Fano threefolds in positive characteristic III

Masaya Asai and Hiromu Tanaka Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 153-8914, JAPAN [email protected] [email protected]

Abstract.

We classify primitive Fano threefolds in positive characteristic whose Picard numbers are at least two. We also classify Fano theefolds of Picard rank two.

Key words and phrases:

Fano threefolds, positive characteristic, classification.

2010 Mathematics Subject Classification:

14G17, 14E30.

1. Introduction

This article is the third part of our series of papers. In the first and second parts [TanI], [TanII], we studied Fano threefolds $X$ in positive characteristic with $\rho(X)=1$ . In this paper, we classify primitive Fano threefolds $X$ in positive characteristic with $\rho(X)\geq 2$ . More precisely, the main theorem is as follows.

Theorem 1.1 (Theorem 5.18, Theorem 5.23, Theorem 5.34, Theorem 6.1).

Let $X$ be a primitive Fano threefold with $\rho(X)\geq 2$ . Then $X$ is isomorphic to one of the following varieties.

No.	$\rho(X)$	$(-K_{X})^{3}$	Description
2-2	$2$	$6$	a split double cover of $\mathbb{P}^{2}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(2,1)$
2-6	$2$	$12$	a split double cover of $W$ with $\mathcal{L}^{\otimes 2}\simeq\omega_{W}^{-1}$
2-6	$2$	$12$	a smooth divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(2,2)$
2-8	$2$	$14$	a split double cover of $V_{7}=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$ with $\mathcal{L}^{\otimes 2}\simeq\omega_{V_{7}}^{-1}$
2-18	$2$	$24$	a split double cover of $\mathbb{P}^{2}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(1,1)$
2-24	$2$	$30$	a smooth divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$
2-32	$2$	$48$	$W$
2-34	$2$	$54$	$\mathbb{P}^{2}\times\mathbb{P}^{1}$
2-35	$2$	$56$	$V_{7}=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$
2-36	$2$	$62$	$\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))$
3-1	$3$	$12$	a split double cover of $\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1,1)$
3-2	$3$	$14$	a smooth member of the complete linear system $\|\mathcal{O}_{P}(2)\otimes\pi^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3)\|$ on the $\mathbb{P}^{2}$ -bundle $\pi\colon P=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})\to\mathbb{P}^{1}\times\mathbb{P}^{1}$
3-27	$3$	$48$	$\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$
3-31	$3$	$52$	$\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))$

In the above table, we use the following notation and terminologies.

(1)

We say that $f:X\to Y$ is a split double cover if $f$ is a finite surjective morphism of projective normal varieites such that $\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}$ splits as an $\mathcal{O}_{Y}$ -module homomorphism and the induced field extension $K(X)\supset K(Y)$ is of degree two.
(2)

For a split double cover $f:X\to Y$ , we set $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ , which is an invertible sheaf on $Y$ (Remark 2.2).
(3)

$W$ is a smooth prime divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $\mathcal{O}(1,1)$ . Note that such a threefold is unique up to isomorphisms (Lemma 5.16).

We also classify Fano threefolds with $\rho(X)=2$ . Although this case has been treated already in [Sai03], their strategies are different.

Theorem 1.2 (Section 5).

Let $X$ be a Fano threefold with $\rho(X)=2$ . Then $X$ is isomorphic to one of the following varieties.

No.	$(-K_{X})^{3}$	Description	Extremal rays
2-1	$4$	blowup of $V_{1}$ along an elliptic curve of degree $1$	$D_{1}+E_{1}$
2-2	$6$	a split double cover of $\mathbb{P}^{2}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}(2,1)$	$C_{1}+D_{1}$
2-3	$8$	blowup of $V_{2}$ along an elliptic curve of degree $2$	$D_{1}+E_{1}$
2-4	$10$	blowup of $\mathbb{P}^{3}$ along a curve of genus $10$ degree $9$	$D_{1}+E_{1}$
2-5	$12$	blowup of $V_{3}$ along an elliptic curve of degree $3$	$D_{1}+E_{1}$
2-6	$12$	a smooth divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(2,2)$ , or a split double cover of $W$ with $\mathcal{L}^{\otimes 2}\simeq\omega_{W}^{-1}$	$C_{1}+C_{1}$
2-7	$14$	blowup of $Q$ along a curve of genus $5$ degree $8$	$D_{1}+E_{1}$
2-8	$14$	a split double cover of $V_{7}=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$ with $\mathcal{L}^{\otimes 2}\simeq\omega_{V_{7}}^{-1}$	$C_{1}+E_{3}\,{\rm or}\,E_{4}$
2-9	$16$	blowup of $\mathbb{P}^{3}$ along a curve of genus $5$ and degree $7$	$C_{1}+E_{1}$
2-10	$16$	blowup of $V_{4}$ along an elliptic curve of degree $4$	$D_{1}+E_{1}$
2-11	$18$	blowup of $V_{3}$ along a line	$C_{1}+E_{1}$
2-12	$20$	blowup of $\mathbb{P}^{3}$ along a curve of genus $3$ and degree $6$	$E_{1}+E_{1}$
2-13	$20$	blowup of $Q$ along a curve of genus $2$ and degree $6$	$C_{1}+E_{1}$
2-14	$20$	blowup of $V_{5}$ along an elliptic curve of degree $5$	$D_{1}+E_{1}$
2-15	$22$	blowup of $\mathbb{P}^{3}$ along a curve of genus $3$ and degree $6$	$E_{1}+E_{3}\,{\rm or}\,E_{4}$
2-16	$22$	blowup of $V_{4}$ along a conic	$C_{1}+E_{1}$
2-17	$24$	blowup of $\mathbb{P}^{3}$ along an elliptic curve curve of degree $5$	$E_{1}+E_{1}$
2-18	$24$	a split double cover of $\mathbb{P}^{2}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(1,1)$	$C_{1}+D_{2}$
2-19	$26$	blowup of $V_{4}$ along a line	$E_{1}+E_{1}$
2-20	$26$	blowup of $V_{5}$ along a cubic rational curve	$C_{1}+E_{1}$
2-21	$28$	blowup of $Q$ along a rational curve of degree $4$	$E_{1}+E_{1}$
2-22	$30$	blowup of $\mathbb{P}^{3}$ along a rational curve of degree $4$	$E_{1}+E_{1}$
2-23	$30$	blowup of $Q$ along an elliptic curve of degree $4$	$E_{1}+E_{3}\,{\rm or}\,E_{4}$
2-24	$30$	a smooth divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$	$C_{1}+C_{2}$
2-25	$32$	blowup of $\mathbb{P}^{3}$ along an elliptic curve of degree $4$	$D_{2}+E_{1}$
2-26	$34$	blowup of $Q$ along a cubic rational curve	$E_{1}+E_{1}$
2-27	$38$	blowup of $\mathbb{P}^{3}$ along a cubic rational curve	$C_{2}+E_{1}$
2-28	$40$	blowup of $\mathbb{P}^{3}$ along an elliptic curve of degree $3$	$E_{1}+E_{5}$
2-29	$40$	blowup of $Q$ along a conic	$D_{2}+E_{1}$
2-30	$46$	blowup of $\mathbb{P}^{3}$ along a conic	$E_{1}+E_{2}$
2-31	$46$	blowup of $Q$ along a line	$C_{2}+E_{1}$
2-32	$48$	$W$	$C_{2}+C_{2}$
2-33	$54$	blowup of $\mathbb{P}^{3}$ along a line	$D_{3}+E_{1}$
2-34	$54$	$\mathbb{P}^{2}\times\mathbb{P}^{1}$	$C_{2}+D_{3}$
2-35	$56$	$V_{7}=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$	$C_{2}+E_{2}$
2-36	$62$	$\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))$	$C_{2}+E_{5}$

In the above table, we use the following notation and terminologies in addition to (1)–(3) from Theorem 1.1.

(4)

$Q$ is the smooth quadric hypersurace in $\mathbb{P}^{4}$ .
(5)

For $1\leq d\leq 7$ with $d\neq 6$ , let $V_{d}$ be a Fano threefold of index $2$ satisfying $(-K_{V_{d}}/2)^{3}=d$ . Note that $V_{d}$ is not determined uniquely by $d$ .
(6)

The centres of all the blowups are smooth curves. The degree of a curve $B$ on a Fano threefold $Y$ is defined as $-\frac{1}{r_{Y}}K_{Y}\cdot B$ , where $r_{Y}$ denotes the index of $Y$ . A line (resp. conic) is a smooth rational curve of degree one (resp. two). Note that if $|-\frac{1}{r_{Y}}K_{Y}|$ is very ample, then a curve of degree one (resp. two) is automatically a line (resp. conic).

For more details, see Table LABEL:table-pic2 in Section 9. The above theorems are positive-characteristic analogues of Mori–Mukai’s results in characteristic zero [MM81], [MM83]. The resulting tables are identical to those of characteristic zero. Our main strategy is the same as in characteristic zero, although we need to overcome some obstructions. For the primitive case, we shall follow the thesis by Ott [Ott], which is extensively detailed. We here point out some differences.

(I) Conic bundles. In characteristic zero, the discriminant divisor $\Delta_{f}$ of a threefold conic bundle $f:X\to Y$ is reduced and normal crossing. However, both the conclusions fail in characteristic two. Because of this, we shall need some minor modifications of the proofs. For example, if $Y=\mathbb{P}^{1}\times\mathbb{P}^{1}$ and $\Delta_{f}\sim\mathcal{O}(2,0)$ , then we immediately get a contradiction in characteristic zero, because $\Delta_{f}$ does not contain any smooth rational curve as a connected component.

(II) Double covers. In characteristic two, double covers are more complicated than those in characteristic $p\neq 2$ . Fortunately, all the double covers $f:X\to Y$ appearing in this paper can be checked to be split, i.e., $\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}$ splits. Although it is hard to control general double covers in characteristic two, split double covers are quite similar to the ones in characteristic $p\neq 2$ , e.g., they are explicitly given, $\kappa(X)\geq\kappa(Y)$ , etc (Section 8).

Acknowledgements: This paper is based on the master thesis [Asa] of the first author, in which the primitive case of odd characteristic is carried out. The second author was funded by JSPS KAKENHI Grant numbers JP22H01112 and JP23K03028.

2. Preliminaries

2.1. Notation

In this subsection, we summarise notation used in this paper.

(1)

We will freely use the notation and terminology in [Har77] and [KM98]. In particular, $D_{1}\sim D_{2}$ means linear equivalence of Weil divisors.
(2)

Throughout this paper, we work over an algebraically closed field $k$ of characteristic $p>0$ unless otherwise specified.
(3)

For an integral scheme $X$ , we define the function field $K(X)$ of $X$ as the local ring $\mathcal{O}_{X,\xi}$ at the generic point $\xi$ of $X$ . For an integral domain $A$ , $K(A)$ denotes the function field of ${\operatorname{Spec}}\,A$ .
(4)

For a scheme $X$ , its reduced structure $X_{{\operatorname{red}}}$ is the reduced closed subscheme of $X$ such that the induced closed immersion $X_{{\operatorname{red}}}\to X$ is surjective.
(5)

We say that $X$ is a variety (over $k$ ) if $X$ is an integral scheme which is separated and of finite type over $k$ . We say that $X$ is a curve (resp. a surface, resp. a threefold) if $X$ is a variety over $k$ of dimension one (resp. two, resp. three).
(6)

We say that $X$ is a Fano threefold if $X$ is a three-dimensional smooth projective variety over $k$ such that $-K_{X}$ is ample. A Fano threefold $X$ is imprimitive if there exists a Fano threefold $Y$ and a smooth curve $B$ on $Y$ such that $X$ is isomorphic to the blowup ${\operatorname{Bl}}_{B}Y$ of $Y$ along $B$ . We say that a Fano threefold $X$ is primitive if $X$ is not imprimitive.

2.2. Split double covers

Definition 2.1.

(1)

We say that a morphism $f:X\to Y$ is a double cover if $f$ is a finite surjective morphism of normal varieties such that the induced field extension $K(Y)\subset K(X)$ is of degree two.
(2)

We say that a double cover $f:X\to Y$ is split if $\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}$ splits as an $\mathcal{O}_{Y}$ -module homomorphism.

Remark 2.2.

Let $f:X\to Y$ be a double cover, where $X$ and $Y$ are smooth varieties.

(1)

It is known that ${\operatorname{Coker}}(\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X})=f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y}$ is an invertible sheaf on $Y$ [Kaw21, Lemma A.1]. We shall frequently use its inverse $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ . We have the following exact sequence:

$0\to\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}\to\mathcal{L}^{-1}\to 0.$
(2)

If $p\neq 2$ , then the branched divisor $D$ satisfies $\mathcal{O}_{Y}(D)\simeq\mathcal{L}^{\otimes 2}$ .

Lemma 2.3.

Let $f:X\to Y$ be a double cover of smooth projective varieties. For the invertible sheaf $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ (Remark 2.2), it holds that $\omega_{X}\simeq f^{*}(\omega_{Y}\otimes\mathcal{L})$ .

Proof.

See [CD89, Proposition 0.1.3]. ∎

The following lemma gives criteria for the splitting of $f$ .

Lemma 2.4.

Let $f:X\to Y$ be a double cover of smooth projective varieties. Set $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ , which is an invertible sheaf (Remark 2.2). Assume that one of the following (1)–(3) holds.

(1)

$p\neq 2$ .
(2)

$H^{1}(X,\mathcal{L})=0$ .
(3)
Both (a) and (b) hold.
1. (a)
  
  $Y$ is $F$ -split, i.e., $\mathcal{O}_{Y}\to F_{*}\mathcal{O}_{Y}$ splits as an $\mathcal{O}_{Y}$ -module homomorphism, where $F:Y\to Y$ denotes the absolute Frobenius morphism.
2. (b)
  
  $H^{1}(Y,\mathcal{O}_{Y}(E))=0$ for any effective Cartier divisor $E$ on $Y$ .

Then $f$ is a split double cover, i.e., $\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}$ splits.

Proof.

It is well known that $f$ is split when (1) holds. Note that $\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}$ splits if and only if

0\to\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}\to\mathcal{L}^{-1}\to 0

splits. Since the extension class corresponding to this exact sequence is contained in $H^{1}(Y,\mathcal{L})$ , (2) implies that $f$ is split.

Assume (3). If $f$ is inseparable, then the absolute Frobenius morphism $F:Y\to Y$ of $Y$ factors through $f:X\to Y$ :

F:Y\to X\xrightarrow{f}Y,

and hence the splitting of $\mathcal{O}_{Y}\to F_{*}\mathcal{O}_{Y}$ implies the splitting of $\mathcal{O}_{Y}\to f_{*}\mathcal{O}_{X}$ . Therefore, we may assume that $f:X\to Y$ is separable. In this case, the trace map ${\operatorname{Tr}}:f_{*}\mathcal{O}_{X}\to\mathcal{O}_{Y}$ is nonzero, because the field-theoretic trace map ${\operatorname{Tr}}:K(X)\to K(Y)$ is given by ${\operatorname{Tr}}(b)=b+\sigma(b)$ for the Galois involution $K(X)\to K(X)$ , and hence ${\operatorname{Tr}}(b)\neq 0$ for $b\in K(X)\setminus K(Y)$ . As the composite $\mathcal{O}_{Y}$ -module homomorphism $\mathcal{O}_{Y}\hookrightarrow f_{*}\mathcal{O}_{X}\xrightarrow{{\operatorname{Tr}}}\mathcal{O}_{Y}$ is zero, we get a nonzero $\mathcal{O}_{Y}$ -module homomorphism $\mathcal{L}^{-1}\simeq f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y}\to\mathcal{O}_{Y}$ . Therefore, $H^{0}(Y,\mathcal{L})\neq 0$ . In this case, (b) implies $H^{1}(Y,\mathcal{L})=0$ , i.e., (2) holds. Therefore, (3) implies that $f$ is split. ∎

2.3. Brauer groups

Definition 2.5 ([CTS21]*Definition 3.2.1).

For a scheme $X$ , the Brauer group of $X$ is defined by

{\operatorname{Br}}(X):=H^{2}_{{\operatorname{\acute{e}t}}}(X,\mathbb{G}_{m}),

where $H^{2}_{{\operatorname{\acute{e}t}}}(-)$ denotes the étale cohomology. For a ring $R$ , we set

{\operatorname{Br}}(R):={\operatorname{Br}}({\operatorname{Spec}}\,R)=H^{2}_{{\operatorname{\acute{e}t}}}({\operatorname{Spec}}\,R,\mathbb{G}_{m}).

Proposition 2.6.

The following hold.

(1)

If $Y$ is a regular noetherian integral scheme, then we have an injective group homorphism ${\operatorname{Br}}(Y)\hookrightarrow{\operatorname{Br}}(K(Y))$ .
(2)

If $C$ is a smooth curve over $k$ , then ${\operatorname{Br}}(C)={\operatorname{Br}}(K(C))=0$ .
(3)

If $Y$ is a smooth projective rational variety over $k$ , then ${\operatorname{Br}}(Y)=0$ .

Proof.

The assertion (1) follows from [Gro95, Corollaire 1.10]. The assertion (2) holds by (1) and [CTS21, Theorem 1.2.14]. The assertion (3) holds by [CTS21, Corollary 6.2.11]. ∎

Proposition 2.7.

Let $f:X\to Y$ be a projective morphism of smooth varieties. Fix $r\in\mathbb{Z}_{>0}$ . Assume that ${\operatorname{Br}}(Y)=0$ and any fibre of $f$ is isomorphic to $\mathbb{P}^{r}$ . Then there exists a vector bundle $E$ of rank $r+1$ such that $X$ is isomorphic to $\mathbb{P}(E)$ over $Y$ .

For the reader’s convenience, we include a sketch of a proof. For more details, we refer to websites [Mur] and [Mus], which the following proof is based on.

Proof.

Fix a closed point $y\in Y$ and set $A:=\mathcal{O}_{Y,y}$ . For the completion $\widehat{A}=\widehat{\mathcal{O}}_{Y,y}$ , we have an isomorphism $X\times_{Y}{\operatorname{Spec}}\,\widehat{A}\simeq\mathbb{P}^{r}_{\widehat{A}}$ over $\widehat{A}$ [Ser06, Corollary 1.2.15]. By Artin’s approxximation theorem, We obtain an $A^{h}$ -isomorphism $X\times_{Y}{\operatorname{Spec}}\,A^{h}\simeq\mathbb{P}^{r}_{A^{h}}$ over the henselisation $A^{h}=\mathcal{O}_{Y,y}^{h}$ . Therefore, there exists an étale surjective morphism $Y^{\prime}\to Y$ such that $X\times_{Y}Y^{\prime}\simeq\mathbb{P}^{r}_{Y^{\prime}}$ .

By the exact sequence

1\to\mathbb{G}_{m}\to{\rm GL}_{r+1}\to{\rm PGL}_{r+1}\to 1

on étale topology, we obtain the following exact sequence:

H^{1}_{{\operatorname{\acute{e}t}}}(Y,{\rm GL}_{r+1})\xrightarrow{\alpha}H^{1}_{{\operatorname{\acute{e}t}}}(Y,{\rm PGL}_{r+1})\to H^{2}_{{\operatorname{\acute{e}t}}}(Y,\mathbb{G}_{m})={\operatorname{Br}}(Y)=0.

This implies that the ${\rm PGL}_{r+1}$ -torsor $f:X\to Y$ comes from ${\rm GL}_{r+1}$ -torsor. Since any ${\rm GL}_{r+1}$ -torsor is nothing but a projective space bundle $\mathbb{P}_{Y}(E)$ , the assertion holds by the following facts.

•

$H^{1}_{{\operatorname{\acute{e}t}}}(X,{\rm GL}_{r+1})$ can be considered as a set of (Zariski) locally free sheaf of rank $r+1$ [SGA1II, Exposé XI, Corollaire 5.3].
•

$H^{1}_{{\operatorname{\acute{e}t}}}(Y,{\rm PGL}_{r+1})$ consists of the isomorphism classes of $\pi:W\to Y$ such that $W$ is a flat proper morphism such that $W\times_{Y}Y^{\prime}\simeq P^{r+1}_{Y^{\prime}}$ for some étale surjective morphism $Y^{\prime}\to Y$ /. $\mathbb{P}^{r+1}$ .
•

$\alpha(E)=\mathbb{P}_{Y}(E)$ .

∎

3. Extremal rays and Mori fibre spaces

3.1. Types of extremal rays

Definition 3.1.

Let $X$ be a smooth projective threefold. Let $R$ be a $K_{X}$ -negative extremal ray of $\overline{{\operatorname{NE}}}(X)$ . By [Kol91, (1.1.1)], there exists a unique morphism $f\colon X\to Y$ , called the contraction (morphism) of $R$ , to a projective normal variety $Y$ such that the following hold.

(1)

$f_{*}\mathcal{O}_{X}=\mathcal{O}_{Y}$ .
(2)

For any curve $C$ on $X$ , $[C]\in R$ if and only if $f(C)$ is a point.

Definition 3.2.

Let $X$ be a smooth projective threefold. Let $R$ be an extremal ray of $\overline{{\operatorname{NE}}}(X)$ and let $f:X\to Y$ be the contraction of $R$ . We set

\mu_{R}:=\min\left\{(-K_{X}\cdot C)_{X}\mid C\mathrm{~{}is~{}a~{}rational~{}curve~{}on~{}}X\mathrm{~{}with~{}}[C]\in R\right\},

which is called the length of an extremal ray $R$ . We say that $\ell$ is an extremal rational curve if $\ell$ is a rational curve on $X$ such that $[\ell]\in R$ and $-K_{X}\cdot\ell=\mu_{R}$ .

Definition 3.3.

Let $X$ be a smooth projective threefold. Let $R$ be a $K_{X}$ -negative extremal ray of $\overline{{\operatorname{NE}}}(X)$ and let $f:X\to Y$ be the contraction of $R$ .

(1)
$R$ and $f$ are called of type $C$ if $\dim Y=2$ .
- •
  
  $R$ and $f$ are called of type $C_{1}$ if $f$ is not smooth.
- •
  
  $R$ and $f$ are called of type $C_{2}$ if $f$ is smooth.
(2)
$R$ and $f$ are called of type $D$ if $\dim Y=1$ . Let $X_{K}$ be the generic fibre of $f$ , which is a regular projective surface over $K:=K(Y)$ .
- •
  
  $R$ and $f$ are called of type $D_{1}$ if $1\leq K_{X_{K}}^{2}\leq 7$ .
- •
  
  $R$ and $f$ are called of type $D_{2}$ if $K_{X_{K}}^{2}=8$ .
- •
  
  $R$ and $f$ are called of type $D_{3}$ if $K_{X_{K}}^{2}=9$ .
(3)
$R$ and $f$ are called of type $E$ if $\dim Y=3$ . Set $D:={\operatorname{Ex}}(f)$ .
- •
  
  $R$ and $f$ are called of type $E_{1}$ if $f(D)$ is a curve.
- •
  
  $R$ and $f$ are called of type $E_{2}$ if $f(D)$ is a point, $D\simeq\mathbb{P}^{2}$ , and $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{D}(-1)$ .
- •
  
  $R$ and $f$ are called of type $E_{3}$ if $f(D)$ is a point, $D$ is isomorphic to a smooth quadric surface $Q$ on $\mathbb{P}^{3}$ , and $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{\mathbb{P}^{3}}(-1)|_{Q}$ .
- •
  
  $R$ and $f$ are called of type $E_{4}$ if $f(D)$ is a point, $D$ is isomorphic to a singular quadric surface $Q$ on $\mathbb{P}^{3}$ , and $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{\mathbb{P}^{3}}(-1)|_{Q}$ .
- •
  
  $R$ and $f$ are called of type $E_{5}$ if $f(D)$ is a point, $D\simeq\mathbb{P}^{2}$ , and $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{D}(-2)$ .
(4)

$R$ and $f$ are called of type $F$ if $\dim Y=0$ .

Remark 3.4.

Let $X$ be a smooth projective threefold. Let $R$ be a $K_{X}$ -negative extremal ray of $\overline{{\operatorname{NE}}}(X)$ and let $f:X\to Y$ be the contraction of $R$ .

(1)
Assume that $R$ is of type $C$ . Then the following hold.
- •
  
  $R$ is of type $C_{1}$ if and only if $\mu_{R}=1$ .
- •
  
  $R$ is of type $C_{2}$ if and only if $\mu_{R}=2$ .
(2)
Assume that $R$ is of type $D$ . Then the following hold [TanII, Proposition 3.17].
- •
  
  $R$ is of type $D_{1}$ if and only if $\mu_{R}=1$ .
- •
  
  $R$ is of type $D_{2}$ if and only if $\mu_{R}=2$ .
- •
  
  $R$ is of type $D_{3}$ if and only if $\mu_{R}=3$ .
(3)
Assume that $R$ is of type $E$ . Then the following hold [TanII, Proposition 3.22].
- •
  
  If $R$ of type $E_{1}$ , then $\mu_{R}=1$ .
- •
  
  If $R$ of type $E_{2}$ , then $\mu_{R}=2$ .
- •
  
  If $R$ of type $E_{3}$ , then $\mu_{R}=1$ .
- •
  
  If $R$ of type $E_{4}$ , then $\mu_{R}=1$ .
- •
  
  If $R$ of type $E_{5}$ , then $\mu_{R}=1$ .

3.2. Del Pezzo fibrations

Proposition 3.5.

Let $X$ be a smooth projective threefold and let $f:X\to Y$ be a contraction of a $K_{X}$ -negative extremal ray of type $D$ . For $K:=K(Y)$ and its algebraic closure $\overline{K}:=\overline{K(Y)}$ , let $X_{K}$ and $X_{\overline{K}}$ be the generic fibre and the geometric generic fibre of $f$ , respectively. Then the following hold.

(1)

For a closed point $y\in Y$ , the effective Cartier divisor $f^{*}y$ is a prime divisor, i.e., any scheme-theoretic fibre of $f$ is geometrically integral.
(2)

$X_{\overline{K}}$ is a projective canonical del Pezzo surface, i.e., $X_{\overline{K}}$ is a projective normal surface such that $V_{\overline{K}}$ has at worst canonical singularities and $-K_{V_{\overline{K}}}$ is ample.
(3)

$1\leq K_{X_{K}}^{2}=K_{X_{\overline{K}}}^{2}\leq 9$ .
(4)

${\operatorname{Pic}}X_{\overline{K}}$ is a finitely generated free abelian groups. Furthermore, ${\operatorname{Pic}}X_{K}\simeq\mathbb{Z}$ .
(5)
Assume that $K_{X_{K}}^{2}=9$ . Then the following hold.
1. (a)
  
  $X_{K}\simeq\mathbb{P}^{2}_{K}$ . In particular, general fibres are $\mathbb{P}^{2}$ .
2. (b)
  
  There exist Cartier divisors $D$ on $X$ and $E$ on $Y$ such that $-K_{X}\sim 3D+f^{*}E$ .
(6)
Assume that $K_{X_{K}}^{2}=8$ . Then the following hold.
1. (a)
  
  If $X_{K}$ is smooth over $K$ and $y$ is a general closed point, then both $X_{\overline{K}}$ and $X_{y}$ are isomorphic to $\mathbb{P}^{1}\times\mathbb{P}^{1}$ .
2. (b)
  
  If $X_{K}$ is not smooth over $K$ and $y$ is a general closed point, then $p=2$ and both $X_{\overline{K}}$ and $X_{y}$ are isomorphic to $\mathbb{P}(1,1,2)$ .
3. (c)
  
  There exist Cartier divisors $D$ on $X$ and $E$ on $Y$ such that $-K_{X}\sim 2D+f^{*}E$ .

Proof.

The assertions (1)–(4) follow from [TanII, Proposition 3.5]. As for (5) and (6), see [TanII, Lemma 3.14 and Lemma 3.15]. ∎

Remark 3.6.

By [FS20, Proposition 14.7], there actually exists a del Pezzo fibration $f:X\to Y$ such that $X$ is smooth and the generic fibre is not smooth.

Proposition 3.7.

Let $X$ be a smooth projective threefold with an extremal ray $R$ of type $D$ . Let $f:X\to Y$ be the contraction of $R$ . For $K:=K(Y)$ , $X_{K}$ denotes the generic fibre of $f$ .

(1)

If $K_{X_{K}}^{2}=9$ , then $f$ is a $\mathbb{P}^{2}$ -bundle, i.e., there exists a locally free $F$ on $Y$ of rank $3$ such that $X$ is isomorphic to $\mathbb{P}_{Y}(F)$ over $Y$ . In particular, any fibre is $\mathbb{P}^{2}$ .
(2)

If $K_{X_{K}}^{2}=8$ , then there exists a $\mathbb{P}^{3}$ -bundle $\mathbb{P}(E)$ over $Y$ and a closed immersion $j:X\hookrightarrow\mathbb{P}(E)$ such that $j_{y}(X_{y})\subset\mathbb{P}^{2}$ is a quadric surface on $\mathbb{P}^{2}_{k}$ for every point $y\in Y$ .

Proof.

Let us show (1). By Proposition 3.5(5), we have $-K_{X}\sim 3D+f^{*}E$ for some Cartier divisors $D$ on $X$ and $E$ on $Y$ , respectively. We use Fujita’s $\Delta$ -genus: $\Delta(V,L)=\dim V+L^{\dim V}-h^{0}(V,L)$ . It holds that

\Delta(X_{\overline{K}},\mathcal{O}_{X}(D)|_{X_{\overline{K}}})=2+\mathcal{O}_{X}(D)^{2}-h^{0}(X_{\overline{K}},\mathcal{O}_{X}(D)|_{X_{\overline{K}}})=0.

Then the assertion (1) follows from the same argument as in [Fuj75, Corollary 5.4]. Similarly, the assertion (2) holds by using Proposition 3.5(6) (cf. [Fuj75, Corollary 5.5]). ∎

3.3. Contraction morphisms

Lemma 3.8.

Let $X$ be a Fano threefold. Then the following hold.

(1)

${\operatorname{Pic}}X\simeq\mathbb{Z}^{\oplus\rho(X)}$ .
(2)

$H^{i}(X,\mathcal{O}_{X})=0$ for all $i>0$ . In particular, $\chi(X,\mathcal{O}_{X})=1$ .
(3)

$-K_{X}\cdot c_{2}(X)=24$ .
(4)

For any effective divisor $D$ on $X$ ,

$D\cdot c_{2}(X)=6\chi(D,\mathcal{O}_{D})+6\chi(D,\mathcal{O}_{X}(D)|_{D})-2D^{3}-(-K_{X})^{2}\cdot D$

Proof.

The assertions (1) and (2) follow from [TanI, Theorem 2.4] (cf. [Kaw21, Corollary 3.7]). The assertion (3) holds by $\chi(X,\mathcal{O}_{X})=1$ and $\chi(X,\mathcal{O}_{X})=\frac{1}{24}(-K_{X}\cdot c_{2}(X))$ .

Let us show (4). By $\chi(X,\mathcal{O}_{X})=1$ , the Riemann–Roch theorem implies

	$\displaystyle\chi(\mathcal{O}_{X}(D))=$	$\displaystyle 1+\frac{1}{12}((-K_{X})^{2}+c_{2}(X))\cdot D+\frac{1}{4}(-K_{X}\cdot D^{2})+\frac{1}{6}D^{3},$
	$\displaystyle\chi(\mathcal{O}_{X}(-D))=$	$\displaystyle 1-\frac{1}{12}((-K_{X})^{2}+c_{2}(X))\cdot D+\frac{1}{4}(-K_{X}\cdot D^{2})-\frac{1}{6}D^{3}.$

Hence

\chi(\mathcal{O}_{X}(D))-\chi(\mathcal{O}_{X}(-D))=\frac{1}{6}((-K_{X})^{2}+c_{2}(X))\cdot D+\frac{1}{3}D^{3}.

On the other hand, by the exact sequences

0\longrightarrow\mathcal{O}_{X}(-D)\longrightarrow\mathcal{O}_{X}\longrightarrow\mathcal{O}_{D}\longrightarrow 0

and

0\longrightarrow\mathcal{O}_{X}\longrightarrow\mathcal{O}_{X}(D)\longrightarrow\mathcal{O}_{D}(D)\longrightarrow 0,

we have

\chi(X,\mathcal{O}_{X}(-D))=\chi(X,\mathcal{O}_{X})-\chi(D,\mathcal{O}_{D}),\quad\chi(X,\mathcal{O}_{X}(D))=\chi(X,\mathcal{O}_{X})+\chi(D,\mathcal{O}_{D}(D)),

where $\mathcal{O}_{D}(D):=\mathcal{O}_{X}(D)|_{D}$ . Therefore, we obtain

\chi(X,\mathcal{O}_{X}(D))-\chi(X,\mathcal{O}_{X}(-D))=\chi(D,\mathcal{O}_{D}(D))+\chi(D,\mathcal{O}_{D}),

which implies

\chi(D,\mathcal{O}_{D}(D))+\chi(D,\mathcal{O}_{D})=\frac{1}{6}((-K_{X})^{2}+c_{2}(X))\cdot D+\frac{1}{3}D^{3},

as required. ∎

Proposition 3.9.

Let $X$ be a smooth projective threefold and let $f:X\to Y$ be the contraction of a $K_{X}$ -negative extremal ray $R$ of $X$ . Let $\ell$ be a curve with $[\ell]\in R$ . Then the following sequence

0\longrightarrow{\operatorname{Pic}}(Y)\xrightarrow{f^{*}}{\operatorname{Pic}}(X)\xrightarrow{(-\cdot\ell)}\mathbb{Z}

is exact, where $(-\cdot\ell)(D):=D\cdot\ell$ for $D\in{\operatorname{Pic}}(X)$ .

Proof.

See [TanII, Proposition 3.12]. ∎

Corollary 3.10.

Let $X$ be a Fano threefold with $\rho(X)=2$ and let $R$ be an extremal ray of ${\operatorname{NE}}(X)$ with the corresponding contraction $f\colon X\to Y$ . Then ${\operatorname{Pic}}\,Y\simeq\mathbb{Z}$ .

Proof.

The assertion follows from ${\operatorname{Pic}}(X)\simeq\mathbb{Z}^{2}$ (Lemma 3.8) and Proposition 3.9. ∎

Proposition 3.11.

Let $X$ be a smooth projective threefold and let $f:X\to Y$ be the contraction of a $K_{X}$ -negative extremal ray $R$ of $X$ . Then the following hold.

(1)

If $\dim Y\neq 1$ , then $R^{i}f_{*}\mathcal{O}_{X}=0$ for any $i>0$ .
(2)

If $X$ is a Fano threefold, then $R^{i}f_{*}\mathcal{O}_{X}=0$ for any $i>0$ . Furthermore, $H^{j}(X,\mathcal{O}_{X})\simeq H^{j}(Y,\mathcal{O}_{Y})$ for any $j\geq 0$ .

Proof.

(1) If $\dim X_{y}\leq 1$ for any closed point $y\in Y$ , then $R^{i}f_{*}\mathcal{O}_{X}=0$ holds by [Tan15, Theorem 0.5]. We may assume that $\dim X_{y}\geq 2$ for some closed point $y\in Y$ . In particular, $\dim Y=0$ or $\dim Y=3$ , as we assume $\dim Y\neq 1$ . If $\dim Y=0$ , then we obtain $R^{i}f_{*}\mathcal{O}_{X}=0$ by Lemma 3.8.

Assume that $\dim Y=3$ . By $\dim X_{y}\geq 1$ , $f(E)$ is a point for $E:={\operatorname{Ex}}(f)$ . Then we have an exact sequence for any $r\in\mathbb{Z}_{\geq 0}$ :

0\to\mathcal{O}_{X}(-(r+1)E)\to\mathcal{O}_{X}(-rE)\to\mathcal{O}_{X}(-rE)|_{E}\to 0.

Since $\mathcal{O}_{X}(-E)|_{E}$ is ample (Definition 3.3) and $E$ is a normal projective toric surface, we have that $H^{i}(E,\mathcal{O}_{X}(-rE)|_{E})=0$ for every $i>0$ . Therefore, we obtain surjections:

R^{i}f_{*}\mathcal{O}_{X}(-mE))\to\cdots\to R^{i}f_{*}\mathcal{O}_{X}(-E))\to R^{i}f_{*}\mathcal{O}_{X}.

By the Serre vanishing theorem, we have $R^{i}f_{*}\mathcal{O}_{X}(-mE))=0$ for some $m>0$ , as required.

(2) By (1), we may assume that $\dim Y=1$ . Pick general closed point $Q_{1},...,Q_{n}$ on $Y$ . Set $S_{1}:=f^{*}Q_{1},...,S_{n}:=f^{*}Q_{n}$ , and $D_{Y}:=Q_{1}+\cdots+Q_{n}$ . By the Serre vanishing theorem

\displaystyle H^{q}(Y,R^{i}f_{*}(f^{*}\mathcal{O}_{Y}(D_{Y})))=H^{q}(Y,R^{i}f_{*}\mathcal{O}_{X}\otimes\mathcal{O}_{Y}(D_{Y}))=0.

for all $q>0$ and $n\gg 0$ . Hence, by the Leray spectral sequence, we have

H^{0}(Y,(R^{i}f_{*}\mathcal{O}_{X})\otimes\mathcal{O}_{Y}(D_{Y}))\simeq H^{i}(X,f^{*}\mathcal{O}_{Y}(D_{Y})).

Hence it is enough to show that $H^{i}(X,f^{*}\mathcal{O}_{Y}(D_{Y}))=H^{i}(X,\mathcal{O}_{X}(S_{1}+\cdots+S_{n}))=0$ for every $n>0$ . We have an exact sequence

0\longrightarrow\mathcal{O}_{X}(-\sum_{\ell=1}^{n}S_{\ell})\longrightarrow\mathcal{O}_{X}\longrightarrow\bigoplus_{\ell=1}^{n}\mathcal{O}_{S_{\ell}}\longrightarrow 0.

Since $Q_{1}+\cdots+Q_{n}\sim D^{\prime}_{Y}$ for some divisor $D^{\prime}_{Y}$ such that $Q_{1},...,Q_{n}\not\in{\operatorname{Supp}}\,D^{\prime}_{Y}$ , it holds that

0\longrightarrow\mathcal{O}_{X}\longrightarrow\mathcal{O}_{X}(f^{*}D_{Y})\longrightarrow\bigoplus_{\ell=1}^{n}\mathcal{O}_{S_{\ell}}\longrightarrow 0.

Since each $S_{i}$ is a canonical del Pezzo surface (Proposition 3.5(2)), we have $H^{i}(S_{\ell},\mathcal{O}_{S_{\ell}})=0$ for every $\ell$ and every $i>0$ . It holds that $H^{i}(X,\mathcal{O}_{X})=0$ for any $i>0$ , which implies $H^{i}(X,\mathcal{O}_{X}(f^{*}D_{Y}))=0$ for any $i>0$ . ∎

Corollary 3.12.

Let $X$ be a Fano threefold with $\rho(X)\geq 2$ . For an extremal ray $R$ of ${\operatorname{NE}}(X)$ , let $f\colon X\to Y$ be the contraction of $R$ . Let $D_{Y}$ be an effective Cartier divisor on $Y$ and set $D:=f^{*}D_{Y}$ . Then we have

\chi(D,\mathcal{O}_{D})=1-\chi(Y,\mathcal{O}_{Y}(-D_{Y}))\quad\text{and}\quad\chi(D,\mathcal{O}_{X}(D)|_{D})=\chi(Y,\mathcal{O}_{Y}(D_{Y}))-1.

Proof.

Set $\mathcal{O}_{D}(D):=\mathcal{O}_{X}(D)|_{D}$ . We have the exact sequence

0\longrightarrow\mathcal{O}_{X}(-D)\longrightarrow\mathcal{O}_{X}\longrightarrow\mathcal{O}_{D}\longrightarrow 0,

which yields

0\longrightarrow\mathcal{O}_{X}\longrightarrow\mathcal{O}_{X}(D)\longrightarrow\mathcal{O}_{D}(D)\longrightarrow 0.

