Algebraic Montgomery-Yang problem: the log del Pezzo surface case

Transcription

1 c 2014 The Matheatical Society of Japan J. Math. Soc. Japan Vol. 66, No. 4 (2014) pp doi: /jsj/ Algebraic Montgoery-Yang proble: the log del Pezzo surface case By DongSeon Hwang and JongHae Keu (Received June 9, 2012) (Revised Sep. 20, 2012) Abstract. We prove that a log del Pezzo surface of Picard nuber one contains at ost 3 singular points if its sooth locus is siply connected. This establishes the algebraic Montgoery-Yang proble for log del Pezzo surfaces. 1. Introduction. The present paper is a continuation of two papers [HK2] and [HK3] on the conjecture called algebraic Montgoery-Yang proble. Conjecture 1.1 ([K]). (Algebraic Montgoery-Yang Proble). Let S be a Q- hoology projective plane with quotient singularities, i.e., a noral projective surface with quotient singularities such that b 2 (S) = 1. Assue that S 0 := S\Sing(S) is siply connected. Then S contains at ost 3 singular points. In previous papers [HK2] and [HK3], we have confired the conjecture when S contains at least one non-cyclic singularity or S is not rational. In this paper we confir the conjecture when K S is aple, or equivalently when S is a log del Pezzo surface. By [HK2], we ay assue that S has cyclic singularities only. Theore 1.2. Let S be a log del Pezzo surface of Picard nuber one with cyclic singularities only. If H 1 (S 0, Z) = 0, then S contains at ost 3 singular points. The condition H 1 (S 0, Z) = 0 is weaker than the condition π 1 (S 0 ) = 1. In fact, there are log del Pezzo surfaces S of Picard nuber one with H 1 (S 0, Z) = 0 but π 1 (S 0 ) 1. Such surfaces have been classified in [HK2], under the assuption that the nuber of singularities is at least 4 and at least one of the singularities is non-cyclic. The ain ingredient of the proof is the classification theory of log del Pezzo surfaces of Picard nuber one developed by Zhang [Z], Gurjar and Zhang [GZ], Belousov [Be] together with the forulas developed in [HK3] for the intersection nubers of divisors on the inial resolution. Conjecture 1.1 is now reduced to the case where S is a rational surface with cyclic singularities such that K S is aple. We do not know any exaple of a rational surface with 4 cyclic singularities such that K S is aple. However, there are infinitely any 2010 Matheatics Subject Classification. Priary 14J26; Secondary 14J17, 14J45. Key Words and Phrases. Montgoery-Yang proble, log del Pezzo surface, quotient singularity.

2 1074 D. Hwang and J. Keu exaples with saller nuber of singularities ([KM], [K] and [HK4]). Throughout this paper, we work over the field C of coplex nubers. 2. Algebraic surfaces with cyclic singularities A singularity p of a noral surface S is called a cyclic singularity if the ger is locally analytically isoorphic to (C 2 /G, O) for soe nontrivial finite cyclic subgroup G of GL 2 (C) without quasi-reflections. Such subgroups are copletely classified by Brieskorn ([Br]). For a cyclic singularity of type 1 q (1, q 1), one can associate a Hirzebruch-Jung continued fraction 1 [n 1, n 2,..., n l ] = n 1 1 n n l = q q 1. Let H be the set of all Hirzebruch-Jung continued fractions [n 1, n 2,..., n l ], H = l 1{[n 1, n 2,..., n l ] all n j are integers 2}. We will use the following notation in this paper. Notation 2.1. Fix w = [n 1, n 2,..., n l ] H and an integer 0 s l + 1. (1) The length of w, denoted by l(w), is the nuber of entries of w. We will write siply l for l(w) if there is no confusion. (2) Let q be the order of the cyclic singularity corresponding to w, i.e., q = w = [n 1, n 2,..., n l ] := det(m( n 1,..., n l )) where n n n 3 0 M( n 1,..., n l ) = n l n l is the intersection atrix corresponding to the singularity [n 1, n 2,..., n l ]. (3) u s := [n 1, n 2,..., n s 1 ] (2 s l + 1), u 0 = 0, u 1 = 1. (4) v s := [n s+1, n s+2,..., n l ] (0 s l 1), v l = 1, v l+1 = 0.

