Algebraic surfaces
Minimal model program on surfaces
The previous result shows that a minimal model for surface always exists and that a surface X is minimal if and only if X has no(−1)-curve.Hence we may and we usually do assume that X is minimal. However,the criterion by(−1)curve only valid for surface.We might prefer to have a criterion which is also valid for higher dimension.
If X has a(−1)-curve E,then K X.E=−1<0.Thus if K X is nef then X has no(−1)-curve.Hence X is minimal.The minimal model program(or sometimes called Mori’s program can be describe as a program tofind minimal models.The important criterion is nefness of K X.Let’s start with a variety X,if K X is nef,then we have a minimal model and stop here.If K X is not nef,then there is a curve C such that C.K X<0.Then one can produce a morphism f:X→Y contracting C(and possibly contracting more curves at the same time). If dim Y<dim X,then the morphism has some special structure which is called Mori’sfibration.And the program stop here.
If dim X=dim Y and f contracts a divisor(we called it divisorial contraction).Thenρ(Y)<ρ(X):=rk(NS(X)).We replace X by Y and start the program over again.By looking atρ(X),it can’t be an i
nfinite loop,that is,the program must stop.
The remaining is the subtle one.If dim X=dim Y and f contracts a subvariety of codimension≥2.Then call it a small contraction.One needsflips to produce another birational model f :X →Y.However, there is no infinite quence offlips.Thus one must stop at somewhere f:˜X→˜Y which doesn’t allowflips.Thus replacing X by˜X then it must go to other cas.
For surface,we don’t need to worry about small contraction.There-fore,by running the minimal model program,the resulting products are surfaces with K X nef and Morifibration over a curve or a point. Theorem0.1.Let X be a minimal surface,then either K X is nef or X is a ruled surface or P2.In fact,X is P2whenρ(X)=1and X is ruled whenρ(X)≥2.
We need the following two highly non-trivial facts:
(1)If K X is not nef,then there exists a rational curve C∼=P1such
that C.K X<0.
(2)Fix an ample divisor H,there is a rational curve C with K X.C<
is maximal.
oxy0such that−K X.C
H.C
The point for thefirst fact is that if K X.C<0then by reduction to characteristic p,one es that the curve C can be deformed(in char p). Thus one has a morphism F:C×A1→X.The morphism extends to a rational map¯F:C×P1 .By rigidity lemma,one shows that
1
2
¯F can’t be a ust have point of indeterminacy.We then eliminate the indeterminacy by blowing-ups to get a morphism
˜F:Y→X.The exceptional curve E∼=P1then maps to a rational curve in X,we denote it by E.Moreover,C≡C +E,thus either E.K X<0or C .K X<0.If K X.E<0then we are done,otherwi, we replace C by C .With arithmetic genus p a(C )<p a(C),we must stop somewhere and get a rational curve.
The idea for proving the cond fact is more subtle,it’s basically the rationality theorem.
Before we get into the proof,we would like to define the arithmetic genus which will be uful in the quel.
Definition0.2.Let D be an effective divisor in a surface X,then we define the arithmetic genus
p a(D):=1
2
(D2+K X.D)+1.
intheafternoonNote that if D is a non-singular curve,then p a(D)=g(D).
Let C⊂X be a possibly singular curve.By blowing-up on X along singularities of C,one has proper transform C⊂ X→X which is non-singular.We leave it as an exerci to show that p a( C)≤p a(C)and< holds if C is singular.Nevertheless,p a( C)=g( C)≥0.Therefore we have:
malpracticeProposition0.3.Let C⊂X be a possibly singular curve.Then p a(C)≥0.
And if p a(C)=0,then C is non-singular and C∼=P1.nothing gonna change my love for you
proof of the theorem.Assume tho facts,we have a a rational curve
C with K X.C<0such that−K X.C
statueoflibertyH.C is maximal,where H is afixed
very ample divisor.Let q
p be the maximal value.Let D:=pK X+qH,supplement
it’s clear that D.C=0and D.C ≥0for any irreducible curve C .In particular,D is nef.
Remark.If D is a nef divisor on a surface,then D.C≥0for all curves and D2≥0.
Wefirst take care of the ca thatρ(X)≥2.
Claim1.h0(X,O X(mD))>0for m 0.
Grant the for the time being,we thenfix an m0 0such that |m0D|is free.We have a morphism
ϕm
0D
:X→ϕ(X)=:Y⊂P n.
Note that the restriction
H0(X,O(m0D))→H0(C,O(m0D|C)=O)∼=C,
3
gives constant functions.One concludes that the morphism ϕmaps C to a point.
Facts we need.Another fact we need is that Y is non-singular for m 0.Then by minimality of X ,dim Y <dim X .If ρ(X )≥2,then one can conclude that there is a curve C with D.C >0.