Then the following hold:

	$\displaystyle\chi(D,\mathcal{O}_{D})$	$\displaystyle=\chi(X,\mathcal{O}_{X})-\chi(X,\mathcal{O}_{X}(-D)),$
	$\displaystyle\chi(D,\mathcal{O}_{D}(D))$	$\displaystyle=\chi(X,\mathcal{O}_{X}(D))-\chi(X,\mathcal{O}_{X}).$

Recall that $\chi(X,\mathcal{O}_{X})=1$ (Lemma 3.8). By the projection formula, we have

R^{i}f_{*}\mathcal{O}_{X}(D)\simeq R^{i}f_{*}(f^{*}\mathcal{O}_{Y}(D_{Y}))\simeq\mathcal{O}_{Y}(D_{Y})\otimes R^{i}f_{*}\mathcal{O}_{X}

for all $i\geq 0$ . We obtain $f_{*}\mathcal{O}_{X}(D)\simeq\mathcal{O}_{Y}(D_{Y})$ , and $R^{i}f_{*}\mathcal{O}_{X}(D)=0$ for all $i>0$ (Proposition 3.11). Therefore, by the Leray spectral sequence, the following holds for every $i\geq 0$ :

H^{i}(X,\mathcal{O}_{X}(D))\simeq H^{i}(Y,f_{*}\mathcal{O}_{X}(D))\simeq H^{i}(Y,\mathcal{O}_{Y}(D_{Y})).

Hence $\chi(X,\mathcal{O}_{X}(D))=\chi(Y,\mathcal{O}_{Y}(D_{Y}))$ , which implies

\chi(D,\mathcal{O}_{D}(D))=\chi(Y,\mathcal{O}_{Y}(D_{Y}))-1.

Similarly, we get $\chi(D,\mathcal{O}_{D})=1-\chi(Y,\mathcal{O}_{Y}(-D_{Y}))$ . ∎

Lemma 3.13.

Let $Y$ be a smooth projective threefold and let $f:X\to Y$ be a blowup along a smooth curve $B$ on $Y$ . Assume that $-K_{X}$ is ample. Then $(-K_{X})^{3}<(-K_{Y})^{3}$ .

Proof.

Set $E:={\operatorname{Ex}}(f)$ . Since $-K_{X}$ is ample, we have $(-K_{X})^{2}\cdot E>0$ . This, together with [TanII, Lemma 3.21], implies the following:

(1)

$(-K_{X})^{3}=(-K_{Y})^{3}+2K_{X}\cdot B+2g(B)-2$ .
(2)

$-K_{X}\cdot B-2g(B)+2=(-K_{X})^{2}\cdot E>0$

By (1), it is enough to show $-K_{X}\cdot B>g(B)-1$ . Since $-K_{X}$ is ample, this holds when $g(B)\leq 1$ . Hence we may assume that $g(B)>1$ . In this case, the required inequality $-K_{X}\cdot B>g(B)-1$ holds by (2): $-K_{X}\cdot B>2g(B)-2>g(B)-1$ . ∎

3.4. Conic bundles

In this subsection, we recall some terminologies and results on conic bundles. For more details, we refer to [Tan-conic] and [TanII, Subsection 3.3].

Definition 3.14.

We say that $f:X\to S$ is a conic bundle if $f:X\to S$ is a flat projective morphism of noetherian schemes such that $X_{s}$ is isomorphic to a conic on $\mathbb{P}^{2}_{\kappa(s)}$ .

If $X$ is a smooth projective threefold and $R$ is an extremal ray of type $C$ , then its contraction $f:X\to Y$ is a conic bundle to a smooth projective surface $Y$ . For the definition of $\Delta_{f}$ , we refer to [TanII, 3.10].

Proposition 3.15.

Let $f:X\to S$ be a conic bundle, where $X$ and $S$ are smooth varieties. Then the following hold.

(1)

$f_{*}\mathcal{O}_{X}=\mathcal{O}_{S}$ .
(2)

$f_{*}\omega_{X}^{-1}$ is a locally free sheaf of rank $3$ .
(3)

$\omega_{X}^{-1}$ is very ample over $S$ , and hence it defines a closed immersion $\iota:X\hookrightarrow\mathbb{P}(f_{*}\omega_{X}^{-1})$ over $S$ .

Proof.

See [Tan-conic, Lemma 2.5 and Proposition 2.7]. ∎

Proposition 3.16.

Let $f:X\to S$ be a conic bundle, where $X$ is a smooth projective threefold and $S$ is a smooth projective surface. Then the following hold.

(1)

$f_{*}K_{X}\sim-2S$ .
(2)

$-f_{*}(K_{X/S}^{2})\equiv\Delta_{f}$ .
(3)

For every divisor $D$ on $S$ , it holds that

$K_{X}^{2}\cdot f^{*}D=-4K_{S}\cdot D-\Delta_{f}\cdot D.$

Proof.

See [TanII, Proposition 3.11]. ∎

4. Basic properties of primitive Fano threefolds

4.1. Extremal rays on primitive Fano threefolds

Theorem 4.1 (Cone theorem).

Let $X$ be a Fano threefold. Then there exist finitely many extremal rays $R_{1},...,R_{n}$ of ${\operatorname{NE}}(X)$ such that

{\operatorname{NE}}(X)=R_{1}+\cdots+R_{n}.

For each $1\leq i\leq n$ , there exists an extremal rational curve $\ell_{i}$ such that $R_{i}=\mathbb{R}_{\geq 0}[\ell_{i}]$ (Definition 3.2).

Proof.

See [Mor82, Theorem 1.2]. ∎

4.1.1. Type C

Theorem 4.2.

Let $X$ be a Fano threefold. Assume that there exists a conic bundle $f:X\to Y$ such that any fibre of $f$ is not smooth. Then the following hold.

(1)

$p=2$ .
(2)
One of the following holds.
1. (a)
  
  $Y\simeq\mathbb{P}^{2}$ and $X$ is isomorphic to a prime divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$ . Furthermore, $\rho(X)=2$ and $X$ is primitive.
2. (b)
  
  $Y\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ and $X$ is isomorphic to a prime divisor on $P:=\mathbb{P}_{Y}(\mathcal{O}(0,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O})$ which is linearly equivalent to $\mathcal{O}_{P}(1)^{\otimes 2}$ , where $\mathcal{O}_{P}(1)$ denotes the tautological bundle with respect to the $\mathbb{P}^{2}$ -bundle structure $P\to Y$ . Furthermore, $\rho(X)=3$ and $X$ is imprimitive.

In particular, if $X$ is a primitive Fano threefold and there exists an extremal ray of type $C$ whose contraction $f:X\to Y$ has no smooth fibres, then (a) of (2) holds.

Proof.

See [MS03, Corollary 8 and Remark 10]. ∎

Lemma 4.3.

Let $X$ be a Fano threefold with an extremal ray $R$ of type $C$ . Let $f:X\to Y$ be the contraction of $R$ . Then $Y$ is a smooth rational surface.

Proof.

By Theorem 4.2, we may assume that $f$ is generically smooth. Note that $Y$ is a smooth projective surface [Kol91, Main Theorem 1.1, (1.1.3.1)]. Since the geometric generic fibre $X_{\overline{\eta}}$ is smooth, $(X_{\overline{\eta}},0)$ is $F$ -pure. Then it follows from [Eji19, Corollary 4.10(2)] that $-K_{Y}$ is big. In particular, the Kodaira dimension of $Y$ is negative, i.e., $Y$ is a smooth ruled surface. On the other hand, $Y$ is rationally chain connected, because so is $X$ [Kol96, Ch. V, Theorem 2.13]. Therefore, $Y$ is a smooth rational surface. ∎

Theorem 4.4.

Let $X$ be a Fano threefold and let $R$ be an extremal ray of type $C$ . Let $f\colon X\to Y$ be the contraction of $R$ . Then $f$ is a conic bundle and $Y$ is a smooth rational surface. Furthermore, if $f$ is generically smooth, then the following hold for the discriminant divisor $\Delta_{f}$ , the length of extremal ray $\mu_{R}$ , and an extremal rational curve $\ell$ of $R$ .

type of $R$	$f$	$\mu_{R}$	$\ell$
$C_{1}$	$\Delta_{f}$ is a nonzero effective divisor, if $f$ is generically smooth	$1$	a curve in a singular fibre
$C_{2}$	$\Delta_{f}=0$ and f is a $\mathbb{P}^{1}$ -bundle	$2$	a fibre of $f$

Proof.

We may assume that $f$ is generically smooth. Then general fibres of $f$ are $\mathbb{P}^{1}$ .

Assume that $f$ is of type $C_{1}$ , i.e., $f$ has a singular fibre. Then $\Delta_{f}\neq 0$ . By $-K_{X}\cdot f^{-1}(y)=2$ , we have that $-K_{X}\cdot\ell=1$ and $\mu_{R}=1$ .

Assume that $f$ is of type $C_{2}$ , i.e., every fibre of $f$ is smooth. Then $\Delta_{f}=0$ and any fibre is isomorphic to $\mathbb{P}^{1}$ . Since $Y$ is a smooth projective rational surface, we have ${\operatorname{Br}}(Y)=0$ (Proposition 2.6(2)). It follows from Proposition 2.7 that $X$ is isomorphic to $\mathbb{P}(E)$ over $Y$ for some vector bundle $E$ of rank $2$ . In this case, we have $\mu_{R}=2$ and $-K_{X}\cdot\ell=2$ for any fibre $\ell$ . ∎

Corollary 4.5.

Let $X$ be a Fano threefold. Let $R=\mathbb{R}_{\geq 0}[\ell]$ be an extremal ray of type $C$ with an extremal rational curve $\ell$ . Let $f\colon X\to Y$ be the contraction of $C$ . Then all the fibres of $f$ are numerically equivalent to $(2/\mu_{R})\ell$ , where $\mu_{R}$ denotes the length of the extremal ray $R$ .

Proof.

Assume that $R$ is of type $C_{1}$ . By Theorem 4.4, we have $\mu_{R}=1$ and $f^{-1}(y)\equiv 2\ell$ for any closed point $y$ . Note that this holds even if no fibre of $f$ is smooth.

Assume that $R$ is of type $C_{2}$ . Then we have $\mu_{R}=2$ and $f^{-1}(y)\equiv\ell$ for any closed point $y$ . ∎

4.1.2. Type D

Theorem 4.6.

Let $X$ be a Fano threefold and let $R$ be an extremal ray of type $D$ . Let $f\colon X\to Y$ be the contraction of $R$ . Fix a closed point $y$ and $X_{y}:=f^{-1}(y)$ denotes the scheme-theoretic fibre over $y$ . Then $Y\simeq\mathbb{P}^{1}$ and $X_{y}$ is a projective Goresntein surface on $X$ such that $\omega_{X_{y}}^{-1}$ is ample. Furthermore, the following hold for the length of extremal ray $\mu_{R}$ and an extremal rational curve $\ell$ of $R$ .

type of $R$	$f$	$\mu_{R}$	$\ell$
$D_{1}$	$1\leq K_{X_{y}}^{2}\leq 7$	$1$
$D_{2}$	$K_{X_{y}}^{2}=8$	$2$	a line on a fibre
$D_{3}$	$K_{X_{y}}^{2}=9$	$3$	a line on a fibre

Proof.

By Proposition 3.11, we obtain $Y\simeq\mathbb{P}^{1}$ . The remaining assertions follow from Remark 3.4 and Proposition 3.5. ∎

4.1.3. Type E

Lemma 4.7.

Let $X$ be a Fano threefold and let $f:X\to Y$ be the contraction of an extremal ray $R$ of type $E_{1}$ . Note that $C:=f(D)$ is a smooth curve on $Y$ , $Y$ is a smooth projective threefold, and $f$ coincides with the blowup along $C$ . Assume that $-K_{Y}$ is not ample. Then the following hold.

(1)

$C\simeq\mathbb{P}^{1}$ .
(2)

$\mathcal{N}^{*}_{C/Y}\simeq\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1)$ .
(3)

$D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ .
(4)

$\mathcal{O}_{X}(K_{X})|_{D}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$ and $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$ .

Proof.

It follows from Proposition 7.2(4) that

(4.7.1)

-K_{Y}\cdot C+2-2g(C)=(-K_{X})^{2}\cdot D>0,

where the inequality holds because $-K_{X}$ is ample. For $\mathcal{E}:=\mathcal{N}^{*}_{C/Y}\otimes\mathcal{O}_{C}(-K_{Y})$ and $D^{\prime}:=\mathbb{P}(\mathcal{E})$ , it follows from [Har77, Ch. II, Lemma 7.9] that

•

$\pi^{\prime}\colon D^{\prime}\xrightarrow{\simeq,\varphi}D\xrightarrow{\pi}C$ , and
•

$\mathcal{O}_{D^{\prime}}(1)\simeq\varphi^{*}\mathcal{O}_{D}(1)\otimes(\pi^{\prime})^{*}\mathcal{O}_{C}(-K_{Y})$ .

It holds that

\mathcal{E}=\mathcal{N}^{*}_{C/Y}\otimes\mathcal{O}_{C}(-K_{Y})\simeq\mathcal{N}^{*}_{C/Y}\otimes\mathcal{O}_{C}(-K_{C})\otimes\bigwedge^{2}\mathcal{N}_{C/Y}\simeq\mathcal{O}_{C}(-K_{C})\otimes\mathcal{N}_{C/Y}.

where the first isomorphism holds by [Har77, Ch. II, Proposition 8.20] and the second one follows from [Har77, Ch. II, Exercise 5.16(b)]. Moreover,

	$\displaystyle\mathcal{O}_{D^{\prime}}(1)$	$\displaystyle\simeq\varphi^{}\mathcal{O}_{D}(1)\otimes(\pi^{\prime})^{}\mathcal{O}_{C}(-K_{Y})$
		$\displaystyle\simeq\varphi^{}\mathcal{O}_{D}(1)\otimes(\varphi^{}\pi^{*}\mathcal{O}_{C}(-K_{Y}))$
		$\displaystyle\simeq\varphi^{}(\mathcal{O}_{D}(1)\otimes\pi^{}\mathcal{O}_{C}(-K_{Y}))$
		$\displaystyle\simeq\varphi^{}((\mathcal{O}_{X}(-D)\otimes f^{}\mathcal{O}_{Y}(-K_{Y}))\|_{D})$
		$\displaystyle\simeq\varphi^{*}(\mathcal{O}_{X}(-K_{X})\|_{D}),$

where the fourth and fifth isomorphisms follow from (2) and (1) of Proposition 7.2, respectively. Since $-K_{X}$ is ample, also $\mathcal{O}_{D^{\prime}}(1)$ is ample. Hence $\mathcal{E}$ is an ample vector bundle on $C$ by [Har70, Ch. III, Theorem 1.1].

Claim.

It holds that $-K_{Y}\cdot C\leq 0$ .

Proof of Claim.

Suppose the contrary, i.e., $-K_{Y}\cdot C>0$ . In order to derive a contraction, it suffices to prove that $-K_{Y}$ is ample. To this end, it is enough to show (i) and (ii) below [Har70, Ch. I, Proposition 4.6].

(i)

$-K_{Y}\cdot Z>0$ for any curve $Z$ on $Y$ .
(ii)

$-K_{Y}$ is semi-ample, i.e., $|-mK_{Y}|$ is base point free for some $m\in\mathbb{Z}_{>0}$ .

Let us show (i). Fix a curve $Z$ on $Y$ . If $Z=C$ , then $-K_{Y}\cdot Z>0$ holds by our assumption. Assume that $Z\neq C$ . Let $\widetilde{Z}$ be the proper transform of $Z$ on $X$ . Since $-K_{X}$ is ample and $D$ is an effective divisor with $\widetilde{Z}\not\subset{\operatorname{Supp}}\,D$ , we have

\displaystyle-K_{Y}\cdot Z=-f^{*}K_{Y}\cdot\widetilde{Z}=(-K_{X}+D)\cdot\widetilde{Z}>0.

Hence (i) holds.

Let us show (ii). By (i), $-K_{Y}$ is nef. By $-K_{Y}\cdot C>0$ , $\mathcal{O}_{Y}(-K_{Y})|_{C}$ is ample. Hence $\pi^{*}(\mathcal{O}_{Y}(-K_{Y})|_{C})\simeq f^{*}(\mathcal{O}_{Y}(-K_{Y}))|_{D}$ is semi-ample. By $f^{*}(-K_{Y})=-K_{X}+D$ , it follows from [CMM14, Theorem 3.2] that $f^{*}(-K_{Y})$ is semi-ample. Since $X$ and $Y$ are normal, $-K_{Y}$ is semi-ample. Thus (ii) holds. This completes the proof of Claim. ∎

By (4.7.1) and Claim, we have $g(C)=0$ . Hence (1) holds. Since $\mathcal{E}=\mathcal{O}_{C}(-K_{C})\otimes\mathcal{N}_{C/Y}$ is a vector bundle of rank 2 on $C\simeq\mathbb{P}^{1}$ , we can write

\mathcal{E}\simeq\mathcal{O}_{\mathbb{P}^{1}}(a)\oplus\mathcal{O}_{\mathbb{P}^{1}}(b).

for some $a,b\in\mathbb{Z}$ . Since $\mathcal{E}$ is ample, we have $a>0$ and $b>0$ [Har70, Ch. III, Corollary 1.8]. By $\mathcal{O}_{D^{\prime}}(1)\simeq\varphi^{*}(\mathcal{O}_{X}(-K_{X})|_{D})$ , we obtain

	$\displaystyle a+b$	$\displaystyle=\deg_{C}(\mathcal{E})$
		$\displaystyle=c_{1}(\mathcal{O}_{D^{\prime}}(1))^{2}$
		$\displaystyle=c_{1}(\varphi^{*}(\mathcal{O}_{X}(-K_{X})\|_{D}))^{2}$
		$\displaystyle=(-K_{X})^{2}\cdot D$
		$\displaystyle=-K_{Y}\cdot C+2-g(C)$
		$\displaystyle\leq 2,$

where the second equality holds by Proposition 7.1(4) and the fifth one follows from Proposition 7.2(4). Hence we get $a=b=1$ and $\mathcal{E}\simeq\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1)$ .

By $\mathcal{E}\simeq\mathcal{O}_{C}(-K_{C})\otimes\mathcal{N}_{C/Y}$ , we have

\displaystyle\mathcal{N}_{C/Y}^{*}\simeq\mathcal{E}^{*}\otimes\omega_{C}^{-1}\simeq(\mathcal{O}_{\mathbb{P}^{1}}(-1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(-1))\otimes\mathcal{O}_{\mathbb{P}^{1}}(2)\simeq\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1).

Thus (2) holds. By Lemma 7.2(2), we obtain

D\simeq\mathbb{P}(\mathcal{N}_{C/Y}^{*})\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1))\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}.

Thus (3) holds.

Let us show (4). By $D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ , we can write

\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(m,n)

for some $m,n\in\mathbb{Z}$ . It holds that

2mn=c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(m,n))^{2}=c_{1}(\mathcal{O}_{X}(D)|_{D})^{2}

=D^{3}=\deg_{C}(\mathcal{N}^{*}_{C/Y})=\deg_{\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1))=2,

where the fourth equality follows from Proposition 7.2(3). Hence we have $(m,n)=(1,1)$ or $(m,n)=(-1,-1)$ . Since $D\cdot\zeta<0$ for an $f$ -exceptional curve $\zeta\subset D$ , we obtain $(m,n)=(-1,-1)$ . By the adjunction formula $\mathcal{O}_{X}(K_{X}+D)|_{D}\simeq\mathcal{O}_{D}(K_{D})\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,-2)$ , we obtain $\mathcal{O}_{X}(K_{X})|_{D}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$ . Thus (4) holds. ∎

Theorem 4.8.

Let $X$ be a primitive Fano threefold and let $R$ be an extremal ray of type $E$ . Let $f\colon X\to Y$ be the contraction of $R$ . Then $f$ is a birational morphism to a projective normal threefold $Y$ . Furthermore, the following hold for the length of extremal ray $\mu_{R}$ and an extremal rational curve $\ell$ of $R$ .

type of $R$	$f$ and $D$	$\mu_{R}$	$\ell$
$E_{1}$	$Y$ is smooth,	$1$	$P\times\mathbb{P}^{1}$
	$C=f(D)\simeq\mathbb{P}^{1}$ ,
	$\mathcal{N}^{*}_{C/Y}\simeq\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1),$
	$D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1},$
	$\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$
$E_{2}$	$Y$ is smooth,	$2$	a line on $D$
	$f(D)$ is a point,
	$D\simeq\mathbb{P}^{2}$ ,
	$\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{2}}(-1)$
$E_{3}$	$f(D)$ is a point,	$1$	$P\times\mathbb{P}^{1}$ or $\mathbb{P}^{1}\times Q$
	$D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ ,		on $D$ $(P,Q\in\mathbb{P}^{1})$
	$\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$ ,
	$P\times\mathbb{P}^{1}$ and $\mathbb{P}^{1}\times Q$ are numerically
	equivalent on $X$ for all $P,Q\in\mathbb{P}^{1}$
$E_{4}$	$f(D)$ is a point,	$1$	a line on $D$
	$D$ is isomorphic to
	the singular quadric surface
	in $\mathbb{P}^{3}$ ,
	$\mathcal{O}_{D}(D)\simeq\mathcal{O}_{D}\otimes\mathcal{O}_{\mathbb{P}^{3}}(-1)$
$E_{5}$	$f(D)$ is a point,	$1$	a line on $D$
	$D\simeq\mathbb{P}^{2}$ ,
	$\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{2}}(-2)$

In particular, $\mathcal{O}_{X}(-D)|_{D}$ is ample.

Proof.

The columns $\mu_{R}$ and $\ell$ can be confirmed by $(K_{X}+D)\cdot\ell=K_{D}\cdot\ell$ . ∎

Corollary 4.9.

Let $X$ be a primitive Fano threefold with an extremal ray $R$ of type $E$ . Let $f:X\to Y$ be the contraction of $R$ and set $D:={\operatorname{Ex}}(f)$ . Then any effective divisor $Z$ on $D$ is semi-ample.

Proof.

If $D\simeq\mathbb{P}^{2}$ or $D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ , then the assertion is clear. Hence we may assume that $D$ is a singular quadric surface in $\mathbb{P}^{3}$ (Theorem 4.8). In this case, the assertion follows from the fact that $D$ is $\mathbb{Q}$ -factorial and $\rho(D)=1$ (it is well known that $D$ is a projective toric surface which is obtained by contracting the $(-2)$ -curve on $\mathbb{P}_{\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}}(2))$ ). ∎

Corollary 4.10.

Let $X$ be a Fano threefold with an extremal ray $R$ of type $E_{2},E_{3},E_{4}$ , or $E_{5}$ . Let $f:X\to Y$ be the contraction of $R$ and set $D:={\operatorname{Ex}}(f)$ . Let $g\colon X\to\mathbb{P}^{1}$ be a morphism. Then $g(D)$ is a point.

Proof.

Assume that $g(D)$ is not a point, i.e., $g(D)=\mathbb{P}^{1}$ . Then there exist curves $\Gamma_{1}$ and $\Gamma_{2}$ on $D$ such that $g(\Gamma_{1})$ is a point and $g(\Gamma_{2})=\mathbb{P}^{1}$ . We have $g^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)\cdot\Gamma_{1}=0$ and $g^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)\cdot\Gamma_{1}>0$ . Since the numerical equivalence classes $[\Gamma_{1}]$ and $[\Gamma_{2}]$ lie on $R\setminus\{0\}$ , we can find $\lambda\in\mathbb{R}_{>0}$ satisfying $\Gamma_{1}\equiv\lambda\Gamma_{2}$ , which is a contradiction. ∎

4.2. Picard groups

Theorem 4.11.

Let $X$ be a Fano threefold with $\rho(X)\geq 2$ . Let $R$ an extremal ray of ${\operatorname{NE}}(X)$ and let $f\colon X\to Y$ be the contraction of $R$ . Pick an extremal rational curve $\ell$ with $R=\mathbb{R}_{\geq 0}[\ell]$ . Then the sequence

0\longrightarrow{\operatorname{Pic}}\,Y\stackrel{{\scriptstyle f^{*}}}{{\longrightarrow}}{\operatorname{Pic}}\,X\stackrel{{\scriptstyle(-\cdot\ell)}}{{\longrightarrow}}\mathbb{Z}\longrightarrow 0

is exact. In particular, $\rho(X)=\rho(Y)+1$ .

Proof.

By Proposition 3.9, it is enough to find a divisor $F$ on $X$ such that $F\cdot\ell=1$ . If $\mu_{R}=1$ , then we get $-K_{X}\cdot\ell=\mu_{R}=1$ . Therefore, we are done for the case when the type of $R$ is one of $C_{1},D_{1},E_{1},E_{3},E_{4},E_{5}$ (Remark 3.4). The remaining cases are $C_{2},D_{2},D_{3}$ , and $E_{2}$ .

Assume that $R$ is of type $C_{2}$ . By Theorem 4.4, $X$ is isomorphic to a $\mathbb{P}^{1}$ -bundle over $Y$ , associated to some locally free sheaf of rank $2$ . Moreover, $\ell$ is a fibre of this bundle (Theorem 4.4). Hence $\mathcal{O}_{X}(1)\cdot\ell=1$ by Lemma 7.1(1).

Assume that $R$ is of type $D_{3}$ . Then there exist Cartier divisors $D$ on $X$ and $E$ on $Y$ such that $K_{X}\sim 3D+f^{*}E$ (Proposition 3.5(5)). Then $3=-K_{X}\cdot\ell=3D\cdot\ell$ , i.e., $D\cdot\ell=1$ . By using Proposition 3.5(6), the same argument works for the case when $R$ is of type $D_{2}$ .

Assume that $R$ is of type $E_{2}$ . Then the exceptional divisor $D$ on $X$ satisfies $D\simeq\mathbb{P}^{2},\mathcal{O}_{D}(D)\simeq\mathcal{O}_{D}(-1)$ , and $\ell$ is a line on $D$ [TanII, Proposition 3.22]. Hence $(-D\cdot\ell)=-(D|_{D}\cdot\ell)_{D}=-(c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(-1))\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(1)))_{\mathbb{P}^{2}}=1$ . ∎

Corollary 4.12.

Let $X$ be a Fano threefold with $\rho(X)=2$ . Then $X$ is primitive if and only if $X$ has no extremal ray of type $E_{1}$ .

Proof.

If $X$ has no extremal ray of type $E_{1}$ , then it is clear that $X$ is primitive. Suppose that $X$ is primitive and there exists an extremal ray $R$ of type $E_{1}$ . It suffices to derive a contradiction. Let $f\colon X\to Y$ be the contraction of $R$ . Then $Y$ is a smooth projective threefold and $f$ is the blowup along a smooth curve $C$ . It suffices to prove that $Y$ is Fano, because $X$ is a primitive Fano threefold.

By Corollary 3.10, there is an ample divisor $L$ on $Y$ which generates ${\operatorname{Pic}}\,Y\simeq\mathbb{Z}$ . Hence we can write $-K_{Y}\sim\alpha L$ for some $\alpha\in\mathbb{Z}$ . By the Bertini theorem, we can find a curve $Z$ on $Y$ with $C\cap Z=\emptyset$ . In particular, its proper transform $\tilde{Z}$ on $X$ is disjoint from $D:={\operatorname{Ex}}(f)$ . Then

\displaystyle(-K_{X})\cdot\tilde{Z}=(f^{*}(-K_{Y})-D)\cdot\tilde{Z}=-K_{Y}\cdot Z=\alpha L\cdot Z.

Since $-K_{X}$ and $L$ are ample, we get $\alpha>0$ . Hence $-K_{Y}\sim\alpha L$ is ample. ∎

4.3. Existence of conic bundle structures

Lemma 4.13.

Let $X$ be a primitive Fano threefold with $\rho(X)\geq 2$ . Then $X$ has an extremal ray of type $C$ or $D$ .

Proof.

Suppose the contrary. Let $R_{1},\dots,R_{n}$ be all the extremal rays. We can find curve $\ell_{1},...,\ell_{n}$ such that

{\operatorname{NE}}(X)=\sum_{i=1}^{n}R_{i}=\sum_{i=1}^{n}\mathbb{R}_{\geq 0}[\ell_{i}].

For each $i\in\{1,...,n\}$ , let $f_{i}:X\to Y_{i}$ be the contraction of $R_{i}$ , which is of type $E$ . Set $D_{i}:={\operatorname{Ex}}(f_{i})$ . Note that each $D_{i}$ is a projective normal toric surface (Theorem 4.8).

Claim.

$D_{1}\cap D_{i}=\emptyset$ for all $2\leq i\leq n$ .

Proof of Claim.

Suppose that $D_{1}\cap D_{i}\neq\emptyset$ for some $2\leq i\leq n$ . Since $D_{i}|_{D_{1}}$ is a nonzero effective Cartier divisor on $D_{1}$ , there exists a curve $Z$ such that $Z\subset D_{1}\cap D_{i}$ .

Since $\mathcal{O}_{X}(-D_{1})|_{D_{1}}$ is ample (Theorem 4.8), we have

D_{1}\cdot Z=\mathcal{O}_{X}(D_{1})\cdot Z=(\mathcal{O}_{X}(D_{1})|_{D_{1}})\cdot Z<0.

On the other hand, it follows from Corollary 4.9 that there exist $m\in\mathbb{Z}_{>0}$ and an effective Cartier divisor $Z^{\prime}$ on $D_{i}$ such that $mZ\sim Z^{\prime}$ and ${\operatorname{Supp}}\,Z^{\prime}$ does not contain any irreducible component of $D_{1}\cap D_{i}$ . Then it holds that

D_{1}\cdot Z=(D_{1}|_{D_{i}})\cdot Z=\frac{1}{m}(D_{1}|_{D_{i}})\cdot Z^{\prime}\geq 0,

which is a contradiction. This completes the proof of Claim. ∎

Fix $s\in\mathbb{Z}_{>0}$ such that $|-sK_{X}|$ is very ample. Pick two general members $H_{1},H_{2}\in|-sK_{X}|$ such that $H_{1}\cap H_{2}$ is a smooth curve. By $\mathrm{NE}(X)=\mathbb{R}_{\geq 0}[\ell_{1}]+\cdots+\mathbb{R}_{\geq 0}[\ell_{n}]$ , we have $H_{1}\cdot H_{2}=H_{1}\cap H_{2}\equiv\sum_{i=1}^{n}a_{i}\ell_{i}$ for some $a_{1},...,a_{n}\in\mathbb{R}_{\geq 0}$ . We then obtain the following contradiction:

0<D_{1}\cdot(-sK_{X})^{2}=D_{1}\cdot H_{1}\cdot H_{2}=a_{1}D_{1}\cdot\ell_{1}+\sum_{i=2}^{n}a_{i}D_{1}\cdot\ell_{i}=a_{1}D_{1}\cdot\ell_{1}\leq 0,

where Claim implies $D_{1}\cdot\ell_{2}=\cdots=D_{1}\cdot\ell_{n}=0$ and the inequality $D_{1}\cdot\ell_{1}<0$ follows from the ampleness of $\mathcal{O}_{X}(-D_{1})|_{D_{1}}$ (Theorem 4.8). ∎

Proposition 4.14.

Let $X$ be a primitive Fano threefold with $\rho(X)\geq 2$ . Then $X$ has an extremal ray of type $C$ .

Proof.

By Lemma 4.13, we may assume that there exists an extremal ray $R_{1}$ is of type $D$ . Then we have $\rho(X)=2$ (Theorem 4.11). Let $R_{2}$ be the other extremal ray. For each $i\in\{1,2\}$ , let $f_{i}:X\to Y_{i}$ be the contraction of $R_{i}$ , where $Y_{1}=\mathbb{P}^{1}$ . Suppose that $R_{2}$ is of type $E$ or $D$ . It suffices to derive a contradiction. If $R_{2}$ is of type $D$ , then Lemma 4.15(1) leads to a contradiction.

The problem is reduced to the case when $R_{2}$ is of type $E$ . By Corollary 4.12, $R_{2}$ is not of type $E_{1}$ , and hence $R_{2}$ is of type $E_{2},E_{3},E_{4},$ or $E_{5}$ . Set $D_{2}:={\operatorname{Ex}}(f_{2})$ . It follows from Corollary 4.10 that $f_{1}(D_{2})$ is a point. Pick a curve $Z$ with $Z\subset D_{2}$ . Then both $f_{1}(Z)$ and $f_{2}(Z)$ are points. Hence we get $[Z]\in R_{1}\cap R_{2}=\{0\}$ , which contradicts $-K_{X}\cdot Z>0$ . ∎

Lemma 4.15.

Let $X$ be a primitive Fano threefold. Let $R_{1}$ and $R_{2}$ be two distinct extremal rays. For each $i\in\{1,2\}$ , let $f_{i}:X\to Y_{i}$ be the contraction of $R_{i}$ . Then the following hold.

(1)

The induced morphism $f_{1}\times f_{2}:X\to Y_{1}\times Y_{2}$ is a finite morphism.
(2)

If $f_{1}$ is of type $C$ and $f_{2}$ is of type $E$ , then the composite morphism

$f_{1}|_{D}:D\overset{j}{\hookrightarrow}X\xrightarrow{f_{1}}Y_{1}$

is a finite surjective morphism, where $D:={\operatorname{Ex}}(f_{2})$ is the $f_{2}$ -exceptional prime divisor and $j:D\hookrightarrow X$ denotes the induced closed immersion.

Proof.

Let us show (1). Suppose that $f_{1}\times f_{2}:X\to Y_{1}\times Y_{2}$ is not a finite morphism. Then there exists a curve $\Gamma$ on $X$ such that $(f_{1}\times f_{2})(\Gamma)$ is a point. Then both $f_{1}(\Gamma)$ and $f_{2}(\Gamma)$ are points. However, this implies $[\Gamma]\in R_{1}\cap R_{2}=\{0\}$ , which is a contradiction. Thus (1) holds.

Let us show (2). By $\dim D=\dim Y_{1}$ , it suffices to show that $f_{1}|_{D}$ is a finite morphism. Suppose that $f_{1}|_{D}:D\to Y_{1}$ is not a finite morphism. Then there exists a curve $\Gamma$ on $X$ such that $\Gamma\subset D$ and $f_{1}(\Gamma)$ is a point. If $R_{2}$ is of type $E_{2},E_{3},E_{4}$ , or $E_{5}$ , then also $f_{2}(\Gamma)$ is a point. This is a contradiction: $[\Gamma]\in R_{1}\cap R_{2}=\{0\}$ . Hence $R_{2}$ is of type $E_{1}$ . Since $\mathcal{O}_{X}(-D)|_{D}$ is ample, we have that $D\cdot\Gamma<0$ . Let $\Gamma^{\prime}$ be a general fibre of $f_{1}:X\to Y_{1}$ , so that $\Gamma^{\prime}\not\subset D$ . Hence we get $D\cdot\Gamma^{\prime}\geq 0$ . On the other hand, we have $\Gamma^{\prime}\equiv\lambda\Gamma$ for some $\lambda\in\mathbb{R}_{>0}$ , which implies $D\cdot\Gamma\geq 0$ . This is a contradiction. ∎

4.4. $\rho(X)\leq 3$

Theorem 4.16.

Let $X$ be a primitive Fano threefold. Let $R$ be an extremal ray of type $C$ and let $f:X\to S$ be the contraction of $R$ . Then $S\simeq\mathbb{P}^{2}$ or $S\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ .

Proof.

If $f$ is a wild conic bundle, then the assertion follows from Theorem 4.2. Hence we may assume that $f$ is generically smooth. Note that $S$ is a smooth projective rational surface (Theorem 4.4). By the classification of smooth projective rational surfaces, it suffices to show that there exists no curve $E$ on $S$ such that $E^{2}<0$ . Suppose that there is a curve $E$ on $S$ such that $E^{2}<0$ . Let us derive a contradiction.

Fix a curve $C$ on $X$ such that $C\subset f^{-1}(E)$ and $f(C)=E$ . We have

{\operatorname{NE}}(X)=\sum_{i=1}^{n}R_{i}=\sum_{i=1}^{n}\mathbb{R}_{\geq 0}[\ell_{i}],\qquad R_{i}=\mathbb{R}_{\geq 0}[\ell_{i}]

for the extremal rays $R_{1}=R,R_{2},...,R_{n}$ and curves $\ell_{1},...,\ell_{n}$ on $X$ . We then have

C\equiv\sum_{i=1}^{n}a_{i}\ell_{i}

for some $a_{1},...,a_{n}\in\mathbb{R}_{\geq 0}$ .

Now, we can write $f_{*}C=bE$ for some $b\in\mathbb{Q}_{>0}$ . By $E^{2}<0$ , we get

\sum_{i=1}^{n}a_{i}E\cdot f_{*}\ell_{i}=E\cdot\sum_{i=1}^{n}a_{i}f_{*}\ell_{i}=E\cdot f_{*}C=bE^{2}<0.

Hence, possibly after permuting the indices, we have $E\cdot f_{*}\ell_{2}<0$ (note that $f_{*}\ell_{1}=0$ ). Let $f_{2}:X\to Y_{2}$ be the contraction of $R_{2}$ . By $E\cdot f_{*}\ell_{2}<0$ and $E^{2}<0$ , we get $f(\ell_{2})=E$ . By the projection formula, we obtain

(4.16.1)

f^{*}E\cdot\ell_{2}=E\cdot(f_{*}\ell_{2})<0.

Assume that $R_{2}$ is of type $C$ or $D$ . Then the fibres of $f_{2}:X\to Y_{2}$ are curves or surfaces. By $f^{-1}(E)\neq X$ , there is a curve $Z$ on $X$ such that

(i)

$Z\not\subset f^{-1}(E)$ and
(ii)

$f_{2}(Z)$ is a point.

By (ii), we can write $Z\equiv c\ell_{2}$ for some $c\in\mathbb{R}_{>0}$ . Therefore, we obtain

f^{*}E\cdot\ell_{2}=c^{-1}f^{*}E\cdot Z\geq 0,

where the inequality holds by (i). This contradicts (4.16.1).

Assume that $R_{2}$ is of type $E$ . Set $D_{2}:={\operatorname{Ex}}(f_{2})$ . Since $D_{2}$ is covered by curves $Z$ with $[Z]\in R_{2}=\mathbb{R}_{\geq 0}[\ell_{2}]$ , it follows from (4.16.1) that $D_{2}\subset f^{-1}(E)$ . By $f(D_{2})\subset f(f^{-1}(E))=E$ , there exists a curve $C^{\prime}$ on $D_{2}$ such that $f(C^{\prime})$ is a point. We get $[C^{\prime}]\in R$ . Since $\mathcal{O}_{X}(-D_{2})|_{D_{2}}$ is ample (Theorem 4.8), we have

D_{2}\cdot C^{\prime}<0.

On the other hand, $D_{2}$ is disjoint from $f^{-1}(s)$ for a general closed point $s\in S$ , and hence we get

D_{2}\cdot f^{-1}(s)=0.

This is a contradiction, because we have $[C^{\prime}]\in R$ and $[f^{-1}(s)]\in R$ . ∎

Theorem 4.17.

Let $X$ be a primitive Fano threefold. Then the following hold.

(1)
If $f:X\to S$ is the contraction of an extremal ray of type $C$ , then the following hold.
- •
  
  $\rho(X)=2$ if and only if $S\simeq\mathbb{P}^{2}$ .
- •
  
  $\rho(X)=3$ if and only if $S\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ .
(2)

$\rho(X)\leq 3.$
(3)

Assume $\rho(X)=2$ . Then $X$ has an extremal ray $R$ of type $C$ and we have $S\simeq\mathbb{P}^{2}$ for the contraction $f:X\to S$ of $R$ .
(4)
Assume $\rho(X)=3$ . Then the following hold.
1. (i)
  
  Let $R$ be an extremal ray of $X$ . Then $R$ is of type $C$ or $E_{1}$ .
2. (ii)
  
  $X$ has two extremal rays $R_{1}$ and $R_{2}$ such that $R_{1}$ is of type $C$ and $R_{2}$ is of type $C$ or $E_{1}$ .
(5)

If $X$ has an extremal ray of type $C$ whose contraction $f:X\to Y$ is not generically smooth, then $p=2$ , $\rho(X)=2$ , and $X$ is a prime divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$ .

