Proof. We have $[K_{{\mathbb {F}}}]=-[D_{t_0} + D_{t_1}+D_{x_0} + D_{x_1} + D_y + D_z]$ by [Reference Cox, Little and SchenckCLS11, Thm 8.2.3].

Proof. Clearly, $F^2=0$ because any two distinct fibres are disjoint. Since the intersection $D_{t_0}\cap D_{x_0} \cap D_y \cap D_z$ is a reduced smooth point, $D_{t_0} D_{x_0} D_y D_z=10H^3F=1$ . Similarly, $D_{x_0} \cap D_{x_1} \cap D_y \cap D_z$ is empty, so

$$\begin{align*}D_{x_0} D_{x_1} D_y D_z=10 H^4+(10 \cdot (-d_0+d_0-3d) - 5 \cdot 2d - 2 \cdot 5d)H^3F=0. \end{align*}$$

Rearranging and substituting $H^3F=\frac 1{10}$ gives $H^4=\frac d2$ .

Proof. By [Reference Cox, Little and SchenckCLS11, Thms 6.3.12 and 6.3.13], $aH+bF$ is nef (resp. ample) if and only if its restriction to any torus invariant irreducible curve is nonnegative (resp. positive). Torus invariant irreducible curves on ${\mathbb {F}}$ are intersections of three of the divisors $D_{\rho }$ .

The Proposition then follows from

$$ \begin{align*} (aH+bF)D_{t_0}D_yD_z&=10aH^3F=a,\\ (aH+bF)D_{x_1}D_yD_z&=10(aH^4+(b-a(5d-d_0)))H^3F=b+ad_0,\\ (aH+bF)D_{x_0}D_{x_1}D_y&=2aH^4+2(b-4ad)H^3F=\tfrac15(b+ad). \end{align*} $$

The other triples do not add any extra conditions.

Proof. We prove part 1 since part 2 is similar. By (1.5), $K_X=(H-2F)|_X$ . By Proposition 1.3, if $\min (d,d_0)\geq 3$ , then $H-2F$ is ample on ${\mathbb {F}}(d;d_0)$ , and therefore, its restriction to X is ample too.

Conversely, consider the curve $\Gamma :=X \cap D_{x_0} \cap D_{x_1}$ which is contained in X. Then, $K_X\Gamma =d-2$ , so $d\leq 2$ implies that $K_X$ is not ample. Finally, if $d_0\leq 2$ and $d\geq 3$ , then $d_0 < \frac 78 d$ , and so ${\mathfrak {s}}_0 \subset X$ by Proposition 1.6. Since $(H-2F){\mathfrak {s}}_0=d_0-2$ , we are done.

Proof. By (1.5) and the vanishing of $H^1({\mathbb {F}},{\mathcal {O}}_{{\mathbb {F}}}(-X+H-2F)) = H^1({\mathbb {F}},K_{{\mathbb {F}}})$ , the canonical system of X is spanned by the following $3d-2$ monomials:

$$\begin{align*}t_0^{d_0-2}x_0,\dots, t_1^{d_0-2}x_0,\ t_0^{3d-d_0-2}x_1,\dots,t_1^{3d-d_0-2}x_1.\end{align*}$$

Thus, X is mapped to the image of the toric variety $D_y \cap D_z \cong {\mathbb {F}}_{e}$ in ${\mathbb {P}}^{3d-3}$ . This is an embedding of ${\mathbb {F}}_e$ because $d_0\ge 3$ .

Proof. Most of the statement follows by Lemma 1.8, Proposition 1.9 and Proposition 1.6, reformulating the inequalities in Proposition 1.6 in terms of e (instead of $d_0$ ) and d. It remains to prove the given formulas for the invariants.

We already showed that $p_g=3d-2$ and $q_1=0$ . Since the Leray spectral sequence of the direct image of ${\mathcal {O}}_{X}$ degenerates at page $2$ , we have $h^2({\mathcal {O}}_{X})=h^0(R^2f_*{\mathcal {O}}_{X})$ . By Grothendieck duality,

$$\begin{align*}R^2f_*{\mathcal{O}}_{X} \cong f_*{\mathcal{O}}_{X}(K_X+2F)^\vee\cong f_*{\mathcal{O}}_{X}(H)^\vee\cong {\mathcal{O}}_{{\mathbb{P}}^1} (-d_0) \oplus {\mathcal{O}}_{{\mathbb{P}}^1}(d_0-3d),\end{align*}$$

and since $3d>d_0>0$ , we get $q_2=0$ . Finally, $K_{X}^3=X(H-2F)^3=10(H^4-(d+6)H^3F)=4d-6$ .

Proof. Note that $ |10(H-dF)|$ is nef but not ample by Proposition 1.3. In particular, we cannot apply directly [Reference Ravindra and SrinivasRS06, Theorem 1].

We consider a desingularisation $\tilde {{\mathbb {F}}} \rightarrow {\mathbb {F}}$ of the singular locus, the curves $\mathfrak {s}_2$ and $\mathfrak {s}_5$ , of ${\mathbb {F}}$ . Let E be the exceptional locus.

The general X is a smooth $3$ -fold that does not intersect $\mathfrak {s}_2$ or $\mathfrak {s}_5$ , so its pull-back is a divisor $\tilde {X}$ in $\tilde {{\mathbb {F}}}$ mapped isomorphically to X. The divisor $\tilde {X}$ is big since $\tilde {X}^4=X^4=10^4(H^4-4dH^3F)=10^3 d>0$ . By the first lines of the proof of Proposition 1.6, since we assumed $d\geq d_0$ , the linear system $|10(H-dF)|$ is base point free and therefore $|\tilde {X}|$ is base point free as well.

