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. 2023 Feb 7;7(1):1–33. doi: 10.1007/s42543-022-00057-x

Boundedness of Complements for Log Calabi–Yau Threefolds

Guodu Chen 1, Jingjun Han 2,, Qingyuan Xue 3
PMCID: PMC10913441  PMID: 38444737

Abstract

In this paper, we study the theory of complements, introduced by Shokurov, for Calabi–Yau type varieties with the coefficient set [0, 1]. We show that there exists a finite set of positive integers N, such that if a threefold pair (X/Zz,B) has an R-complement which is klt over a neighborhood of z, then it has an n-complement for some nN. We also show the boundedness of complements for R-complementary surface pairs.

Keywords: Complements, Log Calabi–Yau pairs, Fano varieties

Introduction

We work over the field of complex numbers C.

The theory of complements (for Fano varieties) was introduced by Shokurov when he proved the existence of flips for threefolds [45]. It originates from his earlier work on anti-canonical systems on Fano threefolds [44]. The boundedness of complements [4, 26, 47] played an important role in various contexts in the study of Fano varieties, including the solution of the Borisov–Alexeev–Borisov conjecture (boundedness of Fano varieties) [4, 5] and the Yau–Tian–Donaldson conjecture (the existence of Kähler–Einstein metrics on log Fano pairs) [8, 40, 49]. We refer the reader to [911, 13, 14, 16, 25, 26, 38] and references therein for more recent progress and applications.

According to the minimal model program, varieties of general type, Fano varieties and Calabi–Yau varieties form three fundamental classes in birational geometry and are building blocks of algebraic varieties. In this paper, we study the theory of complements for Calabi–Yau type varieties with the coefficient set [0, 1] in dimensions 2 and 3. Note that Calabi–Yau type varieties form a large class of varieties which includes both Fano varieties and Calabi–Yau varieties. For Calabi–Yau varieties, since the boundedness of complements implies the boundedness of the non-vanishing index of KX, we expect that the theory of complements will play an important role in the study of Calabi–Yau varieties, including the boundedness of Calabi–Yau varieties. We also remark that replacing a coefficient set which satisfies the descending chain condition (DCC) with the set [0, 1] is considered as a very hard problem in the theory of complements.

Our first main result is the boundedness of complements for threefold pairs.

Theorem 1.1

Let l be a positive integer. Then, there exists a finite set of positive integers N depending only on l satisfying the following.

Assume that (X/Zz,B) is a threefold pair which has an R-complement that is klt over a neighborhood of z. Then, (X/Zz,B) has an n-complement for some nN such that ln.

Theorem 1.1 fails if we remove the assumption “klt over a neighborhood of z”; see [47, Example 11]. However, if we require the coefficients of the boundaries to lie in a set Γ[0,1] such that ΓQ is DCC, then we can remove the klt assumption.

Theorem 1.2

Let l be a positive integer, and Γ[0,1] a set such that ΓQ is DCC. Then, there exists a finite set of positive integers N depending only on l and Γ satisfying the following.

Assume that (X/Zz,B) is an R-complementary threefold pair such that X is of Calabi–Yau type over a neighborhood of z and BΓ. Then, (X/Zz,B) has an n-complement for some nN such that ln.

Here, we say that X is of Calabi–Yau type over a neighborhood of z, if there exists a boundary C on X such that (XC) is klt and KX+CR,Z0 over a neighborhood of z; see Definition 7.1.

Our last main result is the boundedness of complements for surface pairs where we do not require the pair has a klt R-complement nor ΓQ is DCC. Theorem 1.3 completely answers a question of Shokurov [46, 1.3 Conjecture on complements] for surfaces.

Theorem 1.3

Let l be a positive integer. Then, there exists a finite set of positive integers N depending only on l satisfying the following.

Assume that (X/Zz,B) is an R-complementary surface pair. Then, (X/Zz,B) has an n-complement for some nN such that ln.

Sketch of proofs. We now sketch the proofs of Theorems 1.1 and 1.3. For convenience, in what follows, we will assume that l=1 and (XB) is a Q-factorial klt log Calabi–Yau pair, that is, Z=z={pt}, (XB) is Q-factorial klt and KX+BR0.

We first sketch the proof of Theorem 1.3. If X is of Fano type, then (XB) is N1-complementary for some finite set of positive integers N1 by Theorem 2.19; here, (XB) being N1-complementary means that (XB) is n-complementary for some nN1 (see Definition 2.12). Thus, we may assume that X is not of Fano type and κ(X,B-BΦ1)1, where Φ1:=Γ(N1,{0,1}) is a hyperstandard set and BΦ1 is a Q-divisor with coefficients in Φ1 such that 0BΦ1B (see Definition 2.1). Suppose that κ(X,B-BΦ1)=κ(X,B-BΦ2)=1, where N2 is a finite set of positive integers given by Theorem 2.20 and Φ2:=Γ(N1N2,{0,1}). In this case, we claim that (XB) is N2-complementary. Indeed, although X is not of Fano type, by Lemma 2.15 we can still run an MMP on -(KX+BΦ2) and get a good minimal model X such that -(KX+BΦ2) is semi-ample and hence defines a contraction π:XZ, where D denotes the strict transform of D on X for any R-divisor D on X. Then, we run an MMP on -(KX+BΦ1) over Z and reach a model X on which -(KX+BΦ1) is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. As κ(X,B-BΦ1)=κ(X,B-BΦ2)=1, the natural morphism π:XZ is the contraction defined by -(KX+BΦ1) over Z. By the similar arguments as in [4, Proposition 6.3] and using Effective Adjunction [43, Conjecture 7.13.3 and Theorem 8.1], there exists a positive integer p which only depends on Φ1 such that p(KX+BΦ1)p(π)(KZ+BZ(1)+Mπ,Z) and pMπ is base point free, where BZ(1) and Mπ are given by the canonical bundle formula for (X,BΦ1) over Z in Proposition 3.3. It follows that p(KX+BΦ2)p(π)(KZ+BZ(2)+Mπ,Z) and pMπ is base point free, where BZ(2) and Mπ are given by the canonical bundle formula for (X,BΦ2) over Z. We may assume that pn for any nN2. As pMπ is base point free, one can find an effective Q-divisor MZ such that pMZpMπ,Z, MZ has no common components with BZ(2), and (Z,BZ(2)+MZ) has an n-complement for some nN2. Then, we can lift this complement to X and get an n-complement of (XB); see Proposition 3.5. If κ(X,B-BΦ2)=0, then we can easily show that n0(KX+B)0 for some positive integer n0 which only depends on Φ2; see Lemma 2.18. Hence, N1N2{n0} has the required property.

Now, we sketch the proof of Theorem 1.1. The main strategy is similar. One of the key steps is to construct a positive integer n0 and finite sets of positive integers Ni (i=1,2,3) such that

  1. if κ(X,B-BΦ1)=3, then (XB) is N1-complementary,

  2. if κ(X,B-BΦ1)=κ(X,B-BΦ2)=2, then (XB) is N2-complementary,

  3. if κ(X,B-BΦ2)=κ(X,B-BΦ3)=1, then (XB) is N3-complementary, and

  4. if κ(X,B-BΦ3)=0, then (XB) is n0-complementary,

where Φi:=Γ(j=1iNj,{0,1}) for any 1i3; see Sect. 6 for the details. However, there are some issues when we construct these finite sets. One issue is that when we apply the canonical bundle formula, Effective Adjunction is still open when the relative dimension is 2. But in our setting we can give a positive answer to Effective Adjunction; see Proposition 3.4 for the details. On the other hand, there is also an issue when we try to lift complements from lower dimensional varieties. More precisely, it may happen that some components of SuppB have images of codimension 2 in Z. Therefore, we must lift complements more carefully; see Proposition 3.5 and Sect. 6 for the details.

Structure of the paper. We outline the organization of the paper. In Sect. 2, we introduce some notation and tools which will be used in this paper, and prove certain basic results. In Sect. 3, we recall the canonical bundle formula, some well-known results, as well as some new results. In Sect. 4, we prove the boundedness of complements for sdlt curves. In Sect. 5, we prove Theorem 1.3. In Sect. 6, we prove Theorem 1.1. In Sect. 7, we prove Theorem 1.2.

Preliminaries

Arithmetic of Sets

Definition 2.1

(1) We say that a set Γ[0,1] satisfies the descending chain condition (DCC) if any decreasing sequence a1a2 in Γ stabilizes. We say that Γ satisfies the ascending chain condition (ACC) if any increasing sequence a1a2 in Γ stabilizes.

(2) Suppose that R[0,1]Q is a finite set. We define

Φ(R):=1-rl|rR,lZ>0[0,1]

to be the set of hyperstandard multiplicities associated to R (cf. [4, 2.2]). We may say that Φ(R) is the hyperstandard set associated to R. When we say Φ[0,1]Q is a hyperstandard set, we mean that Φ=Φ(R) for some finite set R[0,1]Q. We usually assume 0,1R without mentioning, so Φ({0,1})Φ(R).

(3) (cf. [47, Page 30]) Let NZ>0, R[0,1]Q be two finite sets, and Φ:=Φ(R). We define

Γ(N,Φ(R)):={1-rl+1lnNmnn+1|rR,lZ>0,mnZ0}[0,1].

By Remark 2.2 (1), Γ(N,Φ(R)) is independent of the choice of R. Hence, we may write Γ(N,Φ) instead of Γ(N,Φ(R)) for convenience. By Remark 2.2 (2), Γ(N,Φ) is a hyperstandard set. In particular, it is a DCC set whose only accumulation point is 1. Then, for any b[0,1], we define

bN_Φ:=maxb|bb,bΓ(N,Φ).

If N={n} (respectively, N=), we may write bn_Φ (respectively, bΦ) rather than bN_Φ.