Proof.

The assertion (1) follows from $\rho(X)=\rho(S)+1$ and Theorem 4.16. Then (1) implies (2) and (3) by Proposition 4.14. The assertion (5) follows from Theorem 4.2.

Let us show (4). Since (i) implies (ii) (Proposition 4.14), it suffices to prove (i). Suppose that there exists an extremal ray $R$ which is not of type $C$ nor $E_{1}$ . Let us derive a contradiction. Since $R$ is not of type $D$ , the type of $R$ is $E_{2},E_{3},E_{4}$ , or $E_{5}$ . Let $f:X\to Y$ be the contraction of $R$ . Set $D:={\operatorname{Ex}}(f)$ . There exists an extremal ray $R^{\prime}$ of type $C$ (Proposition 4.14). Let $f^{\prime}:X\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ be the contraction of $R^{\prime}$ . For each $i\in\{1,2\}$ , consider

g_{i}:X\xrightarrow{f^{\prime}}\mathbb{P}^{1}\times\mathbb{P}^{1}\xrightarrow{{\rm pr}_{i}}\mathbb{P}^{1}.

By Corollary 4.10, we have that $g_{i}(D)$ is a point for each $i\in\{1,2\}$ . Hence also $f^{\prime}(D)$ is a point. However, this is a contradiction, because any fibre of $f^{\prime}$ is one-dimensional. ∎

5. Fano threefolds with $\rho(X)=2$

In this section, we classify Fano threefolds with $\rho(X)=2$ . If $X$ has an extremal ray of type C and its contraction is not generically smooth (i.e., a wild conic bunlde), then $X$ is a prime divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$ (Theorem 4.2). In what follows, we always assume that the contraction of type C is generically smooth.

Notation 5.1.

Let $X$ be a Fano threefold with $\rho(X)=2$ . By Theorem 4.1, $\mathrm{NE}(X)$ has exactly two extremal rays $R_{1}$ and $R_{2}$ . For each $i\in\{1,2\}$ , let

f_{i}:X\to Y_{i}

be the contraction of $R_{i}$ and let $\ell_{i}$ be an extremal rational curve with $R_{i}=\mathbb{R}_{\geq 0}[\ell_{i}]$ . Set $\mu_{i}:=-K_{X}\cdot\ell_{i}$ , which is the length of $R_{i}$ . We have

{\operatorname{NE}}(X)=R_{1}+R_{2}=\mathbb{R}_{\geq 0}[\ell_{1}]+\mathbb{R}_{\geq 0}[\ell_{2}].

It follows from Corollary 3.10 that ${\operatorname{Pic}}(Y_{i})\simeq\mathbb{Z}$ for each $i\in\{1,2\}$ . Let $L_{i}$ be an ample Cartier divisor on $Y_{i}$ such that $\mathcal{O}_{Y_{i}}(L_{i})$ generates ${\operatorname{Pic}}\,Y_{i}$ . Set $H_{i}:=f_{i}^{*}L_{i}$ .

•

If $R_{i}$ is of type $C$ , then we assume that $f_{i}$ is generically smooth and $\Delta_{f_{i}}$ denotes its discriminant divisor (Subsection 3.4).
•

If $R_{i}$ is of type $D$ , then $d_{i}:=(-K_{X})^{2}\cdot H_{i}$ . Note that $1\leq d_{i}\leq 9$ (Theorem 4.6).
•

Assume that $R_{i}$ is of type $E$ . Set $D_{i}:={\operatorname{Ex}}(f_{i})$ . If $R_{i}$ is of type $E_{1}$ , then $B_{i}:=f_{i}(D_{i})$ . If $R_{i}$ is of type $E_{1},E_{3},$ or $E_{4}$ (resp. $E_{2}$ , resp. $E_{5}$ ), then let $r_{i}$ be the largest positive integer which divides $-K_{X}+D_{i}$ (resp. $-K_{X}+2D_{i}$ , resp. $-2K_{X}+D_{i}$ ). Note that $r_{i}$ coincides with the index of the Fano threefold $Y_{i}$ when $R_{i}$ is of type $E_{1}$ or $E_{2}$ .

5.1. The Picard group and the canonical divisor

Lemma 5.2.

We use Notation 5.1. Then, for each $i\in\{1,2\}$ , $H_{i}$ is a primitive element in ${\operatorname{Pic}}\,X$ , i.e., there exists no pair $(s,H)$ such that $s\in\mathbb{Z}_{\geq 2}$ , $H$ is a Cartier divisor on $X$ , and $H_{i}\sim sH$ .

Proof.

Suppose that there exist a Cartier divisor $H$ on $X$ and an integer $s\geq 2$ such that $H_{i}\sim sH$ . By $H_{i}\cdot\ell_{i}=f_{i}^{*}L_{i}\cdot\ell_{i}=0$ , we obtain $H\cdot\ell_{i}=0$ . By Theorem 4.11, there exists a Cartier divisor $M$ on $Y_{i}$ such that $H\sim f_{i}^{*}M$ . Then we get $0\sim H_{i}-sH\sim f_{i}^{*}L_{i}-sf_{i}^{*}M$ , which implies $L_{i}\sim sM$ . This contradicts the fact that $\mathcal{O}_{Y_{i}}(L_{i})$ is a primitive element in ${\operatorname{Pic}}\,Y_{i}$ . ∎

Lemma 5.3.

We use Notation 5.1. Fix $i\in\{1,2\}$ and assume that $R_{i}$ is of type $C$ . Then

H_{i}^{2}\equiv\frac{2}{\mu_{i}}\ell_{i}.

Proof.

By $Y_{i}\simeq\mathbb{P}^{2}$ (Theorem 4.17(1)), we have $\mathcal{O}_{X}(L_{i})\simeq\mathcal{O}_{\mathbb{P}^{2}}(1)$ . Hence $L_{i}^{2}$ is a point on $\mathbb{P}^{2}$ . Then $H_{i}^{2}=f_{i}^{*}L_{i}^{2}$ is a fibre of $f_{i}$ , which is numerically equivalent to $\frac{2}{\mu_{i}}\ell_{i}$ (Corollary 4.5). ∎

We now compute $c_{2}(X)\cdot H_{i}$ for each $i\in\{1,2\}$ .

Lemma 5.4.

We use Notation 5.1. Fix $i\in\{1,2\}$ . Then the following hold.

(1)

Assume that $R_{i}$ is of type $C$ . Recall that $\Delta_{f_{i}}$ denotes the discriminant divisor of the generically smooth conic bundle $f_{i}:X\to Y_{i}=\mathbb{P}^{2}$ . Then $c_{2}(X)\cdot H_{i}$ satisfies the following.

type of $R_{i}$ $C_{1}$ $C_{2}$

$c_{2}(X)\cdot H_{i}$ $6+\deg\Delta_{f_{i}}$ $6$

Moreover, $\deg\Delta_{f_{i}}\geq 1$ if $R_{i}$ is of type $C_{1}$ .
(2)

Assume that $R_{i}$ is of type $D$ . Then $c_{2}(X)\cdot H_{i}$ satisfies the following.

type of $R_{i}$ $D_{1}$ $D_{2}$ $D_{3}$

$c_{2}(X)\cdot H_{i}$ $12-(-K_{X})^{2}\cdot H_{i}$ $4$ $3$

(3)

Assume that $R_{i}$ is of type $E$ . Set $D_{i}:={\operatorname{Ex}}(f_{i})$ . Recall that if $R_{i}$ is of type $E_{1},E_{3}$ , or $E_{4}$ (resp. $E_{2}$ , resp. $E_{5}$ ), then $r_{i}$ is defined as the largest positive integer that divides $-K_{X}+D_{i}$ (resp. $-K_{X}+2D_{i}$ , resp. $-2K_{X}+D_{i}$ ) in ${\operatorname{Pic}}\,X$ . Then $c_{2}(X)\cdot H_{i}$ satisfies the following:

type of $R_{i}$	$E_{1}$	$E_{2}$	$E_{3}$ or $E_{4}$	$E_{5}$
$c_{2}(X)\cdot H_{i}$	$\frac{24}{r_{i}}+\deg B_{i}$	$\frac{24}{r_{i}}$	$\frac{24}{r_{i}}$	$\frac{45}{r_{i}}$

where we set $B_{i}:=f_{i}(D_{i})$ and $\deg B_{i}:=\frac{-K_{Y_{i}}\cdot B}{r_{i}}$ for the case when $f_{i}$ is of type $E_{1}$ .

Proof.

Let us show (1). Assume that $R_{i}$ is of type $C$ . Then $f_{i}\colon X\to Y_{i}=\mathbb{P}^{2}$ is a conic bundle. Since $L_{i}$ is an ample generator of ${\operatorname{Pic}}\,\mathbb{P}^{2}$ , we may assume that $L_{i}$ is a line on $Y_{i}=\mathbb{P}^{2}$ . Since $H_{i}=f_{i}^{*}L_{i}$ is an effective divisor, it follows from Lemma 3.8(4) that

c_{2}(X)\cdot H_{i}=6\chi(H_{i},\mathcal{O}_{H_{i}})+6\chi(H_{i},\mathcal{O}_{X}(H_{i})|_{H_{i}})-2H_{i}^{3}-(-K_{X})^{2}\cdot H_{i}.

By Corollary 3.12, it holds that

\chi(H_{i},\mathcal{O}_{H_{i}})=1-\chi(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(-L_{i}))=1-\chi(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(-1))=1

and

\chi(H_{i},\mathcal{O}_{X}(H_{i})|_{H_{i}})=\chi(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(L_{i}))-1=\chi(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(1))-1=2.

By $L_{i}^{3}=0$ , we have $H_{i}^{3}=f^{*}L_{i}^{3}=0$ . Since $L_{i}$ is a line on $\mathbb{P}^{2}$ , it follows from Proposition 3.16 that

(-K_{X})^{2}\cdot H_{i}=-4K_{\mathbb{P}^{2}}\cdot L_{i}-\Delta_{f_{i}}\cdot L_{i}=12-\deg(\Delta_{f_{i}}).

To summarise, it holds that

	$\displaystyle c_{2}(X)\cdot H_{i}$	$\displaystyle=6+12-0-(12-\deg(\Delta_{f_{i}}))$
		$\displaystyle=6+\deg(\Delta_{f_{i}}).$

Thus (1) holds.

Let us show (2). Assume that $R_{i}$ is of type $D$ . Then we have $f_{i}\colon X\to Y_{i}=\mathbb{P}^{1}$ . Since $L_{i}$ is an ample generator of ${\operatorname{Pic}}\,\mathbb{P}^{1}$ , we may assume that $L_{i}$ is a point on $Y_{i}=\mathbb{P}^{1}$ . Applying Lemma 3.8(4), we get

c_{2}(X)\cdot H_{i}=6\chi(H_{i},\mathcal{O}_{H_{i}})+6\chi(H_{i},\mathcal{O}_{X}(H_{i})|_{H_{i}})-2H_{i}^{3}-(-K_{X})^{2}\cdot H_{i}.

By Corollary 3.12, it holds that

\chi(H_{i},\mathcal{O}_{H_{i}})=1-\chi(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(-L_{i}))=1-\chi(\mathcal{O}_{\mathbb{P}^{1}}(-1))=1

and

\chi(H_{i},\mathcal{O}_{X}(H_{i})|_{H_{i}})=\chi(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(L_{i}))-1=\chi(\mathcal{O}_{\mathbb{P}^{1}}(1))-1=1.

We have $H_{i}^{3}=(f_{i}^{*}L_{i})^{3}=0$ . To summarise, we obtain

\displaystyle c_{2}(X)\cdot H_{i}=6+6-0-(-K_{X})^{2}\cdot H_{i}=12-(-K_{X})^{2}\cdot H_{i}.

Thus (2) holds.

Let us show (3). Assume that $R_{i}$ is of type $E_{1}$ . By $K_{X}=f_{i}^{*}K_{X_{i}}+D_{i}$ , we obtain

-K_{X}+D_{i}\sim-f_{i}^{*}K_{X_{i}}\sim f_{i}^{*}(r_{i}L_{i})=r_{i}H_{i}.

By Lemma 3.8(3), we obtain

	$\displaystyle c_{2}(X)\cdot H_{i}$	$\displaystyle=\frac{1}{r_{i}}(c_{2}(X)\cdot(-K_{X})+c_{2}(X)\cdot D_{i})$
		$\displaystyle=\frac{1}{r_{i}}(24+c_{2}(X)\cdot D_{i}).$

Let $g(B_{i})$ be the genus of $B_{i}$ . We have $d_{i}:=\deg B_{i}=(-\frac{1}{r_{i}}K_{X_{i}})\cdot B_{i}$ . By [TanII, Lemma 3.21(2)], we get

•

$(-K_{X})^{2}\cdot D_{i}=r_{i}d_{i}-2g(B_{i})+2$ ,
•

$(-K_{X})\cdot D_{i}^{2}=2g(B_{i})-2$ , and
•

$D_{i}^{3}=-r_{i}d_{i}+2-2g(B_{i})$ .

It holds that

\chi(D_{i},\mathcal{O}_{D_{i}})=1-g(B_{i}),

and

\chi(D_{i},\mathcal{O}_{X}(D_{i})|_{D_{i}})=\chi(D_{i},\mathcal{O}_{D_{i}})+\frac{1}{2}(D_{i}|_{D_{i}})\cdot(D_{i}|_{D_{i}}-K_{D_{i}})

=1-g(B_{i})+\frac{1}{2}(-K_{X})\cdot D_{i}^{2}=0.

It follows from Lemma 3.8(4) that

	$\displaystyle c_{2}(X)\cdot D_{i}$	$\displaystyle=6\chi(D_{i},\mathcal{O}_{D_{i}})+6\chi(D_{i},\mathcal{O}_{X}(D_{i})\|_{D_{i}})-2D_{i}^{3}-(-K_{X})^{2}\cdot D_{i}$
		$\displaystyle=6(1-g(B_{i}))+0-2(-r_{i}d_{i}+2-2g(B_{i}))-(r_{i}d_{i}-2g(B_{i})+2)=r_{i}d_{i}.$

To summarise, we obtain

c_{2}(X)\cdot H_{i}=\frac{24}{r_{i}}+d_{i},

which completes the proof for the case when $R_{i}$ is of type $E_{1}$ .

Assume that $R_{i}$ is of type $E_{2}$ . By [TanII, Proposition 3.22], we get $\mu_{i}=2,D_{i}\simeq\mathbb{P}^{2},D_{i}^{3}=1,(-K_{X})^{2}\cdot D_{i}=4$ , and $\mathcal{O}_{X}(D_{i})|_{D_{i}}\simeq\mathcal{O}_{D_{i}}(-1)$ . Moreover, $\ell_{i}$ is a line on $D_{i}$ . By $K_{X}\sim f_{i}^{*}K_{Y_{i}}+2D_{i}$ , it holds that

-K_{X}+2D_{i}\sim f_{i}^{*}K_{Y_{i}}\sim f_{i}^{*}(r_{i}L_{i})=r_{i}H_{i}.

By Lemma 3.8(3), we obtain

	$\displaystyle c_{2}(X)\cdot H_{i}$	$\displaystyle=\frac{1}{r_{i}}(c_{2}(X)\cdot(-K_{X})+2c_{2}(X)\cdot D_{i})$
		$\displaystyle=\frac{1}{r}(24+2c_{2}(X)\cdot D_{i}).$

By $D_{i}\simeq\mathbb{P}^{2}$ and $\mathcal{O}_{X}(D_{i})|_{D_{i}}\simeq\mathcal{O}_{\mathbb{P}^{2}}(-1)$ , it holds that $\chi(D_{i},\mathcal{O}_{D_{i}})=1$ and $\chi(D_{i},\mathcal{O}_{D_{i}}(D_{i}))=0$ . It follows from Lemma 3.8(4) that

	$\displaystyle c_{2}(X)\cdot D_{i}$	$\displaystyle=6\chi(D_{i},\mathcal{O}_{D_{i}})+6\chi(D_{i},\mathcal{O}_{X}(D_{i})\|_{D_{i}})-2D_{i}^{3}-(-K_{X})^{2}\cdot D_{i}$
		$\displaystyle=6+0-2-4=0.$

Thus we obtain $c_{2}(X)\cdot H_{i}=\frac{24}{r_{i}}$ , which completes the proof for the case when $R_{i}$ is of type $E_{2}$ .

Assume that $R_{i}$ is of type $E_{3}$ or $E_{4}$ . It follows from [TanII, Proposition 3.22] that $\mu_{i}=1$ , $(-K_{X})^{2}\cdot D_{i}=2$ , $D_{i}^{3}=2$ , $D_{i}$ is isomorphic to a possibly singular quadric surface in $\mathbb{P}^{3}$ , $\mathcal{O}_{X}(D_{i})|_{D_{i}}\simeq\mathcal{O}_{\mathbb{P}^{3}}(-1)|_{D_{i}}$ , and $\ell_{i}$ is a line on $\mathbb{P}^{3}$ contained in $D_{i}$ (Theorem 4.8). In particular, $D_{i}\cdot\ell_{i}=-1$ . We have $-K_{X}+D_{i}\sim-f_{i}^{*}K_{Y_{i}}\sim f_{i}^{*}(r_{i}L_{i})=r_{i}H_{i}$ . We then get

c_{2}(X)\cdot H_{i}=\frac{1}{r_{i}}(24+c_{2}(X)\cdot D_{i}).

Let us compute $c_{2}(X)\cdot D_{i}$ . By Lemma 3.8(4), we have

c_{2}(X)\cdot D_{i}=6\chi(D_{i},\mathcal{O}_{D_{i}})+6\chi(D_{i},\mathcal{O}_{X}(D_{i})|_{D_{i}})-2D_{i}^{3}-(-K_{X})^{2}\cdot D_{i}.

It holds that $\chi(D_{i},\mathcal{O}_{D_{i}})=h^{0}(D_{i},\mathcal{O}_{D_{i}})=1$ and $\chi(D_{i},\mathcal{O}_{D_{i}}(D_{i}))=\chi(D_{i},\mathcal{O}_{\mathbb{P}^{3}}(-1)|_{D_{i}})=0$ . Therefore, we obtain

c_{2}(X)\cdot D_{i}=6+0-4-2=0,

which implies $c_{2}(X)\cdot H_{i}=\frac{24}{r_{i}}$ . This completes the proof for the case when $R_{i}$ is of type $E_{3}$ or $E_{4}$ .

Assume that $R_{i}$ is of type $E_{5}$ . It follows from [TanII, Proposition 3.22] that $\mu_{i}=1$ , $D_{i}\simeq\mathbb{P}^{2}$ , $D_{i}^{3}=4$ , $(-K_{X})^{2}\cdot D_{i}=1$ , $\mathcal{O}_{X}(D_{i})|_{X_{i}}\simeq\mathcal{O}_{\mathbb{P}^{2}}(-2)$ , and $\ell_{i}$ is a line on $D_{i}$ . Hence

\displaystyle(-2K_{X}+D_{i})\cdot\ell_{i}=-2K_{X}\cdot\ell_{i}+D_{i}\cdot\ell_{i}=2-2=0,

which implies $-2K_{X}+D_{i}\sim r_{i}H_{i}$ . As before, we can write

c_{2}(X)\cdot H_{i}=\frac{1}{r_{i}}(2\cdot 24+c_{2}(X)\cdot D_{i}).

By Lemma 3.8(4), we have

c_{2}(X)\cdot D_{i}=6\chi(D_{i},\mathcal{O}_{D_{i}})+6\chi(D_{i},\mathcal{O}_{X}(D_{i})|_{D_{i}})-2D_{i}^{3}-(-K_{X})^{2}\cdot D_{i}.

It holds that $\chi(D_{i},\mathcal{O}_{D_{i}})=1$ and $\chi(D_{i},\mathcal{O}_{X}(D_{i})|_{D_{i}})=0$ . We obtain

\displaystyle c_{2}(X)\cdot D_{i}=6+0-8-1=-3,

and hence $c_{2}(X)\cdot H_{i}=\frac{45}{r_{i}}$ . Thus (3) holds. ∎

Lemma 5.5.

We use Notation 5.1. Fix $i\in\{1,2\}$ .

(1)

If $R_{i}$ is of type $C_{1}$ , then $7\leq c_{2}(X)\cdot H_{i}\leq 17$ .
(2)

If $R_{i}$ is not of type $E_{1}$ , then $1\leq c_{2}(X)\cdot H_{i}\leq 24$ or $c_{2}(X)\cdot H_{i}=45$ .
(3)

If $R_{i}$ is neither of type $E_{1}$ nor $E_{5}$ , then $1\leq c_{2}(X)\cdot H_{i}\leq 24$ .

Proof.

Let us show (1). Assume that $R_{i}$ is of type $C_{1}$ . By the same argument as in the proof of Lemma 5.4(1),

c_{2}(X)\cdot H_{i}=18-(-K_{X})^{2}\cdot H_{i}.

Since $-K_{X}$ is ample, we get $c_{2}(X)\cdot H_{i}\leq 17$ . On the other hand, it follows from Lemma 5.4(1) that

c_{2}(X)\cdot H_{i}=6+\deg\Delta_{f_{i}}

and $\deg\Delta_{f_{i}}>0$ . In particular, $(c_{2}(X)\cdot H_{i})\geq 7$ . Thus (1) holds.

Let us show (2). If $R_{i}$ not of type $D_{1}$ , then these assertions follow from (1) and Lemma 5.4. Assume that $R_{i}$ is of type $D_{1}$ . Then Lemma 5.4 implies

c_{2}(X)\cdot H_{i}=12-(-K_{X})^{2}\cdot H_{i}.

Since $1\leq(-K_{X})^{2}\cdot H_{i}\leq 7$ , we have $5\leq c_{2}(X)\cdot H_{i}\leq 11$ . Thus (2) holds. Similarly, (3) holds. ∎

Lemma 5.6.

We use Notation 5.1. Set

a:=|{\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})|,

which is the cardinality of the set ${\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})$ . Then the following hold.

(1)

$H_{1}$ and $H_{2}$ are linearly equivalent over $\mathbb{R}$ in $({\operatorname{Pic}}\,X)\otimes_{\mathbb{Z}}\mathbb{R}/\equiv$ . Moreover, $a\in\mathbb{Z}_{>0}$ .
(2)

${\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})\simeq\mathbb{Z}/(H_{2}\cdot\ell_{1})\mathbb{Z}\simeq\mathbb{Z}/(H_{1}\cdot\ell_{2})\mathbb{Z}$ .
(3)

$a=H_{2}\cdot\ell_{1}=H_{1}\cdot\ell_{2}$ .
(4)

$-aK_{X}\sim\mu_{2}H_{1}+\mu_{1}H_{2}.$
(5)

$24a=\mu_{2}H_{1}\cdot c_{2}(X)+\mu_{1}H_{2}\cdot c_{2}(X)$ .

Proof.

Let us show (1). Assume

\lambda_{1}H_{1}+\lambda_{2}H_{2}\equiv 0

for $\lambda_{1},\lambda_{2}\in\mathbb{R}$ . By $H_{1}=f^{*}_{1}L_{1}$ , we have $H_{1}\cdot\ell_{1}=0$ . Hence we obtain $\lambda_{2}H_{2}\cdot\ell_{1}=0$ . Since $f_{2}(\ell_{1})$ is a curve, we have $H_{2}\cdot\ell_{1}>0$ . We then get $\lambda_{2}=0$ , which implies $\lambda_{1}=0$ . Therefore, $H_{1}$ and $H_{2}$ are linearly independent over $\mathbb{R}$ . Since $\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2}$ is a free subgroup of rank $2$ of ${\operatorname{Pic}}\,X\simeq\mathbb{Z}^{\oplus 2}$ , the quotient group

{\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})

is a finite abelian group. Thus (1) holds.

Let us show (2) and (3). By $H_{2}\cdot\ell_{1}>0$ and $H_{1}\cdot\ell_{2}>0$ , (2) implies (3). We now prove (2). The composite group homomorphism

\varphi:{\operatorname{Pic}}\,X\stackrel{{\scriptstyle(-\cdot\ell_{1})}}{{\longrightarrow}}\mathbb{Z}\longrightarrow\mathbb{Z}/(H_{2}\cdot\ell_{1})\mathbb{Z}

is surjective, because so is each map (Theorem 4.11). It suffices to show that ${\operatorname{Ker}}(\varphi)=\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2}$ . The inclusion ${\operatorname{Ker}}(\varphi)\supset\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2}$ is clear. It is enough to show the opposite inclusion. Take $H\in{\operatorname{Ker}}(\varphi)$ . Then $H\cdot\ell_{1}\in(H_{2}\cdot\ell_{1})\mathbb{Z}$ , and hence there exists $b\in\mathbb{Z}$ such that

H\cdot\ell_{1}=b(H_{2}\cdot\ell_{1}).

Thus we obtain $H-bH_{2}\in\mathrm{Ker}(-\cdot\ell_{1})=\mathrm{Im}(f^{*}_{1})=\mathbb{Z}H_{1}$ by Theorem 4.11. Then we can write $H\sim cH_{1}+bH_{2}$ for some $c\in\mathbb{Z}$ . Thus (2) and (3) holds.

Let us show (4). Since $H_{1}$ and $H_{2}$ are linearly independent over $\mathbb{R}$ by (1), $\{H_{1},H_{2}\}$ is a basis of $\mathrm{N}^{1}(X)_{\mathbb{R}}$ . Then we can write

-K_{X}\equiv\beta_{1}H_{1}+\beta_{2}H_{2}

for some $\beta_{1},\beta_{2}\in\mathbb{R}$ . By $\mu_{1}=-K_{X}\cdot\ell_{1}$ and $H_{2}\cdot\ell_{1}=a$ , we get

\displaystyle\mu_{1}=-K_{X}\cdot\ell_{1}=\beta_{1}(H_{1}\cdot\ell_{1})+\beta_{2}(H_{2}\cdot\ell_{1})=\beta_{2}a.

By symmetry, we obtain $\mu_{2}=\beta_{1}a$ . Hence (4) holds. Finally, (5) follows from (4) and $(-K_{X})\cdot c_{2}(X)=24$ (Lemma 3.8(3)). ∎

Lemma 5.7.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ . Then

\deg B_{1}\leq\left(r_{1}-\frac{\mu_{2}}{a}\right)^{2}\cdot L_{1}^{3}

for $a:=|{\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})|\in\mathbb{Z}_{>0}$ (cf. Lemma 5.6(1)).

Proof.

Since $H_{1}$ and $H_{2}$ are nef, we get $H_{1}\cdot H_{2}^{2}\geq 0$ . By $-aK_{X}\sim\mu_{2}H_{1}+H_{2}$ (Lemma 5.6(4)) and $K_{X}-D_{1}=f_{1}^{*}K_{Y_{1}}\sim-r_{1}H_{1}$ , the following holds:

$\displaystyle 0\leq H_{1}\cdot H_{2}^{2}$	$\displaystyle=$	$\displaystyle H_{1}\cdot(-aK_{X}-\mu_{2}H_{1})^{2}$
	$\displaystyle=$	$\displaystyle H_{1}\cdot((ar_{1}-\mu_{2})H_{1}-aD_{1})^{2}$
	$\displaystyle=$	$\displaystyle(ar_{1}-\mu_{2})^{2}H^{3}_{1}-2a(ar_{1}-\mu_{2})H^{2}_{1}\cdot D_{1}+a^{2}H_{1}\cdot D_{1}^{2}.$

Then the required inequality $\deg B_{1}\leq\left(r_{1}-\frac{\mu_{2}}{a}\right)^{2}\cdot L_{1}^{3}$ follows from $H_{1}^{3}=L_{1}^{3},H_{1}^{2}\cdot D_{1}=0,$ and $H_{1}\cdot D_{1}^{2}=-\deg B_{1}$ . ∎

Proposition 5.8.

Let $X$ be a Fano threefold with $\rho(X)=2$ . Let $R$ be an extremral ray of type $E_{1}$ and let $f:X\to Y$ be the contraction of $R$ . Then $Y$ is a Fano threefold of index $r_{Y}\geq 2$ .

Proof.

Set $R_{1}:=R$ and $a:=|{\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})|\in\mathbb{Z}_{>0}$ (Lemma 5.6(1)). We use Notation 5.1. By $-K_{Y}=f_{*}(-K_{X})$ , $-K_{Y}$ is big. This, together with $\rho(Y)=1$ , implies that $-K_{Y}$ is ample. Suppose $r_{Y}=1$ . It is enough to derive a contradiction. Recall that $|-K_{Y}|$ is not very ample [TanII, Theorem 1.1]. By [TanI, Theorem 1.1], we obtain $(-K_{Y})^{3}\in\{2,4\}$ . It follows from $(-K_{X})^{3}\in 2\mathbb{Z}$ and $0<(-K_{X})^{3}<(-K_{Y})^{3}$ (Lemma 3.13) that $(-K_{Y})^{3}=4$ and $(-K_{X})^{3}=2$ . Let $\varphi_{|-K_{X}|}:X\dashrightarrow\mathbb{P}^{h^{0}(X,-K_{X})-1}$ be the rational map induced by the complete linear system $|-K_{X}|$ . If $\dim({\operatorname{Im}}\,\varphi_{|-K_{X}|})\geq 2$ , then our condition $\rho(X)=2$ contradicts [TanI, Proposition 6.8]. Hence we may assume that $\dim({\operatorname{Im}}\,\varphi_{|-K_{X}|})=1$ . In this case, the mobile part $M$ of $|-K_{X}|$ is base point free and the induced morphism $\varphi_{|M|}:X\to\mathbb{P}^{1}$ is the contraction of the extremal ray $R_{2}$ of type D with $d_{2}:=(-K_{X})^{2}\cdot H_{2}=1$ [TanI, Proposition 4.1]. In particular, $\mu_{2}=1$ . Then Lemma 5.4 and Lemma 5.6 imply

24a=\frac{24}{r_{1}}+\deg B_{1}+(12-d_{2})=35+\deg B_{1}.

By $\deg B_{1}\leq(r_{1}-\frac{\mu_{2}}{a})^{2}\cdot L_{1}^{3}=(1-\frac{1}{a})^{2}\cdot 4<4$ (Lemma 5.7), we obtain $24a=35+\deg B_{1}<35+4$ , which implies $a=1$ . However, this leads to the following contradiction: $\deg B_{1}\leq(1-\frac{1}{a})^{2}\cdot 4=0$ . ∎

Proposition 5.9.

We use Notation 5.1. Then the following hold.

(1)

$\{H_{1},H_{2}\}$ is a $\mathbb{Z}$ -linear basis of ${\operatorname{Pic}}\,X\simeq\mathbb{Z}^{\oplus 2}$ .
(2)

$\{\ell_{1},\ell_{2}\}$ is a dual basis of $\mathrm{N}_{1}(X)_{\mathbb{Z}}$ , i.e., $H_{1}\cdot\ell_{1}=H_{2}\cdot\ell_{2}=0$ and $H_{1}\cdot\ell_{2}=H_{2}\cdot\ell_{1}=1$ . Here we set $\mathrm{N}_{1}(X)_{\mathbb{Z}}:={\rm Z}_{1}(X)/\equiv$ , where ${\rm Z}_{1}(X):=\bigoplus_{C}\mathbb{Z}C$ is the free $\mathbb{Z}$ -module that is freely generated by all the curves $C$ on $X$ and $\equiv$ denotes the numerical equivalence.
(3)

$-K_{X}\sim\mu_{2}H_{1}+\mu_{1}H_{2}.$
(4)

$24=\mu_{2}H_{1}\cdot c_{2}(X)+\mu_{1}H_{2}\cdot c_{2}(X)$ .

Proof of Proposition 5.9 for the case when $X$ is primitive.

Set $a:=|{\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})|\in\mathbb{Z}_{>0}$ (Lemma 5.6(1)). Since $X$ is primitive, there is an extremal ray of type $C$ (Proposition 4.14). In what follows, we assume that $R_{1}$ is of type $C$ . The proof consists of 7 steps.

Step 1.

In order to complete the proof of Proposition 5.9, it is enough to show that $a=1$ .

Proof of Step 1.

Assume $a=1$ . Then the assertion (1) follows from Lemma 5.6(1)(2). The assertions (3) and (4) hold by Lemma 5.6(4)(5). The assertion (2) follows from Lemma 5.6(3) and $H_{1}\cdot\ell_{1}=H_{2}\cdot\ell_{2}=0$ . This completes the proof of Step 1. ∎

Step 2.

If $(\mu_{1},\mu_{2})=(1,1)$ , then $a=1$ .

Proof of Step 2.

By Lemma 5.6(5), we obtain

24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X).

Since $R_{1}$ is of type $C_{1}$ , the following holds (Lemma 5.5(1)):

7\leq c_{2}(X)\cdot H_{1}\leq 17.

If $c_{2}(X)\cdot H_{2}=45$ , then $52\leq 24a\leq 62$ , which is a contradiction. Hence $1\leq c_{2}(X)\cdot H_{2}\leq 24$ (Lemma 5.5(2)), which implies

24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 17+24=41.

Thus $a=1$ . This completes the proof of Step 2. ∎

Step 3.

If $(\mu_{1},\mu_{2})=(1,2)$ , then $a=1$ .

Proof of Step 3.

By Lemma 5.6(5), we obtain

24a=2H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X).

In this case, $a$ is odd. Indeed, if $a$ is even, then the linear equivalence

a(-K_{X})\sim 2H_{1}+H_{2}

would imply that $H_{2}$ is not a primitive element, which is a contradiction (Lemma 5.2). Since $R_{1}$ is of type $C_{1}$ and $R_{2}$ is not of type $E_{5}$ by $\mu_{2}=2$ , we get $c_{2}(X)\cdot H_{1}\leq 17$ and $c_{2}(X)\cdot H_{2}\leq 24$ (Lemma 5.5), which imply

24a=2H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 2\cdot 17+24=58.

As $a$ is odd, we get $a=1$ . This completes the proof of Step 3. ∎

Step 4.

If $(\mu_{1},\mu_{2})=(1,3)$ , then $a=1$ .

Proof of Step 4.

In this case, $R_{1}$ is of type $C_{1}$ and $R_{2}$ is of type $D_{3}$ . By Lemma 5.6(5), we obtain

24a=3H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X).

We now show that $a$ is odd. By $-aK_{X}\sim 3H_{1}+H_{2}$ (Lemma 5.6(4)), we get

	$\displaystyle a^{3}(-K_{X})^{3}$	$\displaystyle=27H_{1}^{3}+27H_{1}^{2}\cdot H_{2}+9H_{1}\cdot H_{2}^{2}+H_{2}^{3}$
		$\displaystyle\overset{{\rm(a)}}{=}27\cdot 2(H_{2}\cdot\ell_{1})$
		$\displaystyle\overset{{\rm(b)}}{=}27\cdot 2a,$

where (b) holds by Lemma 5.6(3) and (a) follows from $H^{3}_{1}=0$ , $H_{2}^{2}\equiv 0$ , and $H_{1}^{2}\equiv 2\ell_{1}$ (Lemma 5.3). Hence we obtain $a^{2}(-K_{X})^{3}=2\cdot 27$ . By $(-K_{X})^{3}\in 2\mathbb{Z}$ , we get $a\not\in 2\mathbb{Z}$ .

We have $c_{2}(X)\cdot H_{1}\leq 17$ (Lemma 5.5) and $c_{2}(X)\cdot H_{2}=3$ (Lemma 5.4), which imply

24a=3H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 3\cdot 17+3=54.

As $a$ is odd, we get $a=1$ . This completes the proof of Step 4. ∎

In what follows, we treat the case when $\mu_{1}=2$ . In this case, $R_{1}$ is of type $C_{2}$ and $H_{1}\cdot c_{2}(X)=6$ (Lemma 5.4).

Step 5.

If $(\mu_{1},\mu_{2})=(2,1)$ , then $a=1$ .

Proof of Step 6.

By Lemma 5.6(5), we obtain

24a=H_{1}\cdot c_{2}(X)+2H_{2}\cdot c_{2}(X)=6+2H_{2}\cdot c_{2}(X).

By the same argument as in the case $(\mu_{1},\mu_{2})=(1,2)$ (Step 3), we see that $a$ is odd. If $c_{2}(X)\cdot H_{2}=45$ , then

24a=6+2\cdot 45=96,

which contradicts the fact that $a$ is odd. Hence $c_{2}(X)\cdot H_{2}\leq 24$ (Lemma 5.5), which implies

24a=6+2H_{2}\cdot c_{2}(X)\leq 6+2\cdot 24=54.

As $a$ is odd, we get $a=1$ . This completes the proof of Step 5. ∎

Step 6.

If $(\mu_{1},\mu_{2})=(2,2)$ , then $a=1$ .

Proof of Step 6.

By Lemma 5.6(5), we obtain

24a=2H_{1}\cdot c_{2}(X)+2H_{2}\cdot c_{2}(X)=12+2H_{2}\cdot c_{2}(X).

Note that the type of $R_{2}$ is $E_{2},C_{2}$ , or $D_{2}$ (Remark 3.4). If $R_{2}$ is of type $E_{2}$ , then $c_{2}(X)\cdot H_{2}=24/r$ for some $r\in\mathbb{Z}_{>0}$ (Lemma 5.4). Hence

12a=6+H_{2}\cdot c_{2}(X)=6+\frac{24}{r}.

By $24/r\in 6\mathbb{Z}$ , we get $r\in\{1,2,4\}$ . Then it is easy to see that $(a,r)=(1,4)$ is a unique solution of $12a=6+\frac{24}{r}$ .

We may assume that $R_{2}$ is of type $C_{2}$ or $D_{2}$ . In this case, we have $c_{2}(X)\cdot H_{2}\leq 6$ (Lemma 5.4), which implies

12a=6+H_{2}\cdot c_{2}(X)\leq 12.

Thus $a=1$ . This completes the proof of Step 6. ∎

Step 7.

If $(\mu_{1},\mu_{2})=(2,3)$ , then $a=1$ .

Proof of Step 7.

In this case, $R_{2}$ is of type $D_{3}$ . By Lemma 5.6(5) and Lemma 5.4, we obtain

24a=3H_{1}\cdot c_{2}(X)+2H_{2}\cdot c_{2}(X)=18+6=24.

Hence $a=1$ . This completes the proof of Step 7. ∎

Step 1–Step 7 complete the proof of Proposition 5.9 for the case when $X$ is primitive. ∎

Proof of Proposition 5.9 for the case when $X$ is imprimitive.

Since $X$ is imprimitive, $X$ has an extremal ray of type $E_{1}$ . In what follows, we assume that $R_{1}$ is of type $E_{1}$ . Set $a:=|{\operatorname{Pic}}(X)/(\mathbb{Z}H_{1}\oplus\mathbb{Z}H_{2})|\in\mathbb{Z}_{>0}$ (Lemma 5.6(1)). It is enough to show $a=1$ (cf. Step 1 of the primitive case).

Claim 5.10.

(1)

$H_{1}\cdot c_{2}(X)\leq 31$ .
(2)

$H_{1}\cdot c_{2}(X)\leq\begin{cases}25\qquad(\text{if }a\leq 3)\\ 23\qquad(\text{if }\text{if }a\leq 2)\\ 17\qquad(\text{if }a=1).\end{cases}$
(3)

$H_{1}\cdot c_{2}(X)\leq 20$ if $a\leq 3$ and $\mu_{2}=2$ .

Proof of Claim 5.10.

