Notation 1.3.1.

1. (1) We denote by $\unicode[STIX]{x1D6FE}_{a}$ and $\unicode[STIX]{x1D6FE}_{b}$ the flopping curves on $B_{a}$ and $B_{b}$ , respectively.

2. (2) It is important to notice that there exist exactly two singular $\unicode[STIX]{x1D70B}_{b}$ -fibers, which are isomorphic to the quadric cone (this follows from the calculation of the topological Euler number of $B_{a}$ and invariance of Euler number under flop). We denote them by $F_{1}$ and $F_{2}$ .

Proof. This follows from [Reference Fujita4, p.166, (si111o) Case (a)]. ◻

Proof. Note that $G=\operatorname{Aut}B_{a}=\operatorname{Aut}(B_{a},\unicode[STIX]{x1D6FE}_{a},L_{\mathsf{j}})$ since $\unicode[STIX]{x1D6FE}_{a}$ and $L_{\mathsf{j}}$ are preserved by $G$ . By universality of blow-ups, a $G$ -action on $\widetilde{B}_{a}$ is naturally induced and then the map $\unicode[STIX]{x1D707}$ is $G$ -equivariant. Thus we have $\operatorname{Aut}(B_{a},\unicode[STIX]{x1D6FE}_{a},L_{\mathsf{j}})=\operatorname{Aut}(\widetilde{B}_{a},E_{a},\widetilde{L}_{\mathsf{j}})$ , where $E_{a}$ is the $\unicode[STIX]{x1D707}$ -exceptional divisor and $\widetilde{L}_{\mathsf{j}}$ is the strict transform of $L_{\mathsf{j}}$ . By a similar reason to the above, a $G$ -action on $B_{c}$ is naturally induced and then the map $\unicode[STIX]{x1D708}$ is $G$ -equivariant. Thus we have $\operatorname{Aut}(\widetilde{B}_{a},E_{a},\widetilde{L}_{\mathsf{j}})=\operatorname{Aut}(B_{c},\mathbb{P}^{1}\times \mathsf{j},\unicode[STIX]{x1D6FE}_{c})$ since the images of $E_{a}$ and $\widetilde{L}_{\mathsf{j}}$ on $B_{c}$ are $\mathbb{P}^{1}\times \mathsf{j}$ and $\unicode[STIX]{x1D6FE}_{c}$ respectively by Proposition 1.4.1. Moreover, since $\mathbb{P}^{1}\times \mathsf{j}$ is uniquely determined from $\unicode[STIX]{x1D6FE}_{c}$ , we have $\operatorname{Aut}(\mathbb{P}^{1}\times \mathbb{P}^{2},\mathbb{P}^{1}\times \mathsf{j},\unicode[STIX]{x1D6FE}_{c})=\operatorname{Aut}(\mathbb{P}^{1}\times \mathbb{P}^{2},\unicode[STIX]{x1D6FE}_{c}).$ Therefore, we have $G=\operatorname{Aut}(\mathbb{P}^{1}\times \mathbb{P}^{2},\unicode[STIX]{x1D6FE}_{c})$ and the first assertion follows. Explicit descriptions of $G$ as a subgroup of $\operatorname{Aut}B_{b}$ can be obtained by elementary calculations.◻

Proof. We only show that the two points $(1:0:0),(0:1:0)\in \mathsf{j}$ correspond to the lines $\ell _{1}$ and $\ell _{2}$ by projective duality. This follows from the orbit decomposition of $\mathbb{P}^{2}$ by the identity component $G_{0}$ of $G$ since the two points $\in \mathbb{P}^{2}$ corresponding to the lines $\ell _{1}$ and $\ell _{2}$ are fixed by $G_{0}$ , and $G_{0}$ has only two fixed points.◻

Proof. As we mention in Section (1.2.b), $M_{B_{a}}$ is the tautological line bundle on $B_{a}$ associated to the bundle ${\mathcal{E}}(2)$ . Therefore, comparing (1.2) and (1.4), we see that $H-L=M_{B_{a}}-2L$ is the tautological line bundle associated to the bundle ${\mathcal{E}}$ . If $\mathsf{m}$ is not equal to the jumping line $\mathsf{j}$ , then ${\mathcal{E}}_{|\mathsf{m}}\simeq {\mathcal{O}}_{\mathbb{P}^{1}}\oplus {\mathcal{O}}_{\mathbb{P}^{1}}(-1)$ and hence $(H-L)_{|L_{\mathsf{m}}}$ is linearly equivalent to the negative section $C_{0}(\mathsf{m})$ of $L_{\mathsf{m}}\simeq \mathbb{F}_{1}$ . If $\mathsf{m}=\mathsf{j}$ , then ${\mathcal{E}}_{|\mathsf{m}}\simeq {\mathcal{O}}_{\mathbb{P}^{1}}(-2)\oplus {\mathcal{O}}_{\mathbb{P}^{1}}(1)$ by Proposition 1.2.2(4) and hence $(H-L)_{|L_{\mathsf{m}}}$ is linearly equivalent to the negative section $\unicode[STIX]{x1D6FE}_{a}$ plus a ruling of $L_{\mathsf{j}}\simeq \mathbb{F}_{3}$ .◻

Proof. The claims follow easily from the exact sequence (1.1) noting that $H-L$ is the tautological line bundle associated to ${\mathcal{E}}$ . Here we only show that $H^{2}({\mathcal{O}}_{B_{a}}(H-2L))\simeq H^{2}(\mathbb{P}^{2},{\mathcal{E}}(-1))=\{0\}$ . By the Serre duality, we have $H^{2}(\mathbb{P}^{2},{\mathcal{E}}(-1))\simeq H^{0}(\mathbb{P}^{2},{\mathcal{E}}^{\ast }(-2))^{\ast }\simeq H^{0}(\mathbb{P}^{2},{\mathcal{E}}(-1))^{\ast }$ , which is zero by (1.1).◻

Notation 2.3.1. Let $b:\widetilde{(\mathbb{P}^{2})^{\ast }}\rightarrow (\mathbb{P}^{2})^{\ast }$ be the blow-up at the point $[\mathsf{j}]$ . Let $E_{0}$ be the $b$ -exceptional curve, and $r$ be a ruling of $\widetilde{(\mathbb{P}^{2})^{\ast }}\simeq \mathbb{F}_{1}$ . The surface $\widetilde{(\mathbb{P}^{2})^{\ast }}$ will be the parameter space of the family of $B_{a}$ -lines which we are going to construct.

For a point $[\mathsf{m}]\in (\mathbb{P}^{2})^{\ast }\setminus [\mathsf{j}]$ , we use the same character $[\mathsf{m}]$ for the corresponding point on $\widetilde{(\mathbb{P}^{2})^{\ast }}$ .