3 Algebraic Montgoery-Yang proble 1075 Now let S be a noral projective surface with cyclic singularities and f : S S be a inial resolution of S. Since cyclic singularities are log-terinal singularities, one can write K S f K S nu p Sing(S) where D p = (a j A j ) is an effective Q-divisor with 0 a j < 1 supported on f 1 (p) = A j for each singular point p. Intersecting the forula with D p, we get D p, K 2 S = K 2 S p D 2 p = K 2 S + p D p K S. When p is a cyclic singularity of order q, the coefficients of D p can be expressed in ters of v j and u j (see Notation 2.1) as follows. Lea 2.2 ([HK3, Lea 3.1]). Let p be a cyclic singular point of S. Assue that f 1 (p) has l coponents A 1,..., A l with A 2 i = n i foring a string of sooth rational curves n1 n2 n l (1) D p K S = D 2 p = (2) D 2 p = 2l l j=1 l j=1 ( 1 v j + u j q. Then n j + 2 q 1 + q l + 2. q ) (n j 2), In particular, if l = 1, then D 2 p = (n 1 2) 2 n The torsion-free part of the second cohoology group, H 2 (S, Z) free := H 2 (S, Z)/(torsion) has a lattice structure which is uniodular. For a cyclic singular point p S, let R p H 2 (S, Z) free be the sublattice of H 2 (S, Z) free spanned by the nuerical classes of the coponents of f 1 (p). Then it is a negative definite lattice. Let R = p Sing(S) R p H 2 (S, Z) free

4 1076 D. Hwang and J. Keu be the sublattice of H 2 (S, Z) free spanned by the nuerical classes of the exceptional curves of f : S S. Here, the order G p of the local fundaental group is equal to the absolute value det(r p ) of the deterinant of the intersection atrix of R p. The following will be also useful in our proof. Lea 2.3 ([HK2, Lea 2.5]). Let S be a log del Pezzo surface of Picard nuber one with cyclic singularities such that H 1 (S 0, Z) = 0. Let f : S S be a inial resolution. Then (1) H 2 (S, Z) is torsion free, i.e., H 2 (S, Z) = H 2 (S, Z) free, (2) R is a priitive sublattice of the uniodular lattice H 2 (S, Z), (3) the orders G p = det(r p ) of the local fundaental groups are pairwise relatively prie, (4) D := det(r + K S ) = det(r) KS 2 and is a nonzero square nuber. The intersection nubers EK S and E 2 can be expressed in ters of the intersection nubers EA j,p of E and the exceptional curves A j,p. See ([HK3, Section 4]) for a ore general description. Proposition 2.4 ([HK3, Proposition 4.2]). Let S be a inial resolution of a log del Pezzo surface of Picard nuber one with cyclic singularities, and E be a divisor on it. Then, for soe positive integer depending on E, the following hold true. (1) EK S = D K 2 S p l p j=1 ( 1 v ) j,p + u j,p EA j,p. q p (2) If EA j,p = 0 for j s p, t p for soe s p and t p with 1 s p < t p l p, then E 2 = 2 D K2 S ( vsp u sp (EA sp ) 2 + v t p u tp (EA tp ) 2 + 2v ) t p u sp (EA sp )(EA tp ). q p p q p q p Theore 2.5 ([HK1, Theore 1.1], [Be, Theore 1.2]). Let S be a Q-hoology projective plane with quotient singularities. If S is rational, then it contains at ost 4 singular points. Proof. This is the result of Belousov ([Be, Theore 1.2]) if K S is aple, and is one of our previous results ([HK1, Theore 1.1]) if K S is nef. 3. Log del Pezzo surfaces of Picard nuber one. Let Throughout this section, S denotes a log del Pezzo surface of Picard nuber one. be its inial resolution. We denote by f : S S F := f 1 (Sing(S))

5 Algebraic Montgoery-Yang proble 1077 the reduced exceptional divisor of f. We review the work of Zhang [Z], Gurjar and Zhang [GZ] and Belousov [Be] on log del Pezzo surfaces of Picard nuber one. Assue that S does not contain any non-cyclic singularities, even though ost of the results in this section hold for general case. S. Lea 3.1. B 2 1 for any irreducible curve B S not contracted by f : S Proof. This is well-known (cf. [HK2, Lea 2.1]). The following lea is given in Lea 4.1 in [Z], and can also be easily derived fro the inequality of Proposition 2.4 (1). Lea 3.2 ([Z, Lea 4.1]). Let E be a ( 1)-curve on S. Let A 1,..., A r exhaust all irreducible coponents of F such that EA i > 0. Suppose that A 2 1 A 2 2 A 2 r. Then the r-tuple ( A 2 1,..., A 2 r) is one of the following: (2,..., 2, n), n 2, (2,..., 2, 3, 3), (2,..., 2, 3, 4), (2,..., 2, 3, 5). An irreducible curve C on S is called a inial curve if C.( f K S ) attains the inial positive value. Lea 3.3 ([Be, Lea 3.2, Lea 4.1]). rational curve. A inial curve C is a sooth Lea 3.4. Let C be a inial curve. Suppose that C + F + K S =. Then there is a unique decoposition F = F + F such that (1) F consists of ( 2)-curves not eeting C + F, (2) C + F + K S 0, (3) F = f 1 (p) for soe singular point p unless F = 0. Furtherore, if F 0, then CF = CF = 2 and one of the following holds: (1) F consists of one irreducible coponent, which C eets in a single point with ultiplicity 2 or in two points, (2) F consists of two irreducible coponents, whose intersection point C passes through, (3) F consists of at least two irreducible coponents, and C eets the two end coponents of F. Proof. The result can be easily derived fro either [GZ, Lea 3.2, Reark 3.4], or [Be, Lea 3.1, Lea 3.2]. Lea 3.5 ([GZ, Proposition 3.6]). Let C be a inial curve. Suppose that C + F + K S =. Then either C is a ( 1)-curve, S = P 2, or S is the Hirzebruch surface with the inial section contracted. Lea 3.6 ([Be, Lea 4.1]). Suppose that S contains a inial curve C with C 2 = 1. Suppose that C + F + K S =. Then CF 1 for any connected coponent

