Volume 296, Issue 5 pp. 2046-2069
ORIGINAL ARTICLE
Open Access

On the fixed part of pluricanonical systems for surfaces

Jihao Liu

Corresponding Author

Jihao Liu

Department of Mathematics, Northwestern University, Evanston, IL, USA

Correspondence

Jihao Liu, Department of Mathematics, Northwestern University, Evanston, IL, USA.

Email: [email protected]

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Lingyao Xie

Lingyao Xie

Department of Mathematics, University of Utah, Salt Lake City, Utah, USA

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First published: 21 February 2023

Abstract

We show that | m K X | $|mK_X|$ defines a birational map and has no fixed part for some bounded positive integer m for any 1 2 $\frac{1}{2}$ -lc surface X such that K X $K_X$ is big and nef. For every positive integer n 3 $n\ge 3$ , we construct a sequence of projective surfaces X n , i $X_{n,i}$ , such that K X n , i $K_{X_{n,i}}$ is ample, mld ( X n , i ) > 1 n ${\rm {mld}}(X_{n,i})>\frac{1}{n}$ for every i, lim i + mld ( X n , i ) = 1 n $\lim _{i\rightarrow +\infty }{\rm {mld}}(X_{n,i})=\frac{1}{n}$ , and for any positive integer m, there exists i such that | m K X n , i | $|mK_{X_{n,i}}|$ has nonzero fixed part. These results answer the surface case of a question of Xu.

1 INTRODUCTION

We work over the field of complex numbers C $\mathbb {C}$ .

Pluricanonical systems are central objects in the study of birational geometry. More precisely, given a normal projective variety X such that K X $K_X$ is effective, we would like to study the behavior of the linear systems | m K X | $|mK_X|$ for any positive integer m.

It is well known that for any sufficiently divisible m 0 $m\gg 0$ , the rational map given by | m K X | $|mK_X|$ is birationally equivalent to the Iitaka fibration of K X $K_X$ . In 2014, Hacon–McKernan–Xu proved that for any lc projective variety X of general type and of fixed dimension, there exists a uniform positive integer m such that | m K X | $|mK_X|$ defines a birational map [8, Theorem 1.3] (see also [7, 15, 16]). In other words, | m K X | $|mK_X|$ defines a birational morphism X Bs ( | m K X | ) P ( | m K X | ) $X\backslash \operatorname{Bs}(|mK_X|)\rightarrow \mathbf {P}(|mK_X|)$ for some uniform positive integer m, where Bs ( | m K X | ) $\operatorname{Bs}(|mK_X|)$ is the base locus of | m K X | $|mK_X|$ .

It is then natural to ask whether the behavior | m K X | $|mK_X|$ can be described more accurately. Since we already have a birational morphism X Bs ( | m K X | ) P ( | m K X | ) $X\backslash \operatorname{Bs}(|mK_X|)\rightarrow \mathbf {P}(|mK_X|)$ for some uniform positive integer m, one would like to focus on the asymptotic behavior of Bs ( | m K X | ) $\operatorname{Bs}(|mK_X|)$ . As the very first step, we have the following question proposed by Prof. C. Xu to the first author in 2018:

Question 1. (Xu)Assume that X is a klt projective variety of fixed dimension such that K X $K_X$ is big and nef. When will we have a uniform positive integer m, such that | m K X | $|mK_X|$ defines a birational map and does not have a fixed part?

Note that it is natural to assume K X $K_X$ to be nef as we can always run an MMP with scaling and reach a minimal model for varieties of general type (cf. [3, Corollary 1.4.2]).

Question 1 naturally arises as a combination of [8, Theorem 1.3] and the effective base-point-freeness theorem [9, 1.1 Theorem]. Note that when the Cartier index is bounded, | m K X | $|mK_X|$ not only defines a birational map but is also base-point-free for some uniform positive integer m. The interesting cases of Question 1 appear when the Cartier index of K X $K_X$ is unbounded, in which case, the uniform base-point-freeness cannot be guaranteed.

Question 1 is trivial in dimension 1 but remained widely open in dimension ≥2. In this paper, we study Question 1 when dim X = 2 $\dim X=2$ . The main theorem of this paper is the following:

Theorem 1.1.There exists a uniform positive integer m satisfying the following. Assume that X is a 1 2 $\frac{1}{2}$ -lc projective surface and K X $K_X$ is big and nef. Then, | m K X | $|mK_X|$ defines a birational map and does not have a fixed part.

The following theorem is a complementary statement for Theorem 1.1, which shows that if the Cartier index of K X $K_X$ is not bounded and X is not 1 2 $\frac{1}{2}$ -lc, then Theorem 1.1 is not expected to hold.

Theorem 1.2.For any integer n 3 $n\ge 3$ , there exists a sequence of projective surfaces { X i } i = 1 + $\lbrace X_i\rbrace _{i=1}^{+\infty }$ , such that

  • 1. mld ( X i ) > 1 n ${\rm {mld}}(X_i)>\frac{1}{n}$ for each i and lim i + mld ( X i ) = 1 n $\lim _{i\rightarrow +\infty }{\rm {mld}}(X_i)=\frac{1}{n}$ ,
  • 2. K X i $K_{X_i}$ is ample, and
  • 3. if m i $m_i$ is the minimal positive integer such that | m i K X i | $|m_iK_{X_i}|$ defines a birational map and has no fixed part, then lim i + m i = + $\lim _{i\rightarrow +\infty }m_i=+\infty$ .

Note that the assumptions on mld(X) in Theorem 1.1 and Theorem 1.2 are natural assumptions: We are only interested in varieties such that the Cartier index of K X $K_X$ is not bounded, and if we consider a family of singularities { ( X x ) } $\lbrace (X\ni x)\rbrace$ such that the index of K X $K_{X}$ is unbounded, then { mld ( X x ) } $\lbrace {\rm {mld}}(X\ni x)\rbrace$ is an infinite set (cf. [4, Proposition 7.4]) and the accumulation points of { mld ( X x ) } $\lbrace {\rm {mld}}(X\ni x)\rbrace$ belong to { 0 } { 1 n | n Z 2 } $\lbrace 0\rbrace \cup \lbrace \frac{1}{n}|\nobreakspace n\in \mathbb {Z}_{\ge 2}\rbrace$ (cf. [1, Corollary 3.4]). The 1 2 $\frac{1}{2}$ accumulation point case is resolved by Theorem 1.1 and the remaining cases are resolved by Theorem 1.2.

It is also interesting to ask whether Question 1 has a positive answer for canonical or terminal threefolds in dimension 3, as 1 is the largest accumulation points of mld ( X x ) ${\rm {mld}}(X\ni x)$ in dimension 3 (cf. [13, Appendix, Theorem]). We will not address this question in this paper, but we will provide a related example (cf. Theorem 5.7).

2 PRELIMINARIES

We adopt the standard notation and definitions in [11], and will freely use them.

Definition 2.1. (Pairs and singularities)A pair ( X , B ) $(X,B)$ consists of a normal quasi-projective variety X and an R $\mathbb {R}$ -divisor B 0 $B\ge 0$ such that K X + B $K_X+B$ is R $\mathbb {R}$ -Cartier. Moreover, if the coefficients of B are ⩽1, then B is called a boundary of X.

Let E be a prime divisor on X and D an R $\mathbb {R}$ -divisor on X. We define mult E D $\operatorname{mult}_ED$ to be the multiplicity of E along D. Let ϕ : W X $\phi :W\rightarrow X$ be any log resolution of ( X , B ) $(X,B)$ and let

K W + B W : = ϕ ( K X + B ) . $$\begin{equation*} K_W+B_W:=\phi ^{*}(K_X+B). \end{equation*}$$
The log discrepancy of a prime divisor D on W with respect to ( X , B ) $(X,B)$ is 1 mult D B W $1-\operatorname{mult}_{D}B_W$ and it is denoted by a ( D , X , B ) $a(D,X,B)$ . For any positive real number ε, we say that ( X , B ) $(X,B)$ is lc (resp. klt, ε-lc, ε-klt) if a ( D , X , B ) 0 $a(D,X,B)\ge 0$ (resp. >0, ε $\ge \epsilon$ , > ε $>\epsilon$ ) for every log resolution ϕ : W X $\phi :W\rightarrow X$ as above and every prime divisor D on W. We say that X is lc (resp. klt, ε-lc, ε-klt) if (X, 0) is lc (resp. klt, ε-lc, ε-klt).

A germ ( X x , B ) $(X\ni x,B)$ consists of a pair ( X , B ) $(X,B)$ and a closed point x X $x\in X$ . ( X x , B ) $(X\ni x,B)$ is called an lc (resp. a klt, an ε-lc) germ if ( X , B ) $(X,B)$ is lc (resp. klt, ε-lc) near x. ( X x , B ) $(X\ni x,B)$ is called ε-lc at x if a ( D , X , B ) ε $a(D,X,B)\ge \epsilon$ for any prime divisor D over X x $X\ni x$ (i.e., center X D = x $\operatorname{center}_{X}D=x$ ).

Definition 2.2.Let I $\mathcal {I}$ be a set of real numbers. We say that I $\mathcal {I}$ satisfies the descending chain condition (DCC) if any decreasing sequence a 1 a 2 a k $a_1\ge a_2 \ge \cdots \ge a_k \ge \cdots$ in I $\mathcal {I}$ stabilizes. We say that I $\mathcal {I}$ satisfies the ascending chain condition (ACC) if any increasing sequence in I $\mathcal {I}$  stabilizes.

Definition 2.3. (Minimal log discrepancies)Let ( X , B ) $(X,B)$ be a pair and x X $x\in X$ a closed point. The minimal log discrepancy of ( X , B ) $(X,B)$ is defined as

mld ( X , B ) : = inf { a ( E , X , B ) E is an exceptional prime divisor over X } . $$\begin{equation*} {\rm {mld}}(X,B):=\inf \lbrace a(E,X,B)\mid E \text{ is an exceptional prime divisor over } X\rbrace . \end{equation*}$$
The minimal log discrepancy of ( X x , B ) $(X\ni x,B)$ is defined as
mld ( X x , B ) : = inf { a ( E , X , B ) E is a prime divisor over X x } . $$\begin{equation*} {\rm {mld}}(X\ni x,B):=\inf \lbrace a(E,X,B)\mid E \text{ is a prime divisor over } X\ni x\rbrace . \end{equation*}$$
If X is Q $\mathbb {Q}$ -Gorenstein, we define mld ( X ) : = mld ( X , 0 ) ${\rm {mld}}(X):={\rm {mld}}(X,0)$ . If X is Q $\mathbb {Q}$ -Gorenstein near x, we define mld ( X x ) : = mld ( X x , 0 ) ${\rm {mld}}(X\ni x):={\rm {mld}}(X\ni x,0)$ . For any positive integer d, we define
mld ( d ) : = { mld ( X x ) ( X x , 0 ) is lc, dim X = d } . $$\begin{equation*} {\rm {mld}}(d):=\lbrace {\rm {mld}}(X\ni x)\mid (X\ni x,0)\text{ is lc, } \dim X=d\rbrace . \end{equation*}$$

Definition 2.4.Let X be a normal projective variety and D an R $\mathbb {R}$ -divisor on X. We define

| D | : = { D 0 D D } . $$\begin{equation*} |D|:=\lbrace D^{\prime }\mid 0\le D^{\prime }\sim \lfloor D\rfloor \rbrace . \end{equation*}$$
For any R $\mathbb {R}$ -divisor D such that | D | $|D|\not=\emptyset$ , the base locus of D is
Bs ( D ) : = D D Supp D , $$\begin{equation*} \operatorname{Bs}(D):=\cap _{D^{\prime }\sim D}\operatorname{Supp}D^{\prime }, \end{equation*}$$
the fixed part of D is the unique R $\mathbb {R}$ -divisor F 0 $F\ge 0$ , such that
  • (1) for any D | D | $D^{\prime }\in |D|$ , D F $D^{\prime }\ge F$ , and
  • (2) Bs ( | D F | ) $\operatorname{Bs}(|D-F|)$ does not contain any divisor,
and the movable part of D is D F $D-F$ . We also say that F is the fixed part of | D | $|D|$ .

We denote by ρ ( X ) $\rho (X)$ the Picard number of X.

Definition 2.5.A surface is a variety of dimension 2. A rational surface is a projective surface that is birational to P 2 $\mathbb {P}^2$ . For ever nonnegative integer k, the Hirzebruch surface F k $\mathbb {F}_k$ is P P 1 ( O P 1 O P 1 ( k ) ) $\mathbb {P}_{\mathbb {P}^1}(\mathcal {O}_{\mathbb {P}^1}\oplus \mathcal {O}_{\mathbb {P}^1}(k))$ .

Definition 2.6.Let n be a nonnegative integer, and C = i = 1 n C i $C=\cup _{i=1}^nC_i$ a collection of proper curves on a smooth surface U. The determinant of C is defined as det ( C ) : = det ( { ( C i · C j ) } 1 i , j n ) $\det (C):=\det (\lbrace -(C_i\cdot C_j)\rbrace _{1\le i,j\le n})$ if C $C\not=\emptyset$ , and we define det ( ) = 1 $\det (\emptyset )=1$ . We define the dual graph DG ( C ) $\mathcal {DG}(C)$ of C as follows.

  • 1. The vertices v i = v i ( C i ) $v_i=v_i(C_i)$ of DG ( C ) $\mathcal {DG}(C)$ correspond to the curves C i $C_i$ .
  • 2. For each i, v i $v_i$ is labeled by the integer e i : = ( C i 2 ) $e_i:=-(C_i^2)$ . e i $e_i$ is called the weight of v i $v_i$ .
  • 3. For i j $i\ne j$ ,the vertices v i $v_i$ and v j $v_j$ are connected by C i · C j $C_i\cdot C_j$ edges.
The determinant of DG ( C ) $\mathcal {DG}(C)$ is defined as det ( C ) $\det (C)$ . For any birational morphism f : Y X $f: Y\rightarrow X$ between normal surfaces, let E = i = 1 n E i $E=\cup _{i=1}^nE_i$ be the reduced exceptional divisor for some nonnegative integer n. We define DG ( f ) : = DG ( E ) $\mathcal {DG}(f):=\mathcal {DG}(E)$ . If f is the minimal resolution of X (resp. the minimal resolution of ( X x , 0 ) $(X\ni x,0)$ for some closed point x X $x\in X$ ), we define DG ( X ) : = DG ( f ) $\mathcal {DG}(X):=\mathcal {DG}(f)$ (resp. DG ( X x ) : = DG ( f ) $\mathcal {DG}(X\ni x):=\mathcal {DG}(f)$ ).

Theorem 2.7. (cf. [[1], Theorem 3.2, Corollary 3.4], [[14]])mld(2) satisfies the ACC, and the set of accumulation points of mld(2) is { 1 n n 2 } { 0 } $\lbrace \frac{1}{n}\mid n\ge 2\rbrace \cup \lbrace 0\rbrace$ .

Proposition 2.8. (cf. [[4], Proposition A.5])Let I 0 [ 0 , 1 ] $\mathcal {I}_0\subset [0,1]$ be a finite set. Then, there exists a positive integer I depending only on I 0 $\mathcal {I}_0$ satisfying the following. Assume that ( X x , 0 ) $(X\ni x,0)$ is an lc surface germ such that mld ( X x ) I 0 ${\rm {mld}}(X\ni x)\in \mathcal {I}_0$ . Then, I K X $IK_X$ is Cartier near x.

Lemma 2.9.Let ε be a positive real number and ( X x , 0 ) $(X\ni x,0)$ an ε-lc (resp. ε-klt) surface germ. Then, for any vertex v of DG ( X x ) $\mathcal {DG}(X\ni x)$ , the weight of v is 2 ε $\le \frac{2}{\epsilon }$ (resp. < 2 ε $&lt;\frac{2}{\epsilon }$ ).

Proof.[1, Corollary 2.19] proves the ε-lc case and the ε-klt case immediately follow. $\Box$

Lemma 2.10. (cf. [[1], Lemma 3.3], [[4], Lemma A.1])Let ε be a positive real number. Then, there exists a finite set G = G ( ε ) $\mathcal {G}=\mathcal {G}(\epsilon )$ of dual graphs and a finite set I 0 = I 0 ( ε ) $\mathcal {I}_0=\mathcal {I}_0(\epsilon )$ of positive integers, such that for any ε-lc germ ( X x , 0 ) $(X\ni x,0)$ , one of the following holds:

  • 1. DG ( X x , 0 ) G $\mathcal {DG}(X\ni x,0)\in \mathcal {G}$ .
  • 2. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ is of the type as in Figure 1
    Details are in the caption following the image
     .
    . Here, e 1 = e 1 ( X x ) , q 1 = q 1 ( X x ) $e_1=e_1(X\ni x),q_1=q_1(X\ni x)$ and e 2 = e 2 ( X x ) , q 2 = q 2 ( X x ) $e_2=e_2(X\ni x),q_2=q_2(X\ni x)$ are the determinants of the subdual graphs, such that e 1 , e 2 , q 1 , q 2 I 0 $e_1,e_2,q_1,q_2\in \mathcal {I}_0$ , and
    min 1 e 1 q 1 , 1 e 2 q 2 ε . $$\begin{equation*} \min {\left\lbrace \frac{1}{e_1-q_1},\frac{1}{e_2-q_2}\right\rbrace} \ge \epsilon . \end{equation*}$$
    Moreover, we may assume that
    • (a) either e 1 = w 1 = 2 $e_1=w_1=2$ and q 1 = 1 $q_1=1$ , or w 1 > 2 $w_1&gt;2$ ; and
    • (b) either e 2 = w 2 = 2 $e_2=w_2=2$ and q 2 = 1 $q_2=1$ , or w 2 > 2 $w_2&gt;2$ .
  • 3. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ is of the type as in Figure 2
    Details are in the caption following the image
     .
    . Here, e 1 = e 1 ( X x ) $e_1=e_1(X\ni x)$ and q 1 = q 1 ( X x ) $q_1=q_1(X\ni x)$ are the determinants of the subdual graphs, such that e 1 , q 1 I 0 $e_1,q_1\in \mathcal {I}_0$ , and
    mld ( X x ) = 1 e 1 q 1 ε . $$\begin{equation*} {\rm {mld}}(X\ni x)=\frac{1}{e_1-q_1}\ge \epsilon . \end{equation*}$$

We remark that each oval in Figures 1 and 2 corresponds to a subdual graph, which is a chain, as shown in [1, Lemma 3.3, 2] and [4, Appendix, Notation].