Proof. By noting the natural isomorphism

$$\begin{eqnarray}\displaystyle & & \displaystyle H^{0}(\widetilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\otimes \widetilde{p}_{2}^{\ast }{\mathcal{O}}(E_{0}+2r))\nonumber\\ \displaystyle & & \displaystyle \quad \simeq H^{0}(\widetilde{p}_{2\ast }\widetilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\otimes {\mathcal{O}}(E_{0}+2r)),\nonumber\end{eqnarray}$$

the assertion follows once we show that

(2.2) $$\begin{eqnarray}\tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\simeq {\mathcal{O}}(-E_{0}-2r).\end{eqnarray}$$

Let $\tilde{\unicode[STIX]{x1D70C}}_{1}:B_{a}\times \widetilde{(\mathbb{P}^{2})^{\ast }}\rightarrow B_{a}$ and $\unicode[STIX]{x1D70C}_{1}:B_{a}\times (\mathbb{P}^{2})^{\ast }\rightarrow B_{a}$ be the first projections, and $\tilde{\unicode[STIX]{x1D70C}}_{2}:B_{a}\times \widetilde{(\mathbb{P}^{2})^{\ast }}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ and $\unicode[STIX]{x1D70C}_{2}:B_{a}\times (\mathbb{P}^{2})^{\ast }\rightarrow (\mathbb{P}^{2})^{\ast }$ the second projections. By Remark 2.1.2, when we consider the variety ${\mathcal{L}}$ as a divisor on $B_{a}\times (\mathbb{P}^{2})^{\ast }$ , it is a member of $|\unicode[STIX]{x1D70C}_{1}^{\ast }L+\unicode[STIX]{x1D70C}_{2}^{\ast }{\mathcal{O}}_{(\mathbb{P}^{2})^{\ast }}(1)|$ . Since ${\mathcal{L}}$ does not contain the fiber of $B_{a}\times (\mathbb{P}^{2})^{\ast }\rightarrow (\mathbb{P}^{2})^{\ast }$ over $[\mathsf{j}]$ , the variety $\widetilde{{\mathcal{L}}}$ is the total pullback of ${\mathcal{L}}$ by $B_{a}\times \widetilde{(\mathbb{P}^{2})^{\ast }}\rightarrow B_{a}\times (\mathbb{P}^{2})^{\ast }$ . Hence $\widetilde{{\mathcal{L}}}$ belongs to the linear system $|\tilde{\unicode[STIX]{x1D70C}}_{1}^{\ast }L+\tilde{\unicode[STIX]{x1D70C}}_{2}^{\ast }(E_{0}+r)|$ since ${\mathcal{O}}(E_{0}+r)=b^{\ast }{\mathcal{O}}_{(\mathbb{P}^{2})^{\ast }}(1)$ .

To compute $\tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)$ , let us consider the following exact sequence:

$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle \widetilde{\unicode[STIX]{x1D70C}}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-2L)\otimes \widetilde{\unicode[STIX]{x1D70C}}_{2}^{\ast }{\mathcal{O}}(-E_{0}-r)\nonumber\\ \displaystyle & \rightarrow & \displaystyle \widetilde{\unicode[STIX]{x1D70C}}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\rightarrow \widetilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\rightarrow 0,\nonumber\end{eqnarray}$$

which is obtained from the natural exact sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{O}}_{B_{a}\times \widetilde{(\mathbb{P}^{2})^{\ast }}}(-\widetilde{{\mathcal{L}}})\rightarrow {\mathcal{O}}_{B_{a}\times \widetilde{(\mathbb{P}^{2})^{\ast }}}\rightarrow {\mathcal{O}}_{\widetilde{{\mathcal{L}}}}\rightarrow 0\end{eqnarray}$$

by tensoring $\widetilde{\unicode[STIX]{x1D70C}}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)$ . By Lemma 2.3.1, the pushforward of the exact sequence by $\widetilde{\unicode[STIX]{x1D70C}}_{2}$ is

$$\begin{eqnarray}\displaystyle 0\rightarrow \tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\rightarrow {\mathcal{O}}(-E_{0}-r)^{\oplus 2}\rightarrow {\mathcal{O}}\rightarrow R^{1}\tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\rightarrow 0. & & \displaystyle \nonumber\end{eqnarray}$$

Note that, for a point $[\mathsf{m}]\not =[\mathsf{j}]$ , it holds that $H^{0}({\mathcal{O}}_{L_{\mathsf{m}}}(H-L))\simeq \mathbb{C}$ and $H^{1}({\mathcal{O}}_{L_{\mathsf{m}}}(H-L))=\{0\}$ by Lemma 2.1.1. Therefore, by Grauert’s theorem, $\tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)$ is an invertible sheaf possibly outside $E_{0}$ , and the support of $R^{1}:=R^{1}\tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)$ is contained in $E_{0}$ .

Let ${\mathcal{I}}$ be the image of the map ${\mathcal{O}}(-E_{0}-r)^{\oplus 2}\rightarrow {\mathcal{O}}$ in the above exact sequence, which is an ideal sheaf. Note that $R^{1}={\mathcal{O}}_{\unicode[STIX]{x1D6E5}}$ , where $\unicode[STIX]{x1D6E5}$ is the closed subscheme defined by ${\mathcal{I}}$ . We show that $\unicode[STIX]{x1D6E5}=E_{0}$ as a closed subscheme. Indeed, $\unicode[STIX]{x1D6E5}$ is a member of $|E_{0}+r|$ , or the intersection of two members of  $|E_{0}+r|$ . In particular, $\unicode[STIX]{x1D6E5}$ is nonempty. Noting the support of ${\mathcal{O}}_{\unicode[STIX]{x1D6E5}}=R^{1}$ is contained in $E_{0}$ , the subscheme $\unicode[STIX]{x1D6E5}$ must be equal to $E_{0}$ .

Therefore, the map ${\mathcal{O}}(-E_{0}-r)^{\oplus 2}\rightarrow {\mathcal{O}}$ is decomposed as ${\mathcal{O}}(-E_{0}-r)^{\oplus 2}\rightarrow {\mathcal{O}}(-E_{0}){\hookrightarrow}{\mathcal{O}}$ and ${\mathcal{O}}(-E_{0}-r)^{\oplus 2}\rightarrow {\mathcal{O}}(-E_{0})$ is surjective. Hence the kernel $\tilde{p}_{2\ast }\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)$ of the map ${\mathcal{O}}(-E_{0}-r)^{\oplus 2}\rightarrow {\mathcal{O}}(-E_{0})$ is isomorphic to ${\mathcal{O}}(-E_{0}-2r)$ . Now we have shown (2.2) and finished the proof of this lemma.◻

Proof. Note that ${\mathcal{U}}_{1}$ is Cohen–Macaulay since it is a divisor on the smooth variety $\widetilde{{\mathcal{L}}}$ . Therefore, the flatness follows from the smoothness of $\widetilde{(\mathbb{P}^{2})^{\ast }}$ and the descriptions of fibers as in (1) and (2). Besides, the irreducibility of ${\mathcal{U}}_{1}$ also follows from the descriptions of fibers, which now we are going to give below.