6 1078 D. Hwang and J. Keu F of F. Lea 3.7 ([Z, Lea 4.4]). Suppose that S contains a inial curve C with C 2 = 1. Suppose that C + F + K S =, and that C eets exactly two coponents F 1, F 2 of F. Then either F1 2 = 2 or F2 2 = 2. The following lea was proved in ([Z, Proof of Lea 5.3]). Lea 3.8. With the sae assuption as in Lea 3.7, assue further that F1 2 = F2 2 = 2. If F 1 is not an end coponent, then one of the following two cases holds: (1) There exists another inial ( 1)-curve C such that C + F + K S =. (2) F 2 = f 1 (p i ) for soe singular point p i. Lea 3.9. Suppose that S contains a inial curve C with C 2 = 1. Suppose that C + F + K S =, and that C eets three coponents F 1, F 2, F 3 of F and possibly ore. Define G := 2C + F 1 + F 2 + F 3 + K S. Then either G 0 or G Γ for soe ( 1)-curve Γ such that CΓ = F i Γ = 0 for i = 1, 2, 3. Furtherore, the following hold true. (1) In the first case, there are 3 singular points p 1, p 2, p 3 such that f 1 (p i ) = F i, and C eets no coponent of F (F 1 + F 2 + F 3 ). (2) In the second case, (a) L = 2 (F F F 2 3 ), where L is the nuber of irreducible coponents of F, (b) each curve in F F 1 F 2 F 3 is a ( 2)- or a ( 3)-curve and there are at ost two ( 3)-curves in F F 1 F 2 F 3, (c) each connected coponent of F contains at ost one ( n)-curve with n 3. Proof. The ain assertion is exactly ([Z, Lea 2.3]). (1) Let F i be an irreducible coponent of f 1 (p i ). Suppose that f 1 (p i ) has at least 2 irreducible coponents. Then there is an irreducible coponent I of f 1 (p i ) such that IF i = 1. By Lea 3.6, IC = 0, hence 0 = IG = I.(2C + F 1 + F 2 + F 3 + K S ) = IF i + IK S = 1 I 2 2. Thus I 2 = 1, a contradiction. Suppose that C eets a coponent J of F (F 1 + F 2 + F 3 ). Then 0 = JG = J.(2C + F 1 + F 2 + F 3 + K S ) = 2 + JK S, so J 2 = 0, a contradiction. (2) By ([GZ, Reark 6.4]), we ay assue that f 1 (p i ) has at least 2 irreducible coponents for i = 1, 2 or 3. Alternatively, by using Proposition 2.4, one can also derive

