# Kawamata–Miyaoka type inequality for canonical $\mathbb{Q}$-Fano varieties
II: terminal $\mathbb{Q}$-Fano threefolds

Haidong Liu  Haidong Liu, Sun Yat-Sen University, Department of Mathematics,
Guangzhou, 510275, China  liuhd35<EMAIL_ADDRESS>https://sites.google.com/view/liuhaidong and  Jie Liu  Jie Liu, Institute of
Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of
Sciences, Beijing, 100190, China<EMAIL_ADDRESS>http://www.jliumath.com

###### Abstract.

We prove an optimal Kawamata–Miyaoka type inequality for terminal
$\mathbb{Q}$-Fano threefolds with Fano index $\geq 3$. As an application, we
obtain that any terminal $\mathbb{Q}$-Fano threefold $X$ satisfies the
following Kawamata–Miyaoka type inequality

$c_{1}(X)^{3}<3c_{2}(X)c_{1}(X).$

###### Key words and phrases:

Kawamata–Miyaoka type inequality, $\mathbb{Q}$-Fano threefolds, second Chern
class

###### 2020 Mathematics Subject Classification:

Primary 14J45; Secondary 14J10, 14J30.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Sarkisov link
  4. 4 Case $q\mathbb{Q}(X)=5$
  5. 5 Case $q\mathbb{Q}(X)=4$
  6. 6 Case $q\mathbb{Q}(X)=8$
  7. A Database for $\mathbb{Q}$-Fano threefolds with large $q\mathbb{Q}$

##  1\. Introduction

A normal projective variety is called _Fano_ if its anti-canonical divisor is
ample. A Fano variety is called _$\mathbb{Q}$ -Fano_ if it is
$\mathbb{Q}$-factorial and its Picard number is one. According to the minimal
model program, $\mathbb{Q}$-Fano varieties with terminal singularities are one
of the building blocks of algebraic varieties and they form a bounded family
(see [Kaw92], [KMMT00] and [Bir21]).

Thanks to Reid’s orbifold Riemann–Roch formula ([Rei87]), terminal
$\mathbb{Q}$-Fano threefolds have been studied intensively in the past two
decades, see [ABR02, Suz04, BS07a, BS07b, Pro10, Pro13, CJ16, BK22, Pro22a,
Pro22b] and the references therein. In particular, all the possibilities of
their numerical types are obtained in the _Graded Ring Database_ (Grdb for
short, [BK09]).

One of the key ingredients in the computation of possible numerical types of
terminal $\mathbb{Q}$-Fano threefolds is an effective version of
_Kawamata–Miyaoka type inequality_ (see [Kaw92, Proposition 1], [Suz04,
Theorem 1.7] and [BK22, Theorem 2.6] for more details). In our previous paper
[LL23], we established such an effective Kawamata–Miyaoka type inequality for
terminal $\mathbb{Q}$-Fano varieties in arbitrary dimension. In particular,
using a refined argument and Reid’s orbifold Riemann–Roch formula, we have
proved the following inequality in dimension three, which significantly
improves the known ones in the literature.

######  Theorem 1.1 ([LL23, Theorem 1.2]).

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold. Then we have

$c_{1}(X)^{3}\leq\frac{25}{8}c_{2}(X)c_{1}(X).$

In this paper, we aim to pursue further in this direction. To give a precise
statement, let us recall the definition of Fano indices. In general, let $X$
be a singular Fano variety. We can define the Fano index of $X$ in two
different ways:

$\begin{split}q\mathbb{Q}(X)&\coloneqq\max\\{q\mid-
K_{X}\sim_{\mathbb{Q}}qA,\quad A\in\operatorname{Cl}X\\},\\\
qW(X)&\coloneqq\max\\{q\mid-K_{X}\sim qB,\quad
B\in\operatorname{Cl}X\\}.\end{split}$

If we assume in addition that $X$ has at worst terminal singularities, then
the Weil divisor class group $\operatorname{Cl}(X)$ is finitely generated and
the numerical equivalence for $\mathbb{Q}$-Cartier Weil divisors coincides
with the $\mathbb{Q}$-linear equivalence. This implies that both
$q\mathbb{Q}(X)$ and $qW(X)$ are positive integers.

The main result of this paper is the following optimal Kawamata–Miyaoka type
inequality for terminal $\mathbb{Q}$-Fano threefolds with large $q\mathbb{Q}$.

######  Theorem 1.2 (Corollary 2.3, Theorems 4.1 and 5.1).

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold such that
$q\mathbb{Q}(X)\geq 3$. Then we have

$c_{1}(X)^{3}\leq\frac{121}{41}c_{2}(X)c_{1}(X)$

and the equality holds if and only if $X$ is isomorphic to
$\mathbb{P}(1,2,3,5)$.

As an immediate application, combining Theorem 1.2 with [LL23, Theorem 4.4]
yields the following generalisation of Theorem 1.1.

######  Theorem 1.3.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold. Then we have

$c_{1}(X)^{3}<3c_{2}(X)c_{1}(X).$

The main idea of the proof of Theorem 1.2 is as follows. Let $X$ be a terminal
$\mathbb{Q}$-Fano threefold with $q\mathbb{Q}(X)\geq 3$ which does not satisfy
the inequality in Theorem 1.2. Using the orbifold Riemann–Roch theorem and
[LL23, Corollary 4.3], we have the following three possibilities for $X$ (see
also [LL23, Theorem 4.4 and Remark 4.5] and Corollary 2.3):

  1. (1.3.1)

$qW(X)\not=q\mathbb{Q}(X)=4$;

  2. (1.3.2)

$qW(X)=q\mathbb{Q}(X)=4$ and $\mathcal{R}_{X}=\\{7,13\\}$;

  3. (1.3.3)

$qW(X)=q\mathbb{Q}(X)=5$ and $\mathcal{R}_{X}=\\{3,7^{2}\\}$;

where $\mathcal{R}_{X}$ denotes the _local index basket_ of $X$ (§ 2.1). The
first case can be excluded by the possible numerical types of $X$ in Table
LABEL:table.qWnot=qQ=4, which is obtained by Y. Prokhorov using computer
computation ([Pro22b, Proposition 3.3]). Our main contribution of this paper
is to exclude the remaining two cases. To this end, we will introduce two
different approaches. In the first one we use the theory of foliations to rule
the last case out and it is relatively simple (§ 4). The second case is much
more difficult and we will follow a second approach developed by Y. Prokhorov
in [Pro10] by using the Sarkisov link (§ 5).

Theorem 1.3 can be directly applied to show that 13559 numerical types in the
Grdb of [BK09] do not actually occur for terminal $\mathbb{Q}$-Fano threefolds
and the Hilbert series of a terminal $\mathbb{Q}$-Fano threefold always lies
in the list $\mathcal{F}_{ss}$ [BK22, Theorem 1.2]. On the other hand, if
$q\mathbb{Q}(X)\geq 9$, then [BS07a, § 4] and [Pro10] show that there are
exactly nine numerical types which can be geometrically realised (№ 23-31 in
Table LABEL:table.qgeq6). However, the situation becomes much more complicated
for smaller $q\mathbb{Q}$. For instance, by Theorem 1.3, if
$q\mathbb{Q}(X)=8$, there are four possibilities (№ 19-22 in Table
LABEL:table.qgeq6) for the numerical type of $X$ with the corresponding local
index baskets as follows:

$\\{3^{2},5\\},\quad\\{3,7\\},\quad\\{5,7\\},\quad\\{3,5,11\\}.$

It is known that the first three cases can be geometrically realised by
appropriate examples [BS07a, Table 1] and we will treat the last case in § 6
(see № 22 in Table LABEL:table.qgeq6 and Theorem 6.1), from which we derive
the following result.

######  Theorem 1.4.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold with $q\mathbb{Q}(X)=8$.
Then its local index basket $\mathcal{R}_{X}$ cannot be $\\{3,5,11\\}$.

Finally, we remark that Theorem 1.3 cannot be generalised to canonical
$\mathbb{Q}$-Fano threefolds as $\mathbb{P}(1,3,7,11)$ shows ([BK22, 4.5]),
and our proof of Theorem 1.2 is not completely independent of previously known
non-existence results in the literature (see the proof of Corollary 2.3 and
Remark 4.4 for more details) and there are also some non-existence results,
which cannot be recovered by Theorem 1.2 (e.g. № 5 and № 8 in [Pro13, Lemma
8.1] and № 5, № 8 and № 9 in [Pro13, Lemma 9.1]).

### Acknowledgements

We would like to thank Y. Prokhorov and C. Jiang for helpful communications
and suggestions. In particular, we thank Y. Prokhorov for allowing us to
include Table LABEL:table.qWnot=qQ=4 in our appendix. H. Liu is supported in
part by the National Key Research and Development Program of China (No.
2023YFA1009800). J. Liu is supported by the National Key Research and
Development Program of China (No. 2021YFA1002300), the CAS Project for Young
Scientists in Basic Research (No. YSBR-033), the NSFC grant (No. 12288201) and
the Youth Innovation Promotion Association CAS.

##  2\. Preliminaries

Throughout this paper, we work over $\mathbb{C}$ and varieties are always
supposed to be irreducible. We will freely use the terminology of [KM98] for
birational geometry, especially the minimal model program (MMP for short).

###  2.1. Orbifold Riemann–Roch formula

Let $X$ be a terminal Fano threefold and $q\coloneqq qW(X)$. According to Reid
[Rei87], there is a basket of orbifold points

$\mathbf{B}_{X}=\\{\left(r_{i},b_{i}\right)\mid
i=1,\cdots,s;0<b_{i}\leq\frac{r_{i}}{2};b_{i}\text{ is coprime to }r_{i}\\}$

associated to $X$, where a pair $\left(r_{i},b_{i}\right)$ corresponds to an
orbifold point $Q_{i}$ of type $\frac{1}{r_{i}}\left(1,-1,b_{i}\right)$.
Denote by $\mathcal{R}_{X}$ the collection of $r_{i}$ (permitting weights)
appearing in $\mathbf{B}_{X}$ and simply write it as a set of integers whose
weights appear in superscripts, say for example,

$\mathcal{R}_{X}=\\{3,7,7\\}=\\{3,7^{2}\\}.$

Note that $r_{X}\coloneqq\operatorname{lcm}(\mathcal{R}_{X})$ coincides with
the Gorenstein index of $X$. Let $A$ be a $\mathbb{Q}$-Cartier Weil divisor on
$X$ such that $-K_{X}\sim qA$. According to [Rei87, (10.3)],

$\chi(tA)=1+\frac{t(q+t)(q+2t)}{12}A^{3}+\frac{t}{12q}c_{2}(X)c_{1}(X)+\sum_{Q\in\mathbf{B}_{X}}c_{Q}(tA)$
(2.0.1)

for $t\in\mathbb{Z}$; here if the orbifold point $Q$ is of type
$\frac{1}{r_{Q}}(1,-1,b_{Q})$ and $0\leq i_{Q,t}<r_{Q}$ is the integer
uniquely determined by $qi_{Q,t}\equiv-t\mod r_{Q}$, then

$c_{Q}(tA)=-\frac{i_{Q,t}(r^{2}_{Q}-1)}{12r_{Q}}+\sum_{j=0}^{i_{Q,t}-1}\frac{\overline{jb_{Q}}(r_{Q}-\overline{jb_{Q}})}{2r_{Q}},$

where the symbol $\overline{\bullet}$ means the smallest residue mod $r_{Q}$
and $\sum_{j=0}^{-1}\coloneqq 0$.

If $t=-q$, then (2.0.1) and Serre’s duality imply that

$1=\frac{1}{24}c_{2}(X)c_{1}(X)+\frac{1}{24}\sum_{Q\in\mathbf{B}_{X}}(r_{Q}-\frac{1}{r_{Q}}).$
(2.0.2)

If $q\geq 3$ and $-q<t<0$, then $\chi(tA)=0$ by the Kawamata–Viehweg vanishing
theorem. Hence (2.0.1) implies that

$1+\frac{t(q+t)(q+2t)}{12}A^{3}+\frac{t}{12q}c_{2}(X)c_{1}(X)+\sum_{Q\in\mathbf{B}_{X}}c_{Q}(tA)=0$
(2.0.3)

for $-q<t<0$. In particular, if $q\geq 3$ and $t=-1$, then we obtain

$A^{3}=\frac{12}{(q-1)(q-2)}\left(1-\frac{c_{2}(X)c_{1}(X)}{12q}+\sum_{Q\in\mathbf{B}_{X}}c_{Q}(-A)\right).$
(2.0.4)

Moreover, the degree $A^{3}$ and the Gorenstein index $r_{X}$ have the
following relations (see [Suz04, Lemma 1.2] for example).

$\gcd(r_{X},q)=1\text{ and }r_{X}\cdot A^{3}\in\mathbb{Z}_{\geq 0}.$ (2.0.5)

###  2.2. Fano indices

We collect in the following some basic facts about the Fano indices of
terminal Fano threefolds. We refer the reader to [Suz04], [Pro10, Lemma 3.2
and Lemma 3.3] and [Pro22b, Proposition 3.3] for more details.

######  Lemma 2.1.

Let $X$ be a terminal Fano threefold.

  1. (2.1.1)

$qW(X)|q\mathbb{Q}(X)$.

  2. (2.1.2)

If $-K_{X}\sim_{\mathbb{Q}}qA$ for some Weil divisor $A$, then
$q|q\mathbb{Q}(X)$.

  3. (2.1.3)

If $-K_{X}\sim qB$ for some Weil divisor $B$, then $q|qW(X)$.

  4. (2.1.4)

If $q\mathbb{Q}(X)\geq 5$, then $qW(X)=q\mathbb{Q}(X)$.

  5. (2.1.5)

$qW(X),q\mathbb{Q}(X)\in\\{1,\dots,9,11,13,17,19\\}$.

###### Proof.

The first three statements follow from [Pro10, Lemma 3.2]. For the last two
statements, without loss of generality we may assume that $q\mathbb{Q}(X)\geq
5$ and let $A$ be a $\mathbb{Q}$-Cartier Weil divisor such that
$-K_{X}\sim_{\mathbb{Q}}q\mathbb{Q}(X)A$. Let $\mu\colon X^{\prime}\rightarrow
X$ be a small $\mathbb{Q}$-factorialization (see [BCHM10, Corollary 1.4.4]).
Then we have $-K_{X^{\prime}}\sim_{\mathbb{Q}}q\mathbb{Q}(X)A^{\prime}$, where
$A^{\prime}=\mu^{*}A$ is a nef and big $\mathbb{Q}$-Cartier Weil divisor.
Running a MMP $g\colon X^{\prime}\dashrightarrow X^{\prime\prime}$ yields a
Mori fibre space $f\colon X^{\prime\prime}\rightarrow Z$. Denote by
$A^{\prime\prime}$ the strict transform of $A^{\prime}$ on $X^{\prime\prime}$.
Then we obtain

$-K_{X^{{}^{\prime\prime}}}\sim_{\mathbb{Q}}q\mathbb{Q}(X)A^{\prime\prime}.$

If $\dim(Z)>0$, then the general fibre $F$ of $f$ is a nonsingular Fano
variety such that
$-K_{F}\sim_{\mathbb{Q}}q\mathbb{Q}(X)A^{\prime\prime}|_{F}$, which is
impossible as $\dim F\leq 2$ and $A^{\prime\prime}|_{F}$ is Cartier. So $Z$ is
a point and $X^{\prime\prime}$ is a terminal $\mathbb{Q}$-Fano threefold.

