On projective threefolds of general type with p_g=2

TL;DR

This paper classifies minimal projective threefolds of general type with geometric genus 2 by analyzing the birationality of their sixth canonical maps, advancing understanding of their geometric structure.

Contribution

It provides a classification of such threefolds based on the birationality of their 6-canonical maps, a novel approach in this context.

Findings

Classification of minimal projective 3-folds with p_g=2

Conditions for birationality of 6-canonical maps

New insights into the structure of general type threefolds

Abstract

We classify minimal projective 3-folds of general type with $p_g = 2$ by studying the birationality of their 6-canonical maps.

Equations

$$\mathcal{F}=\{\hat{F}\subset W|\hat{F}={\tau }_{\ast }^{-1}(F),F\text{is a fiber of}g\}$$

$$|M_1|\supseteq |M_2|+\text{(fixed effective divisor)}.$$

$$h^0(X^{\prime },M)=h^0({\mathcal{O}}_X(K_{X^{\prime }}))$$

$${\pi }^{\ast }(K_X){\sim }_{\mathbb{Q}}M+E^{\prime },$$

$$d_1=\dim \overline{{\phi }_1(X)}=\dim (\Gamma ).$$

$$\zeta =\{\begin{array}{ll}1,& \text{if}{d}_1\ge 2;\\ a,& \text{if}{d}_1=1.\end{array}$$

$${\pi }^{\ast }(K_X){\sim }_{\mathbb{Q}}{\mu }^{\prime }S+E_S^{\prime }$$

$$|D|=\text{Mov}|D|+\text{Fix}|D|$$

$$\xi =\xi (|G|)$$

$$\alpha (m)={\alpha }_{(|G|)}(m)$$

$${\alpha }_0(m)$$

$$m\xi \ge \mathrm{deg}(K_C)+{\alpha }_0(m).$$

$$\xi \ge \frac{\mathrm{deg}(K_C)}{1+\frac{1}{\mu }+\frac{1}{\beta }}.$$

$$K_X^3\ge \mu \beta \xi .$$

$$\overline{{\Phi }_{|L|}(W_1)}\overline{{\Phi }_{|L|}(W_2)}\text{and}\overline{{\Phi }_{|L|}(W_2)}\overline{{\Phi }_{|L|}(W_1)}.$$

$$H^0(Z,m(K_Z+D))⟶H^0(D,m{K}_D)$$

$$|n(\mu +1)K_{X^{\prime }}|{|}_S\succcurlyeq |n\mu (K_{X^{\prime }}+S)|{|}_S=|n\mu K_S|$$

$$n(\mu +1){\pi }^{\ast }(K_X)\ge M_{n(\mu +1)}$$

$${\pi }^{\ast }(K_X){|}_S\ge \frac{\mu }{\mu +1}{\sigma }^{\ast }(K_{S_0})\ge \frac{\zeta }{1+\zeta }{\sigma }^{\ast }(K_{S_0}).$$

$${\chi }_m(\mathbb{B}(X))=P_m(X)$$

$$H^0(X^{\prime },M_{m_1}-jF)\stackrel{{\theta }_{m_1,-j}}{⟶}{H}^0(F,M_{m_1}{|}_F),$$

$$H^0(F,M_{m_1}{|}_F-jC)\stackrel{{\psi }_{m_1,-j}}{⟶}{H}^0(C,M_{m_1}{|}_C).$$

$$M_{m_1}\ge jF+S_1.$$

$${\phi }_{\lfloor \frac{1}{\xi (|G|)}(2-\frac{\tilde{\delta }}{j})+\frac{m_1}{j}+\frac{1}{\beta }\rfloor +2,X}$$

$${\phi }_{\lfloor \frac{2}{\xi (|G|)}(1-\frac{\tilde{\delta }}{j})+\frac{2m_1}{j}\rfloor +3,X}$$

$${\phi }_{\lfloor \frac{2m_1}{\tilde{\delta }}+\frac{1}{\beta }+\frac{1}{\mu }\cdot (1-\frac{2j}{\tilde{\delta }})\rfloor +2,X}$$

$${\phi }_{\lceil \frac{m_1}{j}+\frac{1}{\beta }\rceil +1,X}$$

$${\pi }^{\ast }(K_X){|}_F\ge \frac{j}{j+m_1}{\sigma }^{\ast }(K_{F_0})+\frac{1}{j+m_1}{S}_1{|}_F.$$

$$(n-\frac{m_1}{j}-\frac{1}{\beta })\xi +\frac{\tilde{\delta }}{j}>1,$$

$$(n+1)\xi \ge \lceil (n-\frac{m_1}{j}-\frac{1}{\beta })\xi +\frac{\tilde{\delta }}{j}\rceil +2.$$

$$n=\{\begin{array}{ll}\lfloor \frac{1}{\xi (|G|)}(2-\frac{\tilde{\delta }}{j})+\frac{m_1}{j}+\frac{1}{\beta }\rfloor +1,& \text{when}\tilde{\delta }\le 2j;\\ \lfloor \frac{2m_1}{\tilde{\delta }}+\frac{1}{\beta }+\frac{1}{\mu }\cdot (1-\frac{2j}{\tilde{\delta }})\rfloor +1,& \text{when}\tilde{\delta }>2j.\end{array}$$

Full text

The first author was supported by National Natural Science Foundation of China (#11571076, #11731004) and Program of Shanghai Subject Chief Scientist (#16XD1400400). The second author was supported by National Researcher Program of National Research Foundation of Korea (Grant No. 2010-0020413). The third author was supported by PRIN 2015 “Geometry of Algebraic Varieties” and by GNSAGA of INdAM

1. Introduction

The study of pluricanonical maps is a fundamental aspect of birational geometry. Let $\varphi_{m}$ be the m-canonical map of a projective variety $X$. It is known, by Hacon-McKernan [H-M], Takayama [Ta] and Tsuji [Tsu], that there exists a constant $r_{n}$ (for any integer $n>0$) such that the pluricanonical map $\varphi_{m}$ is birational onto its image for all $m\geq r_{n}$ and for all smooth projective $n$-folds of general type. Despite the great efforts of several authors, $r_{n}$ is not explicily given except for $n\leq 3$. By now it is a classical result for curves and surfaces that $r_{1}=3$ and $r_{2}=5$ (see [Bom]). In addition, very recently Chen and the first author proved the bound $r_{3}\leq 57$ (see [CC1, CC2, CC3, Che16]).

Provided that the property we are studying is birationally invariant, the $3$-dimensional MMP allows us to work with any minimal model $X$ (${\mathbb{Q}}$-factorial with at worst terminal singularities) of a nonsingular projective $3$-fold of general type. The aim of this paper is to give a more precise bound for $r_{3}$ adding some extra information on the nature of $X$. We have already studied the birational geomentry of projective $3$-folds of general type with geometric genus $p_{g}=1$ and $3$ in [CHP17]. Here, we shall assume that the geometric genus $p_{g}$ of $X$ is equal to 2. Under this hypothesis, we are able to classify minimal projective $3$-folds of general type with $p_{g}=2$ by studying only the birationality of their $6$-canonical maps.

We need to introduce some terminology in order to state main result of this paper: Theorem 1.2 which was announced in [CHP17].

By Chen-Chen’s series of works in [CC1, CC2, CC3], there exists a positive integer $m_{0}\leq 18$ such that $P_{m_{0}}(X)=h^{0}(X,m_{0}K_{X})\geq 2$. Hence it is possible to investigate the birational geometry of $X$ by studying the behavior of the $m_{0}$-canonical map $\varphi_{m_{0},X}$. This strategy proves to be very effective.

Definition 1.1**.**

Let $W$ be a ${\mathbb{Q}}$-factorial normal projective variety of dimension $n$. Assume that the two maps $\tau:W\dashrightarrow W^{\prime}$ and $g:W^{\prime}\longrightarrow S$ satisfy the following properties:

• (1)

$W^{\prime}$ is a nonsingular projective variety and $S$ is normal projective of dimension $s<n$;

• (2)

$\tau$ is a dominant birational map and $g$ is a fibration.

Then we say that the set

[TABLE]

forms an $(n-s)$-fold class of $W$, where $\tau_{*}^{-1}(\cdot)$ denotes the strict transform. In particular, if $n-s=1$ ($=2$), we call it a curve class (a surface class). The number $(K_{W}^{n-s}\cdot\tau_{*}^{-1}(F))$ ($F$ a general fiber of $g$) is called the canonical degree of $\mathcal{F}$. Such degree is also denoted as “$\deg_{c}(\mathcal{F})$”.

Especially, when $\varphi_{m_{0},X}$ is of fiber type (i.e. $\dim\overline{\varphi_{m_{0},X}(X)}<\dim X$), the induced fibration (obtained by taking the Stein factorization of $\varphi_{m_{0},X}$) automatically forms either a curve class $\mathcal{C}$ or a surface class $\mathcal{S}$ of $X$. We also say that $X$ is $m_{0}$-canonically fibred by a curve class $\mathcal{C}$ (or a surface class $\mathcal{S}$). Note that in our case $m_{0}=1$, so we can simply say canonically fibred. We can now state the main theorem.

Theorem 1.2**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)=2$. Then one of the following statements is true:

• (1)

$\varphi_{6,X}$ is birational onto its image;

• (2)

$X$ is canonically fibered by a $(2,3)$-surface class of canonical degree $\frac{1}{2}$, in which case $\varphi_{6,X}$ is non-birational;

• (3)

$X$ is canonically fibered by a $(1,2)$-surface class (denote by $\mathcal{C}$ the genus $2$ curve class which is naturally induced from $\mathcal{S}$) and one of the following holds:

1. (i)

$\deg_{c}(\mathcal{C})=\frac{2}{3}$; 2. (ii)

$\deg_{c}(\mathcal{C})=\frac{4}{5}$; 3. (iii)

$P_{2}(X)=5$, $\deg_{c}(\mathcal{C})=1$ and $\deg_{c}(\mathcal{S})=\frac{1}{2}$; 4. (iv)

$P_{4}(X)=14$, $\deg_{c}(\mathcal{C})=1$ and $\deg_{c}(\mathcal{S})=\frac{1}{2}$.

In this case $\varphi_{6,X}$ is non-birational.

• (4)

There is an explicit finite set ${\mathbb{S}}_{2}$ such that $X$ is canonically fibered by a $(1,2)$-surface class and ${\mathbb{B}}(X)\in{\mathbb{S}}_{2}$, in which case $\varphi_{6,X}$ is non-birational. (see Subsection 2.6 for the definition of ${\mathbb{B}}(X)$, the weighted basket of $X$)

Remark 1.3**.**

The existence of threefolds described in Theorem 1.2 (4) are provided by the following examples. Denote by $X_{d}$ a general weighted hypersurface of degree $d$ in the sense of Fletcher (see [Flet]).

1. (1)

The 3-fold $X_{16}\subset{\mathbb{P}}(1,1,2,3,8)$ has $K^{3}=\frac{1}{3}$, $p_{g}=2$ and $\varphi_{7}$ is non-birational; 2. (2)

The 3-fold $X_{14}\subset{\mathbb{P}}(1,1,2,2,7)$ has $K^{3}=\frac{1}{2}$, $p_{g}=2$ and $\varphi_{6}$ is non-birational.

Moreover,

1. (1)

We do not know whether any threefold with properties described in Theorem 1.2 (2) might exist, nor if all those encoded by ${\mathbb{S}}_{2}$ exist (most likely not). 2. (2)

A complete list of the 263 elements of the set ${\mathbb{S}}_{2}$ can be found at the following webpage.

http://www.dima.unige.it/~penegini/publ.html

The plan of the paper is the following:

In Section 2, we describe the set up of the work. We recall some key theorems for the study of the pluricanonical maps for 3-folds of general type and some necessary inequalities in a general frame work. Moreover we introduce the notion of weighted basket.

Section 3 contains the core technical theorems of the paper, which will be effectively used to do the classification. These theorems concern 3-folds with $p_{g}\geq 2$ and canonically fibered by a $(1,2)$-surface class.

Theorem 1.2 is proved in several steps in Section 4, which is the longest section of the paper. Subsection 4.1 takes care of Theorem 1.2 cases (1) and (2). Theorem 1.2 cases (3)(i) and (3)(ii) are proved in Subsection 4.2. Most of Section 4 (Subsection 4.3 and 4.4) is then devoted to constructing effective numerical constraints on $P_{2}(X)$, $P_{3}(X)$, $P_{4}(X)$, $P_{5}(X)$ and $P_{6}(X)$. This is done by repeatedly applying the theorems of Section 3 in a rigorous case by case analysis. These constrains on the plurigenera will be used to produce (by computer aided computation) the set ${\mathbb{S}}_{2}$ that proves Theorem 1.2 case (4) (see Subsection 4.5). Finally, in Subsection 4.3, Theorem 1.2 cases (3)(iii) and (3)(iv) (See Propositions 4.11, 4.13) are proved. This section provides also more details and insights on the computations done in [CHP17], where the tedious calculations are omitted (see Proposition X [CHP17]).

Notation and conventions. We work over the field ${\mathbb{C}}$ of complex numbers. A minimal threefold of general type $X$ is a ${\mathbb{Q}}$-factorial 3-fold with at worst terminal singularities such that the canonical divisor $K_{X}$ is a nef and big ${\mathbb{Q}}$-Cartier divisor. Moreover, let $\omega_{X}=\mathcal{O}_{X}(K_{X})$ be the canonical sheaf. Throughout the paper we use the following symbols.

• $\diamond$

“$\sim$” denotes linear equivalence or ${\mathbb{Q}}$-linear equivalence when specified “$\sim_{{\mathbb{Q}}}$”;

• $\diamond$

“$\equiv$” denotes numerical equivalence;

• $\diamond$

“$|M_{1}|\succcurlyeq|M_{2}|$” (or, equivalently, “$|M_{2}|\preccurlyeq|M_{1}|$”) means, for linear systems $|M_{1}|$ and $|M_{2}|$ on a variety,

[TABLE]

2. Preliminaries

2.1. Set up

Let $X$ be a minimal projective 3-fold of general type and we assume that $p_{g}(X)=h^{0}(X,{\mathcal{O}_{X}}(K_{X}))\geq 2$. So we may consider the canonical map $\varphi_{1}:X\dashrightarrow{\mathbb{P}}^{p_{g}(X)-1}$, which is a non-constant rational map.

From the very beginning we fix an effective Weil divisor $K_{1}\sim K_{X}$. Take successive blow-ups $\pi:X^{\prime}\rightarrow X$, which exists by Hironaka’s big theorem, such that:

(i) $X^{\prime}$ is nonsingular and projective;

(ii) the moving part of $|K_{X^{\prime}}|$ is base point free;

(iii) the union of supports of both $\pi^{*}(K_{1})$ and exceptional divisors of $\pi$ is simple normal crossing.

Denote by $\tilde{g}$ the composition $\varphi_{1}\circ\pi$. So $\tilde{g}:X^{\prime}\rightarrow\Sigma\subseteq{\mathbb{P}}^{p_{g}(X)-1}$ is a non-constant morphism by the above assumption. Let $X^{\prime}\overset{f}{\rightarrow}\Gamma\overset{s}{\rightarrow}\Sigma$ be the Stein factorization of $\tilde{g}$. We get the following commutative diagram:

X$$X^{\prime}$$\Sigma$$\Gamma-----------f$$s$$\pi$$\varphi_{1}$$\tilde{g}

We may write $K_{X^{\prime}}=\pi^{*}(K_{X})+E_{\pi}$, where $E_{\pi}$ is an effective ${{\mathbb{Q}}}$-divisor which is a sum of distinct exceptional divisors with positive rational coefficients. By definition, for any positive integer $m$, we have $\lceil{m\pi^{*}(K_{X})}\rceil\leq mK_{X^{\prime}}$. Set $|M|=\text{Mov}|K_{X^{\prime}}|$. Since

[TABLE]

and $X$ has at worst terminal singularities, we may also write

[TABLE]

where $E^{\prime}$ is another effective ${\mathbb{Q}}$-divisor. Set

[TABLE]

Clearly one has $1\leq d_{1}\leq 3$.

If $d_{1}=2$, a general fiber of $f$ is a smooth projective curve of genus $\geq 2$. We say that $X$ is canonically fibred by curves.

If $d_{1}=1$, a general fiber $F$ of $f$ is a smooth projective surface of general type. We say that $X$ is canonically fibred by surfaces with invariants $(c_{1}^{2}(F_{0}),p_{g}(F_{0})),$ where $F_{0}$ is the minimal model of $F$ via the contraction morphism $\sigma:F\rightarrow F_{0}$. We may write $M\equiv aF$ where $a=\deg f_{*}{\mathcal{O}}_{X^{\prime}}(M)$.

Just to fix the notion, a generic irreducible element $S$ of $|M|$ means either a general member of $|M|$ in the case of $d_{1}\geq 2$ or, otherwise, a general fiber $F$ of $f$.

For any positive integer $m$, $|M_{m}|$ denotes the moving part of $|mK_{X^{\prime}}|$. Let $S_{m}$ be a general member of $|M_{m}|$ whenever $m>1$.

Set

[TABLE]

Naturally one has $\pi^{*}(K_{X})\sim_{\mathbb{Q}}\zeta S+E^{\prime}.$ In practice we need such a real number $\mu=\mu(S)$ which is defined to be the supremum of those rational numbers $\mu^{\prime}$ satisfying the following property:

[TABLE]

for certain effective ${\mathbb{Q}}$-divisor $E_{S}^{\prime}$. Clearly we have $\mu(S)\geq\zeta$.

2.2. Convention

For any linear system $|D|$ of positive dimension on a normal projective variety $Z$, we may write

[TABLE]

and consider the rational map $\Phi_{|D|}=\Phi_{\text{Mov}|D|}$. We say that $|D|$* is not composed of a pencil if $\dim\overline{\Phi_{|D|}(Z)}\geq 2$. A generic irreducible element of $|D|$ means a general member of $\text{Mov}|D|$ when $|D|$ is not composed of a pencil or, otherwise, an irreducible component in a general member of $\text{Mov}|D|$. For a nonsingular projective surface $S$ of general type, we say that $S$ is a $(u,v)$

• surface* if $K_{S_{0}}^{2}=u$ and $p_{g}(S_{0})=v$ where $S_{0}$ is the minimal model of $S$.

2.3. Known inequalities

Pick a generic irreducible element $S$ of $|M|$. Clearly, $S$ is a nonsingular projective surface of general type. Assume that $|G|$ is a base point free linear system on $S$. Denote by $C$ a generic irreducible element of $|G|$. Since $\pi^{*}(K_{X})|_{S}$ is nef and big, there is a rational number $\beta>0$ such that $\pi^{*}(K_{X})|_{S}\geq\beta C$. Granted the existence of such $\beta$, we may assume from now on that $\beta=\beta(|G|)$ is the supremum satisfying the above property.

For any integer $m>0$, we define

[TABLE]

When no confusion arises as it is likely in the context, we will simply use the simple notation $\zeta$, $\mu$, $\beta$, $\xi$ and $\alpha(m)$. According to [CC2, Theorem 2.11], whenever $\alpha(m)>1$, one has

[TABLE]

In particular, as $m$ is sufficiently large so that $\alpha(m)>1$, Inequality (2.2) implies

[TABLE]

Moreover, by [Che07, Inequality (2.1)] one has

[TABLE]

2.4. Birationality principle

We refer to [CC2, 2.7] for birationality principle. Recall the following concept for point separations.

