LEMMA 4.3.9 The K¨hler structure on Hg is Sp(2g, R)-invariant.

a

AB

Proof. Assume that a g — g matrix γ = lies in the symplectic

CD

0 ’Id

group Sp(2g, R), then it preserves the matrix S = , i.e.

Id 0

t

γSγ = S. If we spell it out, it amounts to the matrices t AC and t BD

being symmetric and the equality t AD ’t CB = Id. Let us notice the

following four subgroups of Sp(2g, R):

A 0 Id B

, ,

t

A’1

0 0 Id

Id 0 D P

, and t ’1 ,

C Id 0 D

where matrices B, C, and D’1 P are symmetric. The last subgroup

mentioned is actually a maximal parabolic subgroup of Sp(2g, R).

AB

As we know, the matrix γ = acts on the Siegel upper-half

CD

space Hg by

γZ = (AZ + B)(CZ + D)’1 .

A K¨hler potential ρ(Z) = log det(Y ), where Y = Im(Z) gives rise to

a √

’1

the symplectic form ω = ’ 2 d d ρ on H. We intend to show that

the action of Sp(2g, R) preserves ω as well as the corresponding K¨hler

a

metric g. Using the formula for the di¬erential of the inverse matrix

dM = ’M ’1 · dM · M ’1 and the facts that det exp(M ) = exp(T r(M ))

√

¯

and 2 ’1Y = Z ’ Z we see that

¯ ¯ ¯

d d log det Y = T r((Z ’ Z)’1 dZ§(Z ’ Z)’1 dZ).

150 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

Thus √

’1 ¯

T r(Y ’1 dZ§Y ’1 dZ) and

ω=

8

1 ¯

g = T r(Y ’1 dZ§Y ’1 dZ).

8

Now let Z1 = γZ = (AZ + B)(CZ + D)’1 , and let Y1 = Im(Z1 ). Since

both Z and Z1 are symmetric, we have

Z1 =t Z1 = (Z t C +t D)’1 (Z t A +t B).

Using this, one gets

√

¯ ¯ ¯

2 ’1Y1 = Z1 ’ Z1 = (AZ + B)(CZ + D)’1 ’ (AZ + B)(C Z + D)’1 =

¯ ¯ ¯

= (Z t C +t D)’1 [(Z t A+t B)(C Z +D)’(Z t C +t D)(AZ +B)](C Z +D)’1 .

If we denote the matrix inside the square brackets by K, then we have

¯

K = Z(t AC ’t CA) + Z(t AD ’t CB) + (t BC ’t DZ)Z + (t BD ’t DB).

We notice that due to the above conditions on γ we have

t

AC ’t CA =t BD ’t DB = 0, t

AD ’t CB =t BC ’t DZ = Id.

√

¯

Therefore K = Z ’ Z = 2 ’1Y , and

¯

Y1 =t (CZ + D)’1 Y (C Z + D)’1 ,

generalizing the well-know formula for the Lobachevsky plane, where

we have y1 = y/(c|z| + d)2 . Passing to the K¨hler potential we get

a

¯

ρ = log det Y1 = log det Y ’ log det(CZ + D) ’ log det(C Z + D).

Since the second term is holomorphic and the third is anti-holomorphic,

we get d d ρ(Y ) = d d ρ(Y1 ) proving the invariance of ω under the

action of Sp(2g, R). We also know that Sp(2g, R) acts transitively on

Hg , and thus it is enough to check that g is positive de¬nite at one

√

point, say Z0 = ’1Id to prove that it is such everywhere. Let a

4.3. EXAMPLES AND SIEGEL SPACE 151

tangent vector at Z0 be represented by a symmetric matrix W = (wij ).

¯

At Z0 we have g = 1 T r(dZ§dZ), and

8

1 1

|wij |2 .

g(W, W ) = wij wji =

¯

8 8

ij ij

Let Σ be a compact Riemann surface of genus g with a ¬xed sym-

plectic basis (a1 , ..., ag , b1 , ..., bg ) of H1 (Σ, Z) (meaning that it has the

standard intersection pairings). Next we take the following discrete

subgroup “ = Sp(2g, Z) ‚ Sp(2g, R). It is clear that “ acts transitively

on the set of symplectic bases of H1 (Σ, Z). It is also not hard to see

that γ ∈ “ by the change of symplectic basis transforms the element

Z ∈ Hg into γ · Z. Therefore, we get a well-de¬ned Torelli map denoted

by T from the set of Riemann surfaces of genus g to the factor-space

H/“. In fact the following is known to be true.