By (2.1.2), we have $q\mathbb{Q}(X)|q\mathbb{Q}(X^{\prime\prime})$ and then it
follows from [Pro10, Theorem 1.4] and [Pro22b, Proposition 3.3] that we must
have

$q\mathbb{Q}(X)=qW(X^{\prime\prime})=q\mathbb{Q}(X^{\prime\prime})\in\\{5,\dots,9,11,13,17,19\\}.$

Let $m$ be the order of $K_{X^{\prime\prime}}+q\mathbb{Q}(X)A^{\prime\prime}$.
By [Pro22b, Proposition 3.4], we have $(m,q\mathbb{Q}(X^{\prime\prime}))=1$.
On the other hand, by the Negativity Lemma, we also have

$g^{*}(K_{X^{\prime\prime}}+q\mathbb{Q}(X)A^{\prime\prime})=K_{X}+q\mathbb{Q}(X)A.$

So we obtain $m(K_{X}+q\mathbb{Q}(X)A)\sim 0$. Thus it follows from [Pro10,
Lemma 3.2 (iv)] that $qW(X)=q\mathbb{Q}(X)$. ∎

###  2.3. Algorithm and numerical types

Let $X$ be a terminal Fano threefold. Then $X$ is nonsingular in codimension
two. If $T$ is a $\mathbb{Q}$-Cartier Weil divisor such that
$T\sim_{\mathbb{Q}}0$, then there exists an étale in codimension two finite
cover $\mu\colon Y\rightarrow X$ such that $\mu^{*}T\sim 0$, which is called
the _index one cover defined by $T$_ (see [Rei87, (3.6)] or [KM98, Definition
2.52]). Since $Y$ is also a terminal Fano threefold by [KM98, Proposition
5.20], it follows from the boundedness of terminal Fano threefolds [KMMT00,
Theorem 1.2] that there exists an étale in codimension two finite cover
$\pi\colon X^{\prime}\rightarrow X$ such that $\operatorname{Cl}(X^{\prime})$
is torsion free. In particular, we have
$qW(X^{\prime})=q\mathbb{Q}(X^{\prime})$. Let $A$ be a $\mathbb{Q}$-Cartier
Weil divisor on $X$ such that $-K_{X}\sim_{\mathbb{Q}}q\mathbb{Q}(X)A$. Then
we get

$-K_{X^{\prime}}=-\pi^{*}K_{X}\sim q\mathbb{Q}(X)\pi^{*}A.$

Let $A^{\prime}$ be a $\mathbb{Q}$-Cartier Weil divisor on $X^{\prime}$ such
that $-K_{X^{\prime}}\sim qW(X^{\prime})A^{\prime}$. By Lemma 2.1, there
exists a positive integer $n$ such that

$\pi^{*}A\sim nA^{\prime}\quad\textup{and}\quad
qW(X^{\prime})=q\mathbb{Q}(X^{\prime})=nq\mathbb{Q}(X).$

Throughout this paper, we are mainly interested in the case where $X$ is a
terminal $\mathbb{Q}$-Fano threefold. In particular, we have an effective
bound from [LL23] for the following number

$b_{X}\coloneqq\frac{c_{1}(X)^{3}}{c_{2}(X)c_{1}(X)}.$

Since $\pi\colon X^{\prime}\rightarrow X$ is étale in codimension two, we have
$b_{X^{\prime}}=b_{X}$. Now with the help of a computer program or using the
Grdb, we get the following result, which is our starting point.

######  Lemma 2.2.

Let $X$ be a terminal Fano threefold such that $qW(X)=q\mathbb{Q}(X)\geq 3$
and let $A$ be a $\mathbb{Q}$-Cartier Weil divisor such that $-K_{X}\sim
qW(X)A$. If

$\frac{121}{41}\leq b_{X}\leq c$

for the constant $c$ in the following table:

$qW(X)$ |  $3$ |  $4$ |  $5$ |  $6$ |  $7$ |  $8$ |  $\geq 9$
---|---|---|---|---|---|---|---
c |  $3$ |  $\frac{64}{21}$ |  $\frac{25}{8}$ |  $\frac{16}{5}$ |  $\frac{49}{15}$ |  $\frac{256}{77}$ |  $4$

then one of the following holds:

  1. (2.2.1)

$qW(X)=3$ and $b_{X}=3$;

  2. (2.2.2)

$qW(X)=4$, $\mathcal{R}_{X}=\\{7,13\\}$ and $b_{X}=3$ (№ 41313 in [BK09]);

  3. (2.2.3)

$qW(X)=5$, $\mathcal{R}_{X}=\\{3,7^{2}\\}$ and $b_{X}=25/8$ (№ 41436 in
[BK09]);

  4. (2.2.4)

$qW(X)=5$, $\mathcal{R}_{X}=\\{4,7\\}$ and $b_{X}=3$ (№ 41449 in [BK09]);

  5. (2.2.5)

$qW(X)=7$, $\mathcal{R}_{X}=\\{2^{2},8\\}$ and $b_{X}=49/15$ (№ 41490 in
[BK09]);

  6. (2.2.6)

$qW(X)=11$, $\mathcal{R}_{X}=\\{2,3,5\\}$ and $b_{X}=121/41$ (№ 41510 in
[BK09] or № 25 in Table LABEL:table.qgeq6).

###### Proof.

We use a computer program written in Python, whose algorithm is sketched as
follows.

Firstly, as $c_{2}(X)c_{1}(X)>0$ by [IJL23, Corollary 7.3] (cf. [Kaw92,
Proposition 1] and [KMMT00, Theorem 1.2]), we can list huge but finitely many
possibilities of $\mathcal{R}_{X}$ and $c_{2}(X)c_{1}(X)$ satisfying (2.0.2).

Secondly, for each $q=qW(X)\geq 3$, we calculate $A^{3}$ by (2.0.4) and pick
up those satisfying (2.0.3) and (2.0.5).

Finally, we find out all candidates satisfying $121/41\leq b_{X}\leq c$, which
are listed as above. ∎

######  Corollary 2.3.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold such that
$q\mathbb{Q}(X)\geq 3$. If $b_{X}\geq 121/41$, then either
$X\cong\mathbb{P}(1,2,3,5)$ or one of the following cases holds:

  1. (2.3.1)

$qW(X)=q\mathbb{Q}(X)=4$, $\mathcal{R}_{X}=\\{7,13\\}$ and $b_{X}=3$ (№ 41313
in [BK09]);

  2. (2.3.2)

$qW(X)=q\mathbb{Q}(X)=5$, $\mathcal{R}_{X}=\\{3,7^{2}\\}$ and $b_{X}=25/8$ (№
41436 in [BK09]).

###### Proof.

Let $\pi\colon X^{\prime}\rightarrow X$ be an étale in codimension two finite
cover such that $\operatorname{Cl}(X^{\prime})$ is torsion free. Then we have
$qW(X^{\prime})=q\mathbb{Q}(X^{\prime})\geq 3$. Moreover, it follows from
[LL23, Theorem 1.1, Corollary 4.3 and Theorem 4.4] that the numerical type of
$X^{\prime}$ appears as one of (2.2.2)-(2.2.6) in Lemma 2.2. In particular, as
$q\mathbb{Q}(X)\geq 3$ and $qW(X^{\prime})=nq\mathbb{Q}(X)$ for some
$n\in\mathbb{Z}_{>0}$, we obtain $n=1$ and

$q\mathbb{Q}(X)=qW(X^{\prime})=q\mathbb{Q}(X^{\prime}).$

If $q\mathbb{Q}(X)=4$ and $qW(X)\not=4$, then it follows from Table
LABEL:table.qWnot=qQ=4 that $b_{X}\leq 8/3$, which is impossible by our
assumption. Hence, we must have $qW(X)=q\mathbb{Q}(X)=4$ in our situation and
then applying Lemma 2.2 to $X$ shows that $X$ appears as (2.2.2) in Lemma 2.2.

If $q\mathbb{Q}(X)\geq 5$, then Lemma 2.1 implies $qW(X)=q\mathbb{Q}(X)$ and
thus applying Lemma 2.2 to $X$ shows that the numerical type of $X$ appears as
one of (2.2.3)-(2.2.6) in Lemma 2.2. Note that the cases (2.2.4) and (2.2.5)
are excluded in [Pro13, 7.5] and [Pro13, Claim 6.5], respectively. Moreover,
if the numerical type of $X$ appears as (2.2.6) in Lemma 2.2, then we have
$\dim|-K_{X}|=23$ by [Pro10, Proposition 3.6] (see also [BK09, № 41510]).
Hence, the variety $X$ is isomorphic to $\mathbb{P}(1,2,3,5)$ by [Pro10,
Theorem 1.4]. ∎

######  Lemma 2.4.

Let $X$ be a terminal Fano threefold such that $qW(X)=q\mathbb{Q}(X)=3$ and
$b_{X}\leq 3$. Let $A$ be a $\mathbb{Q}$-Cartier Weil divisor on $X$ such that
$-K_{X}\sim 3A$. Then we have $\dim|A|\leq 4$.

###### Proof.

Using the same computer program as that in the proof of Lemma 2.2 or the Grdb,
we can derive that the remaining case satisfying $qW(X)=q\mathbb{Q}(X)=3$,
$b_{X}\leq 3$ and $\dim|A|\geq 4$ is either $\mathcal{R}_{X}=\\{4\\}$ or $X$
has at worst Gorenstein singularities. In both cases, we have $\dim|A|=4$. ∎

###  2.4. Torsion part of $\operatorname{Cl}(X)$

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold. If $q\mathbb{Q}(X)\geq 8$,
then we have $qW(X)=q\mathbb{Q}(X)\geq 8$ by Lemma 2.1 and so
$\operatorname{Cl}(X)$ is torsion free by [Pro10, Lemma 3.5]. On the other
hand, if $5\leq q\mathbb{Q}(X)\leq 7$ and the torsion subgroup
$\operatorname{Cl}(X)_{t}$ of $\operatorname{Cl}(X)$ is non-trivial, then the
possibilities of the numerical types of $X$ are given in [Pro22b, Propoistion
3.4]. In particular, combining it with Corollary 2.3 yields the following
result.

######  Proposition 2.5 ([Pro22b, Proposition 3.4]).

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold such that
$q\mathbb{Q}(X)\geq 5$ and $\operatorname{Cl}(X)_{t}$ is non-trivial. Then
$\operatorname{Cl}(X)_{t}$ is cyclic of order $d$.

Moreover, let $T$ be the generator of $\operatorname{Cl}(X)_{t}$ and let
$\pi\colon X^{\prime}\rightarrow X$ be the index one cover defined by $T$.
Then $q\mathbb{Q}(X^{\prime})=q\mathbb{Q}(X)$ and one of the following cases
holds:

Table 2. $q\mathbb{Q}(X)\geq 5$ with $\operatorname{Cl}(X)_{t}\not=\\{e\\}$ |  |  |  |  |  |  |  |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---
№ |  $q\mathbb{Q}$ |  $d$ |  $\mathcal{R}_{X}$ |  $\mathcal{R}_{X^{\prime}}$ |  $|A|$ |  $|2A|$ |  $|3A|$ |  $|4A|$ |  $|A^{\prime}|$ |  $|2A^{\prime}|$ |  $|3A^{\prime}|$ |  $|4A^{\prime}|$
1 |  $5$ |  $2$ |  $\\{2,4,14\\}$ |  $\\{4^{2},7\\}$ |  $0$ |  $0$ |  $0$ |  $1$ |  $0$ |  $0$ |  $1$ |  $3$
2 |  $5$ |  $3$ |  $\\{2,9^{2}\\}$ |  $\\{2^{3},3^{2}\\}$ |  $0$ |  $0$ |  $0$ |  $1$ |  $0$ |  $2$ |  $4$ |  $7$
3 |  $5$ |  $2$ |  $\\{4^{2},12\\}$ |  $\\{2,4^{2},6\\}$ |  $0$ |  $0$ |  $1$ |  $3$ |  $0$ |  $1$ |  $3$ |  $7$
4 |  $5$ |  $2$ |  $\\{2^{2},4,8\\}$ |  $\\{2,4\\}$ |  $0$ |  $1$ |  $2$ |  $5$ |  $1$ |  $3$ |  $6$ |  $11$
5 |  $5$ |  $2$ |  $\\{2,4^{2},6\\}$ |  $\\{2^{2},3\\}$ |  $0$ |  $1$ |  $3$ |  $7$ |  $1$ |  $4$ |  $8$ |  $14$
6 |  $7$ |  $2$ |  $\\{2,6,10\\}$ |  $\\{2^{2},3,5\\}$ |  $0$ |  $0$ |  $0$ |  $1$ |  $0$ |  $1$ |  $2$ |  $4$
7 |  $7$ |  $2$ |  $\\{2^{2},3,4,8\\}$ |  $\\{2,3^{2},4\\}$ |  $-1$ |  $0$ |  $1$ |  $2$ |  $0$ |  $1$ |  $3$ |  $5$

where $A$ (resp. $A^{\prime}$) is a $\mathbb{Q}$-Cartier Weil divisor on $X$
(resp. $X^{\prime}$) such that

$-K_{X}\sim qW(X)A\quad(resp.-K_{X^{\prime}}\sim qW(X^{\prime})A^{\prime})$

###### Proof.

If the numerical type of $X$ appears as one of № 4 and № 5 in [Pro22b], then
we can easily derive the values of $b_{X}$ as following (see for instance
[BK09, № 41424 and № 41431])

№ |  $q\mathbb{Q}(X)$ |  $\mathcal{R}_{X}$ |  $c_{1}^{3}$ |  $c_{2}c_{1}$ |  $b_{X}\approx$
---|---|---|---|---|---
4 |  $5$ |  $\\{2^{2},3,14\\}$ |  $\frac{625}{42}$ |  $\frac{185}{42}$ |  $3.3783$
5 |  $5$ |  $\\{2,3,4,12\\}$ |  $\frac{125}{6}$ |  $\frac{25}{6}$ |  $5$

In particular, it follows that $b_{X}>25/8$ in both cases, which contradicts
to Corollary 2.3 (see also [LL23, Corollary 4.3]). Finally the dimensions of
$|kA|$ and $|kA^{\prime}|$ can be derived directly from the Grdb in [BK09] or
Table LABEL:table.qgeq6 and Table LABEL:table.qQ=5. ∎

######  Remark 2.6.

Note that $\pi^{*}A\sim A^{\prime}$ in all the cases above. In particular, for
any positive integer $k$, it follows from the projection formula that we have

$\dim|kA^{\prime}|=\sum_{s=0}^{d-1}\dim|kA+sT|+d-1.$ (2.6.1)

######  Corollary 2.7.

Let $X$ be a terminal Fano threefold such that $q\mathbb{Q}(X)\geq 5$. If the
torsion subgroup $\operatorname{Cl}(X)_{t}$ of $\operatorname{Cl}(X)$ is non-
trivial, then $\operatorname{Cl}(X)_{t}$ is cyclic of order $d$ and the pair
$(q\mathbb{Q}(X),d)$ is one of the following:

$(5,3),\quad(5,2),\quad(7,2).$

###### Proof.

Let $\mu\colon X^{\prime}\rightarrow X$ be a small
$\mathbb{Q}$-factorialization. As in the proof of Lemma 2.1, running a MMP
$g\colon X^{\prime}\dashrightarrow X^{\prime\prime}$ yields a terminal
$\mathbb{Q}$-Fano threefold $X^{\prime\prime}$ such that

$qW(X)=q\mathbb{Q}(X)=qW(X^{\prime\prime})=q\mathbb{Q}(X^{\prime\prime})\geq
5.$

Moreover, we have the following induced injection of groups

$\iota\colon\operatorname{Cl}(X)_{t}\xrightarrow[\cong]{\mu^{*}}\operatorname{Cl}(X^{\prime})_{t}\xrightarrow{g_{*}}\operatorname{Cl}(X^{\prime\prime})_{t}.$

So we can conclude by applying Proposition 2.5 to $X^{\prime\prime}$ and
$\operatorname{Cl}(X^{\prime\prime})_{t}$. ∎

##  3\. Sarkisov link

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold such that
$\operatorname{Cl}(X)$ is torsion free. In particular, we have
$qW(X)=q\mathbb{Q}(X)$. Set $q\coloneqq qW(X)=q\mathbb{Q}(X)$. Let $A$ be an
ample Weil divisor that generates the group
$\operatorname{Cl}(X)\simeq\mathbb{Z}$. So $-K_{X}\sim qA$. Following [Ale94],
let $\scr{M}$ be a movable linear system on $X$ and let
$c\coloneqq\operatorname{ct}(X,\scr{M})$ be the canonical threshold of
$(X,\scr{M})$. Then the pair $(X,c\scr{M})$ is canonical but not terminal.
Assume that $-(K_{X}+c\scr{M})$ is ample. Then there exists a
$(K+c\scr{M})$-crepant blowup $f\colon\tilde{X}\to X$ such that

$K_{\tilde{X}}+c\tilde{\scr{M}}=f^{*}(K_{X}+c\scr{M}),$

where $\tilde{\scr}{M}$ is the strict transform of $\scr{M}$ and $\tilde{X}$
has only terminal $\mathbb{Q}$-factorial singularities.

