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Complex Manifolds

Lectures by Julius Ross

Notes by Tony Feng

Lent 2014

Preface

These are lecture notes for a course taught in Cambridge during Lent2014 by Julius Ross, on complex manifolds. There are likely to be er-rors, which are the fault of the scribe. If you find any, please let me knowat [email protected].

Contents

Preface i

1 Complex Manifolds 1

1.1 Several complex variables . . . . . . . . . . . . . . . . . . . . 1

1.2 Complex manifolds . . . . . . . . . . . . . . . . . . . . . . . 3

2 Almost Complex Structures 6

2.1 Linear algebra preliminaries . . . . . . . . . . . . . . . . . . 6

2.2 Almost complex structures . . . . . . . . . . . . . . . . . . . 7

3 Differential Forms 10

3.1 Differential forms on Complex Manifolds . . . . . . . . . . . 10

3.2 Dolbeault Cohomology . . . . . . . . . . . . . . . . . . . . . 12

3.3 The Mittag-Leffler Problem . . . . . . . . . . . . . . . . . . 13

4 The ∂ Poincare Lemma 15

4.1 Holomorphic Functions on C . . . . . . . . . . . . . . . . . . 15

4.2 Holomorphic functions of several complex variables . . . . . 17

4.3 The Poincare Lemma . . . . . . . . . . . . . . . . . . . . . . 18

5 Sheaves and Cohomology 21

5.1 Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.2 Cech Cohomology . . . . . . . . . . . . . . . . . . . . . . . . 24

5.3 Properties of Cech cohomology . . . . . . . . . . . . . . . . 27

6 More on Several Complex Variables 31

6.1 Hartog’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 31

7 Holomorphic Vector Bundles 34

7.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . 34

7.1.1 Bundle Constructions . . . . . . . . . . . . . . . . . . 35

7.2 Holomorphic Line Bundles . . . . . . . . . . . . . . . . . . . 35

ii

Contents iii

7.3 Line bundles on Projective Space . . . . . . . . . . . . . . . 37

7.4 Ample Line Bundles . . . . . . . . . . . . . . . . . . . . . . 38

8 Kahler Manifolds 40

8.1 Kahler metrics . . . . . . . . . . . . . . . . . . . . . . . . . 40

8.2 The Kahler Identities . . . . . . . . . . . . . . . . . . . . . . 44

9 Hodge Theory 52

9.1 The Hodge Decomposition . . . . . . . . . . . . . . . . . . . 52

10 Lefschetz Theorems 57

10.1 Lefschetz, I . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

10.2 Lefschetz, II . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

11 Hermitian Vector Bundles 61

11.1 Hermitian metrics . . . . . . . . . . . . . . . . . . . . . . . . 61

11.2 The Chern connection . . . . . . . . . . . . . . . . . . . . . 62

11.3 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

11.4 Kodaira’s Theorems . . . . . . . . . . . . . . . . . . . . . . 66

12 Example Sheet 1 68



Chapter 1

Complex Manifolds

1.1 Several complex variables

Identify C ∼= R2 in the standard way, x+ iy ↔ (x, y). Similarly, Cn ∼= R2n

by(x1 + y1i, . . . , xn + yni)↔ (x1, y1, . . . , xn, yn).

Definition 1.1. A smooth function f : Cn → C is holomorphic if it isholomorphic in each variable.

Remark 1.2. There is a more natural definition multivariable holomorphicfunction (you can probably guess what it is), but it is equivalent to thisone, and this is more useful for applications.

Writing f = u + iv, the theory of single-variable complex functionsimplies that being holomorphic is equivalent to satisfying the Cauchy-Riemann equations: for all j,

∂u

∂xj=

∂v

∂yj∂u

∂yj= − ∂v

∂xj.

Formally define the differential operators

∂

∂zj=

1

2

(∂

∂xj− i ∂

∂yj

)∂

∂zj=

1

2

(∂

∂xj+ i

∂

∂yj

)Then the Cauchy-Riemann equations are equivalent to

∂f

∂zj= 0 j = 1, . . . , n.

1

Complex Manifolds 2

Again, this is probably familiar from single-variable complex analysis.

Proposition 1.3 (Maximum Principle). Let U ⊂ Cn be open and con-nected. Let f be holomorphic on a region D such that D ⊂ U . Then

maxD

f(z) = max∂D

f(z)

and f achieves this maximum at an interior point if and only if f is con-stant.

Proof. For n = 1, this is the usual maximum principle. Apply the single-variable principle repeatedly to deduce it in higher dimensions.

Proposition 1.4 (Identity Principle). Let U ⊂ Cn be open and connectedand f : U → C be a holomorphic function. If f vanishes on an open subsetof U , then f ≡ 0 on U .

Proof. This follows from the single-variable identity principle, applied re-peatedly.

Lemma 1.5. Let U, V be open sets in Cn. A smooth map f : U → V isholomorphic if and only if df is C-linear.

Proof. Picking frames ∂∂xk

and ∂∂yk

, we have that

df =

...

...

. . .∂uj∂xk

∂uj∂yk

. . .

. . .∂vj∂xk

∂vj∂yk

. . ....

...

.

Also,

J =

0 −11 0

. . .

0 −11 0

Now, df is C-linear if and only if Jdf = (df)J , and we see that

Jdf =

...

...

. . . − ∂vj∂xk

− ∂vj∂yk

. . .

. . .∂uj∂xk

∂uj∂yk

. . ....

...

and (df)J =

...

...

. . .∂uj∂yk

− ∂uj∂xk

. . .

. . .∂vj∂jk

− ∂vj∂xk

. . ....

...

The equality of these two matrices is precisely the content of the Cauchy-Riemann equations.

Complex Manifolds 3

Theorem 1.6 (Inverse function theorem). Let U, V be open sets in Cn andf : U → V a holomorphic function. Suppose that z0 ∈ U is such that

det JC(f)(z0) 6= 0.

Then there exists an open subset U ′ containing z0 such that f |U ′ : U ′ →f(U ′) is a biholomorphism.

Proof. By the real inverse function theorem, there is a local smooth inversef−1. Furthermore, df−1f(x) = (df)−1x , which is C-linear because df is, so f−1

is holomorphic by Lemma 1.5.

Theorem 1.7 (Implicit function theorem). Let U ⊂ Cn and f : U → Cm aholomorphic function such that ( ∂f

∂zi(x))mi=1 has full rank. Then there exists

a holomorphic function g defined in a neighborhood of U ′ of x such that(zm+1, . . . , zn) → (g(zm+1, . . . , zn), zm+1, . . . , zn) is a biholomorphism ontoU ′ ∩ f−1(0).

Proof. This is deduced from the (holomorphic) inverse function theoremin the usual way. Let w = (z1, . . . , zm) and z = (zm+1, . . . , zn). Then wedefine the map Cn → Cn by

(w, z) 7→ (f(w, z), z).

By the hypothesis, this is a diffeomorphism at x, so there exists a localholomorphic inverse. In particular, the set (0, z) maps via this inverse to

(0, z) 7→ (g(z), z).

1.2 Complex manifolds

Complex manifolds lie at the intersection of several different mathemat-ical areas: several several complex variables, differential geometry, andalgebraic geometry. This makes their theory very rich and naturally inter-esting.

Let X be a smooth manifold of real dimension 2n. We now developthe basic theory of complex manifolds, which is completely analogous tosmooth manifolds except that one demands all functions to be holomorphicinstead of smooth.

Definition 1.8. A holomorphic atlas for X is a collection of charts (Uα, ϕα)where

ϕα : Uα → Vα ⊂︸︷︷︸open

R2n ∼= Cn

such that

Complex Manifolds 4

1. X =⋃α Uα, and

2. the transition functions

ϕαβ = ϕα ϕ−1β

are holomorphic.

Definition 1.9. Two holomorphic atlases (Uα, ϕα) and (Uα, ϕα) are saidto be equivalent if ϕα ϕ−1β is holomorphic for all α, β.

Definition 1.10. A complex manifold X is a smooth manifold with achoice of equivalence class of holomorphic atlases.

This is also called a “complex structure” on X. Note that if X hasdimension 2n as a real manifold, then it has dimension n as a complexmanifold.

Example 1.11. We gives some examples of complex manifolds.

• Cn is a complex manifold.

• Mn(C) ∼= Cn2is a complex manifold. Hence GLn(C) is a complex

manifold, as it is an open subset of a complex manifold.

• Complex projective space CPn. As a set, this consists of one-dimensionalsubspaces of Cn+1. A point can be written as [z0, . . . , zn]. A holo-morphic atlas is given by taking Ui = zi 6= 0, and

ϕi : Ui → Cn

[z0, . . . , zn] 7→ (z0zi, . . . ,

zizi, . . . ,

znzi

).

The transition function on Ui ∩ Uj is

ϕi ϕ−1j (w0, . . . , wj, . . . , wn) = (w0

wi, . . . ,

wjwi, . . . ,

wnwi

).

In contrast to our previous examples, CPn is compact.

Definition 1.12. A function f : X → C is holomorphic if for all holomor-phic charts (U,ϕ), the composition

f ϕ−1 : ϕ(U)→ C is holomorphic.

Definition 1.13. A morphism F : X → Y between complex manifolds isholomorphic if for all holomorphic charts (U,ϕ) on X and (V, ψ) on Y ,

ψ F ϕ−1 is holomorphic.

Almost Complex Structures 5

We say that X and Y are isomorphic (or biholomorphic) complex man-ifolds if there exists a holomorphic function F : X → Y with holomorphicinverse.

An important theme is that there are fewer holomorphic functions thansmooth functions. In the smooth theory, functions can be constructed lo-cally and patched together using bump functions (partitions of unity). Inthe analytic theory, everything is much more rigid: for instance, a func-tion is determined globally by its local behavior (the Identity Theorem).Holomorphic functions on complex manifolds are much more like regularfunctions on algebraic varieties.

Proposition 1.14. Let X be compact and connected. Then any holomor-phic function on X is constant.

Proof. Since X is connected, it suffices to show that f is locally constant.Suppose otherwise. Then ||f || has a local maximum at some x ∈ X. Com-posing with a chart, we get a function f ϕ−1 : ϕ(U) → C assuminga local maximum in the interior. By the maximum principle, f ϕ−1 isconstant.

Corollary 1.15. A compact complex manifold cannot embed in CN .

Proof. Indeed, otherwise one could obtain nonconstant holomorphic func-tions by pulling back holomorphic functions on CN (e.g. the coordinatefunctions).

This contrasts with the Whitney embedded theorem for smooth man-ifolds, which says that any real manifold embeds in a finite-dimensionalEuclidean space.

Proposition 1.16. Let X be a connected complex manifold. If f : X → Cis holomorphic vanishes on an open subset of X, then f is identically zero.

Proof. Let ϕ be a chart around such an accumulation point, and apply theidentity theorem for complex functions.

In particular, this prohibits a holomorphic analogue of partitions ofunity. This means that it is much hard to pass from local to global, as weare accustomed to do in the theory of smooth manifolds.

Definition 1.17. Let Y ⊂ X be a smooth submanifold of dimension 2k.We say that Y is a closed complex submanifold if there exist holomorphiccharts (Uα, ϕα) for X where Y ⊂

⋃Uα and ϕα : Uα∩Y → ϕα(Uα)∩Ck with

the canonical inclusion of Ck ⊂ Cn as (z1, . . . , zk) 7→ (z1, . . . , zk, 0, . . . , 0).

Definition 1.18. We say that X is projective if it is isomorphic to a closedcomplex submanifold of PN for some N .

Chapter 2

Almost Complex Structures

We might ask, to what extent can the complex structure of a manifold becaptured by linear data? In particular, if X is a complex manifold, weshould get some extra structure on TX, the tangent bundle of X. Thisleads to the notion of almost-complex structures.

A model case is X = R2n ∼= Cn, with real coordinates x1, y1, . . . , xn, yn.Consider the linear map R2n → R2n

(x1, y1, . . . , xn, yn) 7→ (−y1, x1, . . . ,−yn, xn).

Under the identification with Cn, this corresponds to “multiplication by i.”Similarly on TX = X ×R2n, we have frames ∂

∂xj, ∂∂yj

and we can define an

endomorphism of the tangent bundle J : TX → TX by

J

(∂

∂xj

)= − ∂

∂yjJ

(∂

∂yj

)=

∂

∂xj.

Note that again, J2 is −1.We want to investigate what extra structure we get from having this

kind of endomorphism of the tangent bundle.

2.1 Linear algebra preliminaries

Let V be a real vector space.

Definition 2.1. A linear map J : V → V such that J2 = −1 is called acomplex structure on V .

On R2n, the endomorphism (xi, yi, . . . , xn, yn) 7→ (y1,−x1, . . . , yn,−xn)is called the standard complex structure.

Given a complex structure J , the relation J2 = −1 implies that itseigenvalues are ±i. As V is real, this means that J has no eigenspaces inV . But if we consider the complexification VC = V ⊗R C, and any real

6


linear map T : V → V extends to a complex linear map T : VC → VC. Inparticular, J extends to an endomorphism of VC satisfying J2 = −1, andit now has two eigenspaces. We let V 1,0 and V 0,1 denote the eigenspaceswith eigenvalues i,−i.

Lemma 2.2. With the notation above,

1. VC = V 1,0 ⊕ V 0,1.

2. V 1,0 = V 0,1.

Proof. 1. This is standard linear algebra. It is worth noting that onecan explicitly write this decomposition as

v =1

2(v − iJv))︸︷︷︸∈V (1,0

+1

2(v + iJv)︸︷︷︸∈V (0,1)

.

2. Straightforward computation. Note that it is obvious from the aboveexplicit representation.

Let J∗ : V ∗ → V ∗ be the dual map, defined by

J∗α(v) = α(J(v)) for all α ∈ V ∗.

This is necessarily an almost-complex structure on V ∗. Therefore, its com-plexification decompses V ∗C into a direct sum of +i and −I eigenspaces.

Lemma 2.3. (VC)∗ = Hom(VC,C) = HomR(V,C), and we have the decom-position

(VC)∗ = (V ∗)1,0 ⊕ (V ∗)0,1.

where(V ∗)1,0 = α ∈ V ∗C : α(Jv) = iα(v)

and similarly for (V ∗)0,1.

2.2 Almost complex structures

This discusion extends to real vector bundles. Given a real vector bundleV on X it makes sense to consider VC, obtained by complexifying fiber-by-fiber.

Definition 2.4. Let V → X be a real vector bundle. A bundle morphismJ : V → V satisfying J2 = −1 is called an almost complex structure on V .


Given such a bundle, we have a decomposition

VC = V 1,0 ⊕ V 0,1

whereJ |V 1,0(v) = iv J |V 0,1(v) = −iv.

This decomposition exists fiber by fiber, of course, but the point is that itvaries smoothly in X. That is clear from the fact that V 1,0 can be viewedas ker J − i id.

Definition 2.5. Let X be a real manifold. An almost complex structureon X is an almost complex structure on TX.

We now again assume that X is a complex manifold. Recall that therewas a standard almost complex structure on TR2n, which we denote byJst. Now, given a holomorphic chart ϕ : U → R2n, we get a bundle mapJ : TU → TU by J = Dϕ−1 Jst Dϕ.

Theorem 2.6. The J defined above is independent of the choice of holo-morphic chart, and gives a well-defined almost complex structure on X.

Proof. Suppose that ϕ, ψ are two different charts defined in neighborhoodsof some point. We must check that

Dϕ−1 Jst Dϕ = Dψ−1 Jst Dψ

or equivalently,

D((ϕ ψ−1)−1) Jst D(ϕ ψ−1) = Jst.

Now, ϕ ψ−1 is a biholomorphism between subsets of Cn, so we are donesince if f holomorphic, then df commutes with Jst.

By the preceding discussion, we get a splitting

(TX)C = T 1,0X ⊕ T 0,1X.

Definition 2.7. T 1,0X is the holomorphic tangent bundle of X.

We have a similar splitting (T ∗X)C = (T ∗X)1,0 ⊕ (T ∗X)0,1.

In terms of local coordinates, suppose that ϕ : U → R2n is a holo-morphic chart. We say that xj + iyj are holomorphic coordinates. Bydefinition,

J

(∂

∂xj

)=

∂

∂yjJ

(∂

∂yj

)= − ∂

∂xj

andJ(dxj) = −dyj J(dyj) = dxj.

Differential Forms 9

Definition 2.8. We define

dzj := dxj + idyj dzj = dxj − idyj.

Then you can check that

J(dzj) = J(dxj) + iJ(dyj) = −dyj + idxj = i(dxj + idyj) = idzj

so dz1, . . . , dzn give a frame for (T ∗X)1,0. Similarly, dz1, . . . , dzn give aframe for (T ∗X)0,1.

Definition 2.9. We define

∂

∂zj=

1

2

(∂

∂xj− i ∂

∂yj

)∂

∂zj=

1

2

(∂

∂xj+ i

∂

∂yj

)These are local frames for TX1,0 and TX0,1, respectively.

Chapter 3

Differential Forms

3.1 Differential forms on Complex Manifolds

Recall that if X is a complex manifold, we have local holomorphic coordi-nates zj = xj + iyj. We showed last time that

• dz1, . . . , dzn is a local frame for (T ∗X)(1,0),

• dz1, . . . , dzn is a local frame for (T ∗X)(0,1),

• ∂∂zj

is a local frame for (TX)(1,0),

• ∂∂zj

is a local frame for (TX)(0,1).

We already showed that if f : X → R2 ∼= C is smooth, then f isholomorphic if and only if the Cauchy-Riemann equations hold, which isequivalent to ∂f

∂zj= 0 for all j.

Let u : X → R be a smooth function. We have a map

dux : TxX → Tu(x)R ∼= R.

Abusing notation, we also let

dux : TxX ⊗ C→ C

be the complexified map. If f : X → C is smooth, say f = u + iv, thendf = du+ idv, which is a smooth section of T ∗X ⊗ C.

Exercise 3.1. In a local frame, we have an equality of the expression

df =∑j

∂f

∂jdxj +

∑j

∂f

∂jyj =

∑j

∂f

∂zjdzj︸︷︷︸

∂f

+∑j

∂f

∂zjdzj︸︷︷︸

∂f

.

10


This shows that we have chosen our definitions well (or at least, we havenot chosen them to be ridiculous).

Remark 3.2. The condition that f be holomorphic is equivalent to ∂f = 0.

Recall that T ∗X⊗C = (T ∗X)(1,0)⊕(T ∗X)(0,1). We now study what hap-pens upon taking exterior powers. We will get summands of the followingform.

Definition 3.3. Let∧p,q(T ∗X) =

∧p(T ∗X)(1,0) ⊗∧q(T ∗X)(0,1).

A smooth section of∧p,q(T ∗X) is called a (p, q)-form, and (p, q) is called

the “bi-degree.” Locally, it loos like∑fα,β(z)dzαdzβ

where |α| = p and |β| = q, i.e. a smooth function times a form with p“holomorphic coordinates” and q “antiholomorphic coordinates.”

We then defineAkC(U) to be the space of smooth sections of Λk(T ∗X⊗C)over U , and Ap,qC =

∧p,q(T ∗X). In particular, A0C(U) is the set of smooth

complex-valued fucntions on U . We will suppress the subscript C when thecontext is clear.

Lemma 3.4. 1. There is a natural identification

k∧(T ∗X ⊗ C) ∼=

⊕p+q=k

p,q∧(T ∗X)

and in particular,

AkC(U) =⊕p+q=k

Ap,qC (U).

2. If α ∈ Ap,q(U) and β ∈ Ap′,q′(U) then

α ∧ β ∈ Ap+p′,q+q′(U).

Proof. At the level of fibers, this is all formal multilinear algebra. If V =A⊕B, with corresponding representations vj = aj + bj, then

v1 ∧ . . . ∧ vk = (a1 + b1) ∧ . . . ∧ (ak + bk) ∈⊕p+q=k

p∧V1 ⊗

q∧V2.

At the level of vector bundles, we can establish it using local frames, or byrecognizing the subspaces as kernels of the relevant operators, which arewedge powers of J and the identity.


What does this discussion look like in local coordinates? Choose localholomorphic coordinates z1, . . . , zn, write

dzI = dzi1 ∧ . . . dzip I = (i1, . . . , ip).

A frame for∧p,q T ∗X is

dzI ∧ dzJ |I| = p, J = |q|.

Example 3.5. Elements of A(1,0)(C) include z dz and z dz. In particular,we allow non-holomorphic functions to multiply the holomorphic differen-tials.

3.2 Dolbeault Cohomology

Definition 3.6. Denote by d : Ak(U) → Ak+1(U) the C-linear extensionof the usual exterior derivative.

In addition, we can define the operators

• ∂ : Ap,q(U) → Ap+1,q(U) which is the composition of d with projec-tion to A(p+1,q).

• ∂ : Ap,q(U) → Ap,q+1(U) which is the composition of d with projec-tion to A(p,q+1).

On A(0,0), these agree with the differential operators ∂ and ∂ that we al-ready defined.

This is very concrete in locally coordinates. Locally, if α = fdzI∧dzJ ∈Ap,q(U), then

dα =∑j

∂f

∂zjdzj ∧ dzI ∧ dzJ︸︷︷︸

∂α

+∑j

∂f

∂zjdzj ∧ dzI ∧ dzJ︸︷︷︸

∂α

.

The following lemma is apparent from this example.

Lemma 3.7. On Ap,q, we have

1.d = ∂ + ∂.