By Lemma 5.4 and Lemma 5.7, we have

(5.10.1)

H_{1}\cdot c_{2}(X)=\frac{24}{r_{1}}+\deg B_{1}\leq\frac{24}{r_{1}}+\left(r_{1}-\frac{\mu_{2}}{a}\right)^{2}\cdot L_{1}^{3}.

In what follows, we shall separate the argument depending on the value of $r_{1}$ . Note that $r_{1}\in\{2,3,4\}$ (Proposition 5.8).

Assume $r_{1}=4$ . In this case, we have $L_{1}^{3}=1$ . This, together with (5.10.1), implies

H_{1}\cdot c_{2}(X)\leq 6+\left(4-\frac{\mu_{2}}{a}\right)^{2}<6+4^{2}=22.

If $a=1$ , then $H_{1}\cdot c_{2}(X)\leq 6+(4-1)^{2}=15$ . If $a\leq 3$ and $\mu_{2}=2$ , then $H_{1}\cdot c_{2}(X)\leq 6+(4-\frac{2}{3})^{2}=6+\frac{100}{9}<6+12=18$ .

Assume $r_{1}=3$ . In this case, we have $L_{1}^{3}=2$ . This, together with (5.10.1), implies

H_{1}\cdot c_{2}(X)\leq 8+2\left(3-\frac{\mu_{2}}{a}\right)^{2}<8+2\cdot 3^{2}=28.

If $a\leq 3$ , then $H_{1}\cdot c_{2}(X)\leq 8+2\left(3-\frac{1}{3}\right)^{2}=8+\frac{128}{9}<8+\frac{135}{9}=8+15=23$ . If $a\leq 2$ , then $H_{1}\cdot c_{2}(X)\leq 8+2\left(3-\frac{1}{2}\right)^{2}=8+\frac{25}{2}<21$ . If $a=1$ , then $H_{1}\cdot c_{2}(X)\leq 8+2\left(3-1\right)^{2}=16$ . If $a\leq 3$ and $\mu_{2}=2$ , then $H_{1}\cdot c_{2}(X)\leq 8+2\left(3-\frac{2}{3}\right)^{2}=8+\frac{98}{9}<8+\frac{99}{9}=19$ .

Assume $r_{1}=2$ . In this case, we have $1\leq L_{1}^{3}\leq 5$ . This, together with (5.10.1), implies

H_{1}\cdot c_{2}(X)\leq 12+5\left(2-\frac{\mu_{2}}{a}\right)^{2}<12+5\cdot 4=32.

If $a\leq 3$ , then $H_{1}\cdot c_{2}(X)\leq 12+5\left(2-\frac{1}{3}\right)^{2}=12+\frac{125}{9}<12+\frac{126}{9}=26$ . If $a\leq 2$ , then $H_{1}\cdot c_{2}(X)\leq 12+5\left(2-\frac{1}{2}\right)^{2}=12+\frac{45}{4}<24$ . If $a=1$ , then $H_{1}\cdot c_{2}(X)\leq 12+5\left(2-1\right)^{2}=17$ . If $a\leq 3$ and $\mu_{2}=2$ , then $H_{1}\cdot c_{2}(X)\leq 12+5\left(2-\frac{2}{3}\right)^{2}=12+\frac{80}{9}<21$ . This completes the proof of Claim 5.10. ∎

We first treat the case when $R_{2}$ is of type $E_{1}$ . In this case, the same conclusion of Claim 5.10 holds for $H_{2}\cdot c_{2}(X)$ . Claim 5.10(1) implies

24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(x)\leq 31+31=62.

Hence we may assume that $a=2$ . By Claim 5.10(2), we get

48=24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(x)\leq 23+23=46.

This is absurd. Thus $R_{2}$ is not of type $E_{2}$ .

We now consider the case when $H_{2}\cdot c_{2}(X)=45$ . Then $R_{2}$ is of type $E_{5}$ (Lemma 5.4, Lemma 5.5), and hence $\mu_{2}=1$ . Then Claim 5.10(1) implies

24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)=H_{1}\cdot c_{2}(X)+45\leq 31+45=76.

Hence $a\leq 3$ . By Claim 5.10(2), we obtain

24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)=H_{1}\cdot c_{2}(X)+45\leq 25+45=70.

Hence we may assume that $a=2$ . We then get $48=24a=H_{1}\cdot c_{2}(X)+45$ , which implies

\frac{24}{r_{1}}<\frac{24}{r_{1}}+\deg B=H_{1}\cdot c_{2}(X)=3.

This contradicts $r_{1}\in\{2,3,4\}$ . Thus $H_{2}\cdot c_{2}(X)\neq 45$ , and hence $1\leq H_{2}\cdot c_{2}(X)\leq 24$ (Lemma 5.5).

Assume that $\mu_{2}=1$ . Then Claim 5.10(1) implies

24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 31+24=55.

Hence we may assume that $a=2$ . By Claim 5.10(2), we get

48=24a=H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 23+24=47,

which is absurd.

Assume that $\mu_{2}=2$ . Then Claim 5.10(1) implies

24a=2H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 2\cdot 31+24=86.

Hence $a\leq 3$ . By Claim 5.10(3), we get

24a=2H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)\leq 2\cdot 20+24=64.

Thus we may assume that $a=2$ . We then get

-2K_{X}\sim 2H_{1}+H_{2},

which contradicts the fact that $H_{2}$ is primitive (Lemma 5.2).

Assume that $\mu_{2}=3$ . Then $R_{2}$ is of type $D_{3}$ and $H_{2}\cdot c_{2}(X)=3$ (Lemma 5.4). Therefore, $24a=3H_{1}\cdot c_{2}(X)+H_{2}\cdot c_{2}(X)$ (Lemma 5.6) implies the following (Claim 5.10(1)):

8a=H_{1}\cdot c_{2}(X)+1\leq 32.

By $-aK_{X}\sim 3H_{1}+H_{2}$ , we obtain $a\not\in 3\mathbb{Z}$ . Hence we may assume that $a\in\{2,4\}$ .

Suppose $a=4$ . Then Lemma 5.4 and Lemma 5.7 imply

31=H_{1}\cdot c_{2}(X)=\frac{24}{r_{1}}+\deg B_{1}\leq\frac{24}{r_{1}}+\left(r_{1}-\frac{3}{4}\right)^{2}L_{1}^{3}.

If $r_{1}=4$ , then $L_{1}^{3}=1$ , and hence $31\leq 6+(4-\frac{3}{a})^{2}<6+16=22$ , which is absurd. If $r_{1}=3$ , then $L_{1}^{3}=2$ , and hence $31\leq 8+(3-\frac{3}{a})^{2}\cdot 2<8+9\cdot 2=26$ , which is impossible. If $r_{1}=2$ , then $1\leq L_{1}^{3}\leq 5$ , and hence $31\leq 12+(2-\frac{3}{4})^{2}\cdot L_{1}^{3}\leq 12+\frac{25}{16}\cdot 5<12+8$ , which is a contradiction. We then obtain $a\neq 4$ .

Suppose $a=2$ . It suffices to derive a contradiction. We get $16=H_{1}\cdot c_{2}(X)+1$ . Hence Lemma 5.4 and Lemma 5.7 imply

15=H_{1}\cdot c_{2}(X)=\frac{24}{r_{1}}+\deg B_{1}\leq\frac{24}{r_{1}}+\left(r_{1}-\frac{3}{2}\right)^{2}\cdot L_{1}^{3}.

If $r_{1}=4$ , then $L_{1}^{3}=1$ and $15\leq 6+\frac{25}{4}\cdot 1<6+7$ , which is absurd. If $r_{1}=3$ , then $L_{1}^{3}=2$ and $15\leq 8+\frac{9}{4}\cdot 2<8+5$ , which is a contradiction. If $r_{1}=2$ , then $1\leq L_{1}^{3}\leq 5$ and $15\leq 12+\frac{1}{4}\cdot L_{1}^{3}\leq 12+\frac{5}{4}<12+2$ , which is absurd. This completes the proof of Proposition 5.9. ∎

Lemma 5.11.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ . Then the following hold.

(1)

$H_{1}^{2}\cdot H_{2}=(r_{1}-\mu_{2})L_{1}^{3}$ .
(2)

$H_{1}\cdot H_{2}^{2}=(r_{1}-\mu_{2})^{2}L_{1}^{3}-\deg B_{1}$ .

Proof.

Let us show (1). Set $D_{1}:={\operatorname{Ex}}(f_{1})$ . We obtain

H_{1}^{2}\cdot H_{2}\overset{{\rm(i)}}{=}H_{1}^{2}\cdot(-K_{X}-\mu_{2}H_{1})\overset{{\rm(ii)}}{=}H_{1}^{2}\cdot((r_{1}-\mu_{2})H_{1}-D_{1})=(r_{1}-\mu_{2})L_{1}^{3},

where (i) follows from $-K_{X}\sim\mu_{2}H_{1}+H_{2}$ (Proposition 5.9(3)) and (ii) holds by $K_{X}=f_{1}^{*}K_{X_{1}}+D_{1}=-r_{1}H_{1}+D_{1}$ . Thus (1) holds. The assertion (2) follows from the displayed equation in the proof of Lemma 5.7 by using $H_{1}^{2}\cdot D_{1}=0$ and $a=1$ (Proposition 5.9). ∎

5.2. Case $C-C$

Notation 5.12.

We use Notation 5.1. Assume that $R_{1}$ is of type $C$ and $R_{2}$ is of type $C$ . In particular, for each $i\in\{1,2\}$ , $f_{i}:X\to Y_{i}=\mathbb{P}^{2}$ is of type $C$ . Set

f:=f_{1}\times f_{2}\colon X\to\mathbb{P}^{2}\times\mathbb{P}^{2}.

Let $\widetilde{f}:X\to f(X)$ be the induced morphism, where $f(X)$ denotes the scheme-theoretic image. For the $i$ -th projection $\pi_{i}:\mathbb{P}^{2}\times\mathbb{P}^{2}\to\mathbb{P}^{2}$ , set $M_{i}:=\pi^{*}L_{i}$ . Note that $\{M_{1},M_{2}\}$ is a free $\mathbb{Z}$ -linear basis of ${\operatorname{Pic}}(\mathbb{P}^{2}\times\mathbb{P}^{2})$ .

Lemma 5.13.

We use Notation 5.12. Then

\deg(\widetilde{f})\cdot f(X)=f_{*}(X)\sim\frac{2}{\mu_{2}}M_{1}+\frac{2}{\mu_{1}}M_{2}.

In particular, $\deg(\widetilde{f})=1$ or $\deg(\widetilde{f})=2$ .

Proof.

Recall that $f:X\to Y_{1}\times Y_{2}=\mathbb{P}^{2}\times\mathbb{P}^{2}$ is a finite morphism (Lemma 4.15). We can write

f_{*}(X)\sim a_{1}M_{1}+a_{2}M_{2}

for some $a_{1},a_{2}\in\mathbb{Z}$ . Pick a curve $\Gamma_{1}=(\mathrm{point}\times\mathrm{line})$ on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ . Since $M_{1}\sim(\mathrm{line}\times\mathbb{P}^{2})$ and $M_{2}\sim(\mathbb{P}^{2}\times\mathrm{line})$ , we have

M_{1}\cdot\Gamma_{1}=0\qquad\text{and}\qquad M_{2}\cdot\Gamma_{1}=1.

Hence we obtain $f_{*}(X)\cdot\Gamma_{1}=a_{2}$ . Now,

	$\displaystyle M_{1}^{2}\cdot M_{2}$	$\displaystyle\equiv(\mathrm{point}\times\mathbb{P}^{2})\cdot(\mathbb{P}^{2}\times\mathrm{line})$
		$\displaystyle\equiv(\mathrm{point}\times\mathrm{line})$
		$\displaystyle\equiv\Gamma_{1}.$

We have $H_{2}\cdot\ell_{1}=1$ (Proposition 5.9(2)) and $H_{1}^{2}\equiv\frac{2}{\mu_{1}}\ell_{1}$ (Lemma 5.3). We obtain

	$\displaystyle f_{*}(X)\cdot\Gamma_{1}$	$\displaystyle=f_{*}(X)\cdot M_{1}^{2}\cdot M_{2}$
		$\displaystyle=(f^{}M_{1})^{2}\cdot f^{}M_{2}$
		$\displaystyle=H_{1}^{2}\cdot H_{2}$
		$\displaystyle=\frac{2}{\mu_{1}}H_{2}\cdot\ell_{1}$
		$\displaystyle=\frac{2}{\mu_{1}}.$

Hence $a_{2}=\frac{2}{\mu_{1}}$ . Similarly, we have $a_{1}=\frac{2}{\mu_{2}}$ . ∎

Lemma 5.14.

We use Notation 5.12. Then $(-K_{X})^{3}=6(\mu_{1}^{2}+\mu_{2}^{2})$ .

Proof.

By Proposition 5.9, we have $-K_{X}\sim\mu_{2}H_{1}+\mu_{1}H_{2}$ . Since $L_{i}$ is a line on $Y_{i}=\mathbb{P}^{2}$ ,

H_{i}^{3}=f_{i}^{*}L_{i}^{3}=0.

By $H_{i}^{2}\equiv\frac{2}{\mu_{i}}\ell_{i}$ (Lemma 5.3) and $H_{1}\cdot\ell_{2}=H_{2}\cdot\ell_{1}=1$ (Proposition 5.9), we obtain

	$\displaystyle(-K_{X})^{3}$	$\displaystyle=(\mu_{2}H_{1}+\mu_{1}H_{2})^{3}$
		$\displaystyle=\mu_{2}^{3}H_{1}^{3}+3\mu_{2}^{2}\mu_{1}H_{1}^{2}\cdot H_{2}+3\mu_{2}\mu_{1}^{2}H_{1}\cdot H_{2}^{2}+\mu_{1}^{3}H_{2}^{3}$
		$\displaystyle=3\mu_{2}^{2}\mu_{1}\cdot\frac{2}{\mu_{1}}(H_{2}\cdot\ell_{1})+3\mu_{2}\mu_{1}^{2}\cdot\frac{2}{\mu_{2}}(H_{1}\cdot\ell_{2})$
		$\displaystyle=6(\mu_{1}^{2}+\mu_{2}^{2}).$

∎

5.2.1. Case $\deg(\widetilde{f})=1$

Lemma 5.15.

We use Notation 5.12. Assume that $\deg\widetilde{f}=1$ . Then the following hold.

(1)

$\widetilde{f}:X\to f(X)$ is an isomorphism.
(2)

If $(\mu_{1},\mu_{2})=(1,1)$ , then $X\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ is of bidegree $(2,2)$ .
(3)

If $(\mu_{1},\mu_{2})=(1,2)$ , then $X\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ is of bidegree $(1,2)$ .
(4)

If $(\mu_{1},\mu_{2})=(2,2)$ , then $X\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ is of bidegree $(1,1)$ .

Proof.

If (1) holds, then (2), (3), and (4) hold. Thus it suffices to show (1).

Set $Y:=f(X)$ . Since $\widetilde{f}:X\to Y$ is a finite birational morphism from a smooth variety $X$ to a variety $Y$ , $\widetilde{f}$ is nothing but the normalisation of $Y$ . As $Y$ is a prime divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ , $Y$ is Gorenstein. By [Rei94, Proposition 2.3], we have

\omega_{X}\otimes\mathcal{O}_{X}(C)\simeq\widetilde{f}^{*}\omega_{Y}

for the conductor $C$ , which is an effective Cartier divisor on $X$ whose support coincides with the non-isomorphic locus ${\operatorname{Ex}}(\widetilde{f})$ of $\widetilde{f}$ . By the adjunction formula, we have

	$\displaystyle\omega_{Y}$	$\displaystyle\simeq(\omega_{\mathbb{P}^{2}\times\mathbb{P}^{2}}\otimes\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(Y))\otimes\mathcal{O}_{Y}$
		$\displaystyle\simeq\left(\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(-3,-3)\otimes\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}\left(\frac{2}{\mu_{2}},\frac{2}{\mu_{1}}\right)\right)\otimes\mathcal{O}_{Y}$
		$\displaystyle\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}\left(\frac{2}{\mu_{2}}-3,\frac{2}{\mu_{1}}-3\right)\otimes\mathcal{O}_{Y},$

where the second isomorphims holds by Lemma 5.13. Therefore, we obtain

\displaystyle K_{X}+C\sim\left(\frac{2}{\mu_{2}}-3\right)H_{1}+\left(\frac{2}{\mu_{1}}-3\right)H_{2}.

It follows from $-K_{X}\sim\mu_{2}H_{1}+\mu_{1}H_{2}$ (Proposition 5.9) that

C\sim\left(\mu_{2}+\frac{2}{\mu_{2}}-3\right)H_{1}+\left(\mu_{1}+\frac{2}{\mu_{1}}-3\right)H_{2}.

As we have $\mu_{i}\in\{1,2\}$ for each $i$ , we obtain $C\sim 0$ , which implies $C=0$ . Hence $\tilde{f}$ is an isomorphism. ∎

5.2.2. Case $\deg(\widetilde{f})=2$

Lemma 5.16.

Let $W$ be a nonzero effective Cartier divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,1)$ . Then the following hold.

(1)
$W$ is isomorphic to one of the following:
- •
  
  $W_{1}:=\{x_{0}y_{0}=0\}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ .
- •
  
  $W_{2}:=\{x_{0}y_{0}+x_{1}y_{1}=0\}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ .
- •
  
  $W_{3}:=\{x_{0}y_{0}+x_{1}y_{1}+x_{2}y_{2}=0\}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ .
Note that $W_{1}$ is not irreducible, $W_{2}$ is not smooth but irreducible, and $W_{3}$ is smooth.
(2)

If $W\simeq W_{2}$ , then there exists a closed point $Q\in\mathbb{P}^{2}$ such that $\dim q_{1}^{-1}(Q)=2$ , where $q_{1}:W\hookrightarrow\mathbb{P}^{2}\times\mathbb{P}^{2}\xrightarrow{{\rm pr}_{1}}\mathbb{P}^{2}$ .

Proof.

Let us show (1). Since $W$ is defined by

f\in H^{0}(\mathbb{P}^{2}\times\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,1))\simeq H^{0}(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(1))\otimes_{k}H^{0}(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(1)),

we have

W=\left\{\sum_{0\leq i,j\leq 2}a_{ij}x_{i}y_{j}=0\right\}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}.

Applying the elementary operations to the $3\times 3$ -matrix $(a_{ij})$ , we may assume that $W$ is one of $W_{1},W_{2}$ , and $W_{3}$ (note that we can apply row and column elementray operations).

Let us show (2). By the proof of (1), $W\simeq W_{i}$ is obtained by applying a suitable ${\rm Aut}(\mathbb{P}^{2})\times{\rm Aut}(\mathbb{P}^{2})$ -action on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ . Hence we may assume that $W=W_{2}$ . Then the assertion (2) holds for $Q:=[0:0:1]$ . ∎

Lemma 5.17.

We use Notation 5.12. Assume that $\deg\widetilde{f}\neq 1$ . Set $W:=f(X)$ . Then the following hold.

(1)

$\deg\widetilde{f}=2$ and $\mu_{1}=\mu_{2}=1$ .
(2)

$(-K_{X})^{3}=12$ .
(3)

$W$ is a smooth prime divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,1)$ .
(4)

For $\mathcal{L}:=({\operatorname{Coker}}(\mathcal{O}_{W}\to\widetilde{f}_{*}\mathcal{O}_{X}))^{-1}$ , it holds that $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,1)|_{W}$ .
(5)

$\widetilde{f}:X\to W$ is a split double cover.

Proof.

By Lemma 5.13, we have

\deg(\tilde{f})\cdot f(X)=f_{*}(X)\sim\frac{2}{\mu_{2}}M_{1}+\frac{2}{\mu_{1}}M_{2}.

We then get $deg(\tilde{f})=2$ and $\mu_{1}=\mu_{2}=1$ . Hence (1) holds. By Lemma 5.14, $(-K_{X})^{3}=6(\mu_{1}^{2}+\mu_{2}^{2})=12.$ Thus (2) holds. Furthermore, we have

\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(W)\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,1).

Then the assertion (3) follows from Lemma 5.16(2).

Let us show (4). Pick a Cartier divisor $L$ on $W$ with $\mathcal{L}\simeq\mathcal{O}_{W}(L)$ . By Lemma 2.3, we have

(5.17.1)

K_{X}\sim\widetilde{f}^{*}(K_{W}+L).

By the adjunction formula, we obtain

	$\displaystyle K_{W}$	$\displaystyle\sim(K_{\mathbb{P}^{2}\times\mathbb{P}^{2}}+W)\|_{W}$
		$\displaystyle\sim(-3M_{1}-3M_{2}+M_{1}+M_{2})\|_{W}$
		$\displaystyle\sim-2(M_{1}+M_{2})\|_{W}.$

By (5.17.1) and $K_{X}\sim-H_{1}-H_{2}$ (Lemma 5.9), we obtain

\widetilde{f}^{*}L\sim K_{X}-\widetilde{f}^{*}K_{W}\sim(-H_{1}-H_{2})-\widetilde{f}^{*}(-2(M_{1}+M_{2})|_{W})=H_{1}+H_{2}.

It holds that $L\equiv c_{1}(\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,1)|_{W})$ . Since $W$ is a smooth Fano threefold, we obtain $L\sim c_{1}(\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,1)|_{W})$ . Thus (4) holds. The assertion (5) follows from (3), (4), and Lemma 2.4(2). ∎

Theorem 5.18.

Let $X$ be a Fano threefold with $\rho(X)=2$ . Set $W:=\{x_{0}y_{0}+x_{1}y_{1}+x_{2}y_{2}=0\}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ , which is a smooth hypersurface on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,1)$ . Let $R_{1}$ and $R_{2}$ be the distinct extremal rays of ${\operatorname{NE}}(X)$ . Assume that each of $R_{1}$ and $R_{2}$ is of type $C$ . Then, possibly after permuting $R_{1}$ and $R_{2}$ , one and only one of the following holds.

(1)
$R_{1}$ is of type $C_{1}$ , $R_{2}$ is of type $C_{1}$ , and one of the following holds.
1. (a)
  
  $X$ is isomorphic to a hypersurface of $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(2,2)$ and $(-K_{X})^{3}=12$ .
2. (b)
  
  There exists a split double cover $f:X\to W$ such that $(f_{*}\mathcal{O}_{X}/\mathcal{O}_{W})^{-1}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,1)|_{W}$ and $(-K_{X})^{3}=12$ .
(2)

$R_{1}$ is of type $C_{1}$ , $R_{2}$ is of type $C_{2}$ , $(-K_{X})^{3}=30$ , and $X$ is isomorphic to a hypersurface of $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$ .
(3)

$R_{1}$ is of type $C_{2}$ , $R_{2}$ is of type $C_{2}$ , $(-K_{X})^{3}=48$ , and $X$ is isomorphic to $W$ .

Proof.

If one of the contractions is not generically smooth, then (2) holds by Proposition 4.17(5). Hence we may assume that each contraction is generically smooth. Then the assertion follows from Lemma 5.14, Lemma 5.15, and Lemma 5.17. ∎

5.3. Case $C-D$

Notation 5.19.

We use Notation 5.1. Assume that $R_{1}$ is of type $C$ and $R_{2}$ is of type $D$ . In particular, $f_{1}:X\to Y_{1}=\mathbb{P}^{2}$ is of type $C$ and $f_{2}:X\to Y_{2}=\mathbb{P}^{1}$ is of type $D$ . If $f_{2}$ is of type $D_{i}$ for $i\in\{1,2,3\}$ , then we have $\mu_{2}=i$ (Remark 3.4). Set

f:=f_{1}\times f_{2}\colon X\to\mathbb{P}^{2}\times\mathbb{P}^{1}.

Lemma 5.20.

We use Notation 5.19. Then $(-K_{X})^{3}=6\mu_{2}^{2}$ .

Proof.

By Proposition 5.9, $-K_{X}\sim\mu_{2}H_{1}+\mu_{1}H_{2}$ . By $L_{1}\simeq\mathcal{O}_{\mathbb{P}^{2}}(1)$ and $L_{2}\simeq\mathcal{O}_{\mathbb{P}^{1}}(1)$ , we have $H_{1}^{3}=f_{1}^{*}L_{1}^{3}=0$ and $H_{2}^{2}=f_{2}^{*}L_{2}^{2}\equiv 0$ . Therefore, we obtain

	$\displaystyle(-K_{X})^{3}$	$\displaystyle=(\mu_{2}H_{1}+\mu_{1}H_{2})^{3}$
		$\displaystyle=\mu_{2}^{3}H_{1}^{3}+3\mu_{2}^{2}\mu_{1}H_{1}^{2}\cdot H_{2}+3\mu_{2}\mu_{1}^{2}H_{1}\cdot H_{2}^{2}+\mu_{2}^{3}H_{2}^{3}$
		$\displaystyle=3\mu_{2}^{2}\mu_{1}\cdot\left(\frac{2}{\mu_{1}}\ell_{1}\right)\cdot H_{2}$
		$\displaystyle=6\mu_{2}^{2},$

where the third equality holds by $H_{1}^{2}\equiv\frac{2}{\mu_{1}}\ell_{1}$ (Lemma 5.3) and the fourth one follows from $H_{2}\cdot\ell_{1}=1$ (Proposition 5.9). ∎

Lemma 5.21.

We use Notation 5.19.

(1)

If $R_{1}$ is of type $C_{1}$ , then $f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ is a double cover.
(2)

If $R_{1}$ is of type $C_{2}$ , then $f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ is an isomorphism and $R_{2}$ is of type $D_{3}$ .

Proof.

By Lemma 4.15(1), $f$ is a finite morphism. It follows from $L_{1}\simeq\mathcal{O}_{\mathbb{P}^{2}}(1)$ and $L_{2}\simeq\mathcal{O}_{\mathbb{P}^{1}}(1)$ that $\deg f=(f_{1}^{*}L_{1})^{2}\cdot f_{2}^{*}L_{2}$ . By Corollary 4.5, any fibre of $f_{1}:X\to Y_{1}=\mathbb{P}^{2}$ is numerically equivalent to $\frac{2}{\mu_{1}}\ell_{1}$ , i.e., $(f_{1}^{*}L_{1})^{2}\equiv\frac{2}{\mu_{1}}\ell_{1}$ . We then obtain

\deg f=(f_{1}^{*}L_{1})^{2}\cdot f_{2}^{*}L_{2}=\frac{2}{\mu_{1}}\ell_{1}\cdot H_{2}=\frac{2}{\mu_{1}},

where the last equality holds by Proposition 5.9.

(1) Assume that $R_{1}$ is of type $C_{1}$ . We then have $\mu_{1}=1$ , and hence $f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ is a double cover.

(2) Assume that $R_{2}$ is of type $C_{2}$ . Then we have $\mu_{1}=1$ and $\deg f=1$ . Hence, $f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ is a finite birational morphism of normal varieties, which is automatically an isomorphism. In particular, $R_{2}$ is of type $D_{3}$ . ∎

Lemma 5.22.

We use Notation 5.19. Assume that $R_{1}$ is of type $C_{1}$ . Set $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}})^{-1}$ . Then the following hold.

(1)

$R_{2}$ is of type $D_{1}$ or $D_{2}$ .
(2)

$\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(3-\mu_{2},1)$ .
(3)

$f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ is a split double cover.

Proof.

By $\mu_{1}=1$ , we have $K_{X}\sim-\mu_{2}H_{1}-H_{2}$ , i.e., $\omega_{X}\simeq f^{*}\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(-\mu_{2},-1)$ . Recall that we have $\omega_{X}\simeq f^{*}(\omega_{\mathbb{P}^{2}\times\mathbb{P}^{1}}\otimes\mathcal{L})$ (Lemma 2.3). It holds that $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(b_{1},b_{2})$ for some $b_{1},b_{2}\in\mathbb{Z}$ . Then

	$\displaystyle\omega_{X}$	$\displaystyle\simeq f^{*}\left(\omega_{\mathbb{P}^{2}\times\mathbb{P}^{1}}\otimes\mathcal{L}\right)$
		$\displaystyle\simeq f^{*}\left(\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(-3,-2)\otimes\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(b_{1},b_{2})\right)$
		$\displaystyle\simeq f^{*}\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(b_{1}-3,b_{2}-2).$

Hence $b_{1}=3-\mu_{2}$ and $b_{2}=1$ . Thus (2) holds.

Let us show (3), i.e.,

0\to\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}\to f_{*}\mathcal{O}_{X}\to\mathcal{L}\to 0

splits. The extension class lies in

{\rm Ext}^{1}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(\mathcal{L},\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}})\simeq H^{1}(\mathbb{P}^{2}\times\mathbb{P}^{1},\mathcal{L}^{-1})\simeq H^{1}(\mathbb{P}^{2}\times\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(\mu_{2}-3,-1))=0,

where the last equality holds by the Künneth formula. Thus (3) holds.

Let us show (1). Suppose that $R_{2}$ is of type $D_{3}$ . In this case, we have that $X\simeq S\times\mathbb{P}^{1}$ for some smooth projective surface $S$ (Lemma 8.4). Then $X$ has an extremal ray of type $C_{2}$ , which is a contraction. Thus (1) holds. ∎

Theorem 5.23.

Let $X$ be a Fano threefold with $\rho(X)=2$ . Let $R_{1}$ and $R_{2}$ be extremal rays of ${\operatorname{NE}}(X)$ . Assume that $R_{1}$ is of type $C$ and $R_{2}$ is of type $D$ . Then one and only one of the following holds.

(1)

$R_{1}$ is of type $C_{1}$ , $R_{2}$ is of type $D_{1}$ , $(-K_{X})^{3}=6$ , and there exists a split double cover $f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ such that $(f_{*}\mathcal{O}_{X}/\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}})^{-1}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(2,1)$ .
(2)

$R_{1}$ is of type $C_{1}$ , $R_{2}$ is of type $D_{2}$ , $(-K_{X})^{3}=24$ , and there exists a split double cover $f:X\to\mathbb{P}^{2}\times\mathbb{P}^{1}$ such that $(f_{*}\mathcal{O}_{X}/\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}})^{-1}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(1,1)$ .
(3)

$R_{1}$ is of type $C_{2}$ , $R_{2}$ is of type $D_{3}$ , $(-K_{X})^{3}=54$ , and $X\simeq\mathbb{P}^{2}\times\mathbb{P}^{1}$ .

Proof.

The assertions follow from Lemma 5.20, Lemma 5.21, and Lemma 5.22. ∎

5.4. Case $C-E$ (primitive)

Notation 5.24.

We use Notation 5.1. Assume that $R_{1}$ is of type $C$ and $R_{2}$ is of type $E_{2},E_{3},E_{4}$ , or $E_{5}$ . Set $D:={\operatorname{Ex}}(f_{2}:X\to Y_{2})$ , which is a prime divisor on $X$ .

Let us consider the induced morphism $f_{1}|_{D}\colon D\to Y_{1}=\mathbb{P}^{2}$ .

Lemma 5.25.

We use Notation 5.24. Then the following hold.

(1)

If $R_{1}$ is of type $C_{1}$ , then $f_{1}|_{D}:D\to Y_{1}=\mathbb{P}^{2}$ is a double cover and $R_{2}$ is of type $E_{3}$ or $E_{4}$ .
(2)

If $R_{1}$ is of type $C_{2}$ , then $f_{1}|_{D}:D\to Y_{1}=\mathbb{P}^{2}$ is an isomorphism and $R_{2}$ is of type $E_{2}$ or $E_{5}$ .

Proof.

By Lemma 4.15(2), $f_{1}|_{D}:D\to Y_{1}=\mathbb{P}^{2}$ is a finite surjective morphism.

Step 1.

It holds that

\deg(f_{1}|_{D})=\frac{2}{\mu_{1}}D\cdot\ell_{1}=\frac{1}{\mu_{2}^{2}}(\omega_{D}\otimes\mathcal{O}_{D}(-D))^{2},

where $\mathcal{O}_{D}(-D):=\mathcal{O}_{X}(-D)|_{D}$ .

Proof of Step 1.

By Corollary 4.5, a fibre of $f_{1}$ is numerically equivalent to $(2/\mu_{1})\ell_{1}$ . We then obtain

\deg(f_{1}|_{D})=\frac{2}{\mu_{1}}D\cdot\ell_{1}=D\cdot H_{1}^{2},

where the latter equality follows from Lemma 5.3. By Proposition 5.9(3), we have

	$\displaystyle H_{1}^{2}$	$\displaystyle\equiv\frac{1}{\mu_{2}^{2}}(-K_{X}-\mu_{1}H_{2})^{2}$
		$\displaystyle=\frac{1}{\mu_{2}^{2}}((-K_{X})^{2}-2\mu_{1}(-K_{X})\cdot H_{2}+\mu_{1}^{2}H_{2}^{2}).$

Since $R_{2}$ is of type $E_{2},E_{3},E_{4}$ or $E_{5}$ , $f_{2}(D)$ is a point on $Y$ . Therefore, we get

H_{2}\cdot D=f^{*}_{2}L_{2}\cdot D\equiv 0.

Hence we obtain

\deg(f_{1}|_{D})=D\cdot H_{1}^{2}=\frac{1}{\mu_{2}^{2}}D\cdot((-K_{X})^{2}-2\mu_{1}(-K_{X})\cdot H_{2}+\mu_{1}^{2}H_{2}^{2})

=\frac{1}{\mu_{2}^{2}}D\cdot(-K_{X})^{2}=\frac{1}{\mu_{2}^{2}}(\omega_{D}\otimes\mathcal{O}_{D}(-D))^{2},

where the last equality holds by the adjunction formula. This completes the proof of Step 1. ∎

Step 2.

The assertion (1) holds.

Proof of Step 2.

Assume that $R_{1}$ is of type $C_{1}$ . We have $\mu_{1}=1$ . Recall that $R_{2}$ is of type $E_{2},E_{3},E_{4},$ or $E_{5}$ .

Assume that $R_{2}$ is of type $E_{2}$ . Then we have $\mu_{2}=2,D\simeq\mathbb{P}^{2}$ , and $\mathcal{O}_{D}(D)\simeq\mathcal{O}_{D}(-1)$ . By Step 1, we get

2D\cdot\ell_{1}=\frac{1}{\mu_{2}^{2}}(\omega_{D}\otimes\mathcal{O}_{D}(-D))^{2}=\frac{1}{4}(-3+1)^{2}=1.

This is absurd.

Assume that $R_{2}$ is of type $E_{5}$ . Then $\mu_{2}=1,D\simeq\mathbb{P}^{2},$ and $\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{2}}(-2)$ . By Step 1, we get

2D\cdot\ell_{1}=\frac{1}{\mu_{2}^{2}}(\omega_{D}\otimes\mathcal{O}_{D}(-D))^{2}=(\mathcal{O}_{\mathbb{P}^{2}}(-3+2))^{2}=1.

This is a contradiction.

Assume that $R_{2}$ is of type $E_{3}$ or $E_{4}$ . Then we have $\mu_{2}=1$ , $D$ is isomorphic to a possibly singular quadric surface in $\mathbb{P}^{3}$ , and $\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{3}}(-1)|_{D}$ . We have

\omega_{D}\simeq(\omega_{\mathbb{P}^{3}}\otimes\mathcal{O}_{\mathbb{P}^{3}}(D))|_{D}\simeq\mathcal{O}_{\mathbb{P}^{3}}(-4+2)|_{D}\\ \simeq\mathcal{O}_{\mathbb{P}^{3}}(-2)|_{D}.

By Step 1, we get

\deg(f_{1}|_{D})=2D\cdot\ell_{1}=\frac{1}{\mu_{2}^{2}}(\omega_{D}\otimes\mathcal{O}_{D}(-D))^{2}=(\mathcal{O}_{\mathbb{P}^{3}}(-2+1)|_{D})^{2}=2.

This completes the proof of Step 2. ∎

Step 3.

The assertion (2) holds.

Proof of Step 3.

Assume that $R_{1}$ is of type $C_{2}$ . We have $\mu_{1}=2$ .

Suppose that $R_{2}$ is of type $E_{3}$ or $E_{4}$ . Then $\mu_{2}=1$ , and hence Proposition 5.9 implies

-K_{X}\sim H_{1}+2H_{2}.

We then get

	$\displaystyle 24$	$\displaystyle=c_{2}(X)\cdot(-K_{X})$
		$\displaystyle=c_{2}(X)\cdot H_{1}+2c_{2}(X)\cdot H_{2}$
		$\displaystyle=6+\frac{48}{r}$

for some $r\in\mathbb{Z}_{>0}$ (Lemma 5.4). This is impossible.

Hence $R_{2}$ is of type $E_{2}$ or $E_{5}$ . It holds that

\deg(f_{1}|_{D})=D\cdot\ell_{1}=\frac{1}{\mu_{2}^{2}}(\omega_{D}\otimes\mathcal{O}_{D}(-D))^{2}=1,

where the first two equalities are guaranteed by Step 1 and the last one follows from the same argument as in Step 2. Then $f_{1}|_{D}:D\to\mathbb{P}^{2}$ is a finite birational morphism of normal varieties, which is automatically an isomorphism. This completes the proof of Step 3. ∎

Step 2 and Step 3 complete the proof of Lemma 5.25. ∎

5.4.1. Case $C_{1}-E$

Notation 5.26.

We use Notation 5.1. Assume that $R_{1}$ is of type $C_{1}$ and $R_{2}$ is of type $E_{2},E_{3},E_{4},$ or $E_{5}$ . Set $D:={\operatorname{Ex}}(f_{2}:X\to Y_{2})$ . In particular, $f_{1}:X\to Y_{1}=\mathbb{P}^{2}$ is of type $C_{1}$ and $f_{2}:X\to Y_{2}$ is of type $E_{3}$ or $E_{4}$ (Lemma 5.25). Recall that $D$ is a (possibly singular) quadric surface in $\mathbb{P}^{3}$ and set $\mathcal{O}_{D}(n):=\mathcal{O}_{\mathbb{P}^{3}}(n)|_{D}$ for $n\in\mathbb{Z}$ . By Lemma 3.15, we have a diagram:

Since $\iota$ is a closed immersion, we identify $X$ with the smooth prime divisor $\iota(X)$ on the $\mathbb{P}^{2}$ -bundle $P$ over $\mathbb{P}^{2}$ .

Lemma 5.27.

We use Notation 5.26. Then $R^{1}(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)=0$ and there exists an exact sequence:

0\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\longrightarrow(f_{1})_{*}\mathcal{O}_{D}(-K_{X})\longrightarrow 0.

Proof.

Since $(-K_{X}-D)-K_{X}$ is $f_{1}$ -ample (Lemma 5.25), the assertion follows from $R^{1}(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)=0$ , which is guaranteed by [Tan15, Theorem 0.5]. ∎

Lemma 5.28.

We use Notation 5.26. Then $(f_{1}|_{D})^{*}\mathcal{O}_{\mathbb{P}^{2}}(n)\simeq\mathcal{O}_{D}(n)$ for all $n\in\mathbb{Z}$ .

Proof.

We may assume $n=1$ .

Assume that $R_{2}$ is of type $E_{3}$ . Then $D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ and we can write

(f_{1}|_{D})^{*}\mathcal{O}_{\mathbb{P}^{2}}(1)\simeq\mathcal{O}_{D}(a,b)

for some $a,b\in\mathbb{Z}$ . Since $f_{1}|_{D}:D\to\mathbb{P}^{2}$ is a finite morphism, this invertible sheaf is ample, which implies $a>0$ and $b>0$ . By the projection formula, we get

	$\displaystyle 2ab$	$\displaystyle=(\mathcal{O}_{D}(a,b))^{2}$
		$\displaystyle=\deg(f_{1}\|_{D})\cdot(\mathcal{O}_{\mathbb{P}^{2}}(1))^{2}$
		$\displaystyle=2.$

Hence we have $ab=1$ . We then get $(a,b)=(1,1)$ and

(f_{1}|_{D})^{*}\mathcal{O}_{\mathbb{P}^{2}}(1)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1)\simeq\mathcal{O}_{\mathbb{P}^{3}}(1)|_{D}=\mathcal{O}_{D}(1).

