof is by reduction to the standard case: If the metric h is constant, then LC = d and Ch = (, ) and R() = d. We only need to show that, at each point, the metric h can be made constant to first order when it is Kähler. So, it suffices to prove the following lemma. Lemma If h is a Kähler metric on then, in a nbh of each point, there are holomorphic coordinates z i so that the matrix of h: ( ) ( ) h ij = h, = h, x i x j z i z j is the identity matrix plus O( z 2 ). Proof. We can choose coordinates which are ortho-normal at the chosen point (z = 0). This makes the constant term of h ij the identity matrix. But we also have linear terms: h ij = δ ij + ɛ ij + ɛ ij + O( z 2 ) where ɛ ij is a linear combination of z k (ɛ ij are holomorphic) ɛ ij = ɛ k ijz k and ɛ ij is a linear combination of z k (ɛ ij are antiholomorphic): Since h is conjugate symmetric we have: ɛ ij = ɛ k ijz k The key property of these numbers is: Claim: If h is Kähler then ɛ ij = ɛ ji ɛ k ij = ɛ i kj Proof: Since δ ij = 0 and z k = 0 ɛ ij = 0, at the point z = 0 we have: 0 = ω = i ɛij dz i dz j = i ɛ k 2 2 ij dz k dz i dz j

23 Then Now let Making MATH 250B: COMPLE ALGEBRAIC GEOMETRY 23 z j = z j + 1 ɛ k 2 ij z i z k = z j ɛk kjzk 2 + ɛ k ijz i z k i<k dz j = dz j + ɛ k ijz k dz i = dz j + ɛ ij dz i dz j = dz j ɛ ij dz i + O( z 2 ) = z j = (δ z i z i ij ɛ ij ) + O( z 2 ) z j z j So, up to terms of second order, we have: ( ) h ij = h, z i z = (δ ik ɛ ik )h kl (δ jl ɛ jl ) j k,l = k,l (δ ik ɛ ik )(δ kl + ɛ kl + ɛ kl)(δ jl ɛ lj) since ɛ jl = ɛ lj. = δ ik δ kl δ jl ɛ ik δ kl δ jl + δ ik ɛ kl δ jl + δ ik ɛ klδ jl δ ik δ kl ɛ jl = δ ij ɛ ij + ɛ ij + ɛ ij ɛ ij = δ ij 3.5. Examples. An easy example is a Riemann surface. This is a complex 1-dimensional and real 2-dimensional manifold. Any Hermitian metric is Kähler since all 2-forms on a real 2-dimensional manifold are closed. The next example was CP n = P n (C). We constructed the Fubini-Study metric on complex projective space P n (C) and showed that it is a Kähler metric. This implies that all smooth projective varieties over C are Kähler manifolds. The outline of the construction is: h L ω h h ω Given any holomorphic line bundle L on a complex manifold, there is an associated 2- form ω L on (giving the Chern class of L). This 2-form ω is associated to a hermitian form h ω (since g, ω, h determine each other) which, if we are lucky, will be positive definite and therefore a Kähler metric. A line bundle L over is the union over open sets U i of U i C. For each U i, take the unit section σ i (v) = (v, 1). These in general don t match. So, there are functions g ij : U i U j C so that σ i (v) = g ij (v)σ j (v) for all v U i U j. Since σ j = g jk σ k we have the equation g ik = g ij g jk on U i U j U k. Conversely, any collection of maps g ij : U i U j C satisfying the equations in the box will uniquely determine a line bundle. If the g ij are holomorphic functions, the line bundle will be a holomorphic bundle.