Definition 2.1**.**

Let $|L|$ be a moving (without fixed part) linear system on a normal projective variety $Z$. We say that the rational map $\Phi_{|L|}$ distinguishes sub-varieties $W_{1},W_{2}\subset Z$ if, set theoretically,

[TABLE]

We say that $\Phi_{|L|}$ separates points $P,Q\in Z$ (for $P,Q\not\in\text{Bs}|L|$), if $\Phi_{|L|}(P)\neq\Phi_{|L|}(Q)$.

We will tacitly and frequently use the following theorem in the context:

Theorem 2.2**.**

(see [CC2, Theorem 2.11]) Keep the same setting and assumption as in Subsection 2.1 and Subsection 2.3. Pick up a generic irreducible element $S$ of $|M|$. For $m>0$, assume that the following conditions are satisfied:

• (i)

$|mK_{X^{\prime}}|$* distinguishes different generic irreducible elements of $|M|$;*

• (ii)

$|mK_{X^{\prime}}||_{S}$* distinguishes different generic irreducible elements of $|G|$;*

• (iii)

$\alpha(m)>2$.

Then $\varphi_{m,X}$ is birational onto its image.

2.5. A weak form of extension theorem

Sometimes we use the following theorem which is a special form of Kawamata’s extension theorem (see [KaE, Theorem A]):

Theorem 2.3**.**

(see [CZ16, Theorem 2.4]) Let $Z$ be a nonsingular projective variety on which $D$ is a smooth divisor such that $K_{Z}+D\sim_{\mathbb{Q}}A+B$ for an ample ${\mathbb{Q}}$-divisor $A$ and an effective ${\mathbb{Q}}$-divisor $B$ and that $D$ is not contained in the support of $B$. Then the natural homomorphism

[TABLE]

is surjective for all $m>1$.

In particular, when $Z$ is of general type and $D$ moves in a base point free linear system, the condition of Theorem 2.3 is automatically satisfied. Taking $Z=X^{\prime}$, $D=S$ and modulo a process of taking the limit (so may assuming $\mu$ to be rational), it holds that

[TABLE]

for some sufficiently large and divisible integer $n$. Noting that

[TABLE]

and that $|n(\mu+1)\sigma^{*}(K_{S_{0}})|$ is base point free, we have

[TABLE]

2.6. The weighted basket of $X$

The weighted basket ($=$ formal basket) $\mathbb{B}(X)$ is defined to be the triple $\{B_{X},P_{2}(X),\chi({\mathcal{O}}_{X})\}$. We keep all the definitions and symbols in [CC1, Sections 2 and 3] such as “basket”, “prime packing”, “the canonical sequence of a basket”, $\Delta^{j}(B)$ ($j>0$), $\sigma$, $\sigma^{\prime}$, $B^{(n)}$ ($n\geq 0$), $\chi_{m}(\mathbb{B}(X))$ ($m\geq 2$), $K^{3}(\mathbb{B}(X))$, $\sigma_{5}$, $\varepsilon$, $\varepsilon_{n}$ ($n\geq 5$) and so on.

As $X$ is of general type, the vanishing theorem and Reid’s Riemann-Roch formula [R87] (see also front lines in [CC1, 4.5]) imply that

[TABLE]

for all $m\geq 2$ and $K^{3}(\mathbb{B}(X))=K_{X}^{3}$. For any $n\geq 0$, $B^{(n)}$ can be expressed in terms of $\chi({\mathcal{O}}_{X})$, $P_{2}$, $P_{3}$, $\cdots$, $P_{n+1}$ (see [CC1, (3.3)$\sim$(3.14)] for more details), which serves as a considerably powerful tool for our classification.

3. Some technical theorems

3.1. Two restriction maps on canonical class of $(1,2)$-surfaces

Within this subsection, we always work under the following assumption:

($\pounds$) *Keep the setting in 2.1. Let $m_{1}>1$ be an integer. Assume that $|M_{m_{1}}|$ is base point free, $d_{1}=1$, $\Gamma\cong{\mathbb{P}}^{1}$ and that $F$ is a $(1,2)$-surface. Take $|G|=\text{Mov}|K_{F}|$, which is assumed to be base point free. Let $C$ be a generic irreducible element of $|G|$. *

Definition 3.1**.**

For any integers $j\geq 0$, define the following restriction maps:

[TABLE]

[TABLE]

Set $U_{m_{1},-j}=\text{Im}(\theta_{m_{1},-j})$, $V_{m_{1},-j}=\text{Im}(\psi_{m_{1},-j})$, $u_{m_{1},-j}=\dim U_{m_{1},-j}$ and $v_{m_{1},-j}=\dim V_{m_{1},-j}$.

In this section we prove three technical theorems that relate the numbers $\beta,\xi,\mu$ and $\alpha$ to the linear systems of the definition above. These three theorems will be used systematically in Section 4 together with [CHP17, Proposition 3.4, 3.5, 3.6, 3.7] while using the setting $m_{0}=1$.

Theorem 3.2**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)\geq 2$. Keep Assumption ($\pounds$). Let $m_{1}>1$ be an integer. Suppose that $|S_{1}|$ is a base point free linear system on $X^{\prime}$ with $h^{0}(S_{1}|_{F})\geq 2$ and that, for some integer $j\geq 2$,

[TABLE]

Denote by $C_{1}$ the generic irreducible element of $|S_{1}|_{F}|$. Assume that $|S_{1}|_{F}|$ and $|G|$ are not composed of the same pencil. Set $\tilde{\delta}=(C_{1}\cdot C)$. Then

• (i)

when $\tilde{\delta}\leq 2j$,

[TABLE]

is birational.

• (i)′

when $\tilde{\delta}\leq j$,

[TABLE]

is birational.

• (ii)

when $\tilde{\delta}>2j$,

[TABLE]

is birational.

• (ii)′

when $\tilde{\delta}>2j$ and $S_{1}|_{F}$ is big,

[TABLE]

is birational.

• (iii)

one has

[TABLE]

• (iv)

For any integer $n>\frac{m_{1}}{j}+\frac{1}{\beta}$ with

[TABLE]

one has

[TABLE]

Proof.

Let $|G_{1}|=|S_{1}|_{F}|$. By assumption, $|G_{1}|$ is also base point free. Set

[TABLE]

Write

[TABLE]

where $E_{m_{1}}$ is an effective ${\mathbb{Q}}$-divisor on $X^{\prime}$.

By Kawamata-Viehweg vanishing theorem ([KaV, V82]), we have

[TABLE]

since

[TABLE]

is simple normal crossing (by our assumption), nef and big.

Since $p_{g}(X)>0$, one sees that $|(n+1)K_{X^{\prime}}|$ distinguishes different general $F$ and $|(n+1)K_{X^{\prime}}||_{F}$ distinguishes different general $C$. What we need to do is to investigate the behavior of $|(n+1)K_{X^{\prime}}||_{C}$.

Recall that we have

[TABLE]

where $H_{1}$ is certain effective ${\mathbb{Q}}$-divisor. The vanishing theorem on $F$ gives

[TABLE]

where $\tilde{D}_{1}=\lceil{n\pi^{*}(K_{X})|_{F}-\frac{1}{j}E_{m_{1}}|_{F}-H_{1}}\rceil|_{C}$ with

[TABLE]

Thus $\varphi_{n+1,X}$ is birational, which implies Item (i).

For Item (ii)’, even if $n=\frac{m_{1}}{j}+\frac{1}{\beta}$ is integral, the ${\mathbb{Q}}$-divisor

[TABLE]

is nef and big since $S_{1}|_{F}$ is nef and big, we still have $\deg(\tilde{D}_{1})>2$. Hence $\varphi_{\lceil{\frac{m_{1}}{j}+\frac{1}{\beta}}\rceil+1,X}$ is birational.

A direct application of the above argument implies that, whenever $\deg(\tilde{D}_{1})>1$, $|K_{C}+\tilde{D}_{1}|$ is base point free and so

[TABLE]

which proves Item (iv).

Finally, modulo a further birational modification, we may and do assume that the linear system

[TABLE]

is base point free. It is clear that $M_{2j-1}^{\prime}$ is big. By the vanishing theorem and Theorem 2.3, we have

[TABLE]

which directly implies the statement in Item (iii).

Statement $(i)^{\prime}$ follows from the similar argument to that for (i). Instead of using the relation (3.2), one may use the statement (iii), namely:

[TABLE]

for an effective ${\mathbb{Q}}$-divisor $H^{\prime\prime}$ on $F$. In fact, it suffices to take

[TABLE]

to obtain (i).

We are left to treat Item (ii). By Kawamata-Viehweg vanishing theorem, we have

[TABLE]

since

[TABLE]

is simple normal crossing (by our assumption), nef and big.

Then the vanishing theorem on $F$ gives

[TABLE]

where $\tilde{D_{n}}=\lceil{n\pi^{*}(K_{X})|_{F}-\frac{2}{\tilde{\delta}}E_{m_{1}}|_{F}-\frac{1}{\mu}\cdot(1-\frac{2j}{\tilde{\delta}})E_{F}^{{}^{\prime}}|_{F}-H_{1}}\rceil|_{C}$ with $\mathrm{deg}\tilde{D_{n}}>2$. Hence

[TABLE]

is birational. ∎

Theorem 3.3**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)\geq 2$. Keep Assumption ($\pounds$). Let $m_{1}$ be a positive integer. Suppose that $M_{m_{1}}\geq j_{1}F+S_{1}$ for some moving divisor $S_{1}$ on $X^{\prime}$, $j_{1}>0$ is an integer and that ${S_{1}}|_{F}\geq j_{2}C+C^{\prime}$ where $C^{\prime}$ is a moving irreducible curve on $F$ with $C^{\prime}\not\equiv C$, $j_{2}>0$ is an integer. Set $\delta_{2}=(C^{\prime}\cdot C)$. The following statements hold:

• (i)

if $j_{1}\geq j_{2}$, then

• (i.1)

For any positive integer $n$ satisfying $n>\frac{m_{1}}{j}+\frac{j_{1}-j_{2}}{j_{1}}\cdot\frac{1}{\beta}$ and

[TABLE]

one has

[TABLE]

• (i.2)

Either

[TABLE]

or

[TABLE]

is birational.

• (ii)

if $j_{2}\geq j_{1}$, then

• (ii.1)

For any positive integer $n>\frac{m_{1}}{j_{2}}+\frac{1}{\mu}\cdot(1-\frac{j_{1}}{j_{2}})$ and

[TABLE]

one has

[TABLE]

• (ii.2)

Either

[TABLE]

or

[TABLE]

is birational.

Proof.

Modulo further birational modification, we may and do assume that $|S_{1}|$ is also base point free. Hence $S_{1}$ is nef. By assumption we may find two effective ${\mathbb{Q}}$-divisor $\tilde{E}^{\prime}_{m_{1}}$ and $\tilde{E}^{\prime\prime}_{m_{1}}$ such that

[TABLE]

[TABLE]

Set

[TABLE]

By the vanishing theorem, one has

[TABLE]

where

[TABLE]

Recall that we have

[TABLE]

where $H_{1}$ is an effective ${\mathbb{Q}}$-divisor. Hence

[TABLE]

By the vanishing theorem once more, we have

[TABLE]

where

[TABLE]

with

[TABLE]

Now, for the similar reason to previous ones, we see that $\varphi_{n+1,X}$ is brational and Statement (i.2) ($\delta_{2}\leq 2j_{1}$) follows. Statement (i.1) follows similarly.

Now turn to (i.2) where $\delta_{2}>2j_{1}$. By Kawamata-Viehweg vanishing theorem, we have

[TABLE]

since $n\pi^{*}(K_{X})-\frac{2}{\delta_{2}}\tilde{E}_{m_{1}}^{{}^{\prime}}-\frac{1}{\mu}(1-\frac{2j_{1}}{\delta_{2}})E_{F}^{{}^{\prime}}$ is simple normal crossing (by definition), nef and big. By vanishing theorem on $F$, we have

[TABLE]

where $\tilde{D}_{1,m_{1}}\equiv(n-\frac{2m_{1}}{\delta_{2}}-\frac{1}{\mu}(1-\frac{2j_{1}}{\delta_{2}})-\frac{1}{\beta}(1-\frac{2j_{2}}{\delta_{2}}))\pi^{*}(K_{X})|_{C}+\frac{2}{\delta_{2}}C^{\prime}|_{C}$ with $\mathrm{deg}(\tilde{D}_{1,m_{1}})>2$. Thus $\varphi_{n+1,X}$ is birational.

One gets Statement (ii) in the similar way, which we leave to interested readers. ∎

Remark 3.4**.**

From the proof of Theorem 3.3, one clearly sees that the variant of Theorem 3.3 with $C^{\prime}=0$ is also true.

Theorem 3.5**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)\geq 2$. Keep Assumption ($\pounds$). Let $m_{1}$ be a positive integer. Suppose that $M_{m_{1}}\geq j_{1}F+S_{1}$ for some moving divisor $S_{1}$ on $X^{\prime}$, $j_{1}>0$ is an integer and that ${S_{1}}|_{F}\geq j_{2}C$ where $j_{2}>0$ is an integer. The following statements hold:

• (i)

if $j_{1}\geq j_{2}$, then

• (i.1)

For any positive integer $n$ satisfying $n>\frac{m_{1}}{j}+\frac{j_{1}-j_{2}}{j_{1}}\cdot\frac{1}{\beta}$ and

[TABLE]

one has

[TABLE]

• (i.2)

The map

[TABLE]

is birational.

• (ii)

if $j_{2}\geq j_{1}$, then

• (ii.1)

For any positive integer $n>\frac{m_{1}}{j_{2}}+\frac{1}{\mu}\cdot(1-\frac{j_{1}}{j_{2}})$ and

[TABLE]

one has

[TABLE]

• (ii.2)

The map

[TABLE]

is birational.

Proof.

This follows directly from the proof of Theorem 3.3 by blinding $C^{\prime}$. We omit the redundant details. ∎

4. Threefolds of general type with $p_{g}=2$

This section is devoted to the classification of 3-folds of general type with $p_{g}(X)=2$. Keep the same notation as in 2.1. We have an induced fibration $f\colon X^{\prime}\longrightarrow\Gamma$ of which the general fiber $F$ is a nonsingular projective surface of general type. Denote by $\sigma\colon F\rightarrow F_{0}$ the contraction onto its minimal model.

By [Che03, Theorem 3.3], it is sufficient to assume that $b=g(\Gamma)=0$, i.e. $\Gamma\cong{\mathbb{P}}^{1}$. Since $p_{g}(X)>0$ and $F$ is a general fiber, we have $p_{g}(F)>0$. By the surface theory, $F$ belongs to one of the $3$ types:

1. (1)

$(K_{F_{0}}^{2},p_{g}(F_{0}))=(1,2)$; 2. (2)

$(K_{F_{0}}^{2},p_{g}(F_{0}))=(2,3)$; 3. (3)

other surfaces with $p_{g}(F_{0})>0$.

Recall that we have

[TABLE]

where $E_{1}^{{}^{\prime}}$ is an effective $\mathbb{Q}$-divisor since $\mu\geq 1$. By Theorem 2.3, the natural restriction map

[TABLE]

is surjective. Since $|6K_{X^{\prime}}|\succcurlyeq|3K_{X^{\prime}}+3F|$, by (4.1) and (4.2), we may write

[TABLE]

where $Q^{{}^{\prime}}$ and $\hat{E}_{F}$ are effective ${\mathbb{Q}}$-divisors.

4.1. Non-$(1,2)$-surface case

Lemma 4.1**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)=2$ and keep the setting in 2.1. Suppose $d_{1}=1$, $\Gamma\cong{\mathbb{P}}^{1}$ and $F$ is neither a $(2,3)$ surface nor a $(1,2)$-surface. Then $\varphi_{6,X}$ is birational.

Proof.

As $|6K_{X^{\prime}}|$ distinguishes different general fibers of $f$, (4.2) implies that $\varphi_{6,X}$ is birational unless $F$ is either a $(1,2)$-surface or a $(2,3)$-surface. ∎

Proposition 4.2**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)=2$, $d_{1}=1$, $\Gamma\cong{\mathbb{P}}^{1}$. Assume that $F$ is a $(2,3)$-surface. Then $\varphi_{6,X}$ is not birational if and only if $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$.

Proof.

We have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{1}{2}$ by (4.3) and by our assumption. By (4.1) and Kawamata-Viehweg vanishing theorem, we have

[TABLE]

If $(\pi^{*}(K_{X})|_{F})^{2}>\frac{1}{2}$, then $|K_{F}+\ulcorner 4\pi^{*}(K_{X})|_{F}\urcorner|$ is birational by [CHP17, Lemma 2.3], [CHP17, Lemma 2.5] and (4.3). So $\varphi_{6,X}$ is birational.

If $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$, we have $\pi^{*}(K_{X})|_{F}\equiv\frac{1}{2}\sigma^{*}(K_{F_{0}})$ by Hodge index theorem and (4.3). It is clear that $|3\sigma^{*}(K_{F_{0}})||_{C}\preccurlyeq|K_{C}+C|_{C}|$. Note that $|C|=|\sigma^{*}(K_{F_{0}})|$ is base point free, we have

[TABLE]

Since $(\pi^{*}(K_{X})\cdot C)=1$, the vanishing theorem and (4.1) implies that ${|M_{6}|}|_{C}=|K_{C}+D|$ with $\deg(D)\geq 2$. Hence ${|M_{6}|}|_{C}=|K_{C}+C|_{C}|$. Since $|C|_{C}|$ gives a $g^{1}_{2}$, $|K_{C}+C|_{C}|$ is clearly non-birational.

Therefore $\varphi_{6,X}$ is non-birational if $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$. ∎

4.2. The $(1,2)$-surface case

From now on, we always assume that $F$ is a $(1,2)$-surface. We have $0\leq\chi(\omega_{X})\leq 1$ by our assumption and [Che04, Lemma 4.5]. It is well known that $|K_{F_{0}}|$ has exactly one base point and that, after blowing up this point, $F$ admits a canonical fibration of genus $2$ with a unique section which we denote by $H$. Denote by $C$ a general member in $|G|=\text{Mov}|{\sigma^{*}(K_{F_{0}})}|$.

Remark 4.3**.**

By [Che14, Theorem 1.1], $\varphi_{7,X}$ is non-birational if and only if $\xi=\frac{2}{3}$. Since $p_{g}(X)>0$, $\varphi_{6,X}$ is non-birational as well if $\xi=\frac{2}{3}$.

Lemma 4.4**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)=2$, $d_{1}=1$, $\Gamma\cong{\mathbb{P}}^{1}$. Assume that $F$ is a $(1,2)$-surface. Keep the setting in 2.1. Then $|6K_{X^{\prime}}|$ distinguishes different generic irreducible elements of $|M|$ and $|6K_{X^{\prime}}||_{F}$ distinguishes generic irreducible elements of $|G|$.

Proof.

