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180 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

Therefore if we identify E R2,19 , then M is a symmetric space for the
group SO(2, 19), i.e. M = SO(2, 19)/SO(2)—SO(19), and dimR (M) =
38. We recall that inside E we have the lattice Λ, and inside SO(2, 19)
we have a discrete arithmetic cocompact subgroup “ which stabilizes Λ.
We arrive to the following correspondence: for a pair (X, [ω]) consisting
of a K3 surface and a choice of polarization we have a well-de¬ned point
of the quotient M/“ = “ \ SO(2, 19)/SO(2) — SO(19) and a theorem
of Piatetski-Shapiro says that it is actually a bijection.
To understand the geometry of M we start with

PROPOSITION 4.8.3 The symmetric space of the group SO(2, n),
n > 1, de¬ned as above, is an open set of Cn and SO(2, n) acts holo-
morphically on it.

Proof. Let us represent R2,n = R1,n’1 • H, where H is the standard
01
hyperbolic plane with pairing given by the matrix . We denote
10
by (, ) the form in R2,n and by , the form in R1,n’1 and we let T be the
positive cone in R1,n’1 , i.e. consisting of those vectors b ∈ R1,n’1 which
satisfy b, b > 0, and b0 > 0, where b0 is the ¬rst coordinate of the

vector b. Then we de¬ne U = R1,n’1 • ’1T ‚ R1,n’1 — C Cn and
we de¬ne a map φ : U ’ M (a subset of the set of lines in R2,n — C)as
1
follows: φ(u) is the line spanned by v = (u; ’ 2 u, u , 1). One easily
checks that
’ 1 u, u
1 01
2 2
||v|| = ||u|| + (’ u, u , 1) = 0,
2
10 1
2

and since v = (¯, ’ 1 u, u , 1) we have
¯ u2

1
(v, v ) = (u, u) ’ [ u, u + u, u ] = u, u ’ Re u, u .
¯ ¯ ¯¯ ¯
2

But if u = a + ’1b then Re u, u = a, a ’ b, b and u, u = ¯
a, a + b, b . Hence (v, v ) = 2 b, b > 0 as desired. To get the converse
¯
statement we need to show that if v = (u, ±, β) ∈ R2,n — C, where ± and
β are complex numbers, satis¬es (v, v) = 0 and (v, v ) > 0 then β = 0.
¯
If we are able to do so, then by a simple rescaling we obtain β = 1 and
this will prove the result. Assume that β = 0. Then (v, v) = ||u||2 =
4.9. COMPACT COMPLEX SURFACES 181

(||a||2 ’ ||b||2 ) ’ 2 ’1 a, b and (v, v ) = u, u = ||a||2 + ||b||2 . This
¯ ¯
2 2
implies easily that ||a|| = ||b|| > 0 and a, b = 0. But this means
that a, b span a 2-dimensional subspace of R1,n’1 on which the form ,
is positive de¬nite, which gives us a contradiction.

Our grand conclusion now is that M is R1,18 + ’1T ‚ C19 as an
open subset of C19 , where T was de¬ned above (it is a tube domain).
The symmetric space M is a so-called hermitian symmetric space.


4.9 Compact complex surfaces
We would like to give the classi¬cation of compact complex analytic
surfaces. This problem was solved in the 60s by Kodaira [39]. Most of
the results and the proofs in this section are due to him.
Let S be a compact complex analytic non-singular surface. We
shall always assume that S is connected. We have several numeri-
cal invariants of S such as the Betti numbers b1 and b2 , de¬ned as
bi = dim H i (S, C) (clearly b3 = b1 , and b0 = b4 = 1). If we identify
H 4 (S, Q) Q, then we have two other numbers: c2 and c2 , where c1
1
and c2 are the ¬rst and the second Chern classes of the holomorphic
tangent bundle T S. We also have Hodge numbers hp,q de¬ned in terms
ˇ
of Cech cohomology as

hp,q = dim H q (S, §p T — S).

Besides, we have the signature I(S) de¬ned as the di¬erence b+ ’ b’
of the numbers of positive and negative eigenvalues of the symmetric
pairing on H 2 (S, Q). We shall use the following simple version of the
Hirzebruch signature theorem [32]:

THEOREM 4.9.1
c2 ’ 2c2
I(S) = 1 .
3
Proof. We shall prove this assertion under the assumption that S has
a K¨hler structure. We recall Proposition 4.2.1, which says that
a

I(S) = 2 ’ h1,1 + 2h0,2 .
182 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

Therefore

I(S) = 2(1 ’ h0,1 + h0,2 ) + (h1,0 ’ h1,1 + h1,2 ),

where we used h1,2 = h0,1 = h1,0 . We also know from Chapter 2 that
the Hirzebruch-Riemann-Roch theorem implies that the holomorphic
Euler characteristic of S is given by
c2 + c2
χ(S) = 1 .
12
On the other hand side, by de¬nition

χ(S) = h0,0 ’ h0,1 + h0,2 = 1 ’ h0,1 + h0,2 .