######  Lemma 3.1 ([Pro10, 4.2]).

Let $P\in X$ be a point of local index $r>1$. Assume that $\scr{M}\sim-tK_{X}$
near $P$, where $0<t<r$. Then $c\leq 1/t$.

By [Ale94, 4.2] (see also [Pro10, 4.3]), there exists a diagram called
_Sarkisov link (of type I or type II)_ as follows

$\textstyle{\tilde{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\chi}$$\textstyle{\bar{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{f}}$$\textstyle{X}$$\textstyle{\hat{X}}$

where $\tilde{X}$ and $\bar{X}$ have only $\mathbb{Q}$-factorial terminal
singularities, $\rho(\tilde{X})=\rho(\bar{X})=2$, the morphism $f$ is a Mori
extremal divisorial contraction, the rational map $\chi$ is a sequence of log
flips and $\hat{f}$ is a Mori extremal contraction (either divisorial or fiber
type).

In what follows, for a divisor $D$ (also a linear system) on $X$, we denote by
$\tilde{D}$ and $\bar{D}$ the strict transforms of $D$ on $\tilde{X}$ and
$\bar{X}$ respectively; if $\hat{f}$ is birational, then we put
$\hat{D}=\hat{f}_{*}\bar{D}$. Vise versa.

Let $\tilde{E}$ be the $f$-exceptional divisor. Set $\scr{S}_{k}\coloneqq|kA|$
for $k\geq 1$. Write

$\begin{split}K_{\tilde{X}}=f^{*}K_{X}+\alpha\tilde{E},\\\
\tilde{\scr}{S}_{k}=f^{*}\scr{S}_{k}-\beta_{k}\tilde{E},\end{split}$ (3.1.1)

where $\alpha\in\mathbb{Q}_{>0}$ and $\beta_{k}\in\mathbb{Q}_{\geq 0}$ for
$k\geq 1$. Since $\operatorname{Cl}(X)$ is torsion free and $-kK_{X}\sim
q\scr{S}_{k}$, the relations (3.1.1) provide that for any $k\geq 1$, we have

$q\beta_{k}-k\alpha\in\mathbb{Z}.$

Assume that the morphism $\hat{f}$ is birational (Sarkisov link of type I).
Then $\hat{X}$ is a terminal $\mathbb{Q}$-Fano threefold. Set
$\hat{q}\coloneqq q\mathbb{Q}(\hat{X})$ and let $A_{\hat{X}}$ be an ample Weil
divisor on $\hat{X}$ that generates the group
$\operatorname{Cl}(\hat{X})/\sim_{\mathbb{Q}}$. Denote by $\bar{F}$ the
$\hat{f}$-exceptional divisor. Write

$F\sim
dA,\quad\hat{E}\sim_{\mathbb{Q}}eA_{\hat{X}},\quad\hat{\scr}{S}_{k}\sim_{\mathbb{Q}}s_{k}A_{\hat{X}},$

where $d,e,s_{k}\in\mathbb{Z}_{\geq 0}$.

Assume that the morphism $\hat{f}$ is not birational (Sarkisov link of type
II). Denote by $\bar{F}$ a general fiber of $\hat{f}$, which is either a
nonsingular rational curve or a nonsingular del Pezzo surface. The image of
the restriction map $\operatorname{Cl}(\bar{X})\to\operatorname{Pic}(\bar{F})$
is isomorphic to $\mathbb{Z}$. Let $A_{\bar{F}}$ be an ample generator of this
image. Write

$-K_{\bar{X}}|_{\bar{F}}=-K_{\bar{F}}\sim\hat{q}A_{\bar{F}},\quad\bar{E}|_{\bar{F}}\sim
eA_{\bar{F}},\quad\bar{\scr}{S}_{k}|_{\bar{F}}\sim s_{k}A_{\bar{F}},$

where $\hat{q},e,s_{k}\in\mathbb{Z}_{\geq 0}$.

######  Proposition 3.2 ([Pro10, § 4]).

Notation are as above.

  1. (3.2.1)

$d$ and $e$ are positive integers. Moreover, if $\hat{f}$ is birational, then
$d/e$ is the order of the torsion subgroup $\operatorname{Cl}(\hat{X})_{t}$ of
$\operatorname{Cl}(\hat{X})$.

  2. (3.2.2)

Assume that $\hat{f}$ is birational. Then $s_{k}=0$ if and only if
$\dim\scr{S}_{k}=0$ and the unique element of $\bar{\scr{S}}_{k}$ is
$\bar{F}$. Moreover, if $s_{k}>0$ and $\operatorname{Cl}(\hat{X})$ is torsion
free, then $\hat{\scr{S}}_{k}\sim s_{k}A_{\hat{X}}$ and thus

$\dim|s_{k}A_{\hat{X}}|\geq\dim\hat{\scr}{S}_{k}=\dim\scr{S}_{k}.$

  3. (3.2.3)

If $\hat{q}\geq 4$, then $\hat{f}$ is birational and if $\hat{f}$ is not
birational and $\hat{q}=3$, then $\bar{F}\cong\mathbb{P}^{2}$ and
$\hat{X}\cong\mathbb{P}^{1}$.

###### Proof.

The first part of (3.2.1) follows directly from [Pro10, Claim 4.6] and the
second part follows from [Pro10, Lemma 4.13]; The assertions of (3.2.2)
follows from the definition. For the assertions of (3.2.3), we assume that
$\hat{f}$ is not birational. Since $\bar{F}$ is either a nonsingular rational
curve or a nonsingular del Pezzo surface, it follows that $\hat{q}\leq 3$
where the equality holds only if $\bar{F}\cong\mathbb{P}^{2}$. ∎

The following very important equality, which indicates the relation between
the various invariants introduced above, will be intensively used in § 5 and §
6.

######  Lemma 3.3 ([Pro22b, 4.8]).

Notation are as above. For any $k\geq 1$, we have

$k\hat{q}=qs_{k}+(q\beta_{k}-k\alpha)e.$ (3.3.1)

For any point $P\in X$ of local index $r_{P}$, recall that the local Weil
divisor class group $\operatorname{Cl}(X,P)$ is a cyclic group of order
$r_{P}$ generated by $-K_{X}$ by [Kaw88, Corollary 5.2]. We denote by
$t_{k}(P)$ the local index of $\scr{S}_{k}$ near $P$. On the other hand, if
$P$ is clear from the context, then we will usually omit $P$ and abbreviate
$t_{k}(P)$ to $t_{k}$ for simplicity. The following fact allows us to
determine the possible value of $\beta_{k}$ once the value of $\alpha$ is
known (see [Kaw96, Kaw05] and [Pro22b, Lemma 2.6]).

######  Lemma 3.4.

If $P\coloneqq f(\tilde{E})$ is a closed point of $X$, then we have

$\beta_{k}-t_{k}\alpha\in\mathbb{Z}.$ (3.4.1)

###### Proof.

This follows from the relations (3.1.1) and the fact that
$t_{k}K_{X}+\scr{S}_{k}$ is Cartier near $P$. ∎

##  4\. Case $q\mathbb{Q}(X)=5$

The goal of this section is to prove the following result.

######  Theorem 4.1.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold such that
$q\mathbb{Q}(X)=5$. Then we have

$c_{1}(X)^{3}<\frac{121}{41}c_{2}(X)c_{1}(X).$

Assume to the contrary that there exists a terminal $\mathbb{Q}$-Fano
threefold $X$ such that $q\mathbb{Q}(X)=5$ and $b_{X}\geq 121/41$. Then the
numerical type of $X$ appears as (2.3.2) in Corollary 2.3. In particular, we
have the numerical invariants of $X$ from [BK09, № 41436] as follows:

$\mathbf{B}_{X}=\\{(3,1),(7,2),(7,3)\\},\,c_{1}(X)^{3}=\frac{500}{21},\,c_{2}(X)c_{1}(X)=\frac{160}{21}\quad\textup{and}\quad
b_{X}=\frac{25}{8}.$

In this case, the tangent sheaf $\scr{T}_{X}$ is not semi-stable by the
Bogomolov–Gieseker inequality and it follows from [LL23, Theorem 4.2 and
Corollary 4.3] that the maximal destabilizing subsheaf $\scr{F}$ of
$\scr{T}_{X}$ is of rank two with $c_{1}(\scr{F})\equiv 4A$, where $A$ is a
generator of $\operatorname{Cl}(X)/\sim_{\mathbb{Q}}$.

###  4.1. Foliation

Let $X$ be a normal variety. A _foliation_ on $X$ is a non-zero coherent
subsheaf $\scr{F}$ of the tangent sheaf $\scr{T}_{X}$ such that

  1. (4.1.1)

$\scr{F}$ is saturated in $\scr{T}_{X}$ (i.e., $\scr{T}_{X}/\scr{F}$ is
torsion free);

  2. (4.1.2)

$\scr{F}$ is closed under the Lie bracket.

The _rank_ of $\scr{F}$ is defined to be the generic rank $r$ of $\scr{F}$ and
the _codimension_ of $\scr{F}$ is defined as $\dim X-r$.

Given a rank $r$ foliation $\scr{F}$ on a normal variety $X$, the inclusion
$\scr{F}\hookrightarrow\scr{T}_{X}$ induces a non-zero map

$\eta\colon\Omega^{r}_{X}\rightarrow\wedge^{r}\scr{T}_{X}^{*}\rightarrow\wedge^{r}\scr{F}^{*}\rightarrow\det(\scr{F}^{*}).$

The _singular locus_ of $\scr{F}$ is the singular scheme of the map $\eta$,
i.e., it is the closed subscheme of $X$ whose ideal sheaf is the image of the
induced map

$(\Omega_{X}^{r}\otimes\det(\scr{F}))^{**}\rightarrow\scr{O}_{X}.$

A point $x\in X$ is called a _regular point_ of $\scr{F}$ if $x$ is not
contained in the singular locus of $\scr{F}$. If $x$ is a regular point of
$X$, then $\scr{F}$ is regular at $x$ if and only if $\scr{T}_{X}/\scr{F}$ is
locally free at $x$. In particular, the singular locus of $\scr{F}$ has
codimension at least two. Moreover, if both $X$ and $\scr{F}$ are regular at
$x$, then by the holomorphic Frobenius theorem, there exists a unique leaf of
$\scr{F}$ passing through $x$.

A foliation $\scr{F}$ is called _algebraically integrable_ if there exists a
dominant rational map $\varphi\colon X\dashrightarrow Y$ to a normal variety
$Y$ such that $\scr{F}=\ker(\scr{T}_{X}\rightarrow\varphi^{*}\scr{T}_{Y})$.

######  Lemma 4.2.

Let $X$ be a normal projective variety of dimension $n$, which is regular in
codimension two. Let $\scr{F}$ be a codimension one algebraically integrable
foliation on $X$. If $X$ is $\mathbb{Q}$-factorial and $\rho(X)=1$, then the
singular locus of $\scr{F}$ has codimension two.

###### Proof.

Assume to the contrary that $\scr{F}$ is regular in codimension two. By
assumption, the foliation $\scr{F}$ is induced by a rational map
$\varphi\colon X\dashrightarrow Y$. Let $F$ and $F^{\prime}$ be the closures
of two general fibres of $\varphi$. Since $X$ is $\mathbb{Q}$-factorial and
$\rho(X)=1$, the divisors $F$ and $F^{\prime}$ are ample and $\dim(F\cap
F^{\prime})=n-2$. Since both of $X$ and $\scr{F}$ are regular in codimension
two, there exists a point $x\in F\cap F^{\prime}$ such that $X$ and $\scr{F}$
are both regular at $x$. In particular, both of $F$ and $F^{\prime}$ are the
leaves of $\scr{F}$ at $x$, which contradicts the uniqueness of leaves. ∎

###  4.2. Proof of Theorem 4.1

Let $X$ be a terminal $\mathbb{Q}$-Fano variety of dimension $n$. Denote by
$r$ the rank of the maximal destabilizing subsheaf $\scr{F}$ of $\scr{T}_{X}$
and by $k$ the length of the Harder–Narasimhan filtration of $\scr{T}_{X}$.
Note that $\scr{T}_{X}$ is semi-stable if and only if $(r,k)=(n,1)$. In this
case, the Bogomolov–Gieseker inequality and [GKP21, Theorem 1.2] imply

$c_{1}(X)^{n}<3c_{2}(X)c_{1}(X)^{n-2}.$

In [LL23, Proposition 3.8], the authors studied the case $(r,k)=(1,k)$ for
$k\geq 2$. Now we study the case $(r,k)=(n-1,2)$ and prove the following
result.

######  Proposition 4.3.

Let $X$ be a terminal $\mathbb{Q}$-Fano variety of dimension $n$. Assume that
the maximal destabilizing subsheaf $\scr{F}$ of $\scr{T}_{X}$ has rank $n-1$.
Then we have

$c_{2}(X)c_{1}(X)^{n-2}>\frac{(2(n-1)c_{1}(X)-nc_{1}(\scr{F}))c_{1}(\scr{F})}{2(n-1)}c_{1}(X)^{n-2}.$

###### Proof.

Since $\scr{T}_{X}$ is generically ample by [LL23, Proposition 3.6], it
follows from [CP19, Lemma 4.10] that $\scr{F}$ is actually a foliation.
Moreover, as $\scr{F}$ is semi-stable and $\det(\scr{F})$ is ample, the
foliation $\scr{F}$ is algebraically integrable by [CP19, Theorem 1.1]
111Though [CP19, Lemma 4.10 and Theorem 1.1] are only stated for nonsingular
projective varieties, it is easy to generalise them to $\mathbb{Q}$-factorial
normal projective varieties by taking a resolution.. Consider the following
short exact sequence of coherent sheaves

$0\rightarrow\scr{F}\rightarrow\scr{T}_{X}\rightarrow\scr{Q}\rightarrow 0.$

By Lemma 4.2, the singular locus of the foliation $\scr{F}$ has codimension
two. In particular, the non locally free locus of the quotient $\scr{Q}$ is
supported on a closed subset of $X$ with codimension two and therefore
$c_{2}(\scr{Q})$ is represented by a non-zero effective codimension two cycle.
So we get

$\displaystyle c_{2}(\scr{T}_{X})c_{1}(X)^{n-2}$
$\displaystyle=(c_{2}(\scr{F})+c_{2}(\scr{Q})+c_{1}(\scr{F})c_{1}(\scr{Q}))c_{1}(X)^{n-2}$
$\displaystyle>c_{2}(\scr{F})c_{1}(X)^{n-2}+c_{1}(\scr{F})c_{1}(X)^{n-1}-c_{1}(\scr{F})^{2}c_{1}(X)^{n-2}$
$\displaystyle\geq
c_{1}(\scr{F})c_{1}(X)^{n-1}-\frac{n}{2n-2}c_{1}(\scr{F})^{2}c_{1}(X)^{n-2},$

where the last inequality follows from the Bogomolov–Gieseker inequality. ∎

###### Proof of Theorem 4.1.

Assume to the contrary that there exists a terminal $\mathbb{Q}$-Fano
threefold $X$ such that $b_{X}\geq 121/41$. By the discussion at the beginning
of this section, the maximal destabilizing subsheaf $\scr{F}$ of $\scr{T}_{X}$
is of rank two and $c_{1}(\scr{F})\equiv 4A$. In particular, as $n=3$ and
$c_{1}(X)\equiv 5A$, it follows from Proposition 4.3 that

$c_{2}(X)c_{1}(X)>8A^{2}c_{1}(X)=\frac{8}{25}c_{1}(X)^{3},$

which contradicts the fact that $b_{X}=25/8$. ∎

######  Remark 4.4.

In Corollary 2.3, we ruled out the case (2.2.5) by [Pro13, Claim 6.5].
However, notice that in this case, the rank of the maximal destabilizing
subsheaf $\scr{F}$ of $\scr{T}_{X}$ is two and $c_{1}(\scr{F})\equiv 6A$ by
[LL23, Theorem 4.2 and Corollary 4.3], where $A$ is the generator of
$\operatorname{Cl}(X)/\sim_{\mathbb{Q}}$. In particular, as $b_{X}=49/15$, we
can also rule it out by the same argument as that in the proof of Theorem 4.1.

