Exceptional del Pezzo hypersurfaces
aa r X i v : . [ m a t h . AG ] A p r EXCEPTIONAL DEL PEZZO HYPERSURFACES
IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
Abstract.
We compute global log canonical thresholds of a large class of quasismooth well-formed del Pezzo weighted hypersurfaces in P ( a , a , a , a ). As a corollary we obtain the exis-tence of orbifold K¨ahler–Einstein metrics on many of them, and classify exceptional and weaklyexceptional quasismooth well-formed del Pezzo weighted hypersurfaces in P ( a , a , a , a ). Contents
Part 1. Introduction
Part 2. Infinite series I = 1 162.2. Infinite series with I = 2 172.3. Infinite series with I = 4 242.4. Infinite series with I = 6 28 Part 3. Sporadic cases I = 1 403.2. Sporadic cases with I = 2 573.3. Sporadic cases with I = 3 853.4. Sporadic cases with I = 4 1003.5. Sporadic cases with I = 5 1193.6. Sporadic cases with I = 6 1263.7. Sporadic cases with I = 7 1283.8. Sporadic cases with I = 8 1293.9. Sporadic cases with I = 9 1313.10. Sporadic cases with I = 10 132 Part 4. The Big Table
Part Introduction
All varieties are always assumed to be complex, algebraic, projective and normal unless oth-erwise stated. 1.1.
Background
The multiplicity of a nonzero polynomial f ∈ C [ z , . . . , z n ] at a point P ∈ C n is the non-negative integer m such that f ∈ m mP \ m m +1 P , where m P is the maximal ideal of polynomialsvanishing at the point P in C [ z , . . . , z n ]. It can be also defined by derivatives: the multiplicityof f at the point P is the numbermult P ( f ) = min (cid:26) m (cid:12)(cid:12)(cid:12) ∂ m f∂ m z ∂ m z . . . ∂ m n z n ( P ) = 0 (cid:27) . On the other hand, we have a similar invariant that is defined by integrations. This invariant,which is called the complex singularity exponent of f at the point P , is given by c P ( f ) = sup n c (cid:12)(cid:12)(cid:12) | f | − c is locally L near the point P ∈ C n o . In algebraic geometry this invariant is usually called a log canonical threshold. Let X be avariety with at most log canonical singularities, let Z ⊆ X be a closed subvariety, and let D bean effective Q -Cartier Q -divisor on the variety X . Then the numberlct Z (cid:0) X, D (cid:1) = sup n λ ∈ Q (cid:12)(cid:12)(cid:12) the log pair (cid:0) X, λD (cid:1) is log canonical along Z o is called a log canonical threshold of the divisor D along Z . It follows from [26] that for apolynomial f in n variables over C and a point P ∈ C n lct P (cid:16) C n , D (cid:17) = c P (cid:0) f (cid:1) , where the divisor D is defined by the equation f = 0 on C n . We can define the log canonicalthreshold of D on X bylct X (cid:0) X, D (cid:1) = inf n lct P (cid:0) X, D (cid:1) (cid:12)(cid:12)(cid:12) P ∈ X o = sup n λ ∈ Q (cid:12)(cid:12)(cid:12) the log pair (cid:0) X, λD (cid:1) is log canonical o . For simplicity, the log canonical threshold lct X ( X, D ) will be denoted by lct(
X, D ). XCEPTIONAL DEL PEZZO HYPERSURFACES 3
Example 1.1.1.
Let D be a cubic curve on the projective plane P . Thenlct (cid:0) P , D (cid:1) = D is a smooth curve , D is a curve with ordinary double points ,
56 if D is a curve with one cuspidal point ,
34 if D consists of a conic and a line that are tangent ,
23 if D consists of three lines intersecting at one point ,
12 if Supp (cid:0) D (cid:1) consists of two lines ,
13 if Supp (cid:0) D (cid:1) consists of one line . Now we suppose that X is a Fano variety with at most log terminal singularities. Definition 1.1.2.
The global log canonical threshold of the Fano variety X is the numberlct (cid:0) X (cid:1) = inf n lct (cid:0) X, D (cid:1) (cid:12)(cid:12)(cid:12) D is an effective Q -divisor on X with D ∼ Q − K X o . The number lct( X ) is an algebraic counterpart of the α -invariant introduced in [44] and [48](see [14, Appendix A]). Because X is rationally connected (see [50]), we havelct (cid:0) X (cid:1) = sup ( λ ∈ Q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) the log pair (cid:16) X, λD (cid:17) is log canonical for everyeffective Q -divisor numerically equivalent to − K X ) . It immediately follows from Definition 1.1.2 thatlct (cid:0) X (cid:1) = sup ε ∈ Q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) the log pair (cid:16) X, εn D (cid:17) is log canonical for everydivisor D ∈ (cid:12)(cid:12) − nK X (cid:12)(cid:12) and every positive integer n . Example 1.1.3 ([14]) . Suppose that P ( a , a , . . . , a n ) is a well-formed weighted projective spacewith a a . . . a n (see [22]). Thenlct (cid:16) P (cid:0) a , a , . . . , a n (cid:1)(cid:17) = a P ni =0 a i . Example 1.1.4.
Let X be a smooth hypersurface in P n of degree m n . The paper [6] showsthat lct (cid:0) X (cid:1) = 1 n + 1 − m if m < n . For the case m = n > − n lct( X ) IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV and the left equality holds if X contains a cone of dimension n −
2. Meanwhile, the papers [13]and [38] show that 1 > lct (cid:0) X (cid:1) > n > , n = 5 , n = 4 ,
34 if n = 3 , if X is general. Example 1.1.5.
Let X be a smooth hypersurface in the weighted projective space P (1 n +1 , d )of degree 2 d >
4. Then lct (cid:0) X (cid:1) = 1 n + 1 − d in the case when d < n (see [8, Proposition 20]). Suppose that d = n . Then the inequalities2 n − n lct (cid:0) X (cid:1) X ) = 1 if X is general and n >
3. Furthermore for the case n = 3 thepapers [13] and [38] prove thatlct (cid:0) X (cid:1) ∈ (cid:26) , , , , , , , , , , , , , , , (cid:27) and all these values are attained. For instance, if the hypersurface X is given by w = x + y + z + t + x y zt ⊂ P (cid:0) , , , , (cid:1) ∼ = Proj (cid:16) C (cid:2) x, y, z, t, w (cid:3)(cid:17) , where wt( x ) = wt( y ) = wt( z ) = wt( t ) = 1 and wt( w ) = 3, then lct( X ) = 1 (see [13]). Example 1.1.6 ([21]) . Let X be a rational homogeneous space such that the Picard group of X is generated by an ample Cartier divisor D and − K X ∼ rD for some positive integer r . Thenlct( X ) = r . Example 1.1.7.
Let X be a quasismooth well-formed (see [22]) hypersurface in P (1 , a , a , a , a ) of degree P i =1 a i with at most terminal singularities, where a . . . a .Then there are exactly 95 possibilities for the quadruple ( a , a , a , a ) (see [22], [24]). For ageneral hypersurface X , it follows from [7], [9], [10] and [13] that1 > lct (cid:0) X (cid:1) > a = a = a = a = 1 ,
79 if ( a , a , a , a ) = (1 , , , ,
45 if ( a , a , a , a ) = (1 , , , ,
67 if ( a , a , a , a ) = (1 , , , , . XCEPTIONAL DEL PEZZO HYPERSURFACES 5
The global log canonical threshold of the hypersurface w = t + z + y + x ⊂ P (cid:0) , , , , (cid:1) ∼ = Proj (cid:16) C (cid:2) x, y, z, t, w (cid:3)(cid:17) is equal to , where wt( x ) = wt( y ) = 1, wt( z ) = 2, wt( t ) = 6, wt( w ) = 9 (see [7]). Example 1.1.8 ([12]) . Let X be a singular cubic surface in P with at most canonical singu-larities. The possible singularities of X are listed in [5]. The global log canonical threshold of X is as follows:lct (cid:0) X (cid:1) =
23 if Sing (cid:0) X (cid:1) = (cid:8) A (cid:9) ,
13 if Sing (cid:0) X (cid:1) ⊇ (cid:8) A (cid:9) , Sing (cid:0) X (cid:1) = (cid:8) D (cid:9) or Sing (cid:0) X (cid:1) ⊇ (cid:8) A , A (cid:9) ,
14 if Sing (cid:0) X (cid:1) ⊇ (cid:8) A (cid:9) or Sing (cid:0) X (cid:1) = (cid:8) D (cid:9) ,
16 if Sing (cid:0) X (cid:1) = (cid:8) E (cid:9) ,
12 otherwise . So far we have not seen any single variety whose global log canonical threshold is irrational.In general, it is unknown whether global log canonical thresholds are rational numbers or not(cf.Question 1 in [46]). Even for del Pezzo surfaces with log terminal singularities the rationality oftheir global log canonical thresholds is unknown. However, we expect more than this as follows:
Conjecture 1.1.9.
There is an effective Q -divisor D on the variety X such that it is Q -linearlyequivalent to − K X and lct (cid:0) X (cid:1) = lct (cid:0) X, D (cid:1) . The following definition is due to [42] (cf. [23], [29], [32], [37]).
Definition 1.1.10.
The Fano variety X is exceptional (resp. weakly exceptional, stronglyexceptional) if for every effective Q -divisor D on the variety X such that D ∼ Q − K X and thepair ( X, D ) is log terminal (resp. lct( X ) >
1, lct( X ) > ⇒ exceptional = ⇒ weakly exceptional . However, if Conjecture 1.1.9 holds for X , then we see that X is exceptional if and only if X isstrongly exceptional. Exceptional del Pezzo surfaces, which are called del Pezzo surfaces withouttigers in [27], lie in finitely many families (see [42], [37]). We expect that strongly exceptionalFano varieties with quotient singularities enjoy very interesting geometrical properties (cf. [41,Theorem 3.3], [35, Theorem 1]).The main motivation for this article is that the global log canonical threshold turns out to playimportant roles both in birational geometry and in complex geometry. We have two significantapplications of the global log canonical threshold of a Fano variety X . The first one is for thecase when lct( X ) >
1. This inequality has serious applications to rationality problems for Fano
IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV varieties in birational geometry. The other is for the case when lct( X ) > dim( X )1+dim( X ) . This hasimportant applications to K¨ahler-Einstein metrics on Fano varieties in complex geometry.For a simple application of the first inequality, we can mention the following. Theorem 1.1.11 ([7] and [38]) . Let X i be birationally super-rigid Fano variety with lct( X i ) > i = 1 , . . . , r . Then the variety X × . . . × X r is non-rational andBir (cid:16) X × . . . × X r (cid:17) = Aut (cid:16) X × . . . × X r (cid:17) . For every dominant map ρ : X × . . . × X r Y whose general fiber is rationally connected,there is a subset { i , . . . , i k } ⊆ { , . . . , r } and a commutative diagram X × . . . × X rπ (cid:15) (cid:15) σ / / ______ X × . . . × X r ρ ) ) RRRRRRRR X i × . . . × X i k ξ / / ___________________ Y, where ξ and σ are birational maps, and π is the natural projection.This theorem may be more generalized so that we could obtain the following Example 1.1.12 ([7]) . Let X i be a threefold satisfying hypotheses of Example 1.1.7 withlct( X i ) = 1 for each i = 1 , . . . r . Suppose, in addition, that each X i is general in its deformationfamily. Then the variety X × . . . × X r is non-rational andBir (cid:16) X × . . . × X r (cid:17) = D r Y i =1 Bir( X i ) , Aut (cid:16) X × . . . × X r (cid:17)E . For every dominant map ρ : X × . . . × X r Y whose general fiber is rationally connected,there is a subset { i , . . . , i k } ⊆ { , . . . , r } and a commutative diagram X × . . . × X rπ (cid:15) (cid:15) σ / / ______ X × . . . × X r ρ ) ) RRRRRRRR X i × . . . × X i k ξ / / ___________________ Y, where ξ and σ are birational maps, and π is the natural projection.The following result that gives strong connection between global log canonical thresholds andK¨ahler-Einstein metrics was proved in [16], [34],[44] (see [14, Appendix A]). Theorem 1.1.13.
Suppose that X is a Fano variety with at most quotient singularities. Thenit admits an orbifold K¨ahler–Einstein metric iflct (cid:0) X (cid:1) > dim (cid:0) X (cid:1) dim (cid:0) X (cid:1) + 1 . Examples 1.1.4, 1.1.5 and 1.1.7 are good examples to which we may apply Theorem 1.1.13.There are many known obstructions for the existence of orbifold K¨ahler–Einstein metrics onFano varieties with quotient singularities (see [17], [19], [31], [33], [40], [47]).
XCEPTIONAL DEL PEZZO HYPERSURFACES 7
Example 1.1.14 ([20]) . Let X be a quasismooth hypersurface in P ( a , . . . , a n ) of degree d < P ni =0 a i , where a . . . a n . Suppose that X is well-formed and has a K¨ahler–Einstein metric.Then d n X i =0 a i − d ! n n n n Y i =0 a i , and P ni =0 a i d + na (see [2], [43]).The problem of existence of K¨ahler–Einstein metrics on smooth del Pezzo surfaces is com-pletely solved by [45] as follows: Theorem 1.1.15. If X is a smooth del Pezzo surface, then the following conditions are equiv-alent: • the automorphism group Aut( X ) is reductive; • the surface X admits a K¨ahler–Einstein metric; • the surface X is not a blow up of P at one or two points. Acknowledgments.
The first author is grateful to the Max Plank Institute for Mathematicsat Bonn for the hospitality and excellent working conditions. The first author was supported bythe grants NSF DMS-0701465 and EPSRC EP/E048412/1, the third author was supported bythe grants RFFI No. 08-01-00395-a, N.Sh.-1987.1628.1 and EPSRC EP/E048412/1. The secondauthor has been supported by the Korea Research Foundation Grant funded by the KoreanGovernment (KRF-2008-313-C00024).The authors thank I. Kim, B. Sea, and J. Won for their pointing out numerous mistakes inthe first version of this paper. 1.2.
Results
Let X d be a quasismooth and well-formed hypersurface in P ( a , a , a , a ) of degree d , where a a a a . Then the hypersurface X d is given by a quasihomogeneous polynomialequation f ( x, y, z, t ) = 0 of degree d . The quasihomogeneous equation f (cid:0) x, y, z, t (cid:1) = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, y, z, t (cid:3)(cid:17) , defines an isolated quasihomogeneous singularity ( V, O ) with the Milnor number Q ni =0 ( da i − O is the origin of C . It follows from the adjunction formula that K X d ∼ Q O P ( a , a , a , a ) (cid:16) d − X i =0 a i (cid:17) , and it follows from [18], [26, Proposition 8.14], [39] that the following conditions are equivalent: • the inequality d P i =0 a i − • the surface X d is a del Pezzo surface; • the singularity ( V, O ) is rational; • the singularity ( V, O ) is canonical.
IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
Blowing up C at the origin O with weights ( a , a , a , a ), we get a purely log terminal blowup of the singularity ( V, O ) (see [28], [36]). The paper [36] shows that the following conditionsare equivalent: • the surface X d is exceptional (weakly exceptional, respectively); • the singularity ( V, O ) is exceptional (weakly exceptional, respectively).From now on we suppose that d P i =0 a i −
1. Then X d is a del Pezzo surface. Put I = P i =0 a i − d . The list of possible values of ( a , a , a , a , d ) with 2 I < a can be foundin [4] and [15]. For the case I = 1, we can obtain the complete list of del Pezzo surfaces X d ⊂ P ( a , a , a , a ) from [25] as follows: • smooth del Pezzo surfaces X ⊂ P (1 , , , , X ⊂ P (1 , , , , X ⊂ P (1 , , , • singular del Pezzo surfaces X n +4 ⊂ P (2 , n + 1 , n + 1 , n + 1), where n is a positive integer, X ⊂ P (1 , , , , X ⊂ P (1 , , , , X ⊂ P (1 , , , , X ⊂ P (2 , , , X ⊂ P (3 , , , , X ⊂ P (3 , , , , X ⊂ P (3 , , , ,X ⊂ P (3 , , , , X ⊂ P (5 , , , , X ⊂ P (5 , , , ,X ⊂ P (5 , , , , X ⊂ P (7 , , , , X ⊂ P (7 , , , X ⊂ P (9 , , , , X ⊂ P (9 , , , , X ⊂ P (11 , , , X ⊂ P (11 , , , , X ⊂ P (13 , , , , X ⊂ P (13 , , , . The global log canonical thresholds of such del Pezzo surfaces have been considered eitherimplicitly or explicitly in [1], [3], [11], [16], [25]. For example, the papers [1], [3], [16] and[25] gives us lower bounds for global log canonical thresholds of singular del Pezzo surfaceswith I = 1. Meanwhile, the paper [11] deals with the exact values of the global log canonicalthresholds of smooth del Pezzo surfaces with I = 1. Theorem 1.2.1.
Suppose that I = 1 and X d is smooth. Thenlct (cid:0) X d (cid:1) = (cid:0) a , a , a , a (cid:1) = (cid:0) , , , (cid:1) and | − K X | contains no cuspidal curves ,
56 if (cid:0) a , a , a , a (cid:1) = (cid:0) , , , (cid:1) and | − K X | contains a cuspidal curve ,
56 if (cid:0) a , a , a , a (cid:1) = (cid:0) , , , (cid:1) and | − K X | contains no tacnodal curves ,
34 if (cid:0) a , a , a , a (cid:1) = (cid:0) , , , (cid:1) and | − K X | contains a tacnodal curve ,
34 if X is a cubic in P with no Eckardt points ,
23 if X is a cubic in P with an Eckardt point . However, for singular del Pezzo surfaces, the exact values of global log canonical thresholdshave not been considered seriously.A singular del Pezzo hypersurface X d ⊂ P ( a , a , a , a ) must satisfy exclusively one of thefollowing properties: For notions of exceptional and weakly exceptional singularities see [36, Definition 4.1], [42], [23].
XCEPTIONAL DEL PEZZO HYPERSURFACES 9 (1) 2 I > a ;(2) 2 I < a and (cid:0) a , a , a , a , d (cid:1) = (cid:0) I − k, I + k, a, a + k, a + k + I (cid:1) for some non-negative integer k < I and some positive integer a > I + k .(3) 2 I < a but (cid:0) a , a , a , a , d (cid:1) = (cid:0) I − k, I + k, a, a + k, a + k + I (cid:1) for any non-negative integer k < I and any positive integer a > I + k .For the first two cases one can check that lct( X d ) (for instance, see [4] and [15]). Allthe quintuples ( a , a , a , a , d ) such that the hypersurface X d is singular and satisfies the lastcondition are listed in Section 4. They are taken from [4] and [15]. Note that we rearrangeda little the quintuples taken from [4] by putting some cases that were contained in the infiniteseries of [4] into the sublist of sporadic cases; on the other hand, we removed two sporadic cases,because they are contained in the additional infinite series found in [15]. The completeness ofthis list is proved in [15] by using [49].We already know the global log canonical thresholds of smooth del Pezzo surfaces. For delPezzo surfaces satisfying one of the first two conditions, their global log canonical thresholdsare relatively too small to enjoy the condition of Theorem 1.1.13. However, the global logcanonical thresholds of del Pezzo surfaces satisfying the last condition have not been investigatedsufficiently. In the present paper we compute all of them and then we obtain the following result. Theorem 1.2.2.
Let X d be a quasismooth well-formed singular del Pezzo surface in the weightedprojective space Proj( C (cid:2) x, y, z, t (cid:3) ) with weights wt( x ) = a wt( y ) = a wt( z ) = a wt( t ) = a such that 2 I < a but ( a , a , a , a , d ) = ( I − k, I + k, a, a + k, a + k + I ) for anynon-negative integer k < I and any positive integer a > I + k , where I = P i =0 a i − d . Then if a = a , then lct( X d ) = min (cid:26) lct (cid:16) X d , Ia C x (cid:17) , lct (cid:16) X d , Ia C y (cid:17) , lct (cid:16) X d , Ia C z (cid:17)(cid:27) , where C x (resp. C y , C z ) is the divisor on X d defined by x = 0 (resp. y = 0, z = 0). If a = a ,then lct( X d ) = lct (cid:16) X d , Ia C (cid:17) , where C is a reducible divisor in |O X d ( a ) | .In particular, we obtain the value of lct( X d ) for every del Pezzo surface X d listed in Section 4.As a result, we obtain the following corollaries. Corollary 1.2.3.
The following assertions are equivalent: • the surface X d is exceptional; • lct( X d ) > • the quintuple ( a , a , a , a , d ) lies in the set (2 , , , , , (3 , , , , , (3 , , , , , (3 , , , , , (3 , , , , , (5 , , , , , (5 , , , , , (5 , , , , , (7 , , , , , (7 , , , , , (9 , , , , , (9 , , , , , (11 , , , , , (11 , , , , , (13 , , , , , (13 , , , , , (3 , , , , , (3 , , , , , (5 , , , , , (5 , , , , , (6 , , , , , (7 , , , , , (7 , , , , , (9 , , , , , (9 , , , , , (9 , , , , , (10 , , , , , (11 , , , , , (11 , , , , , (11 , , , , , (11 , , , , , (13 , , , , , (13 , , , , , (13 , , , , , (13 , , , , , (14 , , , , , (5 , , , , , (5 , , , , , (11 , , , , , (11 , , , , , (13 , , , , , (13 , , , , , (15 , , , , , (9 , , , , , (10 , , , , , (11 , , , , , (11 , , , , , (11 , , , , , (13 , , , , , (13 , , , , , (13 , , , , , (11 , , , , , (11 , , , , , (13 , , , , , (11 , , , , , (13 , , , , , (11 , , , , . Corollary 1.2.4.
The following assertions are equivalent: • the surface X d is weakly exceptional and not exceptional; • lct( X d ) = 1; • one of the following holds – the quintuple ( a , a , a , a , d ) lies in the set (2 , n + 1 , n + 1 , n + 1 , n + 4) , (3 , n, n + 1 , n + 1 , n + 3) , (3 , n + 1 , n + 2 , n + 2 , n + 6) , (3 , n + 1 , n + 2 , n + 1 , n + 5) , (3 , n + 1 , n + 1 , n, n + 3) , (3 , n + 1 , n + 1 , n + 3 , n + 6) , (4 , n + 1 , n + 2 , n + 1 , n + 6) , (4 , n + 3 , n + 3 , n + 4 , n + 12) , (6 , n + 3 , n + 5 , n + 5 , n + 15) , (6 , n + 5 , n + 8 , n + 9 , n + 24) , (6 , n + 5 , n + 8 , n + 15 , n + 30) , (8 , n + 5 , n + 7 , n + 9 , n + 23) , (9 , n + 8 , n + 11 , n + 13 , n + 35) , (1 , , , , , (2 , , , , , (5 , , , , , (5 , , , , , where n is a positive integer, – ( a , a , a , a , d ) = (1 , , , ,
6) and the pencil |− K X | does not have cuspidal curves, – ( a , a , a , a , d ) = (1 , , , ,
10) and C x = { x = 0 } has an ordinary double point, – ( a , a , a , a , d ) = (1 , , , ,
15) and the defining equation of X contains yzt , – ( a , a , a , a , d ) = (2 , , , ,
12) and the defining equation of X contains yzt . XCEPTIONAL DEL PEZZO HYPERSURFACES 11
Corollary 1.2.5.
In the notation and assumptions of Theorem 1.2.2, the surface X d has anorbifold K¨ahler–Einstein metric with the following possible exceptions: X ⊂ P (7 , , , X ⊂ P (7 , , , X ⊂ P (7 , , , X ⊂ P (7 , , , X ⊂ P (7 , , , X ⊂ P (1 , , ,
7) whose defining equation does not contain yzt , and X ⊂ P (2 , , ,
5) whosedefining equation does not contain yzt .Corollary 1.2.3 illustrates the fact that exceptional del Pezzo surfaces lie in finitely manyfamilies (see [42], [37]). On the other hand, Corollary 1.2.3 shows that weakly-exceptional delPezzo surfaces do not enjoy this property. Note also that Corollary 1.2.3 follows from [29].1.3.
Preliminaries
For the basic definitions and properties concerning singularities of pairs we refer the readerto [26]. To prove Theorem 1.2.2 we need to compute the log canonical thresholds of individ-ual effective divisors. The following two lemmas are rather basic properties of log canonicalthresholds but will be useful to compute them. For the proofs the reader is referred to [26] and[30].
Lemma 1.3.1.
Let f ∈ C [ x , . . . , x n ] and D = ( f = 0). Suppose that the polynomial f vanishesat the origin O in C n . Set d = mult O ( f ) and let f d denote the degree d homogeneous part of f . Let T D = ( f d = 0) ⊂ C n be the tangent cone of D and P ( T D ) = ( f d = 0) ⊂ P n − be theprojectivised tangent cone of D . Then(1) d lct O ( C n , D ) nd .(2) The log pair ( P n − , nd P ( T D )) is log canonical if and only if lct O ( C n , D ) = nd .(3) If P ( T D ) is smooth (or even log canonical) then lct O ( C n , D ) = min { , nd } . Lemma 1.3.2.
Let f be a polynomial in C [ z , z ]. Suppose that the polynomial defines anirreducible curve C passing through the origin O in C . We then havelct O ( C , C ) = min (cid:18) , m + 1 n (cid:19) , where ( m, n ) is the first pair of Puiseux exponents of f . We also havelct O (cid:0) C , ( z n z n ( z m + z m ) = 0) (cid:1) = min (cid:18) n , n , m + m m m + m n + m n (cid:19) , where n , n , m , m are non-negative integers.Throughout the proof of Theorem 1.2.2, Inversion of Adjunction that enables us to computelog canonical thresholds on lower dimensional varieties will be frequently utilized. Let X be anormal (but not necessarily projective) variety. Let S be a smooth Cartier divisor on X and B be an effective Q -Cartier Q -divisor on X such that K X + S + B is Q -Cartier and S Supp( B ). Theorem 1.3.3.
The log pair (
X, S + B ) is log canonical along S if and only if the log pair( S, B | S ) is log canonical.In the case when X is a surface, Theorem 1.3.3 can be stated in terms of local intersectionnumbers. Lemma 1.3.4.
Suppose that X is a surface. Let P be a smooth point of X such that it is alsoa smooth point of S . Then the log pair ( X, S + B ) is log canonical at the point P if and only ifthe local intersection number of B and S at the point P is at most 1. In particular, if the logpair ( X, mS + B ) is not log canonical at the point P for m
1, then B · S > Lemma 1.3.5.
Let D be an effective Q -divisor such that K X + D is Q -Cartier. For a smoothpoint P of X , the log pair ( X, D ) is log canonical at the point P if mult P ( D ) Lemma 1.3.6.
Let D and D be effective Q -divisors on Y with D ∼ Q D . Suppose that thepair ( X, D ) is not log canonical at a point P ∈ Y but the pair ( X, D ) is log canonical at thepoint P . Then there is an effective Q -divisor D on Y such that • D ∼ Q D ; • at least one irreducible component of D is not contained in the support of D ; • the pair ( X, D ) is not log canonical at the point P . Proof.
Write D = P ri =1 b i C i where b i ’s are positive rational numbers and C i ’s are distinctirreducible and reduced divisors. Also, we write D = ∆ + P ri =1 e i C i where e i ’s are non-negative rational numbers and ∆ is an effective Q -divisor whose support contains none of C i ’s.Suppose that e i > i . If not, then we put D = D . Let α = min (cid:26) e i b i (cid:12)(cid:12)(cid:12) i = 1 , , . . . , r (cid:27) . Then the positive rational number α is less than 1 since D ∼ Q D . Put D = 11 − α D − α − α D = 11 − α ∆ + r X i =1 (cid:18) e i − αb i − α (cid:19) C i . It is easy to see that the divisor D satisfies the first two conditions. If the pair ( X, D ) is logcanonical at the point P , then the pair ( X, D ) = ( X, (1 − α ) D + αD ) must be log canonicalat the point P . Therefore, the divisor D also satisfies the last condition. (cid:3) In the present paper, we deal with surfaces with at most quotient singularities. However, thestatements mentioned so far require smoothness of the ambient space for us to utilize them tothe fullest. Fortunately, the following proposition enables us to apply the statements with easesince we have a natural finite morphism of a germ of the origin in C to a germ of a quotientsingularity that is ramified only at a point. Proposition 1.3.7 ([26]) . Let f : Y → X be a finite morphism between normal varieties andassume that f is unramified outside a set of codimension two. Let D be an effective Q -Cartier Q -divisor. Then a log pair ( X, D ) is log canonical (resp. Kawamata log terminal) if and only ifthe log pair (
Y, f ∗ D ) is log canonical (resp. Kawamata log terminal). XCEPTIONAL DEL PEZZO HYPERSURFACES 13
The following two lemmas will be useful for this paper. The first lemma is just a reformulationof Lemma 1.3.4 mixed with Proposition 1.3.7 that we can apply to our cases immediately.Suppose that X is a quasismooth well-formed hypersurface in P = P ( a , a , a , a ) of degree d . Lemma 1.3.8.
Let C be a reduced and irreducible curve on X and D be an effective Q -divisor on X . Suppose that for a given positive rational number λ we have λ mult C ( D )
1. If λ ( C · D − (mult C ( D )) C )
1, then the pair (
X, λD ) is log canonical at each smooth point P of C not in Sing( X ). Furthermore, if the point P of C is a singular point of X of type r ( a, b ) and rλ ( C · D − (mult C ( D )) C )
1, then the pair (
X, λD ) is log canonical at P . Proof.
We may write D = mC + Ω, where Ω is an effective divisor whose support does notcontain the curve C . Suppose that the pair ( X, λD ) is not log canonical at a smooth point P of C not in Sing( X ). Since λm
1, the pair (
X, C + λ Ω) is not log canonical at the point P .Then by Lemma 1.3.4 we obtain an absurd inequality1 < λ Ω · C = λC · ( D − mC ) . Also, if the point P is a singular point of X , then we obtain from Lemma 1.3.4 and Proposi-tion 1.3.7 1 r < λ Ω · C = λC · ( D − mC ) r . This proves the second statement. (cid:3)
Let D be an effective Q -divisor on X such that D ∼ Q O P ( a , a , a , a ) (cid:0) I (cid:1)(cid:12)(cid:12)(cid:12) X . The next lemma will be applied to show that the log pair (
X, D ) is log canonical at some smoothpoints on X . Lemma 1.3.9.
Let k be a positive integer. Suppose that H ( P , O P ( k )) contains • at least two different monomials of the form x α y β , • at least two different monomials of the form x γ z δ .For a smooth point P of X in the outside of C x ,mult P (cid:0) D (cid:1) Ikda a a a if either H ( P , O P ( k )) contains at least two different monomials of the form x µ t ν or the point P is not contained in a curve contracted by the projection ψ : X P ( a , a , a ). Here, α , β , γ , δ , µ and ν are non-negative integers. Proof.
The first case follows from [1, Lemma 3.3]. Arguing as in the proof of [1, Corollary 3.4],we can also obtain the second case. (cid:3)
Let us conclude this section by mentioning two results that are never used in this paper, butnevertheless can be used to give shorter proofs of Corollaries 1.2.3 and 1.2.5. Suppose that X isgiven by a quasihomogeneous equation f (cid:0) x, y, z, t (cid:1) = 0 ⊂ P (cid:0) a , a , a , a (cid:1) ∼ = Proj (cid:16) C (cid:2) x, y, z, t (cid:3)(cid:17) , where wt( x ) = a wt( y ) = a wt( z ) = a wt( t ) = a . Lemma 1.3.10.
Suppose that I = P i =0 a i − d >
0. Thenlct (cid:0) X (cid:1) > a a dI ,a a dI if f (0 , , z, t ) = 0 ,a a dI if f (0 , , , t ) = 0 . Proof.
See [4, Corollary 5.3] (cf. [25, Proposition 11]). (cid:3)
Lemma 1.3.11.
Suppose that I = P i =0 a i − d >
0, the curve C x = { x = 0 } is irreducible andreduced. Then lct (cid:0) X (cid:1) > min (cid:18) a a dI , lct (cid:18) X, Ia C x (cid:19)(cid:19) , min (cid:18) a a dI , lct (cid:18) X, Ia C x (cid:19)(cid:19) if f (0 , , , t ) = 0 . Proof.
Arguing as in the proof of [25, Proposition 11] and using Lemma 1.3.6, we obtain therequired assertion. (cid:3)
Notation
We reserve the following notation that will be used throughout the paper: • P ( a , a , a , a ) denotes the well-formed weighted projective space Proj( C (cid:2) x, y, z, t (cid:3) ) withweights wt( x ) = a , wt( y ) = a , wt( z ) = a , wt( t ) = a , where we always assume theinequalities a a a a . We may use simply P instead of P ( a , a , a , a ) whenthis does not lead to confusion. • X denotes a quasismooth and well-formed hypersurface in P ( a , a , a , a ) (see Defini-tions 6.3 and 6.9 in [22], respectively). • O x is the point in P ( a , a , a , a ) defined by y = z = t = 0. The points O y , O z and O t are defined in the similar way. • C x is the curve on X cut out by the equation x = 0. The curves C y , C z and C t aredefined in the similar way. • L xy is the curve in P ( a , a , a , a ) defined by x = y = 0. The curves L xz , L xt , L yz , L yt and L zt are defined in the similar way. • Let D be a divisor on X and P ∈ X . Choose an orbifold chart π : ˜ U → U for someneighborhood P ∈ U ⊂ X . We put mult P ( D ) = mult Q ( π ∗ D ), where Q is a point on ˜ U with π ( Q ) = P , and refer to this quantity as the multiplicity of D at P .1.5. The scheme of the proof
We have 83 families of del Pezzo hypersurfaces in The Big Table. In the present section we ex-plain the methods to compute the global log canonical thresholds of the del Pezzo hypersurfacesin The Big Table. By family we mean either one-parameter series (which actually gives rise to an infinite number of deformationfamilies) or a sporadic case. We hope that this would not lead to a confusion.
XCEPTIONAL DEL PEZZO HYPERSURFACES 15
Let X ⊂ P ( a , a , a , a ) be a del Pezzo surface of degree d in one of the 83 families (actually,one infinite series has been treated in [15], so we will omit the computations in this case). Set I = a + a + a + a − d . There are two exceptional cases where a = a . The method forthese two cases is a bit different from the other cases. Both cases will be individually dealt with(Lemmas 2.2.4 and 3.1.5).If a = a , then we will take steps as follows: Step 1.
Using Lemmas 1.3.1 and 1.3.2 with Proposition 1.3.7, we compute the log canonicalthresholds lct( X, Ia C x ), lct( X, Ia C y ), lct( X, Ia C z ) and lct( X, Ia C t ). Set λ = min (cid:26) lct( X, Ia C x ) , lct( X, Ia C y ) , lct( X, Ia C z ) , lct( X, Ia C t ) (cid:27) . Then the global log canonical threshold lct( X ) is at most λ . Step 2.
We claim that the global log canonical threshold lct( X ) is equal to λ . To provethis assertion, we suppose lct( X ) < λ . Then there is an effective Q -divisor D equivalent to theanticanonical divisor − K X of X such that the log pair ( X, λD ) is not log canonical at somepoint P ∈ X . In particular, we obtainmult P ( λD ) > P is a smooth point of X ,1 r if the point P is a singular point of X of type 1 r ( a, b ).from Lemma 1.3.5 and Proposition 1.3.7. Step 3.
We show that the point P cannot be a singular point of X using the followingmethods. Method 3.1. (Multiplicity)
We may assume that a suitable irreducible component C of C x , C y , C z , and C t is not contained in the support of the divisor D . We derive a possiblecontradiction from the inequality C · D > mult P ( C ) · mult P ( D ) r > mult P ( C ) rλ , where r is the index of the quotient singular point P . The last inequality follows from theassumption that ( X, λD ) is not log canonical at P . This method can be applied to exclude asmooth point. Method 3.2. (Inversion of Adjunction)
We consider a suitable irreducible curve C smooth at P . We then write D = µC + Ω, where Ω is an effective Q -divisor whose support doesnot contain C . We check λµ
1. If so, then the log pair (
X, C + λ Ω) is not log canonical atthe point P either. By Lemma 1.3.8 we have λ ( D − µC ) · C = λC · Ω > r . We try to derive a contradiction from this inequality. The curve C is taken usually from anirreducible component of C x , C y , C z , or C t . This method can be applied to exclude a smoothpoint. Method 3.3. (Weighted Blow Up)
Sometimes we cannot exclude a singular point P onlywith the previous two methods. In such a case, we take a suitable weighted blow up π : Y → X at the point P . We can write K Y + D Y ∼ Q π ∗ ( K X + λD ) , where D Y is the log pull-back of λD by π . Using method 3.1 we obtain that D Y is effective.Then we apply the previous two methods to the pair ( Y, D Y ), or repeat this method until weget a contradictory inequality. Step 4.
We show that the point P cannot be a smooth point of X . To do so, we first applyLemma 1.3.9. However, this method does not work always. If the method fails, then we try tofind a suitable pencil L on X . The pencil has a member F which passes through the point P .We show that the pair ( X, λF ) is log canonical at the point P . Then, we may assume that thesupport of D does not contain at least one irreducible component of F . If the divisor D itselfis irreducible, then we use Method 3.1 to exclude the point P . If F is reducible, then we useMethod 3.2. Part Infinite series
Infinite series with I = 1 Lemma 2.1.1.
Let X be a quasismooth hypersurface of degree 8 n +4 in P (2 , n +1 , n +1 , n +1)for a natural number n . Then lct( X ) = 1. Proof.
