# Noether inequality for irregular threefolds of general type

Yong Hu  and  Tong Zhang  School of Mathematical Sciences, Shanghai Jiao Tong
University, 800 Dongchuan Road, Shanghai 200240, People’s Republic of China
<EMAIL_ADDRESS>School of Mathematical Sciences, Ministry of Education Key
Laboratory of Mathematics and Engineering Applications & Shanghai Key
Laboratory of PMMP, East China Normal University, Shanghai 200241, China
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $X$ be a smooth irregular $3$-fold of general type over $\mathbb{C}$. We
prove that the optimal Noether inequality

${\mathrm{vol}}(X)\geq\frac{4}{3}p_{g}(X)$

holds if $p_{g}(X)\geq 16$ or if $X$ has a Gorenstein minimal model. Moreover,
when $X$ attains the equality and $p_{g}(X)\geq 16$, its canonical model can
be explicitly described.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Irregular $3$-folds of general type with $q\geq 2$
  4. 4 Irregular $3$-folds of general type with $q=1$: Part I
  5. 5 Irregular $3$-folds of general type with $q=1$: Part II
  6. 6 Proof of main theorems

##  1\. Introduction

Throughout this paper, all varieties are projective over the complex numbers
$\mathbb{C}$.

###  1.1. Background

Classifying algebraic varieties is a central problem in algebraic geometry.
One way to attack this problem, especially for varieties of general type, is
to study the relation among their birational invariants first, and then to
obtain an explicit description using these numerical information. This is
often refered as the geography of algebraic varieties in the literature.

It is via this approach that many classification results for varieties of
general type have been proved. As a typical example, it was proved by Noether
[27] that every smooth surface $S$ of general type satisfies the following
optimal inequality:

(1.1) ${\mathrm{vol}}(S)\geq 2p_{g}(S)-4,$

which is now called the Noether inequality. In his celebrated paper [14],
Horikawa completely classified surfaces of general type which attain the
equality in (1.1). The Noether inequality problem in dimension three also has
attracted lots of attentions (e.g. [21, 8, 2, 6, 9]). Recently, J. Chen, M.
Chen and C. Jiang [7] proved that every smooth $3$-fold $X$ of general type,
either having a Gorenstein minimal model or with $p_{g}(X)\geq 11$, satisfies
the following optimal Noether inequality:

(1.2) ${\mathrm{vol}}(X)\geq\frac{4}{3}p_{g}(X)-\frac{10}{3}.$

Parallel to Horikawa’s work [14], they asked [Question 1.5, loc. cit] whether
$3$-folds attaining the equality can be classified, and this question has been
solved by the authors in [17].

Motivated by the explicit classification of irregular varieties of general
type, it is a fundamental question to what extent does the irregularity affect
the distribution of birational invariants (see [30, Introduction] for a
detailed description). In fact, if a smooth surface $S$ of general type
satisfies the Noether equality in (1.1), then $q(S)=0$ [1, Theorem 10]. Thus
there must be a sharper Noether inequality for irregular surfaces of general
type. This problem was studied extensively by Bombieri [1, §10] and later
solved by Debarre [11] who proved that every irregular surface $S$ of general
type satisfies the following optimal Noether inequality:

(1.3) ${\mathrm{vol}}(S)\geq 2p_{g}(S).$

Moreover, Debarre proved that if the equality holds, then $1\leq q(S)\leq 4$.
Based on the results of Horikawa, Debarre, Catanese, Ciliberto, Mendes Lopes
and Pardini [15, 11, 3, 10], a complete classification of irregular surfaces
of general type satisfying the equality in (1.3) has been established. We
refer to [10, §1] for the history and details regarding this result.

Very similar to the surface case, it is proved recently that if a smooth
$3$-fold $X$ of general type with $p_{g}(X)\geq 11$ satisfies the Noether
equality in (1.2), then $q(X)=0$ [17]. Thus it is natural to ask:

######  Question.

What is the optimal Noether inequality for irregular $3$-folds of general
type? Once the desired inequality is obtained, can one classify the $3$-folds
attaining the equality?

However, unlike the Noether inequality (1.2) which was known already two
decades ago for smooth canonically polarized $3$-folds [8], this question is
unknown even for irregular $3$-folds with smooth canonical models.

###  1.2. Main results

The main purpose of this paper is to give an answer to the above question for
“almost all” irregular $3$-folds of general type. Our first main result is the
following.

######  Theorem 1.1.

Let $X$ be a smooth irregular $3$-fold of general type. Then we have the
following optimal Noether inequality:

(1.4) ${\mathrm{vol}}(X)\geq\frac{4}{3}p_{g}(X),$

provided one of the following conditions holds:

  * (1)

$X$ has a Gorenstein minimal model;

  * (2)

$q(X)=1$ and $p_{g}(X)\geq 16$;

  * (3)

$q(X)\geq 2$.

Moreover, if $X$ satisfies the equality in (1.4) as well as any of the above
three conditions (1)–(3), then $1\leq q(X)\leq 2$.

Note that minimal models of $3$-folds $X$ with
${\mathrm{vol}}(X)<\frac{4}{3}p_{g}(X)\leq 20$ form a bounded family [24,
Corollary 2]. Thus Theorem 1.1 shows that the Noether inequality (1.4) holds
for all minimal irregular $3$-folds of general type except possibly finitely
many families.

We would like to remark on the optimality of the Noether inequality (1.4).
Indeed, the authors in [18] described the canonical model of all smooth
irregular $3$-folds $X$ of general type with
${\mathrm{vol}}(X)=\frac{4}{3}\chi(\omega_{X})$ and showed that $q(X)=1$ and
$h^{2}(X,\mathcal{O}_{X})=0$ for such $X$. Thus we deduce that

${\mathrm{vol}}(X)=\frac{4}{3}\left(p_{g}(X)-h^{2}(X,\mathcal{O}_{X})+q(X)-1\right)=\frac{4}{3}p_{g}(X)$

for such $X$. A notable feature is that the general Albanese fibre of such $X$
is a $(1,2)$-surface. Here and throughout the paper, a $(1,2)$-surface always
denotes a smooth surface $S$ with ${\mathrm{vol}}(S)=1$ and $p_{g}(S)=2$.

Our second theorem shows that “almost all” irregular $3$-folds attaining the
equality in (1.4) are exactly of the above form.

######  Theorem 1.2.

Let $X$ be a smooth irregular $3$-fold of general type with
${\mathrm{vol}}(X)=\frac{4}{3}p_{g}(X)$ and $p_{g}(X)\geq 16$. Then
${\mathrm{vol}}(X)=\frac{4}{3}\chi(\omega_{X})$. Thus the canonical model of
$X$ can be explicitly described as in [18, Theorem 1.3].

Interestingly, a similar phenomenon actually holds also in the surface case.
Namely, let $S$ be a smooth irregular surface of general type with
$p_{g}(S)\geq 5$. Then ${\mathrm{vol}}(S)=2p_{g}(S)$ if and only if
${\mathrm{vol}}(S)=2\chi(\omega_{S})$ [10, Theorem 1.1].

Results in [21, 7, 17] suggest that a $3$-fold $X$ of general type with small
${{\mathrm{vol}}(X)}/{p_{g}(X)}$ admits a pencil of $(1,2)$-surfaces. Thus
there should be a sharper Noether inequality for irregular $3$-folds of
general type without pencils of $(1,2)$-surfaces. Our third theorem provides a
strong evidence that the desired inequality may be just like Debarre’s
inequality (1.3).

######  Theorem 1.3.

Let $X$ be a smooth $3$-fold of general type with $q(X)\geq 2$ whose general
Albanese fibre is not a $(1,2)$-surface. Then we have the following optimal
Noether inequality:

(1.5) ${\mathrm{vol}}(X)\geq 2p_{g}(X),$

provided one of the following conditions holds:

  * (1)

$X$ has a Gorenstein minimal model;

  * (2)

$p_{g}(X)\neq 2$ or $3$.

The optimality of (1.5) can be seen from the following example. Take a minimal
surface $S$ of general type with $K_{S}^{2}=2$, $p_{g}(S)=3$ and $q(S)=0$.
Such a surface $S$ exists and has been classified by Horikawa [14]. Take a
smooth curve $C$ of genus two. Let $X=S\times C$ be the product $3$-fold. Then
it is easy to compute that $q(X)=2$, $p_{g}(X)=6$ and
${\mathrm{vol}}(X)=K_{X}^{3}=12$. In particular, we have
${\mathrm{vol}}(X)=2p_{g}(X)$.

###  1.3. Sketch of the proof

The whole proof of the Noether inequality is subject to the irregularity, and
is divided into two main cases. Since the canonical volume and the geometric
genus are birational invariants, we may assume that $X$ is a minimal irregular
$3$-fold of general type.

Case 1: $q(X)\geq 2$. In this case, when ${\rm alb}\dim X=1$, we apply the
Noether-Severi inequality and the slope inequality in [16, 18] to deduce the
inequalities (1.4) and (1.5). When ${\rm alb}\dim X\geq 2$, up to a birational
modification, we choose a smooth member $S\in{\mathrm{Mov}}|K_{X}|$ and deduce
a relation of the form ${\mathrm{vol}}(X)\geq\alpha{\mathrm{vol}}(S)$ from
Kawamata’s extension theorem (here $\alpha$ depends on the canonical image of
$X$). Since $S$ is an irregular surface of general type, we apply the Severi
inequality in [29] or the slope inequality in [34] to get another relation of
the form ${\mathrm{vol}}(S)\geq\beta\chi(\omega_{S})$ (here $\beta$ depends on
the Albanese dimension of $S$). Moreover, by a detailed analysis on the
canonical image of $X$, we also prove that $p_{g}(X)$ is bounded from above by
$\chi(\omega_{S})$ and $\chi(\omega_{X})$. Combining the above three
inequalities with the Severi inequality in [35], the inequality (1.5) in this
case can be proved.

Case 2: $q(X)=1$. In this case, the Albanese map of $X$ induces a fibration
$f:X\to B$ over a genus one curve $B$. Let $F$ be a general fibre of $f$. When
$F$ is not a $(1,2)$-surface, we may slightly modify the proof in Case 1 to
get the inequality (1.4). However, dealing with the case when $F$ is a
$(1,2)$-surface and $h^{2}(X,\mathcal{O}_{X})>0$ requires new insight, mainly
because the Noether-Severi inequality in [18] only provides a weaker Noether
inequality. Our new input here is to use the positivity of the direct image
sheaf $f_{*}\mathcal{O}_{X}(mK_{X})$ to get more positivity of $K_{X}$. More
precisely, we prove that $K_{X}-aF$ is nef for some $a>\frac{4}{3}$ when
$p_{g}(X)\geq 16$. If $X$ is Gorenstein, we prove a weaker result that
$K_{X}-\frac{3}{4}F$ is nef when $p_{g}(X)\geq 4$. These two positivity
results plus a known estimate in [18] help us establish the inequality (1.4)
in the corresponding cases.

As has been pointed out by Debarre [11, §6], Bombieri [1, Lemma 14] proved
that ${\mathrm{vol}}(S)\geq 2\chi(\omega_{S})$ holds for every irregular
surface $S$ of general type. When $q(S)=1$, we have
$p_{g}(S)=\chi(\omega_{S})$ and Debarre’s inequality (1.3) follows directly
from Bombieri’s result. As a dramatic difference, for an irregular $3$-fold
$X$ with $q(X)=1$, we only have
$p_{g}(X)=\chi(\omega_{X})+h^{2}(X,\mathcal{O}_{X})$. Thus the above
implication in dimension two fails in dimension three once
$h^{2}(X,\mathcal{O}_{X})>0$, which is the main difficulty in this paper.

###  1.4. Notation and conventions

In this paper, we adopt the following notation and conventions.

#### Varieties and divisors

Let $V$ be a normal variety of dimension $d$. We say that $V$ is _minimal_ ,
if $V$ has at worst $\mathbb{Q}$-factorial terminal singularities and the
canonical divisor $K_{V}$ is nef. The _geometric genus_ $p_{g}(V)$ of $V$ is
defined as

$p_{g}(V):=h^{0}(V,K_{V}).$

For a Weil divisor $L$ on $V$, the _volume_ ${\mathrm{vol}}(L)$ of $L$ is
defined as

${\mathrm{vol}}(L):=\limsup\limits_{n\to\infty}\frac{h^{0}(V,nL)}{n^{d}/d!}.$

The volume ${\mathrm{vol}}(K_{V})$ is called the _canonical volume_ of $V$,
and is denoted by ${\mathrm{vol}}(V)$. If $V$ is birational to a smooth
variety $V^{\prime}$ with ${\mathrm{vol}}(V^{\prime})>0$, we say that $V$ is
of general type. Note that if $V$ is minimal, then
${\mathrm{vol}}(V)=K_{V}^{d}$.

For a linear system $\Lambda$ on $V$, we denote by ${\mathrm{Mov}}\Lambda$ and
${\mathrm{Bs}}\Lambda$ the movable part and the base locus of $\Lambda$,
respectively. If $p_{g}(V)\geq 2$, the rational map

$\phi_{K_{V}}:V\dasharrow\mathbb{P}^{p_{g}(V)-1}$

induced by the linear system $|K_{V}|$ is called the _canonical map_ of $V$,
and $\phi_{K_{V}}(V)$ is called the _canonical image_.

#### Irregular varieties

Let $V$ be a normal variety with at worst rational singularities. We say that
$V$ is _irregular_ , if $q(V):=h^{1}(V,\mathcal{O}_{V})>0$. Note that $V$ has
a well-defined _Albanese map_

$a:V\to\mathrm{Alb}(V),$

where $A:=\mathrm{Alb}(V)$ is an abelian variety referred as the Albanese
variety of $V$. The number $\dim a(V)$, denoted by ${\rm alb}\dim V$, is
called the _Albanese dimension_ of $V$. Let

$V\stackrel{{\scriptstyle f}}{{\rightarrow}}W\stackrel{{\scriptstyle
g}}{{\rightarrow}}A$

be the Stein factorization of $a$. Then $f$ is called the _Albanese fibration_
of $V$, and a fibre of $f$ is called an _Albanese fibre_. If
$\phi:V^{\prime}\to V$ is a resolution of singularities of $V$ and
$a^{\prime}:V^{\prime}\to\mathrm{Alb}(V^{\prime})$ is the Albanese map of
$V^{\prime}$, then $\mathrm{Alb}(V^{\prime})\cong\mathrm{Alb}(V)$ and
$a^{\prime}=a\circ\phi$.

### Acknowledgments

Both authors would like to thank Professors Jungkai Alfred Chen and Zhi Jiang
for their interests in this paper.

Y.H. is supported by National Key Research and Development Program of China
#2023YFA1010600 and the National Natural Science Foundation of China (Grant
No. 12201397). T.Z. is supported by the National Natural Science Foundation of
China (Grant No. 12071139), Science and Technology Commission of Shanghai
Municipality (No. 22JC1400700, No. 22DZ2229014), and Fundamental Research
Funds for the Central Universities.

##  2\. Preliminaries

In this section, we list some preliminary results used in the later proof.

###  2.1. Linear systems on $3$-folds

Let $V$ be a normal $3$-fold. Let $L$ be a Cartier divisor on $V$ with
$h^{0}(V,L)\geq 2$. Let

$\phi_{L}:V\dasharrow\mathbb{P}^{h^{0}(V,L)-1}$

be the rational map induced by the linear system $|L|$. Denote
$\Sigma_{L}=\phi_{L}(V)$.

We first recall a lemma due to Reid.

######  Lemma 2.1.

[31, Corollaries (i)] Let $W\subseteq\mathbb{P}^{N}$ be an irreducible and
non-degenerate variety of dimension $w$. Then

$h^{0}\left(\mathbb{P}^{N},\mathcal{I}_{W}(2)\right)\leq\binom{N-w+2}{2}-\min\\{\deg
W,2(N-w)+1\\}.$

######  Lemma 2.2.

If $\dim\Sigma_{L}=3$, then

$\displaystyle h^{0}(V,2L)$ $\displaystyle\geq
3h^{0}(V,L)-3+\min\\{\deg\Sigma_{L},2h^{0}(V,L)-7\\}\geq 4h^{0}(V,L)-6.$

If $\dim\Sigma_{L}=2$, then

$\displaystyle h^{0}(V,2L)\geq
2h^{0}(V,L)-1+\min\\{\deg\Sigma_{L},2h^{0}(V,L)-5\\}\geq 3h^{0}(V,L)-3.$

###### Proof.

Up to a birational modification, we may assume that $V$ is smooth and $|L|$ is
base point free. Write $N=h^{0}(V,L)-1$. By the assumption,
$\Sigma_{L}\subseteq\mathbb{P}^{N}$ is non-degenerate.