##  5\. Case $q\mathbb{Q}(X)=4$

The aim of this section is to prove the following result.

######  Theorem 5.1.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold such that
$q\mathbb{Q}(X)=4$. Then we have

$c_{1}(X)^{3}<\frac{121}{41}c_{2}(X)c_{1}(X).$

Assume to the contrary that there exists a terminal $\mathbb{Q}$-Fano
threefold $X$ such that $q\mathbb{Q}(X)=4$ and $b_{X}\geq 121/41$. Then the
numerical type of $X$ appears as (2.3.1) in Corollary 2.3. In particular, we
have the numerical invariants of $X$ from [BK09, № 41313] as follows:

$\mathbf{B}_{X}=\\{(7,2),(13,6)\\},\quad c_{1}(X)^{3}=\frac{1152}{91},\quad
c_{2}(X)c_{1}(X)=\frac{384}{91}\quad\textup{and}\quad b_{X}=3.$

###  5.1. Geometry of $X$

We collect in the following some basic geometric properties of $X$, which will
be used later in the proof of Theorem 5.1.

######  Lemma 5.2.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold with numerical type (2.3.1)
in Corollary 2.3.

  1. (5.2.1)

Every non-Gorenstein point of $X$ is a cyclic quotient singularity.

  2. (5.2.2)

The Weil divisor class group $\operatorname{Cl}(X)$ is torsion free.

  3. (5.2.3)

Let $A$ be a $\mathbb{Q}$-Cartier Weil divisor on $X$ such that $-K_{X}\sim
4A$. Then we have

$\mathcal{R}_{X}$ |  $A^{3}$ |  $|A|$ |  $|2A|$ |  $|3A|$ |  $|4A|$ |  $|5A|$ |  $|6A|$
---|---|---|---|---|---|---|---
$\\{7,13\\}$ |  $\frac{18}{91}$ |  $-1$ |  $1$ |  $3$ |  $6$ |  $10$ |  $15$
  4. (5.2.4)

Every divisor contained in the linear system $|2A|$ or $|3A|$ is reduced and
irreducible.

###### Proof.

The first statement follows from the form of the basket $\mathbf{B}_{X}$ and
[Mor85] (see [Rei87, (6.1) and (6.4)] for more details).

For the second statement, let $\pi\colon X^{\prime}\rightarrow X$ be an étale
in codimension two finite cover such that $\operatorname{Cl}(X^{\prime})$ is
torsion free. In particular, we have $qW(X^{\prime})=q\mathbb{Q}(X^{\prime})$
and $b_{X^{\prime}}=b_{X}=3$. On the other hand, by Lemma 2.1, the index
$qW(X^{\prime})=q\mathbb{Q}(X^{\prime})$ is divided by
$qW(X)=q\mathbb{Q}(X)=4$. So the numerical type of $X^{\prime}$ also appears
as (2.2.2) in Lemma 2.2 and in particular we have
$c_{1}(X)^{3}=c_{1}(X^{\prime})^{3}$ by (2.0.4) (see also [BK09, № 41313]).
Thus $\pi$ is actually an isomorphism and hence $\operatorname{Cl}(X)$ is
torsion free.

The dimensions of the linear systems $|kA|$ can be derived from the orbifold
Riemann–Roch formula (2.0.1) (see also [BK09, № 41313]).

Finally, as $\dim|A|=-1$ and $\operatorname{Cl}(X)$ is torsion free, any
element in $|2A|$ and $|3A|$ is reduced and irreducible. ∎

We will use the Sarkisov link introduced in § 3 to prove Theorem 5.1. This
idea was initiated by Y. Prokhorov in [Pro10]. To be more precise, following
the notation in § 3, we consider the movable linear system
$\scr{M}=\scr{S}_{3}=|3A|$ and the corresponding Sarkisov link:

$\textstyle{\tilde{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\chi}$$\textstyle{\bar{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{f}}$$\textstyle{X}$$\textstyle{\hat{X}.}$

######  Lemma 5.3.

$d\geq 2$ and $\beta_{3}\geq 6\alpha$.

###### Proof.

We have $d\geq 2$ as $\dim|A|=-1$. Now let $P\in X$ be the point of local
index 7. Then $\scr{S}_{3}\sim-6K_{X}$ near the point $P$, hence $c\leq 1/6$
by Lemma 3.1. Since $c=\alpha/\beta_{3}$ by (3.1.1), our inequality follows. ∎

The following observation can be easily derived from the definition and Lemma
5.2. It will play a key role in the proof of Theorem 5.1.

######  Lemma 5.4.

Any element in $\bar{\scr{S}}_{2}$ (resp. $\bar{\scr{S}}_{3}$) is of the form
$\bar{\Delta}+a\bar{E}$, where $a$ is a non-negative integer and
$\bar{\Delta}$ is the strict transform of an element $\Delta$ in $\scr{S}_{2}$
(resp. $\scr{S}_{3}$). In particular, the divisor $\bar{\Delta}$ is reduced
and irreducible. Moreover, if $\hat{f}$ is birational and
$\hat{f}_{*}\bar{\Delta}=0$, then $\bar{\Delta}=\bar{F}$ and $\Delta=F$.

The remaining part of this section is devoted to prove Theorem 5.1 case by
case according to the singular type of $f(\tilde{E})$ and finally we conclude
by showing that every possibility for $f(\tilde{E})$ does not actually happen.

###  5.2. The point $f(\tilde{E})$ is a curve or a Gorenstein point.

If $f(\tilde{E})$ is either a curve or a Gorenstein point of $X$, then
$\alpha$ and $\beta_{3}$ are integers and then it follows from (3.3.1) and
Lemma 5.3 that

$3\hat{q}=4s_{3}+(4\beta_{3}-3\alpha)e\geq 4s_{3}+21\alpha e.$

In particular, we get $\hat{q}\geq 7\alpha e\geq 7$ and therefore it follows
from Proposition 3.2 that the morphism $\hat{f}$ is birational. Moreover, as
$\dim\hat{\scr{S}}_{3}=3$, we get $s_{3}>0$ by Proposition 3.2 and hence
$\hat{q}\geq 9$. By Proposition 2.5, the group $\operatorname{Cl}(\hat{X})$ is
torsion free and so it follows from Proposition 3.2 that $e=d\geq 2$. Hence
$\hat{q}\geq 16$. As $\dim|s_{3}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3$, it
follows from Table LABEL:table.qgeq6 that $s_{3}\geq 9$, which is impossible
as $\hat{q}\leq 19$.

###  5.3. The point $f(\tilde{E})$ is of local index 13.

If $f(\tilde{E})$ is a point of local index $13$, then $\alpha=1/13$ by
[Kaw96]. Moreover, as $A\sim-10K_{X}$ near the point $f(\tilde{E})$, we obtain
$t_{1}=10$, $t_{2}=7$ and $t_{3}=4$. Then it follows from (3.4.1) and Lemma
5.3 that $\beta_{2}=7/13+m_{2}$ for some $m_{2}\in\mathbb{Z}_{\geq 0}$ and
$\beta_{3}=4/13+m_{3}$ for some $m_{3}\in\mathbb{Z}_{>0}$. Hence (3.3.1)
provides

$3\hat{q}=4s_{3}+(4m_{3}+1)e=4(s_{3}+m_{3}e)+e.$ (5.4.1)

If $\hat{q}\geq 4$, by Proposition 3.2, the morphism $\hat{f}$ is birational
and $s_{3}>0$ as $\dim\hat{\scr{S}}_{3}=3$. Then we derive from (5.4.1) that
$\hat{q}\not=4$ or $5$ and the possibilities for $\hat{q}=6$ and $7$ are
listed as below. If $\hat{q}\geq 8$, then $\operatorname{Cl}(\hat{X})$ is
torsion free by Proposition 2.5 and thus $e=d\geq 2$. Moreover, as
$\dim|s_{3}A_{\hat{X}}|\geq\dim\hat{\scr}{S}_{3}=3$, combining this with
(5.4.1) and Table LABEL:table.qgeq6 yields the corresponding possible values
of $s_{3}$ for each $\hat{q}$. In summary, the possibilities for
$(\hat{q},e,s_{3})$ are one of the following:

  * •

$\hat{q}=3$, $e=1$, and $s_{3}\leq 1$;

  * •

$\hat{q}=6$, $e=2$, and $s_{3}=2$;

  * •

$\hat{q}=7$, $e=1$, and $s_{3}\leq 4$;

  * •

$\hat{q}=13$, $e=3$, and $s_{3}=6$;

  * •

$\hat{q}=17$, $e=3$, and $s_{3}=9$.

We treat in the following case by case according to the value of $\hat{q}$.

####  5.3.1. The case $\hat{q}=3$ and $e=1$.

In this case we have $s_{3}\leq 1$ and $s_{2}=1-m_{2}\leq 1$ by (3.3.1). Now
we divide the proof into two subcases according to the type of $\hat{f}$.

#####  5.3.1.1 The subcase where $\hat{f}$ is birational.

Then $s_{3}=1$ and $s_{2}=1$ by Proposition 3.2 and so every element in
$\hat{\scr{S}}_{2}$ and $\hat{\scr{S}}_{3}$ is reduced and irreducible. Let
$\pi\colon\hat{X}^{\prime}\rightarrow\hat{X}$ be an étale in codimension two
finite cover such that $\operatorname{Cl}(\hat{X}^{\prime})$ is torsion free.
Then we have

$\pi^{*}\hat{E}\sim\pi^{*}\hat{\scr{S}}_{2}\sim\pi^{*}\hat{\scr{S}}_{3}.$

We claim that $\dim|\pi^{*}\hat{E}|\geq 5$. Indeed, note that both
$\pi^{*}\hat{\scr{S}}_{2}$ and $\pi^{*}\hat{\scr{S}}_{3}$ are linear subspaces
of $|\pi^{*}\hat{E}|$ and it follows from Lemma 5.4 that the divisor
$\pi^{*}\hat{E}$ is the only possible element contained in
$\pi^{*}\hat{\scr{S}}_{2}\cap\pi^{*}\hat{\scr{S}}_{3}$. However, if this
happens, then we must have $\hat{E}\in\hat{\scr{S}}_{2}\cap\hat{\scr{S}}_{3}$
and thus Lemma 5.4 implies $F\in\scr{S}_{2}\cap\scr{S}_{3}$, which is
impossible. So the subspaces $\pi^{*}\hat{\scr{S}}_{2}$ and
$\pi^{*}\hat{\scr{S}}_{3}$ are disjoint and we have

$\dim|\pi^{*}\hat{E}|\geq\dim\hat{\scr{S}}_{2}+\dim\hat{\scr{S}}_{3}+1=\dim\scr{S}_{2}+\dim\scr{S}_{3}+1=5.$

On the other hand, since $q\mathbb{Q}(\hat{X}^{\prime})$ is divided by
$q\mathbb{Q}(\hat{X})=3$, it follows from Lemma 2.1 that
$q\mathbb{Q}(\hat{X}^{\prime})\in\\{3,6,9\\}$. Moreover, since
$-K_{\hat{X}^{\prime}}\sim 3\pi^{*}\hat{E}$ and the group
$\operatorname{Cl}(\hat{X}^{\prime})$ is torsion free, we get

$\pi^{*}\hat{E}\sim\frac{q\mathbb{Q}(\hat{X}^{\prime})}{3}A_{\hat{X}^{\prime}},$

where $A_{\hat{X}^{\prime}}$ is a Weil divisor on $\hat{X}^{\prime}$ such that
$-K_{\hat{X}^{\prime}}\sim q\mathbb{Q}(\hat{X}^{\prime})A_{\hat{X}^{\prime}}$.
Then we obtain a contradiction from Table LABEL:table.qgeq6 and Lemma 2.4.

#####  5.3.1.2 The subcase where $\hat{f}$ is not birational.

Then the general fibre $\bar{F}$ of $\hat{f}$ is isomorphic to
$\mathbb{P}^{2}$ and $\hat{X}\cong\mathbb{P}^{1}$ by Proposition 3.2.
Moreover, since $\dim\bar{\scr{S}}_{3}=3$ and the general elements in
$\bar{\scr{S}}_{3}$ are irreducible, we obtain $s_{3}>0$ and so $s_{3}=1$.
Consider the following restriction map

$\bar{\scr{S}}_{3}\rightarrow\bar{\scr{S}}_{3}|_{\bar{F}}\subset|\scr{O}_{\mathbb{P}^{2}}(1)|.$

As $\dim|\scr{O}_{\mathbb{P}^{2}}(1)|=2$, $\dim\bar{\scr{S}}_{3}=3$ and
$e=s_{3}=1$, we get $\bar{F}+\bar{E}\in\bar{\scr{S}}_{3}$ by Lemma 5.4. In
particular, we have

$\hat{f}^{*}|\scr{O}_{\mathbb{P}^{1}}(1)|+\bar{E}\subset\bar{\scr{S}}_{3}.$

Then it follows from Lemma 5.4 that the (cycle-theoretic) fibres of $\hat{f}$
is reduced and irreducible.

Next we show that $s_{2}=1$. Assume to the contrary that we have $s_{2}=0$.
Since $\dim\bar{\scr{S}}_{2}=1$ and its general elements are reduced and
irreducible, we have $\bar{F}\in\bar{\scr{S}}_{2}$. This implies that
$F\in\scr{S}_{2}\cap\scr{S}_{3}$, which is absurd.

Finally we claim that $\bar{\scr{S}}_{3}-\bar{\scr{S}}_{2}\sim c\bar{F}$ for
some $c\in\mathbb{Z}$. Assuming this for the time being, then pushing
$\bar{\scr{S}}_{3}-\bar{\scr{S}}_{2}$ to $X$ yields

$A\sim\scr{S}_{3}-\scr{S}_{2}\sim cF\sim 3cA,$

which is a contradiction. To prove the claim, let $\bar{S}_{2}$ (resp.
$\bar{S}_{3}$) be a general element in $\bar{\scr{S}}_{2}$ (resp.
$\bar{\scr{S}}_{3}$). Then the divisor $\bar{S}_{3}-\bar{S}_{2}$ is Cartier in
a neighbourhood of the general fibre $\bar{F}$ and
$(\bar{S}_{3}-\bar{S}_{2})|_{\bar{F}}\sim 0$. In particular, the push-forward
$\hat{f}_{*}\scr{O}_{\bar{X}}(\bar{S}_{3}-\bar{S}_{2})$ is a torsion free
sheaf of rank one over $\hat{X}\cong\mathbb{P}^{1}$ and thus there exists an
integer $b$ such that

$\hat{f}_{*}\scr{O}_{\bar{X}}(\bar{S}_{3}-\bar{S}_{2})\simeq\scr{O}_{\mathbb{P}^{1}}(b).$

Since the induced morphism

$\hat{f}^{*}\scr{O}_{\mathbb{P}^{1}}(b)\simeq\hat{f}^{*}\hat{f}_{*}\scr{O}_{\bar{X}}(\bar{S}_{3}-\bar{S}_{2})\rightarrow\scr{O}_{\bar{X}}(\bar{S}_{3}-\bar{S}_{2})$

is an isomorphism in the neighbourhood of $\bar{F}$, there exists a
$\hat{f}$-vertical effective Weil divisor $\bar{\Delta}$ such that

$\bar{S}_{3}-\bar{S}_{2}\sim b\bar{F}+\bar{\Delta}.$

Since any fibre of $\hat{f}$ is reduced and irreducible, there exists a non-
negative integer $b^{\prime}$ such that $\bar{\Delta}\sim b^{\prime}\bar{F}$
and then we get $\bar{S}_{3}-\bar{S}_{2}\sim(b+b^{\prime})\bar{F}$, which
proves the claim.

