˜

singular variety X. The operation of blowing up one point can be

analyzed in our local model: it transforms the cone into a cylinder, and

the vertex of the cone corresponds to a projective conic in the cylinder.

˜ ˜

This conic is called an exceptional line in X. In fact X is K¨hler . We

a

will show that

˜ ˜

H 0 (X, C) = C = H 4 (X, C)

˜ ˜

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

˜

H 2 (X, C) = H 2 (X, C) • C16 ,

where C16 is spanned by the cohomology classes of those exceptional

lines.

LEMMA 4.3.2 Let M be a manifold and let G be a ¬nite group of

automorphisms of M. Then

H — (M/G, C) = H — (M, C)G .

144 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

Thus the cohomology H — (X, C) is computed by the complex (A— (M )G , d),

where A— (M )G is the space of complex-valued G-invariant di¬erential

forms on M. Thus H — (X, C) = H — (A, C)ι . On H 1 (A, C) = Hom(Λ, C)

the map ι also acts as ’Id. Hence on H j (A, C) = §j H 1 (A, C) it acts

as (’1)j . From here we are able to conclude that

H j (X, C) = H j (A, C), j is even;

H j (X, C) = 0, j is odd.

It means that we have zero cohomology for X in dimensions 1 and

3, besides H 2 (X, C) = H 2 (A, C) has complex dimension 6. Using the

Hodge decomposition

H 2 (A, C) = H 2,0 (A) • H 1,1 (A) • H 0,2 (A)

˜ ˜

we see immediately that h2,0 (X) = h2,0 (A) = 1, h1,1 (X) = h1,1 (A) +

˜

16 = 20, and h0,2 (X) = 1. One can ask what is the meaning of

˜

h2,0 (X) = 1 ? It means that there exists unique holomorphic 2-form β

˜

on X which comes from the holomorphic 2-form dz1 §dz2 on A. Further-

more, local computations at ¬xed points show that β never vanishes.

˜

(In the late 70s Yau showed that X has a hyper-K¨hler structure.)

a

The Kummer surface that we considered is an example of a K3 surface,

which all enjoy the property of having locally non-vanishing holomor-

phic 2-form.

Now we go one dimension down and consider a compact Riemann

surface Σ of genus g. As we did before, we pick a nice basis of inte-

ger homology: a1 , ..., ag , b1 , ..., bg , such that the intersection pairing on

H1 (Σ, Z) gives

(ai , aj ) = (bi , bj ) = 0, (ai , bj ) = δij .

The linear span < a1 , ..., ag > is a lagrangian subspace of H1 (Σ, Z). We

saw before the symplectic pairing on cohomology H 1 (Σ, C) given by

Q([±], [β]) = ±§β.

Σ

4.3. EXAMPLES AND SIEGEL SPACE 145

In terms of periods it is given by

g

Q([±], [β]) = ( ±)( β) ’ ( ±)( β).

ai bi bi ai

i=1

We want to use the complex structure on Σ in a signi¬cant way. We

know that

H 1 (Σ, C) = H 1,0 (Σ) • H 0,1 (Σ) = „¦1 (Σ) • „¦1 (Σ).

We also know that „¦1 (Σ) is a lagrangian subspace of H 1 (Σ, C).

LEMMA 4.3.3 There exists a unique (normalized) basis (ω1 , ..., ωg )

of „¦1 (Σ) such that aj ωi = δij .

Proof. Let us take any basis (·1 , ..., ·g ) of „¦1 (Σ). We will show that

the matrix ( ai ·j ) is non-singular. After that one can apply a linear

transformation to make it the identity matrix. This amounts to showing

that if ω ∈ „¦1 (Σ) is such that ai ω = 0 for any i then ω = 0. We

remember that the space V of γ ∈ H 1 (Σ, C) such that ai γ = 0 for

any i is a lagrangian subspace. This subspace is de¬ned over the real

numbers, i.e. V is stable under complex conjugation. Now ω ∈ V

implies that ω ∈ V , and since V is lagrangian we have

¯

ω§¯ = 0,

ω

Σ

which contradicts to our previous knowledge.

Let us consider in detail the genus 1 case. We identify Σ = C/Λ,

where Λ is a lattice spanned by complex numbers w1 and w2 satisfying

Im(w2 /w1 ) > 0. As a basis in H1 (Σ, Z) we can take loops a and b

obtained from the line segments connecting the origin with the points

w1 and w2 respectively. The periods of the holomorphic one-form dz

are

dz = w1 , dz = w2 .

a b

The invariant of the Riemann surface Σ is the ratio „ = w2 /w1 , which

satis¬es Im(„ ) > 0. One can also rescale the lattice Λ in such a way

that w1 = 1.

146 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

Let us denote

Z=( ωi )

bj

the g — g period matrix.

THEOREM 4.3.4 (Riemann-Siegel) (i) Z is a symmetric matrix

(ii) Im(Z) is positive-de¬nite.

