Proof. For $7\le r\le 9$, the proposition is trivial because S is smooth. The case r = 3 is the main result of [Reference Brevik9]. Assume r > 3. Fix a general $A\subset S$ such that $\#A=r-3$ and let S ^ʹ be the blowing-up of S at all points of A. Let $M\subset {\mathbb{P}}^r$ be the $(r-4)$-dimensional linear space spanned by A. For a general A, the surface S ^ʹ has the same number of singularities as S and it is isomorphic to the cubic surface $S^{\prime\prime}\subset {\mathbb{P}}^3$ which is the closure in ${\mathbb{P}}^3$ of the image of $S\setminus A$ by the linear projection ${\ell} : {\mathbb{P}}^r\setminus M\rightarrow {\mathbb{P}}^3$; [Reference Dolgachev15, p. 389]. Since A is general we know $C\cap A=\emptyset$. Since the linear projection $\ell$ restricted to C is an embedding, $C\cong \ell(C)$ and $\deg(\ell(C)) =\deg (C)$. By [Reference Brevik9], there is a smoothing family of Sʹʹ on which the curve $\ell(C)$ is a flat limit of a family of curves on nearby smooth cubic surface S_t; $t\in T\setminus \{0\}$, T an integral quasi-projective curve, $0\in T$, $S_0= S^{\prime\prime}$, and S_t smooth for all $t\in T\setminus \{0\}$. Obviously, $\ell(C)\cap \ell_i=\emptyset$ for all r − 3 exceptional divisors $\ell _i$ of Sʹʹ associated with the exceptional divisors of the blowing up of the r − 3 points of A. By [Reference Brevik9, proposition 4.7], we may assume that the curve $\ell (C)$ is a limit of curves $C_t\subset S_t$, $t\in T\setminus \{0\}$, which have empty intersection with the r − 3 exceptional divisors $\ell_i(t)$ of S_t, $t\in T\setminus \{0\}$ specializing to the r − 3 exceptional divisors $\ell _i$ of Sʹʹ. For $t\in T\setminus \{0\}$, let $\tilde{S}_t$ the surface obtained from S_t by blowing down the r − 3 divisors $\ell_i(t)$. Each surface $\tilde{S}_t$ is a smooth del Pezzo surface in ${\mathbb{P}}^r$. Since $C_t\cap \ell_i(t)=\emptyset$, each curve C_t is isomorphic to a curve $\tilde{C}_t\subset \tilde{S}_t$. The flat family over T gives rise to a flat family of smooth del Pezzo surfaces $\tilde{S}_t$, $t\in T\setminus \{0\}$ with S as its flat limit together with a flat family curves $\tilde{C}_t$, $t\in T\setminus \{0\}$ with C as its flat limit.

Proof. By [Reference Martens28, corollary 1], X is a-gonal with a unique pencil $g^1_a$ induced by the pencil $|{\mathcal{O}}_Q(0,1)|$. By [Reference Paxia, Raciti and Ragusa32, theorem 2.1], there is no base-point-free $g^1_c$ with $a \lt c \lt b$. Take any $g^1_b$ and take a general $A\in g^1_b$. Since X has no base-point-free $g^1_{b-1}$ (or if $b=a+1$ a unique $g^1_a$), the linear series $g^1_b$ is complete. By adjunction, we have $h^1(Q,{\mathcal{I}}_A(a-2,b-2)) \gt 0$. To prove the lemma, it is sufficient to find $R\in |{\mathcal{O}}_Q(1,0)|$ containing A.