####  5.3.2. The case $\hat{q}=6$, $e=2$ and $s_{3}=2$.

In this case the numerical type of $\hat{X}$ appears as № 1 in Table
LABEL:table.qgeq6 and $\operatorname{Cl}(\hat{X})$ is torsion free by
Proposition 2.5. In particular, we have
$\dim\scr{S}_{3}=\dim\hat{\scr}{S}_{3}=\dim|2A_{\hat{X}}|=3$ and
$\dim|A_{\hat{X}}|=1$. So there exists an element in $\hat{\scr{S}}_{3}$ of
the form $2G$ for some $G\in|A_{\hat{X}}|$. This contradicts to Lemma 5.4 as
$e=s_{3}=2$.

####  5.3.3. The case $\hat{q}=7$, $e=1$ and $s_{3}\leq 4$.

Since $d/e=d\geq 2$ is the order of the torsion subgroup of
$\operatorname{Cl}(\hat{X})$, the numerical type of $\hat{X}$ appears as
either № 6 or № 7 in Table LABEL:table.qgeq5 by Proposition 2.5. Let $T$ be
the unique $2$-torsion element in $\operatorname{Cl}(\hat{X})$. Then we have

$\textup{either}\quad\hat{\scr{S}}_{3}\sim
s_{3}A_{\hat{X}}\quad\textup{or}\quad\hat{\scr{S}}_{3}\sim
s_{3}A_{\hat{X}}+T.$

Then one can easily derive a contradiction from Table LABEL:table.qgeq5 and
(2.6.1) as $\dim\hat{\scr{S}}_{3}=3$.

####  5.3.4. The case $\hat{q}=13$, $e=3$ and $s_{3}=6$.

In this case the numerical type of $\hat{X}$ appears as № 28 in Table
LABEL:table.qgeq6 and $\operatorname{Cl}(\hat{X})$ is torsion free by
Proposition 2.5. As $e=3$ and $\dim|A_{\hat{X}}|=1$, the unique element
$D\in|A_{\hat{X}}|$ is a prime divisor which is different from $\hat{E}$.
Moreover, as

$6=\dim|5A_{\hat{X}}|+\dim\hat{\scr{S}}_{3}\geq\dim|s_{3}A_{\hat{X}}|=\dim|6A_{\hat{X}}|=4,$

there exists an element in $\hat{\scr{S}}_{3}$ of the form $D+G$ for some
$G\in|5A_{\hat{X}}|$. This contradicts to Lemma 5.4 as $e=3$.

####  5.3.5. The case $\hat{q}=17$, $e=3$ and $s_{3}=9$.

In this case the numerical type of $\hat{X}$ appears as № 30 in Table
LABEL:table.qgeq6 and $\operatorname{Cl}(\hat{X})$ is torsion free by
Proposition 2.5. As $e=3$ and $\dim|A_{\hat{X}}|=-1$, the unique element
$D\in|2A_{\hat{X}}|$ is a prime divisor which is different from $\hat{E}$.
Moreover, as $\dim\hat{\scr{S}}_{3}=3$, we must have
$\hat{\scr{S}}_{3}=|9A_{\hat{X}}|$. In particular, as $\dim|7A_{\hat{X}}|=2$,
there exists an element in $\hat{\scr{S}}_{3}$ of the form $D+G$ for some
$G\in|7A_{\hat{X}}|$. This contradicts to Lemma 5.4 as $e=3$.

###  5.4. The point $f(\tilde{E})$ is of local index $7$.

If $f(\tilde{E})$ is a point of local index $7$, then $\alpha=1/7$ by [Kaw96].
Moreover, as $A\sim-2K_{X}$ near the point $f(\tilde{E})$, we obtain
$t_{1}=2$, $t_{2}=4$ and $t_{3}=6$. Then it follows from (3.4.1) that
$\beta_{2}=4/7+m_{2}$ for some $m_{2}\in\mathbb{Z}_{\geq 0}$ and
$\beta_{3}=6/7+m_{3}$ for some $m_{3}\in\mathbb{Z}_{\geq 0}$. Hence (3.3.1)
provides

$3\hat{q}=4s_{3}+(4m_{3}+3)e=4(s_{3}+m_{3}e)+3e.$

As $\hat{q}\leq 19$ by Lemma 2.1, we obtain $\hat{q}-e=4l$ for some integer
$0\leq l\leq 4$. If $\hat{q}\geq 8$, then $\operatorname{Cl}(\hat{X})$ is
torsion free by Proposition 2.5 and thus $e=d\geq 2$. So the possibility for
$(\hat{q},e,s_{3})$ is one of the following:

  * •

$\hat{q}=e$, $s_{3}=0$;

  * •

$\hat{q}=4+e$, $e\geq 1$ and $s_{3}\leq 3$;

  * •

$\hat{q}=8+e$, $e\geq 2$ and $s_{3}\leq 6$;

  * •

$\hat{q}=12+e$, $e\geq 2$ and $s_{3}\leq 9$;

  * •

$\hat{q}=16+e$, $e\geq 2$ and $s_{3}\leq 12$.

We treat in the following case by case according to the value of $\hat{q}$.

####  5.4.1. The case $\hat{q}=e$ and $s_{3}=0$.

As $\dim\scr{S}_{3}=3$ and $s_{3}=0$, it follows from Proposition 3.2 that
$\hat{f}$ is not birational and hence $\hat{q}=e\leq 3$. As
$\bar{\scr{S}}_{3}$ is $\hat{f}$-vertical, the elements in $\bar{\scr{S}}_{3}$
are prime divisors by Lemma 5.4. So we get

$\bar{\scr{S}}_{3}=\hat{f}^{*}|bA_{\hat{X}}|$

for some $b\in\mathbb{Z}_{>0}$. However, as $\dim|A_{\hat{X}}|\geq 0$ by
[Pro13, 2.3], we must have $b=1$ and $\hat{X}\cong\mathbb{P}^{2}$. On the
other hand, note that we also have $s_{2}=0$ by (3.3.1). In particular, as
$\dim\bar{\scr{S}}_{2}=1$ and the elements in $\bar{\scr{S}}_{2}$ are prime
divisors, we obtain

$\bar{\scr{S}}_{2}=\hat{f}^{*}|A_{\hat{X}}|$

similarly. This is impossible as $\dim|A_{\hat{X}}|=3$.

####  5.4.2. The case $\hat{q}=4+e$ and $s_{3}\leq 3$.

In this case Proposition 3.2 provides that the morphism $\hat{f}$ is
birational and thus $s_{3}\geq 1$ as $\dim\scr{S}_{3}=3$. If $\hat{q}\geq 8$,
then $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 2.5 and thus
we have

$\dim|s_{3}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3.$

Then it follows from Table LABEL:table.qgeq6 that $\hat{q}\leq 8$ and hence
$e\leq 4$. We divide the proof into four subcases according to the value of
$e$.

#####  5.4.2.1 The subcase $e=4$.

Then $\hat{q}=8$, $s_{3}=3$ and the numerical type of $\hat{X}$ appears as №
19 in Table LABEL:table.qgeq6. In particular, we have
$\dim|3A_{\hat{X}}|=\dim\hat{\scr{S}}_{3}=3$ and thus
$|3A_{\hat{X}}|=\hat{\scr{S}}_{3}$. As $e=4$ and $\dim|A_{\hat{X}}|=0$, the
unique element $D\in|A_{\hat{X}}|$ is a prime divisor which is different from
$\hat{E}$. So we have $3D\in\hat{\scr{S}}_{3}$, which contradicts to Lemma 5.4
as $e=4$.

#####  5.4.2.2 The subcase $e=3$.

Then $\hat{q}=7$ and $s_{3}=3$. As $\dim\hat{\scr{S}}_{3}=3$, the same
argument as in § 5.3.3 shows that the group $\operatorname{Cl}(\hat{X})$ is
torsion free and thus $\dim|3A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3$. It
follows that the numerical type of $\hat{X}$ appears as one of № 7, № 8 and №
9 in Table LABEL:table.qgeq6. In all these three cases, we have
$\dim|A_{\hat{X}}|\geq 0$. In particular, as $e=3$, there exists a prime
divisor $D\in|A_{\hat{X}}|$ which is different from $\hat{E}$. Moreover, in
all these three cases we have

$\dim|2A_{\hat{X}}|+\dim\hat{\scr{S}}_{3}\geq\dim|3A_{\hat{X}}|.$

So there exists an element in $\hat{\scr{S}}_{3}$ of the form $D+G$ with
$G\in|2A_{\hat{X}}|$, which contradicts to Lemma 5.4 as $e=3$.

#####  5.4.2.3 The subcase $e=2$.

Then $\hat{q}=6$ and $s_{3}=1$ or $3$. By Proposition 2.5, the group
$\operatorname{Cl}(\hat{X})$ is torsion free and thus
$\dim|s_{3}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3$. It follows that $s_{3}=3$
and the numerical type of $\hat{X}$ appears as № 1 in Table LABEL:table.qgeq6.
On the other hand, we have $s_{2}=2$ by (3.3.1). As $\dim|A_{\hat{X}}|=1$, the
image of the following map

$\Phi\colon|A_{\hat{X}}|\times|A_{\hat{X}}|\rightarrow|2A_{\hat{X}}|,\quad(D,D^{\prime})\mapsto
D+D^{\prime}$

is a projective plane in $|2A_{\hat{X}}|\cong\mathbb{P}^{3}$. In particular,
since $\hat{\scr{S}}_{2}$ is a one-dimensional subspace of $|2A_{\hat{X}}|$,
the intersection $\hat{\scr{S}}_{2}\cap\operatorname{Image}(\Phi)$ is non-
empty; that is, there exists an element in $\hat{\scr{S}}_{2}$ of the form
$D+D^{\prime}$ with $D$, $D^{\prime}\in|A_{\hat{X}}|$. This contradicts to
Lemma 5.4 as $e=2$.

#####  5.4.2.4 The subcase $e=1$.

Then $\hat{q}=5$, $1\leq s_{2}\leq 2$ and $1\leq s_{3}\leq 3$ by (3.3.1). As
$d\geq 2$, the group $\operatorname{Cl}(\hat{X})$ is not torsion free by
Proposition 3.2. Let $T$ be the generator of $\operatorname{Cl}(\hat{X})_{t}$
and let $\pi\colon\hat{X}^{\prime}\rightarrow\hat{X}$ be the index one cover
defined by $T$ (see Proposition 2.5). By (2.6.1), we have

$\textup{either}\quad\dim|s_{k}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{k}\quad\textup{or}\quad\dim|s_{k}A_{\hat{X}}|+\dim\hat{\scr{S}}_{k}\leq\dim|s_{k}A_{\hat{X}^{\prime}}|-1,$

where $A_{\hat{X}}$ (resp. $A_{\hat{X}^{\prime}}$) is a $\mathbb{Q}$-Cartier
Weil divisor on $\hat{X}$ (resp. $\hat{X}^{\prime}$) such that

$-K_{\hat{X}}\sim 5A_{\hat{X}}\quad(\textup{resp.}\,-K_{\hat{X}^{\prime}}\sim
5A_{\hat{X}^{\prime}}).$

In particular, as $\dim\hat{\scr{S}}_{2}=1$ and $\dim\hat{\scr{S}}_{3}=3$, it
follows from Proposition 2.5 that $s_{2}=2$, $s_{3}=3$ and the numerical type
of $\hat{X}$ appears as one of № 2, № 4 and № 5 in Table LABEL:table.qgeq5,
and therefore $d=s_{d}$. On the other hand, since $\operatorname{Cl}(X)$ is
torsion free, we have $F\in\scr{S}_{d}$. So there exists a non-negative
integer $a$ such that

$\bar{F}+a\bar{E}\in\bar{\scr{S}}_{d}.$

Pushing it forward to $\hat{X}$ shows that $a=s_{d}=d$ as $e=1$, i.e.,
$d\hat{E}\in\hat{\scr{S}}_{d}$. This yields $\hat{\scr{S}}_{d}\sim
dA_{\hat{X}}$ as $d\hat{E}\sim dA_{\hat{X}}$. Therefore, the numerical type of
$\hat{X}$ cannot be № 2 in Table LABEL:table.qgeq5. Otherwise, we have $d=3$
and $\dim|3A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3$, which is impossible. In
the remaining two cases, we always have $d=2$ and
$\dim\hat{\scr{S}}_{2}=\dim|2A_{\hat{X}}|=1$ and thus
$\hat{\scr{S}}_{2}=|2A_{\hat{X}}|$. Moreover, by (2.6.1) and Proposition 2.5,
we also have

$\dim|A_{\hat{X}}|=\dim|A_{\hat{X}}+T|=0.$

Let $D$ and $D^{\prime}$ be the unique element in $|A_{\hat{X}}|$ and
$|A_{\hat{X}}+T|$, respectively. Then $D$ and $D^{\prime}$ are distinct prime
divisors. However, since $2D$ and $2D^{\prime}$ are contained in
$\hat{\scr{S}}_{2}$, it follows from Lemma 5.4 that we must have
$D=D^{\prime}=\hat{E}$, which is impossible.

####  5.4.3. The case $\hat{q}=8+e$, $e\geq 2$ and $s_{3}\leq 6$.

Since $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 2.5, we have
$\dim|s_{3}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3$ by Proposition 3.2.
Moreover, as $s_{3}+m_{3}e=6$ and $e\geq 2$, it follows from Table
LABEL:table.qgeq6 that the numerical type of $\hat{X}$ appears as one of the
following possibilities in Table LABEL:table.qgeq6:

  * •

№ 25 with $\hat{q}=11$, $e=3$ and $s_{3}=6$;

  * •

№ 26 with $\hat{q}=11$, $e=3$ and $s_{3}=6$;

  * •

№ 28 with $\hat{q}=13$, $e=5$ and $s_{3}=6$.

In the first two cases, we have $\dim|A_{\hat{X}}|=0$. As $e=3$, the unique
element $D\in|A_{\hat{X}}|$ is a prime divisor which is different from
$\hat{E}$. Moreover, in these two cases we have

$\dim|5A_{\hat{X}}|+\dim\hat{\scr{S}}_{3}>\dim|6A_{\hat{X}}|.$

So there exists an element in $\hat{\scr{S}}_{3}$ of the form $D+G$ with
$G\in|5A_{\hat{X}}|$, which contradicts to Lemma 5.4 as $e=3$.

In the last case, we have $s_{2}=4$ by (3.3.1). As $e=5$ and
$\dim|A_{\hat{X}}|=0$, the unique element $D\in|A_{\hat{X}}|$ is a prime
divisor which is different from $\hat{E}$. Moreover, we have

$2=\dim|3A_{\hat{X}}|+\dim\hat{\scr{S}}_{2}=\dim|4A_{\hat{X}}|.$

So there exists an element in $\hat{\scr{S}}_{2}$ of the form $D+G$ for some
$G\in|3A_{\hat{X}}|$, which contradicts to Lemma 5.4 as $e=5$.

####  5.4.4. The case $\hat{q}=12+e$, $e\geq 2$ and $s_{3}\leq 9$.

Since $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 2.5, we have
$\dim|s_{3}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{3}=3$ by Proposition 3.2. Then
it follows from Table LABEL:table.qgeq6 that the numerical type of $\hat{X}$
appears as № 30 with $\hat{q}=17$, $e=5$ and $s_{3}=9$. In particular, we have
$\dim|9A_{\hat{X}}|=\dim\hat{\scr{S}}_{3}=3$ and thus
$|9A_{\hat{X}}|=\hat{\scr{S}}_{3}$. As $e=5$ and $\dim|A_{\hat{X}}|=-1$, the
unique element $D\in|2A_{\hat{X}}|$ is a prime divisor which is different from
$\hat{E}$. As $\dim|7A_{\hat{X}}|=2$, there exists an element in
$\hat{\scr{S}}_{3}=|9A_{\hat{X}}|$ of the form $D+G$ for some
$G\in|7A_{\hat{X}}|$, which contradicts to Lemma 5.4 as $e=5$.