Remark 2.2

(1) If R[0,1]Q is a finite set such that Φ(R)=Φ(R), then Γ(N,Φ(R))=Γ(N,Φ(R)). Indeed, for any rR, there exist rR and lZ>0 such that r=r/l. Thus, Γ(N,Φ(R))Γ(N,Φ(R)), and the converse inclusion follows similarly.

(2) Γ(N,Φ) is the hyperstandard set associated to the following finite set:

R:={r-nNmnn+1|rR,mnZ0}[0,1].

Indeed,

Φ(R)={1-rl|rR,lZ>0}[0,1]={1-rl+1lnNmnn+1|rR,lZ>0,mnZ0}[0,1]=Γ(N,Φ).

(3) If N1 and N2 are two finite sets of positive integers, then

Γ(N1N2,Φ)=Γ(N2,Γ(N1,Φ)).

Indeed, let

R1:={r-nN1mnn+1|rR,mnZ0}[0,1].

Then, Γ(N1,Φ)=Φ(R1) by (2). Therefore,

Γ(N2,Γ(N1,Φ))=Γ(N2,Φ(R1))={1-r1l+1lnN2mnn+1|r1R1,lZ>0,mnZ0}[0,1]={1-1l(r-nN1mn+1)+1lnN2mnn+1|rR,lZ>0,m,mnZ0}[0,1]={1-rl+1l(nN1mn+1+nN2mnn+1)|rR,lZ>0,mn,mZ0}[0,1]=Γ(N1N2,Φ).

The following lemma was observed by the second named author. It will play an important role in the proof of the main theorems.

Lemma 2.3

Assume that R[0,1]Q is a finite set, Φ:=Φ(R), and n is a positive integer such that nRZ. Then, for any b[0,1], we have

(n+1){b}n+bbn_Φ.

Proof

Without loss of generality, we may assume that 1>b=bn_Φ=1-rl+ml(n+1)Γ({n},Φ) for some rR,lZ>0 and mZ0. It suffices to show that

1-(n+1)rl+ml-nrl+mnl(n+1).

Suppose on the contrary that there exists an integer k such that

1-(n+1)rl+ml<kand-nrl+mnl(n+1)>k-1.

The first inequality above gives us that l-lk+m<(n+1)r, and thus l-lk+mnr as nrZ. Therefore, we have

mnn+1>(k-1)l+nr(k-1)l+l-lk+m=m,

a contradiction.

Lemma 2.4

Let N be a finite set of positive integers, Φ a hyperstandard set, and nN. Suppose that b,b+[0,1] such that nb+Z and

nb+(n+1){bN_Φ}+nbN_Φ.

Then, nb+(n+1){b}+nb.

Proof

If b=1, then b+=bN_Φ=1 and there is nothing to prove.

If b<1, then bN_Φb<1. It suffices to show that (n+1)b=(n+1)bN_Φ. Let

b:=max{ln+1|ln+1b,lZ0}Γ(N,Φ).

By the construction, (n+1)b=(n+1)b, which implies that (n+1)b=(n+1)bN_Φ as bbN_Φb.

Divisors

Let F be either the rational number field Q or the real number field R. Let X be a normal variety and WDiv(X) the free abelian group of Weil divisors on X. Then, an F-divisor is defined to be an element of WDiv(X)F:=WDiv(X)F.

A b-divisor on X is an element of the projective limit

WDiv(X)=limYXWDiv(Y),

where the limit is taken over all the pushforward homomorphisms f:WDiv(Y)WDiv(X) induced by proper birational morphisms f:YX. In other words, a b-divisor D on X is a collection of Weil divisors DY on higher models Y of X that are compatible with respect to pushforward. The divisor DY is called the trace of D on the birational model Y. A b-F-divisor is defined to be an element of WDiv(X)F, and the trace of a b-F-divisor is defined similarly.

The Cartier closure of an F-Cartier F-divisor D on X is the b-F-divisor D¯ with trace D¯Y=fD for any proper birational morphism f:YX. A b-F-divisor D on X is b-F-Cartier if D=DY¯ where DY¯ is an F-Cartier F-divisor on a birational model Y of X; in this situation, we say D descends to Y. Moreover, if DY is a Cartier divisor, then we say D is b-Cartier. Let XZ be a projective morphism. Then, a b-F-divisor is nef (respectively, base point free) over Z if it descends to a nef (respectively, base point free) over Z F-divisor on some birational model of X.

Assume that Γ[0,1] is a set, and B:=biBi, B:=biBi are two R-divisors on X, where Bi are prime divisors. By BΓ we mean biΓ for any i. We define B:=biBi,{B}:={bi}Bi,B:=max{|bi|}, and BB:=min{bi,bi}Bi. Assume that N is a finite set of positive integers and Φ is a hyperstandard set. We define

BN_Φ:=(bi)N_ΦBi.

If N={n} (respectively, N=), we may write Bn_Φ (respectively, BΦ) instead of BN_Φ.

Definition 2.5

(1) We say π:XZ is a contraction if X and Z are normal quasi-projective varieties, π is a projective morphism, and πOX=OZ.

(2) We say that a birational map ϕ:XY is a birational contraction if ϕ is projective and ϕ-1 does not contract any divisors.

Lemma 2.6

Suppose that τ:ZZ and ZZ are contractions. Suppose that H (respectively, H) is an R-Cartier R-divisor on Z (respectively, Z) which is ample over Z (respectively, Z). Then, ϵH+τH is ample over Z for any 0<ϵ1.

Proof

Pick any closed point zZ. Let Zz:=Z×Z{z}, Zz:=Z×Z{z}, Hz:=H|Zz, and Hz:=H|Zz. By assumption Hz is ample over Zz and Hz is ample. According to [36, Proposition 1.45], ϵzHz+(τ|Zz)Hz is ample for any 0<ϵz1. In particular, ϵzH+τH is ample over z. It follows that ϵzH+τH is ample over some neighborhood of z by [37, Theorem 1.2.17]. Since Z is quasi-compact, the lemma follows.

Generalized Pairs and Singularities

In this paper, we usually discuss the (sub-)pair in the relative setting (X/Zz,B); we refer the reader to [11, §2] (cf. [6, 36]). Moreover, if the (sub-)pair (X/Zz,B) is (sub-)lc over z for any zZ, then we say (X/ZB) is (sub-)lc.

Here, we briefly discuss the analogous concepts for generalized pairs, and refer the reader to [7, 22, 24] for further details.

Definition 2.7

A generalized pair (g-pair for short) (X/Z,B+M) consists of a contraction XZ, an effective R-divisor B on X, and a nef/Z b-R-divisor M on X, such that KX+B+MX is R-Cartier.

Let (X/Z,B+M) be a g-pair and f:WX a log resolution of (X,SuppB) to which M descends. We may write

KW+BW+MW=f(KX+B+MX)

for some R-divisor BW on W. Let E be a prime divisor on W. The log discrepancy of E with respect to (X,B+M) is defined as

a(E,X,B+M):=1-multEBW.

We say (X/Z,B+M) is generalized lc or glc (respectively, generalized klt or gklt) if a(E,X,B+M)0 (respectively, >0) for any prime divisor E over X.

We say that two g-pairs (X/Z,B+M) and (X/Z,B+M) are crepant if X is birational to X, M=M, and f(KX+B+MX)=(f)(KX+B+M) for some common resolution f:WX and f:WX. We also call (X/Z,B+M) a crepant model of (X/Z,B+M).

Lemma 2.8

Let d be a positive integer and Γ[0,1] a DCC set. Then, there is a positive real number ϵ depending only on d and Γ satisfying the following. Assume that (XB) is a projective klt pair of dimension d such that KX+BR0 and BΓ. Then, (XB) is ϵ-lc.

Proof

The lemma follows from [4, Lemma 2.48].

Definition 2.9

Let XZ be a contraction and D an R-Cartier R-divisor on X. We denote by κ(X,D) and κ(X/Z,D) the Iitaka dimension and relative Iitaka dimension of D respectively; see [41, II, §3.b and §3.c].

Definition 2.10

Let XZ be a contraction, D an R-Cartier R-divisor on X, and ϕ:XY a birational contraction of normal quasi-projective varieties over Z. We say that Y is a good minimal model of D over Z, if ϕ is D-negative, DY is semi-ample over Z, and Y is Q-factorial, where DY is the strict transform of D on Y.

Lemma 2.11

Let XZ be a contraction, and D2D1 two effective R-Cartier R-divisors on X. Suppose that XX is a sequence of steps of the D1-MMP over Z. Let D2 be the strict transform of D2 on X. Then, κ(X/Z,D2)=κ(X/Z,D2).

Proof

Pick a positive real number ϵ such that XX is also a sequence of steps of the (D1+ϵD2)-MMP over Z. As Supp(D1+ϵD2)=SuppD2, one can see that

κ(X/Z,D2)=κ(X/Z,D1+ϵD2)=κ(X/Z,D1+ϵD2)=κ(X/Z,D2),

where D1 is the strict transform of D1 on X.

Complements

Definition 2.12

We say that a pair (X/Zz,B+) is an R-complement of (X/Zz,B) if (X,B+) is lc, B+B, and KX+B+R0 over a neighborhood of z. In addition if (X,B+) is klt over a neighborhood of z, then we call (X/Zz,B+) a klt R-complement of (X/Zz,B).

Let n be a positive integer. An n-complement of (X/Zz,B) is a pair (X/Zz,B+), such that over some neighborhood of z,  we have

  1. (X,B+) is lc,

  2. n(KX+B+)0, and

  3. nB+nB+(n+1){B}.

We say that (X/Zz,B+) is a monotonic n-complement of (X/Zz,B) if we additionally have B+B.

Let N be a non-empty set of positive integers. We say that (X/Zz,B) is N-complementary (respectively, n-complementary, R-complementary) if (X/Zz,B) has an m-complement (respectively, n-complement, R-complement) for some mN. If (X/Zz,B) has an R-complement (respectively, n-complement) for any zZ, then we say that (X/ZB) is R-complementary (respectively, n-complementary).