2. ∂2 = 0, ∂2

= 0, and ∂∂ = −∂∂.

3. If α ∈ Ap,q(U), then

∂(α ∧ β) = ∂α ∧ β + (−1)p+qα ∧ ∂β


and∂(α ∧ β) = ∂α ∧ β + (−1)p+qα ∧ ∂β.

Proof. The first part follows from the above local calculation. The secondfollows from the fact that d2 = 0. The third follows from the correspondingidentity for d.

Remark 3.8. The operators ∂ and ∂ commute with restriction because ddoes. This will show that they induce maps of sheaves.

Definition 3.9. The (p, q) Dolbeault cohomology group is

Hp,q

∂(X) =

ker ∂ : Ap,q → Ap,q+1(X)

im ∂ : Ap,q−1 → Ap,q(X).

Remark 3.10. One could make the obvious analogous definition for ∂. How-ever, this gives no extra information, since ∂ and ∂ are essentially the same(up the conjugation). This is more natural because the kernel of ∂ corre-sponds to holomorphic forms.

Example 3.11. The convention is that negative degree groups are zero,so H0,0

∂(X) is the space of holomorphic functions on X.

Lemma 3.12 (Functoriality). Let F : X → Y be holomorphic. Then themap F ∗ : Ak(Y ) → Ak(X) respect the bigrading and commutes with ∂,hence induces a map

f ∗ : Hp,q

∂(Y )→ Hp,q

∂(X).

3.3 The Mittag-Leffler Problem

Let S be a Riemann surface (a one-dimensional complex manifold). Aprincipal part at a point x ∈ S is a Laurent series

P =n∑k=1

akz−k

where z is a holomorphic coordinate centered at x. The Mittag-Leffler prob-lem is: given distinct points x1, . . . , xn ∈ S and principal parts P1, . . . ,Pnat these points, does there exist a meromorphic function on S, which isholomorphic away from the xi, and has precisely these principal parts atthe xi?

This is, in some sense, answered by Dolbeault cohomology. Clearly,locally the answer is yes, i.e. there exist neighborhoods Uj around xjand meromorphic functions fj on Uj with principal part Pj. Considera partition of unity (ρj) subordinate to the Uj, such that ρj = 1 near xj

The ∂ Poincare Lemma 14

and supp ρj ⊂ Uj. Then∑

j ρjfj is a function which is smooth except atthe xj, with the correct principal parts.

The smooth form g =∑

j ∂(ρjfj) satisfies ∂g = 0 and is identically zeroin a neighborhood of xα. (There is actually something subtle here - fj is nota smooth function on Uj. But since fj is locally meromorphic, ∂fj ≡ 0 onthe punctured neighborhood of xj.) So g represents a cocycle in Dolbeaultcohomology. If it were zero in cohomology, then g = ∂h and the functionf =

∑ρjfj − h would be a solution to the Mittag-Leffler problem. So the

Dolbeault cohomology groups are obstructions to Mittag-Leffler problems.

Chapter 4

The ∂ Poincare Lemma

Our goal is to show that the Dolbeault groups Hp,q

∂(X) vanish when X is

a product of discs in Cn. This is essentially an analytical problem, likePoincaree’s Lemma for de Rham cohomology.

4.1 Holomorphic Functions on CDefinition 4.1. A disc in C is a set BR = |z| < R for some R ∈ [0,∞].

Suppose that B is a bounded disc, u is a smooth complex-valued func-tion on B. Then Stokes’ formula gives

∫∂B

u(z) dz =

∫B

du ∧ dz =

∫B

∂u

∂zdz ∧ dz = 2i

∫B

∂u

∂zdx ∧ dy.

If u is holomorphic, then ∂u∂z

= 0 and we recover Cauchy’s formula.

Theorem 4.2 (Cauchy Integral formula). If u is smooth on B, then

u(w) =1

2πi

(∫∂B

u(z)

z − wdz +

∫B

∂u

∂z

1

z − wdzdz

).

If u is holomorphic, we recover the usual Cauchy integral formula.

Proof. Apply Stokes’ theorem B \Bε(w):∫B\Bε(w)

∂u∂z

(z)

z − wdzdz =

∫∂B

u(z)

z − wdz −

∫∂Bε(w)

u(z)

z − wdz.

We compute the second term by explicit parametrization, and let ε tend to0: ∫

u(w + εeiθ) idθ → 2πiu(w).

15


We now give a converse to this formula, which you can think of as the∂-lemma in one variable.

Theorem 4.3. Let B be a bounded disc in C such that B ⊂ B ⊂ U , forsome open set U . Let g ∈ C∞(U). Then the integral

u(w) =1

2πi

∫B

g(z)

z − wdzdz

is well-defined, smooth, and satisfies ∂u∂w

(w) = g(w) on B.

This says that if α = g dz ∈ A0,1(B) where g has compact support,then there exists u such that ∂u = α.

Remark 4.4. We cannot just differentiate under the integral sign here,which would give

∂

∂w

g(z)

z − w= 0,

since the integrand is not continuous on the domain of integration. Asabove, we get around this by analyzing the contribution in a small neigh-borhood of w.

Proof. Let ρ be a smooth bump function such that ρ = 1 on a neighborhoodV of B such that supp ρ ⊂ U . Then ρg is smooth.

1

2πi

∫B

g(z)

z − wdzdz =

1

2πi

∫B

ρ(z)g(z)

z − wdzdz +

1

2πi

∫B

(1− ρ(z))g(z)

z − wdzdz.

The second integrand has supported away from w, so differentiating by wkills it. Therefore, we reduce to the case where g is smooth and compactlysupported in C. We then reparametrize

1

2πi

∫C

g(z)

z − wdzdz =

1

2πi

∫C

g(z + w)

zdzdz

=1

2πi

∫ 2π

0

∫ ∞0

g(w + reiθ)2irdrdθ

reiθ.

Written in this way, we see that the integral is well-defined and the inte-grand is smooth on all of C, so we can differentiate under the integral signand trace our steps backwards to deduce that

1

2πi

∫C

g(z)

z − wdzdz =

1

2πi

∫C

∂g(z)

∂z

1

z − wdzdz.

Then Cauchy’s integral formula says that this is g(w).


4.2 Holomorphic functions of several com-

plex variables

Definition 4.5. A polydisc in Cn is a set of the form B = BR1× . . .×BRn ,where the Ri ∈ [0,∞].

For a polydisc B, we define ∂0B =∏∂jBRj .

Theorem 4.6. If B is a bounded polydisc, u is a smooth function on Band holomorphic in B, then

u(w) =1

(2πi)n

∫B

u(z)

(z1 − w1) . . . (zn − wn)dz1 . . . dzn.

Proof. This is essentially by applying the Cauchy-formula several times.

Notation: If α = (α1, . . . , αn), with αj ∈ N, then we use the usualmulti-index notation

zα = zα11 . . . zαnn

∂α

∂zα=

∂α1

∂zα11

. . .∂αn

∂zαnn

andα! = α1! . . . αn!.

Corollary 4.7. We have the Taylor series expansion

u(w) =∑α

aαwα,

where

aα =1

α!

∂α

∂zαu|0.

Proof. We write

1

z1 − w1

=1

z1(1− w1/z1)=

1

z1

∑α1≥0

wα11

zα11

for |w1| ≥ |z1|.

So1

(z1 − w1) . . . (zn − wn)=

1

z1 . . . zn

∑α

wα

zα.

Then

u(w) =∑α

1

(2πi)n

(∫∂0B

u(z)

z1 . . . znzαdz1 . . . dzn

)wα.

Again, differentiating the Cauchy integral formula gives aα = ∂αu∂zα|0.


4.3 The Poincare Lemma

Theorem 4.8. Let B be a polydisc. Assume B ⊂ B ⊂ U for some openU ⊂ Cn. Suppose that α ∈ Ap,q(U) with ∂α = 0. Then there existsβ ∈ Ap,q−1(B) with α = ∂β on B.

Proof. We induct on the highest index dzn that appears in this sum. Wecan write

α = β ∧ dzn + α′

where α′ involves only dz1, . . . , dzn−1. If β =∑gIdzI .

Write

uI(w1, . . . , wn) =

∫gI(w1, . . . , wn−1, zn) dzn

zn − wn.

Then∂uI∂wn

(w1, . . . , wn) = gI(w1, . . . , wn).

Setting

δ = (−1)q−1∑I

uIdzI

we obtain

∂δ =∑I

∂uI∂zn

dzI ∧ dzn + (−1)q−1∑i 6=n

∂uI∂zi

dzi ∧ dzI

=∑I

gIdzI ∧ dzn + (−1)q−1∑i 6=n

∂uI∂zi

dzi ∧ dzI

Then we can modify by the coboundary ∂δ: the cocycle α − ∂δ is closedand doesn’t involve dzn. We claim that it doesn’t involve dzi for i > n,since α is closed, ∂gI

∂zidzi ∧ dzI ∧ dzn = for any i > n.

This reduces to the case of fdz1∧ . . .∧dzq, and then the same argumentshows that this is a coboundary.

Lemma 4.9. Let B be a bounded polydisc, B ⊂ B ⊂ B′ ⊂ U where B′ is alarger bounded polydisc. Then for any multi-index α there exists a constantCα such that for all holomorphic functions u on U ,

||∂αu||C0(B) ≤ Cα

∫B′|u(z)| dz ∧ dz.

Proof. This follows from the multivariate Cauchy integral formula, as inthe one-variable case.

Corollary 4.10. Let un be holomorphic on U , say un → u uniformly oncompact sets. Then u is holomorphic.


Proof. By the previous lemma,

∂un∂z− ∂um

∂z

converges uniformly on compact sets. Also, ∂un∂z

= 0. So u is C1 andsatisfies the Cauchy-Riemann equations, hence is holomorphic.

Theorem 4.11. If B is a (possibly unbounded) polydisc in Cn, then

Hp,q

∂(B) = 0 for q > 0.

In other words, if α ∈ Ap,q(B) with ∂α = 0, then there exists β ∈Ap,q−1(B) with ∂β = α.

Proof. The case p = 0, q = 1 is the most difficult. The others reduce easilyto this one, so let’s study it first.

Say B = Br1 × . . .×Brn . We approximate this region by an increasingsequence of bounded polydiscs. Letting εi(m) increase to ri as i→∞, and

Bm = Bε1 × . . . Bεn .

Then B1 ⊂ B2 ⊂ . . . is an ascending chain of open polydiscs whose unionis B.

Let α be a cocycle; we want to write it as a coboundary. We alreadyknow that we can do this after restricting to a smaller open set, so we filterour space as an increasing union of open sets, and patch together solutions.Specifically, we claim that we can find a sequence βm ∈ A0,1(Bm) such that

1. ∂βm = α on Bm, and

2. ||βm+1 − βm||C0(Bm−1) < 2−m.

Assuming this for now, we obtain that βm form a Cauchy sequence,hence converge uniformly to some smooth function γ on B. Moreover,∂(βm+1 − βm) = 0 on Bm, so βm − βm0 → γ − βm0 , and each βm+1 − βm0

is holomorphic on Bm0 . So γ − Bm0 is a uniform limit of holomorphicfunctions, hence is holomorphic on Bm0 . In particular, γ is smooth onBm0 , but since this is independent of m0 we get that γ is smooth on B and∂γ = α on B.

Now it suffices to prove the claim. Suppose that β1, . . . , βm have beendefined, and we want to construct βm+1. By the ∂-Poincare Lemma appliedto Bm+1 ⊂ Bm+2, there exists a smooth β ∈ A0,0(Bm+2) such that ∂β = αon Bm+1. After multiplying by some smooth bump function ρ which isidentically 1 on Bm+1 and with support contained in Bm+2, we can assumethat β is smooth on B. In order to get the norm bound, we modify β by aholomorphic function on B; since β − βm is holomorphic on Bm, it suffices

Sheaves and Cohomology 20

to take a sufficiently high degree Taylor polynomial approximation to theirdifference.

The case for higher p, q is completed in Example Sheet 1, Question8.

In fact, the same proof shows the following slightly stronger result. If

∆ε = z ∈ C : |z| < ε and

∆ε = ∆ε − 0.

Theorem 4.12. If q ≥ 1 and p ≥ 0, then

Hp,q

∂(∆ε1 × . . .×∆εR ×

∆εk+1

× . . .

∆ε`) = 0.

Chapter 5

Sheaves and Cohomology

5.1 Sheaves

Let X be a topological space.

Definition 5.1. A presheaf F of groups consists of a group F(U) for allopen subsets U ⊂ X and restriction homomorphisms rV U : F(U)→ F(V )for each inclusion of open subset V ⊂ U .

Remark 5.2. In other words, a presheaf is a contravariant functor from theposet of open subsets of X to Grp.

You should think of F(U) as being some class (e.g. continuous, smooth,holomorphic, etc.) functions on U , and the restriction maps as being re-striction of functions to a smaller domain. In keeping with this intuition,for s ∈ F(U) we write s|V for rV U(s).

Example 5.3. C0(U) is the group of continuous functions on U , and therestriction maps are restriction of functions.

Remark 5.4. One can similarly define a presheaf of sets, rings, vector spaces,etc. If O is a pre-sheaf of rings, then one can define pre-sheaves of O-modules by asking each F(U) to be an O(U)-module, in a compatible way.

Definition 5.5. A presheaf F is a sheaf if

1. For all s ∈ F(U), if U =⋃Ui is an open cover and s|Ui = 0 for all i,

then s = 0.

2. If U =⋃Ui is an open cover and si ∈ F(Ui) is such that

si|Ui∩Uj = sj|Ui∩Uj for all i, j

then there exists s ∈ F(U) such that S|Ui = si for all i.

Example. C0(U) is a sheaf. More generally, the presheaf functions satis-fying some nice purely local property will be a sheaf, e.g.:

21


• Z(U) = locally constant functions f : U → Z,

• R(U) = locally constant functions f : U → R.

If X is a smooth manifold, we have sheaves

• C∞(U) = smooth functions on U ,

• Ap(U) = smooth p-forms on U ,

• For E a vector bundle on X, C∞(E)(U) = smooth sections of E|U .

If X is a complex manifold, then we have further sheaves

• O(U) = holomorphic functions on U ,

• O∗(U) = nowhere vanishing holomorphic functions on U ,

• Ωp(U) = holomorphic p-forms on U .

Definition 5.6. Let U ⊂ Cn. A meromorphic function on U is a functionf : U \ S → C where S ⊂ U is nowhere dense, such that U has an opencover U =

⋃Ui and there exists gi, fi ∈ O(Ui) with

f |Ui\S =fi|Ui\Sgi|Ui\S

.

Exercise 5.7. Define a sheaf M of meromorphic functions on X, andshow that it is a sheaf. If X is a complex manifold, we say that f ismeromorphic on X if there exists a nowhere-dense S ⊂ X such that f |X−Sis meromorphic on some open cover by holomorphic charts.

Definition 5.8. A morphism α : F → G between (pre-)sheaves on Xconsists of homomorphisms αU : F(U)→ G(U) for all open subset U ⊂ X,such that

F(U)αU //

rV U

G(U)

rV U

F(V ) αV// G(V )

Remark 5.9. If we view sheaves as functor, then this is just the notion ofnatural transformations, i.e. morphisms in the functor category.

Definition 5.10. We say that

0→ F α−→ G β−→ H → 0

is exact if for all U , the sequence

0→ F(U)αU−→ G(U)

βU−→ H(U)


is exact and if s ∈ H(U) and x ∈ U , there exists an open neighborhoodV ⊂ U of x and t ∈ G(V ) such that βV (t) = s|V .

These are the notions of exactness in the functor category. In particular,this is not the same as the short exact sequence of abelian groups beingexact at the level of each open set U .

Definition 5.11 (Kernels). Let α : F → G be a morphism of sheaves. Wedefine the kernel sheaf by

(kerα)(U) = ker(αU : F → G(U)).

Cokernels are trickier, since the naıve definition does not necessarilysatisfy the sheaf axiom.

Definition 5.12 (Cokernels). If

0→ F α−→ G → H → 0

is a short exact sequence of sheaves, we define cokerα = H.

This does not imply that G(U)→ H(U) is surjective. In general, thereis a sheaf which is the cokernel of any sheaf morphism F → G (in the cat-egory theoretic sense), but it is not defined as the object-wise cokernel, asthis is a pre-sheaf but not necessarily a sheaf. One has to take the sheafi-fication of this naıve definition.

Example 5.13. If X is a complex manifold

0→ 2πiZ ι−→ OXexp−−→ O∗X → 0

is a short exact sequence of sheaves. Given any non-vanishing holomorphicfunction, we can locally take its logarithm (but not globally). We will callthis the “fundamental short exact sequence.”

Definition 5.14. A complex of sheaves

F1α1−→ F2

α2−→→ . . .

is exact if0→ kerαn → Fn → kerαn+1 → 0

is short exact for all n.

If X is a complex manifold, we have sheaves Ap,q. The ∂ maps give riseto a complex of sheaves

. . .∂q−2−−→ Ap,q−1 ∂q−1−−→ Ap,q ∂q−→ Ap,q+1 ∂q+1−−→ . . .


The ∂-Poincare Lemma says that this is exact. Indeed, consider the se-quence

0→ ker ∂q → Ap,q → ker ∂q+1 → 0.

The ∂-Poincare Lemma said that we can locally find a lift of any form inker ∂q+1 (in particular, over any polydisc).

Recall that we also defined the sheaf Ωp(U) = σ ∈ Ap,q(U) : ∂σ = 0,the holomorphic (p, 0 forms on U . In our new language,

Ωp = ker ∂ : Ap,0 → Ap,1.

So

0→ Ωp ι−→ Ap,0 ∂−→ Ap,1 ∂−→ . . .

is a resolution of Ωp. We will later see that the sheafs Ap,q are fine, henceacyclic, and can therefore be used to compute the cohomology of Ωp.

5.2 Cech Cohomology

We consider a toy example to illustrate the idea. Let X be a topologicalspace with open cover X = U ∪ V . Suppose that F is a sheaf on X.Say we have sections sU ∈ F(U) and sV ∈ F(V ) - when do they comefrom a global section? By the sheaf axiom, this is the case if and only ifsU |U∩V = sV |U∩V .

Define a map δ : F(U)⊕F(V )→ F(U ∩ V ) by sending

(sU , sV ) 7→ sU |U∩V − sV |U∩V .

Then the gluing condition is precisely δ(sU , sV ) = 0, so F(X) = ker δ. Sowe have identified an algebraic obstruction to patching local sections intoa global one.

Let X be a topological space and F a sheaf on X. Let U = Uα bean open cover of X, indexed by a subset of N (though this isn’t reallynecessary). We introduce the notation

Uα0...αp = Uα0 ∩ . . . ∩ Uαp .

DefineC0(U ,F) =

∏α

F(Uα)

andC1(U ,F) =

∏α<β

F(Uαβ).

More generally,


Definition 5.15.

Cp(U ,F) =∏

α0<...<αp

F(Uα0α1...αp).

By convention, for any multiindex α0, . . . , αp we set

σα0,...,αi,αi+1,...,αp = −σα0,...,αi+1,αi,...,αp .

If α = (α0, . . . , αp), we say that σ = (σα) ∈ Cp(U ,F) is a p-cochain.

Definition 5.16. Define the boundary map δ : Cp(U ,F) → Cp+1(U ,F)by

(δσ)α0,...,αp+1 =

p+1∑j=0

(−1)jσα0,...,αj ,...,αp+1|Uα0 ,...,αp+1 .

Lemma 5.17. The composition δ δ : Cp(U ,F)→ Cp+2(U ,F) is the zeromap.

Proof. If σ ∈ Cp(U ,F), the δ δ(σ)α1,...,αp+2 is∑i,j

[(−1)i(−1)j−1 + (−1)i(−1)j]σα1,...,αi,...αj ,...,αp+2

where the two opposite signs depend on whether i or j is deleted first.

Example 5.18. Let U = U, V,W be an open cover of X. Then a σ ∈C1(U ,F) looks like

σ = (σUV , σUW , σVW ).

Thenδσ = σUV − σUW + σVW .

Definition 5.19. We say that σ ∈ Cp(U ,F) is

• a cocycle if δσ = 0,

• a coboundary if σ = δτ

Define

Hq(U ,F) =ker δ : Cq(U ,F)→ Cq+1(U ,F)

im δ : Cq−1(U ,F)→ Cq(U ,F).

In other words, this is the group of cocycles mod coboundaries, or the co-homology of the chain complex described earlier. Obviously, this dependson the open cover and is not our final definition of Cech cohomology.


Example 5.20. Let X = P1 and F = O. We pick the open cover U =[z : 1] and V = [1 : w]. Both are isomorphic to C, and their intersection isC∗. So C0(U ,O) = O(U)⊕O(V ) and C1(U ,O) = O(U ∩ V ). The map toC1(U,O) is

δ(f + g)(z) = f(z)− g(1/z).

The kernel consists of (f, g) where f = g is constant. The image of δconsists of all holomorphic functions on C∗, since all holomorphic funtionson C∗ have Laurent series. So H0(U ,O) = C and H i(U ,O) = 0 for i > 0.

Definition 5.21. Given two covers U and V , we say that V refines U , andwrite V ≤ U , if there exists α : N→ N such that for all β, Vβ ⊂ Uα(β).

If V ≤ U , we have natural restriction maps ρV,U : Cp(U ,F)→ Cp(V ,F)such that

ρV,U(σβ1,...,βp) = σα(β1),...,α(βp).