####  5.4.5. The case $\hat{q}=16+e$, $e\geq 2$ and $s_{3}\leq 12$.

Then we have $\hat{q}=19$, $e=3$ and $s_{2}=8$ or $s_{2}\leq 5$ by (3.3.1).
Since $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 2.5, we have
$\dim|s_{2}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{2}=1$ by Proposition 3.2. This
implies that $s_{2}=8$ from Table LABEL:table.qgeq6. In particular, we have
$\dim|8A_{\hat{X}}|=\dim\hat{\scr{S}}_{2}=1$ and thus
$|8A_{\hat{X}}|=\hat{\scr{S}}_{2}$. As $e=3$ and $\dim|kA_{\hat{X}}|=-1$ for
$k\leq 2$, the unique element $D\in|4A_{\hat{X}}|$ is a prime divisor which is
different from $\hat{E}$. So we have $2D\in\hat{\scr{S}}_{2}$, which
contradicts to Lemma 5.4 as $e=3$.

##  6\. Case $q\mathbb{Q}(X)=8$

This section is devoted to prove the following result.

######  Theorem 6.1.

The case № 22 in Table LABEL:table.qgeq6 does not occur for terminal
$\mathbb{Q}$-Fano threefolds.

Assume to the contrary that there exists a terminal $\mathbb{Q}$-Fano
threefold $X$ whose numerical type appears as № 22 in Table LABEL:table.qgeq6.
Then we have the numerical invariants of $X$ from [BK09, № 41495] as follows:

$\mathbf{B}_{X}=\\{(3,1),(5,2),(11,4)\\},\quad
c_{1}(X)^{3}=\frac{2048}{165},\quad c_{2}(X)c_{1}(X)=\frac{928}{165},\quad
b_{X}=\frac{64}{29}.$

###  6.1. Geometry of $X$

We collect in the following some geometric properties of $X$ which will be
used latter in the proof of Theorem 6.1.

######  Lemma 6.2.

Let $X$ be a terminal $\mathbb{Q}$-Fano threefold with numerical type № 22 in
Table LABEL:table.qgeq6.

  1. (6.2.1)

Every non-Gorenstein point of $X$ is a cyclic quotient singularity.

  2. (6.2.2)

The Weil divisor class group $\operatorname{Cl}(X)$ is torsion free.

  3. (6.2.3)

Let $A$ be a $\mathbb{Q}$-Cartier Weil divisor on $X$ such that $-K_{X}\sim
8A$. Then we have

$\mathcal{R}_{X}$ |  $A^{3}$ |  $|A|$ |  $|2A|$ |  $|3A|$ |  $|4A|$ |  $|5A|$ |  $|6A|$
---|---|---|---|---|---|---|---
$\\{3,5,11\\}$ |  $\frac{4}{165}$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$
  4. (6.2.4)

Let $S_{2}$ be the unique prime divisor in $|2A|$. Then any element
$\Delta\in|4A|$ is either equal to $2S_{2}$ or a prime divisor.

###### Proof.

The first statement follows from the form of $\mathbf{B}_{X}$ and [Mor85]. The
second and the third statements follow directly from Proposition 2.5 and Table
LABEL:table.qgeq6. The last statement follows from the facts that
$\dim|A|=-1$, $\dim|2A|=0$ and $\operatorname{Cl}(X)$ is torsion free. ∎

We will always denote by $S_{i}$ the unique prime divisor in $|iA|$ for $i=2$
and $3$. As in § 3 and § 5, we consider the movable linear system
$\scr{M}=\scr{S}_{4}=|4A|$ and the corresponding Sarkisov link:

$\textstyle{\tilde{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\chi}$$\textstyle{\bar{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{f}}$$\textstyle{X}$$\textstyle{\hat{X}.}$

######  Lemma 6.3.

$d\geq 2$ and $\beta_{4}\geq 6\alpha$.

###### Proof.

As $\dim|A|=-1$, we get $d\geq 2$. Let $P\in X$ be the point of local index
$11$. Then we have $\scr{M}=\scr{S}_{4}\sim-6K_{X}$ near the point $P$, hence
$c\leq 1/6$ by Lemma 3.1 and then we conclude by (3.1.1). ∎

The following simple but useful observation follows from (6.2.4) of Lemma 6.2.

######  Lemma 6.4.

Any element in $\bar{\scr{S}}_{4}$ is of the form $\bar{\Delta}+a\bar{E}$,
where $a$ is a non-negative integer and $\bar{\Delta}$ is the strict transform
of an element $\Delta$ in $\scr{S}_{4}$. In particular, if
$\bar{\Delta}\not=2\bar{S}_{2}$, then $\bar{\Delta}$ is a prime divisor, where
$\bar{S}_{2}$ is the strict transform of $S_{2}$.

Similar to § 5, the remaining part of this section is devoted to prove Theorem
6.1 case by case according to the singular type of $f(\tilde{E})$.

###  6.2. The point $f(\tilde{E})$ is a curve or a Gorenstein point.

Then both $\alpha$ and $\beta_{4}$ are integers and it follows from (3.3.1)
and Lemma 6.3 that

$4\hat{q}=8s_{4}+(8\beta_{4}-4\alpha)e\geq 8s_{4}+44\alpha e\geq 44.$ (6.4.1)

So we have $\hat{q}\geq 11$. In particular, the morphism $\hat{f}$ is
birational by Proposition 3.2 and $\operatorname{Cl}(\hat{X})$ is torsion free
by Proposition 2.5. Therefore, we have $e=d\geq 2$ by Proposition 3.2. Then
(6.4.1) implies $\hat{q}\geq 22$, which contradicts to Lemma 2.1.

###  6.3. The point $f(\tilde{E})$ is of local index $3$.

Then we have $\alpha=1/3$ by [Kaw96]. Moreover, as $A\sim-2K_{X}$ near the
point $f(\tilde{E})$, we obtain $t_{4}=2$. In particular, by (3.4.1) and Lemma
6.3, there exists an integer $m_{4}\geq 2$ such that $\beta_{4}=2/3+m_{4}$.
Then applying (3.3.1) to $k=4$ yields

$4\hat{q}=8s_{4}+(8m_{4}+4)e.$

So $\hat{q}=2s_{4}+(2m_{4}+1)e$. As $e\geq 1$ and $m_{4}\geq 2$, we have
$\hat{q}\geq 5$. In particular, the morphism $\hat{f}$ is birational and thus
$s_{4}\geq 1$ by Proposition 3.2. Therefore, we have $\hat{q}\geq 7$.
Moreover, if $e=1$, then by Proposition 3.2, the group
$\operatorname{Cl}(\hat{X})$ is not torsion free as $d\geq 2$. It follows from
Proposition 2.5 that $\hat{q}=7$, $s_{4}=1$ and the numerical type of
$\hat{X}$ appears as № 6 or № 7 in Table LABEL:table.qgeq5. Let $T$ be the
unique $2$-torsion element in $\operatorname{Cl}(\hat{X})$. Then we have

either $\hat{\scr{S}}_{4}\sim A_{\hat{X}}$ or $\hat{\scr{S}}_{4}\sim
A_{\hat{X}}+T$.

However, as $\dim\hat{\scr{S}}_{4}=1$, we get a contradiction from Proposition
2.5 and (2.6.1). Hence we have $e\geq 2$ and $\hat{q}\geq 12$. In particular,
the group $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 2.5 and
we have

$\dim|s_{4}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{4}=1.$

Then one can easily derive from Table LABEL:table.qgeq6 that $s_{4}\geq 3$ and
therefore $\hat{q}\geq 16$. Then $s_{4}\geq 5$ by Table LABEL:table.qgeq6
again and hence $\hat{q}\geq 20$, which contradicts to Lemma 2.1.

###  6.4. The point $f(\tilde{E})$ is of local index $5$.

Then we have $\alpha=1/5$ by [Kaw96]. Moreover, as $A\sim-2K_{X}$ near the
point $f(\tilde{E})$, we obtain $t_{2}=4$, $t_{3}=1$ and $t_{4}=3$. Then it
follows from (3.4.1) and Lemma 6.3 that $\beta_{2}=4/5+m_{2}$ for some
$m_{2}\in\mathbb{Z}_{\geq 0}$, $\beta_{3}=1/5+m_{3}$ for some
$m_{3}\in\mathbb{Z}_{\geq 0}$ and $\beta_{4}=3/5+m_{4}$ for some
$m_{4}\in\mathbb{Z}_{>0}$. In particular, applying (3.3.1) to $k=4$ yields

$4\hat{q}=8s_{4}+(8m_{4}+4)e.$

So $\hat{q}=2s_{4}+(2m_{4}+1)e$. As $e\geq 1$ and $m_{4}\geq 1$, we have
$\hat{q}\geq 3$. On the other hand, if $\hat{q}\geq 4$, then $\hat{f}$ is
birational and $s_{4}\geq 1$ by Proposition 3.2. So $\hat{q}\not=4$ or $6$. If
$\hat{q}\geq 8$, then $\operatorname{Cl}(\hat{X})$ is torsion free by
Proposition 2.5 and thus $e=d\geq 2$ by Proposition 3.2. In particular, we
have

$\dim|s_{4}A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{4}=1.$

Then one can easily derive from Table LABEL:table.qgeq6 that there is no
solution for $\hat{q}=2s_{4}+(2m_{4}+1)e$ in the case $\hat{q}\geq 8$. In
conclusion, the possibility for $(\hat{q},e,s_{4})$ is one of the following:

  * •

$\hat{q}=3$, $e=1$ and $s_{4}=0$;

  * •

$\hat{q}=5$, $e=1$ and $s_{4}=1$;

  * •

$\hat{q}=7$, $e=1$ and $1\leq s_{4}\leq 2$.

We treat in the following case by case according to the value of $\hat{q}$.

####  6.4.1. The case $\hat{q}=3$, $e=1$ and $s_{4}=0$.

By Proposition 3.2, the morphism $\hat{f}$ is not birational and
$\hat{X}\cong\mathbb{P}^{1}$. In particular, $\bar{\scr{S}}_{4}$ is
$\hat{f}$-vertical as $s_{4}=0$ and hence
$\bar{\scr{S}}_{4}=\hat{f}^{*}|\scr{O}_{\mathbb{P}^{1}}(1)|$ as
$\dim\bar{\scr{S}}_{4}=1$. Thanks to Lemma 6.4, the divisor
$2\bar{S}_{2}\in\bar{\scr{S}}_{4}$ is the only (cycle-theoretic) fibre of
$\hat{f}$ which is not a prime divisor on $\bar{X}$. Applying (3.3.1) to $k=3$
yields

$s_{3}+m_{3}=1.$

If $s_{3}=0$, then $\bar{S}_{3}$ is $\hat{f}$-vertical. In particular, there
exists a positive integer $a$ such that $a\bar{S}_{3}\in\bar{\scr{S}}_{4}$.
Pushing it forward to $X$ yields $aS_{3}\in\scr{S}_{4}$, which is a
contradiction. So we have $s_{3}=1$ and $m_{3}=0$. Moreover, as $e=s_{3}=1$,
applying the same argument as that in (5.3.1.2) shows that there exists an
integer $b$ and an effective $\hat{f}$-vertical Weil divisor $\bar{\Delta}$
such that

$\bar{S}_{3}-\bar{E}\sim b\bar{F}+\bar{\Delta},$

where $\bar{F}$ is a general fibre of $\hat{f}$. However, note that there
exists an integer $b^{\prime}$ such that $b\bar{F}+\bar{\Delta}\sim
b^{\prime}\bar{S}_{2}$. In particular, pushing forward to $X$ shows that
$S_{3}\sim b^{\prime}S_{2}$, which is impossible.

####  6.4.2. The case $\hat{q}=5$, $e=1$ and $s_{4}=1$.

Applying (3.3.1) to $k=2$ yields

$10=2\hat{q}=8s_{2}+(8m_{2}+6)e=8s_{2}+8m_{2}+6,$

which has no solution of integers.

####  6.4.3. The case $\hat{q}=7$, $e=1$ and $1\leq s_{4}\leq 2$.

Applying (3.3.1) to $k=3$ yields

$21=3\hat{q}=8s_{3}+(8m_{3}+1)e=8s_{3}+8m_{3}+1,$

which again has no solution of integers.

###  6.5. The point $f(\tilde{E})$ is of local index $11$.

Then we have $\alpha=1/11$ by [Kaw96]. Moreover, as $A\sim-7K_{X}$ near the
point $f(\tilde{E})$, we obtain $t_{2}=3$, $t_{3}=10$, $t_{4}=6$, $t_{5}=2$
and $t_{6}=9$. In particular, by (3.4.1), we have $\beta_{k}=t_{k}/11+m_{k}$
for some $m_{k}\in\mathbb{Z}_{\geq 0}$ ($2\leq k\leq 6$). Applying (3.3.1) to
$k=3$ yields

$3\hat{q}=8s_{3}+(8m_{3}+7)e.$ (6.4.2)

Then one can easily derive that $3\hat{q}\geq 14$ and hence $\hat{q}\geq 5$.
In particular, the morphism $\hat{f}$ is birational by Proposition 3.2. Now
applying (3.3.1) to $k=2$ yields

$2\hat{q}=8s_{2}+(8m_{2}+2)e.$ (6.4.3)

So $\hat{q}=4s_{2}+(4m_{2}+1)e$. In particular, as $\hat{q}\leq 19$, we have
$\hat{q}-e=4l$ for some integer $0\leq l\leq 4$. On the other hand, applying
(3.3.1) to $k=4$ yields

$4\hat{q}=8s_{4}+(8m_{4}+4)e.$ (6.4.4)

So $\hat{q}=2s_{4}+(2m_{4}+1)e$. In particular, if $\hat{q}=e$, then we have
$s_{4}=m_{4}=0$, which contradicts to Proposition 3.2 as $\hat{f}$ is
birational. Hence $l\geq 1$. If $\hat{q}\geq 8$, then we have $e=d\geq 2$ by
Proposition 2.5. In summary, the possibility for $(\hat{q},e,s_{4})$ is one of
the following:

  * •

$\hat{q}=4+e$ and $1\leq s_{4}\leq 2$;

  * •

$\hat{q}=8+e$, $e\geq 2$ and $1\leq s_{4}\leq 4$;

  * •

$\hat{q}=12+e$, $e\geq 2$ and $1\leq s_{4}\leq 6$;

  * •

$\hat{q}=16+e$, $e\geq 2$ and $1\leq s_{4}\leq 8$.

We treat in the following case by case according to the value of $\hat{q}$.

####  6.5.1. The case $\hat{q}=4+e$ and $1\leq s_{4}\leq 2$.

Then it follows from (6.4.2) that $e=1$ or $3$. We divide the proof into two
subcases according to the value of $e$.

#####  6.5.1.1 The subcase $e=1$.

As $d\geq 2$, the group $\operatorname{Cl}(\hat{X})$ is not torsion free by
Proposition 3.2. By (6.4.2) and (6.4.3), we also have

$s_{2}+m_{2}=1\quad\textup{and}\quad s_{3}+m_{3}=1.$

Note that one of $s_{2}$ and $s_{3}$ is non-zero by Proposition 3.2. So we
assume that $s_{i}=1$ for $i=2$ or $3$. Then we have

$\hat{S}_{i}\sim_{\mathbb{Q}}\hat{E}\sim_{\mathbb{Q}}A_{\hat{X}}.$

As $\hat{S}_{i}\not=\hat{E}$, it follows from Table LABEL:table.qgeq5 and
(2.6.1) that the numerical type of $\hat{X}$ appears as one of № 4 and № 5 in
Table LABEL:table.qgeq5 and $s_{4}=2$. In particular, we have $d=2$ and
$S_{2}=F$. This implies $s_{3}=1$ and $s_{2}=0$ by Proposition 3.2. As
$m_{4}=0$ by (6.4.4), we obtain $\beta_{4}=6/11$; as $m_{2}=1$, we obtain
$\beta_{2}=14/11$. Since $2S_{2}\in\scr{S}_{4}$, we have
$2\bar{F}+2\bar{E}=2\bar{S}_{2}+(2\beta_{2}-\beta_{4})\bar{E}\in\bar{\scr}{S}_{4}$.
Pushing it forward to $\hat{X}$ yields

$2\hat{E}\in\hat{\scr{S}}_{4}.$

Then we have $2\hat{S}_{3}\sim 2A_{\hat{X}}\sim 2\hat{E}\in\hat{\scr{S}}_{4}$
as $\hat{S}_{3}\sim_{\mathbb{Q}}A_{\hat{X}}\sim_{\mathbb{Q}}\hat{E}$ and
$\operatorname{Cl}(\hat{X})_{t}$ has order two. It follows that
$|2A_{\hat{X}}|=\hat{\scr{S}}_{4}$ as
$\dim|2A_{\hat{X}}|=\dim\hat{\scr{S}}_{4}=1$. Hence
$2\hat{S}_{3}\in\hat{\scr}{S}_{4}$, which contradicts to Lemma 6.4 as
$\hat{S}_{3}\not=\hat{E}$.