Note that if zz¯ is a closed point and (X/Zz,B+) is an R-complement (respectively, an n-complement) of (X/Zz,B), then (X/Zz,B+) is an R-complement (respectively, an n-complement) of (X/Zz,B). Hence, when proving the existence of complements we may assume that zZ is a closed point.

The following lemma is well-known to experts. We will use the lemma frequently without citing it in this paper.

Lemma 2.13

(cf. [4, 6.1)], [11, Lemma 2.11]) Let n be a positive integer. Assume that (X/Zz,B) is a pair, ψ:XX is a birational contraction over Z, and B is the strict transform of B on X.

  1. If (X/Zz,B) is R-complementary (respectively, n-complementary), then so is (X/Zz,B).

  2. Suppose ψ is -(KX+B)-non-positive. If (X/Zz,B) has an R-complement (respectively, a monotonic n-complement), then so does (X/Zz,B).

The following lemma is an easy consequence of Lemmas 2.3 and 2.4.

Lemma 2.14

Let NZ>0, R[0,1]Q be two finite sets, Φ:=Φ(R), and n a positive integer such that nRZ0. Assume that (X/Zz,B) is a pair.

  1. If (X/Zz,BN_Φ) is N-complementary, then so is (X/Zz,B).

  2. Any n-complement of (X/Zz,B) is a monotonic n-complement of (X/Zz,Bn_Φ).

We will use the following lemma frequently in this paper.

Lemma 2.15

Let Φ[0,1]Q be a hyperstandard set. Assume that (X/Zz,B) is a pair which is an R-complement of itself. If either

  • dimX=2 and X is Q-factorial, or

  • dimX=3 and (XB) is dlt over a neighborhood of z,

then -(KX+BΦ) has a good minimal model over a neighborhood of z.

Proof

According to [11, Lemma 4.2], possibly shrinking Z near z, there exist a positive real number u and a surface (respectively, threefold) pair (X,Δ), such that the coefficients of Δ are at most 1 (respectively, (X,Δ) is dlt) and -u(KX+BΦ)R,ZKX+Δ. In both cases, we can run an MMP on KX+Δ over Z and reach a good minimal model X over Z by [18, 32, 48]. It is clear that X is a good minimal model of -(KX+BΦ) as -u(KX+BΦ)R,ZKX+Δ. This finishes the proof.

Boundedness of Complements

We propose a conjecture on the boundedness of complements and collect some useful results.

For a positive integer l and a non-empty set NZ>0, we say N is divisible by l, denoted by lN, if ln for any nN.

Conjecture 2.16

Let dl be two positive integers and Φ[0,1]Q a hyperstandard set. Then, there exists a finite set of positive integers N divisible by l depending only on dl and Φ satisfying the following.

Assume that (X/Zz,B) is a pair of dimension d such that (X/Zz,BN_Φ) has an R-complement which is klt over a neighborhood of z. Then, (X/Zz,B) is N-complementary.

Remark 2.17

  1. In Conjecture 2.16, we do not assume (X/Zz,B) is lc.

  2. One can not remove the klt assumption in Conjecture 2.16 when d3; see [47, Example 11]. However, we will show Conjecture 2.16 for R-complementary surface pairs without the klt assumption; see Theorem 1.3.

Lemma 2.18

Let Φ[0,1]Q be a hyperstandard set. Then, there exists a positive integer n depending only on Φ satisfying the following.

Assume that (XB) is a projective Q-factorial dlt pair of dimension 3 such that KX+BR0 and κ(X,B-BΦ)=0. Then, n(KX+B)0.

Proof

By Lemma 2.15, we may run an MMP on -(KX+BΦ)RB-BΦ which terminates with a good minimal model X. Let B be the strict transform of B on X. Since κ(X,B-BΦ)=0 and B-BΦ is semi-ample, we see that B=BΦ and, therefore, KX+BΦR0. By [11, Proposition 6.4, Theorem 1.1] and [12, Theorem 2.14], there is a positive integer n depending only on Φ such that

n(KX+B)=n(KX+BΦ)0.

It follows that n(KX+B)0 as (XB) and (X,B) are crepant.

We will use the following results on the boundedness of complements.

Theorem 2.19

(cf. [47, Theorem 16]) Let dl be two positive integers and Φ[0,1]Q a hyperstandard set. Then, there exists a finite set of positive integers N divisible by l depending only on dl and Φ satisfying the following.

Assume that (X/Zz,B) is a pair of dimension d such that X is of Fano type over Z and (X/Zz,BN_Φ) has a klt R-complement. Then, (X/Zz,B) is N-complementary.

Theorem 2.20

([11, Theorem 1.3]) Let l be a positive integer. Then, there exists a finite set of positive integers N divisible by l depending on l satisfying the following. If (X/Zz,B) is an R-complementary curve pair, then (X/Zz,B) is N-complementary.

Canonical Bundle Formulas

Canonical Bundle Formulas

For the definition and basic properties of the canonical bundle formula, we refer the reader to [4, 15, 27, 30]. Briefly speaking, suppose that (X/ZB) is a sub-pair and ϕ:XT is a contraction over Z, such that (XB) is lc over the generic point of T and KX+BR,T0. Then, there exist a uniquely determined R-divisor BT and a nef over Z b-R-divisor Mϕ which is determined only up to R-linear equivalence, such that (T/Z,BT+Mϕ) is a g-sub-pair and

KX+BRϕ(KT+BT+Mϕ,T).

Here, B (respectively, Mϕ) is called the discriminant part (respectively, a moduli part) of the canonical bundle formula for (X/ZB) over T. Moreover, if (X/ZB) is an lc (respectively, klt) pair, then (T/Z,BT+Mϕ) is a glc (respectively, gklt) g-pair.

It is worthwhile to point out that Mϕ only depends on (XB) over the generic point of T (cf. [4, 3.4 (2)]), and there are many choices of Mϕ, some of which could behave badly. But we can always choose one with the required properties, e.g., Propositions 3.3 and 3.4.

Lemma 3.1

Notation as above.

  1. Assume that (XB) is a klt pair. Then, there exists a crepant model (T~,BT~+Mϕ)(T,BT+Mϕ) such that for any prime divisor PSuppB which is vertical over T, the image of P on T~ is a prime divisor.

  2. Suppose that there is an R-divisor GT on T such that (T,BT+GT+Mϕ) is a sub-glc g-sub-pair. If we let G:=ϕGT, then (X,B+G) is sub-lc.

Proof

(1) According to [29, Theorem B.6] (cf. [1, Theorem 0.3], [31, Theorem 2], and [22, Theorem 2.8]), there exist birational morphisms XX and TT such that XT is an equidimensional contraction. In particular, for any prime divisor PSuppB which is vertical over T, the image Q of P on T is a prime divisor. Moreover, by the canonical bundle formula, a(Q,T,BT+Mϕ)<1 as a(P,X,B)<1. Since (XB) is a klt pair, (T,BT+Mϕ) is a gklt g-pair. Therefore, (1) holds by [7, Lemma 4.6].

(2) Suppose on the contrary that (X,B+G) is not sub-lc. Let P be a non-sub-lc place of (X,B+G), i.e., a(P,X,B+G)<0. It is clear that CenterX(P)SuppG which is vertical over T. We can find birational morphisms f:XX and g:TT such that XT is a contraction, P is a prime divisor on X, and the image Q of P on T is a prime divisor (cf. [34, VI, Theorem 1.3]). We may write KX+Δ:=f(KX+B+G)andKT+ΔT+Mϕ,T:=g(KT+BT+GT+Mϕ,T) for some R-divisors Δ and ΔT. Then, ΔT is the discriminant part of the canonical bundle formula for (X/Z,Δ) over T; see [43, Lemma 7.4 (ii)]. Since (T,BT+GT+Mϕ) is sub-glc, multQΔT1. By the definition of the canonical bundle formula, (X,Δ) is sub-lc over the generic point of Q. In particular, multPΔ=1-a(P,X,B+G)1, a contradiction.

Lemma 3.2

Let p be a positive integer, (XB) and (X,B) two lc pairs, and ϕ:XT a contraction, such that BB, KX+BR,T0, and KX+BR,T0. Let BT (respectively, BT) and Mϕ be the discriminant part and a moduli part of the canonical bundle formula for (XB) (respectively, (X,B)) over T. If p(KX+B)pϕ(KT+BT+Mϕ,T), then p(KX+B)pϕ(KT+BT+Mϕ,T).

Proof

Since B-B0 and B-BR,T0, B-B=ϕHT for some R-Cartier R-divisor HT on T by [12, Lemma 2.5]. Then, BT=BT+HT by [43, Lemma 7.4 (ii)]. Therefore,

p(KX+B)=p(KX+B)+p(B-B)pϕ(KT+BT+Mϕ,T)+pϕHT=pϕ(KT+BT+HT+Mϕ,T)=pϕ(KT+BT+Mϕ,T).

Proposition 3.3

Let R[0,1]Q be a finite set, and Φ:=Φ(R). Then, there exist a positive integer p and a hyperstandard set Φ[0,1]Q depending only on Φ satisfying the following.

Assume that (X/ZB) is an lc pair of dimension 3 and ϕ:XT is a contraction over Z such that dimT>0, BΦ, and KX+BQ,T0. Then, we can choose a moduli part Mϕ of the canonical bundle formula for (XB) over T, such that BTΦ, pMϕ is b-Cartier, and

p(KX+B)pϕ(KT+BT+Mϕ,T),

where BT is the discriminant part of the canonical bundle formula for (XB) over T. Moreover, if dimT=dimX-1, then pMϕ is base point free over Z.

Proof

The result follows from [12, Proposition 3.1] and [16, Theorem 5.5].