7 Algebraic Montgoery-Yang proble 1079 a contradiction for the case when f 1 (p i ) consists of only one irreducible coponent for each i = 1, 2 and 3, but it needs lengthy coputation. Now (2-b) and (2-c) directly follows fro ([GZ, Lea 6.6]). (2-a) We note that G 2 = (2C + F 1 + F 2 + F 3 + K S ) 2 = 1 L ( F1 2 + F2 2 + F3 2 ) where L denotes the nuber of irreducible coponents of F. Since G 2 = Γ 2 = 1, we have L = 2 (F1 2 + F2 2 + F3 2 ). The following lea was proved in ([Z, Proof of Lea 5.2]). Lea With the sae assuption as in Lea 3.9, assue further that 2C +F 1 +F 2 +F 3 +K S Γ for soe ( 1)-curve Γ, and that at least two of F 1, F 2, F 3 are ( 2)-curves. Then there exists another inial ( 1)-curve C such that C +F +K S =. The first reduction results shown in [HK3] can be reforulated, in the case of log del Pezzo surfaces, as follows: Lea 3. ([HK3, Lea 5.2, Lea 5.3, Lea 5.4, Lea 5.6]). Let S be a log del Pezzo surface of Picard nuber one containing exactly 4 cyclic singular points p 1, p 2, p 3, p 4 of orders (q 1, q 2, q 3, q 4 ). Let E be a ( 1)-curve on S. Then E.F 2, and one of the following cases occurs. (1) The orders are (2, 3, 5, q) where q 7 and gcd(q, 30) = 1. Moreover, the order 3 singularity ust be of type 1 3 (1, 1). In this case, E.F = 2 if and only if E.f 1 (p i ) = 0 for i = 1, 2, 3 and E.f 1 (p 4 ) = 2. (2) The orders are either (2, 3, 7, q) where q 41 or (2, 3,, 13). Moreover, the singularity type of S is precisely one of the 24 cases in Table 1. In this case, if E.F = 2, then E does not eet an end coponent of f 1 (p i ) for any i = 1, 2, 3, Proof of Theore 1.2. Throughout this section, S denotes a log del Pezzo surface of Picard nuber one such that H 1 (S 0, Z) = 0. Then S contains at ost 4 singular points by Theore 2.5. Suppose that S contains 4 cyclic singular points p 1, p 2, p 3, p 4. By Lea 3., it reains to consider the following cases: Let A (1, 1) (1, 1) + 1 q (1, q 1), q 7, gcd(q, 30) = 1; A (1, 1) (1, 2) + 1 q (1, q 1), q 7, gcd(q, 30) = 1; A (1, 1) + A q (1, q 1), q 7, gcd(q, 30) = 1; the 24 cases in Table 1. F = f 1 (Sing(S))

8 1080 D. Hwang and J. Keu Table 1. No. Type of R orders K 2 S 1 A 1 + A 2 + [7] + [13] (2, 3, 7, 13) A 1 + A 2 + [7] + [3, 2, 2, 2, 2, 2, 2, 2, 2] (2, 3, 7, 19) A 1 + A 2 + [7] + [5, 4] (2, 3, 7, 19) A 1 + A 2 + [7] + [3, 4, 2] (2, 3, 7, 19) A 1 + A 2 + [4, 2] + [2, 2, 4, 2, 2, 2] (2, 3, 7, 31) A 1 + A 2 + [4, 2] + [6, 2, 2, 2, 2, 2] (2, 3, 7, 31) A 1 + [3] + [3, 2, 2] + [4, 2, 2, 2, 3] (2, 3, 7, 29) A 1 + A 2 + [3, 2, 2] + [7, 2, 2, 2] (2, 3, 7, 25) A 1 + A 2 + [7] + [2, 2, 3, 2, 2, 2, 2, 2, 2] (2, 3, 7, 31) A 1 + [3] + [4, 2] + [3, 3, 2, 2, 3] (2, 3, 7, 41) A 1 + A 2 + [3, 2, 2] + [7, 2, 2, 2, 2, 2] (2, 3, 7, 37) A 1 + A 2 + [4, 2] + [, 2, 2] (2, 3, 7, 31) A 1 + [3] + A 6 + [2, 6, 2, 2] (2, 3, 7, 29) A 1 + [3] + [3, 2, 2] + [4, 3] (2, 3, 7, ) A 1 + [3] + [3, 2, 2] + [3, 2, 2, 2, 2] (2, 3, 7, ) A 1 + [3] + [3, 2, 2] + [4, 2, 2, 3] (2, 3, 7, 23) A 1 + [3] + [3, 2, 2] + [6, 5] (2, 3, 7, 29) A 1 + A 2 + [3, 2, 2] + [3, 5, 2] (2, 3, 7, 25) A 1 + A 2 + [3, 2, 2] + [13, 2] (2, 3, 7, 25) A 1 + A 2 + [4, 2] + [4, 2, 2, 2] (2, 3, 7, 13) A 1 + A 2 + [4, 2] + [5, 2, 2] (2, 3, 7, 13) A 1 + A 2 + [4, 2] + [4, 2, 2, 2, 2, 2] (2, 3, 7, 19) A 1 + [3] + [3, 2, 2, 2, 2] + [4, 2, 2, 2] (2, 3,, 13) A 1 + [3] + [3, 2, 2, 2, 2] + [5, 2, 2] (2, 3,, 13) be the reduced exceptional divisor of the inial resolution f : S S, and L be the nuber of irreducible coponents of F. Let C be a (fixed) inial curve on S Step 1. C + F + K S =. Proof. Suppose that C + F + K S =. By Lea 3.4 (1) and (3), we see that S contains at least 3 rational double points. In the case of (2, 3, 5, q), by Lea 3. (1) we see that S contains at least 3 rational double points, only if the singularities are of type A 1 + [3] + A 4 + A q 1. In this case, by Lea 2.2, L = q + 5 and K 2 S = 9 (q + 5) < 0,