Note that, by the uniqueness of ${\mathcal{U}}_{1}$ , the group $G$ acts on ${\mathcal{U}}_{1}$ , where $G$ acts on ${\mathcal{L}}$ and hence on $\widetilde{{\mathcal{L}}}$ by Remark 2.1.2. Let $x$ be a point of $\widetilde{(\mathbb{P}^{2})^{\ast }}$ . Set $[\mathsf{m}]:=b(x)\in (\mathbb{P}^{2})^{\ast }$ . Note that the fiber of $\widetilde{{\mathcal{L}}}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ over $x$ is $L_{\mathsf{m}}$ .

Proof of (1). If $x\not \in E_{0}$ , then $L_{\mathsf{m}}\subset {\mathcal{U}}_{1}$ or ${{\mathcal{U}}_{1}}_{|L_{\mathsf{m}}}$ is the negative section of $L_{\mathsf{m}}\simeq \mathbb{F}_{1}$ by Lemma 2.1.1 since ${\mathcal{U}}_{1}\in |\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\otimes \tilde{p}_{2}^{\ast }{\mathcal{O}}(E_{0}+2r)|$ . We show that the latter occurs for any $x\not \in E_{0}$ , which implies the assertion (1). If $L_{\mathsf{m}}\subset {\mathcal{U}}_{1}$ for an $x\not \in E_{0}$ such that $[\mathsf{m}]\not \in \ell _{1}\cup \ell _{2}$ , then, by the description of the group action of $G$ (Corollary 1.4.3), $L_{\mathsf{m}}\subset {\mathcal{U}}_{1}$ hold for all such $x$ ’s, which implies that ${\mathcal{U}}_{1}=\widetilde{{\mathcal{L}}}$ , a contradiction. If $L_{\mathsf{m}}\subset {\mathcal{U}}_{1}$ for an $x\not \in E_{0}$ such that $[\mathsf{m}]\in \ell _{i}$ for $i=1~\text{or}~2$ , then, again by the group action of $G$ , we have $L_{\mathsf{m}}\subset {\mathcal{U}}_{1}$ for all such $x$ ’s, which implies that ${\mathcal{U}}_{1}$ contains the pullback of the strict transform $\ell _{i}^{\prime }\subset \widetilde{(\mathbb{P}^{2})^{\ast }}$ of $\ell _{i}$ . Since $\ell _{i}^{\prime }$ is a ruling of $\widetilde{(\mathbb{P}^{2})^{\ast }}$ , this implies that $H^{0}(\tilde{p}_{1}^{\ast }{\mathcal{O}}_{B_{a}}(H-L)\otimes \tilde{p}_{2}^{\ast }{\mathcal{O}}(E_{0}+r))\not =0$ , which is impossible by the proof of Lemma 2.3.2.

Proof of (2). Now assume that $x\in E_{0}$ . By a similar argument to the above one using the group action, we see that ${{\mathcal{U}}_{1}}_{|L_{\mathsf{j}}\times \{x\}}$ is the negative section $\unicode[STIX]{x1D6FE}_{a}$ plus a ruling if $x$ is not contained in the strict transforms $\ell _{i}^{\prime }$ of $\ell _{i}$ ( $i=1,2$ ). Therefore, ${{\mathcal{U}}_{1}}_{|L_{\mathsf{j}}\times E_{0}}$ is a member of the linear system $|{\mathcal{O}}_{L_{\mathsf{j}}}(H-L)\boxtimes {\mathcal{O}}_{E_{0}}(1)|$ on $L_{\mathsf{j}}\times E_{0}$ . Suppose by contradiction that ${{\mathcal{U}}_{1}}_{|L_{\mathsf{j}}\times \{x\}}=L_{\mathsf{j}}\times \{x\}$ for $x=\ell _{1}^{\prime }\cap E_{0}$ or $\ell _{2}^{\prime }\cap E_{0}$ . Then, since the group action interchanges $\ell _{1}^{\prime }\cap E_{0}$ and $\ell _{2}^{\prime }\cap E_{0}$ , ${{\mathcal{U}}_{1}}_{|L_{\mathsf{j}}\times \{x\}}=L_{\mathsf{j}}\times \{x\}$ for both $x=\ell _{1}^{\prime }\cap E_{0}$ and $\ell _{2}^{\prime }\cap E_{0}$ . This would imply that $|{\mathcal{O}}_{L_{\mathsf{j}}}(H-L)\boxtimes {\mathcal{O}}_{E_{0}}(-1)|$ is nonempty, which is absurd. Therefore, the assertion (2) follows.◻

Notation 2.4.1.

1. (1) We recall that the two singular $\unicode[STIX]{x1D70B}_{b}$ -fibers are denoted by $F_{1}$ and $F_{2}$ , which are singular quadrics. We denote by $v_{i}$ the vertex of $F_{i}$ ( $i=1,2$ ).

2. (2) We denote by $F_{i}^{\prime }$ the strict transform on $B_{a}$ of $F_{i}$ ( $i=1,2$ ).

3. (3) By Corollary 1.2.3, we have $L\cdot \unicode[STIX]{x1D6FE}_{a}=1$ , and, by a standard property of flop, we have $L\cdot \unicode[STIX]{x1D6FE}_{b}=-1$ . This and the equality (1.4) imply that $\unicode[STIX]{x1D6FE}_{b}$ is a $\unicode[STIX]{x1D70B}_{b}$ -section. Therefore, $\unicode[STIX]{x1D6FE}_{b}$ does not pass through $v_{1}$ nor $v_{2}$ and so $B_{b}{\dashrightarrow}B_{a}$ is isomorphic near $v_{1}$ and $v_{2}$ . We denote by $v_{i}^{\prime }$ the point on $B_{a}$ corresponding to $v_{i}$ $(i=1,2)$ .

Proof. (1) We use the group actions of $G$ on $B_{a}$ and $B_{b}$ . The action of $G$ on $B_{b}$ fixes or interchanges $F_{1}$ and $F_{2}$ , and hence $v_{1}$ and $v_{2}$ . Since $B_{b}{\dashrightarrow}B_{a}$ is isomorphic near $v_{1}$ and $v_{2}$ as we noted in Notation 2.4.1(3), the group action on $B_{a}$ fixes or interchanges $v_{1}^{\prime }$ and $v_{2}^{\prime }$ . By Corollary 1.4.3, this implies that the images of $v_{1}^{\prime }$ and $v_{2}^{\prime }$ correspond to the lines $\ell _{1}$ and $\ell _{2}$ by projective duality.