Proof.The statement on the structure of the dual graphs are explained both in [1, Lemma 3.3] and in [4, Lemma A.1]. By taking the coefficient set Γ = { 0 } $\Gamma =\lbrace 0\rbrace$ , the inequality min { 1 e 1 q 1 , 1 e 2 q 2 } ε $\min \lbrace \frac{1}{e_1-q_1},\frac{1}{e_2-q_2}\rbrace \ge \epsilon$ in (2) follows from the moreover part of [4, Lemma A.1(2)], and the inequality 1 e 1 q 1 ε $\frac{1}{e_1-q_1}\ge \epsilon$ follows from the moreover part of [4, Lemma A.1(3)].

For the moreover part of (2), note that if w 1 2 $w_1\le 2$ , then we may add the vertex corresponding to w1 to the 2-chains and repeat this process unless this vertex is the tail of the chain. This implies (2.a), and (2.b) is similar to (2.a). $\Box$

Lemma 2.11. ([[10], 3.1.11])Let ( X x , 0 ) $(X\ni x,0)$ be a klt surface germ such that DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ is a chain. Then, X x $X\ni x$ is a cyclic quotient singularity. Moreover, if the dual graph of X x $X\ni x$ is

image
then X x $X\ni x$ is a cyclic quotient singularity of form 1 r ( 1 , a ) $\frac{1}{r}(1,a)$ , such that r a = a 1 1 a 2 1 a 3 1 a n $\frac{r}{a}=a_1-\frac{1}{a_2-\frac{1}{a_3-\frac{1}{\dots a_n}}}$ and gcd ( r , a ) = 1 $\gcd (r,a)=1$ .

Lemma 2.12. (cf. [[12], Lemma 2.11], [[2], Theorem 1])Let X x $X\ni x$ be a cyclic quotient singularity of form 1 r ( 1 , a ) $\frac{1}{r}(1,a)$ such that gcd ( r , a ) = 1 $\gcd (r,a)=1$ . Then,

mld ( X x ) = min k r + k a r 1 k r 1 , k N + . $$\begin{equation*} {\rm {mld}}(X\ni x)=\min {\left\lbrace \frac{k}{r}+{\left\lbrace \frac{ka}{r}\right\rbrace} \mid 1\le k\le r-1,k\in \mathbb {N}^+\right\rbrace} . \end{equation*}$$

Lemma 2.13.Let ( X x , 0 ) $(X\ni x,0)$ and ( Y y , 0 ) $(Y\ni y,0)$ be two klt surface germs such that DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ is a subgraph of DG ( Y y , 0 ) $\mathcal {DG}(Y\ni y,0)$ . Then, mld ( X x ) mld ( Y y , 0 ) ${\rm {mld}}(X\ni x)\ge {\rm {mld}}(Y\ni y,0)$ .

Proof.Let f : W Y $f: W\rightarrow Y$ be a partial resolution, which extracts all divisors corresponding to vertices contained in DG ( Y y , 0 ) DG ( X x , 0 ) $\mathcal {DG}(Y\ni y,0)\backslash \mathcal {DG}(X\ni x,0)$ . Then, ( X x ) ( W w ) $(X\ni x)\cong (W\ni w)$ for some w W $w\in W$ . Since f K Y = K W + B W $f^*K_Y=K_W+B_W$ for some B W 0 $B_W\ge 0$ , we have

mld ( Y y , 0 ) mld ( W w , B W ) mld ( W w , 0 ) = mld ( X x , 0 ) . $$\begin{equation*} {\rm {mld}}(Y\ni y,0)\le {\rm {mld}}(W\ni w,B_W)\le {\rm {mld}}(W\ni w,0)={\rm {mld}}(X\ni x,0). \end{equation*}$$
$\Box$

Lemma 2.14.Let ( X x , 0 ) $(X\ni x,0)$ be a 2 5 $\frac{2}{5}$ -klt surface singularity. Then, either ( X x ) 1 7 ( 1 , 2 ) $(X\ni x)\cong \frac{1}{7}(1,2)$ , or ( X x ) 1 4 ( 1 , 1 ) $(X\ni x)\cong \frac{1}{4}(1,1)$ , or the weight of any vertex of DG ( X x ) $\mathcal {DG}(X\ni x)$ is ⩽3.

Proof.By Lemma 2.9, the weight of any vertex of DG ( X x ) $\mathcal {DG}(X\ni x)$ is ⩽4. By [11, Theorem 4.7], DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ is connected and contains no cycle. We may assume that DG ( X x ) $\mathcal {DG}(X\ni x)$ contains a vertex of weight 4. We have the following cases.

Case 1. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ only contains one point. Then, ( X x ) 1 4 ( 1 , 1 ) $(X\ni x)\cong \frac{1}{4}(1,1)$ and we are done.

Case 2. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ contains the subgraph G n $\mathcal {G}_n$ :

image
for some n 3 $n\ge 3$ . By Lemma 2.11, the singularity corresponding to the dual graph G n $\mathcal {G}_n$ is a cyclic quotient singularity of type 1 4 n 1 ( 1 , 4 ) $\frac{1}{4n-1}(1,4)$ . By Lemma 2.12, when n 4 $n\ge 4$ ,
mld 1 4 n 1 ( 1 , 4 ) 5 4 n 1 1 3 < 2 5 , $$\begin{equation*} {\rm {mld}}{\left(\frac{1}{4n-1}(1,4)\right)}\le \frac{5}{4n-1}\le \frac{1}{3}&lt;\frac{2}{5}, \end{equation*}$$
and when n = 3 $n=3$ ,
mld 1 4 n 1 ( 1 , 4 ) = mld 1 11 ( 1 , 4 ) = 4 11 < 2 5 . $$\begin{equation*} {\rm {mld}}{\left(\frac{1}{4n-1}(1,4)\right)}={\rm {mld}}{\left(\frac{1}{11}(1,4)\right)}=\frac{4}{11}&lt;\frac{2}{5}. \end{equation*}$$
We get a contradiction to Lemma 2.13.

Case 3. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ contains the subgraph G 2 $\mathcal {G}_2$ :

image
but does not contain the subgraph G n $\mathcal {G}_n$ as in Case 1.2 for any n 3 $n\ge 3$ . We have the following cases.

Case 3.1. DG ( X x , 0 ) = G 2 $\mathcal {DG}(X\ni x,0)=\mathcal {G}_2$ . By Lemma 2.11, ( X x ) $(X\ni x)$ is a cyclic quotient singularity of type 1 7 ( 1 , 2 ) $\frac{1}{7}(1,2)$ and we are done.

Case 3.2. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ contains a subgraph H $\mathcal {H}$ :

image
By Lemma 2.11, the singularity corresponds to H , $\mathcal {H,}$ which is a cyclic quotient singularity of type 1 12 ( 1 , 7 ) $\frac{1}{12}(1,7)$ . Since mld 1 12 ( 1 , 7 ) = 1 3 < 2 5 ${\rm {mld}}\left(\frac{1}{12}(1,7)\right)=\frac{1}{3}&lt;\frac{2}{5}$ , we get a contradiction to Lemma 2.13.

Case 3.3. DG ( X x , 0 ) $\mathcal {DG}(X\ni x,0)$ contains a subgraph L n $\mathcal {L}_n$ :

image
for some integer n 2 $n\ge 2$ . By Lemma 2.11, singularity corresponds to L n $\mathcal {L}_n$ , which is a cyclic quotient singularity of type 1 7 n 4 ( 1 , 7 ) $\frac{1}{7n-4}(1,7)$ . By Lemma 2.12, when n 4 $n\ge 4$ ,
mld 1 7 n 4 ( 1 , 7 ) 8 7 n 4 1 3 < 2 5 . $$\begin{equation*} {\rm {mld}}{\left(\frac{1}{7n-4}(1,7)\right)}\le \frac{8}{7n-4}\le \frac{1}{3}&lt;\frac{2}{5}. \end{equation*}$$
When n = 3 $n=3$ ,
mld 1 7 n 4 ( 1 , 7 ) = mld 1 17 ( 1 , 7 ) = mld 1 17 ( 5 , 1 ) = 6 17 < 2 5 . $$\begin{equation*} {\rm {mld}}{\left(\frac{1}{7n-4}(1,7)\right)}={\rm {mld}}{\left(\frac{1}{17}(1,7)\right)}={\rm {mld}}{\left(\frac{1}{17}(5,1)\right)}=\frac{6}{17}&lt;\frac{2}{5}. \end{equation*}$$
When n = 2 $n=2$ ,
mld 1 7 n 4 ( 1 , 7 ) = mld 1 10 ( 1 , 7 ) = mld 1 10 ( 3 , 1 ) = 2 5 . $$\begin{equation*} {\rm {mld}}{\left(\frac{1}{7n-4}(1,7)\right)}={\rm {mld}}{\left(\frac{1}{10}(1,7)\right)}={\rm {mld}}{\left(\frac{1}{10}(3,1)\right)}=\frac{2}{5}. \end{equation*}$$
We get a contradiction to Lemma 2.13. $\Box$

Lemma 2.15.Let ( X x ) $(X\ni x)$ be a 2 5 $\frac{2}{5}$ -klt surface germ and f : Y X $f: Y\rightarrow X$ the minimal resolution of X x $X\ni x$ . Suppose that

K Y + i = 1 n a i E i = f K X , $$\begin{equation*} K_Y+\sum _{i=1}^na_iE_i=f^*K_X, \end{equation*}$$
where E 1 , , E n $E_1,\dots ,E_n$ are the prime exceptional divisors of f. Then, K Y · i = 1 n a i E i n $K_Y\cdot \sum _{i=1}^na_iE_i\le n$ .

Proof.By Lemma 2.14, there are three cases.

Case 1. E i 2 3 $E_i^2\ge -3$ for each i. Since ( K Y + E i ) · E i = 2 $(K_Y+E_i)\cdot E_i=-2$ for each i, K Y · E i 1 $K_Y\cdot E_i\le 1$ for each i. Since a i < 1 $a_i&lt;1$ for each i, the lemma follows.

Case 2. n = 1 $n=1$ and E 1 2 = 4 $E_1^2=-4$ . Then, K Y · E 1 = 2 $K_Y\cdot E_1=2$ and ( K Y + a 1 E 1 ) · E 1 = 0 $(K_Y+a_1E_1)\cdot E_1=0$ , hence a 1 = 1 2 $a_1=\frac{1}{2}$ . We have K Y · i = 1 n a i E i = 1 2 K Y · E 1 = 1 = n $K_Y\cdot \sum _{i=1}^na_iE_i=\frac{1}{2}K_Y\cdot E_1=1=n$ .

Case 3. n = 2 $n=2$ , and possibly reordering indices, E 1 2 = 2 $E_1^2=-2$ and E 2 2 = 4 $E_2^2=-4$ . Then, K Y · E 1 = 0 $K_Y\cdot E_1=0$ , K Y · E 2 = 2 $K_Y\cdot E_2=2$ , ( K Y + a 1 E 1 + a 2 E 2 ) · E 1 = 0 $(K_Y+a_1E_1+a_2E_2)\cdot E_1=0$ , and ( K Y + a 1 E 1 + a 2 E 2 ) · E 2 = 0 $(K_Y+a_1E_1+a_2E_2)\cdot E_2=0$ . Thus, a 1 = 2 7 $a_1=\frac{2}{7}$ and a 2 = 4 7 $a_2=\frac{4}{7}$ , hence K Y · i = 1 n a i E i = 8 7 < 2 = n $K_Y\cdot \sum _{i=1}^na_iE_i=\frac{8}{7}&lt;2=n$ . $\Box$

Lemma 2.16.Let X x $X\ni x$ be a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ for some positive integer k. Then, DG ( X x ) $\mathcal {DG}(X\ni x)$ is the following graph, where there are k 1 $k-1$ “ 2” in the graph.

image
Moreover, let E 1 , , E k $E_1,\dots ,E_k$ be the divisors corresponding to the vertices of DG ( X x ) $\mathcal {DG}(X\ni x)$ , such that E i 2 = 2 $E_i^2=-2$ , when 1 i k 1 $1\le i\le k-1$ , E k 2 = 3 $E_k^2=-3$ , and E i · E j 0 $E_i\cdot E_j\not=0$ if and only if | i j | 1 $|i-j|\le 1$ . Then, a ( E i , X , 0 ) = 2 k + 1 i 2 k + 1 $a(E_i,X,0)=\frac{2k+1-i}{2k+1}$ for each i.

Proof.It is clear that the cyclic quotient singularity is uniquely determined by its dual graph. Since 2 k + 1 k = 3 1 2 1 2 1 2 $\frac{2k+1}{k}=3-\frac{1}{2-\frac{1}{2-\frac{1}{\dots 2}}}$ where there are k 1 $k-1$ “2” in the fraction, the first part of the lemma follows from Lemma 2.11. For the remaining part of the lemma, let a i : = 1 a ( E i , X , 0 ) $a_i:=1-a(E_i,X,0)$ for each i. Since K Y · E i = 2 E i 2 $K_Y\cdot E_i=-2-E_i^2$ for each i and ( K Y + i = 1 k a i E i ) · E i = 0 $(K_Y+\sum _{i=1}^ka_iE_i)\cdot E_i=0$ for each i, when k = 1 $k=1$ , a ( E 1 , x , 0 ) = 2 3 $a(E_1,x,0)=\frac{2}{3}$ and we are done, and when k 2 $k\ge 2$ , we have

  • (1) 2 a 1 = a 2 $2a_1=a_2$ ,
  • (2) 2 a i = a i 1 + a i + 1 $2a_i=a_{i-1}+a_{i+1}$ for any 2 i k 1 $2\le i\le k-1$ , and
  • (3) 3 a k = a k 1 + 1 $3a_k=a_{k-1}+1$ .
Thus, a i = i a 1 $a_i=ia_1$ for each i, and we have
3 k a 1 = 3 a k = a k 1 + 1 = ( k 1 ) a 1 + 1 , $$\begin{equation*} 3ka_1=3a_k=a_{k-1}+1=(k-1)a_1+1, \end{equation*}$$
hence a 1 = 1 2 k + 1 $a_1=\frac{1}{2k+1}$ and a i = i 2 k + 1 $a_i=\frac{i}{2k+1}$ for each i. The lemma follows. $\Box$

3 GLOBAL GEOMETRY OF SMOOTH SURFACES

3.1 Some elementary lemmas

Lemma 3.1.Let X be a smooth projective surface, D a pseudo-effective R $\mathbb {R}$ -divisor on X, and C an irreducible curve on X. If D · C < 0 $D\cdot C&lt;0$ , then C 2 < 0 $C^2&lt;0$ .

Proof.Let D = P + N $D=P+N$ be the Zariski decomposition of D such that P is the positive part and N is the negative part. Since D · C < 0 $D\cdot C&lt;0$ and P is nef, N · C < 0 $N\cdot C&lt;0$ . Since N 0 $N\ge 0$ , C Supp N $C\subset \operatorname{Supp}N$ and C 2 < 0 $C^2&lt;0$ . $\Box$

Lemma 3.2.Let X be a smooth projective surface such that K X $K_X$ is pseudo-effective. Let C be an irreducible curve on X such that K X · C < 0 $K_X\cdot C&lt;0$ . Then, C 2 = K X · C = 1 $C^2=K_X\cdot C=-1$ . In particular, C is a smooth rational curve.

Proof.By Lemma 3.1, C 2 < 0 $C^2&lt;0$ . Since X is smooth, K X · C 1 $K_X\cdot C\le -1$ and C 2 1 $C^2\le -1$ . Thus, ( K X + C ) · C 2 $(K_X+C)\cdot C\le -2$ , which implies that ( K X + C ) · C = 2 $(K_X+C)\cdot C=-2$ , C 2 = K X · C = 1 $C^2=K_X\cdot C=-1$ , and C is a smooth rational curve. $\Box$

Lemma 3.3.Let X be a smooth projective surface such that K X $K_X$ is pseudo-effective, and C a smooth rational curve on X. Then, C 2 1 $C^2\le -1$ .

Proof.If not, then C 2 0 $C^2\ge 0$ . Since ( K X + C ) · C = 2 $(K_X+C)\cdot C=-2$ , K X · C 2 < 0 $K_X\cdot C\le -2&lt;0$ . Since K X $K_X$ is pseudo-effective, C 2 < 0 $C^2&lt;0$ , a contradiction. $\Box$

Lemma 3.4.Let X be a smooth projective surface, C an irreducible curve on X, f : Y X $f: Y\rightarrow X$ a blow-up of a closed point, E the exceptional divisor of f, and C Y $C_Y$ the strict transform of C on Y. If C Y · E 1 $C_Y\cdot E\le 1$ and C Y $C_Y$ is a smooth rational curve, then C is a smooth rational curve.

Proof.Since X is smooth, Y is smooth. Thus, C Y · E { 0 , 1 } $C_Y\cdot E\in \lbrace 0,1\rbrace$ . If C Y · E = 0 $C_Y\cdot E=0$ , then f is an isomorphism near a neighborhood of C Y $C_Y$ and hence C is a smooth rational curve. If C Y · E = 1 $C_Y\cdot E=1$ , then K X · C = K Y · C Y 1 $K_X\cdot C=K_Y\cdot C_Y-1$ and C 2 = C Y 2 + 1 $C^2=C_Y^2+1$ , and hence ( K X + C ) · C = ( K Y + C Y ) · C Y = 2 $(K_X+C)\cdot C=(K_Y+C_Y)\cdot C_Y=-2$ . Thus, C is a smooth rational curve. $\Box$

Lemma 3.5.Let X be a smooth projective surface such that K X $K_X$ is pseudo-effective, and E 1 , E 2 $E_1,E_2$ two different smooth rational curves on X such that E 1 2 = E 2 2 = 1 $E_1^2=E_2^2=-1$ . Then, E 1 · E 2 = 0 $E_1\cdot E_2=0$ .