Proposition 3.4

Let Φ[0,1]Q be a hyperstandard set. Then, there exists a positive integer p depending only on Φ satisfying the following.

Assume that (X/ZB) is a klt threefold pair and ϕ:XT is a contraction over Z, such that dimT=1, BΦ, and KX+BQ,Z0. Then, we can choose a moduli part Mϕ of the canonical bundle formula for (XB) over T, such that

p(KX+B)pϕ(KT+BT+Mϕ,T),

and pMϕ is base point free over Z, where BT is the discriminant part of the canonical bundle formula for (XB) over T.

Proof

If dimZ=1, then T=Z. It follows that p1Mϕ,T is Cartier and thus p1Mϕ,TZ0, where p1 is given by Proposition 3.3 depending only on Φ, and Mϕ is a moduli part chosen as in Proposition 3.3. Therefore, in what follows, we may assume that dimZ=0, i.e., Z is a point.

If KT0, then T=P1. Let Mϕ be a moduli part chosen as in Proposition 3.3. Then, p1Mϕ,T is base point free.

Now, assume that KT0 and in particular, BT=0. Let F be a general fiber of XT and KF+BF:=(KX+B)|FQ0. According to [11, Proposition 6.4 and Theorem 1.1], r(KF+BF)0 for some positive integer r depending only on Φ. Then, there exist a rational function αK(X) and an R-Cartier R-divisor L on T such that KX+B+1r(α)=ϕL. Let Mϕ,T=L-KT-BT. Then, r(KX+B)rϕ(KT+BT+Mϕ,T). Let bF be the second Betti number of a smooth model of the index one cover of F. By Lemma 2.8, there exists a positive real number ϵ which only depends on Φ such that (F,BF) is ϵ-lc. If BF0, then F belongs to a bounded family by [2, Theorem 6.9], and hence bF has an upper bound. If BF=0, then KFQ0, and hence bF22 by the classification of surfaces. Therefore, by [17, Theorem 1.2], there exists a positive integer p2 depending only on bF such that p2Mϕ,T0.

We conclude that p:=p1p2r has the required property.

Lifting Complements

Now, we turn to the following technical statement on lifting complements via the canonical bundle formula.

Proposition 3.5

Let p and n be two positive integers such that pn. Let (X/ZB) be an lc pair and ϕ:XT a contraction over Z such that dimT>0 and KX+BR,T0. Let BT and Mϕ be the discriminant part and a moduli part of the canonical bundle formula for (XB) over T, such that p(KX+B)pϕ(KT+BT+Mϕ,T) and pMϕ is b-Cartier. Let (T,BT+Mϕ)(T,BT+Mϕ) be a crepant model and MT an effective Q-divisor on T, such that

  1. for any prime divisor PSuppB which is vertical over T, the image of P on T is a prime divisor,

  2. pMTZpMϕ,T and MTBT=0, and

  3. (T/Zz,BT+MT) is n-complementary for some zZ.

Then, (X/Zz,B) is also n-complementary.

Proof

Let X be the normalization of the main component of X×TT. Denote by f:XX and ϕ:XT the induced morphisms. We may write KX+B=f(KX+B) for some R-divisor B. Note that by our assumption, we have

p(KX+B)pϕ(KT+BT+Mϕ,T)Zpϕ(KT+BT+MT).

Let (T/Zz,BT++MT) be an n-complement of (T/Zz,BT+MT). We remark that as pn, B+0. Possibly shrinking Z near z, we may assume that

n(KT+BT++MT)Z0.

Let B+:=B+ϕ(BT+-BT) and B+:=fB+. We claim that (X/Zz,B+) is an n-complement of (X/Zz,B). Indeed, we have

n(KX+B+)=n(KX+B)+n(B+-B)Znϕ(KT+BT+MT)+nϕ(BT+-BT)=nϕ(KT+B++MT)Z0.

Hence, n(KX+B+)Z0. According to Lemma 3.1 (2), the sub-pair (X/Zz,B+) is sub-lc, and thus (X/Zz,B+) is also sub-lc. It suffices to prove that

nB+(n+1){B}+nB.

Let PSuppB+ be a prime divisor. If P is horizontal over T, then multPB+=multPB and there is nothing to prove. Therefore, we may assume that Q, the image of P on T, is a prime divisor. Let bP:=multPB, bP+:=multPB+,bQ:=multQBT, bQ+:=multQBT+, and mQ:=multPϕQ over the generic point of Q. It is clear that bQ+0 as BT+0. By construction,

bP+=bP+(bQ+-bQ)mQ.

Hence,

rPQ:=bP+(1-bQ)mQ=bP++(1-bQ+)mQ1nZ0.

Moreover, as 1-bQ is the lc threshold of ϕQ with respect to (X,B) over the generic point of Q, we know rPQ1. If bQ=1, then bQ+=1 and thus bP=bP+. If bP=1, then rPQ=bQ=1 and thus bP+=1. Hence, we may assume that bQ<1 and bP<1. Since bP=bP+-(bQ+-bQ)mQ and nbQ+(n+1)bQ, we can see that

(n+1)bP=(n+1)bP++(n+1)(bQ-bQ+)mQ=nbP++bP++((n+1)bQ-nbQ+)mQ-bQ+mQnbP++bP++{(n+1)bQ}mQ-bQ+mQ=nbP+,

where the last equality holds as

bP++{(n+1)bQ}mQ-bQ+mQ<bP++mQ-bQ+mQ=rPQ1.

We finish the proof.

Boundedness of Complements for sdlt Curves

Definition 4.1

We say X is a semismooth curve if X is a reduced scheme of dimension 1, every irreducible component of X is normal, and all of its singularities are simple normal crossing points.

Let X be a semismooth curve, and let B0 be an R-divisor on X. We say (XB) is sdlt if B is supported in the smooth locus of X and B1.

Definition 4.2

Let X be a semismooth curve, and B0 an R-divisor on X, such that (XB) is sdlt. We say that (X,B+) is an n-semi-complement of (XB), if

  1. (X,B+) is sdlt,

  2. nB+nB+(n+1){B}, and

  3. n(KX+B+)0.

Moreover, we say (X,B+) is monotonic if we additionally have B+B.

The following theorem is a generalization of [45, 5.2.2] and [33, 19.4 Theorem] where the case l=1 is proved.

Theorem 4.3

Let l be a positive integer. Then, there exists a finite set of positive integers Nsdlt divisible by l depending only on l satisfying the following.

Assume that X is a semismooth curve, connected but not necessarily complete, and B0 is an R-divisor on X, such that

  1. (XB) is sdlt,

  2. X has at least one complete component,

  3. each incomplete component of X does not meet any other incomplete component of X,

  4. the union of the complete components of X is connected, and

  5. -(KX+B) is nef on each complete component of X.

Then, there exists an n-semi-complement (X,B+) of (XB) in a neighborhood of the union of the complete components of X for some nNsdlt.

Proof

Let X0 be a complete component of X, and let {P1,,Pk}:=X0SingX. Then, deg(KX|X0)=2g-2+k, where g is the genus of X0. Since deg(KX|X0)0, there are four possibilities:

  • (i)

    g=1,k=0,deg(KX|X0)=0,

  • (ii)

    g=0,k=2,deg(KX|X0)=0,

  • (iii)

    g=0,k=1,deg(KX|X0)=-1,

  • (iv)

    g=0,k=0,deg(KX|X0)=-2.

We remark that B could not meet the components of type (i) or (ii) as deg(KX|X0)=0.

If X0 is of type (i), then X=X0 and B=0. In this case, (XB) is l-complementary.

If X0 is of type (iv), then X=X0 and XP1. By Theorem 2.20, there exists a finite set of positive integers N divisible by l depending only on l, such that (XB) is N-complementary.

Now, suppose that any complete component of X is either of type (ii) or of type (iii). Note that each component of type (ii) (respectively, type (iii)) can only meet other components at two points (respectively, one point). By assumptions (3) and (4), the entire curve X must form a chain or a cycle. If X is a cycle, then B=0 and KX0 by Lemma 4.4. Otherwise, by Lemma 4.5, it suffices to construct B+ such that (X,B+) is an n-complement of (XB) on each component of X. Note that possibly shrinking X near the union of the complete components, for any positive integer n, (XB) is an n-complement of itself on each incomplete component and each complete component of type (ii). Since X has at most two complete components of type (iii), by Lemma 4.6 there exists a finite set of positive integers N divisible by l depending only on l, such that (XB) has an n-complement on each complete component of type (iii) for some nN. Therefore, by Lemma 4.5, (XB) has an n-semi-complement for some nN.

Let Nsdlt:=NN and we are done.

Lemma 4.4

Let X=i=1mXi be a semismooth curve which is a cycle of irreducible curves Xi. Suppose that XiP1 for any 1im. Then, KX0.

Proof

For each integer mj2, we construct a semismooth curve Yj in a smooth projective surface Sj such that Yj is a cycle of j complete rational curves and KSj+Yj0, in particular, KYj=(KSj+Yj)|Yj0. Let Y2P2=:S2 be the union of a line and a conic which is semismooth. Then, KS2+Y20 and thus KY2=(KS2+Y2)|Y20. Suppose that we have constructed a semismooth curve Yj-1 contained in a smooth projective surface Sj-1, such that Yj-1 is a cycle of j-1 complete rational curves and KSj-1+Yj-10. Let πj:SjSj-1 be the blow-up of Sj-1 at one snc point of Yj-1, and Ej the exceptional divisor of πj. Let Yj=(πj)-1Yj-1Ej. Then, we get a semismooth curve YjSj, which is a cycle of j complete rational curves, such that KSj+Yj0. Since X is analytically isomorphic to Ym, by [28, Appendix B, Theorem 2.1], KYm0 implies KX0.

Lemma 4.5

Let X=i=1mXi be a semismooth curve which is a chain of irreducible curves Xi. Suppose that D is an R-divisor on X, supported in the smooth locus of X, such that D|Xi0 for any 1im. Then, D0.