We factor the restriction map $\rho \colon \operatorname {\mathrm {Pic}}(\tilde {{\mathbb {F}}}) \rightarrow \operatorname {\mathrm {Pic}}(\tilde {X})$ through $\operatorname {\mathrm {Pic}}(\tilde {{\mathbb {F}}}\setminus E)$ as follows:

$$\begin{align*}\operatorname{\mathrm{Pic}}(\tilde{{\mathbb{F}}}) \xrightarrow{\rho_1} \operatorname{\mathrm{Pic}}(\tilde{{\mathbb{F}}}\setminus E) \xrightarrow{\rho_2} \operatorname{\mathrm{Pic}}(\tilde{X}). \end{align*}$$

Following [Reference Ravindra and SrinivasRS06, Section 1], we have isomorphisms

$$ \begin{align*}\operatorname{\mathrm{Cl}}({\mathbb{F}}) &\cong \operatorname{\mathrm{Pic}}({\tilde{{\mathbb{F}}}} \setminus E)& \operatorname{\mathrm{Cl}}(X) &\cong \operatorname{\mathrm{Pic}}(X)=\operatorname{\mathrm{Pic}}(\tilde{X}),& \end{align*} $$

so our claim is that $\rho _2$ is an isomorphism.

By a standard argument (detailed in [Reference Ravindra and SrinivasRS06, Section 1]), $\rho _1$ is surjective with kernel isomorphic to the free abelian group generated by the classes of the irreducible divisorial components of E.

Since $\tilde {X}$ is big and base point free (and $\dim \tilde {{\mathbb {F}}}=4\geq 3$ ), we can apply the Grothendieck–Lefschetz Theorem for big linear systems [Reference Ravindra and SrinivasRS06, Theorem 2].

Part a) of the G–L Theorem shows that the kernel of $\rho $ is generated by the classes of the irreducible divisors of $\tilde {{\mathbb {F}}}$ contracted to a point by the map induced by the linear system $|\tilde {X}|$ . They are exactly the divisors supported on E, since no irreducible Weil divisor of ${\mathbb {F}}$ is contracted to a point by $|10(H-dF)|$ . So, $\ker \rho =\ker \rho _1$ which, since $\rho _1$ is surjective, implies that $\rho _2$ is injective.

Finally, since $\dim \tilde {{\mathbb {F}}}=4$ , part c) of the G–L Theorem shows that in our situation, $\rho $ is surjective, and therefore $\rho _2$ is surjective too.

Proof. Using the notation of [Reference Arzhantsev, Derenthal, Hausen and LafaceAD+15], let $\bar X$ be the affine hypersurface $\{f=0\}$ in ${\mathbb {C}}^6$ and let $\hat X = \bar X\smallsetminus \{t_0=t_1=0\}\cup \{x_0=x_1=y=z=0\}$ be the subset of $\bar X$ obtained by removing the irrelevant locus. The only relevant component of $\bar X \smallsetminus \hat X$ is $\{z^2+y^5=t_0=t_1=0\}$ , which has codimension $2$ in $\bar X$ . Hence, the last assumption of [Reference Arzhantsev, Derenthal, Hausen and LafaceAD+15, Corollary 4.1.1.5] is fulfilled.

Proof. First, consider the following natural map of ${\mathcal {S}}$ -modules:

$$\begin{align*}\tilde\varphi\colon\bigoplus_{k=1}^n{\mathcal{S}}_{a_k}\otimes{\mathcal{S}}(-a_k)\to{\mathcal{S}},\end{align*}$$

which is defined on each direct summand by multiplication of sections $s\otimes t\mapsto \deg s\cdot st$ and then extending by linearity. If we choose local isomorphisms

$$\begin{align*}{\mathcal{S}}_{a_k}|_U\cong{\mathcal{O}}_U[x_1,\dots,x_r]_{a_k}\cong\bigoplus_{i=1}^{N}{\mathcal{O}}_U\cdot \textrm{d} m_{k_i},\end{align*}$$

where $\textrm {d} m_{k_i}$ , $i=1,\dots ,N$ are the free generators corresponding to the monomials $m_{k_i}$ in ${\mathcal {S}}_{a_k}|_U$ of degree $a_{k}$ , then $\tilde \varphi $ maps $\textrm {d} m_{k_i}$ to $a_k m_{k_i}$ . The degree shift ensures that $\deg \textrm {d} m_{k_i}=\deg m_{k_i}=a_k$ .

Let $\tilde \Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ be the kernel of $\tilde \varphi $ and $\tilde \Omega _{{\mathbb {F}}/B}$ its associated sheaf on ${\mathbb {F}}$ . Then, the induced maps of associated sheaves on ${\mathbb {F}}$ form a short exact sequence

$$\begin{align*}0\to\tilde\Omega_{{\mathbb{F}}/B}\to\bigoplus_{k=1}^n\pi^*{\mathcal{S}}_{a_k}\otimes{\mathcal{O}}_{{\mathbb{F}}}(-a_k)\to{\mathcal{O}}_{{\mathbb{F}}}\to0,\end{align*}$$

where the exactness on the right follows because $\tilde \varphi $ is surjective in degrees $\ge 1$ .