This commutes with δ, so it descends to a map on cohomology:

ρV,U : H(U ,F)→ H(V ,F).

Note that the definition of ρ depends on α. However, you can check thaton the level of cohomology, it is actually independent of α.

Definition 5.22. The Cech cohomology of a sheaf F on X is

Hp(X,F) = lim−→UHp(U ,F)

For X a complex manifold and F = O, this holds if each intersection isa polydisc. So the cover we used for P1 in Example 5.20 was in fact acyclic.

Example 5.23. Trivially from the definition, H0(U ,F) = F(X) for anyopen cover of X. Therefore, H0(X,F) = F(X).

Example 5.24. Hq(X,Ar,s) = 0 for all q ≥ 1. The point is that thesheaf Ar,s is a module over the smooth functions on X, making it muchflabbier than holomorphic sheaves like O. To see this, let σ ∈ Hq(X,Ar,s)be represented by some σ ∈ Cp(U ,Ar,s). We have δσ = 0, so let ρα be apartition of unity subordinate to Uα (assuming that the cover is locallyfinite). Define

τα0,...,αq−1 =∑β

ρβσβ,α0,...,αq−1︸︷︷︸extend by zero

.

Here, since ρβ is supported in Uβ, multiplying by it allows us to extend tothe complement of Uβ, so that σβ,α0,...,...,αq−1 is defined on Uα0,...,αq−1 . As aguiding example, consider U = U, V,W. Then by hypothesis,

0 = δσ = σUV − σUW + σVW .


We have τU =∑

w ρWσWU . Then

(δτ)UV = τV − τU= ρUσUV + ρWσWV − ρV σV U − ρWσWU

= ρUσUV + ρV σUV + ρW (σWV − σWU)

= (ρU + ρV + ρW )σUV .

Now we tackle the general case. From the definition, we have

(δτ)α0,...,αp =

p∑j=0

(−1)jτα0,...,αj ,...,αp

=

p∑j=0

(−1)j∑β

ρβσβα0,...,αj ,...,αp .

Now, since σ is a cocycle we have

0 = (δσ)βα0...αp

= σα0...αp +

p∑j=0

(−1)j+1σβα0,...αj ,...,αp .

Substituting this above, we find that

(δτ)α0,...,αp =∑β

ρβ

p∑j=0

σβα0,...,αj ,...,αp

=∑β

ρβσα0...αp

= σα0...αp .

This kind of sheaf, with partitions of unity, is said to be fine.

5.3 Properties of Cech cohomology

Let f : F → G be a morhpism of sheaves. This induces f : Cp(U ,F) →Cp(U ,G) for any open cover U . Furthermore, these maps commutes withδ, and hence induce a map on cohomology:

f∗ : Hp(X,F)→ Hp(X,G).

Theorem 5.25. Suppose that

0→ E g−→ F f−→ G → 0.


is a short exact sequece of sheaves on a manifold X. Then there existnatural maps δ∗ : Hp(X,G) → Hp+1(X,E ) such that there is a long exactsequence in cohomology,

. . .→ Hp−1(X,G)δ∗−→ Hp(X, E)

f∗−→ Hp(X,F)g∗−→ Hp(X,G)→ . . .

Proof. First we define δ∗. Suppose that σ ∈ Hp(X,G) represented by someσ ∈ Cp(U ,G). The short exactness implies that there is a refinement V ≤ Usuch that ρVUσ = f(τ), for some τ ∈ Cp(V ,F). Consider δτ :

f(δτ) = δf(τ) = δρVUσ = ρVUδσ = 0.

By exactness, there exists some u ∈ Cp+1(V , E) such that g(µ) = δτ . Nowg(δµ) = δg(µ) = δ2τ = 0. But injectivity of g implies that δµ = 0, so wecan define δ∗[σ] = [µ] ∈ Hp+1(X, E).

We will prove this under the assumption that there exists arbitrarilyfine covers U such that

0→ Cp(U , E)→ Cp(U ,F)→ Cp(U ,G)→ 0

are exact for all p. This is an exact sequence of chain complexes, so bythe Snake Lemma we obtain boundary homomorphisms δ∗ : Hp(U ,G) →Hp+1(U , E) such that

. . .→ Hp−1(U ,G)δ∗−→ Hp(U , E)

f∗−→ Hp(U ,F)g∗−→ Hp(U ,G)→ . . .

is exact. Moreover, these commute with the restriction maps ρU ,V to finishoff the proof.

Theorem 5.26. Let X be a complex manifold. Then there exist naturalidentifications

Hp,q

∂(X) = Hq(X,Ωp).

Recall that Ωq denoted the sheaf of holomorphic q-forms.

Remark 5.27. We have shown that

0→ Ωp → Ap,0 → Ap,1 → . . .

is an acyclic resolution for the sheaf Ωp. The derived functor cohomologythen defines H∗(X,Ωp) to be the cohomology of the complex

0→ Ap,0 → Ap,1 → . . .

which is a high-level explanation of the theorem. In nice situations, thederived functor cohomology will agree with Cech cohomology. The proofwe give here basically reworks the fact that in derived functor cohomology,one can take an acyclic resolution (instead of an injective resolution).


Proof. Let Zp,q∂

= ker ∂ : Ap,q → Ap,q+1. By the ∂-Lemma, we have shortexact sequences of sheaves

0→ Ωp → Ap,0 ∂−→ Zp,1∂→ 0

and

0→ Zp,q−1∂

→ Ap,q−1 ∂−→ Zp,q∂→ 0 for all q.

Of course, the content of the ∂-Lemma is the surjectivity. The long exactsequence for the first short exact sequence reads

. . .→ Hr(Ap,q)→ Hr(Zp,1∂

)→ Hr+1(Ωp)→ Hr+1(Ap,0)→ . . .

but since Ap,q is fine, the middle map is an isomorphism. That shows

Hr(Zp,10 ) ∼= Hr+1(Ωp).

if r ≥ 1, and the long exact sequence for the second implies similarly that

Hr+1(Zp,q∂

) ∼= Hr(Zp,q+1

∂).

if q ≥ 1. This says that we can increase r by dropping q, and if we do thisdropping q to 1, and then using the first row of the long exact sequence inthe last step, we find:

Hq(X,Ωp) ∼= Hq−1(X,Zp,10 )

∼=...

∼= H1(X,Zp,q−1∂

)

∼=H0(Zp,q

∂)

im ∂ : H0(Ap,q−1)→ H0(Zp,q∂

)

∼= Hp,q

∂(X).

Theorem 5.28. Let X be a complex manifold. Suppose that U is an opencover of X such that

Hq(Uα1 ∩ . . . ∩ Uαs ,O) = 0 for q ≥ 1.

Then Hq(X,O) = Hq(U ,O).

Proof. We have Hr(Uα1 ∩ . . . ∩ Uαs ,O) = H0,r

∂(Uα1 ∩ . . . ∩ Uαs) = 0 by the

Dolbeault theorem. This implies that for r ≥ 1, we have an exact sequence

0→ Z0,r−1∂

(Uα1∩. . .∩Uαs)→ A0,r−1(Uα1∩. . .∩Uαs)→ Z0,r

∂(Uα1∩. . .∩Uαs)→ 0

More on Several Complex Variables 30

As this is true for all multi-intersections, the exactness passes to the levelof cochains.

0→ Cp(U ,Z0,r−1∂

)→ Cp(U ,A0,r−1)→ Cp(U ,Z0,r

∂)→ 0 is exact.

This gives a long exact sequence in cohomology, with the middle termsvanishing, giving

Hp(U ,Z0,r

∂) ∼= Hp+1(U ,Z0,r−1

∂).

We then repeat the proof of the above theorem for this open cover, whichgive an isomorphism to Dolbeault cohomology.

Hp(U ,O) = Hp(U,Z0,0

∂) ∼= . . . ∼= H1(U ,Z0,p−1

∂).

As before,

H1(U ,Z0,p−1∂

) ∼=H0(U ,Z0,p

∂)

im ∂= H0,p

∂(X) = Hp(X,O).

Remark 5.29. The same proof works for Ωp. Actually, the following is true:if Hp(Uα,O) = 0 for all p ≥ 1 (no higher intersections), then H1(U ,O) ∼=H1(X,O).

Corollary 5.30. If dimX = n, then Hq(X,O) = H0,q

∂(X) = 0 if q > n.

Example 5.31. Hq(Cn,O) ∼= H0,q(Cn) = 0 for all q ≥ 1 (this is the ∂-Lemma). In addition, Hq(Ck × (C∗)`,O) = 0 if k + ` ≥ 2, by an extensionof the ∂-Lemma.

It is a fact that if X is contractible, then Hq(X,Z) = 0 for all q ≥1. Indeed, the Cech cohomology with coeficients in a constant sheaf isisomorphic to the singular cohomology with the corresponding coefficients.Recall the exact sequence

0→ Z→ O → O∗ → 0.

So then if X is contractible, its higher Z-cohomoogy vanishes, and we getthat

Hq(X,O) ∼= Hq(X,O∗) for q ≥ 1.

Example 5.32.

Hp(Pn,Ωq) =

C p = q ≤ n

0 otherwise.

The idea of the proof is to cover Pn by the standard charts and computeeverything explicitly.

Chapter 6

More on Several ComplexVariables

6.1 Hartog’s Theorem

There are two useful result on extending holomorphic functions of severalcomplex manifolds, both of which have been called Hartog’s theorem inthe literature. The intuition is that the “singularities” of a multivariableanalytic function must look like an analytic hypersurface. This tells us acouple of things: if the singular locus has codimension at least 2, then itshould be removable. Also, if the singular locus is compact, then in higherdimensions we also expect the function to extend. As a simple example,consider 1

z1: we get a pole along z1 = 0, which defines a noncompact

hypersurface.Contrast with this with the one-dimensional case. There are no non-

empty (complex) codimension 2 subsets of C. On the other hand, weknow that there are holomorphic functions in a punctured disk that do notextend, e.g. a pole or essential singularity.

Theorem 6.1 (Hartog’s Theorem 1). Let Bε = Bε1×. . .×Bεn be a polydiscin Cn for n ≥ 2. Suppose that 0 < ε′ < ε for all i so that B′ε ⊂ Bε. Letf : Bε − Bε′ → C be holomorphic. Then there exists a unique extensionf : Bε → C.

Proof. Without loss of generality, let εi = 1. For δ > 0, let

Vi = z ∈ Bε : 1− δi < |zi| < 1

and V = Vi. So for sufficiently small δ, V ⊂ Bε−Bε′ , hence f is holomorphicon V . We want to formalize our intuition that f cannot have a polar part,which would cut out a non-compact singular locus.

31

More on Several Complex Variables 32

For fixed w = (z2, . . . , zn), |zi| < 1, consider the complex function ofone variable

gw(z1) = f(z1, . . . , zn) defined on A = 1− δ < |z1| < 1.

Therefore, this has a Laurent expansion

gw(z1) =∞∑

n=−∞

an(w)zn1 .

We claim that the an(w) are holomorphic in w. One way to see this is touse Cauchy’s integral formula

an(w) =1

2πi

∫|ξ|=1− δ

2

gw(ξ)

ξn+1dξ

and note that gw is holomorphic on this region.If we choose w so that 1− δ < |z2| < 1, then we see that gw is holomor-

phic on the entire disc |z1| < 1. This implies that an(w) = 0 for n < 0, forall such w, i.e. 1 − δ < |z2| < 1. So by the identity theorem, an(w) = 0for all |z2| < 1. This implies that f(z) =

∑n=0 an(w)zn1 is holomorphic

in z1 over the entire polydisc (it converges absolutely since it does when|z2| = 1, and the an are holomorphic so the maximum modulus occurs onthe boundary).

Theorem 6.2 (Hartog’s Theorem 2). Let U be an open subset of Cn forn > 2, and f a holomorphic function on U − U ∩ z1 = z2 = 0. Then fextends to a holomorphic function on all of U .

Proof. Let D = (z1, . . . , zn) : |zi| ≤ ri ⊂ U be a polydisc and D′ =D−D ∩ z1 = z2 = 0. We show that for each z1, . . . , zn) ∈ D′, we have aCauchy formula

f(z1, . . . , zn) =1

(2πi)n

∫∂D

f(ζ1, . . . , ζn) dz1 . . . dzn(ζ1 − z1) . . . (ζn − zn)

.

Since the right hand side is holomorphic in D, it defines a holomorphicextension.

If Dε = (ζ1, . . . , ζn) : |z1|, |z2| ≤ ε, |ζi| ≤ ri for i > 1 is a smallneighborhood of (z1, . . . , zn) contained in D′, then Cauchy’s formula implies

f(z1, . . . , zn) =1

(2πi)n

∫∂Dε

f(ζ1, . . . , ζn) dz1 . . . dzn(ζ1 − z1) . . . (ζn − zn)

.

Therefore, we just have to show that ∂Dε can be homotoped to ∂D throughD′. This is where the codimension 2 hypothesis comes in.

Holomorphic Vector Bundles 33

We choose a homotopy through regions of the form

D(t) = (ζ1, . . . , ζn) : |ζi − tzi| ≤ αi(t) i = 1, 2; |zi| = ri i > 2.

We require αi(0) = ri so that D(0) = ∂D, and αi(1) = εi so D(1) = ∂Dε.In order for D(t) to stay in D, we must have αi(t) < ri − |tzi|. For

D(t) to stay in D′, it also has to avoid (z1, . . . , zn) and z0 = z1 = 0.The first requirement is equivalent to (1 − t)|zi| 6= αi(t) for any t. Sinceri − |tzi| > (1− t)|zi| at t = 0 and t = 1, we can choose α so that

(1− t)|zi| < α(t) < ri − |tzi|.

To avoid z0 = z1 = 0, we cannot simultaneously have α1(t) = t|z1|and α2(t) = t|z2|. That is fine, since we have two degrees of freedomin choosing α1(t), α2(t) and we impose only one condition. (This comesdown to choosing a curve (t, α1(t), α(t)) in a cube that avoids the curve(t, t|z1|, t|z2|)).

Corollary 6.3. If X is a complex manifold and f a holomorphic functionon X − Z, where Z is a complex submanifold of codimension at least 2,then f extends to a holomorphic function on all of X.

Chapter 7

Holomorphic Vector Bundles

7.1 Basic definitions

Let X be a differentiable manifold. Recall that a complex vector bundle Eof rank r on X is a smooth manifold E → X with a smooth projectionmap π : E → X such that each fiber Ex := π−1(x) has the structure of aC-vector space of dimension r, which is locally free: X has an open cover⋃Uα with trivialization ϕα : π−1(Uα) ∼= Uα × Cr.If r = 1, we call E a complex line bundle.

Definition 7.1. If X is a complex manifold, we say that E is a holomorphicvector bundle if E is a complex manifold, π is holomorphic, and the ϕα areholomorphic.

Note that holomorphic vector bundles have stricter demands than com-plex vector bundles.

Recall that a vector bundle is determined by its transition functions

ϕαβ : ϕα ϕ−1β : (Uα ∩ Uβ)× Cr → (Uα ∩ Uβ)× Cr.

Note that ϕαβ can be considered as a map Uα∩Uβ → GLn(C). The bundleE is holomorphic if and only if the ϕαβ can be taken to be holomorphic.

From the definition, the ϕαβ satisfy the cocycle conditions

ϕαβ ϕβγ = ϕαγ

ϕαβ = ϕαγϕγα

for all α, β, γ.Let GLr(C) denote the sheaf GLr(C)(U) = g : U → GLr(C) holomorphic.

(This notation is sligtly confusing, since it would be also used to denote thelocally constant sheaf.) If E is holomorphic, then the data ϕαβ forms acochain in C1(U ,GLr(C)). The cocycle condition says that δ : C1(U ,GLr(C))→C2(U ,GLr(C)) has δϕαβ = 0, i.e. this cochain is actually a cocycle (hence

34


the name), so data of the transition functions descends to an element ofH1(X,GLr(C)).

Note that we have technically only defined cohomology for sheaves ofabelian groups, but we will soon specialize to r = 1.

Definition 7.2. Let πE : E → X and πF : F → X be holomorphic vectorbundles. A morphism f : E → F is a holomorphic map such that πFf = πEand such that the induced map fx : Ex → Fx is C-linear and rank(fx) islocally constant in x.

Remark 7.3. The constancy of rank is required in order to ensure that wecan take kernels and cokernels of morphisms. Be warned that this isn’thow vector bundle morphisms are usually defined in differential geometry.

Definition 7.4. We write O(E) for the sheaf of local sections of E:

O(E)(U) = s : U → E : πE s = id.

Example 7.5. Locally (for small enough opens) we have

O(E)(U) ' O(U)r

since the vector bundle is trivial.

So there is an essentially surjective embedding from the category ofholomorphic vector bundles to that of locally free sheaves, but this is notan equivalence of categories, since we imposed the constant rank condition.

7.1.1 Bundle Constructions

If E and F are holomorphic vector bundles, we can form holomorphic vectorbundles E⊕F,E⊗F,

∧iE, SkE,E∗, etc. This can be easily checked usingtransition functions: if E has transition functions ϕαβ and F has transitionfunctions ψαβ, then E ⊕ F has transition functions ϕαβ ⊕ ψαβ, etc.

7.2 Holomorphic Line Bundles

Let O denote the trivial line bundle X×C (we are sort of abusing notationhere; O evokes the sheaf of sections of X × C). Observe that if L1, L2 areholomorphic line bundles, then so is L1 ⊗ L2. Also, L1 ⊗ L∗1 ∼= O.

Definition 7.6. The Picard group Pic(X) is the group of holomorphic linebundles on X up to isomorphism, with the operation of tensor product.

Given a holomorphic line bundle L and an open cover U = Uα withtrivializations ϕα : L|Uα ' Uα × C a biholomorphism, the collection of


transition functions for L gαβ = ϕα ϕ−1β ∈ C1(U ,GL1(C)) = C1(U ,O∗)satisfy the cocycle condition, and hence descend to an element [gL] ∈H1(X,O∗).

Proposition 7.7. The map Γ: L 7→ [gL] induces an isomorphism of groupsPic(X) ∼= H1(X,O∗).

Proof. First we must show that this is actually well-defined. If L ∼= M areholomorphic line bundles, choose an open cover U trivializing both L andM , so we have

ϕα : L|Uα ' Uα × C and σα : M |Uα ' Uα × C

This gives cocycles gLαβ = ϕαϕ−1β and gMαβ = σασ

−1β .

We have an isomorphism f : L → M commuting with projections toX, giving fα : L|Uα ' M |Uα . Define hα = σαfαϕ

−1α : Uα × C → Uα × C.

This is multiplication by some (non-vanishing) constant on each fiber, socan be regarded as a section of O∗(Uα). Moreover,

(δh)αβ = hαh−1β

= σαfαϕ−1α (σβfβϕ

−1β )−1

= (gLαβ)−1gMαβ (fαf−1β )︸︷︷︸

=1 since f comes from a global section

Clearly Γ is surjective, since any set of transition functions satisfyingthe cocycle condition can be patched to form a holomorphic line bundle Lwith those transition functions. We have to establish that it is injective(or said differently, this inverse is well-defined). Suppose [L], [M ] ∈ Pic(X)such that [gL] = [gM ]. That means that there exists h = hα ∈ C0(U ,O∗)such that

hαh−1β = (gLαβ)−1(gMαβ).

We can use the hα to define an isomorphism of L and M . Let fα : L|Uα →M |Uα be fα = σ−1α hα ϕα. We have to verify that these glue properly,but of course that is precisely encapsulated by the above equation (detailsleft as exercise).

That Γ is a homomorphism is evident from the definitions.

Recall the exact sequence of sheaves

0→ 2πiZ i−→ O exp−−→ O∗ → 0.

This gives an exact sequence on cohomology,

. . .→ H1(X,O)→ H1(X,O∗) δ∗−→ H2(X,Z)→ . . .


We have just made the identification H1(X,O∗) ∼= Pic(X).

Definition 7.8. Given L ∈ Pic(X), the first Chern class of L is

c1(L) = δ∗([gL]) ∈ H2(X,Z).

This gives a discrete isomorphism invariant of a line bundle. The imageof H1(X,O) is the kernel, by exactness; this parametrizes the ambiguity ofthis invariant.

7.3 Line bundles on Projective Space

Recall that Pn is the set of lines in Cn+1. For each point ` ∈ Pn, we can as-sociate the complex line consisting of that line. This gives a (holomorphic)line bundle on Pn:

L = (`, w) ∈ Pn × Cn+1 : w ∈ `= ([z0, . . . , zn], (w0, . . . , wn)) ∈ Pn × Cn+1 : (w0, . . . , wn) = λ(z0, . . . , zn).

To see this, let’s compute the transition functions. Let Ui = [z0, . . . , zn] : zi 6=∅. Let ϕi : L|Ui → Ui×C send ϕi(`, w) = (`, wi). The transition functionsare

ϕij(z, w) =zizjw on Ui ∩ Uj.

Definition 7.9. L is the tautological line bundle on Pn and is denoted byOPn(−1).

We defineOPn(1) = OPn(−1)∗, OPn(a) = OPn(1)a if a > 0, andOPn(a) =OPn(−1)−a if a < 0. Notice that this is the subgroup of Pic(Pn) generatedby the tautological line bundle.

Suppose f : X → Y is holomorphic and E is a complex vector bundleon Y . Then there is a pullback bundle f ∗E → X such that (f ∗E)x =E|f(x). Given transition functions gEαβ over U = Uα, the bundle f ∗E hastransition functions f ∗gEαβ = gEαβ f on f ∗(Uα).