Proof

Let X(i):=j=1iXj for 1im, and Pi:=XiXi+1=X(i)Xi+1 for 1im-1. We will prove by induction that D|X(i)0 for any 1im. Suppose that D|X(i-1)0 for some integer i2. Then, there exist a rational function αi-1 on X(i-1) and a rational function βi on Xi, such that D|X(i-1)=(αi-1) and D|Xi=(βi). Since Pi-1 is not contained in the support of D, αi-1 and βi are non-zero regular functions near Pi-1. Replacing βi by αi-1(Pi-1)βi(Pi-1)βi, we may assume that αi-1(Pi-1)=βi(Pi-1). Then, there exists a rational function αi on X(i) such that αi|X(i-1)=αi-1 and αi|Xi=βi. Hence, D|X(i)=(αi), and thus D|X(i)0. Therefore, by induction we see that D0.

Lemma 4.6

Let l be a positive integer. Then, there exists a finite set of positive integers N divisible by l depending only on l satisfying the following.

Assume that {ai}i=1k and {bi}i=1k are two sequences of non-negative real numbers, such that i=1kai1 and i=1kbi1. Then, there exist positive integers nN and kk, and two sequences of non-negative real numbers {ai+}i=1 and {bi+}i=1, such that

  1. i=1ai+=i=1bi+=1, and

  2. nai+nai+(n+1){ai} and nbi+nbi+(n+1){bi} for any 1ik.

Proof

Without loss of generality, we may assume that ai,bi<1 for any i. Then, it suffices to prove

n-i=1k(n+1)ai0andn-i=1k(n+1)bi0. 4.1

For any positive integer n and non-negative real numbers cd, we have

(n+1)(c+d)(n+1)c+(n+1)d.

Thus, possibly replacing (ai,aj) by (ai+aj,0) (respectively, (bi,bj) by (bi+bj,0)), we may assume that ai+aj1 (respectively, bi+bj1) for any ij. In particular, we may assume that k=2 and a1+a2=b1+b2=1.

By Dirichlet prime number theorem, there exist three distinct prime numbers qj such that lqj-1 for any j{1,2,3}. Let nj:=qj-1, and N:={n1,n2,n3}. We claim that there exists nN satisfying (4.1). It suffices to show that both (nj+1)a1 and (nj+1)b1 are not integers for some j{1,2,3}. Otherwise, by the pigeonhole principle, we may assume that a11nj+1Z[0,1)=1qjZ[0,1) for two indices j{1,2,3}, which is absurd.

Proposition 4.7

Let l be a positive integer. Then, there exists a finite set of positive integers N divisible by l depending only on l satisfying the following.

Assume that (X/Zz,B) is a surface pair such that z is a closed point, (XB) is dlt, S:=B0, B-S is big over Z and KX+BR,Z0. Then, over a neighborhood of z, and (S,BS) has an n-semi-complement for some nN, where KS+BS:=(KX+B)|S.

Proof

Let Nsdlt and N1 be finite sets of positive integers divisible by l given by Theorem 4.3 and Theorem 2.20, respectively, which only depend on l. We will show that N:=NsdltN1 has the required property.

It is clear that S is a semismooth curve, and (S,BS) is sdlt. We first show that S is connected over a neighborhood of z. Otherwise, there exists a contraction ϕ:XT to a curve T such that the general fiber F of ϕ is P1 and each connected component of S is horizontal over T; see Shokurov’s connectedness lemma [33, 17.4 Theorem] and [42, Propositions 3.3.1 and 3.3.2] (see also [45, 5.7 Connectedness lemma], [21, Corollary 1.3]). Note that if dimZ=1, then we take T=Z. As B-S is big over Z, B-S is horizontal over T and (B-S)|F. It follows that (KX+B)|F=(KX+S+B-S)|FR0, a contradiction. Thus, S is connected over a neighborhood of z. Possibly shrinking Z near z, we may assume that KX+BR0 and thus KS+BS is trivial on each complete component of S. If S has two irreducible incomplete components S1 and S2 that S1S2 over any neighborhood of z, then by assumption, we have B=S=S1+S2 over a neighborhood of z. In this case, BS=0 and KS0 over a neighborhood of z. Now, we assume that each irreducible incomplete component of S does not meet any other irreducible incomplete component of S. By the classification of dlt surface pairs (cf. [36, Corollary 5.55]), over a neighborhood of z, either the support of BS lies in the union of the complete components of S or S is irreducible and its image on Z is also a curve. In the former case, (S,BS) has an n-semi-complement in a neighborhood of the union of complete component of S for some nNsdlt by Theorem 4.3. Therefore, over a neighborhood of z, (S,BS) has an n-semi-complement. In the latter case, the morphism from S to its image on Z is a contraction, then (S,BS) has an n-complement over a neighborhood of z for some nN1. This finishes the proof.

Boundedness of Complements for Surfaces

Conjecture 2.16 for Surfaces

In this subsection, we confirm Conjecture 2.16 for surfaces. For convenience, by (Theorem )d we mean Theorem in dimension d.

Notation (). Let Φ1[0,1]Q be a hyperstandard set. Let p=p(Φ1) be a positive integer given by (Proposition 3.3)2 which only depends on Φ1. Let N2=N2(p) be a finite set of positive integers divisible by p given by Theorem 2.20 which only depends on p, and let Φ2:=Γ(N2,Φ1).

Proposition 5.1

Under Notation (), assume that (X/ZB) is a Q-factorial lc surface pair such that KX+BR,Z0 and κ(X/Z,B-BΦ2)+dimZ=κ(X/Z,B-BΦ1)+dimZ=1. Then, (X/Zz,B) is N2-complementary for any closed point zZ.

Proof

By Lemma 2.15 we can run an MMP on -(KX+BΦ2)R,ZB-BΦ2 over Z and reach a good minimal model ψ:XX over Z, such that -(KX+BΦ2) is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. Let π:XZ be the contraction defined by -(KX+BΦ2) over Z. By assumption, dimZ=1. Let BZ(2) and Mπ be the discriminant and moduli parts of the canonical bundle formula for (X,BΦ2) over Z in Proposition 3.3.

Claim 5.2

pMπ is base point free over Z and

p(KX+BΦ2)p(π)(KZ+BZ(2)+Mπ,Z).

Assume Claim 5.2. Then,

p(KX+B(2)):=pψ(KX+B2)p(πψ)(KZ+BZ(2)+Mπ,Z).

Note that since ψ is -(KX+BΦ2)-negative, B(2)BΦ2. Since KX+BR,Z0 and BBΦ2, there exists a boundary BZ on Z such that the g-pair (Z,BZ+Mπ) is glc, BZBZ(2), and KZ+BZ+Mπ,ZR,Z0. As pMπ is base point free over Z, we can pick an effective Q-divisor MZ on Z such that

pMZZpMπ,Z,MZBZ=0,and(Z,BZ+MZ)is lc.

In particular, (Z/Zz,BZ+MZ) is an R-complement of (Z/Zz,BZ(2)+MZ) for any zZ. Now, by our choice of N2, (Z/Zz,BZ(2)+MZ) is N2-complementary. According to Proposition 3.5, (X/Zz,B(2)) is N2-complementary, and hence (X/Zz,BΦ2) is also N2-complementary as B(2)BΦ2. Thus, (X/Zz,B) is N2-complementary by Lemma 2.14. Therefore, it suffices to prove Claim 5.2.

Proof of Claim 5.2

According to Lemma 2.15 again, we may run an MMP on -(KX+B1)R,ZB2-B1 over Z and reach a good minimal model XX over Z, such that B2-B1 is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. One can pick a positive real number ϵ, such that g:XX is also an MMP on B-BΦ2+ϵ(BΦ2-BΦ1) over Z. Furthermore, we may assume that B-BΦ2+ϵ(BΦ2-BΦ1) is semi-ample over Z by Lemma 2.6. graphic file with name 42543_2022_57_Figa_HTML.jpg By assumption,

κ(X/Z,B-BΦ2)=κ(X/Z,B-BΦ2+ϵ(BΦ2-BΦ1)),

and B-B2+ϵ(B2-B1)R,Zϵ(B2-B1). Hence, the natural morphism π:XZ is the contraction defined by B2-B1 over Z. In particular, we have

KX+B1R,Z0andKX+B2R,Z0.

By Lemma 3.2 and Proposition 3.3, we see that pMπ is base point free, and

p(KX+BΦ2)p(π)(KZ+BZ(2)+Mπ,Z).

Since XX is (KX+BΦ2)-trivial, (X,BΦ2) and (X,BΦ2) are crepant. Therefore,

p(KX+BΦ2)p(π)(KZ+BZ(2)+Mπ,Z).

We complete the proof.

Theorem 5.3

Let l be a positive integer and Φ[0,1]Q a hyperstandard set. Then, there exists a finite set of positive integers N divisible by l depending only on l and Φ satisfying the following.

Assume that (X/Zz,B) is a surface pair such that (X/Zz,BN_Φ) has a klt R-complement. Then, (X/Zz,B) is N-complementary.

Proof

Let N1=N1(l,Φ) be a finite set of positive integers divisible by l given by (Theorem 2.19)2 which only depends on l and Φ, and let Φ1:=Γ(N1,Φ). Let p=p(l,Φ1) be a positive integer divisible by l given by (Proposition 3.3)2 which only depends on l and Φ1. Let N2=N2(p) be a finite set of positive integers divisible by p given by Theorem 2.20 which only depends on p, and let Φ2:=Γ(N1N2,Φ). Let nCY=nCY(l,Φ2) be a positive integer divisible by l given by Lemma 2.18 which only depends on l and Φ2. We will show that the finite set N:=N1N2{nCY} has the required property.

