MATH 250B: COMPLEX ALGEBRAIC GEOMETRY Contentspeople.brandeis.edu/~igusa/ComplexNotes.pdf · MATH...

MATH 250B: COMPLEX ALGEBRAIC GEOMETRY

Contents

1. Complex manifolds vs almost complex manifolds 31.1. Definition 31.2. Complex linear structure 31.3. Almost complex manifolds 51.4. Distributions and Frobenius Theorem 61.5. Newlander-Nirenberg Theorem 82. Complex differential forms 92.1. Definitions 92.2. Coordinate invariant definitions 112.3. Local exactness of the ∂ complex 122.4. One complex variable 132.5. Back to differential forms 152.6. Dolbeault complex 163. Kahler manifolds 173.1. Kahler forms 173.2. Kahler metrics 183.3. Kahler manifolds 193.4. Connections 203.5. Examples 234. Harmonic forms 264.1. L2 forms and the Hodge star operator 264.2. Adjoint operators 294.3. Laplacian 314.4. Differential operators 314.5. Symbols and elliptic operators 334.6. Fundamental Theorem 354.7. Serre duality 385. Lefschetz operator and applications 415.1. Hodge decomposition 415.2. Lefschetz decomposition 445.3. Lefschetz decomposition in cohomology 485.4. Representations of sl2(C) 496. Polarization 526.1. Outline of Kodaira’s theorem 526.2. Construction of L 557. Hypercohomology and spectral sequences 597.1. Definition of hypercohomology 59

1

2 MATH 250B: COMPLEX ALGEBRAIC GEOMETRY

Introduction 62References 62

MATH 250B: COMPLEX ALGEBRAIC GEOMETRY 3

1. Complex manifolds vs almost complex manifolds

The first step is to understand the difference between complex manifolds and realmanifolds.

1.1. Definition.

Definition 1.1.1. A (topological) n-manifold is a Hausdorff space X together with acollection of pairs called charts (U,ϕ) where U are open subsets of X which cover Xand ϕ : U → Rn is a homeomorphism of U onto an open subset V of Rn.

The structure of a manifold is given by the transition mappings. These are

ϕij = ϕj ϕ−1i : ϕi(Ui ∩ Uj)→ Vj

where (Ui, ϕi : Ui ∼= Vi) and (Uj, ϕj : Uj ∼= Vj) are two charts which intersect. If thesetransition maps are Ck (k times continuously differentiable) then M is a Ck-manifold.If they are analytic (given locally by converging power series) then M is an analyticmanifold.

Given a chart (U,ϕ) we get local coordinates which are function x1, · · · , xn : U → Rgiven by composition:

xi = pi ϕ : U → Rn → R

Definition 1.1.2. A complex n-manifold is a real 2n manifold X with charts (U,ϕ :U → Cn) so that the transition maps are holomorphic (complex differentiable and thuscomplex analytic).

This formal definition does not help understand what we are talking about. Wewill later go through the proof that complex differentiable functions are analytic. Butthese are more formalities which do not help to visualize it. There is some beautifulmathematics here that I want to show you.

1.2. Complex linear structure. The first point to understand is that complex struc-ture is locally defined. A complex manifold X locally looks like Cn. For a coordinatechart (U,ϕ) we get n complex coordinate functions zi : U → C given by 2n real coordi-nate functions xi, yi : U → R given by zi = xi + iyi.

The tangent plane at one point TX,x is a complex vector space of dimension n:

TX,x ∼= Cn

Recall the definition of the tangent plane TX,x.

Definition 1.2.1. A tangent vector to X at x0 is an R-linear function:

χ : Ck(X)→ Rsatisfying the Leibnitz rule: χ(fg) = g(x0)χ(f) +f(x0)χ(g). Tangent vectors are “pointderivations” on Ck(X), the ring of Ck functions X → R.

Since X is a real 2n-manifold, we learned in elementary school that the tangentvectors form a 2n-dimensional real vector space spanned by the 2n tangent vectors

∂

∂x1

|x0 , · · ·∂

∂xn|x0 ,

∂

∂y1

|x0 , · · · ,∂

∂yn|x0


Call this vector space TX,x0,R∼= R2n. When X is a complex manifold, this vector space

has a complex structure given by

∂

∂yj|x0 = i

∂

∂xj|x0

Definition 1.2.2. Take the complexified tangent space

TX,x0,R ⊗ C ∼= C2n

This has a complex basis of the 2n vectors

∂

∂zj=

1

2

(∂

∂xj− i ∂

∂yj

)∂

∂zj=

1

2

(∂

∂xj+ i

∂

∂yj

)Definition 1.2.3. A (real-) differentiable (C1) function f : X, x → C (notation meansf is only defined in a nbh U of x) is holomorphic if

∂f

∂zj= 0

for j = 1, · · · , n. This is equivalent to saying that the derivative of f is a complex linearmap:

f∗ : TX → Cat each point in the domain of f .

Example: f(z) = azm1 = a(x1 + iy1)m. Then, by the chain rule,

f∗

(∂

∂x1

)=∂a(x1 + iy1)m

∂x1

= am(x1 + iy1)m−1

f∗

(∂

∂y1

)=∂a(x1 + iy1)m

∂y1

= ami(x1 + iy1)m−1

So,

f∗

(∂

∂y1

+ i∂

∂y1

)= amzm−1

1 − amzm−11 = 0 = 2

∂f

∂z1

More generally, if f∗ is C-linear then

f∗

(∂

∂yj

)= f∗

(i∂

∂xj

)= if∗

(∂

∂xj

)f∗

(i∂

∂yj

)= f∗

(i2

∂

∂xj

)= −f∗

(∂

∂xj

)(1.1) fC

∗

(∂

∂xj+ i

∂

∂yj

)= 0

Note that ∂∂xj

+ i ∂∂yj

is a vector field over an open subset U ⊆ X on the complexified

tangent bundle TX ⊗RC which is a complex 2n dimensional vector bundle over X. Here

fC∗ : TX,x ⊗ C→ C


is the C-linear extension of f∗ : TX,x → C considered as an R-linear map.

1.3. Almost complex manifolds.

Definition 1.3.1. An almost complex manifold is a Ck real 2n-manifold together witha Ck endomorphism I of its tangent bundle

Ix : TX,x → TX,x

so that I2 = −id.

This is a local structure. A neighborhood of each point is (equivalent to) an opensubset U ⊆ R2n with a complex structure on the tangent plane (also R2n) at each point.An almost complex structure is equivalent to an honest complex manifold structure onX if, for each such nbh U , there is a differentiable embedding

ϕ : U → Cn

so that, at each point x ∈ U , the derivative ϕ∗ : TU,x → Cn is complex linear. I.e.,

(1.2) ϕ∗(Iv) = iϕ∗(v)

(and a linear isomorphism). We will rephrase this condition to resemble (1.1).Since U is a real manifold, the derivative ϕ∗ : TU,x → Cn is apriori only linear

over R. Tensoring this R linear map with C (and composing with the C-linear mapCn ⊗R C→ Cn) will always give a C-linear map:

ϕC∗ : TU,x ⊗R C→ Cn

Condition (1.2) is equivalent to:

(1.3) ϕC∗ (v + iIv) = 0

Proof: Since ϕC∗ is C-linear,

ϕC∗ (v + iIv) = ϕC

∗ (v) + iϕC∗ (Iv) = ϕC

∗ (v) + i2ϕC∗ (v) = 0

The condition that ϕ∗ is a linear isomorphism translates into ϕC∗ being an epimorphism.

We will rephrase Equation (1.3) in a fancy language and derive necessary and sufficientconditions for an almost complex manifold to be equivalent to a complex manifold.

Remark 1.3.2. For any C-vector space V and any C-linear map J : V → V with J2 =−idV , V will decompose as

V = V 1,0 ⊕ V 0,1

where V 1,0 is the i-eigenspace of J and V 0,1 is the −i-eigenspace of J . For each v ∈ V ,the components are:

v1,0 =1

2(v − iJv) , v0,1 =

1

2(v + iJv)

Definition 1.3.3. Let X be an almost complex manifold (this includes complex mani-folds). Define T 1,0

X , T 0,1X to be the subbundles of the complexified tangent bundle

TX,C := TX ⊗R Cwhich are the i and −i eigenspaces of the C-linear endomorphism IC.


Equation (1.3) says that ϕC∗ : TX ⊗ C → Cn is a complex linear epimorphism (one

each fiber) with kernel equal to T 0,1X . We need conditions under which such a map exists.

1.4. Distributions and Frobenius Theorem. For simplicity we assume all structuresare C∞.

Definition 1.4.1. A k-distribution on a real n-manifoldX is defined to be a k-dimensionalsubbundle F ⊆ TX of the tangent bundle of X. The distribution is called integrable if,in a nbh U of each point x, there is a submersion

ϕ : U → V ⊆ Rn−k

So that the kernel of ϕ∗|U is F |U .

Theorem 1.4.2 (Frobenius). A distribution F on X is integrable if and only if

[F, F ] ⊆ F

I.e., for any two C1 vector fields σ, τ on X with values in F , [σ, τ ] is in F .

Hiding a ton of technicalities under the rug this will give:

Theorem 1.4.3 (Newlander-Nirenberg). An almost complex manifold X is equivalentto a complex manifold if and only if the distribution T 0,1

X ⊆ TX ⊗ C is integrable. I.e.,

iff [T 0,1X , T 0,1

X ] ⊆ T 0,1X .

1.4.1. Review of bracket [σ, τ ]. A Ck-vector field is defined to be a derivation

σ : C`(X)→ Ck(X)

where ` > k. In other words,

σ(fg) = σ(f)g + fσ(g).

It is a standard fact that all Ck-derivations are given locally by

σ =∑

σi∂

∂xi

where σi is a Ck function on the coordinate chart U . Conversely, any such σ gives aderivation Ck+1(X)→ Ck(X). Thus, even if σ is only defined on C`(X) ⊂ Ck+1(X), itwill extend uniquely to the larger domain.

The bracket [σ, τ ] is defined to be the Ck−1 derivation given by

[σ, τ ] = στ − τσ : Ck+1(X)→ Xk−1(X)

Proposition 1.4.4. If σ =∑σi ∂

∂xiτ =

∑τ j ∂

∂xjthen

[σ, τ ] =∑(

σi∂τ j

∂xi− τ i∂σ

j

∂xi

)∂

∂xj

Proof.

στ =∑

σi∂τ j

∂xi

∂

∂xj+∑

σiτ j∂

∂xi

∂

∂xj


Switch σ, τ :

τσ =∑

τ i∂σj

∂xi

∂

∂xj+∑

τ iσj∂

∂xi

∂

∂xjSubtract to get the Proposition.

Example 1.4.5. If σ is the “vertical” vector field σ = ∂∂xn

then

[σ, τ ] =

[∂

∂xn, τ

]=∑j

∂τ j

∂xn

∂

∂xj

This is the “vertical component” of the derivative of the section τ of TX in the directionof the vector σ in X. (The horizontal component is just σ.)

The proof of Frobenius’ Theorem uses the following basic fact.

Theorem 1.4.6 (Constant Rank Theorem). Let f : M → N be a C1 map between C1

manifolds. Suppose that the derivative f∗ : TM → TN has rank k at every point in M .Then the image f(M) of f is (locally) a k-submanifold of N .

“Locally” refers to the case when f is an immersion which has self-crossings.

Proof of Frobenius’ Theorem. The bracket condition is necessary:Given U ⊆ X open and ϕ : U → V ⊆ Rn−k a submersion, we can use ϕ to give the

last n−k local coordinates xk+1, · · · , xn by xk+j = pj ϕ : U → Vpj−→ R. Then a section

of F is σ =∑k

i=1 σi ∂∂xi

and

[σ, τ ] =k∑j=1

(σiτ jxi − τ

iσjxi) ∂

∂xj

also lies in F where we use the notation fx = ∂f∂x

. This “trivial” proof hides an importantconcept that we need to come back to later.

Sufficiency is by induction on k.For k = 1, choose a C1 vector field σ tangent to F . Then σ generates a flow ψ :

V × (−ε, ε)→ U so that∂

∂tψ(x, t) = σ(ψ(x, t))

where V ⊆ U is an n − 1 dimensional set transverse to the vector field σ. By makingV, U smaller we may assume that

ψ : V ⊗ (−ε, ε) ∼= U

and that V is an open subset of Rn−1. Taking the projection to V gives the integrationof the distribution F spanned by σ:

ϕ : U ∼= V ⊗ (−ε, ε) V ⊆ Rn−1

For k ≥ 2, first choose a nonzero vector field σ in F defined in a nbd of any chosenpoint x. Then we get a smaller open nbd U ∼= V ⊗ (−ε, ε) as above with last coordinatetangent to σ:

σ =∂

∂xn


Let F be the k − 1 distribution on V given by F y = Fy ∩ TV for all y ∈ V . Since Fyis spanned by F y and σ and σ is in the kernel of ϕ∗,

ϕ∗(Fy) = F y

Claim 1 For any (y, t) ∈ ϕ−1(y), ϕ∗(F(y,t)) = F y. This implies ϕ−1∗ (F ) = F .

Pf: Let K be the restriction of F to ϕ−1(y) = y × (−ε, ε). (So, K ∼= Rk × (−ε, ε) hasdimension k + 1.) Then the condition [σ, τ ] ∈ F for all sections τ of F implies that, forall (y, t) ∈ ϕ−1(y) and all w = (v, t) ∈ K over (y, t), TK,w ∼= Rk+1 we have:

ϕ∗θ∗(TK,w) = ϕ∗(F(y,t)) = ϕ∗(F (y,t))

where F (y,t) = F(y,t) ∩ TV×t and θ : y × (−ε, ε) → U since TK,w is spanned by F(y,t) andthe vector τ∗(σ) and

θ∗τ∗(σ) =

(∂

∂xn(τ), σ

)= ([σ, τ ], 1) ∈ F(y,t) × R

Thus ϕ∗ has constant rank k − 1 on K. Therefore, by the Constant Rank Theorem,ϕ∗(K) is a k − 1 dimensional manifold (possibly by making ε smaller). But ϕ∗(K)contains ϕ∗(Fy) = F y

∼= Rk−1. So, ϕ∗(K) = F y.Claim 2 F satisfies the bracket condition [F , F ] ⊆ F .Pf: Since F and TV satisfy the bracket condition, so does their intersection.By induction on k, the k − 1 distribution F on V can be integrated giving a map

ρ : V, x→ Rn−k

with ker ρ∗ = F . So ρ ϕ : U, x→ Rn−k is the integral of F .

1.5. Newlander-Nirenberg Theorem. We need the following complex version ofFrobenius’ Theorem. See [Voisin] for explanation.

Theorem 1.5.1. Suppose that X is a complex m manifold and E is a k dimensionalholomorphic subbundle of the tangent bundle of X. Then, E can be integrated, i.e., neareach x ∈ X, E = kerϕ∗ for some locally defined holomorphic X, x→ Cm−k if and onlyif [E,E] ⊆ E.

This is used for m = 2n and k = n.

Comments on proof of Newlander-Nirenberg Theorem. The important part of the prooffor us is the “trivial” part which states that, if X is a complex manifold, then[

T 0,1X , T 0,1

X

]⊆ T 0,1

X

It is also important to note that this is equivalent to the condition[T 1,0X , T 1,0

X

]⊆ T 1,0

X

(See Remark 2.2.4.)


2. Complex differential forms

Some approaches to this topic obscure the difficulties using deceptively simple nota-tion. I will explain why the equation (∂+∂)2 = 0 is equivalent to the bracket conditions[T 0,1X , T 0,1

X

]⊆ T 0,1

X and[T 1,0X , T 1,0

X

]⊆ T 1,0

X .

2.1. Definitions. We assume that (X, I) is an almost complex manifold which satisfiesthe bracket conditions. Thus X is a complex manifold. But it is easier to keep track ofwhat we are doing if we use the notation of an almost complex manifolds and we haveactions of I = I ⊗ 1 and i = 1⊗ i on TX ⊗ C = T 1,0

X ⊕ T0,1X .