9 Algebraic Montgoery-Yang proble 1081 a contradiction. We also see that each of the 24 cases fro Table 1 contains at ost 2 rational double points Step 2. (1) C is a ( 1)-curve. (2) CF = 3, and C eets three distinct coponents F 1, F 2, F 3 of F. Proof. (1) It iediately follows fro Lea 3.5 since S contains 4 singularities. (2) By Lea 3.6, CF 4. Since C 2 = 1 < 0 and the lattice R is negative definite, CF 1. Assue that CF = 1. Blowing up the intersection point, then contracting the proper transfor of C and the proper transfors of all irreducible coponents of F, we obtain a Q-hoology projective plane with 5 quotient singularities, which ay not be a log del Pezzo surface, i.e., whose canonical class ay be nef. Even this case contradicts Theore 2.5. Assue that CF = 4. By Lea 3.6, C eets four coponents F 1, F 2, F 3, F 4 of F, where F i f 1 (p i ). Then G Γ by Lea 3.9 (1). By Lea 3.2, at least two of F 1, F 2, F 3, F 4 have self-intersection 2. Thus, by Lea 3.10, there exists another inial ( 1)-curve C such that C + F + K S =. This is ipossible by Step 1. Assue that CF = 2. (a) Suppose that the case (2, 3, 5, q) occurs for soe q 7 with gcd(q, 30) = 1. By Lea 3. (1), C.f 1 (p 4 ) = 2. But, by Lea 3.6, C.f 1 (p 4 ) 1, a contradiction. (b) Now suppose that one of the 24 cases of Table 1 occurs. By Lea 3.6, there are two coponents F 1 and F 2 of F with CF 1 = CF 2 = 1. By Lea 3.7, we ay assue that F1 2 = 2. Moreover, by Lea 3. (2), C does not eet an end coponent of f 1 (p i ) for any i, i.e., both F 1 and F 2 are iddle coponents. Thus F2 2 2 by Lea 3.8 and Step 1. After contracting the ( 1)-curve C, by contracting the proper transfors of all irreducible coponents of F F 1, we obtain a Q-hoology projective plane with 5 quotient singularities, again contradicting Theore Step 3. 2C + F 1 + F 2 + F 3 + K S Γ for soe ( 1)-curve Γ. Proof. Suppose that 2C + F 1 + F 2 + F 3 + K S 0. Then, by Lea 3.9 (1), each F i is equal to the inverse iage of a singular point of S. By Table 1 and Lea 3., only the following cases satisfy this condition: A 1 + A 2 + [7] + [13] (Case 1, Table 1), A 1 + [3] + [2, 2, 2, 2] + [q], A 1 + [3] + [3, 2] + [q], A 1 + [3] + [5] + 1 q (1, q 1).

10 1082 D. Hwang and J. Keu Thus, ( F 2 1, F2 2, F3 2 ) = (2, 7, 13), (2, 3, q), (2, 5, q), (3, 5, q), (2, 3, 5). Then Lea 3.2 rules out the first four possibilities, since q 7. In the last case ( F 2 1, F 2 2, F 2 3 ) = (2, 3, 5), F i = f 1 (p i ) for i = 1, 2, 3. In this case we consider the sublattice C, F 1, F 2, F 3 H 2 (S, Z) generated by C, F 1, F 2, F 3. It is of rank 4 and has as its intersection atrix. It has deterinant 1, hence the orthogonal copleent of C, F 1, F 2, F 3 in H 2 (S, Z) is uniodular. The orthogonal copleent is an over-lattice of the lattice R p4 generated by the coponents of f 1 (p 4 ). Since R p4 is a priitive sublattice of H 2 (S, Z), it ust be uniodular, hence q = 1, a contradiction Step 4. If one of the cases (2, 3, 5, q), q 7, gcd(q, 30) = 1, occurs, then C.f 1 (p 4 ) = 1. Proof. Suppose that the case (2, 3, 5, q) occurs for soe q 7 with gcd(q, 30) = 1. By Lea 3. (1), p 2 is of type [3]. By Lea 3.6, C.f 1 (p i ) 1 for i = 1, 2, 3, 4. Suppose on the contrary that C.f 1 (p 4 ) = 0. Then, C.f 1 (p 1 ) = C.f 1 (p 2 ) = C.f 1 (p 3 ) = 1. Let F i f 1 (p i ) be the coponent with CF i = 1 for i = 1, 2, 3. Assue that p 3 is of type [5]. Then ( F 2 1, F 2 2, F 2 3 ) = (2, 3, 5) and the sublattice C, F 1, F 2, F 3 H 2 (S, Z) has deterinant 1, leading to the sae contradiction as above, since the orthogonal copleent of C, F 1, F 2, F 3 in H 2 (S, Z) is R p4. Assue that p 3 is of type [2, 3]. Then ( F 2 1, F 2 2, F 2 3 ) = (2, 3, 2) or (2, 3, 3). Let f 1 (p 3 ) = F 3 + F 3. If F 2 3 = 2, then det C, F1, F 2, F 3, F 3 = 13, and by Lea 3.9 (2-a) L = = 9, so l = 5. The orthogonal copleent of C, F 1, F 2, F 3, F 3 in H 2 (S, Z) is R p4, hence det(r p4 ) = q = 13.