Assume that $R_{2}$ is of type $E_{4}$ . By [Har77, Ch. II, Exercise 6.3 and 6.5], the restriction homomorphism ${\operatorname{Pic}}(\mathbb{P}^{3})\to{\operatorname{Pic}}(D)$ is an isomorphism. Hence we have

(f_{1}|_{D})^{*}\mathcal{O}_{\mathbb{P}^{2}}(1)\simeq\mathcal{O}_{D}(a)

for some $a\in\mathbb{Z}$ . Since this invertible sheaf is ample, we get $a>0$ . By the projection formula,

a^{2}c_{1}(\mathcal{O}_{D}(1))^{2}=c_{1}(\mathcal{O}_{D}(a))^{2}=\deg(f_{1}|_{D})\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(1))^{2}=2.

Hence $a=1$ . ∎

Proposition 5.29.

We use Notation 5.26. Then the following hold.

(1)

$(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\simeq\mathcal{O}_{\mathbb{P}^{2}}(2).$
(2)

$(f_{1})_{*}\mathcal{O}_{D}(-K_{X})\simeq\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)$ .
(3)

$(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\simeq\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2)$ .

Proof.

Let us show (1). For a closed point $t\in\mathbb{P}^{2}$ and the fibre $X_{t}$ of $f_{1}:X\to Y_{1}=\mathbb{P}^{2}$ over $t$ , we have $-K_{X}\cdot X_{t}=D\cdot X_{t}=2$ . Hence $h^{0}(X_{t},\mathcal{O}_{X_{t}}(-K_{X}-D))=1$ for all $t\in Y_{1}=\mathbb{P}^{2}$ . By Grauert’s theorem [Har77, Ch. III, Theorem 12.9], $(f_{1})_{*}(\mathcal{O}_{X}(-K_{X}-D))$ is an invertible sheaf on $\mathbb{P}^{2}$ . Hence we can write

(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\simeq\mathcal{O}_{\mathbb{P}^{2}}(a)

for some $a\in\mathbb{Z}$ . Again by [Har77, Ch. III, Theorem 12.9], we get

\mathcal{O}_{X}(-K_{X}-D)\simeq f^{*}_{1}(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\simeq f^{*}_{1}\mathcal{O}_{\mathbb{P}^{2}}(a).

By Lemma 5.28 and adjunction formula, we obtain

\mathcal{O}_{D}(a)\simeq(f_{1}|_{D})^{*}\mathcal{O}_{\mathbb{P}^{2}}(a)\simeq\mathcal{O}_{D}(-K_{X}-D)\simeq\omega_{D}^{-1}\simeq\mathcal{O}_{\mathbb{P}^{3}}(2)|_{D}=\mathcal{O}_{D}(2),

which implies $a=2$ . Thus (1) holds.

Let us show (2). By $\mathcal{O}_{D}(-K_{X}-D)\simeq\mathcal{O}_{D}(2)$ and $\mathcal{O}_{D}(D)\simeq\mathcal{O}_{D}(-1)$ , we obtain $\mathcal{O}_{D}(-K_{X})\simeq\mathcal{O}_{D}(1)$ . Then

	$\displaystyle(f_{1}\|_{D})_{*}\mathcal{O}_{D}(-K_{X})$	$\displaystyle\simeq(f_{1}\|_{D})_{*}\mathcal{O}_{D}(1)$
		$\displaystyle\simeq(f_{1}\|_{D})_{}(\mathcal{O}_{D}\otimes(f_{1}\|_{D})^{}\mathcal{O}_{\mathbb{P}^{2}}(1))$
		$\displaystyle\simeq(f_{1}\|_{D})_{*}\mathcal{O}_{D}\otimes\mathcal{O}_{\mathbb{P}^{2}}(1).$

Since $f_{1}|_{D}\colon D\to\mathbb{P}^{2}$ is a double cover (Lemma 5.25), we have an exact sequence

0\to\mathcal{O}_{\mathbb{P}^{2}}\to(f_{1}|_{D})_{*}\mathcal{O}_{D}\to\mathcal{O}_{\mathbb{P}^{2}}(-b)\to 0

for some $b\in\mathbb{Z}$ . Since this sequence splits, we get $(f_{1}|_{D})_{*}\mathcal{O}_{D}\simeq\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(-b)$ . Furthermore, $\omega_{D}\simeq(f_{1}|_{D})^{*}(\omega_{\mathbb{P}^{2}}\otimes\mathcal{O}_{\mathbb{P}^{2}}(b))$ (Lemma 2.3). By $\omega_{D}\simeq\mathcal{O}_{D}(-2)$ , we obtain

	$\displaystyle\mathcal{O}_{D}(-2)$	$\displaystyle\simeq\omega_{D}$
		$\displaystyle\simeq(f_{1}\|_{D})^{*}(\mathcal{O}_{\mathbb{P}^{2}}(-3)\otimes\mathcal{O}_{\mathbb{P}^{2}}(b))$
		$\displaystyle\simeq(f_{1}\|_{D})^{*}(\mathcal{O}_{\mathbb{P}^{2}}(b-3))$
		$\displaystyle\simeq\mathcal{O}_{D}(b-3),$

where the last isomorphism holds by Lemma 5.28. Thus $b=1$ and (2) holds.

Let us show (3). By (1), (2), and Lemma 5.27, we have the following exact sequence:

0\longrightarrow\mathcal{O}_{\mathbb{P}^{2}}(2)\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\longrightarrow\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\longrightarrow 0.

By $\mathrm{Ext}^{1}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1),\mathcal{O}_{\mathbb{P}^{2}}(2))=0$ , this exact sequence splits, and hence

(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\simeq\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2).

Thus (3) holds. ∎

Lemma 5.30.

We use Notation 5.26. Then $\mathcal{O}_{P}(X)\simeq\mathcal{O}_{P}(2)$ , where $\mathcal{O}_{P}(1)$ denotes the tautological line bundle of the $\mathbb{P}^{2}$ -bundle structure $\pi:P=\mathbb{P}((f_{1})_{*}\mathcal{O}_{X}(-K_{X}))\to\mathbb{P}^{2}$ and we set $\mathcal{O}_{P}(2):=\mathcal{O}_{P}(1)^{\otimes 2}$ .

Proof.

Recall that we have the following diagram (Notation 5.26, Proposition 5.29):

By ${\operatorname{Pic}}\,P\simeq\pi^{*}{\operatorname{Pic}}\,\mathbb{P}^{2}\oplus\mathbb{Z}\mathcal{O}_{P}(1)$ , we can write

(5.30.1)

\mathcal{O}_{P}(X)\simeq\pi^{*}\mathcal{O}_{\mathbb{P}^{2}}(a)\otimes\mathcal{O}_{P}(n)

for some $a,n\in\mathbb{Z}$ . It follows from the adjunction formula that

(5.30.2)

\mathcal{O}_{X}(K_{X})\simeq\mathcal{O}_{P}(K_{P}+X)|_{X}.

By Proposition 7.1(2), it holds that

(5.30.3)

\mathcal{O}_{P}(K_{P})\simeq\mathcal{O}_{P}(-3)\otimes\pi^{*}(\omega_{\mathbb{P}^{2}}\otimes\det(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2)))\simeq\mathcal{O}_{P}(-3).

Since the closed immersion $\iota\colon X\to P$ is induced by the surjection $(f_{1})^{*}(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\to\mathcal{O}_{X}(-K_{X})$ , it follows from [Har77, Ch. II, the proof of Proposition 7.12] that

(5.30.4)

\mathcal{O}_{X}(-K_{X})\simeq\iota^{*}\mathcal{O}_{P}(1)\simeq\mathcal{O}_{P}(1)|_{X}.

By (5.30.1), (5.30.2), (5.30.3) and (5.30.4), we have

\mathcal{O}_{P}(1)|_{X}\simeq\mathcal{O}_{X}(-K_{X})\simeq\mathcal{O}_{P}(-K_{P}-X)|_{X}\simeq(\pi^{*}\mathcal{O}_{\mathbb{P}^{2}}(-a)\oplus\mathcal{O}_{P}(3-n))|_{X}.

Taking the intersection number of this equation with a general fibre of $\pi|_{X}\colon X\to\mathbb{P}^{2}$ , we have $1=3-n$ , i.e., $n=2$ . Hence we obtain $\mathcal{O}_{X}\simeq\pi^{*}\mathcal{O}_{\mathbb{P}^{2}}(a)|_{X}$ . We see that $a=0$ by taking the intersection number of this equation with a curve $C$ on $X$ such that $\pi(C)$ is a curve. ∎

Lemma 5.31.

We use Notation 5.26. Then $(-K_{X})^{3}=14$ .

Proof.

It follows from Proposition 7.1(5) that

	$\displaystyle(-K_{X})^{3}=$	$\displaystyle 2c_{1}(\mathcal{E})^{2}-2c_{2}(\mathcal{E})+4c_{1}(\mathcal{E})\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{2}})+6c_{1}(\mathcal{E})\cdot K_{\mathbb{P}^{2}}$
		$\displaystyle+9c_{1}(\mathcal{O}_{\mathbb{P}^{2}})\cdot K_{\mathbb{P}^{2}}+6K_{\mathbb{P}^{2}}^{2}+3c_{1}(\mathcal{O}_{\mathbb{P}^{2}})^{2}.$

Moreover, we have

	$\displaystyle c_{1}(\mathcal{E})=c_{1}(\mathcal{O}_{\mathbb{P}^{2}})+c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(1))+c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(2))=c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(3))$
	$\displaystyle c_{2}(\mathcal{E})=c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(1))\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(2))=2$
	$\displaystyle c_{1}(\mathcal{O}_{\mathbb{P}^{2}})=0$

by [Har77, Appendix A, §3, C3 and C.5]. Hence we obtain

(-K_{X})^{3}=2\cdot 3^{2}-2\cdot 2+4\cdot 0+6\cdot 3\cdot(-3)+9\cdot 0+6\cdot(-3)^{2}+3\cdot 0\\ =18-4-54+54=14.

∎

Proposition 5.32.

We use Notation 5.26. Then there exist a split double cover

f:X\to V_{7}

such that $K_{X}\sim f^{*}(\frac{1}{2}K_{V_{7}})$ , where $V_{7}$ is the blowup of $\mathbb{P}^{3}$ at a point and $\frac{1}{2}K_{V_{7}}$ is a Weil divisor on $V_{7}$ satisfying $2\cdot(\frac{1}{2}K_{V_{7}})\sim K_{V_{7}}$ (note that $\frac{1}{2}K_{V_{7}}$ is unique up to linear equivalence). In particular, for $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{V_{7}})^{-1}$ , we have $\mathcal{L}\simeq\mathcal{O}_{V_{7}}(-\frac{1}{2}K_{V_{7}})$ .

Proof.

Set $S\subset P$ to be the section of $\pi:P=\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))\to Z:=\mathbb{P}^{2}$ corresponding to the natural projection:

\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2)\to\mathcal{O}_{\mathbb{P}^{2}}.

Let $\mu:\widetilde{P}\to P$ be the blowup along $S$ .

Step 1.

There exists the following commutative diagram

(5.32.1)

such that for any closed point $z\in\mathbb{P}^{2}=Z$ , the base change of (5.32.1) by $z\hookrightarrow\mathbb{P}^{2}=Z$ can be written as follows

(5.32.2)

where $\mu_{z}$ is the blowup at the closed point $S_{z}$ and $\alpha_{z}:\mathbb{F}_{1}\to\mathbb{P}^{1}$ is the $\mathbb{P}^{1}$ -bundle.

Proof of Step 1.

We have the canonical injective graded $\mathcal{O}_{\mathbb{P}^{2}}$ -algebra homomorhism

\iota:{\rm Sym}(\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))\hookrightarrow{\rm Sym}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2)).

Then $\iota$ induces a dominant rational map $\varphi$ over $\mathbb{P}^{2}$ as follows:

Note that this construction commutes with base changes by open immersions $Z^{\prime}\hookrightarrow Z=\mathbb{P}^{2}$ . Then the indeterminacies of $\varphi$ are resolved by the blowup along $S$ , because this can be checked after taking an affine open cover of $Z=\mathbb{P}^{2}$ trivialising $\mathcal{O}_{\mathbb{P}^{2}}(n)$ . Therefore, we get a commutative diagram (5.32.1). This diagram fibrewisely induces the diagram (5.32.2), because we have the corresponding diagram over an arbitrary open subset of $Z=\mathbb{P}^{2}$ which trivialises $\mathcal{O}_{\mathbb{P}^{2}}(n)$ . This completes the proof of Step 1. ∎

Step 2.

It holds that $X\cap S=\emptyset$ . In particular, $\mu^{-1}(X)\simeq X$ .

Proof of Step 2.

Suppose $X\cap S\neq\emptyset$ . Note that $P=\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1)\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))$ is covered by $\mathbb{A}^{4}$ . Then it follows from [Har77, Ch. I, Proposition 7.1] that $\dim(X\cap S)\geq 3+2-4=1$ . Hence there exists a curve $C$ with $C\subset X\cap S$ . By $\mathcal{O}_{X}(-K_{X})\simeq\mathcal{O}_{P}(1)|_{X}$ and $\mathcal{O}_{P}(1)|_{S}\simeq\mathcal{O}_{S}$ , we obtain

-K_{X}\cdot C=(\mathcal{O}_{P}(1)|_{X})\cdot C=\mathcal{O}_{P}(1)\cdot C=(\mathcal{O}_{P}(1)|_{S})\cdot C=\mathcal{O}_{S}\cdot C=0.

This contradicts the ampleness of $-K_{X}$ . This completes the proof of Step 2. ∎

Step 3.

The induced morphism

f:X\xrightarrow{\simeq}\mu^{-1}(X)\xrightarrow{\alpha|_{\mu^{-1}(X)}}V_{7}

is a double cover.

Proof of Step 3.

Fix a closed point $v\in V_{7}$ . Set $z:=\beta(v)\in Z=\mathbb{P}^{2}$ . Take the base change by $z\hookrightarrow\mathbb{P}^{2}$ . We then obtain the above diagram (5.32.2) and the fibre $X_{z}$ is a conic in $\mathbb{P}^{2}=P_{z}$ . It holds by Step 2 that $X_{z}$ is disjoint from the $(-1)$ -curve on the blowup $\widetilde{P}_{z}=\mathbb{F}_{1}$ of $P_{z}=\mathbb{P}^{2}$ lying over the point $S_{z}$ . We then have $X_{z}\cdot\zeta=2$ for a fibre $\zeta$ of $\alpha_{z}:\widetilde{P}_{z}\to(V_{7})_{z}$ , because $\zeta$ is nothing but the proper transform of a line on $\mathbb{P}^{2}=P_{z}$ passing through $S_{z}$ . Since $f^{-1}(v)$ consists of at most two points, $f:X\to V_{7}$ is a finite surjective morphism. By $X_{z}\cdot\zeta=2$ , $f$ is a double cover. This completes the proof of Step 3. ∎

Step 4.

It holds that $K_{X}\sim f^{*}(\frac{1}{2}K_{V_{7}}+\beta^{*}M)$ for some Cartier divisor $M$ on $\mathbb{P}^{2}$ .

Proof of Step 4.

The above diagram (5.32.1) consists of smooth projective toric varieties and toric morphisms. Furthermore, each morphism is of relative Picard number one.

Set

N:=-2(K_{\widetilde{P}}+\mu^{*}X)+\alpha^{*}K_{V_{7}}.

Fix a closed point $z\in Z=\mathbb{P}^{2}$ . We then have the diagram (5.32.2). Take a fibre $\zeta\simeq\mathbb{P}^{1}$ of the $\mathbb{P}^{1}$ -bundle $\alpha_{z}:\mathbb{F}_{1}=\widetilde{P}_{z}\to\mathbb{P}^{1}=(V_{7})_{z}$ over a closed point. Note that $\mu_{z}^{*}X_{z}\cdot\zeta=2$ (cf. the proof of Step 3), and hence we get

(5.32.3)

N\cdot\zeta=(-2(K_{\widetilde{P}}+\mu^{*}X)+\alpha^{*}K_{V_{7}})\cdot\zeta=-2(-2+2)+0=0

and

(5.32.4)

N\cdot\mu^{-1}(X_{z})=(-2(K_{\widetilde{P}}+\mu^{*}X)+\alpha^{*}K_{V_{7}})\cdot\mu^{-1}(X_{z})=-2\deg\omega_{X_{z}}+2\deg K_{(V_{7})_{z}}=4-4=0.

By (5.32.3), there exists a Cartier divisor $N^{\prime}$ on $V_{7}$ such that $N\sim\alpha^{*}N^{\prime}$ . It follows from (5.32.4) that we can find a Cartier divisor $N^{\prime\prime}$ on $\mathbb{P}^{2}=Z$ such that $N^{\prime}\sim\beta^{*}N^{\prime\prime}$ . For $N^{\prime\prime\prime}:=-N^{\prime\prime}$ , we get $2(K_{\widetilde{P}}+\mu^{*}X)\sim\alpha^{*}(K_{V_{7}}+\beta^{*}N^{\prime\prime\prime})$ . Taking the restriction to $\mu^{-1}(X)(\simeq X)$ , we have $2K_{X}\sim f^{*}(K_{V_{7}}+\beta^{*}N^{\prime\prime\prime})$ . Then $K_{X}-f^{*}(\frac{1}{2}K_{V_{7}})$ is numerically trivial over $Z$ . Therefore, we can find a Cartier divisor $M$ on $Z$ such that $K_{X}=f^{*}(\frac{1}{2}K_{V_{7}}+\beta^{*}M)$ . This completes the proof of Step 4. ∎

Step 5.

It holds that $K_{X}\sim f^{*}(\frac{1}{2}K_{V_{7}})$ .

Proof of Step 5.

By Step 4, we have $K_{X}\sim f^{*}(\frac{1}{2}K_{V_{7}}+\beta^{*}M)$ .

We now show that $\deg M\leq 0$ , i.e., $-M$ is nef. For the exceptional divisor $E\simeq\mathbb{P}^{2}$ of the blowup $V_{7}\to\mathbb{P}^{3}$ , it holds that

\mathcal{O}_{V_{7}}\left(\frac{1}{2}K_{V_{7}}\right)\Big{|}_{E}\simeq\mathcal{O}_{E}(-1).

Pick a line $\ell\subset E$ . Since $-(\frac{1}{2}K_{V_{7}}+\beta^{*}M)$ is ample, we obtain

0<-\left(\frac{1}{2}K_{V_{7}}+\beta^{*}M\right)\cdot\ell=1-\beta^{*}M\cdot\ell.

Since $E\to Z=\mathbb{P}^{2}$ is a finite surjective morphism, we get $\deg M\leq 0$ .

By $K_{X}\sim f^{*}(\frac{1}{2}K_{V_{7}}+\beta^{*}M)$ , we have

8K_{X}^{3}=(f^{*}(K_{V_{7}}+2\beta^{*}M))^{3}=2(K_{V_{7}}^{3}+6K_{V_{7}}^{2}\cdot\beta^{*}M+12K_{V_{7}}\cdot(\beta^{*}M)^{2}),

where the last equality holds by $(\beta^{*}M)^{3}=0$ . It follows from $K_{X}^{3}=-14$ (Lemma 5.31) and $K_{V_{7}}^{3}=K_{\mathbb{P}^{3}}^{3}+8=-56$ that

K_{V_{7}}^{2}\cdot\beta^{*}M+2K_{V_{7}}\cdot(\beta^{*}M)^{2}=0.

Suppose that $M\not\sim 0$ . By $\deg M\leq 0$ , we have $\deg M<0$ , i.e., $-M$ is ample. We then have

K_{V_{7}}^{2}\cdot\beta^{*}M=-(-K_{V_{7}})^{2}\cdot(-\beta^{*}M)<0\quad{\rm and}\quad K_{V_{7}}\cdot(\beta^{*}M)^{2}=-(-K_{V_{7}})\cdot(-\beta^{*}M)^{2}<0,

which is a contraction. Therefore, we obtain $M\sim 0$ . This completes the proof of Step 5. ∎

Step 6.

The induced morphism $f:X\to V_{7}$ is a split double cover.

Proof of Step 6.

For a Cartier divisor $L$ satisying

0\to\mathcal{O}_{V_{7}}\to f_{*}\mathcal{O}_{X}\to\mathcal{O}_{V_{7}}(-L)\to 0,

it suffices to show that ${\operatorname{Ext}}^{1}(\mathcal{O}_{V_{7}}(-L),\mathcal{O}_{V_{7}})=0$ , which is equivalent to $H^{1}(V_{7},\mathcal{O}_{V_{7}}(L))=0$ . We have $K_{X}\sim f^{*}(K_{V_{7}}+L)$ (Lemma 2.3). By Step 5, we get $L\sim-\frac{1}{2}K_{V_{7}}$ . Since $V_{7}$ is toric and $L$ is ample, we get $H^{1}(V_{7},\mathcal{O}_{V_{7}}(L))=0$ [Fuj07, Theorem 1.6]. This completes the proof of Step 6. ∎

Step 5 and Step 6 complete the proof of Proposition 5.32. ∎

5.4.2. Case $C_{2}-E$

Lemma 5.33.

We use Notation 5.1. Assume that $R_{1}$ is of type $C_{2}$ and $R_{2}$ is of type $E_{2},E_{3},E_{4}$ , or $E_{5}$ . Then the following hold.

(1)

$R_{2}$ is of type $E_{2}$ or $E_{5}$ .
(2)

If $R_{2}$ is of type $E_{2}$ , then $X\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$ and $(-K_{X})^{3}=56$ .
(3)

If $R_{2}$ is of type $E_{5}$ , then $X\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))$ and $(-K_{X})^{3}=62$ .

Proof.

Set $D:={\operatorname{Ex}}(f_{2}:X\to Y_{2})$ . Since $f_{1}\colon X\to\mathbb{P}^{2}$ is a $\mathbb{P}^{1}$ -bundle, we can write

f_{1}\colon X=\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{E})\to\mathbb{P}^{2}

for some locally free sheaf $\mathcal{E}$ of rank 2 on $\mathbb{P}^{2}$ . By Lemma 5.25, $f_{1}|_{D}\colon D\to\mathbb{P}^{2}$ is an isomorphism and $R_{2}$ is of type $E_{2}$ or $E_{5}$ . In particular, (1) holds. We have $\mathcal{O}_{X}(D)\simeq\mathcal{O}_{X}(n)\otimes(f_{1})^{*}\mathcal{L}$ for some $n\in\mathbb{Z}$ and $\mathcal{L}\in{\operatorname{Pic}}\,\mathbb{P}^{2}$ . Since $D\cdot F=1$ for a fibre $F\simeq\mathbb{P}^{1}$ of $f_{1}\colon X=\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{E})\to\mathbb{P}^{2}$ , we obtain $n=1$ , i.e.,

(5.33.1)

\mathcal{O}_{X}(D)\simeq\mathcal{O}_{X}(1)\otimes(f_{1})^{*}\mathcal{L}.

We have the following exact sequence

0\longrightarrow\mathcal{O}_{X}\longrightarrow\mathcal{O}_{X}(D)\longrightarrow j_{*}(\mathcal{O}_{X}(D)|_{D})\longrightarrow 0,

where $j:D\hookrightarrow X$ denotes the induced closed immersion. By $R^{1}(f_{1})_{*}\mathcal{O}_{X}=0$ (Proposition 3.11), we obtain the exact sequence

(5.33.2)

0\longrightarrow(f_{1})_{*}\mathcal{O}_{X}\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(D)\longrightarrow(f_{1})_{*}j_{*}(\mathcal{O}_{X}(D)|_{D})\longrightarrow 0.

We have $(f_{1})_{*}\mathcal{O}_{X}=\mathcal{O}_{\mathbb{P}^{2}}$ . By (5.33.1) and $(f_{1})_{*}\mathcal{O}_{X}(1)\simeq\mathcal{E}$ [Har77, Ch. II, Proposition 7.11], it holds that $(f_{1})_{*}\mathcal{O}_{X}(D)\simeq\mathcal{E}\otimes\mathcal{L}$ . If $R_{2}$ is of type $E_{2}$ (resp. $E_{5}$ ), then we set $e:=1$ (resp. $e:=2$ ). By Theorem 4.8, we have $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{\mathbb{P}^{2}}(-e)$ . Since $f_{1}\circ j:D\to\mathbb{P}^{2}$ is an isomorphism, we get $(f_{1})_{*}j_{*}(\mathcal{O}_{X}(D)|_{D})\simeq\mathcal{O}_{\mathbb{P}^{2}}(-e)$ . Hence, the exact sequence (5.33.2) can be written as follows:

0\longrightarrow\mathcal{O}_{\mathbb{P}^{2}}\longrightarrow\mathcal{E}\otimes\mathcal{L}\longrightarrow\mathcal{O}_{\mathbb{P}^{2}}(-e)\longrightarrow 0.

By $\mathrm{Ext}^{1}(\mathcal{O}_{\mathbb{P}^{2}}(-e),\mathcal{O}_{\mathbb{P}^{2}})\simeq H^{1}(\mathbb{P}^{2},\mathcal{O}_{\mathbb{P}^{2}}(e))=0$ , we obtain $\mathcal{E}\otimes\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(-e)$ . Therefore, it hold that

X=\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{E})\simeq\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{E}\otimes\mathcal{L}\otimes\mathcal{O}_{\mathbb{P}^{2}}(e))\simeq\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\otimes\mathcal{O}_{\mathbb{P}^{2}}(e)).

What is remaining is to compute $(-K_{X})^{3}$ . By Proposition 7.1(3), we have

(-K_{X})^{3}=2c_{1}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(e))^{2}-8c_{2}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(e))+6K_{\mathbb{P}^{2}}^{2}.

By [Har77, Appendix A], it holds that

\displaystyle c_{1}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(e))=c_{1}(\mathcal{O}_{\mathbb{P}^{2}})+c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(e))=c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(e)),

\displaystyle c_{2}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(e))=c_{1}(\mathcal{O}_{\mathbb{P}^{2}})\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(e))=0.

Therefore, we obtain

	$\displaystyle(-K_{X})^{3}$	$\displaystyle=2c_{1}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(e))^{2}-8c_{2}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(e))+6K_{\mathbb{P}^{2}}^{2}$
		$\displaystyle=2e^{2}-8\cdot 0+6\cdot 9$
		$\displaystyle=2e^{2}+54.$

∎

Theorem 5.34.

Let $X$ be a Fano threefold with $\rho(X)=2$ . Let $R_{1}$ and $R_{2}$ be extremal rays of ${\operatorname{NE}}(X)$ . Assume that $R_{1}$ is of type $C$ and $R_{2}$ is of type $E_{2},E_{3},E_{4}$ , or $E_{5}$ . Then one and only one of the following holds.

(1)

$R_{1}$ is of type $C_{1}$ , $R_{2}$ is of type $E_{3}$ or $E_{4}$ , $(-K_{X})^{3}=14$ , and there exists a split double cover $f:X\to V_{7}$ such that $f_{*}\mathcal{O}_{X}/\mathcal{O}_{V_{7}}\simeq\mathcal{O}_{V_{7}}(-\frac{1}{2}K_{V_{7}})^{-1}$ .
(2)

$R_{1}$ is of type $C_{2}$ , $R_{2}$ is of type $E_{2}$ , $(-K_{X})^{3}=56$ , and $X\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$ .
(3)

$R_{1}$ is of type $C_{2}$ , $R_{2}$ is of type $E_{5}$ , $(-K_{X})^{3}=62$ , and $X\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))$ .

Proof.

The assertion follows from Lemma 5.25, Proposition 5.32, and Lemma 5.33. ∎

5.5. Case $E_{1}-C$

Lemma 5.35.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ . Then

g(B_{1})=\frac{(-K_{X})^{3}}{2}-\frac{(-K_{Y_{1}})^{3}}{2}+r_{1}\deg B_{1}+1

Proof.

See, e.g., [TanII, Lemma 3.21(2)(a)]. ∎

Lemma 5.36.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $C$ . Then the following hold.

(A)

$18=\mu_{2}\left(\frac{24}{r_{1}}+\deg B_{1}\right)+\deg\Delta_{f_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-\mu_{2})L_{1}^{3}=\frac{1}{\mu_{2}^{2}}(8-\deg\Delta_{f_{2}})$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-\mu_{2})^{2}L_{1}^{3}-\deg B_{1}=\frac{2}{\mu_{2}}$ .

Proof.

The assertion (A) follows from Lemma 5.4 and Proposition 5.9(4). The assertion (C) holds by Lemma 5.11 and

2=(-K_{X})\cdot H_{2}^{2}=(\mu_{2}H_{1}+H_{2})\cdot H_{2}^{2}=\mu_{2}H_{1}\cdot H_{2}^{2}.

The assertion (B) follows from Lemma 5.11 and

12-\deg\Delta_{f_{2}}=(-K_{X})^{2}\cdot H_{2}=(\mu_{2}H_{1}+H_{2})^{2}\cdot H_{2}=\mu_{2}^{2}H_{1}^{2}\cdot H_{2}+2\mu_{2}H_{1}\cdot H_{2}^{2}=\mu_{2}^{2}H_{1}^{2}\cdot H_{2}+4,

where the first equality is guaranteed by Proposition 3.16(3). ∎

Proposition 5.37.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $C_{1}$ . Then one of the following holds.

(1)

$(-K_{X})^{3}=16,r_{1}=4,\deg B_{1}=7,g(B_{1})=5,\deg\Delta_{f_{2}}=5$ (No. 2-9).
(2)

$(-K_{X})^{3}=20,r_{1}=3,\deg B_{1}=6,g(B_{1})=2,\deg\Delta_{f_{2}}=4$ (No. 2-13).
(3)

$(-K_{X})^{3}=18,r_{1}=2,(-K_{X_{1}})^{3}=24,\deg B_{1}=1,g(B_{1})=0,\deg\Delta_{f_{2}}=5$ (No. 2-11).
(4)

$(-K_{X})^{3}=22,r_{1}=2,(-K_{X_{1}})^{3}=32,\deg B_{1}=2,g(B_{1})=0,\deg\Delta_{f_{2}}=4$ (No. 2-16).
(5)

$(-K_{X})^{3}=26,r_{1}=2,(-K_{X_{1}})^{3}=40,\deg B_{1}=3,g(B_{1})=0,\deg\Delta_{f_{2}}=3$ (No. 2-20).

Proof.

We have $\mu_{2}=1$ . By Lemma 5.36, the following hold.

(A)

$18=\frac{24}{r_{1}}+\deg B_{1}+\deg\Delta_{f_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-1)L_{1}^{3}=8-\deg\Delta_{f_{2}}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-1)^{2}L_{1}^{3}-\deg B_{1}=2$ .

Assume $r_{1}=4$ , and hence $L_{1}^{3}=1$ . Then (B) and (C) imply $\deg\Delta_{f_{2}}=5$ and $\deg B_{1}=7$ , respectively. Hence the following hold (Lemma 5.35):

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=H_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+H_{2}^{3}=L_{1}^{3}+9+6+0=16

g(B_{1})=8-32+4\cdot 7+1=5.

Assume $r_{1}=3$ , and hence $L_{1}^{3}=2$ . Then (B) and (C) imply $\deg\Delta_{f_{2}}=4$ and $\deg B_{1}=6$ , respectively. Hence the following hold (Lemma 5.35):

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=H_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+H_{2}^{3}=2+12+6+0=20

g(B_{1})=10-27+3\cdot 6+1=2.

Assume $r_{1}=2$ , and hence $1\leq L_{1}^{3}\leq 5$ . Then

(A)

$6=\deg B_{1}+\deg\Delta_{f_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=L_{1}^{3}=8-\deg\Delta_{f_{2}}$ .
(C)

$H_{1}\cdot H_{2}^{2}=L_{1}^{3}-\deg B_{1}=2$ .

By (C), we get $L_{1}^{3}=2+\deg B_{1}\geq 3$ .

Assume $(r_{1},L_{1}^{3})=(2,3)$ . Then (B) and (C) imply $\deg\Delta_{f_{2}}=5$ and $\deg B_{1}=1$ , respectively. In particular, $g(B_{1})=0$ . Hence the following hold:

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=H_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+H_{2}^{3}=3+9+6+0=18

Assume $(r_{1},L_{1}^{3})=(2,4)$ . Then (B) and (C) imply $\deg\Delta_{f_{2}}=4$ and $\deg B_{1}=2$ , respectively. In particular, $g(B_{1})=0$ . Hence the following hold:

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=H_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+H_{2}^{3}=4+12+6+0=22.

Assume $(r_{1},L_{1}^{3})=(2,5)$ . Then (B) and (C) imply $\deg\Delta_{f_{2}}=3$ and $\deg B_{1}=3$ , respectively. Hence the following hold (Lemma 5.35):

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=H_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+H_{2}^{3}=5+15+6+0=26.

g(B_{1})=13-20+2\cdot 3+1=0.

∎

Proposition 5.38.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $C_{2}$ . Then one of the following holds.

(1)

$(-K_{X})^{3}=38,r_{1}=4,\deg B_{1}=3,g(B_{1})=0$ (No. 2-27).
(2)

$(-K_{X})^{3}=46,r_{1}=3,\deg B_{1}=1,g(B_{1})=0$ (No. 2-31).

Proof.

We have $\mu_{2}=2$ and $\Delta_{f_{2}}=0$ . By Lemma 5.36, the following hold.

(A)

$18=\frac{48}{r_{1}}+2\deg B_{1}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-2)L_{1}^{3}=2$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-2)^{2}L_{1}^{3}-\deg B_{1}=1$ .

By (B), we obtain $r_{1}\in\{3,4\}$ . Moreover, (B) and (C) imply

(-K_{X})^{3}=(2H_{1}+H_{2})^{3}=8H_{1}^{3}+12H_{1}^{2}\cdot H_{2}+6H_{1}\cdot H_{2}^{2}+H_{2}^{3}=8L_{1}^{3}+12\cdot 2+6\cdot 1+0=8L_{1}^{3}+30.

Assume $r_{1}=4$ . Then $L_{1}^{3}=1$ and $(-K_{X})^{3}=8L_{1}^{3}+30=38$ . Moreover, (A) implies $\deg B_{1}=3$ . Hence the following hold (Lemma 5.35):

g(B_{1})=19-32+4\cdot 3+1=0.

Assume $r_{1}=3$ . Then $L_{1}^{3}=2$ and $(-K_{X})^{3}=8L_{1}^{3}+30=46$ . Moreover, (A) implies $\deg B_{1}=1$ . Hence the following hold (Lemma 5.35):

g(B_{1})=23-27+3\cdot 1+1=0.

∎

5.6. Case $E_{1}-D$

Lemma 5.39.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $D$ . Set $d_{2}:=(-K_{X})^{2}\cdot H_{2}$ . Then the following hold.

(A)

$12=\mu_{2}\left(\frac{24}{r_{1}}+\deg B_{1}\right)-d_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-\mu_{2})L_{1}^{3}=\frac{d_{2}}{\mu_{2}^{2}}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-\mu_{2})^{2}L_{1}^{3}-\deg B_{1}=0$ .

Proof.

The assertion (A) follows from Lemma 5.4 and Proposition 5.9(4). The assertion (C) holds by Lemma 5.11 and $H_{2}^{2}\equiv 0$ . The assertion (B) follows from Lemma 5.11 and

d_{2}=(-K_{X})^{2}\cdot H_{2}=(\mu_{2}H_{1}+H_{2})^{2}\cdot H_{2}=\mu_{2}^{2}H_{1}^{2}\cdot H_{2}.

∎

Proposition 5.40.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $D_{3}$ . Then $(-K_{X})^{3}=54,r_{1}=4,\deg B_{1}=1,g(B_{1})=0$ (No. 2-33).

Proof.

We have $d_{2}=(-K_{X})^{2}\cdot H_{2}=9$ . By Lemma 5.39, the following hold:

(A)

$7=\frac{24}{r_{1}}+\deg B_{1}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-3)L_{1}^{3}=1$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-3)^{2}L_{1}^{3}-\deg B_{1}=0$ .

By (B), we get $r_{1}=4$ and $L_{1}^{3}=1$ . Then (C) implies $\deg B_{1}=1$ . Hence the following holds:

(-K_{X})^{3}=(3H_{1}+H_{2})^{3}=27H_{1}^{3}+27H_{1}^{2}\cdot H_{2}+9H_{1}\cdot H_{2}^{2}+H_{3}^{3}=27+27+0+0=54.

∎

Proposition 5.41.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $D_{2}$ . Then one of the following holds.

(1)

$(-K_{X})^{3}=32,r_{1}=4,\deg B_{1}=4,g(B_{1})=1$ (No. 2-25).
(2)

$(-K_{X})^{3}=40,r_{1}=3,\deg B_{1}=2,g(B_{1})=0$ (No. 2-29).

Proof.

We have $d_{2}=8$ . By Lemma 5.39, the following hold:

(A)

$10=\frac{24}{r_{1}}+\deg B_{1}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-2)L_{1}^{3}=2$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-2)^{2}L_{1}^{3}-\deg B_{1}=0$ .

By (B), $r_{1}\geq 3$ . Moreover, we get

(-K_{X})^{3}=(2H_{1}+H_{2})^{3}=8H_{1}^{3}+12H_{1}^{2}\cdot H_{2}+6H_{1}\cdot H_{2}^{2}+H_{2}^{3}=8L_{1}^{3}+24.

Assume $r_{1}=4$ . We then get $L_{1}^{3}=1$ and $(-K_{X})^{3}=8+24=32$ . Moreover, (A) implies $\deg B_{1}=4$ . Hence the following hold (Lemma 5.35):

g(B_{1})=16-32+4\cdot 4+1=1.

Assume $r_{1}=3$ . We then get $L_{1}^{3}=2$ and $(-K_{X})^{3}=16+24=40$ . Moreover, (A) implies $\deg B_{1}=2$ . In particular, $g(B_{1})=0$ . ∎

Proposition 5.42.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $D_{1}$ . Then one of the following holds.

(1)

$(-K_{X})^{3}=10,r_{1}=4,\deg B_{1}=9,g(B_{1})=10,(-K_{X})^{2}\cdot H_{2}=3$ (No. 2-4).
(2)

$(-K_{X})^{3}=14,r_{1}=3,\deg B_{1}=8,g(B_{1})=5,(-K_{X})^{2}\cdot H_{2}=4$ (No. 2-7).
(3)

$(-K_{X})^{3}=4,r_{1}=2,(-K_{Y_{1}})^{3}=8,\deg B_{1}=1,g(B_{1})=1,(-K_{X})^{2}\cdot H_{2}=1$ (No. 2-1).
(4)

$(-K_{X})^{3}=8,r_{1}=2,(-K_{Y_{1}})^{3}=16,\deg B_{1}=2,g(B_{1})=1,(-K_{X})^{2}\cdot H_{2}=2$ (No. 2-3).
(5)

$(-K_{X})^{3}=12,r_{1}=2,(-K_{Y_{1}})^{3}=24,\deg B_{1}=3,g(B_{1})=1,(-K_{X})^{2}\cdot H_{2}=3$ (No. 2-5).
(6)

$(-K_{X})^{3}=16,r_{1}=2,(-K_{Y_{1}})^{3}=32,\deg B_{1}=4,g(B_{1})=1,(-K_{X})^{2}\cdot H_{2}=4$ (No. 2-10).
(7)

$(-K_{X})^{3}=20,r_{1}=2,(-K_{Y_{1}})^{3}=40,\deg B_{1}=5,g(B_{1})=1,(-K_{X})^{2}\cdot H_{2}=5$ (No. 2-14).

Proof.