Set $A_0:= A$. Take any $L_1\in |{\mathcal{O}}_Q(0,1)|$ such that $e_1:= \#(L_1\cap A_0)$ is maximal and set $A_1:= A_0\setminus A_0\cap L_1$. Take any $L_2\in |{\mathcal{O}}_Q(0,1)|$ such that $e_2:= \#(L_2\cap A_1)$ is maximal and set $A_2:= A_1\setminus A_1\cap L_2$. We define recursively the sets $A_3,\ldots,A_b$, the elements $L_3,\ldots ,L_b\in |{\mathcal{O}}_Q(0,1)|$, and the integers $e_3,\ldots ,e_b$ with $e_i:= \#(A_{i-1}\cap L_i)$ maximal and $A_i:= A_{i-1}\setminus A_{i-1}\cap L_i$. Since at each step we require the maximality of the integer i and $A_{i-1}\supseteq A_i$, we have $e_i\ge e_{i+1}$ for all $1\le i\le b-1$. Note that if $e_i=0$, then $A_{i-1}=\emptyset$ and $A_j=\emptyset$ for all $j \gt i-1$ with $j\le b$. Thus $A_b=\emptyset$.

Let c be the first integer such that $\#A_c\le 2$. We saw that $c\le b-2$ and $c=b-2$ if and only if $e_1=1$. Set $T:= L_1\cup \cdots \cup L_c\in |{\mathcal{O}}_Q(0,c)|$. Consider the exact sequence

(5)\begin{equation} 0\rightarrow {\mathcal{I}}_{A_c}(a-2,b-2-c) \rightarrow {\mathcal{I}}_A(a-2,b-2)\rightarrow {\mathcal{I}}_{T\cap A,T}(a-2,b-2)\rightarrow 0. \end{equation}

First assume $e_1\ge 2$ and hence $c\le b-3$. Since ${\mathcal{O}}_Q(a-2,b-2-c)$ is very ample and $\#A_c\le 2$, we have $h^1(Q,{\mathcal{I}}_{A_c}(a-2,b-2-c))=0$. Thus, the long cohomology exact sequence of (5) gives $h^1(T,{\mathcal{I}}_{A\cap T}(a-2,b-2)) \gt 0$. The set T has c connected components, $L_1,\dots ,L_c$, with $L_i\cong {\mathbb{P}}^1$. Since $ \displaystyle{h^1(T,{\mathcal{I}}_{A\cap T,T}(a-2,b-2))= \sum_{i=1}^{c} h^1(L_i,{\mathcal{I}}_{A\cap T,T}(a-2,b-2)_{|L_i}) \gt 0,} $ we get $h^1(L_i,{\mathcal{I}}_{A\cap T,T}(a-2,b-2)_{|L_i}) \gt 0$ and $e_i\ge a$ for some i; by $\deg {\mathcal{O}}_{L_i}(a-2,b-2) =a-2$ for all i, we have $h^1(L_i,{\mathcal{I}}_{A\cap T,T}(a-2,b-2)_{|L_i})=0$ if $e_i\le a-1$ for all i. But then A contains a subset A ^ʹ such that $\#A^{\prime}=a$, $A^{\prime}\subset L_i$ and $|A^{\prime}|$ is the $g^1_a$ on X. Since the $g^1_b$ is base-point-free and complete, we get a contradiction.

Now assume $e_1=1$. Thus $c=b-2$ and $\#(A\cap L)\le 1$ for all $L\in |{\mathcal{O}}_Q(0,1)|$. Thus $h^1(T,{\mathcal{I}}_{T\cap A, T}(a-2,b-2)) =0$. Hence the long cohomology exact sequence of (5) gives $h^1(Q,{\mathcal{I}}_{A_c}(a-2,0)) \gt 0$, i.e. two points in $A_c=\{p,s\}$ fail to impose independent conditions on $|{\mathcal{O}}_Q(a-2,0)|$. Therefore, there is $R\in |{\mathcal{O}}_Q(1,0)|$ such that $A_c\subset R$. Fix any $q\in A\setminus A_c$. Since $\#(A\cap L)\le 1$ for all $L\in |{\mathcal{O}}_Q(0,1)|$, there are $D_1,\ldots,D_{b-2}\in |{\mathcal{O}}_Q(0,1)|$ such that $T^{\prime}:= D_1\cup \cdots \cup D_{b-2}$ contains $A\setminus \{p,q\}$. Using T ^ʹ instead of T, we get the existence of $R^{\prime}\in |{\mathcal{O}}_Q(1,0)|$ such that $\{p,q\}\subset R^{\prime}$. Since R is the unique element of $|{\mathcal{O}}_Q(1,0)|$ containing p, $R^{\prime}=R$. Thus $A\subset R$.