Definition 2.1.1. Let

ΩX,C := HomR(TX ,C) = HomC(TX ⊗ C,C)

This is a C2n bundle over X. On a coordinate chart U , let

dzj = dxj + idyj

This is the homomorphism TU → C given by:

dzj :∂

∂xj7→ 1

dzj :∂

∂yj7→ i

and dzj sends the other basis elements ∂∂xk

, ∂∂yk

, k 6= j to 0. Similarly,

dzj = dxj − idyjNote that dzj is the composition of dzj with complex conjugation c which is an

R-linear automorphism of C:

dzj = c dzjAs in the case of TX ⊗ C, let

ΩX,C = Ω1,0X ⊕ Ω0,1

X

be the decomposition of ΩX,C into the i and −i eigenspaces of I. Then dzj ∈ Ω1,0X since

I ∂∂xj

= ∂∂yj

and the dzj form a basic for Ω1,0X in the sense that every α ∈ Ω1,0

X can be

written as

α =n∑j=1

αjdzj

where αj are C∞ maps X → C (differentiable over R). Similarly, dzj forms a basis of

Ω0,1X .

Proposition 2.1.2. If f : X → C is a (real) differentiable function (f = g + ih,g, h : X → R) then df = dg + idh ∈ ΩX,C is given in local coordinates by

df =∑ ∂f

∂zjdzj +

∂f

∂zjdzj


Proof. Expand it out:∑ 1

2

(∂f

∂xj− i ∂f

∂yj

)(dxj + idyj) +

1

2

(∂f

∂xj+ i

∂f

∂yj

)(dxj − idyj)

=∑ 1

2

(∂f

∂xjdxj − i

∂f

∂yjidyj

)+ same

=∑ ∂f

∂xjdxj +

∂f

∂yjdyj = df

Definition 2.1.3. Ωp,qX ⊂ Ωp+q

X,C is the space of p + q-forms on X with coefficients in Cwhich are given locally by

α =∑I,J

αI,J dzi1 ∧ dzi2 ∧ · · · ∧ dzip︸︷︷︸dzI

∧ dzj1 ∧ dzj2 ∧ · · · ∧ dzjq︸︷︷︸dzJ

=∑I,J

αI,JdzI ∧ dzJ

where the sum is over all pairs of multi-indices I = (i1, i2, · · · , ip) and J = (j1, j2, · · · , jq)and αI,J ∈ C∞(X,C).

We need to show that this is independent of the choice of local coordinates. Giventhat, the following are obvious.

Proposition 2.1.4.

ΩkX,C =

∑p+q=k

Ωp,qX

dΩp,qX ⊂ Ωp+1,q

X ⊕ Ωp,q+1X

d = ∂ + ∂ where

∂ : Ωp,qX → Ωp+1,q

X

∂ : Ωp,qX → Ωp,q+1

X

∂2 = 0, ∂∂ + ∂∂ = 0, ∂2

= 0

Proof. The key part is the second line. Since dαI,J =∑

∂αI,J

∂zjdzj + ∂αI,J

∂zjdzj,

dα =∑ ∂αI,J

∂zjdzj ∧ dzI ∧ dzJ +

∂αI,J

∂zjdzj ∧ dzI ∧ dzJ

∈ Ωp+1,qX ⊕ Ωp,q+1

X

The important properties of the boundary operators ∂, ∂ have been trivialized! Thekey step is independence of coordinates which I need to explain.


2.2. Coordinate invariant definitions. Suppose that X is a real n-manifold and ΩkX

is the space of k-forms on X. Elements are alternating C∞(X)-multilinear maps

ω : ΓTX ⊗ · · · ⊗ ΓTX → C∞(X)

Alternating means ω(σ1, · · · , σk) = 0 if the σi are not distinct (equivalently, the signchanges if we switch two of them). C∞(X)-multilinear means

ω(f1σ1, · · · , fnσk) = f1f2 · · · fnω(σ1, · · · , σk)for fi : X → R in C∞(X). This second condition is equivalent to saying that ω isdetermined by its value at each point. I.e., it is a section of the k-th tensor power of thecotangent bundle.

Given k vector fields σ1, · · · , σk on X we have ω(σ1, · · · , σk) : X → R. In localcoordinates we have

ω =∑I

ωIdxI

where the sum is over all multi-indices I = (i1, i2, · · · , ik) and dxI = dxi1∧dxi2∧· · ·∧dxik .The differential d : Ωk

X → Ωk+1X in the deRham complex of X has the following

coordinate free definition.

dω(σ0, · · · , σk) =k∑i=0

(−1)iσi (ω(σ0, · · · , σi, · · · , σk))

+∑

0≤i<j≤k

(−1)i+jω([σi, σj], σ0, · · · , σi, · · · , σj, · · · , σk)

For example, when k = 1 and α ∈ Ω1X we have

dα(σ, τ) = σ(α(τ))− τ(α(σ))− α([σ, τ ])

Proposition 2.2.1. Suppose that ω =∑

I ωIdxI ∈ Ωk

X . Then

dω =∑I

∑i

∂ωI

∂xidxi ∧ dxI

Proof. dω =∑cJdxJ . To find cJ for J = (j0, · · · , jk):

cj = dω(σ0, · · · , σk) = dω

(∂

∂xj0, · · · , ∂

∂xjk

)where σi = ∂

∂xji. But, [σi, σj] = 0 for these vector fields (because their coefficients are

constant functions). So, we only get the first sum in the definition of dω. But

ω(σ0, · · · , σi, · · · , σk) = ωI

where I = (j0, · · · , ji, · · · , jn). Call this I = δiJ . So,

dω =∑J

cJdxJ =∑J

∑i

(−1)i∂ωδiJ

∂xjidxJ

which is the same as the other equation since (−1)idxJ = dxji ∧ dxδiJ

Ωp,qX , already defined, has the following coordinate free characterization.


Lemma 2.2.2. A k-form on X with coefficients in C: ω ∈ ΩkX,C lies in Ωp,q

X wherep+ q = k if and only if it has the following property.

Suppose that σ1, · · · , σk are sections of either T 1,0X or T 0,1

X . Then

ω(σ1, · · · , σk) = 0

unless exactly p of the σi lie in T 1,0X and q of them lie in T 0,1

X .

Note that this characterization makes sense for any almost complex manifold. Inother words, Ωp,q

X is well-defined for any almost complex X. The following observationfrom [Voisin] is the crucial concept underlying the construction of the double complexΩ∗,∗X which was hidden in the other approach (which is also correct).

Theorem 2.2.3. The differential d : ΩkX → Ωk+1

X sends Ωp,qX into Ωp+1,q

X ⊕ Ωp,q+1X if and

only if the bracket conditions holds: [T 0,1X , T 0,1

X ] ⊆ T 0,1X and similarly for T 1,0

X .

Proof. Look at the definition of dω. Assume the bracket conditions. In the first sum,one of the σi is deleted. So, in order to have p holomorphic and q anti-holomorphicsections we needed to start with p+ 1, q or p, q + 1.

In the second sum, if σi, σj are in T 1,0X then so is [σi, σj]. So, the number of terms in

T 1,0X decreases by one in that case. Similarly, if σi, σj are in T 0,1

X then so is [σi, σj].

Finally, if one of σi, σj is in T 1,0X and the other in T 0,1

X then [σi, σj] is in TX,C =

T 1,0X ⊕ T

0,1X .

Conversely, suppose the bracket condition does not hold. Then, e.g, there might beσi, σj in T 1,0

X so that [σi, σj] has a T 0,1X component. Such a component will be linearly

independent from all other terms, so there is a ω ∈ Ωp,q+1X so that dω is nonzero on that

term and zero on all other terms giving dω a Ωp+2,qX component.

Remark 2.2.4. (1) The equation d2 = 0 which implies that ∂2 = 0, ∂2

= 0, ∂∂ + ∂∂ = 0follows from the definition since any differential form on X with coefficients in C is thesum of two real forms ω = α + iβ and dω = dα + idβ. So, d2ω = d2α + id2β = 0.

(2) The bracket condition holds for T 1,0X if and only if it holds for T 0,1

X because theseare related by complex conjugation (which acts on the second factor of TX ⊗ C) and

[σ, τ ] = [σ, τ ]

So, conjugating both sides of [T 0,1X , T 0,1

X ] ⊆ T 0,1X gives

[T 0,1X , T 0,1

X ] = [T 0,1X , T 0,1

X ] = [T 1,0X , T 1,0

X ] ⊆ T 0,1X = T 1,0

X

2.3. Local exactness of the ∂ complex. Review: We have a complex manifold X.This is an almost complex manifold (X, I) satisfying the bracket condition [T 0,1

X , T 0,1X ] ⊆

T 0,1X .

Complexified differential forms are:

ΩkX,C = Ωk

X ⊗ C =⊕p+q=k

Ωp,qX


In local coordinates, elements have the form

α =∑I,J

αI,JdzI ∧ dzJ

where I = (i1, · · · , ip), J = (j1, · · · , jq). We proved that the standard differential d :ΩkX → Ωk+1

X only has terms in degree (1, 0) and (0, 1):

d = ∂ + ∂ : Ωp,qX → Ωp+1,q

X ⊕ Ωp,q+1X

where

∂α =∑I,J

∂αI,JdzI ∧ dzJ =∑I,J,i

∂αI,J

∂dzidzi ∧ dzI ∧ dzJ

∂α =∑I,J

∂αI,JdzI ∧ dzJ =∑I,J,i

∂αI,J

∂dzidzi ∧ dzI ∧ dzJ

It follows immediately that

∂∂ = 0, ∂∂ + ∂∂ = 0, ∂∂ = 0

Today, we will prove the local exactness of ∂:

Theorem 2.3.1. Suppose α ∈ Ωp,qX with q > 0 so that ∂α = 0 in a nbh of some point

x ∈ X. Then there is a β ∈ Ωp,q−1X so that ∂β = α in a nbh of x.

To prove this we have to go back to Chapter I.

2.4. One complex variable. We start with Stokes’ Theorem:∫Mn

dα =

∫∂M

α

if M is a (real) n-manifold α ∈ Ωn−1M and either

(1) M is compact or(2) α has compact support

We apply this to the case n = 2 and M ⊆ C.(1) M is compact in C. Suppose α = f(z)dz where f is complex differentiable

(∂f∂z

= 0). Then

dα =

(∂f

∂zdz +

∂f

∂zdz

)∧ dz = 0

since dz ∧ dz = 0 and ∂f∂z

= 0. So,

0 =

∫∂M

f(z)dz = 0

This implies:

Theorem 2.4.1 (Cauchy’s formula). If f is C1 and ∂f∂z

= 0 then, for all |z| < 1,

f(z) =1

2πi

∫∂D1

f(ζ)

ζ − zdζ


Proof. By Stokes’ thm this integral is equal to:

(2.1)1

2πi

∫∂Dε

f(ζ)

ζ − zdζ

where Dε is the ε-disk around z. But (as we will see in a minute) this integral givesthe average value of f(ζ) on ∂Dε. So, as ε → 0, this converges to f(z). Since it isindependent of ε, it must be equal to f(z) for all ε.

This formula shows that f(z) is analytic since the integrand is given by a converging(geometric) series which can be differentiated term by term by the dominated conver-gence theorem.

(2) 2nd example. Suppose f is not holomorphic but has compact support. Then (2.1)is not equal to f(z), but it is still equal to the average value of f(ζ) as we now verify:

Let ζ ′ = ζ − z. Then |ζ ′| = ε. So, ζ ′ = εeiθ, dζ ′ = iζ ′dθ and

(∗) =1

2πi

∫ 2π

0

f(z + ζ ′)

ζ ′iζ ′dθ =

1

2π

∫ 2π

0

f(z + ζ ′)dθ = ave f(ζ)

This converges to f(z) as ε→ 0.By Stokes’ thm,

(∗) = − 1

2πi

∫C−Dε(z)

∂

∂ζ′

(f(z + ζ ′)

ζ ′

)dζ′ ∧ dζ ′ → f(z)

f(z) =1

2πi

∫C

1

ζ ′∂f(z + ζ ′)

∂ζ′ dζ ′ ∧ dζ ′

f(z) =1

2πi

∫C

1

ζ ′∂f(z + ζ ′)

∂zdζ ′ ∧ dζ ′

f(z) =∂

∂z

1

2πi

∫C

1

ζ ′f(z + ζ ′) dζ ′ ∧ dζ ′︸︷︷︸

=g(z)

So, f(z) = ∂g∂z

where

g(z) =1

2πi

∫C

f(ζ)

ζ − zdζ ∧ dζ

If f is Ck then this integral can be (partially) differentiated k times under the integralsign (integral over C−Dε). So, g is Ck.

Theorem 2.4.2. Any Ck function f : C→ C is locally equal to ∂g∂z

for some Ck functiong : C→ C.

We need the multi-variable version of this:

Lemma 2.4.3. Suppose that U, V are open subsets of Cp,Cq and f : U × V → C isholomorphic in the first p variables. I.e., ∂f

∂zi= 0 for i ≤ p. Then, locally,

f =∂g

∂zp+1

where g : U × V → C is also holomorphic in the first p variables.


Proof. Replace V with Cq and multiply f with C∞ function on V with compact support.Then let

g(u, zp+1, · · · , zn) =1

2πi

∫C

f(u, ζ, zp+2, · · · , zn)

ζ − zp+1

dζ ∧ dζ

2.5. Back to differential forms. We are proving Theorem 2.3.1. Suppose that α ∈Ωp,qX with q > 0 so that ∂α = 0. If

α =∑I,J

αI,JdzI ∧ dzJ

then

∂α =∑I,J,k

∂αI,J

∂zkdzk ∧ dzI ∧ dzJ

For this to be zero, the sum of all terms with the same dzI must be zero. So, we mayassume p = 0 and

α =∑J

αJdzJ ∈ Ω0,qX

Let t be the smallest index which occurs in any J . Then• Every αJ is holomorphic in every zi where i < t since the term

∂αJ

∂zidzi ∧ dzJ

must be zero (it cannot be cancelled with another term in ∂α = 0.)•

α = dzt ∧ β + α′

where the smallest index that occurs in either α′ or β is t′ > t and

β =∑K

βKdzK

where each βK is holomorphic in z1, · · · , zt−1.• By Lemma 2.4.3, there are gK holomorphic in z1, · · · , zt−1 so that

∂gK

∂zt= βK

Then,

∂∑K

gKdzK =∑K

βKdzt ∧ dzK + terms with smallest index > t

So, α − ∂∑

K gKdzK has smallest index > t. By downward induction on t, α = ∂γ for

some γ ∈ Ω0,qX .

For p > 0 we have α =∑αIdzI and ∂αI = 0 for each I. So, αI = ∂γI for some

γI ∈ Ω0,q−1X and

α = ∂∑I

γIdzI


2.6. Dolbeault complex. Although Ωp,qX is a bicomplex with vertical and horizontal

boundary maps ∂, ∂, we will see that, for a holomorphic bundle, only ∂ gives a well-defined boundary map.

Let E be a holomorphic bundle over a complex manifold X. By definition this is

locally trivial, we have local trivalizations ϕi : EUi

≈−→ Ui × Ck whose transition mapsϕij = ϕ−1

j ϕi give a k × k matrix of holomorphic maps

Ui ∩ Uj →Mk(C) ∼= Ck2

let A0,q(E) = Ω0,qX ⊗ E be the space of (0, q)-forms on X with coefficients in E. In

local coordinates an element α ∈ A0,q(E) is given by

α = (α1, · · · , αk)where each αi ∈ Ω0,q

U . Let ∂α be given by

∂α = (∂α1, · · · , ∂αk)Given two coordinate charts (Ui, ϕi), (Uj, ϕj), we need to know that ∂α defined by this

formula agree on Ui ∩ Uj. Then we can conclude that ∂ is a well defined operator

∂ : A0,q(E)→ A0,q+1(E)

Proposition 2.6.1. The boundary ∂α = (∂α1, · · · , ∂αk) is well defined. I.e., if α ischanged to α′ = (α′1, · · · , α′k) by holomorphic change of coordinates

α′t =∑

(ϕij)stαs

then ∂α′ is related to ∂α by the same change in coordinates.

Proof. This follows from the Leibnitz rule and the fact that ϕij is holomorphic. So,

∂α′t = ∂∑

(ϕij)stαs =∑

(ϕij)st∂αs

Definition 2.6.2. The Dolbeault complex of a homomorphic bundle E is

· · · ∂−→ A0,q−1(E)∂−→ A0,q(E)

∂−→ A0,q+1(E)∂−→ · · ·

I forgot to mention the key point:

Proposition 2.6.3. The kernel of

∂ : A0,0(E) = Ω∞(E)→ A0,1(E)

is the space of holomorphic sections of E.


