2Surfaces
III Positivity in Algebraic Geometry
2.2 Blow ups
Let
X
be a smooth surface, and
p ∈ X
. There is then a standard way to “replace”
p
with a copy of
P
1
. The result is a surface
¯
X
=
Bl
p
X
together with a map
ε
:
¯
X → X
such that
ε
−1
(
p
) is a smoothly embedded copy of
P
1
, and
ε
is an
isomorphism outside of this P
1
.
To construct
¯
X
, pick local coordinates
x, y
in a neighbourhood
U
of
¯
X
such
that
V
(
x, y
) =
p
. Work on
U × P
1
with coordinates [
X
:
Y
] on
P
1
. We then set
¯
U = V (xY − yX).
There is a canonical projection of
¯
U
onto
U
, and this works, since above the
point (0
,
0), any choice of
X, Y
satisfies the equation, and everywhere else, we
must have [X : Y ] = [x : y].
Definition
(Blow up)
.
The blow up of
X
at
p
is the variety
¯
X
=
Bl
p
X
that
we just defined. The preimage of
p
is known as the exceptional curve, and is
denoted E.
We would like to understand
Pic
(
¯
X
) in terms of
Pic
(
X
), and in particular
the product structure on
Pic
(
¯
X
). This will allow us to detect whether a surface
is a blow up.
The first step to understanding Pic(
¯
X) is the localization sequence
Z → Pic(
¯
X) → Pic(
¯
X \ P
1
) → 0.
We also have the isomorphism
Pic(
¯
X \ P
1
) = Pic(X \ {p})
∼
=
Pic(X).
Finally, pulling back along
ε
gives a map
Pic
(
X
)
→ Pic
(
¯
X
), and the push-pull
formula says
π
∗
(π
∗
L) = L ⊗ π
∗
(O
¯
X
).
Hartog’s lemma says
π
∗
O
¯
X
= O
X
.
So in fact
ε
∗
is a splitting. So
Pic
(
¯
X
) is the direct sum of
Pic
(
X
) with a quotient
of
Z
, generated by [
E
]. We will show that in fact
Pic
(
¯
X
)
∼
=
ε
∗
Pic
(
X
)
⊕ Z
[
E
].
This will be shown by calculating E
2
= −1.
In general, if C is a curve in X, then this lifts to a curve
˜
C ⊆
¯
X, called the
strict transform of C. This is given by
˜
C = ε
−1
(C \ {p}).
There is also another operation, we can do, which is the pullback of divisors
π
∗
C.
Lemma.
π
∗
C =
˜
C + mE,
where m is the multiplicity of C at p.
Proof.
Choose local coordinates
x, y
at
p
and suppose the curve
y
= 0 is not
tangent to any branch of
C
at
p
. Then in the local ring
ˆ
O
X,p
, the equation of
C
is given as
f = f
m
(x, y) + higher order terms,
where
f
m
is a non-zero degree
m
polynomial. Then by definition, the multiplicity
is
m
. Then on
¯
U ⊆
(
U × P
1
), we have the chart
U × A
1
where
X 6
= 0, with
coordinates
(x, y, Y/X = t).
Taking (
x, t
) as local coordinates, the map to
U
is given by (
x, t
)
7→
(
x, xt
).
Then
ε
∗
f = f(x, tx) = x
m
f
m
(1, t) + higher order terms = x
m
· h(x, t),
with
h
(0
,
0)
6
= 0. But then on
¯
U
x6=0
, the curve
x
= 0 is just
E
. So this equation
has multiplicity m along E.
Proposition.
Let
X
be a smooth projective surface, and
x ∈ X
. Let
¯
X
=
Bl
x
X
π
→ X. Then
(i) π
∗
Pic(X) ⊕ Z[E] = Pic(
¯
X)
(ii) π
∗
D · π
∗
F = D · F , π
∗
D · E = 0, E
2
= −1.
(iii) K
¯
X
= π
∗
(K
X
) + E.
(iv) π
∗
is defined on NS(X). Thus,
NS(
¯
X) = NS(X) ⊕ Z[E].
Proof.
(i) Recall we had the localization sequence
Z → Pic(
¯
X) → Pic(X) → 1
and there is a right splitting. To show the result, we need to show the
left-hand map is injective, or equivalently,
mE 6∼
0. This will follow from
the fact that E
2
= −1.
(ii)
For the first part, it suffices to show that
π
∗
D · π
∗
F
=
D · F
for
D, F
very ample, and so we may assume
D, F
are curves not passing through
x
.
Then their pullbacks is just the pullback of the curve, and so the result is
clear. The second part is also clear.
Finally, pick a smooth curve
C
passing through
E
with multiplicity 1.
Then
π
∗
C =
˜
C + E.