Proof. Take any isomorphism $\tilde{X}\stackrel{u}{\cong}{X}$. The line bundles $R_1:= u^\ast ({\mathcal{O}}_X(0,1))$ and $R_2:= u^\ast ({\mathcal{O}}_X(1,0))$ are the unique base-point-free line bundles on $\tilde{X}$ of degrees a and b, respectively by proposition 2.3, and hence $R_1= {\mathcal{O}}_{\tilde{X}}(0,1)$ and $R_2 ={\mathcal{O}}_{\tilde{X}}(1,0)$. Thus, u induces isomorphisms $|{\mathcal{O}}_{{X}}(1,0)|={\mathbb{P}}^1\rightarrow {\mathbb{P}}^1=|{\mathcal{O}}_{\tilde{X}}(1,0)|$ and $|{\mathcal{O}}_{{X}}(0,1)| \rightarrow |{\mathcal{O}}_{\tilde{X}}(0,1)|$ and the corresponding pair in ${\operatorname{Aut}}({\mathbb{P}}^1)\times {\operatorname{Aut}}({\mathbb{P}}^1)$ induces v.

Proof. Note that a general element in any component of $\mathcal{H}_{15,15,5}$ is linearly normal since $\pi (15,6)=13 \lt g=15$ and hence $\mathcal{H}_{15,15,5}=\mathcal{H}^\mathcal{L}_{15,15,5}$.

Since $\pi_1(15,5)=16 \gt g$, X is not extremal or nearly extremal—a curve with $p_a(X) \gt \pi_1(d,r)$—and hence $X\subset{\mathbb{P}}^5$ may not sit on a surface of small degree. Instead, we look at the residual series $|K_X(-1)|$ and study the curve induced by it. This technique—if one may call this ‘a technique’—is not new and has been used before, e.g. in [Reference Keem26] or possibly in works by other authors.

For a general $X\in\mathcal{H}_{15,15,5}$, set $\mathcal{E}:=g^4_{13}=|K_X(-1)|$ and let $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ be the dual curve of X, which is by definition the image curve induced by the base-point-free part of $\mathcal{E}$. We first claim that $\mathcal{E}$ is birationally very ample, possibly with non-empty base locus.

Claim 1: $|K_X(-1)|$ is birationally very ample.

Suppose $|K_X(-1)|$ is compounded. Since $\deg |K_X(-1)|=13$ is prime, $|K_X(-1)|$ has non-empty base locus Δ and we have the following four possibilities:

1. (i) $|K_X(-1)|=g^4_{12}+\Delta$, $\deg\Delta=1$,

2. (ii) $|K_X(-1)|=g^4_{10}+\Delta$, $\deg\Delta=3$,

3. (iii) $|K_X(-1)|=g^4_{9}+\Delta$, $\deg\Delta=4$; this can be excluded since the dual curve $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ induced by the compounded $g^4_{9}$ is non-degenerate.

4. (iv) $|K_X(-1)|=g^4_{8}+\Delta$, $\deg\Delta=5$; in this case, X is hyperelliptic and does not carry a very ample special linear series.

(i) Suppose $|K_X(-1)|=g^4_{12}+\Delta$, $\deg\Delta=1$, then one of the following holds;

\begin{equation*} \begin{cases} \mathrm{(ia)} ~X \mathrm{~is ~~trigonal ~~with~} |K_X(-1)(-\Delta)|=g^4_{12}=4g^1_3 \mathrm{~or~~ } \\ \mathrm{(ib)} ~\exists ~~X\stackrel{\varphi}{\rightarrow}C_{\mathcal{E}}\subset{\mathbb{P}}^4, \deg\varphi=2, \mathrm{~~genus}(C_{\mathcal{E}})=2, g^4_{12}= \varphi^*(g^4_6). \end{cases} \end{equation*}

(ia) Set $\Delta=p$ and let $p+q+r\in g^1_3$. We have

\begin{equation*}|K_X-(4g^1_3+p)|=|K_X-5g^1_3+q+r|\end{equation*}

and hence

\begin{align*} |\mathcal{O}_X(1)-q-r|&=|K_X-K_X(-1)-q-r|\\ &=|K_X-(4g^1_3+p)-q-r|=|K_X-5g^1_3|=g^{s}_{13}, s\ge 4 \end{align*}

from which it follows that $|\mathcal{O}_X(1)|$ is not very ample.

