2Surfaces
III Positivity in Algebraic Geometry
2.1 The intersection product
We now focus on the case of smooth projective surfaces over an algebraically
closed field. In this case, there is no need to distinguish between Weil and Cartier
divisors.
Let
X
be such a surface. Let
C, D ⊆ X
be smooth curves intersecting
transversely. We want to count the number of points of intersection. This is
given by
|C ∩ D| = deg
C
(O
C
(D)) = h
0
(O
C∩D
) = χ(O
C∩D
),
using that
O
C∩D
is a skyscraper sheaf supported at
C ∩ D
. To understand this
number, we use the short exact sequences
0 O
C
(−D|
C
) O
C
O
C∩D
0
0 O
X
(−C − D) O
X
(−D) O
C
(−D|
C
) 0
0 O
X
(−C) O
X
O
C
0
This allows us to write
χ(O
C∩D
) = −χ(C, O
C
(−D|
C
)) + χ(C, O
C
)
= χ(O
X
(−C − D)) − χ(O
X
(−D)) + χ(C, O
C
)
= χ(O
X
(−C − D)) − χ(O
X
(−D)) − χ(O
X
(−C)) + χ(O
X
).
Thus suggests for for general divisors D
1
, D
2
, we can define
Definition
(Intersection product)
.
For divisors
D
1
, D
2
, we define the intersec-
tion product to be
D
1
· D
2
= χ(O
X
) + χ(−D
1
− D
2
) − χ(−D
1
) − χ(−D
2
).
Proposition.
(i) The product D
1
· D
2
depends only on the classes of D
1
, D
2
in Pic(X).
(ii) D
1
· D
2
= D
2
· D
1
.
(iii) D
1
· D
2
= |D
1
∩ D
2
| if D
1
and D
2
are curves intersecting transversely.
(iv) The intersection product is bilinear.
Proof. Only (iv) requires proof. First observe that if H is a very ample divisor
represented by a smooth curve, then we have
H · D = deg
H
(O
H
(D)),
and this is linear in D.
Next, check that
D
1
·
(
D
2
+
D
3
)
−D
1
·D
2
−D
1
·D
3
is symmetric in
D
1
, D
2
, D
3
.
So
(a)
Since this vanishes when
D
1
is very ample, it also vanishes if
D
2
or
D
3
is
very ample.
(b) Thus, if H is very ample, then D · (−H) = −(D · H).
(c) Thus, if H is very ample, then (−H) · D is linear in D.
(d)
If
D
is any divisor, write
D
=
H
1
− H
2
for
H
1
, H
2
very ample and smooth.
Then D · D
0
= H
1
· D
0
− H
2
· D
0
by (a), and thus is linear in D
0
.
In fact, bilinearity and (iii) shows that these properties uniquely characterize
the intersection product, since every divisor is a difference of smooth very ample
divisors.
Example.
Take
X
=
P
2
. We saw that
Pic
(
P
2
)
∼
=
Z
, generated by a hyperplane
`
. So all we need to understand is what
`
2
is. But two transverse lines intersect
at a point. So
`
2
= 1. Thus, the intersection product on
Pic
(
P
2
) is just ordinary
multiplication. In particular, if
D
1
, D
2
are curves that intersect transversely, we
find that
D
1
· D
2
= deg(D
1
) deg(D
2
).
This is B´ezout’s theorem!
Exercise. Let C
1
, C
2
⊆ X be curves without common components. Then
C
1
· C
2
=
X
p∈C
1
∩C
2
`(O
C
1
∩C
2
),
where for p ∈ C
1
∩ C
2
, if C
1
= V (x) and C
2
= V (y), then
`(O
C
1
∩C
2
) = dim
k
(O
X,p
/(x, y))
counts multiplicity.
Example. Take X = P
1
× P
1
. Then
Pic(X) = Z
2
= Z[p
∗
1
O(1)] ⊕ Z[p
∗
2
O(1)].
Let A = p
∗
1
O(1), B = p
∗
2
O(1). Then we see that
A
2
= B
2
= 0, A · B = 1
by high school geometry.
The Riemann–Roch theorem for curves lets us compute
χ
(
D
) for a divisor
D. We have an analogous theorem for surfaces:
Theorem (Riemann–Roch for surfaces). Let D ∈ Div(X). Then
χ(X, O
X
(D)) =
D · (D − K
X
)
2
+ χ(O
X
),
where K
X
is the canonical divisor.
To prove this, we need the adjunction formula:
Theorem
(Adjunction formula)
.
Let
X
be a smooth surface, and
C ⊆ X
a
smooth curve. Then
(O
X
(K
X
) ⊗ O
X
(C))|
C
∼
=
O
C
(K
C
).
Proof.
Let
I
C
=
O
X
(
−C
) be the ideal sheaf of
C
. We then have a short exact
sequence on C:
0 → O
X
(−C)|
C
∼
=
I
C
/I
2
C
→ Ω
1
X
|
C
→ Ω
1
C
→ 0,
where the left-hand map is given by d. To check this, note that locally on affine
charts, if
C
is cut out by the function
f
, then smoothness of
C
implies the kernel
of the second map is the span of df.
