We now prove the Cone Theorem (Theorem 4): if X is a smooth projective variety,
NE(X) = NE(X)KX≥0+X
i∈I
R+[Γi]
where Γi are rational curves on X such that 0<−KX ·Γi ≤dim(X) + 1.
The idea is quite simple: if NE(X) is not equal to the closure of the right-hand side, there exists a divisor M on X which is nonnegative on NE(X) (hence nef), positive on the closure of the right-hand side, and vanishes at some nonzero point z of NE(X), which must therefore satisfy KX ·z < 0.
We approximate M by an ample divisor, z by an effective 1-cycle and use the bend-and-break Theorem 19 to get a contradiction. In the third and last step, we prove that the right-hand side is closed by a formal argument with no geometric content.
Proof. There are only countably many families of, hence classes of, rational curves on X. Pick a representative Γi for each such class zi that satisfies 0<−KX ·zi ≤dim(X) + 1.
First step: the raysR+zi are locally discrete in the half-space(N1(X)R)KX<0. LetHbe an ample divisor onX. It is enough to show that for eachε >0, there are only finitely many classes zi in the half-space (N1(X)R)KX+εH<0, since the union of these half-spaces is (N1(X)R)KX<0. If (KX+εH)·Γi <0, we have
H·Γi <−1
εKX ·Γi ≤ 1
ε(dim(X) + 1) and there are finitely many such classes of curves on X.13 Second step: NE(X) is equal to the closure of
V = NE(X)KX≥0+X
i
R+zi
13For any ample divisorH and any integerk, the set{z∈NE(X)|H·z≤k}contains only finitely many classes of irreducible curves. Indeed, letD1, . . . , Drbe Cartier divisors on X such that ([D1], . . . ,[Dr]) is a basis for N1(X)Q. There exists an integer m such that mH±Di is ample for each i in {1, . . . , r}. For any z in NE(X), we then have (mH±Di)·z≥0 hence|Di·z| ≤mH·z. IfH·z≤k, this bounds the coordinates ofz and defines a compact set. It contains at most finitely many classes of irreducible curves, because the set of this classes is by construction discrete inN1(X)R.
If this is not the case, there exists by Lemma 21.d) (since NE(X) contains no lines) an R-divisor M on X which is nonnegative on NE(X) (it is in particular nef), positive onV {0}and which vanishes at some nonzero point z of NE(X). This point cannot be inV, henceKX ·z <0.
Choose a norm onN1(X)R such thatk[C]k ≥1 for each irreducible curve C (this is possible since the set of classes of irreducible curves is discrete).
We may assume, upon replacingM with a multiple, thatM·v ≥2kvkfor all v inV. Since the class [M] is a limit of classes of ample Q-divisors, and z is a limit of classes of effective rational 1-cycles, there exist an ample Q-divisor H and an effective 1-cycleZ such that
2 dim(X)(H·Z)<−KX ·Z and H·v ≥ kvk (5) for all v in V. We may further assume, by throwing away the other compo-nents, that each component C of Z satisfies−KX ·C > 0.
Since X is smooth, the bend-and-break Theorem 19 implies that, for every component C of Z, there exists a rational curve Γ on X such that
2 dim(X) H·C
−KX ·C ≥H·Γ
and −KX ·Γ≤dim(X) + 1. But [Γ] is then in V, henceH·[Γ]≥ k[Γ]k ≥1 by (5). This contradicts the first inequality in (5) and finishes the proof of the second step.
Third step: for any set J of indices, the cone VJ = NE(X)KX≥0+X
j∈J
R+zj
is closed.
By Lemma 21.b), it is enough to show that any extremal rayR+r in VJ
satisfying KX·r <0 is inVJ. LetH be an ample divisor onX and letεbe a positive number such that (KX+εH)·r <0. By the first step, there are only finitely many classes zj1, . . . , zjq, withjα ∈J, such that (KX+εH)·zjα <0.
Writer as the limit of a sequence (rm+sm), whererm ∈NE(X)KX+εH≥0
and sm =Pq
α=1λα,mzjα. SinceH·rm and H·zjα are positive, the sequences (H·rm) and (λα,m) are bounded, hence we may assume, after taking subse-quences, that all sequences (rm) and (λα,m) have limits.14 Because r spans
14See footnote 13.
an extremal ray in VJ, the limits must be nonnegative multiples of r, and since (KX +εH)·r < 0, the limit of (rm) must vanish. Moreover, r is a multiple of one the zjα, hence is in VJ.