3. Kahler manifolds

Having a Hermitian metric on a complex manifold puts strong restrictions on itscohomology.

3.1. Kahler forms. We started with the basic concept of a Kahler form. Suppose thatV is a vector space over C: V ∼= Cn and WR = Hom(V,R),

WC = WR ⊗ C = W 1,0 ⊕W 0,1

W 1,1 = W 1,0 ⊗W 0,1. We want to look at W 1,1R = W 1,1 ∩ ∧2WR.

Lemma 3.1.1. ω ∈ W 1,1R if and only if ω : V × V → R is a skew symmetric R-bilinear

form so that

(3.1) ω(Iu, Iv) = ω(u, v)

for all u, v ∈ V .

Proof. First note that the condition ω(Iu, Iv) = ω(u, v) is equivalent to the condition

(3.2) ω(u, Iv) + ω(Iu, v) = 0

since I2 = −1. By definition, W 1,1R is the set of skew-symmetric forms ω on V so that

ωC = ω ⊗ C lies in W 1,1. This is equivalent to the condition that ωC vanishes on pairsof vectors from V 1,0 or from V 0,1. But V 1,0 is the set of all vectors of the form

u = u− iIu

where u ∈ V and

ω(u, v) = ω(u− iIu, v − iIv)

= ω(u, v)− ω(Iu, Iv)− i(ω(u, Iv) + ω(Iu, v))

which is zero if and only if (3.1) and (3.2) hold.

Note that (3.2) implies that

g(u, v) := ω(u, Iv)

is a symmetric bilinear pairing g : V × V → R since

g(v, u) = ω(v, Iu) = −ω(Iu, v) = ω(u, Iv) = g(u, v)

Definition 3.1.2. A hermitian form on a complex vector space V is defined to be anmap h : V × V → C so that

(1) h(u, v) is C-linear in u(2) h(u, v) is C-anti-linear in v

(3) h(v, u) = h(u, v)

Note that (3) implies that h(v, v) ∈ R. The form h is said to be positive definite ifh(v, v) > 0 for all v 6= 0. A positive definite hermitian form on V is called a Hermitianmetric on V .


Proposition 3.1.3. There is a 1 − 1 correspondence between hermitian forms h on Vand ω ∈ W 1,1

R given byω = −=h

andh(u, v) = g(u, v)− iω(u, v)

where g : V × V → R is given by g(u, v) = ω(u, Iv).

Exercise 3.1.4. Show that there is a 1-1 correspondence between hermitian forms hand symmetric forms g satisfying g(Iu, Iv) = g(u, v).

3.2. Kahler metrics. Suppose that (M, I) is a complex manifold. Then a Hermitianmetric h on M is a Hermitian metric hx on the tangent space TM,x at each point whichvaries smoothly with x ∈M . Associated to h we have:

ω = −=hwhich is a 2-form on M which is also in Ω1,1

M which is equivalent to the equation

ω(Iu, Iv) = ω(u, v)

for any two vector vectors u, v ∈ TM,x at each point x ∈M .

Proposition 3.2.1. The real part of a Hermitian metric h is a Riemannian metric gon M which is also invariant under I:

g(u, v) = ω(u, Iv) = g(Iu, Iv)

Proof. We know that g is a symmetric real form. If v 6= 0 ∈ TM,x is a nonzero vector,

h(v, v) = h(v, v) is a positive real number. So,

g(v, v) = h(v, v) > 0

So, g is a Riemannian metric on M .

Note that M is oriented since any complex vector space has a natural real orientation.

Theorem 3.2.2. The volume form on M associated to g = <h is equal to ωn/n!.

To prove this, we need the matrices for h, ω in local coordinates. Take a coordinatechart U with holomorphic coordinates centered at x0 ∈ U (z(x0) = (0, 0, · · · , 0))

z = (z1, · · · , zn) : U → Cn

and z = (z1, · · · , zn). Then dzj, dzj form bases for A1,0, A0,1. These are the bundleswith fibers W 1,0,W 0,1 for V = TM,x. So h ∈ A1,1 is given by

h =∑

αij dzi ⊗ dzj

where αij : M → C (notation: αij ∈ ΩM). These are given by:

αij = h

(∂

∂xi,∂

∂xj

)Since h(u, v) = h(v, u), αji = αji.


Since the negative imaginary part of a complex number z is given by i2(z − z), we

have:

ω =i

2

∑(αij dzi ⊗ dzj − αji dzi ⊗ dzj) =

i

2

∑αij dzi ∧ dzj

Proof of Theorem 3.2.2. We show equality at the point x0 using complex ortho-normalcoordinates at that point. By a C-linear change of the coordinates z = (z1, · · · , zn), wecan arrange for ∂

∂xito be ortho-normal at the point x0. In other words,

αij(x0) = hx0

(∂

∂xi,∂

∂xj

)= δij

Then, at x0,

ωx0 =i

2

∑dzj ∧ dzj =

∑dxj ∧ dyj

So,

ωn =∑π∈Sn

dxπ(1) ∧ dyπ(1) ∧ dxπ(2) ∧ dyπ(2) ∧ · · · ∧ dxπ(n) ∧ dyπ(n)

which is n! times the volume form dx1 ∧ dy1 ∧ · · · ∧ dxn ∧ dyn at the point x0. Since thishold for every point, ωn/n! is the volume form at each point.

3.3. Kahler manifolds.

Definition 3.3.1. A Kahler manifold is a complex manifold with a Hermitian metric hso that ω is closed (dω = 0).

Corollary 3.3.2. On a compact Kahler manifold M , for every 1 ≤ k ≤ n, the form ωk

is closed but not exact. I.e., [ωk] 6= 0 ∈ H2k(M).

Proof. ωk is clearly closed:

dωk = kωk−1dω = 0

If ωk = dα then

d(α ∧ ωn−k) = dα ∧ ωn−k = ωn

which is impossible since the volume form is nonzero in H2n(M) when M is an orientedcompact manifold.

Corollary 3.3.3. A compact complex k-submanifold N of a Kahler manifold M cannotbe the boundary of a (real) submanifold of M .

This follows easily from Stokes’ Theorem if we take the Kahler form on N to be thepull-back of the Kahler form on M . See [Voisin].


3.4. Connections. Recall that if E is a real C∞ k-dimensional vector bundle on a realmanifold X, a connection on E is an R-linear map:

∇ : C∞(E)→ A1(E) = ΩX,R ⊗ E = HomR(TX , E)

satisfying the Leibnitz equation

∇(fσ) = df ⊗ σ + f∇σ

Note: if we had another one ∇′ then

(∇−∇′)(fσ) = f(∇−∇′)σ

In other words, the difference ϕ = ∇ − ∇′ is a homomorphism of ΩX-modules. Suchmorphisms are given by matrices: Suppose that e1, · · · , ek is a basis of local sections ofE. Then

ϕ(ei) =∑

ϕij ⊗ ejwhere ϕij ∈ ΩX,R. Any section of E is given by

∑fiei where fi ∈ ΩX . Then

ϕ(∑

fiei) =∑

fiϕ(ei) =∑

fiϕij ⊗ ejor:

ϕ(f1, · · · , fk) = (f1, · · · , fk)[ϕij]In local coordinates, d is a connection. So, an arbitrary connection is given by ∇ = d+ϕor:

∇(f1, · · · , fk) = (df1, · · · , dfk) + (f1, · · · , fk)Mwhere M is a k × k matrix with entries in ΩX,R.

If M has a Riemannian metric g then recall that the Levi-Civita connection ∇ is theunique connection on TM having the properties:

(1) dg(σ, τ) = g(∇σ, τ) + g(σ,∇τ) (∇ is compatible with g)(2) ∇(τ)σ −∇(σ)τ = [σ, τ ] (usually written with the notation ∇σ(τ) = ∇(τ)σ)

for any two vector fields σ, τ .Suppose that E is a holomorphic bundle on a complex manifold M . Then any con-

nection on E has two components:

∇ : C∞(E)→ A1(E) = A1,0(E)⊗ A0,1(E)

We write ∇1,0,∇0,1 for these two components. Last time we showed that

∂E : C∞(E)→ A0,1(E)

given in local coordinates by ∂U(f1, · · · , fk) = (∂f1, · · · , ∂fk) is well defined.

Proposition 3.4.1. Given a Hermitian metric h on a holomorphic bundle E, there isa unique connection ∇ on E so that

(1) dh(σ, τ) = h(∇σ, τ) + h(σ,∇τ) for all σ, τ ∈ C∞(E) (∇ is compatible with h)(2) ∇0,1 = ∂E

This unique connection is called the Chern connection on E.


Proof. ∇ = ∇1,0 +∇0,1 where ∇0,1 = ∂E and ∇1,0 is uniquely determined by:

dh(σ, τ) = h(∇1,0σ, τ) + h(σ, ∂τ)

since h(∆0,1σ, τ) = 0 and h(σ,∆1,0τ) = 0.

TX,Rv

iTX,RT 1,0X

v = v − iIv

<

Since iv = iv + Iv = Iv, <(iv) = I<(v) making < : T 1,0X → TX,R an isomorphism of

complex vector spaces with i acting as i on T 1,0X and as I on TX,R.

Theorem 3.4.2. Let (M, I) be a complex manifold with a Hermitian metric h. Thenthe following are equivalent.

(1) h is a Kahler metric (dω = 0).(2) ∇LC(Iσ) = I∇LC(σ) for every real vector field σ on X.(3) The holomorphic part of the Chern connection is equal to the Levi-Civita con-

nection:<(∇1,0) = ∇LC

(So, the Chern connection is “ (∇LC , ∂) ”.)

Proof. (3) ⇒ (2). The Chern connection is complex linear and the holomorphic part isthe part where i = I:

<(iσ) = I<(σ)

for σ a section of T 1,0X . So,

∇LC(Iσ) = <(∇1,0(Iσ)) = <(i∇1,0(σ)) = I<(∇1,0(σ)) = I∇LC(σ)

(2)⇒ (1). Since ∇ = ∇LC is compatible with g = <(h), we have

d(g(σ, τ)) = g(∇σ, τ) + g(σ,∇τ)

Since g(σ, τ) ∈ Ω∞(X), all three terms are 1-forms on X. For example, if ∇σ =∑ξiαi

where ξi are vector fields and αi are 1-forms on X, then

g(∇σ, τ) =∑

g(ξi, τ)αi

Applying both sides to the vector field φ we get:

g(∇σ, τ)(φ) =∑

g(ξi, τ)αi(φ) =∑

g(ξiαi(φ), τ) = g(∇σ(φ), τ) = g(∇φσ, τ)


Replacing τ with Iτ we get g(σ, Iτ) = ω(σ, τ) and

d(ω(σ, τ)) = ω(∇σ, τ) + ω(σ,∇τ)

Apply both to the vector field φ (and use the rule df(φ) = φ(f)):

φ(ω(σ, τ)) = ω(∇φσ, τ) + ω(σ,∇φτ)

Cyclically permute the three vector fields:

σ(ω(τ, φ)) = ω(∇στ, φ) + ω(τ,∇σφ)

τ(ω(φ, σ)) = ω(∇τφ, σ) + ω(φ,∇τσ)

Now use the coordinate invariant definition of dω:

dω(φ, σ, τ) = φ(ω(σ, τ))− σ(ω(φ, τ)) + τ(ω(φ, σ))

−ω([φ, σ], τ) + ω([φ, τ ], σ)− ω([σ, τ ], φ)

This is zero since [φ, σ] = ∇φσ −∇σφ.(1)⇒ (3). The proof is by reduction to the standard case: If the metric h is constant,

then ∇LC = d and ∇Ch = (∂, ∂) and <(∂) = d. We only need to show that, at eachpoint, the metric h can be made constant to first order when it is Kahler. So, it sufficesto prove the following lemma.

Lemma 3.4.3. If h is a Kahler metric on X then, in a nbh of each point, there areholomorphic coordinates zi so that the matrix of h:

hij = h

(∂

∂xi,∂

∂xj

)= h

(∂

∂zi,∂

∂zj

)is the identity matrix plus O(|z|2).

Proof. We can choose coordinates which are ortho-normal at the chosen point (z = 0).This makes the constant term of hij the identity matrix. But we also have linear terms:

hij = δij + εij + ε′ij +O(|z|2)

where εij is a linear combination of zk (εij are holomorphic)

εij =∑

εkijzk

and ε′ij is a linear combination of zk (ε′ij are antiholomorphic):

ε′ij =∑

ε′kijzk

Since h is conjugate symmetric we have:

ε′ij = εji

The key property of these numbers is:Claim: If h is Kahler then

εkij = εikj

Proof: Since ∂δij = 0 and ∂zk = 0⇒ ∂ε′ij = 0, at the point z = 0 we have:

0 = ∂ω =i

2

∑∂εijdzi ∧ dzj =

i

2

∑εkijdzk ∧ dzi ∧ dzj


Now let

z′j = zj +1

2

∑εkijzizk = zj +

1

2εkkjz

2k +

∑i<k

εkijzizk

Thendz′j = dzj +

∑εkijzkdzi = dzj +

∑εijdzi

Making

dzj = dz′j −∑

εijdz′i +O(|z|2)

∂

∂z′i=∑ ∂zj

∂z′i

∂

∂zj=∑

(δij − εij)∂

∂zj+O(|z|2)

So, up to terms of second order, we have:

h′ij = h

(∂

∂z′i,∂

∂z′j

)=∑k,`

(δik − εik)hk`(δj` − εj`)

=∑k,`

(δik − εik)(δk` + εk` + ε′k`)(δj` − ε′`j)

=∑

δikδk`δj` − εikδk`δj` + δikεk`δj` + δikε′k`δj` − δikδk`ε′j`

= δij − εij + εij + ε′ij − ε′ij = δij

since εj` = ε′`j.

3.5. Examples. An easy example is a Riemann surface. This is a complex 1-dimensionaland real 2-dimensional manifold. Any Hermitian metric is Kahler since all 2-forms on areal 2-dimensional manifold are closed.

The next example was CP n = Pn(C). We constructed the Fubini-Study metric oncomplex projective space Pn(C) and showed that it is a Kahler metric. This implies thatall smooth projective varieties over C are Kahler manifolds.

The outline of the construction is:

hL 7→ ωh ↔ hω

Given any holomorphic line bundle L on a complex manifold X, there is an associated 2-form ωL on X (giving the Chern class of L). This 2-form ω is associated to a hermitianform hω (since g, ω, h determine each other) which, if we are lucky, will be positivedefinite and therefore a Kahler metric.

A line bundle L over X is the union over open sets Ui of Ui × C. For each Ui, takethe unit section σi(v) = (v, 1). These in general don’t match. So, there are functionsgij : Ui ∩ Uj → C× so that

σi(v) = gij(v)σj(v)

for all v ∈ Ui ∩ Uj. Since σj = gjkσk we have the equation

gik = gijgjk

on Ui ∩ Uj ∩ Uk. Conversely, any collection of maps gij : Ui ∩ Uj → C× satisfying theequations in the box will uniquely determine a line bundle. If the gij are holomorphicfunctions, the line bundle will be a holomorphic bundle.


For example, g∗ij := 1gij

is another collection of functions satisfying the boxed equa-

tions. So, g∗ij gives another holomorphic line bundle L∗ which one can show is thedual bundle to L.

Let h be a Hermitian metric on L. Let hi : Ui → R+ be the positive function givenby hi(v) = h(σi(v), σi(v)). Since σi = gijσj we get:

hi = h(σi, σi) = gijgijh(σj, σj) = gijgijhj

Lemma 3.5.1. Conversely, any family of functions hi : Ui → R× satisfying the equa-tions hi = gijgijhj gives a Hermitian metric on L.

Proof. Let h′i be another collections of functions so that h′i = gijgijh′j. On each Ui let

fi = h′i/hi. Then fj = h′j/hj = gjigjih′i/gjigjihi = h′i/hi = fi. So, f = fi = fj is a

globally defined function on X and h′ = fh is another metric on L.

For example, h∗i = 1hi

satisfies

h∗i = g∗ijgij∗h∗j

Therefore, h∗i gives a metric on L∗.Let

ωi =1

2πi∂∂ log hi

Note thatlog hi = log gij + log gij + log hj

Since gij is holomorphic, ∂ log gij = 0. Since gij is anti-holomorphic, ∂ log gij = 0. So,

ωi =1

2πi∂∂ log hi =

1

2πi∂∂ log hj = ωj

So, ω = ωi is a well-defined 2-form on all of X. This form is closed since it is by definitionexact on each Ui. This is called the Chern form of L.