By Lemma 5.39, the following hold:

(A)

$12+d_{2}=\frac{24}{r_{1}}+\deg B_{1}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-1)L_{1}^{3}=d_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{1}-1)^{2}L_{1}^{3}-\deg B_{1}=0$ .

It holds that

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=H_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+H_{3}^{3}=L_{1}^{3}+3d_{2}.

Assume $r_{1}=4$ , and hence $L_{1}^{3}=1$ . Then (B) and (C) imply $d_{2}=3$ and $\deg B_{1}=9$ , respectively. In particular, $(-K_{X})^{3}=1+3\cdot 3=10$ . Hence the following hold (Lemma 5.35):

g(B_{1})=5-32+4\cdot 9+1=10.

Assume $r_{1}=3$ , and hence $L_{1}^{3}=2$ . Then (B) and (C) imply $d_{2}=4$ and $\deg B_{1}=8$ , respectively. In particular, $(-K_{X})^{3}=2+3\cdot 4=14$ . Hence the following hold (Lemma 5.35):

g(B_{1})=7-27+3\cdot 8+1=5.

Assume $r_{1}=2$ , and hence $1\leq L_{1}^{3}\leq 5$ . Then

(A)

$d_{2}=\deg B_{1}$ .
(B)

$H_{1}^{2}\cdot H_{2}=L_{1}^{3}=d_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=L_{1}^{3}-\deg B_{1}=0$ .

We get $(-K_{X})^{3}=L_{1}^{3}+3d_{2}=4L_{1}^{3}$ . Then the following hold (Lemma 5.35):

g(B_{1})=2L_{1}^{3}-4L_{1}^{3}+2\cdot L_{1}^{3}+1=1.

∎

Lemma 5.43.

Let $\sigma:X\to Y$ be a blowup along a smooth curve $\Gamma$ on a Fano threefold $Y$ . Let $\pi:X\to\mathbb{P}^{1}$ be a morphism with $\pi_{*}\mathcal{O}_{X}=\mathcal{O}_{\mathbb{P}^{1}}$ . Take a Cartier divisor $D$ on $Y$ . Assume that (i) and (ii) hold.

(i)

$(-K_{Y})\cdot D^{2}=(-K_{Y})\cdot\Gamma$ .
(ii)

$D\sim\sigma_{*}F$ for a fibre $F$ of $\pi:X\to\mathbb{P}^{1}$ .

Then $\Gamma$ is a complete intersection of two members of $|D|$ .

Proof.

Take two fibres $F_{1}$ and $F_{2}$ of $\pi:X\to\mathbb{P}^{1}$ . Set $D_{1}:=\sigma_{*}F_{1}$ and $D_{2}:=\sigma_{*}F_{2}$ . For each $i\in\{1,2\}$ , we get $D_{i}=\sigma_{*}F_{i}\sim\sigma_{*}F\overset{{\rm(ii)}}{\sim}D$ . We then obtain $\Gamma\supset(D_{1}\cap D_{2})_{{\operatorname{red}}}$ . Since $D_{1}\cap D_{2}$ is of pure one-dimensional, we get $(D_{1}\cap D_{2})_{{\operatorname{red}}}=\Gamma$ or $(D_{1}\cap D_{2})_{{\operatorname{red}}}=\emptyset$ . We then obtain $(D_{1}\cap D_{2})_{{\operatorname{red}}}=\Gamma$ , as otherwise the equality $(D_{1}\cap D_{2})_{{\operatorname{red}}}=\emptyset$ would lead to the following contradiction:

0=(-K_{Y})\cdot(D_{1}\cap D_{2})=(-K_{Y})\cdot D_{1}\cdot D_{2}=(-K_{Y})\cdot D^{2}\overset{{\rm(i)}}{=}(-K_{Y})\cdot\Gamma>0.

It is enough to show the scheme-theoretic equality $D_{1}\cap D_{2}=\Gamma$ . Recall that we have the equality $(D_{1}\cap D_{2})_{{\operatorname{red}}}=\Gamma$ . Since $D_{1}\cap D_{2}$ is Cohen-Macaulay, it suffices to prove that $D_{1}\cap D_{2}$ is $R_{0}$ , i.e., $\mathcal{O}_{D_{1}\cap D_{2},\xi}$ is reduced for the generic point $\xi$ of $D_{1}\cap D_{2}$ . This follows from $(-K_{Y})\cdot(D_{1}\cap D_{2})=-K_{Y}\cdot\Gamma$ , because the effective $1$ -cycle associated with $D_{1}\cap D_{2}$ is determined by its length at the generic point [Bad01, Lemma 1.18]. ∎

Proposition 5.44.

Let $X$ be a Fano threefold with $\rho(X)=2$ which has extremal rays of type $D$ and $E_{1}$ . Let $\sigma:X\to Y$ be the contraction of the extremal ray of type $E_{1}$ . Then the blowup centre $\sigma({\operatorname{Ex}}(\sigma))$ of $\sigma$ is a complete intersection $D_{1}\cap D_{2}$ of two members $D_{1}$ and $D_{2}$ of $|D|$ for some Cartier divisor $D$ on $Y$ .

In the following proof, we use No. of $X$ given in Proposition 5.40, Proposition 5.41, and Proposition 5.42. In particular, No. of $X$ is one of 2-1, 2-3, 2-4, 2-5, 2-7, 2-10, 2-14, 2-25, 2-29, 2-33.

Proof.

Set $\Gamma:=\sigma({\operatorname{Ex}}(\sigma))$ . Let $\pi:X\to\mathbb{P}^{1}$ be the contraction of the extremal ray of type $D$ . Take a fibre $F$ of $\pi$ . Set $\mu$ to be the length of $\pi$ . Then $\pi:X\to\mathbb{P}^{1}$ is of type $D_{\mu}$ (Remark 3.4(2)). For each case, it is enough to find a Cartier divisor $D$ on $Y$ satisfying (i) and (ii) of Lemma 5.43. In what follows, we treat the following three cases separately.

(1)

2-4, 2-25, 2-33.
(2)

2-7, 2-29.
(3)

2-1, 2-3, 2-5, 2-10, 2-14.

(1) In these cases, we have $Y=\mathbb{P}^{3}$ and $\deg\Gamma=s^{2}$ for some $1\leq s\leq 3$ (Proposition 5.40, Proposition 5.41, Proposition 5.42). Let $D$ be a Cartier divisor satisfying $\mathcal{O}_{\mathbb{P}^{3}}(D)\simeq\mathcal{O}_{\mathbb{P}^{3}}(s)$ . Then Lemma 5.43(i) holds by $(-K_{Y})\cdot D^{2}=4s^{2}=(-K_{Y})\cdot\Gamma$ . It holds that $-K_{X}\sim\sigma^{*}\mathcal{O}_{\mathbb{P}^{3}}(\mu)+F$ (Proposition 5.9(3)). Hence $\sigma_{*}F\sim-K_{Y}-\mathcal{O}_{\mathbb{P}^{3}}(\mu)\sim\mathcal{O}_{\mathbb{P}^{3}}(4-\mu)$ . In order to prove Lemma 5.43(ii), it suffices to show $s+\mu=4$ . By using the fact that $\pi$ is of type $D_{\mu}$ , this can be checked for all the cases 2-4, 2-25, 2-33 (Proposition 5.40, Proposition 5.41, Proposition 5.42).

(2) In these cases, we have $Y=Q$ and $\deg\Gamma=2s^{2}$ for some $1\leq s\leq 2$ (Proposition 5.41, Proposition 5.42). Let $D$ be a Cartier divisor satisfying $\mathcal{O}_{Q}(D)\simeq\mathcal{O}_{Q}(s)$ . Then Lemma 5.43(i) holds by $(-K_{Y})\cdot D^{2}=6s^{2}=(-K_{Y})\cdot\Gamma$ . It holds that $-K_{X}\sim\sigma^{*}\mathcal{O}_{Q}(\mu)+F$ (Proposition 5.9(3)). Hence $\sigma_{*}F\sim-K_{Y}-\mathcal{O}_{Q}(\mu)\sim\mathcal{O}_{Q}(3-\mu)$ . In order to prove Lemma 5.43(ii), it suffices to show $s+\mu=3$ . By using the fact that $\pi$ is of type $D_{\mu}$ , this can be checked for both cases 2-7, 2-29 (Proposition 5.41, Proposition 5.42).

(3) In these cases, $Y$ is a Fano threefold of index $r_{Y}=2$ and $\Gamma$ is an elliptic curve such that $d=D^{3}$ and $1\leq d\leq 5$ for $d:=\deg\Gamma$ , where $D$ is a Cartier divisor satisfying $-K_{Y}\sim 2D$ (Proposition 5.42). Recall that the degree $\deg\Gamma$ is defined as $\frac{1}{2}(-K_{Y})\cdot\Gamma$ . Then Lemma 5.43(i) holds by $(-K_{Y})\cdot D^{2}=2D^{3}=2d=(-K_{Y})\cdot\Gamma$ . Since $\pi$ is of type $D_{1}$ for all the cases 2-1, 2-3, 2-5, 2-10, 2-14 (Proposition 5.42), it holds that $-K_{X}\sim\sigma^{*}(\mu D)+F\sim\sigma^{*}D+F$ (Proposition 5.9(3)). Hence $\sigma_{*}F\sim-K_{Y}-D\sim D$ . Thus Lemma 5.43(ii) holds. ∎

5.7. Case $E_{1}-E$

Lemma 5.45.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $E_{2},E_{3},$ or $E_{4}$ . Then the following hold.

(A)

$24=\mu_{2}(\frac{24}{r_{1}}+\deg B_{1})+\frac{24}{r_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(\frac{r_{2}-1}{\mu_{2}})^{2}L_{2}^{3}=(r_{1}-\mu_{2})L_{1}^{3}$ .
(C)

$H_{1}\cdot H_{2}^{2}=\frac{r_{2}-1}{\mu_{2}}L_{2}^{3}=(r_{1}-\mu_{2})^{2}L_{1}^{3}-\deg B_{1}$ .

Proof.

The assertion (A) follows from Lemma 5.4(3) and Proposition 5.9(4). For each $i\in\{1,2\}$ , let $a_{i}$ be the positive integer satisfying

K_{X}\sim f_{i}^{*}K_{Y_{i}}+a_{i}D_{i}.

Specifically, we set $a_{1}:=1$ and $a_{2}:=2$ (resp. $a_{2}:=1)$ if $R_{2}$ is of type $E_{2}$ (resp. $E_{3}$ or $E_{4}$ ) [TanII, Proposition 3.22]. For each $i\in\{1,2\}$ . we have

\mu_{2}H_{1}+H_{2}\sim-K_{X}=-f^{*}_{i}K_{X_{i}}-a_{i}D_{i}=r_{i}H_{i}-a_{i}D_{i}.

Then the assertion (B) holds by $H_{1}^{2}\cdot H_{2}=(r_{1}-\mu_{2})L_{1}^{3}$ (Lemma 5.11) and

\mu_{2}^{2}H_{1}^{2}\cdot H_{2}=((r_{2}-1)H_{2}-a_{2}D_{2})^{2}\cdot H_{2}=(r_{2}-1)^{2}L_{2}^{3}.

The assertion (C) follows from $H_{1}\cdot H_{2}^{2}=(r_{1}-\mu_{2})^{2}L_{1}^{3}-\deg B$ (Lemma 5.11) and

\mu_{2}H_{1}\cdot H_{2}^{2}=((r_{2}-1)H_{2}-a_{2}D_{2})\cdot H_{2}^{2}=(r_{2}-1)L_{2}^{3}.

∎

5.7.1. Case $E_{1}-E_{2}$

Proposition 5.46.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $E_{2}$ . Then $(-K_{X})^{3}=46,r_{1}=4,\deg B_{1}=2,g(B_{1})=0,r_{2}=3$ (No. 2-30).

Proof.

For each $i\in\{1,2\}$ , $Y_{i}$ is smooth and $r_{i}$ is the index of the Fano threefold $Y_{i}$ . In particular, $r_{i}\in\{1,2,3,4\}$ . We have $\mu_{2}=2$ . Hence Lemma 5.45 implies the following:

(A)

$24=\frac{48}{r_{1}}+2\deg B_{1}+\frac{24}{r_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(\frac{r_{2}-1}{2})^{2}L_{2}^{3}=(r_{1}-2)L_{1}^{3}$ .
(C)

$H_{1}\cdot H_{2}^{2}=\frac{r_{2}-1}{2}L_{2}^{3}=(r_{1}-2)^{2}L_{1}^{3}-\deg B_{1}$ .

Since each of $H_{1}$ and $H_{2}$ is nef and big, we have $H_{1}^{2}\cdot H_{2}>0$ and $H_{1}\cdot H_{2}^{2}>0$ . By (B) and (C), we obtain $r_{1}>2$ and $r_{2}>1$ . In particular, $r_{1}\in\{3,4\}$ and $r_{2}\in\{2,3,4\}$ .

Assume $r_{1}=4$ , and hence $L_{1}^{3}=1$ . By (B) and (C), the following hold:

r_{2}=3,\quad L_{2}^{3}=2,\quad H_{1}^{2}\cdot H_{2}=2,\quad H_{1}\cdot H_{2}^{2}=2,\quad\deg B_{1}=2.

Then $g(B_{1})=0$ and

(-K_{X})^{3}=(2H_{1}+H_{2})^{3}=8+24+12+2=46.

Assume $r_{1}=3$ , and hence $L_{1}^{3}=2$ . By (B) and (C), the following hold:

r_{2}=3,\quad L_{2}^{3}=2,\quad H_{1}^{2}\cdot H_{2}=2,\quad H_{1}\cdot H_{2}^{2}=2,\quad\deg B_{1}=0.

This is absurd. ∎

5.7.2. Cases $E_{1}-E_{3}$ and $E_{1}-E_{4}$

Proposition 5.47.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $E_{3}$ or $E_{4}$ . Then one of the following holds.

(1)

$(-K_{X})^{3}=22,r_{1}=4,\deg B_{1}=6,g(B_{1})=4,r_{2}=2,L_{2}^{3}=3$ (No. 2-15).
(2)

$(-K_{X})^{3}=30,r_{1}=3,\deg B_{1}=4,g(B_{1})=1,r_{2}=2,L_{2}^{3}=4$ (No. 2-23).

Proof.

Note that $Y_{2}$ is not smooth. We have $\mu_{2}=1$ . Hence Lemma 5.45 implies the following:

(A)

$24=\frac{24}{r_{1}}+\deg B_{1}+\frac{24}{r_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{2}-1)^{2}L_{2}^{3}=(r_{1}-1)L_{1}^{3}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{2}-1)L_{2}^{3}=(r_{1}-1)^{2}L_{1}^{3}-\deg B_{1}$ .

Since each of $H_{1}$ and $H_{2}$ is nef and big, we have $H_{1}^{2}\cdot H_{2}>0$ and $H_{1}\cdot H_{2}^{2}>0$ . By (B) and (C), we obtain $r_{1}\geq 2$ and $r_{2}\geq 2$ .

Assume $r_{1}=4$ , and hence $L_{1}^{3}=1$ . By (B) and (C), the following hold:

r_{2}=2,\quad L_{2}^{3}=3,\quad H_{1}^{2}\cdot H_{2}=3,\quad H_{1}\cdot H_{2}^{2}=3,\quad\deg B_{1}=6.

It follows from Lemma 5.35 that

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=1+9+9+3=22,

g(B_{1})=11-32+24+1=4.

Assume $r_{1}=3$ , and hence $L_{1}^{3}=2$ . By (B), we get $(r_{2}-1)^{2}L_{2}^{3}=4$ . Thus $(r_{2},L_{2}^{3})\in\{(2,4),(3,1)\}$ . Suppose that $(r_{2},L_{2}^{3})=(3,1)$ . Then (B) and (C) imply $H_{1}^{2}\cdot H_{2}\in 2\mathbb{Z}$ and $H_{1}\cdot H_{2}^{2}\in 2\mathbb{Z}$ . We get the following contradiction: $(-K_{X})^{3}=(H_{1}+H_{2})^{3}\equiv L_{1}^{3}+L_{2}^{3}\not\equiv 0\mod 2$ . Thus $(r_{2},L_{2}^{3})=(2,4)$ . Then (B) and (C) imply the following:

H_{1}^{2}\cdot H_{2}=4,\quad H_{1}\cdot H_{2}^{2}=4,\quad\deg B_{1}=4.

Then the following hold (Lemma 5.35):

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=2+12+12+4=30,

g(B_{1})=15-27+12+1=1.

Assume $r_{1}=2$ , and hence $1\leq L_{1}^{3}\leq 5$ . Then (A)-(C) can be rewritten as follows:

(A)

$12=\deg B_{1}+\frac{24}{r_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{2}-1)^{2}L_{2}^{3}=L_{1}^{3}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{2}-1)L_{2}^{3}=L_{1}^{3}-\deg B_{1}$ .

By $1\leq L_{1}^{3}\leq 5$ and (B), we obtain $r_{2}\in\{2,3\}$ . Since the case $r_{2}=2$ contradicts (A), we get $r_{2}=3$ . Then (B) and $1\leq L_{1}^{3}\leq 5$ imply $L_{2}^{3}=1$ and $L_{1}^{3}=4$ . By (B) and (C), we get $H_{1}^{2}\cdot H_{2}\in 2\mathbb{Z}$ and $H_{1}\cdot H_{2}^{2}\in 2\mathbb{Z}$ , respectively. We then obtain the following contradiction: $(-K_{X})^{3}=(H_{1}+H_{2})^{3}\equiv L_{1}^{3}+L_{2}^{3}\not\equiv 0\mod 2$ . ∎

5.7.3. Case $E_{1}-E_{5}$

Proposition 5.48.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $E_{5}$ . Then $(-K_{X})^{3}=40,r_{1}=4,\deg B_{1}=3,g(B_{1})=1,r_{2}=3,L_{2}^{3}=12$ (No. 2-28).

Proof.

By Lemma 5.4, we obtain $\mu_{2}=1$ and $H_{2}\cdot c_{2}(X)=45/r_{2}$ . It holds that

H_{1}+H_{2}\sim-K_{X}=-f^{*}K_{X_{1}}-D_{1}=r_{1}H_{1}-D_{1},

H_{1}+H_{2}\sim-K_{X}=-f^{*}K_{X_{2}}-\frac{1}{2}D_{2}=\frac{r_{2}}{2}H_{2}-\frac{1}{2}D_{2}.

We then get the following:

H_{1}^{2}\cdot H_{2}=H_{1}^{2}\cdot((r_{1}-1)H_{1}-D_{1})=(r_{1}-1)L_{1}^{3}.

H_{1}^{2}\cdot H_{2}=((\frac{r_{2}}{2}-1)H_{2}-\frac{1}{2}D_{2})^{2}\cdot H_{2}=(\frac{r_{2}}{2}-1)^{2}L_{2}^{3}.

H_{1}\cdot H_{2}^{2}=((\frac{r_{2}}{2}-1)H_{2}-\frac{1}{2}D_{2})\cdot H_{2}^{2}=(\frac{r_{2}}{2}-1)L_{2}^{3}.

H_{1}\cdot H_{2}^{2}=H_{1}\cdot((r_{1}-1)H_{1}-D_{1})^{2}=(r_{1}-1)^{2}L_{1}^{3}+H_{1}\cdot D_{1}^{2}=(r_{1}-1)^{2}L_{1}^{3}-\deg B.

Therefore, the following hold, where (A) is guaranteed by Lemma 5.4(3) and Proposition 5.9(4):

(A)

$24=\frac{24}{r_{1}}+\deg B_{1}+\frac{45}{r_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(\frac{r_{2}}{2}-1)^{2}L_{2}^{3}=(r_{1}-1)L_{1}^{3}$ . In particular, $L_{2}^{3}=\frac{4(r_{1}-1)}{(r_{2}-2)^{2}}L_{1}^{3}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(\frac{r_{2}}{2}-1)L_{2}^{3}=(r_{1}-1)^{2}L_{1}^{3}-\deg B_{1}$ .

By (A), we have $r_{1}\geq 2$ and $r_{2}\geq 3$ .

Assume $r_{1}=4$ , and hence $L_{1}^{3}=1$ . By (B), we get $\mathbb{Z}\ni L_{2}^{3}=\frac{4(r_{1}-1)}{(r_{2}-2)^{2}}L_{1}^{3}=\frac{12}{(r_{2}-2)^{2}}$ . Hence $r_{2}\in\{3,4\}$ . Since $r_{2}\neq 4$ by (A), we obtain $r_{2}=3$ . Then (A)-(C) imply the following:

L_{2}^{3}=12,\quad\deg B_{1}=3,\quad H_{1}^{2}\cdot H_{2}=3,\quad H_{1}\cdot H_{2}^{2}=6.

Then the following hold (Lemma 5.35):

(-K_{X})^{3}=(H_{1}+H_{2})^{3}=1+9+18+12=40

g(B_{1})=20-32+12+1=1.

Assume $r_{1}=3$ , and hence $L_{1}^{3}=2$ . By (A), we have $r_{2}\in\{3,5,9,15,45\}$ . This, together with $\mathbb{Z}_{>0}\ni L_{2}^{3}=\frac{4(r_{1}-1)}{(r_{2}-2)^{2}}L_{1}^{3}=\frac{16}{(r_{2}-2)^{2}}$ , implies $r_{2}=3$ and $L_{2}^{3}=16$ . By (C), we get the following contradiction:

\deg B_{1}=(r_{1}-1)^{2}L_{1}^{3}-(\frac{r_{2}}{2}-1)L_{2}^{3}=8-8=0.

Assume $r_{1}=2$ , and hence $1\leq L_{1}^{3}\leq 5$ . Then (A)-(C) can be rewritten as follows:

(A)

$12=\deg B_{1}+\frac{45}{r_{2}}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(\frac{r_{2}}{2}-1)^{2}L_{2}^{3}=L_{1}^{3}$ . In particular, $L_{2}^{3}=\frac{4}{(r_{2}-2)^{2}}L_{1}^{3}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(\frac{r_{2}}{2}-1)L_{2}^{3}=L_{1}^{3}-\deg B_{1}$ .

By (A), we have $r_{2}\in\{5,9,15,45\}$ . However, this contradicts $\mathbb{Z}_{>0}\ni L_{2}^{3}=\frac{4}{(r_{2}-2)^{2}}L_{1}^{3}$ and $1\leq L_{1}^{3}\leq 5$ . ∎

5.7.4. Case $E_{1}-E_{1}$

Proposition 5.49.

We use Notation 5.1. Assume that $R_{1}$ is of type $E_{1}$ and $R_{2}$ is of type $E_{1}$ . Moreover, suppose $r_{1}\geq r_{2}$ . Then one of the following holds possibly after permuting $R_{1}$ and $R_{2}$ .

(1)

$(-K_{X})^{3}=20,Y_{1}\simeq Y_{2}\simeq\mathbb{P}^{3},g(B_{1})=g(B_{2})=3,\deg B_{1}=\deg B_{2}=6$ (No. 2-12).
(2)

$(-K_{X})^{3}=24,Y_{1}\simeq\mathbb{P}^{3},g(B_{1})=1,\deg B_{1}=5,Y_{2}\simeq Q,g(B_{2})=1,\deg B_{2}=5$ (No. 2-17).
(3)

$(-K_{X})^{3}=26,Y_{1}\simeq\mathbb{P}^{3},g(B_{1})=2,\deg B_{1}=5,r_{2}=2,L_{2}^{3}=4,g(B_{2})=0,\deg B_{2}=1$ (No. 2-19).
(4)

$(-K_{X})^{3}=30,Y_{1}\simeq\mathbb{P}^{3},g(B_{1})=0,\deg B_{1}=4,r_{2}=2,L_{2}^{3}=5,g(B_{2})=0,\deg B_{2}=2$ (No. 2-22).
(5)

$(-K_{X})^{3}=28,Y_{1}\simeq Y_{2}\simeq Q,g(B_{1})=g(B_{2})=0,\deg B_{1}=\deg B_{2}=4$ (No. 2-21).
(6)

$(-K_{X})^{3}=34,Y_{1}\simeq Q,g(B_{1})=0,\deg B_{1}=3,r_{2}=2,L_{2}^{3}=5,g(B_{2})=0,\deg B_{2}=1$ (No. 2-26).

Proof.

We may apply Lemma 5.11 for both the extremal rays $R_{1}$ and $R_{2}$ , although we need to permute the indices when we apply it for $R_{2}$ . Then the following hold (Lemma 5.4(3), Proposition 5.9(4)):

(A)

$24=\frac{24}{r_{1}}+\deg B_{1}+\frac{24}{r_{2}}+\deg B_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}=(r_{1}-1)L_{1}^{3}=(r_{2}-1)^{2}L_{2}^{3}-\deg B_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=(r_{2}-1)L_{2}^{3}=(r_{1}-1)^{2}L_{1}^{3}-\deg B_{1}$ .

Since each of $H_{1}$ and $H_{2}$ is nef and big, we have $H_{1}^{2}\cdot H_{2}>0$ and $H_{1}\cdot H_{2}^{2}>0$ . By (B) and (C), we get $r_{1}\geq 2$ and $r_{2}\geq 2$ , respectively. By $r_{1}\geq r_{2}$ and (A), we obtain $r_{1}\geq 3$ . In particular,

(r_{1},r_{2})\in\{(4,4),(4,3),(4,2),(3,3),(3,2)\}.

In what follows, we shall use

•

$(-K_{X})^{3}=(H_{1}+H_{2})^{3}=L_{1}^{3}+3H_{1}^{2}\cdot H_{2}+3H_{1}\cdot H_{2}^{2}+L_{2}^{3}$ , and
•

$g(B_{i})=\frac{(-K_{X})^{3}}{2}-\frac{(-K_{Y_{i}})^{3}}{2}+r_{i}\deg B_{i}+1$ (Lemma 5.35).

(1) Assume $(r_{1},r_{2})=(4,4)$ . Then $(L_{1}^{3},L_{2}^{3})=(1,1)$ and the following hold:

(A)

$12=\deg B_{1}+\deg B_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}=3=9-\deg B_{1}$ .
(C)

$H_{1}\cdot H_{2}^{2}=3=9-\deg B_{2}$ .

Hence $\deg B_{1}=\deg B_{2}=6$ ,

(-K_{X})^{3}=1+9+9+1=20,\qquad g(B_{1})=g(B_{2})=10-32+24+1=3.

(2) Assume $(r_{1},r_{2})=(4,3)$ . Then $(L_{1}^{3},L_{2}^{3})=(1,2)$ and the following hold:

(A)

$10=\deg B_{1}+\deg B_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}^{2}=3=8-\deg B_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=4=9-\deg B_{1}$ .

Hence $(\deg B_{1},\deg B_{2},H_{1}^{2}\cdot H_{2},H_{1}\cdot H_{2}^{2})=(5,5,3,4)$ ,

(-K_{X})^{3}=1+9+12+2=24,

g(B_{1})=12-32+20+1=1,\qquad g(B_{2})=12-27+15+1=1.

(3), (4) Assume $(r_{1},r_{2})=(4,2)$ . Then $L_{1}^{3}=1$ , $1\leq L_{2}^{3}\leq 5$ , and the following hold:

(A)

$6=\deg B_{1}+\deg B_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}^{2}=3=L_{2}^{3}-\deg B_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=L_{2}^{3}=9-\deg B_{1}$ .

By $1\leq L_{2}^{3}\leq 5$ , $L_{2}^{3}-\deg B_{2}=3$ , and $\deg B_{2}>0$ , we have two solutions: $(\deg B_{2},L_{2}^{3})\in\{(1,4),(2,5)\}$ .

Assume $(\deg B_{2},L_{2}^{3})=(1,4)$ . Then we get $(\deg B_{1},H_{1}^{2}\cdot H_{2},H_{1}\cdot H_{2}^{2})=(5,3,4)$ . Then

(-K_{X})^{3}=1+9+12+4=26,

g(B_{1})=13-32+20+1=2,\qquad g(B_{2})=0.

Assume $(\deg B_{2},L_{2}^{3})=(2,5)$ . Then we get $(\deg B_{1},H_{1}^{2}\cdot H_{2},H_{1}\cdot H_{2}^{2})=(4,3,5)$ . Then

(-K_{X})^{3}=1+9+15+5=30,

g(B_{1})=15-32+16+1=0,\qquad g(B_{2})=0.

(5) Assume $(r_{1},r_{2})=(3,3)$ . Then $L_{1}^{3}=L_{2}^{3}=2$ , and the following hold:

(A)

$8=\deg B_{1}+\deg B_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}^{2}=4=8-\deg B_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=4=8-\deg B_{1}$ .

Hence $\deg B_{1}=\deg B_{2}=H_{1}^{2}\cdot H_{2}=H_{1}\cdot H_{2}^{2}=4$ ,

(-K_{X})^{3}=2+12+12+2=28,\qquad g(B_{1})=g(B_{2})=14-27+12+1=0.

(6) Assume $(r_{1},r_{2})=(3,2)$ . Then $L_{1}^{3}=2$ , $1\leq L_{2}^{3}\leq 5$ , and the following hold:

(A)

$4=\deg B_{1}+\deg B_{2}$ .
(B)

$H_{1}^{2}\cdot H_{2}^{2}=4=L_{2}^{3}-\deg B_{2}$ .
(C)

$H_{1}\cdot H_{2}^{2}=L_{2}^{3}=8-\deg B_{1}$ .

By $1\leq L_{2}^{3}\leq 5$ , $L_{2}^{3}-\deg B_{2}=4$ , and $\deg B_{2}>0$ , we obtain $\deg B_{2}=1$ and $L_{2}^{3}=5$ . Hence we get $(\deg B_{1},H_{1}^{2}\cdot H_{2},H_{1}\cdot H_{2}^{2})=(3,4,5)$ . Then

(-K_{X})^{3}=2+12+15+5=34,

g(B_{1})=17-27+9+1=0,\qquad g(B_{2})=17-20+2+1=0.

∎

6. Primitive Fano threefolds with $\rho(X)=3$

In this section, we classify primitive Fano threefolds with $\rho(X)=3$ . The goal of this section is to prove the following theorem.

Theorem 6.1 (Theorem 6.7, Theorem 6.17).

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Then one and only one of the following holds.

(1)

$(-K_{X})^{3}=12$ and there exists a split double cover $f:X\to\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ such that $f_{*}\mathcal{O}_{X}/\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1,1)^{-1}$ .
(2)

$(-K_{X})^{3}=14$ and $X$ is isomorphic to a prime divisor $X^{\prime}$ on $P:=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})$ such that $\mathcal{O}_{P}(X^{\prime})\simeq\mathcal{O}_{P}(2)\otimes\pi^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3)$ , where $\pi:P=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ denotes the natural projection.
(3)

$(-K_{X})^{3}=48$ and $X\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ .
(4)

$(-K_{X})^{3}=52$ and $X\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))$ .

By Theorem 4.17, there are the following two cases, which we shall treat separately:

•

All the extremal rays are of type $C$ (Subsection 6.1).
•

There exist extremal rays $R_{1}$ and $R_{2}$ such that $R_{1}$ is of type $C$ and $R_{2}$ is of type $E_{1}$ (Subsection 6.2).

6.1. Case $C-C-C$

Lemma 6.2.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Assume that all the extremal rays are of type $C$ . Let $F$ be a two-dimensional extremral face generated by extremal rays $R_{1}$ and $R_{2}$ . Let $f_{1}:X\to Y_{1}$ , $f_{2}:X\to Y_{2}$ , and $\pi:X\to B$ be the contractions of $R_{1}$ , $R_{2}$ , and $F$ , respectively. Then we have the following commutative diagram

such that the following hold.

(1)

$Y_{1}=B\times C_{1}$ .
(2)

$Y_{2}=B\times C_{2}$ .
(3)

$B=C_{1}=C_{2}=\mathbb{P}^{1}$ .
(4)

$Y_{1}\to B$ and $Y_{2}\to B$ are the first projections.

Proof.

Each extremal ray $R$ is an intersection of two two-dimensional extremal faces $F$ and $F^{\prime}$ . This implies that there are exactly two non-trivial contractions $X\to B_{1}$ and $X\to B_{2}$ that factor through the contraction $X\to Y$ of $R$ . By $Y=\mathbb{P}^{1}\times\mathbb{P}^{1}$ , these must coinside the ones induced by the projections. ∎

Remark 6.3.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Assume that all the extremal rays are of type $C$ . Let $F$ be an extremal face of ${\operatorname{NE}}(X)$ and let $h:X\to Z$ be the contraction of $F$ . Then the following hold.

(1)
The following are equivalent.
- •
  
  $\dim F=1$ , i.e., $F$ is an extremal ray.
- •
  
  $\dim Z=2$ .
- •
  
  $Z\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ .
(2)
The following are equivalent.
- •
  
  $\dim F=2$ .
- •
  
  $\dim Z=1$ .
- •
  
  $Z\simeq\mathbb{P}^{1}$ .

Proposition 6.4.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Assume that all the extremal rays are of type $C$ . Then the number of the extremal rays of $X$ is three.

Proof.

Suppose that $X$ has at least four extrmemal rays. Then we can find two two-dimensional extremal faces $F$ and $F^{\prime}$ such that $F\cap F^{\prime}=\{0\}$ . Let $\pi:X\to B$ and $\pi^{\prime}:X\to B^{\prime}$ be the contractions of $F$ and $F^{\prime}$ , respectively. By Remark 6.3, we get $\dim B=\dim B^{\prime}=1$ . Then there exists a curve $C$ on $X$ such that $(\pi\times\pi^{\prime})(C)$ is a point for $\pi\times\pi^{\prime}:X\to B\times B^{\prime}$ . This implies that $\pi(C)$ and $\pi^{\prime}(C)$ are points, i.e., $[C]\in F\cap F^{\prime}=\{0\}$ , which is absurd. ∎

Notation 6.5.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Assume that all the extremal rays are of type $C$ . Then there exist exactly three two-dimensional extremal faces $F_{1},F_{2},F_{3}$ (Proposition 6.4). We have also exactly three extremal rays $R_{12},R_{23},R_{13}$ , which are given as follows:

R_{12}:=F_{1}\cap F_{2},\qquad R_{23}:=F_{2}\cap F_{3},\qquad R_{13}:=F_{1}\cap F_{3}.

Corresponding to these extremal faces, we have the following contraction morphisms:

•

$\pi^{X}_{i}:X\to B_{i}:=\mathbb{P}^{1}$ for all $1\leq i\leq 3$ .
•

$f_{ij}:X\to Y_{ij}:=B_{i}\times B_{j}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ for all $1\leq i<j\leq 3$ .

Note that $f_{ij}$ and $\pi_{i}$ are compatible, i.e.,

\pi^{X}_{i}:X\xrightarrow{f_{ij}}Y_{ij}=B_{i}\times B_{j}\xrightarrow{{\rm pr}_{1}}B_{i},\qquad\pi^{X}_{j}:X\xrightarrow{f_{ij}}Y_{ij}=B_{i}\times B_{j}\xrightarrow{{\rm pr}_{2}}B_{j}.

Set

f:=\pi_{1}^{X}\times\pi_{2}^{X}\times\pi_{3}^{X}:X\to Z:=B_{1}\times B_{2}\times B_{3}=\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}.

Let $\pi_{i}^{Z}:Z\to B_{i}=\mathbb{P}^{1}$ be the $i$ -th projection. For each $i\in\{1,2,3\}$ , we set

•

$H_{i}:=(\pi^{X}_{i})^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)$ , and
•

$H^{Z}_{i}:=(\pi^{Z}_{i})^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)$ .

For $1\leq i<j\leq 3$ , let $\ell_{ij}$ be an extremal rational curve of the extremral ray $R_{ij}$ . Note that we have $H_{i}\cdot\ell_{ij}=H_{j}\cdot\ell_{ij}=0$ .

Lemma 6.6.

We use Notation 6.5. Then $f:X\to Z=\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ is a finite surjective morphism.

Proof.

It suffices to show that $f$ is a finite morphism. Suppose that $f$ is not a finite morphism. Then there exists a curve $C$ on $X$ such that $f(C)$ is a point. By $f=\pi^{X}_{1}\times\pi^{X}_{2}\times\pi^{X}_{3}$ , all of $\pi^{X}_{1}(C),\pi^{X}_{2}(C)$ , and $\pi^{X}_{3}(C)$ are points. Then $[C]\in F_{1}\cap F_{2}\cap F_{3}=\{0\}$ , which is a contradiction. ∎

Theorem 6.7.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Assume that all the extremal rays of ${\operatorname{NE}}(X)$ are of type $C$ . Let $R_{1}$ and $R_{2}$ be distinct extremal rays of ${\operatorname{NE}}(X)$ . Then one and only one of the following holds.

(1)

There exists a split double cover $f:X\to\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ such that $f_{*}\mathcal{O}_{X}/\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1,1)^{-1}$ . Furthermore, $\mathcal{O}_{X}(-K_{X})\simeq f^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1,1)$ , $(-K_{X})^{3}=12$ , there are exactly three extremal rays of ${\operatorname{NE}}(X)$ , all the extremal rays are of type $C_{1}$ , and $\Delta_{\varphi}$ is of bidegree $(4,4)$ for the contraction $\varphi:X\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ of an arbitrary extremal ray.
(2)

$X\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ . Furthermore, $(-K_{X})^{3}=48$ , there are exactly three extremal rays $R_{1},R_{2},R_{3}$ of ${\operatorname{NE}}(X)$ , and all the extremal rays are of type $C_{2}$ .

Proof.

We use Notation 6.5. Set $d:=\deg f$ . For each $i\in\{1,2,3\}$ , we have morphisms:

\pi^{X}_{i}:X\xrightarrow{f}Z=\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}\xrightarrow{\pi_{i}^{Z}}\mathbb{P}^{1}.

H_{1}^{Z}\cdot H_{2}^{Z}\cdot H_{3}^{Z}=(\pi_{1}^{Z})^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)\cdot(\pi_{2}^{Z})^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)\cdot(\pi_{3}^{Z})^{*}\mathcal{O}_{\mathbb{P}^{1}}(1)=1,

we have

d=\deg f=(f^{*}H_{1}^{Z})\cdot(f^{*}H_{2}^{Z})\cdot(f^{*}H_{3}^{Z})=H_{1}\cdot H_{2}\cdot H_{3}.

Since $H_{1}\cdot H_{2}$ is a fibre of $f_{12}:X\to Y_{12}=B_{1}\times B_{2}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ , we obtain $H_{1}\cdot H_{2}\equiv\frac{2}{\mu_{12}}\ell_{12}$ , where $\mu_{12}$ denotes the length of $R_{12}$ . Therefore, we get

d=H_{1}\cdot H_{2}\cdot H_{3}=\frac{2}{\mu_{12}}H_{3}\cdot\ell_{12},

which implies

-K_{X}\cdot\ell_{12}=\mu_{12}=\frac{2}{d}H_{3}\cdot\ell_{12}.

By symmetry, we have that $-dK_{X}\cdot\ell_{12}=2H_{3}\cdot\ell_{12},-dK_{X}\cdot\ell_{23}=2H_{1}\cdot\ell_{23},-dK_{X}\cdot\ell_{13}=2H_{2}\cdot\ell_{13}.$ By the exact sequence

0\to{\operatorname{Pic}}\,Y_{12}\to{\operatorname{Pic}}\,X\xrightarrow{\cdot\ell_{12}}\mathbb{Z},

$H_{1},H_{2},H_{3}$ form a $\mathbb{Q}$ -linear basis of $({\operatorname{Pic}}\,X)\otimes_{\mathbb{Z}}\mathbb{Q}$ . Hence we obtain

-dK_{X}\sim 2H_{1}+2H_{2}+2H_{3}.

For the positive integer $g\in\mathbb{Z}_{>0}$ satisying $(-K_{X})^{3}=2g-2$ , we get

d^{3}(2g-2)=d^{3}(-K_{X})^{3}=8(H_{1}+H_{2}+H_{3})^{3}=48\deg f=48d.