The surface X is singular at the point O t , which is of type n +1 (1 , O , O , O , O , which are cut out on X by L xt . Each O i is a singular point oftype n +1 (1 , n ) on the surface X .The curve C x is reducible. We see C x = L + L + L + L , where L i is a smooth rational curves such that − K X · L i = 1(2 n + 1)(4 n + 1) , and L ∩ L ∩ L ∩ L = { O t } . Then L i · L j = 14 n + 1for i = j . Also, we have L i = C x · L i − n + 1 = 2(2 n + 1)(4 n + 1) − n + 1 = − n + 1(2 n + 1)(4 n + 1) . It is easy to see lct( X, C x ) = 1. Therefore, lct( X )
1. Suppose that lct( X ) <
1. Thenthere is an effective Q -divisor D ∼ Q − K X such that the log pair ( X, D ) is not log canonical atsome point P ∈ X . XCEPTIONAL DEL PEZZO HYPERSURFACES 17
Since (4 n + 2)(8 n + 4)2(2 n + 1) (4 n + 1) = 44 n + 1 < H ( P , O P (4 n + 2)) contains x n +1 , y and z , Lemma 1.3.9 implies that P ∈ C x .It follows from Lemma 1.3.6 that we may assume that L i Supp( D ) for some i . Also, P ∈ L j for some j . Put D = mL j + Ω, where Ω is an effective Q -divisor such that L j Supp(Ω). Since1(2 n + 1)(4 n + 1) = D · L i = (cid:0) mL j + Ω (cid:1) · L i > mL i · L j = m n + 1 , we have 0 m n +1 . Since(2 n + 1)Ω · L j = (2 n + 1)( D − mL j ) · L j = (2 n + 1) 1 + m (6 n + 1)(2 n + 1)(4 n + 1) n + 1) < P must be O t . Note that the inequalitymult O t (cid:0) D (cid:1) (4 n + 1) D · L i = 12 n + 1 , shows that the point P cannot be the point O t . This is a contradiction. (cid:3) Infinite series with I = 2 Lemma 2.2.1.
Let X be a quasismooth hypersurface of degree 8 n + 12 in P (4 , n + 3 , n +3 , n + 4) for a natural number n . Then lct( X ) = 1. Proof.
The only singularities of X are a singular point O t of index 4 n + 4, two singular points P , P of index 4 on L yz , and four singular points Q , Q , Q , Q of index 2 n + 3 on L xt .The curve C x is reduced and splits into four irreducible components L , . . . , L . Each L i passes through Q i . They intersect each other at O t . One can easily see that lct( X, C x ) = 1,and hence lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6 we may assume that L i Supp ( D ) for some i . Since(4 n + 4) L i · D = 4 n + 4(2 n + 2)(2 n + 3) < n >
1, the point P cannot belong to the curve L i .For j = i , put D = µL j + Ω, where Ω is an effective Q -divisor such that L j Supp (Ω). Since µ n + 4 = µL i · L j D · L i = 12( n + 1)(2 n + 3) , we have µ n + 3 . Note that L j = C x · L j − L i · L j = 22( n + 1)(2 n + 3) − n + 1) = − n + 54( n + 1)(2 n + 3) . By Lemma 1.3.4 the inequality(2 n + 3)Ω · L j = (2 n + 3)( D − µL j ) · L j = 2 + (6 n + 5) µ n + 1) n + 3 < n > P cannot be contained in L j . Consequently, the point P is located inthe outside of C x .By a suitable coordinate change we may assume that P = O x . Then, the curve C t is reducedand splits into four irreducible components L ′ , . . . , L ′ . Each L ′ i passes through the point Q i .They intersect each other at O x . We can easily see that the log pair ( X, n +4 C t ) is log canonical.By Lemma 1.3.6 we may assume that L ′ i Supp ( D ). Sincemult O x ( D ) L ′ i · D = 22 n + 3 < n >
1, the point P cannot be O x . The point P can be excluded in a similar way.Therefore, P is a smooth point of X \ C x . Applying Lemma 1.3.9, we see that1 < mult P ( D ) n + 12) n + 3) (4 n + 4) n > H ( P , O P (8 n + 12)) contains x n +3 , y and z . The obtained contradictioncompletes the proof. (cid:3) Lemma 2.2.2.
Let X be a quasismooth hypersurface of degree 18 n + 6 in P (3 , n + 1 , n +1 , n + 3) for a natural number n >
1. Then lct( X ) = 1. Proof.
The only singularities of X are a singular point O z of index 6 n + 1, two singular points P , P of index 3 on L yz , and two singular points Q , Q of index 3 n + 1 on L xz .The curve C x is reduced and splits into two components L and L that intersect at O z . It iseasy to see that lct( X, C x ) = 1. Therefore, lct( X )
1. Note that L · L = 36 n + 1 and L = L = − n − n + 1)(6 n + 1) . Suppose that lct( X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .We may assume that L is not contained the support of D . The inequality D · L = 2(3 n + 1)(6 n + 1) n + 1shows that the point P cannot belong to the curve L . Put D = µL + Ω, where Ω is an effective Q -divisor whose support does not contain the curve L . Since3 µ n + 1 = µL · L D · L = 2(3 n + 1)(6 n + 1) , we have 0 µ n + 1) . XCEPTIONAL DEL PEZZO HYPERSURFACES 19
Lemma 1.3.8 and the inequalityΩ · L = ( D − µL ) · L = 2 + µ (9 n − n + 1)(6 n + 1) < n + 1)(6 n + 1)show that the point P is located in the outside of L . Therefore, P C x .The curve C y is irreducible. It is easy to see that the log pair ( X, n +1 C y ) is log canonical.Therefore, we may assume that the support of D does not contain the curve C y . Note that P , P ∈ C y . The inequality 3 D · C y = 46 n + 1 P not P can be the point P .Hence P is a smooth point of X \ C x . Applying Lemma 1.3.9, we get an absurd inequality1 < mult P ( D ) n + 6)(18 n + 3)3(3 n + 1)(6 n + 1)(9 n + 3) H ( P , O P (18 n + 3)) contains x n +1 , x n y and z . The obtained contradiction completesthe proof. (cid:3) Lemma 2.2.3.
Let X be a quasismooth hypersurface of degree 18 n + 3 in P (3 , n + 1 , n + 1 , n )for a natural number n >
1. Then lct( X ) = 1. Proof.
The singularities of X are a singular point O y of index 3 n + 1, a singular point O t ofindex 9 n , and two singular points Q , Q of index 3 on L yz .The curve C x is reduced and irreducible and has the only singularity at O t . It is easy to seethat lct( X, C x ) = 1, and hence lct( X )
1. The curve C y is quasismooth. Therefore, the logpair ( X, n +1 C y ) is log canonical.Suppose that lct( X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assume thatneither C x nor C y is contained in Supp ( D ).The inequalities C x · D < (3 n + 1) C x · D = 23 n < , mult O t ( D ) = mult O t ( C x )mult O t ( D )3 nC x · D n + 1 < P must be located in the outside of C x .Also, the inequality 3 C y · D = 23 n < Q not Q can be the point P . Hence P is a smooth point of X \ C x . Wesee that H ( P , O P (9 n + 3)) contains x n +1 , y and xt . Also, the projection of X from the point O z has only finite fibers. Therefore, Lemma 1.3.9 implies a contradictory inequality1 < mult P ( D ) n + 3)(9 n + 3)3(3 n + 1)(6 n + 1) · n = 23 n < . The obtained contradiction completes the proof. (cid:3)
Lemma 2.2.4.
Let X be a quasismooth hypersurface of degree 12 in P (3 , , , X ) =1. Proof.
The surface X can be defined by the quasihomogeneous equation Y i =1 ( α i x + β i y ) = Y j =1 ( γ j z + δ j t ) , where [ α i : β i ] define four distinct points and [ γ j : δ j ] define three distinct points in P .Let P i be the point in X given by z = t = α i x + β i y = 0. These are singular point of X oftype (1 , Q j be the point in X that is given by x = y = γ j z + δ j t = 0. Then each ofthem is a singular point of X of type (1 , L ij be the curve in X defined by α i x + β i y = γ j z + δ j t = 0, where i = 1 , . . . , j = 1 , . . . , C i cut out by the equation α i x + β i y = 0 consists of three smooth curves L i , L i , L i . These divisors C i , i = 1 , , ,
4, are the only reducible members in the linear system |O X (3) | . Meanwhile, the divisor B j cut out by γ j z + δ j t = 0 consists of four smooth curves L j , L j , L j , L j . Note that L i ∩ L i ∩ L i = { P i } and L j ∩ L j ∩ L j ∩ L j = { Q j } . We have L ij · L ik = and L ji · L ki = if k = j . But L ij = − .Since lct (cid:0) X, C i (cid:1) = lct (cid:0) X, B j (cid:1) = 1 , we have lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . For every i = 1 , . . . ,
4, we may assume thatthe support of the divisor D does not contain at least one curve among L i , L i , L i . Suppose L ik Supp ( D ). Then the inequalitymult P i ( D ) D · L ik = 12implies that none of the points P i can be the point P . For every j = 1 , ,
3, we may also assumethat the support of the divisor D does not contain at least one curve among L j , L j , L j , L j .Suppose L lj Supp ( D ). Then the inequalitymult Q j ( D ) D · L lj = 23implies that none of the points Q j can be the point P . Therefore, the point must be a smoothpoint of X .Write D = µL ij + Ω, where Ω is an effective Q -divisor whose support does not contain L ij .If µ >
0, then we have µL ij · L ik D · L ik , and hence µ . SinceΩ · L ij = 2 + 5 µ < , Lemma 1.3.4 implies the point P cannot be on the curve L ij . Consequently, P [ i =1 3 [ j =1 L ij . XCEPTIONAL DEL PEZZO HYPERSURFACES 21
There is a unique curve C ⊂ X cut out by λx + µy = 0, where [ λ : µ ] ∈ P , passing throughthe point P . Then the curve C is irreducible and quasismooth. Thus, we may assume that C isnot contained in the support of D . Then1 < mult P ( D ) D · C = 12 . This is a contradiction. (cid:3)
Lemma 2.2.5.
Let X be a quasismooth hypersurface of degree 9 n + 3 in P (3 , n, n + 1 , n + 1)for n >
2. Then lct( X ) = 1. Proof.
We may assume that the surface X is defined by the equation xy ( y − ax n )( y − bx n ) + zt ( z − ct ) = 0 , where a , b , c are non-zero constants and b = c . The point O y is a singular point of of index 3 n on X . The three points O x , P a = [1 : a : 0 : 0], P b = [1 : b : 0 : 0] are singular points of index 3on X . Also, X has three singular points O z , O t , P c = [0 : 0 : c : 1] of index 3 n + 1 on L xy .The curve C x consists of three irreducible components L xz , L xt and L c = { x = z − ct = 0 } .These three components intersect each other at O y . It is easy to check lct( X, C x ) = 1. Thus,lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6 we may assume that at least one of the components of C x is not containedin Supp ( D ). Then, the inequality3 nL xz · D = 3 nL xt · D = 3 nL c · D = 23 n + 1 < P cannot be the point O y .Put D = µL xz + Ω, where Ω is an effective Q -divisor whose support does not contain thecurve L xz . We claim that µ n + 1 . Indeed, if the inequality fails, one of the curves L xt and L c is not contained in Supp ( D ). Theneither µ n = µL xz · L xt D · L xt = 23 n (3 n + 1) , or µ n = µL xz · L c D · L c = 23 n (3 n + 1)holds. This is a contradiction. Note that L xz = − n − n (3 n + 1) . The inequality Ω · L xz = 2 + (6 n − µ n (3 n + 1) < n + 1holds for all n >
2. Therefore, Lemma 1.3.8 implies the point P cannot belong to L xz . By thesame way, we can show that P L xt ∪ L c . Let C be the curve on X cut out by the equation z − αt = 0, where α is non-zero constantdifferent from c . Then the curve C is quasismooth and hence lct( X, n +1 C ) >
1. Therefore, wemay assume that the support of D does not contain the curve C . Thenmult O x ( D ) , mult P a ( D ) , mult P b ( D ) D · C = 2 n n >
2. Therefore, P cannot be a singular point of X . Hence P is a smooth point of X \ C x .Applying Lemma 1.3.9, we get an absurd inequality1 < mult P ( D ) n + 3) · n (3 n + 1)(3 n + 1) n > H ( P , O P (9 n + 3)) contains x n +1 , xy and z . The obtained contradictioncompletes the proof. (cid:3) Lemma 2.2.6.
Let X be a quasismooth hypersurface of degree 9 n +6 in P (3 , n +1 , n +2 , n +2)for n >
1. Then lct( X ) = 1. Proof.
The only singularities of X are a singular point O y of index 3 n + 1, and three singularpoints P i , i = 1 , ,
3, of index 3 n + 2 on L xy .The divisor C x consists of three distinct irreducible and reduced curves L , L , L , where each L i contains the singular point P i . Then L ∩ L ∩ L = { O y } . It is obvious that lct( X, C x ) = 1,and hence lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assume that L is not contained in Supp ( D ).Since L · D < (3 n + 1) L · D = 23 n + 2 < n >
1, we see that P L . In particular, we see that P = O y .Put D = µL + Ω, where Ω is an effective Q -divisor such that L Supp (Ω). Then theinequality µ n + 1 = µL · L D · L = 2(3 n + 1)(3 n + 2) , implies that µ n +2 . The intersection number L = − n + 1(3 n + 1)(3 n + 2)shows (3 n + 2)Ω · L = (3 n + 2)( D − µL ) · L = 2 + µ (6 n + 1)(3 n + 1) n + 2)for all n >
1. Therefore, Lemma 1.3.8 excludes all the smooth point on L in the case where n > P in the case where n >
2. For the case n = 1, let C be the uniquecurve in the pencil |O X (5) | that passes through the point P . Then the divisor C consists oftwo distinct irreducible and reduced curve L and R . The curve R is singular at the point XCEPTIONAL DEL PEZZO HYPERSURFACES 23 P . Moreover, the log pair ( X, C ) is log canonical at the point P . By Lemma 1.3.6, we mayassume that R Supp ( D ). Then the inequality2mult P ( D ) mult P ( D )mult P ( R ) D · R = 2excludes the point P in the case where n = 1. By the same method, we can show P L .Hence the point P must be a smooth point in X \ C x . For the case n >
2, we can useLemma 1.3.9 to get a contradiction1 < mult P ( D ) n + 6) n + 1)(3 n + 2)(3 n + 2) = 63 n + 1 < , since H ( P , O P (9 n + 6)) contains x n +2 , y x and z . For the case n = 1, let R P be the uniquecurve in the pencil |O X (5) | that passes through the point P . The log pair ( X, R P ) is logcanonical at the point P . By Lemma 1.3.6, we may assume that Supp ( D ) does not contain atleast one irreducible component of R P . Note that either R P is irreducible or P k ∈ R P for some k = 1 , ,
3. If R P is irreducible, then we can obtain a contradiction1 < mult P ( D ) D · R P = 12 . Thus, P k ∈ R P . Then R P consists of two distinct irreducible curves L k and Z . Since we alreadyshowed that P is located in the outside of L k , the point P must belong to the curve Z . We have L k = − , L k · Z = 35 , Z = 25 . Put D = mZ + ∆, where ∆ is an effective Q -divisor such that Z Supp (∆). If m >
0, then3 m mZ · L k D · L k = 110 , and hence µ . Then Lemma 1.3.8 gives us a contradiction1 < ∆ · Z = 2 − m < . (cid:3) Lemma 2.2.7.
Let X be a quasismooth hypersurface of degree 12 n + 6 in P (4 , n + 1 , n +2 , n + 1) for n >
1. Then lct( X ) = 1. Proof.
We may assume that the surface X is defined by the equation xt + x n +1 z + ax n +1 y − ( z − a y )( z − a y )( z − a y ) = 0 , where a , a , a are distinct constants and a is a constant.The only singularities of X are a singular point O x of index 4, a singular point O t of index6 n + 1, a singular point Q = [1 : 0 : 1 : 0] of index 2, and three singular points P = [0 : 1 : a : 0], P = [0 : 1 : a : 0], P = [0 : 1 : a : 0] of index 2 n + 1.The divisor C x consists of three distinct irreducible curves L i = { x = z − a i y = 0 } , i = 1 , , L i passes through the point P i and L ∩ L ∩ L = { O t } . We can easily checklct( X, C x ) = 1, and hence lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assume that L is not contained in Supp ( D ). Since(6 n + 1) L · D = 22 n + 1 < P is located in the outside of L .Put D = µL + Ω, where Ω is an effective Q -divisor such that L Supp (Ω). Then2 µ n + 1 = µL · L D · L = 2(2 n + 1)(6 n + 1) , and hence µ n +1 . Since L = − n (2 n + 1)(6 n + 1)we have (2 n + 1)Ω · L = (2 n + 1)( D − µL ) · L = 2 + 8 nµ n + 1 n + 1 < n >
1. Then Lemma 1.3.8 excludes all the points on L . Furthermore, the same methodworks for L .The curve C y is quasismooth. Thus the log pair ( X, n +1 C y ) is log canonical. By Lemma 1.3.6we may assume that C y is not contained in Supp ( D ). Then the inequality4 C y · D = 66 n + 1 < P is neither O x nor Q . Hence P is a smooth point of X \ C x . However,Lemma 1.3.9 gives us mult P ( D ) n (2 n + 1)8(2 n + 1) (6 n + 1) < H ( P , O P (12 n )) contains x n , y x n − and z x n − . This is a contradiction. (cid:3) Infinite series with I = 4 Lemma 2.3.1.
Let X be a quasismooth hypersurface of degree 18 n + 15 in P (6 , n + 3 , n +5 , n + 5) for n >
1. Then lct( X ) = 1. Proof.
We may assume that the surface X is defined by the equation( z − a t )( z − a t )( z − a t ) + xy ( y − x n +1 ) = 0 , where a , a , a are distinct constants. The only singularities of X are a singular point O x ofindex 6, a singular point O y of index 6 n + 3, a singular point Q = [1 : 1 : 0 : 0] of index 3, andthree singular points P i = [0 : 0 : a i : 1], i = 1 , ,
3, of index 6 n + 5.The divisor C x consists of three distinct irreducible curves L i = { x = z − a i t = 0 } , i = 1 , , L i passes through the point P i and L ∩ L ∩ L = { O y } . We can easily checklct( X, C x ) = 1, and hence lct( X ) C y consists of three distinct irreducible curves L ′ i = { y = z − a i t = 0 } , i = 1 , , L ′ i passes through the point P i and L ′ ∩ L ′ ∩ L ′ = { O x } . The log pair ( X, n +3 C y ) is logcanonical. XCEPTIONAL DEL PEZZO HYPERSURFACES 25
Suppose that lct( X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .For a general member C in |O X (6 n + 5) | , we havemult Q ( D ) D · C = 66 n + 3 < . Therefore the point P cannot be the point Q .By Lemma 1.3.6 we may assume that L and L ′ are not contained in Supp ( D ). The twoinequalities (6 n + 5) D · L = n +3 < D · L ′ = n +5 < P is locatedin the outside of L ∪ L ′ .Write D = µL + Ω, where Ω is an effective Q -divisor such that L Supp (Ω). Then µ n + 3 = µL · L D · L = 4(6 n + 3)(6 n + 5) , and hence µ n +5 . Note that L = − n + 4(6 n + 3)(6 n + 5) . Therefore, we have(6 n + 5)Ω · L = (6 n + 5)( D − µL ) · L = 4 + (12 n + 4) µ n + 3 n + 5 . Therefore, Lemma 1.3.8 excludes all the smooth point on L in the case where n > P in the case where n >
2. For the case n = 1, let C be the unique curve in thepencil |O X (11) | that passes through the point P . Then the divisor C consists of three distinctirreducible and reduced curve L , L ′ and R . The log pair ( X, C ) is log canonical at thepoint P . If µ = 0, then the inequality above immediately excludes the point P for the case n = 1. Therefore we may assume that either L ′ Supp ( D ) or R Supp ( D ). In the formercase, the intersection number D · L ′ = 233shows that the point P cannot be P . In the latter case, the intersection number D · R = 111excludes the point P . By the same method, we can show P L .Hence the point P must be a smooth point in X \ C x . For the case n >
2, we can useLemma 1.3.9 to get a contradiction1 < mult P ( D ) n + 15) · n + 5)6(6 n + 3)(6 n + 5)(6 n + 5) = 42 n + 1 < , since H ( P , O P (6(6 n + 5)) contains x n +5 , y x and z . For the case n = 1, let R P be the uniquecurve in the pencil |O X (11) | that passes through the point P . The log pair ( X, R P ) is logcanonical at the point P . By Lemma 1.3.6, we may assume that Supp ( D ) does not contain at least one irreducible component of R P . Note that either R P is irreducible or P k ∈ R P for some k = 1 , ,
3. However, if R P is irreducible, then we can obtain a contradiction1 < mult P ( D ) D · R P = 29 . Thus, P k ∈ R P . Then R P consists of three distinct irreducible curves L k , L ′ k and Z . We have D · L ′ k = 233 , D · Z = 433 , L ′ k = − , Z = − . Put D = m Z + m L ′ k + ∆, where ∆ is an effective Q -divisor whose support contains neither Z nor L ′ k . Since the pair ( X, D ) is log canonical at the point P k , we have m , m
1. Since wealready showed that P is located in the outside of L k , the point P must belong to either L ′ k or Z . However, Lemma 1.3.8 shows that the pair ( X, D ) is log canonical at the point P since( D − m Z ) · Z = 4 + 4 m < , ( D − m L ′ k ) · L ′ k = 4 + 13 m < . This is a contradiction. (cid:3)
Lemma 2.3.2.
Let X be a quasismooth hypersurface of degree 36 n + 24 in P (6 , n + 5 , n +8 , n + 9) for n >
1. Then lct( X ) = 1. Proof.
We may assume that the surface X is defined by the equation z + y t + xt − x n +4 + ax n +1 y z = 0 , where a is a constant. The only singularities of X are a singular point O y of index 6 n + 5, asingular point O t of index 18 n + 9, a singular point Q = [1 : 0 : 0 : 1] of index 3, and a singularpoint Q ′ = [1 : 0 : 1 : 0] of index 2.The curve C x is reduced and irreducible with mult O t ( C x ) = 3. Clearly, lct( X, C x ) = 1,and hence lct( X )
1. The curve C y is quasismooth, and hence the log pair ( X, n +5 C y ) is logcanonical.Suppose that lct( X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .Since H ( P , O P (36 n + 30)) contains x n +5 , y and z x , Lemma 1.3.9 impliesmult P ( D ) n + 24)(36 n + 30)6(6 n + 5)(12 n + 8)(18 n + 9) < . Therefore, the point P cannot be a smooth point in the outside of C x .By Lemma 1.3.6 we may assume that neither C x nor C y is contained in Supp ( D ). Then theinequality 3 D · C y = 26 n + 3 P is neither Q nor Q ′ . One the other hand, the inequality(6 n + 5) D · C x = 46 n + 3 < XCEPTIONAL DEL PEZZO HYPERSURFACES 27 shows that the point P can be neither a smooth point on C x nor the point O y . Therefore, itmust be O t . However, this is a contradiction sincemult O t ( D ) = mult O t ( D )mult O t ( C x )3 n + 93 D · C x = 46 n + 5 < . The obtained contradiction completes the proof. (cid:3)
Lemma 2.3.3.
Let X be a quasismooth hypersurface of degree 36 n + 30 in P (6 , n + 5 , n +8 , n + 15) for n >
1. Then lct( X ) = 1. Proof.
We may assume that the surface X is defined by the equation( t − a y )( t − a y ) + xz − x n +5 + ax n +1 y z = 0 , where a = a and a are constants. The only singularities of X are a singular point O z of index12 n + 8, a singular point Q = [1 : 0 : 1 : 0] of index 2, a singular point Q ′ = [1 : 0 : 0 : 1] of index3, and two singular points P = [0 : 1 : 0 : a ], P = [0 : 1 : 0 : a ] of index 6 n + 5.The curve C x consists of two distinct irreducible curves L i = { x = t − a i y = 0 } , i = 1 , L i passes through the point P i . These two curves meet each other at the point O z . It iseasy to see lct( X, C x ) = 1.Suppose that lct( X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6 we may assume that L is not contained in Supp ( D ). Then the inequality(12 n + 8) D · L = 46 n + 5 < P must be located in the outside of L .Write D = µL + Ω, where Ω is an effective Q -divisor such that L Supp (Ω). Then, theinequality 3 µ n + 8 = µL · L D · L = 1(3 n + 2)(6 n + 5) , implies µ n + 5) . Note that L = − n + 9(12 n + 8)(6 n + 5) . Since (6 n + 5)Ω · L = 4 + (18 n + 9) µ n + 8 < n + 5 , Lemma 1.3.8 excludes all the points of L \ { O z } . Consequently, the point P is in the outsideof C x .Meanwhile, the curve C y is quasismooth, and hence the log pair ( X, n +5 C y ) is log canonical.Lemma 1.3.6 enables us to assume that C y is not contained in Supp ( D ). Then the inequality3 C y · D = 13 n + 2 , excludes the singular points Q and Q ′ . Hence P is a smooth point of X \ C x . Applying Lemma 1.3.9, we see that1 < mult P ( D ) n + 30)(3(12 n + 8) + 6)6(6 n + 5)(12 n + 8)(18 n + 15) < , because H ( P , O P (3(12 n + 8) + 6)) contains x n +5 , y and z x . The obtained contradictioncompletes the proof. (cid:3) Infinite series with I = 6 Lemma 2.4.1.
Let X be a quasismooth hypersurface of degree 12 n + 23 in P (8 , n + 5 , n +7 , n + 9) for n >
3. Then lct( X ) = 1. Proof.
The surface X can be given by the equation z t + yt + xy + x n +2 z = 0 . The surface X is singular only at O x , O y , O z and O t .The curve C x (resp. C y , C z , C t ) consists of the irreducible curve L xt (resp. L yz , L yz , L xt ) and aresidual curve R x = { x = z + yt = 0 } (resp. R y = { y = x n +2 + zt = 0 } , R z = { z = t + xy = 0 } , R t = { t = y + x n +1 z = 0 } ) . These two curves intersect each other at O y (resp. O t , O x , O z ).We can easily see thatlct( X, C x ) = 1 , lct( X, n + 5 C y ) = ( n + 3)(4 n + 5)12( n + 2) , lct( X, n + 7 C z ) = 4 n + 79 , lct( X, n + 9 C t ) = (4 n + 9)( n + 4)6(3 n + 6) . Therefore, lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .We have the following intersection numbers: L xt · D = 6(4 n + 5)(4 n + 7) , L yz · D = 68(4 n + 9) , R x · D = 12(4 n + 5)(4 n + 9) ,R y · D = 6( n + 2)(4 n + 7)(4 n + 9) , R z · D = 128(4 n + 5) , R t · D = 188(4 n + 7) ,L xt · R x = 24 n + 5 , L xt · R t = 34 n + 7 , L yz · R y = n + 24 n + 9 , L yz · R z = 14 ,L xt = − n + 6(4 n + 5)(4 n + 7) , L yz = − n + 118(4 n + 9) , R x = − n + 2(4 n + 5)(4 n + 9) ,R y = − n + 4(4 n + 7)(4 n + 9) , R z = 12(4 n + 5) , R t = 12 n + 38(4 n + 7) . By Lemma 1.3.6 we may assume that either L xt Supp ( D ) or R x Supp ( D ). Then at leastone of the inequalitiesmult O y ( D ) (4 n + 5) L xt · D = 64 n + 7 , mult O y ( D ) (4 n + 5) R x · D = 124 n + 9 XCEPTIONAL DEL PEZZO HYPERSURFACES 29 holds. Therefore, the point P cannot be the point O y . We also may assume that either L yz Supp ( D ) or R z Supp ( D ). Then at least one of the inequalitiesmult O x ( D ) L yz · D = 64 n + 9 , mult O x ( D ) R z · D = 64 n + 5holds. Note that the curve R z is singular at the point O x . Therefore, the point P cannot be thepoint O x . We also may assume that either L xt Supp ( D ) or R t Supp ( D ). Then at leastone of the inequalitiesmult O z ( D ) (4 n + 7) L xt · D = 64 n + 5 , mult O z ( D ) n + 73 R t · D = 34holds. Note that the curve R t has multiplicity 3 at the point O z if n >
2. Therefore, the point P cannot be the point O z .Write D = m L xt + m L yz + m R x + m R y + m R z + m R t + Ω, where Ω is an effective Q -divisor whose support contains none of L xt , L yz , R x , R y , R z , R t .If m >
0, then m = 0. Therefore, the inequality2 m n + 5 = m L xt · R x D · R x = 12(4 n + 5)(4 n + 9)shows 0 m n +9 . By Lemma 1.3.8 the inequality( D − m L xt ) · L xt = 6 + m (8 n + 6)(4 n + 5)(4 n + 7) n + 7)(4 n + 9) < P cannot be a smooth point on L xt .If m >
0, then R z Supp ( D ). Therefore, the inequality m m L yz · R z D · R z = 32(4 n + 5)shows 0 m n +5 . By Lemma 1.3.8 the inequality( D − m L yz ) · L yz = 6 + m (4 n + 11)8(4 n + 9) n + 2)(4 n + 5)(4 n + 9) < P cannot be a smooth point on L yz .If m >
0, then m = 0, and hence2 m n + 5 = m L xt · R x D · L xt = 6(4 n + 5)(4 n + 7) . Therefore, 0 m n +7 . The inequality( D − m R x ) · R x = 12 + m (8 n + 2)(4 n + 5)(4 n + 9) n + 7)(4 n + 9) < P cannot be a smooth point on R x . Moreover, this inequality shows thatthe point P cannot be the point O t since n > m >
0, then we may assume that m = 0. We then obtain m ( n + 2)4 n + 9 = m R y · L yz D · L yz = 34(4 n + 9) . Therefore, 0 m n +2) . The inequality( D − m R y ) · R y = 6( n + 2) + 2 m ( n + 2)(4 n + 7)(4 n + 9) n + 7) < P cannot be a smooth point on R y .Since the pair ( X, D ) is log canonical at the point O x and the curve R z contains the point O x , we have m
1. By Lemma 1.3.8, the inequality( D − m R z ) · R z D · R z = 32(4 n + 5) < P cannot be a smooth point on R z .The pair ( X, D ) is log canonical at the point O x and the curve R t contains the point O x .Thus m
1. By Lemma 1.3.8, the inequality( D − m R t ) · R t D · R t = 94(4 n + 7) < P cannot be a smooth point of R t .Consider the pencil L defined by the equations λxy + µt = 0, [ λ : µ ] ∈ P . Note thatthe curve L xt is the only base component of the pencil L . There is a unique divisor C α in L passing through the point P . This divisor must be defined an equation xy + αt = 0, where α is a non-zero constant, since the point P is located in the outside of C x ∪ C y ∪ C z ∪ C t . Notethat the curve C y does not contain any component of C α . Therefore, to see all the irreduciblecomponents of C α , it is enough to see the affine curve ( x + αt = 0 z t + t + x + x n +2 z = 0 ) ⊂ C ∼ = Spec (cid:16) C (cid:2) x, z, t (cid:3)(cid:17) . This is isomorphic to the plane affine curve defined by the equation t { z + (1 − α ) t + ( − α ) n +2 t n +1 z } = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) z, t (cid:3)(cid:17) . Thus, if α = 1, then the divisor C α consists of two reduced and irreducible curves L xt and Z α .If α = 1, then it consists of three reduced and irreducible curves L xt , R z , R . Moreover, Z α and R contain the point P and they are smooth at the point P .Suppose that α = 1. It is easy to check D · Z α = 3(12 n + 19)2(4 n + 5)(4 n + 7) . We also see that Z α = C α · Z α − L xt · Z α > C α · Z α − ( L xt + R x ) · Z α = 4 n + 53 D · Z α > Z α is different from the curve R x . Put D = ǫZ α + Ξ, where Ξ is an effective Q -divisorsuch that Z α Supp(Ξ). Since the pair (
X, D ) is log canonical at the point O y and the curve Z α passes through the point O y , we have ǫ
1. But( D − ǫZ α ) · Z α D · Z α = 3(12 n + 19)2(4 n + 5)(4 n + 7) < XCEPTIONAL DEL PEZZO HYPERSURFACES 31 and hence Lemma 1.3.8 implies that the point P cannot belong to the curve Z α .Suppose that α = 1. Then we have D · R = 6(2 n + 3)(4 n + 5)(4 n + 7) . Since R is different from L yz and R x , R = C α · R − L xt · R − R z · R > C α · R − ( L xt + R x ) · R − ( L yz + R z ) · R > n + 36 D · R > D = ǫ R + Ξ ′ , where Ξ ′ is an effective Q -divisor such that R Supp(Ξ ′ ). Since the curve R passes through the point O y at which the pair ( X, D ) is log canonical, we have ǫ
1. Since( D − ǫ R ) · R D · R = 6(2 n + 3)(4 n + 5)(4 n + 7) < . Lemma 1.3.8 implies that the point P cannot belong to R . (cid:3) Lemma 2.4.2.
Let X be a quasismooth hypersurface of degree 47 in P (8 , , , X ) = 1. Proof.
If we exclude the point O t , then the proof of Lemma 2.4.1 works for this case. Thus wesuppose that P = O t . Then L yz Supp( D ); otherwise we would have a contradictory inequality34 ·
17 = D · L yz > mult P ( D ) > . By Lemma 1.3.6, we may assume that R y Supp( D ). Put D = mL yz + cR x + Ω , where m > c >
0, and Ω is an effective Q -divisor whose support contains neither L yz nor R x . Then 2415 ·
17 = D · R y = (cid:0) mL yz + cR x + Ω (cid:1) · R y > m
17 + mult O t ( D ) − m > m + 117 , and hence m < . Then it follows from Lemma 1.3.8 that6 + 19 m ·
17 = ( D − mL yz ) · L yz > , and hence 219 < m. On the other hand, if c >
0, then613 ·
15 = D · L xt > cR x · L xt = 2 c . Therefore, 0 c . Let π : ¯ X → X be the weighted blow up at the point O t with weights (6 , E be theexceptional curve of π . Also we let ¯Ω, ¯ L yz and ¯ R x be the proper transforms of Ω, L yz and R x ,respectively. Then K ¯ X ∼ Q π ∗ ( K X ) − E, ¯ L yz ∼ Q π ∗ ( L yz ) − E, ¯ R x ∼ Q π ∗ ( R x ) − E, ¯Ω ∼ Q π ∗ (Ω) − a E, where a is a non-negative rational number.The curve E contains two singular points Q and Q of ¯ X . The point Q is a singular point oftype (1 ,
3) and the point Q is a singular point of type (1 , Q is containedin ¯ R x but not in ¯ L yz , on the other hand, Q is contained in ¯ L yz but not in ¯ R x . We also see that¯ L yz ∩ ¯ R x = ∅ . The log pull back of the log pair ( X, D ) is the log pair (cid:18) ¯ X, ¯Ω + m ¯ L yz + c ¯ R x + 4 + a + 7 m + 6 c E (cid:19) . This pair must have non-log canonical singularity at some point Q ∈ E . Then0 ¯ R x · ¯Ω = R x · Ω + 6 a E = 12 − m + 18 c · − a · , ¯ L yz · ¯Ω = L yz · Ω + 7 a E = 6 + 19 m − c · − a · , and hence 0 − a + 126 c − m and 0 − a − c + 57 m . In particular, we see that a . Then 4 + a + 7 m + 6 c <
17 since m and c .Suppose that the point Q is neither Q nor Q . Then the point Q must be located in theoutside of ¯ L yz and ¯ R x . By Lemma 1.3.8, we have a
42 = − a E = ¯Ω · E > , and hence a >
42. This is a contradiction since a < . Therefore, either Q = Q or Q = Q .Suppose that Q = Q . Then Q ¯ L yz . Hence, it follows from Lemma 1.3.8 that17 (cid:18) ¯Ω + m ¯ L yx + 4 + a + 7 m + 6 c E (cid:19) · ¯ R x = 136 + 204 c · · , and hence c > . But c . This is a contradiction.Finally, we suppose that Q = Q . Then Q ¯ R x . It follows from Lemma 1.3.8 that16 (cid:18) ¯Ω + c ¯ R x + 4 + a + 7 m + 6 c E (cid:19) · ¯ L yz = 34 + 85 m · · , and hence m > . This contradiction completes the proof. (cid:3) Lemma 2.4.3.
Let X be a quasismooth hypersurface of degree 35 in P (8 , , , X ) = 1. Proof.