(2) Let $C_{0}^{\prime }(\mathsf{m})$ be the strict transform of $C_{0}(\mathsf{m})$ on $B_{b}$ . Note that $H\cdot C_{0}(\mathsf{m})=0$ by Lemma 2.1.1. If $C_{0}(\mathsf{m})\cap \unicode[STIX]{x1D6FE}_{a}\not =\emptyset$ , then $H\cdot C_{0}^{\prime }(\mathsf{m})<H\cdot C_{0}(\mathsf{m})=0$ by a standard property of flop, which is a contradiction since $H$ is nef on $B_{b}$ .◻

Proof. We show the first assertions of (1) and (2). Since the proofs are similar, we only show (2), which is more difficult. We also only prove the only if part since the if part follows by reversing the argument. Recall that $\unicode[STIX]{x1D6FE}_{a}+f\sim (H-L)_{|L_{\mathsf{j}}}$ by Lemma 2.1.1. Thus $H\cdot f=1$ . Since $f$ intersects $\unicode[STIX]{x1D6FE}_{a}$ transversely at one point, and $H\cdot \unicode[STIX]{x1D6FE}_{a}=-1$ , we have $H\cdot f^{\prime }=0$ , where $f^{\prime }$ is the strict transform of $f$ on $B_{b}$ . Hence $f^{\prime }$ is contained in a $\unicode[STIX]{x1D70B}_{b}$ -fiber $F$ . By the equality (1.4), we have $-K_{F}=-K_{B_{b}}|_{F}=2L|_{F}$ . Therefore, $f^{\prime }$ is a ruling of $F$ since $L\cdot f^{\prime }=L\cdot f+1=1$ .

The latter assertions of (1) and (2) follows from Lemma 2.4.1(1). ◻

Proof. The assertions follow from Proposition 2.4.2(1) and (2), and the description of rulings on quadric surfaces. ◻

Condition 3.2.1. Let $\mathsf{m}\subset \mathbb{P}^{2}$ be a line and $R\subset L_{\mathsf{m}}$ a member of the linear system $|(H+gL)_{|L_{\mathsf{m}}}|$ . We consider the following conditions for $\mathsf{m}$ and  $R$ :

1. (a) $[\mathsf{m}]\not \in \ell _{1}\cup \ell _{2}$ . In particular, $v_{1}^{\prime },v_{2}^{\prime }\not \in R$ by Lemma 2.4.1(1).

2. (b) $R$ is smooth.

3. (c) $R\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ .

4. (d) $R$ intersects $F_{1}^{\prime }$ and $F_{2}^{\prime }$ transversely at $g+1$ points, respectively (note that, by $R\sim (H+gL)_{|L_{\mathsf{m}}}$ , we have $F_{i}^{\prime }\cdot R=H\cdot R=g+1$ ).

   Note that the condition (c) implies that $R\cap F_{1}^{\prime }\cap F_{2}^{\prime }=R\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ .

It is easy to see that, if $g\geqslant 0$ , then general $\mathsf{m}$ and $R$ satisfy these conditions by Proposition 2.1.2.

Proof. Under the assumption that $[\mathsf{m}]\not \in \ell _{1}\cup \ell _{2}$ , the strict transform $C_{0}^{\prime }(\mathsf{m})$ of $C_{0}(\mathsf{m})$ on $B_{b}$ is a ruling of a $\unicode[STIX]{x1D70B}_{b}$ -fiber which is different from $F_{1}$ and $F_{2}$ by Proposition 2.4.2(1). In particular, $C_{0}(\mathsf{m})$ is not contained in $F_{i}^{\prime }$ . We see that ${F_{i}^{\prime }}_{|L_{\mathsf{m}}}$ is linearly equivalent to $C_{0}(\mathsf{m})+L_{|L_{\mathsf{m}}}$ by (1.4) since $F_{i}^{\prime }\sim H$ . Therefore, if ${F_{i}^{\prime }}_{|L_{\mathsf{m}}}$ were reducible, then $C_{0}(\mathsf{m})\subset {F_{i}^{\prime }}_{|L_{\mathsf{m}}}\,\!$ , a contradiction.◻

Proof. (1) The assumption $f\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ is equivalent to that $\unicode[STIX]{x1D70B}_{a}(f)\in \mathbb{P}^{2}$ belongs to the open orbit of $G$ . Therefore, the assertion (1) follows from Corollary 1.4.3 and Lemma 2.4.1(1).

(2) We show that $f$ intersects $F_{i}^{\prime }$ $(i=1,2)$ at one point. By (1.4), we have $F_{i}^{\prime }\cdot f=H\cdot f=1$ since $-K_{B_{a}}\cdot f=2$ and $L\cdot f=0$ . Therefore, we have only to show that $f$ is not contained in $F_{i}^{\prime }$ . If $f\subset F_{i}^{\prime }$ , then the strict transform $f^{\prime }$ of $f$ is contained in $F_{i}$ . Then, however, $-K_{F_{i}}\cdot f^{\prime }=2L_{|F_{i}}\cdot f^{\prime }=0$ , a contradiction.

The assumption implies that $f\cap F_{1}^{\prime }\cap F_{2}^{\prime }=f\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ . Therefore, we have the assertion (2).◻

Proof. We use the notation in Section 2 freely.

(1) By definition, $C_{R}\rightarrow R$ is flat and finite of degree two (note that $R$ satisfies the conditions of Definition 3.1.1 by Condition 3.2.1(a) and (c)). Moreover, by Condition 3.2.1(d), the branch locus of $C_{R}\rightarrow R$ is smooth and of degree $2g+2$ . Therefore, by Condition 3.2.1(b), the scheme $C_{R}$ is a smooth hyperelliptic curve of genus $g$ .

(2) The assertions for $C_{f}$ can be proved similarly to (1) by Lemma 3.2.2. We show the assertion for $M_{f}$ . Since $f\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ , a $B_{a}$ -line intersecting $f$ satisfies Proposition 2.3.3(1), namely, such a $B_{a}$ -line is the negative section of $L_{\mathsf{m}}$ with a line $\mathsf{m}\not =\mathsf{j}$ containing $\unicode[STIX]{x1D70B}_{a}(f)$ . Therefore, the support of $M_{f}$ is the line of $(\mathbb{P}^{2})^{\ast }$ corresponding to the point $\unicode[STIX]{x1D70B}_{a}(f)$ by projective duality. Moreover, $C_{f}\rightarrow \operatorname{Supp}M_{f}$ is one to one since $f$ intersects a $B_{a}$ -line at most at one point. Thus $M_{f}$ is actually the line.