Possibly replacing z by a closed point of z¯, we may assume that z is a closed point. Suppose that (X/Zz,B+) is a klt R-complement of (X/Zz,BN_Φ). Possibly replacing (XB) by a small Q-factorialization of (X,B+) and shrinking Z near z, we may assume that (XB) is Q-factorial klt and KX+BR,Z0. Since BBΦ2BΦ1,

0κ(X/Z,B-BΦ2)+dimZκ(X/Z,B-BΦ1)+dimZ2.

Therefore, we only need to consider the following three cases:

  1. κ(X/Z,B-BΦ1)+dimZ=2,

  2. κ(X/Z,B-BΦ2)+dimZ=κ(X/Z,B-BΦ1)+dimZ=1, and

  3. κ(X/Z,B-BΦ2)+dimZ=0.

If κ(X/Z,B-BΦ1)+dimZ=2, then X is of Fano type over Z. In this case (X/Zz,B) is N1-complementary by the choice of N1 (see Theorem 2.19). If κ(X/Z,B-BΦ2)+dimZ=κ(X/Z,B-BΦ1)+dimZ=1, then (X/Zz,B) is N2-complementary by Proposition 5.1. If κ(X/Z,B-BΦ2)+dimZ=0, that is, dimZ=0 and κ(X,B-BΦ2)=0, then one has

nCY(KX+B)0

by the choice of nCY (see Lemma 2.18). We finish the proof.

Proof of Theorem 1.3

Proposition 5.4

(cf. [33, 16.7 Corollary]) Let R[0,1]Q be a finite set and Φ:=Φ(R). Then, there exists a hyperstandard set Φ~ depending only on Φ satisfying the following.

Assume that (XB) is a dlt pair and S:=B. Let KS+BS:=(KX+B)|S. If BΦ, then BSΦ~, and if BΓ({n},Φ) for some positive integer n, then BSΓ({n},Φ~).

Proof

Let

R~:={1-(1-ri)0|riR},R1:={r-mn+10|rR,mZ0},

and

R~1:={1-(1-ri)0|riR1}={1-(1-ri)-mn+10|riR,mZ0}.

Let Φ~:=Φ(R~) and Φ~1:=Φ(R~1). It is clear that Φ(R1)=Γ({n},Φ) and Φ~1=Γ({n},Φ~). By [33, 16.7 Corollary], if BΦ, then BSΦ~, and if BΓ({n},Φ), then BSΓ({n},Φ~). Therefore, Φ~ has the required property.

Proposition 5.5

Let (X/Zz,B) be a surface pair such that (XB) is Q-factorial dlt and -(KX+B) is nef and big over a neighborhood of z. Let S:=B and KS+BS:=(KX+B)|S. Suppose that S intersects Xz, the fiber of XZ over z, and (S,BS) has a monotonic n-semi-complement (S,BS+) over a neighborhood of z. Then, (X/Zz,B) is n-complementary.

Proof

Possibly replacing z by a closed point of z¯ and shrinking Z near z, we may assume that z is a closed point, (XB) is Q-factorial dlt, -(KX+B) is nef and big over Z, and n(KS+BS+)0.

Let g:WX be a log resolution of (XB) such that g is an isomorphism over the snc locus of (XB) (cf. [35, Theorem 10.45]), and let SW be the strict transform of S on W. Then, the induced morphism gSW:=g|SW:SWS is an isomorphism. We define

KW+BW:=g(KX+B),n(KSW+BSW+):=gSW(n(KS+BS+))0,

and

LW:=-(n+1)(KW+BW).

Let ΔW:=BW-SW. Then,

KSW+BSW:=(KW+BW)|SW=KSW+ΔW|SW=gSW(KS+BS),

and BSW<1 as ΔW<1.

Since -(n+1)(KW+BW) is nef and big over Z, R1h(OW(KW+LW))=0 by the relative Kawamata–Viehweg vanishing theorem for R-divisors (cf. [19, Theorem 3.2.9]), where h is the induced morphism WZ. From the exact sequence,

0OW(KW+LW)OW(KW+SW+LW)OSW(KSW+LW|SW)0,

we deduce that the induced map

H0(W,KW+SW+LW)H0(SW,KSW+LW|SW)

is surjective. Since nBSW+Z, BSW<1, and BSW+-BSW0, we see that

GSW:=nBSW+-(n+1)BSW

is an effective integral divisor. We have

KSW+LW|SW=KSW+-(n+1)(KSW+BSW)=-nKSW-(n+1)BSWGSW0.

Thus, there exists GW0 on W such that GW|SW=GSW and

GWKW+SW+LW=-nKW-nSW-(n+1)ΔW.

Let G:=gGW, and

B+:=S+1n((n+1){B}+G).

Then, we have

n(KX+B+)=n(KX+S)+(n+1){B}+G0.

It remains to show that (X,B+) is lc over a neighborhood of z. Let V be the non-lc locus of (X,B+). There exists a real number a(0,1), such that the non-klt locus of (X,aB++(1-a)B) is equal to SV.

Since g(KX+B+)|SW=gSW(KS+BS+), we have (KX+B+)|S=KS+BS+ and (KX+aB++(1-a)B)|S=KS+aBS++(1-a)BS. By inversion of adjunction (cf. [20, Theorem 1.4]), (X,aB++(1-a)B) is lc near S. In particular, S is disjoint from V. Since

-(KX+aB++(1-a)B)=-a(KX+B+)-(1-a)(KX+B)

is nef and big over Z, by Shokurov–Kollár connectedness principle (cf. [33, 17.4 Theorem]), (SV)Xz is connected. Recall that by assumption, SXz. Hence, VXz= and (X,B+) is lc over a neighborhood of z.

Theorem 5.6

Let l be a positive integer and Φ[0,1]Q a hyperstandard set. Then, there exists a finite set of positive integers N divisible by l depending only on l and Φ satisfying the following.

Assume that (X/Zz,B) is a surface pair such that (X/Zz,BN_Φ) is R-complementary. Then, (X/Zz,B) is N-complementary.

Proof

Let Φ~:=Φ(R~) be the hyperstandard set associated to the finite set R~[0,1]Q given by Proposition 5.4 which only depends on Φ. Possibly replacing l by a multiple, we may assume that lR~Z0. Let N0=N0(l,Φ) be a finite set of positive integers divisible by l given by Theorem 5.3 which only depends on l and Φ. Let N1=N1(l) be a finite set of positive integers divisible by l given by Proposition 4.7 which only depends on l, and let Φ1:=Γ(N1,Φ). Let p=p(l,Φ1) be a positive integer divisible by l given by (Proposition 3.3)2 which only depends on l and Φ1. Let N2=N2(p) be a finite set of positive integers divisible by p given by Theorem 2.20 which only depends on p, and let Φ2:=Γ(N1N2,Φ). Let nCY=nCY(l,Φ2) be a positive integer divisible by l given by Lemma 2.18 which only depends on l and Φ2. We will show that the finite set N:=N0N1N2{nCY} has the required property.

Possibly replacing z by a closed point of z¯, we may assume that z is a closed point. If (X/Zz,BN_Φ) has a klt R-complement, then so does (X/Zz,BN0_Φ), and hence (X/Zz,B) is N0-complementary by the choice of N0. Therefore, we may assume that (X/Zz,BN_Φ) has an R-complement (X/Zz,BN_Φ+G) which is not klt. Possibly replacing (X/Zz,B) by a Q-factorial dlt modification of (X/Zz,BN_Φ+G) and shrinking Z near z, we may assume that (XB) is Q-factorial dlt, KX+BR,Z0, and SXz, where S:=B0 and Xz is the fiber of XZ over z. Since BBΦ2BΦ1BΦ, we have

0κ(X/Z,B-BΦ2)+dimZκ(X/Z,B-BΦ1)+dimZκ(X/Z,B-BΦ)+dimZ2.

Therefore, we only need to consider the following three possibilities:

  1. κ(X/Z,B-BΦ2)+dimZ=0,

  2. κ(X/Z,B-BΦ2)+dimZ=κ(X/Z,B-BΦ1)+dimZ=1, and

  3. κ(X/Z,B-BΦ1)+dimZ=κ(X/Z,B-BΦ)+dimZ=2.

If κ(X/Z,B-BΦ2)+dimZ=0, then nCY(KX+B)0 by our choice of nCY. If κ(X/Z,B-BΦ2)+dimZ=κ(X/Z,B-BΦ1)+dimZ=1, then (X/Zz,B) is N2-complementary by Proposition 5.1. Hence, in what follows we assume that κ(X/Z,B-BΦ1)+dimZ=κ(X/Z,B-BΦ)+dimZ=2. We will show that (X/Zz,B) is N1-complementary.

In this case, both B-BΦ and B-BΦ1 are big over Z. Let KS+BS:=(KX+B)|SR,Z0. By Lemma 2.14 and the choice of N1, (S,(BS)n_Φ~) has a monotonic n-semi-complement over a neighborhood of z for some nN1. Note that Bn_ΦΓ({n},Φ)Γ(N1,Φ), Bn_ΦBΦ1, and B-Bn_Φ is big over Z. According to Lemma 2.15, we may run an MMP on -(KX+Bn_Φ)R,ZB-Bn_Φ over Z and reach a minimal model ψ:XX over Z, such that B-Bn_Φ is nef and big over Z, where D denotes the strict transform of D on X for any R-divisor D on X. No component of S is contracted by ψ and ψS:=ψ|S:SS is an isomorphism as SBn_ΦB and ψ is (KX+B)-trivial.

Since -(KX+B)-ψ(-(KX+Bn_Φ)) is nef over X, and -B+Bn_Φ0, by the negativity lemma, -(KX+B)ψ(-(KX+Bn_Φ)). Let

KS+Bn_Φ,S:=(KX+Bn_Φ)|S.