24 24 MATH 250B: COMPLE ALGEBRAIC GEOMETRY For example, gij := 1 g ij is another collection of functions satisfying the boxed equations. So, {gij} gives another holomorphic line bundle L which one can show is the dual bundle to L. Let h be a Hermitian metric on L. Let h i : U i R + be the positive function given by h i (v) = h(σ i (v), σ i (v)). Since σ i = g ij σ j we get: h i = h(σ i, σ i ) = g ij g ij h(σ j, σ j ) = g ij g ij h j Lemma Conversely, any family of functions h i : U i R satisfying the equations h i = g ij g ij h j gives a Hermitian metric on L. Proof. Let h i be another collections of functions so that h i = g ij g ij h j. On each U i let f i = h i/h i. Then f j = h j/h j = g ji g ji h i/g ji g ji h i = h i/h i = f i. So, f = f i = f j is a globally defined function on and h = fh is another metric on L. For example, h i = 1 h i satisfies h i = gijg ij h j Therefore, h i gives a metric on L. Let ω i = 1 2πi log h i Note that log h i = log g ij + log g ij + log h j Since g ij is holomorphic, log g ij = 0. Since g ij is anti-holomorphic, log g ij = 0. So, ω i = 1 2πi log h i = 1 2πi log h j = ω j So, ω = ω i is a well-defined 2-form on all of. This form is closed since it is by definition exact on each U i. This is called the Chern form of L. Now let = P n (C). Recall that this is the quotient space of C n+1 \0 modulo the relation (z 0, z 1,, z n ) (λz 0,, λz n ) for any λ 0 C. The equivalence class is denoted [z 0,, z n ]. Another interpretation is that P n (C) is the set of one dimensional subspaces of C n+1. Each such is uniquely determined by any nonzero vector (z 0,, z n ) and we make the identification = [z 0,, z n ]. Let S be the tautological line bundle over P n (C) given by S = {(, v) P n (C) and v } This is tautological since the fiber over is. Let U i be the open subset of P n given by Let σ i be the section of S over U i given by σ i ( ) = σ i ([z 0,, z n ]) = U i = {[z] z i 0} ( z0,, z i = 1,, z ) n z i z i z i

25 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 25 Thus σ i ( ) is the unique element of with ith coordinate equal to 1. Compare this with ( z0 σ j ([z]) =,, z i,, z ) n z j z j z j We see that σ i = z j σ j z i So, g ij = z j /z i with dual gij = z i /z j. Since the line bundle S is a subbundle of the trivial bundle P n C n+1 it gets a metric by restricting the standard metric on C n+1 given by h(z) = z j 2. Since the ith coordinate of σ i is 1 we get: On the dual bundle S we have h(σ i ) = 1 + j i h (σ i ) = z j z j 2 Using z j 2 = z j z j, the Chern form of S on U i is ω i = 1 ( ) 1 log 2πi 1 + z j z j which we calculated step by step: ( ) 1 log 1 + = z j dz j z j z j 1 + z j 2 where we used the formula f = f z j dz j ( ) ( 1 (1 + log 1 + zj 2 ) dz j dz j ) z i z j dz i dz j = z j z j (1 + z j 2 ) 2 where we used the formula (fdz j ) = f z i dz i dz j. At the origin z = 0 we get ω = i dzj dz j 2π which is the standard form corresponding to (a scalar multiple of) the standard metric with matrix equal to the identity matrix (divided by π). So, the metric h ω is positive definite at the point z = 0. However, the space P n (C) is homogeneous (the same at every point). This is easier to see if we use a vector space without a basis: Let V be any n + 1 dimensional vector space over C and let P(V ) be the space of 1-dimensional subspaces of V. Then it is clear that every point is the same as every other point. The tautological bundle S and its dual are also defined without choice of coordinates. So, we can choose coordinates to make any point the center point z = 0. So, the canonically defined metric h ω is positive definite at every point.