If $\dim\Sigma_{L}=3$, then $h^{0}(V,L)\geq 4$ and $\deg\Sigma_{L}\geq
N-2=h^{0}(V,L)-3$. By Lemma 2.1, we deduce that

$\displaystyle h^{0}\left(\Sigma_{L},\mathcal{O}_{\Sigma_{L}}(2)\right)$
$\displaystyle\geq
h^{0}\left(\mathcal{O}_{\mathbb{P}^{N}},\mathcal{O}_{\mathbb{P}^{N}}(2)\right)-h^{0}\left(\mathbb{P}^{N},\mathcal{I}_{\Sigma_{L}}(2)\right)$
$\displaystyle\geq\binom{N+2}{2}-\binom{N-1}{2}+\min\\{\deg\Sigma_{L},2N-5\\}$
$\displaystyle=3h^{0}(V,L)-3+\min\\{\deg\Sigma_{L},2h^{0}(V,L)-7\\}$
$\displaystyle\geq 4h^{0}(V,L)-6.$

If $\dim\Sigma_{L}=2$, then $h^{0}(V,L)\geq 3$ and $\deg\Sigma_{L}\geq
N-1=h^{0}(V,L)-2$. By a similar argument via Lemma 2.1, we deduce that

$h^{0}\left(\Sigma_{L},\mathcal{O}_{\Sigma_{L}}(2)\right)\geq
2h^{0}(V,L)-1+\min\\{\deg\Sigma_{L},2h^{0}(V,L)-5\\}\geq 3h^{0}(V,L)-3.$

Note that in either case, we have $h^{0}(V,2L)\geq
h^{0}\left(\Sigma_{L},\mathcal{O}_{\Sigma_{L}}(2)\right)$. Thus the proof is
completed. ∎

######  Lemma 2.3.

Suppose that $V$ is minimal of general type and $\pi:V^{\prime}\to V$ be a
resolution of singularities of $V$. Let $S$ be a semi-ample divisor on
$V^{\prime}$ which itself is a smooth surface of general type. If
$\lambda\pi^{*}K_{V}-S$ is $\mathbb{Q}$-linear equivalent to an effective
$\mathbb{Q}$-divisor for some $\lambda>0$, then

$\lambda(1+\lambda)^{2}K_{V}^{3}\geq K_{S_{0}}^{2},$

where $\sigma:S\to S_{0}$ is the contraction onto the minimal model $S_{0}$ of
$S$.

###### Proof.

Since $K_{V}$ is semi-ample, by [7, Corollary 2.3] which was deduced from
Kawamata’s extension theorem, the $\mathbb{Q}$-divisor
$(1+\lambda)\pi^{*}K_{V}|_{S}-\sigma^{*}K_{S_{0}}$ is pseudo-effective. Thus

$\lambda(1+\lambda)^{2}K_{V}^{3}\geq(1+\lambda)^{2}\left((\pi^{*}K_{V})^{2}\cdot
S\right)\geq(\sigma^{*}K_{S_{0}})^{2}=K_{S_{0}}^{2}.$

The proof is completed. ∎

###  2.2. Positivity of divisors on fibred varieties

Let

$f:V\to B$

be a fibration from an $n$-dimensional normal variety $V$ onto a smooth curve
$B$. Denote by $F$ a general fibre of $f$. Let $L$ be a $\mathbb{Q}$-Cartier
Weil divisor on $V$ such that $L$ is $f$-nef and that $L|_{F}$ is Cartier.
Suppose that $f_{*}\mathcal{O}_{V}(L)$ is nonzero. Let

$0=\mathcal{E}_{0}\subsetneq\mathcal{E}_{1}\subsetneq\cdots\subsetneq\mathcal{E}_{l}=f_{*}\mathcal{O}_{V}(L)$

be the Harder-Narasimhan filtration of $f_{*}\mathcal{O}_{V}(L)$. For any
$1\leq i\leq l$, set

$r_{i}={\mathrm{rank}\,}\mathcal{E}_{i},\quad\mu_{i}=\frac{\deg(\mathcal{E}_{i}/\mathcal{E}_{i-1})}{{\mathrm{rank}\,}(\mathcal{E}_{i}/\mathcal{E}_{i-1})}.$

Then we have $\mu_{1}>\mu_{2}>\cdots>\mu_{l}$.

Consider the evaluation morphism

${\mathrm{ev}}:f^{*}f_{*}\mathcal{O}_{V}(L)\to\mathcal{O}_{V}(L).$

It induces a rational map

$\psi:V\dashrightarrow\mathbb{P}_{B}(f_{*}\mathcal{O}_{V}(L)).$

Note that $\psi$ is actually defined by the linear system $|L+tF|$ for a
sufficiently large $t\in\mathbb{N}$. Denote
$\Delta=\mathrm{Sing}(V)\cup\mathrm{Supp}(\mathcal{O}_{V}(L)/\mathrm{Im}({\mathrm{ev}}))$.

######  Lemma 2.4.

Let $C$ be an integral curve on $V$ not contained in the $f$-horizontal part
of $\Delta$. Then $\left((L-\mu_{l}F)\cdot C\right)\geq 0.$

###### Proof.

Suppose that $C$ is $f$-vertical. Since $L$ is $f$-nef, we have

$\left((L-\mu_{l}F)\cdot C\right)=(L\cdot C)\geq 0.$

Suppose that $C$ is $f$-horizontal now. Choose a birational modification
$\pi:V^{\prime}\to V$ such that $V^{\prime}$ is smooth, $\pi$ is isomorphic
over $V\backslash\Delta$, and
$|M|={\mathrm{Mov}}|{\left\lfloor\pi^{*}(L+tF)\right\rfloor}|$ is base point
free for a sufficiently large $t\in\mathbb{N}$. Write

$\pi^{*}(L+tF)=M+Z.$

Then we have $\pi(Z)\subseteq\Delta$. Since $C$ is not contained in $\Delta$,
the strict transform $C^{\prime}$ of $C$ under $\pi$ is not contained in
${\mathrm{Supp}}(Z)$. Now by [25, Corollary 3.5], the $\mathbb{Q}$-divisor
$\pi^{*}(L-\mu_{l}F)-Z$ is nef. It follows that

$\left((L-\mu_{l}F)\cdot C\right)=\left(\pi^{*}(L-\mu_{l}F)\cdot
C^{\prime}\right)\geq(Z\cdot C^{\prime})\geq 0.$

Thus the proof is completed. ∎

###  2.3. Vector bundles on elliptic curves

Let $E$ be an elliptic curve. For a vector bundle $\mathcal{E}$ on $E$, we may
write

$\mathcal{E}=\bigoplus_{i=1}^{n}\mathcal{E}_{i},$

where each $\mathcal{E}_{i}$ is indecomposable, i.e., not a direct sum of
subbundles. By [32, Page 3, Fact], every $\mathcal{E}_{i}$ is semistable. Thus
it is easy to obtain the Harder-Narasimhan filtration of $\mathcal{E}$.
Indeed, we may further assume that

$\mu(\mathcal{E}_{1})=\cdots=\mu(\mathcal{E}_{i_{1}})>\cdots>\mu(\mathcal{E}_{i_{k-1}+1})=\cdots=\mu(\mathcal{E}_{i_{k}}),$

where
$\mu(\mathcal{E}_{i})=\deg\mathcal{E}_{i}/{\mathrm{rank}\,}\mathcal{E}_{i}$
for each $i$ and $i_{k}=n$. Then the Harder-Narasimhan filtration of
$\mathcal{E}$ is just

$0=\mathcal{F}_{0}\subsetneq\mathcal{F}_{1}\subsetneq\cdots\subsetneq\mathcal{F}_{k}=\mathcal{E},$

where $\mathcal{F}_{l}=\mathcal{E}_{1}\oplus\cdots\oplus\mathcal{E}_{i_{l}}$
for any $1\leq l\leq k$.

######  Definition 2.5.

For a vector bundle $\mathcal{E}$ on $E$ as above, the minimal (resp. maximal)
slope $\mu_{\min}(\mathcal{E})$ (resp. $\mu_{\max}(\mathcal{E})$) of
$\mathcal{E}$ is defined to be $\min\\{\mu(\mathcal{E}_{i})|1\leq i\leq n\\}$
(resp. $\max\\{\mu(\mathcal{E}_{i})|1\leq i\leq n\\}$).

######  Lemma 2.6.

Let $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ be two vector bundles on $E$. Then

$\mu_{\min}(\mathcal{V}_{1}\otimes\mathcal{V}_{2})=\mu_{\min}(\mathcal{V}_{1})+\mu_{\min}(\mathcal{V}_{2}).$

###### Proof.

See [19, Page 29]. ∎

######  Lemma 2.7.

Let $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ be two vector bundles on $E$ such
that $\mathcal{V}_{2}$ is indecomposable. If
${\mathrm{Hom}}(\mathcal{V}_{1},\mathcal{V}_{2})\neq 0$, then
$\mu(\mathcal{V}_{2})\geq\mu_{\min}(\mathcal{V}_{1}).$

###### Proof.

Since $\mathcal{V}_{2}$ is indecomposable, it is semistable, and
$\mu_{\max}(\mathcal{V}_{2})=\mu(\mathcal{V}_{2})$. Thus this lemma follows
from [19, Lemma 1.3.3]. ∎

######  Lemma 2.8.

If there is a generically surjective morphism
$\phi:\mathcal{V}_{1}\to\mathcal{V}_{2}$ between vector bundles
$\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ on $E$, then
$\mu_{\min}(\mathcal{V}_{2})\geq\mu_{\min}(\mathcal{V}_{1}).$

###### Proof.

Let $\mathcal{F}$ be a direct summand of $\mathcal{V}_{2}$ with
$\mu(\mathcal{F})=\mu_{\min}(\mathcal{V}_{2})$. Since $\phi$ is generically
surjective, the composition $\mathcal{V}_{1}\to\mathcal{F}$ of $\phi$ and the
natural projection $\mathcal{V}_{2}\to\mathcal{F}$ is nonzero. Then this lemma
follows from [19, Lemma 1.3.3]. ∎

######  Lemma 2.9.

[5, Lemma 4.3] Let $\mathcal{F}$ be a vector bundle on $E$. Then $\mathcal{F}$
is ample if and only if $H^{1}(E,\mathcal{F}\otimes P)=0$ for all
$P\in{\mathrm{Pic}}^{0}(E)$.

######  Corollary 2.10.

Let $f:V\to E$ be a fibration from a smooth variety $V$ to $E$, and let $L$ be
a nef and big $\mathbb{Q}$-divisor with its fractional part being simple
normal crossing. Then the sheaf $f_{*}\mathcal{O}_{V}(K_{V}+{\left\lceil
L\right\rceil})$ is ample.

###### Proof.

By the Kawamata-Viehweg vanishing theorem, for any
$P\in{\mathrm{Pic}}^{0}(E)$, we have
$h^{1}\left(V,\mathcal{O}_{V}\left(K_{V}+{\left\lceil
L\right\rceil}\right)\otimes f^{*}P\right)=0$. By the Leray spectral sequence,
we deduce that $h^{1}(E,f_{*}\mathcal{O}_{V}(K_{V}+{\left\lceil
L\right\rceil})\otimes P)=0$. By Lemma 2.9,
$f_{*}\mathcal{O}_{V}(K_{V}+{\left\lceil L\right\rceil})$ is ample. ∎

##  3\. Irregular $3$-folds of general type with $q\geq 2$

Throughout this section, we assume that $X$ is a minimal $3$-fold of general
type with $q(X)\geq 2$. The main result in this section is the following
theorem.

######  Theorem 3.1.

Let $X$ be as above. Then

$K_{X}^{3}\geq\frac{4}{3}p_{g}(X),$

and the equality holds only when ${\rm alb}\dim X=2$ and $q(X)=2$. Suppose
furthermore that the general Albanese fibre of $X$ is not a $(1,2)$-surface.
Then

$K_{X}^{3}\geq 2p_{g}(X)$

if $X$ is Gorenstein or $p_{g}(X)\neq 2,3$,

The rest of this section is devoted to the proof.

###  3.1. The case when ${\rm alb}\dim X=1$

Suppose first that ${\rm alb}\dim X=1$. Let

$f:X\to B$

be the Albanese fibration of $X$, where $B$ is a smooth curve of genus
$g(B)=q(X)\geq 2$. Let $F$ be a general fibre of $f$.

######  Proposition 3.2.

Suppose that ${\rm alb}\dim X=1$. Then $K^{3}_{X}>\frac{4}{3}p_{g}(X)$ if $F$
is a $(1,2)$-surface, and $K_{X}^{3}\geq 2p_{g}(X)$ otherwise. In particular,
Theorem 3.1 holds.

###### Proof.

We first assume that $F$ is a $(1,2)$-surface. If
$h^{2}(X,\mathcal{O}_{X})\leq 1$, by [18, Theorem 1.1], we have

$K_{X}^{3}>\frac{4}{3}\chi(\omega_{X})=\frac{4}{3}p_{g}(X)+\frac{4}{3}\left(q(X)-h^{2}(X,\mathcal{O}_{X})-1\right)\geq\frac{4}{3}p_{g}(X).$

Now suppose that $h^{2}(X,\mathcal{O}_{X})\geq 2$. Since $q(F)=0$ and
$R^{1}f_{*}\omega_{X}$ is torsion free by [22, Theorem 2.1], we know that
$R^{1}f_{*}\omega_{X}=0$. By the Leray spectral sequence, we have
$h^{2}(X,\mathcal{O}_{X})=h^{1}(X,\omega_{X})=h^{1}(B,f_{*}\omega_{X})$. Since
$f_{*}\omega_{X/B}$ is of rank two and $h^{1}(B,f_{*}\omega_{X})\geq 2$,
Fujita’s decomposition [12, Theorem 3.1] forces
$f_{*}\omega_{X/B}=\mathcal{O}_{B}^{\oplus 2}$. Thus
$f_{*}\omega_{X}=\omega_{B}^{\oplus 2}$. In particular, $p_{g}(X)=2g(B)\geq
4$. Note that $K_{X/B}$ is nef. It follows that

$K_{X}^{3}=K_{X/B}^{3}+6\left(g(B)-1\right)\geq
6\left(g(B)-1\right)=3p_{g}(X)-6>\frac{4}{3}p_{g}(X).$

We now assume that $F$ is not a $(1,2)$-surface. By [16, Theorem 1.6],
$K_{X/B}^{3}\geq 2\deg f_{*}\omega_{X/B}$. Thus it follows that

$\displaystyle K^{3}_{X}$
$\displaystyle=K_{X/B}^{3}+6\left(g(B)-1\right)K_{F}^{2}$ $\displaystyle\geq
2\deg f_{*}\omega_{X/B}+6\left(g(B)-1\right)K_{F}^{2}$ $\displaystyle=2\deg
f_{*}\omega_{X}+\left(6K_{F}^{2}-4p_{g}(F)\right)\left(g(B)-1\right)$
$\displaystyle=2p_{g}(X)-2h^{1}(B,f_{*}\omega_{X})+\left(6K_{F}^{2}-2p_{g}(F)\right)\left(g(B)-1\right),$

where the last equality follows from the Riemann-Roch theorem. By [12, Theorem
3.1], $h^{1}(B,f_{*}\omega_{X})\leq{\mathrm{rank}\,}f_{*}\omega_{X}=p_{g}(F)$.
Moreover, since $K_{F}^{2}\geq 2p_{g}(F)-4$, it is easy to see that
$3K_{F}^{3}\geq 2p_{g}(F)$ unless $F$ is a $(1,2)$-surface. Combine the above
inequalities together, and we deduce that

$K^{3}_{X}\geq 2p_{g}(X)-2p_{g}(F)+6K_{F}^{2}-2p_{g}(F)\geq 2p_{g}(X).$

Thus the proof is completed. ∎

###  3.2. The case when ${\rm alb}\dim X\geq 2$

We start from the following result.