Note that Bn_Φ,SΓ({n},Φ~) by Proposition 5.4, and the support of -(KX+B)-ψ(-(KX+Bn_Φ)) does not contain any component of S. Then,

-(KS+BS)-ψS(-(KS+Bn_Φ,S))=(-(KX+B)-ψ(-(KX+Bn_Φ)))|S0.

Let BS be the strict transform of BS on S. Since (BS)n_Φ~,Bn_Φ,SΓ({n},Φ~), and BSBn_Φ,S, we deduce that (BS)n_Φ~Bn_Φ,S. Hence, (S,Bn_Φ,S) has a monotonic n-semi-complement over a neighborhood of z. By Proposition 5.5 and Lemma 2.14, (X/Zz,Bn_Φ) has a monotonic n-complement. Since ψ is -(KX+Bn_Φ)-negative, (X/Zz,Bn_Φ) has a monotonic n-complement (X/Zz,B+). By Lemma 2.14, (X/Zz,B+) is an n-complement of (X/Zz,B).

Proof of Theorem 1.3

The theorem follows by Theorem 5.6.

Boundedness of Complements for Threefolds

We will prove the following theorem which is stronger than Theorem 1.1.

Theorem 6.1

Let l be a positive integer and Φ[0,1]Q a hyperstandard set. Then, there exists a finite set of positive integers N divisible by l depending only on l and Φ satisfying the following.

Assume that (X/Zz,B) is a threefold pair such that (X/Zz,BN_Φ) has a klt R-complement. Then, (X/Zz,B) is N-complementary.

Proof

Let N1=N1(l,Φ) be a finite set of positive integers divisible by l given by (Theorem 2.19)3 which only depends on l and Φ, and set Φ1:=Γ(N1,Φ). Let p1=p1(l,Φ1) be a positive integer divisible by l given by (Proposition 3.3)3 and Proposition 3.4 which only depends on l and Φ1. Let N2=N2(p1) be a finite set of positive integers divisible by p1 given by Theorems 2.20 and 1.3 which only depends on p1, and set Φ2:=Γ(N1N2,Φ). Let p2=p2(p1,Φ2) be a positive integer divisible by p1 given by (Proposition 3.3)3 and Proposition 3.4 which only depends on p1 and Φ2. Let N3=N3(p2) be a finite set of positive integers divisible by p2 given by Theorems 2.20 and 1.3 which only depends on p2, and set Φ3:=Γ(N1N2N3,Φ). Let nCY=nCY(l,Φ3) be a positive integer divisible by l given by Lemma 2.18 which only depends on l and Φ3. We will show that N:=N1N2N3{nCY} has the required property.

Replacing z by a closed point of z¯, we may assume that z is a closed point. Possibly replacing (X/Zz,B) by a small Q-factorialization of a klt R-complement of (X/Zz,BN_Φ) and shrinking Z near z, we may assume that (XB) is Q-factorial klt and KX+BR,Z0.

If κ(X/Z,B-BΦ1)+dimZ=3, then X is of Fano type over Z. In this case (X/Zz,B) is N1-complementary by the choice of N1 (see Theorem 2.19). If κ(X/Z,B-BΦ3)+dimZ=0, then nCY(KX+B)0 by the choice of nCY (see Lemma 2.18). Therefore, in the following, we may assume that

1κ(X/Z,B-BΦ3)+dimZκ(X/Z,B-BΦ2)+dimZκ(X/Z,B-BΦ1)+dimZ2.

In particular, there exist integers i,k{1,2} such that

κ(X/Z,B-BΦi)+dimZ=κ(X/Z,B-BΦi+1)+dimZ=k.

We will show that (X/Zz,BΦi+1) is Ni+1-complementary and thus finish the proof by Lemma 2.14.

By Lemma 2.15, we can run an MMP on -(KX+BΦi+1)R,ZB-BΦi+1 over Z and reach a good minimal model XX over Z, such that B-BΦi+1 is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. Let π:XZ be the contraction defined by -(KX+BΦi+1) over Z. By assumption, dimZ=k. Let BZ(i+1) and Mπ be the discriminant and moduli parts of the canonical bundle formula for (X,BΦi+1) over Z in Proposition 3.3 (respectively, Proposition 3.4) if k=2 (respectively, k=1).

Claim 6.2

piMπ is base point free over Z, and

pi(KX+BΦi+1)pi(π)(KZ+BZ(i+1)+Mπ,Z).

Assume Claim 6.2. As XX is an MMP on -(KX+BΦi+1) over Z, for any prime divisor P on X which is exceptional over X, we have

a(P,X,BΦi+1)<a(P,X,BΦi+1)1.

Thus, we can find a crepant model (X~,B~(i+1)) of (X,BΦi+1) such that X~ and X are isomorphic in codimension one. graphic file with name 42543_2022_57_Figb_HTML.jpg It is clear that if (X~/Zz,B~(i+1)) is Ni+1-complementary then so is (X/Zz,BΦi+1). By Lemma 3.1, we may find a crepant model (Z~,BZ~(i+1)+Mπ)(Z,B(i+1)+Mπ) such that for any prime divisor PSuppB~(i+1) which is vertical over Z, the image of P on Z~ is a prime divisor. As (XB) is klt and KX+BR,Z0, we may find a boundary BZ~BZ~(i+1) on Z~ such that (Z~,BZ~+Mπ) is gklt and KZ~+BZ~+Mπ,Z~R,Z0. Since piMπ is base point free over Z, we can pick an effective Q-divisor MZ~ such that piMZ~ZpiMπ,Z~, MZ~BZ~(i+1)=0, and (Z~,BZ~+MZ~) is klt. By our choice of Ni+1, (Z~/Zz,BZ~(i+1)+MZ~) is n-complementary for some nNi+1. By Proposition 3.5, (X~/Zz,B~(i+1)) is also n-complementary. Therefore, it suffices to prove Claim 6.2.

Proof of Claim 6.2

By Lemma 2.15, we may run an MMP on -(KX+BΦi)R,ZBi+1-Bi over Z and reach a good minimal model XX over Z, such that Bi+1-Bi is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. Let π:XZ be the contraction defined by Bi+1-Bi over Z, and τ:ZZ the induced morphism. graphic file with name 42543_2022_57_Figc_HTML.jpg We claim that τ is a birational morphism. In fact, one can pick a positive real number ϵ, such that B-BΦi+1+ϵ(BΦi+1-BΦi) is semi-ample over Z (see Lemma 2.6) and that XX is also an MMP on B-BΦi+1+ϵ(BΦi+1-BΦi) over Z. By assumption,

κ(X/Z,B-BΦi+1)=κ(X/Z,B-BΦi+1+ϵ(BΦi+1-BΦi)).

Hence, we can see that τ:ZZ is birational.

By Lemma 3.2, Proposition 3.3 and the choice of pi, there exists a gklt g-pair (Z,BZ(i+1)+Mπ) induced by the canonical bundle formula, such that piMπ is base point free over Z and

pi(KX+BΦi+1)pi(π)(KZ+BZ(i+1)+Mπ,Z).

Moreover, it is clear that

KZ+BZ(i+1)+Mπ,Z=τ(KZ+BZ(i+1)+Mπ,Z).

Therefore,

pi(KX+BΦi+1)pi(π)(KZ+BZ(i+1)+Mπ,Z)

as (X,BΦi+1) is crepant to (X,BΦi+1). We finish the proof.

Proof of Theorem 1.2

Strictly Lc Calabi–Yau Pairs

Definition 7.1

We say that X is of Calabi–Yau type over Z, if XZ is a contraction, and there is a boundary C such that (XC) is klt and KX+CR,Z0.

Lemma 7.2

Suppose that X is of Calabi–Yau type over Z. Assume that (X,B+) is lc, KX+B+R,Z0 for some boundary B+, and f:YX is a projective birational morphism from a normal quasi-projective variety Y, such that a(Ei,X,B+)<1 for any prime exceptional divisor Ei of f. Then, Y is of Calabi–Yau type over Z.

Proof

Since X is of Calabi–Yau type over Z, there exists a klt pair (XC) such that KX+CR,Z0. Let Dt:=tB++(1-t)C. Then, (X,Dt) is klt and KX+DtR,Z0 for any t[0,1). We have

KY+BY++i(1-ai+)Ei=f(KX+B+)

and

KY+CY+i(1-ai)Ei=f(KX+C),

where BY+ and CY are the strict transforms of B+ and C on Y, respectively, ai+:=a(Ei,X,B+)<1, and ai:=a(Ei,X,C) for any i. Then,

KY+Dt,Y:=f(KX+Dt)=KY+tBY++(1-t)CY+i(t(1-ai+)+(1-t)(1-ai))Ei.

Pick 0<t0<1 such that t0(1-ai+)+(1-t0)(1-ai)0 for any i. Then, (Y,Dt0,Y) is klt and KY+Dt0,YR,Z0. Therefore, Y is of Calabi–Yau type over Z.

Definition 7.3

(cf. [47, §11]) A pair (X/Zz,B) is called strictly lc Calabi–Yau if

  1. (X/Zz,B) is an R-complement of itself, and

  2. for any R-complement (X/Zz,B+) of (X/Zz,B), B+=B over some neighborhood of z.

Remark 7.4

When dimZ=0, (XB) is strictly lc Calabi–Yau if and only if (1) holds.

Lemma 7.5

Assume that (X/Zz,B) is an R-complement of itself. Then, (X/Zz,B) is strictly lc Calabi–Yau if and only if either dimz=dimZ or z¯ is the image of an lc center of (XB) on Z.

Proof

First assume that (X/Zz,B) is strictly lc Calabi–Yau. Suppose that dimz<dimZ and z¯ is not the image of any lc center of (XB). Possibly shrinking Z near z, we may find an ample divisor H0 such that zSuppH and H does not contain the image of any lc center of (XB). Pick a positive real number ϵ, such that (X/Zz,B+ϵπH) is lc and thus an R-complement of (X/Zz,B). However, B+ϵπHB over any neighborhood of z, a contradiction.