Proof.Assume that E 1 · E 2 0 $E_1\cdot E_2\not=0$ , then E 1 · E 2 = n 1 $E_1\cdot E_2=n\ge 1$ for some positive integer n. Let f : X Y $f: X\rightarrow Y$ be the contraction of E1 and E 2 , Y : = f E 2 $E_{2,Y}:=f_*E_2$ . Then, E 2 , Y 2 = 1 + n 2 0 $E_{2,Y}^2=-1+n^2\ge 0$ and K Y · E 2 , Y = 1 n < 0 $K_Y\cdot E_{2,Y}=-1-n&lt;0$ . Since K X $K_X$ is pseudo-effective, K Y $K_Y$ is pseudo-effective, which contradicts Lemma 3.1. $\Box$

Lemma 3.6.Let X be a smooth projective surface such that K X $K_X$ is pseudo-effective, and E 1 , E 2 , E 3 $E_1,E_2,E_3$ three different smooth rational curves on X. If E 1 2 = E 3 2 = 2 $E_1^2=E_3^2=-2$ and E 2 2 = 1 $E_2^2=-1$ , then either E 1 · E 2 = 0 $E_1\cdot E_2=0$ or E 2 · E 3 = 0 $E_2\cdot E_3=0$ .

Proof.Assume that E 1 · E 2 = n 1 > 0 $E_1\cdot E_2=n_1&gt;0$ and E 2 · E 3 = n 3 > 0 $E_2\cdot E_3=n_3&gt;0$ for some positive integers n1 and n3. Let f : X Y $f: X\rightarrow Y$ be the contraction of E2. Then, Y is smooth and K Y $K_Y$ is pseudo-effective. Let E 1 , Y : = f E 1 $E_{1,Y}:=f_*E_1$ , and E 3 , Y : = f E 3 $E_{3,Y}:=f_*E_3$ . Then, E 1 , Y 2 = 2 + n 1 2 $E_{1,Y}^2=-2+n_1^2$ , E 3 , Y 2 = 2 + n 3 2 $E_{3,Y}^2=-2+n_3^2$ , K Y · E 1 , Y = n 1 $K_Y\cdot E_{1,Y}=-n_1$ , and K Y · E 3 , Y = n 3 $K_Y\cdot E_{3,Y}=-n_3$ . Thus, by Lemma 3.1, n 1 = n 3 = 1 $n_1=n_3=1$ , which implies that E 1 , Y 2 = E 3 , Y 2 = 1 $E_{1,Y}^2=E_{3,Y}^2=-1$ and E 1 , Y · E 3 , Y > 0 $E_{1,Y}\cdot E_{3,Y}&gt;0$ . By Lemma 3.4, E 1 , Y $E_{1,Y}$ and E 3 , Y $E_{3,Y}$ are smooth rational curves, which contradicts Lemma 3.5. $\Box$

Lemma 3.7.Let X be a smooth rational surface. Then, K X 2 = 10 ρ ( X ) $K_X^2=10-\rho (X)$ .

Proof.We may run a K X $K_X$ -MMP f : X : = X 0 f 1 X 1 f 2 f n X n $f: X:=X_0\xrightarrow {f_1}X_1\xrightarrow {f_2}\dots \xrightarrow {f_n}X_n$ such that either X n = F k $X_n=\mathbb {F}_k$ for some nonnegative integer k or X n = P 2 $X_n=\mathbb {P}^2$ . For any i { 0 , 1 , 2 , , n 1 } $i\in \lbrace 0,1,2,\dots ,n-1\rbrace$ , we have K X i 2 = K X i + 1 2 1 $K_{X_i}^2=K_{X_{i+1}}^2-1$ and ρ ( X i ) = ρ ( X i + 1 ) + 1 $\rho (X_i)=\rho (X_{i+1})+1$ . Thus, K X 2 + ρ ( X ) = K X n 2 + ρ ( X n ) $K_X^2+\rho (X)=K_{X_n}^2+\rho (X_n)$ . If X n = F k $X_n=\mathbb {F}_k$ for some nonnegative integer k, then K X n 2 + ρ ( X n ) = 8 + 2 = 10 $K_{X_n}^2+\rho (X_n)=8+2=10$ . If X n = P 2 $X_n=\mathbb {P}^2$ , then K X n 2 + ρ ( X n ) = 9 + 1 = 10 $K_{X_n}^2+\rho (X_n)=9+1=10$ . Thus, K X 2 = 10 ρ ( X ) $K_X^2=10-\rho (X)$ . $\Box$

3.2 Zariski decomposition

Lemma 3.8.Let X be a smooth projective surface, and D, D $\tilde{D}$ two Q $\mathbb {Q}$ -divisors on X, such that D D $D\ge \tilde{D}$ and D $\tilde{D}$ is nef. Let D = P + N $D=P+N$ be the Zariski decomposition of D, where P is the positive part and N is the negative part. Then, P D $P\ge \tilde{D}$ .

Proof.Assume that N = i = 1 n a i C i $N=\sum _{i=1}^na_iC_i$ and D D = i = 1 n b i C i + D 0 $D-\tilde{D}=\sum _{i=1}^nb_iC_i+D_0$ , where n is a nonnegative integer, C i $C_i$ are distinct irreducible curves, D 0 0 $D_0\ge 0$ , and for each i, a i > 0 $a_i&gt;0$ , b i 0 $b_i\ge 0$ , and C i Supp D 0 $C_i\not\subset \operatorname{Supp}D_0$ . Then, for every j { 1 , 2 , , n } $j\in \lbrace 1,2,\dots ,n\rbrace$ ,

i = 1 n a i ( C i · C j ) = N · C j = D · C j = D · C j + ( D D ) · C j ( D D ) · C j = i = 1 n b i ( C i · C j ) + D 0 · C j i = 1 n b i ( C i · C j ) , $$\begin{align*} \sum _{i=1}^na_i(C_i\cdot C_j)&=N\cdot C_j=D\cdot C_j=\tilde{D}\cdot C_j+(D-\tilde{D})\cdot C_j\\ &\ge (D-\tilde{D})\cdot C_j=\sum _{i=1}^nb_i(C_i\cdot C_j)+D_0\cdot C_j\ge \sum _{i=1}^nb_i(C_i\cdot C_j), \end{align*}$$
which implies that i = 1 n ( a i b i ) ( C i · C j ) 0 $\sum _{i=1}^n(a_i-b_i)(C_i\cdot C_j)\ge 0$ for every j. Since the intersection matrix { ( C i · C j ) } 1 i , j n $\lbrace (C_i\cdot C_j)\rbrace _{1\le i,j\le n}$ is negative definite, a i b i $a_i\le b_i$ for each i. Thus, D D N $D-\tilde{D}\ge N$ , hence P D $P\ge \tilde{D}$ . $\Box$

Lemma 3.9.Let X be a smooth projective surface, D a big Weil divisor on X, D $\tilde{D}$ an nef Weil divisor on X, and E a Weil divisor on X, such that

  • (1) D = P + N $D=P+N$ is the Zariski decomposition of D, where P is the positive part and N 0 $N\ge 0$ is the negative part,
  • (2) E = D D 0 $E=D-\tilde{D}\ge 0$ , and
  • (3) | D | $|D|$ defines a birational map.
Then, there exist a big Weil divisor D1 on X and a Weil divisor E1 on X, such that
  • 1. D 1 = P $D_1=\lfloor P\rfloor$ ,
  • 2. E E 1 = D 1 D 0 $E\ge E_1=D_1-\tilde{D}\ge 0$ ,
  • 3. | D 1 | $|D_1|$ defines a birational map, and
  • 4. either N = 0 $N=0$ and D = P $D=P$ , or there exists at least one irreducible component F of Supp E $\operatorname{Supp}E$ such that mult F ( E E 1 ) 1 $\operatorname{mult}_F(E-E_1)\ge 1$ .

Proof.We let D 1 : = P $D_1:=\lfloor P\rfloor$ , then (1) holds. Let E 1 : = D 1 D $E_1:=D_1-\tilde{D}$ . Since D $\tilde{D}$ is nef and D D $D\ge \tilde{D}$ , by Lemma 3.8, P D $P\ge \tilde{D}$ . Thus, P D 0 $P-\tilde{D}\ge 0$ , and hence

E 1 = D 1 D = P D = P D 0 . $$\begin{equation*} E_1=D_1-\tilde{D}=\lfloor P\rfloor -\tilde{D}=\lfloor P-\tilde{D}\rfloor \ge 0. \end{equation*}$$
Since
E E 1 = D D 1 = P + N P = { P } + N 0 , $$\begin{equation*} E-E_1=D-D_1=P+N-\lfloor P\rfloor =\lbrace P\rbrace +N\ge 0, \end{equation*}$$
we deduce (2). Since | D 1 | = | P | = | P | | D | $|D_1|=|\lfloor P\rfloor |=|P|\cong |D|$ , | D 1 | $|D_1|$ defines a birational map, hence (3). Finally, if E E 1 0 $E-E_1\not=0$ , then we are done; otherwise, E E 1 = 0 $E-E_1=0$ , hence { P } + N = 0 $\lbrace P\rbrace +N=0$ . Thus, N = 0 $N=0$ , which implies that D = P $D=P$ , hence (4). $\Box$

Proposition 3.10.Let X be a smooth projective surface, D a big Weil divisor on X, and D $\tilde{D}$ an nef Weil divisor on X, such that

  • (1) D = P + N $D=P+N$ is the Zariski decomposition of D, where P is the positive part and N 0 $N\ge 0$ is the negative part,
  • (2) D D 0 $D-\tilde{D}\ge 0$ , and
  • (3) | D | $|D|$ defines a birational map.
Then, there exists a Weil divisor D $D^{\prime }$ on X, such that
  • 1. D D D $D\ge D^{\prime }\ge \tilde{D}$ ,
  • 2. D $D^{\prime }$ defines a birational map, and
  • 3. D $D^{\prime }$ is big and nef.

Proof.Let D 0 : = D $D_0:=D$ , P 0 : = P , N 0 : = N $P_0:=P,N_0:=N$ , and E 0 : = D D $E_0:=D-\tilde{D}$ , and let r0 be the sum of all the coefficients of E0. Then, r0 is a nonnegative integer.

For any nonnegative integer k, assume that there exist big Weil divisors D 1 , , D k $D_1,\dots ,D_k$ on X, Weil divisors E 1 , , E k $E_1,\dots ,E_k$ on X, and nonnegative integers r 1 , , r k $r_1,\dots ,r_k$ , such that for every i { 0 , 1 , , k } $i\in \lbrace 0,1,\dots ,k\rbrace$ ,

  • (1) D i = P i + N i $D_i=P_i+N_i$ is the Zariski decomposition of D i $D_i$ , where P i $P_i$ is the positive part and N i 0 $N_i\ge 0$ is the negative part;
  • (2) E 0 E i = D i D 0 $E_0\ge E_i=D_i-\tilde{D}\ge 0$ ;
  • (3) | D i | $|D_i|$ defines a birational map;
  • (4) r k $r_k$ is the sum of all the coefficients of the components of E i $E_i$ such that 0 r k r 0 k $0\le r_k\le r_0-k$ ; and
  • (5) if i 1 $i\ge 1$ , then D i = P i 1 $D_i=\lfloor P_{i-1}\rfloor$ .
It is clear that these assumptions hold when k = 0 $k=0$ . By Lemma 3.9, there are two cases:

Case 1. N k = 0 $N_k=0$ and D k = P k $D_k=P_k$ . In this case, by our assumptions,

  • (1) D k D 0 $D_k-\tilde{D}\ge 0$ , hence D k D $D_k\ge \tilde{D}$ ;
  • (2) E 0 D k D $E_0\ge D_k-\tilde{D}$ , hence D D k $D\ge D_k$ ;
  • (3) D k $D_k$ is big and defines a birational map; and
  • (4) D k = P k $D_k=P_k$ is nef.
Thus, we may let D : = D k $D^{\prime }:=D_k$ .

Case 2. There exists a big Weil divisor D k + 1 $D_{k+1}$ on X, a Weil divisor E k + 1 $E_{k+1}$ on X, and a nonnegative integer r k + 1 $r_{k+1}$ , such that

  • (1) D k + 1 = P k $D_{k+1}=\lfloor P_{k}\rfloor$ ,
  • (2) E 0 E k + 1 = D k + 1 D 0 $E_0\ge E_{k+1}=D_{k+1}-\tilde{D}\ge 0$ ,
  • (3) | D k + 1 | $|D_{k+1}|$ defines a birational map, and
  • (4) 0 r k + 1 r k 1 $0\le r_{k+1}\le r_k-1$ .
In this case, we may replace k with k + 1 $k+1$ and apply induction on k. Since 0 r k r 0 k $0\le r_k\le r_0-k$ , we have k r 0 $k\le r_0$ . Thus, this process must terminate and we are done. $\Box$

3.3 Effective birationality and existence of special nef Q $\mathbb {Q}$ -divisors

Lemma 3.11.Let X be a klt projective surface such that K X $K_X$ is big and nef, f : Y X $f: Y\rightarrow X$ the minimal resolution of X, and E 1 , , E n $E_1,\dots ,E_n$ the prime f-exceptional divisors. Assume that K Y + i = 1 n a i E i = f K X $K_Y+\sum _{i=1}^na_iE_i=f^*K_X$ . Then, for any positive integer m, if there exist integers r 1 , , r n $r_1,\dots ,r_n$ , such that

  • 1. 0 r i m a i $0\le r_i\le \lfloor ma_i\rfloor$ , and
  • 2. K Y + i = 1 n r i m E i $K_Y+\sum _{i=1}^n\frac{r_i}{m}E_i$ is big and nef,
then | 192 m K X | $|192mK_X|$ does not have a fixed part.

Proof.Let Δ : = i = 1 n r i m E i $\Delta :=\sum _{i=1}^n\frac{r_i}{m}E_i$ and L : = m ( K Y + Δ ) $L:=m(K_Y+\Delta )$ . Then, L is big and nef and Cartier. In particular, 2 L ( K Y + Δ ) Q 2 1 m L $2L-(K_Y+\Delta )\sim _{\mathbb {Q}}\left(2-\frac{1}{m}\right)L$ is big and nef. By [6, Theorem 1.1, Remark 1.2] (see also [9, 1.1 Theorem]), 192L is base-point-free, which implies that the fixed part of 192 m ( K Y + i = 1 n a i E i ) $192m(K_Y+\sum _{i=1}^na_iE_i)$ is supported on i = 1 n E i $\cup _{i=1}^nE_i$ . Thus, | 192 m K X | $|192mK_X|$ does not have a fixed part. $\Box$

Theorem 3.12. (cf. [[8], Theorem 1.3])There exists a uniform positive integer m1, such that for any lc surface X such that K X $K_X$ is big, | m 1 K X | $|m_1K_X|$ defines a birational map.

4 1 3 $\frac{1}{3}$ -KLT SURFACES

In this section, we will prove Theorem 1.1. The structure of this section is as follows. In Section 4.1, we give a detailed classification of 1 3 $\frac{1}{3}$ -klt surface singularities. In Section 4.2, we consider the intersection numbers of the form K X · C $K_X\cdot C$ where X is 1 3 $\frac{1}{3}$ -klt, K X $K_X$ is big and nef, and C is a curve satisfying special properties. For some lemmas and propositions, we need to restrict ourselves to 2 5 $\frac{2}{5}$ -klt surfaces. With a good description of these intersection numbers and with the help of the results on Zariski decomposition in Section 3, in Section 4.3, we will construct special nef Q $\mathbb {Q}$ -divisors on the minimal resolution of 2 5 $\frac{2}{5}$ -klt surfaces. We will prove our main theorem in Section 4.4.

4.1 Classification of ( 1 3 + ε ) $(\frac{1}{3}+\epsilon )$ -lc singularities

Lemma 4.1.Let ε be a positive real number. Then, there exists a positive integer n 0 = n 0 ( ε ) $n_0=n_0(\epsilon )$ depending only on ε satisfying the following. Assume that ( X x , 0 ) $(X\ni x,0)$ is a ( 1 3 + ε ) $(\frac{1}{3}+\epsilon )$ -lc surface germ. Then,

  • 1. either n 0 K X $n_0K_X$ is Cartier near x, or
  • 2. X x $X\ni x$ is a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ for some positive integer k 10 $k\ge 10$ . In particular, DG ( X x ) $\mathcal {DG}(X\ni x)$ is the following graph, where there are k 1 $k-1$ “ 2” in the graph.
    image

Proof.Assume that the lemma does not hold. Then, there exists a sequence of ( 1 3 + ε ) $(\frac{1}{3}+\epsilon )$ -lc surface germs ( X i x i , 0 ) $(X_i\ni x_i,0)$ , and a strictly increasing sequence of positive integer n i $n_i$ , such that

  • (1) n K X i $nK_{X_i}$ is not Cartier near x i $x_i$ for any positive integer n n i $n\le n_i$ , and
  • (2) X i x i $X_i\ni x_i$ is not a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ for any i and any positive integer k.
We consider the set A : = { mld ( X i x i ) } i = 1 + $\mathcal {A}:=\lbrace {\rm {mld}}(X_i\ni x_i)\rbrace _{i=1}^{+\infty }$ . Since mld ( X i x i ) 1 3 + ε ${\rm {mld}}(X_i\ni x_i)\ge \frac{1}{3}+\epsilon$ , by Theorem 2.7, the only possible accumulation point of A $\mathcal {A}$ is 1 2 $\frac{1}{2}$ . If A $\mathcal {A}$ is a finite set, it contradicts Proposition 2.8. Thus, possibly passing to a subsequence and replacing A $\mathcal {A}$ , we may assume that mld ( X i x i ) ${\rm {mld}}(X_i\ni x_i)$ is strictly decreasing and lim i + mld ( X i x i ) = 1 2 $\lim _{i\rightarrow +\infty }{\rm {mld}}(X_i\ni x_i)=\frac{1}{2}$ .