11 Algebraic Montgoery-Yang proble 1083 This leads to a contradiction since there is no continued fraction of length 5 with q = 13. If F 2 3 = 3, then det C, F 1, F 2, F 3, F 3 = 7, hence det(r p4 ) = q = 7. By Lea 3.9 (2), L = = 10, so l = 6. Thus p 4 is of type A 6. But, then K 2 S = 9 L D 2 p 2 D 2 p 3 = < 0, a contradiction. Assue that p 3 is of type A 4 = [2, 2, 2, 2]. Then ( F 2 1, F 2 2, F 2 3 ) = (2, 3, 2). Let f 1 (p 3 ) = H 1 + H 2 + H 3 + H 4. If F 3 is an end coponent of f 1 (p 3 ), say H 1, then det C, F1, F 2, H 1, H 2, H 3, H 4 = 19, and by Lea 3.9 (2-a) L = = 9, so l = 3. Thus det(r p4 ) = q = 19 and rank(r p4 ) = 3. Aong all Hirzebruch-Jung continued fractions of order 19, only two, [7, 2, 2] and [3, 4, 2], have length 3. In each of these two cases, f 1 (p 4 ) contains an irreducible coponent with self-intersection 4. Since f 1 (p 4 ) F F 1 F 2 F 3, we have a contradiction by Lea 3.9 (2-b). If F 3 is a iddle coponent of f 1 (p 3 ), say H 2, then det C, F 1, F 2, H 1, H 2, H 3, H 4 = 31, and by Lea 3.9 (2-a) L = = 9, so l = 3. Thus q = 31 and p 4 is of type [, 2, 2], [3, 6, 2], or [5, 2, 4]. In each of these three cases, f 1 (p 4 ) contains an irreducible coponent with self-intersection 4, a contradiction by Lea 3.9 (2-b). This proves that C.f 1 (p 4 ) = Step 5. None of the cases (2, 3, 5, q), q 7, gcd(q, 30) = 1, occurs. Proof. Suppose that the case (2, 3, 5, q) occurs for soe q 7 with gcd(q, 30) = 1. By Lea 3. (1), p 2 is of type [3]. By Step 2, CF = 3 and C eets the three coponents F 1, F 2, F 3 of F. By Step 3, 2C + F 1 + F 2 + F 3 + K S Γ for soe ( 1)-curve Γ. By Step 4, we ay assue that F 3 f 1 (p 4 ). Let f 1 (p 4 ) = n1 D 1 n2 D 2 n l D l

12 1084 D. Hwang and J. Keu and F 3 = D j for soe 1 j l. Note first that by Lea 3.9 (2-b), n k 3 for all k j. Assue that p 3 is of type [5]. By Lea 3.9 (2-b), C ust eet f 1 (p 3 ), so we ay assue that F 2 = f 1 (p 3 ). Since F 1 = f 1 (p 1 ) or F 1 = f 1 (p 2 ), by Lea 3.2, ( F 2 1, F2 2, F3 2 ) = (2, 5, 2), (3, 5, 2), (2, 5, 3). By Lea 3.9 (2-a), we have hence By Lea 3.9 (2-b) and (2-c), (L, n j ) = (, 2), (12, 2), (12, 3), (l, n j ) = (8, 2), (9, 2), (9, 3). [n 1,..., n l ] = [3, 2, 2, 2, 2, 2, 2, 2], [2, 2, 2, 2, 2, 2, 2, 2]; [3, 2, 2, 2, 2, 2, 2, 2, 2], [2, 2, 2, 2, 2, 2, 2, 2, 2] up to perutation of n 1,..., n l. As you can see in Table 2, none of these cases satisfies the following three conditions: (#1) gcd(q, 30) = 1, (#2) KS 2 > 0, (#3) D = det(r) KS 2 is a positive square integer. Table 2. Type of p 4 q gcd(q, 30) K 2 D A [3, 2, 2, 2, 2, 2, 2, 2] [2, 3, 2, 2, 2, 2, 2, 2] [2, 2, 3, 2, 2, 2, 2, 2] 27 1 [2, 2, 2, 3, 2, 2, 2, 2] A [3, 2, 2, 2, 2, 2, 2, 2, 2] [2, 3, 2, 2, 2, 2, 2, 2, 2] 26 1 [2, 2, 3, 2, 2, 2, 2, 2, 2] [2, 2, 2, 3, 2, 2, 2, 2, 2] 34 1 [2, 2, 2, 2, 3, 2, 2, 2, 2] 35 1