######  Proposition 3.3.

Suppose that ${\rm alb}\dim X\geq 2$ and $p_{g}(X)=1$. Then $K_{X}^{3}\geq 2$.

###### Proof.

By [20, Theorem 1.4 and Proposition 4.3], we have $K_{X}^{3}\geq 12$ when
${\rm alb}\dim X=3$ and $K_{X}^{3}\geq 2$ when ${\rm alb}\dim X=2$. Thus the
result follows. ∎

In the rest of this subsection, we assume that ${\rm alb}\dim X\geq 2$ and
$p_{g}(X)\geq 2$. By [35, Corollary 1.9], we have the following Severi
inequality

(3.1) $K_{X}^{3}\geq 4\chi(\omega_{X})$

which will be useful in the proof. Let $\pi:X^{\prime}\to X$ be a birational
modification such that $X^{\prime}$ is smooth and
$|M|={\mathrm{Mov}}|{\left\lfloor\pi^{*}K_{X}\right\rfloor}|$ is base point
free. Then we have the following commutative diagram:

$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{\psi}$$\scriptstyle{\phi_{M}}$$\textstyle{\Sigma^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{K_{X}}}$$\textstyle{\Sigma}$

where $\Sigma\subset\mathbb{P}^{p_{g}(X)-1}$ is the canonical image of $X$,
$\phi_{M}:X^{\prime}\to\Sigma$ is the morphism induced by $|M|$, and
$X^{\prime}\stackrel{{\scriptstyle\psi}}{{\rightarrow}}\Sigma^{\prime}\stackrel{{\scriptstyle\tau}}{{\rightarrow}}\Sigma$
is the Stein factorization of $\phi_{M}$. Moreover, if $\dim\Sigma\geq 2$, by
Bertini’s theorem, we may take a smooth general member $S\in|M|$. Let
$\sigma:S\to S_{0}$ be the contraction onto the minimal model of $S$. By Lemma
2.3 for $\lambda=1$, we have

(3.2) $4K_{X}^{3}\geq K_{S_{0}}^{2}.$

The notation here will be used throughout this subsection.

####  3.2.1. $\dim\Sigma=3$

We first treat the case when $\dim\Sigma=3$. Thus $p_{g}(X)\geq 4$.

######  Proposition 3.4.

Suppose that ${\rm alb}\dim X\geq 2$ and $\dim\Sigma=3$. Then $K_{X}^{3}\geq
2p_{g}(X)$.

###### Proof.

Since $\phi_{M}$ is generically finite, we know that $S$ is a nef and big
divisor. By the Kawamata-Viehweg vanishing theorem,
$h^{i}(X^{\prime},K_{X^{\prime}}+S)=0$ for all $i>0$. Together with the short
exact sequence

(3.3)
$0\to\mathcal{O}_{X^{\prime}}(K_{X^{\prime}})\to\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+S)\to\mathcal{O}_{S}(K_{S})\to
0,$

it follows that

$\chi(\omega_{X})+\chi(\omega_{S})=\chi\left(\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+S)\right)=h^{0}(X^{\prime},K_{X^{\prime}}+S)\geq
h^{0}(X^{\prime},2S).$

By Lemma 2.2, $h^{0}(X^{\prime},2S)\geq 4h^{0}(X^{\prime},S)-6=4p_{g}(X)-6$.
Thus

(3.4) $\chi(\omega_{X})+\chi(\omega_{S})\geq 4p_{g}(X)-6.$

Since $S$ is a big divisor, we know that ${\rm alb}\dim S=2$ and the Severi
inequality $K_{S_{0}}^{2}\geq 4\chi(\omega_{S})$ holds [29, Theorem 2.1].
Together with (3.2), we have

(3.5) $K_{X}^{3}\geq\chi(\omega_{S}).$

Combine the above inequalities with (3.1). It follows that

$K_{X}^{3}\geq\frac{4}{5}\left(\chi(\omega_{X})+\chi(\omega_{S})\right)\geq\frac{16}{5}p_{g}(X)-\frac{24}{5}\geq
2p_{g}(X),$

where the last inequality holds because $p_{g}(X)\geq 4$. Thus the proof is
completed. ∎

####  3.2.2. $\dim\Sigma=2$

Here we treat the case when $\dim\Sigma=2$. Thus $p_{g}(X)\geq 3$. Denote by
$C$ a general fibre of $\psi$. Then $C$ is a smooth curve of genus $g(C)\geq
2$. Up to a birational modification, we may assume that both $X^{\prime}$ and
$\Sigma^{\prime}$ are smooth. Now $S=\phi_{M}^{*}H$ for a general hyperplane
section $H$ on $\Sigma$, and it is no longer a big divisor. Write
$H^{\prime}=\tau^{*}H$. We may assume $H^{\prime}$ is smooth. Let
$d=(\deg\tau)\cdot(\deg\Sigma)$. Then $M|_{S}\equiv dC$. Since $\Sigma$ is
non-degenerate, we have

$\deg\Sigma\geq p_{g}(X)-2.$

######  Lemma 3.5.

We have
$h^{i}(\Sigma^{\prime},R^{j}\psi_{*}\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+S))=0$
for all $i>0$ and $j\geq 0$. Moreover, $h^{2}(X^{\prime},K_{X^{\prime}}+S)=0$.

###### Proof.

Since $H^{\prime}$ is nef and big, the first vanishing follows from [23,
Corollary 10.15]. Moreover, it implies that
$h^{2}(X^{\prime},K_{X^{\prime}}+S)=h^{0}(\Sigma^{\prime},R^{2}\psi_{*}\omega_{X^{\prime}}\otimes\mathcal{O}_{\Sigma^{\prime}}(H^{\prime}))$
by the Leray spectral sequence. Since $R^{2}\psi_{*}\omega_{X^{\prime}}=0$
[22, Theorem 2.1], the second result follows. ∎

######  Lemma 3.6.

We have $q(S)\geq q(X)$.

###### Proof.

From (3.3), we obtain the following exact sequence:

$H^{1}(S,K_{S})\to H^{2}(X^{\prime},K_{X^{\prime}})\to
H^{2}(X^{\prime},K_{X^{\prime}}+S).$

Note that $q(S)=h^{1}(S,K_{S})$ and $q(X)=h^{2}(X^{\prime},K_{X^{\prime}})$.
It follows from Lemma 3.5 that $q(S)\geq q(X)$. ∎

######  Lemma 3.7.

We have

$\chi(\omega_{X})+\chi(\omega_{S})\geq
2p_{g}(X)+\min\\{\deg\Sigma,2p_{g}(X)-5\\}-h^{0}(\Sigma^{\prime},K_{\Sigma^{\prime}}+H^{\prime})-1.$

###### Proof.

As in the proof of Proposition 3.4, we have

$\chi(\omega_{X})+\chi(\omega_{S})=\chi\left(\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+S)\right).$

By [22, Proposition 7.6],
$R^{1}\psi_{*}\omega_{X^{\prime}}=\omega_{\Sigma^{\prime}}$. Applying the
Leray spectral sequence and Lemma 3.5, we deduce that

$\displaystyle\chi\left(\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+S)\right)$
$\displaystyle=h^{0}(X^{\prime},K_{X^{\prime}}+S)-h^{1}(X^{\prime},K_{X^{\prime}}+S)$
$\displaystyle=h^{0}(X^{\prime},K_{X^{\prime}}+S)-h^{0}\left(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}}\otimes\mathcal{O}_{\Sigma^{\prime}}(H^{\prime})\right)$
$\displaystyle\geq
h^{0}(X^{\prime},2S)-h^{0}(\Sigma^{\prime},K_{\Sigma^{\prime}}+H^{\prime}).$

Moreover, since $p_{g}(X)=h^{0}(X^{\prime},S)$, by Lemma 2.2, we have

$h^{0}(X^{\prime},2S)\geq 2p_{g}(X)-1+\min\\{\deg\Sigma,2p_{g}(X)-5\\}.$

Thus the proof is completed. ∎

######  Lemma 3.8.

Suppose that ${\rm alb}\dim X\geq 2$ and $\dim\Sigma=2$. Then

$K_{X}^{3}\geq\min\left\\{2p_{g}(X),4p_{g}(X)-8,3p_{g}(X)-{\left\lfloor\frac{p_{g}(X)-1}{2}\right\rfloor}-3\right\\}$

provided one of the following holds:

  * (i)

$g(H^{\prime})=0$;

  * (ii)

$g(H^{\prime})=1$ and $\deg\Sigma=p_{g}(X)-1\geq 3$.

###### Proof.

Suppose that $K_{X}^{3}<2p_{g}(X)$. By (3.1), we may assume that
$2\chi(\omega_{X})+1\leq p_{g}(X)$. Since $H^{\prime}$ is nef and big with
$g(H^{\prime})\leq 1$, by the adjunction, we have $(K_{\Sigma^{\prime}}\cdot
H^{\prime})<0$. Thus $p_{g}(\Sigma^{\prime})=0$. By the following short exact
sequence

$0\to\mathcal{O}_{\Sigma^{\prime}}(K_{\Sigma^{\prime}})\to\mathcal{O}_{\Sigma^{\prime}}(K_{\Sigma^{\prime}}+H^{\prime})\to
K_{H^{\prime}}\to 0,$

we deduce that $h^{0}(\Sigma^{\prime},K_{\Sigma^{\prime}}+H^{\prime})\leq
g(H^{\prime})$. Thus by Lemma 3.7, we deduce that under both assumptions (i)
and (ii),

$\chi(\omega_{S})\geq 3p_{g}(X)-\chi(\omega_{X})-3\geq
3p_{g}(X)-{\left\lfloor\frac{p_{g}(X)-1}{2}\right\rfloor}-3.$

Suppose that ${\rm alb}\dim S=2$. Then (3.5) also holds here. Thus we have

$K_{X}^{3}\geq\chi(\omega_{S})\geq
3p_{g}(X)-{\left\lfloor\frac{p_{g}(X)-1}{2}\right\rfloor}-3.$

Suppose that ${\rm alb}\dim S<2$. By Lemma 3.6, $q(S)\geq 2$. Let $h:S\to B$
be the Albanese fibration of $S$ with general fibre $D$. Then $g(B)\geq 2$. In
particular, $h$ descends to a fibration $h_{0}:S_{0}\to B$ via $\sigma$. Since
$g(H^{\prime})\leq 1$ and $g(D)\geq 2$, $D$ is neither a fibre nor a section
of $\psi|_{S}:S\to H^{\prime}$. Thus $(D\cdot C)\geq 2$. Note that
$M|_{S}\equiv dC$. We deduce that

$\left((\sigma^{*}K_{S_{0}})\cdot D\right)=(K_{S}\cdot
D)\geq\left((K_{X^{\prime}}+M)|_{S}\cdot D\right)\geq 2(M|_{S}\cdot D)\geq
4d.$

Moreover, since $K_{S_{0}/B}$ is nef, we deduce that

$K_{S_{0}}^{2}=K_{S_{0}/B}^{2}+4\left(g(B)-1\right)\left((\sigma^{*}K_{S_{0}})\cdot
D\right)\geq 16d.$

Thus by (3.2), we have

$K_{X}^{3}\geq\frac{1}{4}K_{S_{0}}^{2}\geq 4d\geq 4p_{g}(X)-8.$

The proof is completed. ∎

######  Lemma 3.9.

Suppose that ${\rm alb}\dim X\geq 2$, $\dim\Sigma=2$ and $g(H^{\prime})\geq
1$. Then $K_{X}^{3}\geq 2d$.

###### Proof.

Since $g(H^{\prime})\geq 1$, the fibration $h=\psi|_{S}:S\to H^{\prime}$
naturally descends to a fibration $h_{0}:S_{0}\to H^{\prime}$ via $\sigma$. We
still denote by $C$ a general fibre of $h_{0}$.

Suppose first that ${\rm alb}\dim S=2$. Then $q(S)>g(H^{\prime})$. By [34,
Corollary 1], we know that

$K_{S_{0}}^{2}-8\left(g(C)-1\right)\left(g(H^{\prime})-1\right)=K_{S_{0}/H^{\prime}}^{2}\geq
4\deg h_{*}\omega_{S/H^{\prime}}.$

On the other hand, by the adjunction, $K_{S}-2M|_{S}$ is an effective divisor
on $S$. Thus $h_{*}\mathcal{O}_{S}(2M|_{S})\otimes\omega_{H^{\prime}}^{-1}$ is
a subsheaf of $h_{*}\omega_{S/H^{\prime}}$. Let $\mathcal{L}$ be the
saturation of $h_{*}\mathcal{O}_{S}(2M|_{S})\otimes\omega_{H^{\prime}}^{-1}$
in $h_{*}\omega_{S/H^{\prime}}$. Then we obtain a short exact sequence

$0\to\mathcal{L}\to h_{*}\omega_{S/H^{\prime}}\to\mathcal{Q}\to 0,$

where $\mathcal{L}$ is a line bundle with

$\deg\mathcal{L}\geq\deg\left(h_{*}\mathcal{O}_{S}(2M|_{S})\otimes\omega_{H^{\prime}}^{-1}\right)=2d-2\left(g(H^{\prime})-1\right).$

By [12, Theorem 2.7], $h_{*}\omega_{S/H^{\prime}}$ is nef. Thus
$\deg\mathcal{Q}\geq 0$. It follows that

$\deg h_{*}\omega_{S/H^{\prime}}\geq\deg\mathcal{L}\geq
2d-2\left(g(H^{\prime})-1\right).$

Combine the above inequalities with (3.2). It follows that

$K_{X}^{3}\geq\frac{1}{4}K_{S_{0}}^{2}\geq
2d+\left(2g(C)-4\right)\left(g(H^{\prime})-1\right)\geq 2d.$

Now suppose that ${\rm alb}\dim S<2$. Since $g(H^{\prime})\geq 1$, by the
universal property of the Albanese map, we know that $h:S\to H^{\prime}$ is
just the Albanese fibration of $S$. Let $a^{\prime}:X^{\prime}\to A$ be the
Albanese map of $X^{\prime}$. Since $a^{\prime}|_{S}$ factors through $h$ and
$S$ is general, we deduce that $\dim a^{\prime}(X^{\prime})=2$, and $C$ is a
general Albanese fibre of $X^{\prime}$. Moreover, $a^{\prime}$ factors through
$\psi$. Thus $\Sigma^{\prime}\to A$ is generically finite onto its image. Let
$E_{\pi}=K_{X^{\prime}}-\pi^{*}K_{X}$ be the $\pi$-exceptional divisor. Since
$E_{\pi}$ is covered by rational curves [13, Corollary 1.5], we know that
$(E_{\pi}\cdot C)=0$. Otherwise, $E_{\pi}\to\Sigma^{\prime}$ would be
dominant, which is absurd. Thus we have $\left((\pi^{*}K_{X})\cdot
C\right)=(K_{X^{\prime}}\cdot C)=2g(C)-2$. It follows that

$K_{X}^{3}\geq\left((\pi^{*}K_{X})\cdot
M^{2}\right)=d\left((\pi^{*}K_{X})\cdot C\right)=d\left(2g(C)-2\right)\geq
2d.$

Thus the whole proof is completed. ∎

######  Proposition 3.10.

Suppose that ${\rm alb}\dim X\geq 2$ and $\dim\Sigma=2$.

  * (1)

If $p_{g}(X)\geq 4$, then $K_{X}^{3}\geq 2p_{g}(X)$;

  * (2)

If $p_{g}(X)=3$, then $K_{X}^{3}\geq 4$.

###### Proof.

If $g(H^{\prime})=0$, then the result follows from Lemma 3.8. If
$g(H^{\prime})\geq 1$ and $d\geq p_{g}(X)$, then the result follows from Lemma
3.9. Thus we only need to treat the case when $g(H^{\prime})\geq 1$ and $d\leq
p_{g}(X)-1$. In this case, we have $p_{g}(X)-2\leq\deg\Sigma\leq p_{g}(X)-1$.