When f = ι : X → Y is an inclusion map, then we write E|X := ι∗E.

Exercise 7.10. If f : X → Y is a holomorphic map, then the pullbackinduces a group homomorphism f ∗ : Pic(Y )→ Pic(X).

After we study line bundles on Pn, this will give useful information onany projective complex manifold.


7.4 Ample Line Bundles

Recall that ifX is a complex manifold, we cannot use holomorphic functionsto embed X in Cn. The idea is to instead use sections of holomorphic linebundles.

Definition 7.11. X is said to be quasiprojective (resp. projective) if it isisomorphic to a (closed) submanifold of PN for some N .

For a holomorphic bundle E → X, write H0(X,E) = H0(X,O(E)) forthe vector space of global sections of E.

We are mostly interested in the special case of holomorphic line bundles.

Definition 7.12. A trivialization of L over an open set U ⊂ X is a ξ ∈O∗(L)(U), i.e. ξ(x) ∈ Lx is non-zero for all x ∈ U .

Given such a ξ, we define a map ϕ : L|U → U×C by ϕ(v) = (π(v), f(v))where v ∈ Lx is v = f(v)ξ(x) for the unique scalar f(v). In this sense, atrivialization is the same as the usual notion of trivialization of bundles.

Any two trivializations ξα over Uα and ξβ over Uβ, the transition func-tion is ξα/ξβ over Uα ∩ Uβ. Furthermore, any s ∈ O(L)(U) can be writtenuniquely as s = fξ for some f ∈ O(U) (so O(L)(U) is a free O(U)-modulegenerated by a trivialization.)

Fix a holomorphic line bundle π : L→ X. Suppose that there are globalsections s0, . . . , sN ∈ H0(X,L) and assume that x ∈ X is such that not allsi(x) are zero. Given a trivialization ξ of U , we can write si = fiξ wherefi ∈ O(U). By the non-vanishing assumption, [f0(x), . . . , fN(x)] ∈ PN . Weclaim that this is independent of ξ. Indeed, if ξ is another trivializationthen xi = gξ for some g ∈ O(U)∗, and this has the effect of multiplying allthe coordinates by g(x), which is just the same point in PN . We denote thispoint by [s0(x), . . . , sN(x)], which we have now established is well-defined.

Definition 7.13. Let V ⊂ H0(X,L). The base locus of V is

Bs(V ) := x ∈ X : s(x) = 0∀s ∈ V .

Let φV be the holomorphic map V \B(V )→ PN defined by

φv(x) = [s0(x), . . . , sN(x)]

(note that this depends on our choice of basis. There is a way of definingit invariantly).

Such a linear subspace V is called a linear series. If V = H0(X,L) thenwe say that V is the “complete linear series” defined by L.

Even if Bs(V ) = ∅, the map φV need not be an embedding. For in-stance, for L = O the induced map is constant.

Kahler Manifolds 39

Definition 7.14. We say L is ample if the map induced by the completelinear series H0(X,L⊗k) for some k ≥ 1 is an embedding.

This is saying that L⊗k has “enough” sections so that:

1. Bs(H0(X,L⊗k)) 6= ∅.

2. The map φL⊗k is injective, i.e. for x 6= y there exists s ∈ H0(X,L⊗k)such that s(x) = 0 and s(y) 6= 0.

3. dφL⊗k to be injective.

Theorem 7.15. Let X = Pn. There is a canonical identification for k ≥ 0

H0(Pn,OPn(k)) ' C[z0, . . . , zn]k = homogeneous polynomials of degree k .

Proof. Let s ∈ C[z0, . . . , zn]k and fix some [z] ∈ Pn, with coordinatesz = (z0, . . . , zn). Recall that O(−1)|[z] = λz : λ ∈ C ⊂ Cn+1. Lets : O(−1)|[z] → C be defined by s(λz) = λks(z0, . . . , zn). This definesa point in (O(−1)|∗[z])⊗k = O(k)|[z]. Exercise: check that this defines aholomorphic global section. Clearly if s 6= 0 then s 6= 0.

We need to show that s 7→ s is surjective. Let t ∈ H0(Pn,O(k)) andpick some 0 6= s ∈ C[z0, . . . , zn]k. Then f := t

sis a meromorphic function

on Pn. Let π : Cn+1 \ 0 → Pn be the natural projection, and consideru(z) = s(z)f([z]). This is a holomorphic function on Cn+1 − 0. ByHartog’s theorem, it extends to a holomorphic function on Cn+1, and it isa polynomial by homogeneity, say u. Then u = t.

Given any line bundle L→ X, there is a natural multiplicationH0(X,L⊗k)⊗H0(X,L⊗r)→ H0(X,L⊗(k+r)). The isomorphism above is compatible withthe multiplicative structure, so we have an isomorphism of rings⊕

k≥0

H0(X,L⊗k) ' C[z0, . . . , zn].

Chapter 8

Kahler Manifolds

8.1 Kahler metrics

We seek to put a metric on a complex manifold X that respects the complexstructure.

Let V be a finite-dimensional real vector space and J : V → V analmost complex structure on V . Let 〈, 〉 be an inner product on V .

Definition 8.1. We say that 〈, 〉 is compatible with J if 〈v, w〉 = 〈Jv, Jw〉for all v, w ∈ V . When this holds, the fundamental form is

ω(v, w) = 〈Jv, w〉.

Remark 8.2. Any two of 〈, 〉, J, ω detemines the third. Note that ω is anantisymmetric form, since

ω(v, w) = 〈Jv, w〉 = 〈J2v, Jw〉 = −〈v, Jw〉 = −ω(w, v).

It also follows that ω is definite, since the inner product is.

Now extend these to VC = V ⊗ C. The inner product extends to aHermitian inner product on VC,

〈λv, µw〉C = λµ〈v, w〉.

Also, ω extends C-linearly to an element of∧2 V ∗C .

Lemma 8.3. Suppose (V, 〈, 〉) is Euclidean and J is a compatible almostcomplex structure.

1. The decomposition VC = V 1,0⊕V 0,1 is orthogonal with respect to 〈, 〉C,

2. We have ω ∈ Λ1,1V ∗C .

Proof. 1. If v ∈ V 1,0 and w ∈ V 0,1, then

〈v, w〉 = 〈Jv, Jw〉 = 〈iv,−iw〉 = −〈v, w〉.

40

Kahler Manifolds 41

Another way to see this is to observe that any element of V 1,0 isv − iJv and any element of V 0,1 is w + iJw.

2. If v, w ∈ V 1,0 then

ω(v, w) = ω(Jv, Jw) = ω(iv, iw) = −ω(v, w) =⇒ ω(v, w) = 0.

A similar calculation works for V 0,1.

Let X be a complex manifold. This gives an almost complex structureJ : TX → TX with J2 = −Id.

Definition 8.4. A Riemannian metric g on X is said to be compatible withthe almost complex structure if for all x ∈ X, the scalar product gx on TxXis compatible with J , i.e.

gx(v, w) = gx(Jxv, Jxw).

Similarly, define the fundamental form ω(v, w) = g(Jv, w). The exten-sion of g gives a Hermitian inner product on TX ⊗ C. Suppose that wehave local holomorphic coordinates z1, . . . , zn on X. Then dz1, . . . , dzn area C-basis for (T ∗X)1,0 (locally). Let

hij = 2〈∂i, ∂j〉g.

Exercise 8.5. Check that

ω =i

2

∑j,k

hjkdzj ∧ dzk.

Definition 8.6. We say g is a Kahler metric if dω = 0.

Example 8.7. On Cn, with coordinates z1, . . . , zn, we get

ω =i

2

n∑i=1

dzi ∧ dzi.

Example 8.8. If dimX = 1 (i.e. X is a Riemann surface), then any(1, 1)-form is closed.

Example 8.9 (Fubini-Study metric on Pn). Let U ⊂ Pn be open andπ : Cn+1 → Pn be the natural projection map. Suppose s : U → Cn+1

is a holomorphic lift of π, i.e. πs(z) = z for all z ∈ U . Let ωFS|U =i2π∂∂ log ||s||2. This is (1, 1)-form, but we need to check that it is well-

defined, closed, and positive definite.

Kahler Manifolds 42

To show that ωFS is well-defined, choose another s. So s = fs for someholomorphic function f , so ||s||2 = |f |2||s||2. Then

∂∂ log ||s||2 = ∂∂ log |f |2 + ∂∂ log ||s||2.

Now,∂∂ log |f |2 = ∂∂ log f + ∂∂ log f = 0

since f is holomorphic. Next, note that

2ωFS =i

2π(∂ + ∂)(∂ − ∂) log ||s||2

=i

2πd(∂ − ∂) log ||s||2 =⇒ dωFS = 0.

Finally, we need to show that if

ωFS =1

2

∑hijdzj ∧ dzj

then the (hij) is positive definite. Note that U(n + 1) acts on Cn+1 andon Pn holomorphically, transitively, and is ωFS-invariant. So it suffices toprove the positive definiteness at a particular point, e.g. [1, 0, . . . , 0]. Thisis a local calculation, left as an exericse.

Lemma 8.10. If f : X → Y is a holomorphic embedding and w is Kahleron X, then f ∗ω is Kahler on X.

Proof. The closedness follows from df ∗ω = f ∗dω = 0, and the positivedefiniteness is obvious.

Corollary 8.11. Any projective manifold is Kahler.

Pick a local unitary frame for the hermitian metric ϕ1, . . . , ϕn. Settingξj = Repϕj and ξj = imϕj, we have locally

h =∑j

ϕj ⊗ ϕj.

Then

g = Rep∑j

(ηj + iξj)⊗ (ηj − iξj) =∑j

ηj ⊗ ηj + ξj ⊗ ξj

and

ω =i

2

∑j

(ηj + iξj) ∧ (ηj − iξj) =∑j

ηj ∧ ξj.

Kahler Manifolds 43

Therefore, we see that 1n!

∧ωn is a volume form on X, and we conclude

that ∫X

ωn > 0.

when the left hand side makes sense (e.g. if X is compact).

Proposition 8.12. If X is compact Kahler, then

dimH2qdR(X,R) > 0.

Proof. Let ω be a Kahler form on X. Let τ = ∧qω. Then dτ = 0 as dω = 0.So [τ ] ∈ H2q

dR(X,R). Suppose for the sake of contradiction that τ = dσ forsome σ. Then∫

X

∧nω =

∫X

ωn−q ∧ τ =

∫X

ωn−q ∧ dσ =

∫X

d(σ ∧ ωn−q) = 0

(the last equality following from the fact that dω = 0) by Stokes’ Theorem(we are assuming a manifold without boundary).

Note that this is a topological result, since the de Rham cohomologyis a topological invariant (of smooth manifolds). So there is a topologicalobstruction to being a Kahler manifold!

Proposition 8.13. Let ω be a (1, 1)-form associated to a Hermitian metricon a complex manifold X. Then dω = 0 if and only if for all x ∈ X, thereexist (holomorphic) coordinates z1, . . . , zn centered at x so that locally

ω =i

2

∑hijdzi ∧ dzj

with hij = δij + O(|z|2). In other words, ω defines a Kahler metric if andonly if ω = ω0 +O(|z|2) locally, where ω0 is the standard form on Cn.

Proof. Say ω = 12

∑hijdzi ∧ dzj. Then

dω =1

2

∑ ∂hij∂zk

dzk ∧ dzi ∧ dzj +1

2

∑ ∂hij∂zk

dzk ∧ dzi ∧ dzj.

If the relation in the proposition holds, we get∑ ∂hij

∂zk=

∂hij∂zk

= 0 so dω = 0.Conversely, suppose that dω = 0. Let z1, . . . , zn be holomorphic coor-

dinates at x so ω = 12

∑i,j hijdzi ∧ dzj. By a linear change of coordinates,

we may assume that hij(0) = δij. The Taylor expansion will look like

hij(z) = δij +∑k

aijkzk +∑k

bijkzk +O(|z|2).

Kahler Manifolds 44

Since hij = hji, we have bijk = ajik. Since dω = 0,

0 =∑i,j,k

aijkdzk ∧ dzi ∧ dzj +∑i,j,k

bijkdzk ∧ dzi ∧ dzj.

This forces aijk = akji and bijk = bikj.Now let wj = zj + 1

2

∑i,k aijkzizk. We claim that this will kill the linear

term. Indeed,

dwj = dzj +1

2

∑i,k

aijk(zidzk + zkdzi)

dwj = dzj +1

2

∑i,k

aijk(zidzk + zkdzi).

Now we compute:

dwj ∧ dwj =∑

dzj ∧ dzj +1

2

∑i,j,k

aijk(zidzj ∧ dzk + zkdzj ∧ dzi)

+1

2

∑i,j,k

aijk(zidzk ∧ dzj + zkdzi ∧ dzj) +O(|z|2)

=∑

dzj ∧ dzj +∑i,j,k

aijkzkdzidzj +∑i,j,k

bijkzkdzjdzi +O(|z|2)

=2

iω +O(|z|2).

8.2 The Kahler Identities

The Kahler condition implies strong interactions between the Riemanniangeometry and complex geometry. To discuss this, we recall some operatorson forms.

1. Consider (X, g) an oriented Riemannian manifold of dimension 2nover R. There is an exterior derivative

d : Ak → Ak+1

satisfying d d = 0.

2. Let σ be an orientation form for X. The Hodge star ∗ : Ak → A2n−k

is defined so thatα ∧ ∗β = 〈α, β〉σ

for all α ∈ Ak.

Kahler Manifolds 45

Set d∗ = − ∗ d∗ : Ak → Ak−1. The Laplacian (or what is sometimescalled the “Laplace-Beltrami operator” in this general setting) is

∆d = d∗d+ dd∗ : Ak → Ak.

3. Now suppose that X is a complex manifold of complex dimension n,and g is a Riemannian metric on X. Extend g and ∗ to the complexstructure, so we get a map ∗ : AkC → A2n−k

C . Again,

α ∧ ∗β = gC(α, β)σ.

Using the complex structure, we write d = ∂ + ∂ where

∂ : Ap,q → Ap+1,q ∂ : Ap,q → Ap,q+1.

We define ∂∗

= − ∗ ∂∗ and ∂∗ = − ∗ ∂∗ and

∆∂ = ∂∗∂ + ∂∂

∗∆∂ = ∂∗∂ + ∂∂∗.

If ω is a Hermitian form, let

L = Ap,q → Ap+1,q+1

be L(α) = α ∧ ω. Set

Λ = ∗−1L∗ : Ap,q → Ap−1,q−1.

Just as in Riemannian geometry, the operators ∂∗ and ∂∗

are con-structed to be the adjoints of ∂, ∂.

Lemma 8.14. Suppose α ∈ Ap−1,q and β ∈ Ap,q. Then

〈∂∗α, β〉 = 〈α, ∂β〉

and〈∂∗α, β〉 = 〈α, ∂β〉.

Proof. Without loss of generality, we can prove only the first identity. Notethat if α ∈ Ak, ∗ ∗ α = (−1)2n−kα. Also, by Stokes’ Theorem

0 =

∫d(α ∧ ∗β) =

∫∂(α ∧ ∗β) =

∫∂α ∧ ∗β + (−1)kα ∧ ∂ ∗ β.

Kahler Manifolds 46

Therefore,

〈∂α, β〉 =

∫∂α ∧ ∗β

= (−1)k+1

∫α ∧ ∂ ∗ β

= (−1)k+1+k(2n−k)∫α ∧ ∗ ∗ ∂ ∗ β

= 〈α, ∂∗β〉.

Theorem 8.15. Assume ω is Kahler. Then

1. ∆d = 2∆∂ = 2∆∂.

2. Let h =∑2n

k=0(n − k)πk where πk is the projection map An → Ak.Then h,Λ, and L commute with ∆d.

3. [Λ, L] = h, [h, L] = −2L, and [h,Λ] = 2Λ.

This theorem says that there are three operators on the space of har-monic forms on X: h,Λ, L which form a representation of sl2. The rest ofthe section is devoted to a proof of this theorem.

Step 1. For now, we assume that X = Cn and ω = i2

∑dzj ∧ dzj is the

standard Kahler form for the Euclidean metric g = 12

∑dzj ⊗ dzj.

First, we establish some notation.

Definition 8.16. For α ∈ Aq, ξ ∈ A1 let ξ ∨ α ∈ Aq−1 be defined by

〈ξ ∨ α, β〉 = 〈α, ξ ∧ β〉∀β ∈ Aq−1.

Since the Hermitian product is non-degenerate, this specifies the oper-ator uniquely. The point is that ∨ is essentially the adjoint to ∧.

Definition 8.17. If α ∈ Aq, say α =∑|I|+|J |=q αIJdzI ∧ dzJ we let

∂jα =∑I,J

∂αIJ∂zj

dzI ∧ dzJ ∂jα =∑I,J

∂αIJ∂zj

dzI ∧ dzJ

Lemma 8.18. With the notation above, dzj ∨ dzk = 0 and dzj ∨ dzk =gjk = 2δjk for all j, k.

Proof. This is obvious from the definitions:

dzj ∨ dzk := 〈dzj, dzk〉 = 0

Kahler Manifolds 47

anddzj ∨ dzk := 〈dzj, dzk〉 = gjk = 2δjk.

Next, we claim the following identities:

Lemma 8.19. 1. ∂α =∑

j dzj ∧ ∂jα.

2. ∂j〈α, β〉 = 〈∂jα, β〉+ 〈α, ∂jβ〉.

3. ∂j(dzk ∨ α) = dzk ∨ ∂jα.

Proof. 1. This is obvious.

2. The important point here is that our metric standard one, so itdoesn’t depend on z:

∂j〈α, β〉 = ∂j(∑I

αIβI)

=∑I

∂j(αIβI)

=∑I

(∂αI)βI + αI∂βI

3. Follows from pairing both sides with β and using the previous part.The point is that ∂j commutes with dzk∨ since it commutes with itsadjoint dzk∧. In full, gory detail:

〈∂j(dzk ∨ α), β〉 = ∂j〈dzk ∨ α, β〉 − 〈dzk ∨ α, ∂jβ〉= ∂j〈α, dzk ∧ β〉 − 〈α, dzk ∧ β〉= 〈∂jα, ∂j(dzk ∧ β)〉= 〈dzk ∨ ∂jα, β〉.

Step 2. We have the following key identity.

Proposition 8.20.

∂∗α = −

∑j

dzj ∨ ∂jα.

Proof. Let α ∈ Aq and β ∈ Aq−1 with compact support. Then we have∫∂j〈dzj ∨ α, β〉dV = 0 by Stokes’ theorem and compact support.

Kahler Manifolds 48

This says that ∂j to adjoint to (−1)k∂j. By construction,

〈∂∗α, β〉 = 〈α, ∂β〉

=∑j

〈α, dzj ∧ ∂jβ〉

=∑j

〈dzj ∨ α, ∂jβ〉

= −∑j

〈dzj ∨ ∂jα, β〉.

Step 3.

Proposition 8.21. On Cn with the standard metric, we have

[∂∗, L] = i∂.

Proof. We compute:

[∂∗, L]α = ∂

∗Lα− L∂∗α

= ∂∗(ω ∧ α)− ω ∧ ∂∗α.

One can explicitly check that

dzk ∨ (α ∧ β) = (dzk ∨ α) ∧ β + (−1)kα ∧ (dzk ∨ β).

Therefore,

∂∗(ω ∧ α) = −

∑j

dzj ∨ ∂j(ω ∧ α)

= −∑j

dzj ∨ (ω ∧ ∂jα)

= − i2

∑j,k

dzj ∨ (dzk ∧ dzk ∧ ∂jα)

= − i2

∑j,k

(dzj ∨ dzk) ∧ dzk ∧ ∂jα

+i

2

∑j,k

dzj ∧ (dzj ∨ dzk) ∧ ∂jα

− i

2

∑j,k

dzk ∧ dzk ∧ (dzj ∨ ∂jα).

Kahler Manifolds 49

The first term vanishes by the earlier calculation. The second term can bewritten as i

∑j dzj ∧ ∂jα = ∂α, and this is what we want to end up with.

The last term will cancel out with stuff from the commutator. Indeed,

−ω ∧ ∂∗α =∑j

ω ∧ dzj ∨ ∂jα

and this cancels the last term.

Proposition 8.22. On any Kahler manifold X,

1. [∂∗, L] = i∂

2. [∂∗, L] = −i∂,

3. [Λ, ∂] = −i∂∗,

4. [Λ, ∂] = i∂∗.

Proof. For (1), recall that as ω is Kahler, around any x there are hol-moorphic coordinates z1, . . . , zn in which ω = ω0 + o(|z|2), where ω0 isthe standard form. Since [∂

∗, L] only involves first order derivatives in the

metric coefficients, the calculation for Cn shows that this is still true.Then (2) follows from conjugating (noting that ω is a real (1, 1)-form),

and (3) and (4) from taking adjoints.

Proposition 8.23. On any Kahler manifold (X,ω) we have

∆d = 2∆∂ = 2∆∂.

Proof. We claim that

∂∗∂ + ∂∂

∗= 0

∂∗∂ + ∂∂∗ = 0.

This is just formal: the first identity is

∂∗∂ + ∂∂

∗= −i[L∗, ∂]∂ − i∂[L∗, ∂]

= −iL∗∂∂ + i∂L∗∂ − i∂L∗∂ + i∂∂L∗

= 0 since ∂∂ = 0.