If we exclude the points O z and O t , then the proof of Lemma 2.4.1 works also for thiscase.Suppose that P = O z . Then L xt ⊂ Supp( D ), since otherwise we would have an absurdinequality 69 ·
11 = D · L xt > . XCEPTIONAL DEL PEZZO HYPERSURFACES 33
We may assume that M t Supp( D ) by Lemma 1.3.6. Put D = mL xt + cM y + Ω , where m > c >
0, and Ω is an effective Q -divisor whose support contains neither L xt nor R y . Then188 ·
11 = D · R t = (cid:0) mL xt + cR y + Ω (cid:1) · R t > m
11 + 2(mult O z ( D ) − m )11 > m + 211 , and hence m < . Note that mult O z ( R t ) = 2. It follows from Lemma 1.3.8 that6 + 14 m ·
11 = (cid:0) D − mL xt (cid:1) · L xt > . Therefore, < m < . On the other hand, if c >
0, then68 ·
13 = D · L yz > cR y · L yz = 3 c , and hence c .Let π : ¯ X → X be the weighted blow up at the point O z with weights (3 , E bethe exceptional curve of π and let ¯Ω, ¯ L xt and ¯ R y be the proper transforms of Ω, L xt and R y ,respectively. Then K ¯ X ∼ Q π ∗ ( K X ) − E, ¯ L xt ∼ Q π ∗ ( L xt ) − E, ¯ R y ∼ Q π ∗ ( R y ) − E, ¯Ω ∼ Q π ∗ (Ω) − a E. where a is a non-negative rational number.The curve E contains two singular points Q and Q of ¯ X . The point Q is a singular point oftype (1 , L xt but not in ¯ R y . On the other hand, the point Q is a singularpoint of type (2 , R y but not in ¯ L xt . But ¯ L xt ∩ ¯ R y = ∅ .The log pull back of the log pair ( X, D ) is the log pair (cid:18) ¯ X, ¯Ω + m ¯ L xt + c ¯ R y + 6 + a + 3 m + 2 c E (cid:19) , which must have non-log canonical singularity at some point Q ∈ E . We have0 ¯Ω · ¯ R y = 18 + 6 c · − m − a , ¯Ω · ¯ L xt = 6 + 14 m · − c − a . Then, a m < since m < . Also, we obtain 6 + a + 3 m + 2 c <
11 since c .Suppose that the point Q is neither Q nor Q . Then Q ¯ L xt ∪ ¯ R y . By Lemma 1.3.4, wehave a · − a E = ¯Ω · E > , and hence a >
6. This contradicts to the inequality a < . Therefore, we see that either Q = Q or Q = Q . Suppose that Q = Q . Then Q ¯ R y . Lemma 1.3.8 shows that12 < (cid:18) ¯Ω + c ¯ R y + 6 + a + 3 m + 2 c E (cid:19) · ¯ L xt = 66 + 55 m · · m > . But m < . This is a contradiction.Thus, the point Q must be Q . Then Q ¯ L xt . It follows from Lemma 1.3.8 that13 < (cid:18) ¯Ω + m ¯ L xt + 6 + a + 3 m + 2 c E (cid:19) · ¯ R y = 132 + 44 c · . Therefore, c > . But we have seen c . The obtained contradiction shows that P = O z . Thepoint P must be the point O t . Then L yz Supp( D ) since otherwise we would have68 ·
13 = D · L yz > . By Lemma 1.3.6, we may assume that R y Supp( D ). Put D = mL yz + cR x + Ω , where m > c >
0, and Ω is an effective Q -divisor whose support contains neither L yz nor R x . Then 1811 ·
13 = D · R y = (cid:0) mL yz + cR x + Ω (cid:1) · R y > m
13 + mult O t ( D ) − m > m + 113 , and hence m < . On the other hand, Lemma 1.3.8 implies6 + 15 m ·
13 = (cid:0) D − mL yz (cid:1) · L yz > , and hence < m < . If c >
0, then69 ·
11 = D · L xt = cR x · L xt > c . Therefore, c .Let π : ¯ X → X be the weighted blow up at the point O t with weights (5 , E bethe exceptional curve of π . Let ¯Ω, ¯ L yz and ¯ R x be the proper transforms of Ω, L yz and R x ,respectively. Then K ¯ X ∼ Q π ∗ ( K X ) − E, ¯ L yz ∼ Q π ∗ ( L yz ) − E, ¯ R x ∼ Q π ∗ ( R x ) − E, ¯Ω ∼ Q π ∗ (Ω) − a E, where a is a non-negative rational number.The curve E contains two singular points Q and Q of ¯ X . The point Q is a singular pointof type (1 , L yz but not to ¯ R x . The point Q is a singular point of type (1 , R x but not to ¯ L yz . Note that ¯ L yz ∩ ¯ R x = ∅ .The log pull back of the log pair ( X, D ) is the log pair (cid:18) ¯ X, ¯Ω + m ¯ L yz + c ¯ R x + 6 + a + 2 m + 5 c E (cid:19) . XCEPTIONAL DEL PEZZO HYPERSURFACES 35
It must have non-log canonical singularity at some point Q ∈ E . We have0 ¯Ω · ¯ R x = 12 + 10 c · − m − a , ¯Ω · ¯ L yz = 6 + 15 m · − c − a . Therefore, 30 + 75 m > c + 8 a and 24 + 20 c > m + 9 a . In particular, we see that a .Then 6 + a + 2 m + 5 c <
13 since c and m .Suppose that Q = Q and Q = Q . Then Q ¯ L yz ∪ ¯ R x . By Lemma 1.3.8, we have a
10 = − a E = ¯Ω · E > , and hence a >
10. This is a contradiction since a < . Therefore, the point Q is either thepoint Q or the point Q .Suppose that Q = Q . Then Q ¯ L yz . It follows from Lemma 1.3.8 that12 < (cid:18) ¯Ω + m ¯ L yz + 6 + a + 2 m + 5 c E (cid:19) · ¯ R x = 78 + 65 c · , and hence c > . However, c . Thus, the point Q must be Q . Then Q ¯ R x . Again,Lemma 1.3.8 shows that15 < (cid:18) ¯Ω + c ¯ R x + 6 + a + 2 m + 5 c E (cid:19) · ¯ L yz = 78 + 91 m · · , < (cid:0) ¯Ω + m ¯ L yz (cid:1) · E = a
10 + m . Therefore, m > and a + 2 m >
2. In particular, < m < .Let ψ : ˜ X → ¯ X be the weighted blow up at the point Q with weights (1 , G be theexceptional curve of ψ and let ˜Ω, ˜ L yz , ˜ R x and ˜ E be the proper transforms of Ω, L yz , R x and E ,respectively. Then K ˜ X ∼ Q ψ ∗ ( K ¯ X ) − G, ˜ L yz ∼ Q ψ ∗ ( ¯ L yz ) − G, ˜ E ∼ Q ψ ∗ ( E ) − G, ˜Ω ∼ Q ψ ∗ ( ¯Ω) − b G, where b is a non-negative rational number.The surface is smooth along G . The log pull back of ( X, D ) is the log pair (cid:18) ˜ X, ˜Ω + m ˜ L yz + c ˜ R x + 6 + a + 2 m + 5 c
13 ˜ E + θG (cid:19) , where θ = 15 m + 45 + a + 13 b + 5 c . Then the log pair is not log canonical at some point O ∈ G . We have0 ˜ E · ˜Ω = a − b , ˜ L yz · ˜Ω = 6 + 15 m · − c − a − b , and hence 30 + 75 m > a + 13 b + 5 c ) and a > b . In particular, we obtain θ = 15 m + 45 + a + 13 b + 5 c m + 3908 · · · · · < m .Suppose that O ˜ E ∪ ˜ L yz . Then it follows from Lemma 1.3.8 that b = − b G = ˜Ω · G > . However, this gives an absurd inequality 104 < b
30 + 75 m − a − c
30 + 75 m < m . Therefore, O ∈ ˜ E ∪ ˜ L yz . Note that ˜ E ∩ ˜ L yz = ∅ .Suppose that O ∈ ˜ L yz . Then it follows from Lemma 1.3.8 that1 < (cid:0) ˜Ω + c ˜ R x + 6 + a + 2 m + 5 c
13 ˜ E + θG (cid:1) · ˜ L yz = (cid:0) ˜Ω + θG (cid:1) · ˜ L yz = 3 m + 68 , and hence m > . But m . Thus, we see that O ∈ ˜ E . Lemma 1.3.8 implies that1 < (cid:18) ˜Ω + 6 + a + 2 m + 5 c
13 ˜ E (cid:19) · G = b + 6 + a + 2 m + 5 c , < (cid:0) ˜Ω + θG (cid:1) · ˜ E = a − b θ. Therefore, we obtain 13 b + a + 2 m + 5 c > a + 2 c + 6 m > φ : ˆ X → ˜ X be the blow up at the point O . Let F be the exceptional curve of φ . Let ˆΩ,ˆ L yz , ˆ R x , ˆ E and ˆ G be the proper transforms of Ω, L yz , R x , E and G , respectively. Then K ˆ X ∼ Q φ ∗ ( K ˜ X ) + F, ˆ G ∼ Q φ ∗ ( G ) − F, ˆ E ∼ Q φ ∗ ( ˜ E ) − F, ˆΩ ∼ Q φ ∗ ( ˜Ω) − dF, where d is a non-negative rational number. The log pull back of ( X, D ) is the log pair (cid:18) ˆ X, ˆΩ + m ˆ L yz + c ˆ R x + 6 + a + 2 m + 5 c
13 ˆ E + θ ˆ G + νF (cid:19) , where ν = 65 d + 25 m + 6 a + 13 b + 30 c + 1065 . It is not log canonical at some point A ∈ F . We have0 ˆ E · ˆΩ = a − b − d, ˆ G · ˆΩ = b − d, XCEPTIONAL DEL PEZZO HYPERSURFACES 37 and hence b > d and a > b + 10 d . In particular, ν = 65 d + 25 m + 6 a + 13 b + 30 c + 1065 == 13(5 d + b ) + 25 m + 6 a + 30 c + 1065 a + 10 m + 12 c + 426 c < c > m + 9 a and c .Suppose that A ˆ E ∪ ˆ G . Then Lemma 1.3.8 shows that d = ˆΩ · F >
1. This is impossiblesince 10 d a − b a . Thus, we see that A ∈ ˆ E ∪ ˆ G . Note that ˆ E ∩ ˆ G = ∅ .Suppose that A ∈ ˆ E . Then it follows from Lemma 1.3.8 that a − b − d + ν = (cid:0) ˆΩ + νF (cid:1) · ˆ E > , which implies that 5 a + 10 m + 12 c >
22. However, this inequality with 24 + 20 c > m + 9 a gives 95 (22 − c ) <
95 (5 a + 10 m )
24 + 20 c, and hence < c . But c . Thus, the point A cannot belong to ˆ E . Then A ∈ ˆ G . ByLemma 1.3.8, we see that b − d + ν = (cid:0) ˆΩ + νF (cid:1) · ˆ G > , and hence 6 a + 25 m + 30 c + 78 b >
55. But55 < m + 6 a + 78 b + 3 c = 25 m + 34 (8 a + 104 b + 40 c ) m + 34 (30 + 75 m ) < a + 104 b + 40 c
30 + 75 m and m . The obtained contradiction completes theproof. (cid:3) Lemma 2.4.4.
Let X be a quasismooth hypersurface of degree 12 n + 35 in P (9 , n + 8 , n +11 , n + 13) for n >
1. Then lct( X ) = 1. Proof.
The surface X can be defined by the equation z t + y z + xt + x n +3 y = 0 . It is singular only at the points O x , O y , O z and O t .The curve C x (resp. C y , C z , C t ) consists of two irreducible and reduced curves L xz (resp. L yt , L xz , L yt ) and R x = { x = zt + y = 0 } (resp. R y = { y = z + xt = 0 } , R z = { z = t + x n +2 y = 0 } , R t = { t = y z + x n +3 = 0 } ). These two curves intersect at the point O t (resp. O x , O y , O z ). It is easy to see that lct( X, C x ) = 1 is less than each of the numberslct( X, n + 8 C y ) , lct( X, n + 11 C z ) , lct( X, n + 13 C t ) . We have the following intersection numbers. − L xz · K X = 6(3 n + 8)(6 n + 13) , − L yt · K X = 23(3 n + 11) , − R x · K X = 18(3 n + 11)(6 n + 13) , − R y · K X = 43(6 n + 13) , − R z · K X = 43(3 n + 8) , − R t · K X = 6( n + 3)(3 n + 8)(3 n + 11) ,L xz · R x = 36 n + 13 , L yt · R y = 29 , L xz · R z = 23 n + 8 , L yt · R t = n + 33 n + 11 ,L xz = − n + 15(3 n + 8)(6 n + 13) , L yt = − n + 149(3 n + 11) , R x = − n + 6(3 n + 11)(6 n + 13) ,R y = − n + 109(6 n + 13) , R z = 6 n + 49(3 n + 8) , R t = ( n + 3)(3 n + 5)(3 n + 8)(3 n + 11) . Now we suppose that lct( X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such thatthe log pair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6 we may assume that Supp ( D ) does not contain either the curve L yt or thecurve R y . Since these two curves intersect at the point O x , the inequalities L yt · D = 23(3 n + 11) < ,R y · D = 43(6 n + 13) < P cannot be the point O x .By Lemma 1.3.6 we may assume that Supp ( D ) does not contain either the curve L xz or thecurve R z . Therefore, one of the following inequalities must hold:mult O y ( D ) (3 n + 8) L xz · D = 66 n + 13 < , mult O y ( D ) n + 82 R z · D = 23 . Therefore, the point P cannot be the point O y .Suppose that P = O z . If L yt Supp ( D ), then we get an absurd inequality69(3 n + 11) = L yt · D > n + 11 . Therefore Supp ( D ) must contain the curve L yt . By Lemma 1.3.6 we may assume that M t Supp ( D ). Put D = µL yt + Ω, where Ω is an effective Q -divisor whose support does not containthe curve L yt . Then6( n + 3)(3 n + 8)(3 n + 11) = D · R t > µL yt · R t + (mult P ( D ) − µ )mult P ( R t )3 n + 11 > µ ( n + 3)3 n + 11 + 2(1 − µ )3 n + 11 , XCEPTIONAL DEL PEZZO HYPERSURFACES 39 and hence µ < n + 8)( n + 1) . On the other hand, Theorem 1.3.3 shows13 n + 11 < Ω · L yt = D · L yt − µL yt = 6 + µ (3 n + 14)9(3 n + 11) . It implies n +14 < µ . Consequently, the point P cannot be the point O z .Suppose that P = O t . Since L xz · D < n +13 , the curve L xz must be contained in Supp ( D ).Then, we may assume that R x Supp ( D ). Put D = µL xz + Ω, where Ω is an effective Q -divisorwhose support does not contain the curve L xz . Then18(3 n + 11)(6 n + 13) = D · R x > µL xz · R x + mult P ( D ) − µ n + 13 > µ n + 13 , and hence µ < − n n + 22 . However, Theorem 1.3.3 implies16 n + 13 < Ω · L xz = D · L xz − µL xz = 6 + (9 n + 15) µ (3 n + 8)(6 n + 13) , and hence n +29 n +15 < µ . This is a contradiction. Therefore, the point P cannot be the point O t .Write D = aL xz + bR x + ∆, where ∆ is an effective Q -divisor whose support contains neither L xz nor R x . Since the log pair ( X, D ) is log canonical at the point O t , we have 0 a, b bR x + ∆) · L xz = ( D − aL xz ) · L xz = 6 + a (9 n + 15)(3 n + 8)(6 n + 13) < , ( aL xz + ∆) · R x = ( D − bR x ) · R x = 18 + b (9 n + 6)(3 n + 11)(6 n + 13) < P cannot belong to the curve C x . By the same way, we can show P C y ∪ C z ∪ C t .Consider the pencil L defined by the equations λxt + µz = 0, [ λ : µ ] ∈ P . Note that thecurve L xz is the only base component of the pencil L . There is a unique divisor C α in L passingthrough the point P . This divisor must be defined an equation xt + αz = 0, where α is anon-zero constant, since the point P is located in the outside of C x ∪ C z ∪ C t . Note that thecurve C t does not contain any component of C α . Therefore, to see all the irreducible componentsof C α , it is enough to see the affine curve ( x + αz = 0 z + y z + x + x n +3 y = 0 ) ⊂ C ∼ = Spec (cid:16) C (cid:2) x, y, z (cid:3)(cid:17) . This is isomorphic to the plane affine curve defined by the equation z { (1 − α ) z + y + ( − α ) n +3 yz n +5 } = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . Thus, if α = 1, then the divisor C α consists of two reduced and irreducible curves L xz and Z α .If α = 1, then it consists of three reduced and irreducible curves L xz , R y , R . Moreover, Z α and R are smooth at the point P .Suppose that α = 1. Then we have D · Z α = 2(24 n + 61)3(3 n + 8)(6 n + 13) . Since Z α is different from R x , Z α = C α · Z α − L xz · Z α > C α · Z α − ( L xz + R x ) · Z α = 6 n + 136 D · Z α > . Put D = ǫZ α + Ξ, where Ξ is an effective Q -divisor such that Z α Supp(Ξ). Since the pair(
X, D ) is log canonical at the point O t and the curve Z α passes through the point O t , we have ǫ
1. But ( D − ǫZ α ) · Z α D · Z α = 2(24 n + 61)3(3 n + 8)(6 n + 13) < P cannot belong to the curve Z α .Suppose that α = 1. We have D · R = 6(2 n + 5)(3 n + 8)(6 n + 13) . Since R is different from R x and L yt , R = C α · R − L xz · R − R y · R > C α · R − ( L xz + R x ) · R − ( L yt + R y ) · R = 3 n + 56 D · D > . Put D = ǫ R + Ξ ′ , where Ξ ′ is an effective Q -divisor such that R Supp(Ξ ′ ). Since the curve R passes through the point O t at which the pair ( X, D ) is log canonical, ǫ
1. Since( D − ǫ R ) · R D · R = 6(2 n + 5)(3 n + 8)(6 n + 13) < , Lemma 1.3.8 implies that the point P cannot belong to R . (cid:3) Part Sporadic cases
Sporadic cases with I = 1 Lemma 3.1.1.
Let X be a quasismooth hypersurface of degree 10 in P (1 , , , (cid:0) X (cid:1) = C x has an ordinary double point ,
710 if C x has a non-ordinary double point . Proof.
The surface X is singular only at the point O z . The curve C x is reduced and irreducible.Moreover, we havelct (cid:0) X, C x (cid:1) = C x has an ordinary double point at the point O z ,
710 if the curve C x has a non-ordinary double point at the point O z . XCEPTIONAL DEL PEZZO HYPERSURFACES 41
Suppose that lct (cid:0) X (cid:1) < lct (cid:0) X, C x (cid:1) . Then there is an effective Q -divisor D ∼ Q − K X such thatthe log pair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assumethat the support of D does not contain the curve C x . Also Lemma 1.3.9 shows that P ∈ C x .However, we obtain absurd inequalities13 = D · C x > mult P (cid:0) D (cid:1) > P = O z , mult P (cid:0) D (cid:1) >
13 if P = O z . Therefore, lct (cid:0) X (cid:1) = lct (cid:0) X, C x (cid:1) . (cid:3) Lemma 3.1.2.
Let X be the quasismooth hypersurface defined by a quasihomogeneous poly-nomial f ( x, y, z, t ) of degree 15 in P (1 , , , (cid:0) X (cid:1) = f ( x, y, z, t ) contains yzt,
815 if f ( x, y, z, t ) does not contain yzt. Proof.
The surface X is singular only at the point O t . The curve C x is reduced and irreducible.It is easy to check lct (cid:0) X, C x (cid:1) = f ( x, y, z, t ) contains yzt,
815 if f ( x, y, z, t ) does not contain yzt. The proof is exactly the same as the proof of Lemma 3.1.1. The contradictory inequalities17 = D · C x > mult P (cid:0) D (cid:1) > P = O t , mult P (cid:0) D (cid:1) >
17 if P = O t . complete the proof. (cid:3) Lemma 3.1.3.
Let X be a quasismooth hypersurface of degree 16 in P (1 , , , X ) =1. Proof.
The surface X is singular only at the points O y and O z . The former is a singular pointof type (1 ,
1) and the latter is of type (1 , C x consists of two distinct irreducible curves L and L . It is easy to see thatlct( X, C x ) = 1.Suppose that lct (cid:0) X (cid:1) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assume that thesupport of D does not contain the curve L without loss of generality. Moreover, Lemma 1.3.9implies P ∈ C x .We have D · L = D · L = 115 , and L ∩ L = { O y , O z } . We also have L = L = − , L · L = 815 . Since 5 D · L = , the point P cannot belong to L . Therefore, the point P is a smooth pointon L . Put D = mL + Ω , where Ω is an effective Q -divisor such that L Supp(Ω). Since the log pair (
X, D ) is logcanonical at O y , we must have m
1. Then it follows from Lemma 1.3.8 that1 < Ω · L = (cid:0) D − mL (cid:1) · L = 1 + 7 m . This gives us m >
2. This is a contradiction. Consequently, lct( X ) = 1. (cid:3) Lemma 3.1.4.
Let X be a quasismooth hypersurface of degree 18 in P (2 , , , (cid:0) X (cid:1) = C y has a tacnodal point ,
116 if C y has no tacnodal points . Proof.
The surface X is singular at the point O z . This is a singular point of type (1 , X also has two singular points O and O that are cut out by the equations x = z = 0.These are of type (1 ,
1) on the surface X .The curves C x and C y are reduced and irreducible. The curve C y is always singular at thepoint O z . We can see lct( X, C x ) = 1 andlct (cid:0) X, C y (cid:1) =
34 if C y has a tacnodal singularity at the point O z , C y has a non-tacnodal singularity at the point O z . Therefore, if C y has a tacnodal singularity at the point O z , then2 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 94 . If C y has a non-tacnodal singularity at the point O z , then2 = lct (cid:18) X, C x (cid:19) > lct (cid:18) X, C y (cid:19) = 116 . Let ǫ = min (cid:8) lct (cid:0) X, C x (cid:1) , lct (cid:0) X, C y (cid:1)(cid:9) . Then lct( X ) ǫ .Suppose that lct( X ) < ǫ . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, ǫD ) is not log canonical at some point P ∈ X . By Lemma 1.3.6, we may assume thatthe support of the divisor D contains neither the curve C x nor the curve C y .The inequalities mult O z ( ǫD ) ǫ O z ( D )mult O z ( C y ) D · C y = 1imply that the point P cannot be the point O z . If the point P is a smooth point on C y , thenwe have obtain a contradictory inequalities15 = D · C y > mult P ( D ) > ǫ > . Therefore, the point P is located in the outside of the curve C y . XCEPTIONAL DEL PEZZO HYPERSURFACES 43
Suppose that P ∈ C x . Then we obtain the following contradictory inequalities215 = D · C x > mult P (cid:0) D (cid:1) >
12 if P ∈ X \ Sing( X ) , mult P (cid:0) D (cid:1) >
16 if P = O or P = O . Therefore, P C x ∪ C y . Then P is a smooth point. There is a unique curve C in the pencil | − K X | passing through the point P . The curve C is a hypersurface in P (1 , ,
3) of degree 6such that the natural projection C −→ P (1 , ∼ = P is a double cover. Thus, we have mult P ( C )
2. In particular, the log pair ( X, ǫ C ) is logcanonical. Thus, it follows from Lemma 1.3.6 that we may assume that the support of thedivisor D does not contain one of the irreducible components of the curve C . Then13 = D · C > mult P (cid:0) D (cid:1) > C is irreducible (but possibly non-reduced). Therefore, the curve C must bereducible and reduced. Then C = C + C , where C and C are irreducible and reduced smooth rational curves such that C = C = − , C · C = 32 . Without loss of generality we may assume that P ∈ C . Put D = mC + Ω , where Ω is an effective Q -divisor such that C Supp(Ω). If m = 0, then C Supp(Ω) and16 = D · C = (cid:0) mC + Ω (cid:1) · C > mC · C = 3 m , and hence m . Thus, it follows from Lemma 1.3.8 that1 + 4 m (cid:0) D − mC (cid:1) · C = Ω · C > ǫ > . Therefore, m > . But m . Consequently, lct( X ) = ǫ . (cid:3) Lemma 3.1.5.
Let X be a quasismooth hypersurface of degree 15 in P (3 , , , X ) =2. Proof.
The surface X has five singular points O , . . . , O of type (1 , z = t = 0. The surface also has three singular points Q , Q , Q of type (1 , x = y = 0.Let C i be the curve in the pencil | − K X | passing through the point O i , where i = 1 , . . . , C i consists of three reduced and irreducible smooth rational curves C i = L i + L i + L i . The curve L ij contains the point Q j . Furthermore, L i ∩ L i ∩ L i = { O i } . We see that − K X · L ij = 115 , (cid:0) L ij (cid:1) = − , L ij · L ik = 13where j = k .Note that lct( X, C i ) = . Thus lct( X ) X ) <
2. Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . Then, mult P ( D ) > .Suppose that P C ∪ C ∪ C ∪ C ∪ C . Then P is a smooth point of X . There is a uniquecurve C ∈ | − K X | passing through point P . Then C is different from the curves C , . . . , C andhence C is irreducible. Furthermore, the log pair ( X, C ) is log canonical. Thus, it follows fromLemma 1.3.6 that we may assume that C Supp( D ). Then we obtain an absurd inequality15 = D · C > mult P (cid:0) D (cid:1) > , since the log pair ( X, D ) is not log canonical at the point P . Therefore, P ∈ C ∪ C ∪ C ∪ C ∪ C .However, we may assume that P ∈ C without loss of generality. Furthermore, by Lemma 1.3.6,we may assume that L i Supp( D ) for some i = 1 , , D · L i > mult O (cid:0) D (cid:1) , the point P cannot be the point O .Without loss of generality, we may assume that P ∈ L .Let Z be the curve in the pencil | − K X | passing through the point Q . Then Z = Z + Z + Z + Z + Z , where Z i is a reduced and irreducible smooth rational curve. The curve Z i contains the point O i . Moreover, Z ∩ Z ∩ Z ∩ Z ∩ Z = { Q } . It is easy to check lct( X, Z ) = . By Lemma 1.3.6,we may assume that Z k Supp( D ) for some k = 1 , . . . ,
5. Then13 = 5 D · Z k > mult Q (cid:0) D (cid:1) , and hence the point P cannot be the point Q .Thus, the point P is a smooth point on L . Put D = mL + Ω , where Ω is an effective Q -divisor such that L Supp(Ω). If m = 0, then115 = D · L i = (cid:0) mL + Ω (cid:1) · L i > mL · L i = m , and hence m . Then it follows from Lemma 1.3.8 that1 + 7 m
15 = (cid:0) D − mL (cid:1) · L = Ω · L > . This implies that m > . But m . The obtained contradiction completes the proof. (cid:3) Lemma 3.1.6.
Let X be a quasismooth hypersurface of degree 25 in P (3 , , , X ) = . XCEPTIONAL DEL PEZZO HYPERSURFACES 45
Proof.
The curve C x is irreducible and reduced. It is easy to see that lct( X, C x ) = . There-fore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . We may assume that the support of D does not contain the curve C x by Lemma 1.3.6.Since H ( P , O P (21)) contains x , x y , z , by Lemma 1.3.9 we havemult P ( D ) · · · · < P is a smooth point in the outside of the curve C x . Thus, either P = O x or P ∈ C x .If P ∈ C x , then we obtain a contradictory inequalities577 = D · C x > mult P ( D )mult P ( C x ) = mult P ( D ) > P ∈ X \ Sing( X ) , mult P ( D )mult P ( C x )7 = mult P ( D )7 > P = O z , mult P ( D )mult P ( C x )11 = 2mult P ( D )11 > P = O t . Therefore, we see that P = O x .Since the curve C y is irreducible and the log pair ( X, C y ) is log canonical at the point O x ,we may assume that the support of D does not contain the curve C y . Then1063 < mult O x ( D )3 D · C y = 25231 < . This is a contradiction. (cid:3)
Lemma 3.1.7.
Let X be a quasismooth hypersurface of degree 28 in P (3 , , , X ) = . Proof.
The surface X is singular at the point O x and the point O y . The former is a singularpoint of type (1 ,
1) and the latter is of type (1 , O and O be the two points cut outon X by the equations x = y = 0. The points O and O are singular points of type (3 ,
5) onthe surface X .The curve C x consists of two reduced and irreducible smooth rational curves L and L . Thesetwo curves intersect each other only at the point O y . Each curve L i contains the singular point O i . We have − K X · L i = 135 , L · L = 25 , L = L = − . Since lct(
X, C x ) = , lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .If P is a smooth point in the outside of C x , thenmult P (cid:0) D (cid:1) < H ( P , O P (21)) contains x , z , x y . Therefore, either P belongs to thecurve C x or P = O x . By Lemma 1.3.6, we may assume that L i Supp( D ) for some i = 1 ,
2. Similarly, we mayassume that C y Supp( D ) since ( X, C y ) is log canonical and the curve C y is irreducible.The inequalities mult O x ( D ) D · C y = 27 < P cannot be the point O x . Therefore, the point P belongs to the curve C x .The inequalities mult O y ( D ) D · L i = 17 < P cannot be the point O y .Without loss of generality, we may assume that P ∈ L . Put D = mL + Ω, where Ω is aneffective Q -divisor such that L Supp(Ω). If m = 0, then135 = D · L = (cid:0) mL + Ω (cid:1) · L > mL · L = 2 m , and hence m . Then Lemma 1.3.8 implies an absurd inequality598 > m
35 = (cid:0) D − mL (cid:1) · L = Ω · L >
49 if P = O ,
463 if P = O . The obtained contradiction completes the proof. (cid:3)
Lemma 3.1.8.
Let X be a quasismooth hypersurface of degree 36 in P (3 , , , X ) = . Proof.
The surface X is singular at the points O y and O z . It is also singular at two points P and P on the curve L yz . These two points P and P are contained in C y .The curve C x is irreducible and reduced. It is easy to see that lct( X, C x ) = . Also, thecurve C y is always irreducible and the pair ( X, · C y ) is log canonical. We see that lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6, we may assume thatthe support of D contains neither the curve C x nor C y .Then the following inequalitiesmult O y ( D ) D · C x = 5 · · · · · < , mult O z ( D ) D · C x = 11 · · · · · < , mult P i ( D ) D · C y = 3 · · · · · < , show that the point P is a smooth point P on X . Furthermore, the first two inequalities alsoshow that the point P cannot belong to the curve C x . Therefore, the point P is a smooth pointin the outside of the curve C x .However, since H ( P , O P (39)) contains x , x y , x z , by Lemma 1.3.9 we have1021 < mult P ( D ) · · · · < . XCEPTIONAL DEL PEZZO HYPERSURFACES 47
The obtained contradiction completes the proof. (cid:3)
Lemma 3.1.9.
Let X be a quasismooth hypersurface of degree 56 in P (5 , , , X ) = . Proof.
The surface X is singular at the points O x , O z and O t . The first point is a singular pointof type (2 , (7 , (5 , O of type (5 ,
3) on L xz that is different from the singular point O t .The curve C x (resp. C y ) consists of two reduced and irreducible curves L xy and R x (resp. R y ). The curve L xy intersects the curve R x at the point O z . The curve R x is singular at thepoint O z . On the other hand, it intersects the curve R y at the point O t . The curve R y is singularat O t . We have L xy = − , L xy · R x = 217 , R x = − , L xy · R y = 17 , R y = 935 . It is easy to check lct(
X, C x ) = and lct( X, C y ) = , and hence lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assume thateither the support of the divisor D does not contain the curve L xy or it contains neither R x nor R y .Suppose that P C x ∪ C y . Then P is a smooth point andmult P (cid:0) D (cid:1) < X P (5 , ,
17) is a finite morphism outsideof the curve C y and H ( P , O P (85)) contains monomials x , z , x y . This is a contradiction.Thus, the point P must belong to C x ∪ C y .The curve C z is irreducible and the log pair ( X, · C z ) is log canonical. By Lemma 1.3.6 wemay assume that C z Supp( D ). Then825 >
421 = 5 D · C z > mult O x (cid:0) D (cid:1) , and hence the point P cannot be O x .Suppose that P ∈ L xy . Put D = mL xy + Ω, where Ω is an effective Q -divisor such that L xy Supp(Ω). If m = 0, then1119 = D · R x = (cid:0) mL xy + Ω (cid:1) · R x > mL xy · R x = 2 m , and hence m . Then it follows from Lemma 1.3.8 that1 + 37 m
357 = (cid:0) D − mL xy (cid:1) · L xy = Ω · L xy > P = O t , P = O z ,
825 if P = O z and P = O t . This implies m > . But m . The obtained contradiction implies that P L xy . Suppose that P ∈ R x . Put D = aR x + Υ, where Υ is an effective Q -divisor such that R x Supp(Υ). If a = 0, then1357 = D · L xy = (cid:0) aR x + Υ (cid:1) · L xy > aL xy · R x = 2 a , and hence a . Then it follows from Lemma 1.3.8 that1 + 9 a
119 = (cid:0) D − aR x (cid:1) · R x = Υ · R x > P = O,
825 if P = O. This is impossible because a . Thus, we see that P C x .We see that P ∈ R y and P ∈ X \ Sing( X ). Put D = bR y + ∆, where ∆ is an effective Q -divisor such that R y Supp(∆). If b = 0, then1357 = D · L xy = (cid:0) bR y + ∆ (cid:1) · L xy > bL xy · R y = b , and hence b . Then it follows from Lemma 1.3.8 that1 + 9 b
35 = (cid:0) D − bR y (cid:1) · R y = ∆ · R y > . This is impossible because b . The obtained contradiction completes the proof. (cid:3) Lemma 3.1.10.
Let X be a quasismooth hypersurface of degree 81 in P (5 , , , X ) = . Proof.
The curve C x is irreducible and reduced. Moreover, the curve C x is smooth outside ofthe singular locus of the surface X . It is easy to see that lct( X, C x ) = . Hence, we havelct( X ) . The curve C y is irreducible and reduced. The log pair ( X, C y ) is log canonical.Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P ∈ X . We may assume that the support of D contains neither C x nor C y by Lemma 1.3.6.The inequality 31 D · C x = 319 < P cannot be on the curve C x . On the other hand, the inequality5 D · C y = 331 < P cannot be on the curve C y . In particular, the point P cannot be thepoint O x .Therefore, the point P must be a smooth point in the outside of C x . However, Lemma 1.3.9implies mult P ( D ) · · · · < H ( P , O P (190)) contains x , x z, y . This is a contradiction. (cid:3) XCEPTIONAL DEL PEZZO HYPERSURFACES 49
Lemma 3.1.11.
Let X be a quasismooth hypersurface of degree 100 in P (5 , , , X ) = . Proof.
The surface X is singular at the points O y and O z . Also, it is singular at two points P and P on L yz . The point O y is a singular point of type (2 ,
3) on X . The point O z is of type (5 , (2 , C x is irreducible and reduced. It is easy to see that lct( X, C x ) = . Therefore,lct( X ) . The curve C z is irreducible and reduced. The log pair ( X, · C z ) is log canonical.Suppose that lct( X ) < . Then it follows from Lemma 1.3.6 that there is an effective Q -divisor D ∼ Q − K X such that C x , C z Supp( D ) and the pair ( X, D ) is not log canonical atsome point P ∈ X .The inequality 27 D · C x = 219 < P cannot be on the curve C x . On the other hand, the inequality5 D · C z = 219 < P cannot be on the curve C z . In particular, the point P can be neitherthe point P nor the point P .Consequently, the point P must be a smooth point in the outside of C x . However, H ( P , O P (270)) contains x , x y , z . Then, Lemma 1.3.9 implies a contradictory inequality625 < mult P ( D ) · · · · < . (cid:3) Lemma 3.1.12.
Let X be a quasismooth hypersurface of degree 81 in P (7 , , , X ) = . Proof.
The surface X is singular only at the points O x , O y and O t .The curve C x is irreducible and reduced. It is easy to see that lct( X, C x ) = , and hencelct( X ) . The curve C y is irreducible and reduced. Moreover, the log pair ( X, · C y ) is logcanonical.Suppose that lct( X ) < . By Lemma 1.3.6, there is an effective Q -divisor D ∼ Q − K X suchthat the support of D contains neither the curve C x nor the curve C y , and the log pair ( X, D )is not log canonical at some point P ∈ X .The three inequalities 11 D · C x = 337 < , D · C y = 337 < , mult O t ( D ) = mult O t ( D )mult O t ( C x )3 D · C x = 111 < P is a smooth point in the outside of C x . However, since H ( P , O P (189)) contains x , x y , z , Lemma 1.3.9 implies an absurd in-equalities 1249 < mult P ( D ) · · · · < . Therefore, lct( X ) = . (cid:3) Lemma 3.1.13.
Let X be a quasismooth hypersurface of degree 88 in P (7 , , , X ) = . Proof.
The surface X is singular at the points O x and O z . The former is a singular point oftype (3 ,
1) and the latter is of type (11 , O and O on L xz . They are of type (7 , C x consists of two smooth rational curves L and L . Each curve L i contains thesingular point O i . The curves L and L intersects each other only at the point O z . We have L = L = − , L · L = 427 . It is easy to check lct( X, C x ) = . Meanwhile, the curve C y is irreducible and reduced. Also,the log pair ( X, C y ) is log canonical.Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assume thatthe support of D contains neither C y nor L without loss of generality.The inequality 27 D · L = 111 < P is located in the outside of L . The inequality7 D · C y = 227 < P cannot be O x . Write D = mL + Ω , where Ω is an effective Q -divisor such that L Supp(Ω). If m = 0, then1297 = D · L = (cid:0) mL + Ω (cid:1) · L > mL · L = 4 m , and hence m . Then ( D − mL ) · L = 1 + 37 m < · , and hence Lemma 1.3.8 implies that the point P cannot be on the curve L . Therefore, thepoint P is a smooth point in the outside of C x . However, Lemma 1.3.9 showsmult P (cid:0) D (cid:1) < H ( P , O P (189)) contains monomials x , z , x y . This is a contradiction. (cid:3) XCEPTIONAL DEL PEZZO HYPERSURFACES 51
Lemma 3.1.14.