(ib) Since $\deg \varphi=2$, $\deg \varphi (X)=\deg C_{\mathcal{E}}=6$ and $g(C_{\mathcal{E}})\le\pi(6,4)\le 2$ by Castelnuovo genus bound. Since $|K_X(-1)(-\Delta)|$ is complete, $g^4_6$ on the normalization of $C_{\mathcal{E}}$ such that $g^4_{12}=|K_X(-1)(-\Delta)|= \varphi^*(g^4_6)$ is also complete. Hence $C_{\mathcal{E}}$ has geometric genus $g(C_{\mathcal{E}})=2$, smooth and $|\mathcal{O}_{C_{\mathcal{E}}}(1)|=g^4_6$ is non-special. However, we may exclude this case by lemma 3.3.

(ii) Suppose $|K_X(-1)|=g^4_{10}+\Delta$, $\deg\Delta=3$. Since $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ is non-degenerate and X is non-hyperelliptic, we have $g^4_{10}=\varphi^*(g^4_5)$ where $\varphi :X\rightarrow C_{\mathcal{E}}$ is a double cover of an elliptic curve $C_{\mathcal{E}}$ with complete, non-special $g^4_5$. Again by lemma 3.3, we may exclude this case, finishing the proof of Claim 1.

Claim 2: No smooth $X\in\mathcal{H}_{15,15,5}$ is trigonal: Suppose there is a trigonal $X\in\mathcal{H}_{15,15,5}$. Recall that on a trigonal curve X of genus $g\ge 5$ with a $g^r_d, ~ d\le g-1$, either $g^r_d$ is compounded with the unique $g^1_3$ or $|K_X-g^r_d|$ is compounded with $g^1_3$ by Maroni theory; cf. [Reference Martens and Schreyer29, proposition 1]. Since $\mathcal{E}$ is birationally very ample by Claim 1, $|K_X-\mathcal{E}|=|{\mathcal{O}}_X(1)|$ is compounded with $g^1_3$, a contradiction. This finishes the proof of Claim 2.

Claim 3: $|K_X(-1)|$ is base-point-free.

For the birationally very ample ${\mathcal{E}}=|K_X(-1)|$, suppose $\Delta=\mathrm{Bs}(\mathcal{E})\neq\emptyset$ and set $\mathcal{E}=\widetilde{\mathcal{E}}+\Delta$. Note that $\deg\Delta=1$ otherwise $\pi(e,4)\le 12 \lt g$ if $e\le 11$. Put $\Delta=p$ and let $C_{\widetilde{\mathcal{E}}}\subset{\mathbb{P}}^4$ be the image of the morphism induced by the moving part $\widetilde{\mathcal{E}}$. Since $\deg C_{\widetilde{\mathcal{E}}}=12$ and $\pi(12,4)=15=g$, $C_{\widetilde{\mathcal{E}}}$ is an (smooth) extremal curve lying on a cubic surface $S\subset{\mathbb{P}}^4$.

(i) Suppose S is a smooth cubic scroll in ${\mathbb{P}}^4$ and let $C_{\widetilde{\mathcal{E}}}\in|aH+bL|$. We solve the degree and genus formula (2) for r = 4 to get $a=4, b=0$. Since $X\cong C_{\widetilde{\mathcal{E}}}$ and $|K_X(-1)|=\mathcal{E}=\widetilde{\mathcal{E}}+\Delta$, we have

\begin{equation*}|K_S+C_{\widetilde{\mathcal{E}}}-H|_{|X}=|H+L|_{|X}=|K_X-\widetilde{\mathcal{E}}|,\end{equation*}