THEOREM 4.3.10 (Torelli). If Σ1 and Σ2 are Riemann surfaces

of genus g and T (Σ1 ) = T (Σ2 ) then there exists a complex-analytic

isomorphism Σ1 Σ2 .

It is still an open problem to give a precise description of the image of

T . However Shiota proved a conjecture of Novikov which gives a set of

algebraic equations for the closure of the image of T . The problem is

that these equations may not de¬ne an irreducible algebraic variety, so

the image of T is in general an open subset of an irreducible component

of the locus de¬ned by the Novikov equations.

Let us consider the case g = 1. In this case H = H1 ‚ C de¬ned by

Im(z) > 0 is the Lobachevsky upper-half plane and Sp(2, Z) = SL(2, Z)

acts on H. (Actually the action of SL(2, Z) is reduced to the action of

P SL(2, Z) = SL(2, Z)/{±1}.) Let F ‚ H be the closed subset de¬ned

by the following set of conditions: |z| ≥ 1 and ’1/2 ¤ Re(z) ¤ 1/2.

We call F a fundamental domain since it satis¬es the following two

conditions:

1). For any z ∈ H there is γ ∈ P SL(2, Z) such that γ · z ∈ F .

2). If z1 , z2 ∈ F , z1 = z2 and z1 = γz2 for some γ ∈ P SL(2, Z) then

z1 , z2 ∈ ‚F .

There are some ¬xed points on the boundary ‚F . For example,

√ 0 ’1

the point ’1 is ¬xed by the element and the point j =

10

152 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

√ 0 ’1

exp(2π ’1/3) is ¬xed by . In this case the image of T is

11

the whole F if we identify the points on the boundary by the action of

P SL(2, Z).

Let us introduce the altitude of a point z ∈ H. It is simply its

imaginary part y = Im(z) > 0. The basic fact about the altitude is

that on each P SL(2, Z)-orbit it is bounded from above.

We need some facts about positive de¬nite quadratic forms, namely

what is called Minkowski theory. Let Pn be the space of positive de¬nite

quadratic forms Q in variables (x1 , ..., xn ). It is the same as the space

of positive de¬nite n — n matrices A. The group GL(n, R) acts on this

space by

X —¦ A =t XAX, X ∈ GL(n, R).

Its discrete subgroup GL(n, Z) acts as well and we introduce the fun-

damental domain Mn for this action.

Mn = {Y = (yij ∈ Pn , Y (a1 , ..., an ) ≥ ykk

if (a1 , ..., an ) ∈ Zn , g.c.d.(ak , ..., an ) = 1, and yk,k+1 ≥ 0}.

Such a matrix Y is called Minkowski reduced and Mn is a fundamental

domain. It has the following properties:

(1) GL(n, Z) —¦ Mn = Pn .

(2) If Y1 , Y2 ∈ Mn and Y2 = g —¦ Y1 , 1 = g ∈ GL(n, Z), then Y1 , Y2 ∈

‚Mn .

(3) M is closed and connected.

Then Minkowski theory follows that

Pn /GL(n, Z) Mn / ∼

for ∼ the equivalence relation for which Y1 ∼ Y2 if they belong to the

same orbit. The set Mn \ ‚Mn is an open dense set in both. The

boundary ‚Mn is a ¬nite union of semi-algebraic sets.

Let us give a geometric explanation of the above de¬nition of Mn .

When k = 1 the condition that g.c.d.(a1 , ..., an ) = 1 is equivalent to the

fact that (a1 , ..., an ) is a part of a basis in Zn . We call such an n-tuple

unimodular. Thus when k = 1 and e1 is the ¬rst vector of the standard

basis, we have

y11 = Y (e1 ) = min Y (a) = min Y (a).

a∈Zn , unimodular

na∈Z \0

4.3. EXAMPLES AND SIEGEL SPACE 153

When k = 2 the condition g.c.d.(a2 , ..., an ) = 1 is equivalent to the fact

that (e1 , a2 , ..., an ) can be completed to a basis of Zn . This generalizes to

an arbitrary k, i.e. the condition that g.c.d.(ak , ..., an ) = 1 is equivalent

to possibility of completing (e1 , ..., ek’1 , ak , ..., an ) to a basis of Zn .