Now let X = Pn(C). Recall that this is the quotient space of Cn+1\0 modulo therelation

(z0, z1, · · · , zn) ∼ (λz0, · · · , λzn)

for any λ 6= 0 ∈ C. The equivalence class is denoted [z0, · · · , zn]. Another interpretationis that Pn(C) is the set of one dimensional subspaces ∆ of Cn+1. Each such ∆ is uniquelydetermined by any nonzero vector (z0, · · · , zn) ∈ ∆ and we make the identification∆ = [z0, · · · , zn].

Let S be the tautological line bundle over Pn(C) given by

S = (∆, v) |∆ ∈ Pn(C) and v ∈ ∆This is “tautological” since the fiber over ∆ is ∆.

Let Ui be the open subset of Pn given by

Ui = [z] | zi 6= 0Let σi be the section of S over Ui given by

σi(∆) = σi([z0, · · · , zn]) =

(z0

zi, · · · , zi

zi= 1, · · · , zn

zi

)


Thus σi(∆) is the unique element of ∆ with ith coordinate equal to 1. Compare thiswith

σj([z]) =

(z0

zj, · · · , zi

zj, · · · , zn

zj

)We see that

σi =zjziσj

So, gij = zj/zi with dual g∗ij = zi/zj.

Since the line bundle S is a subbundle of the trivial bundle Pn × Cn+1 it gets ametric by restricting the standard metric on Cn+1 given by h(z) =

∑|zj|2. Since the

ith coordinate of σi is 1 we get:

h(σi) = 1 +∑j 6=i

|zj|2

On the dual bundle S∗ we have

h∗(σ∗i ) =1

1 +∑|zj|2

Using |zj|2 = zjzj, the Chern form of S∗ on Ui is

ωi =1

2πi∂∂ log

(1

1 +∑zjzj

)which we calculated step by step:

∂ log

(1

1 +∑zjzj

)=−∑zjdzj

1 +∑|zj|2

where we used the formula ∂f =∑ ∂f

∂zjdzj

∂∂ log

(1

1 +∑zjzj

)= −

((1 +

∑|zj|2)

∑dzj ∧ dzj −

∑zizjdzi ∧ dzj

(1 +∑|zj|2)2

)where we used the formula ∂(fdzj) =

∑ ∂f∂zidzi ∧ dzj. At the origin z = 0 we get

ω =i

2π

∑dzj ∧ dzj

which is the standard form corresponding to (a scalar multiple of) the standard metricwith matrix equal to the identity matrix (divided by π). So, the metric hω is positivedefinite at the point z = 0. However, the space Pn(C) is homogeneous (the same atevery point). This is easier to see if we use a vector space without a basis: Let V beany n + 1 dimensional vector space over C and let P(V ) be the space of 1-dimensionalsubspaces of V . Then it is clear that every point is the same as every other point. Thetautological bundle S and its dual are also defined without choice of coordinates. So, wecan choose coordinates to make any point the center point z = 0. So, the canonicallydefined metric hω is positive definite at every point.


4. Harmonic forms

The main theorem is that the k-th cohomology of a complex manifold is isomorphicto the space of Harmonic forms of degree k and these are constructed using the formaladjoint of the boundary operator d in the de Rham complex of X.

4.1. L2 forms and the Hodge star operator. Recall that the L2-norm of a contin-uous function f on a compact manifold X is

|f |2 :=

√∫X

|f |2dµ

where dµ is a measure on X.Recall the standard definition of an L2-form. If V is a vector bundle over X with a

Riemannian metric g and σ, τ are sections of V then

(σ, τ)L2 :=

∫X

g(σ(x), τ(x))xdµ

Note that g(σ(x), τ(x))x : X → R is a continuous function (if σ, τ, g are continuous).

The L2-norm is |σ|2 =√

(σ, σ)L2 .

4.1.1. L2-norm on forms. Suppose now that X is a oriented Riemannian manifold. Ide-ally the volume form on X should be given by dx1 ∧ dx2 ∧ · · · ∧ dxn. However, thisformula will only work at the single point x = 0 as we saw last time. We use a differentapproach.

Let e1, · · · , en be a positively oriented ortho-normal (o-n) basis for the tangent spaceof X. We can choose these to vary smoothly with x in some open nbh U of any givenpoint. In other words, this is a smoothly varying o-n frame for the tangent space. Let

e∗1, · · · , e∗n ∈ T ∗X,xbe the dual basis in the cotangent space T ∗X,x. We define this to be an ortho-normal basisfor the vector space T ∗X,x

∼= Rn. This defines a metric on the (real) cotangent bundle

Ω1X,R = T ∗X,R. Recall the notation:

ΩkX,R is the bundle over X with fiber ∧kT ∗X,x ∼= R(n

k).

Ak(X) is the space of k-forms on X which are the sections of the bundle ΩkX,R.

The vector space ∧kT ∗X,x ∼= R(nk) has a basis of k-forms

e∗I = e∗ii ∧ · · · ∧ e∗ik

where I = (i1 < i2 < · · · < ik) is multi-index notation. We define the vectors e∗I ,|I| = k to be an ortho-normal basis for ∧kT ∗X,x. In other words, given any two k-forms

α =∑|I|=k α

Ie∗I and β =∑|I|=k β

Ie∗I on X,

(α, β) :=∑

αIβI : X → R

Proposition 4.1.1. The induced norm on ∧kT ∗X,x is independent of the choice of theoiginal basis e1, · · · , en.


Proof. We can define the metric without choosing a basis:

(α1 ∧ · · · ∧ αk, β1 ∧ · · · ∧ βk) :=∑π∈Sk

sgn π∏

(απ(i), βi)

It is easy to see that this defines a metric on the space of k-forms and that e∗I areortho-normal with respect to this metric.

Of particular importance is the case k = n where ∧nT ∗X,x ∼= R is one-dimensional withunit vector

Vol := e∗1 ∧ · · · ∧ e∗nThis is called the volume form on X. This is a well-defined globally defined n-form onX since X is oriented.

The L2-norm on Ak(X) is defined to be

(α, β)L2 :=

∫X

(α, β) Vol

where we note that (α, β) : X → R is the smooth mapping sending x to (α(x), β(x))x.

4.1.2. Hodge ∗ operator. ∗ : ∧kT ∗X,x≈−→ ∧n−kT ∗X,x is defined to be the unique R-linear

map determined by:

α ∧ ∗β = (α, β) Vol

This linear map is the composition of two isomorphisms:

m : ∧kT ∗X,x≈−→ HomR(∧kT ∗X,x,R)

≈−→ HomR(∧kT ∗X,x,∧kT ∗X,x)β 7→ (−, β)x 7→ (−, β)x Vol

and the inverse ofp : ∧n−kT ∗X,x

≈−→ HomR(∧kT ∗X,x,∧kT ∗X,x)γ 7→ − ∧ γ

To see that p is an isomorphism, look at what it does to the o-n basis e∗I of ∧n−kT ∗X,x:

e∗J ∧ e∗I =

±Vol if I

∐J = 1, 2, · · · , n

0 otherwise

Thus, p takes o-n basis to o-n basis. It is an isometry. The formula for ∗ is γ = ∗β =p−1m(β) or

− ∧ ∗β = (−, β)x Vol

This gives an isomorphism of vector bundles

∗ : ΩkX,R

≈−→ Ωn−kX,R

and on forms:∗ : Ak(X)

≈−→ An−k(X)

Furthermore, substituting the boxed equation into the definition of (α, β)L2 gives:

(α, β)L2 =

∫X

α ∧ ∗β

And this determines ∗β uniquely (outside a set of measure zero).


Lemma 4.1.2. ∗ : ∧kT ∗X,x ∼= ∧n−kT ∗X,x is an isometry.

Lemma 4.1.3. ∗2 = (−1)k(n−k) = (−1)k(n−1) on Ak(X).

Proof. For any α, β ∈ Ak(X) we have:

α ∧ ∗β = (α, β) Vol definition of ∗= (∗α, ∗β) Vol ∗ is an isometry= (∗β, ∗α) Vol (·, ·) is symmetric= ∗β ∧ ∗2α definition of ∗= (−1)k(n−k) ∗2 α ∧ ∗β

Since this is true for all β we get α = (−1)k(n−k) ∗2 α for all α.

Note that

(−1)k(n−k) =

(−1)k if n is even

+1 if n is odd

4.1.3. Extension to complex case. I used a really simple example to explain what is meantby saying “Extend ∗ by C-linearity and (·, ·) to a hermitian pairing” on T ∗X,C = T ∗X,R⊗C.

The example is k = 1 and V = C. This has real basis 1, I and W = ∧1V =Hom(V,R) = R2 with basis 1∗, I∗. You know these maps as

1∗ = < : V → RI∗ = = : V → R

By definition, these form an o-n basis for ∧V = a11∗ + a2I∗ : a1, a2 ∈ R: If α =

a11∗ + a2I∗ and β = b11∗ + b2I

∗ then (α, β) = a1b1 + a2b2.The volume form on V is Vol = 1∗ ∧ I∗ ∈ ∧V .Claim: ∗(b11∗ + b2I

∗) = −b21∗ + b1I∗.

1∗

I∗

β

∗β

Verification of claim:

α ∧ ∗β = (a11∗ + a2I∗) ∧ (−b21∗ + b1I

∗)= (a1b1 + a2b2)(1∗ ∧ I∗) = (α, β) Vol

Now extend ∗ by C-linearity to

WC = HomR(V,C) = W ⊗ C = C2

which has the same basis 1∗, I∗ over C. Elements of WC are β = b11∗+b2I∗ where b1, b2 ∈

C. “Extending by C-linearity” means ∗β = −b21∗ + b1I∗. Then ∗β = −b21∗ + b1I

∗.

α ∧ ∗β = (a1b1 + a2b2)(1∗ ∧ I∗)


which illustrates the general formula:

α ∧ ∗β = (α, β)C Vol

where the volume form is

1∗ ∧ I∗ = dx ∧ dy =i

2dz ∧ dz

The last form explains two things: the factor of 2 when comparing the different versionsof the metric and the fact that the volume form is in W 1,1.

We verified that Lemma 4.1.3 holds in the complex case:

α ∧ ∗β = (α, β) Vol definition of ∗ in complex case= (∗α, ∗β) Vol ∗ is an isometry

= (∗β, ∗α) Vol (·, ·) is conjugate symmetric= ∗β ∧ ∗2α definition of ∗= (−1)k(2n−k) ∗2 α ∧ ∗β ∗β ∈ A2n−k(X)

Therefore, in the complex case we get:

∗2α = (−1)|α|

4.2. Adjoint operators. The formal definition of the adjoint of d is given by:

(α, d∗β)L2 = (dα, β)L2

This determines d∗β uniquely (outside a set of measure zero) if it exists. Existence inthe real case is given by the following formula.

Lemma 4.2.1. d∗β = (−1)|β| ∗−1 d(∗β)

Proof. Note that, if β ∈ Ak then (dα, β) makes sense only when α ∈ Ak−1.

(dα, β)L2 =

∫X

dα ∧ ∗β

But, d(α ∧ ∗β) = dα ∧ ∗β + (−1)k−1α ∧ d(∗β). So,

(dα, β)L2 = (−1)k∫X

α ∧ d(∗β)

= (−1)k(α, ∗−1d ∗ β)L2

So, (−1)|β| ∗−1 d ∗ β = d∗β.

This gives:

d∗ =

(−1)k ∗ d ∗ if n is odd

− ∗ d ∗ if n is even


4.2.1. Complex adjoints. In the complex case we have d∗ = − ∗ d ∗. But we also havethe decomposition d = ∂ + ∂.

Lemma 4.2.2. The formal adjoints of ∂ and ∂ are ∂∗ = − ∗ ∂ ∗ and ∂∗

= − ∗ ∂ ∗.

Proof. Very similar to the real case:

(∂α, β)L2 =

∫X

∂α ∧ ∗β

But,∫X∂φ = 0 for any 2n − 1 form φ since φ must be an (n, n − 1) form. So, ∂φ = 0

because An+1,n−1(X) = 0 making∫X∂φ =

∫Xdφ =

∫∂Xφ = 0 since ∂X = ∅. Since

∂(α ∧ ∗β) = ∂α ∧ ∗β + (−1)|α|α ∧ ∂(∗β)

and |α|+ 1 = |β| we get

(∂α, β)L2 = (−1)|β|∫X

α ∧ ∂(∗β)

= (−1)k(α, ∗−1∂(∗β))L2 = −(α, ∗∂(∗β))L2

So, ∂∗

= − ∗ ∂ ∗ and similarly for ∂∗.

Given a holomorphic vector bundle E over X so that E (and X) have Hermitianmetrics, we also have a Hodge ∗ operator

∗E : Ω0,qX ⊗ E → Ωn,n−q

X ⊗ E∗ = Ω0,n−qX ⊗KX ⊗ E∗

where KX = Ωn,0X is the “canonical line bundle” over X. There is a contraction map

c : E ⊗ E∗ → C given by evaluation. Then ∗E is uniquely determined by:

c(α ∧ ∗Eβ) = (α, β) Vol ∈ An,n(X)

But, we take contraction c to be understood and we don’t write it. ∗E is a conjugatelinear isomorphism. So, both sides are conjugate linear in β.

We define the L2-norm on A0,q(E) by

(α, β)L2 :=

∫X

(α, β) Vol =

∫X

α ∧ ∗Eβ

Lemma 4.2.3. The operator ∂E : A0,q(E)→ A0,q−1(E) has formal adjoint given by

∂∗E = (−1)q ∗−1

E ∂KX⊗E∗ ∗EI.e., the following diagram (with all arrow isomorphisms) commutes up to sign.

A0,q(E)∗E //

∂∗E

An,n−q(E∗) A0,n−q(KX ⊗ E∗)∂KX⊗E∗

A0,q−1(E)∗E// An,n−q+1(E∗) A0,n−q−1(KX ⊗ E∗)

The proof, using Stokes’, is the same as before.


4.3. Laplacian. For a Riemannian manifold M we have the Laplace operator: ∆d =dd∗ + d∗d. For a complex manifold X with Hermitian metric we have

∆∂ = ∂∂∗ + ∂∗∂, ∆∂ = ∂ ∂∗

+ ∂∗∂

If E is a holomorphic bundle over X with a Hermitian metric we also have

∆E = ∂E∂∗E + ∂

∗E∂E

Theorem 4.3.1. ker ∆d = ker d ∩ ker d∗ and similarly for ∆∂,∆∂,∆E.

Proof. By definition of adjoint we have:

(α,∆dα)L2 = (dα, dα)L2 + (d∗α, d∗α)L2

But the metric is positive definite. So, both terms are ≥ 0 and equal to zero iff dα, d∗αare both zero.

Definition 4.3.2. Elements of ker ∆d are called harmonic forms. We also have ∆∂-harmonic forms, ∆∂-harmonic forms. ∆E-harmonic forms are also called harmonic(0, q)-forms with coefficients in E.

We need to verify that the Laplace operators are “elliptic differential operators” andwe will quote some theorems about such operators. Our goal is to understand thedefinitions and statements.

4.4. Differential operators. We start with an elementary example. Hook’s law saysthat, for a spring, there is a force which is negative the displacement (times a constant):

d2f(t)

dt2= −f(t)

Or, f ∈ C∞(R) is in the kernel of P where

P (f) =d2f(t)

dt2+ f(t)

The key point is that P is linear.

P (af + bg) = aP (f) + bP (g)

So, kerP , the space of solutions f of Hook’s law, is a vector subspace of C∞(R). P is asecond order linear differential operator in one variable t and one function f .

Now, suppose we have several functions f1, · · · , fp in several variables:

fj(x1, · · · , xn) ∈ C∞(Rn)

and several equations P1, · · · , Pq of the form

Pi =∑

PI,i,j∂fj∂xI

where, for I = (i1, i2, · · · , ik), ∂∂xI

is multi-index notation for

∂

∂xI=

∂k

∂xi1∂xi2 · · · ∂xik=

∂

∂xi1· · · ∂

∂xik


(When k = 0, I = ∅ and ∂∂x∅

is the identity map ∂f∂x∅

= f .) Since ∂∂xi, ∂∂xj

commute, the

indices is can be permuted. If each PI,i,j = PI,i,j(x) ∈ C∞(Rn) is a function only of x,P = (P1, · · · , Pq) is called a (linear) differential operator of order k. So,

kerP = (f1, · · · , fp) : P (f) = 0is a vector subspace of C∞(Rn)p.