Let X be a quasismooth hypersurface of degree 60 in P (9 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation xz + x y − y + t = 0 . Note that the surface X is singular at O x and O z . It is also singular at the point P = [1 : 1 : 0 : 0]and the point P = [0 : 1 : 0 : 1].The curves C x , C y , and C z are irreducible and reduced. We havelct( X, C x ) = 214 , lct( X, C y ) = 10 , lct( X, C z ) = 17 . The curve C x is singular at the point O z with multiplicity 3.Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D contains none of the curves C x , C y , C z .The three inequalities 173 D · C x = 17 · · · · · · < , D · C y = 9 · · · · · < , D · C z = 3 · · · · · < P is located in the outside of C x ∪ C y ∪ C z .Let L be the pencil on X that is cut out by the equations λz + µx y = 0 , where [ λ : µ ] ∈ P . Then the base locus of the pencil L consists of the points P and O x . Let C be the unique curve in L that passes through the point P . Then C is cut out on X by anequation x y = αz , where α is a non-zero constant, since the point P is located in the outside of C x ∪ C y ∪ C z .The curve C is smooth outside of the points P and O x by the Bertini theorem because C isisomorphic to a general curve in the pencil L unless α = −
1. In the case when α = −
1, thecurve C is smooth outside the points P and O x as well.We claim that the curve C is irreducible. If so, then we may assume that the support of D does not contain the curve C and hence we obtain a contradiction421 < mult Q ( D ) D · C = 51 · · · · < . For the irreducibility of the curve C , we may consider the curve C as a surface in C definedby the equations t + y + (1 + α ) xz = 0 and x y = αz . This surface is isomorphic to thesurface in C defined by the equations t + y + βxz = 0 and x y = z , where β = 1 or 0. Then,we consider the surface in P defined by the equations t w + y + βxz = 0 and x y = z w .We take the affine piece defined by t = 1. This affine piece is isomorphic to the surface definedby the equation x y + z ( y + βxz ) = 0 in C . If β = 1, the surface is irreducible. If β = 0, then it has an extra component defined by y = 0. However, this component originates from thehyperplane w = 0 in P . Therefore, the surface in C defined by the equations t + y = 0 and x y = z is also irreducible. (cid:3) Lemma 3.1.15.
Let X be a quasismooth hypersurface of degree 69 in P (9 , , , X ) = 6. Proof.
We may assume that the surface X is defined by the quasihomogeneous equation zt ( z − t ) + xy ( y − x ) = 0 . The surface X is singular at three distinct points O x , O y , P = [1 : 1 : 0 : 0]. Also, it is singularat three distinct points O z , O t , Q = [0 : 0 : 1 : 1].The curve C x consists of three distinct curves L xz , L xt and R x = { x = z − t = 0 } thatintersect altogether at the point O y . Similarly, the curve C y consists of three curves L yz , L yt and R y = { y = z − t = 0 } that intersect altogether at the point O x . The curve C z consists of L xz , L yz , and R z = { z = y − x = 0 } . The curve R z is singular at the point O t with multiplicity3. The curve C t consists of L xt , L yt and R t = { t = y − x = 0 } . The curve R t is singular atthe point O z with multiplicity 3.Note that lct( X, C x ) = 6. The log pairs ( X, C y ), ( X, C z ) and ( X, C t ) are log canonical.Suppose that lct( X ) <
6. Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P ∈ X . Lemma 1.3.6 implies that we may assumethat the support of D contains neither R x nor R y by a linear coordinate change. Furthermore,we may assume that the support of D does not contain at least one component of C z . Also, itmay be assumed not to contain at least one component of C t .The inequalities 15 D · R x = 15 · · · · ·
23 = 123 < , D · R y = 23 · · · · ·
23 = 19 < P is located in the outside of R x ∪ R y .Then the inequalities23 D · L xz = 115 < , D · L yz = 19 < , D · R z = 19 < O t ( D ) < , and hence the point P cannot be the point O t . By the same way, wecan show that P = O z .Write D = mR z + Ω, where Ω is an effective Q -divisor such that R z Supp(Ω). Then m since ( X, D ) is log canonical at O t . We have R z · ( L xz + L yz ) = 823 , R z · D = 169 , and hence R z = − . ThenΩ · R z = D · R z − mR z = 1 + m · · · < · . Lemma 1.3.8 implies that the point P cannot belong to R z . In particular, the point P cannotbe the point P . XCEPTIONAL DEL PEZZO HYPERSURFACES 53
Write D = aL xz + ∆, where ∆ is an effective Q -divisor whose support does not contain thecurve L xz . Then a . ThenΩ · L xz = 6( D · L xz − aL xz ) = 6 · (1 + 37 a )345 < , because L xz = − . Thus, we see that P L xz . Similarly, we can show that P L yz . Thus,we see that P C z . In the same way, we can see that P is not contained in the curves C t and { z − t = 0 } .Therefore, the point P is a smooth point in the outside of C z ∪ C t ∪ { z − t = 0 } . Let E bethe unique curve on X such that E is given by the equation z = λt and P ∈ E , where λ is anon-zero constant different from 1. Then E is quasismooth and hence irreducible. Therefore,we may assume that the support of D does not contain the curve E . Thenmult P ( D ) D · E = 23 · · · · < . This is a contradiction. (cid:3)
Lemma 3.1.16.
Let X be a quasismooth hypersurface of degree 127 in P (11 , , , X ) = . Proof.
We may assume that the hypersurface X is defined by the equation z t + yt + xy + x z = 0 . The singularities of X consist of a singular point of type (7 ,
5) at O x , a singular point of type (1 ,
2) at O y , a singular point of type (11 ,
29) at O z , and a singular point of type (11 , O t .The curve C x (resp. C y , C z , C t ) consists of two irreducible curves L xt (resp. L yz , L yz , L xt )and R x = { x = z + yt = 0 } (resp. R y = { y = x + zt = 0 } , R z = { z = t + xy = 0 } , R t = { t = y + x z = 0 } ). We can see that L xt ∩ R x = { O y } , L yz ∩ R y = { O t } , L yz ∩ R z = { O x } , L xt ∩ R t = { O z } . It is easy to check lct( X, C x ) = . The log pairs ( X, · C y ), ( X, · C z ) and ( X, · C t )are log canonical.Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6, we may assume that the support of D does not contain L xt or R x . Thenone of the following two inequalities must hold:433 >
139 = 29 L xt · D > mult O y ( D ) , >
249 = 29 R x · D > mult O y ( D ) . Therefore, the point P cannot be the point O y . For the same reason, one of two inequalities433 >
149 = 11 L yz · D > mult O x ( D ) , >
229 = 11 R z · D > mult O x ( D ) must hold, and hence the point P cannot be the point O x . Since R t is singular at the point O z with multiplicity 4, we can apply the same method to C t , i.e., one of the following inequalitiesmust be satisfied: 433 >
129 = 39 L xt · D > mult O z ( D ) , >
111 = 394 R t · D >
14 mult O z ( D )mult O z ( R t ) = mult O z ( D ) . Thus, the point P cannot be O z .Write D = µR x + Ω, where Ω is an effective Q -divisor such that R x Supp (Ω). If µ > L xt is not contained in the support of D . Thus,229 µ = µR x · L xt D · L xt = 129 · , and hence µ . We have49Ω · R x = 49( D · R x − µR x ) = 2 + 76 µ < . Then Lemma 1.3.8 shows that the point P cannot belong to R x . In particular, the point P cannot be O t .Put D = ǫL xt +∆, where ∆ is an effective Q -divisor such that L xt Supp (∆). Since ( X, D )is log canonical at the point O y , ǫ and hence∆ · L xt = D · L xt − ǫL xt = 1 + 67 ǫ · < . Then Lemma 1.3.8 implies that the point P cannot belong to L xt .Consequently, the point P must be a smooth point in the outside of C x . Then an absurdinequality 433 < mult P ( D ) · · · · < H ( P , O P (539)) contains x y , x , x z and t . The ob-tained contradiction completes the proof. (cid:3) Lemma 3.1.17.
Let X be a quasismooth hypersurface of degree 256 in P (11 , , , X ) = . Proof.
The curve C x is irreducible and reduced. Moreover, it is easy to see lct( X, C x ) = .The curve C y is also irreducible and reduced and the log pair ( X, C y ) is log canonical.Suppose that lct( X ) < . By Lemma 1.3.6, there is an effective Q -divisor D ∼ Q − K X suchthat C x , C y Supp( D ) and the log pair ( X, D ) is not log canonical at some point P ∈ X .The inequalities 69 D · C x = 69 · · · · · < , D · C y = 11 · · · · · < , XCEPTIONAL DEL PEZZO HYPERSURFACES 55 imply that the point P is a smooth point in the outside of C x . However, since H ( P , O P (759))contains x , x y , z , we obtainmult P ( D ) · · · · < (cid:3) Lemma 3.1.18.
Let X be a quasismooth hypersurface of degree 127 in P (13 , , , X ) = . Proof.
We may assume that the hypersurface X is given by the equation z t + y z + xt + x y = 0 . The only singularities of X are a singular point of type (9 ,
5) at O x , a singular point of type (13 ,
11) at O y , a singular point of type (13 ,
23) at O z , and a singular point of type (23 , O t .The curve C x (resp. C y , C z , C t ) consists of two irreducible curves L xz (resp. L yt , L xz , L yt )and R x = { x = y + zt = 0 } (resp. R y = { y = z + xt = 0 } , R z = { z = t + x y = 0 } , R t = { t = y z + x = 0 } ). We can see that L xt ∩ R x = { O t } , L yz ∩ R y = { O x } , L yz ∩ R z = { O y } , L xt ∩ R t = { O z } . It is easy to check lct( X, C x ) = . The log pairs ( X, · C y ), ( X, · C z ) and ( X, · C t )are log canonical.Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6, we may assume that the support of D does not contain L xz or R x . Thenone of the following two inequalities must hold:865 >
123 = 57 L xz · D > mult O t ( D ) , >
435 = 57 R x · D > mult O t ( D ) . Therefore, the point P cannot be the point O t . For the same reason, one of two inequalities865 >
135 = 13 L yt · D > mult O x ( D ) , >
257 = 13 R y · D > mult O x ( D )must hold, and hence the point P cannot be the point O x .To apply the same method to C z and C t , we note that R z is singular at O y with multiplicity2 and R t is singular at O z with multiplicity 3. Then we can see that one inequality from eachof the pairs 865 >
113 = 35 L yt · D > mult O z ( D ) , > · R t · D >
13 mult O z ( D )mult O z ( R t ) = mult O z ( D ); >
157 = 23 L xz · D > mult O y ( D ) , >
113 = 232 R z · D >
12 mult O y ( D )mult O y ( R z ) = mult O y ( D )must be satisfied. Therefore, the point P can be neither O y nor O z .To apply Lemma 1.3.8 to L xz and R x , we compute L xz = − · , R x = − · . Put D = aL xz + bR x + Ω, where Ω is an effective Q -divisor such that L xz , R x Supp (Ω). Then a, b since the log pair ( X, D ) is log canonical at the point O t . Therefore, D · L xz − aL xz = 1 + 79 a · < ,D · R x − bR x = 4 + 88 b · < . Then, Lemma 1.3.8 implies that the point P is a smooth point in the outside of C x .Applying Lemma 1.3.9, we see that865 < mult P ( D ) · · · · < , since H ( P , O P (455)) contains x , x y , z and the point P is in the outside of L xz . Theobtained contradiction completes the proof. (cid:3) Lemma 3.1.19.
Let X be a quasismooth hypersurface of degree 256 in P (13 , , , X ) = . Proof.
We may assume that the surface X is given by the equation t + y z + xz + x y = 0 . It has a singular point of type (3 ,
11) at O x , a singular point of type (13 ,
23) at O y , and asingular point of type (35 ,
47) at O z .The curve C x is reduced and irreducible. The curve is singular at the point O z . It is easyto check that lct( X, C x ) = . Therefore, lct( X ) . The curve C y is also reduced andirreducible. The curve C y is singular only at O x . Moreover, the log pair ( X, · C y ) is logcanonical.Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thelog pair ( X, D ) is not log canonical at some point P ∈ X . By Lemma 1.3.6 we may assumeneither C x nor C y is contained in Supp ( D ).The following two inequalities show that the point P is located in the outside of C x ∪ C y :812 C x · D = 135 < , C x · D = 181 < . XCEPTIONAL DEL PEZZO HYPERSURFACES 57
However, applying Lemma 1.3.9, we can obtainmult P ( D ) · · · · < , since H ( P , O P (1053)) contains x , x y and z . This is a contradiction. (cid:3) Sporadic cases with I = 2 Lemma 3.2.1.
Let X be the quasismooth hypersurface defined by a quasihomogeneous poly-nomial f ( x, y, z, t ) of degree 12 in P (2 , , , (cid:0) X (cid:1) = f ( x, y, z, t ) contains the term yzt,
712 if f ( x, y, z, t ) does not contain the term yzt. Proof.
We may assume f ( x, y, z, t ) = z ( z − x )( z − ǫx ) + y + xt + ayzt + bxy z + cx yt + dx y , where ǫ ( = 0 , a , b , c , d are constants. Note that X is singular at the point O t and threepoints Q = [1 : 0 : 0 : 0], Q = [1 : 0 : 1 : 0], Q = [1 : 0 : ǫ : 0]. The curve C x always isirreducible and reduced. We can easily check thatlct (cid:0) X, C x (cid:1) = a = 0 ,
712 if a = 0 . Suppose that lct( X ) < λ := lct( X, C x ). Then there is an effective Q -divisor D ∼ Q − K X suchthat the log pair ( X, λD ) is not log canonical at some point P ∈ X . We may assume that thecurve C x is not contained in the support of D .First, we consider the case where a = 0. Since H ( P , O P (6)) contains x , y , and xz ,Lemma 1.3.9 implies that for a smooth point O ∈ X \ C x mult O ( D ) < · · · · · < . Therefore, the point P cannot be a smooth point in X \ C x . Since the curve C x is not containedin the support of D and it is singular at O t with multiplicity 3, the inequality53 D · C x = 5 · · · · · · · < P is located in the outside of C x . Thus, the point P must be one of thepoint Q , Q , Q . The curve C y is quasismooth. Therefore, we may assume that the support of D does not contain the curve C y . Then the inequalitymult Q i ( D ) D · C y = 2 · · · · · · < a = 0. Note that the curve C x is not contained in thesupport of D and it is singular at O t with multiplicity 2. Since52 D · C x = 5 · · · · · · · the point P is located in the outside of C x .The curve C z is irreducible and the log pair ( X, C z ) is log canonical. Therefore, we mayassume that the support of D does not contain the curve C z . The curve C z is singular at thepoint Q . The inequality mult Q ( D ) D · C z = 2 · · · · · < P cannot be the point Q . We consider the curves C z − x defined by z = x and C z − ǫx defined by z = ǫx . Then by coordinate changes we can see that they have the sameproperties as that of C z . Moreover, we can see that the point P can be neither Q nor Q .Therefore, the point P must be located in the outside of C x ∪ C z ∪ C z − x ∪ C z − ǫx .Let L be the pencil on X defined by λx + µz = 0, where [ λ : µ ] ∈ P . Let C the curve in L that passes through the point P . Then it is cut out by z = αx , where α = 0 , , ǫ . The curve C is isomorphic to the curve in P (2 , ,
5) defined by x + y + xt + βx yt + γx y = 0 , where β and γ are constants. We can easily see that the curve C is irreducible. Moreover, wecan check mult P ( C ) X, C ) is log canonical. Therefore, we mayassume that the support of D does not contain the curve C . Then, the inequalitymult P ( D ) D · C = 2 · · · · · < (cid:3) Lemma 3.2.2.
Let X be a quasismooth hypersurface of degree 14 in P (2 , , , X ) =1. Proof.
We may assume that X is defined by the quasihomogeneous equation t − y z + x ( z − β x )( z − β x )( z − β x ) + ǫxy ( y − γx )where ǫ = 0, β , β , β , γ are constants. Note that X is singular at the points O y , O z and threepoints Q = [1 : 0 : β : 0], Q = [1 : 0 : β : 0], Q = [1 : 0 : β : 0]. The constants β , β and β are distinct since X is quasismooth. The curve C x consists of two irreducible reduced curves C − and C + . However, the curves C y and C z are irreducible. We can easily see that lct( X, C x ) = 1,lct( X, C y ) = and lct( X, C z ) > X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that thelog pair ( X, D ) is not log canonical at some point P ∈ X . Since H ( P , O P (6)) contains x , y and xz , Lemma 1.3.9 implies that the point P is either a singular point of X or a point of C x .Furthermore, C y is irreducible and hence we may assume that the support of D does not containthe curve C y . Hence the equality 2 C y · D = 2 · · · · · · P = Q i for each i = 1 , ,
3. In particular, the point P must belong to C x .We have the following intersection numbers: C x · C − = C x · C + = 16 , C − · C + = 712 , C − = C = − . XCEPTIONAL DEL PEZZO HYPERSURFACES 59
We may assume that the support of D cannot contain either C − or C + . If D does not containthe curve C + , then we obtain mult O y ( D ) D · C + = 23 < , mult O z ( D ) D · C + = 23 < . On the other hand, if D does not contain the curve C − , then we obtainmult O y ( D ) D · C − = 23 < , mult O z ( D ) D · C − = 23 < . Therefore, the point P must be in C x \ Sing( X ).We write D = mC + + Ω, where the support of Ω does not contain the curve C + . Then m > since D · C − > mC + · C − . Then we see C + · D − mC <
1. By the same method, we also obtain C − · D − mC − <
1. Then Lemma 1.3.8 completes the proof. (cid:3)
Lemma 3.2.3.
Let X be a quasismooth hypersurface of degree 20 in P (3 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t = y + z + x z + ǫ xy z + ǫ x yz + ǫ x y , where ǫ i ∈ C . Note that the surface X is singular only at the point O x , O = [0 : 1 : 0 : 1], P = [0 : 0 : 1 : 1] and P = [0 : 0 : 1 : − C x , C y and C z are irreducible. Moreover, we have32 = lct( X, C x ) < lct( X, C y ) = 2 , and hence lct( X ) . We also see that lct( X, C z ) > .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D does not contain the curves C x , C y and C z .Suppose that P C x ∪ C y ∪ C z . Then we consider the pencil L on X cut out by the equations λy + µxz = 0, [ λ : µ ] ∈ P . There is a unique member Z in the pencil L with P ∈ Z . Thecurve Z is cut out by an equation of the form αy + xz , where α is a non-zero constant. Thereis a natural double cover ω : Z → C , where C is the curve in P (3 , ,
5) given by the equation αy + xz . The curve C is quasismooth and ω ( P ) is a smooth point of P (3 , , P ( Z )
2, the curve Z consists of at most 2 components, each component of Z is asmooth rational curve. In particular, ( X, Z ) is log canonical. Therefore, we may assume thatSupp ( D ) does not contain at least one irreducible component of Z . Thus, if Z is irreducible,then we obtain an absurd inequality815 = D · Z > mult P ( D ) > . So, we see that Z = Z + Z , where Z and Z are smooth irreducible rational curves. Then Z = Z = − , Z · Z = 43 . Without loss of generality we may assume that P ∈ Z . Put D = mZ + Ω, where Ω is aneffective Q -divisor such that Z Supp(Ω). If m = 0, then415 = D · Z > mZ · Z = 4 m , and hence m . On the other hand, Lemma 1.3.8 shows that4 + 4 m
15 = (cid:0) D − mZ (cid:1) · Z = Ω · Z > , and hence m > . This is a contradiction. Therefore, P ∈ C x ∪ C y ∪ C z .The inequalities D · C x = 15 < , D · C y = 415 < , D · C z = 13 < P must be a singular point of X .The curve C z is singular at the point O x . Thus, we have12 = 32 D · C z > mult O x ( D )mult O x ( C z )2 = mult O x ( D ) . Therefore, the point P cannot be O x .Also, we have 25 = 2 D · C x > mult O ( D ) . This inequality shows that the point P cannot be the point O . Consequently, the point P mustbe either P or P .Without loss of generality we may assume that P = P . Note that C x ∩ C y = { P , P } .Let π : ¯ X → X be the weighted blow up at the point P with weights (3 , E bethe exceptional curve of π and let ¯ D , ¯ C x and ¯ C y be the proper transforms of D , C x and C y ,respectively. Then K ¯ X ∼ Q π ∗ ( K X ) + 25 E, ¯ C x ∼ Q π ∗ ( C x ) − E, ¯ C y ∼ Q π ∗ ( C y ) − E, ¯ D ∼ Q π ∗ ( D ) − a E, where a is a non-negative rational number. The curve E contains one singular point Q of type (1 ,
1) and one singular point of Q of type (1 ,
1) on the surface ¯ X . The point Q is containedin ¯ C y but not in ¯ C x . On the other hand, the point Q is contained in ¯ C x but not in ¯ C y . Theintersection ¯ C x ∩ ¯ C y consists of a single point that dominates the point P .The log pull back of the log pair ( X, D ) is the log pair (cid:18) ¯ X,
32 ¯ D + 3 a − E (cid:19) . This is not log canonical at some point Q ∈ E . We see that0 ¯ C x · ¯ D = C x · D + 3 a E = 15 − a , XCEPTIONAL DEL PEZZO HYPERSURFACES 61 and hence a
4. In particular, 3 a − < . This implies that the log pull back of the log pair ( X, D ) is log canonical in a puncturedneighborhood of the point Q .If a , then the log pair ( ¯ X, ¯ D ) is not log canonical at Q as well. We then obtain a
12 = ¯ D · E >
23 if Q = Q and Q = Q , ·
13 if Q = Q , ·
14 if Q = Q . In particular, we have a >
2. This contradicts the assumption a . Therefore, a > and thelog pull back of the log pair ( X, D ) is effective. Thenmult Q ( ¯ D ) > (cid:18) − a − (cid:19) = 14 − a . Since ¯ D · E = a , Lemma 1.3.8 implies that the point Q cannot be a smooth point. Therefore,the point Q is either Q or Q . However, two inequalities45 − a D · ¯ C x > mult Q ( ¯ D ) > − a , − a D · ¯ C y > mult Q ( ¯ D ) > − a (cid:3) Lemma 3.2.4.
Let X be a quasismooth hypersurface of degree 30 in P (3 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t = z − y z − x + ǫ x yz + ǫ x y + ǫ x y z + ǫ x y , where ǫ i ∈ C . The surface X is singular at the points O y , O = [0 : 1 : 1 : 0], O = [0 : 0 : 1 : 1], P = [1 : 0 : 0 : 1] and P = [1 : 0 : 0 : − C x and C y are irreducible. Moreover, we have32 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 2 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y .Since H ( P , O P (20)) contains the monomials y , y x , z , it follows from Lemma 1.3.9 thatthe point P is either a singular point of X or a smooth point in C y . However, the point P cannot belong to C y since = 5 D · C y . Therefore, the point P must be either the point O y or O . On the other hand, we have 4 D · C x = . This means that the pair ( X, D ) is log canonicalat the points O y and O . Consequently, lct( X ) = . (cid:3) Lemma 3.2.5.
Let X be a quasismooth hypersurface of degree 57 in P (5 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z + yt + xy + x t + ǫx yz = 0 , where ǫ ∈ C . The surface X is singular only at the points O x , O y and O t .The curves C x and C y are irreducible. Moreover, we have2512 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 6521 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y .Since H ( P , O P (110)) contains the monomials x y , x and t , it follows from Lemma 1.3.9that the point P is either a singular point of X or a smooth point on C x . However, this isimpossible since 22 D · C x = < and 5 D · C y = < . (cid:3) Lemma 3.2.6.
Let X be a quasismooth hypersurface of degree 70 in P (5 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + yz + xy − x + ǫx y z = 0 , where ǫ ∈ C . The surface X is singular at the points O y and O z . It is also singular at two points P = [1 : 0 : 0 : 1] and P = [1 : 0 : 0 : − C x is irreducible. On the other hand, the curve C y consists of two smooth curves C = { y = x − t = 0 } and C = { y = x + t = 0 } . Moreover, we have2512 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 267 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D dose not contains C x . Also, we may assume that the support of D doesnot contain either C or C .Since 19 D · C x = < , the point P cannot belong to C x .We put m C + m C + Ω, where Ω is an effective Q -divisor whose support contains neither C nor C . Since the pair ( X, D ) is log canonical at the point O z , we see that m i . Since5( D − m i C i ) · C i = 2 − m i < i , Lemma 1.3.8 implies that the point P can be neither P nor P . Therefore, thepoint P is a smooth point of X in the outside of C x . However, since H ( P , O P (95)) contains the XCEPTIONAL DEL PEZZO HYPERSURFACES 63 monomials x y , x and z , it follows from Lemma 1.3.9 that the point P is either a singularpoint of X or a smooth point on C x . This is a contradiction. (cid:3) Lemma 3.2.7.
Let X be a quasismooth hypersurface of degree 36 in P (6 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation zt + y + xz + x + ǫx y = 0 , where ǫ is a constant different from ±
2. The surface X is singular at the points O z and O t . Itis also singular at two points P and P on L zt . The surface X is also singular at one point Q on L yt .The curves C x and C y are irreducible and reduced. However, the curve C z consists of twoirreducible and reduced curves C and C . The curve C contains the point P but not P . Onthe other hand, C contains the point P but not P . We also see C = C = − , C · C = 613 . It is easy to check2512 = lct (cid:18) X, C z (cid:19) <
94 = lct (cid:18) X, C x (cid:19) <
92 = lct (cid:18) X, C y (cid:19) . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y . In addition, we may assume that it cannotcontain either C or C .Since H ( P , O P (30)) contains the monomials x y , x and z , it follows from Lemma 1.3.9that P ∈ Sing( X ) ∪ C x ∪ C z . However, 2 D · C y = < and hence the point P cannot bethe point Q . Note that the curve C x passes through the point O z with multiplicity 2. Then theinequality 5 D · C x = < shows that the point P cannot be a point on C x \ { O t } .Put D = mC + Ω, where Ω is an effective Q -divisor such that C Supp(Ω). If m = 0, then239 = D · C = (cid:0) mC + Ω (cid:1) · C > mC · C = 6 m , and hence m . Then 3 (cid:0) D − mC (cid:1) · C = 2 + 8 m . Therefore, it follows from Lemma 1.3.8 that the point P cannot be a point on C \ { O t } . Bythe same method, we can show that the point P cannot be a point on C \ { O t } . Therefore, thepoint P must be the point O t .Let π : ¯ X → X be the weighted blow up at the point O t with weights (2 , E bethe exceptional curve of π and let ¯ D , ¯ C x and ¯ C y be the proper transforms of D , C x and C y ,respectively. Then K ¯ X ∼ Q π ∗ ( K X ) − E, ¯ C x ∼ Q π ∗ ( C x ) − E, ¯ C y ∼ Q π ∗ ( C y ) − E, ¯ D ∼ Q π ∗ ( D ) − a E, where a is a non-negative rational number. The curve E contains one singular point Q of type (1 ,
1) and one singular point of Q of type (1 ,
1) on the surface ¯ X . The point Q is containedin ¯ C y but not in ¯ C x . On the other hand, the point Q is contained in ¯ C x but not in ¯ C y .The log pull back of the log pair ( X, D ) is the log pair (cid:18) ¯ X, D + 25 a + 9612 · E (cid:19) . This is not log canonical at some point Q ∈ E . We see that0 ¯ C y · ¯ D = C y · D + 3 a E = 65 · − a · , and hence a . In particular, 25 a + 9612 · . This implies that the log pull back of the log pair ( X, D ) is log canonical in a puncturedneighborhood of the point Q . Thenmult Q ( ¯ D ) > (cid:18) − a + 9612 · (cid:19) = 125 · − a . Since ¯ D · E = a , Lemma 1.3.8 implies that the point Q cannot be a smooth point.Therefore, the point Q is either Q or Q . However, two inequalities125 · − a
13 = 3 ¯ D · ¯ C x > mult Q ( ¯ D ) > · − a , · − a
13 = 2 ¯ D · ¯ C y > mult Q ( ¯ D ) > · − a (cid:3) Lemma 3.2.8.
Let X be a quasismooth hypersurface of degree 57 in P (7 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z + y t + xt + x y + ǫx y z = 0 , where ǫ ∈ C . The surface X is singular at the points O x , O y and O t . The curves C x , C y and C z are irreducible. We have4924 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 103 < lct (cid:18) X, C z (cid:19) = 192 . Thus, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains none of the curves C x , C y and C z . The curve C x is singularat the point O t . Since D · C x = < , 7 D · C y = < and D · C z = < , the point P cannot belong to the set C x ∪ C y ∪ C z . XCEPTIONAL DEL PEZZO HYPERSURFACES 65
Consider the pencil L on X defined by the equations λy z + µx = 0, [ µ, λ ] ∈ P . Then thereis a unique curve Z in the pencil L passing through the point P . Then the curve Z is definedby an equation of the form y z − αx = 0, where α is a non-zero constant.We see that C y Supp( Z ). But the open subset Z \ C y of the curve Z is a Z -quotient ofthe affine curve z − αx = z + t + xt + x + ǫx z = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, z, t (cid:3)(cid:17) that is isomorphic to the plane affine curve defined by the equation α x + t + xt + x + ǫαx = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, t (cid:3)(cid:17) . This curve is irreducible and hence the curve Z is also irreducible. Thus mult P ( Z )
14. Wemay assume that Supp( D ) does not contain the curve Z by Lemma 1.3.6. Then we obtain anabsurd inequality 320 = D · Z > mult P (cid:0) D (cid:1) > . (cid:3) Lemma 3.2.9.
Let X be a quasismooth hypersurface of degree 64 in P (7 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t − y + xz + x y + ǫx y z, where ǫ ∈ C . Note that X is singular at the points O x and O z . The surface X also has twosingular points P = [0 : 1 : 0 : 1] and P = [0 : 1 : 0 : −
1] of type (7 , C x is reducible. We have C x = C + C , where C and C are irreducible andreduced curves. The curve C contains the point P but not the point P . On the other hand,the curve C contains the point P but not the point P . However, these two curves meet eachother only at the point O z . We also have C = C = − · , C · C = 419 . The curve C y is irreducible. It is easy to checklct (cid:18) X, C x (cid:19) = 3516 < lct (cid:18) X, C y (cid:19) = 103 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D does not contain the curve C y . Moreover, we may assume that the support of D does not contain either the curve C or the curve C .Since C i Supp( D ) for either i = 1 or 2, we havemult O z ( D ) D · C i = 14 < , and hence P = O z . Meanwhile, the inequality 7 D · C y = < implies that the point P cannot belong to C y .Suppose that P ∈ C . Then we write D = mC + Ω, where Ω is an effective Q -divisor suchthat C Supp(Ω). If m = 0, then14 ·
19 = D · C = (cid:0) mC + Ω (cid:1) · C > mC · C = 4 m , and hence m . Then it follows from Lemma 1.3.8 that2 + 25 m ·
19 = (cid:0) D − mC (cid:1) · C = Ω · C > P = P , ·
18 if P = P . This is impossible since m . Thus, P C . Similarly, we can show that P C .Consequently, the point P is located in the outside of C x ∪ C y . In particular, it is a smoothpoint of X . But H ( P , O P (64)) contains monomials y , x y , y t and t . This is impossible byLemma 1.3.9. The obtained contradiction completes the proof. (cid:3) Lemma 3.2.10.
Let X be a quasismooth hypersurface of degree 48 in P (9 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t − y + xz + x y = 0 . The surface X is singular at the points O x , O z , Q = [0 : 1 : 0 : 1] and Q = [1 : 1 : 0 : 0].The curves C x , C y , C z and C t are irreducible and reduced. We have6324 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 4 < lct (cid:18) X, C z (cid:19) = 132 < lct (cid:18) X, C t (cid:19) = 162 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains none of the curves C x , C y , C z and C t .Note that the curve C x is singular at O z with multiplicity 3 and the curve C y is singular at O x with multiplicity 3. Then the inequalities133 D · C x = 16 < , D · C y = 213 < , D · C z = 16 < , D · C t = 89 · < P must be located in the outside of C x ∪ C y ∪ C z ∪ C t .Consider the pencil L on X defined by the equations λxt + µyz = 0, [ µ, λ ] ∈ P . Then there isa unique curve Z in the pencil L passing through the point P . Then the curve Z is defined by anequation of the form xt − αyz = 0, where α is a non-zero constant. We see that C x Supp( Z ).But the open subset Z \ C x of the curve Z is a Z -quotient of the affine curve t − αyz = t + y + z + y = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z, t (cid:3)(cid:17) , XCEPTIONAL DEL PEZZO HYPERSURFACES 67 which is isomorphic to the plane affine curve given by the equation α y z + y + z + y = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . Then, it is easy to see that the curve Z is irreducible and mult P ( Z )
4. Thus, we may assumethat Supp( D ) does not contain the curve Z by Lemma 1.3.6. However,2518 ·
13 = D · Z > mult P (cid:0) D (cid:1) > . Consequently, lct( X ) = . (cid:3) Lemma 3.2.11.
Let X be a quasismooth hypersurface of degree 57 in P (9 , , , X ) = 3. Proof.
We may assume that the surface X is defined by the quasihomogeneous equation zt ( z − t ) − xy + x y = 0 . The surface X is singular at three distinct points O x , O y , Q = [1 : 1 : 0 : 0] . Also, it is singularat three distinct points O z , O t , Q = [0 : 0 : 1 : 1].The curve C x consists of three distinct curves L xz , L xt and R x = { x = z − t = 0 } thatintersect altogether at the point O y . Similarly, the curve C y consists of three curves L yz , L yt and R y = { y = z − t = 0 } that intersect altogether at the point O x . The curve C z consistsof three distinct curves L xz , L yz and R z = { z = x − y = 0 } that intersect altogether at thepoint O t . The curve C t consists of three distinct curves L xt , L yt and R t = { t = x − y = 0 } that intersect altogether at the point O z . Let C z − t be the curve cut out on X by the equation z = t . Then C z − t consists of three distinct curves R x , R y and R z − t = { z − t = x − y = 0 } that intersect altogether at the point Q .We have the following intersection numbers: L xz = L xt = R x = − · , L yz = L yt = R y = − · , R z = R t = R z − t = − · − K X · L xz = − K X · L xt = − K X · R x = 119 · , − K X · L yz = − K X · L yt = − K X · R y = 219 · , − K X · R z = − K X · R t = − K X · R z − t = 219 · . Since lct( X, C x ) = 3, we have lct( X )
3. Suppose that lct( X ) <
3. Then there is aneffective Q -divisor D ∼ Q − K X such that the pair ( X, D ) is not log canonical at some point P ∈ X .The pairs ( X, C x ) and ( X, C y ) are log canonical. By Lemma 1.3.6, we may assume thatthe support of D does not contain at least one component of C x . Then one of the inequalitiesmult O y ( D ) D · L xz = 657 < , mult O y ( D ) D · L xt = 657 < , mult O y ( D ) D · R x = 657 < must hold, and hence the point P cannot be the point O y . Also, we may assume that thesupport of D does not contain at least one component of C y . By the same reason, the point P cannot be the point O x .We have lct (cid:18) X, C z (cid:19) = lct (cid:18) X, C t (cid:19) = lct (cid:18) X, C (cid:19) = 72 . By Lemma 1.3.6, we may assume that the support of D does not contain at least one componentof each curve C z , C t and C z − t . Since the curve R z is singular at the point O t with multiplicity3, Then one of the inequalities mult O t ( D ) D · L xz = 16 < , mult O t ( D ) D · L yz = 29 < , mult O t ( D ) D · R z = 29 < P cannot be the point O t . By applying the same method to C t and C z − t , we see that the point P can neither O z not Q .The three curves R z , R t , and R z − t intersects only at the point Q . The log pair (cid:18) X, (cid:16) R z + R t + R z − t (cid:17)(cid:19) is log canonical at Q , and R z + R t + R z − t ∼ − K X . By Lemma 1.3.6, we may assume thatthe support of D does not contain at least one curve among R z , R t and R z − t . Without loss ofgenerality, we may assume that the support of D does not contain the curve R z . Thenmult Q ( D ) D · R z = 219 < , and hence the point P cannot be Q .Write D = m L xz + m L yz + m R z + ∆, where ∆ is an effective Q -divisor whose supportcontains none of the curves L xz , L yz , R z . Since the pair ( X, D ) is log canonical at the point O t , we have m i for each i = 1, 2, 3. By Lemma 1.3.8, the inequalities( D − m L xz ) · L xz = 2 + 29 m · < , ( D − m L yz ) · L yz = 2 + 26 m · < , ( D − m R z ) · R z = 2 + 2 m · < P cannot belong to C z . By the same way, we can show that the point P isnot contained in C t ∪ C z − t . Therefore, the point P is a smooth point of X in the outside of theset C z ∪ C t ∪ C z − t . Then there is a unique quasismooth irreducible curve E ⊂ X passing throughthe point P and defined by the equation z = λt , where λ is a non-zero constant different from 1.By Lemma 1.3.6, we may assume that the support of D does not contain the curve E . Then13 < mult P ( D ) D · E = 118 . XCEPTIONAL DEL PEZZO HYPERSURFACES 69
This is a contradiction. (cid:3)
Lemma 3.2.12.
Let X be a quasismooth hypersurface of degree 81 in P (9 , , , X ) = 3. Proof.
The surface X can be defined by the quasihomogeneous equation yt + y z + xz − x = 0 . It is singular at the point O y , O z and O t . The surface X is also singular at the point Q = [1 :0 : 1 : 0].The curve C x (resp. C y ) consists of two irreducible curves L xy and R x = { x = t + y z = 0 } (resp. R y = { y = z − x = 0 } ). The curve L xy intersects R x (resp. R y ) only at the point O z (resp. O t ). We have the following intersection numbers: − K X · L xy = 112 · , − K X · R x = 16 · , − K X · R y = 23 · , L xy · R x = 112 ,L xy · R y = 331 , L xy = − · , R x = − · , R y = 103 · . Meanwhile, the curve C z is irreducible. We see that lct( X ) (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 20954 < lct (cid:18) X, C z (cid:19) = 223 . Suppose that lct( X ) <
3. Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P . We may assume that the support of D does notcontain at least one component of each of C x and C y by Lemma 1.3.6. One of the inequalitiesmult O z D D · L xy = 231 < , mult O z D D · R x = 419 < P cannot be the point O z . Since the curve R y is singular at thepoint O t with multiplicity 3, one of the inequalitiesmult O t D D · L xy = 112 < , mult O t D D · R y = 29 < P cannot be the point O t .By Lemma 1.3.6, we may also assume that the curve C z is not contained in the support of D . The curve C z is singular at the point O y . Then the inequality192 D · C z = 931 < P cannot be the point O y .Write D = m L xy + m R x + m R y + Ω, where Ω is an effective Q -divisor whose supportcontains none of L xy , R x , R y . If m = 0, then we obtain16 ·
19 = D · R x > m L xy · R x = m , and hence m . Similarly, we see that m and m . Since we have( D − m L xy ) · L xy = 2 + 53 m · < , ( D − m R x ) · R x = 1 + 5 m · < , D − m R y ) · R y = 2 − m < , it follows from Lemma 1.3.8 that the point P is located in the outside of C x and C y . Therefore,the point P is a smooth point in the outside of C x and C y . However, since H ( P , O P (171))contains the monomials y , x , x z and x z , it follows from Lemma 1.3.9 that the point P must be either a singular point of X or a point in C x ∪ C y . This is a contradiction. (cid:3) Lemma 3.2.13.