Hence $d^{2}(g-1)=24=2^{3}\cdot 3$ , which implies $d\in\{1,2\}$ .

Assume $d=1$ . Then $f:X\to Z=\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ is a finite birational morphism (Lemma 6.6). Since $Z$ is normal, $f$ is an isomorphism. We have $(-K_{X})^{3}=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,2,2)^{3}=48$ . Furthermore, each projection $X\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ is clearly of type $C_{2}$ . Hence (2) holds.

Assume $d=2$ . Then $-K_{X}\sim H_{1}+H_{2}+H_{3}$ and $f:X\to Z=\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ is a double cover (Lemma 6.6). This double cover is split (Lemma 2.4), because $\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ is $F$ -split and $H^{1}(\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(E))=0$ for any effective divisor $E$ on $\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ . We have $(-K_{X})^{3}=\deg f\cdot\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1,1)^{3}=12$ .

Let us show that $f_{12}:X\to B_{1}\times B_{2}$ is of type $C_{1}$ . We can write

\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\Delta_{f_{12}})=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(a,b)

for some $a,b\in\mathbb{Z}$ . We obtain

4=2d=2H_{1}\cdot H_{2}\cdot H_{3}=(-K_{X})^{2}\cdot H_{1}\overset{(\star)}{=}-4K_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\cdot\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)-\Delta_{f_{12}}\cdot\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)

=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(8-a,8-b)\cdot\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)=8-b.

where ( $\star$ ) follows from Proposition 3.16. Therefore, $b=4$ . Similarly, $a=4$ . Hence $f_{12}$ is of type $C_{1}$ . By symmetry, $f_{ij}:X\to B_{i}\times B_{j}$ is of type $C_{1}$ for all $1\leq i<j\leq 3$ . Thus (1) holds. ∎

Remark 6.8.

By the above proof, the split double cover $f:X\to\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ in Theorem 6.7(1) can be chosen to be $f$ as in Notation 6.5.

6.2. Case $C-E$

What is remaining is the following case (cf. Theorem 4.17):

Notation 6.9.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Let $R_{1}$ and $R_{2}$ be two distinct extremal rays. Assume that

•

$R_{1}$ is of type $C$ , and
•

$R_{2}$ is of type $E_{1}$ .

For each $i\in\{1,2\}$ , let

f_{i}:X\to Y_{i}

be the contraction of $R_{i}$ , where $Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ . Set $D:={\operatorname{Ex}}(f_{2})$ , which is a prime divisor on $X$ such that $D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$ (Theorem 4.8). Let $\ell_{i}$ be an extremal rational curve with $R_{i}=\mathbb{R}_{\geq 0}[\ell_{i}]$ . Set $\mu_{i}:=-K_{X}\cdot\ell_{i}$ , which is the length of $R_{i}$ .

Lemma 6.10.

We use Notation 6.9. For the induced morphism

f_{1}|_{D}\colon D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}\to\mathbb{P}^{1}\times\mathbb{P}^{1}=Y_{1},

the following hold.

(1)

If $R_{1}$ is of type $C_{1}$ , then $f_{1}|_{D}$ is a double cover.
(2)

If $R_{2}$ is of type $C_{2}$ , then $f_{1}|_{D}$ is an isomorphism.

Proof.

By Lemma 4.15(2), $f_{1}|_{D}:D\to Y_{1}$ is a finite surjective morphism. Fix a fibre $\zeta$ of $f_{1}:X\to Y_{1}$ over a closed point of $Y_{1}$ . By $\zeta\equiv\frac{2}{\mu_{1}}\ell_{1}$ (Corollary 4.5), it holds that

\deg(f_{1}|_{D})=D\cdot\zeta=\frac{2}{\mu_{1}}D\cdot\ell_{1}.

Recall that we have the following exact sequence (Theorem 4.11):

0\longrightarrow{\operatorname{Pic}}\,(\mathbb{P}^{1}\times\mathbb{P}^{1})\stackrel{{\scriptstyle f_{1}^{*}}}{{\to}}{\operatorname{Pic}}\,X\stackrel{{\scriptstyle\cdot\ell_{1}}}{{\to}}\mathbb{Z}\longrightarrow 0.

Fix a Cartier divisor $E$ on $X$ with $E\cdot\ell_{1}=1$ , whose existence is guaranteed by the surjectivity of the last map $E\mapsto E\cdot\ell_{1}$ . Then there exist $a\in\mathbb{Z}$ and a Cartier divisor $L$ on $Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ such that

D\sim f_{1}^{*}L+aE.

By $a=D\cdot\ell_{1}$ , we have $a\in\mathbb{Z}_{>0}$ . Since $L$ is a divisor on $\mathbb{P}^{1}\times\mathbb{P}^{1}$ , we have $L^{3}=0$ and $\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(L)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(b,c)$ for some $b,c\in\mathbb{Z}$ . We have $L^{2}\equiv 2bcQ$ for a closed point $Q$ on $\mathbb{P}^{1}\times\mathbb{P}^{1}$ , which implies $f_{1}^{*}L^{2}\equiv 2bc\zeta\equiv\frac{4bc}{\mu_{1}}\ell_{1}$ and $f_{1}^{*}L^{2}\cdot E=\frac{4bc}{\mu_{1}}$ . Therefore, we obtain

	$\displaystyle D^{3}$	$\displaystyle=(f^{*}_{1}L+aE)^{3}$
		$\displaystyle=f_{1}^{}L^{3}+3f_{1}^{}L^{2}\cdot(aE)+3f^{*}_{1}L\cdot(a^{2}E^{2})+a^{3}E^{3}$
		$\displaystyle=3a\cdot\frac{4bc}{\mu_{1}}+3a^{2}(f^{*}_{1}L\cdot E^{2})+a^{3}E^{3}.$

On the other hand, for $C:=f_{2}(D)$ , it holds that

D^{3}=\deg_{C}(\mathcal{N}^{*}_{C/Y_{2}})=\deg_{\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}}(1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(1))=2,

where the first and second equalities follow from Lemma 7.2(3) and Lemma 4.7(2), respectively. We then obtain

(6.10.1)

2=D^{3}=a\left(\frac{12bc}{\mu_{1}}+3a(f_{1}^{*}L\cdot E^{2})+a^{2}E^{3}\right).

By $a\in\mathbb{Z}_{>0},b,c\in\mathbb{Z}$ , and $\mu_{1}\in\{1,2\}$ , it holds that $a=1$ or $a=2$ . If $a=2$ , then the right hand side of (6.10.1) would be contained in $4\mathbb{Z}$ , which is absurd. Then we obtain $a=1$ , which implies $D\cdot\ell_{1}=1$ and $\deg(f_{1}|_{D})=\frac{2}{\mu_{1}}$ .

(1) Assume that $R_{1}$ is of type $C_{1}$ . Then we have $\deg(f_{1}|_{D})=2$ , i.e., $f_{1}|_{D}$ is a double cover. Hence (1) holds.

(2) Assume that $R_{1}$ is of type $C_{2}$ . Then we have $\deg(f_{1}|_{D})=1$ , i.e., $f_{1}|_{D}$ is a finite birational morphism. Since $f_{1}|_{D}:D=\mathbb{P}^{1}\times\mathbb{P}^{1}\to Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ is a finite birational morphism of normal varieties, $f_{1}|_{D}$ is an isomorphism. Thus (2) holds. ∎

6.2.1. Case $C_{2}-E_{1}$

Proposition 6.11.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Assume that there exist two extremal rays $R_{1}$ and $R_{2}$ such that $R_{1}$ is of type $C_{2}$ and $R_{2}$ is of type $E_{1}$ . Then

X\simeq\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))

and $(-K_{X})^{3}=52$ .

Proof.

We use Notation 6.9. Recall that we have $X\simeq\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{E})$ and $\mathcal{E}=(f_{1})_{*}\mathcal{O}_{X}(1)$ . Set $\mathcal{O}_{D}(D):=\mathcal{O}_{X}(D)|_{D}$ .

Step 1.

There is the following exact sequence:

0\to(f_{1})_{*}\mathcal{O}_{X}\to(f_{1})_{*}\mathcal{O}_{X}(D)\to(f_{1})_{*}\mathcal{O}_{D}(D)\to 0.

Proof of Step 1.

We have an exact sequence

0\to\mathcal{O}_{X}\to\mathcal{O}_{X}(D)\to\mathcal{O}_{D}(D)\to 0,

which induces another exact sequence

0\to(f_{1})_{*}\mathcal{O}_{X}\to(f_{1})_{*}\mathcal{O}_{X}(D)\to(f_{1})_{*}\mathcal{O}_{D}(D)\to R^{1}(f_{1})_{*}\mathcal{O}_{X}.

By Proposition 3.11, $R^{1}(f_{1})_{*}\mathcal{O}_{X}=0$ . This completes the proof of Step 1. ∎

Step 2.

The following hold.

(1)

$(f_{1})_{*}\mathcal{O}_{X}(D)\simeq\mathcal{E}\otimes\mathcal{L}$ for some $\mathcal{L}\in{\operatorname{Pic}}(\mathbb{P}^{1}\times\mathbb{P}^{1})$ .
(2)

$(f_{1})_{*}\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$ .

Proof of Step 2.

Let us show (1). Since $f_{1}:X\to Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ is a $\mathbb{P}^{1}$ -bundle, it holds that

\mathcal{O}_{X}(D)\simeq\mathcal{O}_{X}(n)\otimes f_{1}^{*}\mathcal{L}

for some $n\in\mathbb{Z}$ and $\mathcal{L}\in{\operatorname{Pic}}(\mathbb{P}^{1}\times\mathbb{P}^{1})$ . Fix a fibre $\zeta$ of $f_{1}:X\to Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ . Since $f_{1}|_{D}:D\to Y_{1}$ is an isomorphism (Lemma 6.10), we have $D\cdot\zeta=1$ . By $\mathcal{O}_{X}(1)\cdot\zeta=1$ , we obtain $n=1$ . It holds that

(f_{1})_{*}\mathcal{O}_{X}(D)\simeq(f_{1})_{*}(\mathcal{O}_{X}(1)\otimes f_{1}^{*}\mathcal{L})\simeq((f_{1})_{*}\mathcal{O}_{X}(1))\otimes\mathcal{L}\simeq\mathcal{E}\otimes\mathcal{L}.

Thus (1) holds.

Let us show (2). Since $f_{1}|_{D}:D\to Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ is an isomorphism, Lemma 4.7(4) implies $(f_{1})_{*}\mathcal{O}_{D}(D)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)$ . This completes the proof of Step 2. ∎

Step 3.

$X\simeq\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))$ .

Proof of Step 3.

By Step 1and Step 2, we have the following exact sequence for some $\mathcal{L}\in{\operatorname{Pic}}\,(\mathbb{P}^{1}\times\mathbb{P}^{1})$ :

0\to\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\to\mathcal{E}\otimes\mathcal{L}\to\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)\to 0.

It holds that

{\operatorname{Ext}}^{1}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1),\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}})\simeq H^{1}(\mathbb{P}^{1}\times\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))=0,

which implies

\mathcal{E}\otimes\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1).

Therefore,

X\simeq\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{E})\simeq\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{E}\otimes\mathcal{L}\otimes\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))\simeq\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1)).

This completes the proof of Step 3. ∎

Step 4.

$(-K_{X})^{3}=52$ .

Proof of Step 4.

By Proposition 7.1(2), we have

(-K_{X})^{3}=2c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))^{2}-8c_{2}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))+6K_{\mathbb{P}^{1}\times\mathbb{P}^{1}}^{2}.

By [Har77, Appendix A, §3, C3 and C5], we obtain

	$\displaystyle c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))$	$\displaystyle=c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}})+c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))=c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1)),$
	$\displaystyle c_{2}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))$	$\displaystyle=c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}})\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))=0.$

Hence it holds that $(-K_{X})^{3}=2\cdot 2-8\cdot 0+6\cdot 8=52.$ This completes the proof of Step 4. ∎

Step 3 and Step 4 complete the proof of Proposition 6.11 ∎

6.2.2. Case $C_{1}-E_{1}$

Notation 6.12.

We use Notation 6.9. Assume that $R_{1}$ is of type $C_{1}$ . In particular, $X$ is a primitive Fano threefold with $\rho(X)=3$ , $R_{1}$ is of type $C_{1}$ , and $R_{2}$ is of type $E_{1}$ . By Lemma 3.15, we have morphisms

where $\mathcal{E}^{\prime}:=(f_{1})_{*}\mathcal{O}_{X}(-K_{X})$ and $\iota^{\prime}$ is a closed immersion. We identify $X$ with the smooth prime divisor $\iota^{\prime}(X)$ on $P^{\prime}$ .

Lemma 6.13.

We use Notation 6.12. For the induced morphism

f_{1}|_{D}:D=\mathbb{P}^{1}\times\mathbb{P}^{1}\to Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1},

we set $\mathcal{L}:=((f_{1}|_{D})_{*}\mathcal{O}_{D}/\mathcal{O}_{Y_{1}})^{-1}$ , which is an invertible sheaf (cf. Remark 2.2, Lemma 6.10(1)). Then, after possibly permuting the direct product factors of $Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ , the following hold.

(1)

$\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(0,1)$ .
(2)

$f_{1}|_{D}$ is a split double cover.
(3)

There exists a double cover $h:\mathbb{P}^{1}\to\mathbb{P}^{1}$ such that $f_{1}|_{D}={\rm id}\times h:D=\mathbb{P}^{1}\times\mathbb{P}^{1}\to Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ .
(4)

$(f_{1}|_{D})_{*}\mathcal{O}_{D}(0,-1)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(0,-1)\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(0,-1)$ .

Proof.

Since $f_{1}|_{D}\colon D\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ is a double cover (Lemma 6.10(1)), Lemma 2.3 implies

\omega_{D}\simeq(f_{1}|_{D})^{*}(\omega_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\otimes\mathcal{L}).

We can write $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(a,b)$ for some $a,b\in\mathbb{Z}$ . We then get

(6.13.1)

\mathcal{O}_{D}(-2,-2)\simeq(f_{1}|_{D})^{*}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(a-2,b-2)).

Hence $a<2$ and $b<2$ . Furthermore, we get

	$\displaystyle 8$	$\displaystyle=(c_{1}(\mathcal{O}_{D}(-2,-2)))^{2}$
		$\displaystyle=\deg(f_{1}\|_{D})\cdot c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(a-2,b-2))^{2}$
		$\displaystyle=4(a-2)(b-2),$

i.e., $(a-2)(b-2)=2$ . Therefore, after possibly permuting the direct product factors of $Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ , we obtain $(a,b)=(0,1)$ and $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(0,1)$ . Thus (1) holds. Lemma 2.4 and Lemma 8.4 imply (2) and (3), respectively.

Let us show (4). It is easy to see that $h_{*}\mathcal{O}_{\mathbb{P}^{1}}(1)$ is a locally free sheaf of rank 2 on $\mathbb{P}^{1}$ . Hence we can write $h_{*}\mathcal{O}_{\mathbb{P}^{1}}(1)\simeq\mathcal{O}_{\mathbb{P}^{1}}(c)\oplus\mathcal{O}_{\mathbb{P}^{1}}(d)$ for some $c,d\in\mathbb{Z}$ . We obtain

	$\displaystyle h^{0}(\mathcal{O}_{\mathbb{P}^{1}}(c))+h^{0}(\mathcal{O}_{\mathbb{P}^{1}}(d))$	$\displaystyle=$	$\displaystyle h^{0}(\mathbb{P}^{1},h_{*}\mathcal{O}_{\mathbb{P}^{1}}(1))=h^{0}(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(1))=2$
	$\displaystyle h^{0}(\mathcal{O}_{\mathbb{P}^{1}}(c-1))+h^{0}(\mathcal{O}_{\mathbb{P}^{1}}(d-1))$	$\displaystyle=$	$\displaystyle h^{0}(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(-1)\otimes h_{*}\mathcal{O}_{\mathbb{P}^{1}}(1))=h^{0}(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(-1))=0.$

Therefore, we get $c=d=0$ and

h_{*}\mathcal{O}_{\mathbb{P}^{1}}(-1)\simeq\mathcal{O}_{\mathbb{P}^{1}}(-1)\otimes h_{*}\mathcal{O}_{\mathbb{P}^{1}}(1)\simeq\mathcal{O}_{\mathbb{P}^{1}}(-1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(-1).

This, together with the flat base change theorem, implies the following:

	$\displaystyle(f_{1}\|_{D})_{*}\mathcal{O}_{D}(0,-1)$	$\displaystyle\simeq({\rm id}\times h)_{*}\mathcal{O}_{D}(0,-1)$
		$\displaystyle\simeq({\rm id}\times h)_{}({\rm pr}_{2})^{}\mathcal{O}_{\mathbb{P}^{1}}(-1)$
		$\displaystyle\simeq({\rm pr}_{2})^{}h_{}\mathcal{O}_{\mathbb{P}^{1}}(-1)$
		$\displaystyle\simeq({\rm pr}_{2})^{*}(\mathcal{O}_{\mathbb{P}^{1}}(-1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(-1))$
		$\displaystyle\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(0,-1)\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(0,-1).$

Thus (4) holds. ∎

Lemma 6.14.

We use Notation 6.12. Then, after possibly permuting the direct product factors of $Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1}$ , the following holds:

(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1)\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}.

Proof.

Step 1.

The following hold.

(1)

$(-K_{X}-D)\cdot\zeta=0$ for a fibre $\zeta$ of $f_{1}:X\to Y_{1}$ .

(2)

There exists the following exact sequence:

0\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\longrightarrow(f_{1})_{*}\mathcal{O}_{D}(-K_{X})\longrightarrow 0.

Proof of Step 1.

The assertion (1) follows from $-K_{X}\cdot\zeta=2$ and $D\cdot\zeta=2$ (Lemma 6.10). Let us show (2). We have an exact sequence

0\longrightarrow\mathcal{O}_{X}(-K_{X}-D)\longrightarrow\mathcal{O}_{X}(-K_{X})\longrightarrow\mathcal{O}_{D}(-K_{X})\longrightarrow 0,

which induces another exact sequence

\displaystyle 0\to(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\to(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\to(f_{1})_{*}\mathcal{O}_{D}(-K_{X})\to R^{1}(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D).

By (1) and [Tan15, Theorem 0.5], we obtain $R^{1}f_{1*}\mathcal{O}_{X}(-K_{X}-D)=0$ . Thus (2) holds. This completes the proof of Step 1. ∎

Step 2.

$(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1)$ .

Proof of Step 2.

By Step 1(1), we have $(-K_{X}-D)\cdot\zeta=0$ . Therefore, we can write

\mathcal{O}_{X}(-K_{X}-D)\simeq f_{1}^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(s,t)

for some $s,t\in\mathbb{Z}$ (Theorem 4.11). We obtain

	$\displaystyle\mathcal{O}_{D}(2,2)$	$\displaystyle\simeq\mathcal{O}_{D}(-K_{D})$
		$\displaystyle\simeq\mathcal{O}_{X}(-K_{X}-D)\|_{D}$
		$\displaystyle\simeq(f_{1}\|_{D})^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(s,t)$
		$\displaystyle\simeq({\rm id}\times h)^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(s,t)$
		$\displaystyle\simeq\mathcal{O}_{D}(s,2t),$

where the fourth and fifth isomorphisms follow from Lemma 6.13. Hence we get $s=2$ and $t=1$ . Then it holds that

(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\simeq(f_{1})_{*}f_{1}^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(s,t)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(s,t)\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1).

This completes the proof of Step 2. ∎

Step 3.

$(f_{1})_{*}\mathcal{O}_{D}(-K_{X})\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}$ .

Proof of Step 3.

Recall that we have

(f_{1})_{*}\mathcal{O}_{D}(-K_{X})=(f_{1}|_{D})_{*}(\mathcal{O}_{X}(-K_{X})|_{D}).

By $\mathcal{O}_{X}(-K_{X}-D)|_{D}\simeq\mathcal{O}_{D}(2,2)$ and $\mathcal{O}_{X}(D)|_{D}\simeq\mathcal{O}_{D}(-1,-1)$ (Theorem 4.8), it holds that

(f_{1}|_{D})_{*}(\mathcal{O}_{X}(-K_{X})|_{D})\simeq(f_{1}|_{D})_{*}\mathcal{O}_{D}(1,1).

By Lemma 6.13, we obtain

	$\displaystyle(f_{1})_{*}\mathcal{O}_{D}(-K_{X})$	$\displaystyle\simeq(f_{1}\|_{D})_{*}\mathcal{O}_{D}(1,1)$
		$\displaystyle\simeq({\rm id}\times h)_{*}(\mathcal{O}_{D}(1,2)\otimes\mathcal{O}_{D}(0,-1))$
		$\displaystyle\simeq({\rm id}\times h)_{}((({\rm id}\times h)^{}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))\otimes\mathcal{O}_{D}(0,-1))$
		$\displaystyle\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1)\otimes({\rm id}\times h)_{*}\mathcal{O}_{D}(0,-1)$
		$\displaystyle\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}.$

This completes the proof of Step 3. ∎

Step 4.

It holds that

(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1)\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}.

Proof of Step 4.

By Step 1, we have the following exact sequence:

0\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X}-D)\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\longrightarrow(f_{1})_{*}\mathcal{O}_{D}(-K_{X})\longrightarrow 0.

By Step 2 and Step 3, this exact sequence can be written as follows:

0\longrightarrow\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1)\longrightarrow(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\longrightarrow\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}\longrightarrow 0.

Hence it suffices to show that this exact sequence splits, which follows from the following computation:

\mathrm{Ext}^{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2},\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1))\simeq H^{1}(\mathbb{P}^{1}\times\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))^{\oplus 2}=0.

This completes the proof of Step 4. ∎

Step 4 completes the proof of Lemma 6.14. ∎

Lemma 6.15.

We use Notation 6.12. Then it holds that

\mathcal{O}_{P^{\prime}}(X)\simeq\mathcal{O}_{P^{\prime}}(2)\otimes\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1).

Proof.

We have the following morphisms:

f_{1}:X\overset{\iota^{\prime}}{\hookrightarrow}P^{\prime}=\mathbb{P}(\mathcal{E}^{\prime})\xrightarrow{\pi^{\prime}}Y_{1}=\mathbb{P}^{1}\times\mathbb{P}^{1},

where $\mathcal{E}^{\prime}=(f_{1})_{*}\mathcal{O}_{X}(-K_{X})=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1)\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}$ (Lemma 6.14). We identify $X$ with $\iota^{\prime}(X)$ , which is a smooth prime divisor on $P^{\prime}$ . Since $\iota^{\prime}\colon X\hookrightarrow P^{\prime}$ is the closed immersion induced by the natural surjection $f_{1}^{*}(f_{1})_{*}\mathcal{O}_{X}(-K_{X})\to\mathcal{O}_{X}(-K_{X})$ , it follows from [Har77, Ch. II, the proof of Propsition 7.12] that

(6.15.1)

\mathcal{O}_{X}(-K_{X})\simeq\iota^{\prime*}\mathcal{O}_{P^{\prime}}(1)=\mathcal{O}_{P^{\prime}}(1)|_{X}.

By Proposition 7.1(2),

	$\displaystyle\mathcal{O}_{P^{\prime}}(-K_{P^{\prime}})$	$\displaystyle\simeq\mathcal{O}_{P^{\prime}}(3)\otimes\pi^{\prime*}(\omega_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\otimes\det\mathcal{E}^{\prime})^{-1}$
		$\displaystyle\simeq\mathcal{O}_{P^{\prime}}(3)\otimes\pi^{\prime*}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,-2)\otimes\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(4,1))^{-1}$
(6.15.2)			$\displaystyle\simeq\mathcal{O}_{P^{\prime}}(3)\otimes\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1).$

By (6.15.1) and (6.2.2), we obtain

\mathcal{O}_{P^{\prime}}(X)|_{X}\simeq\mathcal{O}_{P^{\prime}}(-K_{P^{\prime}})|_{X}\otimes\mathcal{O}_{X}(K_{X})\simeq\mathcal{O}_{P^{\prime}}(2)|_{X}\otimes\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)|_{X}.

Set

\mathcal{N}:=\mathcal{O}_{P^{\prime}}(X)^{-1}\otimes\mathcal{O}_{P^{\prime}}(2)\otimes\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1).

We have $\mathcal{N}|_{X}\simeq\mathcal{O}_{X}$ . For a fibre $\zeta$ of $f_{1}:X\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ , we get $\mathcal{N}\cdot\zeta=(\mathcal{N}|_{X})\cdot\zeta=0$ , which implies $\mathcal{N}\simeq\pi^{\prime*}\mathcal{N}^{\prime}$ for some $\mathcal{N}^{\prime}\in{\operatorname{Pic}}(\mathbb{P}^{1}\times\mathbb{P}^{1})$ . Since the pullback of $\mathcal{N}^{\prime}$ to $X$ is trivial: $f_{1}^{*}\mathcal{N}^{\prime}\simeq\mathcal{N}|_{X}\simeq\mathcal{O}_{X}$ , it holds that $\mathcal{N}^{\prime}\simeq\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}$ . Therefore, $\mathcal{N}\simeq\pi^{\prime*}\mathcal{N}^{\prime}\simeq\mathcal{O}_{P^{\prime}}$ , as required. ∎

Proposition 6.16.

We use Notation 6.12. Then there exists a closed immersion

\iota:X\hookrightarrow P:=\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})

such that

\mathcal{O}_{P}(\iota(X))\simeq\mathcal{O}_{P}(2)\otimes\pi^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3),

where $\pi:\mathbb{P}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ denotes the induced projection. Furthermore, it holds that $(-K_{X})^{3}=14$ .

Proof.

By Lemma 6.14 and Lemma 6.15, we may assume that

•

$\mathcal{E}^{\prime}=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,1)\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,0)^{\oplus 2}$ ,
•

$P^{\prime}=\mathbb{P}(\mathcal{E}^{\prime})$ ,
•

$X$ is a smooth prime divisor on $P^{\prime}$ , and
•

$\mathcal{O}_{P^{\prime}}(X)\simeq\mathcal{O}_{P^{\prime}}(2)\otimes\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$ .

Set $\mathcal{M}:=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,-1)$ and

\mathcal{E}:=\mathcal{E}^{\prime}\otimes\mathcal{M}=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2}.

We have the induced morphisms:

\pi:P\xrightarrow{\psi,\simeq}P^{\prime}\xrightarrow{\pi^{\prime}}\mathbb{P}^{1}\times\mathbb{P}^{1}

By $\mathcal{O}_{P}(1)\simeq\psi^{*}\mathcal{O}_{P^{\prime}}(1)\otimes\pi^{*}\mathcal{M}$ [Har77, Ch. II, Lemma 7.9], it holds that

	$\displaystyle\mathcal{O}_{P}(\iota(X))$	$\displaystyle\simeq\psi^{*}\mathcal{O}_{P^{\prime}}(X)$
		$\displaystyle\simeq\psi^{}\mathcal{O}_{P^{\prime}}(2)\otimes\psi^{}\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq(\psi^{}\mathcal{O}_{P^{\prime}}(1))^{\otimes 2}\otimes\psi^{}\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq(\mathcal{O}_{P}(1)\otimes\pi^{}\mathcal{M}^{-1})^{\otimes 2}\otimes\psi^{}\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq\mathcal{O}_{P}(2)\otimes\pi^{}\mathcal{M}^{-2}\otimes\pi^{}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq\mathcal{O}_{P}(2)\otimes\pi^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3).$

What is remaining is to compute $(-K_{X})^{3}$ . By Lemma 7.1(5), we have

	$\displaystyle(-K_{X})^{3}=2c_{1}(\mathcal{E})^{2}-$	$\displaystyle 2c_{2}(\mathcal{E})+4(c_{1}(\mathcal{E})\cdot c_{1}(\mathcal{F}))+6(c_{1}(\mathcal{E})\cdot K_{\mathbb{P}^{1}\times\mathbb{P}^{1}})$
		$\displaystyle+9(c_{1}(\mathcal{F})\cdot K_{\mathbb{P}^{1}\times\mathbb{P}^{1}})+6K_{\mathbb{P}^{1}\times\mathbb{P}^{1}}^{2}+3c_{1}(\mathcal{F})^{2},$

where $\mathcal{F}:=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3)$ . By [Har77, Appendix A, §3, C.3 and C.5], it holds that

	$\displaystyle c_{1}(\mathcal{E})$	$\displaystyle=2c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1))\quad{\rm and}$
	$\displaystyle c_{2}(\mathcal{E})$	$\displaystyle=c_{1}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1))^{2}=2.$

Hence we obtain

\displaystyle(-K_{X})^{3}=2\cdot 8-2\cdot 2+4\cdot(-10)+6\cdot 8+9\cdot(-10)+6\cdot 8+3\cdot 12=14.

∎

Theorem 6.17.

Let $X$ be a primitive Fano threefold with $\rho(X)=3$ . Let $R_{1}$ and $R_{2}$ be extremal rays of ${\operatorname{NE}}(X)$ . Assume that $R_{1}$ is of type $C$ and $R_{2}$ is of type $E$ . Then one and only one of the following holds.

(1)

$R_{1}$ is of type $C_{1}$ , $R_{2}$ is of type $E_{1}$ , $(-K_{X})^{3}=14$ , and $X$ is isomorphic to a prime divisor $X^{\prime}$ on $P:=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})$ such that $\mathcal{O}_{P}(X^{\prime})\simeq\mathcal{O}_{P}(2)\otimes\pi^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3)$ , where $\pi:P=\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-1,-1)^{\oplus 2})\to\mathbb{P}^{1}\times\mathbb{P}^{1}$ denotes the natural projection.
(2)

$R_{1}$ is of type $C_{2}$ , $R_{2}$ is of type $E_{1}$ , $(-K_{X})^{3}=52$ , and $X\simeq\mathbb{P}(\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(1,1))$ .

Proof.

See Proposition 6.11 and Proposition 6.16. ∎

7. Appendix: Computation of Chern classes

In this section, we compute some intersection numbers on projective space bundles (Proposition 7.1) and blowups (Proposition 7.2). We include the proofs for the sake of completeness, although both results are obtained just by applying standard results on Chern classes.

Proposition 7.1.

Let $Y$ be an $n$ -dimensional smooth variety and let $\mathcal{E}$ be a locally free sheaf of rank $r$ on $Y$ . Let $\pi\colon X=\mathbb{P}(\mathcal{E})\to Y$ be the projection induced by the $\mathbb{P}^{r-1}$ -bundle structure. Then the following hold.

(1)

If $r=2$ , then it holds that $\mathcal{O}_{X}(1)\cdot F=1$ for any fibre $F$ of $\pi:X\to Y$ over a closed point.
(2)

The following holds:

$\omega_{X}\simeq\mathcal{O}_{X}(-r)\otimes\pi^{*}(\omega_{Y}\otimes\det\mathcal{E}).$
(3)

If $n=2$ and $r=2$ , then

$(-K_{X})^{3}=2c_{1}(\mathcal{E})^{2}-8c_{2}(\mathcal{E})+6K_{Y}^{2}.$
(4)

If $n=1$ and $r=2$ , then

$c_{1}(\mathcal{O}_{X}(1))^{2}=\deg_{Y}(\mathcal{E}).$

(5)

If $n=2$ , $r=3,$ and $D$ is a smooth prime divisor on $X$ such that $\mathcal{O}_{X}(D)\simeq\mathcal{O}_{X}(2)\otimes\pi^{*}\mathcal{F}$ for some invertible sheaf $\mathcal{F}$ on $Y$ , then

	$\displaystyle(-K_{D})^{3}=$	$\displaystyle 2c_{1}(\mathcal{E})^{2}-2c_{2}(\mathcal{E})+4c_{1}(\mathcal{E})\cdot c_{1}(\mathcal{F})+6c_{1}(\mathcal{E})\cdot K_{Y}$
		$\displaystyle+9c_{1}(\mathcal{F})\cdot K_{Y}+6K_{Y}^{2}+3c_{1}(\mathcal{F})^{2}.$

Proof.

Fix a Cartier divisor $\xi$ on $X$ such that $\mathcal{O}_{X}(1)\simeq\mathcal{O}_{X}(\xi)$ . The assertion (1) holds by

\mathcal{O}_{X}(1)\cdot F=\deg_{F}(\mathcal{O}_{X}(1)|_{F})=\deg_{\mathbb{P}^{1}}(\mathcal{O}_{\mathbb{P}^{1}}(1))=1.

Let us show (2). Since $\pi$ is locally trivial, it follows from [Har77, Ch. II, Proposition 8.11 and Exercise 8.3(a)] that we have an exact sequence

0\to\pi^{*}\Omega_{Y}\longrightarrow\Omega_{X}\longrightarrow\Omega_{X/Y}\longrightarrow 0.

By [Har77, Ch. II, Exercise 5.16(d),(e)], it holds that

	$\displaystyle\omega_{X}$	$\displaystyle\simeq\bigwedge^{n+r-1}\Omega_{X}$
		$\displaystyle\simeq\bigwedge^{n}\pi^{*}\Omega_{Y}\otimes\bigwedge^{r-1}\Omega_{X/Y}$
		$\displaystyle\simeq\pi^{*}\left(\bigwedge^{n}\Omega_{Y}\right)\otimes\bigwedge^{r-1}\Omega_{X/Y}$
		$\displaystyle\simeq\pi^{*}\omega_{Y}\otimes\omega_{X/Y}.$

Moreover, we have $\omega_{X/Y}\simeq\mathcal{O}_{X}(-r)\otimes\pi^{*}\det\mathcal{E}$ by [Har77, Ch. III, Exercise 8.4(b)]. Thus (2) holds.

Let us show (3). Set $\mathcal{L}:=\omega_{Y}\otimes\det\mathcal{E}$ . It follows from (2) that $\omega_{X}\simeq\mathcal{O}_{X}(-2)\otimes\pi^{*}\mathcal{L}$ . We then obtain

(7.1.1)

\displaystyle(-K_{X})^{3}=(2\xi-c_{1}(\pi^{*}\mathcal{L}))^{3}=8\xi^{3}-12\xi^{2}\cdot\pi^{*}c_{1}(\mathcal{L})+6\xi\cdot\pi^{*}c_{1}(\mathcal{L})^{2}.

By [Har77, Appendix A, §3, page 429], we have an equation

\pi^{*}c_{0}(\mathcal{E})\cdot\xi^{2}-\pi^{*}c_{1}(\mathcal{E})\cdot\xi+\pi^{*}c_{2}(\mathcal{E})=0.

By $c_{0}(\mathcal{E})=1$ , this can be written as

(7.1.2)

\xi^{2}=\pi^{*}c_{1}(\mathcal{E})\cdot\xi-\pi^{*}c_{2}(\mathcal{E}).

We then get

	$\displaystyle\xi^{3}$	$\displaystyle=\pi^{}c_{1}(\mathcal{E})\cdot\xi^{2}-\pi^{}c_{2}(\mathcal{E})\cdot\xi$
		$\displaystyle=\pi^{}c_{1}(\mathcal{E})\cdot(\pi^{}c_{1}(\mathcal{E})\cdot\xi-\pi^{}c_{2}(\mathcal{E}))-\pi^{}c_{2}(\mathcal{E})\cdot\xi$
(7.1.3)			$\displaystyle=(\pi^{}c_{1}(\mathcal{E}))^{2}\cdot\xi-\pi^{}c_{1}(\mathcal{E})\cdot\pi^{}c_{2}(\mathcal{E})-\pi^{}c_{2}(\mathcal{E})\cdot\xi$
		$\displaystyle=\pi^{}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi-\pi^{}(c_{1}(\mathcal{E})\cdot c_{2}(\mathcal{E}))$
		$\displaystyle=\pi^{*}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi,$

where the first and second equalities follows from (7.1.2) and the last one holds, because $\dim Y=2$ implies $c_{1}(\mathcal{E})\cdot c_{2}(\mathcal{E})=0$ . By (7.1.1), (7.1.2), and (7), we obtain

	$\displaystyle(-K_{X})^{3}$	$\displaystyle=8\pi^{}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi-12(\pi^{}c_{1}(\mathcal{E})\cdot\xi-\pi^{}c_{2}(\mathcal{E}))\cdot\pi^{}c_{1}(\mathcal{L})+6\xi\cdot\pi^{*}c_{1}(\mathcal{L})^{2}$
		$\displaystyle=8\pi^{}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi-12\pi^{}(c_{1}(\mathcal{E})\cdot c_{1}(\mathcal{L}))\cdot\xi+6\pi^{*}c_{1}(\mathcal{L})^{2}\cdot\xi,$
		$\displaystyle=8(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))-12c_{1}(\mathcal{E})\cdot c_{1}(\mathcal{L})+6c_{1}(\mathcal{L})^{2}.$

where the second equality holds by $c_{2}(\mathcal{E})\cdot c_{1}(\mathcal{L})=0$ and the third one follows from $\pi^{*}Z\cdot\xi=\deg Z$ for any $0$ -cycle $Z\in\mathrm{A}^{2}(Y)$ (note that $\deg$ is dropped by abuse of notation). By $c_{1}(\mathcal{L})=c_{1}(\omega_{Y})+c_{1}(\det\mathcal{E})=K_{Y}+c_{1}(\mathcal{E})$ , we obtain

	$\displaystyle(-K_{X})^{3}$	$\displaystyle=8(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))-12c_{1}(\mathcal{E})\cdot(K_{Y}+c_{1}(\mathcal{E}))+6(K_{Y}+c_{1}(\mathcal{E}))^{2}$
		$\displaystyle=2c_{1}(\mathcal{E})^{2}-8c_{2}(\mathcal{E})+6K_{Y}^{2}.$

Thus (3) holds.

Let us show (4). It follows from [Har77, Appendix A, §3, page 429] that

\xi^{2}=\pi^{*}c_{1}(\mathcal{E})\cdot\xi-\pi^{*}c_{2}(\mathcal{E}).

By $\dim Y=1$ , we have $c_{2}(\mathcal{E})=0$ , which implies

c_{1}(\mathcal{O}_{X}(1))^{2}=\xi^{2}=\pi^{*}c_{1}(\mathcal{E})\cdot\xi=\deg_{Y}(\mathcal{E}).

Thus (4) holds.

Let us show (5). By the adjunction formula,

\displaystyle(-K_{D})^{3}=((-K_{X}-D)|_{D})^{3}=(-K_{X}-D)^{3}\cdot D.

By (2) and our assumption, we get $K_{X}\sim-3\xi+\pi^{*}c_{1}(\omega_{Y}\otimes\det\mathcal{E})$ and $D\sim 2\xi+\pi^{*}c_{1}(\mathcal{F})$ , respectively. Thus

K_{X}+D\sim-\xi+\pi^{*}c_{1}(\omega_{Y}\otimes\det\mathcal{E}\otimes\mathcal{F})=-\xi+\pi^{*}c_{1}(\mathcal{G})

\text{for}\qquad\mathcal{G}:=\omega_{Y}\otimes\det\mathcal{E}\otimes\mathcal{F}

Then

	$\displaystyle(-K_{D})^{3}$	$\displaystyle=(-K_{X}-D)^{3}\cdot D$
		$\displaystyle=(\xi-\pi^{}c_{1}(\mathcal{G}))^{3}\cdot(2\xi+\pi^{}c_{1}(\mathcal{F}))$
(7.1.4)			$\displaystyle=(\xi^{3}-3\pi^{}c_{1}(\mathcal{G})\cdot\xi^{2}+3\pi^{}c_{1}(\mathcal{G})^{2}\cdot\xi-(\pi^{}c_{1}(\mathcal{G}))^{3})\cdot(2\xi+\pi^{}c_{1}(\mathcal{F}))$
		$\displaystyle=(\xi^{3}-3\pi^{}c_{1}(\mathcal{G})\cdot\xi^{2}+3\pi^{}c_{1}(\mathcal{G})^{2}\cdot\xi)\cdot(2\xi+\pi^{*}c_{1}(\mathcal{F}))$
		$\displaystyle=2\xi^{4}+(-6\pi^{}c_{1}(\mathcal{G})+\pi^{}c_{1}(\mathcal{F}))\cdot\xi^{3}+(6\pi^{}c_{1}(\mathcal{G})^{2}-3\pi^{}(c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{F})))\cdot\xi^{2},$

where the fourth equality holds by $c_{1}(\mathcal{G})^{3}=0$ and the last one follows from $c_{1}(\mathcal{G})^{2}\cdot c_{1}(\mathcal{F})=0$ . By [Har77, Appendix A, §3, page 429], we have an equation

\pi^{*}c_{0}(\mathcal{E})\cdot\xi^{3}-\pi^{*}c_{1}(\mathcal{E})\cdot\xi^{2}+\pi^{*}c_{2}(\mathcal{E})\cdot\xi-\pi^{*}c_{3}(\mathcal{E})=0.