Then we have
0 = π
∗
C · E =
˜
C · E + E
2
,
But
˜
C and E intersect at exactly one point. So
˜
C · E = 1.
(iii) By the adjunction formula, we have
(K
¯
X
+ E) · E = deg(K
P
1
) = −2.
So we know that
K
¯
X
· E
=
−
1. Also, outside of
E
,
K
¯
X
and
K
X
agree. So
we have
K
¯
X
= π
∗
(K
X
) + mE
for some m ∈ Z. Then computing
−1 = K
¯
X
· E = π
∗
(K
X
) · E + mE
2
gives m = 1.
(iv)
We need to show that if
D ∈ Num
0
(
X
), then
π
∗
(
D
)
∈ Num
0
(
¯
X
). But this
is clear for both things that are pulled back and things that are
E
. So we
are done.
One thing blow ups are good for is resolving singularities.
Theorem
(Elimination of indeterminacy)
.
Let
X
be a sooth projective surface,
Y
a projective variety, and
ϕ
:
X → Y
a rational map. Then there exists a
smooth projective surface X
0
and a commutative diagram
X
0
X Y
p
q
ϕ
where p : X
0
→ X is a composition of blow ups, and in particular is birational.
Proof.
We may assume
Y
=
P
n
, and
X 99K P
n
is non-degenerate. Then
ϕ
is
induced by a linear system
|V | ⊆ H
0
(
X, L
). We first show that we may assume
the base locus has codimension > 1.
If not, every element of
|V |
is of the form
C
+
D
0
for some fixed
C
. Let
|V
0
|
be the set of all such
D
0
. Then
|V |
and
|V
0
|
give the same rational map. Indeed,
if
V
has basis
f
0
, . . . , f
n
, and
g
is the function defining
C
, then the two maps
are, respectively,
[f
0
: · · · : f
n
] and [f
0
/g : · · · : f
n
/g],
which define the same rational maps. By repeating this process, replacing
V
with V
0
, we may assume the base locus has codimension > 1.
We now want to show that by blowing up, we may remove all points from
the base locus. We pick
x ∈ X
in the base locus of
|D|
. Blow it up to get
X
1
= Bl
x
X → X. Then
π
∗
|D| = |D
1
| + mE,
where
m >
0, and
|D
1
|
is a linear system which induces the map
ϕ ◦ π
1
. If
|D
1
|
has no basepoints, then we are done. If not, we iterate this procedure. The
procedure stops, because
0 ≤ D
2
1
= (π
∗
1
D
2
+ m
2
E
2
) < D
2
.
Exercise.
We have the following universal property of blow ups: If
Z → X
is
a birational map of surfaces, and suppose
f
−1
is not defined at a point
x ∈ X
.
Then f factors uniquely through the blow up Bl
x
(X).
Theorem.
Let
g
:
Z → X
be a birational morphism of surfaces. Then
g
factors
as
Z
g
0
→ X
0
p
→ X
, where
p
:
X
0
→ X
is a composition of blow ups, and
g
0
is an
isomorphism.
Proof.
Apply elimination of indeterminacy to the rational inverse to
g
to obtain
p
:
X
0
→ X
. There is then a lift of
g
0
to
X
0
by the universal property, and these
are inverses to each otehr.
Birational isomorphism is an equivalence relation. So we can partition the
set of smooth projective surfaces into birational isomorphism classes. If
X, X
0
are in the same birational isomorphism class, we can say
X ≤ X
0
if
X
0
is a blow
up of
X
. What we have seen is that any two elements have an upper bound.
There is no global maximum element, since we can keep blowing up. However,
we can look for minima elements.
Definition
(Relatively minimal)
.
We say
X
is relatively minimal if there does
not exist a smooth
X
0
and a birational morphism
X → X
0
that is not an
isomorphism. We say
X
is minimal if it is the unique relative minimal surface
in the birational equivalence class.
If
X
is not relatively minimal, then it is the blow up of another smooth
projective surface, and in this case the exceptional curve
C
satisfies
C
2
=
−
1. It
turns out this criterion itself is enough to determine minimality.
Theorem
(Castelnuovo’s contractibility criterion)
.
Let
X
be a smooth projective
surface over
K
=
¯
K
. If there is a curve
C ⊆ X
such that
C
∼
=
P
1
and
C
2
=
−
1,
then there exists
f
:
X → Y
that exhibits
X
as a blow up of
Y
at a point with
exceptional curve C.
Exercise.
Let
X
be a smooth projective surface. Then any two of the following
three conditions imply the third:
(i) C
∼
=
P
1
(ii) C
2
= −1
(iii) K
X
· C = −1
When these are satisfied, we say C is a (−1) curve.
For example, the last follows from the first two by the adjunction formula.
Corollary.
A smooth projective surface is relatively minimal if and only if it
does not contain a (−1) curve.
Proof.