Let X be a quasismooth hypersurface of degree 105 in P (10 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z + yt + xy − x z = 0 . The surface X is singular at the points O x , O y , O t and Q = [1 : 0 : 1 : 0].The curve C x is irreducible. However, the curve C y consists of two irreducible curves L yz and R y = { y = z − x = 0 } . The curve L yz intersects R y at the point O t . We have L yz = − · , R y = − · , L yz · R y = 743 . We also have lct( X ) since5714 = lct (cid:18) X, C y (cid:19) < lct (cid:18) X, C x (cid:19) = 256 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D does not contain the curve C x . Similarly, we may assume that thesupport of the divisor D does not contain either L yz or R y .Since the support of the divisor D does not contain either L yz or R y and the curve R y issingular at the point O t , one of the inequalitiesmult O t ( D ) D · L yz = 15 < , mult O t ( D ) D · R y = 15 < P cannot be O t .We write D = m L yz + m R y + Ω, where Ω is an effective Q -divisor whose support containsneither L yz nor R y . If m = 0, then m = 0 and hence25 ·
43 = D · R y > m L yz · R y = 7 m . Therefore, m . Similarly, we have m . Since10( D − m L yz ) · L yz = 2 + 51 m < , D − m R y ) · R y = 2 + 16 m < , it follows from Lemma 1.3.8 that the point P is located in the outside of C y . XCEPTIONAL DEL PEZZO HYPERSURFACES 71
Since the divisor D does not contain the curve C x , mult O y ( D ) D · C x = < , andhence the point P cannot belong to the curve C x . Therefore, the point P is a smooth pointin the outside of C x ∪ C y . However, since H ( P , O P (190)) contains x , y , x z and x z , itfollows from Lemma 1.3.9 that the point P must be either a singular point of X or a point in C x ∪ C y . This is a contradiction. (cid:3) Lemma 3.2.14.
Let X be a quasismooth hypersurface of degree 105 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation yz − y + xt + x z = 0 . The surface X is singular at the point O x , O z , O t and Q = [0 : 1 : 1 : 0].The curve C x (resp. C y ) consists of two irreducible curves L xy and R x = { x = z − y = 0 } (resp. R y = { y = t + x z = 0 } . The curve L xy intersects R x (resp. R y ) only at the point O t (resp. O z ). We have the following intersection numbers: − K X · L xy = 114 · , − K X · R x = 27 · , − K X · R y = 17 · , L xy · R x = 347 ,L xy · R y = 114 , L xy = − · , R x = − · , R y = 57 · . We see that lct( X ) since7730 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 6 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D does not contain at least one component of each of C x and C y . Note that the curve R x is singular at the point O t with multiplicity 3 and the curve R y is singular at the point O z .Then one of two inequalitiesmult O t ( D ) D · L xy = 114 < , mult O t ( D ) D · R x = 221 < P cannot be O t . Applying the same method to C y , we showthat the point P cannot be the point O z .Write D = m L xy + m R x + m R y + Ω, where Ω is an effective Q -divisor whose supportcontains none of L xy , R x , R y . If m = 0, then we obtain27 ·
47 = D · R x > m L xy · R x = 3 m , and hence m . Similarly, we see that m and m . Since we have( D − m L xy ) · L xy = 2 + 73 m · < , D − m R x ) · R x = 2 + 10 m < , D − m R y ) · R y = 1 − m < , it follows from Lemma 1.3.8 that the point P is located in the outside of C x and C y . Therefore,the point P is a smooth point in the outside of C x . However, since H ( P , O P (517)) contains x y , x y , x , x z , x , t , it follows from Lemma 1.3.9 that the point P must be eithera singular point of X or a point in C x . This is a contradiction. (cid:3) Lemma 3.2.15.
Let X be a quasismooth hypersurface of degree 107 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation yt + y z + xz + x t = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consistsof L xy (resp. L xy , L zt , L zt ) and R x = { x = t + y z = 0 } (resp. R y = { y = z + x t = 0 } , R z = { z = x + yt = 0 } , R t = { t = y + xz = 0 } ). Also, we see that L xy ∩ R x = { O z } , L xy ∩ R y = { O t } , L zt ∩ R z = { O y } , L zt ∩ R t = { O x } . We have the following intersection numbers: − K X · L xy = 116 · , − K X · L zt = 211 · , − K X · R x = 18 · , − K X · R y = 611 · , − K X · R z = 1225 · , − K X · R t = 311 · , L xy · R x = 116 , L xy · R y = 341 , L zt · R z = 625 L xy = − · , L zt = − · , R x = − · , R y = 4211 · X ) since113 = lct (cid:18) X, C x (cid:19) <
509 = lct (cid:18) X, C y (cid:19) <
283 = lct (cid:18) X, C z (cid:19) < (cid:18) X, C t (cid:19) . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that eitherSupp( D ) does not contain at least one irreducible component of each of C x , C y , C z and C t .Since the curve R y is singular at the point O t with multiplicity 3, one of the inequalitiesmult O t ( D ) D · L xy = 116 < , mult O t ( D ) D · R y = 211 < P cannot be the point O t . Applying the same method to eachof C x and C t , we can show that the point P can be neither O z nor O x .Since H ( P , O P (352)) contains the monomials x y , x and z , it follows from Lemma 1.3.9that the point P is either the point O t or a smooth point on C x .Write D = m L xy + m R x + m L zt + Ω, where Ω is an effective Q -divisor whose supportcontains none of L xy , R x , R z . If m = 0, then m = 0 and hence we obtain18 ·
25 = D · R x > m L xy · R x = m . XCEPTIONAL DEL PEZZO HYPERSURFACES 73
Therefore, m . Similarly, we get m . Since we have( D − m L xy ) · L xy = 2 + 71 m · < , ( D − m R x ) · R x = 1 + 7 m · < P must be the point O y .Suppose that m = 0. Then the inequalitymult O y ( D ) D · L zt = 211 < m = 0 and hence the curve R z is not contained in thesupport of D . Then1225 ·
41 = D · R z > m L zt · R z + mult O y ( D ) − m > m
25 + 311 · , and hence m < · · . Since25( D − m L zt ) · L zt = 2 + 34 m < X, D ) is log canonical at the point O y by Lemma 1.3.8. This is a contradiction. (cid:3) Lemma 3.2.16.
Let X be a quasismooth hypersurface of degree 111 in P (11 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + tz + xy + x z = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consistsof L xt (resp. L yz , L yz , L xt ) and R x = { x = yt + z = 0 } (resp. R y = { y = zt + x = 0 } , R z = { z = xy + t = 0 } , R t = { t = y + x z = 0 } ). Also, we see that L xt ∩ R x = { O y } , L yz ∩ R y = { O t } , L yz ∩ R z = { O x } , L xt ∩ R t = { O z } . The intersection numbers among the divisors D , L xt , L yz , R x , R y , R z , R t are as follows: − K X · L xt = 117 · , − K X · R x = 425 · , − K X · R y = 717 · , − K X · L yz = 211 · , − K X · R z = 411 · , − K X · R t = 411 · ,L xt · R x = 225 , L yz · R y = 743 , L yz · R z = 211 , L xt · R t = 217 ,L xt = − · , R x = − · , R y = − · ,L yz = − · , R z = 1811 · , R t = 6411 · . We can easily see that lct( X, C x ) = is less than each of the numbers lct( X, C y ),lct( X, C z ) and lct( X, C t ). Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . The inequalities25 D · L xt = 117 < , D · R x = 443 < P = O y . The inequalities11 D · L yz = 243 < , D · R z = 425 < P = O x . Since the curve R t is singular at the point O z , the inequalities34 D · L xt = 3417 · < , D · R t = 211 < P = O z .We write D = a L xt + a L yz + a R x + a R y + a R z + a R t + Ω, where Ω is an effectivedivisor whose support contains none of the curves L xt , L yz , R x , R y , R z , R t . Since the pair( X, D ) is log canonical at the points O x , O y , O z , the numbers a i are at most . Then byLemma 1.3.8 the following inequalities enable us to conclude that either the point P is in theoutside of C x ∪ C y ∪ C z ∪ C t or P = O t :338 D · L xt − L xt = 2618 · · < , D · L yz − L xt = 2414 · · < , D · R x − R x = 1612 · · < , D · R y − R y = 4834 · · < , D · R z − R z D · R z = 32 · < , D · R t − R t D · R t = 334 < . Suppose that P = O t . Then we consider the pencil L defined by λyt + µz = 0, [ λ : µ ] ∈ P .The base locus of the pencil consists of the curve L yz and the point O y . Let E be the uniquedivisor in L that passes through the point P . Since P C x ∪ C y ∪ C z ∪ C t , the divisor E isdefined by the equation z = αyt , where α = 0.Suppose that α = −
1. Then the curve E is isomorphic to the curve defined by the equations yt = z and t y + xy + x z = 0. Since the curve E is isomorphic to a general curve in L ,it is smooth at the point P . The affine piece of E defined by t = 0 is the curve given by z ( z + xz + x ) = 0. Therefore, the divisor E consists of two irreducible and reduced curves L yz and C . We have D · C = D · E − D · L yz = 39411 · · . Also, we see C = E · C − C · L yz > E · C − ( L yz + R y ) · C = 432 D · C > . By Lemma 1.3.8 the inequality D · C < gives us a contradiction.
XCEPTIONAL DEL PEZZO HYPERSURFACES 75
Suppose that α = −
1. Then divisor E consists of three irreducible and reduced curves L yz , R x , and M . Note that the curve M is different from the curves R y and L xt . Also, it is smoothat the point P . We have D · M = D · E − D · L yz − D · R x = 1411 · ,M = E · M − L yz · M − R x · M > E · M − C y · M − C x · M > . By Lemma 1.3.8 the inequality D · M < gives us a contradiction. Therefore, P = O t .Put D = bR x + ∆, where ∆ is an effective divisor whose support does not contain R x . ByLemma 1.3.6, we may assume that R x Supp(∆) if b >
0. Thus, if b >
0, then225 ·
34 = D · L xt > bR x · L xt = 2 b , and hence b . On the other hand, it follows from Lemma 1.3.8 that4 + 64 b ·
43 = ∆ · R x > · . Therefore, b > . Since > , this is a contradiction. (cid:3) Lemma 3.2.17.
Let X be a quasismooth hypersurface of degree 226 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + yz + xy + x z = 0 . The surface X is singular at the points O x , O y and O z . The curves C x and C y are irreducible.We have 5512 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 17 · . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y . Then the inequalities61 D · C x = 443 < , D · C y = 461 < P must be a smooth point of X in the outside of C x . However, since H ( P , O P (671)) contains the monomials x y , x and z , it follows from Lemma 1.3.9 thatthe point P is either a singular point of X or a point on C x . This is a contradiction. (cid:3) Lemma 3.2.18.
Let X be a quasismooth hypersurface of degree 135 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z − y z + xt + x y = 0 . The surface X is singular at the points O x , O y , O t , Q = [0 : 1 : 1 : 0]. The curve C x consists of two irreducible and reduced curves L xz and R x = { x = z − y = 0 } .The curve L xz intersects R x at the point O t . It is easy to check L xz = − · , R x = − · , L xz · R x = 561 . Meanwhile, the curve C y is irreducible. We have9130 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 152 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D does not contain the curve C y . Similarly, we may assume that either L xz Supp( D ) or R x Supp( D ).Since the support of D cannot contain either L xz or R x one of the inequalitiesmult O t ( D ) D · L xz = 19 < , mult O t ( D ) D · R x = 29 < P cannot be O t . Also, the inequality13 D · C y = 661 < P cannot be O x .We write D = m L xz + m R x + Ω, where Ω is an effective Q -divisor whose support containsneither L xz nor R x . If m = 0, then we obtain29 ·
61 = D · R x > m L xz · R x = 5 m m . By the same way, we get m . Since18( D − m L xz ) · L xz = 2 + 77 m < , D − m R x ) · R x = 2 + 32 m < P is a smooth point in the outside of C x . However,since H ( P , O P (585)) contains x , x y , z , this is impossible by Lemma 1.3.9. (cid:3) Lemma 3.2.19.
Let X be a quasismooth hypersurface of degree 107 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation yz + y t + xt + x z = 0 . The surface X is singular at the points O x , O y , O z and O t . Each of the divisors C x , C y , C z ,and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t )consists of L xy (resp. L xy , L zt , L zt ) and R x = { x = z + y t = 0 } (resp. R y = { y = t + x z = 0 } , R z = { z = y + xt = 0 } , R t = { t = x + yz = 0 } ). The curve L xy intersects R x (resp. R y )only at the point O t (resp. O z ). Also, the curve L zt intersects R z (resp. R t ) only at the point O x (resp. O y ). It is easy to check XCEPTIONAL DEL PEZZO HYPERSURFACES 77 − K X · L xy = 229 · , − K X · L zt = 113 · , − K X · R x = 310 · , − K X · R y = 413 · , − K X · R z = 613 · , − K X · R t = 35 · ,L xy = − · , R x = − · , L xy · R x = 347 . We see lct( X ) since6518 = lct (cid:18) X, C x (cid:19) < (cid:18) X, C y (cid:19) <
293 = lct (cid:18) X, C z (cid:19) <
949 = lct (cid:18) X, C t (cid:19) . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D does not contain at least one irreducible component of each of the curves C x , C y , C z and C t . The curve R x (resp. R y , R t ) is singular at the point O t (resp. O z , O y ). Then ineach of the following pairs of inequalities, at least one of two must hold:mult O t ( D ) D · L xy = 229 < , mult O t ( D ) D · R x = 320 < O z ( D ) D · L xy = 247 < , mult O z ( D ) D · R y = 213 < O x ( D ) D · L zt = 110 < , mult O x ( D ) D · R z = 647 < O y ( D ) D · L zt = 213 < , mult O y ( D ) D · R t = 629 < . Therefore, the point P must be a smooth point of X .We write D = m L xy + m R x + Ω, where Ω is an effective Q -divisor whose support containsnone of L xy , R x . If m = 0, then m = 0 and hence we obtain310 ·
47 = D · R x > m L xy · R x = 3 m . Therefore, m . Similarly, we get m . Since( D − m L xy ) · L xy = 2 + 74 m · < , ( D − m R x ) · R x = 6 + 21 m · < P is a smooth point in the outside of C x . How-ever, since H ( P , O P (377)) contains the monomials x y , x and z , this is impossible byLemma 1.3.9. (cid:3) Lemma 3.2.20.
Let X be a quasismooth hypersurface of degree 111 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z t + y z + xt + x y = 0 . It is singular at the point O x , O y , O z and O t . Each of the divisors C x , C y , C z , and C t consistsof two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consists of L xz (resp. L yt , L xz , L yt ) and R x = { x = y + zt = 0 } (resp. R y = { y = z + xt = 0 } , R z = { z = t + x y = 0 } , R t = { t = x + y z = 0 } ). The curve L xz intersects R x (resp. R z )only at the point O t (resp. O y ). Also, the curve L yt intersects R y (resp. R t ) only at the point O x (resp. O z ). It is easy to check − K X · L xz = 110 · , − K X · L yt = 213 · , − K X · R x = 831 · , − K X · R y = 413 · , − K X · R z = 15 · , − K X · R t = 710 · ,L xz = − · , R x = − · , L xz · R x = 449 . We have lct( X ) since6516 = lct (cid:18) X, C x (cid:19) <
304 = lct (cid:18) X, C y (cid:19) < (cid:18) X, C t (cid:19) <
627 = lct (cid:18) X, C z (cid:19) . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D does not contain at least one irreducible component of each of the curves C x , C y , C z and C t . The curve R z is singular at the point O y . The curve R t is singular at O z withmultiplicity 3. Then in each of the following pairs of inequalities, at least one of two must hold:mult O x ( D ) D · L yt = 231 < , mult O x ( D ) D · R y = 449 < O y ( D ) D · L xz = 249 < , mult O y ( D ) D · R z = 213 < O z ( D ) D · L yt = 213 < , mult O z ( D ) D · R t = 730 < . Therefore, the point P can be none of O x , O y , O z .Since H ( P , O P (403)) contains the monomials x y , x and z , it follows from Lemma 1.3.9that the point P is either the point O t or a smooth point of X in C x .Write D = m L xz + m R x + Ω, where Ω is an effective Q -divisor whose support contains noneof L xz , R x . If m = 0, then m = 0 and hence we obtain831 ·
49 = D · R x > m L xz · R x = 4 m . Therefore, m . Similarly, we get m . Since we have( D − m L xz ) · L xz = 2 + 67 m · < , ( D − m R x ) · R x = 8 + 72 m · < XCEPTIONAL DEL PEZZO HYPERSURFACES 79 it follows from Lemma 1.3.8 that the point P must be the point O t .Suppose that m = 0. Then the inequalitymult O t ( D ) D · L xz = 110 < m = 0 and hence the curve R x is not contained in thesupport of D . Then831 ·
49 = D · R x > m L xz · R x + mult O t ( D ) − m > m
49 + 1665 · , and hence m < · . Since49( D − m L xz ) · L xz = 2 + 67 m < X, D ) is log canonical at the point O t by Lemma 1.3.8. This is a contradiction. (cid:3) Lemma 3.2.21.
Let X be a quasismooth hypersurface of degree 226 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + y z + xz + x y = 0 . It is singular at the points O x , O y and O z . The curves C x and C y are irreducible. We have9120 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 15512 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y . Then the inequalities71 D · C x = 431 < , D · C y = 471 < P is a smooth point in the outside of C x . However, since H ( P , O P (923))contains x , y x , y x and z , it follows from Lemma 1.3.9 that the point P is either asingular point of X or a point on C x . This is a contradiction. (cid:3) Lemma 3.2.22.
Let X be a quasismooth hypersurface of degree 99 in P (14 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + tz + xy + x z = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consistsof L xt (resp. L yz , L yz , L xt ) and R x = { x = yt + z = 0 } (resp. R y = { y = zt + x = 0 } , R z = { z = xy + t = 0 } , R t = { t = y + x z = 0 } ). Also, we see that L xt ∩ R x = { O y } , L yz ∩ R y = { O t } , L yz ∩ R z = { O x } , L xt ∩ R t = { O z } . We can easily check that lct( X, C y ) = is less than each of the numbers lct( X, C y ),lct( X, C z ) and lct( X, C t ). Therefore, lct( X ) . Suppose lct( X ) < . Then, there isan effective Q -divisor D ∼ Q − K X such that the log pair ( X, D ) is not log canonical at somepoint P ∈ X .The intersection numbers among the divisors D , L xt , L yz , R x , R y , R z , R t are as follows: D · L xt = 217 · , D · R x = 417 · , D · R y = 1029 · ,D · L yz = 17 · , D · R z = 27 · , D · R t = 57 · ,L xt · R x = 217 , L yz · R y = 541 , L yz · R z = 17 , L xt · R t = 529 ,L xt = − · , R x = − · , R y = − · ,L yz = − · , R z = 127 · , R t = 13514 · . By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . The inequalities17 D · L xt = 229 < , D · R x = 441 < P = O y . The inequalities14 D · L yz = 241 < , D · R z = 217 < P = O x . The curve R z is singular at the point O x . The inequalities29 D · L xt = 217 < , D · R t = 528 < P = O z . The curve R t is singular at the point O z .We write D = m L xt + m L yz + m R x + m R y + m R z + m R t + Ω, where Ω is an effectivedivisor whose support contains none of the curves L xt , L yz , R x , R y , R z , R t . Since the pair( X, D ) is log canonical at the points O x , O y , O z , the numbers m i are at most . Then byLemma 1.3.8 the following inequalities enable us to conclude that either the point P is in theoutside of C x ∪ C y ∪ C z ∪ C t or P = O t :( D − m L xt ) · L xz = 2 + 44 m · , ( D − m L yz ) · L yt = 2 + 53 m · , ( D − m R x ) · R x = 4 + 54 m · , ( D − m R y ) · R y = 10 + 60 m · , ( D − m R z ) · R z = 2 − m · , ( D − m R t ) · R t = 10 − m · . Suppose that P = O t . Then we consider the pencil L on X defined by λyt + µz = 0,[ λ : µ ] ∈ P . The base locus of the pencil L consists of the curve L yz and the point O y . Let E be the unique divisor in L that passes through the point P . Since P C x ∪ C y ∪ C z ∪ C t , thedivisor E is defined by the equation z = αyt , where α = 0. XCEPTIONAL DEL PEZZO HYPERSURFACES 81
Suppose that α = −
1. Then the curve E is isomorphic to the curve defined by the equations yt = z and t y + xy + x z = 0. Since the curve E is isomorphic to a general curve in L ,it is smooth at the point P . The affine piece of E defined by t = 0 is the curve given by z ( z + xz + x ) = 0. Therefore, the divisor E consists of two irreducible and reduced curves L yz and C . We have the intersection number D · C = D · E − D · L yz = 1817 · · . Also, we see C = E · C − C · L yz > E · C − C y · C > C is different from R y . By Lemma 1.3.8 the inequality D · C < gives us a contradiction.Suppose that α = −
1. Then divisor E consists of three irreducible and reduced curves L yz , R x , and M . Note that the curve M is different from the curves R y and L xt . Also, it is smoothat the point P . We have D · M = D · E − D · L yz − D · R x = 1537 · · ,M = E · M − L yz · M − R x · M > E · M − C y · M − C x · M > . By Lemma 1.3.8 the inequality D · M < gives us a contradiction. Therefore, the log pair( X, D ) is not log canonical at the point O t .Put D = aL yz + bR x + ∆, where ∆ is an effective Q -divisor whose support contains neither L yz nor R x . Then a > D · L yz > mult O t ( D ) > . Therefore, we may assume that R y Supp(∆) by Lemma 1.3.6. Similarly, we may assume that L xt Supp(∆) if b > b >
0, then 217 ·
29 = D · L xt > bR x · L xt = 2 b , and hence b . Similarly, we have1029 ·
41 = D · R y > a
41 + b
41 + mult O t ( D ) − a − b > a
41 + 421 · . Therefore, a < · · .Let π : ¯ X → X be the weighted blow up at the point O t with weights (9 ,
4) and let F be theexceptional curve of the morphism π . Then F contains two singular points Q and Q of ¯ X such that Q is a singular point of type (1 , Q is a singular point of type (3 , K ¯ X ∼ Q π ∗ ( K X ) − F, ¯ L yz ∼ Q π ∗ ( L yz ) − F, ¯ R x ∼ Q π ∗ ( R x ) − F, ¯∆ ∼ Q π ∗ (∆) − c F, where ¯ L yz , ¯ R x and ¯∆ are the proper transforms of L yz , R x and ∆ by π , respectively, and c is anon-negative rational number. Note that F ∩ ¯ R x = { Q } and F ∩ ¯ L yz = { Q } . The log pull-back of the log pair ( X, D ) by π is the log pair (cid:18) ¯ X, a
10 ¯ L yz + 51 b
10 ¯ R x + 5110 ¯∆ + θ F (cid:19) , where θ = 280 + 51(4 a + 9 b + c )10 · . This is not log canonical at some point Q ∈ F .We have 0 ¯∆ · ¯ R x = 4 + 54 b · − a − c · . This inequality shows 4 a + c (4 + 54 b ). Since b , we obtain θ = 280 + 51(4 a + 9 b + c )10 · b )10 · · < . Suppose that Q ¯ R x ∪ ¯ L yz . Then the log pair ( F, ¯∆ | F ) is not log canonical at the point Q , and hence 17 c
120 = 5110 ¯∆ · F > c > . However, since b , we obtain c a + c
417 (4 + 54 b ) < . Therefore, the point Q must be either Q or Q .Suppose that Q = Q . The pair ( ¯ R x , ( ¯∆ + θ F ) | ¯ R x ) is not log canonical at Q . It thenfollows from Lemma 1.3.8 that1 < (cid:18) θ F (cid:19) · ¯ R x = 4 · (cid:18) b · − a − c · (cid:19) + θ . However, 4 · (cid:18) b · − a − c · (cid:19) + θ = 4760 + 51(16 + 369 b )10 · · < . This is a contradiction. Consequently, the point Q must be Q .Let ψ : ˜ X → ¯ X be the blow up at the point Q and let E be the exceptional curve of themorphism ψ . The surface ˜ X is smooth along the exceptional divisor E . Then K ˜ X ∼ Q ψ ∗ ( K ¯ X ) − E, ˜ L yz ∼ Q ψ ∗ ( ¯ L yz ) − E, ˜ F ∼ Q ψ ∗ ( F ) − E, ˜∆ ∼ Q ψ ∗ ( ¯∆) − d E, where ˜ L yz , ˜ F and ˜∆ are the proper transforms of ¯ L yz , F and ¯∆ by ψ , respectively, and d is anon-negative rational number.The log pull-back of the log pair ( X, D ) by π ◦ ψ is the log pair (cid:18) ˜ X, a
10 ˜ L yz + 51 b
10 ˜ R x + 5110 ˜∆ + θ ˜ F + θ E (cid:19) , XCEPTIONAL DEL PEZZO HYPERSURFACES 83 where ˜ R x is the proper transform of ¯ R x by ψ and θ = 70 + 51( a + d ) + 10 θ
90 = 3150 + 51(45 a + 9 b + c + 41 d )90 · . This is not log canonical at some point O ∈ E .We have 0 ˜∆ · ˜ L yx = ¯∆ · ¯ L yz − d a · − b − c · − d , and hence 9 b + c + 41 d (2 + 53 a ). Therefore, this inequality together with a < · · givesus θ = 3150 + 51(45 a + 9 b + c + 41 d )90 ·
41 == 3150 + 2295 a ·
41 + 51(9 b + c + 41 d )90 · a · · < . Suppose that the point O is in the outside of ˜ L yz and ˜ F . Then the log pair ( E, ˜∆ | E ) is notlog canonical at the point O and hence1 < · E = 51 d . However, 41 d b + c + 41 d
914 (2 + 53 a ) < · a < · · . This is a contradiction.Suppose that the point O belongs to ˜ L yz Then the log pair ( E, ( a ˜ L yz + ˜∆) | E ) is not logcanonical at the point O and hence1 < ( 51 a
10 ˜ L yz + 5110 ˜∆) · E = 5110 ( a + d ) . However, 5110 ( a + d ) (cid:18) a + 914 ·
41 (2 + 53 a ) (cid:19) < a < · · . This is a contradiction. Therefore, the point O is the intersection point of ˜ F and E .Let ξ : ˆ X → ˜ X be the blow up at the point O and let H be the exceptional divisor of ξ .We also let ˆ L yz , ˆ R x , ˆ∆, ˆ E , and ˆ F be the proper transforms of ˜ L yz , ˜ R x , ˜∆, E and ˜ F by ξ ,respectively. We have K ˆ X ∼ Q ξ ∗ ( K ˜ X ) + H, ˆ E ∼ Q ξ ∗ ( E ) − H, ˆ F ∼ Q ξ ∗ ( ˜ F ) − H, ˆ∆ ∼ Q ξ ∗ ( ˜∆) − eH, where e is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ ◦ ξ is (cid:18) ˆ X, a
10 ˆ L yz + 51 b
10 ˆ R x + 5110 ˆ∆ + θ ˆ F + θ ˆ E + θ H (cid:19) , where θ = θ + θ + 51 e − a + 90 b + 10 c + 41 d + 369 e )90 · . This log pair is not log canonical at some point A ∈ H . We have c · − d − e = ˆ∆ · ˆ F > . Therefore, 4 d + 36 e c . Then θ = 1980 + 51(81 a + 90 b + 10 c )90 ·
41 + 51( d + 9 e )90 a + 360 b + 81 c )4 · ·
41 == 22 + 51 b
41 + 51 · a + c )4 · ·
22 + 51 b
41 + 9 · b )5 · · < b and 4 a + c (4 + 54 b ).Suppose that A ˆ F ∪ ˆ E . Then the log pair (cid:16) ˆ X, ˆ∆ + θ H (cid:17) is not log canonical at the point A . Applying Lemma 1.3.4, we get 1 < · H = 51 e . However, e
136 (4 d + 36 e ) c
136 (4 a + c ) b · < . Therefore, the point A must be either in ˆ F or in ˆ E .Suppose that A ∈ ˆ F . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ F + θ H (cid:17) is not log canonical at thepoint A . Applying Lemma 1.3.4, we get1 < (cid:18) θ H (cid:19) · ˆ F = 5110 (cid:18) c · − d − e (cid:19) + θ = 7920 + 51(324 a + 360 b + 81 c )4 · · . However, 7920 + 51(324 a + 360 b + 81 c )4 · · < A is the intersection point of H and ˆ E . Then thelog pair (cid:16) ˆ X, ˆ∆ + θ ˆ E + θ H (cid:17) is not log canonical at the point A . From Lemma 1.3.4, weobtain 1 < (cid:18) θ H (cid:19) · ˆ E = 5110 ( d − e ) + θ = 1980 + 51(81 a + 90 b + 10 c + 410 d )90 · . XCEPTIONAL DEL PEZZO HYPERSURFACES 85
However, 1980 + 51(81 a + 90 b + 10 c + 410 d )90 ·
41 = 220 + 459 a ·
41 + 51(9 b + c + 41 d )9 ·
220 + 459 a ·
41 + 51(2 + 53 a )14 · < b + c + 41 d (2 + 53 a ) and a < · · . The obtained contradiction completes theproof. (cid:3) Sporadic cases with I = 3 Lemma 3.3.1.
Let X be a quasismooth hypersurface of degree 33 in P (5 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z + yt + xy + x t + ǫx yz = 0 , where ǫ ∈ C . Note that the surface X is singular at O x , O y and O t .The curves C x , C y are irreducible. Moreover, we have2518 = lct( X, C x ) > lct( X, C y ) = 4936 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D contains neither C x nor C y . Since the curve C y is singular at the point O t , thethree inequalities 5 D · C y = 913 < , D · C x = 6391 < P is located in the outside of the set C x ∪ C y .Let L be the pencil on X that is cut out by the equations λx + µy = 0 , where [ λ : µ ] ∈ P . Then the base locus of the pencil L consists of the point O t . Let C be theunique curve in L that passes through the point P . Since the point P is in the outside of the set C x ∪ C y , the curve C is defined by an equation of the form y − αx = 0, where α is a non-zeroconstant. Suppose that C is irreducible and reduced. Then mult P ( C ) C isa triple cover of the curve y − αx = 0 ⊂ P (cid:0) , , (cid:1) ∼ = Proj (cid:16) C (cid:2) x, y, t (cid:3)(cid:17) . In particular, lct( X, C ) > . Thus, we may assume that the support of D does not containthe curve C and hence we obtain3649 < mult P ( D ) D · C = 913 < . This is a contradiction. Thus, to conclude the proof it suffices to prove that the curve C isirreducible and reduced. Let S ⊂ C be the affine variety defined by the equations y − αx = z + yt + xy + x t + ǫx yz = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, y, z, t (cid:3)(cid:17) . To conclude the proof, it is enough to prove that the variety S is irreducible.Consider the projectivised surface ¯ S of S defined by the homogeneous equations y w − αx = z w + yt w + xy + x t + ǫx yz = 0 ⊂ P ∼ = Proj (cid:16) C (cid:2) x, y, z, t, w (cid:3)(cid:17) . Then we consider the affine piece S ′ of ¯ S defined by y = 0. The affine surface S ′ is defined bythe equations w − αx = z w + t w + x + x t + ǫx z = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, z, t, w (cid:3)(cid:17) . This is isomorphic to the affine hypersurface defined by x ( αx z + αx t + 1 + x t + ǫx z ) = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, z, t (cid:3)(cid:17) . This affine hypersurface has two irreducible components. However, the component defined by x = 0 originates from the hyperplane section of ¯ S by w = 0. Therefore, the original affinesurface S must be irreducible and reduce. (cid:3) Lemma 3.3.2.
Let X be a quasismooth hypersurface of degree 40 in P (5 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t ( t − x ) + yz + xy + ǫx y z, where ǫ ∈ C . Note that X is singular at the points O x , O y , O z and Q = [1 : 0 : 0 : 1].The curve C x is irreducible. We havelct( X, C x ) = 2518 . Therefore, lct( X ) . Meanwhile, the curve C y is reducible. It consists of two irreduciblecomponents L yt and R y = { y = t − x = 0 } . The curve L yt intersects R y only at the point O z .It is easy to see L yt = R y = − , L yt · R y = 411 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D does not contain the curve C x . Moreover, we may assume that the support of D does not contain either L yt or R y sincelct( X, C y ) = 3524 > . Then one of the inequalitiesmult O z ( D ) D · L yt = 35 < , mult O z ( D ) D · R y = 35 < XCEPTIONAL DEL PEZZO HYPERSURFACES 87 must hold, and hence the point P cannot be the point O z . Also, since 7 D · C x = < , thepoint P cannot belong to the curve C x .We write D = aL yt + bR y + Ω, where Ω is an effective Q -divisor whose support containsneither L yt nor R y . If a = 0, then we have355 = D · R y > aL yt · R y = 4 a . Therefore, a . By the same way, we also obtain b .Since we have5( D − aL yt ) · L yt = 3 + 13 a < , D − bR y ) · R y = 3 + 13 a < P is in the outside of C y . Consequently, the point P is locatedin the outside of C x ∪ C y . However, since H ( P , O P (40)) contains monomials x , xy , x t and thenatural projection X P (5 , ,
20) is a finite morphism outside of the curve C y , Lemma 1.3.9shows that the point P must belong to the set C x ∪ C y . This is a contradiction. (cid:3) Lemma 3.3.3.
Let X be a quasismooth hypersurface of degree 95 in P (11 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + tz + xy + x z = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consistsof L xt (resp. L yz , L yz , L xt ) and R x = { x = yt + z = 0 } (resp. R y = { y = zt + x = 0 } , R z = { z = xy + t = 0 } , R t = { t = y + x z = 0 } ). Also, we see that L xt ∩ R x = { O y } , L yz ∩ R y = { O t } , L yz ∩ R z = { O x } , L xt ∩ R t = { O z } . It is easy to check that lct( X, C x ) = is less than each of the numbers lct( X, C y ),lct( X, C z ) and lct( X, C t ). Therefore, lct( X ) .Suppose that lct( X ) < . Then, there is an effective Q -divisor D ∼ Q − K X such that the logpair ( X, D ) is not log canonical at some point P ∈ X .The intersection numbers among the divisors D , L xt , L yz , R x , R y , R z , R t are as follows: D · L xt = 17 · , D · R x = 27 · , D · R y = 1829 · ,D · L yz = 311 · , D · R z = 27 · , D · R t = 1211 · ,L xt · R x = 221 , L yz · R y = 637 , L yz · R z = 211 , L xt · R t = 429 ,L xt = − · , R x = − · , R y = − · ,L yz = − · , R z = 1611 · , R t = 10411 · . By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . The inequalities21 D · L xt = 329 < , D · R x = 637 < P = O y . The inequalities11 D · L yz = 337 < , D · R z = 27 < P = O x . The inequalities29 D · L xt = 17 < , D · R t = 311 < P = O z . The curve R t is singular at the point O z with multiplicity 4.We write D = m L xt + m L yz + m R x + m R y + m R z + m R t + Ω, where Ω is an effectivedivisor whose support contains none of the curves L xt , L yz , R x , R y , R z , R t . Since the pair( X, D ) is log canonical at the points O x , O y , O z , the numbers m i are at most . Then byLemma 1.3.8 the following inequalities enable us to conclude that either the point P is in theoutside of C x ∪ C y ∪ C z ∪ C t or P = O t :( D − m L xt ) · L xt = 3 + 47 m · , ( D − m L yz ) · L yz = 3 + 45 m · , ( D − m R x ) · R x = 6 + 52 m · , ( D − m R y ) · R y = 18 + 48 m · , ( D − m R z ) · R z = 6 − m · , ( D − m R t ) · R t = 12 − m · . Suppose that P = O t . Then we consider the pencil L defined by λyt + µz = 0, [ λ : µ ] ∈ P .The base locus of the pencil L consists of the curve L yz and the point O y . Let E be the uniquedivisor in L that passes through the point P . Since P C x ∪ C y ∪ C z ∪ C t , the divisor E isdefined by the equation z = αyt , where α = 0.Suppose that α = −
1. Then the curve E is isomorphic to the curve defined by the equations yt = z and t y + xy + x z = 0. Since the curve E is isomorphic to a general curve in L ,it is smooth at the point P . The affine piece of E defined by t = 0 is the curve given by z ( z + xz + x ) = 0. Therefore, the divisor E consists of two irreducible and reduced curves L yz and C . We have the intersection number D · C = D · E − D · L yz = 1697 · · . Also, we see C = E · C − C · L yz > E · C − C y · C > C is different from R y . The multiplicity of D along the curve C is at most since theintersection number C · C t is positive and the pair ( X, D ) is log canonical along the curve C t .Then by Lemma 1.3.8 the inequality D · C < gives us a contradiction.