By $\dim Y=2$ , we have $c_{3}(\mathcal{E})=0$ . It holds by $c_{0}(\mathcal{E})=1$ that

(7.1.5)

\xi^{3}=\pi^{*}c_{1}(\mathcal{E})\cdot\xi^{2}-\pi^{*}c_{2}(\mathcal{E})\cdot\xi.

Moreover, we obtain

(7.1.6)	$\displaystyle\xi^{4}$	$\displaystyle=\pi^{}c_{1}(\mathcal{E})\cdot\xi^{3}-\pi^{}c_{2}(\mathcal{E})\cdot\xi^{2}$
	$\displaystyle=\pi^{}c_{1}(\mathcal{E})\cdot(\pi^{}c_{1}(\mathcal{E})\cdot\xi^{2}-\pi^{}c_{2}(\mathcal{E})\cdot\xi)-\pi^{}c_{2}(\mathcal{E})\cdot\xi^{2}$
	$\displaystyle=\pi^{*}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi^{2},$

where we used $c_{1}(\mathcal{E})\cdot c_{2}(\mathcal{E})=0$ , which is guaranteed by $\dim Y=2$ . By (7), (7.1.5) and (7.1.6),

	$\displaystyle(-K_{D})^{3}$	$\displaystyle=2\xi^{4}+(-6\pi^{}c_{1}(\mathcal{G})+\pi^{}c_{1}(\mathcal{F}))\cdot\xi^{3}+(6\pi^{}c_{1}(\mathcal{G})^{2}-3\pi^{}(c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{F}))\cdot\xi^{2}$
		$\displaystyle=2\pi^{*}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi^{2}$
		$\displaystyle+(-6\pi^{}c_{1}(\mathcal{G})+\pi^{}c_{1}(\mathcal{F}))\cdot(\pi^{}c_{1}(\mathcal{E})\cdot\xi^{2}-\pi^{}c_{2}(\mathcal{E})\cdot\xi)$
(7.1.7)			$\displaystyle+(6\pi^{}c_{1}(\mathcal{G})^{2}-3\pi^{}(c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{F}))\cdot\xi^{2}$
		$\displaystyle=2\pi^{*}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi^{2}$
		$\displaystyle+(-6\pi^{}c_{1}(\mathcal{G})+\pi^{}c_{1}(\mathcal{F}))\cdot\pi^{*}c_{1}(\mathcal{E})\cdot\xi^{2}$
		$\displaystyle+(6\pi^{}c_{1}(\mathcal{G})^{2}-3\pi^{}(c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{F}))\cdot\xi^{2}$

where the last equality holds by $c_{1}(\mathcal{G})\cdot c_{2}(\mathcal{E})=c_{1}(\mathcal{F})\cdot c_{2}(\mathcal{E})=0$ .

Since we have $\mathcal{O}_{X}(1)|_{F}=\mathcal{O}_{\mathbb{P}^{2}}(1)$ for every fibre $F\simeq\mathbb{P}^{2}$ of $\pi$ , we have $\xi^{2}\cdot F=(\xi|_{F})^{2}=c_{1}(\mathcal{O}_{\mathbb{P}^{2}}(1))^{2}=1$ , which implies $\pi^{*}Z\cdot\xi^{2}=\deg Z$ for any $0$ -cycle $Z\in\mathrm{A}^{2}(Y)$ . It follows from (7) that

\displaystyle(-K_{D})^{3}

\displaystyle=2c_{1}(\mathcal{E})^{2}-2c_{2}(\mathcal{E})+(-6c_{1}(\mathcal{G})+c_{1}(\mathcal{F}))\cdot c_{1}(\mathcal{E})+(6c_{1}(\mathcal{G})^{2}-3c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{F}))

By $c_{1}(\mathcal{G})=K_{Y}+c_{1}(\mathcal{E})+c_{1}(\mathcal{F})$ , we obtain

		$\displaystyle-6c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{E})+6c_{1}(\mathcal{G})^{2}-3c_{1}(\mathcal{G})\cdot c_{1}(\mathcal{F})$
	$\displaystyle=$	$\displaystyle c_{1}(\mathcal{G})\cdot(-6c_{1}(\mathcal{E})+6c_{1}(\mathcal{G})-3c_{1}(\mathcal{F}))$
	$\displaystyle=$	$\displaystyle(K_{Y}+c_{1}(\mathcal{E})+c_{1}(\mathcal{F}))\cdot(6K_{Y}+3c_{1}(\mathcal{F})).$

Therefore,

	$\displaystyle(-K_{D})^{3}$	$\displaystyle=2c_{1}(\mathcal{E})^{2}-2c_{2}(\mathcal{E})+c_{1}(\mathcal{F})\cdot c_{1}(\mathcal{E})$
		$\displaystyle+(K_{Y}+c_{1}(\mathcal{E})+c_{1}(\mathcal{F}))\cdot(6K_{Y}+3c_{1}(\mathcal{F}))$
		$\displaystyle=2c_{1}(\mathcal{E})^{2}-2c_{2}(\mathcal{E})+4c_{1}(\mathcal{F})\cdot c_{1}(\mathcal{E})$
		$\displaystyle+6K_{Y}^{2}+9K_{Y}\cdot c_{1}(\mathcal{F})+6K_{Y}\cdot c_{1}(\mathcal{E})+3c_{1}(\mathcal{F})^{2}.$

Thus (5) holds. ∎

Proposition 7.2.

Let $Y$ be a smooth projective threefold and let $C$ be a smooth curve on $Y$ . Let $f\colon X\to Y$ be the blowup along $C$ . Set $D:={\operatorname{Ex}}(f)$ . Then the following hold:

(1)

$K_{X}\sim f^{*}K_{Y}+D$ .
(2)

The induced morphism $f|_{D}\colon D\to C$ coincides with the $\mathbb{P}^{1}$ -bundle $\mathbb{P}(\mathcal{N}^{*}_{C/Y})$ over $C$ , and $\mathcal{O}_{X}(-D)|_{D}\simeq\mathcal{O}_{\mathbb{P}(\mathcal{N}^{*}_{C/Y})}(1)$ .
(3)

$D^{3}=\deg_{C}(\mathcal{N}^{*}_{C/Y})$ .
(4)

$(-K_{X})^{2}\cdot D=-K_{Y}\cdot C+2-2g(C)$ .

Proof.

The assertions (1) and (2) follow from [Har77, Ch. II, Exercise 8.5(b)] and and [Har77, II, Theorem 8.24(b),(c)], respectively. Let us show (3). By (2), we have $D\simeq\mathbb{P}(\mathcal{N}^{*}_{C/Y})$ . We then get

	$\displaystyle D^{3}$	$\displaystyle=(\mathcal{O}_{X}(-D)\|_{D})\cdot(\mathcal{O}_{X}(-D)\|_{D})$
		$\displaystyle=c_{1}(\mathcal{O}_{D}(1))^{2}$
		$\displaystyle=\deg_{C}(\mathcal{N}^{*}_{C/Y}),$

where the second equality holds by (2) and the last one follows from Proposition 7.1(4). Thus (3) holds.

Let us show (4). Set $\pi:=f|_{D}\colon D=\mathbb{P}(\mathcal{N}^{*}_{C/Y})\to C$ and $\mathcal{L}:=(\omega_{C}\otimes\det\mathcal{N}^{*}_{C/Y})^{-1}$ . Fix a divisor $\xi$ on $D$ such that $\mathcal{O}_{D}(\xi)=\mathcal{O}_{D}(1)$ . Then

(-K_{X})^{2}\cdot D=(\mathcal{O}_{X}(-K_{X})|_{D})^{2}.

It holds that

	$\displaystyle\mathcal{O}_{X}(-K_{X})\|_{D}$	$\displaystyle\simeq\mathcal{O}_{D}(-K_{D})\otimes\mathcal{O}_{X}(D)\|_{D}$
		$\displaystyle\simeq\mathcal{O}_{D}(2)\otimes\pi^{}(\omega_{C}\otimes\det\mathcal{N}^{}_{C/Y})^{-1}\otimes\mathcal{O}_{D}(-1)$
		$\displaystyle\simeq\mathcal{O}_{D}(1)\otimes\pi^{*}\mathcal{L},$

where the first isomorphism holds by the adjunction formula and the second one follows from (2) and Proposition 7.1(2). Hence

	$\displaystyle(-K_{X})^{2}\cdot D$	$\displaystyle=(\xi+c_{1}(\pi^{*}\mathcal{L}))^{2}$
		$\displaystyle=\xi^{2}+2\xi\cdot\pi^{}c_{1}(\mathcal{L})+\pi^{}c_{1}(\mathcal{L})^{2}$
		$\displaystyle=\xi^{2}+2\pi^{*}c_{1}(\mathcal{L})\cdot\xi,$

where $c_{1}(\mathcal{L})^{2}=0$ holds by $\dim C=1$ . By [Har77, Appendix A, §3, page 429], we have $\xi^{2}=\pi^{*}c_{1}(\mathcal{N}^{*}_{C/Y})\cdot\xi$ . Therefore, we get

	$\displaystyle(-K_{X})^{2}\cdot D$	$\displaystyle=\pi^{}c_{1}(\mathcal{N}^{}_{C/Y})\cdot\xi+2\pi^{*}c_{1}(\mathcal{L})\cdot\xi$
		$\displaystyle=\pi^{}(c_{1}(\mathcal{N}^{}_{C/Y})+2c_{1}(\mathcal{L}))\cdot\xi$
		$\displaystyle=\deg_{C}(c_{1}(\mathcal{N}^{*}_{C/Y})+2c_{1}(\mathcal{L}))$
		$\displaystyle=\deg_{C}(-c_{1}(\mathcal{N}^{*}_{C/Y})-2K_{C})$
		$\displaystyle=\deg_{C}(-c_{1}(\omega_{Y}\|_{C}))-K_{C})$
		$\displaystyle=-K_{Y}\cdot C+2-2g(C),$

where the fourth equality holds by $\mathcal{L}=(\omega_{C}\otimes\det\mathcal{N}^{*}_{C/Y})^{-1}$ and the fifth equality follows from $\omega_{C}\simeq\omega_{Y}|_{C}\otimes(\det\mathcal{N}_{C/Y})$ [Har77, Ch. II, Proposition 8.20]. ∎

8. Appendix: Description of split double covers

Definition 8.1.

Let $Y$ be a normal variety and let $\mathcal{L}$ be an invertible sheaf. Set

\mathbb{V}:={\operatorname{Spec}}_{Y}\bigoplus_{i\geq 0}\mathcal{L}^{\otimes-i},

where $\bigoplus_{i\geq 0}\mathcal{L}^{\otimes-i}$ denotes the natural graded $\mathcal{O}_{Y}$ -algebra. Let $\pi:\mathbb{V}\to Y$ be the projection. Take the tautological section $z\in H^{0}(\mathbb{V},\pi^{*}\mathcal{L})$ , which corresponds to the element $(0,1,0,0,...)\in\bigoplus_{i\geq 0}H^{0}(Y,\mathcal{L}^{1-i})$ via

H^{0}(\mathbb{V},\pi^{*}\mathcal{L})\simeq H^{0}(Y,\pi_{*}\pi^{*}\mathcal{L})\simeq\bigoplus_{i\geq 0}H^{0}(Y,\mathcal{L}^{1-i}).

(1)

For $s\in H^{0}(Y,\mathcal{L}^{\otimes 2})$ , the closed subscheme

$X:=\{z^{2}-\pi^{*}(s)=0\}\subset\mathbb{V}$

and the induced morphism $X\to Y$ are called the simple $\mu_{2}$ -cover associated to $(\mathcal{L},s)$ .
(2)

Assume $p=2$ . For $s\in H^{0}(X,\mathcal{L})$ and $t\in H^{0}(X,\mathcal{L}^{\otimes 2})$ , the closed subscheme

$X:=\{z^{2}-\pi^{*}(s)\cdot z+\pi^{*}(t)=0\}\subset\mathbb{V}$

and the induced morphism $X\to Y$ are called the split $\alpha_{\mathcal{L},s}$ -torsor associated to $(\mathcal{L},s,t)$ . Note that the split $\alpha_{\mathcal{L},0}$ -torsor associated to $(\mathcal{L},0,t)$ coincides with the simple $\mu_{2}$ -cover associated to $(\mathcal{L},-t)$ .

Lemma 8.2.

Assume that $p\neq 2$ . Let $f:X\to Y$ be a double cover of smooth projective varieties. Set $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ , which is an invertible sheaf (Remark 2.2). Then $f$ is a simple $\mu_{2}$ -cover associated to $(\mathcal{L},s)$ for some $s\in H^{0}(Y,\mathcal{L}^{\otimes 2})$

Proof.

See [CD89, Page 10]. ∎

Lemma 8.3.

Assume $p=2$ . Let $f:X\to Y$ be a double cover of smooth projective varieties. Set $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ , which is an invertible sheaf (Remark 2.2). Then $f$ is a spit double cover if and only if $f$ is a split $\alpha_{\mathcal{L},s}$ -torsor associated to some $(\mathcal{L},s,t)$ .

Proof.

If $f$ is the split $\alpha_{\mathcal{L},s}$ -torsor associated to some $(\mathcal{L},s,t)$ , then it is clear that $f$ is a split double cover. Conversely, assume that $f$ is a split double cover. By [CD89, Page 10-11], $f$ is given by a splittable admissible triple in the sense of [CD89, Page 11], which is nothing but a split $\alpha_{\mathcal{L},s}$ -torsor associated to some $(\mathcal{L},s,t)$ . ∎

Lemma 8.4.

Let $f:X\to Y$ be a split double cover of smooth projective varieties. Set $\mathcal{L}:=(f_{*}\mathcal{O}_{X}/\mathcal{O}_{Y})^{-1}$ , which is an invertible sheaf (Remark 2.2). Let $\pi:Y\to Y^{\prime}$ be a morphism of smooth projective varieties such that $g_{*}\mathcal{O}_{Y}=\mathcal{O}_{Y^{\prime}}$ and $\mathcal{L}\simeq g^{*}\mathcal{L}^{\prime}$ for some invertible sheaf $\mathcal{L}^{\prime}$ on $Y^{\prime}$ . Then there exist a split double cover $f^{\prime}:X^{\prime}\to Y^{\prime}$ and a morphism $\pi_{X}:X\to X^{\prime}$ such that the following diagram is cartesian:

In particular, if $Y=Y^{\prime}\times_{k}Y^{\prime\prime}$ and $\pi:Y\to Y^{\prime}$ is the first projection in addition to the above assumptions, then $X\simeq X^{\prime}\times_{k}Y^{\prime\prime}$ and $\pi_{X}:X\to X^{\prime}$ coincides with the first projection.

Proof.

Assume $p=2$ . By Lemma 8.3, there exist $s\in H^{0}(Y,\mathcal{L})$ and $t\in H^{0}(Y,\mathcal{L}^{\otimes 2})$ such that $f$ is the split $\alpha_{\mathcal{L},s}$ -torsor associated to $(\mathcal{L},s,t)$ . By $\pi_{*}\mathcal{O}_{Y}=\mathcal{O}_{Y^{\prime}}$ and $\mathcal{L}\simeq g^{*}\mathcal{L}^{\prime}$ , we have the following corresponding elements:

•

$H^{0}(Y,\mathcal{L})\simeq H^{0}(Y^{\prime},\mathcal{L}^{\prime}),\qquad s\mapsto s^{\prime}$ .
•

$H^{0}(Y,\mathcal{L}^{\otimes 2})\simeq H^{0}(Y^{\prime},\mathcal{L}^{\prime\otimes 2}),\qquad t\mapsto t^{\prime}$ .

Set $f^{\prime}:X^{\prime}\to Y^{\prime}$ to be the split $\alpha_{\mathcal{L}^{\prime},s^{\prime}}$ torsor associated to $(\mathcal{L}^{\prime},s^{\prime},t^{\prime})$ .

Fix an affine open cover $Y^{\prime}=\bigcup_{i\in I}Y^{\prime}_{i}$ such that $\mathcal{L}^{\prime}|_{Y^{\prime}_{i}}\simeq\mathcal{O}_{Y^{\prime}_{i}}$ for every $i\in I$ . Then $X^{\prime}$ is given as follows:

X^{\prime}=\bigcup_{i\in I}{\operatorname{Spec}}\,(\mathcal{O}_{Y^{\prime}_{i}}(Y^{\prime}_{i})[z^{\prime}_{i}]/(z^{\prime 2}_{i}+s^{\prime}_{i}z^{\prime}_{i}+t^{\prime}_{i})),

where $s^{\prime}_{i}:=s^{\prime}|_{Y^{\prime}_{i}}$ and $t^{\prime}_{i}:=t^{\prime}|_{Y^{\prime}_{i}}$ . By taking an affine open cover of $Y$ which refines $Y=\bigcup_{i\in I}\pi^{-1}(Y^{\prime}_{i})$ , we can directly check that the induced morphism $X\to X^{\prime}\times_{Y^{\prime}}Y$ is an isomorphism (more explicitly, if $\widetilde{Y}$ is an affine open subset contained in $\pi^{-1}(Y^{\prime}_{i})$ for some $i$ , then both $X$ and $X^{\prime}\times_{Y^{\prime}}Y$ are given by $\mathcal{O}_{\widetilde{Y}}(\widetilde{Y})[z]/(z^{2}+(s|_{\widetilde{Y}})z+(t|_{\widetilde{Y}})$ ). If $p\neq 2$ , the same argument works by using Lemma 8.2 instead of Lemma 8.3. ∎

9. Classification table of Fano threefolds with $\rho=2$

Let $X$ be a Fano threefold with $\rho(X)=2$ . Then one of No. 2-1, …, No. 2-36 holds. In addition to (1)–(6) of Theorem 1.2, we use the follwoing notation and terminologies:

(7)

If an extremal ray is of type $C$ , then $\deg\Delta$ denotes the degree of the discriminant bundle $\Delta$ , which is an invertible sheaf on $\mathbb{P}^{2}$ . When the conic bundle is generically smooth, then $\Delta$ coincides with the invertible sheaf associated with the discriminant divisor.
(8)

If an extremal ray is of type $D$ , then $X_{t}$ denotes a fibre over a closed point $t\in\mathbb{P}^{1}$ . In particular, $(-K_{X})^{2}\cdot X_{t}$ coincides with $(-K_{X_{K}})^{2}$ for the generic fibre $X_{K}$ .

Table 5. Fano threefolds with

\rho(X)=2

No.	$(-K_{X})^{3}$	descriptions and extremal rays
2-1	$4$	$D_{1}:(-K_{X})^{2}\cdot X_{t}=1$
		$E_{1}:$ blowup of $V_{1}$ along an elliptic curve of degree $1$ which is a complete intersection of two members of $\|-\frac{1}{2}K_{V_{1}}\|$
2-2	$6$	$X$ is a split double cover of $\mathbb{P}^{2}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}(2,1)$
		$C_{1}:\deg\Delta=8$ , $X\xrightarrow{{\rm 2:1}}\mathbb{P}^{2}\times\mathbb{P}^{1}\xrightarrow{{\rm pr}_{1}}\mathbb{P}^{2}$
		$D_{1}:(-K_{X})^{2}\cdot X_{t}=2$
2-3	$8$	$D_{1}:(-K_{X})^{2}\cdot X_{t}=2$
		$E_{1}:$ is a blowup of $V_{2}$ along an elliptic curve of degree $2$ which is a complete intersection of two members of $\|-\frac{1}{2}K_{V_{2}}\|$
2-4	$10$	$D_{1}:(-K_{X})^{2}\cdot X_{t}=3$
		$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a curve of genus $10$ and degree $9$ which is a complete intersection of two cubic surfaces
2-5	$12$	$D_{1}:(-K_{X})^{2}\cdot X_{t}=3$
		$E_{1}:$ blowup of $V_{3}$ along an elliptic curve of degree $3$ which is a complete intersection of two members of $\|-\frac{1}{2}K_{V_{3}}\|$
2-6	$12$	a smooth divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(2,2)$ , or a split double cover of $W$ with $\mathcal{L}^{\otimes 2}\simeq\omega_{W}^{-1}$
		$C_{1}:\deg\Delta=6$ , $X\to\mathbb{P}^{2}\times\mathbb{P}^{2}\xrightarrow{{\rm pr_{1}}}\mathbb{P}^{2}$
		$C_{1}:\deg\Delta=6$ , $X\to\mathbb{P}^{2}\times\mathbb{P}^{2}\xrightarrow{{\rm pr_{2}}}\mathbb{P}^{2}$
2-7	$14$	$D_{1}:(-K_{X})^{2}\cdot X_{t}=4$
		$E_{1}:$ blowup of $Q$ along a curve of genus $5$ and degree $8$ which is a complete intersection of two members of $\|\mathcal{O}_{Q}(2)\|$
2-8	$14$	a split double cover of $V_{7}$ with $\mathcal{L}^{\otimes 2}\simeq\omega_{V_{7}}^{-1}$
		$C_{1}:\deg\Delta=6$ , $X\xrightarrow{{\rm 2:1}}V_{7}=\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))\xrightarrow{{\rm pr}}\mathbb{P}^{2}$
		$E_{3}\,{\rm or}\,E_{4}$
2-9	$16$	$C_{1}:\deg\Delta=5$
		$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a curve of genus $5$ and degree $7$
2-10	$16$	$D_{1}:$ $(-K_{X})^{2}\cdot X_{t}=4$
		$E_{1}:$ blowup of $V_{4}$ along an elliptic curve of degree $4$ which is a complete intersection of two members of $\|-\frac{1}{2}K_{V_{4}}\|$
2-11	$18$	$C_{1}:$ $\deg\Delta=5$
		$E_{1}:$ blowup of $V_{3}$ along a line
2-12	$20$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a curve of genus $3$ and degree $6$
		$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a curve of genus $3$ and degree $6$
2-13	$20$	$C_{1}:$ $\deg\Delta=4$
		$E_{1}:$ blowup of $Q$ along a curve of genus $2$ and degree $6$
2-14	$20$	$D_{1}:$ $(-K_{X})^{2}\cdot X_{t}=5$
		$E_{1}:$ blowup of $V_{5}$ along an elliptic curve of degree $5$ which is a complete intersection of two members of $\|-\frac{1}{2}K_{V_{5}}\|$
2-15	$22$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a curve of genus $3$ and degree $6$
		$E_{3}\,{\rm or}\,E_{4}$
2-16	$22$	$C_{1}:\deg\Delta=4$
		$E_{1}:$ blowup of $V_{4}$ along a conic
2-17	$24$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along an elliptic curve of degree $5$
		$E_{1}:$ blowup of $Q$ along an elliptic curve of degree $5$
2-18	$24$	a split double cover of $\mathbb{P}^{2}\times\mathbb{P}^{1}$ with $\mathcal{L}\simeq\mathcal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{1}}(1,1)$
		$C_{1}:\deg\Delta=4$ , $X\xrightarrow{{\rm 2:1}}\mathbb{P}^{2}\times\mathbb{P}^{1}\xrightarrow{{\operatorname{pr}}_{1}}\mathbb{P}^{2}$
		$D_{2}:(-K_{X})^{2}\cdot X_{t}=8$
2-19	$26$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a curve of genus $2$ and degree $5$
		$E_{1}:$ blowup of $V_{4}$ along a line
2-20	$26$	$C_{1}:\deg\Delta=3$
		$E_{1}:$ blowup of $V_{5}$ along a cubic rational curve
2-21	$28$	$E_{1}:$ blowup of $Q$ along a rational curve of degree $4$
	$28$	$E_{1}:$ blowup of $Q$ along a rational curve of degree $4$
2-22	$30$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a rational curve of degree $4$
		$E_{1}:$ blowup of $V_{5}$ along a conic
2-23	$30$	$E_{1}:$ blowup of $Q$ along an elliptic curve of degree $4$
		$E_{3}\,{\rm or}\,E_{4}$
2-24	$30$	a divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,2)$
		$C_{1}:\deg\Delta=3$
		$C_{2}$
2-25	$32$	$D_{2}:(-K_{X})^{2}\cdot X_{t}=8$
		$E_{1}:$ blowup of $\mathbb{P}^{3}$ along an elliptic curve of degree $4$ which is a complete intersection of two quadric surfaces
2-26	$34$	$E_{1}:$ blowup of $Q$ along a cubic rational curve
		$E_{1}:$ blowup of $V_{5}$ along a line
2-27	$38$	$C_{2}$
		$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a cubic rational curve
2-28	$40$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along an elliptic curve of degree $3$
		$E_{3}\,{\rm or}\,E_{4}$
2-29	$40$	$D_{2}:(-K_{X})^{2}\cdot X_{t}=8$
		$E_{1}:$ blowup of $Q$ along a conic which is a complete intersection of two members of $\|\mathcal{O}_{Q}(1)\|$
2-30	$46$	$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a conic
		$E_{2}:$ blowup of $Q$ at a point
2-31	$46$	$C_{2}$
		$E_{1}:$ blowup of $Q$ along a line
2-32	$48$	$W$ , a divisor on $\mathbb{P}^{2}\times\mathbb{P}^{2}$ of bidegree $(1,1)$
		$C_{2}:W\hookrightarrow\mathbb{P}^{2}\times\mathbb{P}^{2}\xrightarrow{{\operatorname{pr}}_{1}}\mathbb{P}^{2}$
		$C_{2}:W\hookrightarrow\mathbb{P}^{2}\times\mathbb{P}^{2}\xrightarrow{{\operatorname{pr}}_{1}}\mathbb{P}^{2}$
2-33	$54$	$D_{3}:(-K_{X})^{2}\cdot X_{t}=9$
		$E_{1}:$ blowup of $\mathbb{P}^{3}$ along a line
2-34	$54$	$\mathbb{P}^{2}\times\mathbb{P}^{1}$
		$C_{2}:$ the projection $\mathbb{P}^{2}\times\mathbb{P}^{1}\to\mathbb{P}^{2}$
		$D_{3}:$ the projection $\mathbb{P}^{2}\times\mathbb{P}^{1}\to\mathbb{P}^{1}$
2-35	$56$	$V_{7}$ , i.e., $\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))$
		$C_{2}:$ the projection $\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(1))\to\mathbb{P}^{2}$
		$E_{2}:$ blowup of $\mathbb{P}^{3}$ at a point
2-36	$62$	$\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))$
		$C_{2}:$ the projection $\mathbb{P}_{\mathbb{P}^{2}}(\mathcal{O}_{\mathbb{P}^{2}}\oplus\mathcal{O}_{\mathbb{P}^{2}}(2))\to\mathbb{P}^{2}$
		$E_{5}:$ blowup at the singular point of the cone over the Veronese surface

Proof.

(of Table LABEL:table-pic2) We use Notation 5.1. Almost all the parts follow from Section 5. It is enough to compute $\deg\Delta$ (type $C$ ) and $(-K_{X})^{2}\cdot X_{t}$ (type $D$ ).

If $f_{1}:X\to\mathbb{P}^{2}$ is of type $C$ , then we have $H_{1}^{3}=0$ and $H_{1}^{2}\cdot H_{2}=(\frac{2}{\mu_{1}}\ell_{1})\cdot H_{2}=\frac{2}{\mu_{1}}$ (Lemma 5.3, Proposition 5.9), which imply the following:

\deg\Delta\overset{{\rm(a)}}{=}12-(-K_{X})^{2}\cdot H_{1}\overset{{\rm(b)}}{=}12-(\mu_{2}H_{1}+\mu_{1}H_{2})^{2}\cdot H_{1}=12-4\mu_{2}-\mu_{1}^{2}H_{1}\cdot H_{2}^{2},

where (a) and (b) follow from Proposition 3.16 and Proposition 5.9, respectively. If $f_{1}:X\to\mathbb{P}^{1}$ is of type $D$ , then we have $H_{1}^{3}=H_{1}^{2}\cdot H_{2}=0$ , which imply the following:

(-K_{X})^{2}\cdot X_{t}=(-K_{X})^{2}\cdot H_{1}=(\mu_{2}H_{1}+\mu_{1}H_{2})^{2}\cdot H_{1}=\mu_{1}^{2}H_{1}\cdot H_{2}^{2}.

Therefore, it is enough to compute one of $(-K_{X})^{2}\cdot H_{1}$ and $H_{1}\cdot H_{2}^{2}$ . If $R_{2}$ is type $D$ , then we have $H_{2}^{2}\equiv 0$ , and hence $H_{1}\cdot H_{2}^{2}=0$ . If $R_{2}$ is of type $C$ , then we get $H_{1}\cdot H_{2}^{2}=\frac{2}{\mu_{2}}$ .

In what follows, we assume that $X$ is a Fano threefold of No. 2-8 in order to explain how to compute $\deg\Delta$ and $(-K_{X})^{2}\cdot H_{1}$ for the case when $R_{1}$ is of type $C$ and $R_{2}$ is of type $E$ . It holds that $-K_{X}\sim H_{1}+H_{2}$ (Proposition 5.9). For $D:={\operatorname{Ex}}(f_{2})$ , we have $K_{X}\sim f_{2}^{*}K_{Y}+D$ and $((-K_{X})^{2}\cdot D,(-K_{X})\cdot D^{2},D^{3})=(2,-2,2)$ [TanII, Proposition 3.22]. By Lemma 5.25(1), we have $D\cdot\ell_{1}=1$ (note that $\zeta\equiv 2\ell_{1}$ for a fibre $\zeta$ of $f_{1}:X\to\mathbb{P}^{2}$ ). Hence we obtain $-f_{2}^{*}K_{Y}\cdot\ell_{1}=-K_{X}\cdot\ell_{1}+D\cdot\ell_{1}=2$ . By $H_{2}\cdot\ell_{1}=1$ (Proposition 5.9), we get $-f_{2}^{*}K_{Y}\equiv 2H_{2}$ . Then

(-K_{X})^{2}\cdot H_{1}=(-K_{X})^{2}\cdot(-K_{X}-H_{2})=(-K_{X})^{2}\cdot\left(-K_{X}-\frac{-K_{X}+D}{2}\right)

=\frac{(-K_{X})^{3}-(-K_{X})^{2}\cdot D}{2}=\frac{14-2}{2}=6.

Thus $\deg\Delta=12-(-K_{X})^{2}\cdot H_{1}=12-6=6$ . ∎

We close this paper with the following proposition. Although this result is not used in this paper, we will need it in the classification of Fano threefolds with $\rho\geq 3$ .

Proposition 9.1.

(1)

Let $X$ be a Fano threefold of No. 2-15. Let $f:X\to\mathbb{P}^{3}$ be a blowup along a smooth curve $B$ of degree $6$ . Then $B$ is contained in a (possibly singular) quadric surface $S$ on $\mathbb{P}^{3}$ .
(2)

Let $X$ be a Fano threefold of No. 2-23. Let $f:X\to Q$ be a blowup along a smooth curve $B$ of degree $4$ . Then $B$ is contained in a (possibly singular) quadric surface $S$ on $Q$ , i.e., $S$ is a hyperplane section of $Q\subset\mathbb{P}^{4}$ .

Proof.

We prove (1) and (2) simultaneously. If $X$ is 2-15 (resp. 2-23), then we set $Y:=\mathbb{P}^{3}$ (resp. $Y:=Q$ ) and $s:=4$ (resp. $s:=3$ ). Then we have a blowup $f:X\to Y$ and $Y$ is a Fano threefold of index $s$ . We have the contraction $g:X\to Z$ of the other extremal ray, which is of type $E_{3}$ or $E_{4}$ . In particular, $T:={\operatorname{Ex}}(g)$ is a (possibly singular) quadric surface. Set $S:=f(T)$ . It is enough to show that $B\subset S$ and the induced composite morphism

f_{T}:T\hookrightarrow X\xrightarrow{f}S

is an isomorphism. Set $H_{f}$ (resp. $H_{g}$ ) to be the pullback on $X$ of the ample generator of $Y$ (resp. $Z$ ). It holds that $-K_{X}\sim H_{f}+H_{g}$ (Proposition 5.9). Let $r$ be the positive integer satisfying $-g^{*}K_{Y}\sim rH_{g}$ .

Step 1.

$B\subset S$ .

Proof of Step 1.

Suppose $B\not\subset S$ . Then $T=f^{*}S$ . Pick a curve $C$ on $S$ disjoint from $S\cap B$ . Set $C_{X}:=f^{-1}(C)\xrightarrow{\simeq}C$ . Since $S$ is ample, we have $S\cdot C>0$ . On the other hand, $\mathcal{O}_{X}(-T)|_{T}$ is ample, and hence $T\cdot C_{X}=\mathcal{O}_{X}(T)|_{T}\cdot C<0$ . By $T=f^{*}S$ , we get the following contradiction:

0<S\cdot C=f^{*}S\cdot C_{X}=T\cdot C_{X}<0.

This completes the proof of Step 1. ∎

Step 2.

$T\cdot\zeta=r-1$ for a one-dimensional fibre $\zeta$ of $f:X\to Y$ .

Proof of Step 2.

We have that

-K_{X}\sim H_{f}+H_{g}\equiv\frac{-K_{X}+E}{s}+\frac{-K_{X}+T}{r},

which implies

-(sr-r-s)K_{X}\sim rE+sT.

Taking the intersection with $\zeta$ , we get $sr-r-s=-r+sT\cdot\zeta$ , which implies $T\cdot\zeta=r-1$ . This completes the proof of Step 2. ∎

Step 3.

It holds that $r=2$ and $T\cdot\zeta=1$ .

Proof of Step 3.

By [TanII, Proposition 3.22], we get

K_{Z}^{3}=(g^{*}K_{Z})^{3}=(K_{X}-T)^{3}=K_{X}^{3}-3K_{X}^{2}\cdot T+3K_{X}\cdot T^{2}-T^{3}=K_{X}^{3}-3\cdot 2+3\cdot 2-2,

and hence $r^{3}H_{g}^{3}=(-K_{Z})^{3}=(-K_{X})^{3}+2$ . If $X$ is of No. 2-15, then we get $r^{3}H_{g}^{3}=22+2=24$ . If $X$ is of No. 2-23, then we get $r^{3}H_{g}^{3}=30+2=32$ . Since the inequality $r\geq 2$ holds by Step 2, we obtain $r=2$ . Again by Step 2, we get $T\cdot\zeta=r-1=1$ . This completes the proof of Step 3. ∎

Step 4.

$f_{T}:T\to S$ is an isomorphism.

Proof of Step 4.

Let us show that $S$ is normal. Consider an exact sequence

0\to\mathcal{O}_{X}(-T)\to\mathcal{O}_{X}\to\mathcal{O}_{T}\to 0.

By $(-T-K_{X})\cdot\zeta=-1+1=0$ (Step 3), we get $R^{1}f_{*}\mathcal{O}_{X}(-T)=0$ [Tan15, Theorem 0.5]. Then $\mathcal{O}_{S}=f_{*}\mathcal{O}_{T}$ , i.e., $S$ is normal.

Note that any curve $\zeta$ on $T$ is not contracted by $f$ , as otherwise we would get a contradiction: $\zeta\in R_{1}\cap R_{2}=\{0\}$ . Hence $T\neq{\operatorname{Ex}}(f)$ , which implies that $f:T\to S$ is birational. Since $T$ is a (possibly singular) quadric surface, $f|_{T}:T\to S$ is an isomorphism. This completes the proof of Step 4. ∎

Step 1 and Step 4 complete the proof of Proposition 9.1. ∎

	$\displaystyle\mathcal{O}_{D^{\prime}}(1)$	$\displaystyle\simeq\varphi^{}\mathcal{O}_{D}(1)\otimes(\pi^{\prime})^{}\mathcal{O}_{C}(-K_{Y})$
		$\displaystyle\simeq\varphi^{}\mathcal{O}_{D}(1)\otimes(\varphi^{}\pi^{*}\mathcal{O}_{C}(-K_{Y}))$
		$\displaystyle\simeq\varphi^{}(\mathcal{O}_{D}(1)\otimes\pi^{}\mathcal{O}_{C}(-K_{Y}))$
		$\displaystyle\simeq\varphi^{}((\mathcal{O}_{X}(-D)\otimes f^{}\mathcal{O}_{Y}(-K_{Y}))\|_{D})$
		$\displaystyle\simeq\varphi^{*}(\mathcal{O}_{X}(-K_{X})\|_{D}),$

	$\displaystyle K_{W}$	$\displaystyle\sim(K_{\mathbb{P}^{2}\times\mathbb{P}^{2}}+W)\|_{W}$
		$\displaystyle\sim(-3M_{1}-3M_{2}+M_{1}+M_{2})\|_{W}$
		$\displaystyle\sim-2(M_{1}+M_{2})\|_{W}.$

	$\displaystyle(f_{1}\|_{D})_{*}\mathcal{O}_{D}(-K_{X})$	$\displaystyle\simeq(f_{1}\|_{D})_{*}\mathcal{O}_{D}(1)$
		$\displaystyle\simeq(f_{1}\|_{D})_{}(\mathcal{O}_{D}\otimes(f_{1}\|_{D})^{}\mathcal{O}_{\mathbb{P}^{2}}(1))$
		$\displaystyle\simeq(f_{1}\|_{D})_{*}\mathcal{O}_{D}\otimes\mathcal{O}_{\mathbb{P}^{2}}(1).$

	$\displaystyle\mathcal{O}_{P}(\iota(X))$	$\displaystyle\simeq\psi^{*}\mathcal{O}_{P^{\prime}}(X)$
		$\displaystyle\simeq\psi^{}\mathcal{O}_{P^{\prime}}(2)\otimes\psi^{}\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq(\psi^{}\mathcal{O}_{P^{\prime}}(1))^{\otimes 2}\otimes\psi^{}\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq(\mathcal{O}_{P}(1)\otimes\pi^{}\mathcal{M}^{-1})^{\otimes 2}\otimes\psi^{}\pi^{\prime*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq\mathcal{O}_{P}(2)\otimes\pi^{}\mathcal{M}^{-2}\otimes\pi^{}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(-2,1)$
		$\displaystyle\simeq\mathcal{O}_{P}(2)\otimes\pi^{*}\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{1}}(2,3).$

	$\displaystyle\xi^{3}$	$\displaystyle=\pi^{}c_{1}(\mathcal{E})\cdot\xi^{2}-\pi^{}c_{2}(\mathcal{E})\cdot\xi$
		$\displaystyle=\pi^{}c_{1}(\mathcal{E})\cdot(\pi^{}c_{1}(\mathcal{E})\cdot\xi-\pi^{}c_{2}(\mathcal{E}))-\pi^{}c_{2}(\mathcal{E})\cdot\xi$
(7.1.3)			$\displaystyle=(\pi^{}c_{1}(\mathcal{E}))^{2}\cdot\xi-\pi^{}c_{1}(\mathcal{E})\cdot\pi^{}c_{2}(\mathcal{E})-\pi^{}c_{2}(\mathcal{E})\cdot\xi$
		$\displaystyle=\pi^{}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi-\pi^{}(c_{1}(\mathcal{E})\cdot c_{2}(\mathcal{E}))$
		$\displaystyle=\pi^{*}(c_{1}(\mathcal{E})^{2}-c_{2}(\mathcal{E}))\cdot\xi,$