Let us indicate why GL(n, Z) —¦ Mn = Pn . Fix some Y ∈ Pn .

There exists a vector v ∈ Zn min 0 which gives the minimum value of

Y (v). In other words, there exists X ∈ GL(n, Z) such that Xv =

e1 , because if we consider Y = X ’1 Y then Y (v) = Y (e1 ) realizes

the minimum. So we can assume that e1 realizes the minimum of

Y (v) among all unimodular v. Similarly, after transforming Y by some

element of GL(n, Z) which ¬xes e1 , we may assume that e2 realizes the

minimum of Y (v) among all vectors v such that (e1 , v) is unimodular,

etc... At the end if the signs of the yk,k+1 are not correct, we apply a

diagonal element of GL(n, Z) to correct them.

Let us also give an explicit description of M2 .

y11 y12

M2 = { ∈ P2 , 0 ¤ 2y12 ¤ y11 ¤ y22 }.

y12 y22

To show that any such matrix satis¬es the necessary conditions we

notice that

y22 = Y [e2 ] ≥ Y [e1 ] = y11

and

Y [e1 ’ e2 ] = y11 ’ 2y12 + y22 .

Since (e1 , e1 ’ e2 ) is a basis of Z2 , one has

Y [e1 ’ e2 ] ≥ Y [e2 ],

whence 2y12 ¤ y11 .

To go in the opposite direction, we assume that 0 ¤ 2y12 ¤ y11 ¤

y22 . Next we take a = (a1 , a2 ) and compute

Y (a) = a2 y11 + 2a1 a2 y12 + a2 y22 ≥ a2 y11 ’ |a1 a2 |y11 + a2 y11 =

1 2 1 2

= y11 (a2 ’ |a1 a2 | + a2 ) = y11 ((|a1 | ’ |a2 |)2 + |a1 a2 |) ≥ y11 .

1 2

When k = 2, a = (a1 , 1), and (e1 , a) is a basis of Z2 we have

Y (a) = a2 y11 + 2a1 y12 + y22 ≥ a2 y11 ’ |a1 |y11 + y22 ≥ y22 .

1 1

154 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

The notion of altitude passes up to higher genera. On Hg the al-

titude function is det(Y ) = exp(ρ(Z)), where ρ is a K¨hler potential.

a

Now we de¬ne the closed subset F ‚ Hg consisiting of all Siegel-reduced

Z. We say that Z is Siegel-reduced if the following three conditions

hold:

1). Z is at a higher (or equal) altitude than any other γ · Z for all

γ ∈ Sp(2g, Z).

2). The symmetric matrix Y is Minkowski-reduced.

3). If X = Re(Z) then |xkl | ¤ 1 .

2

LEMMA 4.3.11 The space F is a fundamental domain for the action

of “ = Sp(2g, Z)/ ± Id on Hg .

Proof. (Sketch.) Let us pick Z ∈ Hg . One checks straightforwardly

that given a > 0 there exist ¬nitely many γ ∈ “ such that ρ(γZ) ≥

a. Then there exists some γZ which realizes the maximum. So the

condition 1) is veri¬ed.

A 0

For 2) we use γ = , A ∈ P GL(g, Z). Then γ —¦ Z =

0 t A’1

AZ t A’1 and Im(γ —¦ Z) = AY t A’1 and we can always pick A such that

Im(γ —¦ Z) is Minkowski-reduced.

Id B

For 3) we make a transformation by an element of the form .

0 Id

Now we must show that our domain is not too big. Let Z and γ —¦ Z

lie in the interior of F , and g = Id. Then, obviously, ρ(Z) = ρ(γ —¦ Z)

and it follows that | det(CZ + D)| = 1, where γ is given by the matrix

AB

. We are able to ¬nd a neighbourhood U of Z such that the

CD

equality | det(CZ + D)| = 1 holds for any Z from U . It can only

happen if C = 0, because det(CZ + D) is a holomorphic function of

A B

Z . Therefore, γ = . Now we see that if Y = Im(Z),

0 t A’1

Y1 = Im(γ —¦ Z) then Y1 = AY t A’1 and if Y, Y1 are in the interior of

Mg then A = I and γ —¦ Z = Z + B. Using the third condition and the

fact that B has to have integer entries, we conclude that B = 0.

In fact, Siegel proved that Hg /Sp(2g, Z) has ¬nite volume equal to

2

ζ(2) · · · ζ(2g).

π