which is birationally very ample. From the standard exact sequence

\begin{equation*}0\rightarrow \mathcal{I}_{C_{\widetilde{\mathcal{E}}}}(H+L)\rightarrow \mathcal{O}(H+L)\rightarrow\mathcal{O}(H+L)\otimes\mathcal{O}_{C_{\widetilde{\mathcal{E}}}}\rightarrow 0\end{equation*}

the restriction map $H^0(S, \mathcal{O}(H+L))\stackrel{\rho}{\longrightarrow} H^0(C_{\widetilde{\mathcal{E}}},\mathcal{O}(H+L)\otimes\mathcal{O}_{C_{\widetilde{\mathcal{E}}}})$ is injective since $H^0(S, \mathcal{I}_{C_{{\widetilde{\mathcal{E}}}}}(H+L))=H^0(S, \mathcal{O}(-3H+L))=0$. ρ is surjective as well by Castelnuovo genus bound; if ρ is not surjective then

\begin{equation*}\mathbb{C}^7\cong H^0(S, \mathcal{O}(H+L))\subsetneq H^0(C_{\widetilde{\mathcal{E}}},\mathcal{O}(H+L)\otimes\mathcal{O}_{C_{\widetilde{\mathcal{E}}}}),\end{equation*}

which would imply that the birationally very ample $|\mathcal{O}(H+L)\otimes\mathcal{O}_{C_{\widetilde{\mathcal{E}}}}|=|K_X-\widetilde{\mathcal{E}}|$ on $C_{\widetilde{\mathcal{E}}}\cong X$ induces a morphism $X\rightarrow{\mathbb{P}}^s, s\ge 7$, a contradiction by $\pi(16,7)=12 \lt g$.

By the surjectivity of the restriction map ρ and by the very ampleness of $|H+L|$ on the cubic scroll S, $|K_X-\widetilde{\mathcal{E}}|$ is very ample. Note that $|\mathcal{O}_X(1)|=|K_X-\mathcal{E}|=|K_X-\widetilde{\mathcal{E}}-\Delta|$ gives rise to the morphism which is the composition of the embedding $X\stackrel{\varphi}{\longrightarrow}{\mathbb{P}}^6$ given by the very ample $|K_X-\widetilde{\mathcal{E}}|$ followed by the projection τ with centre at p;

However, the four-secant line through Δ, i.e. the line though Δ in the ruling $|L|$, produces a singularity and hence $|\mathcal{O}_X(1)|$ is not very ample if $\Delta\neq \emptyset$.

(ii) Suppose $S\subset{\mathbb{P}}^4$ is a cone over a twisted cubic in ${\mathbb{P}}^3$. Let $\widetilde{C_{\widetilde{\mathcal{E}}}}\subset \mathbb{F}_3$ be the strict transformation of $C_{\widetilde{\mathcal{E}}}\subset{\mathbb{P}}^4$ under $\mathbb{F}_{3}\stackrel{|h+3f|}{\longrightarrow} S$ and set $\widetilde{C_{\widetilde{\mathcal{E}}}}\in |ah+bf|$. Recall that $C_{\widetilde{\mathcal{E}}}$ is smooth (extremal) and so is $\widetilde{C_{\widetilde{\mathcal{E}}}}\cong X$. From

\begin{equation*}\widetilde{C_{\widetilde{\mathcal{E}}}}\cdot (h+3f)=(ah+bf)\cdot (h+3f)=b=\deg\widetilde{{\mathcal{E}}}=12\end{equation*}

\begin{equation*}\widetilde{C_{\widetilde{\mathcal{E}}}}\cdot(\widetilde{C_{\widetilde{\mathcal{E}}}}+K_{\mathbb{F}_3})=(ah+bf)\cdot((a-2)h+(b-5)f)=2g-2=28\end{equation*}

we get a = 4 and $\widetilde{C_{\widetilde{\mathcal{E}}}}\in |4h+12f|$. Note that

\begin{equation*} |K_{\mathbb{F}_3}+\widetilde{C_{\widetilde{\mathcal{E}}}}-(h+3f)|=|h+4f| \end{equation*}