The other one follows by conjugation.Now we claim that ∆d = ∆∂ + ∆∂. Write

∆d = dd∗ + d∗d = (∂ + ∂)(∂∗ + ∂∗) + (∂∗ + ∂

∗)(∂ + ∂)

and expand and collect terms, canceling the cross-terms.

Kahler Manifolds 50

Finally, one computes that ∆∂ = ∆∂ in a similar fashion: the righthand side is

−i∂[L∗, ∂]− i[L∗, ∂]∂ = −i∂L∗∂ + i∂∂L∗ − iL∗∂∂ + i∂L∗∂

and the right hand side is

i∂[L∗, ∂] + i[L∗, ∂]∂ = i∂L∗∂ − i∂∂L∗ + iL∗∂∂ − i∂L∗∂.

These are equal since ∂∂ = −∂∂.

Now, we are ready to prove the remaining Kahler identities.

Theorem 8.24. Let (X,ω) be Kahler. Define L : A∗ → A∗ by L(α) =α ∧ ω, and Λ = L∗. Let πk : A∗ → Ak be the projection, and define

h =2n∑p=0

(n− k)πk.

Then

1. Then h,Λ, and L commute with ∆d.

2. [Λ, L] = h, [h, L] = −2L, and [h,Λ] = 2Λ.

Proof. First we study the commutators with h. By linearity, suffices tocheck the identities on α ∈ Ap,q where p+ q = k. Then it is clear that

[h,∆∂]α = (n− k)∆∂α−∆∂(n− k)α = 0.

Also,

[h, L]α = hLα− Lhα= (n− (k + 2))Lα− L(n− k)α

= −2Lα.

Taking the adjoint (and noting that h∗ = h since π∗k = πk) we obtain thethird commutation relation.

We show that [L,∆d] = 0. (Note that this is equivalent to showingthat ω is harmoinc). Using d(ω ∧ α) = ω ∧ dα, we have [L, d] = 0. Then

Hodge Theory 51

[L, ∂] = 0 and [L, ∂] = 0. Therefore,

[L,∆∂] = [L, ∂∂∗ + ∂∗∂]

= ∂[L, ∂∗] + [L, ∂∗]∂

= −i(∂∂ + ∂∂)

= 0.

Taking adjoints, we get that Λ commutes with ∆d (since ∆d is self-adjoint).

The only thing left to see is that [Λ, L] = h. We can recast this assaying that if α ∈ Ap,q then

[Λ, L]α = (n− p− q)α.

This has no derivatives, so it certainly follows if it is shown on (Cn, ω0),which can be checked explicitly. In the case n = 1, we have Λ(g(z)dzdz) =g(z), and the commutator identity follows in that case. In general, write

L =∑k

Lk where Lkα = dzk ∧ dzk ∧ α

and Λ =∑

k Λk, where Λk = L∗k just removes dzk ∧ dzk if it exists and killsthe form otherwise (we are ignoring issues with constants). Furthermore,it is clear that [Lk,Λj] = 0 if j 6= k, so this reduces to the one-variablecase.

Chapter 9

Hodge Theory

9.1 The Hodge Decomposition

Our aim is to use the Kahler identities to get info on Hp,q

∂(X). We do this

indirectly, by considering the harmonic forms.

Definition 9.1. Given an oriented Riemannian manifold (X, g), we definethe space of harmonic forms of degree k to be

Hk(X, g) = α ∈ Ak(X) : ∆dα = 0.

Remark 9.2. On Rn with the standard Riemannian metric, ∆d reducesto the usual Laplace operator, so functions lying in the kernel of ∆d areharmonic functions in the usual sense.

Lemma 9.3. We have

∆∂α = 0 ⇐⇒ ∂α = ∂∗α = 0.

Proof. The direction ⇐= is obvious since ∆∂ = ∂∂∗

+ ∂∗∂. For the other

direction, if ∆∂α = 0 then

0 = (∆∂α, α) = (∂∂∗

+ ∂∗∂, α) = ||∂∗α||2 + ||∂α||2.

If (X, g) is Kahler, then the Kahler identities imply that Hk(X, g) isalso the set of forms annihilated by ∆∂ (or ∆∂, since they are the same).Let

Hp,q

∂(X, g) = α ∈ Ap,q(X) : ∆∂α = 0.

Recall that for a compact oriented Riemannian manifold (X, g), we havethe following Hodge decomposition:

52

Hodge Theory 53

Theorem 9.4 (Hodge Decomposition for Riemannian manifolds). If (X, g)is a compact Riemannian manifold, then there is a natural orthogonal com-position

Ak(X) ∼= Hk(X)⊕ dAk−1(X)⊕ d∗Ak−1(X).

Note that another way to write this is

Ak(X) ∼= Hk(X)⊕∆d(Ak(X))

since ∆dAk(X) = dd∗Ak(X)⊕d∗dAk(X) = dAk−1(X)⊕d∗Ak+1(X) by theabove.

If (X, g) is in addition Kahler (keeping the compactness hypothesis),then we have an inner product

(α, β) =

∫X

α ∧ ∗β =

∫X

gC(α, β)dV.

This is obviously positive definite, so it induces a norm on |Ak(X)|. (Itmight be worth warning that the forms are not complete with respect to thisnorm.) The analogous Hodge decomposition for complex Kahler manifoldsis:

Theorem 9.5 (Hodge Decomposition for Kahler manifolds). If (X, g) is acompact Kahler manifold, then there are natural orthogonal decompositions

Ap,q(X) ∼= Hp,q(X)⊕ ∂Ap−1,q(X)⊕ ∂∗Ap+1,q(X)

∼= Hp,q(X)⊕ ∂Ap,q−1(X)⊕ ∂∗Ap,q+1(X).

Moreover, the spaces Hp,q(X) are finite dimensional.

Again, this could be written as

Ap,q(X) ∼= Hp,q(X)⊕∆∂(Ap,q(X))∼= Hp,q(X)⊕∆∂(Ap,q(X))

As mentioned just above we have ∆d = 2∆∂ = 2∆∂, so we may identifyHp,q = Hp,q

∂ = Hp,q

∂.

Corollary 9.6. The map Hp,q

∂(X)→ Hp,q

∂(X) is an isomorphism, i.e. each

class in Hp,q(X) is represented by a unique harmonic form.

Proof. Lemma 9.3 shows that the map α 7→ [α] from Hp,q → Hp,q

∂is well-

defined.Now let’s show surjectivity. Let α ∈ Ap,q(X) with ∂α = 0. By the

Hodge Decomposition Theorem, we may write

α = β1 + ∂β2 + ∂∗β3 where β1 is harmonic.

Hodge Theory 54

Applying ∂ to both sides, we see that 0 = ∂∂∗β3 =⇒ ∂

∗β3 = 0 (by a

similar argument as before, consider (∂∂∗β3, β3) = 0). Therefore, [α] = [β1]

in Hp,q

∂(X).

Now for injectivity, suppose that α ∈ Hp,q(X, g) and [α] = 0 in Hp,q

∂(X).

Then α = ∂β for some β, so 0 = ∂∗∂β implies ∂β = 0.

By the same argument in the Riemannian case, we have

Corollary 9.7. The map Hk(X)→ HkdR(X) is an isomorphism, i.e. each

class in HkdR(X) is represented by a unique harmonic form.

Some operations on Hp,q(X, g)

1. Conjugation α 7→ α sends harmonic forms to harmonic forms since∂α = ∂α = 0, hence induces an isomorphism

Hp,q

∂(X, g) ∼= Hq,p

∂(X, g).

Note that we are really using the Kahler identities here, since a prioriwe would end up in H∂.

2. The Hodge star α 7→ ∗α sends harmonic forms to harmonic formssince ∂∗ ∗ α = ± ∗ ∂α = 0, hence gives an isomorphism

Hp,q

∂(X, g) ∼= Hn−q,n−p

∂(X, g).

3. Serre duality: consider the pairing

Hp,q

∂(X, g)×Hn−p,n−q

∂(X, g)→ C

where (α, β) 7→∫Xα∧β. If α 6= 0, then (α, ∗α) pairs to

∫α∧∗α > 0,

giving an isomorphism

Hp,q

∂(X, g) ∼= Hn−p,n−q

∂(X, g)∗.

Since the harmonic forms are canonically identified with Dolbeaultcohomology (by above), we get a similar pairing on the Dolbeaultcohomology.

4. There is a Lefschetz operator

L : Ap,q(X)→ Ap+1,q+1(X)

sending L(α) = α ∧ ω. So as [L,∆∂] = 0 we get an induced map

L : Hp,q

∂(X, g)→ Hp+1,q+1

∂(X, g).

Hodge Theory 55

Define hp,q = dimHp,q

∂(X) on any complex manifold X. These are

certainly finite if X is compact Kahler (since the harmonic forms are finitedimensional). On a compact Kahler X, the Hodge diamond is the array

h0,0

h0,1 h1,0

h0,2 h1,1 h2,0

h0,3 h1,2 h2,1 h3,0

......

hn−1,n hn,n−1

hn,n

Note that the rows are symmetric by conjugation, the columns are sym-metric by the Hodge star, and their composition is the symmetry of Serreduality.

Theorem 9.8. Let (X, g) be compact Kahler. Then there exists a decom-position

Hk(X,C) ∼=⊕p+q=k

Hp,q

∂(X)

independent of the chosen Kahler structure.

This is in accordance with what we got earlier, but the point is thatthis doesn’t depend on the metric.

Proof. The decomposition is induced by the Hodge decomposition

HkdR(X,C) ∼= Hk

∂(X,C) ∼=

⊕p+q=k

Hp,q

∂(X, g) ∼=

⊕p+q=k

Hp,q

∂(X).

We must show that the composition is independent of the choice of g. Sincewe know that the composition is actually an isomorphism, it suffices to showthat if g1, g2 are two Kahler metrics, and α1 ∈ Hp,q(X, g1), α2 ∈ Hp,q(X, g2)such that [α1] = [α2] ∈ Hp,q

∂(X), then [α1] = [α2] in Hk

dR(X,C).

Lefschetz Theorems 56

To see this, suppose [α1] = [α2] in Hp,q

∂(X), so α1 = α2 + ∂γ for some

γ. So d(∂γ) = d(α1 − α2) = 0 as ∆d(α1 − α2) = 0. Moreover, ∂γ isorthogonal to Hk(X, g) by the (Kahler) Hodge decomposition theorem,so ∂γ ∈ d(Ak−1C (X)) by the (Riemannian) Hodge decomposition theorem.That implies [α1]− [α2] in Hk

dR(X,C).

Chapter 10

Lefschetz Theorems

10.1 Lefschetz, I

Throughout this chapter, X is compact Kahler. The basic question is: howmuch of H2 can we see from line bundles on X? To make this more precise,recall that from the fundamental exact sequence we had

Pic(X) ∼= H1(X,O∗) c1−→ H2(X,Z).

Since Z ⊂ C, we have H2(X,Z) ⊂ H2(X,C). Last time, we proved that ifX is compact Kahler then we have a Hodge decomposition

H2(X,C) ∼= H2,0

∂(X)⊕H1,1

∂(X)⊕H0,2

∂(X).

Definition 10.1. Let H1,1(X,Z) = H2(X,Z) ∩H1,1

∂(X).

This is the image of the map H2(X,Z) → H2(X,C) in H1,1

∂(X). The

Lefschetz theorem on (1, 1) classes identifies the image of c1 : Pic(X) →H2(X,Z) with H1,1(X,Z).

The inclusion C ⊂ OX induces

f1 : H2(X,C)→ H2(X,OX) ∼=︸︷︷︸Dolbeault

H0,2

∂(X).

Here and throughout, we are identifying H∗(X,C) with H∗dR(X,C). Wealso have a projection f2 : H2(X,C)→ H0,2

∂(X) by Hodge decomposition.

Lemma 10.2. With the notation above, f1 = f2.

Proof. Recall the complexes of sheaves we used to prove the Dolbeaulttheorem:

C

// A0 //

=

A1 //

π1

A2

π2

OX // A0 // A0,1 // A0,2

57


By chasing through the Dolbeault isomorphism, the map f1 : H2(X,C)→H0,2

∂(X) can be given as follows. Let [α] ∈ H2(X,C) be represented by

some α ∈ A2C(X), then f1([α]) = [π2(α)] in H0,2

∂(X). ♠♠♠ Tony: [The rest

of the argument is a pedantic way of saying that the result is automatic fromconstruction of the Hodge Decomposition.]

We have to show that f2[α] = [π2(α)]. To see this, we can take α to bethe (unique) harmonic representative with respect to some Kahler metric g.Then π2(α) is the (unique) harmonic representative of [π2(α)] in H0,2

π1(X).

On the other hand, π2(α) also represents f2([α]) by our construction ofthe Hodge decomposition in Theorem 9.8, so we conclude that p([α]) =[π2(α)].

Theorem 10.3. The image of the first Chern class H1(X,O∗) → H2(X,C)is H(1,1)(X,Z).

Proof. Let α ∈ H2(X,Z) ⊂ H2(X,C). Say α = α2,0 + α1,1 + α0,2 whereαp,q ∈ Hp,q

∂(X). As α ∈ H2(X,Z) is real, we have α2,0 = α0,2. So α ∈

H1,1

∂⇐⇒ α2,0 = 0.

By the preceding lemma, [α2,0] is precisely the class obtained by fromthe composition

Pic(X) // H2(X,Z)f //

s

&&

H2(X,O) ∼= H0,2

∂(X)

H2(X,C)

β 7→β0,266

So α0,2 = 0 ⇐⇒ α ∈ ker f , which is what the exactness tells us.

Let X be a complex manifold. The Jacobian of X is Pic0(X) =ker c1 = L ∈ Pic(X) : c1(L) = 0. So understanding Pic(X) comes downto understanding the discrete group H1,1(X,Z) and Pic0(X). Note thatPic(X)0(X) = H1(X,O)/H1(X,Z) from the long exact sequence. If X iscompact Kahler then one can show that H1(X,Z) is a full-rank lattice inH1(X,O). So it is not hard to then get that Pic0(X) is a complex manifold(in fact, a complex torus).

Remark 10.4. It is a fact that given a smooth hypersurface D ⊂ X onecan associate a line bundle O(D) with a unique s ∈ H0(O(D)) such thatD = s = 0. If X is projective then one can show that H1,1(X,Z) isspanned over Q by c1(OD) for such divisors.

Definition 10.5. Let Z ⊂ X be a complex submanifold of complex di-mension n− p. We define [Z] ∈ Hp,p(X,C) by requiring∫

X

α ∧ [Z] =

∫Z

α|Z ∀α ∈ H2n−2p(X,Q).


Naıvely, one could hope that all classes in Hp,p(X,C) ∩ H∗(X,Z) arelinear combinations of the classes corresponding to submanifolds [Z]. Wesaw above that this is actually true for H1,1, but it cannot be true ingeneral. The Hodge conjecture (a Millenium Prize Problem) predicts thatthis is so for rational classes, i.e. any class in Hp,p(X) ∩ H∗(X,Z) is aQ-linear combination of classes of the form [Z] for Z ⊂ X.

10.2 Lefschetz, II

Recall that if X is Kahler and ω is the Kahler form, we considered theoperator L(α) = α∧ω. By the Kahler identities we have [L,∆] = 0 so thisinduces

L : Hp,q(X,ω)→ Hp+1,q+1(X,ω)

hence alsoL : Hp,q

∂(X)→ Hp+1,q+1

∂(X)

andL : Hk(X,C)→ Hk+2(X,C).

Theorem 10.6 (Hard Lefschetz). Let X be compact Kahler of complexdimension n. Then

Lk : Hn−k(X,C)→ Hn+k(X,C)

is an isomorphism.

This result (and improvements) follow from the general theory aboutfinite-dimensional representations of sl2(C) and the Kahler identities. Forcompleteness, we give a discussion of representations of sl2. As a vectorspace, this consists of traceless 2 × 2 complex matrices. A standard basisis

X =

(0 10 0

)H =

(1 00 −1

)Y =

(0 01 0

)and these satisfy the commutation relations

[X, Y ] = H, [H,X] = 2X, [H,Y ] = −2Y.

A representation of sl2 is a Lie algebra homomorphism sl2 → gl(V ). Wesay that ρ is irreducible if it has no proper subrepresentations. It is a factthat any finite dimensional representation of sl2 is semisimple, i.e. a directsum of irreducibles.

In any representation, H has eigenspaces Vλ called weight spaces. Ob-serve that if v ∈ Vλ, then Xv ∈ Vλ+2 and Y v ∈ Vλ−2. We say that v ∈ Vis a highest weight vector if it’s an eigenvector for H and Xv = 0.

Hermitian Vector Bundles 60

Lemma 10.7. If V is an irreducible finite dimensional representation ofsl2, then V ∼= Vn ⊕ Vn−2 ⊕ . . .⊕ V−n where XVλ = Vλ+2 and Y Vλ = Vλ−2.

Proof. Let v ∈ Vλ be a highest weight vector with weight λ (it is true butnot trivial that such a vector exists). We claim that v, Y v, Y 2v, . . . spansV . The elements are linearly independent, since they are eigenvectors withdistinct eigenvalues: we can inductively show that H(Y kv) = (λ − 2k)v.You can check that the subspace V ′ generated by this set is subrepresen-tation: it is obviously preserved by Y,H and the result for X follows byusing the commutation relations.

If V is finite dimensional, then for some n we have Y nv 6= 0 but Y n+1v =0. We calculate

Xv = 0

XY v = Y Xv +Hv = λv

XY 2v = (λ+ λ− 2)Y v

......

XY nv = Y XY n−1v +HY n−1v = (nλ− n2 + n)Y n−1v.

Therefore, λ = n.

In particular, a byproduct of the proof is that Xm : V−m → Vm is anisomorphism. If V is not an irreducible representation, then it is a directsum of irreducibles of this form, so Xm is still an isomorphism.

Now, let’s return to studying a compact Kahler manifold (X,ω). TheKahler identities show that H = h,X = L, Y = Λ give a representationon H∗(X, g). The eigenspace of h with eigenvalue n − p is Hp(X, g) ∼=Hp(X,C). So the preceding discussion proves the Hard Lefschetz theorem.

Chapter 11

Hermitian Vector Bundles

11.1 Hermitian metrics

Let E → X be a complex vector bundle over a smooth manifold X. Wedefine

Ak(E)(U) = Ak(U)⊗ C∞(E)(U) (the sheaf of smooth sections of E).

If X is a complex manifold, then we get a splitting

Ak(E) =⊕p+q=n

Ap,q(E)

just from the usual splitting on Ak.

Definition 11.1. A Hermitian metric on E is a smoothly varying Hermi-tian metric on the fibers Ex, for x ∈ X.

If e1, . . . , er is a local frame for E, then [hij = h(ei, ej)] is a Hermitianmatrix varying smoothly with x. We call (E, h) a Hermitian vector bundle.It is convenient to think of h as giving an isomorphism h : E ' E∗. (Notethat we have to choose the conjugate complex structure on E∗, or allowthis to be conjugate linear (so not a morphism of complex vector bundles.)

Exercise 11.2. If E,F are given Hermitian metrics, then the bundlesE ⊕ F,E ⊗ F,E∗,

∧iE etc have naturallly induced Hermitian metrics.

Now assume that E is a holomorphic vector bundle.

Proposition 11.3. There is a natural map C-linear map ∂ = ∂E : Ap,q(E)→Ap,q+1(E) satisfying

∂E(α⊗ s) = ∂α⊗ s+ α⊗ ∂Es for all α ∈ Ap,q(U), s ∈ C∞(E)(U).

61


Proof. In a local holomorphic frame (e1, . . . , en), we define

∂E(α⊗ ei) = ∂α⊗ ei.

To see that this is well-defined, let e′i = ψijej be a new choice of holomorphicframe (we are using Einstein summation convention), so that the ψij areholomorphic functions. Then

∂E(α⊗ ψijej) = ∂E(ψijα⊗ ej) = ψij∂α⊗ ej = ∂α⊗ ψijej.

11.2 The Chern connection

Definition 11.4. A connection on a complex vector bundle is a sheaf mapD : A0(E)→ A1(E) such that

D(fs) = df ⊗ s+ fDs for f ∈ C∞(U), s ∈ A0(E)(U).

If e1, . . . , en is a local frame for E then this gives a connection matrix

Dei =∑j

Θijej

where Θ = (Θij) is a matrix of 1-forms.

Definition 11.5. Le E be a holomorphic vector bundle and D be a con-nection on E. Then we can write D = D′ +D′′, where

D′ : A0(E)→ A1,0(E) D′′ : A0(E)→ A0,1(E).

We say that D is compatible with the holomorphic structure on E if D′′ =∂E : A0(E)→ A0,1(E).

Here is an alternate characterization which is useful for local analysis.

Proposition 11.6. Let E be a holomorphic vector bundle and D a con-nection on E. Then D is compatible with the holomorphic structure on Eif and only if in every local holomorphic frame (s1, . . . , sn), the connectioncoefficients D = (Θij) are all (1, 0) forms.

Proof. The (0, 1) part of Dsj =∑

ij Θijsj vanishes since sj is holomorphic.Since vanishing can be checked locally, this shows that if D is compatiblewith the holomorphic structure then the (0, 1) part of D must vanish inevery holomorphic frame.


Conversely, if that is the case then for any local holomorphic frame(s1, . . . , sn) over U and αj ∈ C∞(U), we have

D(αjsj) = dαj ⊗ sj + αjDsj.

Projecting to the (0, 1)-part shows that

D′′(αjsj) = ∂αj ⊗ sj,

which is the same as the local description of ∂E.