Since $|M|$ is composed of a rational pencil, it is clear that $|6K_{X^{\prime}}|$ distinguishes different generic irreducible elements of $|M|$. Notice that $|G|$ is composed of a rational pencil. The surjectivity of (4.2) implies that $|6K_{X^{\prime}}||_{F}$ distinguishes generic irreducible elements of $|G|$. ∎

Lemma 4.5**.**

(see [Che07, 2.15]) Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)=2$, $d_{1}=1$, $\Gamma\cong{\mathbb{P}}^{1}$. Assume that $F$ is a $(1,2)$-surface. Then $\beta\geq\frac{1}{2}$ and $\xi\geq\frac{2}{3}.$

Lemma 4.6**.**

Under the same condition as that of Lemma 4.5. Assume that $\xi=1$ and $(\pi^{*}(K_{X})|_{F})^{2}>\frac{1}{2}$. Then $\varphi_{6,X}$ is birational.

Proof.

Consider the Zariski decomposition of the following $\mathbb{Q}$-divisor:

[TABLE]

where

1. (1)

both $N^{+}$ and $N^{-}$ are effective $\mathbb{Q}$-divisors with $N^{+}+N^{-}=2\hat{E}_{F}$; 2. (2)

the $\mathbb{Q}$-divisor $(2\pi^{*}(K_{X})|_{F}+N^{+})$ is nef; 3. (3)

$((2\pi^{*}(K_{X})|_{F}+N^{+})\cdot N^{-})=0$.

Step 1. $(\pi^{*}(K_{X})|_{F})^{2}>\frac{1}{2}$ implies $(N^{+}\cdot C)>0$.

Since $C$ is nef, we see $(N^{+}\cdot C)\geq 0$. Assume the contrary that $(N^{+}\cdot C)=0$. Then $(N^{+})^{2}\leq 0$ as $C$ is a fiber of the canonical fibration of $F$. Notice that

[TABLE]

implies $(\pi^{*}(K_{X})|_{F}\cdot\hat{E}_{F})>0$. We clearly have $(\pi^{*}(K_{X})|_{F}\cdot N^{+})>0$ by the definition of Zariski decomposition. Hence

[TABLE]

a contradiction.

Step 2. $(N^{+}\cdot C)>0$ implies the birationality of $\varphi_{6,X}$.

By Kawamata-Viehweg vanishing theorem, we have

[TABLE]

Noting that

[TABLE]

and that $2\pi^{*}(K_{X})|_{F}+N^{+}$ is nef and big, the vanishing theorem gives

[TABLE]

where $\deg(D^{+})\geq 2\xi+(N^{+}\cdot C)>2$. By Lemma 4.4, (4.5) and (4.6), $\varphi_{6,X}$ is birational. ∎

Lemma 4.7**.**

Under the same condition as that of Lemma 4.5. If $\beta>\frac{2}{3}$, then $\varphi_{6,X}$ is birational.

Proof.

Since

[TABLE]

we have $\xi\geq\frac{4}{5}$ by (2.2). Now, as $\alpha(6)>2$, $\varphi_{6,X}$ is birational by Theorem 2.2. ∎

Lemma 4.8**.**

Under the same condition as that of Lemma 4.5. If $\mu>\frac{4}{3}$, then $\varphi_{6,X}$ is birational.

Proof.

By assumption and (2.5), one has $\beta\geq\frac{4}{7}$. So

[TABLE]

which implies $\xi\geq\frac{4}{5}$. Since $\alpha(6)>\frac{5}{2}\cdot\xi\geq 2$, $\varphi_{6,X}$ is birational for the similar reason. ∎

Lemma 4.9**.**

Let $X$ be a minimal projective $3$-fold of general type with $p_{g}(X)=2$, $d_{1}=1$ and $\Gamma\cong{\mathbb{P}}^{1}$. Keep the notation in 2.1. Assume that $F$ is a $(1,2)$-surface. Let $m_{1}\geq 2$ be any integer. Then $\varphi_{6,X}$ is birational provided that one of the following holds:

1. (i)

$u_{m_{1},0}=h^{0}(F,m_{1}K_{F})$; 2. (ii)

$h^{0}(M_{m_{1}}-jF)\geq\lfloor{\frac{4}{3}m_{1}}\rfloor-j+2\geq 2$* and $u_{m_{1},-j}\leq 1$ for some integer $j\geq 0$.*

Proof.

(i). Since $\theta_{m_{1},0}$ is surjective, we have $m_{1}\pi^{*}(K_{X})|_{F}\geq m_{1}C$, which means $\beta=1$. Hence $\varphi_{6,X}$ is birational by Lemma 4.7.

(ii). By assumption, $|M_{m_{1}}-jF|$ and $|F|$ are composed of same pencil. Hence we have $\mu>\frac{4}{3}$ and $\varphi_{6,X}$ is birational by Lemma 4.8. ∎

Lemma 4.10**.**

Under the same condition as that of Lemma 4.5. Suppose that $\xi=\frac{4}{5}$. Then $\varphi_{6,X}$ is not birational.

Proof.

Write $|M_{6}|=\mathrm{Mov}|6K_{X^{\prime}}|$. One may assume that $|M_{6}|$ is base point free. Denote by $G_{6,0}$ a general member of $|M_{6}|_{F}|$. Then

[TABLE]

Since $\alpha(6)>1$, we have $(6\pi^{*}(K_{X})|_{F}\cdot C)\geq(G_{6,0}\cdot C)\geq 4$ by (2.2). Therefore $(G_{6,0}\cdot C)=4$ by the assumption. On the other hand, the vanishing theorem gives

[TABLE]

where $\deg(\hat{D})\geq 2$, which forces $|S_{6,0}||_{F}|_{C}=|G_{6,0}|_{C}|$. Take a general effective divisor $K\in|K_{X}|$. Then $\mathrm{supp}(\pi^{*}(K)|_{F})|_{C}$ consists of just one point $P\in C$ since $\deg(\sigma^{*}(K_{F_{0}})|_{C})=1$. Here the divisor $P$ satisfies $2P\in|K_{C}|$. So $|4P|=|\llcorner(6\pi^{*}K|_{F})|_{C}\lrcorner|\succcurlyeq|G_{6,0}|_{C}|$. Therefore the restriction of the linear system $|S_{6,0}|$ on $C$ is just $|2K_{C}|$, which implies that $\varphi_{6,X}$ is not birational. ∎

4.3. Effective constraints on $P_{2}$, $P_{3}$, $P_{4}$, $P_{5}$ and $P_{6}$

This subsection is devoted to link some numerical constrains on plurigenera $P_{i}(X)$ ($i=1,\ldots,6$) to the birationality of $\varphi_{6}$.

The following proposition is the prototype for Propositions 4.12, 4.13, 4.14 and 4.15.

Proposition 4.11**.**

Under the same condition as that of Lemma 4.5, then

1. (1)

when $P_{2}(X)\geq 6$, $\varphi_{6,X}$ is birational; 2. (2)

when $P_{2}(X)=5$, $\varphi_{6,X}$ is not birational if and only if

[TABLE]

Proof.

Set $m_{1}=2$. By Lemma 4.9, we may assume

[TABLE]

Case 1. $u_{2,0}\leq 3$ and $u_{2,-1}=3$.

There is a moving divisor $S_{2,-1}$ on $X^{\prime}$ such that

[TABLE]

and $h^{0}(F,S_{2,-1}|_{F})\geq 3$. Modulo further birational modification, we may assume that $|S_{2,-1}|$ is base point free. Denote by $C_{2,-1}$ a generic irreducible element of $|S_{2,-1}|_{F}|$. Then $|C_{2,-1}|$ is moving since $q(F)=0$.

If $|S_{2,-1}|_{F}|$ and $|C|$ are composed of same pencil, then

[TABLE]

which means $\beta\geq 1$ and $\varphi_{6,X}$ is birational by Lemma 4.7.

If $|S_{2,-1}|_{F}|$ and $|C|$ are not composed of the same pencil, then $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.6] (1).

Case 2. $u_{2,0}\leq 3$ and $u_{2,-1}=2$.

If $|S_{2,-1}|_{F}|$ and $|C|$ are not composed of the same pencil, then $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.6](1).

If $|S_{2,-1}|_{F}|$ and $|C|$ are composed of the same pencil, then we have $\xi\geq\frac{4}{5}$ by [CHP17, Proposition 3.6] (2.1) (as $2\xi(|G|)\geq\frac{4}{3}>1$). By [CHP17, Proposition 3.6] (2.2) (taking $n=3$), $\varphi_{6,X}$ is birational.

Case 3. $u_{2,0}\leq 3$, $u_{2,-1}\leq 1$ and $P_{2}(X)\geq 6$.

We have $h^{0}(M_{2}-F)\geq 3$. Hence $\varphi_{6,X}$ is birational by Lemma 4.9 (ii).

Suppose $P_{2}(X)=5$.

We first assume that $\varphi_{6,X}$ is not birational. By the arguments in Case 1-Case 3, we have $u_{2,0}\leq 3$ and $u_{2,-1}=1$. If $u_{2,0}\leq 2$, one has $h^{0}(M_{2}-F)\geq 3$ by our assumption. Thus we have $\mu\geq\frac{3}{2}$. Lemma 4.8 implies that $\varphi_{6,X}$ is birational, which is a contradiction. So we have $u_{2,0}=3$. If $|M_{2}|_{F}|$ and $|C|$ are composed of the same pencil, we get $\beta\geq 1$. Lemma 4.7 implies that $\varphi_{6,X}$ is birational. So $|M_{2}|_{F}|$ and $|C|$ are not composed of the same pencil. Thus we have

[TABLE]

Lemma 4.6 implies that $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$.

Conversely, assume that $\xi=1$ and $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$. Observing that $\beta>\frac{1}{2}$ induces $(\pi^{*}(K_{X})|_{F})^{2}>\frac{1}{2}$. Since we assume that $\xi=1$, by the argument in Case 1-Case 3, one has $u_{2,0}\leq 3$ and $u_{2,-1}=1$. If $u_{2,0}\leq 2$, we have $h^{0}(F,M_{2}-F)\geq 3$. By [CHP17, Proposition 3.5], one has $\beta>\frac{3}{5}$, which contradicts to our assumption. So we have $u_{2,0}=3$. Similarly $|M_{2}|_{F}|$ induces a generically finite morphism. Pick a generic irreducible element $C_{2}$ in $|M_{2}|_{F}|$.

Since

[TABLE]

we have $C_{2}^{2}=2$ and $(\pi^{*}(K_{X})|_{F}\cdot C_{2})=1$. Since $|C_{2}|_{C_{2}}|$ is generically finite, $|C_{2}|_{C_{2}}|$ is a $g^{1}_{2}$. By Kawamata-Viehweg vanishing theorem, we have

[TABLE]

and

[TABLE]

So $|M_{6}||_{C_{2}}\succcurlyeq|K_{C_{2}}+C_{2}|_{C_{2}}|$. Since $6=(6\pi^{*}(K_{X})|_{F}\cdot C_{2})\geq\mathrm{deg}(M_{6}|_{C_{2}})$, we have

[TABLE]

$|C_{2}|_{C_{2}}|$ is $g^{1}_{2}$ implies that $\varphi_{6,X}$ is not birational. ∎

The interested reader should read carefully the proof of the next following four Propositions . Otherwise, one can possibly skip the proofs knowing only that they on the same lines as the previous one, with an increasing order of cases and difficulties.

Proposition 4.12**.**

Under the same condition as that of Lemma 4.5, if $P_{3}(X)\geq 9$, then $\varphi_{6,X}$ is birational.

Proof.

Set $m_{1}=3$. By Lemma 4.9, we may assume that $u_{3,0}\leq h^{0}(3K_{F})-1=5$. Since $|3K_{X^{\prime}}|\succcurlyeq|K_{X^{\prime}}+F_{1}+F|$ for two distinct general fibers of $f$, it follows from [Che04, Lemma 4.6] that

[TABLE]

which means $u_{3,0}\geq 2$.

Case 1. $u_{3,0}=5$.

We have $h^{0}(M_{3}|_{F}-C)\geq 3$ since $v_{3,0}\leq 2$ by [CHP17, Proposition 3.4].

If $v_{3,-1}\geq 2$, we have $M_{3}|_{F}\geq C+C_{3,-1}$ where $C_{3,-1}$ is a moving curve on $F$ with $(C_{3,-1}\cdot C)\geq 2$. Hence $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7](i).

If $v_{3,-1}\leq 1$, then $|M_{3}|_{F}-C|$ and $|C|$ are composed of the same pencil. Then $M_{3}|_{F}\geq 3C$ which means $\beta\geq 1$ and so $\varphi_{6,X}$ is birational by Lemma 4.7.

Case 2. $u_{3,0}=4$

We have $h^{0}(M_{3}|_{F}-C)\geq 2$.

If $v_{3,-1}\geq 2$, we get $M_{3}|_{F}\geq C+C_{3,-1}$ where $C_{3,-1}$ is a moving curve on $F$ with $(C_{3,-1}\cdot C)\geq 2$. Hence $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7](i).

If $v_{3,-1}\leq 1$, we see that $|M_{3}|_{F}-C|$ and $|C|$ are composed of the same pencil. Then $M_{3}|_{F}\geq 2C$ which implies $\beta\geq\frac{2}{3}$.

Subcase 2.1. $u_{3,-1}\geq 3$.

If $|S_{3,-1}|_{F}|$ and $|C|$ are not composed of the same pencil, then $(S_{3,-1}|_{F}\cdot C)\geq 2$ and so $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.6](1.2) ($m_{1}=3$, $j=1$, $\beta\geq\frac{2}{3}$, $\mu=1$, $\tilde{\delta}=2$).

If $|S_{3,-1}|_{F}|$ and $|C|$ are composed of the same pencil, then $S_{3,-1}|_{F}\geq 2C$. We get $\beta\geq\frac{3}{4}$ by [CHP17, Proposition 3.5] ($n_{1}=3$, $j_{1}=1$, $l_{1}=2$) and so $\varphi_{6,X}$ is birational by Lemma 4.7.

Subcase 2.2. $u_{3,-1}\leq 2$ and $u_{3,-2}=2$.

If $|S_{3,-2}|_{F}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.2(i).

If $|S_{3,-2}|_{F}|$ and $|C|$ are composed of the same pencil, we hope to use Theorem 3.5 with $j_{1}=2$ and $j_{2}=1$. Recall that $\beta\geq\frac{2}{3}$, $\xi\geq\frac{2}{3}$ and $\mu=1$. By taking $n=7$ and applying Inequality (2.2), we get $\xi\geq\frac{5}{7}$. Similarly, one gets $\xi\geq\frac{4}{5}$ by one more step optimization. Finally Theorem 3.5(i) implies the birationality of $\varphi_{6,X}$.

Subcase 2.3. $u_{3,-1}\leq 2$, $u_{3,-2}=1$ and $P_{3}(X)\geq 9$.

One has $h^{0}(M_{3}-2F)\geq 3$. As $u_{3,-2}=1$, one has $M_{3}\geq 4F$, which implies $\mu\geq\frac{4}{3}$. As $\alpha(5)>1$, we get $\xi\geq\frac{4}{5}$ by (2.2). Since $\alpha(6)>2$, $\varphi_{6,X}$ is birational.

Case 3. $u_{3,0}\leq 3$, $u_{3,-1}\leq 3$ and $u_{3,-2}\geq 2$.

We have $M_{3}\geq 2F+S_{3,-2}$ for a moving divisor $S_{3,-2}$ with $h^{0}(S_{3,-2}|_{F})\geq 2$. Clearly $\beta\geq\frac{2}{3}$.

If $|S_{3,-2}|_{F}|$ and $|C|$ are not composed of the same pencil, then $\xi\geq\frac{4}{5}$ by Theorem 3.2(iv) ($n=4$). Hence $\varphi_{6,X}$ is birational by Theorem 3.2(i).

If $|S_{3,-2}|_{F}|$ and $|C|$ are composed of the same pencil, we get $\xi\geq\frac{4}{5}$ by Theorem 3.5(i.1) ($n=4$) and $\varphi_{6,X}$ is birational by Theorem 3.5(i.2).

Case 4. $u_{3,0}\leq 3$, $u_{3,-1}=3$, $u_{3,-2}=1$ and $P_{3}(X)\geq 9$.

We have $h^{0}(M_{3}-2F)\geq 3$. Since $u_{3,-2}=1$, we have $M_{3}\geq 4F$. Thus, by [CHP17, Proposition 3.5], we have $\beta\geq\frac{4}{7}$.

If $|S_{3,-1}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(S_{3,-1}|_{F}\cdot C)\geq 2$. [CHP17, Proposition 3.6](1.2) implies the birationality of $\varphi_{6,X}$ ($m_{1}=3$, $j=1$, $\delta=2$, $\mu=\frac{4}{3}$).

If $|S_{3,-1}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{3}{4}$ by [CHP17, Proposition 3.5]. Thus $\varphi_{6,X}$ is birational by Lemma 4.7.

Case 5. $u_{3,0}\leq 3$, $u_{3,-1}\leq 2$, $u_{3,-2}=1$ and $P_{3}(X)\geq 9$.

Clearly we have $h^{0}(M_{3}-2F)\geq 4$, which implies that $\mu\geq\frac{5}{3}$. Hence $\varphi_{6,X}$ is birational by Lemma 4.8. ∎

Proposition 4.13**.**

Under the same condition as that of Lemma 4.5, then

1. (1)

when $P_{4}(X)\geq 15$, $\varphi_{6,X}$ is birational; 2. (2)

when $P_{4}(X)=14$, $\varphi_{6,X}$ is non-birational if and only if one of the following holds:

1. (2.1)

$\xi=\frac{4}{5}$; 2. (2.2)

$\xi=1$* and $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$.*

Proof.

Set $m_{1}=4$. By Lemma 4.7, we may and do assume that $u_{4,0}\leq h^{0}(4K_{F})-1=8$.

By [CHP17, Proposition 3.4](1), we know that $v_{4,0}\leq 3$. We claim that $v_{4,-1}=3$ is impossible. Otherwise, we have $M_{4}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(C,C_{-1}|_{C})\geq 3$. In particular, one has $(C_{-1}\cdot C)\geq 4$ by Riemann-Roch formula. Now we have

[TABLE]

which is a contradiction. Hence we have $v_{4,-1}\leq 2$.

In the proof we will always apply a setting such that, for some integer $j\geq 0$,

[TABLE]

for a moving divisor $S_{4,-j}$ with $h^{0}(F,S_{4,-j}|_{F})\geq 2$. Modulo further birational modifications, we may and do assume that $|S_{4,-j}|$ is base point free.

Case 1. $u_{4,0}\geq 7$.

Since $v_{4,0}\leq 3$ and $v_{4,-1}\leq 2$, we have $h^{0}(F,M_{4}|_{F}-2C)\geq 2$. There is a moving curve $C_{-2}$ such that $M_{4}|_{F}\geq 2C+C_{-2}$. When $|C|$ and $|C_{-2}|$ are composed of the same pencil, we get $M_{4}|_{F}\geq 3C$ and so $\beta\geq\frac{3}{4}$. Lemma 4.7 implies the birationality of $\varphi_{6,X}$. When $|C|$ and $|C_{-2}|$ are not composed of the same pencil, [CHP17, Proposition 3.7](iii) implies that we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=4$, $j=2$, $\delta_{1}=2$, $\xi\geq\frac{2}{3}$, $\mu=1$). Therefore $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7](i) ($m_{1}=4$, $j=2$, $\delta_{1}=2$, $\xi\geq\frac{4}{5}$, $\mu=1$).

Case 2. $u_{4,0}=6$.

The argument is organised according to the value of $v_{4,0}$.

Subcase 2.1. $u_{4,0}=6$ and $v_{4,0}\leq 2$.

