Algebraic surfaces
Blowing-up and Blowing-down
Remark 0.1. The construction of blowing up can be found almost in any book.
(Some called it σ-process however). We refer [Beauville, complex algebraic surfaces, chap. II]. However, Beauville only proved that the map h is a bijective morphism.
It would be a good exercise to prove that h indeed an isomorphism.
In this section, we introduce the important notion of blowing-up. This process is essential in studying singularities and hence birational geometry in general.
We first introduce the local version. Let Anbe the affine space with coordinates z0, ..., zn−1 and 0 ∈ An be the ”origin”. We construct a variety Y ⊂ An× Pn−1 by {ziXj = zjXi}i6=j, where X0, ..., Xn are the homogeneous coordinates of Pn−1. There is a natural morphism π : Y → An by projection. One sees that π−1(0) ∼= Pn−1and π : Y − π−1(0) ∼= An− {0}. We say Y is the blowing-up of An at 0 and denoted Bl0(Y ).
In general, let x ∈ X be a point in a variety X. Pick an open affine neighborhood U of x. We identify (U, x) with an open set (U0, 0) ⊂ An. Then one has eU :=
π−1(U0) → U0 which is the blowing-up of U0 at 0. Glue X − U and eU together, we get πX : eX → X. Which is called the blowing-up of X at x. Note that one has similarly that πX−1(x) ∼= Pn−1and πX : eX − πX−1(x) ∼= X − {x}. The divisor π−1X (x) is called the exceptional divisor, and usually denoted E.
Exercise 0.2. Let π : X = Blx(P2) → P2be the blowing-up of P2at a point x ∈ P2. Prove that
KX = π∗KP2+ E by local coordinate computation.
In fact, if dimX = 2, π : eX → X is a blowing-up at a point x ∈ X, then KX˜ = π∗KX+ E.
More generally, if dimX = n, and π : eX = Blx(X) → X is the blowing-up at x, then
KX˜ = π∗KX+ (n − 1)E.
Let’s play a little bit around the blowing-ups. Let’s restrict ourselves to surfaces.
One might expect that there are similar higher-dimensional formulation. Let X be a surface, and C ⊂ X be a curve. Let f be the local equation of C around x. By fixing local coordinates z1, z2, we can write
f = f (z1, z2) = fm+ fm+1+ ...
with fm6= 0. We define the multiplicity of C at x to be mx(C) := m.
One can have an equivalent definition by vanishing order of partial differentials.
Hence one can check the mx(C) is well-defined.
We consider π : eX = Blx(X) → X. And let C be a curve passing through x ∈ X. Then π−1(C) consists of irreducible components, E and the other part maps onto C. The part maps onto C can be defined as
C := πe −1(C − {x}),
which is called the proper transform of C. Thus we have π−1(C) = eC ∪ E. More precisely, by computing the equations, one has
π∗C = eC + mx(C)E,
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this is called the total transform of C.
We here collect some properties regarding the blowing-up on surface.
Proposition 0.3. Let π : eX = Blx(X) → X be the blowing-up at x ∈ X. Then one has:
(1) There is a natural isomorphism Div(X) ⊕ ZE → Div( eX) by (D, nE) 7→
π∗D + nE. And the isomorphism induces an isomorphism PicX ⊕ ZE → Pic eX.
(2) Let D, D0 ∈ Div(X), then (π∗D).(π∗D0) = D.D0. (3) Let D ∈ Div(X), then (π∗D).E = 0.
(4) E.E = −1.
Proof. It’s easy to check the isomorphism given in (1).
For (2) and (3), it follows by choosing ∆, ∆0 which are linear equivalent to D, D0 respectively but not passing through x.
For (4), by adjunction formula and the fact the E ∼= P1,
−2 = deg(KE) = (KX˜+ E).E = (π∗KX+ 2E).E = 2E.E.
¤ The blowing-up gives the first example of binational morphism.
Definition 0.4. By a rational map f : X 99K Y from X to Y , we mean a regular function on a dense Zariski open (or simply non-empty Zariski open) set U ⊂ X.
More precisely, a rational map can be written as (U, f ) where U ⊂ X is a dense Zariski-open set and f : U → Y is regular.
We say (U, f ) ∼ (V, g) if f = g on U ∩ V . In fact, a precise definition of rational map should be the equivalent class of the pairs (U, f ). However, we usually abuse the notation if no confusion is likely.
Definition 0.5. A rational map φ : X 99K Y is said to be birational if it admits an inverse. That is, there is an ψ : Y 99K X such that ψ ◦ φ = idX, φ ◦ ψ = idY
Example 0.6. Let π : Y = Bl0(A2) → A2, take ψ : A2− {0} → Y ⊂ A2× P1 such that ψ(x, y) = ((x, y), [x, y]). Then ψ ◦ π = idY, π ◦ ψ = idX. Hence π is a birational morphism.
Exercise 0.7. The following are equivalent:
(1) X and Y are birationally equivalent.
(2) there are non-empty open subset U ⊂ X and V ⊂ Y such that U, V are isomorphic.
(3) K(X) ∼= K(Y ) as k-algebra.
Given a variety X, one can obtain various birational equivalent varieties ... → Xn → Xn−1→ ... → X1→ X
by successive blowing-ups. It’s also a natural question to ask if X is obtained by blowing-ups? Another way to put it is if X minimal or not? The precise formulation of minimal model in any dimension is quite subtle.
We start by working on contraction on surfaces. In order to produce a minimal object, we need to tell whether a surface X is obtained from blowing-ups.