13 Algebraic Montgoery-Yang proble 1085 Assue that p 3 is of type [2, 3]. Then, by Lea 3.2, ( F 2 1, F2 2, F3 2 ) = (2, 3, nj ), n j 5, or (3, 3, 2), or (2, 2, n j ). The last case can be ruled out by Lea 3.10 and Step 1. Now, by Lea 3.9 (2), we have and (l, n j ) = (5, 2), (6, 3), (7, 4), (8, 5), (6, 2), [n 1,..., n l ] = [3, 2, 2, 2, 2], [2, 2, 2, 2, 2]; [3, 2, 2, 2, 2, 2]; [4, 2, 2, 2, 2, 2, 2]; [5, 2, 2, 2, 2, 2, 2, 2]; [2, 2, 2, 2, 2, 2], up to perutation of n 1,..., n l. None of the 16 cases satisfies the three conditions (#1), (#2), (#3). Table 3 suarizes the coputation. Table 3. Type of p 4 q gcd(q, 30) K 2 D A [3, 2, 2, 2, 2] [2, 3, 2, 2, 2] 14 1 [2, 2, 3, 2, 2] 15 1 A [3, 2, 2, 2, 2, 2] [2, 3, 2, 2, 2, 2] [2, 2, 3, 2, 2, 2] [4, 2, 2, 2, 2, 2, 2] 22 1 [2, 4, 2, 2, 2, 2, 2] 32 1 [2, 2, 4, 2, 2, 2, 2] 38 1 [2, 2, 2, 4, 2, 2, 2] 40 1 [5, 2, 2, 2, 2, 2, 2, 2] 33 1 [2, 5, 2, 2, 2, 2, 2, 2] 51 1 [2, 2, 5, 2, 2, 2, 2, 2] 63 1 [2, 2, 2, 5, 2, 2, 2, 2] 69 1 Assue that p 3 is of type [2, 2, 2, 2]. Then, by Lea 3.2, ( F 2 1, F2 2, F3 2 ) = (2, 3, nj ), n j 5, or (2, 2, n j ).

14 1086 D. Hwang and J. Keu The last case can be ruled out by Lea 3.10 and Step 1. Now, by Lea 3.9 (2), we have and (l, n j ) = (3, 2), (4, 3), (5, 4), (6, 5), [n 1,..., n l ] = [3, 2, 2], [2, 2, 2]; [3, 2, 2, 2]; [4, 2, 2, 2, 2]; [5, 2, 2, 2, 2, 2], up to perutation of n 1,..., n l. None of the cases satisfies the three conditions (#1), (#2), (#3). Table 4 suarizes the coputation. Table 4. Type of p 4 q gcd(q, 30) K 2 D A [3, 2, 2] [2, 3, 2] 8 1 [3, 2, 2, 2] 9 1 [2, 3, 2, 2] [4, 2, 2, 2, 2] 16 1 [2, 4, 2, 2, 2] 22 1 [2, 2, 4, 2, 2] 24 1 [5, 2, 2, 2, 2, 2] 25 1 [2, 5, 2, 2, 2, 2] [2, 2, 5, 2, 2, 2] Next, we will show that none of the cases (2, 3, 7, q), q 41, gcd(q, 42) = 1, and (2, 3,, 13) occurs. To do this, it is enough to consider the 24 cases of Table Step 6. None of the 24 cases of Table 1 occurs. Proof. By Step 2, CF = 3 in each of the 24 cases of Table 1. Each of Cases (1), (2), (3), (4), (6), (8), (9), (), (12), (13), (17), and (19), contains an irreducible coponent F with self-intersection 6. Lea 3.9 (2-b) iplies that C eets F. Thus C eets two coponents of F with self-intersection 2 by Lea 3.2. Thus we get a contradiction for those cases by Lea 3.10 and Step 1. By Lea 3.9 (2-c), we get a contradiction iediately for Cases (7), (10), (14), (16), (18), since each of these cases contains a connected coponent of F with at least two irreducible coponents of self-intersection 3. By Lea 3.2 and Lea 3.9 (2-b), we get a contradiction iediately for Cases (5), (20), (21), (22), since each of these cases contains at least two irreducible coponents with self-intersection 4.