To show the remaining assertions (3) and (4), we investigate fibers of $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ induced by $\tilde{p}_{2}$ . For this, we note that the fiber over a point $s\in \widetilde{(\mathbb{P}^{2})^{\ast }}$ is the intersection between $L_{\mathsf{m}}$ and the $B_{a}$ -line corresponding to $s$ . Therefore, the fiber over $[\mathsf{m}]\in \widetilde{(\mathbb{P}^{2})^{\ast }}$ can be identified with the negative section $C_{0}(\mathsf{m})$ of $L_{\mathsf{m}}$ . Recall that $E_{0}$ is as in Notation 2.3.1. Let $t$ be the point of $E_{0}$ over which the fiber of ${\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ is the union $\unicode[STIX]{x1D6FE}_{a}\cup \unicode[STIX]{x1D70B}_{a}^{-1}(\mathsf{j}\cap \mathsf{m})$ . Then the fiber of $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ over $t$ is $\unicode[STIX]{x1D70B}_{a}^{-1}(\mathsf{j}\cap \mathsf{m})$ . Besides, over $\widetilde{(\mathbb{P}^{2})^{\ast }}\setminus ([\mathsf{m}]\cup t)$ , the map $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ is one to one, hence is an isomorphism by the Zariski main theorem.

We denote by $E_{\mathsf{m}}$ and $E_{t}$ the exceptional curves of $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ over $[\mathsf{m}]$ and $t$ , respectively. By Lemmas 2.4.1(2), 3.2.1 and Corollary 2.4.3, the map $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ is an étale double cover near $C_{0}(\mathsf{m})$ . Hence $\tilde{p}_{1}^{-1}(C_{0}(\mathsf{m}))\cap {\mathcal{U}}_{1}$ consists of two disjoint smooth rational curves, and $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}$ is smooth near $\tilde{p}_{1}^{-1}(C_{0}(\mathsf{m}))\cap {\mathcal{U}}_{1}$ . Since $E_{\mathsf{m}}\subset \tilde{p}_{1}^{-1}(C_{0}(\mathsf{m}))\cap {\mathcal{U}}_{1}$ , we see that $E_{\mathsf{m}}$ is one of its components. Therefore, $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}\rightarrow \widetilde{(\mathbb{P}^{2})^{\ast }}$ is the blow-up at $[\mathsf{m}]$ around $[\mathsf{m}]$ .

(3) We show that $M_{R}$ is reduced. Indeed, since $R\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ , a $B_{a}$ -line intersecting $R$ satisfies Proposition 2.3.3(1) or is the $B_{a}$ -line $\unicode[STIX]{x1D6FE}_{a}$ plus the $\unicode[STIX]{x1D70B}_{a}$ -fiber $\unicode[STIX]{x1D70B}_{a}^{-1}(\mathsf{j}\cap \mathsf{m})$ . Therefore, such a $B_{a}$ -line intersects $R$ at one point except the negative section $C_{0}(\mathsf{m})$ which intersects $R$ at $g$ points counted with multiplicity. Thus $C_{R}\rightarrow \operatorname{Supp}M_{R}$ is one to one outside the point $[\mathsf{m}]$ , which in particular shows that $M_{R}$ is reduced.

We show that $M_{R}$ has a $g$ -ple point at $[\mathsf{m}]$ . Note that $\widetilde{M}_{R}\rightarrow M_{R}$ is isomorphic outside $[\mathsf{j}]\in M_{R}$ since $b:\widetilde{(\mathbb{P}^{2})^{\ast }}\rightarrow (\mathbb{P}^{2})^{\ast }$ is the blow-up at $[\mathsf{j}]$ . Therefore, to show $M_{R}$ has a $g$ -ple point at $[\mathsf{m}]$ , it suffices to show that $\widetilde{M}_{R}$ has a $g$ -ple point at $[\mathsf{m}]$ , where we denote by the same letter the point of $\widetilde{M}_{R}$ corresponding to $[\mathsf{m}]$ . Note that $R\cdot C_{0}(\mathsf{m})=g$ in $L_{\mathsf{m}}$ . Therefore, by the results of the paragraph before the proof of (3), we have $C_{R}\cdot E_{\mathsf{m}}=g$ in $\tilde{p}_{1}^{-1}(L_{\mathsf{m}})\cap {\mathcal{U}}_{1}$ . Hence we see that $\widetilde{M}_{R}$ has a $g$ -ple point at $[\mathsf{m}]$ by blowing down $E_{\mathsf{m}}$ .

Now we compute $\deg M_{R}$ . Let us take a general fiber $f$ of $L_{\mathsf{m}}\rightarrow \mathsf{m}$ such that $f\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ and $R\cap f\not \in F_{1}^{\prime }\cup F_{2}^{\prime }\cup C_{0}(\mathsf{m})$ . Then, since $R\cap f\not \in F_{1}^{\prime }\cup F_{2}^{\prime }$ , we see that $C_{R}$ and $C_{f}$ intersect transversely at two points, which is the inverse image of one point $R\cap f$ . Since $R\cap f\not \in C_{0}(\mathsf{m})$ , we see that $C_{R}$ and $C_{f}$ does not intersect on $E_{\mathsf{m}}$ . Therefore, the intersection multiplicity of $M_{R}$ and $M_{f}$ at $[\mathsf{m}]$ is $g$ . Thus we conclude that $\deg M_{R}=M_{R}\cdot M_{f}=g+2$ since $M_{f}$ is a line by generality of $f$ and the assertion (2).

It remains to show that $M_{R}$ is smooth outside $[\mathsf{m}]$ . Let $M_{R}^{\prime }$ be the strict transform of $M_{R}$ by the blow-up of $\mathbb{P}^{2}$ at $[\mathsf{m}]$ . Since $\deg M_{R}=g+2$ and $M_{R}$ has a $g$ -ple point at $[\mathsf{m}]$ , the arithmetic genus of $M_{R}^{\prime }$ is $g$ . Since we have shown that $C_{R}\rightarrow M_{R}$ is birational above, $C_{R}$ is the normalization of $M_{R}^{\prime }$ . Since $p_{a}(M_{R}^{\prime })=g(C_{R})=g$ , we see that $M_{R}^{\prime }$ is smooth, hence $M_{R}$ is smooth outside $[\mathsf{m}]$ .