Now we replace Rn with an n-dimensional manifold M . The functions fj will bereplaced with sections of a bundle E, and Pi will be sections of another bundle F . Inthis setting, P is a (linear) morphism of sheaves:

P : C∞(E)→ C∞(F )

Here C∞(E) is the sheaf of sections of E. This is the functor which assigns to each opensubset U ⊆M the space C∞(EU) of sections of EU . When we say P is a “morphism ofsheaves” we mean it is locally defined, such as a differential operator.

A morphism P : C∞(E) → C∞(F ) is called a differential operator of order k if, oneach coordinate chart U with local coordinates x1, · · · , xn for U , α1, · · · , αp : EU → Rgiving an isomorphism α : EU ∼= U × Rp and β = (β1, · · · , βq) : FU ∼= U × Rq, there aresmooth functions PI,i,j : U → R so that, for any section γ of E,

βi(P (γ)) =∑

PI,i,j∂αj(γ)

∂xI

where αj(γ), βi(P (γ)) are the smooth functions on U given by Uγ−→ EU

αj−→ R and

UP (γ)−−→ FU

βi−→ R. When k = 0, P ∈ Hom(E,F ):

βi =∑

Pijαj

(P (x) ∈ HomR(Ex, Fx) for each point x when k = 0.)What happens when we change the choice of coordinates? When we change the

βi nothing much happens. We just get a linear combination of the previous differentialoperators. When we change the αj we use the Leibnitz rule. A simple example will showthe idea. Suppose p = q = 1 and the equation is the first order differential operator:

β(γ) = φ(x)∂α(γ)

∂x

we change α = α1 by α(γ) = ψ(x)α′(γ). (The change is linear, so it is the same on theentire fiber.) Then

β(γ) = φ(x)∂ψ(x)α′(γ)

∂x= φ(x)ψ(x)

∂α′(γ)

∂x+ φ(x)

∂ψ(x)

∂x︸︷︷︸∈C∞(U)

α′(γ)︸︷︷︸order 0

The second term is an order 0 differential operator. This is still a linear differentialoperator of order 1. If we ignore the lower order terms, the leading (degree k) term islinear with respect to the change of coordinates of E and F .

Proposition 4.4.1. The definition of a linear differential operator of order k is inde-pendent of choice of coordinates for E,F,M .


4.5. Symbols and elliptic operators. The key point is that the highest order termin a differential operator actually does not differentiate anything! This leads to the“symbol” σP of P .

For each I of maximal order |I| = k, the matrix

(PI,i,j)i,j : E → F

is a well-defined linear map of vector bundles since it commutes with the action ofC∞(M). For example, given any φ : M → R and with k = 1 we have:

∂φ(x)αj(γ)

∂xi= φ(x)

∂αj(γ)

∂xi+ αj(γ)

∂φ

∂xi

As in the change of variables case, the second term has order 0 < k = 1 since αj(γ) isnot differentiated.

Definition 4.5.1. The highest order term of P , as a differential operator, is the symbolof P

σP =∑|I|=k

PI,i,j∂

∂xI∈ Hom(E,F )⊗ SkTM

As I pointed out, (PI,i,j) ∈ Hom(E,F ) and (as an example for k = 2)

∂2

∂x∂y=

∂

∂x

∂

∂y=

∂

∂y

∂

∂x∈ S2TM

which is the second symmetric tensor power of TM .Given any 1-form α ∈ A1(M) we can evaluate the symbol on α =

∑aidxi and get

σP (α) =∑

PI,i,jaI ∈ Hom(E,F )

where aI is the product of the functions ais for I = (is). The example below (computingσ∆(α)) should make the notation clear.

Definition 4.5.2. The differential operator P is elliptic if

σP (α)x : Ex → Fx

is injective at every point where α(x) 6= 0.

For example, for Hook’s law, σP = id⊗ ∂∂t2

and σP (adt) = a2 6= 0 : R → R. So, thisoperator is elliptic.

4.5.1. Symbol of the Laplacian. We went through the calculation for n = 2 using localcoordinates x1 = x, x2 = y. We took an arbitrary metric:

(dxi, dxj) = gij

This is a symmetric positive definite n×n matrix (2×2 when n = 2). The volume formis some scalar multiple of dx ∧ dy:

Vol = V dx ∧ dyfor some function V on M . Our first task is to find ∗dx and ∗dy. These are given by:

(adx+ bdy, dx) Vol = (adx+ bdy, dx) ∧ ∗dx


(adx+ bdy, dx)V dx ∧ dy = (adx+ bdy, dx) ∧ (cdx+ ddy)

(ag11 + bg21)V dx ∧ dy = (ad− bc)dx ∧ dySo,

∗dx = −g21V dx+ g11V dy

Similarly,

∗cy = −g22V dx+ g12V dy

We want to show that the symbol of ∆ = dd∗ + d∗d = −d ∗ d ∗ − ∗ d ∗ d evaluated onany 1-form α is

σ∆(α) = −||α||2idFirst we computed the norm of the 1-form α = adx+ bdy:

(4.1) ||α||2 = (adx+ bdy, adx+ bdy) = a2g11 + 2abg12 + b2g22

Then we computed ∗ d ∗ d(fdx):

d(fdx) = −fydx ∧ dy

∗ d(fdx) = − 1

Vfy

d ∗ d(fdx) = − 1

Vfxydx−

1

Vfyydy + fyd (−1/V )

but we ignore the term fyd(−1/V ) since it has order 1 (f is differentiated only once).

∗ d ∗ d(fdx) = g21fxydx− g11fxydy + g22fyydx− g12fyydy + lower

Next we computed d ∗ d ∗ (fdx):

∗(fdx) = −g21V fdx+ g11V fdy

d ∗ (fdx) = g21V fydx ∧ dy + g11V fxdx ∧ dy + lower order terms

∗ d ∗ (fdx) = g21fy + g11fx + lower

d ∗ d ∗ (fdx) = g21fxydx+ g21fyydy + g11fxxdx+ g11fxydy + lower

Since d∗ = − ∗ d ∗, the sum is:

−∆(fdx) = (−dd∗ − d∗d)(fdx)

= g11fxxdx+ 2g21fxydx+ g22fyydx+ lower

−∆(fdx) =

(g11 ∂

2

∂x2+ 2g21 ∂2

∂x∂y+ g22 ∂

2

∂y2

)(fdx) + lower

The blue part is the (negative) symbol −σ∆ of ∆. The value on α = adx+ bdy is

−σ∆(α) =(g11a2 + 2g21ab+ g22b2

)id = ||α||2id

since, e.g.,

α

(∂2

∂x∂y

)= α

(∂

∂x

)α

(∂

∂y

)= ab


Theorem 4.5.3. The symbols of ∆,∆∂,∆∂,∆E are given by

σ∆(α) = −||α||2id

σ∆∂(α) = −1

2||α||2id

σ∆∂(α) = −1

2||α||2id

σ∆E(α) = −||α||2id on Ω0,q(E)

Since nonzero scalar multiples of the identity map are monomorphisms we get:

Corollary 4.5.4. The Laplacians ∆,∆∂,∆∂,∆E are elliptic differential operators.

4.6. Fundamental Theorem. Any differential operator P : C∞(E) → C∞(F ) has aformal adjoint (assuming E,F,M have metrics) P ∗ : C∞(F )→ C∞(E) given by

(α, P ∗β)L2 = (Pα, β)L2

Lemma 4.6.1.

(φ(x)α, β)L2 = (α, φ(x)β)L2 for any φ : M → R(∂

∂xiα, β

)L2

= −(α,

∂

∂xiβ

)L2

Proof. The first identity is trivial:

(φ(x)α, β)L2 =

∫M

(φ(x)α, β) Vol =

∫M

φ(x)(α, β) Vol =

∫M

(α, φ(x)β) Vol = (α, φ(x)β)L2

The second identity follows from Stokes’ Theorem:(∂

∂xiα, β

)L2

+

(α,

∂

∂xiβ

)L2

=

∫M

∂

∂xi(α, β) Vol

= (−1)i+1

∫M

d(

(α, β)V dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn)

= (−1)i+1

∫∂M

(...) = 0

Lemma 4.6.2. The symbol of the adjoint P ∗ is (up to sign) the adjoint of the symbolof P .

Proof. If P =∑φI ∂

∂xIthen the adjoint of each term is given by(

φI∂

∂xIα, β

)L2

= (−1)k(α,

∂

∂xI

(φIβ

))L2

= (−1)k(α, φI

∂

∂xIβ

)L2

+ lower terms

since, if we differentiate φ even once, β would not be differentiated k-times. If we keeptrack of the indices φIi,j (PI,i,j) we see that, for each I, the matrix is transposed (since jindexes the basis of E and i indexes the basis of F ).

Proposition 4.6.3. If E,F have the same dimension, the adjoint P ∗ of an ellipticoperator P : C∞(E)→ C∞(F ) is also elliptic.


Proof. If the matrix of P has full rank then so does the matrix of P ∗ (being the trans-pose). If the matrix is square, both are invertible and thus monomorphisms.

In our case this statement is obvious since ∆ is self-adjoint :

∆∗ = (dd∗ + d∗d)∗ = d∗d+ dd∗ = ∆

Next we quote the following without proof.

Theorem 4.6.4. If M is compact and P : C∞(E) → C∞(F ) is an elliptic operatorwhere E,F have the same dimension then:

(1) kerP ⊂ C∞(E) is finite dimensional.(2) imP ∗ ⊂ C∞(E) is orthogonal to kerP and

C∞(E) = kerP ⊕ imP ∗

Orthogonality is easy: If α ∈ kerP and P ∗β ∈ imP ∗ then

(α, P ∗β)L2 = (Pα, β)L2 = 0

Corollary 4.6.5. Let Hk = ker ∆ be the space of harmonic k-forms on M . Then themap

Hk ≈−→ Hk(M ;R)

sending α to [α] is an isomorphism.

Recall (Thm 4.3.1) that ker ∆ = ker d ∩ ker d∗. Thus any α ∈ Hk = ker ∆ is closed.

Proof. Since ∆ is an elliptic self-adjoint operator the theorem gives:

Ak(M) = Hk ⊕∆(Ak(M))

If β ∈ Ak(M) is closed then

β = α + ∆γ = α + dd∗γ + d∗dγ

where α ∈ Hk. Then β, α, dd∗γ are all closed. So, d∗dγ ∈ ker d∩ im d∗ = 0. So, [β] = [α]proving that Hk Hk(M ;R) is onto.

To show the map is 1-1, suppose β ∈ Hk and β = dγ. Then β ∈ im d ∩ ker d∗ = 0.So, Hk → Hk(M ;R), and it is an isomorphism.

4.6.1. Comment on finite dimensional case. In the finite dimensional case, Corollary4.6.5 is trivial: Suppose we have a cochain complex · · · → Ck−1 → Ck → Ck+1 → · · ·where Ck are finite dimensional complex vector spaces. Let d∗k : Ck+1 → Ck be theadjoints:

Ck−1

dk−1,, Ck

d∗k−1

mmdk --

Ck+1d∗k

ll

Choosing orthonormal bases, the matrix of d∗k is the conjugate transpose of the matrixof dk. In particular, they have the same rank, say rk. If the dimension of Ck is nk thendim ker dk = nk − rk and dim im dk−1 = rk−1. So,

dimHk(C∗) = nk − rk − rk−1


We know from general principles that im dk−1∩ker d∗k−1 = 0. Since these have dimensionsnk − rk−1 and rk−1 we have Ck = im dk−1 ⊕ ker d∗k−1. Since im dk−1 ⊆ ker dk we haveker dk + ker d∗k−1 = Ck. This gives a short exact sequence:

0→ ker ∆→ ker dk ⊕ ker d∗k−1 → Ck → 0

where ker ∆ = ker dk∩ker d∗k−1. Counting dimensions we see that dim ker ∆ = dim ker dk+dim ker d∗k−1 − dimCk = nk − rk − rk−1. So, ker ∆ ∼= Hk(C∗).

4.6.2. Poincare duality.

Corollary 4.6.6 (Poincare duality).

Hp(M) ∼= Hn−p(M)∗

This follows from:

Lemma 4.6.7. ∆∗ = ∗∆

Proof. When n is even we have ∗2k = (−1)kid and ∆k = −∗n−k d ∗k+1 d− d ∗k+1 d∗k. So,

∗∆k = (−1)k+1d ∗ d− ∗d ∗ d∗

∆n−k∗ = − ∗ d ∗ d ∗+(−1)n−k+1d ∗ dwhich are equal for n even. (The lemma also holds for n odd.)

Lemma 4.6.8. α ∈ Ak(M) is harmonic if and only if ∗α ∈ An−k(M) is harmonic.

Proof. If α is harmonic then ∆α = 0. So, ∗∆α = ∆(∗α) = 0. So, ∗α is harmonic.

Corollary 4.6.9. Hk ∗−→ Hn−k, α 7→ ∗α is an isomorphism.

The isomorphism given by ∗ depends on the metric. So, it is not natural. The naturalstatement is that there is a pairing:

Ak(M)⊗ An−k(M)→ R, α⊗ β 7→∫M

α ∧ β

which induces a duality in cohomology:

Theorem 4.6.10. The pairing above induces a perfect pairing

Hk(M ;R)⊗Hn−k(M ;R)→ R

showing that Hk(M ;R) ∼= Hn−k(M ;R)∗.

Proof. For any α ∈ Hk, ∗α ∈ Hn−k by the lemma. This maps to∫α ∧ ∗α =

∫(α, α) Vol = (α, α)L2 = ||α||2 6= 0

So, the left kernel of the pairing is zero. Similarly, the right kernel is zero and the pairingis perfect.


4.7. Serre duality. Let E be a holomorphic bundle over a complex manifold X. ThenSerre duality is a statemen about the cohomology of X with coefficients in the sheafof holomorphic sections of E. So, we need to review parts of the chapter on sheavesthat we skipped. We need definitions of flasque sheaves, fine sheaves, exact sequences ofsheaves. All sheaves will be sheaves of vector spaces over R or C.

Definition 4.7.1. An exact sequence of sheaves is a sequence:

F0d0−→ F1

d1−→ F2d2−→ · · ·

of morphism of sheaves over X so that the kernel of each morphism dk is the sheafificationof the image of dk−1. (The kernel of a morphism is a sheaf but the image may not beas we will seen in a simple example below.) An exact sequence of sheaves, as above, iscalled a resolution of a sheaf F if F = ker d0.

The cohomology of a sheaf F is defined in terms of an “injective” resolution of F .Such resolutions always exist and

(4.2) Hk(X,F) =ker ΓFk → ΓFk+1

im ΓFk−1 → ΓFkwhere ΓF is the space of global sections of F . It is not easy to find injective resolutionsof sheaves. So, we use other kinds of resolutions which are just as good.

Definition 4.7.2. A sheaf F is acyclic if

Hk(X;F) =

ΓF if k = 0

0 otherwise

Theorem 4.7.3. Let F∗ be any resolution of F by acyclic sheaves Fk. Then (4.2) holds.

Thus, acyclic sheaves can be used instead of injective sheaves to compute sheaf co-homology. Examples of acyclic sheaves are flasque sheaves and fine sheaves.

Definition 4.7.4. A sheaf F over X is flasque if for any V ⊆ U open subsets of X, therestriction map F(U)→ F(V ) is an epimorphism.

Definition 4.7.5. A fine sheaf is a sheaf F of modules over a ring sheaf R over Xwhich admits partitions of unity.

I used an example to explain this definition. Given a holomorphic bundle E over X,the sheaf C∞(E) of C∞ sections of E is a fine sheaf. This is a sheaf of modules over thering sheaf C∞(X) of all C∞ maps X → C. This ring sheaf admits partitions of unity inthe following sense.

Let Uα be a locally finite covering of X with open sets Uα. Then a partition ofunity subordinate to this covering is defined to be a collection of global sections

φα ∈ C∞(X)

whose support (the closure of the set of all points x ∈ X so that φα(x) 6= 0) is containedin Uα so that

∑φα = 1. In sheaf-theoretic terms the support of φα is the complement

of the union of all open sets W so that φα|W = 0.


Example, the constant sheaf is not acyclic. The sheaf E is not acyclic. But the sheafC∞(E) of C∞ sections of E is fine and therefore acyclic. This implies the following.