Definition 0.8. Let C ⊂ X be a curve on X, we say that C is a (−1)-curve if C ∼= P1 and C2= −1
We seen that we can have a (−1)-curve by blowing-up. In fact we will prove that any (−1)-curve comes from blowing-ups.
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Theorem 0.9 (Caltelnuovo). Let X be a surface (non-singular complex projective surface) with E ⊂ X a (−1)-curve. Then there is a morphism π : X → X0 with X0 non-singular such that π is the blowing-up of X0 with exceptional divisor E.
Proof. The idea is to construct a morphism which is identical at E but isomorphic outside E.
First pick H0 any very ample divisor on X, Let k := H.E > 0. We consider H0= H + kE, then H0.E = 0. Notice that the restriction map
H0(X, O(H0)) → H0(E, O(H0|E) = OE) ∼= H0(P1, O) ∼= C.
Hence the map ϕH0 produce by |H0| is constant on E. We need to refine H so that ϕH0 is isomorphic outside E.
To this end, we first pick any very ample H0. It’s clear that nH0 is very ample for all n > 0. On the other hand, H0 is ample, one can arrange that H := nH0 is very ample with H1(X, O(H)) = 0.
Consider the exact sequence
0 → OX(H + (i − 1)E) → OX(H + iE) → OE(H + iE|E) = OE(k − i) → 0.
Claim. H1(X, O(H + iE)) = 0 for all 1 ≤ i ≤ k.
Grant this for the time being, then one has an exact sequence
0 → H0(X, OX(H + (i − 1)E)) → H0(X, OX(H + iE)) → H0(E, OE(k − i)) → 0.
Note that H0(E, OE(k − i)) is of dimension k − i + 1, let ai,0, ..., ai,k−i ∈ H0(X, O(H + iE)) be the lifting of a basis in H0(E, O(k − i)).
Remark. Before we move on, we would like to remark the difference between H0(X, O(D)) and L(D). It actually comes from two possible definition of O(D).
If we define the sheaf O(D) as O(D)(U ) = {f ∈ K(X)|div(f ) + D|U ≥ 0 on U }.
Then H0(X, O(D)) = L(D). However, another way to look at the sheaf O(D) is to consider it as the sheaf of sections line bundle associate to D. Then under this consideration, for s ∈ H0(X, O(D)), div(s) gives an effective divisor Ds linearly equivalent to D. To view it as L(D) is the classical treatment. The Modern viewpoint tends to think it as section of line bundles. We take the convention that H0(X, O(D)) represents the global section of line bundle of D from now on.
Let me describe the correspondence in more detail. Given a divisor D, one has a system of local equations (Uifi). The basic idea behind the notion of line bundle is instead of looking at functions, we look at local functions satisfying given patching conditions. The correspondence is given as
L(D) → H0(X, O(D)), f 7→ (Ui, f fi) = s.
And the correspondence between their divisor is given by div(s) = div(f ) + D, which is an effective divisor Ds∈ |D|.
Turning back to the proof, let s ∈ H0(X, O(E)) be a section such that div(s) = E. Then the map H0(X, OX(H + (i − 1)E)) → H0(X, OX(H + iE)) is given by multiplying s. Therefore, by working on the sequence inductively, one can have a basis of H0(X, O(H + kE)), given as
{s0sk, ..., snsk, a1,0sk−1, ..., a1,k−1sk−1, ..., ak−1,0s, ak−1,1s, ak}.
We consider the map ϕH0 : X → PN given by the above basis. Note that ak ∈ H0(X, O(H0)) whose restriction to E is a non-zero constant. Hence one has
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ϕ is well-defined along E and ϕ(E) = [0, ..., ak] = [0, ..., 1]. Moreover, for x 6∈ E, s(x) 6= 0, hence
[s0sk(x), ..., snsk(x)] = [s0(x), ..., sn(x)] = ϕH(x).
Since H is very ample, ϕH defines an embedding on X and hence on X − E. One sees that the first n + 1 coordinate of ϕH0 gives an embedding on X − E already, so it follows that ϕH0 gives an embedding on X − E.
It remains to show that X0 := ϕH0(X) is non-singular. Let U ⊂ X be the open subset defined by ak 6= 0. It’s clear that E ⊂ U . We want to identify U with an open set V ⊂ fA2⊂ A2× P1. This can be achieved by considering
h : U → A2× P1, x 7→¡
(ak−1,0s
ak (x),ak−1,1s
ak (x)), [ak−1,0(x), ak−1,1(x)]¢ .
We might need to shrink U so that ak−1,0(x) and ak−1,1(x) are not simultaneously vanishing. It’s obvious that h factor through fA2. Let V = h(U ) ⊂ fA2. Moreover, one has the commutative diagram
U −−−−→ fh A2
ϕH0
y π
y ϕH0(U ) −−−−→ Ah¯ 2, where ¯h = (ak−1,0a s
k ,ak−1,1a s
k ) is a rational map on PN defined on ϕH0(U ). Another remark is that h clearly map E ⊂ U onto E ⊂ fA2. It suffices to show that h : U → V is an isomorphism. Because, the induced map ¯h is an isomorphism . Therefore, ϕH0(U ) is non-singular at ϕH0(E), which is the only possible singularity.
However, to show that h is an isomorphism is not trivial. One can first prove that it’s a hemeomorphism, hence in particular, bijective. Then one prove the h induces isomorphism on all local rings. (cf. [Ha. Ex I.3.2, I.3.3]) ¤