Definition 11.7. If (E, h) is a Hermitian vector bundle, a connection Don E is compatible with h if

d(α, β)h = (Dα, β)h + (α,Dβ)h for all α, β ∈ A0(E).

Again, there is a local characterization of this condition.

Proposition 11.8. Let (E, h) be a Hermitian vector bundle. A connec-tion D on E is compatible with h if and only if for every unitary frame(e1, . . . , en), the connection coefficients D = (Θij) form a skew-Hermitianmatrix, i.e. Θij = −Θji.

Proof. If (e1, . . . , en) is a unitary frame, then (ei, ej)h = δij. Therefore,

0 = d(ei, ej) = (Dei, ej)+(ei, Dej) = (∑

Θikek, ej)+(ei,∑k

Θjkek) = Θij+Θji.

Conversely, suppose that (Θij) is skew-Hermitian in any unitary frame.Since the equality of forms can be checked locally, it suffices to prove thatfor any α, β ∈ A0(U) we have

d(α, β) = (Dα, β) + (α,Dβ).

By the above computation, this is satisfied when α, β are among e1, . . . , en.Both sides are linear in the arguments, so it suffices to show that they havethe same C∞(U) dependence, i.e. if the above identity holds then

d(fα, β) = (D(fα), β)h + (fα,Dβ)

The left hand side expands out as

d(fα, β) = df ⊗ (α, β) + fd(α, β)

= df ⊗ (α, β) + f(Dα, β) + f(α,Dβ).


The right hand side expands as

(D(fα), β) + (fα,Dβ)h = (df ⊗ α, β) + (fDα, β) + (fα,Dβ)

= df ⊗ (α, β) + f(Dα, β) + f(α,Dβ).

There are many connections, but a unique one which is compatiblewith a holomorphic and Hermitian structure (which is the Levi-Civita con-nection in the real case). This is called the metric connection or Chernconnection.

Proposition 11.9. If (E, h) is a holomorphic, Hermitian vector bundlethen there exists a unique connection D compatible with the holomorphicstructure and with h.

Proof. Let’s start with uniqueness. Let e1, . . . , en be a local frame andhij = h(ei, ej). Write

Dei =∑

Θijej.

Then

dhij = d(ei, ej)h

=

(∑k

Θikek, ej

)+

(ei,∑k

Θjkek

)=∑k

Θikhk,j +∑k

Θjkhik

As D is compatible with the complex structure, the Θij are (1, 0) forms.So

∂hij =∑k

Θikhkj and ∂hij =∑k

Θjkhik.

Therefore, Θ = (∂h)h−1. This gives uniqueness. On the other hand, thisformula satisfies the conditions necessary, which also gives existence.

A connection D extends to

D : Ap(E)→ Ap+1(E)

byD(α⊗ s) = dα⊗ s+ (−1)pα ∧Ds α ∈ Ap(U), s ∈ O(E).


11.3 Curvature

Definition 11.10. The curvature of D is the map

FD : D2 : A0(E)→ A2(E).

Lemma 11.11. FD is linear over C∞.

Proof. We have

FD(fs) = D2(fs) = D(df ⊗ s+ fDs)

= d2f ⊗ s− df ⊗Ds+ df ⊗Ds+ f ⊗D2s.

The tricky sign swap in the second term comes from the definition of howwe extend D. Since d2 = 0, this is exactly what we want.

Therefore, this map FD looks locally like multiplication by a matrix oftwo-forms:

Corollary 11.12. FD is induced by an element FD ∈ A2(End(E)).

Given a local frame e1, . . . , er, the connection matrix is

Θei =∑j

Θijej

where the Θij are one-forms. Given any local section s =∑siei,

Ds =∑

dsi ⊗ ei +∑

siΘijej.

We can write this as D = d+ Θ.In this notation, the curvature is

D2s = (d+ Θ)(d+ Θ)s

= (d+ Θ)(ds+ Θs)

= d2s+ (dΘ)s−Θ(ds) + Θds+ ΘΘs

= (dΘ + Θ ∧Θ)s.

Lemma 11.13. 1. If (E, h) is hermitian and D is compatible with h,then h(FDsi, sj) + h(si, FDsj) = 0.

2. If E is holomorphic and D is compatible with the holomorphic struc-ture, then FD has no (0, 2)-part, i.e. FD ∈ A2,0(End(E))⊕A1,1(End(E,E)).

3. If D is the metric connection on holomorphic (E, h) then FD is askew-hermitian form in A1,1(End(E)) (i.e. the identity in 1 holds).


Proof. 1. Since the statement is local, we can assume that E = X ×Cr

and pick a unitary local frame e1, . . . , er. Therefore, we can writeD = d+ Θ where Θ∗ = −Θ as D is compatible with h.

We have (dΘ + Θ ∧ Θ)∗ = dΘ∗ − Θ∗ ∧ Θ∗ (the − sign comes fromflipping the factors in the wedge), and this is

= −dΘ− (−Θ ∧ −Θ)

= −dΘ−Θ ∧Θ

= −FD

Here’s another way to see this. We have dd(si, sj)h = d(Dsi, sj)h +d(si, Dsj)h = 0. On the other hand,

dh(Dsi, sj) = h(DDsi, sj)− h(Dsi, Dsj)

(the sign coming again from the tricky sign swap) and

dh(si, Dsj) = h(Dsi, Dsj) + h(si, DDsj).

2. We have D : Ak(E) → Ak+1(E) which restricts to D : Ap,q(E) →Ap+1,q(E) ⊕ Ap,q+1(E). In the notation from before, D = D′ ⊕ D′′where D′′ = ∂E as D is compatible with the holomorphic structure,so

D2 = (D′ +D′′)2

= (D′)2 +D′∂E + ∂E + ∂2

E

so we see that the (0, 2) part vanishes.

Alternatively, if D = d+Θ and D is compatible with the holomorphicstructure, then Θ has type (1, 0). So

FD = dΘ + Θ ∧Θ

= ∂Θ︸︷︷︸2,0

+ ∂Θ︸︷︷︸1,1

+ Θ ∧Θ︸︷︷︸2,0

3. Follows from the first two parts.

11.4 Kodaira’s Theorems

We write Fn for the curvature of the Chern connection on a holomorphic,Hermitian vector bundle.

Example Sheet 1 67

Example 11.14. If L is a holomorphic, hermitian line bundle with metrich, we have

FD ∈ A1,1(End(E)) skew-hermitian

hence is a real (1, 1)-form. If z is a local unitary frame for L, then h(z) =||z||2. In this local frame, Θ = ∂ log h and F = ∂∂ log h.

In particular, if X = Pn and L = OPn(1), we have a natural hermitianmetric on L∗ = OPn(−1) since the total space minus zero section is Cn+1−0,and we can pull back the standard metric on Cn+1. This gives a Fubini-Study Hermitian metric on L∗, hence on L. By definition, the Fubini-StudyKahler metric on Pn is

ωFS =i

2π∂∂ log(1 + ||w||2) on [1 : w].

We have just seen that this coincides with i2πF |hFS .

We have just seen an example of a metric whose curvature is a Kahlerform. This is important enough to make it into a definition.

Definition 11.15. We say a holomorphic line bundle L is positive if thereexists a Hermitian metric h such that i

2πFh is a Kahler form, i.e.

iFh2π

=∑

hijdzidzj

where hij are positive-definite.

If f : X → Y is holomorphic, and E is holomorphic and hermitianon Y , then F ∗E → X is holomorphic and f ∗h is Hermitian on F ∗E, andFf∗h = f ∗Fh.

So if f : X → Pn is holomorphic, then L := f ∗OPn(1) for some holo-morphic L, then f ∗hFS is a positive Hermitian metric on L so L is positive.In particular, we can take kth roots to deduce that any ample divisor ispositive.

Theorem 11.16 (Kodaira Embedding). L is ample if and only if L ispositive.

Theorem 11.17 (Kodaira Vanishing). Let L be a positive line bundle ona compact Kahler manifold of complex dimension n. Then

Hq(X,Ωp ⊗ L) = 0 if p+ q > n.

Chapter 12

Example Sheet 1

1. Let U be an open subset of Cm = R2m and f : U → Cn = R2n bea smooth function. The complex Jacobian of f is defined to be thematrix

JC(f) =

(∂fi∂zj

)where zj = xj + iyj are the standard coordinates on U . With thestandard coordinates wj = uj + ivj on Cn, compute the matrix ofdf : TRm → TR2n. Find also the matrix of the induced map

dfC : TR2m ⊗ C→ TR2n ⊗ C

in terms of the frames ∂∂zj, ∂∂zj and ∂

∂wj, ∂∂wj and related this to JC

when f is holomorphic. Conclude that det df = | det JC|2. Use this toprove that a complex structure induces a natural orientation on theunderlying manifold.

As a real function, f is (u1, v1, u2, v2, . . . , un, vn). Therefore,

df =

...

...

. . .∂uj∂xk

∂uj∂yk

. . .

. . .∂vj∂xk

∂vj∂yk

. . ....

...

.

In terms of the frames ∂∂zj, ∂∂zj and ∂

∂wj, ∂∂wj, JC has matrix

dfC =

...

...

. . .∂fj∂zk

∂fj∂zk

. . .

. . .∂fj∂zk

∂fj∂zk

. . ....

...

.

68

Example Sheet 1 69

If f is holomorphic, then∂fj∂zk

=∂fj∂zk

= 0 and∂fj∂zk

=∂fj∂zk

. In particular,if n = m then

det df = | det JC|2.

Here is a another, more conceptual perspective on the isomorphismdfC ∼ JC⊕JC. Let V ∼= R2m be a real vector space. As we have seen,a complex structure on V is the same thing as a map J : V → Vsatisfying J2 = −1. (Given such a map, we can define a linear actionof C on V by having i act as multiplication by J .) We denote theassociated complex vector space by (V, J). Complexifying, we have amap JC : VC → VC satisfying J2

C = −1 and we denote by V 1,0 the +ieigenspace.

Lemma 12.1. There is a natural isomorphism (V, J)→ V 1,0.

Proof. The map can be described explicitly as v 7→ v−iJv2

. Then

(a+ Jb)v 7→ (a+ Jb)v − iJ(a+ Jb)v

2= (a+ ib)

v − iJv2

.

This describes VC = V 1,0 + V 0,1 as (V, J)⊕ (V, J).

Lemma 12.2. The conjugation map z⊗v 7→ z⊗V : C⊗V → C⊗Vis an R-linear involution of VC interchanging V 1,0 and V 0,1.

Proof. Clear from our explicit description of V 1,0 as v − iJv.

Now, if T : (V, J) → (V ′, J ′) is a C-linear map, then we get also getan R-linear map TR : V → V ′. The complexification TC : VC → V ′C isthe direct sum of T and T , by the previous two lemmas. In particular,if (V ′, J ′) = (V, J) then we have detTC = | detT |2.Back to the setting of complex manifolds, TU is endowed with a com-plex structure by the complex structure on Cm. If f is holomorphic,then df is C-linear (see next question), and JC is just the map in-duced on T 1,0U since ∂

∂zj is a frame for T 1,0. So the result follows

from our preceding dscussion.

In particular, we see that the Jacobians of the transition functions forany frame ∂

∂zj, ∂∂zj respecting the natural decomposition of TX ⊗

C have positive determinant, implying that a complex manifold isorientable.

2. Included as Lemma 1.5 and Theorem 1.6.

Example Sheet 1 70

3. Show that a closed complex submanifold X of a complex manifoldY is naturally a complex manifold and the inclusion ι : X → Y isholomorphic.

By hypothesis, there exist open sets Uα in Y covering X, and anatlas ϕα : Uα ∼= Vα ⊂ Cn such that ϕα|Uα∩X ∼= Vα ∩ zn = 0,the latter being an open subset of Cn−1. The transition functions areholomorphic, since a holomorphic map to V landing in V ′ is the sameas a holomorphic map to V ′.

4. Let f : Cn → C be holomorphic and 0 ∈ C be a regular value (i.e. thecomplex Jacobian JC is surjective at any point in f−1(0)). Show thatZ := f−1(0) is a complex submanifold of C.

(ii) Let F be a homogeneous polynomial in n+1 variables. Show thatthe set

Z = [z0, . . . , zn] ∈ Pn : F (z0, . . . , zn) = 0

is well defined. By considering the map f : Cn+1 − 0 → C inducedby F , show that if 0 is a regular value of f then Z is a complexsubmanifold of Pn.

(i) For any point p ∈ Z, the implicit function theorem 1.7 im-plies that an open neighborhood (in Z) of p can be described as(z1, . . . , zn−1, φ(z1, . . . , zn−1)) where φ is holomorphic. The chart

(z1, . . . , zn) 7→ (z1, . . . , zn−1, zn − φ(z1, . . . , zn−1)

sends a neighborhood of p to the subset zn = 0 of an open subset ofCn, exhibiting Z as a closed submanifold in this chart.

(ii) The set is well-defined by homogeneity. The previous part appliedto f : Cn+1 − 0 shows that if 0 is a regular value of f then Z(f) ⊂Cn+1 is a complex submanifold of Cn+1, so the same is true afterintersecting with the standard open (say) Ui = zi 6= 0, and we canchoose coordinates for Z(f) that look like (z0, . . . , t︸︷︷︸

i

, . . . , zn = 0).

Let Ui be the image of this open in Pn. Then Z(F ) is described in Uiby (z0, . . . , 1, . . . , zn = 0).

5. Prove that a closed submanifold D ⊂ Cn of complex dimension n− 1is given by D = Z(f) for some holomorphic f : Cn → C.

By hypothesis, there is an open cover Uα of Cn such that in eachopen set, D is cut out by a local equation fα ∈ O(Uα). (This is stilltrue if D ∩ Uα = ∅, since we can take fα to be non-vanishing).

An important and subtle point is that we can choose fα to not bedivisible by the square of any non-unit in O(Uα), possibly after refin-ing our cover. (This is related to the fact that the logarithm cannot

Example Sheet 1 71

be defined in an open neighborhood of the origin, or else we couldtake f

1/nα for each n.) Otherwise, fα would be divisible by arbitrar-

ily high powers of non-units. Consider the germ of fα in p ∈ D.Any holomorphic germ gp ∈ Op is a unit in a neighborhood of p if itdoesn’t vanish at p. If it does vanish at p, then its power series hasno constant term, so gnp has terms all of order at least n.

Remark 12.3. More generally, the Auslander-Buchsbaum theoremguarantees that a regular local ring is a UFD, so we can apply thisanalysis to any smooth variety.

Now, on Uα∩Uβ the hypersurface D is cut out by squarefree equationsfα and fβ, so

gαβ :=fαfβ∈ O∗(Uα ∩ Uβ).

We can then view (gαβ) ∈ C1(Uα,O∗). This is a cocycle, since

gαβgβδgδα =fαfβ

fβfδ

fδfα

= 1.

Therefore, it descends to a cohomology class g ∈ H1(Cn,O∗). If thisis the 0 class, then g is a coboundary, so (after possibly refining thecover again) there is an open cover where

gαβ =hαhβ

hα ∈ O∗(Uα)∀α.

Now we claim that fαhα

defines a global section cutting out D (sincehα is nonvanishing, it cuts out the same local subset as fα, so theonly thing to check is that it is well-defined). Indeed,

fαhα

=gαβfβgαβhβ

=fβhβ.

Finally, we show that the class g is 0 by showing that the entirecohomology group H1(Cn,O∗) vanishes. By the exponential shortexact sequence of sheaves, we have a long exact sequence

. . .→ H1(Cn,O)→ H1(Cn,O∗)→ H2(Cn,Z)→ 0

By the Dolbeault theorem, H1(Cn,O) ∼= H0,1(Cn) = 0 by the ∂-Poincare Lemma. Also, H2(Cn,Z) = H2(Cn,Z) = 0. Therefore, themiddle group vanishes as well.

6. Let f : X → Y be a holomorphic map between complex manifolds.Prove that if α is a (p, q)-form on Y then f ∗α is a (p, q) form on X.

Example Sheet 1 72

Using this, show that f induces a homomorphism

f ∗ : Hp,q

∂(Y )→ Hp,q

∂(X)

given byf ∗[α] = [f ∗α].

There are two ways to see this: using the invariant definitons, orusing local coordinates. From the invariant perspective, note thatthe pullback is defined by

f ∗α(X) = α(df(X)) for X ∈ T kCX.

Therefore, it suffices to show that if X ∈ T p,qX, then df(X) ∈ T p,qX.Since T p,qC =

∧p T 1,0⊗∧q T 0,1, it suffices to establish this in the special

case (1, 0) and (0, 1), and since those spaces is conjugate it sufficesto show that df preserves T 1,0. Since T 1,0 is the +i eigenspace ofJ , this follows from the fact that df commutes with J because f isholomorphic (Lemma 1.5).

In local coordinates (w1, . . . , wn) on Y and (z1, . . . , zm) on X, we have

f ∗dwj = d(fj) =∑k

∂fj∂zk

dzk +∂fj∂zk

dzk

but the second term vanishes because f is holomorphic.

To show that f descends to a map on Dolbeault cohomology, we mustshow that f commutes with ∂, since this implies f ∗ preserves cocyclesand coboundaries. Since f ∗ preserves the bigrading by the previouspart, this follows from the fact in the smooth case that f ∗ commuteswith d; recall that this can be checked on functions and 1-forms.

7. Let F be a sheaf on a topological space X. Define the stalk of F atx ∈ X by

Fx = lim←−U3xF(U).

Observe that any sheaf morphism α : F → G induces a morphismαx : Fx → Gx. Prove that

0→ F α−→ G β−→ H → 0

is short exact if and only if

0→ Fxαx−→ Gx

βx−→ Hx → 0

Example Sheet 1 73

is short exact.

Lemma 12.4. F(U) →∏

x∈U Fx.

Proof. If s ∈ F(U) maps to 0, then around each x ∈ U there is a Vxsuch that s|Vx = 0. By the identity axiom, s = 0.

This shows that if Fxαx−→ Gx is injective for all x, then so is F α−→ G.

Conversely, if fx ∈ Fx maps to 0 in GX , then this is witnessed byrepresentatives in some V , so if F(U) → G(U) for all U then fx = 0.

The equivalence of surjectivity is obvious from our definition of asurjective map of sheaves.

Finally, we consider exactness at the middle term. First suppose thatthe sequence of sheaves is short exact. If g ∈ Gx maps to 0 in Hx,then there is a representative g ∈ G(U) for gx mapping to 0 in H(U),so it is the image of f ∈ F(U), hence gx = αx(fx).

Conversely, suppose the sequence of stalks is exact for all x. If gx ∈G(U) maps to 0 in H(U), then βx(gx) = 0 for all x ∈ U . Therefore,gx = αx(fx) for fx ∈ Fx. For each x, there is some Vx with a liftfx ∈ F(Vx) restricting to fx. Let gx = α(fx). We know that the stalkof gx at x agrees with gx, so by shrinking Vx further if necessary wemay assume that gx = g|Vx .We claim that the sections fx are compatible. Indeed, for each z ∈ Vxwe have α(fx)z = gz, so for each z ∈ Vx ∩ Vy we see that α(fx)z =α(fy)z. Since αx is injective, fx and fy agree on all stalks, and thenLemma 12.4 implies that fx = fy. Now the sheaf axiom glues all thefx to a section f ∈ F(U). We must have α(U)(f) = g, since equalityholds after mapping to each stalk and by Lemma 12.4 again.

Remark 12.5. The argument above may seem rather involved, but itis guided by the simple intuition that a section of a sheaf is a “bundleof compatible stalks.” Having produced a bundle of stalks fx for thecandidate section f , we have to show that they are compatible, whichis true because their images under the injective map α are compatible.

8. Complete the proof of the ∂-Poincare Lemma 4.11 (in the remainingcases q > 1).

We proceed by induction on q, having established the base case q = 1.We suppose that ∂α = 0. As before, we filter B by an increasingsequence of bounded polydiscs B1, B2, . . . such that

⋃Bi = B.

(a) We claim that there exist smooth p, q forms β1, β2, . . . such that∂βm|Bm = α.

Example Sheet 1 74

Indeed, since α is ∂-closed Lemma 4.3 guarantees that thereexists βm defined on Bm+1 such that ∂βm|Bm = α. Then we canmultiply βm by a smooth bump function which is ≡ 1 on Bm

and supported in Bm to extend it to all of B.

(b) Next, we claim that we may arrange that βm|Bm−1 = βm+1|Bm−1 .Indeed, we have ∂(βm+1−βm)|Bm = 0, so the induction hypoth-esis furnishes us with some γ on Bm such that βm+1 − βm = ∂γon Bm. Multiplying by a bump function ≡ 1 on Bm−1 and sup-ported in Bm, we may extend γ to B, so βm+1 − γ is defined onB and restricts to βm on Bm−1.

(c) Obviously, the βm converge to β satisfying ∂β = γ.

9. (i) Let X = C∗. Consider the open cover U0 = C − R+ × 0 andU2 = C− R− × 0. Compute Hq(U ,Z).

C0(U ,Z) consists of pairs (a, b) where a ∈ Z(U0) = Z and b ∈ Z(U1) =Z. The boundary map is is (a, b) 7→ (a− b, a− b), so the cocyles arethe pairs where a = b, i.e. H0(U ,Z) ∼= Z.

C1(U ,Z) consists of an element of Z(U0∩U1) = Z⊕Z. The cobound-aries are the diagonal, and the cocycles are everything since there areno higher intersections, so H1(U ,Z) ∼= Z.