Theorem 4.7.6. Cohomology of X with coefficients in the sheaf E of holomorphic sec-tions of E is isomorphic to the cohomology of the Dolbeault complex

A0,0(E)∂−→ A0,1(E)

∂−→ · · · ∂−→ A0,n(E)

Proof. The Dolbeault complex is the sequence of global sections of the sequence ofsheaves

Ω0,0(E)∂−→ Ω0,1(E)

∂−→ · · · ∂−→ Ω0,n(E)

We already showed in Theorem 2.3.1 that this sequence is locally exact and that the

kernel of Ω0,0(E)∂−→ Ω0,1(E) is the sheaf of holomorphic sections of E. So this is a

resolution of the sheaf E . But Ωp,q(E) are sheaves of modules over C∞(X). So they arefine and can be used to compute the cohomology of E .

Another example: Compute the cohomology of the circle S1 with constant coefficientsF = R by giving a resolution with fine sheaves.

0→ F → C∞(S1)∂∂t−→ C∞(S1)→ 0

This is an exact sequence of sheaves. It is also an example of a morphism C∞(S1)∂∂t−→

C∞(S1) which is not surjective but the sheafification of its image is all of C∞(S1). Takingglobal sections we see that ΓC∞(S1) = C∞(S1) is the space of smooth functions S1 → R.We easily see that the kernel and cokernel of the derivative map ∂

∂t: C∞(S1)→ C∞(S1)

are R. So, H0(S1;R) = H1(S1;R) = R as expected.

Theorem 4.7.7 (Serre duality). Let X be a compact complex n manifold, E a holomor-phic bundle over X, E the sheaf of holomorphic sections of E. Then

Hq(X; E) ∼= Hn−q(X; E∗ ⊗KX)∗

where we recall that KX = Ωn,0(X) is the canonical line bundle over X.

The proof is very similar to the proof of Poincare duality and we only went over thedifferences. The fundamental theorem applied to the Dolbeault complex gives us:

Hq(X; E) ∼= H0,q = ker ∆0,qE .

We used a version of the lemma ∗∆ = ∆∗:∗E∗⊗KX

∆E = ∆E∗⊗KX∗E

to show that

∗E : H0,q(E)→ H0,n−q(E∗ ⊗KX)

is an isomorphism.To get the natural statement we use the pairing:

A0,q(E)⊗ A0,n−q(KX ⊗ E∗)→ C, α⊗ β 7→∫c(α ∧ β)


where c : E ⊗ E∗ → C is contraction. This gives the perfect pairing:

H0,q(E)⊗H0,n−q(E∗ ⊗KX)→ C


5. Lefschetz operator and applications

We use the Lefschetz operator to prove the Hodge decomposition and Lefschetz de-composition theorems.

5.1. Hodge decomposition.

Definition 5.1.1. Let X be a Kahler manifold. Then the Lefschetz operator

L : Ak(X)→ Ak+2(X)

is wedge with the Kahler form: Lα = ω ∧ α = α ∧ ω (since ω is even). Let

Λ : Ak(X)→ Ak−2(X)

be the formal adjoint of L. I.e.

(Lα, β) = (α,Λβ)

using the Riemannian metric g uniquely determined by ω.

Lemma 5.1.2.

Λ = (−1)k ∗ L ∗

Proof. For k-form α and k + 2-form β we have

(Lα, β) Vol = Lα ∧ ∗β = α ∧ ω ∧ ∗β = (α, ∗−1ω ∧ ∗β) Vol

So, Λ = ∗−1(ω∧)∗ = (−1)k ∗ L ∗ (since ∗2 = (−1)kid).

The key property of Λ is the following

Lemma 5.1.3.

[Λ, ∂] = −i∂∗, [Λ, ∂] = i∂∗

Assume this lemma for a moment.

5.1.1. Proof of Hodge decomposition. The key theorem is:

Theorem 5.1.4. For a Kahler manifold we have:

∆d = 2∆∂ = 2∆∂

This theorem immediately implies the Hodge decomposition.

Corollary 5.1.5. ∆d(Ap,q(X)) ⊆ Ap,q(X).

Proof. ∆∂ and ∆∂ clearly satisfy this.

Theorem 5.1.6 (Hodge decomposition). Let Hp,q = Hp+q ∩ Ap,q(X) be the space of(p, q)-forms which are Kahler. Then

Hk ∼=⊕p+q=k

Hp,q.


Proof. Clearly ⊇. We show ⊆. Let β ∈ Hk. Since Hk ⊂ Ak(X) =⊕

p+q=k Ap,q(X), we

have β =∑βp,q where βp,q ∈ Ap,q(X). So,

0 = ∆dβ =∑p+q=k

∆βp,q

By the corollary, each term ∆βp,q ∈ Ap,q(X). So, they must all be zero. So, βp,q ∈ Hp,q

and these generate Hk.

Proof that Lemma implies Theorem. We assume Lemma 5.1.3. The first consequence ofLemma 5.1.3 is that ∂ and ∂

∗anti-commute:

∂∂∗

+ ∂∗∂ = −i∂[Λ, ∂]− i[Λ, ∂]∂ = −i∂Λ∂ + i∂Λ∂

since ∂∂ = 0. So,

∆d = (∂ + ∂)(∂∗ + ∂∗) + (∂∗ + ∂

∗)(∂ + ∂)

= ∂∂∗ + ∂∂∗

+ ∂∗∂ + ∂∗∂

= ∆∂ + ∆∂

But, ∆∂ = ∆∂ by the following calculations.

∆∂ = i∂[Λ, ∂] + i[Λ, ∂]∂ = i∂Λ∂ − i∂∂Λ + iΛ∂∂ − i∂Λ∂

∆∂ = −i∂[Λ, ∂]− i[Λ, ∂]∂ = −i∂Λ∂ + i∂∂Λ− iΛ∂∂ + i∂Λ∂

Since ∂∂ = −∂∂, these are equal.

It remains to prove the key Lemma 5.1.3.

5.1.2. Proof of key lemma. We will show that [Λ, ∂] = −i∂∗. The other is the complexconjugate of this formula.

Since ∂∗ = − ∗ ∂∗ and Λ = (−1)k ∗ L ∗ we have:

−i∂∗ = i ∗ ∂ ∗

[Λ, ∂] = (−1)k+1 ∗ L ∗ ∂ − (−1)k∂ ∗ L ∗Multiply on both sides by the Hodge star operator ∗:

∗(−i∂∗)∗ = −i∂

∗[Λ, ∂]∗ = L ∗ ∂ ∗ − ∗ ∂ ∗ L = [∂∗, L]

So, it suffices to show:

[∂∗, L] = L ∗ ∂ ∗ − ∗ ∂ ∗ L = −i∂By Lemma 3.4.3 we may assume we have holomorphic coordinates zi centered at any

point so that

ω =i

2

∑dzj ∧ dzj

up to terms of order |z|2.Next, we note that, i∂ and [∂∗, L] are both first order differential operators which, in

the chosen coordinates take any k-form with constant coefficients to k + 1 forms which


are zero at z = 0. This means that we only need to show that the homogeneous degree1 parts are equal. Equivalently, the symbols of the two operators are equal.

σ∂ =∑

dzj ⊗∂

∂zj

In the symbols, the operator ∂∂zj

is formal and can be pulled out from both sides. So,

we need to show:

(5.1) L ∗ (dzj∧) ∗ − ∗ (dzj∧) ∗ L = −idzj∧

Lemma 5.1.7.

∗(dzj∧)∗ = 2int

(∂

∂zj

)Proof. We decompose dzj = dxj − idyj.

(dxj∧)∗ = (−1)k+1 ∗ int(

∂

∂xj

)(Apply this to α = dxj ∧ β where β has degree k − 1: ∗int(∂/∂xj)α = dxj ∧ γ whereβ ∧ dxj ∧ γ = Vol. Then Vol = (−1)k−1dxj ∧ β ∧ γ. So, ∗(α) = (−1)k−1γ makingdxj ∧ ∗α = (−1)k−1 ∗ int(∂/∂xj).) Thus:

∗(dxj∧)∗ = int

(∂

∂xj

)∗(dyj∧)∗ = int

(∂

∂yj

)So,

∗(dzj)∗ = ∗(dxj∧) ∗ −i ∗ (dyj∧)∗ = int

(∂

∂xj− i ∂

∂yj

)= int

(2∂

∂zj

)

Since ∗(dzj)∗ is a derivation,

[∗(dzj)∗, ω∧] = (∗(dzj) ∗ ω)∧ = idzj∧proving (5.1) and thus Lemma 5.1.3 and completing the proof of the Hodge decomposi-tion theorem.


5.1.3. Independence of metric.

Theorem 5.1.8. The Hodge decomposition Hk(X;C) ∼=⊕

p+q=kHp,q(X;C) is indepen-

dent of the choice of metrics. (Assuming a Kahler metric exists.)

Proof. Define Hp,q(X;C) to be the space of cohomology classes represented by closed(p, q)-forms. These include the harmonic (p, q) forms. So, we have an inclusion Hp,q ⊆Hp,q(X;C). We need to show the inclusion the other way.

Let α ∈ Ap,q(X) be a closed (p, q) form on X. By the Fundamental Theorem we have

α = β + ∆γ

for a harmonic form β. Since ∆ preserves the bidegree (p, q), we can project β, γ toAp,q(X) to get

α = βp,q + dd∗γp,q + d∗dγp,q

But α, β, dd∗γp,q lie in ker d. So d∗dγp,q ∈ ker d ∩ im d∗ = 0. So, [α] = [βp,q] andHp,q(X) ∼= Hp,q.

Corollary 5.1.9. Hp,q(X) ∼= Hq,p(X). In particular, Hp,q(X) and Hq,p(X) have thesame rank.

Please see the book for other easy applications.

5.2. Lefschetz decomposition. The Lefschetz decomposition follows from the follow-ing important lemma.

Lemma 5.2.1. Let X be a compact Kahler manifold of complex dimension n. Then

[L,Λ] = (k − n)id on Ak(X)

Equivalently, [L,Λ] = (k − n)id on ΩkX,R.

The number k − n comes from the Pigeon Hole Principle: If there are k pigeons in kholes, if one of the holes is empty then another has at least two pigeons. Our case: Ifthere are k pigeons in n holes and each hole has 0,1 or 2 pigeons then

k − n = number of holes with two pigeons − number of empty holes.

By induction on k: It is true if k = 0. Each additional pigeon increases the RHS by 1.

Proof of Lemma 5.2.1. As before we can assume that we have standard coordinates andω = i

2

∑dzj ∧ dzj. Then Λ : Ak(X)→ Ak−2(X) is i/2 times the sum over all j of

∗−1(dzj ∧ dzj∧)∗ : Ak(X)→ Ak−2(X)

Recall (Lemma 5.1.7) that

∗−1(dzj∧)∗ = 2(−1)k+1int

(∂

∂zj

): Ak(X)→ Ak−1(X)

∗−1(dzj∧)∗ = 2(−1)kint

(∂

∂zj

): Ak−1(X)→ Ak−2(X)


So,

∗−1(dzj∧)(dzj∧)∗ = −4int

(∂

∂zj

)int

(∂

∂zj

): Ak(X)→ Ak−2(X)

(5.2) ∗−1 (dzj ∧ dzj∧)∗ = −4int

(∂

∂zj∧ ∂

∂zj

)L = ω∧ is i/2 times the sum of the operators dzi ∧ dzi∧ which commutes with theoperator (5.2) when i 6= j. (Note: when we put back the two factors of i/2 we cancelthe −4 since (i/2)2 = −1/4. So, we can ignore these three terms.)

We apply these operators to k-forms

α =∑

αI,JdzI ∧ dzJwhere |I| + |J | = k. These are our k “pigeons”. The “pigeon holes” are the numbers1, 2, · · · , n. For each index j ∈ I ∩ J we have the pair dzj ∧ dzj and the operator

(dzj ∧ dzj∧)int

(∂

∂zj∧ ∂

∂zj

)removes this pair and puts it back in (giving back exactly what we started with). Whenj /∈ I ∩ J , this operator kills the term αI,JdzI ∧ dzJ .

If j /∈ I ∪ J (empty pigeon hole) then the operator

int

(∂

∂zj∧ ∂

∂zj

)(dzj ∧ dzj∧)

sticks on the pair dzj ∧ dzj then removes it. This operator gives zero on terms wherej ∈ I ∪ J . So, [L,Λ] multiplies each summand of α by the same number: k − n.

5.2.1. Primitive and almost primitive forms. The Lefschetz decomposition theorem isan easy computational consequence of Lemma 5.2.1. However, I prefer the conceptualproof (from [W]]). So, this will be slightly different from the book.

Lemma 5.2.2. Suppose X is a compact Kahler manifold of complex dimension n and0 ≤ k ≤ n. Then

Ln−k : Ak(X)→ A2n−k(X)

is an isomorphism. Equivalently, Ln−k : ΩkX,R(X) ∼= Ω2n−k

X,R .

We will prove this first for “primitive” forms.

Definition 5.2.3. A k-form α is primitive if Λα = 0. (We will see that this is equivalentto the book definition: α is primitive if Ln−k+1α = 0.)

I will say that α is “almost primitive” (not standard terminology) if LΛα is a scalarmultiple of α:

LΛα = cα

with scalar c ∈ C. (Λα = 0 ⇒ LΛα = 0α. So, primitive implies almost primitive.)Since LΛα− ΛLα = (k − n)α, LΛα is a multiple of α iff ΛLα is a multiple of α:

ΛLα = [Λ, L]α + LΛα = (n− k + c)α


Lemma 5.2.4. If α is almost primitive then so are Lα and Λα.

Proof. Suppose LΛα = cα. Then ΛLα = (n− k + c)α. So,

(5.3) LΛ(Lα) = (n− k + c)Lα

ΛLΛα = cΛα

Since Λα has degree k − 2 this implies

LΛ(Λα) = (k − 2− n+ c)Λα

Proposition 5.2.5. If α 6= 0 ∈ Ak(X) is primitive then

(1) Ljα is almost primitive for all j ≥ 0 and

ΛLjα = j(n− k + 1− j)Lj−1α

(2) Ljα 6= 0 for all 0 ≤ j ≤ n− k.(3) Ln−k+1α = 0.(4) k ≤ n

Proof. (1) follows from (5.3) by induction on j (start with j = 0, c = 0).(2) follows from (1) by induction on j.(4) If k > n then j(n − k − j + 1) 6= 0 for all j > 0. So, Ljα 6= 0 for all j ≥ 0. But

that is impossible since A2n+1(X) = 0.(3): Equation (1) tells us that ΛLn−k+1α = 0. In other words, Ln−k+1α is primitive.

But Ln−k+1α ∈ A2n−k+2(X) has degree 2n − k + 2 ≥ n + 2 which is impossible by (4).Therefore, Ln−k+1α = 0.

As I explained after the class, this proposition is usually drawn as follows.

•n−k

0

__ •1

aa

n−k−1

•

2

aa

n−k−2

3

aa · · ·1 •

n−k

__

0

Here the bullets (•) represent the one dimensional vector spaces generated by α,Lα, L2α,etc. The upper arrows are L. The lower arrows are Λ. The sum of the two arrows whichare above each other is always n−k+1 and the product is j(n−k−j+1) which appearsin formula (1) in the proposition.

Proposition 5.2.6. If k ≤ n, every α ∈ Ak(X) can be written uniquely as a sumα =

∑Lrαr where αr ∈ Ak−2r(X) is primitive.

We will show later that this holds for k > n. Part of the proof is used in the nexttheorem. So, we separate it as a lemma.

Lemma 5.2.7. Let α =∑Lrαr. Let R be the maximum value of r so that αR 6= 0.

Then Ln−k+Rα 6= 0. (In particular, α 6= 0.)


Proof. We have

Ln−k+Rα =∑

Ln−k+r+Rαr =(3) Ln−k+2RαR 6=(2) 0

where Ln−k+r+Rαr = 0 for r < R by (3) in the previous proposition and Ln−k+2RαR 6= 0by (2) in the previous proposition.

Proof. (Uniqueness) Given two such expressions, take the difference. Then 0 =∑Lr(αr−

α′r). By the lemma, αr = α′r for all r.(Existence) The proof of existence is by induction s ≥ 1 minimal so that Λsα = 0. If

s = 1 then α is primitive and we are done. So, suppose s ≥ 2. Then

β = Λs−1α ∈ Ak−2s+2(X)

is primitive. So, β = cΛs−1Ls−1β for some constant c. This implies that

Λs−1(α− cLs−1β) = 0

So, α−cLs−1β =∑Lrαr for αr primitive by induction on s and α =

∑Lrαr+Ls−1(cβ)

where cβ is primitive.