We have $M_{4}|_{F}\geq 2C+C_{-2}$, where $h^{0}(C_{-2})=h^{0}(M_{4}|_{F}-2C)\geq 2$. If $|C_{-2}|$ and $|C|$ are composed of the same pencil, then we have $\beta\geq\frac{3}{4}$ and $\varphi_{6,X}$ is birational by Lemma 4.7. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $\xi\geq\frac{4}{5}$ by [CHP17, Proposition 3.7](iii) ($n=4$, $m_{1}=4$, $j=2$, $\delta_{1}=2$, $\xi\geq\frac{2}{3}$, $\mu=1$). Then $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7](i) ($m_{1}=4$, $j=2$, $\delta_{1}=2$, $\xi\geq\frac{4}{5}$, $\mu=1$).

Subcase 2.2. $u_{4,0}=6$, $u_{4,-1}\geq 4$ and $v_{4,0}=3$.

Since $v_{4,0}=3$, we have $\xi=1$. Clearly we have $h^{0}(F,S_{4,-1}|_{F})\geq 4$ by assumption.

If $|S_{4,-1}|_{F}|$ and $|C|$ are not composed of the same pencil and $(S_{4,-1}|_{F}\cdot C)\geq 4$. We have

[TABLE]

by Theorem 3.2(iii). So $\varphi_{6,X}$ is birational by Lemma 4.6.

If $|S_{4,-1}|_{F}|$ and $|C|$ are not composed of the same pencil and $(S_{4,-1}|_{F}\cdot C)\leq 3$. Then we have

[TABLE]

where $C_{-1}$ is a moving curve on $F$. If $|C_{-1}|$ and $|C|$ are not composed of the same pencil, noting that

[TABLE]

we still have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{3}{5}$ by Theorem 3.2(iii) and above inequality. Hence $\varphi_{6,X}$ is birational by Lemma 4.6. If $|C_{-1}|$ and $|C|$ are composed of the same pencil. We have $\beta\geq\frac{3}{5}$ by [CHP17, Proposition 3.5] ($n_{1}=4$, $j_{1}=1$, $l_{1}=2$). Since

[TABLE]

$\varphi_{6,X}$ is birational by Lemma 4.6.

If $|S_{4,-1}|_{F}|$ and $|C|$ are composed of the same pencil. We have $S_{4,-1}|_{F}\geq 3C$. By [CHP17, Proposition 3.5], we get $\beta\geq\frac{4}{5}$ and so $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{5}$. Hence $\varphi_{6,X}$ is birational by Lemma 4.7.

Subcase 2.3. $u_{4,0}=6$, $u_{4,-1}\leq 3$, $u_{4,-2}\leq 3$, $u_{4,-3}\geq 2$ and $v_{4,0}=3$.

$v_{4,0}=3$ implies that $\xi=1$.

If $|S_{4,-3}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{7}>\frac{1}{2}$ by Theorem 3.2(iii) and $\varphi_{6,X}$ is birational by Lemma 4.6.

If $|S_{4,-3}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. Similarly, one has $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{7}>\frac{1}{2}$. Hence $\varphi_{6,X}$ is birational by Lemma 4.6.

Subcase 2.4. $u_{4,0}=6$, $u_{4,-1}\leq 3$, $u_{4,-2}\leq 3$, $u_{4,-3}=1$, $v_{4,0}=3$ and $P_{4}(X)\geq 15$.

$v_{4,0}=3$ implies that $\xi=1$.

As $|M_{4}-3F|$ and $|F|$ are composed of the same pencil and $h^{0}(M_{4}-3F)\geq 3$, we have $\mu\geq\frac{5}{4}$ and so $\beta\geq\frac{5}{9}$ by [CHP17, Proposition 3.5]. Hence $\varphi_{6,X}$ is birational by Lemma 4.6 since $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{9}>\frac{1}{2}$.

Case 3. $u_{4,0}\leq 5$ and $u_{4,-1}\geq 4$.

If $|S_{4,-1}|_{F}|$ and $|C|$ are composed of the same pencil, then $\beta\geq\frac{4}{5}$ by [CHP17, Proposition 3.5]. Hence $\varphi_{6,X}$ is birational by Lemma 4.7.

Assume that $|S_{4,-1}|_{F}|$ and $|C|$ are not composed of the same pencil. For the case $(S_{4,-1}|_{F}\cdot C)=4$, $\varphi_{6,X}$ is birational by Theorem 3.2(ii) ($m_{1}=4$, $j=1$, $\tilde{\delta}=4$, $\beta=\frac{1}{2}$, $\mu=1$). For the case $(S_{4,-1}|_{F}\cdot C)\leq 3$, we have $S_{4,-1}|_{F}\geq C+C_{-1}$ where $C_{-1}$ is a moving curve. When $|C_{-1}|$ and $|C|$ are not composed of the same pencil, then $\varphi_{6,X}$ is birational by Theorem 3.3(i.2) ($m_{1}=4$, $j_{1}=j_{2}=1$, $\mu=1$, $\beta=\frac{1}{2}$, $\delta_{2}=2$).

When $|C_{-1}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{3}{5}$ by [CHP17, Proposition 3.5]. Since $\alpha(7)>2$, we have $\xi\geq\frac{5}{7}$.

Subcase 3.1. $u_{4,0}\leq 5$, $u_{4,-1}=4$ and $u_{4,-2}=4$.

Since $|S_{4,-2}|_{F}|=|S_{4,-1}|_{F}|$, we have $S_{4,-2}|_{F}\geq 2C$. Hence $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2)($\xi=\frac{5}{7}$, $m_{1}=4$, $j_{1}=j_{2}=2$, $\beta=\frac{3}{5}$).

Subcase 3.2. $u_{4,0}\leq 5$, $u_{4,-1}=4$, $u_{4,-2}\leq 3$ and $u_{4,-3}\geq 2$.

If $|S_{4,-3}|_{F}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=4$, $j=3$, $\xi=\frac{5}{7}$, $\tilde{\delta}=2$, $\beta=\frac{3}{5}$). If $|S_{4,-3}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\xi\geq\frac{4}{5}$ by Theorem 3.5(i.1)($n=4$, $m_{1}=4$, $j_{1}=3$, $j_{2}=1$, $\beta=\frac{3}{5}$, $\xi=\frac{5}{7}$). Hence $\varphi_{6,X}$ is birational by Theorem 3.5(i.2) ($m_{1}=4$, $j_{1}=3$, $j_{2}=1$, $\xi=\frac{4}{5}$, $\beta=\frac{3}{5}$).

Subcase 3.3. $u_{4,0}\leq 5$, $u_{4,-1}=4$, $u_{4,-2}\leq 3$, $u_{4,-3}=1$ and $P_{4}(X)\geq 15$.

We have $\zeta(4)\geq 5$ by our assumption. Recall that we have $\beta\geq\frac{3}{5}$ in this case. We get $\xi\geq\frac{4}{5}$ since $\alpha(5)\geq\frac{46}{45}>1$. Hence $\varphi_{6,X}$ is birational as $\alpha(6)\geq\frac{152}{75}>2$.

Case 4. $u_{4,0}\leq 5$, $u_{4,-1}\leq 3$ and $u_{4,-3}\geq 3$.

If $|S_{4,-3}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{5}{7}>\frac{2}{3}$ by [CHP17, Proposition 3.5]. So $\varphi_{6,X}$ is birational by Lemma 4.7.

Assume that $|S_{4,-3}|_{F}|$ and $|C|$ are not composed of the same pencil. When $(S_{4,-3}|_{F}\cdot C)\geq 3$, we have $\xi\geq\frac{6}{7}$ by Theorem 3.2(iii). Thus Theorem 3.2(i) ($\beta=\frac{1}{2}$) implies the birationality of $\varphi_{6,X}$. When $(S_{4,-3}|_{F}\cdot C)\leq 2$, we have $S_{4,-3}|_{F}\geq C$ and so $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. Also we have $\xi\geq\frac{5}{7}$ by Theorem 3.2(iii) ($j=3$, $m_{1}=4$). Hence $\varphi_{6,X}$ is birational by Theorem 3.2(i) ($m_{1}=4$, $j=3$, $\tilde{\delta}=2$, $\xi=\frac{5}{7}$, $\beta=\frac{4}{7}$).

Case 5. $u_{4,0}\leq 5$, $u_{4,-1}\leq 3$, $u_{4,-3}\leq 2$ and $u_{4,-4}\geq 2$.

If $|S_{4,-4}|_{F}|$ and $|C|$ are composed of the same pencil, we get $\beta\geq\frac{5}{8}$ by [CHP17, Proposition 3.5]. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($m_{1}=4$, $j_{1}=4$, $j_{2}=1$, $\beta=\frac{5}{8}$, $\xi=\frac{2}{3}$, $n=4$). Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($m_{1}=4$, $j_{1}=4$, $j_{2}=1$, $\beta=\frac{5}{8}$, $\xi=\frac{4}{5}$).

If $|S_{4,-4}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $\xi\geq\frac{3}{4}$ by Theorem 3.2(iii). Furthermore one gets $\xi\geq\frac{7}{9}$ by Theorem 3.2(iv) ($n=8$). Finally $\varphi_{6,X}$ is birational by Theorem 3.2(i) ($m_{1}=4$, $j=4$, $\beta=\frac{1}{2}$, $\xi=\frac{7}{9}$, $\tilde{\delta}=2$).

Case 6. $u_{4,0}\leq 5$, $u_{4,-1}\leq 3$, $u_{4,-3}\leq 2$, $u_{4,-4}=1$ and $P_{4}(X)\geq 15$.

In any case, the assumption implies that $\mu\geq\frac{5}{4}$.

If $u_{4,0}\leq 4$, we get $h^{0}(M_{4}-4F)\geq 3$. One has $M_{4}\geq 6F$ which means $\mu\geq\frac{3}{2}>\frac{4}{3}$. Hence $\varphi_{6,X}$ is birational by Lemma 4.8.

If $u_{4,0}=5$, then either $(M_{4}|_{F}\cdot C)=4$ or $M_{4}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve with $h^{0}(C_{-1})\geq 3$. We have $\xi=1$ in first case and, since $\beta\geq\frac{5}{9}$ by (2.5), one has $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{9}>\frac{1}{2}$. Hence $\varphi_{6,X}$ is birational by Lemma 4.6. Now turn to the later case. When $|C_{-1}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{3}{4}$ and $\varphi_{6,X}$ is birational by Lemma 4.7. If $|C_{-1}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7] ($j=1$, $\delta_{1}=2$, $\mu=\frac{5}{4}$).

Now suppose $P_{4}(X)=14$.

We first assume that $\varphi_{6,X}$ is not birational. By the arguments in Case 1-Case 6, one of the following holds.

1. (a)

$u_{4,0}=6$, $u_{4,-1}\leq 3$, $u_{4,-2}\leq 3$, $u_{4,-3}=1$, $v_{4,0}=3$; 2. (b)

$u_{4,0}\leq 5$, $u_{4,-1}\leq 3$, $u_{4,-2}\leq 3$ and $u_{4,-3}=2$; 3. (c)

$u_{4,0}\leq 5$, $u_{4,-1}\leq 3$, $u_{4,-2}\leq 3$ and $u_{4,-3}=1$.

For (a), we have $\xi=1$ since $v_{4,0}=3$. Lemma 4.6 implies that we have $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$. Thus (2.2) holds.

For (b). If $|S_{4,-3}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(S_{4,-3}|_{F}\cdot C)\geq 2$. By Theorem 3.2 (iv), we have $\xi\geq\frac{4}{5}$ ($n=4$). Theorem 3.2 (i) implies that $\varphi_{6,X}$ is birational when $\xi>\frac{4}{5}$ ($\tilde{\delta}=2$, $j=3$, $\xi>\frac{4}{5}$, $\beta=\frac{1}{2}$). So $\xi=\frac{4}{5}$ holds.

If $|S_{4,-3}|_{F}|$ and $|C|$ are composed of the same pencil. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{4}{7}$. Since

[TABLE]

we have $\xi\geq\frac{5}{7}$. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($m_{1}=4$, $j_{1}=3$, $j_{2}=1$, $\beta=\frac{4}{7}$, $\xi=\frac{5}{7}$). Theorem 3.5 (i.2) implies that $\varphi_{6,X}$ is birational when $\xi>\frac{4}{5}$ ($j_{1}=3$, $j_{2}=1$, $\xi>\frac{4}{5}$, $\beta=\frac{4}{7}$). Thus we have $\xi=\frac{4}{5}$.

We are left to treat (c). We claim that $u_{4,0}=5$. Otherwise, one has $u_{4,0}\leq 4$. Thus we have $h^{0}(M_{4}-3F)\geq 4$. Then $\varphi_{6,X}$ is birational by Lemma 4.9 (ii) ($m_{1}=4$, $j=3$), which contradicts to our assumption. So we have $u_{4,0}=5$. By our assumption, we have $h^{0}(M_{4}-3F)\geq 3$. Thus one has $\mu\geq\frac{5}{4}$ and $\beta\geq\frac{5}{9}$ by [CHP17, Proposition 3.5].

If $v_{4,0}\leq 2$, one has $M_{4}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve satisfying $h^{0}(F,C_{-1})\geq 3$. If $|C_{-1}|$ and $|C|$ are composed of the same pencil, one gets $\beta\geq\frac{3}{4}$. Lemma 4.7 implies that $\varphi_{6,X}$ is birational, which contradicts to our assumption. Thus we have $(C_{-1}\cdot C)\geq 2$. [CHP17, Proposition 3.7] (i) and (ii) implies that $\varphi_{6,X}$ is birational ($m_{1}=4$, $j=1$, $\mu=\frac{5}{4}$, $\beta=\frac{5}{9}$), which is a contradiction. So we have $v_{4,0}=3$. In particular, one has $\xi=1$. Lemma 4.6 implies that we have $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$.

Now we consider the other direction. Lemma 4.10 implies that we only need to consider the case where $\xi=1$ and $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$. Observing that $\beta>\frac{1}{2}$ implies that $(\pi^{*}(K_{X})|_{F})^{2}>\frac{1}{2}$. Thus we get $\mu=1$ and $\beta=\frac{1}{2}$. By the argument in Case 1-Case 6, one of the following holds:

1. (i)

$u_{4,0}=6$, $u_{4,-1}\leq 3$, $u_{4,-2}\leq 3$, $u_{4,-3}=1$ and $v_{4,0}=3$; 2. (ii)

$u_{4,0}\leq 5$, $u_{4,-1}\geq 4$, $S_{4,-1}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve. Moreover, $|C_{-1}|$ and $|C|$ are not composed of the same pencil; 3. (iii)

$u_{4,0}\leq 5$, $u_{4,-1}\leq 3$, $u_{4,-3}=2$; 4. (iv)

$u_{4,0}\leq 5$, $u_{4,-1}\leq 3$, $u_{4,-3}=1$.

We first consider (i). Our assumption gives $M_{4}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1})\geq 3$. Since $q(F)=0$, $\xi=1$ and $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$, $|C_{-1}|$ is not composed of pencil and we have $(\pi^{*}(K_{X})|_{F}\cdot C_{-1})=1$. We may and do assume that $|C_{-1}|$ is base point free. Take a general member $C_{-1}\in|C_{-1}|$. One has $g(C_{-1})\geq 3$ and $|C|_{C_{-1}}|$ is $g^{1}_{2}$. By Kawamata-Viehweg vanishing theorem, we have

[TABLE]

By Ramanujam vanishing theorem, one has $h^{1}(F,K_{F}+C)=0$. Thus we have

[TABLE]

So

[TABLE]

Since

[TABLE]

we have

[TABLE]

Since $|C|_{C_{-1}}|$ is $g^{1}_{2}$, $\varphi_{6,X}$ is non-birational.

For (ii). By Theorem 3.2 (iii) ($j=1$), we have

[TABLE]

Since $\xi=1$ and $\beta\geq\frac{1}{2}$, we have

[TABLE]

which contradicts to our assumption.

For (iii). If $|S_{4,-3}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5], which contradicts to our assumption. Thus $|S_{4,-3}|_{F}|$ and $|C|$ are not composed of the same pencil. In particular, we have $(S_{4,-3}|_{F}\cdot C)\geq 2$. By Theorem 3.2 (iii) ($j=3$), we have

[TABLE]

One can gets $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{7}$, which is a contradiction.

We are left to treat (iv). By our assumption, we have $\mu\geq\frac{5}{4}$, which is a contradiction. ∎

Proposition 4.14**.**

Under the same condition as that of Lemma 4.5, then

1. (1)

when $P_{5}(X)\geq 24$, then $\varphi_{6,X}$ is birational; 2. (2)

when $22\leq P_{5}(X)\leq 23$, $\varphi_{6,X}$ is non-birational if and only if $\xi=\frac{4}{5}$.

Proof.

Set $m_{1}=5$. By Lemma 4.9, we may and do assume that $u_{5,0}\leq h^{0}(5K_{F})-1=12$.

By [CHP17, Proposition 3.4], we know that $v_{5,0}\leq 4$. When $\mathrm{dim}\psi_{5,0}(U_{5,0})=4$, we have $\mathrm{deg}(M_{5}|_{C})=5$, which implies that $\xi=1$. By Riemann-Roch formula, we have $h^{0}(C,M_{5}|_{C})=4$. So $|M_{5}||_{C}$ is the complete linear system $|{M_{5}}|_{C}|=|K_{C}+D_{1}|$ with $\mathrm{deg}(D_{1})=3$. Thus $\varphi_{5,X}$ is birational which implies that $\varphi_{6,X}$ is birational. So we may assume that $\mathrm{dim}\psi_{5,0}(U_{5,0})\leq 3$.

Suppose $v_{5,-1}\geq 3$. Then $M_{5}|_{F}\geq C+C_{-1}$ for some moving divisor $C_{-1}$ satisfying $h^{0}(C,C_{-1}|_{C})\geq 3$. By Riemann-Roch formula, we have $(C_{-1}\cdot C)\geq 4$. Then $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7] (ii) ($\mu=1$, $m_{1}=5$, $j=1$, $\delta_{1}=4$, $\beta=\frac{1}{2}$). From now on, we may assume that $v_{5,-1}\leq 2$.

Case 1. $u_{5,0}\geq 8$.

Since $\mathrm{dim}\psi_{5,0}(U_{5,0})\leq 3$, we have $h^{0}(M_{5}|_{F}-C)\geq 5$. Because $v_{5,-1}\leq 2$, we have $M_{5}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$ satisfying $h^{0}(C_{-2})\geq 3$. If $|C_{-2}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{4}{5}$ and $\varphi_{6,X}$ is birational by Lemma 4.7. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}\cdot C)\geq 2$. We have

[TABLE]

By [CHP17, Proposition 3.7](iii), we have $\xi\geq\frac{4}{5}$ ($n=4$, $\mu=1$, $m_{1}=5$, $j=2$, $\xi(1,|C|)=\frac{2}{3}$). Then $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7] (i) ($\mu=1$, $m_{1}=5$, $j=2$, $\delta_{1}=2$, $\xi=\frac{4}{5}$).

Case 2. $u_{5,0}=7$.

If $\mathrm{dim}\psi_{5,0}(U_{5,0})\leq 2$, we have $h^{0}(M_{5}|_{F}-C)\geq 5$. The same argument as in Case 1 implies that $\varphi_{6,X}$ is birational. So we may assume that $\mathrm{dim}\psi_{5,0}(U_{5,0})=3$, which implies that $\xi\geq\frac{4}{5}$.