Note that $\tilde \Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ is a locally free ${\mathcal {S}}$ -module and using the isomorphism ${\mathcal {S}}|_U\cong {\mathcal {O}}_U[x_1,\dots ,x_r]$ , it has local generators

$$\begin{align*}a_i \cdot x_i\textrm{d} x_j-a_j\cdot x_j\textrm{d} x_i \ \ \text{ and }\ \ \textrm{d} m_k-\sum_{i=1}^r\dfrac{\partial m_k}{\partial x_i}\textrm{d} x_i,\end{align*}$$

where here we write $a_i$ for the degree of $x_i$ . However, $\Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ is locally generated as an ${\mathcal {S}}|_U$ -module by the generators of the first type – that is, $a_i \cdot x_i\textrm {d} x_j-a_j\cdot x_j\textrm {d} x_i$ .

Let ${\mathcal {K}}:=\bigoplus _{k=1}^n{\mathcal {S}}[a_k-1]_{a_k}\otimes {\mathcal {S}}(-a_k)$ . We construct a map $\alpha \colon {\mathcal {K}}\to \tilde \Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ whose cokernel is $\Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ . Locally, we define $\alpha _U\colon {\mathcal {K}}|_U\to \tilde \Omega _{{\mathcal {S}}/{\mathcal {O}}_B}|_U$ by

$$\begin{align*}m\otimes 1\mapsto\textrm{d} m-\sum_{i=1}^r\tfrac{\partial m}{\partial x_i}\textrm{d} x_i,\end{align*}$$

where $m=m(x_1,\dots ,x_r)$ is a section of ${\mathcal {S}}|_U[a_k-1]_{a_k}$ . Next, we show that $\alpha $ is well-defined. Suppose that ${\mathcal {S}}|_V\cong {\mathcal {O}}_V[x_1',\dots ,x_r']$ . The transition function on $U\cap V$ is an isomorphism $v\colon {\mathcal {O}}_{U\cap V}[x_1',\dots ,x_r']\to {\mathcal {O}}_{U\cap V}[x_1,\dots ,x_r]$ , $x_i=v_i(x^{\prime }_1,\dots ,x^{\prime }_r)$ . We denote the induced isomorphisms on ${\mathcal {K}}$ and $\tilde \Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ by v as well. Then,

$$\begin{align*}\alpha_U(m(x))=\alpha_U(m(v(x')))=\textrm{d} m(v(x')))-\sum_{i=1}^r \dfrac{\partial m(v(x'))}{\partial v_i}\textrm{d} v_i(x'). \end{align*}$$

Now, by the chain rule, we have

$$ \begin{align*} \left(\tfrac{\partial m(v(x'))}{\partial x_1'},\dots,\tfrac{\partial m(v(x'))}{\partial x_r'}\right)&=\left(\tfrac{\partial m(v)}{\partial v_1},\dots,\tfrac{\partial m(v)}{\partial v_r}\right)\cdot D_{x'}v(x')\ \ \text{and }\\ (\textrm{d} v_1,\dots,\textrm{d} v_r)^t&=D_{x'}v(x')\cdot(\textrm{d} x_1',\dots,\textrm{d} x_r')^t. \end{align*} $$

Multiplying the first equation on the right by $D_{x'}v(x')^{-1}$ and combining them, we get

$$\begin{align*}\sum_{i=1}^r \dfrac{\partial m(v(x'))}{\partial v_i}\textrm{d} v_i(x')=\left(\tfrac{\partial m(v(x'))}{\partial x_1'},\dots,\tfrac{\partial m(v(x'))}{\partial x_r'}\right)\cdot(\textrm{d} x_1',\dots,\textrm{d} x_r')^t.\end{align*}$$

It follows that

$$\begin{align*}\alpha_U(m(x))=\textrm{d} m(v(x'))-\sum_{i=1}^r \dfrac{\partial m(v(x'))}{\partial x^{\prime}_i}\textrm{d} x_i'=\alpha_V(m(v(x'))), \end{align*}$$

and thus, $\alpha $ is well-defined. Hence, we have a short exact sequence

$$\begin{align*}0\to\mathcal{K}\xrightarrow{\alpha}\tilde\Omega_{{\mathcal{S}}/{\mathcal{O}}_B}\to\Omega_{{\mathcal{S}}/{\mathcal{O}}_B}\to0.\end{align*}$$

The two short exact sequences involving $\tilde \Omega _{{\mathcal {S}}/{\mathcal {O}}_B}$ fit together as the middle row respectively first column of the following commutative diagram:

The composition $\tilde \iota \circ \alpha \colon {\mathcal {K}}\to \bigoplus _{k=1}^n{\mathcal {S}}_{a_k}\otimes {\mathcal {S}}(-a_k)$ is injective. We call the cokernel ${\mathcal {V}}_{{\mathcal {S}}}$ and fill in the third row using a diagram chasing argument. Since $\tilde \varphi $ is surjective in degrees $\ge 1$ , $\tilde \varphi =\varphi \circ \beta $ and $\beta $ is surjective, it follows that $\varphi $ is surjective in degrees $\ge 1$ . Thus, the sequence of sheaves of ${\mathcal {O}}_{{\mathbb {F}}}$ -modules associated to the bottom row is the relative Euler sequence (3.2) on ${\mathbb {F}}$ .

We can now prove statements (1) and (2).

(1) is proved by taking the exact sequence of sheaves associated to the middle column of the above diagram.

(2) Since ${\mathbb {F}}$ is well-formed, the singular locus of ${\mathbb {F}}$ has codimension $\ge 2$ . Hence, it suffices to prove that the two sheaves are isomorphic on the nonsingular locus $W\subset {\mathbb {F}}$ . We restrict the relative Euler sequence (3.2) to W. Then, this is an exact sequence of vector bundles and since $\omega _{W/B}=\det \Omega _{W/B}$ , we deduce that $\omega _{W/B}=\det {\mathcal {V}}$ . By part (1), we conclude that $\omega _{W/B}$ is the restriction of $\det \left (\bigoplus _{k}\pi ^*{\mathcal {E}}_{a_k}(-a_k)\right )\cong \pi ^*\left ( \bigotimes _k \det {\mathcal {E}}_{a_k} \right )(-\sum _k r_ka_k)$ to W.