XCEPTIONAL DEL PEZZO HYPERSURFACES 89
Suppose that α = −
1. Then divisor E consists of three irreducible and reduced curves L yz , R x , and M . Note that the curve M is different from the curves R y and L xt . Also, it is smoothat the point P . We have D · M = D · E − D · L yz − D · R x = 1477 · · ,M = E · M − L yz · M − R x · M > E · M − C y · M − C x · M > . The multiplicity of D along the curve M is at most since the intersection number M · C t is positive and the pair ( X, D ) is log canonical along the curve C t . By Lemma 1.3.8 theinequality D · M < gives us a contradiction. Therefore, P = O t .Put D = aL yz + bR x + ∆, where ∆ is an effective Q -divisor whose support contains neither L yz nor R x . Then a > D · L yz > mult O t ( D ) > . Therefore, we may assume that R y Supp(∆) by Lemma 1.3.6.If b >
0, the curve L xt is not contained in the support of D , and hence321 ·
29 = D · L xt > bR x · L xt = 2 b . Therefore, b . Similarly, we have1829 ·
37 = D · R y > a
37 + b
37 + mult O t ( D ) − a − b > a
37 + 411 · , and hence a < · · .Let π : ¯ X → X be the weighted blow up of the point O t with weights (13 ,
4) and let F bethe exceptional curve of the morphism π . Then F contains two singular points Q and Q of¯ X such that Q is a singular point of type (1 ,
2) and Q is a singular point of type (3 , K ¯ X ∼ Q π ∗ ( K X ) − F, ¯ L yz ∼ Q π ∗ ( L yz ) − F, ¯ R x ∼ Q π ∗ ( R x ) − F, ¯∆ ∼ Q π ∗ (∆) − c F, where ¯ L yz , ¯ R x and ¯∆ are the proper transforms of L yz , R x and ∆ by π , respectively, and c is anon-negative rational number.The log pull-back of the log pair ( X, D ) by π is the log pair (cid:18) ¯ X, a L yz + 11 b R x + 114 ¯∆ + θ F (cid:19) , where θ = 11(4 a + 13 b + c ) + 804 · . This pair is not log canonical at some point Q ∈ F . We have0 ¯∆ · ¯ R x = 6 + 52 b · − a − c · . This inequality shows 4 a + c (6 + 52 b ). Then θ = 11(4 a + c )4 ·
37 + 143 b ·
37 + 2037 ·
37 (6 + 52 b ) + 143 b ·
37 + 2037 = 1944 + 5291 b · · < b . Note that F ∩ ¯ R x = { Q } and F ∩ ¯ L yz = { Q } .Suppose that the point Q is neither Q nor Q . Then the pair (cid:0) ¯ X, ¯∆ + F (cid:1) is not logcanonical at the point Q . Then 11 c ·
13 = 114 ¯∆ · F > c a + c (6 + 52 b ). This is a contradiction since b .Therefore, the point Q is either Q or Q .Suppose that the point Q is the point Q . Then the log pair (cid:0) ¯ X, b ¯ R x + ¯∆ + θ F (cid:1) is notlog canonical at the point Q . It then follows from Lemma 1.3.4 that1 < (cid:18)
114 ¯∆ + θ F (cid:19) · ¯ R x = 11 (cid:18) b · − a − c · (cid:19) + θ . However, 11 (cid:18) b · − a − c · (cid:19) + θ = 1944 + 5291 b · · < . Therefore, the point Q must be the point Q .Let φ : ˜ X → ¯ X be the weighted blow up at the point Q with weights (1 , G be theexceptional divisor of the morphism φ . Then G contains one singular point Q of the surface ˜ X that is a singular point of type (1 , L yz , ˜ R x , ˜∆ and ˜ F be the proper transforms of L yz , R x , ∆ and F by φ , respectively. We have K ˜ X ∼ Q φ ∗ ( K ¯ X ) − G, ˜ L yz ∼ Q φ ∗ ( ¯ L yz ) − G, ˜ F ∼ Q φ ∗ ( F ) − G, ˜∆ ∼ Q φ ∗ ( ¯∆) − d G, where d is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ is (cid:18) ˜ X, a L yz + 11 b R x + 114 ˜∆ + θ ˜ F + θ G (cid:19) , where θ = 114 ·
13 (2 a + d ) + θ
13 + 1013 = 1560 + 11(78 a + 13 b + c + 37 d )4 · · . This log pair is not log canonical at some point O ∈ G . We have0 ˜∆ · ˜ L yz = 3 + 45 a · − b − c · − d . We then obtain 13 b + c + 37 d (3 + 45 a ). Since a < · · , we see θ = 1560 + 11(78 a + 13 b + c + 37 d )4 · · a · ·
37 + 3 + 45 a · < . XCEPTIONAL DEL PEZZO HYPERSURFACES 91
Note that ˜ F ∩ G = Q and Q ˜ L yz . Suppose that O ˜ F ∪ ˜ L yz . The log pair (cid:16) ˜ X, ˜∆ + G (cid:17) is not log canonical at the point O . Applying Lemma 1.3.4, we get1 <
114 ˜∆ · G = 11 d · , and hence d > . However, d (13 b + c + 37 d ) · (3 + 45 a ). This is a contradictionsince a < · · . Therefore, the point O is either the point Q or the intersection point of G and ˜ L yz . In the latter case, the pair (cid:16) ˜ X, a ˜ L yz + ˜∆ + θ G (cid:17) is not log canonical at the point O . Then, applying Lemma 1.3.4, we get1 < (cid:18)
114 ˜∆ + θ G (cid:19) · ˜ L yz = 114 (cid:18) a · − b − c · − d (cid:19) + θ . However, 114 (cid:18) a · − b − c · − d (cid:19) + θ = 114 (cid:18) a · (cid:19) + 1560 + 858 a · · < a < · · . Therefore, the point O must be the point Q .Let ξ : ˆ X → ˜ X be the blow up at the point Q and let H be the exceptional divisor of ξ .We also let ˆ L yz , ˆ R x , ˆ∆, ˆ G , and ˆ F be the proper transforms of ˜ L yz , ˜ R x , ˜∆, G and ˜ F by ξ ,respectively. Then ˆ X is smooth along the exceptional divisor H . We have K ˆ X ∼ Q ξ ∗ ( K ˜ X ) , ˆ G ∼ Q ξ ∗ ( G ) − H, ˆ F ∼ Q ξ ∗ ( ˜ F ) − H, ˆ∆ ∼ Q ξ ∗ ( ˜∆) − e H, where e is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ ◦ ξ is (cid:18) ˆ X, a L yz + 11 b R x + 114 ˆ∆ + θ ˆ F + θ ˆ G + θ H (cid:19) , where θ = θ + θ e a + 182 b + 14 c + 37 d + 481 e )8 · · . This log pair is not log canonical at some point A ∈ H . We have c · − d · − e · ˆ F > . Therefore, 2 d + 26 e c . Then θ = 2600 + 11(130 a + 182 b + 14 c )8 · ·
37 + 11( d + 13 e )8 · a + 364 b + 65 c )16 · ·
37 == 5200 + 4004 b · ·
37 + 11 · a + c )16 · ·
100 + 77 b ·
37 + 5 · b )4 · ·
37 = 2430 + 4477 b · · < b and 4 a + c (6 + 52 b ). Suppose that A ˆ F ∪ ˆ G . Then the log pair (cid:16) ˆ X, ˆ∆ + θ H (cid:17) is not log canonical at the point A . Applying Lemma 1.3.4, we get 1 <
114 ˆ∆ · H = 11 e . However, e
126 (2 d + 26 e ) c
126 (4 a + c ) b )21 · . Therefore, the point A must be either in ˆ F or in ˆ G .Suppose that A ∈ ˆ F . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ F + θ H (cid:17) is not log canonical at thepoint A . Applying Lemma 1.3.4, we get1 < (cid:18)
114 ˆ∆ + θ H (cid:19) · ˆ F = 114 (cid:18) c · − d · − e (cid:19) + θ = 5200 + 11(260 a + 364 b + 65 c )16 · · . However,5200 + 11(260 a + 364 b + 65 c )16 · ·
37 = 400 + 11 · b ·
37 + 11 · a + c )16 · b · · < . Therefore, the point A is the intersection point of H and ˆ G . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ G + θ H (cid:17) is not log canonical at the point A . From Lemma 1.3.4, we obtain1 < (cid:18)
114 ˆ∆ + θ H (cid:19) · ˆ G = 114 (cid:18) d − e (cid:19) + θ = 2600 + 11(130 a + 182 b + 14 c )8 · ·
37 + 77 d · . However,2600 + 11(130 a + 182 b + 14 c )8 · ·
37 + 77 d ·
13 = 100 + 55 a ·
37 + 77(13 b + c + 37 d )4 · ·
121 + 370 a · < a < · · and 13 b + c + 37 d (3 + 45 a ). The obtained contradiction completes theproof. (cid:3) Lemma 3.3.4.
Let X be a quasismooth hypersurface of degree 196 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + yz + xy + x z = 0 . It is singular at the points O x , O y and O z . The curves C x and C y are irreducible. We have5518 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 37 · , and hence lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y . Then the inequalities53 D · C x = 637 < , D · C y = 653 < XCEPTIONAL DEL PEZZO HYPERSURFACES 93 show that the point P is a smooth point in the outside of C x . However, since H ( P , O P (583))contains the monomials x , y x and z , it follows from Lemma 1.3.9 that the point P iseither a singular point of X or a point on C x . This is a contradiction. (cid:3) Lemma 3.3.5.
Let X be a quasismooth hypersurface of degree 95 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z t + y z + xt + x y = 0 . The surface X is singular at the point O x , O y , O z and O t . Each of the divisors C x , C y , C z ,and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t )consists of L xz (resp. L yt , L xz , L yt ) and R x = { x = y + zt = 0 } (resp. R y = { y = z + xt = 0 } , R z = { z = t + x y = 0 } , R t = { t = x + y z = 0 } ). The curve L xz intersects R x (resp. R z )only at the point O t (resp. O y ). Also, the curve L yt intersects R y (resp. R t ) only at the point O x (resp. O z ).It is easy to check − K X · L xz = 317 · , − K X · L yt = 19 · , − K X · R x = 49 · , − K X · R y = 613 · , − K X · R z = 613 · , − K X · R t = 23 · ,L xz = − · , L yt = − · , R x = − · , R y = − · , R z = 2813 · R t = 163 · , L xz · R x = 441 , L yt · R y = 213 , L xz · R z = 217 , L yt · R t = 29 . We have lct( X ) since6524 = lct (cid:18) X, C x (cid:19) < (cid:18) X, C y (cid:19) <
418 = lct (cid:18) X, C t (cid:19) <
214 = lct (cid:18) X, C z (cid:19) . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D does not contain at least one irreducible component of each of the curves C x , C y , C z and C t . The curve R z is singular at the point O y . The curve C t is singular at O z withmultiplicity 3. Then in each of the following pairs of inequalities, at least one of two must hold:mult O x ( D ) D · L yt = 19 < , mult O x ( D ) D · R y = 641 < O y ( D ) D · L xz = 341 < , mult O y ( D ) D · R z = 313 < O z ( D ) D · L yt = 313 < , mult O z ( D ) D · R t = 617 < P can be none of O x , O y , O t . Put D = m L xz + m L yt + m R x + m R y + m R z + m R t + Ω, where Ω is an effective Q -divisor whose support contains none of L xz , L yt , R x , R y , R z , R t . Since the pair ( X, D ) is logcanonical at the points O x , O y , O z , we have m i for each i . Since( D − m L xz ) · L xz = 3 + 55 m · , ( D − m L yt ) · L yt = 3 + 37 m · , ( D − m R x ) · R x = 12 + 56 m · , ( D − m R y ) · R y = 6 + 48 m · , ( D − m R z ) · R z = 6 − m · , ( D − m R t ) · R t = 2 − m · P cannot be a smooth point of X on C x ∪ C y ∪ C z ∪ C t .Therefore, the point P is either a point in the outside of C x ∪ C y ∪ C z ∪ C t or the point O t .Suppose that the point P is not the point O t . We consider the pencil L on X defined by theequations λxt + µz = 0, [ λ : µ ] ∈ P . Then there is a unique curve Z α in the pencil L passingthrough the point P . Since the point P is located in the outside of C x ∪ C z ∪ C t , the curve Z α is defined by an equation of the form xt + αz = 0 , where α is a non-zero constant. Note that any component of C t is not contained in Z α . Theopen subset Z α \ C t is a Z -quotient of the affine curve x + αz = z + y z + x + x y = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, y, z (cid:3)(cid:17) that is isomorphic to the plane affine curve defined by the equation z ( y + (1 − α ) z + α z y ) = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . Therefore, if α = 1, then the curve Z α consists of two irreducible components L xz and C α . Onthe other hand, if α = 1, then the curve Z α consists of three irreducible components L xz , R y ,and C . Since P C x ∪ C y ∪ C z ∪ C t , the point P must be contained in C α (including α = 1).Also, the curve C α is smooth at the point P . By Lemma 1.3.6, we may assume that Supp( D )does not contain at least one irreducible component of the curve Z α .Write D = mC α + Γ, where Γ is an effective Q -divisor whose support contains C α . Supposethat m = 0. If α = 1, then we obtain317 ·
41 = D · L xz > mC α · L xz = 109 m · m . If α = 1, then one of the inequalities317 ·
41 = D · L xz > mC · L xz = 92 m · , ·
41 = D · R y > mC · R y = 11 m m · . We also see that D · C α = D · ( Z α − L xz ) = 53113 · ·
41 if α = 1 ,D · ( Z α − L xz − R y ) = 3317 ·
41 if α = 1 . Also, if α = 1, then XCEPTIONAL DEL PEZZO HYPERSURFACES 95 C α = Z α · C α − L xz · C α > Z α · C α − ( L xz + R x ) · C α = 413 D · C α . If α = 1, C = Z α · C − ( L xz + R y ) · C > Z α · C − ( L xz + R x + L yt + R y ) · C = 8 D · C . In both cases, we have C α >
0. Since( D − mC α ) · C α D · C α < P must be the point O t .If L xz is not contained in the support of D , then the inequalitymult O t ( D ) D · L xz = 317 < L xz must be contained in the supportof D , and hence the curve R x is not contained in the support of D . Put D = aL xz + bR y + ∆,where ∆ is an effective Q -divisor whose support contains neither L xz nor R y . Then49 ·
41 = D · R x > aL xz · R x + mult O t ( D ) − a > a
41 + 2441 · a . If b = 0, then L yt is not contained in the support of D . Therefore,19 ·
13 = D · L yt > bR y · L yt = 2 b , and hence b .Let π : ¯ X → X be the weighted blow up at the point O t with weights (1 ,
4) and let F be theexceptional curve of the morphism π . Then F contains one singular point Q of ¯ X such that Q is a singular point of type (3 , K ¯ X ∼ Q π ∗ ( K X ) − F, ¯ L xz ∼ Q π ∗ ( L xz ) − F, ¯ R y ∼ Q π ∗ ( R y ) − F, ¯∆ ∼ Q π ∗ (∆) − c F, where ¯ L xz , ¯ R y and ¯∆ are the proper transforms of L xz , R y and ∆ by π , respectively, and c is anon-negative rational number. Note that F ∩ ¯ R y = { Q } .The log pull-back of the log pair ( X, D ) by π is the log pair (cid:18) ¯ X, a
24 ¯ L xz + 65 b
24 ¯ R y + 6524 ¯∆ + θ F (cid:19) , where θ = 864 + 65(4 a + b + c )24 · . This is not log canonical at some point Q ∈ F . We have0 ¯∆ · ¯ L xz = 3 + 55 a · − b − c . This inequality shows b + c (3 + 55 a ). Since a , we obtain θ = 864 + 260 a ·
41 + 65( b + c )24 ·
864 + 260 a ·
41 + 65(3 + 55 a )17 · ·
41 = 121 + 65 a · < . Suppose that the point Q is neither Q nor the intersection point of F and ¯ L xz . Then, thepoint Q is not in ¯ L xz ∪ ¯ R y . Therefore, the pair (cid:0) ¯ X, ¯∆ + F (cid:1) is not log canonical at the point Q , and hence 1 < · F = 65 c · . But c b + c (3 + 55 a ) < · since a . Therefore, the point Q is either Q or theintersection point of F and ¯ L xz .Suppose that the point Q is the intersection point of F and ¯ L xz . Then the point Q is in ¯ L xz but not in ¯ R y . Therefore, the pair (cid:0) ¯ X, ¯ L xz + ¯∆ + θ F (cid:1) is not log canonical at the point Q .Then 1 < (cid:18) θ F (cid:19) · ¯ L xz = 6524 (cid:18) a · − b + c (cid:19) + θ = 121 + 65 a · . However, this is impossible since a . Therefore, the point Q must be the point Q .Let ψ : ˜ X → ¯ X be the weighted blow up at the point Q with weights (3 ,
1) and let E be theexceptional curve of the morphism ψ . The exceptional curve E contains one singular point O of ˜ X . This singular point is of type (1 , K ˜ X ∼ Q ψ ∗ ( K ¯ X ) , ˜ R y ∼ Q ψ ∗ ( ¯ R y ) − E, ˜ F ∼ Q ψ ∗ ( F ) − E, ˜∆ ∼ Q ψ ∗ ( ¯∆) − d E, where ˜ R y , ˜ F and ˜∆ are the proper transforms of ¯ R y , F and ¯∆ by ψ , respectively, and d is anon-negative rational number.The log pull-back of the log pair ( X, D ) by π ◦ ψ is the log pair (cid:18) ˜ X, a
24 ˜ L xz + 65 b
24 ˜ R y + 6524 ˜∆ + θ ˜ F + θ E (cid:19) , where θ = 65(3 b + d )4 ·
24 + 14 θ . This is not log canonical at some point O ∈ E .We have 0 ˜∆ · ˜ R y = ¯∆ · ¯ R y − d b · − a + c · − d , and hence 4 a + c + 41 d (6 + 48 b ). Therefore, this inequality together with b gives us θ = 65(3 b + d )4 ·
24 + 864 + 65(4 a + b + c )4 · ·
41 == 864 + 8060 b + 65(4 a + c + 41 d )4 · · b < . Suppose that the point O is in the outside of ˜ R y and ˜ F . Then the log pair ( E, ˜∆ | E ) is notlog canonical at the point O , and hence1 < · E = 65 d . XCEPTIONAL DEL PEZZO HYPERSURFACES 97
However, 41 d a + c + 41 d
413 (6 + 48 b ) < · b . This is a contradiction.Suppose that the point O belongs to ˜ R y Then the log pair (cid:16) E, (cid:16) b ˜ R y + ˜∆ (cid:17) (cid:12)(cid:12)(cid:12) E (cid:17) is not logcanonical at the point O , and hence1 < (cid:18) b
24 ˜ R y + 6524 ˜∆ (cid:19) · E = 6524 (cid:18) b + d (cid:19) . However, 6524 (cid:18) b + d (cid:19) (cid:18) b + 43 · ·
41 (6 + 48 b ) (cid:19) < b . This is a contradiction. Therefore, the point O is the point O which is theintersection point of E and ˜ F .Let ξ : ˆ X → ˜ X be the weighted blow up at the point O with weights (1 ,
2) and let H be theexceptional divisor of ξ . The exceptional divisor H contains a singular point of ˆ X . This singularpoint is of type (1 , K ˆ X ∼ Q ξ ∗ ( K ˜ X ) , ˆ E ∼ Q ξ ∗ ( E ) − H, ˆ F ∼ Q ξ ∗ ( ˜ F ) − H, ˆ∆ ∼ Q ξ ∗ ( ˜∆) − e H, where ˆ E , ˆ F , ˆ∆, be the proper transforms of E , ˜ F , ˜∆ by ξ , respectively, and e is a non-negativerational number. The log pull-back of the log pair ( X, D ) via π ◦ ψ ◦ ξ is (cid:18) ˆ X, a
24 ˆ L xz + 65 b
24 ˆ R y + 6524 ˆ∆ + θ ˆ F + θ ˆ E + θ H (cid:19) , where ˆ L xz and ˆ R y are the proper transforms of ˜ L xz and ˜ R y by ξ , respectively, and θ = 13 (2 θ + θ ) + 65 e · . This log pair is not log canonical at some point A ∈ H . We have0 ˆ∆ · ˆ F = ¯∆ · F − d − e c − d − e , and hence d + 4 e c . Then θ = 13 (2 θ + θ ) + 65 e ·
24 = = 34 θ + 65(3 b + d )3 · ·
24 + 65 e · = 2592 + 65(12 a + 44 b + 3 c )3 · ·
41 + 65( d + 4 e )3 · · a + 44 b + 3 c )3 · ·
41 + 65 c ·
24 == 2592 + 65(12 a + 44 b + 44 c )3 · ·
216 + 65 a ·
41 + 65 · a )3 · · ·
41 = 321 + 1040 a · · < since b + c (3 + 55 a ) and a .Suppose that A ˆ F ∪ ˆ E . Then the log pair (cid:16) ˆ X, ˆ∆ + θ H (cid:17) is not log canonical at the point A . Applying Lemma 1.3.4, we get 1 < · H = 65 e . However, e
14 ( d + 4 e ) c
34 ( b + c ) a )4 · < . Therefore, the point A must be either in ˆ F or in ˆ E .Suppose that A ∈ ˆ F . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ F + θ H (cid:17) is not log canonical at thepoint A . Applying Lemma 1.3.4, we get1 < (cid:18) θ H (cid:19) · ˆ F = 6524 (cid:18) c − d − e (cid:19) + θ = 2592 + 65(12 a + 44 b + 44 c )3 · · . However, 2592 + 65(12 a + 44 b + 44 c )3 · ·
321 + 1040 a · · < . Therefore, the point A is the intersection point of H and ˆ E . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ E + θ H (cid:17) is not log canonical at the point A . From Lemma 1.3.4, we obtain1 < (cid:18) θ H (cid:19) · ˆ E = 6524 (cid:18) d − e (cid:19) + θ = 2592 + 65(12 a + 44 b + 3 c )3 · ·
41 + 65 d a + 44 b + 3 c )3 · ·
41 + 65 d
32 = 648 + 715 b · ·
41 + 65(4 a + c + 41 d )32 · b · · < b and 4 a + c + 41 d (6 + 48 b ). The obtained contradiction completes the proof. (cid:3) Lemma 3.3.6.
Let X be a quasismooth hypersurface of degree 196 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + y z + xz + x y = 0 . The surface X is singular at the points O x , O y and O z . The curves C x and C y are irreducible.We have 9130 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 152 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither the curve C x nor the curve C y . Then the inequalities61 D · C x = 29 < , D · C y = 661 < XCEPTIONAL DEL PEZZO HYPERSURFACES 99 show that the point P is a smooth point in the outside of C x . However, H ( P , O P (793)) contains x , y x , y x and z , it follows from Lemma 1.3.9 that the point P must be a singularpoint of X or a point on C x . This is a contradiction. (cid:3) Lemma 3.3.7.
Let X be a quasismooth hypersurface of degree 148 in P (15 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + yz + xy + x z = 0 . The surface X is singular at the points O x , O y and O z . The curves C x , C y and C z are irreducible.We can see thatlct (cid:18) X, C y (cid:19) = 5714 < lct (cid:18) X, C x (cid:19) = 256 < lct (cid:18) X, C z (cid:19) = 12914 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains none of C x , C y , C z . Note that the curve C y is singular at thepoint O z . The inequalities19 D · C x = 643 < , D · C y = 15 < , D · C z = 295 < P is located in the outside of C x ∪ C y ∪ C z .Now we consider the pencil L on X defined by the equations λz + µxy = 0, [ λ : µ ] ∈ P .Then there is a unique member C in L passing through the point P . Since the point P is locatedin the outside of C x ∪ C y ∪ C z , the curve C is cut out by the equation of the form xy + αz = 0,where α is a non-zero constant. Since the curve C is a double cover of the curve defined by theequation xy + αz = 0 in P (15 , , P ( C )
2. Therefore, we may assume thatthe support of D does not contain at least one irreducible component. If α = 1, then the curve C is irreducible, and hence the inequalitymult P ( D ) D · C = 65 · < α = 1, then the curve C consists of two distinct irreducible and reducedcurve C and C . We have D · C = D · C = 35 · , C = C = 1119 . Put D = a C + a C + ∆, where ∆ is an effective Q -divisor whose support contains neither C nor C . Since the pair ( X, D ) is log canonical at O x , both a and a are at most . Then acontradiction follows from Lemma 1.3.8 since( D − a i C i ) · C i D · C i = 35 · < i . (cid:3)
00 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
Sporadic cases with I = 4 Lemma 3.4.1.
Let X be a quasismooth hypersurface of degree 24 in P (5 , , , X ) =1. Proof.
The surface X can be defined by the quasihomogeneous equation z + yt − y + ǫx yz + x t = 0 , where ǫ ∈ C . The surface X is singular at the points O x , O t , Q = [0 : 1 : 1 : 0] and Q = [0 : 1 : 0 : 1].The curves C x , C y , C z and C t are all irreducible. We have1 = lct (cid:18) X, C y (cid:19) < lct (cid:18) X, C x (cid:19) = 54 < lct (cid:18) X, C z (cid:19) = 2and lct (cid:0) X, C t (cid:1) >
1. Therefore, lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that the supportof the divisor D contains none of the curves C x , C y , C z and C t . Also, the curve C y is singular atthe point O t with multiplicity 3 and the curve C t is singular at the point O x . Then the followingintersection numbers show that the point P is located in the outside of the set C x ∪ C y ∪ C z ∪ C t :3 D · C x = 23 < , D · C y = 45 < , D · C z = 1645 < , D · C t = 1 . Now we consider the pencil L on X defined by the equations λxt + µyz = 0, where [ λ : µ ] ∈ P . There is a unique member Z in the pencil L passing through the point P . Since P C x ∪ C y ∪ C z ∪ C t , the divisor Z is defined by an equation of the form xt = αyz, where α is non-zero constant. Note that the curve C x is not contained in the support of Z . Theopen subset Z \ C x of the curve Z is a Z -quotient of the affine curve t − αyz = z + yt + y + ǫyz + t = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z, t (cid:3)(cid:17) , that is isomorphic to the plane affine quintic curve Z ′ given by the equation z + α y z + y + ( ǫ + α ) yz = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . This affine plane curve Z ′ is irreducible and hence the curve Z is also irreducible. The multiplicityof Z at the point P is at most 3 since the quintic Z ′ is singular at the origin. This implies thatthe log pair ( X, Z ) is log canonical at the point P . Thus, we may assume that Supp( D ) doesnot contain the curve Z by Lemma 1.3.6. Then we obtain a contradictory inequality2845 = D · Z > mult P (cid:0) D (cid:1) > . (cid:3) Lemma 3.4.2.
Let X be a quasismooth hypersurface of degree 30 in P (5 , , , X ) = 1. XCEPTIONAL DEL PEZZO HYPERSURFACES 101
Proof.
The surface X can be defined by the quasihomogeneous equation t ( t − x ) − y + yz + ǫx y z = 0 . The surface X is singular at the points O x , O z , Q = [1 : 0 : 0 : 1], Q = [0 : 1 : 0 : 1] and Q = [0 : 1 : 1 : 0].The curve C x is irreducible. However, the curve C y consists of two irreducible curves L yt and L = { y = t − x = 0 } . It is easy to check1 = lct (cid:18) X, C y (cid:19) < lct (cid:18) X, C x (cid:19) = 54 . Therefore, lct( X ) X ) <
1. Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that the supportof the divisor D does not contain the curve C x . Similarly, we may assume that the support of D does not contain either L yt or L .We have the following intersection numbers for L yt and L : L yt = L = − , L yt · L = 38 . Since H ( P , O P (30)) contains the monomials y , yz and t , it follows from Lemma 1.3.9 thatthe point P is either a singular point of X or a point on C y . However, since 3 D · C x = < P must belong to the curve C y .Since the support of D does not contain either L yt or L , one of the inequalitiesmult O z ( D ) D · L yt = 45 < , mult O z ( D ) D · L = 45 < P cannot be the point O z .We put D = kL + mL yt + ∆, where ∆ is an effective Q -divisor whose support contains neither L nor L yt . If k = 0, then m = 0 and110 = D · L yt > kL · L yt = 3 k . Therefor, k . By the same way, we can also obtain m . Then, by Lemma 1.3.8, theinequalities 5( D − kL ) · L = 4 + 9 k < , D − mL yt ) · L yt = 4 + 9 m < P cannot belong to the curve C y . This is a contradiction. (cid:3) Lemma 3.4.3.
Let X be a quasismooth hypersurface of degree 45 in P (9 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + y z + xz + x = 0 . It is singular at the points O y , O z , O t , and the point Q = [1 : 0 : − C x consistsof two irreducible and reduced curves L xy and R x = { x = t + y z = 0 } . The curve C y consistsof two irreducible and reduced curves L xy and R y = { y = z + x = 0 } . The curves C z and C t
02 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV are irreducible and reduced. It is easy to check that lct( X, C y ) = is less than each of thenumbers lct( X, C x ), lct( X, C z ) and lct( X, C t ).Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6 we may assume that thesupport of D contains neither C z nor C t . Similarly, we may assume that the support of D doesnot contain either L xy or R x . Also, we may assume that the support of D does not containeither L xy or R y . Then in each of the following pairs of inequalities, at least one of two musthold: mult O z ( D ) D · L xy = 417 < , mult O z ( D ) D · R x = 811 < O t ( D ) D · L xy = 13 < , mult O t ( D ) D · R y = 49 < . Therefore, the point P can be neither O z nor O t . The curve C z is singular at the point O y .Then the inequalities 112 D · C z = 1017 < , D · C t = 511 < P cannot belong to C z ∪ C t .We can see that L xy · D = 117 · , R x · D = 233 , R y · D = 43 · , L xy · R x = 16 ,L xy · R y = 317 , L xy = − · , R x = − , R y = 23 · . If we write D = nL xy + ∆, where ∆ is an effective Q -divisor whose support does not containthe curve L xy , then we can see that n since D · R x > nR x · L xy for n = 0. By Lemma 1.3.8the inequality ( L xy · D − nL xy ) = 4 + 25 n · < P cannot belong to the curve L xy . By the same method, we see that thepoint P must be in the outside of R x .If we write D = mR y + Ω, where Ω is an effective Q -divisor whose support does not containthe curve R y , then we can see that 0 m since D · L xy > mR y · L xy for m = 0. ByLemma 1.3.8 the inequality ( R y · D − mR y ) R y · D < P cannot belong to the curve R y .Now we consider the pencil L on X cut out by λt + µy z = 0. The base locus of the pencil L consists of three points O y , O z , and Q . Let F be the member in L defined by t + y z = 0. Thedivisor F consists of two irreducible and reduced curves R x and E = { t + y z = x + z = 0 } .The curve E is smooth in the outside of the base points. We have E · D = 833 . Since E = F · E − R x · E > F · E − ( L xy + R x ) · E = 254 D · E, XCEPTIONAL DEL PEZZO HYPERSURFACES 103 the self-intersection E is positive. We write D = kE + Γ, where Γ is an effective Q -divisorwhose support does not contain the curve E . Since ( X, D ) is log canonical at the point O y ,the non-negative number k is at most . By Lemma 1.3.8, the inequality( E · D − kE ) E · D = 833 < P cannot belong to the curve E .So far we have seen that the point P must lie in the outside of C x ∪ C y ∪ C z ∪ C t ∪ E . Inparticular, it is a smooth point. There is a unique member C in L which passes through thepoint P . Then the curve C is cut out by t = αy z where α is a constant different from 0 and −
1. The curve C is isomorphic to the curve defined by y z + xz + x = 0 and t = y z . Thecurve C is smooth in the outside of the base points and the singular locus of X by the Bertinitheorem, since it is isomorphic to a general curve in the pencil L . We claim that the curve C isirreducible. If so then we may assume that the support of D does not contain the curve C andhence we obtain mult P ( D ) C · D = 1033 < . This is a contradiction.For the irreducibility of the curve C , we may consider the curve C as a surface in C definedby the equations y z + xz + x = 0 and t = y z . Then, we consider the surface in P definedby the equations y zw + xz w + x = 0 and t w = y z . We take the affine piece defined by t = 0.This affine piece is isomorphic to the surface defined by the equation y zw + xz w + x = 0 and w = y z in C . It is isomorphic to the irreducible hypersurface y z + xy z + x = 0 in C .Therefore, the curve C is irreducible. (cid:3) Lemma 3.4.4.
Let X be a quasismooth hypersurface of degree 75 in P (10 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + z + xy + x z = 0 . It has singular points at O x , O y , O t and Q = [ − C x and C t areirreducible and reduced. The curve C y (resp. C z ) consists of two irreducible reduced curves L yz and R y = { y = z + x = 0 } (resp. R z = { y = t + xy = 0 } ). It is easy to see thatlct( X, C y ) = 9160 < lct( X, C x ) < lct( X, C z ) < lct( X, C t ) . Also, we have the following intersection numbers: − K X · L yz = 25 · , − K X · R y = 45 · , − K X · R z = 45 · ,L yz · R y = 531 , L yz · R z = 15 , L yz = − · , R y = − · , R z = 125 · . Suppose that lct( X ) < . Then, there is an effective Q -divisor D ∼ Q − K X such that thelog pair ( X, D ) is not log canonical at some point P ∈ X . Since the curves C x and C t are
04 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV irreducible we may assume that the support of D contains none of them. The inequalities13 D · C x < , D · C t < P must lie in the outside of C x ∪ C t \ { O x , O t } .By Lemma 1.3.6, we may assume that the support of D does not contain either L yz or R y . Ifthe support of D does not contain L yz , then the inequality31 D · L yz = 25 < P cannot be O t . On the other hand, if the support of D does not contain R y ,then the inequality 312 D · R y = 25 < P cannot be O t . Note that the curve R y is singular at the point O t . Weuse the same method for C z = R z + L yz so that we can see that the point P cannot be O x .We write D = mR y + Ω, where Ω is an effective Q -divisor whose support does not containthe curve R y . Then we see m since the support of D does not contain either L yz or R y and D · L yz > mR y · L yz . Since R y · D − mR y < , Lemma 1.3.8 implies that the point P islocated in the outside of R y . Using the same argument for L yz , we can also see that the point P is located in the outside of L yz . Also, the same method shows that the point P is located inthe outside of R z . Consequently, the point P must lie in the outside of C x ∪ C y ∪ C z ∪ C t .Now we consider the pencil L on X cut out by λt + µxy = 0. The base locus of the pencil L consists of three points O x , O y , and Q . Let F be the member of L defined by t + xy = 0. Thedivisor F consists of two irreducible and reduced curves R z and E = { t + xy = z + x = 0 } .The curve E is smooth in the outside of Sing( X ). We have E · D = 85 · . Since E = F · E − R z · E > F · E − ( L yz + R z ) · E = 374 D · E, the self-intersection E is positive. We write D = kE + Γ, where Γ is an effective Q -divisorwhose support does not contain the curve E . Since ( X, D ) is log canonical at the point O y ,the non-negative number k is at most . By Lemma 1.3.8, the inequality( E · D − kE ) E · D = 85 · < P cannot belong to the curve E .So far we have seen that the point P must lie in the outside of C x ∪ C y ∪ C z ∪ C t ∪ E . Inparticular, it is a smooth point. There is a unique member C in L which passes through thepoint P . Then the curve C is cut out by t = αxy where α is a constant different from 0and −
1. The curve C is isomorphic to the curve defined by xy + z + x z = 0 and t = xy .The curve C is smooth in the outside of the base points and the singular locus of X by Bertinitheorem, since it is isomorphic to a general curve in the pencil L . We claim that the curve C is XCEPTIONAL DEL PEZZO HYPERSURFACES 105 irreducible. If so then we may assume that the support of D does not contain the curve C andhence we obtain mult P ( D ) C · D = 125 · < . This is a contradiction.For the irreducibility of the curve C , we may consider the curve C as a surface in C definedby the equations xy + z + x z = 0 and t = xy . Then, we consider the surface in P definedby the equations xy + w z + x z = 0 and t w = xy . We then take the affine piece defined by y = 0. This affine piece is isomorphic to the surface defined by the equation x + w z + x z = 0and t w = x in C . It is isomorphic to the hypersurface defined by t w + w z + t w z = 0in C . It has two irreducible components w = 0 and t + z + t w z = 0. The former componentoriginates from the hyperplane at infinity in P . Therefore, the curve C must be irreducible. (cid:3) Lemma 3.4.5.