If $\deg\Sigma=p_{g}(X)-1$, then $\tau$ is birational and $p_{g}(X)\geq 4$. By
[26, Theorem 8], we know that $g(H^{\prime})=1$. Thus by Lemma 3.8, we have

$K_{X}^{3}\geq 2p_{g}(X).$

If $\deg\Sigma=p_{g}(X)-2$, since $g(H^{\prime})\geq 1$, by [26, Theorem 7],
we know that $\deg\tau\geq 2$. Now $p_{g}(X)-1\geq
d=(\deg\tau)\cdot(\deg\Sigma)\geq 2p_{g}(X)-4$. We deduce that $p_{g}(X)=3$
and $d=2$. By Lemma 3.9, we have

$K_{X}^{3}\geq 2d=4.$

The proof is completed. ∎

####  3.2.3. $\dim\Sigma=1$

Finally, we treat the case when $\dim\Sigma=1$.

######  Proposition 3.11.

Suppose that ${\rm alb}\dim X\geq 2$ and $\dim\Sigma=1$.

  * (1)

If $p_{g}(X)\geq 4$, then $K_{X}^{3}\geq 2p_{g}(X)$;

  * (2)

If $p_{g}(X)=3$, then $K_{X}^{3}\geq\frac{16}{3}$.

###### Proof.

Since $\dim\Sigma=1$, we have $M\equiv bS,$ where $S$ is a general fibre of
$\psi$.

We first consider the case when $g(\Sigma^{\prime})>0$. In this case, every
fibre of $\pi$ is rationally chain connected by [13, Corollary 1.5]. Thus
$\psi$ naturally descends to a fibration $h:X\to\Sigma^{\prime}$. By the
adjunction, the general fibre of $h$ is smooth and minimal. Thus we may assume
that $S_{0}$ is a general fibre of $h$. Since ${\rm alb}\dim X\geq 2$, we know
that $q(S_{0})>0$. By [1, Lemma 14], $K_{S_{0}}^{2}\geq 2$. On the other hand,
$g(\Sigma^{\prime})>0$ implies that $b\geq p_{g}(X)$. Thus it follows that

$K_{X}^{3}\geq b(K_{X}^{2}\cdot S_{0})=bK_{S_{0}}^{2}\geq 2p_{g}(X).$

From now on, we assume that $g(\Sigma^{\prime})=0$. Then every vector bundle
over $\Sigma^{\prime}$ splits. In this case, $M\sim(p_{g}(X)-1)S$. By Lemma
2.3 for $\lambda=\frac{1}{p_{g}(X)-1}$, when $K_{S_{0}}^{2}\geq 6$ and
$p_{g}(X)\geq 3$, we always have

$K_{X}^{3}\geq\frac{\left(p_{g}(X)-1\right)^{3}}{\left(p_{g}(X)\right)^{2}}K_{S_{0}}^{2}\geq\frac{6\left(p_{g}(X)-1\right)^{3}}{\left(p_{g}(X)\right)^{2}}>2p_{g}(X),$

except the case when $K_{S_{0}}^{2}=6$ and $p_{g}(X)=3$ in which we have
$K_{X}^{3}\geq\frac{16}{3}$.

It remains to treat the case when $K_{S_{0}}^{2}\leq 5$. In this case, by [11,
Théorème 6.1], $q(S)\leq p_{g}(S)\leq 2$. On the other hand, by the Leray
spectral sequence, we have

$q(X)=h^{2}(X^{\prime},\omega_{X^{\prime}})=h^{1}(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}})+h^{0}(\Sigma^{\prime},\omega_{\Sigma^{\prime}})=h^{1}(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}}).$

Since $R^{1}\psi_{*}\omega_{X^{\prime}/\Sigma^{\prime}}$ splits into nef line
bundles [28, Lemma 2.5], we deduce that

$q(S)={\mathrm{rank}\,}R^{1}\psi_{*}\omega_{X^{\prime}}\geq
h^{1}(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}})=q(X)\geq 2.$

It follows that $p_{g}(S)=q(S)=q(X)=2$. By [12, Theorem 2.7] $p_{g}(S)=2$
implies that $\psi_{*}\omega_{X^{\prime}/\Sigma^{\prime}}$ splits into two nef
line bundles. Since
$h^{0}(\Sigma^{\prime},\psi_{*}\omega_{X^{\prime}})=p_{g}(X)\geq 3$, we deduce
that $h^{1}(\Sigma^{\prime},\psi_{*}\omega_{X^{\prime}})\leq 1$. Furthermore,
$q(S)=q(X)=2$ implies that
$R^{1}\psi_{*}\omega_{X^{\prime}}=\omega_{\Sigma^{\prime}}^{\oplus 2}$. By the
Leray spectral sequence, we have

$h^{2}(X,\mathcal{O}_{X})=h^{1}(X^{\prime},\omega_{X^{\prime}})=h^{1}(\Sigma^{\prime},\psi_{*}\omega_{X^{\prime}})+h^{0}(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}})\leq
1.$

Thus by (3.1), it follows that

$K_{X}^{3}\geq 4\left(p_{g}(X)-h^{2}(X,\mathcal{O}_{X})+q(X)-1\right)\geq
4p_{g}(X).$

The proof is completed. ∎

######  Proposition 3.12.

Suppose that ${\rm alb}\dim X\geq 2$ and $p_{g}(X)=2$. Then
$K_{X}^{3}\geq\frac{8}{3}$.

###### Proof.

By [20, Theorem 1.4], if ${\rm alb}\dim X=3$, then $K_{X}^{3}\geq 12$. Thus to
prove the proposition, we may assume that ${\rm alb}\dim X=2$. Let $a:X\to A$
be the Albanese map of $X$, and let

$h^{0}_{a}(X,K_{X}):=\min\\{h^{0}(X,\omega_{X}\otimes\alpha)|\alpha\in{\mathrm{Pic}}^{0}(X)\\}.$

If $h^{0}_{a}(X,K_{X})\geq 1$, by [35, Theorem 1.3], we have

$K_{X}^{3}\geq 4h^{0}_{a}(X,K_{X})\geq 4.$

Thus in the following, we assume that $h^{0}_{a}(X,K_{X})=0$. By the proof of
[20, Proposition 4.3], under this assumption, $a$ is surjective. In
particular, $A$ is an abelian surface. By the proof of [20, Proposition 4.3],
we have

$a_{*}\omega_{X}=\mathcal{E}_{0}\oplus\left(\bigoplus_{i=1}^{l}\mathcal{E}_{i}\right),$

where $\mathcal{E}_{0}$ is either zero or a direct sum of torsion line bundles
on $A$, and
$\mathcal{E}_{i}=\bigoplus_{j=1}^{n_{i}}(p_{i}^{*}\mathcal{F}_{ij}\otimes\mathcal{P}_{ij})$
is non-zero for each $i>0$. Here $p_{i}:A\to E_{i}$ is a fibration from $A$ to
a genus one curve $E_{i}$, each $\mathcal{F}_{ij}$ is an ample vector bundle
on $E_{i}$, each $\mathcal{P}_{ij}\in{\mathrm{Pic}}^{0}(A)$ is a torsion line
bundle and we may assume that $\mathcal{P}_{ij}\notin
p_{i}^{*}{\mathrm{Pic}}^{0}(E_{i})$ when it is non-trivial. Moreover, the
$p_{i}$’s are all distinct and $l\geq 2$.

Case 1: $h^{0}(A,\mathcal{E}_{0})<2$. We first consider the case when
$h^{0}(A,\mathcal{E}_{0})<2$. Since $p_{g}(X)=2$, we have
$h^{0}(A,\bigoplus_{i=1}^{l}\mathcal{E}_{i})>0$.

Without loss of generality, we assume that $h^{0}(A,\mathcal{E}_{1})>0$. Then
at least one $\mathcal{P}_{1j}$ must be trivial. Say $\mathcal{P}_{11}$ is
trivial. Let $f:X\to B$ be the Stein factorization of the composition
$p_{1}\circ a:X\to E_{1}$. Denote by $F$ the general fibre of $f$. Then $F$ is
an irregular surface of general type. If $g(B)\geq 2$, since $K_{X/B}$ is nef,
we have

$K_{X}^{3}\geq K_{X/B}^{3}+\left(6g(B)-6\right)K_{F}^{2}\geq 6K_{F}^{2}\geq
6.$

Thus we only need to treat the case when $g(B)=1$. In this case, by the
universal property of the Albanese morphism, $p_{1}$ factors through the
morphism $A\to B$. Thus we may further assume that $B=E_{1}$ and $f=p_{1}\circ
a$. Now $\mathcal{F}_{11}$ is a direct summand of $f_{*}\omega_{X}$. Since
$f_{*}\omega_{X}$ is nef [12, Theorem 2.7], we know that

$\deg f_{*}\omega_{X}\geq\deg\mathcal{F}_{11}\geq 1.$

Suppose that $\deg f_{*}\omega_{X}\geq 2$. By [16, Theorem 1.6], we have

$K_{X}^{3}\geq 2\deg f_{*}\omega_{X}\geq 4.$

Thus in the following, we assume that $\deg f_{*}\omega_{X}=1$. Then
$\deg\mathcal{F}_{11}=1$. If $K_{F}^{2}\geq 8$, by [16, Theorem 1.7], we have

$K_{X}^{3}\geq\left(\frac{4K_{F}^{2}}{K_{F}^{2}+4}\right)\deg
f_{*}\omega_{X}=\frac{4K_{F}^{2}}{K_{F}^{2}+4}\geq\frac{8}{3}.$

Now assume that $K_{F}^{2}<8$. By [11, Théorème 6.1], $K_{F}^{2}\geq
2p_{g}(F)$. Since $\mathcal{F}_{11}$ is ample, we know that
$h^{1}(E_{1},\mathcal{F}_{11})=0$. Thus by the Riemann-Roch theorem,
$h^{0}(E_{1},\mathcal{F}_{11})=1$. Note that
$h^{0}(E_{1},f_{*}\omega_{X})=p_{g}(X)=2$. Thus
${\mathrm{rank}\,}\mathcal{F}_{11}<{\mathrm{rank}\,}f_{*}\omega_{X}=p_{g}(F)$.
On the other hand, by [25, Corollary 3.5], we know that
$K_{X}-\mu(\mathcal{F}_{11})F$ is pseudo-effective. We deduce that

$K_{X}^{3}\geq\mu(\mathcal{F}_{11})K_{F}^{2}\geq\frac{K_{F}^{2}}{p_{g}(F)-1}\geq\frac{2K_{F}^{2}}{K_{F}^{2}-2}>\frac{8}{3}$

Case 2: $h^{0}(A,\mathcal{E}_{0})=2$. We now treat the case when
$h^{0}(A,\mathcal{E}_{0})=2$. In this case, $\mathcal{E}_{0}$ contains
$\mathcal{O}_{A}^{\oplus 2}$ as a direct summand.

Since $p_{1}$ and $p_{2}$ are distinct, $p_{1}^{*}{\mathrm{Pic}}^{0}(E_{1})$
and $p_{2}^{*}{\mathrm{Pic}}^{0}(E_{2})$ generate ${\mathrm{Pic}}^{0}(A)$. We
may assume that each $\mathcal{P}_{1j}\in p_{2}^{*}{\mathrm{Pic}}^{0}(E_{2})$
and each $\mathcal{P}_{2j}\in p_{1}^{*}{\mathrm{Pic}}^{0}(E_{1})$. Since
$h^{0}(A,\mathcal{E}_{0})=p_{g}(X)=2$, every $\mathcal{P}_{1j}$ and
$\mathcal{P}_{2j}$ is a non-trivial torsion. Applying the same argument as in
Case 1, we see that the proof is reduced again to the case when both
$f_{i}=p_{i}\circ a:X\to E_{i}$ have connected fibres for $i=1,2$. Now
$\mathcal{F}_{i1}$ is a direct summand of $(f_{i})_{*}(\omega_{X}\otimes
a^{*}\mathcal{P}_{i1}^{-1})$. Since $\mathcal{P}_{i1}$ is a torsion, by [33,
Corollary 3.4] (take $\mathcal{L}=a^{*}\mathcal{P}_{11}^{-1}$ and $D=0$), we
know that $(f_{i})_{*}(\omega_{X}\otimes a^{*}\mathcal{P}_{i1}^{-1})$ is also
nef. Thus

$\deg(f_{i})_{*}(\omega_{X}\otimes
a^{*}\mathcal{P}_{i1}^{-1})\geq\deg\mathcal{F}_{i1}\geq 1.$

Let $F_{i}$ be a general fibre of $f_{i}$ for $i=1,2$. If $K_{F_{i}}^{2}\geq
8$ for any $i$, applying the same proof of [16, Theorem 1.7] for
$\omega_{X}\otimes\mathcal{P}_{i1}^{-1}$, we deduce that

$K_{X}^{3}\geq\left(\frac{4K_{F_{i}}^{2}}{K_{F_{i}}^{2}+4}\right)\deg(f_{i})_{*}(\omega_{X}\otimes
a^{*}\mathcal{P}_{i1}^{-1})=\frac{4K_{F_{i}}^{2}}{K_{F_{i}}^{2}+4}\geq\frac{8}{3}.$

From now on, we assume that $K_{F_{i}}^{2}\leq 7$ for $i=1,2$. By the
assumption on $a_{*}\omega_{X}$, we have $\mathcal{O}_{E_{1}}^{\oplus
2}\oplus(p_{1})_{*}\mathcal{E}_{2}\subset(f_{1})_{*}\omega_{X}$. Note that
$(p_{1})_{*}\mathcal{E}_{2}\neq 0$. Thus
$p_{g}(F_{1})={\mathrm{rank}\,}(f_{1})_{*}\omega_{X}\geq 3$. By the proof of
[20, Proposition 4.3], an étale cover of $F_{1}$ is of maximal Albanese
dimension. Thus the Severi inequality $K_{F_{1}}^{2}\geq
4\chi(\omega_{F_{1}})$ [29, Theorem 2.1] implies that
$\chi(\omega_{F_{1}})=1$. Since $K_{F_{1}}^{2}\geq 2p_{g}(F_{1})$ [11,
Théorème 6.1], we deduce that $p_{g}(F_{1})=q(F_{1})=3$. By [3], it follows
that $K_{F_{1}}^{2}=6$ and $F_{1}$ is isomorphic to a smooth theta divisor in
a principally polarized abelian $3$-fold $V$. The same holds also for $F_{2}$.

We claim that $a$ has connected fibres. Otherwise, let $h:X\to W$ be the Stein
factorization of $a$. Then $W$ is a normal surface whose smooth model has
Kodaira dimension at least one. Note that both $h_{i}:W\to E_{i}$ have
connected fibres for $i=1,2$. Thus there exists an $i$ such that the general
fibre of $h_{i}$ has genus at least two. Say $i=1$, and let $B$ be a general
fibre of $h_{1}$. Then $g(B)\geq 2$. Moreover, the morphism
$h|_{F_{1}}:F_{1}\to B$ is just the Stein factorization of $a|_{F_{1}}$. Let
$C$ be a general fibre of $h|_{F_{1}}$. Since $K_{F_{1}/B}$ is nef, we have

$K_{F_{1}}^{2}=K_{F_{1}/B}^{2}+8(g(B)-1)(g(C)-1)\geq 8(g(B)-1)(g(C)-1)\geq 8.$

This is absurd. Thus the claim is verified.

By the claim, $a|_{F_{1}}:F_{1}\to E^{\prime}_{2}$ is a fibration, where
$E^{\prime}_{2}$ is a general fiber of $p_{1}$. It is clear that
$p_{2}|_{E^{\prime}_{2}}:E^{\prime}_{2}\to E_{2}$ is an isogeny. Since
$\mathcal{P}_{11}\in p_{2}^{*}{\mathrm{Pic}}^{0}(E_{2})$ is non-trivial,
$\mathcal{P}:=(a^{*}\mathcal{P}_{11}^{-1})|_{F_{1}}$ is a non-trivial torsion
line bundle on $F$. Since $F_{1}$ is a smooth theta divisor in $V$ and
${\mathrm{Pic}}^{0}(F_{1})\simeq{\mathrm{Pic}}^{0}(V)$, we may assume that
$\mathcal{P}\in{\mathrm{Pic}}^{0}(V)$ is non-trivial. Consider the long exact
sequence

$H^{0}(V,\mathcal{P})\to H^{0}(V,\mathcal{O}_{V}(F_{1})\otimes\mathcal{P})\to
H^{0}(F_{1},\omega_{F_{1}}\otimes\mathcal{P})\to H^{1}(V,\mathcal{P}).$

The non-triviality of $\mathcal{P}$ implies that
$h^{0}(V,\mathcal{P})=h^{1}(V,\mathcal{P})=0$. It follows that
$h^{0}(F_{1},\omega_{F_{1}}\otimes\mathcal{P})=h^{0}(V,\mathcal{O}_{V}(F_{1})\otimes\mathcal{P})=1$.
As a result, ${\mathrm{rank}\,}(f_{1})_{*}(\omega_{X}\otimes
a^{*}\mathcal{P}_{11}^{-1})=1$. In particular,
${\mathrm{rank}\,}\mathcal{F}_{11}=1$. Thus $\mu(\mathcal{F}_{11})\geq 1$. By
[25, Corollary 3.5] again, $K_{X}-F_{1}$ is pseudo-effective. It follows that

$K_{X}^{3}\geq\mu(\mathcal{F}_{11})K_{F_{1}}^{2}\geq K_{F_{1}}^{2}=6.$

The whole proof is completed. ∎

###  3.3. The case when $X$ is Gorenstein

We now treat the case when $X$ is Gorenstein.