Proof. Throughout this proof, we use implicitly the formula $\operatorname {\mathrm {rank}} {\mathcal {R}}_n=h^0(S,K_S)$ , where S is a $(1,2)$ -surface. Hence, $\operatorname {\mathrm {rank}} {\mathcal {R}}_1=2$ and $\operatorname {\mathrm {rank}}{\mathcal {R}}_n=3+\frac 12n(n-1)$ for $n\ge 2$ .

We construct ${\mathcal {S}}(X)$ as follows: First, consider the weighted symmetric algebra $\operatorname {\mathrm {Sym}} {\mathcal {R}}_1$ with weights $(1^2)$ . Since any fibre S is mapped to ${\mathbb {P}}^1$ by $|K_S|$ , there are no relations involving only variables of degree $1$ . Hence, the natural map $\operatorname {\mathrm {Sym}} {\mathcal {R}}_1 \to {\mathcal {R}}$ is injective and an isomorphism in degree $1$ .

The multiplication map $\sigma _2 \colon \operatorname {\mathrm {Sym}}({\mathcal {R}}_1)_2 \rightarrow {\mathcal {R}}_2$ has cokernel ${\mathcal {E}}_2$ which is locally free of rank $1$ because S is mapped onto ${\mathbb {P}}(1,1,2)$ by $|2K_S|$ . Hence, we can construct (cf. §3.1) a weighted symmetric algebra ${\mathcal {S}}'$ with weights $(1^2,2)$ with an injective morphism ${\mathcal {S}}'\hookrightarrow {\mathcal {R}}$ , which is an isomorphism in degrees $1$ , $2$ , $3$ and $4$ .

The cokernel ${\mathcal {E}}_5$ of the inclusion ${\mathcal {S}}_5'\hookrightarrow {\mathcal {R}}_5$ is locally free of rank $1$ , so we get a weighted symmetric algebra ${\mathcal {S}} \supset {\mathcal {S}}'$ with weights $(1^2,2,5)$ such that ${\mathcal {S}}' \cong {\mathcal {S}}[2]$ . There is a morphism ${\mathcal {S}} \rightarrow {\mathcal {R}}$ that is an isomorphism in degrees $\le 9$ and thereafter surjective, so inducing an inclusion

$$\begin{align*}X\cong {\mathbf{P}\mathrm{roj}}_B {\mathcal{R}} \subset {\mathbb{F}}:={\mathbf{P}\mathrm{roj}}_B{\mathcal{S}} \end{align*}$$

of X as divisor in a ${\mathbb {P}}(1,1,2,5)$ -bundle over B. The relative degree of X is then $10$ , the degree of the single equation defining its general fibre.

Proof. (1) Suppose $p\in X\cap {\mathfrak s}_5$ . Then, in a neighbourhood of p, ${\mathbb {F}}(X)$ has a singular point that is a quotient singularity of type $\frac 15(1,1,2,0)$ . Since X is a Cartier divisor, P is a noncanonical singular point of X, at best a $\frac 15(1,1,2)$ point, which is a contradiction (see also [Reference Franciosi, Pardini and RollenskeFPR17, Rmk 4.6]).

(2) Suppose ${\mathfrak {s}}_2\subset X$ . For a general point p in B, there is a local analytic neighbourhood $p\in V\subset B$ such that the equation of X has the form $z^2=q(x_0,x_1)y^4+\dots $ , where q is (at best) a relative quadratic form over V. Thus, in a neighbourhood of ${\mathfrak {s}}_2$ , X looks like $V\times \{(z^2=q)\subset \frac 12(1,1,1)\}$ . This is at best a curve of singularities $V\times \frac 14(1,1)$ [Reference HackingHac16], which is not canonical, a contradiction.

Proof. The indeterminacy locus of g is ${\mathfrak {s}}_5$ , so by Proposition 4.9(1), the restriction $g_{|X} \colon X \to {{\mathbb {Q}}(X)}$ is a finite morphism of degree $2$ . The involution on X which swaps the two sheets of this covering can be lifted to ${\mathcal {S}}$ .

Indeed, on an open subset $U\subset B$ , we have ${\mathcal {S}}_{|U}\cong {\mathcal {O}}_U[x_0,x_1,y,z]$ with $\deg x_j=1$ , $\deg y=2$ , $\deg z=5$ . Proposition 4.9(1), implies that the coefficient of $z^2$ in the equation of X never vanishes; completing the square, we may assume that the equation has the form $z^2=f(x_0,x_1,y)$ . Then, the involution may be lifted to ${\mathcal {S}}_{|U}$ as the involution fixing $x_0,x_1,y$ and mapping $z \mapsto -z$ . These glue in the obvious way to give an involution on ${\mathcal {S}}$ and a splitting into invariant and anti-invariant parts: ${\mathcal {S}}={\mathcal {S}}^+ \oplus {\mathcal {S}}^-$ . By construction, the involution preserves X, so we get a splitting ${\mathcal {R}}={\mathcal {R}}^+ \oplus {\mathcal {R}}^-$ , and clearly, ${\mathcal {R}}^+={\mathcal Q}$ .