is very ample by [Reference Hartshorne22, V.Cor. 2.18, p. 380]. We consider the restriction map $H^0(\mathbb{F}_3, \mathcal{O}(h+4f))\stackrel{\rho}{\longrightarrow} H^0(\widetilde{C_{\widetilde{\mathcal{E}}}}, \mathcal{O}(h+4f)\otimes\mathcal{O}_{\widetilde{C_{\widetilde{\mathcal{E}}}}})=H^0(\widetilde{C_{\widetilde{\mathcal{E}}}}, |K_X-\widetilde{\mathcal{E}}|).$ By the same routine as we did for the previous case (smooth cubic scroll case), we may claim that the restriction map ρ is surjective.

By the surjectivity of the restriction map ρ and the very ampleness of $|h+4f|$ on ${\mathbb{F}}_3$, we see that $|K_X-\widetilde{\mathcal{E}}|$ is very ample. $|\mathcal{O}_X(1)|=|K_X-\widetilde{\mathcal{E}}-\Delta|$ induces the composition of two maps $X\stackrel{|h+4f|_{|X}}{\longrightarrow}{\mathbb{P}}^6\stackrel{\pi_p}{\longrightarrow}{\mathbb{P}}^5$ and the projection map π_p is not an embedding; since $f\cdot (h+4f)=1, f\cdot \widetilde{C_{\widetilde{\mathcal{E}}}}=4$, the image of f under $|h+4f|$ in ${\mathbb{P}}^6$ is a four-secant line to the smooth image of $X\cong\widetilde{C_{\widetilde{\mathcal{E}}}}$. Therefore, $|\mathcal{O}_X(1)|$ is not very ample, finishing the proof of Claim 3.

Since $\mathcal{E}=|K_X(-1)|$ is base-point-free and birationally very ample, we have

\begin{equation*}g=\pi_1(13,4)=15\le p_a(C_{\mathcal{E}})\le\pi(13,4)=18.\end{equation*}

Thus, $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ lies on a surface S, $3\le\deg S\le4$; cf. [Reference Harris19, theorem 3.15, p. 99].

(A) Suppose $\deg S=3$ and S is smooth. Let $C_{\mathcal{E}}\in|aH+bL|$. Solving (2), i.e. $13=C_{\mathcal{E}}\cdot H=3a+b, ~p_a(C_{\mathcal{E}})=\frac{3(a-1)(a-2)}{2}+(2+b)(a-1),$ the following pairs $(a,b)\in\mathbb{Z}\times\mathbb{Z}$ are possible for $15\le p_a(C_{\mathcal{E}})\le 18$:

\begin{equation*} \begin{cases} p_a(C_{\mathcal{E}})=18: (5,-2), (4,1), \textrm{cases (A1), (A2) below} \\ p_a(C_{\mathcal{E}})=17, 16: \mathrm{no ~~ solution }\\ p_a(C_{\mathcal{E}})=15: (3,4), (6-5), \textrm{cases (A3), (A4) below}. \ \end{cases} \end{equation*}

Before proceeding, we recall some standard notation concerning linear systems and divisors on a blown up projective plane. Let ${\mathbb{P}}^2_s$ be the rational surface ${\mathbb{P}}^2$ blown up at s general points. Let e_i be the class of the exceptional divisor E_i and l be the class of a line L in ${\mathbb{P}}^2$. For integers $b_1\ge b_2\ge\cdots\ge b_s$, let $(a;b_1,\cdots, b_i, \cdots,b_s)$ denote class of the linear system $|aL-\sum b_i E_i|$ on ${\mathbb{P}}^2_s$. By abuse of notation, we use the expression $(a;b_1,\cdots, b_i, \cdots,b_s)$ for the divisor $aL-\sum b_i E_i$ and $|(a;b_1,\cdots, b_i, \cdots,b_s)|$ for the linear system $|aL-\sum b_i E_i|$. We use the convention

\begin{equation*}(a;b_1^{s_1},\cdots,b_j^{s_j},\cdots,b_t^{s_t}), ~ \sum s_j=s\end{equation*}

when b_j appears s_j times consecutively in the linear system $|aL-\sum b_i E_i|$.