(4) We have already shown that $C_{R}\rightarrow M_{R}$ is the normalization. A general line in $(\mathbb{P}^{2})^{\ast }$ through $[\mathsf{m}]$ intersects $M_{R}$ at two points outside $[\mathsf{m}]$ since $M_{R}$ has a $g$ -ple point at $[\mathsf{m}]$ . Therefore, we obtain the description of the unique $g_{2}^{1}$ of the hyperelliptic curve $C_{R}$ .◻

Notation 3.3.1. (The marked point $[\mathsf{j}]_{R}$ on $C_{R}$ ) Assume that $R$ satisfies Condition 3.2.1(a)–(d). Then, since $R\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ , the $\unicode[STIX]{x1D70B}_{a}$ -fiber $\unicode[STIX]{x1D70B}_{a}^{-1}(\mathsf{m}\cap \mathsf{j})$ is the only one which intersects both $R$ and $\unicode[STIX]{x1D6FE}_{a}$ . Therefore, $\unicode[STIX]{x1D6FE}_{a}\cup \unicode[STIX]{x1D70B}_{a}^{-1}(\mathsf{m}\cap \mathsf{j})$ is the unique $B_{a}$ -line intersecting $R$ of the form $\unicode[STIX]{x1D6FE}_{a}$ plus a $\unicode[STIX]{x1D70B}_{a}$ -fiber. Since $C_{R}\rightarrow M_{R}$ is an isomorphism outside $[\mathsf{m}]$ by Proposition 3.3.1(1) and (4), we denote by $[\mathsf{j}]_{R}$ the point of the hyperelliptic curve $C_{R}$ corresponding to this $B_{a}$ -line (note that this point $[\mathsf{j}]_{R}$ is mapped to $[\mathsf{j}]\in M_{R}\subset (\mathbb{P}^{2})^{\ast }$ ).

Proof. Let $F$ be one of the two singular $\unicode[STIX]{x1D70B}_{b}$ -fibers and $F^{\prime }$ its strict transform on $B_{a}$ . By Condition 3.2.1(d), $R$ intersects $F^{\prime }$ transversely at $g+1$ points, which we denote by $s_{1},\ldots ,s_{g+1}$ . By Proposition 3.3.1(1), these points are contained in the branched locus of the hyperelliptic double cover $C_{R}\rightarrow R$ . We denote by $t_{1},\ldots ,t_{g+1}$ the inverse images on $C_{R}$ of $s_{1},\ldots ,s_{g+1}$ , and by $u_{1},\ldots ,u_{g+1}$ the images on $M_{R}$ of $t_{1},\ldots ,t_{g+1}$ . By Proposition 3.3.1(1), $C_{0}(\mathsf{m})$ is not the strict transform of a ruling of $F$ since $[\mathsf{m}]\not \in \ell _{1}\cup \ell _{2}$ . Therefore, $u_{1},\ldots ,u_{g+1}\not =[\mathsf{m}]$ . By Proposition 3.3.1(2), $\unicode[STIX]{x1D70B}_{a}^{-1}(\mathsf{m}\cap \mathsf{j})$ is not the strict transform of a ruling of $F$ since $[\mathsf{m}]\not \in \ell _{1}\cup \ell _{2}$ . Therefore, $u_{1},\ldots ,u_{g+1}\not =[\mathsf{j}]$ . By Proposition 2.4.2(1), $u_{1},\ldots ,u_{g+1}$ are contained in $\ell :=\ell _{1}$ or $\ell _{2}$ . Therefore, since $\ell$ and $M_{R}$ contain $[\mathsf{j}]$ , and $\deg M_{R}=g+2$ , we have $\ell _{|M_{R}}=u_{1}+\cdots +u_{g+1}+[\mathsf{j}]$ . Then, by the definition of $\unicode[STIX]{x1D703}_{R}$ , we have $\unicode[STIX]{x1D703}_{R}=t_{1}+\cdots +t_{g+1}-h_{R}$ . Now the assertion follows from [Reference Arbarello, Cornalba, Griffiths and Harris1, p. 288, Exercise 32].◻

Proof. (1) We show that the linear system $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ has no base points. By (4.2), we see that $\operatorname{Bs}|\unicode[STIX]{x1D703}+g_{2}^{1}+p|\subset \{p\}$ . By the Serre duality, we have

$$\begin{eqnarray}H^{1}(\unicode[STIX]{x1D703}+g_{2}^{1}+p)\simeq H^{0}(K_{C}-\unicode[STIX]{x1D703}-g_{2}^{1}-p)^{\ast }=H^{0}(\unicode[STIX]{x1D703}-g_{2}^{1}-p)=0\end{eqnarray}$$

since $\unicode[STIX]{x1D703}$ is ineffective. Similarly, we have $H^{1}(\unicode[STIX]{x1D703}+g_{2}^{1})=\{0\}$ . Therefore, by the Riemann–Roch theorem,

$$\begin{eqnarray}h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p)-h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1})=\unicode[STIX]{x1D712}(\unicode[STIX]{x1D703}+g_{2}^{1}+p)-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D703}+g_{2}^{1})=1,\end{eqnarray}$$

which implies that $p\not \in \operatorname{Bs}|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ .

By the above argument, we see that $h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p)=\deg (\unicode[STIX]{x1D703}+g_{2}^{1}+p)+1-g=3$ . Therefore, $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ gives a morphism $\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}+g_{2}^{1}+p|}:C\rightarrow \mathbb{P}(V)\simeq \mathbb{P}^{2}$ with $V=H^{0}(C,{\mathcal{O}}_{C}(\unicode[STIX]{x1D703}+g_{2}^{1}+p))^{\ast }$ . Let $M:=\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}+g_{2}^{1}+p|}(C)$ be the image of $C$ . We show that $C\rightarrow M$ is birational. Note that by the Riemann–Roch theorem and $h^{1}(\unicode[STIX]{x1D703}+p)=h^{0}(K-\unicode[STIX]{x1D703}-p)=0$ , we have $h^{0}(\unicode[STIX]{x1D703}+p)=1$ . Therefore, the hyperelliptic double cover $\unicode[STIX]{x1D711}_{|g_{2}^{1}|}:C\rightarrow \mathbb{P}^{1}$ factors through the map $\unicode[STIX]{x1D711}_{|g_{2}^{1}+\unicode[STIX]{x1D703}+p|}$ . So we have only to show that $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ separates the two points in a member of $|g_{2}^{1}|$ . This is equivalent to $h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p-g_{2}^{1})=h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p)-2$ , which follows from the above computations. Therefore, we have shown that $C\rightarrow M$ is birational. This implies that the degree of $M$ is $g+2$ .

(2) Since $h^{0}(\unicode[STIX]{x1D703}+p)=1$ as in the proof of (1), the linear system $|\unicode[STIX]{x1D703}+p|$ has a unique member $D$ . We see that $D$ is mapped to a point since $\unicode[STIX]{x1D703}+g_{2}^{1}+p$ is the pullback of ${\mathcal{O}}_{\mathbb{P}^{2}}(1)|_{M}$ and $h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p-(\unicode[STIX]{x1D703}+p))=h^{0}(g_{2}^{1})=2$ .

(3) The assertion clearly follows from (2).