Proof. For all $d\le 4$ , we get ${\mathcal {S}}_d^+={\mathcal Q}_d$ and thus, ${\mathcal {S}}_d^-=0$ . In degree $5$ , ${\mathcal {S}}_5^+={\mathcal Q}_5=\ker \epsilon _5$ so that $(\epsilon _5)_{|{\mathcal {S}}_5^-}$ is an isomorphism, whose inverse is a right inverse for $\epsilon _5$ . Hence, ${\mathcal {R}}_5^- \cong {\mathcal {S}}_5^- \cong {\mathcal {E}}_5$ .

Now as locally free ${\mathcal {O}}_B$ -modules, ${\mathcal {R}}_d^+$ and ${\mathcal {R}}_d^-$ are generated by the monomials $x_0^ax_1^by^c$ with $a+b+2c=d$ (resp. $x_0^ax_1^by^cz$ with $a+b+2c=d-5$ ). Thus, all multiplication maps

$$\begin{align*}{{\mathcal{R}}}^+_d \otimes {\mathcal{R}}_5^- \rightarrow {{\mathcal{R}}}^-_{d+5} \end{align*}$$

are isomorphisms, completing the proof.

Proof. Since the involution of ${\mathbb {F}}(X)$ preserves X, the image of ${\mathcal {L}}$ is contained in the invariant part ${\mathcal {S}}^+_{10}$ (or in ${\mathcal {S}}^-_{10}$ , but this would contradict Proposition 4.9(1)). By Proposition 4.14, ${\mathcal {E}}_5 \cong {\mathcal {R}}_5^- \cong {\mathcal {S}}_5^-$ . Then, ${\mathcal {S}}_{10}^+$ splits as ${\mathcal Q}_{10} \oplus {\mathcal {E}}_5^2$ . This corresponds locally to separating the polynomials in $x_0,x_1,y$ from those involving $z^2$ .

Consider the induced projection ${\mathcal {S}}_{10}^+ \to {\mathcal {E}}_5^2$ . Then, Proposition 4.9(1) says that the composition of maps $ {\mathcal {L}} \to {\mathcal {S}}_{10}^+ \to {\mathcal {E}}_5^2 $ is surjective. Therefore, since both ${\mathcal {L}}$ and ${\mathcal {E}}_5$ are line bundles, ${\mathcal {L}} \cong {\mathcal {E}}_5^2$ .

Proof. (1) By Proposition 3.19,

$$\begin{align*}\omega_{{\mathbb{F}}/B}={\mathcal{O}}_{{\mathbb{F}}}(-9) \otimes \pi_{\mathbb{F}}^*\det({\mathcal{E}}_1 \oplus {\mathcal{E}}_2 \oplus {\mathcal{E}}_5) \end{align*}$$

and since $X \in |{\mathcal {O}}_{\mathbb {F}}(10) \otimes \pi ^* {\mathcal {E}}_5^{-2}|$ , the adjunction formula gives

$$\begin{align*}\omega_{X/B}={\mathcal{O}}_{X}(1) \otimes \pi^*\det({\mathcal{E}}_1 \oplus {\mathcal{E}}_2 \oplus {\mathcal{E}}_5^{-1}). \end{align*}$$

Finally, by Remark 4.7, we have $\omega _{X/B}\cong {\mathcal {O}}_{X}(1)$ , so the thesis follows immediately.

(2) This proof is inspired by [Reference PignatelliPig12, Definition 2.4 and Proposition 2.5]). The map $\operatorname {\mathrm {Sym}}^5(\epsilon _2) \colon \operatorname {\mathrm {Sym}}^5 ({\mathcal {S}}_2) \to {\mathcal {E}}_2^5$ induced by $\epsilon _2 \colon {\mathcal {S}}_2 \to {\mathcal {E}}_2$ factors through ${\mathcal Q}_{10}$ giving a map $\alpha \colon {\mathcal Q}_{10}\to {\mathcal {E}}_2^5$ whose kernel consists of those elements of ${\mathcal Q_{10}}$ vanishing along ${\mathfrak {s}}_2$ .

By Corollary 4.15, X is defined by a map ${\mathcal {E}}_5^2 \rightarrow {\mathcal {S}}_{10}^+\cong {\mathcal Q}_{10} \oplus {\mathcal {E}}_5^2$ , where locally, the factor ${\mathcal {E}}_5^2$ gives the multiples of $z^2$ , and therefore, ${\mathcal {E}}_5^2$ is in the ideal sheaf of ${\mathfrak {s}}_2$ . Hence, ${\mathfrak {s}}_2 \not \subset X$ if and only if the composition of the first component of this map ${\mathcal {E}}_5^2 \to {\mathcal Q}_{10}$ with $\alpha \colon {\mathcal Q}_{10}\to {\mathcal {E}}_2^5$ is not the zero map. Thus, $\operatorname {\mathrm {Hom}}_{{\mathcal {O}}_B}({\mathcal {E}}_5^2,{\mathcal {E}}_2^5)\neq 0$ . Substituting ${\mathcal {E}}_5 \cong (\det {\mathcal {E}}_1) \otimes {\mathcal {E}}_2$ , we obtain the result.

Proof. Since the fibres are hypersurfaces in weighted projective space, we have

$$ \begin{align*} \pi_* {\mathcal{O}}_{X}& \cong {\mathcal{O}}_B;& R^1\pi_*{\mathcal{O}}_{X}(dK_{X/B}) =0& \text{ for all }d. \end{align*} $$

Thus, in combination with the projection formula

$$\begin{align*}R^i \pi_* ({\mathcal{O}}_{X}(dK_{X/B}) \otimes \pi^*{\mathcal{L}}) \cong \left( R^i \pi_* {\mathcal{O}}_{X}(dK_{X/B}) \right) \otimes {\mathcal{L}}, \end{align*}$$

we see that the Leray spectral sequence of the direct image of ${\mathcal {O}}_{X}(dK_{X/B}) \otimes \pi ^*{\mathcal {L}}$ degenerates at page $2$ for each d and ${\mathcal {L}}$ .