######  Proposition 3.13.

Suppose that $X$ is Gorenstein. Then Theorem 3.1 holds.

###### Proof.

By Proposition 3.2, we may assume that ${\rm alb}\dim X\geq 2$. Since $X$ is
Gorenstein, we know that $\chi(\omega_{X})\geq 1$ [4, §2.1]. By (3.1), we have
$K_{X}^{3}\geq 4$. Thus we may further assume that $p_{g}(X)\geq 3$. Moreover,
by Proposition 3.4, 3.10 and 3.11, the proof reduces to the case when
$p_{g}(X)=3$ and $\chi(\omega_{X})=1$, and we only need to show that
$K_{X}^{3}\geq 6$.

Let $\Sigma$ be the canonical image of $X$. Since $p_{g}(X)=3$, we have
$\dim\Sigma\leq 2$. If $\dim\Sigma=2$, by Lemma 2.2, $h^{0}(X,2K_{X})\geq
3p_{g}(X)-3=6$. On the other hand, by the Riemann-Roch formula, we have

$h^{0}(X,2K_{X})=\frac{1}{2}K_{X}^{3}+3\chi(\omega_{X})=\frac{1}{2}K_{X}^{3}+3.$

Thus we deduce that $K_{X}^{3}\geq 2h^{0}(X,2K_{X})-6\geq 6$. If
$\dim\Sigma=1$, by Proposition 3.11 and the fact that $K_{X}^{3}$ is even [4,
§2.2], we see that $K_{X}^{3}\geq 6$. Thus the proof is completed. ∎

###  3.4. Proof of Theorem 3.1

It is clear that the two inequalities in Theorem 3.1 follow from Proposition
3.2, 3.4, 3.10, 3.11, 3.12 and 3.13.

Suppose that $K_{X}^{3}=\frac{4}{3}p_{g}(X)$. Then the above propositions
imply that ${\rm alb}\dim X\geq 2$ and $p_{g}(X)=2,3$. Thus $K_{X}^{3}\leq 4$.
Let $a:X\to A$ be the Albanese map of $X$. As in the proof of Proposition
3.12, we have $\dim a(X)=2$. Consider

$h^{0}_{a}(X,K_{X}):=\min\\{h^{0}(X,\omega_{X}\otimes\alpha)|\alpha\in{\mathrm{Pic}}^{0}(X)\\}.$

If $h^{0}_{a}(X,K_{X})\geq 1$, since $K_{X}^{3}\leq 4$, we know that
$K_{X}^{3}=4h^{0}_{a}(X,K_{X})$ and $A$ is an abelian surface by [35, Theorem
1.3]. If $h^{0}_{a}(X,K_{X})=0$, by the proof of Proposition 3.12, we know
that $A$ is also an abelian surface. In both cases, $q(X)=2$. Thus the proof
is completed.

##  4\. Irregular $3$-folds of general type with $q=1$: Part I

Let $X$ be a minimal $3$-fold of general type with $q(X)=1$. Let

$f:X\to B$

be the Albanese fibration of $X$, where $B$ is a smooth curve of genus
$g(B)=1$. Throughout this section, we always assume that the general fibre $F$
of $f$ is a $(1,2)$-surface. The main result in this section is the following
theorem.

######  Theorem 4.1.

Let $X$ be as above. Suppose that either $X$ is Gorenstein or $p_{g}(X)\geq
16$. Then

$K_{X}^{3}\geq\frac{4}{3}p_{g}(X).$

Moreover, if $p_{g}(X)\geq 16$ and $h^{2}(X,\mathcal{O}_{X})>0$, then

$K_{X}^{3}\geq\frac{4}{3}p_{g}(X)+\frac{1}{33}.$

The rest of this section is devoted to the proof.

###  4.1. General settings

Suppose that $p_{g}(X)\geq 2$. Take a birational modification
$\pi:X^{\prime}\to X$ such that $X^{\prime}$ is smooth and
$|M|={\mathrm{Mov}}|{\left\lfloor\pi^{*}K_{X}\right\rfloor}|$ is base point
free. Then we have the following commutative diagram

$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{f^{\prime}}$$\scriptstyle{\psi}$$\scriptstyle{\phi_{M}}$$\textstyle{\Sigma^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{B}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{K_{X}}}$$\scriptstyle{f}$$\textstyle{\Sigma}$

where $\Sigma$ is the canonical image of $X$, $\phi_{M}:X^{\prime}\to\Sigma$
is the morphism induced by $|M|$,
$X^{\prime}\stackrel{{\scriptstyle\psi}}{{\rightarrow}}\Sigma^{\prime}\stackrel{{\scriptstyle\tau}}{{\rightarrow}}\Sigma$
is the Stein factorization of $\phi_{M}$, and $f^{\prime}=f\circ\pi$. Denote
by $F^{\prime}$ a general fibre of $f^{\prime}$, and write
$d=(\deg\tau)\cdot(\deg\Sigma)$.

We start from the following lemma.

######  Lemma 4.2.

If $h^{2}(X,\mathcal{O}_{X})>0$, then

$f_{*}\omega_{X}=\mathcal{O}_{B}\oplus\mathcal{O}_{B}(L),$

where $L$ is an effective divisor on $B$. In particular, if $p_{g}(X)\geq 3$,
then $h^{2}(X,\mathcal{O}_{X})=1$. Moreover, if $p_{g}(X)\geq 3$, then
$\Sigma$ is a surface.

###### Proof.

By the assumption, we have $h^{1}(X,\omega_{X})>0$. Since $h^{1}(F,K_{F})=0$,
we know that $R^{1}f_{*}\omega_{X}=0$ by the torsion freeness of
$R^{1}f_{*}\omega_{X}$. Thus by the Leray spectral sequence,
$h^{1}(B,f_{*}\omega_{X})=h^{1}(X,\omega_{X})=h^{2}(X,\mathcal{O}_{X})>0$.
Since ${\mathrm{rank}\,}f_{*}\omega_{X}=p_{g}(F)=2$, by Fujita’s decomposition
[12, Theorem 3.1], we may write

$f_{*}\omega_{X}=\mathcal{O}_{B}\oplus\mathcal{O}_{B}(L),$

where $L$ is an effective divisor on $B$. If $p_{g}(X)\geq 3$, by the Riemann-
Roch theorem, $\deg L=p_{g}(X)-1\geq 2$. Thus

$h^{2}(X,\mathcal{O}_{X})=h^{1}(B,f_{*}\omega_{X})=h^{1}(B,\mathcal{O}_{B})=1.$

Moreover, if $p_{g}(X)\geq 3$, by the projection formula, we deduce that

$h^{0}(X,K_{X}-F)=h^{0}(B,f_{*}\mathcal{O}_{X}(K_{X}-F))=h^{0}(B,L-P)=p_{g}(X)-2,$

where $P=f(F)$. Therefore, the natural restriction $H^{0}(X,K_{X})\to
H^{0}(F,K_{F})$ is surjective. Thus $\dim\phi_{K_{X}}(F)=1$. Since
$h^{0}(B,L)\geq 2$ and $h^{0}(B,K_{X}-f^{*}L)>0$, we know that $\Sigma$ is a
surface. ∎

Since $F$ is a minimal $(1,2)$-surface, $|K_{F}|$ has a unique base point.
Therefore, the horizontal part of the base locus of the relative canonical map
of $X$ with respect to $f$ is a section $\Gamma$ whose intersection
$\Gamma\cap F$ with $F$ is just the base point of $|K_{F}|$.

######  Proposition 4.3.

Suppose that $p_{g}(X)\geq 4$ and $\dim\Sigma=2$. Then the following
inequalities hold:

  * (1)

$d\geq\min\\{2p_{g}(X)-4,\,p_{g}(X)-1\\}=p_{g}(X)-1\geq 3$;

  * (2)

$\displaystyle{(K_{X}\cdot\Gamma)\geq\frac{1}{3}d}$

  * (3)

$\displaystyle{K_{X}^{3}\geq\frac{4}{3}d}$.

###### Proof.

The first inequality follows from [18, Lemma 2.3 (1)] and the assumption that
$p_{g}(X)\geq 4$ and $g(B)=1$. The second and the third one follow from [18,
Proposition 2.5 (2)]. ∎

######  Proposition 4.4.

[18, Remark 2.12] Suppose that $p_{g}(X)\geq 4$, $\dim\Sigma=2$,
$d=p_{g}(X)-1$ and $K_{X}-cF$ is nef for some $c\geq 0$. Then

$K_{X}^{3}\geq\frac{4}{3}\left(p_{g}(X)-1\right)+c.$

###  4.2. The special case

In this subsection, we assume that

$f_{*}\omega_{X}=\mathcal{O}_{B}\oplus\mathcal{O}_{B}(L),$

where $e:=\deg L=p_{g}(X)-1\geq 3$. Under this assumption, we have

$K_{X}\sim f^{*}L+D\equiv eF+D.$

where $D$ is an effective divisor. Since $X$ has terminal singularities, we
may write

$K_{X^{\prime}}=\pi^{*}K_{X}+E_{\pi},$

where $E_{\pi}$ is an effective $\pi$-exceptional $\mathbb{Q}$-divisor. Up to
a birational modification of $X^{\prime}$, we may further assume that
$\pi^{*}D+E_{\pi}$ is simple normal crossing.

####  4.2.1. $X$ is Gorenstein

We first consider the case when $X$ is Gorenstein. In this case, $D$ is a
Cartier divisor. Write $D^{\prime}=\pi^{*}D$. Then we have the following
proposition in which the Gorenstein assumption is essential.

######  Proposition 4.5.

[2, Theorem 3.2] For any $t>\frac{3}{10}$, we have

$h^{0}\left(F^{\prime},K_{F^{\prime}}+{\left\lceil
tD^{\prime}|_{F^{\prime}}\right\rceil}\right)\geq 3.$

######  Lemma 4.6.

The vector bundle

$\mathcal{E}:=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}\left(K_{X^{\prime}}+{\left\lceil\pi^{*}K_{X}-2F^{\prime}-\frac{2}{e}D^{\prime}\right\rceil}\right)$

on $B$ is ample with ${\mathrm{rank}\,}\mathcal{E}\geq 3$. In particular,
$\mu_{\min}(\mathcal{E})>0$.

###### Proof.

Note that the $\mathbb{Q}$-divisor

$\pi^{*}K_{X}-2F^{\prime}-\frac{2}{e}D^{\prime}\equiv\left(1-\frac{2}{e}\right)\pi^{*}K_{X}$

is nef and big with simple normal crossing fractional part. By Corollary 2.10,
$\mathcal{E}$ is ample. Thus $\mu_{\min}(\mathcal{E})>0$. Note that

$\displaystyle{\mathrm{rank}\,}\mathcal{E}$
$\displaystyle=h^{0}\left(F^{\prime},K_{F^{\prime}}+\left.{\left\lceil\pi^{*}K_{X}-2F^{\prime}-\frac{2}{e}D^{\prime}\right\rceil}\right|_{F^{\prime}}\right)$
$\displaystyle\geq
h^{0}\left(F^{\prime},K_{F^{\prime}}+{\left\lceil\left(1-\frac{2}{e}\right)D^{\prime}|_{F^{\prime}}\right\rceil}\right)$

Since $1-\frac{2}{e}\geq\frac{1}{3}>\frac{3}{10}$, by Proposition 4.5, we
conclude that ${\mathrm{rank}\,}\mathcal{E}\geq 3$. ∎

As a simple corollary of Lemma 4.6, we have

######  Corollary 4.7.

Let

$\mathcal{F}=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}\left(K_{X^{\prime}}+{\left\lceil\pi^{*}K_{X}-\frac{2}{e}D^{\prime}\right\rceil}\right).$

Then ${\mathrm{rank}\,}\mathcal{F}\geq 3$ and $\mu_{\min}(\mathcal{F})>2$.

######  Lemma 4.8.

One of the following two cases occurs:

  * (1)

We have $\mu_{\min}(f_{*}\mathcal{O}_{X}(2K_{X}))\geq 2.$ In this case,
$K_{X}-F$ is nef.

  * (2)

We have $f_{*}\mathcal{O}_{X}(2K_{X})=\mathcal{V}_{1}\oplus\mathcal{L}_{1},$
where $\mathcal{V}_{1}$ is of rank three with $\mu_{\min}(\mathcal{V}_{1})>2$
and $\mathcal{L}_{1}$ is a line bundle of degree one. In this case, $2K_{X}-F$
is nef.

###### Proof.

Since $f^{\prime}_{*}$ is left exact, we have the following injection

$\mathcal{F}:=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}\left(K_{X^{\prime}}+{\left\lceil\pi^{*}K_{X}-\frac{2}{e}D^{\prime}\right\rceil}\right)\to
f^{\prime}_{*}\mathcal{O}_{X^{\prime}}(2K_{X^{\prime}})=f_{*}\mathcal{O}_{X}(2K_{X}).$

By Corollary 4.7, ${\mathrm{rank}\,}\mathcal{F}\geq 3$ and
$\mu_{\min}(\mathcal{F})>2$. Let $\mathcal{V}_{1}$ be the saturation of
$\mathcal{F}$ in $f_{*}\mathcal{O}_{X}(2K_{X})$. Then $\mathcal{V}_{1}$ is a
subbundle of $f_{*}\mathcal{O}_{X}(2K_{X})$ with
${\mathrm{rank}\,}\mathcal{V}_{1}\geq 3$ and $\mu_{\min}(\mathcal{V}_{1})>2$.

Consider the evaluation morphism

$\mathrm{ev}_{2}:f^{*}f_{*}\mathcal{O}_{X}(2K_{X})\to\mathcal{O}_{X}(2K_{X}).$

Let
$\Delta_{2}=\mathrm{Sing}(X)\cup\mathrm{Supp}(\mathcal{O}_{X}(2K_{X})/\mathrm{Im}({\mathrm{ev}}_{2}))$.
Since $F$ is a minimal $(1,2)$-surface, $|2K_{F}|$ is base point free. We
deduce that $\Delta_{2}$ is $f$-vertical.

Suppose that $\mu_{\min}(f_{*}\mathcal{O}_{X}(2K_{X}))\geq 2$. By Lemma 2.4,
$2K_{X}-2F$ is nef. Thus (1) is proved. Suppose that
$\mu_{\min}(f_{*}\mathcal{O}_{X}(2K_{X}))<2$. Since
${\mathrm{rank}\,}f_{*}\mathcal{O}_{X}(2K_{X})=h^{0}(F,2K_{F})=4$, we deduce
that ${\mathrm{rank}\,}\mathcal{V}_{1}=3$, and $f_{*}\mathcal{O}_{X}(2K_{X})$
has a direct summand line bundle $\mathcal{L}_{1}$ with
$\deg\mathcal{L}_{1}<2$. Write

$f_{*}\mathcal{O}_{X}(2K_{X})=\mathcal{E}_{1}\oplus\mathcal{L}_{1}.$

Since $\mu_{\min}(\mathcal{V}_{1})>2$, by Lemma 2.7,
${\mathrm{Hom}}(\mathcal{V}_{1},\mathcal{L}_{1})=0$. It follows that
$\mathcal{V}_{1}\subset\mathcal{E}_{1}$. Since $\mathcal{V}_{1}$ is a
subbundle and ${\mathrm{rank}\,}\mathcal{E}_{1}=3$, we deduce that
$\mathcal{V}_{1}=\mathcal{E}_{1}$. By Corollary 2.10,
$f_{*}\mathcal{O}_{X}(2K_{X})=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+\pi^{*}K_{X})$
is ample. As a result, $\deg\mathcal{L}_{1}=1$. Thus
$\mu_{\min}(f_{*}\mathcal{O}_{X}(2K_{X}))=1$. By Lemma 2.4 again, we deduce
that $2K_{X}-F$ is nef. This proves (2). ∎

######  Lemma 4.9.

We have $\mu_{\min}(f_{*}\mathcal{O}_{X}(3K_{X}))\geq 2.$ Moreover,
$3K_{X}-2F$ is nef.