Since $v_{5,-1}\leq 2$, we have $M_{5}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve satisfying $h^{0}(C_{-2})\geq 2$. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $2\leq(C_{-2}\cdot C)\leq((M_{5}|_{F}-2C)\cdot\sigma^{*}(K_{F_{0}}))\leq 3$. [CHP17, Proposition 3.7] (i) implies that $\varphi_{6,X}$ is birational ($\mu=1$, $m_{1}=5$, $j=2$, $\delta_{1}=2$, $\xi=\frac{4}{5}$).

Suppose $|C_{-2}|$ and $|C|$ are composed of the same pencil. We get $\beta\geq\frac{3}{5}$ by our assumption. So we may assume that $\xi\geq\frac{4}{5}$ and $\beta\geq\frac{3}{5}$ in this case.

Subcase 2.1. $u_{5,-1}\geq 6$.

If $\mathrm{dim}\psi_{5,0}(U_{5,-1})=3$, we have $(S_{5,-1}|_{F}\cdot C)\geq 4$. By Theorem 2.3, we have

[TABLE]

[CHP17, Lemma 3.1] implies that $M_{4}$ is a big divisor. So $S_{5,-1}$ is nef and big. Kawamata-Viehweg vanishing theorem yields $|6K_{X^{\prime}}||_{F}\succcurlyeq|K_{F}+S_{5,-1}|_{F}|$. Thus $M_{6}|_{F}\geq C+S_{5,-1}|_{F}$. Then $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7] (ii) ($\mu=1$, $m_{1}=5$, $j=1$, $\delta_{1}=4$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

If $\mathrm{dim}\psi_{5,0}(U_{5,-1})\leq 2$, we have $S_{5,-1}|_{F}\geq 2C+C^{\prime}$ where $C^{\prime}$ is a moving divisor on $F$. If $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, we have $2\leq(C^{\prime}\cdot C)\leq(5\pi^{*}(K_{X})|_{F}-2C)\cdot\sigma^{*}(K_{F_{0}})\leq 3$. Theorem 3.3 (ii) implies that $\varphi_{6,X}$ is birational ($\mu=1$, $m_{1}=5$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\xi=\frac{4}{5}$). If $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $S_{5,-1}|_{F}\geq 3C$. By Theorem 3.5 (ii.2), $\varphi_{6,X}$ is birational ($\mu=1$, $m_{1}=5$, $j_{1}=1$, $j_{2}=3$, $\xi=\frac{4}{5}$).

Subcase 2.2. $u_{5,-1}\leq 5$, $u_{5,-2}=5$.

If $(S_{5,-2}|_{F}\cdot C)=5$, we have $\xi=1$. Since $\beta\geq\frac{3}{5}$, $\varphi_{6,X}$ is birational by Lemma 4.6. So we may assume $(S_{5,-2}|_{F}\cdot C)\leq 4$.

If $(S_{5,-2}|_{F}\cdot C)=4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=5$, $j=2$, $\tilde{\delta}=4$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

We are left to treat the case when $(S_{5,-1}|_{F}\cdot C)\leq 3$. We have $S_{5,-2}|_{F}\geq C+C^{\prime}$, where $C^{\prime}$ is a moving curve satisfying $h^{0}(F,C^{\prime})\geq 3$. If $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{5}{7}>\frac{2}{3}$ by [CHP17, Proposition 3.5]. By Lemma 4.7, $\varphi_{6,X}$ is birational. If $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2) ($m_{1}=5$, $j_{1}=2$, $j_{2}=1$, $\delta_{2}=2$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$ ).

Subcase 2.3. $u_{5,-1}\leq 5$, $u_{5,-2}\leq 4$ and $u_{5,-3}\geq 4$.

If $(S_{5,-3}|_{F}\cdot C)\geq 4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=5$, $j=3$, $\tilde{\delta}=4$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

If $(S_{5,-3}|_{F}\cdot C)\leq 3$, we have $S_{5,-3}|_{F}\geq C+C^{\prime}$, where $C^{\prime}$ is a moving curve. If $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{5}{8}$ and $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($m_{1}=5$, $j_{1}=3$, $j_{2}=2$, $\beta=\frac{5}{8}$, $\xi=\frac{4}{5}$). If $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, we have $(C^{\prime}\cdot C)\geq 2$ and $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2) ($m_{1}=5$, $j_{1}=3$, $j_{2}=1$, $\delta_{2}=2$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

Subcase 2.4. $u_{5,-1}\leq 5$, $u_{5,-2}\leq 4$, $u_{5,-3}\leq 3$ and $u_{5,-4}\geq 3$.

If $|S_{5,-4}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{2}{3}$ by [CHP17, Proposition 3.5]. Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($m_{1}=5$, $j_{1}=4$, $j_{2}=2$, $\beta=\frac{2}{3}$, $\xi=\frac{4}{5}$).

If $|S_{5,-4}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $2\leq(S_{5,-4}|_{F}\cdot C)\leq 5$. Then $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=5$, $j=4$, $\tilde{\delta}=2$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

Subcase 2.5. $u_{5,-1}\leq 5$, $u_{5,-2}\leq 4$, $u_{5,-3}\leq 3$, $u_{5,-4}\leq 2$ and $u_{5,-5}\geq 2$.

If $|S_{5,-5}|_{F}|$ and $|C|$ are composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($m_{1}=5$, $j_{1}=5$, $j_{2}=1$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

If $|S_{5,-5}|_{F}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=5$, $j=5$, $\tilde{\delta}=2$, $\beta=\frac{3}{5}$, $\xi=\frac{4}{5}$).

Subcase 2.6. $u_{5,-1}\leq 5$, $u_{5,-2}\leq 4$, $u_{5,-3}\leq 3$, $u_{5,-4}\leq 2$, $u_{5,-5}=1$ and $P_{5}(X)\geq 24$.

Since $h^{0}(M_{5}-5F)\geq 3$, we have $\mu\geq\frac{7}{5}>\frac{4}{3}$. Thus $\varphi_{6,X}$ is birational by Lemma 4.8.

Case 3. $u_{5,0}\leq 6$, $u_{5,-1}\leq 6$, $u_{5,-2}\geq 5$.

If $(S_{5,-2}|_{F}\cdot C)\geq 5$, $\varphi_{6,X}$ is birational by Theorem 3.2 (ii) ($m_{1}=5$, $j=2$, $\beta=\frac{1}{2}$, $\tilde{\delta}=5$).

If $(S_{5,-2}|_{F}\cdot C)=4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=5$, $j=2$, $\xi=\frac{4}{5}$, $\tilde{\delta}=4$, $\beta=\frac{1}{2}$).

The remaining case is $(S_{5,-2}|_{F}\cdot C)\leq 3$. By Riemann-Roch formula, we have $\mathrm{dim}\psi_{5,0}(U_{5,-2})\leq 2$. So we have $S_{5,-2}|_{F}\geq 2C$ and $S_{5,-2}|_{F}\geq C+C^{\prime}$, where $C^{\prime}$ is a moving curve satisfying $h^{0}(F,C^{\prime})\geq 3$. The former implies $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. If $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{5}{7}>\frac{2}{3}$. Lemma 4.7 implies that $\varphi_{6,X}$ is birational. If $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, we have

[TABLE]

By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($\xi=\frac{2}{3}$, $\delta_{2}=2$, $j_{1}=2$, $j_{2}=1$, $\beta=\frac{4}{7}$).

Case 4. $u_{5,0}\leq 6$, $u_{5,-1}\leq 6$, $u_{5,-2}\leq 4$, $u_{5,-3}\geq 4$.

If $(S_{5,-3}|_{F}\cdot C)\geq 4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($\xi=\frac{4}{5}$, $\tilde{\delta}=4$, $j=3$, $\beta=\frac{1}{2}$, $m_{1}=5$). So we may assume $(S_{5,-3}|_{F}\cdot C)\leq 3$. By Riemann-Roch formula, we have $S_{5,-3}|_{F}\geq C+C^{\prime}$ where $C^{\prime}$ is a moving curve on $F$.

If $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{5}{8}$ by [CHP17, Proposition 3.5]. We have $\alpha(7)\geq\frac{34}{15}>2$, which implies $\xi\geq\frac{5}{7}$. Since $\alpha(8)\geq\frac{22}{7}>3$, we have $\xi\geq\frac{3}{4}$. So $\alpha(5)\geq\frac{21}{20}>1$ and $\xi\geq\frac{4}{5}$ follows. By Theorem 3.5 (i.2), $\varphi_{6,X}$ is birational ($\xi=\frac{4}{5}$, $m_{1}=5$, $j_{1}=3$, $j_{2}=2$, $\beta=\frac{5}{8}$).

If $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, we have $2\leq(C^{\prime}\cdot C)\leq 4$. By Theorem 3.3 (i.1), we have $\xi\geq\frac{5}{7}$ ($n=6$, $\xi=\frac{2}{3}$, $\delta_{2}=2$, $j_{1}=3$, $j_{2}=1$, $m_{1}=5$, $\beta=\frac{1}{2}$). So $\varphi_{6,X}$ is birational by Theorem 3.3(i.2) ($\xi=\frac{5}{7}$, $\delta_{2}=2$, $j_{1}=3$, $j_{2}=1$, $m_{1}=5$, $\beta=\frac{1}{2}$).

Case 5. $u_{5,0}\leq 6$, $u_{5,-1}\leq 6$, $u_{5,-2}\leq 4$, $u_{5,-3}\leq 3$, $u_{5,-4}\geq 3$.

If $(S_{5,-4}|_{F}\cdot C)\geq 4$, we have $\xi\geq\frac{4}{5}$. By Theorem 3.2 (i), $\varphi_{6,X}$ is birational ($\tilde{\delta}=4$, $\xi=\frac{4}{5}$, $j=4$, $m_{1}=5$, $\beta=\frac{1}{2}$). We may assume that $(S_{5,-4}|_{F}\cdot C)\leq 3$. By Riemann-Roch formula, we have $S_{5,-4}|_{F}\geq C$. We get $\beta\geq\frac{5}{9}$ by [CHP17, Proposition 3.5].

If $|S_{5,-4}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{2}{3}$ by [CHP17, Proposition 3.5]. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $\xi=\frac{2}{3}$, $j_{1}=4$, $m_{1}=5$, $j_{2}=2$, $\beta=\frac{2}{3}$ ). Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($\xi=\frac{4}{5}$, $j_{1}=4$, $m_{1}=5$, $j_{2}=2$, $\beta=\frac{2}{3}$).

If $|S_{5,-4}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(S_{5,-4}|_{F}\cdot C)\geq 2$. By Theorem 3.2 (iv), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=5$, $j=4$, $\beta=\frac{5}{9}$, $\tilde{\delta}=2$). Hence $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($\xi=\frac{4}{5}$, $\tilde{\delta}=2$, $j=4$, $m_{1}=5$, $\beta=\frac{5}{9}$).

Case 6. $u_{5,0}\leq 6$, $u_{5,-1}\leq 6$, $u_{5,-2}\leq 4$, $u_{5,-3}\leq 3$, $u_{5,-4}\leq 2$, $u_{5,-5}\geq 2$.

If $|S_{5,-5}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{3}{5}$ by [CHP17, Proposition 3.5]. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=5$, $j_{1}=5$, $j_{2}=1$, $\beta=\frac{3}{5}$, $\xi=\frac{2}{3}$). By Theorem 3.5 (i.2), $\varphi_{6,X}$ is birational ($\xi=\frac{4}{5}$, $m_{1}=5$, $j_{1}=5$, $j_{2}=1$, $\beta=\frac{3}{5}$).

If $|S_{5,-5}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(S_{5,-5}|_{F}\cdot C)\geq 2$. By Theorem 3.2 (iv) , we have $\xi\geq\frac{4}{5}$. By Theorem 3.2 (i)′, $\varphi_{6,X}$ is birational ($m_{1}=5$, $j=5$, $\tilde{\delta}=2$, $\xi=\frac{4}{5}$).

Case 7. $u_{5,0}\leq 6$, $u_{5,-1}\leq 6$, $u_{5,-2}\leq 4$, $u_{5,-3}\leq 3$, $u_{5,-4}\leq 2$, $u_{5,-5}\leq 1$ and $P_{5}(X)\geq 24$.

We have $h^{0}(M_{5}-5F)\geq 3$ by our assumption. Since $u_{5,-5}=1$, we have $\mu\geq\frac{7}{5}>\frac{4}{3}$ by [CHP17, Proposition 3.5]. Then $\varphi_{6,X}$ is birational by Lemma 4.8.

Now we prove the second statement. Assume that $22\leq P_{5}(X)\leq 23$.

By Lemma 4.10, it suffices to consider the direction by assuming that $\varphi_{6,X}$ is not birational.

By the arguments in Case 1$\sim$Case 7, it suffices to consider one of the following situations:

1. (i)

$u_{5,-2}=4$; 2. (ii)

$u_{5,0}=7$, $u_{5,-1}\leq 5$, $u_{5,-2}\leq 3$, $u_{5,-3}\leq 3$, $u_{5,-4}\leq 2$, $\mathrm{dim}\psi_{5,0}(U_{5,0})=3$, $\xi\geq\frac{4}{5}$, $\beta\geq\frac{3}{5}$; 3. (iii)

$u_{5,0}\leq 6$, $u_{5,-1}=6$, $u_{5,-2}\leq 3$, $u_{5,-4}\leq 2$, $u_{5,-5}=1$; 4. (iv)

$u_{5,0}\leq 6$, $u_{5,-1}\leq 5$, $u_{5,-2}\leq 3$, $u_{5,-4}\leq 2$, $u_{5,-5}=1$.

We first consider (i). If $(S_{5,-2}|_{F}\cdot C)\geq 4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) and (ii) ($\tilde{\delta}\geq 4$, $m_{1}=5$, $j=2$, $\beta=\frac{1}{2}$), which contradicts to our assumption. So we have $(S_{5,-2}|_{F}\cdot C)\leq 3$. Then we have $S_{5,-2}|_{F}\geq C+C^{\prime}$ for a moving curve $C^{\prime}$ on $F$. When $|C|$ and $|C^{\prime}|$ are not composed of the same pencil, Theorem 3.3 (i.1) implies $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=5$, $j_{1}=2$, $j_{2}=1$, $\delta_{2}=2$, $\beta=\frac{1}{2}$). Then $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2), a contradiction. Otherwise, we have $S_{5,-2}|_{F}\geq 2C$. Thus $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. Since $\alpha(7)\geq\frac{13}{6}>2$, we have $\xi\geq\frac{5}{7}$. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=5$, $j_{1}=2$, $j_{2}=2$, $\beta=\frac{4}{7}$, $\xi\geq\frac{5}{7}$). When $\xi>\frac{4}{5}$, by Theorem 3.5 (i.2), $\varphi_{6,X}$ is birational ($m_{1}=5$, $j_{1}=2$, $j_{2}=2$), which is a contradiction. So the only possibility is $\xi=\frac{4}{5}$.

For (ii), the condition $P_{5}(X)\geq 22$ and Lemma 4.9 (ii) imply that $u_{5,-4}=2$. If $|S_{5,-4}|_{F}|$ and $|C|$ are not composed of the same pencil, Theorem 3.2 (i) implies that $\varphi_{6,X}$ is birational ($m_{1}=5$, $j=4$, $\tilde{\delta}=2$, $\xi\geq\frac{4}{5}$, $\beta=\frac{3}{5}$), a contradiction. If $|S_{5,-4}|_{F}|$ and $|C|$ are composed of the same pencil. When $\xi>\frac{4}{5}$, Theorem 3.5 (i.2) implies that $\varphi_{6,X}$ is birational ($m_{1}=5$, $j_{1}=4$, $j_{2}=1$, $\beta=\frac{3}{5}$). Thus the only possibility is $\xi=\frac{4}{5}$.

For (iii), we have $\mu\geq\frac{6}{5}$ and $\beta\geq\frac{6}{11}$ by our assumption. Since $\alpha(7)>2$, we have $\xi\geq\frac{5}{7}$. If $(S_{5,-1}\cdot C)\geq 4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (ii) ($m_{1}=5$, $\tilde{\delta}=4$, $\mu\geq\frac{6}{5}$, $\beta\geq\frac{6}{11}$). So we may assume that $(S_{5,-1}|_{F}\cdot C)\leq 3$. Thus we have $S_{5,-1}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.3 (ii.2) ($m_{1}=5$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\mu=\frac{6}{5}$). If $|C_{-2}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{2}{3}$ by [CHP17, Proposition 3.5]. Since $\alpha(5)>1$, we have $\xi\geq\frac{4}{5}$. Since $\alpha(6)>2$, $\varphi_{6,X}$ is birational, which is a contradiction. Thus (iii) does not occur.

We are left to treat (iv). Since $h^{0}(M_{5}-5F)\geq 3$, Lemma 4.9 (ii) implies that $\varphi_{6,X}$ is birational, which is a contradiction.

Therefore we have $\xi=\frac{4}{5}$.∎

Proposition 4.15**.**

Under the same assumption as that of Lemma 4.5, then

1. (1)

when $P_{6}(X)\geq 35$, $\varphi_{6,X}$ is birational; 2. (2)

when $32\leq P_{6}(X)\leq 34$, $\varphi_{6,X}$ is non-birational if and only if $\xi=\frac{4}{5}$.

Proof.

Set $m_{1}=6$. By Lemma 4.9 (i), we may and do assume that $u_{6,0}\leq P_{6}(F)-1=17$.

Reduction to: $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 4$, $v_{6,-1}\leq 3$, $v_{6,-2}\leq 2$ and $v_{6,-3}\leq 1$.

By [CHP17, Proposition 3.4], we have $v_{6,0}\leq 5$. If $\mathrm{dim}\psi_{6,0}(U_{6,0})=5$, the Riemann-Roch formula implies that $\mathrm{deg}(M_{6}|_{C})\geq 6$. Noting that $\mathrm{deg}(M_{6}|_{C})\leq 6$, $|M_{6}||_{C}$ must be complete. So we can write $|M_{6}||_{C}=|K_{C}+D|$ where $\mathrm{deg}(D)=4$. Thus $\varphi_{6,X}$ is birational. Hence we may assume that $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 4$.

Suppose $v_{6,-1}\geq 4$. Then $M_{6}|_{F}\geq C+C_{-1}$ for some moving curve $C_{-1}$ on $F$ satisfying $h^{0}(C,C_{-1}|_{C})\geq 4$. In particular, we have $(C_{-1}\cdot C)\geq 5$. By [CHP17, Proposition 3.7] (ii), $\varphi_{6,X}$ is birational ($\mu=1$, $m_{1}=6$, $\delta_{1}=5$, $\beta=\frac{1}{2}$, $j=1$). We may assume that $v_{6,-1}\leq 3$.

Suppose $v_{6,-2}\geq 3$. We have $M_{6}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve satisfying $h^{0}(C,C_{-2}|_{C})\geq 3$. By Riemann-Roch formula, one has $(C_{-2}\cdot C)\geq 4$. We also have

[TABLE]

So $(C_{-2}\cdot C)=4$. By [CHP17, Proposition 3.7] (i), $\varphi_{6,X}$ is birational ($\xi=\frac{2}{3}$, $\delta_{1}=4$, $\mu=1$, $j=2$, $m_{1}=6$). Thus we may assume that $v_{6,-2}\leq 2$.