Whence for $i=0,1$ and any $d\ge 1$ , we have

$$\begin{align*}h^i({\mathcal{O}}_{X}(dK_{X/B})\otimes \pi^*{\mathcal{L}})= h^i(\pi_*{\mathcal{O}}_{X}(dK_{X/B})\otimes{\mathcal{L}})=h^i({\mathcal{R}}_d \otimes {\mathcal{L}}).\end{align*}$$

When $i=2,3$ , we get

$$\begin{align*}h^i({\mathcal{O}}_{X}(dK_{X/B})\otimes \pi_{\mathbb{F}}^*{\mathcal{L}})= h^{i-2}(R^2\pi_*{\mathcal{O}}_{X}(dK_{X/B})\otimes{\mathcal{L}}).\end{align*}$$

If $d\geq 2$ , then $R^2\pi _* {\mathcal {O}}_{X}(dK_{X/B})=0$ by the base change theorem because the fibres are canonically polarised. Thus, $h^i({\mathcal {O}}_X(dK_{X/B})\otimes \pi ^*{\mathcal {L}})=0$ for $d\ge 2$ . When $d=1$ , $R^2\pi _*{\mathcal {O}}_X(K_{X/B})\cong (\pi _*{\mathcal {O}}_X)^{\vee }$ by Grothendieck duality, so $h^i({\mathcal {O}}_X(K_{X/B})\otimes \pi ^*{\mathcal {L}})=h^{i-2}(\pi _*{\mathcal {O}}_X\otimes {\mathcal {L}})=h^{i-2}({\mathcal {L}})$ .

Proof. The first three equalities follow from Lemma 4.19 with ${\mathcal {L}}=\omega _B$ . For the last, note that $\chi (\omega _X)=p_g-q_2+q_1-1=\chi ({\mathcal {E}}_1\otimes \omega _B)-\chi ({\mathcal {O}}_B)$ . Then, by the Riemann–Roch Theorem for curves, $\chi ({\mathcal {E}}_1\otimes \omega _B)=\chi ({\mathcal {E}}_1)+2\deg (\omega _B)$ , and the result follows.

The inequality $q_2 \leq \operatorname {\mathrm {rank}}({\mathcal {E}}_1)= 2$ follows then by the semipositivity of ${\mathcal {E}}_1$ ([Reference ViehwegVie83, Thm III]).

Proof. By Lemma 4.19, $\chi (\omega _X^2)=\chi ({\mathcal {R}}_2\otimes \omega _B^2)$ ; twisting the exact sequence $ 0\to \operatorname {\mathrm {Sym}}^2({\mathcal {R}}_1)\to {\mathcal {R}}_2 \to {\mathcal {E}}_2 \to 0$ by $\omega _B^2$ , we get

$$ \begin{align*} \chi(\omega_X^2)&=\chi({\mathcal{R}}_2\otimes\omega_B^2)\\ &=\chi(S^2({\mathcal{R}}_1\otimes\omega_B))+\chi({\mathcal{E}}_2\otimes\omega_B^2)\\ &=3\chi({\mathcal{R}}_1\otimes\omega_B)-3\chi({\mathcal{O}}_B)+\chi({\mathcal{E}}_2)+\deg\omega_B^2. \end{align*} $$

By Proposition 4.20, this last line is equivalent to

$$\begin{align*}\chi(\omega_X^2)=-3\chi({\mathcal{O}}_X)+\chi({\mathcal{E}}_2)-4\chi({\mathcal{O}}_B).\end{align*}$$

However, the Riemann–Roch formula [Reference ReidRei87, Cor. 10.3] gives

$$\begin{align*}\chi(\omega_X^2)=\frac12 K_{X}^3-3\chi({\mathcal{O}}_X)+\frac N4. \end{align*}$$

Combining these two expressions to eliminate $\chi (\omega _X^2)$ and simplifying gives $K^3_{X} =2\chi ({\mathcal {E}}_2)-8\chi ({\mathcal {O}}_B)-\frac {N}2$ . Finally, substituting $\chi ({\mathcal {E}}_2)= \frac 13 \left ( N + 2 \chi ({\mathcal {E}}_1)-\chi ({\mathcal {O}}_B) \right ) $ and then $\chi ({\mathcal {E}}_1)=\chi (\omega _X)+5\chi ({\mathcal {O}}_B)$ , we obtain the result.

Proof. Since X is regular, we know that $B={\mathbb {P}}^1$ . Moreover, X is Gorenstein, so $N=q_2=0$ . By definition of N and Corollary 4.16, there is a d such that $\det {\mathcal {E}}_1={\mathcal {O}}_{{\mathbb {P}}^1}(3d)$ , ${\mathcal {E}}_2={\mathcal {O}}_{{\mathbb {P}}^1}(2d)y$ , ${\mathcal {E}}_5={\mathcal {O}}_{{\mathbb {P}}^1}(5d)z$ . Moreover, there is a unique $d_0\le 3d-d_0$ such that ${\mathcal {E}}_1={\mathcal {O}}_{{\mathbb {P}}^1}(d_0)x_0\oplus {\mathcal {O}}_{{\mathbb {P}}^1}(3d-d_0)x_1$ .

Now consider the short exact sequence

$$\begin{align*}0\to \operatorname{\mathrm{Sym}}^2{\mathcal{E}}_1\xrightarrow{\sigma_2} {\mathcal{S}}_2\xrightarrow{\epsilon_2}{\mathcal{E}}_2\to0.\end{align*}$$

The main point of the proof is to show that $\epsilon _2$ has a right inverse.