Claim 3.We may assume that the restriction
H 0(X,O (mD ))→H 0(C ,O (mD |C ))
is non-constant.
As a result,the restriction of ϕto C is not constant and so ϕis not constant.Hence dim Y ≥1.So dim Y =1.We may assume that ϕ:X →Y is a fisurjective with connected fibers.Moreover,a general fiber is a non-singular curve.
It remains to analyze the structure of ϕ.Especially,we wish to prove that the fiber is ∼=P 1.We need the famous
Zariski Lemma.Let π:X →B be a fibration
from a surface to a curve.Let F s = i n i C i be a fiber and D = i m i D i with m i ≥0for all i .Then D 2≤0.In particular,C 2i ≤0for all i .
Let’s look at the fibration ϕ:X →Y .We hope to prove that every fiber F s ∼=P 1.Let C 0⊂F s be an irreducible component.As we have en,the curve C 0contains in a fiber if and only if D.C 0=0.Hence
K X .C 0<0.Moreover,by Zariski Lemma,C 20≤0.By adjunction
formula,
−2≤2p a (C 0)−2=K X .C 0+C 0.C 0<0.
The only possibility is C 20=0,K X .C 0=−2since X has no (−1)-curve.
Let F s :=ϕ∗(s )= i n i C i be a fiber of ϕ.It’s clear that F 2s =0.And we have en that C 2i =0.It follows that 0=F 2s =2
i =jown
n i n j C i C j ≥0.Since F s is connected,if there are more than two components in F s ,then C i .C j >0for some i =j which is a contradiction.Therefore F s is say F s =n s C s .
For s =t ∈B ,
−2n s =F s .K X =F t .K X =−2n t .
It turns out that n s =n t for all s,t ∈B .However,for general fiber F is a non-singular curve.One has n s =1for all s .This completes the proof of the ca that ρ(X )≥2.
proof of the claims.In order to prove the claims,we need
Kodaira Vanishing Theorem.Let X be a non-singular projective variety over C .Let L be an ample divisor,then
charteredH i (X,O X (K X +L ))=0∀i >0.
4
To prove the Claim1,we consider
mD−K X=(mp−1)K X+mqH≡mp−1
p
D+(mq−
(mp−1)q
p
)H.
Since D is nef and H is ample.It’s clear that”nef+ample is ample”. Hence mD−K X is ample for all m>0.By Kodaira vanishing theorem, one has
χ(X,O(mD))=h0(X,O(mD)).
dajieBy Riemann-Roch,
h0(X,O(mD))=χ(X,O X)+1
2
(mD−K X).mD.
It suffices to prove that D.H>0for any ample divisor since mD−K X
is ample.
Suppo on the contrary that D.H=0,(recall that D.C >0for some C ,so D≡0.)By Hodge Index Theorem,D2<0.This contra-dicts to D being nef.(D is nef implies that D2≥0.)This completes
the proof of Claim1.
To prove the Claim2.We remark that the following conditions are equivalent.
(1)x is a ba point of|D|.
(2)Every ction of H0(X,O(D))vanishing at x.
(3)The evaluation map H0(X,O(D))→C(p)is zero.
(4)The natural mao H0(X,O(D)⊗I x)→H0(X,O(D))is an iso-
morphism.
(5)H1(X,O(D)⊗I x)=0
Where I x denotes the ideal sheaf of x and O(D)⊗I x is obtained by considering ctions in O(D)vanishing along x.Therefore,in order to prove the ba point freeness,it’s enough to prove that H1(X,O(mD)⊗I x)= 0.One might want to apply Kodaira vanishing theorem to prove
H1=0,however,it only works for divisor.Therefore,we consider
π:X =Bl x(X)→X.It’s not too difficult(but not trivial)to e
that
H1(X ,O(π∗mD−E))∼=H1(X,O(mD)⊗I x).
Consider now L m:=π∗mD−E−K X =π∗(mD−K X)−2E.We leave it as an exerci to show that L m is
ample for m 0.Then by Kodaira vanishing theorem,we are done.
To prove the last claim,it suffices to prove that|mD|parate two general points on C .To this end,wefirstfixed x∈C ⊂X.We con-sider the linear ries|mD⊗I x|which is a subries of|mD|consisting
of tho divisors passing through x.As long as dim|mD⊗I x|≥1then
mondialBs|mD⊗I x|isfinite.We pick any y∈Bs|mD⊗I x|.Therefore,a general member D ∈|mD⊗I x|passing through x but not y.Hence
the corresponding ction s∈H0(X,O(mD))has the property that
5 s(x)=0,s(y)=0.In particular,we have proved that Claim3.It follows thatϕ(x)=ϕ(y). The remaining ca is to show that a minimal surface withρ(X)=1 is P2.This might require some characterization of P2which we will prove later.