Let X be a quasismooth hypersurface of degree 71 in P (11 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + y z + xz + x t = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consistsof L xy (resp. L xy , L zt , L zt ) and R x = { x = y z + t = 0 } (resp. R y = { y = x t + z = 0 } , R z = { z = x + yt = 0 } , R t = { t = y + xz = 0 } ). Also, we see that L xy ∩ R x = { O z } , L xy ∩ R y = { O t } , L zt ∩ R z = { O y } , L zt ∩ R t = { O x } . One can easily check that lct( X, C x ) = is less than each of the numbers lct( X, C y ),lct( X, C z ) and lct( X, C t ). Therefore, lct( X ) . Suppose lct( X ) < . Then, there isan effective Q -divisor D ∼ Q − K X such that the log pair ( X, D ) is not log canonical at somepoint P ∈ X .The intersection numbers among the divisors D , L xy , L zt , R x , R y , R z , R t are as follows: D · L xy = 15 · , D · R x = 25 · , D · R y = 49 · ,D · L zt = 411 · , D · R z = 1617 · , D · R t = 35 · ,L xy · R x = 110 , L xy · R y = 19 , L zt · R z = 417 , L zt · R t = 311 ,L xy = − · , R x = − · , R y = 23 · ,L zt = − · , R z = − · , R t = 2120 · . By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . Since the curve R t is singular at the point O x and the curve R y
06 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV is singular at the point O t with multiplicity 3, in each of the following pairs of inequalities, atleast one of two must hold:mult O x ( D ) D · L zt = 417 < , mult O x ( D ) D · R t = 310 <
611 ;mult O z ( D ) D · L xy = 427 < , mult O z ( D ) D · R x = 817 <
611 ;mult O t ( D ) D · L xy = 15 < , mult O t ( D ) D · R y = 411 < . Therefore, the point P can be none of O x , O z , O t .Suppose that the point P is the point O y . We then put D = mL zt + ∆, where ∆ is an effective Q -divisor whose support does not contain the curve L zt . If m = 0, thenmult O y ( D ) D · L zt = 411 < . This is a contradiction. Therefore, m >
0, and hence the support of D does not contain thecurve R z . Since 1617 ·
27 = D · R z > m
17 + mult O y ( D ) − m > m
17 + 611 · m < · · . However, we obtain17( D − mL zt ) · L zt = 4 + 24 m > P is a smooth point of X .We write D = a L xy + a L zt + a R x + a R y + a R z + a R t + Ω, where Ω is an effective Q -divisor whose support contains none of the curves L xy , L zt , R x , R y , R z , R t . Since the pair( X, D ) is log canonical at the points O x , O z , O t , the numbers a i are at most . Then byLemma 1.3.8 the following inequalities enable us to conclude that the point P is in the outsideof C x ∪ C y ∪ C z ∪ C t :( D − a L xy ) · L xy = 4 + 43 a · , ( D − a L zt ) · L zt = 4 + 24 a · , ( D − a R x ) · R x = 2 + 3 a · , ( D − a R y ) · R y = 4 − a · , ( D − a R z ) · R z = 16 + 28 a · , ( D − a R t ) · R t = 12 − a · . We consider the pencil L defined by λty + µx = 0, [ λ : µ ] ∈ P . The base locus of the pencil L consists of the curve L xy and the point O y . Let E be the unique divisor in L that passesthrough the point P . Since P C x ∪ C y ∪ C z ∪ C t , the divisor E is defined by the equation ty = αx , where α = 0.Suppose that α = −
1. Then the curve E is isomorphic to the curve defined by the equations ty = x and x t + y z + xz = 0. Since the curve E is isomorphic to a general curve in L ,it is smooth at the point P . The affine piece of E defined by t = 0 is the curve given by XCEPTIONAL DEL PEZZO HYPERSURFACES 107 x ( x + x z + z ) = 0. Therefore, the divisor E consists of two irreducible and reduced curves L xy and C . We have D · C = D · E − D · L xy = 2675 · · ,C = E · C − L xy · C > E · C − L xy · C − R x · C = 334 D · C > . By Lemma 1.3.8 the inequality D · C < gives us a contradiction.Suppose that α = −
1. Then divisor E consists of three irreducible and reduced curves L xy , R z , and M . Note that the curve M is different from the curves R x and L zt . Also, it is smoothat the point P . We have D · M = D · E − D · L xy − D · R z = 115 · ,M = E · M − L xy · M − R z · M > E · M − C x · M − C z · M = 134 D · M > . By Lemma 1.3.8 the inequality D · M < gives us a contradiction. (cid:3)
Lemma 3.4.6.
Let X be a quasismooth hypersurface of degree 79 in P (11 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation t y + tz + xy + x z = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consistsof L xt (resp. L yz , L yz , L xt ) and R x = { x = yt + z = 0 } (resp. R y = { y = zt + x = 0 } , R z = { z = xy + t = 0 } , R t = { t = y + x z = 0 } ). Also, we see that L xt ∩ R x = { O y } , L yz ∩ R y = { O t } , L yz ∩ R z = { O x } , L xt ∩ R t = { O z } . One can easily check that lct( X, C x ) = is less than each of the numbers lct( X, C y ),lct( X, C z ) and lct( X, C t ). Therefore, lct( X ) . Suppose lct( X ) < . Then, there isan effective Q -divisor D ∼ Q − K X such that the log pair ( X, D ) is not log canonical at somepoint P ∈ X .The intersection numbers among the divisors D , L xt , L yz , R x , R y , R z , R t are as follows: D · L xt = 16 · , D · R x = 817 · , D · R y = 56 · ,D · L yz = 411 · , D · R z = 811 · , D · R t = 23 · ,L xt · R x = 217 , L yz · R y = 531 , L yz · R z = 211 , L xt · R t = 16 ,L xt = − · , R x = − · , R y = − · ,L yz = − · , R z = 1411 · , R t = 103 · .
08 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . The inequalities17 D · L xt = 16 < , D · R x = 831 < P = O y . The inequalities11 D · L yz = 431 < , D · R z = 817 < P = O x . Since the curve R t is singular at the point O z with multiplicity 4 theinequalities 24 D · L xt = 246 · < , D · R t = 411 < P = O z .We write D = a L xt + a L yz + a R x + a R y + a R z + a R t + Ω, where Ω is an effective Q -divisor whose support contains none of the curves L xt , L yz , R x , R y , R z , R t . Since the pair( X, D ) is log canonical at the points O x , O y , O z , the numbers a i are at most . Then byLemma 1.3.8 the following inequalities enable us to conclude that either the point P is in theoutside of C x ∪ C y ∪ C z ∪ C t or P = O t :3316 D · L xt − L xt = 1813 · · < , D · R x − R x = 1132 · · < , D · R y − R y = 253 · < , D · L yz − L xt = 1854 · · < , D · R z − R z = 52 · · < , D · R t − R t = − · · < . Suppose that P = O t . Then we consider the pencil L defined by λyt + µz = 0, [ λ : µ ] ∈ P .The base locus of the pencil L consists of the curve L yz and the point O y . Let E be the uniquedivisor in L that passes through the point P . Since P C x ∪ C y ∪ C z ∪ C t , the divisor E isdefined by the equation z = αyt , where α = 0.Suppose that α = −
1. Then the curve E is isomorphic to the curve defined by the equations yt = z and t y + xy + x z = 0. Since the curve E is isomorphic to a general curve in L ,it is smooth at the point P . The affine piece of E defined by t = 0 is the curve given by z ( z + xz + x ) = 0. Therefore, the divisor E consists of two irreducible and reduced curves L yz and C . We have the intersection numbers D · C = D · E − D · L yz = 56411 · · , C · L yz = E · L yz − L yz = 211 . Also, we see C = E · C − C · L yz > . By Lemma 1.3.8 the inequality D · C < gives us a contradiction.Suppose that α = −
1. Then divisor E consists of three irreducible and reduced curves L yz , R x , and M . Note that the curve M is different from the curves R y and L xt . Also, it is smoothat the point P . We have D · M = D · E − D · L yz − D · R x = 4 · · · ,M = E · M − L yz · M − R x · M > E · M − C y · M − C x · M = 5 D · M > . XCEPTIONAL DEL PEZZO HYPERSURFACES 109
By Lemma 1.3.8 the inequality D · M < gives us a contradiction. Therefore, P = O t .We write D = aL yz + bR x + ∆, where ∆ is an effective divisor whose support contains neither L yz nor R x . Note that we already assumed that the support of D cannot contain either L yz or R y . If the support of D contains R y , then it does not contain L yz . However, the inequality31 D · L yz = < shows that P = O t . Therefore, the support of D does not contain the curve R y . The inequality D · L xt > bR x · L xt implies b . On the other hand, we have56 ·
31 = D · R y > a
31 + b
31 + mult O t ( D ) − a − b > a
31 + 1631 · , and hence a < · .Let π : ¯ X → X be the weighted blow up of O t with weights (7 ,
4) and let F be the exceptionalcurve of π . Then K ¯ X ∼ Q π ∗ ( K X ) − F, ¯ L yz ∼ Q π ∗ ( L yz ) − F, ¯ R x ∼ Q π ∗ ( R x ) − F, ¯∆ ∼ Q π ∗ (∆) − c F, where ¯∆, ¯ L yz , ¯ R x are the proper transforms of ∆, L yz , R x , respectively, and c is a non-negativerational number. The curve F contains two singular points Q and Q of ¯ X . The point Q is asingular point of type (1 ,
1) and the point Q is of type (1 , R x passesthrough the point Q but not the point Q . The curve ¯ L yz passes through the point Q but notthe point Q .The log pull-back of the log pair ( X, D ) by π is the log pair (cid:18) ¯ X, a
16 ¯ L yz + 33 b
16 ¯ R x + 3316 ¯∆ + θ F (cid:19) , where θ = 33(4 a + 7 b + c ) + 32016 · . This pair is not log canonical at some point Q ∈ F . We have0 ¯∆ · ¯ R x = 8 + 40 b · − a − c · . This inequality shows 4 a + c (8 + 40 b ). Then θ = 33(4 a + c ) + 231 b + 32016 · b · · < b .Suppose that the point Q is neither the point Q nor the point Q . Then the log pair (cid:0) ¯ X, ¯∆ + F (cid:1) is not log canonical at the point Q . Then33 c ·
28 = 3316 ¯∆ · F > c a + c (8 + 40 b ). This is a contradiction since b .Therefore, the point Q is either the point Q or the point Q .
10 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
Suppose that the point Q is the point Q . This point is the intersection point of F and ¯ R x .Then the log pair (cid:0) ¯ X, b ¯ R x + ¯∆ + θ F (cid:1) is not log canonical at the point Q . It then followsfrom Lemma 1.3.4 that1 < (cid:18) θ F (cid:19) · ¯ R x = 33 · (cid:18) b · − a − c · (cid:19) + θ . However, 33 · (cid:18) b · − a − c · (cid:19) + θ = 6496 + 9207 b · · < . Therefore, the point Q is the point Q . This point is the intersection point of F and ¯ L yz .Let φ : ˜ X → ¯ X be the blow up at the point Q . Let G be the exceptional divisor of themorphism φ . The surface ˜ X is smooth along the exceptional divisor G . Let ˜ L yz , ˜ R x , ˜∆ and ˜ F be the proper transforms of L yz , R x , ∆ and F by φ , respectively. We have K ˜ X ∼ Q φ ∗ ( K ¯ X ) − G, ˜ L yz ∼ Q φ ∗ ( ¯ L yz ) − G, ˜ F ∼ Q φ ∗ ( F ) − G, ˜∆ ∼ Q φ ∗ ( ¯∆) − d G, where d is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ is (cid:18) ˜ X, a
16 ˜ L yz + 33 b
16 ˜ R x + 3316 ˜∆ + θ ˜ F + θ G (cid:19) , where θ = 337 ·
16 ( a + d ) + θ a + 7 b + c + 31 d )7 · · . This log pair is not log canonical at some point O ∈ G . We have0 ˜∆ · ˜ L yz = 4 + 38 a · − b − c · − d . We then obtain 7 b + c + 31 d (4 + 38 a ). Since a , we see θ = 2800 + 33(35 a + 7 b + c + 31 d )7 · · a · · < . Suppose that O ˜ F ∪ ˜ L yz . The log pair (cid:16) ˜ X, ˜∆ + G (cid:17) is not log canonical at the point O .Applying Lemma 1.3.4, we get 1 < · G = 33 d , and hence d > . However, d (7 b + c + 31 d ) · (4 + 38 a ). This is a contradiction since a . Therefore, the point O is either the intersection point of G and ˜ F or the intersectionpoint of G and ˜ L yz . In the latter case, the pair (cid:16) ˜ X, a ˜ L yz + ˜∆ + θ G (cid:17) is not log canonicalat the point O . Then, applying Lemma 1.3.4, we get1 < (cid:18) θ G (cid:19) · ˜ L yz = 3316 (cid:18) a · − b − c · − d (cid:19) + θ . However, 3316 (cid:18) a · − b − c · − d (cid:19) + θ = 4532 + 3069 a · · < . XCEPTIONAL DEL PEZZO HYPERSURFACES 111
Therefore, the point O must be the intersection point of G and ˜ F .Let ξ : ˆ X → ˜ X be the blow up at the point O and let H be the exceptional divisor of ξ .We also let ˆ L yz , ˆ R x , ˆ∆, ˆ G , and ˆ F be the proper transforms of ˜ L yz , ˜ R x , ˜∆, G and ˜ F by ξ ,respectively. Then ˆ X is smooth along the exceptional divisor H . We have K ˆ X ∼ Q ξ ∗ ( K ˜ X ) − H, ˆ G ∼ Q ξ ∗ ( G ) − H, ˆ F ∼ Q ξ ∗ ( ˜ F ) − H, ˆ∆ ∼ Q ξ ∗ ( ˜∆) − eH, where e is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ ◦ ξ is (cid:18) ˆ X, a
16 ˆ L yz + 33 b
16 ˆ R x + 3316 ˆ∆ + θ ˆ F + θ ˆ G + θ H (cid:19) , where θ = θ + θ + 33 e − a + 56 b + 8 c + 31 d + 217 e )7 · · . This log pair is not log canonical at some point A ∈ H . We have c − d − e = ˆ∆ · ˆ F > . Therefore, 4 d + 28 e c .Then θ = 1568 + 33(63 a + 56 b + 8 c )7 · ·
31 + 33 · d + 7 e )7 · · a + 224 b + 63 c )4 · · ·
31 == 6272 + 7392 b · · ·
31 + 33 · a + c )4 · · ·
28 + 33 b ·
31 + 9 · b )2 · ·
31 = 773 + 2046 b · · < b and 4 a + c (8 + 40 b ). In particular, θ is a positive number.Suppose that A ˆ F ∪ ˆ G . Then the log pair (cid:16) ˆ X, ˆ∆ + θ H (cid:17) is not log canonical at the point A . Applying Lemma 1.3.4, we get 1 < · H = 33 e . However, e
128 (4 d + 28 e ) c
128 (4 a + c ) b )17 · . Therefore, the point A must be either in ˆ F or in ˆ G .Suppose that A ∈ ˆ F . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ F + θ H (cid:17) is not log canonical at thepoint A . Applying Lemma 1.3.4, we get1 < (cid:18) θ H (cid:19) · ˆ F = 3316 (cid:18) c − d − e (cid:19) + θ = 6272 + 33(252 a + 224 b + 63 c )4 · · · .
12 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
However, 6272 + 33(252 a + 224 b + 63 c )4 · · ·
773 + 2046 b · · < . Therefore, the point A is the intersection point of H and ˆ G . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ G + θ H (cid:17) is not log canonical at the point A . From Lemma 1.3.4, we obtain1 < (cid:18) θ H (cid:19) · ˆ G = 3316 ( d − e ) + θ = 1568 + 33(63 a + 56 b + 8 c + 248 d )7 · · . However,1568 + 33(63 a + 56 b + 8 c + 248 d )7 · ·
31 = 224 + 297 a ·
31 + 33(7 b + c + 31 d )2 · ·
320 + 1209 a · < a < · and 7 b + c + 31 d (4 + 38 a ). The obtained contradiction completes theproof. (cid:3) Lemma 3.4.7.
Let X be a quasismooth hypersurface of degree 166 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + yz + xy + x z = 0 . The surface X is singular only at the points O x , O y and O z . The curves C x and C y areirreducible. We have 5524 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 13 · . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y . Then the inequalities45 D · C x = 831 < , D · C y = 845 < P is a smooth point in the outside of C x . However, H ( P , O P (495)) containsthe monomials x , y x and z , it follows from Lemma 1.3.9 that the point P is either asingular point of X or a point on C x . This is a contradiction. (cid:3) Lemma 3.4.8.
Let X be a quasismooth hypersurface of degree 71 in P (13 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation ty + yz + xt + x z = 0 . The surface X is singular at the points O x , O y , O z , O t . Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components. The divisor C x (resp. C y , C z , C t ) consists XCEPTIONAL DEL PEZZO HYPERSURFACES 113 of L xy (resp. L xy , L zt , L zt ) and R x = { x = z + ty = 0 } (resp. R y = { y = x z + t = 0 } , R z = { z = y + xt = 0 } , R t = { t = x + yz = 0 } ). Also, we see that L xy ∩ R x = { O t } , L xy ∩ R y = { O z } , L zt ∩ R z = { O x } , L zt ∩ R t = { O y } . One can easily check that lct( X, C x ) = is less than each of the numbers lct( X, C y ),lct( X, C z ) and lct( X, C t ). Therefore, lct( X ) . Suppose lct( X ) < . Then, there isan effective Q -divisor D ∼ Q − K X such that the log pair ( X, D ) is not log canonical at somepoint P ∈ X .The intersection numbers among the divisors D , L xy , L zt , R x , R y , R z , R t are as follows: D · L xy = 419 · , D · R x = 67 · , D · R y = 813 · ,D · L zt = 27 · , D · R z = 1213 · , D · R t = 87 · ,L xy · R x = 329 , L xy · R y = 219 , L zt · R z = 313 , L zt · R t = 27 ,L xy = − · , R x = − · , R y = 213 · ,L zt = − · , R z = − · , R t = 207 · . By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . Since the curve R t is singular at the point O y and the curve R y is singular at the point O z , in each of the following pairs of inequalities, at least one of two musthold: mult O x ( D ) D · L zt = 27 < , mult O x ( D ) D · R z = 1229 < O y ( D ) D · L zt = 413 < , mult O y ( D ) D · R t = 819 < O z ( D ) D · L xy = 429 < , mult O z ( D ) D · R y = 413 < O t ( D ) D · L xy = 419 < , mult O t ( D ) D · R x = 37 < . Therefore, the point P can be none of O x , O y , O z , O t .We write D = a L xy + a L zt + a R x + a R y + a R z + a R t + Ω, where Ω is an effective Q -divisor whose support contains none of the curves L xy , L zt , R x , R y , R z , R t . Since the pair( X, D ) is log canonical at the points O x , O y , O z , O t , the numbers a i are at most . Then byLemma 1.3.8 the following inequalities enable us to conclude that the point P must be locatedin the outside of C x ∪ C y ∪ C z ∪ C t :( D − a L xy ) · L xy = 4 + 44 a · , ( D − a L zt ) · L zt = 4 + 23 a · , ( D − a R x ) · R x = 12 + 3 a · , ( D − a R y ) · R y = 8 − a · , ( D − a R z ) · R z = 12 + 30 a · , ( D − a R t ) · R t = 8 − a · .
14 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
We consider the pencil L defined by λtx + µy = 0, [ λ : µ ] ∈ P . The base locus of the pencilconsists of the curve L xy and the point O x . Let E be the unique divisor in L that passes throughthe point P . Since P C x ∪ C y ∪ C z ∪ C t , the divisor E is defined by the equation tx = αy ,where α = 0.Suppose that α = −
1. Then the curve E is isomorphic to the curve defined by the equations tx = y and xt + yz + x z = 0. Since the curve E is isomorphic to a general curve in L ,it is smooth at the point P . The affine piece of E defined by t = 0 is the curve given by y ( y + y z + z ) = 0. Therefore, the divisor E consists of two irreducible and reduced curves L xy and C . We have D · C = D · E − D · L xy = 80013 · · . Also, we see C = E · C − C · L xy > E · C − C x · C > . By Lemma 1.3.8 the inequality D · C < gives us a contradiction.Suppose that α = −
1. Then divisor E consists of three irreducible and reduced curves L xy , R z , and M . Note that the curve M is different from the curves R x and L zt . Also, it is smoothat the point P . We have D · M = D · E − D · L xy − D · R z = 57213 · · ,M = E · M − L xy · M − R z · M > E · M − C x · M − C z · M = 52 D · M > . By Lemma 1.3.8 the inequality D · M < gives us a contradiction. (cid:3)
Lemma 3.4.9.
Let X be a quasismooth hypersurface of degree 79 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z t + y z + xt + x y = 0 . The surface X is singular at O x , O y , O z and O t . We havelct (cid:18) X, C x (cid:19) = 6532 < lct (cid:18) X, C x (cid:19) = 218 < lct (cid:18) X, C t (cid:19) = 3310 < lct (cid:18) X, C z (cid:19) = 6920 . In particular, lct( X ) .Each of the divisors C x , C y , C z , and C t consists of two irreducible and reduced components.The divisor C x (resp. C y , C z , C t ) consists of L xz (resp. L yt , L xz , L yt ) and R x = { x = y + zt = 0 } (resp. R y = { y = z + xt = 0 } , R z = { z = x y + t = 0 } , R t = { t = x + y z = 0 } ). The curve L xz intersects R x (resp. R z ) only at the point O t (resp. O y ). The curve L yt intersects R y (resp. R t ) only at the point O x (resp. O z ).We suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thelog pair ( X, D ) is not log canonical at some point P ∈ X .The intersection numbers among the divisors D , L xz , L yt , R x , R y , R z , R t are as follows: L xz = − · , R x = − · , L xz · R x = 433 , D · L xz = 414 · , D · R x = 1623 · , XCEPTIONAL DEL PEZZO HYPERSURFACES 115 L yt = − · , R y = − · , L yt · R y = 213 , D · L yt = 413 · , D · R y = 813 · ,R z = 2013 · , L xz · R z = 214 , D · R z = 813 · ,R t = 9514 · , L yt · R t = 523 , D · R t = 2014 · . By Lemma 1.3.6 we may assume that the support of D does not contain at least one componentof each divisor C x , C y , C z , C t . Since the curve R t is singular at the point O z with multiplicity3 and the curve R z is singular at the point O y , in each of the following pairs of inequalities, atleast one of two must hold:mult O x ( D ) D · L yt = 423 < , mult O x ( D ) D · R y = 833 < O y ( D ) D · L xz = 433 < , mult O y ( D ) D · R z = 413 < O z ( D ) D · L yt = 413 < , mult O z ( D ) D · R t = 1021 < . Therefore, the point P can be none of O x , O y , O z .Put D = m L xz + m L yt + m R x + m R y + m R z + m R t + Ω, where Ω is an effective Q -divisor whose support contains none of L xz , L yt , R x , R y , R z , R t . Since the pair ( X, D ) is logcanonical at the points O x , O y , O z , we have m i for each i . Since( D − m L xz ) · L xz = 4 + 43 m · , ( D − m L yt ) · L yt = 4 + 32 m · , ( D − m R x ) · R x = 16 + 40 m · , ( D − m R y ) · R y = 8 + 38 m · , ( D − m R z ) · R z = 8 − m · , ( D − m R t ) · R t = 20 − m · P cannot be a smooth point of X on C x ∪ C y ∪ C z ∪ C t .Therefore, the point P is either a point in the outside of C x ∪ C y ∪ C z ∪ C t or the point O t .Suppose that P C x ∪ C y ∪ C z ∪ C t . Then we consider the pencil L on X defined by theequations λxt + µz = 0, [ λ : µ ] ∈ P . There is a unique curve Z α in the pencil passing throughthe point P . This curve is cut out by xt + αz = 0 , where α is a non-zero constant.The curve Z α is reduced. But it is always reducible. Indeed, one can easily check that Z α = C α + L xz where C α is a reduced curve whose support contains no L xy . Let us prove that C α is irreducibleif α = 1.Any component of the curve C t is not contained in the curve Z α . The open subset Z α \ C t ofthe curve Z α is a Z -quotient of the affine curve x + αz = z + y z + x + x y = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, y, z (cid:3)(cid:17) ,
16 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV that is isomorphic to a plane affine curve defined by the equation z (cid:0) ( α − z + y − α yz (cid:1) = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . Thus, if α = 1, then the curve Z α consists of two irreducible and reduced curves L xz and C α . If α = 1, then the curve Z α consists of three irreducible and reduced curves L xz , R y , and C . Inboth cases, the curve C α (including α = 1) is smooth at the point P . By Lemma 1.3.6, we mayassume that Supp( D ) does not contain at least one irreducible component of the curve Z α .If α = 1, then D · C α = 813 · ,C α = Z α · C α − L xz · C α > Z α · C α − ( R x + L xz ) · C α = 334 D · C α > . If α = 1, then D · C = 15213 · · ,C = Z · C − ( L xz + R y ) · C > Z · C − ( R x + L xz ) · C − ( L yt + R y ) · C = 194 D · C > . We put D = mC α + ∆ α , where ∆ α is an effective Q -divisor such that C α Supp(∆ α ). Since C α intersects the curve C t and the pair ( X, D ) is log canonical along the curve C t , we obtain m . Then, the inequality ( D − mC α ) · C α D · C α < X, D ) is log canonical at the point P by Lemma 1.3.8. The obtainedcontradiction conclude that the point P must be the point O t .If L xz is not contained in the support of D , then the inequalitymult O t ( D ) D · L xz = 27 < L xz must be contained in the support of D , and hencethe curve R x is not contained in the support of D . Put D = aL xz + bR y + ∆, where ∆ is aneffective Q -divisor whose support contains neither L xz nor R y . Then1623 ·
33 = D · R x > aL xz · R x + mult O t ( D ) − a > a
33 + 3233 · a < · · . If b = 0, then L yt is not contained in the support of D . Therefore,413 ·
23 = D · L yt > bR y · L yt = 2 b , and hence b .Let π : ¯ X → X be the weighted blow up at the point O t with weights (13 ,
19) and let F bethe exceptional curve of the morphism π . Then F contains two singular points Q and Q of¯ X such that Q is a singular point of type (1 , Q is a singular point of type (3 , K ¯ X ∼ Q π ∗ ( K X ) − F, ¯ L xz ∼ Q π ∗ ( L xz ) − F, ¯ R y ∼ Q π ∗ ( R y ) − F, ¯∆ ∼ Q π ∗ (∆) − c F, XCEPTIONAL DEL PEZZO HYPERSURFACES 117 where ¯ L xz , ¯ R y and ¯∆ are the proper transforms of L xz , R y and ∆ by π , respectively, and c is anon-negative rational number. Note that F ∩ ¯ R y = { Q } and F ∩ ¯ L xz = { Q } .The log pull-back of the log pair ( X, D ) by π is the log pair (cid:18) ¯ X, a
32 ¯ L xz + 65 b
32 ¯ R y + 6532 ¯∆ + θ F (cid:19) , where θ = 32 + 65(19 a + 13 b + c )32 · . This is not log canonical at some point Q ∈ F . We have0 ¯∆ · ¯ L xz = 4 + 43 a · − b − c · . This inequality shows 13 b + c (4 + 43 a ). Since a · · , we obtain θ = 32 + 1235 a ·
33 + 65(13 b + c )32 ·
32 + 1235 a ·
33 + 13 · a )14 · · < . Suppose that the point Q is neither Q nor Q . Then, the point Q is not in ¯ L xz ∪ ¯ R y .Therefore, the pair (cid:0) ¯ X, ¯∆ + F (cid:1) is not log canonical at the point Q , and hence1 < · F = 65 c · · . But c b + c (4 + 43 a ) < · · since a · · . Therefore, the point Q is either Q or Q .Suppose that the point Q is Q . Then the point Q is in ¯ L xz but not in ¯ R y . Therefore, thepair (cid:0) ¯ X, ¯ L xz + ¯∆ + θ F (cid:1) is not log canonical at the point Q . However, this is impossible since13 (cid:18) θ F (cid:19) · ¯ L xz = 13 · (cid:18) a · − b − c · (cid:19) + θ == 32 + 1235 a ·
33 + 13 · a )14 · · < . Therefore, the point Q must be the point Q .Let ψ : ˜ X → ¯ X be the weighted blow up at the point Q with weights (3 ,
7) and let E be theexceptional curve of the morphism ψ . The exceptional curve E contains two singular points O and O of ˜ X . The point O is of type (1 ,
2) and the point O is of type (4 , K ˜ X ∼ Q ψ ∗ ( K ¯ X ) − E, ˜ R y ∼ Q ψ ∗ ( ¯ R y ) − E, ˜ F ∼ Q ψ ∗ ( F ) − E, ˜∆ ∼ Q ψ ∗ ( ¯∆) − d E, where ˜ R y , ˜ F and ˜∆ are the proper transforms of ¯ R y , F and ¯∆ by ψ , respectively, and d is anon-negative rational number.The log pull-back of the log pair ( X, D ) by π ◦ ψ is the log pair (cid:18) ˜ X, a
32 ˜ L xz + 65 b
32 ˜ R y + 6532 ˜∆ + θ ˜ F + θ E (cid:19) ,
18 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV where ˜ L xz is the proper transform of ¯ L xz by ψ and θ = 65(3 b + d )19 ·
32 + 719 θ + 919 = 9728 + 65(133 a + 190 b + 7 c + 33 d )19 · · . This is not log canonical at some point O ∈ E .We have 0 ˜∆ · ˜ R y = ¯∆ · ¯ R y − d ·
19 = 8 + 38 b · − a + c · − d · , and hence 133 a + 7 c + 33 d (8 + 38 b ). Therefore, this inequality together with b < givesus θ = 9728 + 65 · b · ·
33 + 65(133 a + 7 c + 33 d )19 · · · b · ·
33 + 65 · b )13 · · < . Suppose that the point O is in the outside of ˜ R y and ˜ F . Then the log pair ( E, ˜∆ | E ) is notlog canonical at the point O and hence1 < · E = 65 d · · . However, d
133 (133 a + 7 c + 33 d ) ·
33 (8 + 38 b ) < · · b . This is a contradiction.Suppose that the point O belongs to ˜ R y . Then the log pair (cid:16) ˜ X, b ˜ R y + ˜∆ + θ E (cid:17) is notlog canonical at the point O and hence1 < (cid:18) θ E (cid:19) · ˜ R x = 7 · (cid:18) b · − a + c · − d · (cid:19) + θ . However,7 · (cid:18) b · − a + c · − d · (cid:19) + θ = 9728 + 65 · b · ·
33 + 65 · b )13 · · < . This is a contradiction. Therefore, the point O is the point O .Suppose that the point O belongs to ˜ F . Then the log pair (cid:16) ˜ X, ˜∆ + θ ˜ F + θ E (cid:17) is not logcanonical at the point O and hence1 < (cid:18) θ E (cid:19) · ˜ F = 3 · (cid:18) c · − d · (cid:19) + θ . XCEPTIONAL DEL PEZZO HYPERSURFACES 119
However, 3 · (cid:18) c · − d · (cid:19) + θ = 3 · c · ·
32 + 9728 + 65(133 a + 190 b + 7 c )19 · ·
33 == 512 + 455 a ·
33 + 65 · b + c )13 · · ·
512 + 455 a ·
33 + 65 · a )14 · · · < b + c (4 + 43 a ) and a · · . This is a contradiction. (cid:3) Lemma 3.4.10.
Let X be a quasismooth hypersurface of degree 166 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + y z + xz + x y = 0 . The surface X is singular only at the points O x , O y and O z . The curves C x and C y areirreducible. We have 9140 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 11524 , and hence lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D contains neither C x nor C y . Then the inequalities51 D · C x = 823 < , D · C y = 851 < P is a smooth point of X in the outside of C x . However, H ( P , O P (663))contains x , y x , y x and z , and hence it follows from Lemma 1.3.9 that the point P iseither a singular point of X or a point on C x . This is a contradiction. (cid:3) Sporadic cases with I = 5 Lemma 3.5.1.
Let X be a quasismooth hypersurface of degree 63 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z t + yt + xy + x z = 0 , and X is singular at O x , O y , O z and O t .The curve C x (resp. C y , C z , C t ) consists of two irreducible and reduced curves L xt (resp L yz , L yz , L xt ) and R x = { x = z + yt = 0 } (resp. R y = { y = x + zt = 0 } , R z = { z = t + xy = 0 } , R t = { t = y + x z = 0 } ). The curve L xt intersects R x (resp. R t ) only at the point O y (resp. O z ). The curve L yz intersects R y (resp. R z ) only at the point O t (resp. O x ).
20 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
We have the following intersection numbers D · L xt = 513 · , D · L yz = 15 · , D · R x = 25 · , D · R y = 45 · , D · R z = 1011 · ,D · R t = 2011 · , L xt · R x = 213 , L xt · R t = 419 , L yz · R y = 425 , L yz · R z = 211 ,L xt = − · , L yz = − · , R x = − · , R y = − · , R z = 1211 · , R t = 5611 · . We havelct (cid:18) X, C y (cid:19) = 138 < lct (cid:18) X, C x (cid:19) = 3320 < lct (cid:18) X, C t (cid:19) = 3516 < lct (cid:18) X, C z (cid:19) = 198 . In particular, we have lct( X ) .We suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thelog pair ( X, D ) is not log canonical at some point P ∈ X .Suppose that the point P is located in the outside of C x ∪ C y ∪ C z ∪ C t . We consider thepencil L on X defined by the equations λx + µzt = 0, where [ λ : µ ] ∈ P . The curve L xt is theunique base component of the pencil L . There is a unique member Z in the pencil L passingthrough the point P . Since the point P is in the outside of C x ∪ C y ∪ C z ∪ C t , the curve Z isdefined by an equation of the form αx + zt = 0 , where α is a non-zero constant.The open subset Z \ C z of the curve Z is a Z -quotient of the affine curve αx + t = t + yt + xy + x = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) x, y, z (cid:3)(cid:17) , that is isomorphic to the affine curve given by the equation x (cid:0) (1 − α ) x + α x y + y (cid:1) = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . If α = 1, the divisor Z consists of two irreducible and reduced curves L xt and Z α . On theother hand, if α = 1, then the divisor Z consists of three irreducible and reduced curves L xt , R y and Z . Since P C x ∪ C y ∪ C z ∪ C t , the point P must be contained in Z α (including α = 1).Also, the curve Z α is smooth at the point P .Write D = nZ α + Γ, where Γ is an effective Q -divisor whose support contains Z α . Since Z α passing through the point O t and the pair ( X, D ) is log canonical at the point O z , we have n . We can easily check D · Z α = D · ( Z − L xt ) = 2275 · ·
19 if α = 1 ,D · ( Z − L xt − R y ) = 3513 ·
19 if α = 1 . Also, if α = 1, then Z α = Z · Z α − L xt · Z α > Z · Z α − ( L xt + R x ) · Z α = 335 D · Z α . XCEPTIONAL DEL PEZZO HYPERSURFACES 121 If α = 1, Z α = Z · Z α − ( L xt + R y ) · Z α > Z · Z α − ( L xt + R x + L yz + R y ) · Z α = 4 D · Z α . In both cases, we have Z α >
0. Since( D − nZ α ) · Z α D · Z α < X, D ) is log canonical at the point P . This is a contradiction.Therefore, the point P must belong to the set C x ∪ C y ∪ C z ∪ C t .It follows from Lemma 1.3.6 that we may assume that Supp( D ) does not contain at leastone irreducible component of the curves C x , C y , C z , C t . Since the curve R t is singular at thepoint O z with multiplicity 3 and the support of D does not contain either L xt or R t , one of theinequalitiesmult O z ( D ) D · L xt = 513 < , mult O z ( D ) D · R t = 203 · < P cannot be the point O z . Similarly, we see that the point P can be neither O x nor O y .Now we write D = m L xt + m L yz + m R x + m R y + m R z + m R t + Ω, where Ω is an effective Q -divisor whose support contains none of L xt , L yz , R x , R y , R z , R t . Since the pair ( X, D ) islog canonical at the points O x , O y , O z , we must have m i . Then the inequalities( D − m L xt ) · L xt = 5 + 27 m · D − m L yz ) · L yz = 5 + 31 m · D − m R x ) · R x = 10 + 28 m · D − m R y ) · R y = 20 + 24 m · D − m R z ) · R z = 10 − m · D − m R t ) · R t = 20 − m · P must be the point O t .Put D = aL yz + bR x + ∆, where ∆ is an effective Q -divisor whose support contains neitherthe curve L yz nor R x . If a = 0, then we obtainmult O t ( D ) D · L yz = 511 < . This is a contradiction. Therefore, a >
0, and hence the support of D dose not contain the curve R y . Since 45 ·
19 = D · R y > aL yz · R y + mult O t ( D ) − a > a
25 + 813 · ,
22 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV and hence a . If b >
0, then513 ·
19 = D · L xt > bR x · L xt = 2 b , and hence b .Let π : ¯ X → X be the weighted blow up of O t with weights (7 ,
3) and let F be the exceptionalcurve of π . Then K ¯ X ∼ Q π ∗ ( K X ) − F, ¯ L yz ∼ Q π ∗ ( L yz ) − F, ¯ R x ∼ Q π ∗ ( R x ) − F, ¯∆ ∼ Q π ∗ (∆) − c F, where ¯∆, ¯ L yz , ¯ R x are the proper transforms of ∆, L yz , R x , respectively, and c is a non-negativerational number. The curve F contains two singular points Q and Q of ¯ X . The point Q is asingular point of type (1 ,
1) and the point Q is of type (2 , R x passesthrough the point Q but not the point Q . The curve ¯ L yz passes through the point Q but notthe point Q .The log pull-back of the log pair ( X, D ) by π is the log pair (cid:18) ¯ X, a L yz + 13 b R x + 138 ¯∆ + θ F (cid:19) , where θ = 13(3 a + 7 b + c ) + 1208 · . This pair is not log canonical at some point Q ∈ F . We have0 ¯∆ · ¯ R x = 10 + 28 b · − a − c · . This inequality shows 3 a + c (10 + 28 b ). Then θ = 13(3 a + c ) + 91 b + 1208 · b < b .Suppose that the point Q is neither the point Q nor the point Q . Then the log pair (cid:0) ¯ X, ¯∆ + F (cid:1) is not log canonical at the point Q . Then13 c ·
21 = 138 ¯∆ · F > c a + c (10 + 28 b ). This is a contradiction since b .Therefore, the point Q is either the point Q or the point Q .Suppose that the point Q is the point Q . This point is the intersection point of F and ¯ R x .Then the log pair (cid:0) ¯ X, b ¯ R x + ¯∆ + θ F (cid:1) is not log canonical at the point Q . It then followsfrom Lemma 1.3.4 that1 < (cid:18)
138 ¯∆ + θ F (cid:19) · ¯ R x = 13 · (cid:18)
10 + 28 b · − a − c · (cid:19) + θ . However, 13 · (cid:18)
10 + 28 b · − a − c · (cid:19) + θ = 6 + 7 b < . XCEPTIONAL DEL PEZZO HYPERSURFACES 123
Therefore, the point Q is the point Q . This point is the intersection point of F and ¯ L yz .Let φ : ˜ X → ¯ X be the blow up at the point Q . Let G be the exceptional divisor of themorphism φ . The surface ˜ X is smooth along the exceptional divisor G . Let ˜ L yz , ˜ R x , ˜∆ and ˜ F be the proper transforms of L yz , R x , ∆ and F by π ◦ φ , respectively. We have K ˜ X ∼ Q φ ∗ ( K ¯ X ) − G, ˜ L yz ∼ Q φ ∗ ( ¯ L yz ) − G, ˜ F ∼ Q φ ∗ ( F ) − G, ˜∆ ∼ Q φ ∗ ( ¯∆) − d G, where d is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ is (cid:18) ˜ X, a L yz + 13 b R x + 138 ˜∆ + θ ˜ F + θ G (cid:19) , where θ = 137 · a + d ) + θ a + 7 b + c + 25 d )7 · · . This log pair is not log canonical at some point O ∈ G . We have0 ˜∆ · ˜ L yz = 5 + 31 a · − b − c · − d . We then obtain 7 b + c + 25 d (5 + 31 a ). Since a , we see θ = 1120 + 13(28 a + 7 b + c + 25 d )7 · ·
511 + 273 a · · < . Suppose that O ˜ F ∪ ˜ L yz . The log pair (cid:16) ˜ X, ˜∆ + G (cid:17) is not log canonical at the point O .Applying Lemma 1.3.4, we get 1 <
138 ˜∆ · G = 13 d , and hence d > . However, d (7 b + c + 25 d ) · (5 + 31 a ). This is a contradiction since a . Therefore, the point O is either the intersection point of G and ˜ F or the intersectionpoint of G and ˜ L yz . In the latter case, the pair (cid:16) ˜ X, a ˜ L yz + ˜∆ + θ G (cid:17) is not log canonicalat the point O . Then, applying Lemma 1.3.4, we get1 < (cid:18)
138 ˜∆ + θ G (cid:19) · ˜ L yz = 138 (cid:18) a · − b − c · − d (cid:19) + θ . However, 138 (cid:18) a · − b − c · − d (cid:19) + θ = 511 + 273 a · · < . Therefore, the point O must be the intersection point of G and ˜ F .Let ξ : ˆ X → ˜ X be the blow up at the point O and let H be the exceptional divisor of ξ .We also let ˆ L yz , ˆ R x , ˆ∆, ˆ G , and ˆ F be the proper transforms of ˜ L yz , ˜ R x , ˜∆, G and ˜ F by ξ ,respectively. Then ˆ X is smooth along the exceptional divisor H . We have K ˆ X ∼ Q ξ ∗ ( K ˜ X ) − H, ˆ G ∼ Q ξ ∗ ( G ) − H, ˆ F ∼ Q ξ ∗ ( ˜ F ) − H, ˆ∆ ∼ Q ξ ∗ ( ˜∆) − eH,
24 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV where e is a non-negative rational number. The log pull-back of the log pair ( X, D ) via π ◦ φ ◦ ξ is (cid:18) ˆ X, a L yz + 13 b R x + 138 ˆ∆ + θ ˆ F + θ ˆ G + θ H (cid:19) , where θ = θ + θ + 13 e − a + 56 b + 8 c + 25 d + 175 e )7 · · . This log pair is not log canonical at some point A ∈ H . We have c − d − e ˆ∆ · ˆ F > . Therefore, 3 d + 21 e c .Then θ = 560 + 13(49 a + 56 b + 8 c )7 · ·
25 + 13( d + 7 e )7 · a + 168 b + 49 c )3 · · ·
25 == 1680 + 1284 b · · ·
25 + 13 · a + c )3 · · ·
140 + 107 b · ·
25 + 7(5 + 14 b )4 ·
25 = 21 + 36 b < b and 3 a + c (10 + 28 b ). In particular, θ is a positive number.Suppose that A ˆ F ∪ ˆ G . Then the log pair (cid:16) ˆ X, ˆ∆ + θ H (cid:17) is not log canonical at the point A . Applying Lemma 1.3.4, we get 1 <
138 ˆ∆ · H = 13 e . However, e
121 (3 d + 21 e ) c
121 (3 a + c ) b )13 · . Therefore, the point A must be either in ˆ F or in ˆ G .Suppose that A ∈ ˆ F . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ F + θ H (cid:17) is not log canonical at thepoint A . Applying Lemma 1.3.4, we get1 < (cid:18)
138 ˆ∆ + θ H (cid:19) · ˆ F = 138 (cid:18) c − d − e (cid:19) + θ = 1680 + 13(147 a + 168 b + 49 c )3 · · · . However, 1680 + 13(147 a + 168 b + 49 c )3 · · ·
21 + 36 b < . Therefore, the point A is the intersection point of H and ˆ G . Then the log pair (cid:16) ˆ X, ˆ∆ + θ ˆ G + θ H (cid:17) is not log canonical at the point A . From Lemma 1.3.4, we obtain1 < (cid:18)
138 ˆ∆ + θ H (cid:19) · ˆ G = 138 ( d − e ) + θ = 560 + 13(49 a + 56 b + 8 c + 200 d )7 · · . XCEPTIONAL DEL PEZZO HYPERSURFACES 125
However,560 + 13(49 a + 56 b + 8 c + 200 d )7 · ·
25 = 80 + 91 a ·
25 + 13(7 b + c + 25 d )7 ·
56 + 169 a · < a < and 7 b + c +25 d (5+31 a ). The obtained contradiction completes the proof. (cid:3) Lemma 3.5.2.