We let G : = G ( 1 3 + ε ) $\mathcal {G}:=\mathcal {G}(\frac{1}{3}+\epsilon )$ be the finite set of dual graphs and I 0 : = I 0 ( 1 3 + ε ) $\mathcal {I}_0:=\mathcal {I}_0(\frac{1}{3}+\epsilon )$ be the finite set of real numbers as in Lemma 2.10. Then, for any ( X x ) $(X\ni x)$ such that DG ( X x , 0 ) G $\mathcal {DG}(X\ni x,0)\in \mathcal {G}$ , mld ( X x ) ${\rm {mld}}(X\ni x)$ belongs to a finite set. Thus, possibly passing to a subsequence, by Lemma 2.10, we may assume that one of the following holds:

  • (1) ( X i x i ) $(X_i\ni x_i)$ satisfies (2) of Lemma 2.10 for each i, and e 1 = e 1 ( X i x i ) , q 1 = q 1 ( X i x i ) $e_1=e_1(X_i\ni x_i),q_1=q_1(X_i\ni x_i)$ , e 2 = e 2 ( X i x i ) , q 2 = q 2 ( X i x i ) I 0 $e_2=e_2(X_i\ni x_i),q_2=q_2(X_i\ni x_i)\in \mathcal {I}_0$ for each i. Since I 0 $\mathcal {I}_0$ is a finite set, possibly passing to a subsequence, we may assume that e 1 , e 2 , q 1 , q 2 $e_1,e_2,q_1,q_2$ are constants for each i.
  • (2) ( X i x i ) $(X_i\ni x_i)$ satisfies (3) of Lemma 2.10 for each i, and e 1 = e 1 ( X i x i ) , q 1 = q 1 ( X i x i ) I 0 $e_1=e_1(X_i\ni x_i),q_1=q_1(X_i\ni x_i)\in \mathcal {I}_0$ for each i. Since I 0 $\mathcal {I}_0$ is a finite set, possibly passing to a subsequence, we may assume that e 1 , e 2 , q 1 , q 2 $e_1,e_2,q_1,q_2$ are constants for each i.
If ( X i x i ) $(X_i\ni x_i)$ satisfies (3) of Lemma 2.10 for each i, and e 1 = e 1 ( X i x i ) , q 1 = q 1 ( X i x i ) $e_1=e_1(X_i\ni x_i),q_1=q_1(X_i\ni x_i)$ are constants for each i, then by Lemma 2.10(3), mld ( X i x i ) = 1 e 1 q 1 ${\rm {mld}}(X_i\ni x_i)=\frac{1}{e_1-q_1}$ is a constant, a contradiction.

If ( X i x i ) $(X_i\ni x_i)$ satisfies (2) of Lemma 2.10 for each i, and e 1 = e 1 ( X i x i ) , q 1 = q 1 ( X i x i ) $e_1=e_1(X_i\ni x_i),q_1=q_1(X_i\ni x_i)$ , e 2 = e 2 ( X i x i ) , q 2 = q 2 ( X i x i ) $e_2=e_2(X_i\ni x_i),q_2=q_2(X_i\ni x_i)$ are constants for each i, then by Lemma 2.10(2),

min 1 e 1 q 1 , 1 e 2 q 2 1 3 + ε . $$\begin{equation*} \min {\left\lbrace \frac{1}{e_1-q_1},\frac{1}{e_2-q_2}\right\rbrace} \ge \frac{1}{3}+\epsilon . \end{equation*}$$
Thus, e 1 q 1 2 $e_1-q_1\le 2$ and e 2 q 2 2 $e_2-q_2\le 2$ . We get a contradiction by enumerating possibilities as follows:

Case 1. q 1 = 1 $q_1=1$ . Then e 1 = 2 $e_1=2$ or 3.

Case 1.1 e 1 = 2 $e_1=2$ .

Case 1.1.1 q 2 = 1 $q_2=1$ . Then e 2 = 2 $e_2=2$ or 3.

Case 1.1.1.1 e 2 = 2 $e_2=2$ . In this case, all the weights in DG ( X i x i ) $\mathcal {DG}(X_i\ni x_i)$ are 2. Thus, mld ( X i x i ) = 1 ${\rm {mld}}(X_i\ni x_i)=1$ for every i, a contradiction.

Case 1.1.1.2 e 2 = 3 $e_2=3$ . In this case, for each i, the dual graph of X i x i $X_i\ni x_i$ is of the following form

image
By Lemma 2.11, X i x i $X_i\ni x_i$ is a cyclic quotient singularity of type 1 2 k i + 1 ( 1 , k i ) $\frac{1}{2k_i+1}(1,k_i)$ for some positive integer k i $k_i$ , and k i + $k_i\rightarrow +\infty$ when i + $i\rightarrow +\infty$ , a contradiction.

Case 1.1.2 q 2 2 $q_2\ge 2$ . In this case, there exist an integer w 2 3 $w_2\ge 3$ and a nonnegative integer d 2 < q 2 $d_2&lt;q_2$ , such that e 2 = w 2 q 2 d 2 $e_2=w_2q_2-d_2$ . Thus,

2 e 2 q 2 = ( w 2 1 ) q 2 d 2 ( w 2 2 ) q 2 + 1 q 2 + 1 3 , $$\begin{equation*} 2\ge e_2-q_2=(w_2-1)q_2-d_2\ge (w_2-2)q_2+1\ge q_2+1\ge 3, \end{equation*}$$
a contradiction.

Case 1.2 e 1 = 3 $e_1=3$ .

Case 1.2.1 q 2 = 1 $q_2=1$ . Then, e 2 = 2 $e_2=2$ or 3.

Case 1.2.1.1 e 2 = 2 $e_2=2$ . In this case, for each i, the dual graph of X i x i $X_i\ni x_i$ is of the following form

image
By Lemma 2.11, X i x i $X_i\ni x_i$ is a cyclic quotient singularity of type 1 2 k i + 1 ( 1 , k i ) $\frac{1}{2k_i+1}(1,k_i)$ for some positive integer k i $k_i$ , and k i + $k_i\rightarrow +\infty$ when i + $i\rightarrow +\infty$ , a contradiction.

Case 1.2.1.2 e 2 = 3 $e_2=3$ . In this case, for each i, the dual graph of X i x i $X_i\ni x_i$ is of the following form:

image
By Lemma 2.11, X i x i $X_i\ni x_i$ is a cyclic quotient singularity of type 1 4 k i + 8 ( 1 , 2 k i + 3 ) $\frac{1}{4k_i+8}(1,2k_i+3)$ for some nonnegative integer k i $k_i$ . By Lemma 2.12, mld ( X i x i ) = 1 2 ${\rm {mld}}(X_i\ni x_i)=\frac{1}{2}$ , a contradiction.

Case 1.2.2 q 2 2 $q_2\ge 2$ . In this case, exactly the same argument as in Case 1.1.2 holds and we get a contradiction.

Case 2. q 1 2 $q_1\ge 2$ . In this case, there exists an integer w 1 3 $w_1\ge 3$ and a nonnegative integer d 1 < q 1 $d_1&lt;q_1$ , such that e 1 = w 1 q 1 d 1 $e_1=w_1q_1-d_1$ . Thus,

2 e 1 q 1 = ( w 1 1 ) q 1 d 1 ( w 1 2 ) q 1 + 1 q 1 + 1 3 , $$\begin{equation*} 2\ge e_1-q_1=(w_1-1)q_1-d_1\ge (w_1-2)q_1+1\ge q_1+1\ge 3, \end{equation*}$$
a contradiction. $\Box$

4.2 Intersection numbers

Lemma 4.2.Let X be a projective klt surface such that K X $K_X$ is nef and f : Y X $f: Y\rightarrow X$ the minimal resolution of X. If X is not rational, then K Y $K_Y$ is pseudo-effective.

Proof.If X is not rational, Y is not rational. If K Y $K_Y$ is not pseudo-effective, then there exists a birational morphism g : Y W $g: Y\rightarrow W$ to a smooth projective surface W and a P 1 $\mathbb {P}^1$ -fibration h : W R $h: W\rightarrow R$ . Since Y is not a rational surface, g ( R ) 0 $g(R)\ge 0$ . Thus, for any exceptional curve F of f, F does not dominate R. Pick a general h-vertical curve Σ and let Σ Y $\Sigma _Y$ , Σ X $\Sigma _X$ be the strict transforms of Σ on Y and X, respectively. Then,

0 K X · Σ X = K Y · Σ Y = K W · Σ = 2 , $$\begin{equation*} 0\le K_X\cdot \Sigma _X=K_Y\cdot \Sigma _Y=K_W\cdot \Sigma =-2, \end{equation*}$$
a contradiction. $\Box$

Lemma 4.3.Let X be a 1 3 $\frac{1}{3}$ -klt surface such that K X $K_X$ is big and nef, C an irreducible curve on X, x C $x\in C$ a closed point, f : Y X $f: Y\rightarrow X$ the minimal resolution of X, and C Y $C_Y$ the strict transform of C on Y. Assume that

  • X is not a rational surface,
  • K Y · C Y < 0 $K_Y\cdot C_Y&lt;0$ ,
  • X x $X\ni x$ is a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ for some integer k 5 $k\ge 5$ , and
  • E 1 , , E k $E_1,\dots ,E_k$ are prime f-exceptional divisors over X x $X\ni x$ , such that
    • 1. E i 2 = 2 $E_i^2=-2$ when 1 i k 1 $1\le i\le k-1$ ,
    • 2. E k 2 = 3 $E_k^2=-3$ , and
    • 3. E i · E j 0 $E_i\cdot E_j\not=0$ if and only if | i j | 1 $|i-j|\le 1$ .
Then,
  • 1. C Y · E i = 0 $C_Y\cdot E_i=0$ when 1 i k 1 $1\le i\le k-1$ , and
  • 2. C Y · E k = 1 $C_Y\cdot E_k=1$ .

Proof.By Lemma 4.2, K Y $K_Y$ is pseudo-effective. By Lemma 3.2, K Y · C Y = 1 $K_Y\cdot C_Y=-1$ and C Y 2 = 1 $C_Y^2=-1$ . Moreover, each E i $E_i$ is a smooth rational curve. Let g : Y W $g: Y\rightarrow W$ be the contraction of C Y $C_Y$ and E i , W : = g E i $E_{i,W}:=g_*E_i$ for each i. Then, W is smooth and K W $K_W$ is pseudo-effective. $\Box$

Claim 4.4. C Y · E j 1 $C_Y\cdot E_j\le 1$ for every j { 1 , 2 , , k } $j\in \lbrace 1,2,\dots ,k\rbrace$ .

Proof of Claim 4.4.Suppose this is not the case, then there exists an integer n 2 $n\ge 2$ and an integer j { 1 , 2 , , k } $j\in \lbrace 1,2,\dots ,k\rbrace$ , such that C Y · E j = n $C_Y\cdot E_j=n$ . We have

E j , W 2 = E j 2 + n 2 3 + 4 1 $$\begin{equation*} E_{j,W}^2=E_j^2+n^2\ge -3+4\ge 1 \end{equation*}$$
and
K W · E j , W = K Y · E j n 1 , $$\begin{equation*} K_W\cdot E_{j,W}=K_Y\cdot E_j-n\le -1, \end{equation*}$$
which contradicts Lemma 3.1 as K W $K_W$ is pseudo-effective. $\Box$

Claim 4.5. E i , W $E_{i,W}$ are smooth rational curves for every i.

Proof of Claim 4.5.It immediately follows from Lemma 3.4 and Claim 4.4. $\Box$

Claim 4.6. C Y · E j = 0 $C_Y\cdot E_j=0$ for every j { 2 , 3 , , k 2 } $j\in \lbrace 2,3,\dots ,k-2\rbrace$ .

Proof of Claim 4.6.Suppose that the claim does not hold. Then, by Claim 4.4, there exists j { 2 , 3 , , k 2 } $j\in \lbrace 2,3,\dots ,k-2\rbrace$ such that C Y · E j = 1 $C_Y\cdot E_j=1$ . There are three cases:

Case 1. C Y · E j + 1 = 1 $C_Y\cdot E_{j+1}=1$ . In this case, E j , W 2 = E j + 1 , W 2 = 1 $E_{j,W}^2=E_{j+1,W}^2=-1$ and E j , W · E j + 1 , W = 2 $E_{j,W}\cdot E_{j+1,W}=2$ . By Claim 4.5, E j , W $E_{j,W}$ and E j + 1 , W $E_{j+1,W}$ are smooth rational curves. Since K W $K_W$ is pseudo-effective, this contradicts Lemma 3.5.

Case 2. C Y · E j 1 = 1 $C_Y\cdot E_{j-1}=1$ . We get a contradiction by the same arguments as Case 1 except that we replace E j + 1 $E_{j+1}$ with E j 1 $E_{j-1}$ .

Case 3. C Y · E j 1 = C Y · E j + 1 = 0 $C_Y\cdot E_{j-1}=C_Y\cdot E_{j+1}=0$ . In this case, E j , W 2 = 1 $E_{j,W}^2=-1$ , E j 1 , W 2 = E j + 1 , W 2 = 2 $E_{j-1,W}^2=E_{j+1,W}^2=-2$ , E j , W · E j 1 , W = E j , W · E j + 1 , W = 1 $E_{j,W}\cdot E_{j-1,W}=E_{j,W}\cdot E_{j+1,W}=1$ , which contradicts Lemma 3.6. $\Box$

Claim 4.7. C Y · E k 1 = 0 $C_Y\cdot E_{k-1}=0$ .

Proof of Claim 4.7.Suppose that the claim does not hold. Then, by Claim 4.4, C Y · E k 1 = 1 $C_Y\cdot E_{k-1}=1$ . By Claim 4.6, C Y · E j = 0 $C_Y\cdot E_j=0$ for every j { 2 , 3 , , k 2 } $j\in \lbrace 2,3,\dots ,k-2\rbrace$ . There are two cases:

Case 1. C Y · E k = 1 $C_Y\cdot E_k=1$ . In this case, E k 1 , W 2 = 1 $E_{k-1,W}^2=-1$ , E k , W 2 = 2 $E_{k,W}^2=-2$ , E k 2 , W 2 = 2 $E_{k-2,W}^2=-2$ , E k 1 , W · E k , W = 2 $E_{k-1,W}\cdot E_{k,W}=2$ , and E k 1 , W · E k 2 , W = 1 $E_{k-1,W}\cdot E_{k-2,W}=1$ . This contradicts Lemma 3.6.

Case 2. C Y · E k = 0 $C_Y\cdot E_k=0$ . In this case, E k 1 , W 2 = 1 $E_{k-1,W}^2=-1$ , E k , W 2 = 3 $E_{k,W}^2=-3$ , E k 2 , W 2 = E k 3 , W 2 = 2 $E_{k-2,W}^2=E_{k-3,W}^2=-2$ , and for every i , j { k 3 , k 2 , k 1 , k } $i,j\in \lbrace k-3,k-2,k-1,k\rbrace$ , E i · E j = 1 $E_i\cdot E_j=1$ if | i j | = 1 $|i-j|=1$ and E i · E j = 0 $E_i\cdot E_j=0$ if | i j | 2 $|i-j|\ge 2$ .

Let h : W Z $h: W\rightarrow Z$ be the contraction of E k 1 , W $E_{k-1,W}$ and E i , Z : = h E i , W $E_{i,Z}:=h_*E_{i,W}$ for any i k 1 $i\not=k-1$ . Then, Z is smooth and K Z $K_Z$ is pseudo-effective. By Lemma 3.4, E k 3 , Z , E k 2 , Z $E_{k-3,Z},E_{k-2,Z}$ , and E k , Z $E_{k,Z}$ are smooth rational curves. Moreover, E k 3 , Z 2 = E k , Z 2 = 2 $E_{k-3,Z}^2=E_{k,Z}^2=-2$ , E k 2 , Z 2 = 1 $E_{k-2,Z}^2=-1$ , and E k 3 , Z · E k 2 , Z = E k 2 , Z · E k , Z = 1 $E_{k-3,Z}\cdot E_{k-2,Z}=E_{k-2,Z}\cdot E_{k,Z}=1$ . This contradicts Lemma 3.6. $\Box$

Claim 4.8. C Y · E 1 = 0 $C_Y\cdot E_1=0$ .

Proof of Claim 4.8.Suppose that the claim does not hold. Then, by Claim 4.4, C Y · E 1 = 1 $C_Y\cdot E_1=1$ . By Claim 4.6 and Claim 4.7, C Y · E j = 0 $C_Y\cdot E_j=0$ for every j { 2 , , k 1 } $j\in \lbrace 2,\dots ,k-1\rbrace$ . By Claim 4.4, there are two cases:

Case 1. C Y · E k = 1 $C_Y\cdot E_k=1$ . In this case, E 1 , W 2 = 1 $E_{1,W}^2=-1$ , E 2 , W 2 = E k , W 2 = 2 $E_{2,W}^2=E_{k,W}^2=-2$ , E 1 , W · E 2 , W = E 1 , W · E k , W = 1 $E_{1,W}\cdot E_{2,W}=E_{1,W}\cdot E_{k,W}=1$ , which contradicts Lemma 3.6.

Case 2. C Y · E k = 0 $C_Y\cdot E_k=0$ . The are two subcases:

Case 2.1. For any closed point y C $y\in C$ such that y x $y\not=x$ , X is smooth near y. In this case, let a : = 1 a ( E 1 , X , 0 ) = 1 2 k + 1 $a:=1-a(E_1,X,0)=\frac{1}{2k+1}$ . Since K X $K_X$ is big and nef,

0 K X · C = f K X · C Y = ( K Y + ( 1 a ) E 1 ) · C Y = 1 + ( 1 a ) = a < 0 , $$\begin{equation*} 0\le K_X\cdot C=f^*K_X\cdot C_Y=(K_Y+(1-a)E_1)\cdot C_Y=-1+(1-a)=-a&lt;0, \end{equation*}$$
a contradiction.

Case 2.2. There exists a closed point y C $y\in C$ such that y x $y\not=x$ and X is not smooth near y. Then, there exists a prime divisor F on Y that is over X y $X\ni y$ , such that C Y F $C_Y\cap F\not=\emptyset$ . Moreover, F is a smooth rational curve. Since X is 1 3 $\frac{1}{3}$ -klt, by Lemma 2.9, F 2 5 $F^2\ge -5$ . Let F W : = g F $F_W:=g_*F$ .