Now, we prove the converse direction. Assume that (X/Zz,B+G) is an R-complement of (X/Zz,B) for some G0. Since GR0 over some neighborhood of z, G=πLZ for some R-Cartier R-divisor LZ on Z by [12, Lemma 2.5]. If dimz=dimZ, then G=0 over a neighborhood of z. If z¯ is the image of some lc center of (XB), then zSuppLZ as (X,B+G) is lc over a neighborhood of z. Therefore, in both cases, (X/Zz,B) is strictly lc Calabi–Yau.

Example 7.6

Let π:X:=P1×P1Z:=P1, zZ a closed point, and L1,L2 two sections. Then, over a neighborhood of z, we have (X,L1+L2) is lc and KX+L1+L2R,Z0. Since KX+L1+L2+πzR,Z0, (X/Zz,L1+L2) is not strictly lc Calabi–Yau.

Lemma 7.7

Suppose that (X/Zz,B) is strictly lc Calabi–Yau and XX is a birational contraction over Z. Let B be the strict transform of B on X. Then, (X/Zz,B) is strictly lc Calabi–Yau.

Proof

It follows from the definition of strictly lc Calabi–Yau and the fact that (X,B)(X,B) is crepant over some neighborhood of z.

Proposition 7.8

Let Γ[0,1]Q be a DCC set. Then, there exists a positive integer I depending only on Γ satisfying the following. If (X/Zz,B) is a strictly lc Calabi–Yau threefold pair such that BΓ, then I(KX+B)0 over some neighborhood of z.

Proof

Possibly replacing (X/Zz,B) by a Q-factorial dlt modification, we may assume that X is Q-factorial. Since (X/Zz,B) is a strictly lc Calabi–Yau pair, by [26, Theorem 5.20], BΓ over a neighborhood of z for some finite subset ΓΓ which only depends on Γ. According to [12, Theorem 2.14], we may find a positive integer I which only depends on Γ such that (X/Zz,B) has a monotonic I-complement (X/Zz,B+G) for some G0. By assumption, G=0 over some neighborhood of z. Thus,

I(KX+B)=I(KX+B+G)0

over some neighborhood of z.

Proof of Theorem 1.2

We first show a special case of Theorem 1.2.

For convenience, we say a pair (XB) is klt over a closed subset Z0Z, if a(E,X,B)>0 for any prime divisor E over X such that π(CenterX(E))Z0, where π:XZ is a contraction. For two R-divisors D1 and D2 on X, by D1D2 (respectively, D1>D2) over Z0, we mean that multED1multED2 (respectively, multED1>multED2) for any prime divisor E on X with π(E)Z0. By D1D2 over an open subset UZ, we mean D1|π-1(U)D2|π-1(U).

Proposition 7.9

Let I be a positive integer. Assume that N is a finite set of positive integers divisible by I given by Theorem 1.1 which only depends on I.

Assume that (X/Zz,B) is an R-complementary threefold pair such that X is of Calabi–Yau type over Z. Assume that there is a contraction π:XZ over Z, and an open subset UZ, such that

  1. IBZ0 over U,

  2. (XB) is klt over Z\U, and

  3. -(KX+B)R(π)H for some R-divisor H which is ample over Z.

Then, (X/Zz,B) is N-complementary.

Proof

Possibly shrinking Z near z, we may assume that (XB) is lc. Set N:=maxnNn. We claim that there exists a boundary B on X such that

  • (X,B) is klt,

  • KX+BR,Z0, and

  • BNN+1B over U and BB over Z\U.

Assume the claim holds. Then,

(n+1)BnB+(n+1){B} 7.1

for any nN. By Theorem 1.1 and the construction of N, (X/Zz,B) is n-complementary for some nN. Thus, (X/Zz,B) is n-complementary by (7.1).

Therefore, it suffices to prove the claim. By assumption, we may find an effective R-Cartier R-divisor H1R,ZH such that Z\USuppH1 and (X,B+H1) is lc, where H1:=(π)H1. In particular, we have

KX+B+H1R,Z0,B+H1>BoverZ\U.

Since X is of Calabi–Yau type over Z, there exists a boundary C such that (XC) is klt and KX+CR,Z0. Let δ(0,1) be a positive real number such that

B:=(1-δ)(B+H1)+δCNN+1BoverU,(1-δ)(B+H1)BoverZ\U.

It is clear that (X,B) is klt and KX+BR,Z0. This completes the proof.

Proof of Theorem 1.2

Let I=I(l,ΓQ) be a positive integer divisible by l given by Proposition 7.8 which only depends on l and ΓQ, and let Φ:=Φ(1IZ[0,1]). Let N=N(I) be a finite set of positive integers divisible by I given by Theorem 1.1 which only depends on I. We will show that N has the required property.

Possibly shrinking Z near z, we may assume that (XB) is lc. By Lemma 7.2, we can replace (XB) by a dlt modification and thus assume that (XB) is Q-factorial dlt. Suppose that (X/Zz,B+) is an R-complement of (X/Zz,B). Possibly replacing z by a closed point of z¯ and shrinking Z near z, we may assume that z is a closed point, (X,B+) is lc, and KX+B+R,Z0. Write

-(KX+BN_Φ)R,ZB+-BN_Φ=F+M,

where F:=Nσ(B+-BN_Φ/Z)0 and M:=B+-BN_Φ-F0 (cf. [41, III, §4], [39, §3]). Note that F is well defined as B+-BN_Φ0. Since X is of Calabi–Yau type over Z, there exists a boundary C such that (XC) is klt and KX+CR,Z0. Choose a positive real number ϵ0 such that (X,C+ϵ0M) is klt. We may run an MMP on KX+C+ϵ0M over Z and reach a good minimal model X, such that KX+C+ϵ0M is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. Since

KX+C+ϵ0MR,Zϵ0MR,Z-ϵ0(KX+BN_Φ+F),

X and X are isomorphic in codimension one by [23, Lemma 2.4]. We also see that -(KX+BN_Φ+F) is semi-ample over Z and thus induces a contraction π:XZ over Z. In particular, there is an effective R-divisor H on Z which is ample over Z such that -(KX+BN_Φ+F)R(π)H. Note that (X,B+) is lc and thus (X,BN_Φ+F) is also lc.

Let η be the generic point of Z, and

Zscy:={η}{zZ|(X/Zz,BN_Φ+F)is strictly lc Calabi--Yau}.

By Lemma 7.5 and [3, Theorem 1.1], Zscy is a non-empty finite set.

Claim 7.10

We have I(KX+BN_Φ+F)0 over a neighborhood of z for any zZscy.

Assume Claim 7.10. By [12, Lemma 2.6], Lemma 7.5 and [3, Theorem 1.1], there exists an open subset UZ such that

I(KX+BN_Φ+F)0over{U}and(X,BN_Φ+F)is klt overZ\U.

In particular, I(BN_Φ+F)Z0 over U. Recall that -(KX+BN_Φ+F)R(π)H where H is ample over Z. By Proposition 7.9, (X/Zz,BN_Φ+F) is N-complementary, and thus (X/Zz,B) is also N-complementary by Lemma 2.14. Moreover, as X and X are isomorphic in codimension one, (X/Zz,B) is N-complementary.

Proof of Claim 7.10

We may pick a positive real number ϵ such that (X,C+ϵF) is klt and that XX is a sequence of steps of the -(KX+BN_Φ+(1-ϵ)F)-MMP over Z. Since

-KX+BN_Φ+(1-ϵ)FR,ZϵFR,ZKX+C+ϵF,

one can run an MMP on -(KX+BN_Φ+(1-ϵ)F) over Z. This MMP terminates with a model X on which -(KX+BN_Φ+(1-ϵ)F) is semi-ample over Z, where D denotes the strict transform of D on X for any R-divisor D on X. graphic file with name 42543_2022_57_Figd_HTML.jpg For any positive real number ϵϵ, we infer that ψ:XX is also an MMP on -(KX+BN_Φ+(1-ϵ)F) over Z, and

-(KX+BN_Φ+(1-ϵ)F)=-(1-ϵϵ)(KX+BN_Φ+F)-ϵϵ(KX+BN_Φ+(1-ϵ)F)

is semi-ample over Z (see Lemma 2.6). In particular, X is a good minimal model of -(KX+BN_Φ+(1-ϵ)F) over Z. Since Nσ(-(KX+BN_Φ+(1-ϵ)F)/Z)=ϵF (cf. [41, III, 4.2 Lemma]), by [23, Lemma 2.4], F is contracted by ψ. Hence,

KX+BN_Φ=ψKX+BN_Φ+F.

If dimX=dimZ and z=ηZscy, then X is smooth and B+=0 over a neighborhood of η. Otherwise, by Lemma 7.7, we know that (X/Zz,BN_Φ) is strictly lc Calabi–Yau for any zZscy, which implies that BN_Φ=B+=BΓQ over a neighborhood of z. We, therefore, see that I(KX+BN_Φ)0 over some neighborhood of z by our choice of I. Since (X,BN_Φ+F) is crepant to (X,BN_Φ), our claim holds.

Acknowledgements

The authors would like to thank Qianyu Chen, Chen Jiang, Jihao Liu, Wenfei Liu, Lingyao Xie and Chenyang Xu for valuable discussions and suggestions. The third named author would also like to thank his advisor Christopher D. Hacon for his constant support and useful suggestions. The first named author was supported by the China Post-doctoral Science Foundation (Grant Nos. BX2021269 and 2021M702925). The second named author was supported by National Key Research and Development Program of China (Grant No. 2020YFA0713200). The second named author is a member of LMNS, Fudan University. The third named author was partially supported by NSF Research Grants (Nos. DMS-1801851 and DMS-1952522) and by a grant from the Simons Foundation (Award Number: 256202). Finally, the authors would like to thank the referees for their careful reading of this paper and the many useful comments.

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