Besides, we saw there that the Euler characteristic of the holomorphic
cotangent bundle is given by
c2 ’ 5c2
χ(T S) = 1

.
6
On the other hand side,

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

= h1,0 ’ h1,1 + h1,2 .
It gives us
c2 + c2 c2 ’ 5c2 c2 ’ 2c2
1 1
=1

I(S) = 2χ(S) + χ(T (S)) = 2 + .
12 6 3
We recall that the number pg := h0,2 = h2,0 is called the geometric
genus of S. The equality h0,2 = h2,0 is a consequence of Serre™s duality
theorem which in general tells us that for a compact complex manifold
M of dimension n one has

H i (M, V ) H n’i (M, V — — §n T — X)

if one takes i = 2 and V a trivial line bundle. We also de¬ne the
irregularity q of S to be the number q := h0,1 . Let us denote by K =
§2 T — S the canonical line bundle on S.
4.9. COMPACT COMPLEX SURFACES 183

LEMMA 4.9.2 Let ± be a holomorphic one-form on S, then d± = 0.

Proof. Locally d± is given by f (z1 , z2 )dz1 §dz2 and if f (z1 , z2 ) is not
identically zero then
d±§d¯ > 0.
±
On the other hand side

d±§d¯ =
± d(±§d¯ ) = 0.
±
S S

If S admits an embedding into a projective space, this surface will
be called algebraic. If S can be mapped properly surjectively and holo-
morphically to a non-singular algebraic curve (called the base curve),
in such a way that a general ¬ber is a non-singular elliptic curve, then
S is called an elliptic surface. A surface S is called regular if q = 0.
LEMMA 4.9.3
1
q ≥ b1 .
2
Proof. Let ±1 , ..., ±k be linearly independent holomorphic 1-forms on S.
We know that d±i = 0 and if we take their complex conjugates, then the
forms ±1 , ..., ±k , ±1 , ..., ±k are linearly independent. To see this, assume
otherwise that

ai ±i + bj ±j = df, ai , bj ∈ C
i j

for some function f . Then simple local arguments show that function f
has to be harmonic (i.e. to satisfy d d f = 0). Therefore, by a version
of the maximum principle, f is constant and hence

ai ±i = 0, and bj ±j = 0,
i j

which contradicts our assumption.
We also notice that each non-zero holomorphic 1-form ± is not exact,
because if ± = df for a holomorphic function f , then by the maximum
principle f has to be constant. It means that if ± = 0 then both [±] = 0
and [±] = 0. Therefore all [±i ] and [±j ] are linearly independent as well.
This shows that h1,0 ¤ b1 /2.
184 CHAPTER 4. COMPLEX ALGEBRAIC VARIETIES

On the other hand side, the exact cohomology sequence

0 ’ H 0 (S, T — S) ’ H 1 (S, C) ’ H 1 (S, 1S ) ’ · · · ,
d
which can be deduced from the exact sequence 0 ’ C ’ OS ’ dOS ’
0, gives us the inequality

q + h1,0 ≥ b1 .
b1
Together with h1,0 ¤ b1 /2 this gives the desired h1,0 ¤ and the
2
Lemma follows.

THEOREM 4.9.4
If b1 is even, then b1 = 2q, and b+ = 2pg + 1.
If b1 is odd, then 2q = b1 + 1, b+ = 2pg .

Proof. Let us choose pg linearly independent holomorphic 2-forms βj ,
1 ¤ j ¤ pg . They all are closed and not exact, because if d± = βj then

¯
0= d(±§β) = β§β > 0.
S S

We sure can choose them is such a way that

¯
βi §βj = δij .
S

Now we construct a new set consisting of 2pg 2-forms ν1 , ..., ν2pg given

¯j , and ν2j = ’1(βj ’ βj ). One easily checks that
¯
by ν2j’1 = βj + β
the intersection pairing on the set {νj } is positive de¬nite and thus
b+ ≥ 2pg .
If one combines the following identities

c2 + c2
pg ’ q + 1 = 1 ,
12
c2 ’ 2c2
b ’b = 1
+ ’
, and
3
c2 = 2 ’ 2b1 + b+ + b’
4.9. COMPACT COMPLEX SURFACES 185

(the ¬rst two appear in the previous Theorem and the last that we saw
before is the statement of the fact that the highest Chern class of the
tangent bundle is equal to the topological Euler characteristic of the
manifold) then one gets

(2q ’ b1 ) + (b+ ’ 2pg ) = 1.

Now the above Lemma and the inequality b+ ≥ 2pg imply that we
have only two choices:
(1) 2q = b1 and b+ = 2pg + 1 or
(2) 2q = b1 + 1 and b+ = 2pg .
The straightforward consequence of the above result and the in-
equalities h1,0 + q ≥ b1 and h1,0 ¤ b1 /2 is that if b1 is even then h1,0 = q,
and if b1 is odd, then h1,0 = q ’ 1. In both cases,

h1,0 + q = b1 .

COROLLARY 4.9.5 The numbers pg , q and h1,0 are homological in-
variants of S.

One of the most important results of [39] is
THEOREM 4.9.6 Surfaces can be classi¬ed into the following seven
classes
(I) b1 is even and pg = 0 (algebraic type)
(II) b1 = 0 and pg = 1 (K3 surfaces)
(III) b1 = 4 and pg = 1 (complex tori)
(IV) b1 is even, pg > 0, and c2 = 0 (elliptic type)
1
2
(V) b1 is even, pg > 0, and c1 > 0 (algebraic type)

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