26 26 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 4. Harmonic forms The main theorem is that the k-th cohomology of a complex manifold is isomorphic to the space of Harmonic forms of degree k and these are constructed using the formal adjoint of the boundary operator d in the de Rham complex of L 2 forms and the Hodge star operator. Recall that the L 2 -norm of a continuous function f on a compact manifold is f 2 := f 2 dµ where dµ is a measure on. Recall the standard definition of an L 2 -form. If V is a vector bundle over with a Riemannian metric g and σ, τ are sections of V then (σ, τ) L 2 := g(σ(x), τ(x)) x dµ Note that g(σ(x), τ(x)) x : R is a continuous function (if σ, τ, g are continuous). The L 2 -norm is σ 2 = (σ, σ) L L 2 -norm on forms. Suppose now that is a oriented Riemannian manifold. Ideally the volume form on should be given by dx 1 dx 2 dx n. However, this formula will only work at the single point x = 0 as we saw last time. We use a different approach. Let e 1,, e n be a positively oriented ortho-normal (o-n) basis for the tangent space of. We can choose these to vary smoothly with x in some open nbh U of any given point. In other words, this is a smoothly varying o-n frame for the tangent space. Let e 1,, e n T,x be the dual basis in the cotangent space T,x. We define this to be an ortho-normal basis for the vector space T,x = R n. This defines a metric on the (real) cotangent bundle Ω 1,R = T,R. Recall the notation: Ω k,r is the bundle over with fiber k T,x = R (n k). A k () is the space of k-forms on which are the sections of the bundle Ω k,r. The vector space k T,x = R (n k) has a basis of k-forms e I = e i i e i k where I = (i 1 < i 2 < < i k ) is multi-index notation. We define the vectors e I, I = k to be an ortho-normal basis for k T,x. In other words, given any two k-forms α = I =k αi e I and β = I =k βi e I on, (α, β) := α I β I : R Proposition The induced norm on k T,x oiginal basis e 1,, e n. is independent of the choice of the

27 MATH 250B: COMPLE ALGEBRAIC GEOMETRY 27 Proof. We can define the metric without choosing a basis: (α 1 α k, β 1 β k ) := π S k sgn π (α π(i), β i ) It is easy to see that this defines a metric on the space of k-forms and that e I ortho-normal with respect to this metric. Of particular importance is the case k = n where n T,x = R is one-dimensional with unit vector Vol := e 1 e n This is called the volume form on. This is a well-defined globally defined n-form on since is oriented. The L 2 -norm on A k () is defined to be (α, β) L 2 := (α, β) Vol where we note that (α, β) : R is the smooth mapping sending x to (α(x), β(x)) x Hodge operator. : k T,x n k T,x is defined to be the unique R-linear map determined by: α β = (α, β) Vol This linear map is the composition of two isomorphisms: m : k T,x Hom R ( k T,x, R) Hom R ( k T,x, k T,x) β (, β) x (, β) x Vol and the inverse of p : n k T,x Hom R ( k T,x, k T,x) γ γ To see that p is an isomorphism, look at what it does to the o-n basis e I of n k T,x { : e J e ± Vol if I J = {1, 2,, n} I = 0 otherwise Thus, p takes o-n basis to o-n basis. It is an isometry. The formula for is γ = β = p 1 m(β) or β = (, β) x Vol This gives an isomorphism of vector bundles : Ω k,r Ω n k,r and on forms: : A k () A n k () Furthermore, substituting the boxed equation into the definition of (α, β) L 2 gives: (α, β) L 2 = α β And this determines β uniquely (outside a set of measure zero). are