Let X be a quasismooth hypersurface of degree 136 in P (11 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation xy + x z + yz + t = 0 . The surface X is singular at the points O x , O y and O z .The curves C x and C y are reduced and irreducible. We have116 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 5518 . Thus, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6 we may assume that thesupport of D contains neither C x nor C y . Then two inequalities37 D · C x = 25 < , D · C y = 1037 < P is neither a singular point of X nor a point on C x . Since H ( P , O P (407))contains x , z and x y , we see that this cannot happen by Lemma 1.3.9. (cid:3) Lemma 3.5.3.
Let X be a quasismooth hypersurface of degree 136 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation x y + xz + y z + t = 0 . The surface X is singular only at the points O x , O y and O z .The curves C x and C y are reduced and irreducible. Also, it is easy to check9150 = lct (cid:18) X, C x (cid:19) < lct( X, C y ) = 196 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6 we may assume that thesupport of D contains neither C x nor C y . Then two inequalities41 D · C x = 1019 < , D · C y = 1041 < P is neither a singular point of X nor a point on C x . However, byLemma 1.3.9 this is impossible since H ( P , O P (533)) contains x , z and x y . (cid:3)
26 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
Sporadic cases with I = 6 Lemma 3.6.1.
Let X be a quasismooth hypersurface of degree 45 in P (7 , , , X ) = . Proof.
The surface X can be defined by the equation z − y z + xt + x y = 0. It is singular atthe points O x , O y , O t and Q = [0 : 1 : 1 : 0].The curve C x consists of two irreducible and reduced curves L xz and R x = { x = z − y = 0 } .These two curves L xz and R x meets each other at the point O t . Also, L xz = − · , R x = − · , L xz · R x = 319 . The curve R x is singular at the point O t . The curve C y is irreducible and3554 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 2518 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D does not contain the curve C y . Similarly, we may assume that either L xz Supp( D ) or R x Supp( D ).Since H ( P , O P (105)) contains the monomials x , y x and z , it follows from Lemma 1.3.9that the point P is either a point on C x or the singular point O x .Since either L xz Supp( D ) or R x Supp( D ), one of the inequalitiesmult O t ( D ) D · L xz = 35 < , mult O t ( D ) D · R x = 35 < P cannot be the point O t . On the other hand, the inequality7 D · C y = < shows that the point P cannot be the point O x .Put D = mL xz + Ω, where Ω is an effective Q -divisor such that L xz Supp(Ω). If m = 0,then 65 ·
19 = D · R x > mL xz · R x = 3 m , and hence m . Then, 10( D − mL xz ) · L xz = 6 + 23 m . Thus it follows from Lemma 1.3.8 that the point P cannot belong to L xz .Now we write D = ǫR x + ∆, where ∆ is an effective Q -divisor such that R x Supp(∆). If ǫ = 0, then 35 ·
19 = D · L xz > ǫR x · L xz = 3 ǫ , and hence ǫ . Then 5( D − ǫR x ) · R x = 3 + 8 ǫ . By Lemma 1.3.8 the point P cannot be contained in R x either. Therefore, the point P is locatednowhere. (cid:3) XCEPTIONAL DEL PEZZO HYPERSURFACES 127
Lemma 3.6.2.
Let X be a quasismooth hypersurface of degree 106 in P (11 , , , X ) = . Proof.
We may assume that the surface X is defined by the quasihomogeneous equation x z + xy + yz + t = 0 . The surface X is singular at O x , O y and O z . The curves C x and C y are irreducible. It is easyto see lct( X, C x ) = 5536 < lct( X, C y ) = 5728 . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical. For a smooth point P ∈ X \ C x , we havemult P ( D ) · · · · · < H ( P , O P (319)) contains the monomials x , z and x y . Therefore,either there is a point P ∈ C x such that mult P ( D ) > or we have mult O x ( D ) > . Since thepairs ( X, · · C x ) and ( X, · · C y ) are log canonical and the curves C x and C y are irreducible,we may assume that the support of D contains neither the curve C x nor the curve C y . Then wecan obtain mult O x ( D ) C y · D · · · · · · < P ∈ C x mult P ( D ) C x · D · · · · · · < . This is a contradiction. Therefore, lct( X ) = . (cid:3) Lemma 3.6.3.
Let X be a quasismooth hypersurface of degree 106 in P (13 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation x y + xz + y z + t = 0 . The surface X is singular at the points O x , O y and O z .The curves C x , C y and C z are reduced and irreducible. We havelct (cid:18) X, C x (cid:19) = 9160 < lct (cid:18) X, C y (cid:19) = 2512 < lct (cid:18) X, C z (cid:19) = 9328 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P . By Lemma 1.3.6 we may assume that the supportof D contains none of C x , C y , C z . Since C y is singular at the point O z and D · C y = < ,the point P must be in the outside of C y . Furthermore, the point P is in the outside of C x ∪ C z since 15 D · C x = < and D · C z = < .Now we consider the pencil L on X defined by the equations λz + µx y = 0, [ λ : µ ] ∈ P .Then there is a unique member C in L passing through the point P . Since the point P is located
28 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV in the outside of C x ∪ C y ∪ C z , the curve C is cut out by the equation of the form x y + αz = 0,where α is a non-zero constant. Since the curve C is a double cover of the curve defined by theequation x y + αz = 0 in P (13 , , P ( C )
2. Therefore, we may assume thatthe support of D does not contain at least one irreducible component. If α = 1, then the curve C is irreducible, and hence the inequalitymult P ( D ) D · C = 1265 < α = 1, then the curve C consists of two distinct irreducible and reducedcurve C and C . We have D · C = D · C = 665 , C = C = 813 . Put D = a C + a C + ∆, where ∆ is an effective Q -divisor whose support contains neither C nor C . Since the pair ( X, D ) is log canonical at O x , both a and a are at most . Then acontradiction follows from Lemma 1.3.8 since( D − a i C i ) · C i D · C i = 1265 < i . (cid:3) Sporadic cases with I = 7 Lemma 3.7.1.
Let X be a quasismooth hypersurface of degree 76 in P (11 , , , X ) = . Proof.
We may assume that the surface X is defined by the equation t + yz + xy + x z = 0.The surface X is singular at O x , O y and O z . The curves C x , C y and C z are irreducible. Wehave 2110 = lct( X, C z ) > X, C x ) > lct( X, C y ) = 1310 . Therefore, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of D contains none of the curves C x , C y and C z .Since the curve C y is singular at the point O z , the inequality 11 D · C y = < shows thatthe point P does not belong to the curve C y . Also, the inequality 13 D · C x = < impliesthat the point P cannot belong to C x either. The inequality D · C z = · < shows that thepoint P cannot belong to C z Consider the pencil L on X defined by the equations λy + µx z = 0, [ λ : µ ] ∈ P . There isa unique member Z in L passing through the point P . Since P C x ∪ C y ∪ C z , the curve Z isdefined by an equation of the form x z = αy , where α is a non-zero constant. The open subset Z \ C x of the curve Z is a Z -quotient of the affine curve z − αy = t + yz + y + z = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z, t (cid:3)(cid:17) XCEPTIONAL DEL PEZZO HYPERSURFACES 129 that is isomorphic to the plane affine curve C ⊂ C defined by the equation t + α y + (1 + α ) y = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . The curve C is irreducible if α = − α = 1. Since the C x is not contained in thesupport of Z , the curve Z is irreducible if α = − α = 1. From the equation of C , we can see that the log pair ( X, Z ) is log canonical at the point P . By Lemma 1.3.6, wemay assume that Supp( D ) does not contain at least one irreducible component of the curve Z .Suppose that α = −
1. Then Z Supp( D ) and1033 = D · Z > mult P (cid:0) D (cid:1) > . This is a contradiction. Thus, α = −
1. Then it follows from the equation of C that the curve Z consists of two irreducible and reduced curves Z and Z . Without loss of generality we mayassume that the point P belongs to the curve Z .Put D = mZ + Ω, where Ω is an effective Q -divisor such that Z Supp(Ω). Since the pair( X, D ) is log canonical at the point O x , one has m . Then (cid:0) D − mZ (cid:1) · Z < D · Z = 533 < . since Z >
0. By Lemma 1.3.8, the log pair ( X, D ) is log canonical at the point P . This is acontradiction. (cid:3) Sporadic cases with I = 8 Lemma 3.8.1.
Let X be a quasismooth hypersurface of degree 46 in P (7 , , , X ) = . Proof.
The surface X can be defined by the equation t + y z + xz + x y = 0. The surface X is singular at the points O x , O y and O z . The curves C x , C y and C z are irreducible. We have3548 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C z (cid:19) = 9180 < lct (cid:18) X, C y (cid:19) = 5548 . In particular, lct( X ) . Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the pair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6,we may assume that the support of the divisor D contains none of the curves C x , C y and C z .Since the curve C x is singular at the point O z , the inequality11 D · C x = 1613 < P cannot belong to C x . Also, the inequality7 D · C y = 1613 < P is not in C y . Since D · C z = 167 · < , the point P cannot be in C z either.
30 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV
Consider the pencil L on X defined by the equations λx y + µz = 0, [ λ : µ ] ∈ P . There isa unique member Z in L passing through the point P . Since P C x ∪ C y ∪ C z , the curve Z isdefined by an equation of the form x y = αz , where α is a non-zero constant. The open subset Z \ C x of the curve Z is a Z -quotient of the affine curve y − αz = t + y z + z + y = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z, t (cid:3)(cid:17) that is isomorphic to the plane affine curve C ⊂ C defined by the equation t + α z + (1 + α ) z = 0 ⊂ C ∼ = Spec (cid:16) C (cid:2) y, z (cid:3)(cid:17) . The curve C is irreducible if α = − α = −
1. Since the C x is not containedin the support of Z , the curve Z is irreducible if α = − α = −
1. Fromthe equation of C , we can see that the log pair ( X, Z ) is log canonical at the point P . ByLemma 1.3.6, we may assume that Supp( D ) does not contain at least one irreducible componentof the curve Z .Suppose that α = −
1. Then Z Supp( D ) and4877 = D · Z > mult P (cid:0) D (cid:1) > . This is a contradiction. Thus, α = −
1. Then it follows from the equation of C that the curve Z consists of two irreducible and reduced curves Z and Z . Without loss of generality we mayassume that the point P belongs to the curve Z .Put D = mZ + Ω, where Ω is an effective Q -divisor such that Z Supp(Ω). Since the pair( X, D ) is log canonical at the point O x , one has m . Then (cid:0) D − mZ (cid:1) · Z < D · Z = 2477 < . since Z >
0. By Lemma 1.3.8, the log pair ( X, D ) is log canonical at the point P . This is acontradiction. (cid:3) Lemma 3.8.2.
Let X be a quasismooth hypersurface of degree 81 in P (7 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation z − y z + xt + x y = 0 . The surface X is singular at the points O x , O y , O t and Q = [0 : 1 : 1 : 0].The curve C x consists of two irreducible and reduced curves L xz and R x = { x = z − y = 0 } .These two curves intersect each other only at the point O t . Also, L xz = − · , R x = − · , L xz · R x = 337 . The curve C y is irreducible and3572 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 158 . XCEPTIONAL DEL PEZZO HYPERSURFACES 131
Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D does not contain the curve C y . Similarly, we may assume that either L xz Supp( D ) or R x Supp( D ).Since either L xz Supp( D ) or R x Supp( D ), one of the inequalitiesmult O t ( D ) D · L xz = 49 < , mult O t ( D ) D · R x = 89 < P cannot be O t . Since mult O x ( D ) D · C y = < , thepoint P cannot be the point O x .Put D = mL xz + Ω, where Ω is an effective Q -divisor such that L xz Supp(Ω). If m = 0,then 1618 ·
37 = D · R x > mL xz · R x = 3 m , and hence m . Since 18 (cid:0) D − mL xz (cid:1) · L xz = 8 + 47 m P cannot belong to L xz .Now we write D = ǫZ x + ∆, where ∆ is an effective Q -divisor such that Z x Supp(∆). If ǫ = 0, then 818 ·
37 = D · L xz > ǫR x · L xz = 3 ǫ , and hence ǫ . Since 9 (cid:0) D − ǫR x (cid:1) · R x = 8 + 20 ǫ P cannot belong to R x . Consequently, the point P must be a smooth point in the outside of C x . However, since H ( P , O P (189)) contains themonomials x , y x and z , it follows from Lemma 1.3.9 that P must be either a singular pointof X or a point on C x . This is a contradiction. (cid:3) Sporadic cases with I = 9 Lemma 3.9.1.
Let X be a quasismooth hypersurface of degree 64 in P (7 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + y z + xz + x y = 0 . The surface X is singular only at the points O x , O y and O z . The curves C x and C y areirreducible, and 3554 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 2518 . In particular, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that the
32 IVAN CHELTSOV, JIHUN PARK, CONSTANTIN SHRAMOV support of the divisor D contains neither the curve C x nor the curve C y . Then two inequalities19 D · C x = < , 7 D · C y = < show that the point P must be a smooth point in theoutside of C x .Note that H ( P , O P (133)) contains the monomials x , y x and z and hence it follows fromLemma 1.3.9 that the point P is either a singular point of X or a point on C x . This is acontradiction. (cid:3) Sporadic cases with I = 10 Lemma 3.10.1.
Let X be a quasismooth hypersurface of degree 82 in P (7 , , , X ) = . Proof.
The surface X can be defined by the quasihomogeneous equation t + y z + xz + x y = 0 . It is singular at the points O x , O y and O z .The curves C x and C y are irreducible. We have712 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 1912 , and hence lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that the pair( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that the supportof the divisor D contains neither the curve C x nor the curve C y . Since 25 D · C x = < and7 D · C y = < , the point P must be a smooth point in the outside of the curve C x . Notethat H ( P , O P (175)) contains the monomials x , x y and z , and hence the point P cannotbe a smooth point in the outside of C x by Lemma 1.3.9. Consequently, lct( X ) = . (cid:3) Lemma 3.10.2.
Let X be a quasismooth hypersurface of degree 117 in P (7 , , , X ) = . Proof.
The surface X can be defined by the equation z − y z + xt + x y = 0. It is singularat the points O x , O y , O t and Q = [0 : 1 : 1 : 0].The curve C x consists of two irreducible curves L xz and R x = { x = z − y = 0 } . These twocurves intersect each other only at the point O t . It is easy to check L xz = − · , R x = − · , L xz · R x = 355 . On the other hand, the curve C y is irreducible. We have718 = lct (cid:18) X, C x (cid:19) < lct (cid:18) X, C y (cid:19) = 136 . In particular, lct( X ) .Suppose that lct( X ) < . Then there is an effective Q -divisor D ∼ Q − K X such that thepair ( X, D ) is not log canonical at some point P . By Lemma 1.3.6, we may assume that thesupport of the divisor D does not contain the curve C y . Similarly, we may assume that either L xz Supp( D ) or R x Supp( D ). XCEPTIONAL DEL PEZZO HYPERSURFACES 133
Since 7 D · C y = < , the point P cannot be the point O x . Meanwhile, since the supportof D does not contain at least one components of C x , one of the inequalitiesmult O t ( D ) D · L xz = 513 < , mult O t ( D ) D · R x = 1013 < P cannot be the point O t .Put D = mL xz + Ω, where Ω is an effective Q -divisor such that L xz Supp(Ω). If m = 0,then 1013 ·
55 = D · R x > mL xz · R x = 3 m , and hence m . Then 26 (cid:0) D − mL xz (cid:1) · L xz = 10 + 71 m < , and hence Lemma 1.3.8 implies that the point P cannot belong to L xz .Now we write D = ǫR x + ∆, where ∆ is an effective Q -divisor such that R x Supp(∆). If ǫ = 0, then 1026 ·
55 = D · L xz > ǫR x · L xz = 3 ǫ , and hence ǫ . Then 13( D − ǫR x (cid:1) · R x = 10 + 32 ǫ < . Thus, Lemma 1.3.4 shows that the point P is not on R x .Therefore, the point P must be a smooth point in the outside of the curve C x . Since H ( P , O P (273)) contains the monomials x , y x and z , it follows from Lemma 1.3.9 that P is either a point on C x or a singular point of X . This is a contradiction. (cid:3) Part The Big Table
The tables contains the following information on del Pezzo surfaces. • The first column: the weights ( a , a , a , a ) of the weighted projective space P . • The second column: the degree of the surface X ⊂ P . • The third column: the self-intersection number K X of an anticanonical divisor of X . • The fourth column: the rank ρ of the Picard group of the surface X . • The fifth column: the global log canonical theeshold lct( X ) of X . • The sixth column: the possible monomials in x , y , z , t in the defining equation f ( x, y, z, t ) = 0 of the surface X . • The seventh column: the information on the singular points of X . We use the standardnotation for cyclic quotient singularities along with the following convention: when wewrite, for instance, O x O y = n × r ( a, b ), we mean that there are n cyclic quotient sin-gularities of type r ( a, b ) cut out on X by the equations z = t = 0 that are differentfrom the point O x in the case when O x ∈ X and O x is not of type r ( a, b ), and that aredifferent from the point O y in the case when O y ∈ X and O y is not of type r ( a, b ). I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 1Weights Degree K X ρ lct Monomials Singular Points(2 , n + 1 , n + 1 , n + 1) 8 n + 4 n +1)(4 n +1) y , y z , y z , yz , z , xt , x n +1 yt , x n +1 zt , x n +1 y , x n +1 yz , x n +1 z , x n +2 O t = n +1 (1 , O y O z == 4 × n +1 (1 , n )(1 , , ,
5) 10 a710 b t , yzt , y z , y , xz , xy t , xy z , x zt , x yz , x y , x yt , x y z , x z , x y , x t , x yz , x y , x z , x y , x O z = (1 , , , ,
7) 15 c815 d z , yzt , y , xt , xy z , x yz , x y t , x zt , x y , x y z , x z , x yt , x y , x yz , x t , x y , x z , x y , x O t = (3 , , , ,
8) 16
10 1 t , yzt , y z , xz , xy , x y t , x y z , x zt , x yz , x y , x yt , x y z , x z , x y , x t , x yz , x y , x z , x y , x O y = (1 , O z = (1 , C x has an ordinary double point, b: if C x has a non-ordinary double point,c: if the defining equation of X contains yzt , d: if the defining equation of X does not contain yzt . X C EP T I O NA L D E L PE ZZ O HY PE R S U R F A C E S Log del Pezzo surfaces with I = 1Weights Degree K X ρ lct Monomials Singular Points(2 , , ,
9) 18 a116 b t , yz , y t , y , xy z , x zt , x y z , x yt , x y , x z , x yz , x y , x O z = (1 , O y O t = 2 × (1 , , , ,
5) 15 t , zt , z t , z , y , xy , x y , x y , x y , x O x O y = 5 × (1 , O z O t = 3 × (1 , , , ,
11) 25 z t , y , xt , xy z , x yz , x yt , x y , x z O x = (1 , O z = (3 , O t = (5 , , , ,
14) 28 t , z t , z , xy , x y z , x yt , x yz , x y , x z O x = (1 , O y = (1 , O z O t = 2 × (3 , , , ,
18) 36 t , y z , xz , xy t , x y , x yz , x y z , x t , x y , x O y = (1 , O z = (5 , O x O t = 2 × (1 , , , ,
21) 56 yt , y , xz , x yz , x t O x = (2 , O z = (7 , O t = (5 , O y O t = 1 × (5 , , , ,
31) 81 z , yt , xy , x yz , x t O x = (2 , O y = (2 , O t = (5 , , , ,
50) 100 t , yz , xy , x y z , x t , x O y = (2 , O z = (5 , O x O t = 2 × (2 , , , ,
37) 81 z , y t , xt , x y z , x y O x = (3 , O y = (7 , O t = (11 , C y has a tacnodal point, b: if C y has no tacnodal points. I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 1Weights Degree K X ρ lct Monomials Singular Points(7 , , ,
44) 88 t , y t , y , xz , x y z , x y O x = (3 , O z = (11 , O y O t = 2 × (7 , , , ,
20) 60 t , y , xz , x y O x = (4 , O z = (5 , O x O y = 1 × (1 , O y O t = 1 × (2 , , , ,
23) 69 t , zt , z t , z , xy , x y O x = (1 , O y = (1 , O x O y = 1 × (1 , O z O t = 3 × (3 , , , ,
49) 127 z t , yt , xy , x z O x = (7 , O y = (1 , O z = (11 , O t = (11 , , , , t , yz , xy , x z O x = (5 , O y = (2 , O z = (11 , , , ,
57) 127 z t , y z , xt , x y O x = (9 , O y = (13 , O z = (13 , O t = (23 , , , , t , y z , xz , x y O x = (3 , O y = (13 , O z = (35 , X C EP T I O NA L D E L PE ZZ O HY PE R S U R F A C E S Log del Pezzo surfaces with I = 2Weights Degree K X ρ lct Monomials Singular Points(3 , n, n + 1 , n + 1) 9 n + 3 n (3 n +1) t , zt , z t , z , xy , x n +1 y , x n +1 y , x n +1 O y = n (1 , O x O y = 3 × (1 , O z O t = 3 × n +1 (1 , n )(3 , n + 1 , n + 2 , n + 2) 9 n + 6 n +1)(3 n +2) t , zt , z t , z , xy , x n +1 yt , x n +1 yz , x n +2 O y = n +1 (1 , O z O t == 3 × n +2 (3 , n + 1)(3 , n + 1 , n + 2 , n + 1) 12 n + 5 n +5)3(3 n +1)(3 n +2)(6 n +1) x n +1 z , y z , z t , t x , x n yz , x n +1 yt , x n +1 y O x = (1 , O y = n +1 (3 , n ) O z = n +2 (3 , n + 1) O t = n +1 (3 n + 1 , n + 2)(3 , n + 1 , n + 1 , n ) 18 n + 3 n (3 n +1) z , y t , xt x n yz , x n y z , x n y , x n +1 t , x n +1 O y = n +1 (1 , n ) O t = n (3 n + 1 , n + 1) O x O t = 2 × (1 , , n + 1 , n + 1 , n + 3) 18 n + 6 n +1)(6 n +1) t , y t , y , xz , x n +1 yz , x n +1 y z , x n +1 t , x n +1 y , x n +2 O z = n +1 (3 n + 1 , n + 2) O x O t = 2 × (1 , O y O t = 2 × n +1 (1 , n )(4 , n + 1 , n + 2 , n + 1) 12 n + 6 n +1)(6 n +1) z , y z , y z , y , xt , x n +1 yt , x n +1 z , x n +1 y O x = (1 , O t = n +1 (2 n + 1 , n + 2) O x O z = 1 × (1 , O y O z = 3 × n +1 (1 , n )(4 , n + 3 , n + 3 , n + 4) 8 n + 12 n +1)(2 n +3) y , y z , y z , yz , z , xt , x n +2 t , x n +3 O t = n +4 (2 , O x O t = 2 × (1 , O y O z == 4 × n +3 (4 , n + 1) I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 2Weights Degree K X ρ lct Monomials Singular Points(2 , , ,
5) 12 a712 b z , yzt , y , xt , xy z , x z , x yt , x y , x z , x O t = (3 , O x O z = 3 × (1 , , , ,
7) 14 t , yzt , y z , xz , xy , x yt , x y z , x z , x y , x z , x O y = (1 , O z = (1 , O x O z = 3 × (1 , , , ,
10) 20 t , z t , z , y , xy z , x yt , x yz , x y , x z O x = (1 , O y O t = 1 × (1 , O z O t = 2 × (3 , , , ,
15) 30 t , z , y z , xy t , x yz , x y , x y z , x t , x y , x O y = (1 , O y O z = 1 × (1 , O z O t = 1 × (3 , O x O t = 2 × (1 , , , ,
22) 57 t y , z , xy , x yz , x t O x = (3 , O y = (2 , O t = (5 , , , ,
35) 70 t , yz , xy , x y z , x t , x O y = (2 , O z = (5 , O x O t = 2 × (3 , , , ,
13) 36 t z , y , xz , x y , x O z = (3 , O t = (2 , O x O y = 2 × (1 , O x O z = 1 × (1 , X contains yzt , b: if the defining equation of X contains no yzt . X C EP T I O NA L D E L PE ZZ O HY PE R S U R F A C E S Log del Pezzo surfaces with I = 2Weights Degree K X ρ lct Monomials Singular Points(7 , , ,
25) 57 ty , z , xt , x y z , x y O x = (5 , O y = (7 , O t = (8 , , , ,
32) 64 t , ty , y , xz , x y z , x y O x = (5 , O z = (8 , O y O t = 2 × (7 , , , ,
16) 48 t , y , xz , x y O x = (4 , O z = (4 , O x O y = 1 × (1 , O y O t = 1 × (1 , , , ,
19) 57 t , t z , tz , z , xy , x y O x = (1 , O y = (1 , O x O y = 1 × (1 , O z O t = 3 × (3 , , , ,
31) 81 t y , y z , xz , x O y = (3 , O z = (19 , O t = (3 , O x O z = 1 × (1 , , , ,
43) 105 t y , z , xy , x z O x = (3 , O y = (16 , O t = (2 , O x O z = 1 × (4 , , , ,
47) 105 y , yz , xt , x z O x = (10 , O y = (11 , O t = (3 , O y O z = 1 × (4 , I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 2Weights Degree K X ρ lct Monomials Singular Points(11 , , ,
41) 107 t y , y z , xz , x t O x = (3 , O y = (11 , O z = (25 , O t = (11 , , , ,
43) 111 t y , z t , xy , x z O x = (3 , O y = (1 , O z = (11 , O t = (11 , , , , t , yz , xy , x z O x = (10 , O y = (2 , O z = (11 , , , ,
61) 135 z , y z , xt , x y O x = (2 , O y = (13 , O t = (2 , O y O z = 1 × (4 , , , ,
47) 107 y t , yz , xt , x z O x = (7 , O y = (13 , O z = (13 , O t = (20 , , , ,
49) 111 z t , y z , xt , x y O x = (1 , O y = (13 , O z = (13 , O t = (20 , , , , t , y z , xz , x y O x = (6 , O y = (13 , O z = (31 , , , ,
41) 99 t y , z t , xy , x z O x = (3 , O y = (12 , O z = (14 , O t = (14 , X C EP T I O NA L D E L PE ZZ O HY PE R S U R F A C E S Log del Pezzo surfaces with I = 3Weights Degree K X ρ lct Monomials Singular Points(5 , , ,
13) 33 t y , z , xy , x yz , x t O x = (2 , O y = (2 , O t = (5 , , , ,
20) 40 t , yz , xy , x y z , x t , x O y = (2 , O z = (1 , O x O t = 2 × (2 , , , ,
37) 95 t y , z t , xy , x z O x = (5 , O y = (1 , O z = (11 , O t = (11 , , , ,
98) 196 t , yz , xy , x z O x = (2 , O y = (2 , O z = (11 , , , ,
41) 95 z t , y z , xt , x y O x = (1 , O y = (13 , O z = (13 , O t = (17 , , , ,
98) 196 t , y z , xz , x y O x = (9 , O y = (13 , O z = (27 , , , ,
74) 148 t , yz , xy , x z O x = (2 , O y = (5 , O z = (15 , I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 4Weights Degree K X ρ lct Monomials Singular Points(6 , n + 3 , n + 5 , n + 5) 18 n + 15 n +3)(6 n +5) t , zt , z t , z , xy , x n +2 y O x = (1 , O y = n +3 (1 , O x O y = 1 × (1 , O z O t == 3 × n +5 (2 , n + 1)(6 , n + 5 , n + 8 , n + 9) 36 n + 24 n +3)(6 n +5) z , y t , xt , x n +1 y z , x n +4 O y = n +5 (2 , n + 1) O t == n +9 (6 n +5 , n +8) O x O z = 1 × (1 , O x O t = 1 × (1 , , n + 5 , n + 8 , n + 15) 36 n + 30 n +2)(6 n +5) t , y t , y , xz , x n +2 y z , x n +5 O z = n +8 (6 n +5 , n +7) O x O z = 1 × (1 , O x O t = 1 × (1 , O y O t == 2 × n +5 (2 , n + 1)(5 , , ,
9) 24 t y , y , z , x yz , x t O x = (1 , O t = (5 , O y O z = 1 × (1 , O y O t = 1 × (1 , , , ,
15) 30 t , y , yz , x y z , x t , x O z = (5 , O x O t = 2 × (1 , O y O t = 1 × (1 , O y O z = 1 × (1 , , , ,
17) 45 t y , y z , xz , x O y = (3 , O z = (11 , O t = (3 , O x O z = 1 × (1 , X C EP T I O NA L D E L PE ZZ O HY PE R S U R F A C E S Log del Pezzo surfaces with I = 4Weights Degree K X ρ lct Monomials Singular Points(10 , , ,
31) 75 t y , z , xy , x z O x = (3 , O y = (12 , O t = (2 , O x O z = 1 × (3 , , , ,
27) 71 t y , y z , xz , x t O x = (2 , O y = (11 , O z = (17 , O t = (11 , , , ,
31) 79 t y , tz , xy , x z O x = (2 , O y = (1 , O z = (11 , O t = (11 , , , ,
83) 166 t , yz , xy , x z O x = (3 , O y = (2 , O z = (11 , , , ,
29) 71 ty , yz , xt , x z O x = (1 , O y = (13 , O z = (13 , O t = (14 , , , ,
33) 79 tz , y z , xt , x y O x = (1 , O y = (13 , O z = (13 , O t = (14 , , , ,
83) 166 t , y z , xz , x y O x = (7 , O y = (13 , O z = (23 , I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 5Weights Degree K X ρ lct Monomials Singular Points(11 , , ,
25) 63 t y , tz , xy , x z O x = (2 , O y = (1 , O z = (11 , O t = (11 , , , ,
68) 136 t , yz , xy , x z O x = (3 , O y = (2 , O z = (11 , , , ,
68) 136 t , y z , xz , x y O x = (2 , O y = (13 , O z = (19 , X C EP T I O NA L D E L PE ZZ O HY PE R S U R F A C E S Log del Pezzo surfaces with I = 6Weights Degree K X ρ lct Monomials Singular Points(8 , n + 5 , n + 7 , n + 9) 12 n + 23 n +23)2(4 n +5)(4 n +7)(4 n +9) z t , yt , xy , x n +2 z O x = (4 n + 5 , n + 9) O y = n +5 (1 , O z = n +7 (8 , n + 5) O t = n +9 (8 , n + 7)(9 , n + 8 , n + 11 , n + 13) 12 n + 35 n +35)(3 n +8)(3 n +11)(6 n +13) z t , y z , xt , x n +3 y O x = (3 n + 11 , n + 13) O y = n +8 (9 , n + 13) O z = n +11 (9 , n + 8) O t == n +13 (3 n +8 , n +11)(7 , , ,
19) 45 z , y z , xt , x y O x = (1 , O y = (7 , O t = (2 , O y O z = 1 × (1 , , , ,
53) 106 t , yz , xy , x z O x = (8 , O y = (2 , O z = (11 , , , ,
53) 106 t , y z , xz , x y O x = (5 , O y = (13 , O z = (15 , I VAN C H E L T S O V , J I HUN P A R K , C O N S T AN T I N S H R A M O V Log del Pezzo surfaces with I = 7Weights Degree K X ρ lct Monomials Singular Points(11 , , ,
38) 76 t , yz , xy , x z O x = (2 , O y = (2 , O z = (11 , I = 8Weights Degree K X ρ lct Monomials Singular Points(7 , , ,
23) 46 t , y z , xz , x y O x = (3 , O y = (7 , O z = (11 , , , ,
37) 81 y z , z , xt , x y O x = (3 , O y = (7 , O t = (2 , O y O z = 1 × (7 , I = 9Weights Degree K X ρ lct Monomials Singular Points(7 , , ,
32) 64 t , y z , xz , x y O x = (5 , O y = (7 , O z = (15 , I = 10Weights Degree K X ρ lct Monomials Singular Points(7 , , ,
41) 82 t , y z , xz , x y O x = (2 , O y = (7 , O z = (19 , , , ,
55) 117 y z , z , xt , x y O x = (2 , O y = (7 , O t = (26 , O y O z = 1 × (7 , XCEPTIONAL DEL PEZZO HYPERSURFACES 147
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Ivan Cheltsov
School of Mathematics, The University of Edinburgh, Edinburgh, EH9 3JZ, UK; [email protected]
Jihun Park
Department of Mathematics, POSTECH, Pohang, Kyungbuk 790-784, Korea; [email protected]
Constantin Shramov