15 Algebraic Montgoery-Yang proble 1087 We need to rule out the reaining three cases: (15), (23), (24). Consider Case (24). Note that L = 10 in this case. On the other hand, by Lea 3.9 (2-b), C ust eet the coponent having self-intersection nuber 5. Thus, we ay assue that F 2 3 = 5. Since F 2 1 2, F 2 2 2, Lea 3.9 (2-a) gives L = 2 (F F F 2 3 ) =, a contradiction. Case (15): Let 2 A 3 B 3 3 C 1 D 1 C 2 C 3 D 2 D 3 D 4 D 5 be the exceptional curves. In this case, K 2 S = 50/231, D = 10. Since L = 10 = 2 (F F F 2 3 ), C eets only two of B, C 1, D 1. If CC 1 = CD 1 = 1, then CA = 1. Applying Proposition 2.4 (1) to C and looking at Table 5, we get KS 2 = 1 3 D 7 5 = 9 77, thus = 27/5, not an integer, a contradiction. Table 5. [2] [3] [3, 2, 2] [3, 2, 2, 2, 2] j vj+uj 1 q If CB = CC 1 = CA = 1, then Γ only eets C 2 and D 1, a contradiction to Lea 3. (2). If CB = CC 1 = CD j = 1 for soe j, then Proposition 2.4 (1) gives hence j = 4, 5. If j = 4, then KS 2 = 1 1 D 3 3 ( 7 1 v j + u j q KS 2 = 1 1 D = , thus = 13/5, a contradiction. If j = 5, then thus = 34/5, a contradiction. If CB = CD 1 = CA = 1, then D K 2 S = = , ) > 0,

16 1088 D. Hwang and J. Keu thus = 49/5, a contradiction. If CB = CD 1 = CC 2 = 1, then a contradiction. If CB = CD 1 = CC 3 = 1, then thus = 16/5, a contradiction. Case (23): Let KS 2 = 1 1 D 3 5 = 7 33, KS 2 = 1 1 D = < 0, KS 2 = 1 1 D = , 2 A 3 B 4 C 1 D 1 3 C 2 C 3 C 4 C 5 D 2 D 3 D 4 be the exceptional curves. Since C eets D 1 and L =, C ust eet only one of B and C 1. If CB = CA = 1, then Γ eets exactly two irreducible coponents C 1, D 2 with ultiplicity 1, a contradiction to Lea 3. (2). If CB = CC j = 1 for soe j 2, then Table 6 gives D K 2 S < 0, a contradiction. If CC 1 = 1, then CA = 1 and Proposition 2.4 (1) together with Table 6 gives a contradiction. D K 2 S = < 0, Table 6. [2] [3] [3, 2, 2, 2, 2] [4, 2, 2, 2] j vj+uj 1 q This copletes the proof of Theore 1.2.

17 Algebraic Montgoery-Yang proble 1089 References [Be] G. N. Belousov, Del Pezzo surfaces with log terinal singularities, Math. Notes, 83 (2008), [Br] E. Brieskorn, Rationale Singularitäten koplexer Flächen, Invent. Math., 4 (1968), [GZ] R. V. Gurjar and D.-Q. Zhang, π 1 of sooth points of a log del Pezzo surface is finite. I, J. Math. Sci. Univ. Tokyo, 1 (1994), [HK1] D. Hwang and J. Keu, The axiu nuber of singular points on rational hoology projective planes, J. Algebraic Geo., 20 (20), [HK2] D. Hwang and J. Keu, Algebraic Montgoery-Yang proble: the non-cyclic case, Math. Ann., 350 (20), [HK3] D. Hwang and J. Keu, Algebraic Montgoery-Yang proble: the non-rational surfacce case, Michigan Math. J., 62 (2013), [HK4] D. Hwang and J. Keu, Construction of singular rational surfaces of Picard nuber one with aple canonical divisor, Proc. Aer. Math. Soc., 140 (2012), [KM] S. Keel and J. McKernan, Rational Curves on Quasi-Projective Surfaces, Me. Aer. Math. Soc., 140, no. 669, Aer. Math. Soc., Providence, RI, [K] J. Kollár, Is there a topological Bogoolov-Miyaoka-Yau inequality?, Pure Appl. Math. Q., 4 (2008), [Z] D.-Q. Zhang, Logarithic del Pezzo surfaces of rank one with contractible boundaries, Osaka J. Math., 25 (1988), DongSeon Hwang Departent of Matheatics Ajou University Suwon , Korea E-ail: dshwang@ajou.ac.kr JongHae Keu School of Matheatics Korea Institute For Advanced Study Seoul , Korea E-ail: jhkeu@kias.re.kr