Now assume that $v_{6,-3}\geq 2$. Then $M_{6}|_{F}\geq 3C+C_{-3}$ for some moving curve $C_{-3}$ on $F$. In particular, we have $(C_{-3}\cdot C)\geq 2$. By [CHP17, Proposition 3.7] (iii), we have $\xi\geq\frac{4}{5}$ ($n=4$, $\mu=1$, $m_{1}=6$, $j=3$, $\xi=\frac{2}{3}$, $\delta_{1}=2$). Thus $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.7] (i) ($\xi=\frac{4}{5}$, $\delta_{1}=2$, $j=3$, $\mu=1$, $m_{1}=6$). So we may assume that $v_{6,-3}\leq 1$.

Case 1. $u_{6,0}\geq 11$.

If $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$, one has $(M_{6}|_{F}\cdot C)\geq 5$ by Riemann-Roch formula. Hence $\xi\geq\frac{5}{6}$. By our assumption, we have $M_{6}|_{F}\geq 4C$. Thus we get $\beta\geq\frac{2}{3}$. Since

[TABLE]

$\varphi_{6,X}$ is birational by Theorem 2.2.

If $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$, we get $M_{6}|_{F}\geq 5C$ by our assumption. In particular, we have $\beta\geq\frac{5}{6}$. By Lemma 4.7, $\varphi_{6,X}$ is birational.

Case 2. $u_{6,0}=10$ and $P_{6}(X)\geq 31$.

If $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$ and $v_{6,-1}=3$, we have $h^{0}(F,M_{6}|_{F}-C)\geq 7$ and $M_{6}|_{F}\geq C+C_{-1}$, where $(C_{-1}\cdot C)\geq 4$. On the other hand, by our assumption ($u_{6,0}=10$, $v_{6,-2}\leq 2$ and $v_{6,-3}\leq 1$), we have $M_{6}|_{F}\geq 4C$. In particular, we have $\beta\geq\frac{2}{3}$. By [CHP17, Proposition 3.7] (ii), $\varphi_{6,X}$ is birational ($\mu=1$, $m_{1}=6$, $\delta_{1}=4$, $j=1$, $\beta=\frac{2}{3}$).

If $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$ and $v_{6,-1}\leq 2$, we have $h^{0}(F,M_{6}|_{F}-C)\geq 8$. By our assumption ($u_{6,0}=10$, $v_{6,-2}\leq 2$ and $v_{6,-3}\leq 1$), we have $M_{6}|_{F}\geq 5C$. In particular, we get $\beta\geq\frac{5}{6}$. Lemma 4.7 implies that $\varphi_{6,X}$ is birational.

So we may and do assume that $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$ throughout this Case. By Riemann-Roch formula, one has $\deg({M_{6}}|_{C})\geq 5$. When $\deg({M_{6}}|_{C})=5$, then $|M_{6}||_{C}$ must be complete and clearly $\varphi_{6,X}$ is birational. Thus we can assume, from now on within this case, that $(M_{6}|_{F}\cdot C)=6$. In particular, $\xi=1$.

Subcase 2.1. $u_{6,-1}\geq 7$.

We first consider the case when $\mathrm{dim}\psi_{6,0}(U_{6,-1})=4$. By our assumption, we have $(S_{6,-1}|_{F}\cdot C)=(M_{6}|_{F}\cdot C)=6$. [CHP17, Proposition 3.6] (1.2) implies that $\varphi_{6,X}$ is birational ($\beta=\frac{1}{2}$, $m_{1}=6$, $\delta=6$, $\mu=1$).

So we may assume that $\mathrm{dim}\psi_{6,0}(U_{6,-1})\leq 3$. Thus we have $S_{6,-1}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1})\geq 4$. If $\mathrm{dim}\psi_{6,-1}(H^{0}(F,C_{-1}))\geq 3$, we have $(C_{-1}\cdot C)\geq 4$ by Riemann-Roch formula. By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($\xi=1$, $j_{1}=j_{2}=1$, $\delta_{2}=4$, $\mu=1$, $\beta=\frac{1}{2}$). We are left to treat the case when $\mathrm{dim}\psi_{6,-1}(H^{0}(F,C_{-1}))\leq 2$. We have $S_{6,-1}|_{F}\geq 2C+C_{-2}$ where $C_{-2}$ is a moving curve on $F$. If $|C_{-2}|$ and $|C|$ are composed of the same pencil, we get $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. Since $\xi=1$, we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{7}$. Lemma 4.6 implies that $\varphi_{6,X}$ is birational. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, $\varphi_{6,X}$ is birational by Theorem 3.3 (ii.2) ($m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\xi=1$, $\mu=1$) .

Subcase 2.2. $u_{6,-1}\leq 6$, $u_{6,-3}\geq 4$.

If $\psi_{6,-3}(U_{6,-3})\geq 3$, we have $(S_{6,-3}|_{F}\cdot C)\geq 4$. Therefore $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{9}$ by Theorem 3.2 (iii) ($j=3$, $m_{1}=6$, $\tilde{\delta}=4$). Then $\varphi_{6,X}$ is birational by Lemma 4.6.

If $\psi_{6,-3}(U_{6,-3})\leq 2$, we have $S_{6,-3}|_{F}\geq C+C^{\prime}$ where $C^{\prime}$ is a moving curve. Thus we still have $(\pi^{*}(K_{X})|_{F}\cdot S_{6,-3}|_{F})\geq 2$. By Theorem 3.2 (iii), we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{9}$. Hence $\varphi_{6,X}$ is birational by Lemma 4.6.

Subcase 2.3. $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}\geq 2$.

If $|S_{6,-5}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{6}{11}$ by [CHP17, Proposition 3.5]. Thus $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{6}{11}$. Lemma 4.6 implies that $\varphi_{6,X}$ is birational.

If $|S_{6,-5}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $((\pi^{*}(K_{X}))|_{F}\cdot S_{6,-5}|_{F})\geq 1$. By Theorem 3.2 (iii), we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{6}{11}$ ($j=5$). Lemma 4.6 implies that $\varphi_{6,X}$ is birational.

Subcase 2.4. $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$ and $P_{6}(X)\geq 31$.

We have $h^{0}(M_{6}-5F)\geq 3$. Since $u_{6,-5}=1$, we have $\beta\geq\frac{7}{13}$ by (2.5). So $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{7}{13}$ and $\varphi_{6,X}$ is birational by Lemma 4.6.

Case 3. $u_{6,0}\leq 9$, $u_{6,-1}\geq 8$.

If $(S_{6,-1}|_{F}\cdot C)=6$, we have $\xi=1$. By [CHP17, Proposition 3.6](1.2) ($m_{1}=6$, $\beta=\frac{1}{2}$, $\mu=1$, $\delta=6$), $\varphi_{6,X}$ is birational.

If $(S_{6,-1}|_{F}\cdot C)\leq 5$ and $\mathrm{dim}\psi_{6,0}(U_{6,-1})\geq 4$, the Riemann-Roch formula on $C$ tells that

[TABLE]

where $\mathrm{deg}(D)=3$. Thus $\varphi_{6,X}$ is birational.

If $\mathrm{dim}(\psi_{6,0}(U_{6,-1}))\leq 3$, we have $S_{6,-1}|_{F}\geq C+C_{-1}$ where $C_{-1}$ is a moving curve satisfying $h^{0}(F,C_{-1})\geq 5$. By our reduction, we have $\mathrm{dim}\psi_{6,-1}(H^{0}(F,C_{-1}))\leq 3$. If $\mathrm{dim}\psi_{6,-1}(H^{0}(F,C_{-1}))=3$, we have $(C_{-1}\cdot C)\geq 4$ by Riemann-Roch formula. By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($j_{1}=1$, $j_{2}=1$, $m_{1}=6$, $\mu=1$, $\beta=\frac{1}{2}$, $\delta_{2}=4$). If $\mathrm{dim}\psi_{6,-1}(H^{0}(F,C_{-1}))\leq 2$, we have $S_{6,-1}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-2})\geq 3$. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}\cdot C)\geq 2$. By Theorem 3.3 (ii.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\xi=\frac{2}{3}$, $\mu=1$). So $\varphi_{6,X}$ is birational by Theorem 3.3 (ii.2) ($j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\mu=1$, $\xi=\frac{4}{5}$). If $|C_{-2}|$ and $|C|$ are composed of the same pencil, we have $S_{6,-1}|_{F}\geq 4C$. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{5}{7}$. Lemma 4.7 implies that $\varphi_{6,X}$ is birational.

Case 4. $u_{6,0}\leq 9$, $u_{6,-1}\leq 7$ and $u_{6,-2}\geq 7$.

Note that $S_{6,-2}|_{F}\geq M_{4}|_{F}\geq 2\sigma^{*}(K_{F_{0}})$. So $S_{6,-2}|_{F}$ is a big divisor.

If $(S_{6,-2}|_{F}\cdot C)\geq 5$, $\varphi_{6,X}$ is birational by Theorem 3.2 (ii)′ ($m_{1}=6$, $j=2$, $\beta=\frac{1}{2}$).

If $(S_{6,-2}|_{F}\cdot C)\leq 4$, we have $S_{6,-2}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve satisfying $h^{0}(F,C_{-1})\geq 4$. If $h^{0}(C,C_{-1}|_{C})\geq 3$, we have $(C_{-1}\cdot C)=4$ by Riemann-Roch formula and our assumption $(S_{6,-2}|_{F}\cdot C)\leq 4$. By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($j_{1}=2$, $j_{2}=1$, $\delta_{2}=4$, $m_{1}=6$, $\xi=\frac{2}{3}$, $\beta=\frac{1}{2}$, $\mu=1$ ) . We may assume that $h^{0}(C,C_{-1}|_{C})\leq 2$. Thus we have $S_{6,-2}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}\cdot C)\geq 2$. By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($j_{1}=2$, $j_{2}=2$, $\xi=\frac{2}{3}$, $\delta_{2}=2$, $m_{1}=6$, $\beta=\frac{1}{2}$). If $|C_{-2}|$ and $|C|$ are composed of the same pencil, we have $S_{6,-2}|_{F}\geq 3C$. By Theorem 3.5 (ii.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=2$, $j_{2}=3$, $\mu=1$, $\xi=\frac{2}{3}$). Thus $\varphi_{6,X}$ is birational by Theorem 3.5 (ii.2) ($j_{1}=2$, $j_{2}=3$, $m_{1}=6$, $\mu=1$, $\xi=\frac{4}{5}$).

Case 5. $u_{6,0}\leq 9$, $u_{6,-1}\leq 7$, $u_{6,-2}\leq 6$ and $u_{6,-3}\geq 5$.

If $(S_{6,-3}|_{F}\cdot C)\geq 4$, we have $\xi\geq\frac{7}{9}$ by Theorem 3.2 (iii) ($j=3$, $m_{1}=6$, $\tilde{\delta}=4$). By Theorem 3.2 (i) and (ii), $\varphi_{6,X}$ is birational ($j=3$, $\tilde{\delta}\geq 4$, $\xi=\frac{7}{9}$, $m_{1}=6$, $\beta=\frac{1}{2}$).

If $(S_{6,-3}|_{F}\cdot C)\leq 3$, we have $S_{6,-3}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve satisfying $h^{0}(F,C_{-1})\geq 3$. If $|C_{-1}|$ and $|C|$ are composed of the same pencil, we have $S_{6,-3}|_{F}\geq 3C$. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{2}{3}$. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=j_{2}=3$, $\xi=\frac{2}{3}$, $\beta=\frac{2}{3}$). Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($j_{1}=3$, $j_{2}=3$, $\xi=\frac{4}{5}$, $m_{1}=6$, $\beta=\frac{2}{3}$). If $|C_{-1}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-1}\cdot C)\geq 2$. By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($j_{1}=3$, $j_{2}=2$, $\xi=\frac{2}{3}$, $m_{1}=6$, $\delta_{2}=2$, $\beta=\frac{1}{2}$).

Case 6. $u_{6,0}\leq 9$, $u_{6,-1}\leq 7$, $u_{6,-2}\leq 6$, $u_{6,-3}\leq 4$ and $u_{6,-4}\geq 4$.

If $(S_{6,-4}|_{F}\cdot C)\geq 4$, we get $\xi\geq\frac{4}{5}$ by Theorem 3.2 (iii) ($j=4$, $\tilde{\delta}=4$, $m_{1}=6$). Hence $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($\tilde{\delta}=4$, $j=4$, $\xi=\frac{4}{5}$, $\beta=\frac{1}{2}$).

If $(S_{6,-4}|_{F}\cdot C)\leq 3$, we get $S_{6,-4}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve. If $|C_{-1}|$ and $|C|$ are composed of the same pencil, we have $S_{6,-4}|_{F}\geq 2C$. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{3}{5}$. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=4$, $j_{2}=2$, $\xi=\frac{2}{3}$, $\beta=\frac{3}{5}$). Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($j_{1}=4$, $j_{2}=2$, $\xi=\frac{4}{5}$, $\beta=\frac{3}{5}$). If $|C_{-1}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-1}\cdot C)\geq 2$. By Theorem 3.3 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=4$, $j_{2}=1$, $\beta=\frac{1}{2}$, $\xi=\frac{2}{3}$, $\delta_{2}=2$). Theorem 3.3 (i.2) implies that $\varphi_{6,X}$ is birational ($j_{1}=4$, $j_{2}=1$, $m_{1}=6$, $\xi=\frac{4}{5}$, $\beta=\frac{1}{2}$).

Case 7. $u_{6,0}\leq 9$, $u_{6,-1}\leq 7$, $u_{6,-2}\leq 6$, $u_{6,-3}\leq 4$, $u_{6,-4}\leq 3$ and $u_{6,-5}\geq 3$.

If $(S_{6,-5}|_{F}\cdot C)\geq 4$. By the same argument as in Case 6, $\varphi_{6,X}$ is birational (Note that we have $S_{6,-4}|_{F}\geq S_{6,-5}|_{F}$).

If $2\leq(S_{6,-5}|_{F}\cdot C)\leq 3$, we have $S_{6,-5}|_{F}\geq C$ by Riemann-Roch formula. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{6}{11}$. By Theorem 3.2 (iv), we have $\xi\geq\frac{4}{5}$ ($n=4$, $\xi=\frac{2}{3}$, $m_{1}=6$, $j=5$, $\beta=\frac{6}{11}$, $\tilde{\delta}=2$). Then $\varphi_{6,X}$ is birational by Theorem 3.2(i) ($j=5$, $m_{1}=6$, $\tilde{\delta}=2$, $\xi=\frac{4}{5}$, $\beta=\frac{6}{11}$ ).

If $|S_{6,-5}|_{F}|$ and $|C|$ are composed of the same pencil, we get $S_{6,-5}|_{F}\geq 2C$. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{7}{11}$. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=5$, $j_{2}=2$, $\beta=\frac{7}{11}$, $\xi=\frac{2}{3}$). Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($j_{1}=5$, $j_{2}=2$, $m_{1}=6$, $\xi=\frac{4}{5}$, $\beta=\frac{7}{11}$).

Case 8. $u_{6,0}\leq 9$, $u_{6,-1}\leq 7$, $u_{6,-2}\leq 6$, $u_{6,-3}\leq 4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=2$.

If $|S_{6,-6}|_{F}|$ and $|C|$ are composed of the same pencil, we get $\beta\geq\frac{7}{12}$ by [CHP17, Proposition 3.5]. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=6$, $j_{2}=1$, $\beta=\frac{7}{12}$, $\xi=\frac{2}{3}$). Then $\varphi_{6,X}$ is birational by Theorem 3.5 (i.2) ($j_{1}=6$, $j_{2}=1$, $m_{1}=6$, $\xi=\frac{4}{5}$, $\beta=\frac{7}{12}$).

If $|S_{6,-6}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(S_{6,-6}|_{F}\cdot C)\geq 2$. By Theorem 3.2 (iv), we have $\xi\geq\frac{5}{7}$ ($n=6$, $m_{1}=6$, $j=6$, $\beta=\frac{1}{2}$, $\xi=\frac{2}{3}$). One has $\xi\geq\frac{4}{5}$ by Theorem 3.2 (iv) ($n=4$, $m_{1}=6$, $j=6$, $\beta=\frac{1}{2}$, $\xi=\frac{5}{7}$). Thus $\varphi_{6,X}$ is birational by Theorem 3.2 (i)′ ($\tilde{\delta}=2$, $j=6$, $\xi=\frac{4}{5}$, $m_{1}=6$, $\beta=\frac{1}{2}$) .

Case 9. $u_{6,0}\leq 9$, $u_{6,-1}\leq 7$, $u_{6,-2}\leq 6$, $u_{6,-3}\leq 4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$ and $P_{6}(X)\geq 35$.

We have $\mu\geq\frac{3}{2}>\frac{4}{3}$ by our assumption. So $\varphi_{6,X}$ is birational by Lemma 4.8.

Now suppose $32\leq P_{6}(X)\leq 34$. Lemma 4.10 implies that we only need to consider the direction by assuming the non-birationality of $\varphi_{6,X}$. By the arguments in Case 1-Case 9, it suffices to consider one of the following cases: Case i $\sim$ Case iii.

Case i. $u_{6,0}=9$, $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 4$.

(${\dagger}$) We first treat the case when $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$. We have $(M_{6}|_{F}\cdot C)\geq 5$ by Riemann-Roch formula. If $(M_{6}|_{F}\cdot C)=5$, Rimann-Roch formula implies that $|M_{6}||_{C}$ is a complete linear system $|K_{C}+D|$, where $\mathrm{deg}D=3$. So $\varphi_{6,X}$ is birational, which is a contradiction. So we have $(M_{6}|_{F}\cdot C)=6$, which implies $\xi=1$. We will prove that this can not happen at all. By Lemma 4.6, we have $\beta=\frac{1}{2}$.

Subcase i.a. $u_{6,-1}=7$, $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$.

If $\dim\psi_{6,0}(U_{6,-1})\geq 4$, we have $(S_{6,-1}|_{F}\cdot C)\geq 5$ by Riemann-Roch formula. For the case $(S_{6,-1}|_{F}\cdot C)\geq 6$, $\varphi_{6,X}$ is birational by [CHP17, Proposition 3.6] (1.2) ($m_{1}=6$, $\delta=6$, $\beta=\frac{1}{2}$, $\mu=1$). For the case $(S_{6,-1}|_{F}\cdot C)=5$, the linear system $|S_{6,-1}||_{C}$ must be the complete one, i.e. $|S_{6,-1}|_{C}|$, due to Riemann-Roch formula as well. In fact, $|S_{6,-1}|_{C}|=|K_{C}+D|$ with $\deg(D)=3$. Clearly, $\varphi_{6,X}$ is birational.

If $\dim\psi_{6,0}(U_{6,-1})\leq 3$, we have $S_{6,-1}|_{F}\geq C+C_{-1}^{{}^{\prime}}$, where $C_{-1}^{{}^{\prime}}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1}^{{}^{\prime}})\geq 4$. When $(C_{-1}^{{}^{\prime}}\cdot C)\geq 4$, then $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2) ($m_{1}=6$, $j_{1}=j_{2}=1$, $\delta_{2}=4$, $\beta=\frac{1}{2}$, $\mu=1$), which is a contradiction. So we have $(C_{-1}^{{}^{\prime}}\cdot C)\leq 3$, which implies that $S_{6,-1}|_{F}\geq 2C+C_{-2}^{{}^{\prime}}$, where $C_{-2}^{{}^{\prime}}$ is a moving curve on $F$. When $|C_{-2}^{{}^{\prime}}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. Then we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{7}>\frac{1}{2}$, which means $\varphi_{6,X}$ is birational by Lemma 4.6 (a contradiction).