We have F W · E i , W = 0 $F_W\cdot E_{i,W}=0$ for every i 1 $i\not=1$ , E 1 , W 2 = 1 $E_{1,W}^2=-1$ , E 2 , W 2 = E 3 , W 2 = E 4 , W 2 = 2 $E_{2,W}^2=E_{3,W}^2=E_{4,W}^2=-2$ , and for every i , j { 1 , 2 , 3 , 4 } $i,j\in \lbrace 1,2,3,4\rbrace$ , E i , W · E j , W = 1 $E_{i,W}\cdot E_{j,W}=1$ when | i j | = 1 $|i-j|=1$ and E i , W · E j , W = 0 $E_{i,W}\cdot E_{j,W}=0$ when | i j | 2 $|i-j|\ge 2$ .

There are two subcases:

Case 2.2.1. C Y · F = 1 $C_Y\cdot F=1$ . In this case, by Lemma 3.4, F W $F_W$ is a smooth rational curve. Moreover, F W 2 4 $F_W^2\ge -4$ and F W · E 1 , W = 1 $F_W\cdot E_{1,W}=1$ ,

Let h : W Z $h: W\rightarrow Z$ be the contraction of E 1 , W $E_{1,W}$ , E i , Z : = h E i , W $E_{i,Z}:=h_*E_{i,W}$ for each i 1 $i\not=1$ , and F Z : = h F W $F_Z:=h_*F_W$ . Then, Z is smooth and K Z $K_Z$ is pseudo-effective. By Lemma 3.4, E 2 , Z , E 3 , Z , E 4 , Z $E_{2,Z},E_{3,Z},E_{4,Z}$ , and F Z $F_Z$ are smooth rational curves. Moreover, E 2 , Z 2 = 1 $E_{2,Z}^2=-1$ , E 3 , Z 2 = E 4 , Z 2 = 2 $E_{3,Z}^2=E_{4,Z}^2=-2$ , E 2 , Z · E 3 , Z = E 3 , Z · E 4 , Z = F Z · E 2 , Z = 1 $E_{2,Z}\cdot E_{3,Z}=E_{3,Z}\cdot E_{4,Z}=F_Z\cdot E_{2,Z}=1$ , F 2 , Z · E 3 , Z = F 2 , Z · E 4 , Z = E 2 , Z · E 4 , Z = 0 $F_{2,Z}\cdot E_{3,Z}=F_{2,Z}\cdot E_{4,Z}=E_{2,Z}\cdot E_{4,Z}=0$ , and F Z 2 3 $F_Z^2\ge -3$ .

Let p : Z T $p: Z\rightarrow T$ be the contraction of E 2 , Z $E_{2,Z}$ , E i , T : = p E i , Z $E_{i,T}:=p_*E_{i,Z}$ for each i 1 , 2 $i\not=1,2$ , and F T : = p F Z $F_T:=p_*F_Z$ . Then, T is smooth and K T $K_T$ is pseudo-effective. By Lemma 3.4, E 3 , T , E 4 , T $E_{3,T},E_{4,T}$ , and F T $F_T$ are smooth rational curves. Moreover, E 3 , T 2 = 1 $E_{3,T}^2=-1$ , E 4 , T 2 = 2 $E_{4,T}^2=-2$ , F T 2 2 $F_T^2\ge -2$ , and E 3 , T · E 4 , T = F T · E 3 , T = 1 $E_{3,T}\cdot E_{4,T}=F_T\cdot E_{3,T}=1$ .

By Lemma 3.3, F T 2 { 1 , 2 } $F_T^2\in \lbrace -1,-2\rbrace$ . By Lemma 3.5, F T 2 = 2 $F_T^2=-2$ . But this contradicts Lemma 3.6.

Case 2.2.2. C Y · F 2 $C_Y\cdot F\ge 2$ . In this case, we let b : = F 2 $b:=F^2$ and c : = C Y · F $c:=C_Y\cdot F$ . Then, F W 2 = b + c 2 $F_W^2=b+c^2$ , K W · F W = K Y · F c = 2 b c $K_W\cdot F_W=K_Y\cdot F-c=-2-b-c$ , and F W · E 1 , W = c $F_W\cdot E_{1,W}=c$ .

Let h : W Z $h: W\rightarrow Z$ be the contraction of E 1 , W $E_{1,W}$ and F Z : = h F W $F_Z:=h_*F_W$ . Then, Z is smooth and K Z $K_Z$ is pseudo-effective. Moreover, F Z 2 = F W 2 + c 2 = b + 2 c 2 $F_Z^2=F_W^2+c^2=b+2c^2$ , and K Z · F Z = K W · F W c = 2 b 2 c $K_Z\cdot F_Z=K_W\cdot F_W-c=-2-b-2c$ . Since b 5 $b\ge -5$ and c 2 $c\ge 2$ , F Z 2 3 > 0 $F_Z^2\ge 3&gt;0$ and K Z · F Z 1 < 0 $K_Z\cdot F_Z\le -1&lt;0$ , which contradicts Lemma 3.1. $\Box$

Proof of Lemma 4.3 continued.By Claim 4.6, Claim 4.7, and Claim 4.8, we get (1). Since x C $x\in C$ , C Y $C_Y$ intersects i = 1 k C i $\cup _{i=1}^kC_i$ , which implies that C Y $C_Y$ intersects E k $E_k$ . Thus, C Y · E k 1 $C_Y\cdot E_k\ge 1$ . (2) follows from Claim 4.4. $\Box$

Lemma 4.9.Let X be a rational 2 5 $\frac{2}{5}$ -klt surface such that K X $K_X$ is big and nef and k 10 $k\ge 10$ an integer. Then, X does not contain a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ .

Proof.Assume not. Then, there exists a closed point x X $x\in X$ such that x is a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ . By Lemma 2.16, we may let f : Y X $f: Y\rightarrow X$ be the minimal resolution of X and write

K Y + i = 1 k i 2 k + 1 E i + i = 1 s b i F i = f K X $$\begin{equation*} K_Y+\sum _{i=1}^k\frac{i}{2k+1}E_i+\sum _{i=1}^sb_iF_i=f^*K_X \end{equation*}$$
where E 1 , , E k , F 1 , , F s $E_1,\dots ,E_k,F_1,\dots ,F_s$ are the prime f-exceptional divisors, where
  • (1) E 1 , , E k $E_1,\dots ,E_k$ are the prime f-exceptional divisors over X x $X\ni x$ such that E i 2 = 2 $E_i^2=-2$ when 1 i k 1 $1\le i\le k-1$ and E k 2 = 3 $E_k^2=-3$ , and
  • (2) for every i { 1 , 2 , , s } $i\in \lbrace 1,2,\dots ,s\rbrace$ , center X F i = x i $\operatorname{center}_XF_i=x_i$ for some closed point x i X $x_i\in X$ , such that x i x $x_i\not=x$ .
In particular, K Y · E i = 0 $K_Y\cdot E_i=0$ when i k $i\not=k$ and K Y · E k = 1 $K_Y\cdot E_k=1$ . Since X is 2 5 $\frac{2}{5}$ -klt, by Lemma 2.15, K Y · i = 1 s b i F i s $K_Y\cdot \sum _{i=1}^sb_iF_i\le s$ . Since f extracts k + s $k+s$ divisors, we have ρ ( Y ) 1 + k + s $\rho (Y)\ge 1+k+s$ . Since K X $K_X$ is big and nef, we have K X 2 > 0 $K_X^2&gt;0$ , which implies that
K Y 2 = K X 2 K Y · i = 1 k i 2 k + 1 E i + i = 1 s b i F i > k 2 k + 1 s > 1 2 s . $$\begin{equation*} K_Y^2=K_X^2-K_Y\cdot {\left(\sum _{i=1}^k\frac{i}{2k+1}E_i+\sum _{i=1}^sb_iF_i\right)}&gt;-\frac{k}{2k+1}-s&gt;-\frac{1}{2}-s. \end{equation*}$$
Since X is rational, Y is rational. By Lemma 3.7, K Y 2 = 10 ρ ( Y ) $K_Y^2=10-\rho (Y)$ . Thus,
1 2 s < K Y 2 = 10 ρ ( Y ) 10 ( 1 + k + s ) = 9 k s , $$\begin{equation*} -\frac{1}{2}-s&lt;K_Y^2=10-\rho (Y)\le 10-(1+k+s)=9-k-s, \end{equation*}$$
which implies that k < 19 2 < 10 $k&lt;\frac{19}{2}&lt;10$ , a contradiction. $\Box$

Lemma 4.10.Then, there exists a positive integer n1, a DCC set I $\mathcal {I}$ of nonnegative real numbers, and a positive real number γ0 satisfying the following. Assume the following:

  • X is a 2 5 $\frac{2}{5}$ -klt surface such that K X $K_X$ is big and nef,
  • C is an irreducible curve on X,
  • f : Y X $f: Y\rightarrow X$ is the minimal resolution of X,
  • C Y $C_Y$ is the strict transform of C on Y, and
  • K Y · C Y < 0 $K_Y\cdot C_Y&lt;0$ ,
then
  • 1. K X · C I $K_X\cdot C\in \mathcal {I}$ ,
  • 2. if K X · C = 0 $K_X\cdot C=0$ , then n 1 K X $n_1K_X$ is Cartier near C, and
  • 3. if K X · C > 0 $K_X\cdot C&gt;0$ , then K X · C γ 0 $K_X\cdot C\ge \gamma _0$ .

Proof.By Lemma 4.1, there exists a positive integer n 0 = n 0 ( 1 15 ) $n_0=n_0(\frac{1}{15})$ , such that for any closed point X x $X\ni x$ , either n 0 K X $n_0K_X$ is Cartier near x, or x is a cyclic quotient singularity of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ for some positive integer k 10 $k\ge 10$ . Now we let

I : = γ γ 0 , γ = 1 + i = 1 m k i 2 k i + 1 + l n 0 , m , l , k 1 , , k m N . $$\begin{equation*} \mathcal {I}:={\left\lbrace \gamma \mid \gamma \ge 0,\gamma =-1+\sum _{i=1}^m\frac{k_i}{2k_i+1}+\frac{l}{n_0},m,l,k_1,\dots ,k_m\in \mathbb {N}\right\rbrace} . \end{equation*}$$
Then, I $\mathcal {I}$ is a DCC set of nonnegative real numbers. Since I $\mathcal {I}$ satisfies the DCC, we may let γ 0 : = min { 1 , γ I γ > 0 } $\gamma _0:=\min \lbrace 1,\gamma \in \mathcal {I}\mid \gamma &gt;0\rbrace$ .

Consider the equation

i = 1 m k i 2 k i + 1 + l n 0 = 1 , $$\begin{equation*} \sum _{i=1}^m\frac{k_i}{2k_i+1}+\frac{l}{n_0}=1, \end{equation*}$$
where m , l , k 1 , , k m N $m,l,k_1,\dots ,k_m\in \mathbb {N}$ . Then, there exists a finite set I 0 N $\mathcal {I}_0\subset \mathbb {N}$ such that k i I 0 $k_i\in \mathcal {I}_0$ for each i: to see this, note that k i 2 k i + 1 $\frac{k_i}{2k_i+1}$ belongs to a DCC set of positive real numbers and the sum i = 1 m k i 2 k i + 1 $\sum _{i=1}^m\frac{k_i}{2k_i+1}$ belongs to the finite set { n 0 l n 0 1 l n 0 } $\lbrace \frac{n_0-l}{n_0}\mid 1\le l\le n_0\rbrace$ , which implies that k i 2 k i + 1 $\frac{k_i}{2k_i+1}$ belongs to a finite set, hence k i $k_i$ belongs to a finite set. We define
n 1 : = n 0 γ I 0 ( 2 γ + 1 ) . $$\begin{equation*} n_1:=n_0\prod _{\gamma \in \mathcal {I}_0}(2\gamma +1). \end{equation*}$$

We show that n 1 , I $n_1,\mathcal {I}$ , and γ0 satisfy our requirements. For any curve C as in the assumption, there exists a nonnegative integer s, such that

  • (1) there are closed points x 1 , , x s $x_1,\dots ,x_s$ on X, such that x i C $x_i\in C$ and x i $x_i$ is a cyclic quotient singularity of type 1 2 k i + 1 ( 1 , k i ) $\frac{1}{2k_i+1}(1,k_i)$ for some positive integer k i 10 $k_i\ge 10$ for each i, and
  • (2) for any closed point y { x 1 , , x s } $y\not\in \lbrace x_1,\dots ,x_s\rbrace$ , n 0 K X $n_0K_X$ is Cartier near y.
By Lemma 4.9, we may assume that X is not rational. By Lemma 4.3, we may write
K Y + i = 1 s j = 1 k i a i , j E i , j + k = 1 t c k n 0 F k = f K X , $$\begin{equation*} K_Y+\sum _{i=1}^s\sum _{j=1}^{k_i}a_{i,j}E_{i,j}+\sum _{k=1}^t\frac{c_k}{n_0}F_k=f^*K_X, \end{equation*}$$
where
  • (1) E i , j $E_{i,j}$ and F k $F_k$ are distinct prime f-exceptional divisors for every i , j , k $i,j,k$ ,
  • (2) for any i , j $i,j$ , center X E i , j = x i $\operatorname{center}_XE_{i,j}=x_i$ ,
  • (3) k i , c k $k_i,c_k$ are positive integers,
  • (4) a i , k i = k i 2 k i + 1 $a_{i,k_i}=\frac{k_i}{2k_i+1}$ for each i, and
  • (5) C Y · E i , u i = 1 $C_Y\cdot E_{i,u_i}=1$ and C Y · E i , j = 0 $C_Y\cdot E_{i,j}=0$ for every j u i $j\not=u_i$ .
By Lemma 3.2, K Y · C Y = 1 $K_Y\cdot C_Y=-1$ . Thus,
f K X · C Y = K Y + i = 1 s j = 1 k i a i , j E i , j + k = 1 t c k n 0 F k · C Y = 1 + i = 1 s k i 2 k i + 1 + l n 0 $$\begin{equation*} f^*K_X\cdot C_Y={\left(K_Y+\sum _{i=1}^s\sum _{j=1}^{k_i}a_{i,j}E_{i,j}+\sum _{k=1}^t\frac{c_k}{n_0}F_k\right)}\cdot C_Y=-1+\sum _{i=1}^s\frac{k_i}{2k_i+1}+\frac{l}{n_0} \end{equation*}$$
for some nonnegative integer l. Moreover, since K X $K_X$ is big and nef,
0 K X · C = f K X · C Y . $$\begin{equation*} 0\le K_X\cdot C=f^*K_X\cdot C_Y. \end{equation*}$$
Thus, K X · C = f K X · C Y I $K_X\cdot C=f^*K_X\cdot C_Y\in \mathcal {I}$ , and we get (1). (3) follows from (1). Moreover, if K X · C = 0 $K_X\cdot C=0$ , then
0 = 1 + i = 1 s k i 2 k i + 1 + l n 0 , $$\begin{equation*} 0=-1+\sum _{i=1}^s\frac{k_i}{2k_i+1}+\frac{l}{n_0}, \end{equation*}$$
which implies that k i I 0 $k_i\in \mathcal {I}_0$ for each i. Thus, n 1 K X $n_1K_X$ is Cartier near C by construction of n1, and we get (2). $\Box$

4.3 Construction of nef Q $\mathbb {Q}$ -divisors

Proposition 4.11.There exists a positive integer m0 satisfying the following. Assume the following:

  • (1) X a 2 5 $\frac{2}{5}$ -klt surface such that K X $K_X$ is big and nef,
  • (2) f : Y X $f: Y\rightarrow X$ is the minimal resolution of X, and
  • (3) K Y + i = 1 s a i E i = f K X $K_Y+\sum _{i=1}^sa_iE_i=f^*K_X$ , where E i $E_i$ are the prime f-exceptional divisors,
then m 0 K Y + i = 1 s c i E i $m_0K_Y+\sum _{i=1}^sc_iE_i$ is nef for some nonnegative integers c 1 , , c s $c_1,\dots ,c_s$ , such that c i m 0 a i $c_i\le \lfloor m_0a_i\rfloor$ for each i.

Proof.Let n1 and γ0 be the numbers given by Lemma 4.10, n0 the number given by Lemma 4.1, n 2 : = max { 10 , n 1 , 1 γ 0 } $n_2:=\max \lbrace 10,n_1,\lceil \frac{1}{\gamma _0}\rceil \rbrace$ , and

m 0 : = n 0 n 1 i = 1 n 2 ( 2 i + 1 ) . $$\begin{equation*} m_0:=n_0n_1\prod _{i=1}^{n_2}(2i+1). \end{equation*}$$
We show that m0 satisfies our requirements.

We classify the singularities on X into three classes:

Class 1. Cyclic quotient singularities of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ where k n 2 $k\ge n_2$ . Let these singularities be x 1 , , x s $x_1,\dots ,x_s$ for some nonnegative integer s. We may assume that x i $x_i$ is a cyclic quotient singularity of type 1 2 k i + 1 ( 1 , k i ) $\frac{1}{2k_i+1}(1,k_i)$ for some integer k i n 2 $k_i\ge n_2$ for every 1 i s $1\le i\le s$ .

Class 2. Singularities of type 1 2 k + 1 ( 1 , k ) $\frac{1}{2k+1}(1,k)$ where 5 k < n 2 $5\le k&lt;n_2$ . Let these singularities be x s + 1 , , x t $x_{s+1},\dots ,x_t$ for some integer t s $t\ge s$ . In particular, by the definition of m0. m 0 K X $m_0K_X$ is Cartier near x i $x_i$ for every s + 1 i t $s+1\le i\le t$ .

Class 3. Other singularities. Let these singularities be x t + 1 , , x r $x_{t+1},\dots ,x_r$ for some integer r t $r\ge t$ . In particular, by Lemma 4.1, and the definition of m0, m 0 K X $m_0K_X$ is Cartier near x i $x_i$ for every t + 1 i r $t+1\le i\le r$ .