(4) The point $p$ is not contained in the support of $D$ since $h^{0}(\unicode[STIX]{x1D703}+p-p)=h^{0}(\unicode[STIX]{x1D703})=0$ . Let us consider points of $S\setminus \{p\}$ . Without loss of generality, we have only to show that $h^{0}(\unicode[STIX]{x1D703}+p-p_{1})=0$ . By the Riemann–Roch theorem, the assertion is equivalent to $h^{1}(\unicode[STIX]{x1D703}+p-p_{1})=0$ . By (4.2), $\unicode[STIX]{x1D703}+p-p_{1}=p_{2}+\cdots +p_{g+1}+p-g_{2}^{1}$ . Therefore, by the Serre duality, we have

$$\begin{eqnarray}h^{1}(\unicode[STIX]{x1D703}+p-p_{1})=h^{0}(g\times g_{2}^{1}-(p_{2}+\cdots +p_{g+1}+p))\end{eqnarray}$$

since $K_{C}=(g-1)g_{2}^{1}$ . Now it is easy to verify this is zero by using the hyperelliptic morphism $C\rightarrow \mathbb{P}^{1}$ .

(5) First we show that $t$ is different from the image of any point $x$ of $C\setminus D$ . Indeed, we have

$$\begin{eqnarray}h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p-(\unicode[STIX]{x1D703}+p)-x)=h^{0}(g_{2}^{1}-x)=1,\end{eqnarray}$$

which means that $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ separates $D$ and $x$ . In particular, we have the former assertion of (5) by (4).

We show that $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ separates any two of $p_{1},\ldots ,p_{g+1}$ . Without loss of generality, we have only to consider the case of $p_{1}$ and $p_{2}$ . It suffices to show that $h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p-p_{1}-p_{2})=h^{0}(\unicode[STIX]{x1D703}+g_{2}^{1}+p)-2=1$ , which is equivalent to $h^{1}(\unicode[STIX]{x1D703}+g_{2}^{1}+p-p_{1}-p_{2})=0$ by the Riemann–Roch theorem. By the presentation (4.2), we have $h^{1}(\unicode[STIX]{x1D703}+g_{2}^{1}+p-p_{1}-p_{2})=h^{1}(p_{3}+\cdots +p_{g+1}+p)$ . By the Serre duality, we have

$$\begin{eqnarray}h^{1}(p_{3}+\cdots +p_{g+1}+p)=h^{0}((g-1)g_{2}^{1}-p_{3}-\cdots -p_{g+1}-p)\end{eqnarray}$$

since $K_{C}=(g-1)g_{2}^{1}$ . Now it is easy to verify the r.h.s. is zero by using the hyperelliptic morphism $C\rightarrow \mathbb{P}^{1}$ .

The same argument shows that $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ separates any two of $p_{1}^{\prime },\ldots ,p_{g+1}^{\prime }$ . Moreover, if $p$ is distinct from a $p_{i}$ or $p_{j}^{\prime }$ , the same proof works for the separation of $p$ and $p_{i}$ or $p_{j}^{\prime }$ .

It remains to show that $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ separates one of $p_{1},\ldots ,p_{g+1}$ and one of $p_{1}^{\prime },\ldots ,p_{g+1}^{\prime }$ . Without loss of generality, we have only to consider the case of $p_{1}$ and $p_{1}^{\prime }$ . If $p=p_{1}$ , then $p\not =p_{1}^{\prime }$ , and hence we have already shown that the images of $p=p_{1}$ and $p_{1}^{\prime }$ are different. Thus we may assume that $p\not =p_{1},p_{1}^{\prime }$ . By (4.2), $D_{1}:=p+p_{1}+\cdots +p_{g+1}$ and $D_{2}:=p+p_{1}^{\prime }+\cdots +p_{g+1}^{\prime }$ are two distinct members of $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ . If the images of $p_{1}$ and $p_{1}^{\prime }$ by the map $\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}+g_{2}^{1}+p|}$ coincides, then the images of $D_{1}$ and $D_{2}$ coincides since they are the line through the images of $p$ and $p_{1}$ , and the line through the images of $p$ and $p_{1}^{\prime }$ . This is a contradiction to a property of the map defined by $|\unicode[STIX]{x1D703}+g_{2}^{1}+p|$ .◻

Proof of Theorem 4.2.1.

We set $V\,=\,H^{0}(C,{\mathcal{O}}_{C}(\unicode[STIX]{x1D703}\,+\,g_{2}^{1}\,+\,p))^{\ast }$ . Let $M\,\subset \,\mathbb{P}(V)$ be the $\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}+g_{2}^{1}+p|}$ -image of $C$ . Let $r_{1},\ldots ,r_{g+1}$ and $r_{1}^{\prime },\ldots ,r_{g+1}^{\prime }\in M$ be the $\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}+g_{2}^{1}+p|}$ -images of the Weierstrass points $p_{1},\ldots ,p_{g+1}$ and $p_{1}^{\prime },\ldots ,p_{g+1}^{\prime }$ of $C$ as in (4.2), respectively. Let $r\in M$ be the image of $p$ and $t\in M$ the image of the unique member of $|\unicode[STIX]{x1D703}+p|$ . By Lemma 4.2.2(4) and (5), $r$ , $t$ , $r_{1},\ldots ,r_{g+1}$ , $r_{1}^{\prime },\ldots ,r_{g+1}^{\prime }$ are distinct points (recall that now we are assuming $p$ is not a Weierstrass point). By (4.2), there are two lines $\ell ,\ell ^{\prime }\subset \mathbb{P}(V)$ such that $\ell _{|M}=r_{1}+\cdots +r_{g+1}+r$ and $\ell _{|M}^{\prime }=r_{1}^{\prime }+\cdots +r_{g+1}^{\prime }+r$ (note also that $\deg M=g+2$ by Lemma 4.2.2(1)).

We then identify the polarized space $(\mathbb{P}(V),\ell \cup \ell ^{\prime })$ with $((\mathbb{P}^{2})^{\ast },\ell _{1}\cup \ell _{2})$ (recall the notation as in Proposition 1.2.2). By this identification, the point $r$ corresponds to $[\mathsf{j}]$ . Let $\mathsf{m}$ be the line of $\mathbb{P}^{2}$ such that $[\mathsf{m}]$ corresponds to the point $t$ .

Condition 3.2.1(a). Since $r\not =t$ , the line $\mathsf{m}$ is not the jumping line $\mathsf{j}$ of the bundle ${\mathcal{E}}$ such that $B_{a}\simeq \mathbb{P}({\mathcal{E}})$ . Moreover, $\mathsf{m}$ is not a jumping line of the second kind of ${\mathcal{E}}$ , equivalently, $[\mathsf{m}]\not \in \ell _{1}\cup \ell _{2}$ since $t$ is distinct from $r$ , $r_{1},\ldots ,r_{g+1}$ , $r_{1}^{\prime },\ldots ,r_{g+1}^{\prime }$ . This shows that $[\mathsf{m}]$ satisfies Condition 3.2.1(a).