We have $\operatorname {\mathrm {Sym}}^2{\mathcal {E}}_1={\mathcal {O}}(2d_0)x_0^2\oplus {\mathcal {O}}(3d)x_0x_1\oplus {\mathcal {O}}(6d-2d_0)x_1^2$ and $d,d_0\ge 0$ by Fujita semipositivity. In this case,

(4.1) $$ \begin{align} \operatorname{\mathrm{Ext}}_{{\mathcal{O}}_{{\mathbb{P}}^1}}^1({\mathcal{E}}_2,\operatorname{\mathrm{Sym}}^2{\mathcal{E}}_1)\cong H^1(\operatorname{\mathrm{Sym}}^2{\mathcal{E}}_1\otimes{\mathcal{E}}_2^\vee)\cong H^1({\mathcal{O}}(2(d_0-d))), \end{align} $$

because $3d\ge 2d$ and $6d-2d_0\ge 2d$ . Standard cohomological arguments allow us to deduce the existence of an inverse when $d_0\ge d$ , so we assume that $d_0\le d-1$ .

We complete the proof arguing by contradiction. We show that if the extension class in (4.1) corresponding to the above short exact sequence is nontrivial, then $\operatorname {\mathrm {Hom}}_{{\mathcal {O}}_{{\mathbb {P}}^1}}({\mathcal {E}}_5^2,\mathcal {Q}_{10})$ is zero. Arguing as in the proof of 4.16(2), this implies that ${\mathfrak {s}}_2\subset X$ , which is a contradiction.

Motivated by this, we define $\mathcal {I}=x_1\mathcal {Q}$ , the sheaf of ideals locally principally generated by ${\mathcal {O}}(3d-d_0)x_1$ . Then, define $\mathcal {T}=\mathcal {Q}/\mathcal {I}$ and note that ${\mathbf {P}\mathrm{roj}}_{{\mathbb {P}}^1}(\mathcal {T})$ is the divisor $(x_1=0)$ in the ${\mathbb {P}}(1,1,2)$ -bundle ${\mathbb {Q}}(X)$ over ${\mathbb {P}}^1$ .

Replacing ${\mathcal {E}}_1$ with $\mathcal {Q}_1$ and ${\mathcal {S}}_2$ with $\mathcal {Q}_2$ in the above short exact sequence and quotienting by $\mathcal {I}$ , the multiplication maps in $\mathcal {T}$ give the following exact sequence

(4.2) $$ \begin{align} 0\to\operatorname{\mathrm{Sym}}^2\mathcal{T}_1\cong{\mathcal{O}}(2d_0)\to\mathcal{T}_2\xrightarrow{\tilde\epsilon_2}{\mathcal{E}}_2\to0, \end{align} $$

and $\tilde \epsilon _2$ has a right inverse if and only if $\epsilon _2$ has, because of (4.1).

Since $\mathcal {T}$ is generated in degree 2, and $\mathcal {T}_1$ has rank 1, the multiplication map $\operatorname {\mathrm {Sym}}^{k}\mathcal {T}_2\to \mathcal {T}_{2k}$ is an isomorphism. Note that $\mathcal {T}_2$ is a direct sum of two line bundles on ${\mathbb {P}}^1$ . Thus, if (4.2) does not split, then the maximal degree in $\mathcal {T}_2$ is $<2d$ , and hence, the maximal degree in $\mathcal {T}_{2k}$ is $<2kd$ .

Since $\mathcal {T}$ is a quotient of $\mathcal {Q}$ , we get a surjective map $\mathcal {Q}_{10}\to \mathcal {T}_{10}$ . Thus, all summands of $\mathcal {Q}_{10}$ are line bundles of degree $<10d$ . Since ${\mathcal {E}}_5^2\cong {\mathcal {O}}(10d)$ , it follows that $\operatorname {\mathrm {Hom}}({\mathcal {E}}_5^2,\mathcal {Q}_{10})$ is zero. Hence, $\epsilon _2$ has a right inverse.

We already know that $\epsilon _5$ has a right inverse by Proposition 4.14. Thus, ${\mathcal {S}}(X)\cong \operatorname {\mathrm {wSym}}_{1,2,5}({\mathcal {E}}_1,{\mathcal {E}}_2,{\mathcal {E}}_5)$ , and hence, ${\mathbb {F}}$ is a toric variety as in Example 3.16. This finishes the proof.

Proof. In [Reference Chen and HuCH17, Thm 1.1], the authors generalise Kobayashi’s construction to exhibit threefolds $Y(a,e)$ on the Noether line with canonical image ${\mathbb {F}}_{e}$ and $p_g(Y)=6a-3e-2$ for all pairs of integers $a,e$ with $a\geq e\geq 0$ excepting $(a,e)=(2,2)$ , $(1,1)$ , $(1,0)$ , $(0,0)$ .

We show that $Y(a,e)$ is the same as our $X(d;d_0)$ with $d=2a-e$ , $d_0=2d-a$ .