If we choose a setI of indices such that (R+zj)j∈I is the set of all (distinct) extremal rays among all R+zi, the proof shows that any extremal ray of NE(X)KX<0 is spanned by a zi, with i ∈ I. This finishes the proof of the
Cone Theorem.
Corollary 22 Let X be a smooth projective variety and let R be a KX -negative extremal ray. There exists a nef divisor MR on X such that
a) R ={z ∈NE(X)|MR·z = 0};
b) the divisor mMR−KX is ample for all integers m 0.
The divisor MR will be called a supporting divisor for R. Property b) is useful in conjunction with Theorem 24: it implies that in characteristic zero, the linear system|mMR|is base-point-free for all integersmsufficiently large, and defines the contraction of R.
Proof. With the notation of the proof of the Cone Theorem, there exists i0 in I such thatR =R+zi0. By the third step of the proof, the subcone
V =VI {i0} = NE(X)KX≥0+ X
i∈I, i6=i0
R+zi
of NE(X) is closed and proper since it does not contain R. By Lemma 21.d), there exists a linear form which is nonnegative on NE(X), positive onV {0}
and which vanishes at some nonzero point of NE(X), hence on R since NE(X) = V + R. The intersection of the interior of V∗ and the ratio-nal hyperplane R⊥ is therefore nonempty, hence contains an integral point:
there exists a divisor MR on X which is positive on V {0} and vanishes on R. It is in particular nef and a) holds.
Choose a norm onN1(X)Rand letabe the (positive) minimum ofMR on the set of elements ofV with norm 1. Ifbis the maximum ofKX on the same compact, the divisor mMR−KX is positive onV {0}form rational greater thanb/a, and positive onR {0}form≥0, hence ample form >max(b/a,0)
by Kleiman’s criterion. This proves b).
10 The Cone Theorem for projective klt pairs
Let (X,∆) be a complex projective klt pair.15 The Cone Theorem takes the following form: the set R of all (KX + ∆)-negative extremal rays of NE(X) is countable and
NE(X) = NE(X)KX+∆≥0+X
R∈R
R
These rays are locally discrete in the half-space (N1(X)R)KX+∆<0. For each R ∈R, there exists a rational curve Γ in X such that
R =R+[Γ] 0<−(KX + ∆)·Γ≤2 dim(X)
As the following exercise shows, even when X is smooth, this theorem has applications that do not follow from Theorem 4. However, the method of proof is totally different, since most bend-and-break results break down on singular varieties.
Exercise 1 LetX be a smooth projective variety of general type.16
1) Prove that there exist an effective divisorDsuch thatKX−Dis ample and ε ∈[0,1) such that the pair (X, εD) is klt.
2) Prove that there are only finitely many KX-negative extremal rays in NE(X).
3) IfX contains no rational curves, prove thatKX is ample. (Hint: show that KX is nef and that, if it is not ample, X contains a curve C and an effective divisor D such that KX ·C = 0 (use Theorem 24).)
15For us, this means the following: X is a projective normal variety, ∆ is aQ-divisor on X with coefficients in [0,1) such that some integral mutliple of KX+ ∆ is a Cartier divisor, and for some (hence all) log resolution π : Y →X (i.e., a projective birational morphism whereY is smooth and Exc(π) +π∗∆ is a divisor with simple normal crossings), if one writes
KY + ∆0 ∼π∗(KX+ ∆) +E
where ∆0is the strict tranform of ∆ andE isπ-exceptional, then the coefficients ofEare all>−1.
16This means thatKX is big.
10.1 The Rationality Theorem
This is the main result that is needed for the proof of the Cone Theorem in the singular case. We will not prove it here. It can be “deduced” from the Base-Point-Free Theorem 24 and the Kawamata–Viehweg Vanishing Theorem (but this is not easy; see [D], §7.8).
Theorem 23 (Rationality Theorem) Let (X,∆) be a complex projective klt pair such that the Q-Cartier Q-divisor KX + ∆ is not nef. Let H be an ample Cartier divisor on X.
The number r= sup{t∈R+ |H+t(KX+ ∆) nef}is rationaland can be written asr = juv , wherej is the smallest positive integer such thatj(KX+∆) is a Cartier divisor, anduandv are positive integers withv ≤j(dim(X)+1).
Exercise 2 Deduce the Rationality Theorem from the Cone Theorem for klt pairs, assuming only that H is nef.