#####  6.5.1.2 The subcase $e=3$.

Then we have $s_{3}=m_{3}=0$ by (6.4.2) and $S_{3}=F$ by Proposition 3.2. In
particular, we have $d=e=3$. Thus $s_{2}>0$ and $\operatorname{Cl}(\hat{X})$
is torsion free by Proposition 3.2. Then we obtain $s_{2}=1$ by (6.4.3) and
$s_{4}=2$ by (6.4.4). As $\beta_{6}=9/11+m_{6}$, applying (3.3.1) to $k=6$
yields

$42=6\hat{q}=8s_{6}+(8m_{6}+6)e=8s_{6}+24m_{6}+18.$

So $3m_{6}+s_{6}=3$. As $\dim\scr{S}_{6}=3$ and $\hat{f}$ is birational, it
follows from Proposition 3.2 that $s_{6}>0$. Therefore $s_{6}=3$ and
$m_{6}=0$. In particular, since $\operatorname{Cl}(\hat{X})$ is torsion free,
we have

$\dim|3A_{\hat{X}}|\geq\dim\hat{\scr{S}}_{6}=3.$

Hence, the numerical type of $\hat{X}$ appears as one of № 7-9 in Table
LABEL:table.qgeq6. In particular, we have $\dim|-K_{\hat{X}}|\geq 15$ and it
follows from [Pro16, Theorem 1.1] that $\hat{X}$ is isomorphic to one of the
following

$\mathbb{P}(1^{2},2,3),\quad X_{6}\subset\mathbb{P}(1,2^{2},3,5),\quad
X_{6}\subset\mathbb{P}(1,2,3^{2},4).$

Firstly we consider the case where $\hat{X}$ is isomorphic to
$\mathbb{P}(1^{2},2,3)$ (№ 7 in Table LABEL:table.qgeq6). As
$\dim|A_{\hat{X}}|=1$ and $e=3$, there exists a prime divisor
$\hat{D}\in|A_{\hat{X}}|$ such that $\hat{D}$ is different from the prime
divisors $\hat{S}_{2}$ and $\hat{E}$. Note that we have

$6=\dim|2A_{\hat{X}}|+\dim\hat{\scr{S}}_{6}=\dim|3A_{\hat{X}}|.$

So there exists an element in $\hat{\scr{S}}_{6}$ of the form
$\hat{D}+\hat{\Delta}$ for some $\hat{\Delta}\in|2A_{\hat{X}}|$. Denote by $D$
and $\Delta$ the strict transforms of $\hat{D}$ and $\hat{\Delta}$ on $X$,
respectively. Then $D$ is different from $S_{2}$ and $S_{3}(=F)$. In
particular, there exists an integer $a\geq 4$ such that $D\sim aA$ because
$\operatorname{Cl}(X)$ is torsion free. Let $\gamma$ be the unique non-
negative rational number such that

$\hat{f}^{*}(\hat{D}+\hat{\Delta})-\gamma\bar{F}\in\bar{\scr{S}}_{6}.$

As $d=3$ and $a\geq 4$, pushing the above effective divisor forward to $X$
shows that $D+\Delta\in\scr{S}_{6}$. However, since $\hat{E}$ is not contained
in the support of $\hat{\Delta}$, we have $\Delta\not=0$. Hence we must have
$\Delta=S_{2}$ and $D\in|4A|$. As $e=3$, we obtain
$\hat{S}_{2}\in|2A_{\hat{X}}|$, which is impossible as $s_{2}=1$.

Next we consider the case where $\hat{X}$ is isomorphic to
$X_{6}\subset\mathbb{P}(1,2^{2},3,5)$ (№ 8 in Table LABEL:table.qgeq6). In
this case, we have

$\dim|3A_{\hat{X}}|=\dim\scr{S}_{6}=3$ and
$2=\dim|2A_{\hat{X}}|>\dim\scr{S}_{4}=1$.

In particular, there exists a non-negative rational number $\gamma$ such that

$\bar{\scr{S}}_{6}=\hat{f}^{*}|3A_{\hat{X}}|-\gamma\bar{F}.$

Consider the $2$-dimensional subspace of $\bar{\scr{S}}_{6}$ defined as
follows:

$\bar{\scr{N}}\coloneqq\hat{f}^{*}(\hat{S}_{2}+|2A_{\hat{X}}|)-\gamma\bar{F}.$

Pushing it forward to $X$ yields a $2$-dimensional subspace $\scr{N}$ of
$\scr{S}_{6}$. Clearly the divisor $S_{2}$ is contained in the fixed part of
$\scr{N}$. So we get

$\scr{N}^{\prime}=\scr{N}-S_{2}\subset\scr{S}_{4}.$

However, as $\dim\scr{N}^{\prime}=\dim\scr{N}=\dim|2A_{\hat{X}}|=2$ and
$\dim\scr{S}_{4}=1$, we get a contradiction.

Finally we consider the case where $\hat{X}$ is isomorphic to
$X_{6}\subset\mathbb{P}(1,2,3^{2},4)$ (№ 9 in Table LABEL:table.qgeq6). Recall
from [Pro22b, 4.9] that we can write

$K_{\bar{X}}\sim_{\mathbb{Q}}\hat{f}^{*}K_{\hat{X}}+b\bar{F},\quad\bar{\scr{S}}_{k}\sim_{\mathbb{Q}}\bar{f}^{*}\hat{\scr{S}}_{k}-\gamma_{k}\bar{F},\quad\bar{E}\sim_{\mathbb{Q}}\hat{f}^{*}\hat{E}-\delta\bar{F}.$

Then it follows from [Pro22b, 4.5] that for any $k\geq 1$, we have

$-8s_{k}+7k=3(bs_{k}-7\gamma_{k}).$ (6.4.5)

Applying it to $k=2$ yields

$\gamma_{2}=\frac{b-2}{7}.$

In particular, we get $b\geq 2$ and thus $\hat{f}(\bar{F})$ must be a point as
$b=1$ if $\hat{f}(\bar{F})$ is a curve. If $b>2$, then we have $\gamma_{4}>0$
and $\gamma_{6}>0$ by (6.4.5). In particular, as
$|2A_{\hat{X}}|=\hat{\scr{S}}_{4}$ and $|3A_{\hat{X}}|=\hat{\scr{S}}_{6}$, the
point $\hat{f}(\bar{F})$ is contained in the base loci of $|2A_{\hat{X}}|$ and
$|3A_{\hat{X}}|$. So we obtain

$\hat{f}(\bar{F})=[0:0:0:0:1]\in\hat{X}=X_{6}\subset\mathbb{P}(1,2,3^{2},4).$

Therefore the point $\hat{f}(\bar{F})$ is a point of local index $4$ and thus
$b=1/4$ by [Pro16, Remark 5.1], which is a contradiction. So we have $b=2$ and
then $\gamma_{4}=0$ by (6.4.5). Since $|2A_{\hat{X}}|=\hat{\scr{S}}_{4}$ is a
pencil, there exists an element $\hat{D}\in\hat{\scr{S}}_{4}$ containing
$\hat{f}(\bar{F})$. In particular, we can find a positive integer $a$ such
that

$\hat{f}^{*}\hat{D}=\bar{D}+a\bar{F}\in\hat{f}^{*}\hat{\scr{S}}_{4}=\hat{f}^{*}|2A_{\hat{X}}|=\bar{\scr{S}}_{4},$

where $\bar{D}$ is the strict transform of $\hat{D}$. This contradicts to
Lemma 6.4 as $e=d=3$.

####  6.5.2. The case $\hat{q}=8+e$, $e\geq 2$ and $1\leq s_{4}\leq 4$.

Then $\hat{q}\geq 10$. In particular, it follows from Lemma 2.1 that $\hat{q}$
is odd and so is $e$. On the other hand, by (6.4.2), we have

$6=2(s_{3}+m_{3}e)+e.$

So $e$ is even, which is a contradiction.

####  6.5.3. The case $\hat{q}=12+e$, $e\geq 2$ and $1\leq s_{4}\leq 6$.

Since $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 2.5, we have

$\dim|s_{4}A_{\hat{X}}|\geq\dim\scr{S}_{4}=1.$

In particular, as $s_{4}\leq 6$ and $e\geq 2$, it follows from Table
LABEL:table.qgeq6 that $\hat{q}=17$, $e=5$ and $s_{4}\geq 5$. Then we get
$s_{4}=6$ by (6.4.4) and so $|6A_{\hat{X}}|=\hat{\scr{S}}_{4}$ by Table
LABEL:table.qgeq6. Let $D$ be the unique element in $|2A_{\hat{X}}|$. Then $D$
is a prime divisor and $3D\in\hat{\scr{S}}_{4}$. This contradicts to Lemma 6.4
as $e=5$.

####  6.5.4. The case $\hat{q}=16+e$, $e\geq 2$ and $1\leq s_{4}\leq 8$.

Then it follows from Lemma 2.1 that $\hat{q}=19$ and $e=3$. On the other hand,
by (6.4.2), we have

$9=2s_{3}+6m_{3},$

which is impossible.

##  Appendix A Database for $\mathbb{Q}$-Fano threefolds with large
$q\mathbb{Q}$

###  A.1. Data for $q\mathbb{Q}(X)\geq 6$

We collect in Table LABEL:table.qgeq6 below the possible numerical invariants
for terminal Fano threefolds with $qW(X)=q\mathbb{Q}(X)\geq 6$ and $b_{X}<3$
(cf. Corollary 2.3). It can be derived from the Grdb in [BK09] or using the
same computer program as that in the proof of Lemma 2.2. We denote by $A$ a
$\mathbb{Q}$-Cartier Weil divisor such that $-K_{X}\sim qW(X)A$.

If we assume in addition that $X$ is a terminal $\mathbb{Q}$-Fano threefold,
then we use the symbol “$+$” as a superscript of its numbering № if it can be
geometrically realised by appropriate examples, “$+!$” if it is completely
described, “$-$” if it cannot occur and “$?$” if it is unknown ([BS07a, Pro10,
Pro13, Pro16, Pro22a]). We also use the symbol “$\dagger$” as a subscript in
the case where $\operatorname{Cl}(X)$ is possibly not torsion free (see
Proposition 2.5). There are only two possibilities for such $X$ (№ 12 and №
14) with the corresponding index one covers $X^{\prime}$ (№ 10 and № 9).

Table 6. Data for $q=qW(X)=q\mathbb{Q}(X)\geq 6$ |  |  |  |  |  |  |  |  |  |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
№ |  $q$ |  $\mathcal{R}_{X}$ |  $c_{1}^{3}$ |  $c_{2}c_{1}$ |  $b_{X}\approx$ | $\dim|kA|$
|  |  |  |  |  |  $|A|$ |  $|2A|$ |  $|3A|$ |  $|4A|$ |  $|5A|$ |  $|6A|$ |  $|7A|$ |  $|8A|$ |  $|9A|$
1+! |  $6$ |  $\\{5\\}$ |  $\frac{216}{5}$ |  $\frac{96}{5}$ |  $2.25$ |  $1$ |  $3$ |  $6$ |  $10$ |  $16$ |  $23$ |  $32$ |  $43$ |  $56$
2? |  $6$ |  $\\{5,7\\}$ |  $\frac{648}{35}$ |  $\frac{432}{35}$ |  $1.5$ |  $0$ |  $1$ |  $2$ |  $4$ |  $7$ |  $10$ |  $14$ |  $18$ |  $24$
3? |  $6$ |  $\\{5,17\\}$ |  $\frac{432}{85}$ |  $\frac{192}{85}$ |  $2.25$ |  $0$ |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$
4? |  $6$ |  $\\{5,11\\}$ |  $\frac{216}{55}$ |  $\frac{456}{55}$ |  $0.4736$ |  $0$ |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$
5? |  $6$ |  $\\{5,7^{2}\\}$ |  $\frac{432}{35}$ |  $\frac{192}{35}$ |  $2.25$ |  $-1$ |  $0$ |  $1$ |  $2$ |  $4$ |  $6$ |  $9$ |  $11$ |  $15$
6? |  $6$ |  $\\{7,11\\}$ |  $\frac{432}{77}$ |  $\frac{480}{77}$ |  $0.9$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $1$ |  $3$ |  $4$ |  $5$ |  $7$
7+! |  $7$ |  $\\{2,3\\}$ |  $\frac{343}{6}$ |  $\frac{119}{6}$ |  $2.8823$ |  $1$ |  $3$ |  $6$ |  $10$ |  $15$ |  $22$ |  $30$ |  $40$ |  $52$
8+! |  $7$ |  $\\{2^{3},5\\}$ |  $\frac{343}{10}$ |  $\frac{147}{10}$ |  $2.3333$ |  $0$ |  $2$ |  $3$ |  $6$ |  $9$ |  $13$ |  $18$ |  $24$ |  $31$
9+! |  $7$ |  $\\{2,3^{2},4\\}$ |  $\frac{343}{12}$ |  $\frac{161}{12}$ |  $2.1304$ |  $0$ |  $1$ |  $3$ |  $5$ |  $7$ |  $11$ |  $15$ |  $20$ |  $26$
10+ |  $7$ |  $\\{2^{2},3,5\\}$ |  $\frac{343}{15}$ |  $\frac{203}{15}$ |  $1.6896$ |  $0$ |  $1$ |  $2$ |  $4$ |  $6$ |  $9$ |  $12$ |  $16$ |  $21$
11? |  $7$ |  $\\{3,6,9\\}$ |  $\frac{343}{18}$ |  $\frac{119}{18}$ |  $2.8823$ |  $0$ |  $0$ |  $1$ |  $2$ |  $4$ |  $7$ |  $10$ |  $13$ |  $17$
12${}^{?}_{\dagger}$ |  $7$ |  $\\{2,6,10\\}$ |  $\frac{343}{30}$ |  $\frac{203}{30}$ |  $1.6896$ |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $4$ |  $6$ |  $8$ |  $10$
13? |  $7$ |  $\\{2,3,13\\}$ |  $\frac{343}{78}$ |  $\frac{539}{78}$ |  $0.6363$ |  $0$ |  $0$ |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$
14${}^{?}_{\dagger}$ |  $7$ |  $\\{2^{2},3,4,8\\}$ |  $\frac{343}{24}$ |  $\frac{161}{24}$ |  $2.1304$ |  $-1$ |  $0$ |  $1$ |  $2$ |  $3$ |  $5$ |  $7$ |  $10$ |  $12$
15? |  $7$ |  $\\{2^{2},3,11\\}$ |  $\frac{343}{33}$ |  $\frac{245}{33}$ |  $1.4$ |  $-1$ |  $0$ |  $1$ |  $1$ |  $2$ |  $4$ |  $5$ |  $7$ |  $9$
16+ |  $7$ |  $\\{2^{3},3,4,5\\}$ |  $\frac{343}{60}$ |  $\frac{497}{60}$ |  $0.6901$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$
17? |  $7$ |  $\\{2^{3},5,8\\}$ |  $\frac{343}{40}$ |  $\frac{273}{40}$ |  $1.2564$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $6$ |  $7$
18? |  $7$ |  $\\{3,8,9\\}$ |  $\frac{343}{72}$ |  $\frac{329}{72}$ |  $1.0425$ |  $-1$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $1$ |  $2$ |  $3$ |  $4$
19+! |  $8$ |  $\\{3^{2},5\\}$ |  $\frac{512}{15}$ |  $\frac{208}{15}$ |  $2.4615$ |  $0$ |  $1$ |  $3$ |  $4$ |  $7$ |  $10$ |  $13$ |  $18$ |  $23$
20+! |  $8$ |  $\\{3,7\\}$ |  $\frac{512}{21}$ |  $\frac{304}{21}$ |  $1.6842$ |  $0$ |  $1$ |  $2$ |  $3$ |  $5$ |  $7$ |  $10$ |  $13$ |  $17$
21+ |  $8$ |  $\\{5,7\\}$ |  $\frac{512}{35}$ |  $\frac{432}{35}$ |  $1.1851$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $6$ |  $8$ |  $10$
22- |  $8$ |  $\\{3,5,11\\}$ |  $\frac{2048}{165}$ |  $\frac{928}{165}$ |  $2.2068$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $6$ |  $8$
23+! |  $9$ |  $\\{2,4,5\\}$ |  $\frac{729}{20}$ |  $\frac{279}{20}$ |  $2.6129$ |  $0$ |  $1$ |  $2$ |  $4$ |  $6$ |  $8$ |  $11$ |  $15$ |  $19$
24+ |  $9$ |  $\\{2^{3},5,7\\}$ |  $\frac{729}{70}$ |  $\frac{549}{70}$ |  $1.3278$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$
25+! |  $11$ |  $\\{2,3,5\\}$ |  $\frac{1331}{30}$ |  $\frac{451}{30}$ |  $2.9512$ |  $0$ |  $1$ |  $2$ |  $3$ |  $5$ |  $7$ |  $9$ |  $12$ |  $15$
26+ |  $11$ |  $\\{2,5,7\\}$ |  $\frac{1331}{70}$ |  $\frac{759}{70}$ |  $1.7536$ |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$ |  $6$
27+ |  $11$ |  $\\{2^{2},3,4,7\\}$ |  $\frac{1331}{84}$ |  $\frac{649}{84}$ |  $2.0508$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$
28+! |  $13$ |  $\\{3,4,5\\}$ |  $\frac{2197}{60}$ |  $\frac{767}{60}$ |  $2.8644$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$ |  $5$ |  $7$ |  $9$
29+! |  $13$ |  $\\{2,3^{2},5,7\\}$ |  $\frac{2197}{210}$ |  $\frac{1157}{210}$ |  $1.8988$ |  $-1$ |  $-1$ |  $0$ |  $0$ |  $0$ |  $1$ |  $1$ |  $1$ |  $2$
30+! |  $17$ |  $\\{2,3,5,7\\}$ |  $\frac{4913}{210}$ |  $\frac{1717}{210}$ |  $2.8613$ |  $-1$ |  $0$ |  $0$ |  $0$ |  $1$ |  $1$ |  $2$ |  $2$ |  $3$
31+! |  $19$ |  $\\{3,4,5,7\\}$ |  $\frac{6859}{420}$ |  $\frac{2489}{420}$ |  $2.7557$ |  $-1$ |  $-1$ |  $0$ |  $0$ |  $0$ |  $0$ |  $1$ |  $1$ |  $1$

