c1 (T X) r

χ(X, E) = (1+ )(r+c1 (E)), [X] = c1 (T X), [X] + c1 (E), [X] .

2 2

Now the number c1 (T X), [X] is the degree of the divisor of a mero-

morphic vector ¬eld on X and is equal to 2 ’ 2g. Thus we have

χ(X, E) = r(1 ’ g) + deg(§r E).

72 CHAPTER 2. COHOMOLOGY OF VECTOR BUNDLES

When E is a line bundle L one has χ(X, L) = 1 ’ g + deg(L). By

ˇ

de¬nition, χ(X, L) = dim H 0 (X, L) ’ dim H 1 (X, L), and let us notice

that by the Serre duality theorem

ˇ

dim H 1 (X, E) = dim H 0 (X, T — X — L—’1 ).

The degree of the bundle T — X — L—’1 is equal to 2g ’ 2 ’ deg(L).

Thus when 2g ’ 2 < deg(L) this bundle has no holomorphic sections

ˇ

and H 1 (X, L) = 0. A simple consequence of this result is

COROLLARY 2.11.1 If deg(L) > 2g ’ 2 then dim H 0 (X, L) = 1 ’

g + deg(L).

For instance if L = (T — X)—m and g ≥ 2, m ≥ 2 then we have

deg(L) = m deg(T — X) = m(2g ’ 2) > 2g ’ 2.

Thus we get dim H 0 (X, (T — X)—m ) = (2m ’ 1)(g ’ 1). When m = 2

then it follows that the dimension of the space of quadratic di¬erentials

is equal to 3g ’ 3.

Now we consider the case d = 2, i.e. the case of a surface. Again,

let E be a vector bundle of rank r over a surface X. Here ch(E) =

1 2 2

r + ·i + 2 ( ·i ), c1 (E) = ·i , c2 (E) = i<j ·i ·j , and ·i =

2

c1 (E) ’ 2c2 (E), ch(E) = r + c1 (E) + [ c1 (E) ’ c2 (E)]. As before we

2

decompose c(T X) = (1 + ξ1 )(1 + ξ2 ) to get

2 2

ξ1 ξ1 ξ2 ξ2

T d(T X) = f (ξ1 )f (ξ2 ) = (1 + + )(1 + + ) =

2 12 2 12

c1 (T X) c2 (T X) c1 (T X)2 ’ 2c2 (T X)

=1+ + + =

2 4 12

c1 (T X) c1 (T X)2 + c2 (T X)

=1+ + .

2 12

Now we apply RRH:

χ(X, E) =

c1 (T X) c1 (T X)2 + c2 (T X)

)(r + c1 (E) + c1 (E)2 ’ c2 (E)), [X] =

(1 + +

2 12

2.11. APPLICATIONS OF RIEMANN-ROCH-HIRZEBRUCH 73

c1 (T X)2 + c2 (T X) c1 (E)2

c1 (T X)c1 (E)

= +r + ’ c2 (E), [X] .

2 12 2

Next we would like to use this formula in a couple of special and im-

portant cases. First let us treat the case when E = 1X - the trivial line

bundle. We have

c1 (T X)2 + c2 (T X)

χ(X, 1X ) = , [X] .

12

On the other hand,

ˇ ˇ

χ(X, 1X ) = dim H 0 (X, 1X ) ’ dim H 1 (X, 1X ) + dim H 2 (X, 1X ).

ˇ

By duality, dim H 2 (X, 1X ) = dim H 0 (X, §2 T — X) - the dimension of the

space of holomorphic two forms. It is usually denoted by pg and is called

the geometric genus of X. Besides, dim H 0 (X, 1X ) = 1 and in the case

ˇ

when X is a K¨hler manifold one has dim H 1 (X, 1X ) = 1 dim H 1 (X, C).

a 2

2

Now suppose that X = CP for which we know that

c(T CP2 ) = c(L)3 = (1 + ξ)3 = 1 + 3ξ + 3ξ 2 .

We also computed that pg (CP2 ) = 0 and hence directly χ(CP2 , 1CP2 ) = 1.

To verify RRH in this case we notice that 32 + 3 = 12 and therefore

c1 (T X)2 + c2 (T X)

= ξ2.

12

This leaves us with

χ(CP2 , 1CP2 ) = ξ 2 , [CP2 ] = 1

as it should.

For an arbitrary surface X we now take E = T — X. Since cj (E) =

(’1)j cj (E — ) we obtain

c1 (T X)2 c1 (T X)2 + c2 (T X) c1 (T X)2

—

χ(X, T X) = ’ + + ’c2 (T X), [X] =

2 6 2

1

c1 (T X)2 ’ 5c2 (T X), [X] .

=

6

74 CHAPTER 2. COHOMOLOGY OF VECTOR BUNDLES

When X is so-called K3 surface ( for which §2 T — X is a trivial line

bundle and H 1 (X, C) = 0) we have c1 (T X) = c1 (§2 T X) = 0 and

furthemore

1

χ(X, 1X ) = c2 (T X), [X] = 1 + pg = 2

12

(here we also work under an assumption that X is K¨hler). So we have

a

the topological Euler characteristic:

(’1)i dim H i (X, C).

c2 (T X), [X] = 24 =

By Poincar´ duality,

e

dim H 3 (X, C) = dim H 1 (X, C) = 0.