When $|C_{-2}^{{}^{\prime}}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}^{{}^{\prime}}\cdot C)\geq 2$. Theorem 3.3 (ii.2) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\xi=1$, $\mu=1$), which contradicts to our assumption.

In a word, Subcase i.a does not occur.

Subcase i.b. $u_{6,-1}\leq 6$, $u_{6,-3}\leq 4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$ and $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$.

Since $P_{6}(X)\geq 32$, we have $M_{6}\geq 7F$ by our assumption. By Inequality (2.5), we have $\beta\geq\frac{7}{13}$ and $(\pi^{*}(K_{X})|_{F})^{2}>\frac{1}{2}$, which is a contradiction by Lemma 4.6. Hence this subcase does not occur either.

(${\ddagger}$) We then treat the case when $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$. Since

[TABLE]

$\varphi_{6,X}$ is generically finite, which implies that $\mathrm{dim}\psi_{6,0}(U_{6,0})\geq 3$. We may and do assume that $\mathrm{dim}\psi_{6,0}(U_{6,0})=3$ throughout the rest of this case. We have $M_{6}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1})\geq 6$.

If $(C_{-1}\cdot C)\leq 3$, we have $C_{-1}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$. If $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}\cdot C)\geq 2$. By [CHP17, Proposition 3.7] (iii), we have $\xi\geq\frac{4}{5}$ ($m_{1}=6$, $j=3$, $\delta_{1}=2$, $\mu=1$, $\xi\geq\frac{2}{3}$, $n=4$). [CHP17, Proposition 3.7] (i) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j=3$, $\delta_{1}=2$, $\mu=1$, $\xi\geq\frac{4}{5}$), which contradicts to our assumption. Thus $|C_{-2}|$ and $|C|$ are composed of the same pencil. We get $M_{6}|_{F}\geq 4C$. In particular, we have $\beta\geq\frac{2}{3}$. Since $\alpha(7)>2$, we have $\xi\geq\frac{5}{7}$. We have $\alpha(5)\geq\frac{15}{14}>1$. So $\xi\geq\frac{4}{5}$. We can get $\alpha(6)>2$ when $\xi>\frac{4}{5}$. Thus we have $\xi=\frac{4}{5}$.

If $(C_{-1}\cdot C)\geq 4$, [CHP17, Proposition 3.7] (ii) implies that $\varphi_{6,X}$ is birational whenever $\beta>\frac{1}{2}$. Thus we need to study the situation with $\beta=\frac{1}{2}$. Taking $n=9$, $10$, $11$, $12$, $13$, respectively, and run [CHP17, Proposition 3.7] (iii), one finally gets $\xi\geq\frac{6}{7}$. So we will work under the constraints: $\xi\geq\frac{6}{7}$ and $\beta=\frac{1}{2}$, throughout the rest of this case.

Subcase i.c. $u_{6,-1}=7$, $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$.

Clearly, one has $\dim\psi_{6,0}(U_{6,-1})\leq 3$ , which is parallel to the second part of Subcase i.a. We have $S_{6,-1}|_{F}\geq C+C_{-1}^{{}^{\prime}}$, where $C_{-1}^{{}^{\prime}}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1}^{{}^{\prime}})\geq 4$. When $(C_{-1}^{{}^{\prime}}\cdot C)\geq 4$, then $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2) ($m_{1}=6$, $j_{1}=j_{2}=1$, $\delta_{2}=4$, $\beta=\frac{1}{2}$, $\mu=1$), which is a contradiction. So we have $(C_{-1}^{{}^{\prime}}\cdot C)\leq 3$, which implies that $S_{6,-1}|_{F}\geq 2C+C_{-2}^{{}^{\prime}}$, where $C_{-2}^{{}^{\prime}}$ is a moving curve on $F$. When $|C_{-2}^{{}^{\prime}}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5], a contradiction to our assumption $\beta=\frac{1}{2}$. When $|C_{-2}^{{}^{\prime}}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}^{{}^{\prime}}\cdot C)\geq 2$. Theorem 3.3 (ii.2) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\xi=\frac{6}{7}$, $\mu=1$), which contradicts to our assumption.

Subcase i.d. $u_{6,-1}\leq 6$, $u_{6,-3}=4$.

If $(S_{6,-3}|_{F}\cdot C)\geq 4$, Theorem 3.2 (i) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j=3$, $\beta=\frac{1}{2}$, $\tilde{\delta}=4$, $\xi=\frac{6}{7}$), which contradicts to our assumption.

If $(S_{6,-3}|_{F}\cdot C)\leq 3$, we have $S_{6,-3}|_{F}\geq C+C^{\prime}$, where $C^{\prime}$ is a moving curve on $F$. When $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{5}{9}>\frac{1}{2}$ by [CHP17, Proposition 3.5], which contradicts to assumption. Then $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, we get $(C^{\prime}\cdot C)\geq 2$. Then $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2) ($m_{1}=6$, $j_{1}=3$, $j_{2}=1$, $\delta_{2}=2$, $\beta=\frac{1}{2}$, $\xi\geq\frac{6}{7}$), which contradicts to our assumption.

Subcase i.e. $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=2$.

The assumption $\beta=\frac{1}{2}$ implies that $|S_{6,-5}|_{F}|$ and $|C|$ are not composed of the same pencil. Then $(S_{6,-5}|_{F}\cdot C)\geq 2$. Theorem 3.2 (i)′ implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j=5$, $\tilde{\delta}=2$, $\xi=\frac{6}{7}$), which contradicts to our assumption. Thus this subcase does not occur.

Subcase i.f. $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$, $P_{6}(X)\geq 30$.

We have $h^{0}(M_{6}-5F)\geq 3$. Thus we have $\beta\geq\frac{7}{13}>\frac{1}{2}$ by Inequality (2.5), which contradicts to our assumption.

Case ii. $u_{6,0}\leq 8$, $u_{6,-1}\leq 7$, $u_{6,-2}=6$, $u_{6,-3}\leq 4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$.

If $(S_{6,-2}|_{F}\cdot C)\leq 3$, we have $S_{6,-2}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$. When $|C_{-2}|$ and $|C|$ are composed of the same pencil, we have $\xi\geq\frac{4}{5}$ by Theorem 3.5 (ii.1) ($n=4$, $m_{1}=6$, $j_{1}=2$, $j_{2}=3$, $\xi=\frac{2}{3}$, $\mu=1$). Thus $\varphi_{6,X}$ is birational by Theorem 3.5 (ii.2) ($m_{1}=6$, $j_{1}=2$, $j_{2}=3$, $\xi=\frac{4}{5}$, $\mu=1$), which is a contradiction. When $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}\cdot C)\geq 2$. By Theorem 3.3 (i.2), $\varphi_{6,X}$ is birational ($m_{1}=6$, $j_{1}=2$, $j_{2}=2$, $\xi=\frac{2}{3}$, $\delta_{2}=2$, $\beta=\frac{1}{2}$), which contradicts to our assumption.

Thus we have $(S_{6,-2}|_{F}\cdot C)\geq 4$. Theorem 3.2 (i) and (ii) imply that $\varphi_{6,X}$ is birational if $\beta>\frac{1}{2}$ ($m_{1}=6$, $j=2$, $\tilde{\delta}\geq 4$, $\xi=\frac{2}{3}$, $\mu=1$). So we have $\beta=\frac{1}{2}$ by our assumption. But our assumption in this case gives $h^{0}(M_{6}-6F)\geq 2$. Since $u_{6,-6}=1$, we have $\beta\geq\frac{7}{13}$, which is a contradiction.

Case iii. $u_{6,0}\leq 8$, $u_{6,-1}\leq 7$, $u_{6,-2}\leq 5$, $u_{6,-3}\leq 4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$.

We have $h^{0}(M_{6}-6F)\geq 3$ since $P_{6}(X)\geq 32$. Since $u_{6,-6}=1$, we have $\mu\geq\frac{4}{3}$. Then one gets $\beta\geq\frac{4}{7}$ by Inequality (2.5). We have $\alpha(7)\geq\frac{7}{3}>2$, so $\xi\geq\frac{5}{7}$. Since $\alpha(5)>1$, we have $\xi\geq\frac{4}{5}$. When $\xi>\frac{4}{5}$, we have $\alpha(6)>2$, which implies that $\varphi_{6,X}$ is birational. So the only possibility is $\xi=\frac{4}{5}$. ∎

4.4. Estimation of the canonical volume

We go on working under the same assumption as that of Lemma 4.5.

Lemma 4.16**.**

Let $\pi\colon X^{\prime}\rightarrow X$ be any birational morphism where $X^{\prime}$ is nonsingular and projective. Assume that $|M|$ is a base point free linear system on $X^{\prime}$. Denote by $S$ a general member of $|M|$. Then the following inequality holds:

[TABLE]

Proof.

Take a sufficiently large and divisible integer $m$ such that the linear system $|\pi^{*}(mK_{X})|$ is base point free. Denote by $S_{m}$ a general member of $|\pi^{*}(mK_{X})|$. By Bertini’s theorem, $S_{m}$ is a smooth projective surface of general type. On the surface $S_{m}$, by Hodge index theorem, we have

[TABLE]

which implies (4.7). ∎

Let $m$ be a positive integer and $l$ be another integer satisfying $0\leq l\leq m$. Assume that $h^{0}(M_{m}-lF)\geq 2$. Denote by $|S_{m,-l}|$ the moving part of $|M_{m}-lF|$. Modulo further blowups, we may assume that $|S_{m,-l}|$ is base point free. Multiplying the following inequality with $\pi^{*}(K_{X})^{2}$:

[TABLE]

while applying (4.7), we have

[TABLE]

For the last inequality, we note that $M_{m}\geq mF$, which implies $S_{m,-l}\geq(m-l)F$.

Proposition 4.17**.**

Keep the same assumption as that of Lemma 4.5. Suppose that $\varphi_{6,X}$ is not birational, $\xi\neq\frac{2}{3}$ and $\xi\neq\frac{4}{5}$. Then the following holds:

1. (1)

$K_{X}^{3}\geq\frac{5}{14}$; 2. (2)

when $P_{6}(X)\geq 26$, $K_{X}^{3}\geq\frac{11}{28}$; 3. (3)

when $P_{6}(X)\geq 27$, $K_{X}^{3}>0.4328$; 4. (4)

when $P_{6}(X)\geq 28$, $K_{X}^{3}>0.4714$; 5. (5)

when $P_{6}(X)\geq 31$, $K_{X}^{3}\geq\frac{8}{15}$.

Proof.

Since $\alpha(7)\geq(7-1-1-\frac{1}{\beta})\cdot\xi>2$, we have $\xi\geq\frac{5}{7}$.

Statement (1) follows from inequality (2.4) with $\beta\geq\frac{1}{2}$ and $\xi\geq\frac{5}{7}$.

By the arguments in Case 1$\sim$Case 9 in the proof of Proposition 4.15, one of the following cases holds:

Case I. $u_{6,0}=10$, $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$ and $P_{6}(X)\leq 30$. ($\Rightarrow K_{X}^{3}\geq\frac{1}{2}$)

We have $(M_{6}|_{F}\cdot C)\geq 5$ by Riemann-Roch formula. If $(M_{6}|_{F}\cdot C)=5$, $|M_{6}||_{C}$ is a complete linear system whose general member has degree $5$, which implies that $\varphi_{6,X}$ is birational, which contradicts to our assumption. Thus we have $(M_{6}|_{F}\cdot C)=6$. We get $\xi=1$. Therefore we have $K_{X}^{3}\geq\frac{1}{2}$.

Case IIa. $u_{6,0}=9$, $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$. ($\Rightarrow K_{X}^{3}\geq\frac{1}{2}$)

As we have seen, one has $(M_{6}|_{F}\cdot C)\geq 5$ and ${}^{\prime\prime}=5^{\prime\prime}$ implies the birationality of $\varphi_{6,X}$. Hence $(M_{6}|_{F}\cdot C)=6$. In particular, $\xi=1$. By Lemma 4.6 and the assumption, we have $\beta=\frac{1}{2}$ and $(\pi^{*}(K_{X})|_{F})^{2}=\frac{1}{2}$. Thus $K_{X}^{3}\geq\frac{1}{2}$.

Claim. We have $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$, $P_{6}(X)\leq 29$.

In fact, $u_{6,-1}\leq 6$ follows from the proof of Proposition 4.15 (see Subcase i.a.).

From the proof of Proposition 4.15, we have $u_{6,-3}\leq 4$ and $u_{6,-5}\leq 2$.

Suppose $u_{6,-3}=4$. If $(S_{6,-3}|_{F}\cdot C)\geq 4$, by Theorem 3.2 (iii) ($m_{1}=6$, $j=3$), we have

[TABLE]

Since $\xi=1$, we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{9}>\frac{1}{2}$, which contradicts to our assumption. So $(S_{6,-3}|_{F}\cdot C)\leq 3$, which gives $S_{6,-3}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$. Using the same argument as above, we can get $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{9}>\frac{1}{2}$, which is a contradiction. So $u_{6,-3}\leq 3$.

By the similar argument as above, we also sees that $u_{6,-5}=1$.

If $P_{6}(X)\geq 30$, we have $h^{0}(M_{6}-5F)\geq 3$. By Inequality (2.5), we have $\beta\geq\frac{7}{13}>\frac{1}{2}$, which is a contradiction.

Case IIb. $u_{6,0}=9$, $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$. ($\Rightarrow K_{X}^{3}\geq\frac{10}{21}$)

One has $M_{6}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1})\geq 6$. By the argument in Case i of Proposition 4.15 and the assumption $\xi\neq\frac{4}{5}$, we know $(C_{-1}\cdot C)\geq 4$, $\xi\geq\frac{6}{7}$, $\beta=\frac{1}{2}$, $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$ and $P_{6}(X)\leq 29$.

We have

[TABLE]

In particular, we have $K_{X}^{3}\geq\frac{10}{21}$.

Case IIIa. $u_{6,0}=8$, $\mathrm{dim}\psi_{6,0}(U_{6,0})=4$. ($\Rightarrow K_{X}^{3}\geq\frac{1}{2}$)

For the similar reason, we have $(M_{6}|_{F}\cdot C)=6$ and so $\xi=1$. By the same argument as in Case IIa, we have $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$, $P_{6}(X)\leq 28$. In particular, we have $K_{X}^{3}\geq\frac{1}{2}$.

Case IIIb. $u_{6,0}=8$, $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$, $M_{6}|_{F}\geq C+C_{-1}$ with $(C_{-1}\cdot C)=4$. ($\Rightarrow K_{X}^{3}\geq\frac{10}{21}$)

By the same argument as in Case i of Proposition 4.15 (Subcase i.c $\sim$ Subcase i.f), we know $\xi\geq\frac{6}{7}$, $\beta=\frac{1}{2}$, $u_{6,-1}\leq 6$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$, $P_{6}(X)\leq 28$. By the same argument as in Case IIb, we have $K_{X}^{3}\geq\frac{10}{21}$.

Case IIIc. $u_{6,0}=8$, $\mathrm{dim}\psi_{6,0}(U_{6,0})\leq 3$, $M_{6}|_{F}\geq C+C_{-1}$ with $(C_{-1}\cdot C)\leq 3$ ($\Rightarrow K_{X}^{3}\geq\frac{5}{12}$).

Since $h^{0}(F,C_{-1})\geq 5$, we have $C_{-1}\geq C+C^{\prime}$, where $C^{\prime}$ is a moving curve on $F$ satisfying $h^{0}(F,C^{\prime})\geq 3$. If $|C^{\prime}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{2}{3}$. Since $\alpha(7)>2$, we have $\xi\geq\frac{5}{7}$. We get $\alpha(5)\geq(5-1-1-\frac{1}{\beta})\cdot\xi\geq\frac{15}{14}>1$. Thus $\xi\geq\frac{4}{5}$. By our assumption, we have $\xi>\frac{4}{5}$, which gives $\alpha(6)>2$. Then $\varphi_{6,X}$ is birational, which contradicts to our assumption. So $|C^{\prime}|$ and $|C|$ are not composed of the same pencil.

Therefore we have $M_{6}|_{F}\geq 2C+C^{\prime}$. By [CHP17, Proposition 3.7] (iii), we have $\xi\geq\frac{3}{4}$ ($m_{1}=6$, $j=2$, $\delta_{1}=2$) by successfully taking $n=6$, $7$. Thus we have

[TABLE]

In particular, we have $K_{X}^{3}\geq\frac{5}{12}$.

Case IVa. $u_{6,-1}=7$, $\mathrm{dim}\psi_{6,0}(U_{6,-1})\leq 3$. ($\Rightarrow K_{X}^{3}>0.4714$)

When $(S_{6,-1}|_{F}\cdot C)\leq 3$, we have $S_{6,-1}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-2})\geq 3$. By the same argument as in the last part of Case 3 of Proposition 4.15, $\varphi_{6,X}$ is birational, a contradiction.

Thus we only need to consider the case when $(S_{6,-1}|_{F}\cdot C)\geq 4$. We have $S_{6,-1}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1})\geq 4$.

If $(C_{-1}\cdot C)\geq 4$, $\varphi_{6,X}$ is birational by Theorem 3.3 (i.2) ($m_{1}=6$, $j_{1}=j_{2}=1$, $\delta_{2}=4$, $\mu=1$, $\beta=\frac{1}{2}$), which is a contradiction.

Thus $(C_{-1}\cdot C)\leq 3$. So we have $C_{-1}\geq C+C^{\prime}$, where $C^{\prime}$ is a moving curve on $F$. When $|C^{\prime}|$ and $|C|$ are not composed of the same pencil, we have $(C^{\prime}\cdot C)\geq 2$. By Theorem 3.3 (ii.1), we have $\xi\geq\frac{5}{7}$ ($n=6$, $m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\mu=1$, $\xi=\frac{2}{3}$, $\delta_{2}=2$). Theorem 3.3 (ii.2) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\delta_{2}=2$, $\mu=1$, $\xi=\frac{5}{7}$), which is a contradiction.

So $C^{\prime}\sim C$ and we have $S_{6,-1}|_{F}\geq 3C$. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{4}{7}$. Since $\alpha(7)>2$, we have $\xi\geq\frac{5}{7}$. Since $\alpha(8)\geq(8-1-1-\frac{1}{\beta})\cdot\xi>3$, we have $\xi\geq\frac{3}{4}$. Denote by $\xi_{6,-1}$ the intersection number $(\pi^{*}(K_{X})|_{F}\cdot S_{6,-1}|_{F})$. We have $\xi_{6,-1}\geq\beta(C\cdot S_{6,-1}|_{F})\geq\frac{16}{7}$. Besides, Kawamata-Viehweg vanishing theorem implies

[TABLE]

which directly implies $7\pi^{*}(K_{X})|_{F}\geq C+S_{6,-1}|_{F}$ since $|C+S_{6,-1}|_{F}|$ is base point free. Noting that a general $S_{6,-1}|_{F}$ can be smooth, nef and big, we may use the the similar method to that of [CHP17, Proposition 3.6] (2.1) to obtain the following inequality, for any $n\geq 8$,

[TABLE]

where one notes that $((K_{F}+S_{6,-1}|_{F})\cdot S_{6,-1}|_{F})\geq 16$. Take $n=8$, we get $\xi_{6,-1}\geq\frac{7}{3}$. Take $n=10$, we get $\xi_{6,-1}\geq\frac{26}{11}$. Take $n=9$, we get $\xi_{6,-1}\geq\frac{12}{5}$. Since $7\pi^{*}(K_{X})|_{F}\geq C+S_{6,-1}|_{F}$ and $\xi\geq\frac{3}{4}$, we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{63}{140}$. By (4.8), we have $K_{X}^{3}>0.4714$.