Now we may write

K Y + i = 1 s j = 1 k i j 2 k i + 1 E i , j + 1 m 0 F = f K X , $$\begin{equation*} K_Y+\sum _{i=1}^s\sum _{j=1}^{k_i}\frac{j}{2k_i+1}E_{i,j}+\frac{1}{m_0}F=f^*K_X, \end{equation*}$$
where
  • (1) for every 1 i s $1\le i\le s$ and 1 j k i $1\le j\le k_i$ , center X E i , j = x i $\operatorname{center}_XE_{i,j}=x_i$ ;
  • (2) for every 1 i s $1\le i\le s$ and 1 j k i 1 $1\le j\le k_i-1$ , E i , j 2 = 2 $E_{i,j}^2=-2$ ;
  • (3) for every 1 i s $1\le i\le s$ , E i , k i 2 = 3 $E_{i,k_i}^2=-3$ ;
  • (4) F 0 $F\ge 0$ is a f-exceptional Weil divisor, such that x i center X F $x_i\not\in \operatorname{center}_XF$ for every 1 i s $1\le i\le s$ .
We show that we may take
i = 1 l c i E i : = i = 1 s j = k i n 2 + 1 k i m 0 ( j ( k i n 2 ) ) 2 n 2 + 1 E i , j + F . $$\begin{equation*} \sum _{i=1}^lc_iE_i:=\sum _{i=1}^s\sum _{j=k_i-n_2+1}^{k_i}\frac{m_0(j-(k_i-n_2))}{2n_2+1}E_{i,j}+F. \end{equation*}$$
Indeed, by our constructions, 0 c i m 0 a i $0\le c_i\le \lfloor m_0a_i\rfloor$ for each i, and we only left to check that ( m 0 K Y + i = 1 l c i E i ) · C Y 0 $(m_0K_Y+\sum _{i=1}^lc_iE_i)\cdot C_Y\ge 0$ for any irreducible curve C Y $C_Y$ on Y. We have the following cases:

Case 1. K Y $K_Y$ is not pseudo-effective. In this case, by Lemma 4.2, X is rational. By Lemma 4.9, s = 0 $s=0$ . Thus, i = 1 l c i E i = F $\sum _{i=1}^lc_iE_i=F$ and

m 0 K Y + i = 1 l c i E i = m 0 f K X $$\begin{equation*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}=m_0f^*K_X \end{equation*}$$
is nef. Thus, ( m 0 K Y + i = 1 l c i E i ) · C Y 0 $(m_0K_Y+\sum _{i=1}^lc_iE_i)\cdot C_Y\ge 0$ for any irreducible curve C Y $C_Y$ on Y.

Case 2. K Y $K_Y$ is pseudo-effective.

Case 2.1. C Y $C_Y$ is not exceptional over X. Let C : = f C Y $C:=f_*C_Y$ .

Case 2.1.1. K Y · C Y 0 $K_Y\cdot C_Y\ge 0$ . In this case, E i , j · C Y 0 $E_{i,j}\cdot C_Y\ge 0$ and F · C Y 0 $F\cdot C_Y\ge 0$ , hence ( m 0 K Y + i = 1 l c i E i ) · C Y 0 $(m_0K_Y+\sum _{i=1}^lc_iE_i)\cdot C_Y\ge 0$ .

Case 2.1.2. K Y · C Y < 0 $K_Y\cdot C_Y&lt;0$ . By Lemma 3.2, K Y · C Y = C Y 2 = 1 $K_Y\cdot C_Y=C_Y^2=-1$ . By Lemma 4.3, C Y · E i , j = 0 $C_Y\cdot E_{i,j}=0$ for every i and every j k i 1 $j\le k_i-1$ , and C Y · E i , k i { 0 , 1 } $C_Y\cdot E_{i,k_i}\in \lbrace 0,1\rbrace$ for every i. By Lemma 4.10, there are two possibilities.

Case 2.1.2.1. n 1 K X $n_1K_X$ is Cartier near C. In this case, since n 2 n 1 $n_2\ge n_1$ , we have 2 k i + 1 2 n 2 + 1 > n 1 $2k_i+1\ge 2n_2+1&gt;n_1$ for every i. Since the Cartier index of K X $K_X$ near x i $x_i$ is 2 k i + 1 $2k_i+1$ and n 1 K X $n_1K_X$ is Cartier near C, C does not pass through x i $x_i$ . Thus, C Y $C_Y$ does not intersect E i , j $E_{i,j}$ for any i , j $i,j$ , and hence

m 0 K Y + i = 1 l c i E i · C Y = ( m 0 K Y + F ) · C Y = m 0 K Y + i = 1 s j = 1 k i m 0 j 2 k i + 1 E i , j + F · C Y = m 0 f K X · C Y 0 . $$\begin{align*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}\cdot C_Y&=(m_0K_Y+F)\cdot C_Y\\ &={\left(m_0K_Y+\sum _{i=1}^s\sum _{j=1}^{k_i}\frac{m_0j}{2k_i+1}E_{i,j}+F\right)}\cdot C_Y=m_0f^*K_X\cdot C_Y\ge 0. \end{align*}$$

Case 2.1.2.2. K X · C γ 0 $K_X\cdot C\ge \gamma _0$ . Possibly reordering indices, we may assume that there exists an integer t { 0 , 1 , 2 , , s } $t\in \lbrace 0,1,2,\dots ,s\rbrace$ , such that C Y · E i , k i = 1 $C_Y\cdot E_{i,k_i}=1$ when 1 i t $1\le i\le t$ and C Y · E i , k i = 0 $C_Y\cdot E_{i,k_i}=0$ when t + 1 i s $t+1\le i\le s$ . There are two cases:

Case 2.1.2.2.1. t 2 $t\le 2$ . In this case, since n 2 1 γ 0 $n_2\ge \frac{1}{\gamma _0}$ , γ 0 > 1 2 n 2 + 1 $\gamma _0&gt;\frac{1}{2n_2+1}$ . Thus,

m 0 K Y + i = 1 l c i E i · C Y = m 0 f K X · C i = 1 t m 0 k i 2 k i + 1 m 0 n 2 2 n 2 + 1 m 0 γ 0 m 0 i = 1 t 1 2 n 2 2 n 2 + 1 m 0 γ 0 m 0 2 n 2 + 1 > 0 . $$\begin{align*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}\cdot C_Y&=m_0f^*K_X\cdot C-\sum _{i=1}^t{\left(\frac{m_0k_i}{2k_i+1}-\frac{m_0n_2}{2n_2+1}\right)}\\ &\ge m_0\gamma _0-m_0\sum _{i=1}^t{\left(\frac{1}{2}-\frac{n_2}{2n_2+1}\right)}\ge m_0\gamma _0-\frac{m_0}{2n_2+1}&gt;0. \end{align*}$$

Case 2.1.2.2.2. t 3 $t\ge 3$ . In this case, we have

m 0 K Y + i = 1 l c i E i · C Y m 0 K Y · C Y + i = 1 t m 0 n 2 2 n 2 + 1 = m 0 1 + i = 1 t n 2 2 n 2 + 1 m 0 1 + 3 n 2 2 n 2 + 1 = m 0 ( n 2 1 ) 2 n 2 + 1 > 0 . $$\begin{align*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}\cdot C_Y&\ge m_0K_Y\cdot C_Y+\sum _{i=1}^t\frac{m_0n_2}{2n_2+1}=m_0{\left(-1+\sum _{i=1}^t\frac{n_2}{2n_2+1}\right)}\\ &\ge m_0{\left(-1+\frac{3n_2}{2n_2+1}\right)}=\frac{m_0(n_2-1)}{2n_2+1}&gt;0. \end{align*}$$

Case 2.2. C Y $C_Y$ is exceptional over X. Then, C Supp i = 1 s j = 1 k i E i , j Supp F $C\subset \operatorname{Supp}\left(\cup _{i=1}^s\cup _{j=1}^{k_i}E_{i,j}\right)\cup \operatorname{Supp}F$ .

Case 2.2.1 C Y Supp F $C_Y\subset \operatorname{Supp}F$ . In this case, C Y · E i , j = 0 $C_Y\cdot E_{i,j}=0$ for every i , j $i,j$ , and hence

m 0 K Y + i = 1 l c i E i · C Y = ( m 0 K Y + F ) · C Y = m 0 K Y + i = 1 s j = 1 k i m 0 j 2 k i + 1 E i , j + F · C Y = m 0 f K X · C Y = 0 . $$\begin{align*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}\cdot C_Y&=(m_0K_Y+F)\cdot C_Y\\ &={\left(m_0K_Y+\sum _{i=1}^s\sum _{j=1}^{k_i}\frac{m_0j}{2k_i+1}E_{i,j}+F\right)}\cdot C_Y=m_0f^*K_X\cdot C_Y=0. \end{align*}$$

Case 2.2.2 C Y Supp i = 1 s j = 1 k i E i , j $C_Y\subset \operatorname{Supp}\left(\cup _{i=1}^s\cup _{j=1}^{k_i}E_{i,j}\right)$ . We may assume that C Y = E i , j 0 $C_Y=E_{i,j_0}$ for some i and some 1 j 0 k i $1\le j_0\le k_i$ . In this case,

m 0 K Y + i = 1 l c i E i · C Y = m 0 K Y + j = k i n 2 + 1 k i m 0 ( j ( k i n 2 ) ) 2 n 2 + 1 E i , j · C Y . $$\begin{equation*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}\cdot C_Y={\left(m_0K_Y+\sum _{j=k_i-n_2+1}^{k_i}\frac{m_0(j-(k_i-n_2))}{2n_2+1}E_{i,j}\right)}\cdot C_Y. \end{equation*}$$
There are four possibilities:

Case 2.2.2.1 j 0 = k i $j_0=k_i$ . In this case, m 0 K Y + i = 1 l c i E i · C Y = m 0 1 + n 2 1 2 n 2 + 1 3 n 2 2 n 2 + 1 = 0 $\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)\cdot C_Y=m_0\left(1+\frac{n_2-1}{2n_2+1}-\frac{3n_2}{2n_2+1}\right)=0$ .

Case 2.2.2.2 k i n 2 + 1 j 0 k i 1 $k_i-n_2+1\le j_0\le k_i-1$ . In this case,

m 0 K Y + i = 1 l c i E i · C Y = m 0 0 + j 0 1 ( k i n 2 ) 2 n 2 + 1 2 ( j 0 ( k i n 2 ) ) 2 n 2 + 1 + j 0 + 1 ( k i n 2 ) 2 n 2 + 1 = 0 . $$\begin{equation*} {\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)}\cdot C_Y=m_0{\left(0+\frac{j_0-1-(k_i-n_2)}{2n_2+1}-\frac{2(j_0-(k_i-n_2))}{2n_2+1}+\frac{j_0+1-(k_i-n_2)}{2n_2+1}\right)}=0. \end{equation*}$$

Case 2.2.2.3 j 0 = k i n 2 $j_0=k_i-n_2$ . In this case, m 0 K Y + i = 1 l c i E i · C Y = m 0 2 n 2 + 1 > 0 $\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)\cdot C_Y=\frac{m_0}{2n_2+1}&gt;0$ .

Case 2.2.2.4 1 j 0 k i n 2 $1\le j_0\le k_i-n_2$ . In this case, m 0 K Y + i = 1 l c i E i · C Y = m 0 K Y · C Y = 0 $\left(m_0K_Y+\sum _{i=1}^lc_iE_i\right)\cdot C_Y=m_0K_Y\cdot C_Y=0$ . $\Box$

Proposition 4.12.There exists a uniform positive integer m2 satisfying the following. Assume that

  • 1. X a 2 5 $\frac{2}{5}$ -klt surface such that K X $K_X$ is big and nef,
  • 2. f : Y X $f: Y\rightarrow X$ is the minimal resolution of X, and
  • 3. K Y + i = 1 s a i E i = f K X $K_Y+\sum _{i=1}^sa_iE_i=f^*K_X$ , where E i $E_i$ are the prime f-exceptional divisors,
then m 2 K Y + i = 1 l r i E i $m_2K_Y+\sum _{i=1}^lr_iE_i$ is big and nef for some nonnegative integers r 1 , , r l $r_1,\dots ,r_l$ , such that r i m 2 a i $r_i\le \lfloor m_2a_i\rfloor$ for each i.

Proof.By Proposition 4.11, there exist a positive integer m 0 = m 0 $m_0=m_0$ which does not depend on X, and non-negative integers c 1 , , c l $c_1,\dots ,c_l$ , such that m 0 K Y + i = 1 l c i E i $m_0K_Y+\sum _{i=1}^lc_iE_i$ is nef and c i m 0 a i $c_i\le \lfloor m_0a_i\rfloor$ for each i. By Theorem 3.12, there exists a uniform positive integer m1 such that | m 1 K X | $|m_1K_X|$ defines a birational map. Let m 2 : = m 0 m 1 $m_2:=m_0m_1$ . Then, | m 2 K X | $|m_2K_X|$ defines a birational map, and hence

m 2 K Y + i = 1 l m 2 a i E i = m 2 K Y + i = 1 l m 2 a i E i = | f ( m 2 K X ) | $$\begin{equation*} {\left|m_2K_Y+\sum _{i=1}^l\lfloor m_2a_i\rfloor E_i\right|}={\left|m_2K_Y+\sum _{i=1}^lm_2a_iE_i\right|}=|f^*(m_2K_X)| \end{equation*}$$
defines a birational map.

Let D : = m 2 K Y + i = 1 l m 2 a i E i $D:=m_2K_Y+\sum _{i=1}^l\lfloor m_2a_i\rfloor E_i$ and D : = m 2 K Y + i = 1 l m 1 c i E i $\tilde{D}:=m_2K_Y+\sum _{i=1}^lm_1c_iE_i$ . Since c i m 0 a i $c_i\le \lfloor m_0a_i\rfloor$ ,

m 1 c i m 1 m 0 a i m 1 m 0 a i = m 2 a i . $$\begin{equation*} m_1c_i\le m_1\lfloor m_0a_i\rfloor \le \lfloor m_1m_0a_i\rfloor =\lfloor m_2a_i\rfloor . \end{equation*}$$
Thus, D D $D\ge \tilde{D}$ . By Proposition 3.10, there exists a Weil divisor D $D^{\prime }$ on X, such that D D D $D\ge D^{\prime }\ge \tilde{D}$ and D $D^{\prime }$ is big and nef. In particular, we may write D = m 2 K Y + i = 1 l r i E i $D^{\prime }=m_2K_Y+\sum _{i=1}^lr_iE_i$ for some integers r 1 , , r l $r_1,\dots ,r_l$ such that 0 c i r i m 2 a i $0\le c_i\le r_i\le \lfloor m_2a_i\rfloor$ for each i. m2 and r 1 , , r l $r_1,\dots ,r_l$ satisfy our requirements. $\Box$

4.4 Proof of the main theorem

Proof of Theorem 1.1.Let f : Y X $f: Y\rightarrow X$ be the minimal resolution of X such that K Y + i = 1 n a i E i = f K X $K_Y+\sum _{i=1}^na_iE_i=f^*K_X$ , where E 1 , , E n $E_1,\dots ,E_n$ are the prime exceptional divisors of f. By Proposition 4.12, there exists a uniform positive integer m2, such that K Y + i = 1 n r i m 2 E i $K_Y+\sum _{i=1}^n\frac{r_i}{m_2}E_i$ is big and nef for some integers r 1 , , r n $r_1,\dots ,r_n$ such that 0 r i m 2 a i $0\le r_i\le \lfloor m_2a_i\rfloor$ for each i. By Lemma 3.11, | 192 m 2 K X | $|192m_2K_X|$ defines a birational map and we may let m : = 192 m 2 $m:=192m_2$ . $\Box$

5 EXAMPLES

In this section, we will provide two theorems where we construct some interesting examples. The first one is Theorem 5.3 (= Theorem 1.2), which shows that the 1 2 $\frac{1}{2}$ -lc assumption in Theorem 1.1 is necessary. The second one is Theorem 5.7. It shows that, even if we only have a very strong control on mld(X) (i.e., when X is a terminal threefold), “ | m K X | $|mK_X|$ has no fixed part” is the best we may expect, as we cannot expect | m K X | $|mK_X|$ to be free in codimension 2 for any bounded m.