By definition of the canonical divisor, we have
O
X
(K
X
) = det(Ω
1
X
).
Restricting to C, we have
O
X
(K
X
)|
C
= det(Ω
1
X
|
C
) = det(O
X
(−C)|
C
) ⊗ det(Ω
1
C
) = O
X
(C)|
∨
C
⊗ O
C
(K
C
).
Proof of Riemann–Roch.
We can assume
D
=
H
1
− H
2
for very ample line
bundles
H
1
, H
2
, which are smoothly irreducible curves that intersect transversely.
We have short exact sequences
0 O
X
(H
1
− H
2
) O
X
(H
1
) O
H
2
(H
1
|
H
2
) 0
0 O
X
O
X
(H
1
) O
H
1
(H
1
) 0
where
O
H
1
(
H
1
) means the restriction of the line bundle
O
X
(
H
1
) to
H
1
. We can
then compute
χ(H
1
− H
2
) = χ(O
X
(H
1
)) − χ(H
2
, O
H
2
(H
1
|
H
2
))
= χ(O
X
) + χ(H
1
, O
H
1
(H
1
)) − χ(H
2
, O
H
2
(H
1
|
H
2
)).
The first term appears in our Riemann–Roch theorem, so we leave it alone. We
then use Riemann–Roch for curves to understand the remaining. It tells us
χ(H
i
, O
H
i
(H
1
)) = deg(O
H
i
(H
1
)) + 1 − g(H
i
) = (H
i
· H
1
) + 1 − g(H
i
).
By definition of genus, we have
2g(H
i
) − 2 = deg(K
H
i
).
and the adjunction formula lets us compute deg(K
H
i
) by
deg(K
H
i
) = H
i
· (K
X
+ H
i
).
Plugging these into the formula and rearranging gives the desired result.
Previously, we had the notion of linear equivalence of divisors. A coarser
notion of equivalence is that of numerical equivalence, which is about what the
intersection product can detect:
Definition
(Numerical equivalence)
.
We say divisors
D, D
0
are numerically
equivalent, written D ≡ D
0
, if
D · E = D
0
· E
for all divisors E.
We write
Num
0
= {D ∈ Div(X) : D ≡ 0}.
Definition (N´eron–Severi group). The N´eron–Severi group is
NS(X) = Div(X)/Num
0
(X).
We will need the following important fact:
Fact. NS(X) is a finitely-generated free module.
Note that
Pic
(
X
) is not be finitely-generated in general. For example, for
an elliptic curve,
Pic
(
X
) bijects with the closed points of
X
, which is in general
very large!
Definition (ρ(X)). ρ(X) = dim NS
R
(X) = rk NS(X).
Tensoring with R, the intersection product gives a map
( · , · ) : NS
R
(X) = NS(X) ⊗ R → R.
By definition of
NS
(
X
), this is non-degenerate. A natural question to ask is then,
what is the signature of this form? Since
X
is projective, there is a very ample
divisor
H
, and then
H
2
>
0. So this is definitely not negative definite. It turns
out there is only one positive component, and the signature is (1, ρ(X) − 1):
Theorem
(Hodge index theorem)
.
Let
X
be a projective surface, and
H
be a
(very) ample divisor on
X
. Let
D
be a divisor on
X
such that
D · H
= 0 but
D 6≡ 0. Then D
2
< 0.
Proof.
Before we begin the proof, observe that if
H
0
is very ample and
D
0
is (strictly) effective, then
H
0
· D
0
>
0, since this is given by the number of
intersections between
D
0
and any hyperplane in the projective embedding given
by H
0
.
Now assume for contradiction that D
2
≥ 0.
–
If
D
2
>
0, fix an
n
such that
H
n
=
D
+
nH
is very ample. Then
H
n
·D >
0
by assumption.
We use Riemann–Roch to learn that
χ(X, O
X
(mD)) =
m
2
D
2
− mK
X
· D
2
+ χ(O
X
).
We first consider the
H
2
(
O
X
(
mD
)) term in the left-hand side. By Serre
duality, we have
H
2
(O
X
(mD)) = H
0
(K
X
− mD).
Now observe that for large m, we have
H
n
· (K
X
− mD) < 0.
Since
H
n
is very ample, it cannot be the case that
K
X
− mD
is effective.
So H
0
(K
X
− D) = 0.
Thus, for m sufficiently large, we have
h
0
(mD) − h
1
(mD) > 0.
In particular,
mD
is (strictly) effective for
m
0. But then
H · mD >
0
since H is very ample and mD is effective. This is a contradiction.
–
If
D
2
= 0. Since
D
is not numerically trivial, there exists a divisor
E
on
X such that D · E 6= 0. We define
E
0
= (H
2
)E − (E · H)H.
It is then immediate that
E
0
·H
= 0. So
D
0
n
=
nD
+
E
0
satisfies
D
0
n
·H
= 0.
On the other hand,
(D
0
n
)
2
= (E
0
)
2
+ 2nD · E
0
> 0
for n large enough. This contradicts the previous part.