28 28 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Lemma : k T,x = n k T,x is an isometry. Lemma = ( 1) k(n k) = ( 1) k(n 1) on A k (). Proof. For any α, β A k () we have: α β = (α, β) Vol definition of = ( α, β) Vol is an isometry = ( β, α) Vol (, ) is symmetric = β 2 α definition of = ( 1) k(n k) 2 α β Since this is true for all β we get α = ( 1) k(n k) 2 α for all α. Note that { ( 1) k(n k) ( 1) k if n is even = +1 if n is odd Extension to complex case. I used a really simple example to explain what is meant by saying Extend by C-linearity and (, ) to a hermitian pairing on T,C = T,R C. The example is k = 1 and V = C. This has real basis 1, I and W = 1 V = Hom(V, R) = R 2 with basis 1, I. You know these maps as 1 = R : V R I = I : V R By definition, these form an o-n basis for V = {a a 2 I : a 1, a 2 R}: If α = a a 2 I and β = b b 2 I then (α, β) = a 1 b 1 + a 2 b 2. The volume form on V is Vol = 1 I V. Claim: (b b 2 I ) = b b 1 I. β I β Verification of claim: α β = (a a 2 I ) ( b b 1 I ) = (a 1 b 1 + a 2 b 2 )(1 I ) = (α, β) Vol Now extend by C-linearity to 1 W C = Hom R (V, C) = W C = C 2 which has the same basis 1, I over C. Elements of W C are β = b 1 1 +b 2 I where b 1, b 2 C. Extending by C-linearity means β = b b 1 I. Then β = b b 1 I. α β = (a 1 b 1 + a 2 b 2 )(1 I )

29 which illustrates the general formula: where the volume form is MATH 250B: COMPLE ALGEBRAIC GEOMETRY 29 α β = (α, β) C Vol 1 I = dx dy = i dz dz 2 The last form explains two things: the factor of 2 when comparing the different versions of the metric and the fact that the volume form is in W 1,1. We verified that Lemma holds in the complex case: α β = (α, β) Vol definition of in complex case = ( α, β) Vol is an isometry = ( β, α) Vol (, ) is conjugate symmetric = β 2 α definition of = ( 1) k(2n k) 2 α β β A 2n k () Therefore, in the complex case we get: 2 α = ( 1) α 4.2. Adjoint operators. The formal definition of the adjoint of d is given by: (α, d β) L 2 = (dα, β) L 2 This determines d β uniquely (outside a set of measure zero) if it exists. Existence in the real case is given by the following formula. Lemma d β = ( 1) β 1 d( β) Proof. Note that, if β A k then (dα, β) makes sense only when α A k 1. (dα, β) L 2 = dα β But, d(α β) = dα β + ( 1) k 1 α d( β). So, (dα, β) L 2 = ( 1) k α d( β) So, ( 1) β 1 d β = d β. = ( 1) k (α, 1 d β) L 2 This gives: { d ( 1) k d = d if n is odd if n is even

30 30 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Complex adjoints. In the complex case we have d = d. But we also have the decomposition d = +. Lemma The formal adjoints of and are = and =. Proof. Very similar to the real case: (α, β) L 2 = α β But, φ = 0 for any 2n 1 form φ since φ must be an (n, n 1) form. So, φ = 0 because A n+1,n 1 () = 0 making φ = dφ = φ = 0 since =. Since and α + 1 = β we get So, = and similarly for. (α β) = α β + ( 1) α α ( β) (α, β) L 2 = ( 1) β α ( β) = ( 1) k (α, 1 ( β)) L 2 = (α, ( β)) L 2 Given a holomorphic vector bundle E over so that E (and ) have Hermitian metrics, we also have a Hodge operator E : Ω 0,q E Ωn,n q E = Ω 0,n q K E where K = Ω n,0 is the canonical line bundle over. There is a contraction map c : E E C given by evaluation. Then E is uniquely determined by: c(α E β) = (α, β) Vol A n,n () But, we take contraction c to be understood and we don t write it. E is a conjugate linear isomorphism. So, both sides are conjugate linear in β. We define the L 2 -norm on A 0,q (E) by (α, β) L 2 := (α, β) Vol = α E β Lemma The operator E : A 0,q (E) A 0,q 1 (E) has formal adjoint given by E = ( 1) q 1 E K E E I.e., the following diagram (with all arrow isomorphisms) commutes up to sign. A 0,q (E) E A n,n q (E ) A 0,n q (K E ) E A 0,q 1 (E) K E E A n,n q+1 (E ) A 0,n q 1 (K E ) The proof, using Stokes, is the same as before.