Lemma 5.1.Let X be an lc projective surface such that K X $K_X$ is big and nef, f : Y X $f: Y\rightarrow X$ the minimal resolution of X, and E 1 , , E n $E_1,\dots ,E_n$ the prime f-exceptional divisors of X. Assume that K Y + i = 1 n a i E i = f K X $K_Y+\sum _{i=1}^na_iE_i=f^*K_X$ , where a i : = 1 a ( E i , X , 0 ) $a_i:=1-a(E_i,X,0)$ . Let m be a positive integer and c 1 , , c n $c_1,\dots ,c_n$ nonnegative integers, such that

  • 0 c 1 , , c n m a i $0\le c_1,\dots ,c_n\le \lfloor ma_i\rfloor$ ,
  • | m K Y + i = 1 n c i E i | $|mK_Y+\sum _{i=1}^nc_iE_i|\not=\emptyset$ ,
  • the fixed part of | m K Y + i = 1 n c i E i | $|mK_Y+\sum _{i=1}^nc_iE_i|$ is supported on i = 1 n E i $\cup _{i=1}^nE_i$ , and
  • m K Y + i = 1 n c i E i $mK_Y+\sum _{i=1}^nc_iE_i$ is big but not nef,
then there exist nonnegative integers c 1 , , c n $c^{\prime }_1,\dots ,c_n^{\prime }$ , such that
  • 1. 0 c i c i $0\le c_i^{\prime }\le c_i$ for each i,
  • 2. there exists j { 1 , 2 , , n } $j\in \lbrace 1,2,\dots ,n\rbrace$ such that c j < c j $c_j^{\prime }&lt; c_j$ ,
  • 3. | m K Y + i = 1 n c i E i | $|mK_Y+\sum _{i=1}^nc_iE_i|\not=\emptyset$ , and
  • 4. the fixed part of | m K Y + i = 1 n c i E i | $|mK_Y+\sum _{i=1}^nc^{\prime }_iE_i|$ is supported on i = 1 n E i $\cup _{i=1}^nE_i$ ,

Proof.Since 0 c 1 , , c n m a i $0\le c_1,\dots ,c_n\le \lfloor ma_i\rfloor$ , Y , i = 1 n c i m E i $\left(Y,\sum _{i=1}^n\frac{c_i}{m}E_i\right)$ is lc. Thus, we may run a K Y + i = 1 n c i m E i $\left(K_Y+\sum _{i=1}^n\frac{c_i}{m}E_i\right)$ -MMP h : Y W $h: Y\rightarrow W$ . Since the fixed part of | m K Y + i = 1 n c i E i | $|mK_Y+\sum _{i=1}^nc_iE_i|$ is supported on i = 1 n E i $\cup _{i=1}^nE_i$ , h only contracts divisors supported on i = 1 n E i $\cup _{i=1}^nE_i$ . Let B : = h K Y + i = 1 n c i m E i $B:=h_*\left(K_Y+\sum _{i=1}^n\frac{c_i}{m}E_i\right)$ , then we have

K Y + i = 1 n c i m E i = h ( K W + B ) + i = 1 n b i E i , $$\begin{equation*} K_Y+\sum _{i=1}^n\frac{c_i}{m}E_i=h^*(K_W+B)+\sum _{i=1}^nb_iE_i, \end{equation*}$$
where b i 0 $b_i\ge 0$ are real numbers. Moreover, since m K Y + i = 1 n c i E i $mK_Y+\sum _{i=1}^nc_iE_i$ is big but not nef, h id Y $h\not=\mathrm{id}_Y$ . Thus, there exists j { 1 , 2 , , n } $j\in \lbrace 1,2,\dots ,n\rbrace$ such that b j > 0 $b_j&gt;0$ . We have
m h ( K W + B ) = m K Y + i = 1 n ( c i m b i ) E i . $$\begin{equation*} mh^*(K_W+B)=mK_Y+\sum _{i=1}^n(c_i-mb_i)E_i. \end{equation*}$$
Since f is the minimal resolution of X, E i 2 2 $E_i^2\le -2$ for every i. Thus, h is the minimal resolution of W, which implies that c i m b i 0 $c_i-mb_i\ge 0$ for every i. Let c i : = c i m b i $c_i^{\prime }:=\lfloor c_i-mb_i\rfloor$ for every i. Then, (1)(2) hold. Since
m K Y + i = 1 n c i E i | m ( K W + B ) | | m h ( K W + B ) | m K Y + i = 1 n c i E i , $$\begin{equation*} {\left|mK_Y+\sum _{i=1}^nc_iE_i\right|}\cong |m(K_W+B)|\cong |mh^*(K_W+B)|\cong {\left|mK_Y+\sum _{i=1}^nc_i^{\prime }E_i\right|}, \end{equation*}$$
(3)(4) hold. $\Box$

Theorem 5.2.Let X be an lc projective surface such that K X $K_X$ is big and nef, f : Y X $f: Y\rightarrow X$ the minimal resolution of X, and E 1 , , E n $E_1,\dots ,E_n$ the prime f-exceptional divisors of X. Assume that K Y + i = 1 n a i E i = f K X $K_Y+\sum _{i=1}^na_iE_i=f^*K_X$ , where a i : = 1 a ( E i , X , 0 ) $a_i:=1-a(E_i,X,0)$ . Then, for any positive integer m, if | m K X | $|mK_X|$ defines a birational map and does not have fixed part, then there exist positive integers r 1 , , r n $r_1,\dots ,r_n$ , such that

  • 1. 0 r i m a i $0\le r_i\le \lfloor ma_i\rfloor$ , and
  • 2. K Y + i = 1 n r i m E i $K_Y+\sum _{i=1}^n\frac{r_i}{m}E_i$ is big and nef.

Proof.The fixed part of

| f ( m K X ) | = m K Y + i = 1 n m a i E i = m K Y + i = 1 n m a i E i $$\begin{equation*} |f^*(mK_X)|={\left|mK_Y+\sum _{i=1}^nma_iE_i\right|}={\left|mK_Y+\sum _{i=1}^n\lfloor ma_i\rfloor E_i\right|} \end{equation*}$$
is supported on i = 1 n E i $\cup _{i=1}^nE_i$ . Since | m K X | $|mK_X|$ defines a birational map, | m K Y + i = 1 n m a i E i | $|mK_Y+\sum _{i=1}^n\lfloor ma_i\rfloor E_i|$ defines a birational map. In particular, m K Y + i = 1 n m a i E i $mK_Y+\sum _{i=1}^n\lfloor ma_i\rfloor E_i$ is big.

We inductively define integers c i j $c^j_i$ for every i { 1 , 2 , , n } $i\in \lbrace 1,2,\dots ,n\rbrace$ for nonnegative integers j in the following way: Let c i 0 : = m a i $c^0_i:=\lfloor ma_i\rfloor$ for every i. If K Y + i = 1 n c i j m E i $K_Y+\sum _{i=1}^n\frac{c^j_i}{m}E_i$ is big and nef, then we let r i : = c i j $r_i:=c^j_i$ for every i and we are done. Otherwise, by Lemma 5.1, there exist integers c i j + 1 $c^{j+1}_i$ for every i, such that 0 c i j + 1 c i j $0\le c^{j+1}_i\le c^j_i$ , c k j + 1 < c k j $c^{j+1}_k&lt;c^j_k$ for some k { 1 , 2 , , n } $k\in \lbrace 1,2,\dots ,n\rbrace$ , m K Y + i = 1 n c i j + 1 E i $\left|mK_Y+\sum _{i=1}^n\lfloor c^{j+1}_i\rfloor E_i\right|\not=\emptyset$ , and the fixed part of m K Y + i = 1 n c i j + 1 E i $\left|mK_Y+\sum _{i=1}^n\lfloor c^{j+1}_i\rfloor E_i\right|$ is supported on i = 1 n E i $\cup _{i=1}^nE_i$ . This process must terminate after finitely many steps, and we get the desired r i $r_i$ for every i. $\Box$

Theorem 5.3. (= Theorem 1.2)There exist normal projective surfaces { X n , k } n 4 , k 2 $\lbrace X_{n,k}\rbrace _{n\ge 4,k\ge 2}$ , such that

  • 1. | m K X n , k | $|mK_{X_{n,k}}|\not=\emptyset$ and has a nonzero fixed part for any positive integers m , n $m,n$ , and k m $k\ge m$ ,
  • 2. K X n , k $K_{X_{n,k}}$ is ample for every n , k $n,k$ , and
  • 3. lim k + mld ( X n , k ) = 1 n 1 $\lim _{k\rightarrow +\infty }{\rm {mld}}(X_{n,k})=\frac{1}{n-1}$ for any n.

Proof.Step 1. In this step, we construct X n , k $X_{n,k}$ for every n 4 $n\ge 4$ and k 2 $k\ge 2$ .

For any positive integer n 4 $n\ge 4$ and positive integer k 2 $k\ge 2$ , we let Y n , k $Y_{n,k}$ be a general hypersurface of degree d n , k : = 2 k ( n 2 ) 2 ( 2 k ( n 1 ) 1 ) $d_{n,k}:=2k(n-2)^2(2k(n-1)-1)$ in the weighted projective space P n , k : = P ( 1 , 1 , 2 k ( n 2 ) , 2 k ( n 2 ) ( n 1 ) + 1 ) $P_{n,k}:=\mathbb {P}(1,1,2k(n-2),2k(n-2)(n-1)+1)$ . Since 2 k ( n 2 ) d n , k $2k(n-2)\mid d_{n,k}$ and

d n , k 1 = ( 2 k ( n 2 ) ( n 1 ) + 1 ) · ( 2 k ( n 2 ) 1 ) , $$\begin{equation*} d_{n,k}-1=(2k(n-2)(n-1)+1)\cdot (2k(n-2)-1), \end{equation*}$$
Y n , k $Y_{n,k}$ is well formed and has a unique singularity o n , k $o_{n,k}$ , which is a cyclic quotient singularity of type 1 2 k ( n 2 ) ( n 1 ) + 1 1 , 2 k ( n 2 ) $\frac{1}{2k(n-2)(n-1)+1}\left(1,2k(n-2)\right)$ . The dual graph of this cyclic quotient singularity is the following:
image
where there are 2 k ( n 2 ) 1 $2k(n-2)-1$ “2” in the chain. Let E 1 = E 1 ( n , k ) , , E 2 k ( n 2 ) = E 2 k ( n 2 ) ( n , k ) $E_1=E_1(n,k),\dots ,E_{2k(n-2)}=E_{2k(n-2)}(n,k)$ be the curves in this dual graph in order, that is,
  • (1) E i 2 = 2 $E_i^2=-2$ when i { 1 , 2 , , 2 k ( n 2 ) 1 } $i\in \lbrace 1,2,\dots ,2k(n-2)-1\rbrace$ ,
  • (2) E 2 k ( n 2 ) 2 = n $E_{2k(n-2)}^2=-n$ , and
  • (3) E i · E j 0 $E_i\cdot E_j\not=0$ if and only if | i j | 1 $|i-j|\le 1$ .
Let h n , k : Z n , k Y n , k $h_{n,k}:Z_{n,k}\rightarrow Y_{n,k}$ be the minimal resolution, then we have
K Z n , k + i = 1 2 k ( n 2 ) i ( n 2 ) 2 k ( n 1 ) ( n 2 ) + 1 E i = h n , k K Y n , k . $$\begin{equation*} K_{Z_{n,k}}+\sum _{i=1}^{2k(n-2)}\frac{i(n-2)}{2k(n-1)(n-2)+1}E_i=h_{n,k}^*K_{Y_{n,k}}. \end{equation*}$$
Now let g n , k : W n , k Z n , k $g_{n,k}:W_{n,k}\rightarrow Z_{n,k}$ be the blow-up of E k ( n 1 ) E k ( n 1 ) + 1 $E_{k(n-1)}\cap E_{k(n-1)+1}$ and C n , k , W $C_{n,k,W}$ the exceptional divisor of g n , k $g_{n,k}$ . Let E i , W = E i , W ( n , k ) $E_{i,W}=E_{i,W}(n,k)$ be the strict transform of E i $E_i$ on W n , k $W_{n,k}$ for each i. Then,
K W n , k + i = 1 2 k ( n 2 ) i ( n 2 ) 2 k ( n 1 ) ( n 2 ) + 1 E i , W + n 3 2 k ( n 1 ) ( n 2 ) + 1 C n , k , W = g n , k K Z n , k + i = 1 2 k ( n 2 ) i ( n 2 ) 2 k ( n 1 ) ( n 2 ) + 1 E i = ( h n , k g n , k ) K Y n , k . $$\begin{align*} &K_{W_{n,k}}+\sum _{i=1}^{2k(n-2)}\frac{i(n-2)}{2k(n-1)(n-2)+1}E_{i,W}+\frac{n-3}{2k(n-1)(n-2)+1}C_{n,k,W}\\ &\quad =\, g_{n,k}^*{\left(K_{Z_{n,k}}+\sum _{i=1}^{2k(n-2)}\frac{i(n-2)}{2k(n-1)(n-2)+1}E_i\right)}=(h_{n,k}\circ g_{n,k})^*K_{Y_{n,k}}. \end{align*}$$
Now, we run a K W n , k + i = 1 2 k ( n 2 ) E i , W + n 3 2 k ( n 1 ) ( n 2 ) + 1 C n , k , W $\left(K_{W_{n,k}}+\sum _{i=1}^{2k(n-2)}E_{i,W}+\frac{n-3}{2k(n-1)(n-2)+1}C_{n,k,W}\right)$ -MMP over Y n , k $Y_{n,k}$ , which induces a birational contraction f n , k : W n , k X n , k $f_{n,k}:W_{n,k}\rightarrow X_{n,k}$ . Then, f n , k $f_{n,k}$ contracts precisely E 1 , W , , E 2 k ( n 2 ) , W $E_{1,W},\dots ,E_{2k(n-2),W}$ . We let C n , k $C_{n,k}$ be the pushforward of C n , k , W $C_{n,k,W}$ on X n , k $X_{n,k}$ and p n , k : X n , k Y n , k $p_{n,k}: X_{n,k}\rightarrow Y_{n,k}$ the induced contraction.

Step 2. In this step, we show the following:

Claim 5.4.For any positive integers m, n 4 $n\ge 4$ and k m $k\ge m$ , if | m K X n , k | $|mK_{X_{n,k}}|\not=\emptyset$ and K X n , k $K_{X_{n,k}}$ is big, then | m K X n , k | $|mK_{X_{n,k}}|$ has nonzero fixed part.

Proof.We let o 1 = o 1 ( n , k ) : = ( f n , k ) i = 1 k ( n 1 ) E i , W $o_1=o_1(n,k):=(f_{n,k})_*\left(\cup _{i=1}^{k(n-1)}E_{i,W}\right)$ and o 2 = o 2 ( n , k ) : = ( f n , k ) i = k ( n 1 ) + 1 2 k ( n 2 ) E i , W $o_2=o_2(n,k):=(f_{n,k})_*\left(\cup _{i=k(n-1)+1}^{2k(n-2)}E_{i,W}\right)$ . Then, o1 is a cyclic quotient singularity of type 1 2 k ( n 1 ) + 1 ( 1 , k ( n 1 ) ) $\frac{1}{2k(n-1)+1}(1,k(n-1))$ with dual graph

image
where there are k ( n 1 ) 1 $k(n-1)-1$ “2” in the chain, and o2 is a cyclic quotient singularity of type 1 ( n 3 ) ( 2 k ( n 1 ) 1 ) ( 1 , 2 k ( n 3 ) 1 ) $\frac{1}{(n-3)(2k(n-1)-1)}(1,2k(n-3)-1)$ with dual graph
image
where there are k ( n 3 ) 2 $k(n-3)-2$ “2” in the chain. Then,
K X n , k + n 3 2 k ( n 1 ) ( n 2 ) + 1 C n , k = p n , k K Y n , k , $$\begin{equation*} K_{X_{n,k}}+\frac{n-3}{2k(n-1)(n-2)+1}C_{n,k}=p_{n,k}^*K_{Y_{n,k}}, \end{equation*}$$
K W n , k + i = 1 2 k ( n 2 ) i ( n 2 ) 2 k ( n 1 ) ( n 2 ) + 1 E i , W + n 3 2 k ( n 1 ) ( n 2 ) + 1 C n , k , W = f n , k K X n , k + n 3 2 k ( n 1 ) ( n 2 ) + 1 C n , k , $$\begin{align*} &K_{W_{n,k}}+\sum _{i=1}^{2k(n-2)}\frac{i(n-2)}{2k(n-1)(n-2)+1}E_{i,W}+\frac{n-3}{2k(n-1)(n-2)+1}C_{n,k,W}\\ &\quad = f_{n,k}^*{\left(K_{X_{n,k}}+\frac{n-3}{2k(n-1)(n-2)+1}C_{n,k}\right)}, \end{align*}$$
and
K W n , k + i = 1 k ( n 1 ) i 2 k ( n 1 ) + 1 E i , W + i = k ( n 1 ) + 1 2 k ( n 2 ) i 1 2 k ( n 1 ) 1 E i , W = f n , k K X n , k . $$\begin{equation*} K_{W_{n,k}}+\sum _{i=1}^{k(n-1)}\frac{i}{2k(n-1)+1}E_{i,W}+\sum _{i=k(n-1)+1}^{2k(n-2)}\frac{i-1}{2k(n-1)-1}E_{i,W}=f_{n,k}^*K_{X_{n,k}}. \end{equation*}$$
We have
f n , k K X n , k · C n , k , W = 1 + k ( n 1 ) 2 k ( n 1 ) + 1 + k ( n 1 ) 2 k ( n 1 ) 1 = 1 4 k 2 ( n 1 ) 2 1 < 1 35 k 2 < 5 12 k . $$\begin{equation*} f_{n,k}^*K_{X_{n,k}}\cdot C_{n,k,W}=-1+\frac{k(n-1)}{2k(n-1)+1}+\frac{k(n-1)}{2k(n-1)-1}=\frac{1}{4k^2(n-1)^2-1}&lt;\frac{1}{35k^2}&lt;\frac{5}{12k}. \end{equation*}$$
Now for any positive even number m = 2 l $m=2l$ , any n 4 $n\ge 4$ and any k l $k\ge l$ , we have
m · k ( n 1 ) 2 k ( n 1 ) + 1 m = 2 l k ( n 1 ) 2 k ( n 1 ) + 1 2 l = l 2 k ( n 1 ) + 1 2 l 5 12 l 5 12 k . $$\begin{equation*} \frac{{\left\lbrace m\cdot \frac{k(n-1)}{2k(n-1)+1}\right\rbrace} }{m}=\frac{{\left\lbrace \frac{2lk(n-1)}{2k(n-1)+1}\right\rbrace} }{2l}=\frac{{\left\lbrace -\frac{l}{2k(n-1)+1}\right\rbrace} }{2l}\ge \frac{5}{12l}\ge \frac{5}{12k}. \end{equation*}$$
Thus, for any positive even number m = 2 l $m=2l$ , any n 4 $n\ge 4$ and any k l $k\ge l$ , K W n , k + i = 1 2 k ( n 2 ) c i m E i , W $K_{W_{n,k}}+\sum _{i=1}^{2k(n-2)}\frac{c_i}{m}E_{i,W}$ is not nef for any integers c 1 , , c 2 k ( n 2 ) $c_1,\dots ,c_{2k(n-2)}$ such that 1
  • (1) 0 c i m i 2 k ( n 1 ) + 1 $0\le c_i\le \lfloor \frac{mi}{2k(n-1)+1}\rfloor$ when 1 i k ( n 1 ) $1\le i\le k(n-1)$ , and
  • (2) 0 c i m ( i 1 ) 2 k ( n 1 ) 1 $0\le c_i\le \lfloor \frac{m(i-1)}{2k(n-1)-1}\rfloor$ when k ( n 1 ) + 1 i 2 k ( n 2 ) $k(n-1)+1\le i\le 2k(n-2)$ .
For any integer n 4 $n\ge 4$ , any positive integer m such that | m K X n , k | $|mK_{X_{n,k}}|\not=\emptyset$ , and any integer k m $k\ge m$ ,
  • (1) if K X n , k $K_{X_{n,k}}$ is not nef, then | m K X n , k | $|mK_{X_{n,k}}|$ has nonzero fixed part, and
  • (2) if K X n , k $K_{X_{n,k}}$ is nef, then by Theorem 5.2, | m K X n , k | $|mK_{X_{n,k}}|$ has nonzero fixed part.

Step 3. In this step, we show that K X n , k $K_{X_{n,k}}$ is ample.

Claim 5.5.For any integers n 4 $n\ge 4$ and k 2 $k\ge 2$ , K Y n , k $K_{Y_{n,k}}$ is ample, | K Y n , k | $|K_{Y_{n,k}}|$ defines a birational map, and | K Y n , k | $|K_{Y_{n,k}}|$ and has no fixed part. In particular, | K Y n , k | $|K_{Y_{n,k}}|$ defines a birational map.