###### Proof.

Given that $\mu_{\min}(f_{*}\mathcal{O}_{X}(3K_{X}))\geq 2$, consider the
evaluation morphism

$\mathrm{ev}_{3}:f^{*}f_{*}\mathcal{O}_{X}(3K_{X})\to\mathcal{O}_{X}(3K_{X}),$

and let
$\Delta_{3}=\mathrm{Sing}(X)\cup\mathrm{Supp}(\mathcal{O}_{X}(3K_{X})/\mathrm{Im}({\mathrm{ev}}_{3}))$.
Since $K_{X}$ is nef, by Lemma 2.4, for any curve $C$ not contained in the
$f$-horizontal part of $\Delta_{3}$, we have

$\left((3K_{X}-2F)\cdot C\right)\geq 0.$

On the other hand, since ${\mathrm{Bs}}|3K_{F}|={\mathrm{Bs}}|K_{F}|$, we see
that the $f$-horizontal part of $\Delta_{3}$ is just the section $\Gamma$. By
Proposition 4.3 (1) and (2), we have

$\left((3K_{X}-2F)\cdot\Gamma\right)=3(K_{X}\cdot\Gamma)-2\geq d-2>0.$

As a result, $3K_{X}-2F$ is nef.

To prove that $\mu_{\min}(f_{*}\mathcal{O}_{X}(3K_{X}))\geq 2$, we may write

$f_{*}\mathcal{O}_{X}(3K_{X})=\mathcal{V}_{2}\oplus\mathcal{V}_{3},$

where $\mu_{\min}(\mathcal{V}_{2})\geq 2$ and $\mu_{\max}(\mathcal{V}_{3})<2$.
Since $H^{0}(F,K_{F})\otimes H^{0}(F,2K_{F})\to H^{0}(F,3K_{F})$ is
surjective, we obtain a generically surjective morphism

$f_{*}\omega_{X}\otimes f_{*}\mathcal{O}_{X}(2K_{X})\to
f_{*}\mathcal{O}_{X}(3K_{X}).$

Recall that $f_{*}\omega_{X}=\mathcal{O}_{B}\oplus\mathcal{L}$ with
$\deg\mathcal{L}\geq 3$. If $\mu_{\min}(f_{*}\mathcal{O}_{X}(2K_{X}))\geq 2$,
then $\mu_{\min}(f_{*}\mathcal{O}_{X}(3K_{X}))\geq 2$ by Lemma 2.8. By Lemma
4.8, the proof reduces to the case when

$f_{*}\mathcal{O}_{X}(2K_{X})=\mathcal{V}_{1}\oplus\mathcal{L}_{1},$

where $\mathcal{V}_{1}$ is of rank three with $\mu_{\min}(\mathcal{V}_{1})>2$
and $\mathcal{L}_{1}$ is a line bundle with $\deg\mathcal{L}_{1}=1$. In this
case, the above generic surjection becomes

$\mathcal{V}_{1}\oplus(\mathcal{V}_{1}\otimes\mathcal{L})\oplus\mathcal{L}_{1}\oplus(\mathcal{L}_{1}\otimes\mathcal{L})\to\mathcal{V}_{2}\oplus\mathcal{V}_{3}.$

By Lemma 2.7,
${\mathrm{Hom}}(\mathcal{V}_{1},\mathcal{V}_{3})={\mathrm{Hom}}(\mathcal{L}_{1}\otimes\mathcal{L},\mathcal{V}_{3})={\mathrm{Hom}}(\mathcal{V}_{1}\otimes\mathcal{L},\mathcal{V}_{3})=0$.
By the generic surjectivity, either $\mathcal{V}_{3}=0$ or $\mathcal{V}_{3}$
is a line bundle with $\deg\mathcal{V}_{3}=1$. However, the latter cannot
occur. Indeed, by Lemma 4.8, $2K_{X}-F$ is nef. By Proposition 4.3 (1) and
(3), we have

$(2K_{X}-F)^{3}=8K_{X}^{3}-6K_{F}^{2}=8K_{X}^{3}-6>0.$

Thus $2K_{X}-F$ is also big. By Corollary 2.10,
$f_{*}\mathcal{O}_{X}(3K_{X}-F)=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+\pi^{*}(2K_{X}-F))$
is ample. In particular, $\mu_{\min}(f_{*}\mathcal{O}_{X}(3K_{X}))>1$. Thus
the proof is completed. ∎

######  Proposition 4.10.

We have $\mu_{\min}(f_{*}\mathcal{O}_{X}(4K_{X}))\geq 3.$ Moreover,
$4K_{X}-3F$ is nef.

###### Proof.

Given that $\mu_{\min}(f_{*}\mathcal{O}_{X}(4K_{X}))\geq 3$, consider the
evaluation map

$\mathrm{ev}_{4}:f^{*}f_{*}\mathcal{O}_{X}(4K_{X})\to\mathcal{O}_{X}(4K_{X}),$

and let
$\Delta_{4}=\mathrm{Sing}(X)\cup\mathrm{Supp}(\mathcal{O}_{X}(4K_{X})/\mathrm{Im}({\mathrm{ev}}_{4}))$.
Since $|4K_{F}|$ is base point free, we deduce that $\Delta_{4}$ is
$f$-vertical. By Lemma 2.4, $4K_{X}-3F$ is nef.

To prove that $\mu_{\min}(f_{*}\mathcal{O}_{X}(4K_{X}))\geq 3$, we may write

$f_{*}\mathcal{O}_{X}(4K_{X})=\mathcal{V}_{4}\oplus\mathcal{V}_{5},$

where $\mu_{\min}(\mathcal{V}_{4})\geq 3$ and $\mu_{\max}(\mathcal{V}_{5})<3$.
Since ${\mathrm{Sym}}^{2}H^{0}(F,2K_{F})\to H^{0}(F,2K_{F})$ is surjective, we
obtain a generically surjective morphism

${\mathrm{Sym}}^{2}f_{*}\mathcal{O}_{X}(2K_{X})\to
f_{*}\mathcal{O}_{X}(4K_{X}).$

Similar to the proof of Lemma 4.9, the proof reduces again to the case when

$f_{*}\mathcal{O}_{X}(2K_{X})=\mathcal{V}_{1}\oplus\mathcal{L}_{1},$

where $\mathcal{V}_{1}$ is of rank three with $\mu_{\min}(\mathcal{V}_{1})>2$
and $\mathcal{L}_{1}$ is a line bundle with $\deg\mathcal{L}_{1}=1$. Now the
above generic surjection becomes

${\mathrm{Sym}}^{2}\mathcal{V}_{1}\oplus(\mathcal{V}_{1}\otimes\mathcal{L}_{1})\oplus\mathcal{L}_{1}^{\otimes
2}\to\mathcal{V}_{4}\oplus\mathcal{V}_{5}.$

By Lemma 2.6 and 2.7,
${\mathrm{Hom}}({\mathrm{Sym}}^{2}\mathcal{V}_{1},\mathcal{V}_{5})={\mathrm{Hom}}(\mathcal{V}_{1}\otimes\mathcal{L}_{1},\mathcal{V}_{5})=0$.
Thus the generic surjectivity implies that either $\mathcal{V}_{5}=0$ or
$\mathcal{V}_{5}$ is a line bundle with $\deg\mathcal{V}_{5}=2$. On the other
hand, by Lemma 4.9, $3K_{X}-2F$ is nef. By Proposition 4.3 (1) and (3), we
have

$(3K_{X}-2F)^{3}=27K_{X}^{3}-54K_{F}^{2}=27K_{X}^{3}-54>0.$

Thus $3K_{X}-2F$ is also big. By Corollary 2.10,
$f_{*}\mathcal{O}_{X}(4K_{X}-2F)=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}(K_{X^{\prime}}+\pi^{*}(3K_{X}-2F))$
is ample. In particular, $\mu_{\min}(f_{*}\mathcal{O}_{X}(4K))>2$. We conclude
that $\mathcal{V}_{5}=0$, and the proof is completed. ∎

####  4.2.2. General case

Now we consider the general case, without assuming $X$ is Gorenstein. We first
recall a result due to Kollár.

######  Proposition 4.11.

[7, Theorem A.1] Let $\Delta\geq 0$ be a $\mathbb{Q}$-divisor on $F$ such that
$\Delta\sim_{\mathbb{Q}}K_{F}$. Then

$\mathrm{lct}(F;\Delta)\geq\frac{1}{10}.$

######  Lemma 4.12.

If $e\geq 11$, then $K_{X}-F$ is nef and big.

###### Proof.

Since the nef cone of $X$ is closed, it is enough to prove that
$nK_{X}-(n-1)F$ is nef for any $n\geq 1$. As long as $K_{X}-F$ is nef, we
deduce that

$(K_{X}-F)^{3}=K_{X}^{3}-3K_{F}^{2}=K_{X}^{3}-3>0$

by Proposition 4.3 (1) and (3). Thus $K_{X}-F$ is big.

We prove by induction on $n$. The case $n=1$ is clear. Suppose that
$mK_{X}-(m-1)F$ is nef. Note that the $\mathbb{Q}$-divisor

$m\pi^{*}(K_{X}-F)-\frac{m}{m(e-1)+1}D^{\prime}\equiv\left(1-\frac{1}{m(e-1)+1}\right)\left(m\pi^{*}K_{X}-(m-1)F^{\prime}\right)$

is nef and big with a simple normal crossing fractional part. Let

$\mathcal{E}_{m}=f^{\prime}_{*}\mathcal{O}_{X^{\prime}}\left(K_{X^{\prime}}+{\left\lceil
m\pi^{*}(K_{X}-F)-\frac{m}{m(e-1)+1}D^{\prime}\right\rceil}\right).$

By Corollary 2.10, $\mathcal{E}_{m}$ is ample. On the other hand, since $e\geq
11$, for any $m>0$, we have

$\frac{m}{m(e-1)+1}<\frac{1}{10}.$

By Proposition 4.11, the pair $(F,\frac{m}{m(e-1)+1}D|_{F})$ is klt. In
particular,

$(\pi|_{F^{\prime}})_{*}\mathcal{O}_{F^{\prime}}\left(K_{F^{\prime}/F}-{\left\lfloor\frac{m}{m(e-1)+1}D^{\prime}|_{F^{\prime}}\right\rfloor}\right)=\mathcal{O}_{F}.$

It follows that

$\displaystyle{\mathrm{rank}\,}\mathcal{E}_{m}$
$\displaystyle=h^{0}\left.\left(F^{\prime},K_{F^{\prime}}+{\left\lceil
m\pi^{*}(K_{X}-F)-\frac{m}{m(e-1)+1}D^{\prime}\right\rceil}\right|_{F^{\prime}}\right)$
$\displaystyle=h^{0}\left(F^{\prime},(m+1)(\pi|_{F^{\prime}})^{*}K_{F}+K_{F^{\prime}/F}-{\left\lfloor\frac{m}{m(e-1)+1}D^{\prime}|_{F^{\prime}}\right\rfloor}\right)$
$\displaystyle=h^{0}\left(F,(m+1)K_{F}\right).$

Since there is an injection $\mathcal{E}_{m}\hookrightarrow
f_{*}\mathcal{O}_{X}\left((m+1)K_{X}-mF\right)$ of vector bundles of the same
rank, we deduce that $f_{*}\mathcal{O}_{X}\left((m+1)K_{X}-mF\right)$ is
ample. Let

${\mathrm{ev}}_{m+1}:f^{*}f_{*}\mathcal{O}_{X}\left((m+1)K_{X}-mF\right)\to\mathcal{O}_{X}\left((m+1)K_{X}-mF\right)$

be the evaluation morphism, and let

$\Delta_{m+1}=\mathrm{Sing}(X)\cup\mathrm{Supp}(\mathcal{O}_{X}((m+1)K_{X}-mF)/\mathrm{Im}({\mathrm{ev}}_{m+1})).$

By Lemma 2.4, $(m+1)K_{X}-mF$ is nef away from the $f$-horizontal part of
$\Delta_{m+1}$. Note that $F$ is a $(1,2)$-surface. If the $f$-horizontal part
of $\Delta_{m+1}$ is non-empty, then $m=2$ and the $f$-horizontal part of
$\Delta_{m+1}$ is the section $\Gamma$. Nevertheless, by Proposition 4.3 (1)
and (2), we have

$\left((3K_{X}-2F)\cdot\Gamma\right)=3(K_{X}\cdot\Gamma)-2\geq d-2>0.$

Thus the induction process works, and the result follows. ∎

######  Lemma 4.13.

Suppose that $e\geq 11$ and that $K_{X}-aF$ is nef and big for some rational
number $a\geq 1$. Then $K_{X}-\frac{9a+e}{20}F$ is nef and big.

###### Proof.

We may write $a=\frac{p}{q}$ with $(p,q)=1$. For any $n>1$, let
$\nu_{n}:B_{n}\to B$ be an étale cover of degree $nq$. Then $g(B_{n})=1$. Let
$X_{n}=X\times_{B}B_{n}$. Then we have the following commutative diagram

$\textstyle{X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\scriptstyle{\mu_{n}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{B_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu_{n}}$$\textstyle{B}$

Denote by $F_{n}$ a general fibre of $f_{n}:X_{n}\to B_{n}$. Then we have

$K_{X_{n}}=\mu_{n}^{*}K_{X}\equiv nqeF_{n}+D_{n},$

where $D_{n}=\mu_{n}^{*}D$. Let $\pi_{n}:X^{\prime}_{n}\to X_{n}$ be
birational modification as in §1.4 with respect to $|K_{X_{n}}|$. Since
$X_{n}$ has terminal singularities, we may write

$K_{X^{\prime}_{n}}=\pi_{n}^{*}K_{X_{n}}+E_{\pi_{n}},$

where $E_{\pi_{n}}\geq 0$ is a $\mathbb{Q}$-divisor. We may further assume
that $\pi_{n}^{*}D_{n}+E_{\pi_{n}}$ is simple normal crossing. By the
assumption,

$(qe-p)nF_{n}+D_{n}\equiv K_{X_{n}}-npF_{n}\equiv\mu_{n}^{*}(K_{X}-aF)$

Set $m={\left\lceil\frac{(qe-p)n}{10}\right\rceil}-1$. Then the
$\mathbb{Q}$-divisor

$\pi_{n}^{*}\left(K_{X_{n}}-(np+m)F_{n}-\frac{m}{(qe-p)n}D_{n}\right)\equiv\left(1-\frac{m}{(qe-p)n}\right)\pi_{n}^{*}(K_{X_{n}}-npF_{n})$

is nef and big with a simple normal crossing fractional part. Note that

$\frac{m}{(qe-p)n}<\frac{1}{10}.$

Thus we may run a similar argument as in the proof of Lemma 4.12 to deduce
that $2K_{X_{n}}-(np+m)F_{n}$ is nef, which implies that
$K_{X}-\left(\frac{a}{2}+\frac{m}{2nq}\right)F$ is nef. By the choice of $m$,
we have

$\lim\limits_{n\to+\infty}\frac{m}{nq}=\frac{e}{10}-\frac{q}{10p}=\frac{e-a}{10}.$

Taking $n\to\infty$, it follows that $K_{X}-\frac{9a+e}{20}F$ is nef. Note
that $K_{X}-aF$ is nef implies that $K_{X}^{3}\geq 3aK_{F}^{2}=3a$, and
Proposition 4.3 (3) implies that $K_{X}^{3}\geq\frac{4}{3}e$. It follows that

$\left(K_{X}-\frac{9a+e}{20}F\right)^{3}=K_{X}^{3}-\frac{27a+3e}{20}\geq\frac{3a}{2}+\frac{2e}{3}-\frac{27a+3e}{20}>0.$

Thus $K_{X}-\frac{9a+e}{20}F$ is also big, and the proof is completed. ∎

######  Proposition 4.14.

Suppose that $e\geq 11$. Then $K_{X}-\frac{e}{11}F$ is nef.