All higher cohomology groups are automatically 0.

(ii) Let M = C2 − 0. Consider the cover U = U0, U1 consistingof U0 = C × C∗ and U1 = C∗ × C. Show that H1(M,O) is infinite-dimensional and Hq(M,O) is trivial for q > 1.

Since U0∩U1 = C∗×C∗, Example 5.31 shows that this cover actuallycomputes Cech cohomology. So it is obvious that Hq(M,O) vanishesfor q ≥ 2.

To compute H1(U ,O), observe that the 0-cochains C0(U ,O) consistof (f, g) where f0 ∈ O(U0)

∗ and f1 ∈ O(U1)∗. The 1-cochains consist

of h ∈ O(U0∩U1)∗, and note that U0∩U1 = C∗. Therefore, H1(U ,O)

consists of holomorphic functions on C∗ × C∗ modulo functions thatcan be written as f − g, where f is holomorphic on C× C∗ and g isholomorphic on C∗ × C.

We claim that 1(z0z1)k

are independent elements of the cokernel. Theindependence will follow if we can show that they are non-zero, sincewe can take a linear combination and multiply through by (z0z1)

k−1,where k is the largest exponent appearing in the linear combination.

To see that they are non-zero suppose that

f0 − f1 =1

(z0z1)k

Example Sheet 1 75

with f0, f1 as above. Then (z0z1)kf0 = 1 + (z0z1)

kf1. The left handside is holomorphic on C×C∗ and the right hand side is holomorphicon C∗×C, so they both extend to holomorphic functions on all of M .But then Hartog’s theorem implies that they extend to holomorphicfunctions on all of C2, say g0 and g1.

Now, g0(z0z1)k

= f0 is holomorphic on C× C∗, so g0z0

is holomorphic onC × C∗. It is also holomorphic on C∗ × C since z0 is nonvanishingthere, so it also extends to a global entire function, say h0. Similarly,g1z1

= h1 is entire. So we have

h0zk1− h1zk0

= 1 =⇒ zk0h0 − zk1h1 = 1.

Setting z0 = 0, we see that z1h1(0, z1) = 1, where h1(0, z1) is entireon C1. This is clearly impossible.

10. Let X = P1 and P,Q be distinct points of X. Let OX(−P−Q) denotethe sheaf of holomorphic functions vanishing at both P and Q. Showthat there is a short exact sequence of sheaves

0→ OX(−P −Q)→ OX → CP ⊕ CQ → 0

where the sheaf on the right should be carefully defined. For any sheafF write Γ(F) = F(X) for the space of sections on all of X. Provethat the map Γ(OM) → Γ(M,Cp ⊕ CQ) is not surjective and thatH1(OX(−P −Q)) 6= 0.

The sheaf CP is a skyscraper sheaf at P . This can be characterizedin a few diferent ways. It is the unique sheaf whose stalk at P is Cand all other stalks vanish. Alternatively,

CP (U) =

C P ∈ U0 otherwise

with restriction maps being identity or 0.

To check exactness, we can check exactness at each stalk. At a stalkat R 6= P,Q, the sequence reads

0→ OX(−P −Q)R → (OX)R → 0→ 0

so we have to show that OX(−P − Q)R ∼= (OX)R. This is obvious,since the open sets containing R but not P or Q is cofinal amongopen sets containing R.

Example Sheet 2 76

Without loss of generality, we check exactness of the stalks at P only.Then the sequence reads

0→ OX(−P −Q)P → (OX)P → C→ 0

For the same reason as before, OX(−P − Q)P ∼= OX(−P )P , andsubtituting this above makes the exactness obvious again.

Note that H0(X,OX) = C and H0(X,CP ⊕CQ) ∼= C⊕C, so the mapis obviously not surjective. Its cokernel injects into H1(OX(−P−Q)),so the latter group is non-zero.

Chapter 13

Example Sheet 2

1. Let X be a differentiable manifold. We let R denote the sheaf oflocally constant real-valued functions, and Ap the sheaf of p-forms onX.

(i) Show that there is an exact sequence of sheaves

0→ R→ A0 → A1 → . . .

(ii) Show that Hq(X,Ap) = 0 for q ≥ 1 and all p ≥ 0.

(iii) Using induced long exact sequences, as in lectures, prove thatHp(X,R) = Hp

dR(M).

(i) The exactness can be checked locally, where it is a consequence ofPoincare’s lemma.

(ii) This is immediate from the fact that Ap is fine.

(iii) One can give a direct argument, completely analogous to what wedid for Dolbeault cohomology. Instead, we will show more generallywhat we claimed there: if

0→ F → I0d1−→ I1 → . . .

is a resolution of F by acyclic sheaves, then Hp(F) = ker dp+1

im dp.

Indeed, we have short exact sequences of sheaves

0→ Zi−1 → Ii−1di−→ Zi → 0

for each i > 0. By the long exact sequence in cohomology,

Hj+1(Zi−1) ∼= Hj(Zi) for j ≥ 1.

77

Example Sheet 2 78

Therefore,

Zp(X)

dpIp−1(X)= H1(Zp−1) ∼= H2(Zp−2) ∼= . . . ∼= Hp−1(Z1) = Hp(F).

2. Let X be a complex manifold, and denote the induced almost complexstructure by J : TX → TX. Explain how multiplication by i gives anR-linear map

J : (T ∗X)0,1 → (T ∗X)0,1

such that J2 = − Id. Prove that there is a natural identification

T ∗X ∼= (T ∗X)1,0

taking J to J .

This was already done in Example Sheet 1, Question 1.

(ii) Suppose now that g is a Riemannian metric on X compatible withJ and let gC be the extension to TX ⊗ C given by

gC(λv, µw) = λµg(v, w) λ, µ ∈ C; v, w ∈ TxX.

Show that under the isomorphism in part (i), we have

g = 2gC on (T ∗X)1,0.

The explicit isomorphism is v 7→ v−iJv2

. We compute

gC

(v − iJv

2,w − iJw

2

)=

1

4(g(v, w)− g(v, iJw)− g(iJV, w) + g(iJv, iJw))

=1

2g(v, w)

(iii) Now suppose we have local coordinates z1, . . . , zn on X so thatdz1, . . . , dzn are a frame for (T ∗X)(1,0). Show that if hij = 2gC(dzi, dzj)then h = (hij) is a hermitian matrix, and the associated fundamentalform is

ω =i

2

∑i,j

hijdzi ∧ dzj.

Example Sheet 2 79

With zj = xj + iyj, we compute

ω(dzi, dzj) = gC(Jdzi, dzj)

= igC(dzi, dzj)

=i

2hij.

That establishes the result.

(iv) Deduce that any hermitian metric is given by a unique such formω with hij hermitian (and that this metric is Kahler if and only if ωis closed.

The computation in (iii) shows that h can be recovered from ω.

3. Let U0 = [1, z] ∈ Pn : z = (z1, . . . , zn) ∈ Cn. Show that the Fubini-Study form on Pn can be written locally on U0 as

ωFS =i

2π

(∑j dzj ∧ dzj1 + ||z||2

−∑

j zjdzj ∧∑

j zjdzj

(1 + ||z||2)2

).

Deduce that ωFS defines a Kahler metric on Pn.

It will be useful to record that

∂(1 + ||z||2) =∑j

zjdzj ∂(1 + ||z||2) =∑j

zjdzj.

Now, 2πiωFS = ∂∂ log(1 + ||z||2), which is

∂∂ log(1 + ||z||2) = ∂

(∑j zjdzj

1 + ||z||2

)=

∑j dzj ∧ dzj1 + ||z||2

−∑

j zjdzj ∧∑

j zjdzj

(1 + ||z||2)2

We already showed that ωFS is well-defined and closed, so it sufficesto show that it is positive-definite. From the preceding computation,we see that hij = (1 +

∑|wi|2)δij − wiwj. Therefore, if in terms of

the standard Hermitian inner product on Cn we have

ut(hij)u = ||u||2 + ||w||2||u||2 − utwwtu.

By Cauchy-Schwarz, the last term is ≤ ||u||2 · ||w||2, so we are done.

4. Let Pn−1 ⊂ Pn be the standard inclusion. Show that the restriction ofthe Fubini-Study metric on Pn gives the Fubini-Study metric on Pn−1.We interpret the standard inclusion to be (z1, . . . , zn−1) 7→ (z1, . . . , zn−1, 0).(We could take any of the other inclusions where zj = 0 as well.) Then

Example Sheet 2 80

clearly ∂|Pn restricts to ∂Pn−1 and likewise for ∂, and also the normon Pn restricts to the norm on Pn−1, so the result follows.

5. Show that any holomorphic line bundle on a disc ∆ ⊂ C is trivial.Deduce that any holomorphic line bundle on P1 is of the form O(n))for some integer n.

Remark 13.1. It is true more generally that any line bundle on Pn isisomorphic to O(m). The easiest way to see this is to prove that

(a) H i(Pn,O) = 0 for i > 0. This can be shown directly using Cechcohomology on the standard open cover.

(b) The long exact sequence of the exponential map shows thatH1(P1,O∗) = Z.

(c) The (first) Chern class c1 : H1(Pn,O∗)→ H2(Pn,Z) takes O(1)to 1. This is the hard part, partly because the Chern classreceives a new definition every time it is mentioned. One jus-tification is that the first Chern class of a complex line bundleof the Euler class of the underlying real vector bundle, whichis Poincare dual to the zero set of transverse section. Since asection of O(1) is a linear polynomial, its zero locus is the classof a hyperplane in Pn, which generates. Therefore, c1(O(1)) isa generator of H2(Pn,Z) ∼= Z.

We will present an ad hoc argument that works for P1. First, theDolbeault theorem implies Hq(∆,O) ∼= H0,q(∆) = 0 for q > 0 bythe ∂-Poincare Lemma. By the long exact sequence associated tothe short exact exponential sequence 0 → Z → O → O∗ → 0, wehave H1(∆,O∗) ∼= H2(X,Z) = 0. This shows that any complex linebundle on ∆ is trivial.

Take a cover of P1 by two discs U0, U1∼= ∆ whose intersection is an

annulus V . Let U = U0, U1.

Lemma 13.2. H1(U ,O) = 0.

Remark 13.3. Our cover is good, so in fact this implies H1(P1,O) = 0.

Proof. This boils down to the assertion that if f ∈ O(V ), we canwrite f = g0 − g1 where g0 ∈ O(U0) and g1 ∈ O(U1). By Cauchy’sformula,

f(w) =1

2πi

∫∂V

f(z) dz

z − w=

1

2πi

∫∂U0

f(z) dz

z − w︸︷︷︸g0

− 1

2πi

∫∂U1

f(z) dz

z − w︸︷︷︸g1

.

Example Sheet 2 81

Now, let L be a line bundle on P1. By the first part, L is trivialover U0 and U1, so let s0 and s1 be trivializations over those twoopen sets. We have a transition function g(z) = s0

s1∈ O∗(U0 ∩ U1).

Let d be the winding number of g, so h(z) = z−dg(z) has a windingnumber zero. Therefore, log h(z) is well-defined and holomorphic(rigorous argument: the integral of d log(z−dg(z)) over any closedloop vanishes).

By the Lemma, we may write log h(z) = h0(z) − h1(z) where hj ∈O(Uj). Then z−dg(z) = h(z) = eh0(z)e−h1(z). Since e−hj(z) is non-vanishing and holomorphic on Uj, we can define new trivializationssj(z) := sj(z)e−hj(z). Then

s0(z) = s0(z)e−h0(z) = g(z)s1(z)zdg−1(z)e−h1(z) = zds1(z)

showing that L is isomorphic to O(d).

6. Let ∆d and ∆∂ denote the d and ∂-Laplacians, respectively, on Cn

with the standard metric. Show that ∆∂(f) = −12

∑j ∂

2f/∂x2j +

∂2f/∂y2j . Verify directly that ∆d = 2∆∂.

Let dV = dz1 ∧ dz1 ∧ . . . ∧ dzn ∧ dzn be the volume form on Cn. Wedenote by dzi the form with only dzi missing, e.g.

dz2 = dz1 ∧ dz1 ∧ dz2 ∧ . . . ∧ dzn ∧ dzn.

Observe that ∗dzi = dzi and ∗dzi = −dzi.

Example Sheet 2 82

Since ∂∗ decreases degrees, ∆∂f = ∂ ∗ f = − ∗ ∂ ∗ ∂f .

∆∂f = − ∗ ∂ ∗∑j

∂f

∂zjdzj

= − ∗ ∂∑j

∂f

∂zjdzj

= − ∗∑i,j

∂2f

∂zi∂zjdzi ∧ dzj

= − ∗∑j

∂2f

∂z2jdV

= −∑j

∂2f

∂z2j

= −1

4

∑j

∂

∂zj

(∂f

∂xj− i ∂f

∂yj

)= −1

4

∑j

∂2f

∂x2j− 2i

∂2f

∂xj∂yj+∂2f

∂y2j

Note that the Cauchy-Riemann equations imply

∂f

∂xj=

∂u

∂xj+ i

∂v

∂xj=

∂v

∂yj− i ∂u

∂yj= −i ∂f

∂yj,

so the above is

−1

2

∑j

∂2f

∂x2j+∂2f

∂y2j

as desired.

7. Prove that any complex manifold admits a Hermitian structure.

Let U = (Uα, ϕα) be a holomorphic atlas for a complex manifoldX. By pulling back the standard Hermitian metric on Cn via ϕα, weobtain a Hermitian metric hα on ϕα. Let ρα be a partition of unitysubordinate to Uα. Then

∑ραhα is a Hermitian metric on X.

Remark 13.4. The Hopf surface is the quotient (C2 − 0)/Z wherek · (z, w) = (λkz, λkw) for some λ ∈ (0, 1) is evidently diffeomorphicto S1 × S3, and does not have a Kahler structure since it is compactwith vanishing H2. We show that it is indeed a complex manifold inExample Sheet 3, Question 4.

8. Let X be a complex manifold. Show that T 1,0X is naturally a holo-morphic vector bundle which we denote by TX . GIven a submanifold

Example Sheet 2 83

Y ⊂ X prove that there is a natural inclusion TY ⊂ TX|Y , which hasquotient NY which is a holomorphic bundle on Y .

In local coordinates (z1, . . . , zn) on Uα ⊂ X, a frame for T 1,0X|Uα is∂∂z1, . . . , ∂

∂zn. Suppose that (z1, . . . , zn)T = ψ(z′1, . . . , z

′n), where ψ is a

holomorphic transition function. The induced transition function onT 1,0X is

∂

∂zj7→∑i

∂z′i∂zj

∂

∂z′i= (ψij)

−1 ∂

∂z′j

and (ψij)−1 is holomorphic by the inverse function theorem.

We have a natural inclusion TY → TX|Y , which induces TCY →TCX|Y . Since the inclusion is holomorphic, its differential commuteswith the almost complex structure J so we have an inclusion of +ieigenspaces T 1,0Y → T 1,0X|Y .

Finally, in Uα we may suppose that in our local coordinates (z1, . . . , zn)Y ∩Uα is cut out by zn = 0, so a local frame for TY |Uα is ∂

∂z1, . . . , ∂

∂zn−1

is a frame for TY . In another frame (z′1, . . . , z′n) for Uβ where Y ∩Uβ

is cut out by z′n = 0, we have

∂

∂zn7→∑i

∂z′i∂zn

∂

∂z′i=∂z′n∂zn

∂

∂z′n+ . . .

so the transition function for NY is the holomorphic function ∂z′n∂zn

.

9. Let L1 and L2 be line bundles on a complex manifold X. Supposethat there exists a complex submanifold Y of codimension at least 2such that L1 is isomorphic to L2 on X \ Y . Prove that L1

∼= L2.

A vector bundle isomorphism is an isomorphism of the total spaces(as complex manifolds) over the identity on the base, restricting to alinear map on the fibers.

Let U = X \ Y and f : L1|U → L2|U be an isomorphism with inverseg : L2|U → L1|U . Since L1|U has codimension 2 in L1, this extends toa holomorphic map f : L1 → L2 by Hartog’s Theorem 6.2. Similarly,g extends to a holomorphic map g : L2|U → L1|U . Since g f =Id |L1|U and vice versa, the same holds for the holomorphic extensionby the identity principle. So we have established that f extends toan isomorphism of total spaces.

Let πi denote the projection maps from Li → X. Then we have

π1 = π2 f on L1|U

hence π1 = π2 f by the identity principle again. Therefore, f sendsfibers to fibers.

Example Sheet 2 84

The only thing left to show is that f restricts to a linear map on eachfiber. This can be checked locally, so we may assume that L1, L2 aretrivial and f is of the form

(x, z) 7→ (x, ϕ(z)).

Consider two maps L1 × C→ L2 defined by

(x, z, λ) 7→ (x, λϕ(z)),

(x, z, λ) 7→ (x, ϕ(λz)).

Both are holomorphic, and they agree on a subset of codimension 2(namely U × C × C), hence they agree everywhere by the identityprinciple. That establishes linearity, and we are done.

Remark 13.5. A different argument using transition functions wassuggested in class. It was stated slightly incorrectly, but we can makeit work as follows.

Let Uα be an open cover of X trivializing both L1 and L2. ThenU ∩ Uα is an open cover of U , and since L1

∼= L2 over U we mayassume that the transition functions for L1, L2 are actually equal.That is, for each Uα we have bundle isomorphisms

(π1, fα π1) : L1|U∩Uα ∼= (U ∩ Uα)× C(π2, gα π2) : L2|U∩Uα ∼= (U ∩ Uα)× C.

Moreover, fα = ταβfβ and gα = ταβgβ on Uα ∩ Uβ.

Since fα and fβ are defined on a codimension-2 subset, they extendto fα and gβ which are non-vanishing since the extension of an iso-morphism is an isomorphism (by the previous argument). Similarly,ταβ extends to ταβ satisfying

fα = ταβ fβ

by the identity axiom. This shows that in fact L1 and L2 have thesame transition functions over all of X.

10. (i) Let KX = det(T ∗X), which is a holomorphic line bundle called thecanonical bundle of X. Show that if Y ⊂ X is a complex submanifold,then KY is isomorphic to KX |Y ⊗ detNY .

We have a short exact sequence of vector bundles

0→ TY → TX → NY → 0.

Example Sheet 2 85

Whenever one has a short exact sequence of vector spaces,

0→ Af−→ B

g−→ C → 0

then one gets an isomorphism by taking top exterior powers

detA⊗ detC ∼= detB

defined on pure tensors by

(u1∧ . . .∧ua)⊗ (g(b1)∧ . . . g(bc)) 7→ f(u1)∧ . . .∧ f(ua)∧ b1∧ . . .∧ bc.

This is well-defined because all lifts of b1, . . . , bc differ by elements inthe image of A, which will be killed after wedging with its top form.By naturality, we can taking top exterior powers of our short exactsequence of vector bundles to deduce

detTX|Y = detTY ⊗ detNY .

Dualizing, we getKX |Y ∼= KY ⊗ detN ∗Y ,

and tensoring with NY gives the result.

(ii) Suppose that L is a holomorphic line bundle on X and 0 6= s ∈H0(X,L). Suppose further that Y = x ∈ X : s(x) = 0 is a complexsubmanifold and assume that ds does not vanish along Y (that is, if sis locally given by a holomorphic function f then df does not vanishon Y ). Prove a relation between NY and L|Y , and use this to find aformula for KY in terms of L and KX .

The statement is geometrically intuitive: locally Y looks like thesubset s = 0, so the line spanned by s is complementary to TY inTX. Led by this intuition, we guess that NY

∼= L|Y , which we provelocally.

Suppose that L is trivialized over Uα and Uβ by sections xα and xβ,with transition function xα = gαβxβ. Let the restrictions of s to Uαand Uβ be sα = fαxα and sβ = fβxβ, respectively. Then

fαxα = fβxβ =fβgαβ

(gαβxβ) =⇒ gαβfα = fβ. (13.1)

Since ds|Y 6= 0, we have that fα is not divisible by the square ofany non-unit in any stalk, since if we had (fα)x = ch2x then d(fα)xwould be divisible by hx, hence vanish at x. Therefore, fα is achart coordinate, so we can find local charts for Y ⊂ X of the form(z1, . . . , zn−1, fα) on Uα (after possibly shrinking Uα).

Example Sheet 3 86

In these local coordinates, a frame for TX|Y is ∂∂z1, . . . , ∂

∂zn−1, ∂∂fα

and

the first n − 1 sections form a frame for TY , so a frame for NY |Y is∂∂fα

. The transition function relating these coordinates to coordinates

(z′1, . . . , z′n−1, fβ) on Uβ is

(z1, . . . , zn−1, fα) = (z′1, . . . , z′n−1, gαβ(z′1, . . . , z

′n, 0)fβ).

Therefore, the transition function for NY |Y is given by (using (13.1))

∂

∂fα=∂fβ∂fα

∂

∂fβ+ . . . = gαβ(z′1, . . . , z

′n, 0)

∂

∂fβ+ . . .

These are precisely the transition functions functions for L|Y .

Chapter 14

Example Sheet 3

1. Prove that a Kahler form is harmonic.

One way is simply to use the Kahler identities. Recall from Theo-rem 8.24 that [Λd, L] = 0. Applying this operator to the 1, we findthat ∆d(ω) − L∆d1 = 0, and the second term obviously vanishes so∆d(ω) = 0.