Recall that there is a double cover $X(d;d_0)\to D_z$ , where $D_z$ is a bundle in quadric cones over ${\mathbb {P}}^1$ (see Remark 1.5). We perform a weighted blowup $D^{\prime }_z\to D_z$ of the index 2 section ${\mathfrak {s}}_2=D_{x_0}\cap D_{x_1}\cap D_z$ and a corresponding blowup $X'\to X$ of the preimage of ${\mathfrak {s}}_2$ in X, to obtain the following diagram:

Recall from §1 that $\delta $ is the positive section and l is the fibre on ${\mathbb {F}}_e$ . We claim that $D^{\prime }_z$ is the following ${\mathbb {P}}^1$ -bundle over ${\mathbb {F}}_e$ :

$$\begin{align*}D_z':= {\mathbb{P}}_{{\mathbb{F}}_e}\left({\mathcal{O}}_{{\mathbb{F}}_e}\oplus{\mathcal{O}}_{{\mathbb{F}}_e}(2\delta+(d+e)l)\right).\end{align*}$$

Indeed, in coordinates, the blowup $D_z'\to D_z$ is given by

$$\begin{align*}t_i\mapsto t_i,\ x_0\mapsto ux_0',\ x_1\mapsto ux_1',\ y\mapsto y,\end{align*}$$

where u is the section defining the exceptional divisor E. Note that $s=t_0^{2(d_0-d)}x_0^2$ is a section of ${\mathcal {O}}_{{\mathbb {F}}_e}(2\delta +(d+e)l)$ because $2(d_0-d)=d+e$ . The rational function $s/y$ on $D_z$ pulls back to $su^2/y$ on $D_z'$ . Since y, $u^2$ are the fibre coordinates of $D_z'\to {\mathbb {F}}_e$ , we get the claimed ${\mathbb {P}}^1$ -bundle over ${\mathbb {F}}_e$ .

Moreover, $X'\to D^{\prime }_z$ is a double cover with branch locus $B+E$ , where $B\in |{\mathcal {O}}_{D_z'}(5)|$ is the strict transform of the branch divisor of $X\to D_z$ and E is the exceptional divisor of $D_z'\to D_z$ . This is the Kobayashi–Chen–Hu construction with $d+e=2a$ . The condition $a\geq e$ is equivalent to the condition $e\leq d$ .

The short list of exclusions $(a,e)=(2,2),\dots $ mentioned above are just the $X(d;d_0)$ , which violate $\min (d,d_0)\geq 3$ (i.e., those with $K_X$ nonample).

Proof. By Propositions 1.6, 2.2, 2.4 and 4.22, all smooth simple fibrations with $e\neq \frac 54d$ are Gorenstein regular of the form $X(d;d_0)$ with $d_0\geq d$ . Moreover, they belong to the same irreducible component of the moduli space of canonical $3$ -folds, whose general element is a $X\left (d;\left \lfloor \frac 32d \right \rfloor \right )$ .

Now, assume d divisible by $8$ and choose $d_0=\frac 78 d$ so that $e=\frac 54d$ . A general scrollar deformation ${\mathcal {X}}$ as in §2 gives a degeneration $X(d;d_0+1)\leadsto X(d;d_0)$ with singular central fibre ${\mathcal {X}}_0$ . Indeed, in the notation of the proofs of Propositions 1.6 and 2.4 in this case, we get first of all

• • $\deg c_{10,0,0} <0 \Rightarrow \mathfrak {s}_0\subset X(d;d_0)$ .

• • $\deg c_{9,1,0} =0$ and $c_{9,1,0} \in (t_0,t_1)^{e-1} \Rightarrow c_{9,1,0}=0$ that implies $\mathfrak {s}_0\subset \operatorname {\mathrm {Sing}} X(d;d_0)$ and then the degeneration ${\mathcal {X}}_0$ can not be smooth.

However, for general ${\mathcal {X}}$ , ${\mathcal {X}}_0$ is canonical. Indeed, $\deg c_{7,3,0} =2e$ . Then, following the argument of the proof of Proposition 2.4, we may get any $c_{7,3,0} \in (t_0,t_1^{e-1})^3$ of degree $2e$ : these are all the multiples of $t_0$ , and, in particular, we may get $c_{7,3,0} $ with distinct roots – that is, the condition we used in Proposition 1.6 to ensure that the general element has canonical singularities. In particular, near $t_0=0$ , ${\mathcal {X}}_0$ looks like $(z^2+y^5+t_0x_1^3=0)$ , which is canonical by §1.4.

We now show that there is no degeneration $X(d;d_0)\leadsto X\left ( d; \frac 78 d \right )$ with $d_0 \ge d$ and all fibres nonsingular. The argument is inspired by Horikawa [Reference HorikawaHor76a, Lemma 7.3], although it is more complicated to set up in our situation. Suppose, by contradiction, that ${\mathcal {X}}\to \Lambda $ is such a degeneration. The relative canonical linear system $|K_{{\mathcal {X}}/\Lambda }|$ gives a rational map ${\mathcal {X}}/\Lambda \to {\mathcal {F}}/\Lambda $ , where ${\mathcal {F}}$ is a degeneration of surfaces ${\mathbb {F}}_{3d-2d_0}\leadsto {\mathbb {F}}_{\frac 54d}$ . If $3d-2d_0\ne 0$ , then the Hirzebruch surface ${\mathbb {F}}_{3d-2d_0}$ admits a unique fibration to ${\mathbb {P}}^1$ . However, if $3d-2d_0=0$ , then $d\ge 3$ and by §1.9, one of the two fibrations ${\mathbb {F}}_0\to {\mathbb {P}}^1$ is distinguished by the canonical linear system of X. Thus, each fibre of ${\mathcal {F}}/\Lambda $ has a unique distinguished fibration to ${\mathbb {P}}^1$ , and hence, ${\mathcal {X}}/\Lambda $ admits a map to ${\mathbb {P}}^1_{\Lambda }={\mathbb {P}}^1\times \Lambda $ factoring through ${\mathcal {F}}/\Lambda $ . Moreover, this map induces the fibration in $(1,2)$ -surfaces ${\mathcal {X}}_\lambda \to {\mathbb {P}}^1$ on each fibre.