Proof.Let d n , k : = d n , k deg ( K P n , k ) $d^{\prime }_{n,k}:=d_{n,k}-\deg (-K_{P_{n,k}})$ . Then,

d n , k ( 2 k ( n 2 ) ( n 1 ) + 1 ) = 4 + 2 k ( n 2 ) ( n 1 ) ( 2 k ( n 2 ) 3 ) 116 > 0 . $$\begin{equation*} d^{\prime }_{n,k}-(2k(n-2)(n-1)+1)=-4+2k(n-2)(n-1)(2k(n-2)-3)\ge 116&gt;0. \end{equation*}$$
Thus, K Y n , k $K_{Y_{n,k}}$ is ample and | K Y n , k | $|K_{Y_{n,k}}|$ defines a birational map. In particular, | K Y n , k | $|K_{Y_{n,k}}|\not=\emptyset$ .

Let x , y , z , w $x,y,z,w$ be the coordinates of P n , k $P_{n,k}$ and since d n , k = d n , k ( 1 + 1 + 2 k ( n 2 ) + ( 2 k ( n 2 ) ( n 1 ) + 1 ) ) $d^{\prime }_{n,k}=d_{n,k}-(1+1+2k(n-2)+(2k(n-2)(n-1)+1))$ . Let A : = x d n , k = 0 $A:=\left(x^{d^{\prime }_{n,k}}=0\right)$ and B : = y d n , k = 0 $B:=\left(y^{d^{\prime }_{n,k}}=0\right)$ . Then, A | Y n , k | K Y n , k | $A|_{Y_{n,k}}\in |K_{Y_{n,k}}|$ and B | Y n , k | K Y n , k | $B|_{Y_{n,k}}\in |K_{Y_{n,k}}|$ . We only need to show that A | Y n , k B | Y n , k $A|_{Y_{n,k}}\not=B|_{Y_{n,k}}$ . This is the same as saying that Y n , k $Y_{n,k}$ does not contain the line ( x = y = 0 ) $(x=y=0)$ in P n , k $P_{n,k}$ . Suppose that Y n , k $Y_{n,k}$ is defined by the homogeneous weighted polynomial q n , k ( x , y , z , w ) $q_{n,k}(x,y,z,w)$ . Since Y n , k $Y_{n,k}$ is general, z ( n 2 ) ( 2 k ( n 1 ) 1 ) q n , k ( x , y , z , w ) $z^{(n-2)(2k(n-1)-1)}\in q_{n,k}(x,y,z,w)$ . Thus, Y n , k $Y_{n,k}$ does not contain the line x = y = 0 $x=y=0$ and we are done. $\Box$

Claim 5.6.For any integers n 4 $n\ge 4$ and k 2 $k\ge 2$ , K X n , k $K_{X_{n,k}}$ is ample.

Proof.For any n , k $n,k$ , by Claim 5.5, the fixed part of | p n , k K Y n , k | $|p_{n,k}^*K_{Y_{n,k}}|$ is supported on C n , k $C_{n,k}$ . Since

p n , k K Y n , k = K X n , k + ( n 3 ) 2 k ( n 1 ) ( n 2 ) + 1 C n , k , $$\begin{equation*} p_{n,k}^*K_{Y_{n,k}}=K_{X_{n,k}}+\frac{(n-3)}{2k(n-1)(n-2)+1}C_{n,k}, \end{equation*}$$
there exists a nonnegative integer r = r n , k $r=r_{n,k}$ such that | K X n , k r n , k C n , k | $|K_{X_{n,k}}-r_{n,k}C_{n,k}|$ defines a birational map and has no fixed part. In particular, K X n , k r n , k C n , k $K_{X_{n,k}}-r_{n,k}C_{n,k}$ is big and nef. If r n , k = 0 $r_{n,k}=0$ , then | K X n , k | $|K_{X_{n,k}}|\not=\emptyset$ and has no fixed part, which contradicts Claim 5.4. Thus, r n , k > 0 $r_{n,k}&gt;0$ .

Since K X n , k + ( n 3 ) 2 k ( n 1 ) ( n 2 ) + 1 C n , k $K_{X_{n,k}}+\frac{(n-3)}{2k(n-1)(n-2)+1}C_{n,k}$ is nef and big, K X n , k + ( n 3 ) 2 k ( n 1 ) ( n 2 ) + 1 C n , k · C n , k = 0 $\left(K_{X_{n,k}}+\frac{(n-3)}{2k(n-1)(n-2)+1}C_{n,k}\right)\cdot C_{n,k}=0$ . Since K X n , k r n , k C n , k $K_{X_{n,k}}-r_{n,k}C_{n,k}$ is nef and big, K X n , k $K_{X_{n,k}}$ is nef and big and K X n , k · C n , k > 0 $K_{X_{n,k}}\cdot C_{n,k}&gt;0$ . In particular, K X n , k 2 > 0 $K^2_{X_{n,k}}&gt;0$ .

For any irreducible curve D n , k $D_{n,k}$ on X n , k $X_{n,k}$ such that D n , k C n , k $D_{n,k}\not=C_{n,k}$ , if D n , k · C n , k > 0 $D_{n,k}\cdot C_{n,k}&gt;0$ , we have that

K X n , k · D n , k = ( K X n , k r n , k C n , k ) · D n , k + r n , k C n , k · D n , k > 0 , $$\begin{equation*} K_{X_{n,k}}\cdot D_{n,k}=(K_{X_{n,k}}-r_{n,k}C_{n,k})\cdot D_{n,k}+r_{n,k}C_{n,k}\cdot D_{n,k}&gt;0, \end{equation*}$$
and if D n , k · C n , k = 0 $D_{n,k}\cdot C_{n,k}=0$ , then
K X n , k · D n , k = K Y n , k · ( p n , k ) D n , k > 0 . $$\begin{equation*} K_{X_{n,k}}\cdot D_{n,k}=K_{Y_{n,k}}\cdot (p_{n,k})_*D_{n,k}&gt;0. \end{equation*}$$
Thus, K X n , k $K_{X_{n,k}}$ is ample. $\Box$

Step 4. Claim 5.4 and Claim 5.6 imply (1)(2). Since

mld ( X n , k ) = mld ( X n , k o 2 ( n , k ) ) = 2 k 2 k ( n 1 ) 1 = 1 n 1 1 2 k , $$\begin{equation*} {\rm {mld}}(X_{n,k})={\rm {mld}}(X_{n,k}\ni o_2(n,k))=\frac{2k}{2k(n-1)-1}=\frac{1}{n-1-\frac{1}{2k}}, \end{equation*}$$
we have lim k + mld ( X n , k ) = 1 n 1 $\lim _{k\rightarrow +\infty }{\rm {mld}}(X_{n,k})=\frac{1}{n-1}$ for any n 4 $n\ge 4$ , which implies (3). $\Box$

Theorem 5.7.For any positive integer m0, there exists a terminal threefold X such that K X $K_X$ is ample but | m 0 K X | $|m_0K_X|$ is not free in codimension 2.

Proof.Step 1. We start with a local construction by using the language of toric varieties.

Let N = Z 3 $N=\mathbb {Z}^3$ , e 1 = ( 1 , 0 , 0 ) ${\bf e}_1=(1,0,0)$ , e 2 = ( 0 , 1 , 0 ) ${\bf e}_2=(0,1,0)$ , e 3 = ( 0 , 0 , 1 ) ${\bf e}_3=(0,0,1)$ , w = ( 1 , 1 , 0 ) ${\bf w}=(1,1,0)$ . Let u = ( m , 1 , b ) ${\bf u}=(m,1,-b)$ and v = ( n , 2 , 1 ) ${\bf v}=(-n,2,1)$ , where m , n , b $m,n,b$ are positive integers such that n b = m + 1 $nb=m+1$ and 2 n $2\nmid n$ . Then, all these vectors above are primitive in N.

Let Σ1 be the fan determined by the single maximal Cone ( e 3 , u , v ) $\text{Cone}({\bf e}_3,{\bf u},{\bf v})$ . Let X Σ 1 $X_{\Sigma _1}$ be the corresponding toric variety. Then, X Σ 1 $X_{\Sigma _1}$ is affine and the cyclic quotient singularity is of the form 1 2 m + n ( 1 , 2 , 2 b + 1 ) $\frac{1}{2m+n}(-1,2,2b+1)$ . Notice that X Σ 1 $X_{\Sigma _1}$ has an isolated singularity.

image

Let Σ 2 = Σ 1 ( e 2 ) $\Sigma _2=\Sigma _1^*({\bf e}_2)$ be the star subdivision of Σ1 at e2 as above (see [5, Chapter 11]) and X Σ 2 $X_{\Sigma _2}$ be the corresponding toric variety, then g : X Σ 2 X Σ 1 $g: X_{\Sigma _2}\rightarrow X_{\Sigma _1}$ is a birational morphism, which is an isomorphism outside the unique torus-invariant point P X Σ 1 $P\in X_{\Sigma _1}$ . Since Cone ( u , e 2 , v ) $\text{Cone}({\bf u},{\bf e}_2,{\bf v})$ is smooth, X Σ 2 $X_{\Sigma _2}$ has only two isolated singularities, which are of type 1 m ( 1 , 1 , b ) $\frac{1}{m}(1,-1,b)$ and 1 n ( 1 , 1 , 2 ) $\frac{1}{n}(1,-1,2)$ . In particular, X Σ 2 $X_{\Sigma _2}$ is terminal. We use D e 2 , D v , D u , D e 3 $D_{{\bf e}_2},\nobreakspace D_{{\bf v}},\nobreakspace D_{{\bf u}},\nobreakspace D_{{\bf e}_3}$ to denote the corresponding torus-invariant divisors. We can see that D e 2 $D_{{\bf e}_2}$ is the only exceptional divisor. Let R denote the proper curve in X Σ 2 $X_{\Sigma _2}$ that corresponds to Cone ( e 2 , e 3 ) Σ 2 $\text{Cone}({\bf e}_2,{\bf e}_3)\in \Sigma _2$ . Then, R D e 2 $R\subset D_{{\bf e}_2}$ . By [5, Proposition 6.4.4], D u · R = 1 m $D_{{\bf u}}\cdot R=\frac{1}{m}$ , D v · R = 1 n $D_{{\bf v}}\cdot R=\frac{1}{n}$ , D e 3 · R = b m 1 n > 0 $D_{{\bf e}_3}\cdot R=\frac{b}{m}-\frac{1}{n}&gt;0$ , and D e 2 · R = ( 2 n + 1 m ) $D_{{\bf e}_2}\cdot R=-(\frac{2}{n}+\frac{1}{m})$ . Therefore,

0 < 2 n b m = K X Σ 2 · R < 1 n $$\begin{equation*} 0&lt;\frac{2}{n}-\frac{b}{m}=K_{X_{\Sigma _2}}\cdot R&lt;\frac{1}{n} \end{equation*}$$
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Let Σ 3 = Σ 2 ( w ) $\Sigma _3=\Sigma _2^*({\bf w})$ be the star subdivision of Σ2 at w as above and X Σ 3 $X_{\Sigma _3}$ be the corresponding toric variety, then f : X Σ 3 X Σ 2 $f: X_{\Sigma _3}\rightarrow X_{\Sigma _2}$ is a birational morphism, which is an isomorphism outside the torus-invariant point Q X Σ 2 $Q\in X_{\Sigma _2}$ that corresponds to the maximal Cone ( w , e 2 , e 3 ) Σ 2 $\text{Cone}({\bf w},{\bf e}_2,{\bf e}_3)\in \Sigma _2$ . We use D e 2 , D v , D u , D e 3 , D w $D^{\prime }_{{\bf e}_2},\nobreakspace D^{\prime }_{{\bf v}},\nobreakspace D^{\prime }_{{\bf u}},\nobreakspace D^{\prime }_{{\bf e}_3},\nobreakspace D^{\prime }_{{\bf w}}$ to denote the corresponding torus-invariant divisors. Notice that D w $D^{\prime }_{{\bf w}}$ is the only exceptional divisor of f and D e 2 , D v , D u , D e 3 $D^{\prime }_{{\bf e}_2},\nobreakspace D^{\prime }_{{\bf v}},\nobreakspace D^{\prime }_{{\bf u}},\nobreakspace D^{\prime }_{{\bf e}_3}$ are the birational transforms of D e 2 , D v , D u , D e 3 $D_{{\bf e}_2},\nobreakspace D_{{\bf v}},\nobreakspace D_{{\bf u}},\nobreakspace D_{{\bf e}_3}$ on X Σ 3 $X_{\Sigma _3}$ . Let R $R^{\prime }$ denote the birational transform of R on X3, then R $R^{\prime }$ corresponds to Cone ( e 2 , e 3 ) Σ 3 $\text{Cone}({\bf e}_2,{\bf e}_3)\in \Sigma _3$ .

Since w = 1 m u + b m e 3 + m 1 m e 2 ${\bf w}=\frac{1}{m}{\bf u}+\frac{b}{m}{\bf e}_3+\frac{m-1}{m}{\bf e}_2$ , we have K X Σ 3 = f K X Σ 2 + 1 m + b m + m 1 m 1 D w $K_{X_{\Sigma _3}}=f^*K_{X_{\Sigma _2}}+\left(\frac{1}{m}+\frac{b}{m}+\frac{m-1}{m}-1\right)D^{\prime }_{{\bf w}}$ , hence

f K X Σ 2 = K X Σ 3 b m D w . $$\begin{equation*} f^*K_{X_{\Sigma _2}}=K_{X_{\Sigma _3}}-\frac{b}{m}D^{\prime }_{{\bf w}}. \end{equation*}$$

By [5, Lemma 6.4.2], D w · R = 1 $D^{\prime }_{\bf w}\cdot R^{\prime }=1$ . Thus, for any positive integer k,

k f K X Σ 2 · R = 2 n b m k m k b m . $$\begin{equation*} \lfloor kf^*K_{X_{\Sigma _2}}\rfloor \cdot R^{\prime }={\left(\frac{2}{n}-\frac{b}{m}\right)}k-{\left\lbrace \frac{m-kb}{m}\right\rbrace} . \end{equation*}$$

Step 2. Next, we will use covering trick to make the canonical divisor ample.

Choose a projective threefold Z with the isolated quotient singularity of type 1 2 m + n ( 1 , 2 , 2 b + 1 ) $\frac{1}{2m+n}(-1,2,2b+1)$ at P, after resolving singularities away from P, we may assume that P is the only singular point on Z. By abuse of notation we continue to use f : Y X $f:Y\rightarrow X$ and g : X Z $g: X\rightarrow Z$ to denote the corresponding toric blow-ups defined in Step 1. Let E be the exceptional divisor of g and R E $R\subset E$ be the proper curve defined in Step 1. Then, E $-E$ is g-ample and we have f K Z a E = K X $f^*K_Z-aE=K_X$ , where

a = 2 n b m 2 n + 1 m > 0 . $$\begin{equation*} a=\frac{\frac{2}{n}-\frac{b}{m}}{\frac{2}{n}+\frac{1}{m}}&gt;0. \end{equation*}$$

Let L be a sufficiently ample Cartier divisor on Z such that g ( L + K Z ) a E $g^*(L+K_Z)-aE$ is ample on X. We can find an effective A 2 L $A\sim 2L$ that is smooth and avoids P. Let h : Z Z $h:Z^{\prime }\rightarrow Z$ be the double cover ramified along A. Then, by Hurwitz's Formula, we have K Z = h K Z + 1 2 A $K_Z^{\prime }=h^*\left(K_Z+\frac{1}{2}A\right)$ and h is étale around P. Let X , Y , E , f , g $X^{\prime }, Y^{\prime },E^{\prime },f^{\prime },g^{\prime }$ be the corresponding base change of h. Then, K X = h X K X + 1 2 g A = h X g ( 1 2 A + K Z ) a E $K_{X^{\prime }}=h_X^*\left(K_X+\frac{1}{2}g^*A\right)=h_X^*\left(g^*(\frac{1}{2}A+K_Z)-aE\right)$ is ample, where h X : X = X × Z Z X $h_X: X^{\prime }=X\times _Z Z^{\prime }\rightarrow X$ is the canonical projection. Since R is a proper curve in one of the components of E $E^{\prime }$ as in Step 1 and R $R^{\prime }$ its birational transform in Y $Y^{\prime }$ , we have

m 0 f K X · R = 2 n b m m 0 m m 0 b m $$\begin{equation*} \lfloor m_0f^*K_{X^{\prime }}\rfloor \cdot R^{\prime }={\left(\frac{2}{n}-\frac{b}{m}\right)}m_0-{\left\lbrace \frac{m-m_0b}{m}\right\rbrace} \end{equation*}$$
for any positive integer m0 since Z $Z^{\prime }$ and X Σ 1 $X_{\Sigma _1}$ are isomorphic around the isolated singular point and the computation is local.

Now we can choose m , n 0 $m,n\gg 0$ such that 2 b m 0 < n < m $2bm_0&lt;n&lt;m$ , then

2 n b m m 0 m m 0 b m < m 0 n 1 2 < 0 . $$\begin{equation*} {\left(\frac{2}{n}-\frac{b}{m}\right)}m_0-{\left\lbrace \frac{m-m_0b}{m}\right\rbrace} &lt;\frac{m_0}{n}-\frac{1}{2}&lt;0. \end{equation*}$$
Therefore, m 0 f K X · R < 0 $\lfloor m_0f^*K_{X^{\prime }}\rfloor \cdot R^{\prime }&lt;0$ , which means any effective divisor in | m 0 f K X | $|m_0f^*K_{X^{\prime }}|$ contains R $R^{\prime }$ . Since f $f^{\prime }$ is isomorphic over the generic point of R, this implies that any effective divisor in | m 0 K X | $|m_0K_{X^{\prime }}|$ contains R. $\Box$

ACKNOWLEDGMENTS

The authors would like to thank Christopher D. Hacon for useful discussions and encouragements. They would like to thank Chenyang Xu for proposing Question 1 to them and sharing useful comments to this question. The authors would like to thank useful discussions with Paolo Cascini, Guodu Chen, Jingjun Han, Junpeng Jiao, Yuchen Liu, Yujie Luo, and Qingyuan Xue. They would like to thank the referees for useful suggestions.

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