You can also do this by explicit computation in local coordinates. Letα1, β1, . . . , αn, βn be an orthonormal basis for T ∗X such that J(αi) =βi. This can be constructed inductively, like what one does to obtaina symplectic basis: first pick a unit vector α1, and set β1 = J(α1).By compatibility,

g(β1, β1) = g(Jα1, Jα1) = g(−α1,−α1) = 1.

Also, g(α1, β1) = g(α1, Jα1) = g(Jα1,−α1) = −g(α1, Jα1) showsorthogonality. The basis is then constructed by induction.

In this local frame, we have

ω =∑

αi ∧ βi.

By definition, ω is harmonic if (dd∗ + d∗d)ω = 0. Since ω is Kahler,dω = 0. Therefore, we need to check that d∗ω is closed. But d∗ = ∗d∗and we can explicitly see that

∗(αj ∧ βj) = α1 ∧ β1 ∧ . . . αj ∧ βj ∧ . . . ∧ αn ∧ βn.

It is clear that this is true up to sign, but since (α1, β1, . . . , αn, βn) isan oriented coframe the sign +1 is also clear. In particular, we seethat ∗ω = 1

(n−1)!ωn−1, so d(∗ω) = 0.

2. Prove that Hp,q(Pn) = 0 except when p = q ≤ n, in which case it isone-dimensional. Use this to find the Picard group of Pn.

87

Example Sheet 3 88

Recall the Hodge decomposition

H∗(X,C) =⊕p+q=n

Hp,q(X)

We know the (singular) cohomology of projective space: H∗(X,C) ∼=C[z]/zn, where |z| = 2. Since hp,q = hq,p, this is only possible ifthe only non-zero groups are in bi-degree (p, p), and these all havedimension one.

3. Find an exact sequence involving the holomorphic tangent bundle ofPn, the trivial bundle and the tautological bundle. Use this to showthat

KPn = OPn(−n− 1).

The exact sequence is

0→ OPn → OPn(1)n+1 → TPn → 0.

Before explaining where this comes from, it may be helpful and clari-fying to consider the question for the tangent bundle rather than theholomorphic tangent bundle.

In the topological category, one in fact has a splitting TPn ⊕ C ∼=L−n−1, where C is the trivial bundle and L is the tautological bundle.Indeed, recall that an open chart for Pn(V ) in the nighborhood ofa particular line ` is HomC(`, V/`). Complexifying and taking the(1, 0) part returns this same space. Choosing a Hermitian metric onPn, this is isomorphic to Hom(`, `⊥), so we can identify the tangentbundle as (`, ϕ) : ϕ ∈ Hom(`, `⊥). On the other hand, the trivialbundle is (`, ϕ) : ϕ ∈ Hom(`, `). Taking the direct sum, we obtain

(`, ϕ) : ϕ ∈ Hom(`, `⊥ ⊕ `) = Hom(`, V ).

Since the tautological bundle is (`, v) : v ∈ `, this is visibly the dualof Ln+1. Dualizing, we find that T ∗Pn⊕C ∼= Ln+1, which shows thatthe canonical bundle is O(−n− 1).

This doesn’t work in the holomorphic (or algebraic) category, sincewe do not have the vector bundle `⊥ (there is no holomorphic splittingof the bundle morphism V → V/`). However, one still has an “Eulerexact sequence” of sheaves whose dual is

0→ OPn → OPn(1)n+1 → TPn → 0.

Example Sheet 3 89

The meaning is as follows. If we think of Pn as the quotient of X =Cn+1 − 0 by C∗, then the vector fields on X are freely generated by∂∂xi

, so a general vector field on X can be written as

V (x) =∑

vi(x0, . . . , xn)∂

∂xi.

The vector fields on Pn should correspond to “C∗-equivariant” vectorfields on X. In other words, vi(tx) ∂

∂txi= vi(x) ∂

∂xi, and from this

it is clear that the vi must be linear functions on X, which is thesame as sections of O(1) over Pn. That furnishes a map O(1)n+1 →TPn. What’s the kernel? Recall Euler’s formula for a homogeneouspolynomial F of degree d:

dF (x0, . . . , xn) =∑

xi∂F

∂xi.

Applying this to a rational function FG

on Pn, we see that (x0, . . . , xn)maps to the trivial derivation. In fact, the kernel is freely generatedby this section. We have now defined the maps in a purported shortexact sequence

0→ O → O(1)n+1 → TPn → 0.

To check exactness, we can work locally. Without loss of generality,we consider the standard open set U0 with coordinates (z1, . . . , zn) =(x1x0, . . . , xn

x0). The tangent bundle is trivialized over this open set with

local frame ∂i := ∂∂zi

.

To study the map O(1)n+1 → TPn , we relate ∂i and ∂∂xi

, observe thatfor i > 1 we have

∂

∂xif

(x1x0, . . . ,

xnx0

)=

1

x0∂if(z1, . . . , zn)

and for i = 0,

∂

∂x0f

(x1x0, . . . ,

xnx0

)=

n∑i=1

−xix20∂if(z1, . . . , zn).

Therefore, the map O(1)n+1 → TPn sends

(f0, . . . , fn) 7→∑

fi∂

∂xi=∑i

(fix0− f0xi

x20

)∂i.

Example Sheet 3 90

From this, we visibly see that the kernel is the free submodule gen-erated by (1, z1, . . . , zn), which is the image of the restriction O →O(1)n+1.

Remark 14.1. Yet another point of view is that there is a canonicalidentification T`P(V ) ∼= Hom(`, V/`) (even in the algebraic category).If you have seen deformation theory, a nice point of view is that P(V )represents inclusions of a line bundle into the trivial bundle tensoredwith V , so the tangent space to P(V ) at a line consists of the spaceof first-order deformations of that line in V , i.e. flat families

`

Voo

Spec k Spec k[ε]/ε2oo

and such things are described by a homomorphisms Hom(`, V/`) (thisis just a little commutative algebra; the quotienting corresponds tothe flatness condition).

Using this, we see that T`P(V ) ∼= L∗` ⊗ V`/L`. Tensoring with L`, weobtain an exact sequence

0→ O` → V ⊗ L` → T`P(V )⊗ L` → 0.

This gives isomorphisms on the fibers of the exact sequence we want.To deduce the global exactness in the algebraic category, we use thefact that the sheaves are locally free and apply Nakayama’s Lemma.

4. The k-th Betti number of a smooth manifold is bk = dimRHk(M,R).

Show that the odd betti numbers b2k+1 of a compact Kahler manifoldare even.

Note that bk = dimCHk(MC). By the Hodge Decomposition Theo-

rem,

Hk(M,C) ∼=⊕p+q=n

Hp,q(X,C).

Using hp,q = hq,p on a Kahler manifold, we deduce the result.

The Hopf surface (discussed in Example Sheet 2, Queston 7) is acomplex manifold homeomorphic to S1×3, which fails this property,so it cannot be given a Kahler structure.

5. Calculate the dimensions of the cohomology groups H i(Pn,O(r)) forall values of n, i, r.

You can compute this directly with Cech cohomology, although it ispretty involved. The advantage of that approach is that it illuminatesSerre duality on Pn, which is then used to deduce it for projective

Example Sheet 3 91

varieties. However, it is relatively easy to obtain the dimensionsdirectly.

We claim that the only non-zero cohomology groups are H0(Pn,O(k))for k ≥ 0 and their Serre duals, namely Hn(Pn,O(−n−1−k)). We’llprove this by induction, so let’s first establish it for P1 by directioncomputation. Choosing the usual cover by U0 and U1, we see thatthe restriction maps for O(−k) are (f(z), g(1/z))→ (zkf(z), g(1/z)),so the cokernel is spanned by z, z2, . . . , zk−1. This verifies our claimwhen n = 1.

In general, we have a short exact of sheaves associated to the inclusionPn−1 → Pn:

0→ OPn(−1)→ OPn → ι∗OPn−1 → 0.

Twisting by k, we deduce

0→ OPn(k − 1)→ OPn(k)→ ι∗OPn−1(k)→ 0.

We claim that H i(ι∗OPn−1(k)) ∼= H i(OPn−1(k)). This generally phe-nomenon is most naturally phrased in algebro-geometric terms, whereit is clear from the Cech complex that Cech cohomology is preservedby pushforward under affine morphisms. The result can be seen inthis case because the standard open cover of Pn, whose intersectionsare all cohomologically trivial, restricts to the standard open coverof Pn−1. With this in hand, the long exact sequence in cohomologyreads

. . .→ H i−1(Pn−1,O(k))→ H i(Pn,O(k))→ H i(Pn,O(k − 1))→ . . .

Assuming our claim on Pn−1 to be established, we see that for 1 <i < n, the claim is automatic by induction on k on Pn. When i = 1,it is clear from the explicit description of H0(O(k)) that the mapH0(Pn,O(k)) → H0(Pn−1,O(k)) is surjective. Inducting on k, the“base case” is established by Kodaira vanishing or explicit computa-tion; the rest is straightforward.

6. Let X be the Hopf manifold Cd−0/ ∼, where (w1, . . . , wd) ∼ (z1, . . . , zd)iff wj = 2szj for all j for some s ∈ Z. For d > 1, show that X is acomplex manifold and h0(X,Ωr

X) = 0 for all 1 ≤ r ≤ d.

A quotient of a manifold by a proper, free group action is again amanifold ♠♠♠ Tony: [explain this?], and it is clear that Z acts prop-erly on Cn+1− 0 since every compact subset gets squished arbitrarilysmall or blown up arbitrarily large.

Example Sheet 3 92

If H0(X,Ωr) 6= 0 for some r > 0, let ω ∈ H0(X,Ωr). Pulling backvia the natural projection map π : Cn+1 − 0 → X, we obtain aholomorphic differential π∗ω on Cn+1 which is invariant under theaction of C∗. We can write

π∗ω(z0, . . . , zn) = f(z0, . . . , zn)dzI .

By Hartog’s extension theorem, f extends to a holomorphic functionon all of C∗. The invariance π∗ω(~z) = π∗ω(2j~z) implies that

f(2−jz0, . . . , 2−jzn) = 2jf(z0, . . . , zn).

Taking the limit as j → ∞ contradicts the fact that f extends con-tinuously to 0.

7. Suppose that L is a holomorphic line bundle. Suppose that s ∈H0(X,L) is non-zero and Y = s = 0 is a complex submanifold withds 6= 0 along Y . Show that the sheaf IY ⊂ OX consisting of functionsvanishing along Y can be identified with the sheaf of sections of thedual bundle L∗. Use this to explain the short exact sequence

0→ L⊗(k−1) → L⊗k → L⊗k|Y → 0.

Suppose now that L is positive and E is a holomorphic line bundle.For all i > 0, show that H i(X,E⊗L⊗k) = 0 for k 0. Assuming theKodaira embedding theorem, and Bertini’s theorem that a general hy-perplane section of a complex projective variety is a complex subman-ifold, show that h0(X,E⊗L⊗r) is a polynomial in r for r 0. Whatis this polynomial when X = Pn and L = OPn(1) and E = OX?

We can check that IY ∼= L∗ by examining the transition functions.Suppose that L is trivialized over Uα and Uβ by sections xα and xβ,with transition function xα = gαβxβ. Let the restrictions of s to Uαand Uβ be sα = fαxα and sβ = fβxβ, respectively. Since ds|Y 6= 0,we have that fα is not divisible by the square of any non-unit in anystalk, since if we had (fα)x = ch2x then d(fα)x would be divisible byhx, hence vanish at x. Therefore, IY is trivialized over Uα and Uβ byfα and fβ, and the identity

fαxα = fβxβ =fβgαβ

(gαβxα) =⇒ fα =fβgαβ

.

shows that the transition function for IY is the inverse of that for L,hence IY ∼= L∗.

Example Sheet 3 93

Therefore, we have the short exact sequence of a closed submanifold

0→ IY ∼= L∗ → OX → ι∗OY → 0.

Tensoring with L⊗k (which is exact because L⊗k is a line bundle), weobtain

0→ L⊗(k−1) → L⊗k → L⊗k|Y → 0.

Here L⊗k|Y = ι∗OY ⊗ L⊗k = ι ∗ (OY ⊗ ι∗L⊗k).For the second part, by Kodaira vanishing it suffices to show thatE⊗Lk is positive. Take any hermitian metric g on E, and a positivemetric h on L. Then g ⊗ h⊗k is a metric on E ⊗ L⊗k, and thecorresponding curvature form is Fg + kFh. If X is compact, thensince Fh is positive-definite we can ensure that Fg + kFh is positivedefinite for k 0. ♠♠♠ Tony: [the compactness isn’t in the hypothesis... how to do without it?]

Next, we use the exact sequence

0→ L⊗k−1 → L⊗k → L⊗k|Y → 0.

Twisting by E and taking Euler characteristics gives

χ(E ⊗ L⊗k)− χ(E ⊗ L⊗(k−1)) = χ(E ⊗ L⊗k|Y ).

Since L is positive, the previous part implies that the higher coho-mology vanishes for k 0. Now the result follows from induction ondimX, provided that we can show it for the zero-dimensional case,which is obvious. Note that we are using a fact about polynomials:if f(n)− f(n− 1) is a polynomial, then so is f .

8. If V ⊂ X is the inclusion of a codimension-one complex manifold,show that there are exact sequences of sheaves (and define the sheavesinvolved)

0→ ΩpX(−V )→ Ωp

X → ΩpX |V → 0

and0→ Ωp−1

V (−V )→ ΩpX |V → (Ωp

V )→ 0.

There is some obvious abuse of notation here, since the same notationΩpX |V is used to denote a sheaf on X and a vector bundle on V .

We have the short exact sequence associated to a closed hypersurface.

0→ OX(−V )→ OX → ι∗OV → 0.

Example Sheet 3 94

Tensoring with ΩpX is exact since Ωp

X is locally free (it is the sheaf ofsections of a vector bundle).

0→ ΩpX(−V )→ Ωp

X → ΩpX |V → 0.

where ΩpX |V = ι∗OV ⊗ Ωp

X∼= ι∗(ι

∗ΩpX).

For the second exact sequence, we dualize the sequence from ExampleSheet 2, Question 10 to obtain

0→ N ∗V/X ∼= OV (−V )→ ΩX |V → ΩV → 0

where now ΩX |V = ι∗ΩX . Note that N ∗V/X ∼= L∗|V ∼= OV (−V ). Wetake p − 1 wedge powers to obtain the exact sequence requested inthe problem

0→ OV (−V )⊗ Ωp−1V → ΩX → Ωp

V → 0.

Where does this come from? In general, given an exact sequence ofvector spaces

0→ Af−→ B

g−→ C → 0

where A is one-dimensional we obtain by taking exterior powers

0→ A⊗p−1∧

Bf−→

p∧B

g−→p∧C → 0.

Here g =∧p g, and f(a, g(b1)∧ . . .∧g(bp−1)) = a∧b1∧ . . .∧bp−1. The

point is that this is well-defined because any two pre-images of g(b1)differ by an element of A, which is killed off by the wedge with a.(More generally, it’s clear that we can do this as long as p > dimA.)By comparing dimensions, we see that this is indeed exact.

9. Let M be a smooth manifold and E a complex vector bundle on M .For ψ a complex 1-form on M , we consider dψ as an alternating 2-form via the natural identification. For complex vector fields X, Y ,show that

2dψ(X, Y ) = Xψ(Y )− Y ψ(X)− ψ([X, Y ]).

Suppose that D : A(E) → A1(E) is a connection on E, with R =D2 : A(E) → A2(E) the curvature of D. With the 2-form part ofR ∈ A(Hom(E,E)) considered as an alternating form, show that

2R(X, Y ) = [DX , DY ]−D[X,Y ].

Example Sheet 3 95

It suffices to check the equality locally. By linearity of both sides inψ and X, Y , it suffices to check the result when X, Y , and ψ are allpure tensors. Without loss of generality (by conjugating), we mayassume X = X i∂i and Y = Y j∂j. Then

[X, Y ] = X i(∂iYj)∂j − Y j(∂jX

i)∂i

If ψ is type (0, 1) then both sides are zero, so we may assume thatψ = f(z)dzk and only ∂ψ contributes to the left hand side. Bothsides also vanish unless k = i or k = j, so we assume that k = jwithout loss of generality.

dψ(X, Y ) =∑`

∂f

∂z`dz` ∧ dzj(X, Y )

=∂f

∂zidzi ∧ dzj(X, Y )

=1

2(X iY j)∂if.

The right hand side is

Xψ(Y )− Y ψ(X)− ψ([X, Y ]) = X i∂i(fYj)− ψ(X i(∂iYj)∂j)

= X i(∂if)Y j +X if(∂iYj)− fX i∂iY

j

= ∂if(X iY j).

The second part is a painful explicit computation. To ease notation,we adopt Einstein summation convention. We can check equalitylocally, so let (e1, . . . , en) be a local frame for E. In this local frame,we have connection coefficients D = (Θij). We check that both sidesare equal as sections of Hom(E,E). Applying the right hand side toei, we find that

([DX , DY ]−D[X,Y ])ei = DXDY e

i −DYDXei −D[X,Y ]e

i

= DX(Θij(Y )ej)−DY (Θij(X)ej)−Θij([X, Y ])ej

= Xd(Θij(Y ))ej + Θij(Y )Θjk(X)ek

− Y d(Θij(X))ej + Θij(X)Θjk(Y )ek

−Θij([X, Y ])ej

= Xd(Θij(Y ))ej − Y d(Θij(X))ej

+ Θ ∧Θ(X, Y )ej −Θij([X, Y ])ej.

Example Sheet 3 96

On the other hand, RD = dΘ + Θ ∧ Θ, and by the first part of theproblem we have

2dΘ(X, Y ) = Xd(Θ)− Y dΘ−Θ([X, Y ])

so the two sides are equal, as desired.

10. Let D be a connection on a complex vector bundle E. We define thedual connection D∗ on E∗ by specifying that for local sections σ of E∗

and s of E, we have the identity

(D∗σ)(s) = d(σ(s))− σ(Ds).

Check that D∗ is a connection

Given a hermitian metric on E, we define a dual hermitian metric onE∗ by specifying that the dual frame to any unitary frame is unitary.If D is a hermitian connection on a hermitian vector bundle E, checkthat D∗ is a hermitian connection on E∗.

Suppose now that E is a hermitian vector bundle over a complexmanifold and D is the associated Chern connection on E. Show thatD∗ is the associated Chern connection on E∗.

By definition,

D∗(fσ)(s) = d(fσ(s))− fσ(Ds)

= σ(s)df + fd(σ(s))− fσ(Ds).

We have to check that this agrees with the Leibniz rule, which says

D∗(fσ)(s) = σ(s)df + fD∗σ(s)

= σ(s)df + fd(σ(s))− fσ(Ds).

Indeed, they do agree.

Next, we suppose (E, h) is hermitian. By Proposition 11.8, it sufficesto check that the coefficients of D∗ are skew-hermitian in any localunitary frame. If (e1, . . . , en) is a local unitary frame for E, thenProposition 11.8 tells us that (Θij) is skew-hermitian. We wish tofind the coefficients of the dual connection, so write

D∗e∗i =∑

Θ∗ije∗j .

To find Θ∗ij, we evaluate on ej:

D∗e∗i (ej) = Θ∗ij.

Example Sheet 3 97

By definition, the left hand side is

d(e∗i (ej))− e∗i (Dej) = 0− e∗i (∑

Θjkek) = Θji.

We have thus found that Θ∗ij = Θji, i.e. the coefficient matrix of thedual is the transpose. Now it is obvious that (Θ∗ij) is skew-hermitianif (Θij) is.

Finally, we must show that if D is compatible with the holomorphicstructure on E, then D∗ is compatible with the holomorphic structureon D∗. By Proposition 11.6, it suffices to check that the coefficientsof D∗ are (1, 0) forms in any local holomorphic frame. If (s1, . . . , sj)is a local holomorphic frame for E, then Proposition 11.6 tells usthat the coefficients Θij in this frame are all (1, 0)-forms. By theprevious computation, for the dual holomorphic frame (s∗1, . . . , s

∗j)

the coefficients Θ∗ij = Θji are also (1, 0)-forms.

11. If E is a holomorphic vector bundle on a complex manifold X andF ⊂ E is a holomorphic sub-bundle, then a hermitian metric on Einduces one on F and we have a direct sum decomposition of complexsmooth bundles E = F ⊕ F⊥. If DE is the associated connectionon E, show that the composite (in obvious notation) πF DF is theassociated connection on F .

We must check that DF := πF DE is compatible with the holomor-phic and hermitian structures on F . For the first, we want to showthat D′′F = ∂F . But it is clear that D′′F = πF D′′E. You can easilycheck that ∂F = πF ∂E either by checking the the right hand sidesatisfies the defining conditions (Leibniz rule and agreement with ∂on A0) or by examining our explicit construction. The result can alsobe interpreted via our local description (Proposition 11.6): in thoseterms, it is equivalent to the assertion that a symmetric submatrix ofa matrix of (1, 0) forms is still pure of type (1, 0), which is obvious.

For the compatibility with the the hermitian structure, observe thatfor α, β ∈ F , we have (α, β)F = (α, β)E, so

d(α, β)F = d(α, β)E

= (DEα, β) + (α,DEβ)

= (πFDEα, β) + (α, πFDEβ)

= (DFα, β) + (α,DFβ).

The result can also be interpreted via our local description (Propo-sition 11.8): in those terms, it is equivalent to the assertion that asubmatrix of a skew-Hermitian matrix is skew-Hermitian, which isobvious.

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