31 MATH 250B: COMPLE ALGEBRAIC GEOMETRY Laplacian. For a Riemannian manifold M we have the Laplace operator: d = dd + d d. For a complex manifold with Hermitian metric we have = +, = + If E is a holomorphic bundle over with a Hermitian metric we also have E = E E + E E Theorem ker d = ker d ker d and similarly for,, E. Proof. By definition of adjoint we have: (α, d α) L 2 = (dα, dα) L 2 + (d α, d α) L 2 But the metric is positive definite. So, both terms are 0 and equal to zero iff dα, d α are both zero. Definition Elements of ker d are called harmonic forms. We also have - harmonic forms, -harmonic forms. E -harmonic forms are also called harmonic (0, q)-forms with coefficients in E. We need to verify that the Laplace operators are elliptic differential operators and we will quote some theorems about such operators. Our goal is to understand the definitions and statements Differential operators. We start with an elementary example. Hook s law says that, for a spring, there is a force which is negative the displacement (times a constant): d 2 f(t) = f(t) dt 2 Or, f C (R) is in the kernel of P where P (f) = d2 f(t) + f(t) dt 2 The key point is that P is linear. P (af + bg) = ap (f) + bp (g) So, ker P, the space of solutions f of Hook s law, is a vector subspace of C (R). P is a second order linear differential operator in one variable t and one function f. Now, suppose we have several functions f 1,, f p in several variables: f j (x 1,, x n ) C (R n ) and several equations P 1,, P q of the form where, for I = (i 1, i 2,, i k ), x I = x I P i = P I,i,j f j x I is multi-index notation for k = x i1 x i2 x ik x i1 x ik

32 32 MATH 250B: COMPLE ALGEBRAIC GEOMETRY (When k = 0, I = and x is the identity map f x = f.) Since x i, x j commute, the indices i s can be permuted. If each P I,i,j = P I,i,j (x) C (R n ) is a function only of x, P = (P 1,, P q ) is called a (linear) differential operator of order k. So, ker P = {(f 1,, f p ) : P (f) = 0} is a vector subspace of C (R n ) p. Now we replace R n with an n-dimensional manifold M. The functions f j will be replaced with sections of a bundle E, and P i will be sections of another bundle F. In this setting, P is a (linear) morphism of sheaves: P : C (E) C (F ) Here C (E) is the sheaf of sections of E. This is the functor which assigns to each open subset U M the space C (E U ) of sections of E U. When we say P is a morphism of sheaves we mean it is locally defined, such as a differential operator. A morphism P : C (E) C (F ) is called a differential operator of order k if, on each coordinate chart U with local coordinates x 1,, x n for U, α 1,, α p : E U R giving an isomorphism α : E U = U R p and β = (β 1,, β q ) : F U = U R q, there are smooth functions P I,i,j : U R so that, for any section γ of E, β i (P (γ)) = P I,i,j α j (γ) x I where α j (γ), β i (P (γ)) are the smooth functions on U given by U γ α j E U R and U P (γ) β i F U R. When k = 0, P Hom(E, F ): β i = P ij α j (P (x) Hom R (E x, F x ) for each point x when k = 0.) What happens when we change the choice of coordinates? When we change the β i nothing much happens. We just get a linear combination of the previous differential operators. When we change the α j we use the Leibnitz rule. A simple example will show the idea. Suppose p = q = 1 and the equation is the first order differential operator: β(γ) = φ(x) α(γ) x we change α = α 1 by α(γ) = ψ(x)α (γ). (The change is linear, so it is the same on the entire fiber.) Then β(γ) = φ(x) ψ(x)α (γ) x = φ(x)ψ(x) α (γ) x + φ(x) ψ(x) }{{ x } C (U) α (γ) }{{} order 0 The second term is an order 0 differential operator. This is still a linear differential operator of order 1. If we ignore the lower order terms, the leading (degree k) term is linear with respect to the change of coordinates of E and F. Proposition The definition of a linear differential operator of order k is independent of choice of coordinates for E, F, M.

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