Case IVb. $u_{6,-1}=6$ and $\mathrm{dim}\psi_{6,0}(U_{6,-1})\leq 3$. ($\Rightarrow K_{X}^{3}\geq\frac{11}{28}$)

If $(S_{6,-1}|_{F}\cdot C)\geq 4$, we have $\xi\geq\frac{3}{4}$ by [CHP17, Proposition 3.6] (1.1) ($n=11$, $\delta=4$, $m_{1}=6$, $\beta=\frac{1}{2}$, $\xi=\frac{2}{3}$). By the similar reason to that in Case IVa, we have $7\pi^{*}(K_{X})|_{F}\geq C+S_{6,-1}|_{F}$. Thus we get $K_{X}^{3}\geq(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{11}{28}$.

If $(S_{6,-1}|_{F}\cdot C)\leq 3$, we have $S_{6,-1}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$. When $|C_{-2}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{4}{7}$ by [CHP17, Proposition 3.5]. Besides, $\alpha(7)>2$ implies $\xi\geq\frac{5}{7}$. Thus we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{20}{49}$. When $|C_{-2}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-2}\cdot C)\geq 2$. By Theorem 3.3 (ii.1), we have $\xi\geq\frac{5}{7}$ ($n=6$, $m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\mu=1$, $\xi=\frac{2}{3}$, $\delta_{2}=2$). Theorem 3.3 (ii.2) implies that $\varphi_{6,X}$ is birational ($n=6$, $m_{1}=6$, $j_{1}=1$, $j_{2}=2$, $\mu=1$, $\xi=\frac{5}{7}$, $\delta_{2}=2$), which contradicts to our assumption.

Case V. $u_{6,-2}=6$. ($\Rightarrow K_{X}^{3}>0.4771$)

If $(S_{6,-2}|_{F}\cdot C)\leq 3$, we have $S_{6,-2}|_{F}\geq 2C+C_{-2}$, where $C_{-2}$ is a moving curve on $F$. By the same argument as in the last part of Case 4 of Proposition 4.15, we conclude that $\varphi_{6,X}$ is birational, a contradiction.

So we have $(S_{6,-2}\cdot C)\geq 4$ in this case. In fact, the case $(S_{6,-2}\cdot C)\geq 5$ has been treated in Case 4 of Proposition 4.15, which shows that $\varphi_{6,X}$ is birational (a contradiction). Hence $(S_{6,-2}|_{F}\cdot C)=4$. Theorem 3.2 (i) implies that $\varphi_{6,X}$ is birational if $\beta>\frac{1}{2}$. Thus we have $\beta=\frac{1}{2}$. By Theorem 3.2 (iii) and (iv), we have $\xi\geq\frac{3}{4}$ and, for any $n\geq 6$,

[TABLE]

Take $n=8$, we get $\xi\geq\frac{7}{9}$. Take $n=9$, we get $\xi\geq\frac{4}{5}$. By our assumption, we have $\xi>\frac{4}{5}$. Take $n=10$ in the above inequality, we have $\xi\geq\frac{9}{11}$. By Theorem 3.2 (iii) ($m_{1}=6$, $j=2$, $(S_{6,-2}|_{F}\cdot C)=4$), we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{11}$. By (4.8), we have $K_{X}^{3}>0.4771$.

Case VI. $u_{6,-3}=4$. ($\Rightarrow K_{X}^{3}>0.4734$)

If $(S_{6,-3}|_{F}\cdot C)\geq 4$, $\varphi_{6,X}$ is birational by Theorem 3.2 (i) ($m_{1}=6$, $j=3$, $\tilde{\delta}=4$, $\beta=\frac{1}{2}$, $\xi=\frac{5}{7}$), which contradicts to our assumption. Thus we have $(S_{6,-3}|_{F}\cdot C)\leq 3$. So $S_{6,-3}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$.

If $|C_{-1}|$ and $|C|$ are not composed of the same pencil, we have $(C_{-1}\cdot C)\geq 2$. By Theorem 3.3 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=3$, $j_{2}=1$, $\beta=\frac{1}{2}$, $\delta_{2}=2$). Our assumption implies that we have $\xi>\frac{4}{5}$. Theorem 3.3 (i.2) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j_{1}=3$, $j_{2}=1$, $\xi>\frac{4}{5}$, $\beta=\frac{1}{2}$), which is a contradiction.

Thus $|C_{-1}|$ and $|C|$ are composed of the same pencil. So $S_{6,-3}|_{F}\geq 2C$. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{5}{9}$. By Theorem 3.5 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=3$, $j_{2}=2$, $\beta=\frac{5}{9}$). So $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{4}{9}$. By (4.8), we have $K_{X}^{3}>0.4734$.

Case VII. $u_{6,-5}=2$. ($\Rightarrow K_{X}^{3}>0.4362$)

If $|S_{6,-5}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(S_{6,-5}|_{F}\cdot C)\geq 2$. By Theorem 3.2 (iv), we get $\xi\geq\frac{5}{7}$ by taking $n=6$ and $\xi\geq\frac{3}{4}$ by taking $n=7$. By Theorem 3.2 (iii), we have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{19}{44}$. By (4.8), we have $K_{X}^{3}>0.4746$.

If $|S_{6,-5}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{6}{11}$. By Theorem 3.5(i.1), we get $\xi\geq\frac{5}{7}$ by taking $n=6$ and $\xi\geq\frac{3}{4}$ by taking $n=7$. We have $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{9}{22}$. By (4.8), we obtain $K_{X}^{3}>0.4362$.

Now we prove (2). By the results of Case I $\sim$ Case VII, we only need to consider the case where $u_{6,0}\leq 7$, $u_{6,-1}\leq 5$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$. We have $\mu\geq\frac{7}{6}$ and $\beta\geq\frac{7}{13}$. As $\alpha(7)>2$, we have $\xi\geq\frac{5}{7}$. So we get $K_{X}^{3}\geq\frac{35}{78}>\frac{11}{28}$.

For (3), by the results of Case I $\sim$ Case VII, we are left to treat the following cases:

1. (3.1)

$u_{6,0}=8$, $\xi\geq\frac{3}{4}$, $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{12}$, $u_{6,-4}=3$; 2. (3.2)

$u_{6,0}=8$, $\xi\geq\frac{3}{4}$, $(\pi^{*}(K_{X})|_{F})^{2}\geq\frac{5}{12}$, $u_{6,-1}\leq 6$, $u_{6,-2}\leq 5$, $u_{6,-3}\leq 3$, $u_{6,-4}\leq 2$, $u_{6,-5}=1$; 3. (3.3)

$u_{6,0}\leq 7$, $u_{6,-1}\leq 6$, $u_{6,-2}\leq 5$, $u_{6,-3}\leq 3$, $u_{6,-4}\leq 2$, $u_{6,-5}=1$; 4. (3.4)

$u_{6,0}\leq 7$, $u_{6,-1}\leq 6$, $u_{6,-2}\leq 5$, $u_{6,-3}\leq 3$, $u_{6,-4}=3$, $u_{6,-5}=1$.

We first treat (3.1). If $|S_{6,-4}|_{F}|$ and $|C|$ are composed of the same pencil, we have $\beta\geq\frac{3}{5}$. We get $K_{X}^{3}\geq\beta\xi\geq\frac{9}{20}>0.4328$. If $|S_{6,-4}|_{F}|$ and $|C|$ are not composed of the same pencil, we have $(\pi^{*}(K_{X})|_{F}\cdot S_{6,-4}|_{F})\geq 1$. We have $K_{X}^{3}>0.4328$ by (4.8). Next we treat $(3.2)$. We have $\mu\geq\frac{7}{6}$. By Inequality (2.5), we have $\beta\geq\frac{7}{13}$. Then we get $K_{X}^{3}\geq\frac{49}{104}>0.4328$. For (3.3), we have $\mu\geq\frac{4}{3}$ and $\beta\geq\frac{4}{7}$. Since $\alpha(5)>1$, we get $\xi\geq\frac{4}{5}$. Our assumption implies that we have $\xi>\frac{4}{5}$. Thus we have $\alpha(6)>2$, which implies that $\varphi_{6,X}$ is birational, which is a contradiction. So $(3.3)$ does not occur. Finally, for $(3.4)$, we have $\mu\geq\frac{7}{6}$ and $\beta\geq\frac{7}{13}$. Since we have $\xi\geq\frac{5}{7}$, so we get $K_{X}^{3}\geq\frac{35}{78}>0.4328$.

Now we consider $(4)$. By the arguments in Case I-Case VII, we only need to treat the following cases:

1. (4.1)

$u_{6,0}=8$, $u_{6,-1}\leq 6$, $u_{6,-2}=5$, $u_{6,-3}\leq 3$, $u_{6,-5}=2$; 2. (4.2)

$u_{6,0}\leq 8$, $u_{6,-1}\leq 6$, $u_{6,-2}\leq 4$, $u_{6,-3}\leq 3$, $u_{6,-5}=2$; 3. (4.3)

$u_{6,0}\leq 8$, $u_{6,-1}\leq 6$, $u_{6,-2}=5$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$; 4. (4.4)

$u_{6,0}\leq 8$, $u_{6,-1}\leq 6$, $u_{6,-2}\leq 4$, $u_{6,-3}\leq 3$, $u_{6,-5}=1$; 5. (4.5)

$u_{6,0}\leq 7$, $u_{6,-1}\leq 6$, $u_{6,-2}=5$, $u_{6,-3}\leq 3$, $u_{6,-5}=2$.

(4.1) By the argument in Case VII, we only need to treat the case when $|S_{6,-5}|_{F}|$ and $|C|$ are composed of the same pencil. In particular, we have $\beta\geq\frac{6}{11}$.

We claim that $(S_{6,-2}|_{F}\cdot C)\leq 3$. Otherwise, Theorem 3.2 (i) and (ii) imply that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j=2$, $\tilde{\delta}\geq 4$, $\xi\geq\frac{2}{3}$, $\beta\geq\frac{6}{11}$), which contradicts to our assumption. Thus we have $S_{6,-2}|_{F}\geq C+C_{-1}$, where $C_{-1}$ is a moving curve on $F$ satisfying $h^{0}(F,C_{-1})\geq 3$.

If $|C_{-1}|$ and $|C|$ are composed of the same pencil, we have $S_{6,-2}|_{F}\geq 3C$. By Theorem 3.5 (ii.1), we have $\xi\geq\frac{4}{5}$ ($n=4$). Theorem 3.5 (ii.2) implies that $\varphi_{6,X}$ is birational ($m_{1}=6$, $j_{1}=2$, $j_{2}=3$, $\xi=\frac{4}{5}$, $\mu=1$), which contradicts to our assumption. Thus $|C_{-1}|$ and $|C|$ are not composed of the same pencil. In particular, we have $(C_{-1}\cdot C)\geq 2$. By Theorem 3.3 (i.1), we have $\xi\geq\frac{4}{5}$ ($n=4$, $m_{1}=6$, $j_{1}=2$, $j_{2}=1$, $\beta=\frac{6}{11}$, $\xi=\frac{5}{7}$, $\delta_{2}=2$) .

By the arguments in Case IIIa-Case IIIc, we have $M_{6}|_{F}\geq 2C+C^{\prime}$, where $C^{\prime}$ is a moving curve on $F$ satisfying $h^{0}(F,C^{\prime})\geq 3$. Moreover, $|C^{\prime}|$ and $|C|$ are not composed of the same pencil. Thus we have

[TABLE]

By (4.8) ($m=6$, $l=5$), we have $K_{X}^{3}>0.4766$. Thus $(4)$ holds under the assumption of $(4.1)$.

$(4.2)$. By the argument in Case VII, we only need to consider the case when $|S_{6,-5}|_{F}|$ and $|C|$ are composed of the same pencil. By [CHP17, Proposition 3.5], we have $\beta\geq\frac{6}{11}$. Since $P_{6}(X)\geq 28$, we have $\mu\geq\frac{7}{6}$ by our assumption. Note that we have $\xi\geq\frac{5}{7}$. Since $\alpha(8)>3$, we get $\xi\geq\frac{3}{4}$. Thus we have $K_{X}^{3}\geq\mu\beta\xi>0.4772$. So $(4)$ holds.

$(4.3)$ & $(4.5)$. We have $\mu\geq\frac{7}{6}$ by our assumption. By Inequality (2.5), we have $\beta\geq\frac{7}{13}$. Since $u_{6,-2}=5$, by the same argument as in $(4.1)$, we get $\xi\geq\frac{4}{5}$. Hence $K_{X}^{3}\geq\mu\beta\xi\geq\frac{98}{195}>\frac{1}{2}$. So $(4)$ holds.

$(4.4)$. We have $\mu\geq\frac{4}{3}$ by our assumption. By Inequality (2.5), we have $\beta\geq\frac{4}{7}$. Since $\alpha(5)>1,$ we have $\xi\geq\frac{4}{5}$. By our assumption, we have $\xi>\frac{4}{5}$. Thus $\alpha(6)>2$, which implies that $\varphi_{6,X}$ is birational, which contradicts to our assumption. So $(4.4)$ does not occur.

For $(5)$, by the arguments in Case V ($\Rrightarrow\beta=\frac{1}{2}$) and Case VI ($\Rrightarrow\beta\geq\frac{5}{9}$), we see that $u_{6,-2}=6$ and $u_{6,-3}=4$ can not hold simultaneously. Combining all the arguments in Case I-Case VII, we only need to consider the following cases:

1. (5.1)

$u_{6,0}\leq 8$, $u_{6,-1}=7$, $\beta\geq\frac{4}{7}$, $u_{6,-2}\leq 5$, $u_{6,-3}=4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$; 2. (5.2)

$u_{6,0}\leq 8$, $u_{6,-1}=7$, $\beta\geq\frac{4}{7}$, $u_{6,-2}\leq 5$, $u_{6,-3}\leq 3$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$; 3. (5.3)

$u_{6,0}\leq 8$, $u_{6,-1}\leq 6$, $u_{6,-2}=6$, $u_{6,-3}\leq 3$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$; 4. (5.4)

$u_{6,0}\leq 8$, $u_{6,-1}\leq 6$, $u_{6,-2}\leq 5$, $u_{6,-3}=4$, $u_{6,-4}\leq 3$, $u_{6,-5}\leq 2$, $u_{6,-6}=1$.

$(5.1)$. By the argument in Case VI, we have $\xi\geq\frac{4}{5}$. By our assumption, we have $\mu\geq\frac{7}{6}$. Thus we have

[TABLE]

$(5.2)$. By the assumption, we have $\mu\geq\frac{4}{3}$. Similar to the case $(4.4)$, we see that $\alpha(5)>1$ and $\alpha(6)>2$, which implies the birationality of $\varphi_{6,X}$ (a contradiction).

$(5.3)$ & $(5.4)$. Similar to the case $(5.2)$, one has $\mu\geq\frac{4}{3}$, which gives a contradiction. ∎

4.5. The classification of ${\mathbb{B}}^{(5)}(X)$

Lemma 4.18**.**

Let $X$ be a minimal projective 3-fold of general type with $p_{g}(X)=2$, $d_{1}=1$, $\Gamma\cong{\mathbb{P}}^{1}$. Assume that $F$ is a $(1,2)$-surface. Then

• (i)

$P_{3}(X)\geq P_{2}(X)+2$;

• (ii)

$P_{4}(X)\geq P_{3}(X)+4$;

• (iii)

$P_{6}(X)\geq P_{5}(X)+7$.

Proof.

Since $|3K_{X^{\prime}}||_{F}\succcurlyeq|K_{X^{\prime}}+F_{1}+F|$ for two distinct general fibers of $f$, it follows from [Che04, Lemma 4.6] that

[TABLE]

which implies that $P_{3}(X)\geq P_{2}(X)+2$.

Since $|4K_{X^{\prime}}|\succcurlyeq|2(K_{X^{\prime}}+F)|$ for a general fiber of $f$, by Theorem 2.3, we have

[TABLE]

which implies that $P_{4}(X)\geq P_{3}(X)+4$.

By the same argument as above, we get $|M_{6}||_{F}\succcurlyeq\mathrm{Mov}|\sigma^{*}(3K_{F_{0}})|$. $\alpha(6)>1$ implies that $(M_{6}\cdot C)\geq 4$. Thus $|M_{6}||_{F}\neq\mathrm{Mov}|\sigma^{*}(3K_{F_{0}})|$, which implies that we have $P_{6}(X)\geq P_{5}(X)+7$. ∎

Collecting all above results of this section, we want to classify those basket ${\mathbb{B}}(X)$ which satisfy the following properties:

1. (1)

$3\leq P_{2}(X)\leq 4$; 2. (2)

$P_{2}(X)+2\leq P_{3}(X)\leq 8$; 3. (3)

$P_{3}(X)+4\leq P_{4}(X)\leq 13$; 4. (4)

$P_{4}(X)+4\leq P_{5}(X)\leq 21$; 5. (5)

$P_{5}(X)+7\leq P_{6}(X)\leq 31$; 6. (6)

$\chi(\mathcal{O}_{X})=0,-1;$ 7. (7)

$K_{X}^{3}\geq\frac{5}{14}$; 8. (8)

If $P_{6}(X)\geq 26$, then $K_{X}^{3}\geq\frac{11}{28}$; 9. (9)

If $P_{6}(X)\geq 27$, we have $K_{X}^{3}>0.4328$; 10. (10)

If $P_{6}(X)\geq 28$, we have $K_{X}^{3}>0.4714$; 11. (11)

If $P_{6}(X)\geq 31$, we have $K_{X}^{3}\geq\frac{8}{15}$.

This situation naturally fits into the hypothesis of [CC1, (3.8)] from which we can list all the possibilities for $B^{(5)}(X)$. To be precise,

[TABLE]

with

[TABLE]

where $\sigma_{5}=\sum_{r\geq 5}n_{1,r}^{0}\geq 0$ and

[TABLE]

Note also that, by our definition, each of the above coefficients satisfies $n_{*,*}^{0}\geq 0$.

Inputing above constraints, our independently written computer programs output a raw list for $\{B_{X}^{(5)},P_{2}(X),\chi({\mathcal{O}}_{X})\}$. Taking into account those possible packings, we finally get the list ${\mathbb{S}}_{2}$ which consists of 263 elements. Being aware of the length of this paper, we do not list the set ${\mathbb{S}}_{2}$, which can be found, however, at

http://www.dima.unige.it/~penegini/publ.html

4.6. Proof of Theorem 1.2

Proof.

Theorem 1.2 follows directly from Lemma 4.1, Proposition 4.2, Remark 4.3, Lemma 4.10, Proposition 4.11, Proposition 4.12, Proposition 4.13, Proposition 4.14, Proposition 4.15 and Proposition 4.17. ∎