###### Proof.

Since $e\geq 11$, by Lemma 4.12, $K_{X}-F$ is nef. Take $a_{1}=1$ and define
$a_{n+1}=\frac{9a_{n}+e}{20}$ recursively. Then it is easy to see that
$\\{a_{n}\\}$ is an increasing sequence and bounded from above. Moreover, we
have

$\lim\limits_{n\to\infty}a_{n}=\frac{e}{11}.$

By Lemma 4.13, $K_{X}-a_{n}F$ is nef for every $n\geq 2$. Thus
$K_{X}-\frac{e}{11}F$ is nef. ∎

###  4.3. Proof of Theorem 4.1

Now we prove Theorem 4.1. By [18, Theorem 1.1], we have

$K_{X}^{3}\geq\frac{4}{3}\chi(\omega_{X})=\frac{4}{3}p_{g}(X)-\frac{4}{3}h^{2}(X,\mathcal{O}_{X}).$

If $h^{2}(X,\mathcal{O}_{X})=0$, then there is nothing to prove. Thus in the
following, we assume that $h^{2}(X,\mathcal{O}_{X})>0$.

We first treat the case when $p_{g}(X)\geq 4$. By Lemma 4.2, we know that
$\dim\Sigma=2$. If $d\geq p_{g}(X)$, then the theorem follows from Proposition
4.3 (3). Thus we only need to consider the case when $d=p_{g}(X)-1$. If
$p_{g}(X)\geq 16$. by Proposition 4.4 and 4.14, we have

$K_{X}^{3}\geq\frac{4}{3}\left(p_{g}(X)-1\right)+\frac{15}{11}=\frac{4}{3}p_{g}(X)+\frac{1}{33}.$

If $X$ is Gorenstein, by Proposition 4.4 and 4.10, we have

$K_{X}^{3}\geq\frac{4}{3}\left(p_{g}(X)-1\right)+\frac{3}{4}=\frac{4}{3}p_{g}(X)-\frac{7}{12},$

i.e., $3K_{X}^{3}\geq 4p_{g}(X)-\frac{7}{4}$. Since now $K_{X}^{3}>0$ is an
even integer [4, §2.2], we deduce that

$3K^{3}_{X}\geq 4p_{g}(X).$

We are left to treat the case when $X$ is Gorenstein and $p_{g}(X)\leq 3$.
Since $K_{X}^{3}>0$ is an even integer, the theorem is trivial when
$p_{g}(X)\leq 1$. We may assume that $2\leq p_{g}(X)\leq 3$, and we only need
to prove that $K_{X}^{3}\neq 2$. By Lemma 4.2, we have

$f_{*}\omega_{X}=\mathcal{O}_{B}\oplus\mathcal{L},$

where $\mathcal{L}$ is a line bundle on $B$ with $\deg\mathcal{L}=p_{g}(X)-1$.
Consider the following commutative diagram

$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{B}$

where $\nu:C\to B$ is an étale cover of degree three, and $Y=X\times_{B}C$. It
is easy to see that $g(C)=1$ and that $Y$ is Gorenstein minimal. Moreover,
$f^{\prime}:Y\to C$ is the Albanese fibration of $Y$ with general fibre
isomorphic to $F$. Note that

$f^{\prime}_{*}\omega_{Y}=\nu^{*}f_{*}\omega_{X}=\mathcal{O}_{C}\oplus\nu^{*}\mathcal{L}.$

Thus $\deg\nu^{*}\mathcal{L}=3\deg\mathcal{L}=3p_{g}(X)-3$ and
$p_{g}(Y)=h^{0}(C,f^{\prime}_{*}\omega_{Y})=3p_{g}(X)-2$.

Suppose that $p_{g}(X)=3$. Then $p_{g}(Y)=7$. Thus
$K_{Y}^{3}\geq\frac{4}{3}p_{g}(Y)=\frac{28}{3}$, which implies that

$K_{X}^{3}=\frac{1}{3}K^{3}_{Y}\geq\frac{28}{9}>3.$

Suppose that $p_{g}(X)=2$. Then $\deg\mathcal{L}=1$ and $p_{g}(Y)=4$. Since
$X$ is Gorenstein, by [4, §2.1], we know that

$p_{g}(X)+q(X)-h^{2}(X,\mathcal{O}_{X})-1=\chi(\omega_{X})>0.$

Thus $h^{2}(X,\mathcal{O}_{X})=1$ and $\chi(\omega_{X})=1$. Since
${\mathrm{Sym}}^{2}H^{0}(F,K_{F})\to H^{0}(F,2K_{F})$ is an injection, pushing
forward by $f$, we obtain an injection

$\mathcal{O}_{B}\oplus\mathcal{L}\oplus\mathcal{L}^{\otimes
2}={\mathrm{Sym}}^{2}f_{*}\omega_{X}\to f_{*}\mathcal{O}_{X}(2K_{X}).$

In particular, $f_{*}\mathcal{O}_{X}(2K_{X})$ admits an indecomposable direct
summand $\mathcal{V}_{2}$, contained in the image of $\mathcal{L}^{\otimes
2}$, such that $\mu(\mathcal{V}_{2})\geq\mu(\mathcal{L}^{\otimes 2})=2$.
Suppose on the contrary that $K_{X}^{3}=2$. By the Riemann-Roch formula, we
have

$h^{0}\left(B,f_{*}\mathcal{O}_{X}(2K_{X})\right)=h^{0}(X,2K_{X})=\frac{1}{2}K_{X}^{3}+3\chi(\omega_{X})=4.$

By Corollary 2.10, $f_{*}\mathcal{O}_{X}(2K_{X})$ is ample. We deduce that
$\mu_{\min}(f_{*}\mathcal{O}_{X}(2K_{X}))>0$ and $\deg
f_{*}\mathcal{O}_{X}(2K_{X})=4$. Moreover,
${\mathrm{rank}\,}f_{*}\mathcal{O}_{X}(2K_{X})=h^{0}(F,2K_{F})=4$. This forces
${\mathrm{rank}\,}\mathcal{V}_{2}=1$. Together with the above injection, it
implies that $f_{*}\mathcal{O}_{X}(2K_{X})$ admits another indecomposable
direct summand $\mathcal{V}_{2}^{\prime}\neq\mathcal{V}_{2}$, contained in the
image of $\mathcal{L}$, such that
$\mu(\mathcal{V}_{2}^{\prime})\geq\mu(\mathcal{L})=1$. The existence of such
$\mathcal{V}_{2}$ and $\mathcal{V}^{\prime}_{2}$ in turn forces

$f_{*}\mathcal{O}_{X}(2K_{X})=\mathcal{V}_{2}\oplus\mathcal{V}^{\prime}_{2}\oplus\mathcal{E},$

where $\deg\mathcal{V}_{2}=2$, $\mathcal{V}_{2}^{\prime}$ is a line bundle
with $\deg\mathcal{V}_{2}^{\prime}=1$, and $\mathcal{E}$ is an indecomposable
bundle of rank two with $\deg\mathcal{E}=1$. In particular, we have
$\mu_{\min}\left(f_{*}\mathcal{O}_{X}(2K_{X})\right)=\frac{1}{2}$. Since
$f^{\prime}_{*}\mathcal{O}_{Y}(2K_{Y})=\nu^{*}f_{*}\mathcal{O}_{X}(2K_{X})$,
we deduce that
$\mu_{\min}\left(f^{\prime}_{*}\mathcal{O}_{Y}(2K_{Y})\right)=\frac{3}{2}$.
However, this cannon occur by Lemma 4.8. As a result, we deduce that
$K_{X}^{3}\neq 2$. Thus the proof is completed.

##  5\. Irregular $3$-folds of general type with $q=1$: Part II

Let $X$ be a minimal $3$-fold of general type with $q(X)=1$. Let

$f:X\to B$

be the Albanese fibration of $X$, where $B$ is a smooth curve of genus
$g(B)=1$. Throughout this section, we always assume that the general fibre $F$
of $f$ is not a $(1,2)$-surface. The main result in this section is the
following theorem.

######  Theorem 5.1.

Let $X$ be as above. Suppose that $X$ is Gorenstein or $p_{g}(X)\geq 6$. Then

$K_{X}^{3}\geq\frac{4}{3}p_{g}(X),$

and the equality holds only when $X$ is Gorenstein and $p_{g}(X)=3$.

Note that the theorem is trivial when $X$ is Gorenstein and $p_{g}(X)\leq 1$,
because under this assumption we always have $K_{X}^{3}\geq 2$ [4, §2.2]. Thus
in the following, we always assume that $p_{g}(X)\geq 2$. As in §4.1, we still
have the following commutative diagram:

$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{f^{\prime}}$$\scriptstyle{\psi}$$\scriptstyle{\phi_{M}}$$\textstyle{\Sigma^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{B}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{K_{X}}}$$\scriptstyle{f}$$\textstyle{\Sigma}$

We still denote by $|M|$ the movable part of
$|{\left\lfloor\pi^{*}K_{X}\right\rfloor}|$ and by $F^{\prime}$ a general
fibre of $f^{\prime}$. Note that by [35, Theorem 1.6 and Corollary 1.9], $X$
satisfies the Severi inequality

(5.1) $K_{X}^{3}\geq 2\chi(\omega_{X})$

and the slope inequality

(5.2) $K_{X}^{3}\geq 2\deg f_{*}\omega_{X}.$

###  5.1. The case when $\dim\Sigma=1$

We start with the following proposition.

######  Proposition 5.2.

Suppose that $\dim\Sigma=1$.

  * (1)

If $p_{g}(X)=3$, then $K_{X}^{3}\geq\frac{32}{9}$.

  * (2)

If $p_{g}(X)=4$, then $K_{X}^{3}\geq\frac{27}{4}$;

  * (3)

If $p_{g}(X)\geq 5$, then $K_{X}^{3}\geq 2p_{g}(X)$.

###### Proof.

As in the proof of Proposition 3.11, we may write $M\equiv bS,$ where $S$ is a
general fibre of $\psi$. Let $\sigma:S\to S_{0}$ be the contraction onto the
minimal model $S_{0}$ of $S$.

We first consider the case when $g(\Sigma^{\prime})>0$. In this case, since
$q(X)=1$, we have $g(\Sigma^{\prime})=1$. Thus the universal property of the
Albanese map implies that $\psi=f^{\prime}$, and we may assume that
$S=F^{\prime}$ and $\Sigma^{\prime}=B$. Moreover, $g(\Sigma^{\prime})=1$
implies that $b=p_{g}(X)$. Thus it follows that

$K_{X}^{3}=(\pi^{*}K_{X})^{3}\geq b\left((\pi^{*}K_{X})^{2}\cdot
F^{\prime}\right)=p_{g}(X)K_{F}^{2}.$

If $K_{F}^{2}\geq 2$, then the above inequalities implies that

$K_{X}^{3}\geq 2p_{g}(X).$

If $K_{F}^{2}=1$, since $F$ is not a $(1,2)$-surface and $p_{g}(X)>0$, we
deduce that $p_{g}(F)=1$. Now $f_{*}\omega_{X}$ is a line bundle on $B$, and
$h^{0}(B,f_{*}\omega_{X})=p_{g}(X)\geq 2$. Thus $h^{1}(B,f_{*}\omega_{X})=0$.
By (5.2) and the Riemann-Roch theorem, we have

$K_{X}^{3}\geq 2\deg f_{*}\omega_{X}=2p_{g}(X).$

We now assume that $g(\Sigma^{\prime})=0$. In this case, $f^{\prime}|_{S}:S\to
B$ is non-trivial. Thus $q(S)>0$. Suppose first that $K_{S_{0}}^{2}\geq 4$. As
in the proof of Proposition 3.11, we have

$K_{X}^{3}\geq\frac{\left(p_{g}(X)-1\right)^{3}}{\left(p_{g}(X)\right)^{2}}K_{S_{0}}^{2}\geq\frac{4\left(p_{g}(X)-1\right)^{3}}{\left(p_{g}(X)\right)^{2}},$

and it is easy to check that the three inequalities in the proposition hold.
It remains to treat the case when $K_{S_{0}}^{2}\leq 3$. By [11, Théorème
6.1], $K_{S_{0}}^{2}\geq 2p_{g}(S)\geq 2q(S)$. This forces $p_{g}(S)=q(S)=1$.
First, $p_{g}(S)=1$ implies that $\psi_{*}\omega_{X^{\prime}}$ is a line
bundle, and $p_{g}(X)\geq 2$ further implies that
$\psi_{*}\omega_{X^{\prime}}$ is ample. Second, $q(S)=1$ implies that
$R^{1}\psi_{*}\omega_{X^{\prime}}$ is a line bundle. By the Leray spectral
sequence,
$h^{1}(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}})=h^{2}(X^{\prime},\omega_{X^{\prime}})-g(\Sigma^{\prime})=1$.
Thus $R^{1}\psi_{*}\omega_{X^{\prime}}=\omega_{\Sigma^{\prime}}$. By the Leray
spectral sequence again, we deduce that

$h^{2}(X,\mathcal{O}_{X})=h^{1}(X^{\prime},\omega_{X^{\prime}})=h^{1}(\Sigma^{\prime},\psi_{*}\omega_{X^{\prime}})+h^{0}(\Sigma^{\prime},R^{1}\psi_{*}\omega_{X^{\prime}})=0.$

By (5.1), it follows that

$K_{X}^{3}\geq 2\chi(\omega_{X})=2p_{g}(X)+2q(X)-2=2p_{g}(X).$

The whole proof is completed. ∎

###  5.2. The case when $\dim\Sigma=2$

We now consider the case when $\dim\Sigma=2$. Thus $p_{g}(X)\geq 3$. Since
$\dim\phi_{K_{X}}(F)\geq 1$, we have $p_{g}(F)\geq 2$. We still denote by $C$
a general fibre of $\psi$, where $C$ is a smooth curve of genus $g(C)\geq 2$.

As in §3.2.2, we may assume that $X^{\prime}$ and $\Sigma^{\prime}$ are
smooth. Let $S\in|M|$ be a smooth general member. Denote by $\sigma:S\to
S_{0}$ the contraction onto its minimal model. Then we have $M|_{S}\equiv dC$,
where $d=(\deg\tau)\cdot(\deg\Sigma)\geq p_{g}(X)-2$. Let $H^{\prime}$ be a
smooth curve on $\Sigma^{\prime}$ such that $S=\psi^{*}H^{\prime}$.

######  Lemma 5.3.

Suppose that $\dim\Sigma=2$ and $C$ is $f^{\prime}$-vertical. Then
$K_{X}^{3}\geq 2p_{g}(X)-2$.

###### Proof.

By the assumption, $C$ is contained in a general fibre of $f^{\prime}$. Thus
we have

$K_{X}^{3}\geq\left((\pi^{*}K_{X})\cdot
M^{2}\right)=d\left((\pi^{*}K_{X})|_{F^{\prime}}\cdot C\right)=d(K_{F}\cdot
C_{0}).$

where $C_{0}=\pi_{*}C$. Note that $(K_{F}\cdot C_{0})\geq 2$. Otherwise, we
would have $(K_{F}\cdot C_{0})=1$. By the Hodge index theorem, we deduce that
$K_{F}^{2}=1$, and $F$ is a $(1,2)$-surface, which is absurd. Now $f^{\prime}$
factors through $\psi$. It implies that $\Sigma^{\prime}$ is an irregular
surface. By [26, Theorem 7], we know that $d\geq p_{g}(X)-1$. Combine the
above inequalities together, and it follows that

$K_{X}^{3}\geq 2p_{g}(X)-2.$

Thus the proof is completed. ∎

######  Lemma 5.4.

Suppose that $\dim\Sigma=2$ and $C$ is not $f^{\prime}$-vertical.