We look for a member $R$ of the linear system $|C_{0}(\mathsf{m})+(g+1)L_{|L_{\mathsf{m}}}|$ on $L_{\mathsf{m}}\subset B_{a}$ with Condition 3.2.1(b)–(d) such that $C=C_{R}$ . Let $\mathsf{m}_{i}$ and $\mathsf{m}_{j}^{\prime }$ be the lines of $\mathbb{P}^{2}$ such that $[\mathsf{m}_{i}]=r_{i}$ and $[\mathsf{m}_{j}^{\prime }]=r_{j}^{\prime }$ ( $1\leqslant i,j\leqslant g+1$ ). As we have seen above, $\mathsf{m}_{i}$ and $\mathsf{m}_{j}^{\prime }$ are different from $\mathsf{j}$ . Therefore, the negative sections $C_{0}(\mathsf{m}_{i})$ and $C_{0}(\mathsf{m}_{j}^{\prime })$ are $B_{a}$ -lines disjoint from $\unicode[STIX]{x1D6FE}_{a}$ by Lemma 2.4.1(2). Note that the condition for an $R\in |C_{0}(\mathsf{m})+(g+1)L_{|L_{\mathsf{m}}}|$ to intersect one fixed $B_{a}$ -line is at most of codimension $1$ . Hence there exists at least one $R\in |C_{0}(\mathsf{m})+(g+1)L_{|L_{\mathsf{m}}}|$ intersecting the $2g+2$ $B_{a}$ -lines $C_{0}(\mathsf{m}_{i})$ and $C_{0}(\mathsf{m}_{j}^{\prime })$ ( $1\leqslant i,j\leqslant g+1$ ) since $\dim H^{0}(C_{0}(\mathsf{m})+(g+1)L_{|L_{\mathsf{m}}})=2g+3$ .

Condition 3.2.1(d). Since $\mathsf{m}_{i}$ and $\mathsf{m}_{j}^{\prime }$ are different from $\mathsf{m}$ , the curve $R$ intersects each one of $C_{0}(\mathsf{m}_{i})$ and $C_{0}(\mathsf{m}_{j}^{\prime })$ at one point. Since $[\mathsf{m}]$ satisfies Condition 3.2.1(a), the points $v_{1}^{\prime }$ , $v_{2}^{\prime }$ are not contained in $R$ . Therefore, $R\cap C_{0}(\mathsf{m}_{i})$ ( $1\leqslant i\leqslant g+1$ ) are different and so are $R\cap C_{0}(\mathsf{m}_{j}^{\prime })$ ( $1\leqslant j\leqslant g+1$ ). Then, by Proposition 2.4.2(1), $R$ intersects $F_{1}^{\prime }$ and $F_{2}^{\prime }$ at least at $g+1$ points $R\cap C_{0}(\mathsf{m}_{i})$ ( $1\leqslant i\leqslant g+1$ ) and $R\cap C_{0}(\mathsf{m}_{j}^{\prime })$ ( $1\leqslant j\leqslant g+1$ ), respectively. Therefore, $R$ intersects $F_{1}^{\prime }$ and $F_{2}^{\prime }$ at $g+1$ points respectively transversely since $F_{1}^{\prime }\cdot R=F_{2}^{\prime }\cdot R=g+1$ . This shows that $R$ satisfies Condition 3.2.1(d).

Condition 3.2.1(c). Assume by contradiction that $R$ intersects $\unicode[STIX]{x1D6FE}_{a}$ . Then $R$ would intersect $F_{1}^{\prime }$ and $F_{2}^{\prime }$ at some points outside $C_{0}(\mathsf{m}_{i})$ and $C_{0}(\mathsf{m}_{j}^{\prime })$ ( $1\leqslant i,j\leqslant g+1$ ) since $\unicode[STIX]{x1D6FE}_{a}\subset F_{1}^{\prime }\cap F_{2}^{\prime }$ and $\unicode[STIX]{x1D6FE}_{a}$ is disjoint from $C_{0}(\mathsf{m}_{i})$ and $C_{0}(\mathsf{m}_{j}^{\prime })$ . This contradicts the argument to show Condition 3.2.1(d). Therefore, $R$ satisfies Condition 3.2.1(c).

Condition 3.2.1(b). It suffices to show that $R$ is irreducible. Assume by contradiction that $R$ is reducible. Then $R$ contains a ruling of $L_{\mathsf{m}}$ , say, $f$ . We have $f\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ since $R\cap \unicode[STIX]{x1D6FE}_{a}=\emptyset$ . Thus we can define the curve $M_{f}$ , which is a line in $(\mathbb{P}^{2})^{\ast }$ by Proposition 3.3.1(2). Besides the line $M_{f}$ contains $t=[\mathsf{m}]$ , and one of $r_{1},\ldots ,r_{g+1}$ and one of $r_{1}^{\prime },\ldots ,r_{g+1}^{\prime }$ corresponding to $F_{1}^{\prime }\cap f$ and $F_{2}^{\prime }\cap f$ , respectively. By reordering the points, we may assume that $r_{1},r_{1}^{\prime }\in M_{f}$ . Therefore, $t,r_{1},r_{1}^{\prime }$ are collinear. This is, however, a contradiction since the line through $t$ and $r_{1}$ touches $M$ only at $t$ and $r_{1}$ by Lemma 4.2.2(3) (recall that $r_{1}$ is the image of a Weierstrass point of $C$ ).

Finally we show $M=M_{R}$ (note that we can define $M_{R}$ since we have checked that $\mathsf{m}$ and $R$ satisfy Condition 3.2.1(a)–(d)). Note that, by the constructions of $M$ and $M_{R}$ as the images of the map $\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}+g_{2}^{1}+p|}$ and $\unicode[STIX]{x1D711}_{|\unicode[STIX]{x1D703}_{R}+h_{R}+[\mathsf{j}]_{R}|}$ respectively, there exists a line through $t$ and touches both $M$ and $M_{R}$ at $r_{i}$ with multiplicity two $(i=1,\ldots ,g+1)$ , and the same is true for $r_{j}^{\prime }$ $(j=1,\ldots ,g+1)$ . Hence the intersection multiplicities of $M_{R}$ and $M$ at $r_{i}$ and $r_{j}^{\prime }$ are at least two. Therefore, the scheme theoretic intersection $M\cap M_{R}$ contains $r$ , the $2(g+1)$ points $r_{i},r_{j}^{\prime }$ , $i,j=1,...,g+1$ with multiplicity ${\geqslant}2$ and we also have a fat point of multiplicity $g^{2}$ at $t$ . This implies that, if $M\not =M_{R}$ , then $M\cdot M_{R}\geqslant 1+4(g+1)+g^{2}=(g+2)^{2}+1$ , which is a contradiction since $\deg M=\deg M_{R}=g+2$ . Now we conclude that $M_{R}=M$ .◻