Proof of Lemma 5.2.2. Since L is a vector bundle morphism, it is defined on each finitedimensional vector space Ln−kx : Ak(X)x → A2n−k(X)x for any x ∈ X. Since these vectorspaces have the same dimension, it suffices to show that Ln−k is a monomorphism. ButLn−kα 6= 0 since Ln−k+Rα 6= 0 by Lemma 5.2.7.


5.3. Lefschetz decomposition in cohomology.

Lemma 5.3.1. ∆d commutes with L (and with Λ).

Proof. By the key Lemma 5.1.3, [Λ, ∂] = i∂∗. Taking the adjoint of both side (recall

that adjoint is conjugate transpose at the matrix level) we get:

[∂∗, L] = −i∂

Since ∂∂ + ∂∂ = 0 (coming from d2 = 0) ∂ anticommutes with [∂∗, L]:

∂[∂∗, L] + [∂∗, L]∂ = 0

∂∂∗L− ∂L∂∗ + ∂∗L∂ − L∂∗∂ = 0

But L, ∂ commute: ∂Lα = ∂(ω ∧ α) = ω ∧ ∂α = L∂α since ∂ω = 0. So,

∂∂∗L− L∂∂∗ + ∂∗∂L− L∂∗∂ = 0

or: [∆∂, L] = 0. But ∆d = 2∆∂ = 2∆∂. So, L also commutes with ∆d and ∆∂.

Since L commutes with ∆, L takes ker ∆ to ker ∆. (If ∆α = 0 then ∆Lα = L∆α = 0.)

Theorem 5.3.2. For X a compact Kahler manifold of complex dimension n and k ≤ n,

Ln−k : Hk(X) ∼= H2n−k(X)

is an isomorphism.

Proof. We have an isomorphism on forms:

Ln−k : Ak(X)≈ // A2n−k(X)

Ln−k : Hk //

⊆OO

H2n−k

⊆OO

So, Ln−k must be a monomorphism on Hk. But Hk ∼= H2n−k by Poincare duality. So,Ln−k : Hk → H2n−k is an isomorphism.

Remark 5.3.3. We already know, by Poincare duality that Hk(X) ∼= H2n−k(X). Butthis theorem is stronger than Poincare duality since Ln−k factors as

Hk(X)→ Hk+2(X)→ Hk+4(X)→ · · · → H2n−k(X)

This implies that all of these intermediate cohomology groups are at least as big asHk(X). For example, when k = 0, we know H0(X) = R. So, every even dimensionalcohomology group of X is nonzero.

Since ω ∈ A1,1(X), the Lefschetz operator L takes Ap,q(X;C) into Ap+1,q+1(X;C).

Lemma 5.3.4. For p+ q ≤ n,

Ln−p−q : Ap,q(X;C)≈−→ An−q,n−p(X;C)

is an isomorphism.


Theorem 5.3.5. For p+ q ≤ n,

Ln−p−q : Hp,q(X;C)≈−→ Hn−q,n−p(X;C)

is an isomorphism.

Proof. We already know the two sides are isomorphic. And L takes Hp,q(X;C) intoHn−q,n−p(X;C). Also, Ln−k is a monomorphism on Ap,q(X;C). So, it is an isomorphismof each summand.

Definition 5.3.6. Let Hp,q(X;C)prim denote the set of all β ∈ Hp,q(X;C) which areprimitive meaning Λβ = 0.

Theorem 5.3.7. For p+ q ≤ n we have:

Hp,q(X;C) ∼=⊕r≥0

Hp−r,q−r(X;C)prim

Proof. This follows from Lemma 5.2.6 using the fact that L takes harmonic (p, q)-formsto harmonic (p+ 1, q + 1)-forms.

Exercise 5.3.8. Show that, for k = p+ q > n,

Hp,q(X;C) ∼=⊕r≥n−k

Hp−r,q−r(X;C)prim

5.4. Representations of sl2(C). I explained representations of sl2(C) with an exam-ple. Then I used the example to find an equation for the coefficient c in the proof ofProposition 5.2.6 and thus give another proof of this important proposition.

5.4.1. Example. Let V the vector space of polynomial in two indeterminants x, y ofdegree ≤ n where n is a fixed positive integer. Let Va,b be the one-dimensional subspaceof V spanned by the single monomial xayb. Let L,Λ : V → V be the linear operatorsgiven by L = x ∂

∂y,Λ = y ∂

∂x. On generators we have:

Lxayb = x∂

∂yxayb = bxa+1yb−1

Λxayb = y∂

∂xxayb = axa−1yb+1

which we represent schematically by:

b+2**Va−1,b+1

L

b+1 ))

a−1gg Va,ba

Λ

kkb

L**Va+1,b−1

Λ

a+1iib−1 &&a+2kk

Then, for any α ∈ Va,b we have:

[L,Λ]α = ((ab+ a)− (ab+ b))α = (a− b)αSo, [L,Λ] = c · id on Vc =

⊕a−b=c Va,c. We have linear maps:

L,,Lc−2

L++

Λ

jj LcΛ

llL

,,Vc+2

Λ

kkL

**

Λ

ll


Proposition 5.4.1. If c ≥ 0 then

Lc : V−c≈−→ Vc

Proof. A basis for Lc is xb+cyb : 0 ≤ b ≤ n − c and a basis for L−c is xbyb+c : 0 ≤b ≤ n − c and each basis element xbyb+c of L−c maps to a nonzero multiple of thecorresponding basis element xb+cyb of Lc.

Next we did the Lefschetz decomposition: For any α =∑γa,bx

ayb, we want to findαr which are primitive so that α =

∑Lrαr. But, the primitive elements are powers of

y. The one corresponding to xayb must be ya+b. So, r = a and

Laya+b =(a+ b)!

b!xayb

To obtain γa,bxayb we divide:

La(

b!

(a+ b)!γa,b

)ya+b = γa,bx

ayb

Change to a = r, b = r − c, a+ b = 2r − c: Let R be the largest value of r:

R∑r=0

Lr(

(r − c)!(2r − c)!

γr,r−c

)y2r−c =

R∑r=0

γr,r−cxryr−c = α

So,

αr =(r − c)!(2r − c)!

γr,r−cy2r−c

Next we want to write αR in terms of ΛRα which is

ΛRα = R!γR,R−cy2R−c

So,

αR =(R− c)!

(2R− c)!R!ΛRα

When we subtract this element we get

α− LRαR =R−1∑r=0

γr,r−cxryr−c

which, by induction on R can be written uniquely as∑R−1

r=0 Lrαr for primitive αr. From

theoretical considerations (the classification of all finite dimensional representations ofsl2(C)) we know that the same patter occurs for A∗(X). Namely, if α ∈ Ak(X) andk ≤ n then we can find the decomposition α =

∑Lrαr as follows. Note that [L,Λ] =

c · id = (k − n)id. So, c = k − n ≤ 0.Let R be maximal so that ΛRα 6= 0. Then ΛR+1α = 0. So, ΛRα is primitive. Let

αR =(R + n− k)!

(2R + n− k)!R!ΛRα

Claim 1:ΛR(α− LRαR) = 0.


This claim implies that we can find the Lefschetz decomposition of α by iterating theprocess. For example,

αR−1 =(R− 1 + n− k)!

(2R− 2 + n− k)!(R− 1)!ΛR−1(α− LRαR)

To prove the claim we need to verify that ΛRα = ΛRLRαR. Equivalently, we need toshow that

ΛRLRΛRα =(2R + n− k)!R!

(R + n− k)!ΛRα

Since ΛRα is a primitive element of Ak−2R(X), this formula follows from another claim(using ` = k − 2r):

Claim 2: If β ∈ A`(X) is primitive and `+ 2r ≤ n then:

ΛrLrβ =(n− `)!r!

(n− `− r)!β

which can be proved by induction on r (using the polynomial ring example as an aid,we realize that β corresponds to yn−`).

5.4.2. The Lie algebra sl2(C).

Definition 5.4.2. sl2(C) is the algebra of all 2 × 2 matrices over C with trace 0. Thestandard generators are:

e =

[0 10 0

], f =

[0 01 0

], h =

[1 00 −1

]with relations:

[e, f ] = h, [h, e] = 2e, [h, f ] = −2f

In our case, e = L and f = Λ. The operator h picks out the degree of an element:h(α) = (k − n)α. Thus, Ak(X) is the k − n eigenspace of h = [L,Λ]. The equations[h, L] = 2L, [h,Λ] = −2Λ are equivalent to the statement that L increases degree by 2and Λ decreases degree by 2. Since these relations hold, L,Λ give an action of sl2(C)on A∗(X). At each point x ∈ X, A∗(X)x is then a finite dimensional representation ofsl2(C) and the Lefschetz decomposition follows from the general theorem that says that,up to isomorphism, all finite dimensional representations of sl2(C) generated by a singleelement α is isomorphic to the example that we went over with α being a polynomial intwo indeterminants x, y.


6. Polarization

This section is about integral structures on Kahler manifolds and the Kodaira Em-bedding Theorem.

6.1. Outline of Kodaira’s theorem. This section states theorems without proof. De-tails will come later.

Theorem 6.1.1. If the cohomology class of the Kahler form is integral: [ω] ∈ H2(X;Z)then ω is the Chern form of a holomorphic line bundle L with a Hermitian metric.

Theorem 6.1.2. Given L, for sufficiently large N , the holomorphic sections of the linebundle L⊗N give an embedding of X into a projective space:

ϕ : X → Pr

The projective space is P(V ∗) where V ∗ is the dual of a finite dimensional vector spaceover C. Elements of P(V ∗) are given by hyperplanes H in V since nonzero elements ofV ∗ are linear maps V → C and such a map is given, up to a scalar multiple, by its kernelwhich is a hyperplane in V . The embedding is given as follows.

Lemma 6.1.3. For any x, y ∈ X there is an N and a holomorphic section σ of L⊗N sothat σ(x) 6= 0 and σ(y) = 0.

Since X is compact, we can find a finite collection of sections σi so that, for anyx, y ∈ X, σi(x) 6= 0 and σi(x) 6= σi(y) for some i. Let V be the finite dimensionalcomplex vector space spanned by these sections. Then V is a space of sections of L⊗N

and the embedding

ϕ : X → P(V ∗)

is given by ϕ(x) = H = σ ∈ V |σ(x) = 0. This is a hyperplane in V since V containsa sections so that σ(x) 6= 0. Also, ϕ(x) 6= ϕ(y) for x 6= y ∈ X since the sections separateany pair of points.

6.1.1. Blow up. In order to show that a line bundle L has a section which is nonzero ata point x0 ∈ X, we blow up the point which means replace the point with a projectivespace.

Recall that the tautological bundle S of Pn−1 is the set of all points

S = (V, x) |V ∈ Pn−1, x ∈ V

Let π : S → Cn be the projection to the second factor π(V, x) = x. Then π−1(0) = Pn−1

and, if x 6= 0, then V = Cx is uniquely determined by x. So, π induces a bijection onthe complement which is biholomorphic:

π : S\Pn−1 ∼=−→ Cn\0

To blow up X at a point x0 we cover X with holomorphic coordinate charts X =⋃Ui

so that only U0 contains the point x0 and φ0 : U0∼= V0 ⊆ Cn maps x0 to 0. Then the


blow up Xx0 is given by replacing (U0, φ0) with (U ′0 = π−1(V0), π) where points y ∈ U ′0are identified with φ−1

i π(y) ∈ Ui when φ−10 (π(y)) ∈ U0 ∩ Ui:

Xx0 =⋃i 6=0

Ui ∪ π−1(V0)

Then Xx0 is a complex manifold and there is a holomorphic map τ : Xx0 → X which isan bijection except at x0 which has inverse image τ−1 = Pn−1 = E. This is called theexceptional divisor.

Lemma 6.1.4. A line bundle L over X has a section which is nonzero at x0 if and onlyif τ ∗(L) has a section which is nonzero on E.

Proof. The bundle τ ∗L is trivial over E. So, any section over E must be constant andinduces a section of L.

We consider the short exact sequence of sheaves:

0→ IEL⊗N → L⊗N → L⊗N |E → 0

where L⊗N is the sheaf of section of τ ∗(L⊗N) over Xx0 and IEL⊗N is the subsheaf ofsections which are zero on E. We need two lemmas:

Lemma 6.1.5. a) L⊗N has a section which is nonzero on x0 if H1(Xx0 ; IEL⊗N) = 0.b) H1(Xx0 ; IEL⊗N) = 0 for sufficiently large N .

The converse of (a) is also stated. But we only need this, the trivial implication whichcomes from the long exact cohomology sequence:

· · · → H0(Xx0 ; L⊗N)→ H0(Xx0 ; L⊗N |E)︸︷︷︸C

→ H1(Xx0 ; IEL⊗N)→ · · ·

Part (b) comes from a difficult theorem that we have to skip:

Theorem 6.1.6. If L is a positive holomorphic line bundle over X which is compactthen, for all q > 0, Hq(X;KX ⊗L) = 0 where L is the sheaf of sections of L and KX isthe canonical line bundle over X.

Positive means the Chern form is positive. The proof of (b) is a calculation whichshows that the Chern form of L⊗N becomes positive for N large.

6.1.2. Line bundles. We went over Theorem 4.49 from [V] which proves that line bundlesare classified by H1(X;A∗) for three cases of A simultaneously.

A =

C constant sheaf,

CX,C sheaf of continuous maps X → COX sheaf of holomorphic maps X → C

In all three cases we consider the short exact sequence of sheaves over X:

0→ Z→ A exp−−→ A∗ → 0

where A∗ is the subsheaf of A of nowhere zero sections.


Theorem 6.1.7.

H1(X;A∗) = iso classes of flat/topological/holomorphic line bundles/X

Proof. This follows from the fact that the cohomology of a sheaf is isomorphic to its Cech cohomology by the following double complex. First choose an acyclic resolution ofA∗:

A∗ → F0d−→ F1

d−→ F2d−→ · · ·

Then H∗(X;A∗) is the cohomology of the sequence ΓXF∗. Now consider the doublecomplex:

ΓX(F1) //

d

OO

⊕i

ΓUi(F1)

D //

d

OO

⊕i,j ΓUij

(F1)D //

d

OO

⊕i,j,k ΓUijk

(F1)D //

d

OO

ΓX(F0) //

d

OO

⊕i

ΓUi(F0)

D //

d

OO

⊕i,j

ΓUij(F0)

D //

d

OO

⊕i,j,k ΓUijk

(F0)D //

d

OO

⊕i ΓUi

(A∗) D //

d

OO

⊕i,j

ΓUij(A∗) D //

dOO

⊕i,j,k ΓUijk

(A∗) D //

d

OO

It is a general fact that, since the rows and columns of this diagram are all exact exceptfor the left column and the bottom row, the cohomology of the left column and bottomrow are isomorphic. The isomorphism is give by the “zig-zag” path which is boxed indegree 1. An element of H1(X;A∗) is represented by a cocycle in ΓX(F1). Restrict thisto each Ui and it must have the form dβi where βi is a section of F0 over Ui. ThenDβ = βi − βj is a section of F0 over Uij = Ui ∩ Uj. This must be dgij where gij isa section of A∗ over Uij. But Dgij = gijgjkg

−1ik = 1. So, gij : Uij → C∗ is gives a

collection of transition maps for a line bundle over X. In the three cases, gij is locallyconstant/continuous/holomorphic giving the line bundle a flat/topological/holomorphicstructure.


6.2. Construction of L. We discussed the construction of the line bundle L with Chernform ω. Recall that the Chern form of a line bundle is given locally by

ω =1

2πi∂∂ log h

For example, we calculated the Chern form for S∗, the dual of the canonical line bundleS over P(V ). We got h∗(σ) = 1

1+∑|zi|2 and, at z = 0, ω = i

2π

∑dzj ∧ dzj.

If the line bundle L has Chern form ω equal to the Kahler form then it turns out thatL⊗N is “very ample” which means that there are enough sections to give an embeddingϕ : X → P(V ∗) as explained last time and, furthermore, that ϕ∗(OP(V ∗)(1)) = L⊗N .

We used the following theorem which we skipped (so, the explanation is below).