###  A.2. Data for $q\mathbb{Q}(X)=5$

We collect in Table LABEL:table.qQ=5 below the possible numerical invariants
for terminal Fano threefolds with $qW(X)=q\mathbb{Q}(X)=5$ and $b_{X}<3$ (cf.
Theorem 4.1). Again it can be derived from the Grdb in [BK09] or using the
computer program as that in the proof of Lemma 2.2. We denote by $A$ a
$\mathbb{Q}$-Cartier Weil divisor such that $-K_{X}\sim 5A$.

If we assume in addition that $X$ is a terminal $\mathbb{Q}$-Fano threefold,
then the number in the column “$\operatorname{Cl}(X)_{t}$” is the order of the
possible torsion subgroup $\operatorname{Cl}(X)_{t}$ of $\operatorname{Cl}(X)$
if it is non-trivial. In this case, the group $\operatorname{Cl}(X)_{t}$ is a
cyclic group generated by some $T$ (see [Pro22b, Proposition 3.4] or
Proposition 2.5). Moreover, we have $qW(X^{\prime})=q\mathbb{Q}(X^{\prime})=5$
for the index one cover $X^{\prime}$ of $X$ defined by $T$ and the numerical
type of $X^{\prime}$ coincides with the one appears in the column
“$X^{\prime}$”.

Table 7. Data for $qW(X)=q\mathbb{Q}(X)=5$ |  |  |  |  |  |  |  |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---
№ |  $\mathcal{R}_{X}$ |  $c_{1}^{3}$ |  $c_{2}c_{1}$ |  $b_{X}\approx$ |  $\operatorname{Cl}(X)_{t}$ |  $X^{\prime}$ | $\dim|kA|$
|  |  |  |  |  |  |  $|A|$ |  $|2A|$ |  $|3A|$ |  $|4A|$ |  $|5A|$ |  $|6A|$
1 |  $\\{2^{2},8,11\\}$ |  $\frac{125}{88}$ |  $\frac{195}{88}$ |  $0.6410$ |  - |  - |  $-1$ |  $-1$ |  $-1$ |  $0$ |  $0$ |  $0$
2 |  $\\{2,3^{2},7^{2}\\}$ |  $\frac{125}{42}$ |  $\frac{145}{42}$ |  $0.8620$ |  - |  - |  $-1$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $2$
3 |  $\\{2,3,4,6,7\\}$ |  $\frac{125}{28}$ |  $\frac{95}{28}$ |  $1.3157$ |  - |  - |  $-1$ |  $-1$ |  $0$ |  $1$ |  $2$ |  $3$
4 |  $\\{2,3,7,11\\}$ |  $\frac{2375}{462}$ |  $\frac{955}{462}$ |  $2.4869$ |  - |  - |  $-1$ |  $-1$ |  $0$ |  $1$ |  $2$ |  $3$
5 |  $\\{2,7,8\\}$ |  $\frac{125}{56}$ |  $\frac{435}{56}$ |  $0.2873$ |  - |  - |  $-1$ |  $0$ |  $0$ |  $0$ |  $2$ |  $1$
6 |  $\\{2^{4},3^{2},8\\}$ |  $\frac{125}{24}$ |  $\frac{115}{24}$ |  $1.0869$ |  - |  - |  $-1$ |  $0$ |  $0$ |  $1$ |  $2$ |  $4$
7 |  $\\{7,11\\}$ |  $\frac{500}{77}$ |  $\frac{480}{77}$ |  $1.0416$ |  - |  - |  $-1$ |  $0$ |  $1$ |  $1$ |  $4$ |  $4$
8 |  $\\{2^{3},3^{3},7\\}$ |  $\frac{125}{14}$ |  $\frac{65}{14}$ |  $1.9230$ |  - |  - |  $-1$ |  $0$ |  $1$ |  $2$ |  $4$ |  $7$
9 |  $\\{2^{3},3^{2},4,6\\}$ |  $\frac{125}{12}$ |  $\frac{55}{12}$ |  $2.2727$ |  - |  - |  $-1$ |  $0$ |  $1$ |  $3$ |  $5$ |  $8$
10 |  $\\{2,4,9\\}$ |  $\frac{125}{36}$ |  $\frac{355}{36}$ |  $0.3521$ |  - |  - |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$
11 |  $\\{2,4,14\\}$ |  $\frac{125}{28}$ |  $\frac{135}{28}$ |  $0.9259$ |  $2$ |  16 |  $0$ |  $0$ |  $0$ |  $1$ |  $2$ |  $3$
12 |  $\\{2,9^{2}\\}$ |  $\frac{125}{18}$ |  $\frac{85}{18}$ |  $1.4705$ |  $3$ |  27 |  $0$ |  $0$ |  $0$ |  $1$ |  $3$ |  $5$
13 |  $\\{4,7\\}$ |  $\frac{125}{28}$ |  $\frac{375}{28}$ |  $0.3333$ |  - |  - |  $0$ |  $0$ |  $1$ |  $2$ |  $3$ |  $4$
14 |  $\\{7,9\\}$ |  $\frac{500}{63}$ |  $\frac{520}{63}$ |  $0.9615$ |  - |  - |  $0$ |  $0$ |  $1$ |  $2$ |  $4$ |  $6$
15 |  $\\{7,14\\}$ |  $\frac{125}{14}$ |  $\frac{45}{14}$ |  $2.7777$ |  - |  - |  $0$ |  $0$ |  $1$ |  $2$ |  $4$ |  $6$
16 |  $\\{4^{2},7\\}$ |  $\frac{125}{14}$ |  $\frac{135}{14}$ |  $0.9259$ |  - |  - |  $0$ |  $0$ |  $1$ |  $3$ |  $5$ |  $7$
17 |  $\\{4^{2},12\\}$ |  $\frac{125}{12}$ |  $\frac{55}{12}$ |  $2.2727$ |  $2$ |  26 |  $0$ |  $0$ |  $1$ |  $3$ |  $5$ |  $7$
18 |  $\\{4,7,9\\}$ |  $\frac{3125}{252}$ |  $\frac{1135}{252}$ |  $2.7533$ |  - |  - |  $0$ |  $0$ |  $1$ |  $3$ |  $6$ |  $9$
19 |  $\\{2^{2},3,4\\}$ |  $\frac{125}{12}$ |  $\frac{175}{12}$ |  $0.7142$ |  - |  - |  $0$ |  $1$ |  $2$ |  $4$ |  $6$ |  $9$
20 |  $\\{2^{2},3,9\\}$ |  $\frac{125}{9}$ |  $\frac{85}{9}$ |  $1.4705$ |  - |  - |  $0$ |  $1$ |  $2$ |  $4$ |  $7$ |  $11$
21 |  $\\{2^{2},4,8\\}$ |  $\frac{125}{8}$ |  $\frac{75}{8}$ |  $1.6666$ |  $2$ |  29 |  $0$ |  $1$ |  $2$ |  $5$ |  $8$ |  $12$
22 |  $\\{2,3,7\\}$ |  $\frac{625}{42}$ |  $\frac{545}{42}$ |  $1.1467$ |  - |  - |  $0$ |  $1$ |  $3$ |  $5$ |  $8$ |  $12$
23 |  $\\{2,11\\}$ |  $\frac{375}{22}$ |  $\frac{255}{22}$ |  $1.4705$ |  - |  - |  $0$ |  $1$ |  $3$ |  $6$ |  $9$ |  $13$
24 |  $\\{2,3,4,7\\}$ |  $\frac{1625}{84}$ |  $\frac{775}{84}$ |  $2.0967$ |  - |  - |  $0$ |  $1$ |  $3$ |  $6$ |  $10$ |  $15$
25 |  $\\{2,7,8\\}$ |  $\frac{1125}{56}$ |  $\frac{435}{56}$ |  $2.5862$ |  - |  - |  $0$ |  $1$ |  $3$ |  $6$ |  $10$ |  $15$
26 |  $\\{2,4^{2},6\\}$ |  $\frac{125}{6}$ |  $\frac{55}{6}$ |  $2.2727$ |  $2$ |  32 |  $0$ |  $1$ |  $3$ |  $7$ |  $11$ |  $16$
27 |  $\\{2^{3},3^{2}\\}$ |  $\frac{125}{6}$ |  $\frac{85}{6}$ |  $1.4705$ |  - |  - |  $0$ |  $2$ |  $4$ |  $7$ |  $11$ |  $17$
28 |  $\\{2^{3},3,8\\}$ |  $\frac{625}{24}$ |  $\frac{215}{24}$ |  $2.9069$ |  - |  - |  $0$ |  $2$ |  $4$ |  $8$ |  $13$ |  $20$
29 |  $\\{2,4\\}$ |  $\frac{125}{4}$ |  $\frac{75}{4}$ |  $1.6666$ |  - |  - |  $1$ |  $3$ |  $6$ |  $11$ |  $17$ |  $25$
30 |  $\\{2,9\\}$ |  $\frac{625}{18}$ |  $\frac{245}{18}$ |  $2.5510$ |  - |  - |  $1$ |  $3$ |  $6$ |  $11$ |  $18$ |  $27$
31 |  $\\{7\\}$ |  $\frac{250}{7}$ |  $\frac{120}{7}$ |  $2.0833$ |  - |  - |  $1$ |  $3$ |  $7$ |  $12$ |  $19$ |  $28$
32 |  $\\{2^{2},3\\}$ |  $\frac{125}{3}$ |  $\frac{55}{3}$ |  $2.2727$ |  - |  - |  $1$ |  $4$ |  $8$ |  $14$ |  $22$ |  $33$
33 |  $\\{2\\}$ |  $\frac{125}{2}$ |  $\frac{45}{2}$ |  $2.7777$ |  - |  - |  $2$ |  $6$ |  $12$ |  $21$ |  $33$ |  $49$

###  A.3. Data for $qW(X)\not=q\mathbb{Q}(X)=4$

We collect in Table LABEL:table.qWnot=qQ=4 below the possible numerical
invariants for terminal $\mathbb{Q}$-Fano threefolds with
$qW(X)\not=q\mathbb{Q}(X)=4$ and $b_{X}\leq 64/21$. It is communicated to us
by Y. Prokhorov and the algorithm to find them is outlined in the proof of
[Pro22b, Proposition 3.3]. In all these cases we have
$\operatorname{Cl}(X)_{t}\cong\mathbb{Z}/2\mathbb{Z}$ and $qW(X)=1$. Let $T$
be the unique $2$-torsion element in $\operatorname{Cl}(X)$ and let $A$ be a
$\mathbb{Q}$-Cartier Weil divisor such that $-K_{X}\sim_{\mathbb{Q}}4A$. After
exchanging $A$ and $A+T$ if necessary, we have the following possibilities for
$X$.

Table 8. Data for $qW(X)\not=q\mathbb{Q}(X)=4$ |  |  |  |  |  |  |  |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---
№ |  $\mathcal{R}_{X}$ |  $c_{1}^{3}$ |  $c_{2}c_{1}$ |  $b_{X}\approx$ | $\dim|kA|$ | $\dim|kA+T|$
|  |  |  |  |  $|A|$ |  $|2A|$ |  $|3A|$ |  $|4A|$ |  $|A+T|$ |  $|2A+T|$ |  $|3A+T|$ |  $|4A+T|$
1 |  $\\{2^{5},6\\}$ |  $\frac{64}{3}$ |  $\frac{32}{3}$ |  $2$ |  $0$ |  $3$ |  $6$ |  $11$ |  $1$ |  $2$ |  $6$ |  $11$
2 |  $\\{2^{5},5,6\\}$ |  $\frac{128}{15}$ |  $\frac{88}{15}$ |  $1.4545$ |  $-1$ |  $1$ |  $2$ |  $4$ |  $0$ |  $0$ |  $2$ |  $4$
3 |  $\\{2^{5},6,7\\}$ |  $\frac{64}{21}$ |  $\frac{80}{21}$ |  $0.8$ |  $-1$ |  $0$ |  $0$ |  $1$ |  $0$ |  $-1$ |  $0$ |  $1$
4 |  $\\{2^{3},10\\}$ |  $\frac{128}{5}$ |  $\frac{48}{5}$ |  $2.6666$ |  $0$ |  $3$ |  $7$ |  $13$ |  $1$ |  $3$ |  $7$ |  $13$
5 |  $\\{2^{3},3,10\\}$ |  $\frac{64}{15}$ |  $\frac{104}{15}$ |  $0.6153$ |  $-1$ |  $0$ |  $1$ |  $2$ |  $0$ |  $0$ |  $1$ |  $2$
6 |  $\\{2^{3},5,10\\}$ |  $\frac{64}{5}$ |  $\frac{24}{5}$ |  $2.6666$ |  $-1$ |  $1$ |  $3$ |  $6$ |  $0$ |  $1$ |  $3$ |  $6$
7 |  $\\{2^{3},7,10\\}$ |  $\frac{256}{35}$ |  $\frac{96}{35}$ |  $2.6666$ |  $-1$ |  $0$ |  $1$ |  $3$ |  $0$ |  $0$ |  $1$ |  $3$

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