Proof. We note that

Q([ωi ], [ωj ]) = 0 = ( ωi )( ωj ) ’ ( ωi )( ωj ) =

ak bk bk ak

k

= ωj ’ ωi = zij ’ zji .

bi bj

√

For the second part we recall that the hermitian form H(u, v) = ’1Q(u, v )

¯

is positive de¬nite. Let us compute the matrix of H in the basis (ωi ).

√ √

Hij = H(ωi , ωj ) = ’1( ωj ’

¯ ωi ) = ’1(¯ij ’ zij ) = 2Im(zij ).

z

bi bj

Here we used ωi = 1.

ai

DEFINITION 4.3.5 The Siegel upper-half space Hg is the space of

complex g — g matrices satisfying conditions (i) and (ii) of the above

Theorem.

We will show that the space Hg is a homogeneous space of symplectic

group.

To see the geometry of Hg let us consider the genus 2 case. Here

H2 = R3 — U , where R3 corresponds to real symmetric 2 — 2 matrices

and U corresponds to real symmetric positive de¬nite 2 — 2 matrices.

In general, Hg is a subset of complex symmetric g — g matrices and

it looks like an open convex cone. Another geometric interpretation of

Hg is given by

PROPOSITION 4.3.6 The space Hg identi¬es with the manifold S

of complex lagrangian subspaces Λ ‚ C2g such that

√

’1Q(v, v ) > 0

¯

4.3. EXAMPLES AND SIEGEL SPACE 147

for any v ∈ Λ, v = 0 . Here C2g has the standard symplectic form given

by

0 ’Id

.

Id 0

Proof. The space S lies inside the grassmanian Gras(g, 2g) of complex

g-dimensional subspaces inside 2g-dimensional complex space. The

map Hg ’ S we are looking for is given by

Z

Z ’ column space of ‚ S.

Id

To get the inverse map S√ Hg we take Λ ∈ C2g - a lagrangian subspace

’

with the property that ’1ω(v, v ) > 0 for any v ∈ Λ. We can view

¯

A

Λ as a column space of some 2g — g matrix , where A and B

B

are invertible as we saw before. This matrix can be replaced by the

equivalent matrix

AB ’1

A Z

B ’1 = = .

B Id Id

The matrix Z automatically satis¬es to the above conditions. For ex-

ample, when Λ is a lagrangian subspace we get

0 ’Id Z

= (t Z ’ Z) = 0,

( tZ Id )

Id 0 Id

implying that Z is symmetric.

There is an action of the complex symplectic group Sp(2g, C) on the

set of lagrangian subspaces, but this action does not preserve positivity

condition. The action of the real symplectic group Sp(2g, R) though

has all the necessary properties. Let us exhibit how γ ∈ Sp(2g, R) acts

AB

on an element of S. Let γ = be represented as four square

CD

blocks. Then we have

(AZ + B)(CZ + D)’1

AB Z AZ + B

= ∼ .

CD Id CZ + D Id

148 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

PROPOSITION 4.3.7 There is a holomorphic action of Sp(2g, R)

on Hg such that

A B

· Z = (AZ + B)(CZ + D)’1 .

C D

√

The action is transitive and the stabilizer of ’1 · Id is the unitary

group U (g) ‚ Sp(2g, R).

We notice that all stabilizers in Sp(2g, R) of all points of Hg have the

same dimension. We identify Hg S and consider an element V ∈ S,

¯

where V is a lagrangian subspace in C2g . We have C2g = V • V . If

we think of C2g as the complexi¬cation of R2g then R2g has a complex

√ √

¯

structure J which acts as ’1 on V and as ’ ’1 on V . The stabilizer

HV of V is then the set of real linear symplectic transormations γ

which commute with J. It happens if and only if γ is a complex-

linear transformation. Thus Hv U (g). When we compute the real

dimension of the orbit of V it is equal to

dimR Sp(2g, R) ’ dimR U (g) = dimR Hg .

This shows that there is only one orbit, which proves the transitivity.

COROLLARY 4.3.8 The space Hg identi¬es with the coset space

Sp(2g, R)/U (g).

We also notice that

U (g) = Sp(2g, R) © SO(2g, R)

is a maximal compact subgroup in Sp(2g, R). Therefore Hg is a so-

called symmetric space, which by de¬nition is a riemannian manifold

such that for any point Z ∈ Hg there is a symmetry σ : Hg ’ Hg

which ¬xes Z and has Z as isolated ¬xed point (hence the di¬erential

√

of σ at Z is ’Id). For example the symmetry for Z = ’1Id is

Z ’ ’Z ’1 . The Lie group Sp(2g, R) is the group of isometries of Hg ,

4.3. EXAMPLES AND SIEGEL SPACE 149

and its action is transitive. In addition, Hg is a K¨hler manifold under

a

a K¨hler potential

a

ρ(Z) = log(det(Im(Z))).

√

For g = 1 and the coordinate z = x + ’1y on the Lobachevsky plane

we have recover ρ(z) = log(y).