For computational reasons, we sometimes identify a smooth rational normal surface scroll $S\subset {\mathbb{P}}^4$ with the Hirzebruch surface $\mathbb{F}_1$ embedded by the very ample linear system $|e+2f|$ on ${\mathbb{P}}^2_1$. We also make obvious identifications among divisor classes such as

\begin{equation*}H\cong e+2f, L\cong f, l-e_1\cong f_1\end{equation*}

where e is the class of the minimal degree self-intersection curve and f is class of the fibre on $\mathbb{F}_1$ (l is the class of a line, e ₁ the exceptional divisor on ${\mathbb{P}}^1_1$, and f ₁ is the proper transformation of the line through the blown up point). By abusing notation, we make no distinction between f and f ₁ (e and e ₁) and use the same letters.

(A1) $C_{\mathcal{E}}\in |5H-2L|$, $p_a(C_{\mathcal{E}})=18$: Note that

(6)\begin{equation} C_{\mathcal{E}}\in |5H-2L|=|5(2l-e_1)-2(l-e_1)|=|8l-3e_1| \end{equation}

and hence $C_{\mathcal{E}}$ has a plane model of degree 8. On a fixed $S\cong{\mathbb{F}}_1$, we consider the Severi variety $\Sigma_{{15},\mathcal{M}}$ of curves of genus g = 15 in the linear system $\mathcal{M}=|5H-2L|$. It is known that $\Sigma_{{15},\mathcal{M}}$ is irreducible,

\begin{equation*}\dim\Sigma_{{15},\mathcal{M}}=\dim\mathcal{M}-\delta , ~~\delta=p_a(C)-g\end{equation*}

and a general element of $\Sigma_{{15},\mathcal{M}}$ is a nodal curve with δ = 3 nodes as its only singularities; cf. [Reference Tyomkin34]. We now take a general $C\in \Sigma_{{15},\mathcal{M}}$, i.e. a curve with three nodes in $|5H-2L|$ on S or equivalently a curve with a plane model of degree 8 with an ordinary triple point and three nodes. Let $\widetilde{C}\subset{\mathbb{P}}^2_4$ be the strict transformation of C under the blowing up ${\mathbb{P}}^2_4 \stackrel{\varphi}{\longrightarrow} S\cong{\mathbb{P}}^2_1$ at three nodal singularities. By abusing notation, we set $H:=\varphi^*({\mathcal{O}}_S(1))$. We have $\widetilde{C}\in |(8;3,2^3)|$ and

(7)\begin{equation} |K_{{\mathbb{P}}^2_4}+\widetilde{C}-(2l-e_1)|=|-(3;1^4)+(8;3,2^3)-(2;1,0^3)|=|(3;1^4)|. \end{equation}

On the other hand, the restriction map

\begin{equation*}H^0({\mathbb{P}}^2_4,\mathcal{O}(K_{{\mathbb{P}}^2_4}+\widetilde{C}-H)\longrightarrow H^0(\widetilde{C}, \mathcal{O}(K_{{\mathbb{P}}^2_4}+\widetilde{C}-H)\otimes\mathcal{O}_{\widetilde{C}})\end{equation*}

is an isomorphism; injective by $H^0({\mathbb{P}}^2_4, \mathcal{O}(K_{{\mathbb{P}}^2_4}-H))=0$ and surjective by the Castelnuovo genus bound $\pi(15,6)=13 \lt g$, as we did in the proof of Claim 3. Therefore, the residual series

\begin{equation*}|K_{{\mathbb{P}}^2_4}+\widetilde{C}-\varphi^*{\mathcal{O}}_S(1)|_{|\widetilde{C}}=|K_{\widetilde{C}}-\varphi^*{\mathcal{O}}_S(1)_{|\widetilde{C}}|\end{equation*}

of $|\varphi^*{\mathcal{O}}_S(1)|_{|\widetilde{C}}$ is completely cut out by the very ample linear system $|(3;1^4)|$ on ${\mathbb{P}}^2_4$ which induces an embedding $\widetilde{C}\rightarrow{\mathbb{P}}^5$ as a smooth curve of degree

\begin{equation*}(3;1^4)\cdot \widetilde{C}=(3;1^4)\cdot (8;3,2^3)=24-3-6=15.\end{equation*}