The idea is to produce a suitable linear system
|L|
, giving a map
f
:
X → P
n
, and then take
Y
to be the image. Note that
f
(
C
) =
∗
is the same as
requiring
O
X
(L)|
C
= O
C
.
After finding a linear system that satisfies this, we will do some work to show
that f is an isomorphism outside of C and has smooth image.
Let
H
be a very ample divisor on
X
. By Serre’s theorem, we may assume
H
1
(X, O
X
(H)) = 0. Let
H · C = K > 0.
Consider divisors of the form H + iC. We can compute
deg
C
(H + iC)|
C
= (H + iC) · C = K − i.
Thus, we know deg
C
(H + KC)|
C
= 0, and hence
O
X
(H + KC)|
C
∼
=
O
C
.
We claim that
O
X
(
H
+
KC
) works. To check this, we pick a fairly explicit
basis of H
0
(H + KC). Consider the short exact sequence
0 → O
X
(H + (i − 1)C) → O
X
(H + iC) → O
C
(H + iC) → 0,
inducing a long exact sequence
0 → H
0
(H + (i − 1)C) → H
0
(H + iC) → H
0
(C, (H + iC)|
C
)
→ H
1
(H + (i − 1)C) → H
1
(H + iC) → H
1
(C, (H + iC)|
C
) → · · ·
We know
O
X
(
H
+
iC
)
|
C
=
O
P
1
(
K − i
). So in the range
i
= 0
, . . . , K
, we have
H
1
(C, (H + iC)|
C
) = 0. Thus, by induction, we find that
H
1
(H + iC) = 0 for i = 0, . . . , K.
As a consequence of this, we know that for
i
= 1
, . . . , K
, we have a short
exact sequence
H
0
(H + (i − 1)C) → H
0
(H + iC) H
0
(C, (H + iC)|
C
) = H
0
(O
P
1
(K − i)).
Thus,
H
0
(
H
+
iC
) is spanned by the image of
H
0
(
H
+ (
i −
1)
C
) plus a lift of a
basis of H
0
(O
P
1
(K − i)).
For each
i >
0, pick a lift
y
(i)
0
, . . . , y
(i)
K−i
of a basis of
H
0
(
O
P
1
(
K − i
)) to
H
0
(
H
+
iC
), and take the image in
H
0
(
H
+
KC
). Note that in local coordinates,
if C is cut out by a function g, then the image in H
0
(H + KC) is given by
g
K−i
y
(i)
0
, . . . , g
K−i
y
(i)
K−i
.
For
i
= 0, we pick a basis of
H
0
(
H
), and then map it down to
H
0
(
H
+
KC
).
Taking the union over all
i
of these elements, we obtain a basis of
H
0
(
H
+
KC
).
Let f be the induced map to P
r
.
For concreteness, list the basis vectors as
x
1
, . . . , x
r
, where
x
r
is a lift of
1 ∈ O
P
1
to H
0
(H + KC), and x
r−1
, x
r−2
are gy
(K−1)
0
, gy
(K−1)
1
.
First observe that x
1
, . . . , x
r−1
vanish at C. So
f(C) = [0 : · · · : 0 : 1] ≡ p,
and C is indeed contracted to a point.
Outside of
C
, the function
g
is invertible, so just the image of
H
0
(
H
) is
enough to separate points and tangent vectors. So
f
is an isomorphism outside
of C.
All the other basis elements of
H
+
KC
are needed to make sure the image is
smooth, and we only have to check this at the point
p
. This is done by showing
that
dim m
Y,p
/m
2
Y,p
≤
2, which requires some “infinitesimal analysis” we will not
perform.
Example.
Consider
X
=
P
2
, and
x, y, z
three non-colinear points in
P
2
given
by [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1].
There are three conics passing through
x, y, z
, given by
x
1
x
2
,
x
0
, x
2
and
x
0
x
1
, which is the union of lines
0
,
1
,
2
, and let
|C
x,y,z
|
be the linear system
generated by these three conics. Then we obtain a birational map
P
2
99K P
2
[x
0
: x
1
: x
2
] 7→ [x
1
x
2
: x
0
x
2
: x
0
x
1
] = [x
−1
0
: x
−1
1
: x
−1
2
]
with base locus {x, y, z}.
Blowing up at the three points
x, y, z
resolves our indeterminacies, and we
obtain a diagram
Bl
x,y,z
P
2
P
2
P
2
C
Then in the blowup, we have
π
∗
i
=
˜
i
+ E
j
+ E
k
for
i, j, k
distinct. We then check that
˜
2
i
=
−
1, and each
˜
i
is isomorphic
to
P
1
. Thus, Castelnuovo tells us we can blow these down. This blow down
map is exactly the right-hand map in the diagram above, which one can check
compresses each line
˜
i
to a point.