Of course, dim H 0 (X, C) = dim H 4 (X, C) = 1 and it provides us with

the classical result that for a K3 surface dim H 2 (X, C) = 22. Applying

Hodge decomposition

H 2 (X, C) = H 2,0 • H 1,1 • H 0,2

and duality

dim H 0,2 = dim H 2,0 = dim H 0 (X, §2 T — X) = 1

ˇ

we get dim H 1,1 = 20. (By de¬nition H p,q = H q (X, §p T — X) and we

also use the fact that the spaces H p,q and H q,p are complex conjugate

to one another.) Now, by RRH

5 5

χ(X, T — X) = ’ c2 (T X), [X] = ’ 24 = ’20,

6 6

which is perfectly consistent with our previous computation because

ˇ ˇ

dim H 0 (X, T — X) ’ dim H 1 (X, T — X) + dim H 2 (X, T — X) =

ˇ

’ dim H 1 (X, T — X) = ’20,

ˇ

since the spaces H 0 (X, T — X) and H 2 (X, T — X) are dual to each other

and dim H 0 (X, T — X) = dim H 0,1 = 0.

2.12. DOLBEAULT COHOMOLOGY 75

2.12 Dolbeault cohomology

Let X be a complex manifold of complex dimension n and let Am (X)

be the space of di¬erential forms of degree m with complex coe¬cients.

There is a natural decomposition

Am (X) = Ap,q (X),

p+q=m

where Ap,q (X) is the space of di¬erential forms of type (p, q) i.e. those

forms ± that in local coordinates (z1 , ..., zn ) look like

±= fi1 ...ip j1 ...jq dzi1 § · · · § dzip d¯j1 § · · · § d¯jq .

z z

i1 <···<ip

j1 <···<jq

The exterior di¬erential d : Am (X) ’ Am+1 (X) may be decomposed

as d± = d ± + d ±, where ± ∈ Ap,q (X), d ± ∈ Ap+1,q (X), and d ± ∈

Ap,q+1 (X). It is clear that we have d d = d d = d d + d d = 0 and

that is why

d d d

Ap,0 (X) ’ Ap,1 (X) ’ · · · ’ Ap,n

is a complex called the Dolbeault complex. The basic fact about the

Dolbeault cohomology groups

Ker[d : Ap,q ’ Ap,q+1 ]

p,q

H (X) = .

Im[d : Ap,q’1 ’ Ap,q ]

is provided by the following analogue of the Poincar´ lemma:

e

LEMMA 2.12.1 (Dolbeault lemma.) If X is a Stein manifold then

the Dolbeault cohomology group

H p,q (X) = 0, q > 0;

H p,0 (X) = „¦p (X),

where „¦p (X) is the space of global holomorphic p-forms on X.

For example, if we take p = 0, q = 0 then f ∈ A0,0 is just a function and

the condition d f = 0 simply means that f is a holomorphic function.

76 CHAPTER 2. COHOMOLOGY OF VECTOR BUNDLES

Similarly if q = 0 with p arbitrary, the condition d ± = 0 for ± ∈ Ap,0

means that ± is holomorphic.

If X is a Stein manifold then we have a similar vanishing result

ˇ

when we consider the Cech cohomology groups with coe¬cients in the

holomorphic vector bundle §p T — X, namely

ˇ

H q (X, §p T — X) = 0, q > 0;

ˇ

H 0 (X, §p T — X) = „¦p (X).

Now we give the proof of Dolbeault Lemma for p = 0 (for p > 0 it

works similarly) and for a polydisk ∆ = {(z1 , ..., zn ) ∈ Cn ; |zi ’ wi | <

ai }.

First, we consider the case n = 1. Let U be an open disk in C.

What we need to show is that any 1-form ± of the type ± = f (z)d¯ z

can actually be written as ± = d g, where g is a smooth function on

U . This is equivalent to ¬nding a solution of the equation ‚g/‚ z = f .

¯

V ‚‚ U assume that f is a C ∞ function in a neighbourhood of the

¯

closure U , so that

¯

√ dξ ‚f dξ § dξ

2π ’1f (z) = f (ξ) + ,

¯

ξ’z ‚ξ ξ ’ z

γ U

where z is a point in U , and γ = ‚U . To see this formula we take a

small disk Dµ z and apply Stokes™ theorem to the di¬erential form

dξ

f (ξ) ξ’z over the region U \ Dµ and we get

¯

dξ ‚f dξ § dξ

f (ξ) = .

¯

ξ’z ‚ξ ξ ’ z

‚(U \Dµ ) U \Dµ

The left hand side of this equality is equal to

dξ dξ

f (ξ) ’ f (ξ) .

ξ’z ξ’z

γ ‚Dµ

Now what is left is to go to the limit µ ’ 0. Since we have assumed

that f is a C ∞ - function in a neighbourhood of U , there exists an open

V such that U ‚‚ V is relatively compact in V . One can multiply our

function f by a smooth function χ which is identically equal to 1 on

2.12. DOLBEAULT COHOMOLOGY 77

¯

U and vanishes on ‚V so we may assume that f vanishes identically

outside V . Then we have

¯ ¯

√ ‚f dξ § dξ ‚f (ξ + z) dξ § dξ

2π ’1f (z) = = =

¯ ¯

C ‚ξ ξ ’ z ξ

‚ξ

C

(we make the change of variables ξ ’ ξ + z)

¯ ¯

‚f (ξ + z) dξ § dξ ‚ dξ § dξ

= = ( f (ξ + z) ).