Theorem 6.2.1. H1(X;O∗X) = Pic(X) = the group of isomorphism classes of holo-morphic line bundles over X (with multiplication given by tensor product).

So, the short exact sequence of sheaves:

0→ Z→ OXexp−−→ O∗X → 0

gives the long exact cohomology sequence:

→ H1(X;OX)→ H1(X;O∗X)c1−→ H2(X;Z)→ H2(X;OX)→

The theorem we proved in class is:

Theorem 6.2.2. Suppose that h is a Hermitian metric on a holomorphic line bundle Lover X and ωL,h is the Chern form of (L, h). Then [ωL,h] is equal to the image of c1(L)in H2(X;R).

Conversely, for any real form ω of type (1, 1) s.t. [ω] is the image of c1(L), thereexists a metric h on L s.t. ω = ωL,h.

Note that, by the long exact sequence above, the statement that [ω] is the image ofc1(L) for some line bundle L is equivalent to the statement that [ω] is in the kernel ofH2(X;Z)→ H2(X;OX).

Proof. Recall that, on a coordinate chart,

ωL,h|Ui=

1

2πi∂∂ log hi

where hi = h(σi) and σi is a nowhere zero section of Ui. Then, on Ui, ωL,h = dβi where

βi = 12πi∂ log hi (since ∂

2= 0, dβi = (∂ + ∂)βi = ∂βi = ωL,h).

Then

βi − βj =1

2πi∂ (log hi − log hj)

on Uij := Ui ∩ Uj. We have transition maps for the sections σi = gijσj making hi =|gij|2hj. We may assume the open sets Uij are contractible (a “good covering”). Then

gij = exp(fij) = e2πifij for some fij : Uij → C. Then βi − βj = −df ij.Since gijgjkg

−1ik = 1 we have fij + fjk − fik ∈ Z.

Claim 1: Up to sign, [ωL,h] ∈ H2(X;R) is represented by the integral Cech cocycleaijk = fij + fjk − fik


Proof: This comes from the “zig-zag” map which gives the standard isomorphismbetween deRham cohomology and Cech cohomology. The double complex is:

Λ2(X) //

d

OO

⊕i Λ

2(Ui)D //

d

OO

⊕i,j Λ2(Uij)

D //

d

OO

⊕i,j,k Λ2(Uijk)

D //

d

OO

Λ1(X) //

d

OO

⊕i Λ

1(Ui)D //

d

OO

⊕i,j Λ1(Uij)

D //

d

OO

⊕i,j,k Λ1(Uijk)

D //

d

OO

Λ0(X) //

d

OO

⊕i Λ

0(Ui)D //

d

OO

⊕i,j Λ0(Uij)

D //

d

OO

⊕i,j,k Λ0(Uijk)

D //

d

OO

⊕iC

D //

OO

⊕i,j C

D //

OO

⊕i,j,k C

D //

OO

We started with a 2-form ωL,h ∈ Λ2(X). Expressed in local coordinates we got a collec-tion of 2-forms (dβi) ∈

⊕i Λ

2(Ui) which lifted to (βi) ∈⊕

i Λ1(Ui). Then D(βi) = βi −

βj ∈⊕

i,j Λ1(Uij) which we expressed as d of (−f ij) ∈⊕

i,j Λ0(Uij). Changing the sign

and ignoring the complex conjugation, D(fij) = fij + fjk − fik = aijk ∈⊕

i,j,k Λ0(Uijk).

Since this is always an integer, its derivative is zero: daijk = 0. So, it comes from a Cechcocycle (aijk) ∈

⊕i,j,k C. Since this is an integer, aijk = aijk. This zig-zag map gives

an isomorphism H2(X;C) ∼= H2(X;C) of deRham cohomology with Cech cohomology

with coefficients in C. This shows that the integral cochain (−aijk) represents the realcohomology class [ωL,h] ∈ H2(X;R) ⊆ H2(X;C).

Claim 2: (aijk) represents c1(L) ∈ H2(X;Z).

Proof: The isomorphism H1(X;O∗X) ∼= Pic(X) = the group of isomorphism classesof holomorphic line bundles over X is given by the correspondence L ↔ (gij) since gijare the transition maps for the line bundle L. The map c1 : H1(X;O∗X)→ H2(X;Z) isgiven (using Cech cocycles) by chasing the diagram:⊕

ijk Z //⊕

ijk ΓUijkOX

exp //⊕

ijk ΓUijkO∗X

⊕ij Z //

D

OO

⊕ij ΓUij

OXexp //

D

OO

⊕ij ΓUij

O∗X

D

OO

ωL,h ↔ (aijk) // (aijk)

(fij)

D

OO

// (gij)↔ L


Conversely, suppose that [ω] = [ωL,H ]. Then ω − ωL,H is ∂ and ∂ closed and exact.

Using the ∂∂ lemma we get ω − ωL,H = 12πi∂∂ϕ for some function ϕ : X → C. The

following calculation shows that ω is the Chern form of L with the new metric h = eϕH:

1

2πi∂∂(log h) =

1

2πi∂∂(ϕ+ logH) = ω − ωL,H + ωL,H = ω.

I explained the proof that [ω] integral implies that it is in the kernel of H2(X;Z)→H2(X;OX). And I repeated the proof after reviewing.

6.2.1. Review.

Definition 6.2.3. a Hodge structure of weight k isVZ a finitely generated free abelian group together with a decomposition:VC = VZ ⊗ C =

⊕p+q=k V

p,q so that

V p,q = V q,p

Definition 6.2.4. A polarization of a Hodge structure of weight k is given by a bilinearform

Q : VZ × VZ → Zwhich is symmetric if k is even and alternating if k is odd and which satisfies (i) and(ii):

Let H(α, β) = ikQ(α, β) on VC. Then

(i) The Hodge decomposition is orthogonal for H, i.e., H(V p,q, V p′,q′) = 0 if p 6= p′.

(ii) ip−q−k(−1)(k2)H(α) > 0 for all α 6= 0 of type (p, q).

Definition 6.2.5. A polarized manifold is a Kahler manifold X whose Kahler class [ω]is integral. In that case,

Q(α, β) =

∫X

ωn−k ∧ α ∧ β =⟨Ln−kα, β

⟩gives a polarization of Hk(X).

The short exact sequence of sheaves:

0→ Z→ OX → O∗X → 0

gives the long exact sequence

(6.1) 0→ ker c1 → H1(X;O∗X)c1−→ H2(X;Z)→ H2(X;OX)→

Proposition 6.2.6. Hk(X;OX) = H0,k(X). Furthermore, the map Z → OX gives themap in cohomology:

Hk(X;Z)→ Hk(X;R) → Hk(X;C) H0,k(X)

where the last map is the projection given by the Hodge decomposition of Hk(X;C).


Proof. Since OX is the sheaf of sections of the trivial line bundle over X, it has an acyclicresolution given by the Dolbeault complex:

A0,0(X)∂−→ A0,1(X)

∂−→ A0,2(X)∂−→ A0,3(X)

∂−→However, as we proved last week, there is a filtration of the bigraded complex Ap,q sothat the forms of degree p ≥ 1 form a subcomplex giving the cohomology ⊕p≥1H

p,q(X).So, we have a map of resolutions:

H∗(X;Z)→ H∗(X;C)

A0(X)

// A1(X)

// A2(X)

//

H∗(X;OX) = H0,∗(X;C) A0,0(X) // A0,1(X) // A0,2(X) //

Given by projection to the (0, k) components.

Corollary 6.2.7. Any element [ω] ∈ H2(X;Z) of type (1, 1) goes to zero in H2(X;OX).

6.2.2. Abelian varieties. Continuing with the long exact sequence 6.1, we have:

H0(X;O∗X)→ H1(X;Z)→ H1(X;OX)→ ker c1 → 0

Recall that H1(X;OX) = H0,1(X). This is a vector space over C and the image ofH1(X;Z) is a lattice of full rank. So,

T := H1(X;OX)/H1(X;Z) ∼= ker c1 = Pic0(X)

is a compact manifold, in fact a torus. ker c1 = Pic0(X) is the group of isomorphismclasses of holomorphic line bundles over X with c1 = 0.

Suppose that H1(X;Z) has rank r. Then H1(X;C) ∼= Cr. This decomposes asH1,0(X)⊕H0,1(X). Since the summands are complex vector spaces, r = 2g is even andH0,1(X) ∼= Cg ∼= R2g. And T ∼= R2g/Z2g is homeomorphic to a product of circles: (S1)2g.

Given any T = V/Γ where V is a complex vector space and Γ is a lattice of full rankin V we have a natural isomorphism

H1(T ;Z) ∼= Γ = H1(X;Z)

given by sending any γ ∈ Γ to the path from 0 to γ in V which goes to a loop inT = V/Γ.

Suppose that X is polarized, i.e.,

Q(α, β) =⟨Ln−1α, β

⟩is an integral form on H1(X;Z) = Γ. Since k = 1 is odd, Q is an alternating form onH1(X;Z).

Q : Λ2(H1(X;Z))→ ZSo,

Q ∈ Λ2(H1(X;Z))∗ = Λ2(H1(X;Z)) = Λ2(H1(T ;Z)) = H2(T ;Z)

Lemma 6.2.8. Q is a Kahler form on the complex manifold T .

By Kodaira’s embedding theorem we get:

Proposition 6.2.9. Pic0(X) is a smooth projective variety.


7. Hypercohomology and spectral sequences

I explained the idea of hypercohomology. This something that we have already seenin different words. Basically, hypercohomology is derived cohomology.

H(X, Y ) = RHom(X•, Y •)

The idea of cohomology works well when either X• is a projective complex or Y • is aninjective complex.

7.1. Definition of hypercohomology. We need either projective objects or injectiveobjects to define hypercohomology. For example, in topology,

Hk(X;Y•) = Hk(S•(X);Y•)

where Si(X) is the free abelian group on the set of singular i simplices σ : ∆i → X.Since Si(X) is free and thus projective, Hk(X;Y•) = Hk(X;Y•).

7.1.1. Injective objects. Sometimes, as in the case of sheaves of holomorphic forms, weuse injective objects since there are not enough projective objects.

Suppose that A is an abelian category. So, it has kernels, cokernels, exact sequencesand the key property of the definition:

Every monomorphism is the kernel of its cokernel. Equivalently, every epimorphismis the cokernel of its kernel. Here “kernel” and “cokernel” are referring to the morphismsour of the kernel object and onto the cokernel object, respectively.

Example 7.1.1. Ab, the category of abelian groups and the full subcategory of finitelygenerated abelian groups are abelian categories. The category of sheaves of OX modulesis also an abelian category.

Definition 7.1.2. An object I in an abelian category A is injective if, for all monomor-phisms j : A → B and any morphism f : A→ I there is a morphism f : B → I so thatf = f j. In other words: “Any morphism of a subobject of B into I can be extended toB.” We say that A has enough injectives if, for all objects A ∈ A, there is an injectiveobjects IA ∈ A and a monomorphism A → IA.

Example 7.1.3. Take A = Z ∈ Ab. Then we have an embedding Z → Q and Q isinjective. (An abelian group is injective if and only if it is “divisible” which means that,for any x ∈ I and any positive n, there is a y ∈ I so that ny = x. Clearly Q has thisproperty.)

Note that Q is not finitely generated. The category of finitely generated abeliangroups does not have enough injectives. But it has enough projectives. So, we are OK.

Theorem 7.1.4. For any ring R and any R-module RM there is an injective R-moduleIM and monomorphism M → IM .

The category of OX-modules has enough injectives. These are used to define hyper-cohomology. But, as did before, we use other kinds of sheaves: acyclic sheaves such asflasque and fine sheaves.


7.1.2. Right derived functor.

Definition 7.1.5. A functor F : A → B between abelian categories (which is understoodto be “additive”: F : Hom(A,A′) → Hom(FA, FA′) is a homomorphism and F (A ⊕A′) ∼= FA⊕ FA′) is left exact if for all short exact sequences in A:

0→ A→ A′ → A′′ → 0

we get an exact sequence in B:

0→ FA→ FA′ → FA′′

(without 0 on the right).

For example, FY = Hom(X, Y ) is a left exact functor on any abelian category (whereX ∈ A is fixed).

Remark 7.1.6. If an abelian category A has enough injectives then every object A ∈ Ahas an injective resolution I• giving a long exact sequence

0→ A→ I0 → I1 → I2 → · · ·This resolution is constructed recursively as follows. First, choose a monomorphismA → I0 where I0 is an injective object ofA. Let C0 = I0/A be the cokernel. Then chooseanother injective morphism C0 → I1 and let C1 = I1/C0, choose a monomorphismC1 → I2, etc. One thing that you need to realize is that a long exact sequence is thesame as a sequence of short exact sequences, in this case:

C0

!!

C2

!!

C4

!!0 // A

>>

// I0

!!

// I1

==

// I2

!!

// I3 //

==

· · ·

0

@@

C1

==

C3

==

In this diagram, the diagonals are short exact sequences and this implies the horizontalis a long exact sequence.

Definition 7.1.7. The right derived functors RiF : A → B is defined by:

RiF (A) = H i(F (I•)) =kerFI i → FI i+1

imFI i−1 → FI i

It is a basic theorem proved by diagram chasing that the injective resolution I• isunique up to a homotopy equivalence of cochain complexes which in turn is also uniqueup to higher homotopies and, therefore, F i(A) is uniquely determine up to canonicalisomorphism.

Example 7.1.8. Let A = Ab, A = Z, F = Hom(Z/p,−) : Ab → Ab. An injectiveresolution of Z is given by

0→ Z→ Q→ Q/Z→ 0

since Q and Q/Z are both injective. Apply F = Hom(Z/p,−) to the injective partI• = (Q→ Q/Z) to get

Hom(Z/p,Q)→ Hom(Z/p,Q/Z)


which is:0→ Z/p

Therefore:R0 Hom(Z/p,Z) = 0

R1 Hom(Z/p,Z) = Z/pIn the category A = R-Mod we get:

Ri HomR(A,B) = ExtiR(A,B)


Introduction

The goal of the course is to go through basic properties of Complex Manifolds andin particular Kahler manifolds. We want to understand complex projective varietiesfrom a differentiable point of view. A smooth projective variety is a Kahler manifold,but not all Kahler manifolds can be embedded in projective space. The first step is tounderstand what is a complex manifold. Then, what is a Kahler manifold.

The first topic will be the Newlander-Nirenberg Theorem which explains the differencebetween complex manifolds and almost complex manifolds (real 2n manifolds havingcomplex structure on their tangent bundles). While going through an outline of theproof, I will explain basic complex analysis which is very intuitive and pretty.

Next is the Hodge Decomposition Theorem on compact Kahler manifolds. I also wantto go over the Lefschetz Decomposition Theorem and the Kodaira Embedding Theorem.Again, during the discussion of difficult topics I will explain the basic preliminary topicswhich may not be so well motivated on their own, such as cohomology of sheaves.

I want to avoid getting stuck in preliminaries and technicalities, especially since thisis the second half of the course. So, I will skip a lot of the preliminary topics and “startin the middle of the book” going over basic topics as needed through the course. I willnot attempt to give complete proofs of major theorems in class. However, there will beelementary exercises to make sure we understand the definitions. My style of teachingthis material will be apparent from the first lecture.

By the end of the course, we should have a feeling of the manifold structure ofsmooth projective varieties over the complex numbers and a good idea of which lemmasand theorems whose proofs we are skipping.

The first topic that I plan to skip is the basic theory of complex variables. Thefundamental fact is that complex differentiable functions are analytic. This is not toodifficult to prove and I will explain it later, but not at the beginning. I will also assumethe basic notions of vector bundles.

We will follow the books of Voisin [V] and Wells [W].

Lectures in Room 226 Goldsmith, 3:30-4:50 MWInstructor: Kiyoshi Igusa, 305 GoldsmithOffice hours: Monday, Wednesday, Thursday 1-2.

References

[V] Voisin, Claire.Hodge theory and complex algebraic geometry. I, volume 76 of Cambridge Studies inAdvanced Mathematics. (2002).

[W] Wells, Raymond O’Neil. Differential analysis on complex manifolds. Vol. 65. Springer Science &Business Media, 2008.

Date post:	01-Jun-2020
Category:	Documents
Upload:	others
View:	5 times
Download:	0 times

MATH 250B: COMPLEX ALGEBRAIC GEOMETRY Contentspeople.brandeis.edu/~igusa/ComplexNotes.pdf · MATH...

Documents