  * (1)

If $g(H^{\prime})\geq 1$, then $K_{X}^{3}\geq 2p_{g}(X)-2$;

  * (2)

If $g(H^{\prime})=0$, then

$K_{X}^{3}\geq\min\left\\{\frac{5}{2}p_{g}(X)-5,2p_{g}(X)-\frac{10}{3}\right\\}.$

###### Proof.

Since $S$ admits a surjective morphism $f^{\prime}|_{S}:S\to B$, we know that
$q(S)>0$. By the assumption, $f^{\prime}|_{S}$ does not factor through
$\psi|_{S}:S\to H^{\prime}$.

If $g(H^{\prime})\geq 1$, then we deduce that ${\rm alb}\dim S=2$. By the same
argument as in the proof of Lemma 3.9, we deduce that $K_{X}^{3}\geq 2d$. By
[26, Theorem 7], we have $d\geq p_{g}(X)-1$. Thus it follows that

$K_{X}^{3}\geq 2p_{g}(X)-2.$

The proof of (1) is completed.

In the following, we assume that $g(H^{\prime})=0$. By [7, Corollary 2.3], the
$\mathbb{Q}$-divisor $2(\pi^{*}K_{X})|_{S}-\sigma^{*}K_{S_{0}}$ is pseudo-
effective. Thus we have

$2\left((\pi^{*}K_{X})\cdot C\right)\geq\left((\sigma^{*}K_{S_{0}})\cdot
C\right)=(K_{S_{0}}\cdot C_{0}),$

where $C_{0}=\sigma_{*}C$. If $(K_{S_{0}}\cdot C_{0})\geq 5$, then
$\left((\pi^{*}K_{X})\cdot C\right)\geq\frac{5}{2}$ and we have

$K_{X}^{3}\geq\left((\pi^{*}K_{X})\cdot
M^{2}\right)=d\left((\pi^{*}K_{X})\cdot C\right)\geq\frac{5}{2}p_{g}(X)-5.$

Thus we may assume that $(K_{S_{0}}\cdot C_{0})\leq 4$. Note that by the
adjunction, $K_{S}-2M|_{S}\equiv K_{S}-2dC$ is pseudo-effective. Thus
$K_{S_{0}}-2dC_{0}$ is pseudo-effective. It follows that

$4\geq(K_{S_{0}}\cdot C_{0})\geq 2dC_{0}^{2}\geq
2C_{0}^{2}\left(p_{g}(X)-2\right)\geq 2C_{0}^{2}.$

In particular, $C_{0}^{2}=0$, $1$ or $2$. We claim that in all cases,

$K_{S_{0}}^{2}\geq 4\chi(\omega_{S})-8.$

Suppose the claim holds. Then as in the proof of Lemma 3.8, we know that

$\chi(\omega_{X})+\chi(\omega_{S})\geq 3p_{g}(X)-3.$

Combine the above inequalities with (3.2) and (5.1). It follows that

$K_{X}^{3}\geq\frac{1}{3}K_{X}^{3}+\frac{1}{6}K_{S_{0}}^{2}\geq\frac{2}{3}\left(\chi(\omega_{X})+\chi(\omega_{S})-2\right)\geq
2p_{g}(X)-\frac{10}{3}.$

To prove the above claim, suppose first that $C_{0}^{2}=0$. Then $\psi|_{S}$
descends to a fibration $h:S_{0}\to H^{\prime}$ with $C_{0}$ as a general
fibre. By the adjunction and parity, $g(C_{0})=2$ or $3$. Since $q(S_{0})>0$,
by [34, Corollary 1], we have

$K_{S_{0}}^{2}+8\left(g(C_{0})-1\right)=K_{S_{0}/H^{\prime}}^{2}\geq 4\deg
h_{*}\omega_{S_{0}/H^{\prime}}=4\chi(\omega_{S})+4\left(g(C_{0})-1\right).$

That is,

$K_{S_{0}}^{2}\geq 4\chi(\omega_{S})-4\left(g(C_{0})-1\right)\geq
4\chi(\omega_{S})-8.$

Suppose that $C_{0}^{2}>0$. Then by parity, either $(K_{S_{0}}\cdot C_{0})=3$
and $C_{0}^{2}=1$, or $(K_{S_{0}}\cdot C_{0})=4$ and $C_{0}^{2}=2$. By the
Hodge index theorem, we always have $K_{S_{0}}^{2}\leq 9$. By [1, Lemma 14],
we know that $\chi(\omega_{S})\leq 4$ and

$K_{S_{0}}^{2}\geq 2\chi(\omega_{S})\geq 4\chi(\omega_{S})-8.$

Thus the claim holds, and the proof of the lemma is completed. ∎

######  Proposition 5.5.

Suppose that $q(X)=1$ and $\dim\Sigma=2$. Then

$K_{X}^{3}\geq\min\left\\{\frac{5}{2}p_{g}(X)-5,2p_{g}(X)-\frac{10}{3}\right\\}.$

In particular, if $p_{g}(X)\geq 6$, then $K_{X}^{3}>\frac{4}{3}p_{g}(X)$.

###### Proof.

This is just a combination of Lemma 5.3 and 5.4. ∎

###  5.3. The case when $\dim\Sigma=3$

We now treat the case when $\dim\Sigma=3$.

######  Proposition 5.6.

Suppose that $\dim\Sigma=3$. Then $K_{X}^{3}\geq 2p_{g}(X)-3$.

###### Proof.

Let $S\in|M|$ be a general member. Then $S$ is a smooth surface of general
type. Since $S$ admits a surjective morphism $f^{\prime}|_{S}:S\to B$, we know
that $q(S)>0$. Let $h:S\to B^{\prime}$ be the Stein factorization of
$f^{\prime}|_{S}$ with a general fibre $C$. Then $g(C)\geq 2$. Let
$\sigma:S\to S_{0}$ be the contraction onto the minimal model of $S$. Since
$g(B^{\prime})>0$, $h$ naturally descends to a fibration $h_{0}:S_{0}\to
B^{\prime}$ with a general fibre $C_{0}=\sigma_{*}C$.

Since $\phi_{M}|_{C}$ is a finite morphism, we know that $h^{0}(C,M|_{C})\geq
2$. In particular, $(M\cdot C)\geq 2$. By the adjunction, we deduce that

$\deg K_{C}=(K_{S}\cdot C)=\left((K_{X^{\prime}}+M)\cdot C\right)\geq 2(M\cdot
C)\geq 4.$

Thus $g(C)\geq 3$. By [34, Theorem 2], we know that

$K_{S_{0}}^{2}\geq\left(\frac{4g(C_{0})-4}{g(C_{0})}\right)\chi(\omega_{S})\geq\frac{8}{3}\chi(\omega_{S}).$

As in the proof of Proposition 3.4, we still have

$\chi(\omega_{X})+\chi(\omega_{S})\geq 4p_{g}(X)-6.$

Combine the above inequalities with (3.2) and (5.1). It follows that

$K_{X}^{3}\geq\frac{1}{4}K_{X}^{3}+\frac{3}{16}K_{S_{0}}^{2}\geq\frac{1}{2}(\chi(\omega_{X})+\chi(\omega_{S}))\geq
2p_{g}(X)-3.$

Thus the proof is completed. ∎

###  5.4. The case when $X$ is Gorenstein

Finally, we treat the case when $X$ is Gorenstein.

######  Lemma 5.7.

Suppose that $X$ is Gorenstein. If the map
${\mathrm{Sym}}^{2}H^{0}(F,K_{F})\to H^{0}(F,2K_{F})$ is injective, then

$K_{X}^{3}\geq\left(2p_{g}(F)-4\right)\deg
f_{*}\omega_{X}+h^{1}(B,f_{*}\omega_{X})\left(h^{1}(B,f_{*}\omega_{X})+1\right).$

###### Proof.

Denote $q_{f}=h^{1}(B,f_{*}\omega_{X})$. By [12, Theorem 3.1], we have

$f_{*}\omega_{X}=\mathcal{O}_{B}^{\oplus q_{f}}\oplus\mathcal{V},$

where ${\mathrm{rank}\,}\mathcal{V}=p_{g}(F)-q_{f}$ and
$h^{1}(B,\mathcal{V})=0$. By the assumption, we have
${\mathrm{Sym}}^{2}f_{*}\omega_{X}\subseteq f_{*}\omega_{X}^{\otimes 2}$.
Denote by $\mathcal{F}$ the saturation of ${\mathrm{Sym}}^{2}f_{*}\omega_{X}$
in $f_{*}\omega_{X}^{\otimes 2}$. Then we have the following short exact
sequence

$0\to\mathcal{F}\to f_{*}\omega_{X}^{\otimes 2}\to\mathcal{Q}\to 0.$

By the Kawamata-Viehweg vanishing theorem and the Leray spectral sequence, we
know that $h^{1}(B,f_{*}\omega_{X}^{\otimes 2})=0$. Via the long exact
sequence, we deduce that $h^{1}(B,\mathcal{Q})=0$ and
$h^{0}(B,\mathcal{Q})\geq h^{1}(B,\mathcal{F})$. Thus by the Riemann-Roch
theorem, we have

$\deg f_{*}\omega_{X}^{\otimes
2}=\deg\mathcal{F}+\deg\mathcal{Q}=h^{0}(B,\mathcal{F})-h^{1}(B,\mathcal{F})+h^{0}(B,\mathcal{Q})\geq
h^{0}(B,\mathcal{F}).$

Note that

$\deg\left({\mathrm{Sym}}^{2}f_{*}\omega_{X}\right)=\left(p_{g}(F)+1\right)\deg
f_{*}\omega_{X}$

and

$h^{1}\left(B,{\mathrm{Sym}}^{2}f_{*}\omega_{X}\right)\geq
h^{1}\left(B,{\mathrm{Sym}}^{2}\mathcal{O}_{B}^{\oplus
q_{f}}\right)=\frac{q_{f}(q_{f}+1)}{2}.$

By the Riemann-Roch theorem, we know that

$h^{0}(B,\mathcal{F})\geq
h^{0}\left(B,{\mathrm{Sym}}^{2}f_{*}\omega_{X}\right)=\left(p_{g}(F)+1\right)\deg
f_{*}\omega_{X}+\frac{q_{f}(q_{f}+1)}{2}.$

On the other hand, since $h^{1}(B,f_{*}\omega_{X}^{\otimes 2})=0$, by the
Riemann-Roch formula, we know that

$\deg f_{*}\omega_{X}^{\otimes 2}=h^{0}(B,f_{*}\omega_{X}^{\otimes
2})=h^{0}(X,2K_{X})=\frac{1}{2}K_{X}^{3}+3\chi(\omega_{X}).$

Moreover, since $q(X)=1$, by the Leray spectral sequence, we have

$\chi(\omega_{X})=p_{g}(X)-h^{2}(X,\mathcal{O}_{X})\leq
h^{0}(B,f_{*}\omega_{X})-h^{1}(B,f_{*}\omega_{X})=\deg f_{*}\omega_{X}.$

Combine the above (in)equalities together, and it follows that

$K_{X}^{3}\geq 2\deg f_{*}\omega_{X}^{\otimes
2}-6\chi(\omega_{X})\geq\left(2p_{g}(F)-4\right)\deg
f_{*}\omega_{X}+q_{f}(q_{f}+1).$

The proof is completed. ∎

######  Lemma 5.8.

Suppose that $X$ is Gorenstein with $p_{g}(X)=2$. Then $K_{X}^{3}\geq 4$.

###### Proof.

Since $K_{X}^{3}$ is an even integer, to prove this lemma, it suffices to
prove that $K_{X}^{3}>2$. Now
$\chi(\omega_{X})=p_{g}(X)-h^{2}(X,\mathcal{O}_{X})>0$. By the Leray spectral
sequence, we know that $h^{1}(B,f_{*}\omega_{X})\leq
h^{2}(X,\mathcal{O}_{X})\leq 1$.

If $h^{1}(B,f_{*}\omega_{X})=0$, by (5.2) and the Riemann-Roch theorem, we
have

$K_{X}^{3}\geq 2\deg f_{*}\omega_{X}=2h^{0}(B,f_{*}\omega_{X})=2p_{g}(X).$

From now on, we assume that $h^{1}(B,f_{*}\omega_{X})=1$. By the Riemann-Roch
theorem, $\deg
f_{*}\omega_{X}=h^{0}(B,f_{*}\omega_{X})-h^{1}(B,f_{*}\omega_{X})=1$. Thus
$f_{*}\omega_{X}$ cannot be a line bundle. In particular,
$p_{g}(F)={\mathrm{rank}\,}f_{*}\omega_{X}\geq 2$. If $p_{g}(F)=2$, by [12,
Theorem 2.1], $f_{*}\omega_{X}=\mathcal{O}_{B}\oplus\mathcal{L}$ where
$\mathcal{L}$ is a line bundle on $B$ with $\deg\mathcal{L}=1$. By [25,
Corollary 3.5], $K_{X}-F$ is pseudo-effective. By [28, Theorem 1.4] and [5,
Lemma 4.1], $K_{X}-\epsilon F$ is nef for some $\epsilon>0$. Thus

$K_{X}^{3}>\left((K_{X}-\epsilon F)^{2}\cdot
K_{X}\right)\geq\left((K_{X}-\epsilon F)^{2}\cdot F\right)=K_{F}^{2}\geq 2,$

where the last inequality holds since $F$ is not a $(1,2)$-surface. If
$p_{g}(F)=3$, then the natural map ${\mathrm{Sym}}^{2}H^{0}(F,K_{F})\to
H^{0}(F,2K_{F})$ is injective. Thus by Lemma 5.7, we have

$K_{X}^{3}\geq 2\deg f_{*}\omega_{X}+2=4.$

If $p_{g}(F)\geq 4$, since $f_{*}\omega_{X}^{\otimes m}$ is ample for any
$m\geq 2$ [5, Lemma 4.1], by [28, Main Theorem 1] and the argument in the
proof of [28, Proposition 2.1] (see the inequality $(*)$ there), we know that

$K_{X}^{3}>\left(\frac{4p_{g}(F)-8}{p_{g}(F)}\right)\deg f_{*}\omega_{X}\geq
2\deg f_{*}\omega_{X}=2.$

As a result, we always have $K_{X}^{3}>2$, and the proof is completed. ∎

######  Proposition 5.9.

Suppose that $X$ is Gorenstein. Then

$K_{X}^{3}\geq\left\\{\begin{array}[]{ll}4,&p_{g}(X)=2;\\\
2p_{g}(X)-2,&p_{g}(X)\geq 3.\end{array}\right.$

###### Proof.

As $X$ is Gorenstein, we know that $K_{X}^{3}>0$ is even [4, §2.2]. Denote by
$\Sigma$ the canonical image of $X$. If $\dim\Sigma=3$, then the result
follows from Proposition 5.6. If $\dim\Sigma=2$, then the result follows from
Proposition 5.5. If $\dim\Sigma=1$, then the result follows from Proposition
5.2 for $p_{g}(X)\geq 3$ and Lemma 5.8 for $p_{g}(X)=2$. Thus the whole proof
is completed. ∎

###  5.5. Proof of Theorem 5.1

Now it is clear that Theorem 5.1 is a combination of Proposition 5.2, 5.5, 5.6
and 5.9.

##  6\. Proof of main theorems

We are now ready to prove the three main theorems in this paper.

###  6.1. Proof of Theorem 1.1

Since the canonical volume and the geometric genus are birational invariants,
we may assume that $X$ is minimal.

If $q(X)\geq 2$, the result follows from Theorem 3.1. If $q(X)=1$, the result
follows from Theorem 4.1 and 5.1.

###  6.2. Proof of Theorem 1.2

As in the proof of Theorem 1.1, we may assume that $X$ is minimal with
$K_{X}^{3}=\frac{4}{3}p_{g}(X)$ and $p_{g}(X)\geq 16$. By Theorem 3.1,
$q(X)=1$. Thus the general Albanese fibre $F$ is a minimal surface of general
type. By Theorem 5.1, $F$ is a $(1,2)$-surface. Then we apply Theorem 4.1 to
deduce that $h^{2}(X,\mathcal{O}_{X})=0$. As a result, we have
$K_{X}^{3}=\frac{4}{3}\chi(\omega_{X})$. By [18, Theorem 1.3], the canonical
model of $X$ has an explicit description.

###  6.3. Proof of Theorem 1.3

We may again assume that $X$ is minimal. Then the result just follows from
Theorem 3.1.

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