Let $\mathcal{F}\subset\mathcal{G}^4_{13}$ be the family of such $\mathcal{E}=g^4_{13}$’s arising this way; i.e. the family of complete linear series $\mathcal{E}=g^4_{13}$’s such that $C_{\mathcal{E}}\in \Sigma_{15,\mathcal{M}}$ on a rational normal scroll $S\subset{\mathbb{P}}^4$, $\mathcal{M}=|5H-2L|$ and $\mathcal{E}=|\varphi^*{\mathcal{O}}_S(1)|_{|\widetilde{C_{\mathcal{E}}}}$. $\mathcal{F}$ is irreducible since $\Sigma_{15,\mathcal{M}}$ is irreducible. By an easy dimension count,

\begin{align*} \dim\mathcal{F}&=\dim\Sigma_{{15},\mathcal{M}}-\dim\mathrm{Aut}(S)\\ &=\dim|5H-2L|-\delta-\dim\mathrm{Aut}(S)=\dim|(8;3)|-3-6=29\\ & \gt 3g-3+\rho(13,15,4)=3g-3+\rho(15,15,5)=27, \end{align*}

hence the family $\mathcal{F}$ is our first candidate which may contribute to a full component of $\mathcal{H}_{15,15,5}$. Explicitly, we have the natural map $\mathcal{F}\stackrel{\psi}{\rightarrow} \mathcal{F}^\vee$ sending $\mathcal{E}=g^4_{13}$ to its residual series $|K-\mathcal{E}|=g^5_{15}$ such that $\psi(\mathcal{E})$ is very ample for a general $\mathcal{E}\in\mathcal{F}$. Thus, there exists an irreducible family of smooth curves of degree d = 15 and g = 15 in ${\mathbb{P}}^5$ which is an ${\operatorname{Aut}}({\mathbb{P}}^5)$-bundle over the family $\mathcal{F}^\vee\subset\mathcal{G}^5_{15}$.

(A2) $C_{\mathcal{E}}\in |4H+L|$, $p_a(C_{\mathcal{E}})=18$: We have $C_{\mathcal{E}}\in|4H+L|=|4(2l-e_1)+l-e_1|=|9l-5e_1|$. Set $\mathcal{L}=|4H+L|$ and let $\Sigma_{{15},\mathcal{L}}$ be the Severi variety $\Sigma_{{15},\mathcal{L}}$ of curves of genus g = 15 in the linear system $\mathcal{L}$. A general element of $\Sigma_{{15},\mathcal{L}}$ is a nodal curve with δ = 3 nodes as its only singularities. We take a general $C\in \Sigma_{{15},\mathcal{L}}$. Let $\widetilde{C}\subset{\mathbb{P}}^2_4$ be the strict transformation of C under the blow up of $S\cong{\mathbb{P}}^2_1$ at three nodal singularities. Hence $\widetilde{C}\in|(9;5,2^3)|$ and

\begin{equation*}|K_{{\mathbb{P}}^2_4}+\widetilde{C}-(2l-e_1)|=|-(3;1^4)+(9;5,2^3)-(2;1,0^3)|=|(4;3,1^3)|.\end{equation*}

The restriction map

\begin{equation*}H^0({\mathbb{P}}^2_4,\mathcal{O}(K_{{\mathbb{P}}^2_4}+\widetilde{C}-H)\longrightarrow H^0(\widetilde{C}, \mathcal{O}(K_{{\mathbb{P}}^2_4}+\widetilde{C}-H)\otimes\mathcal{O}_{\widetilde{C}})\end{equation*}

is an isomorphism; injective by $H^0({\mathbb{P}}^2_4, \mathcal{O}(K_{{\mathbb{P}}^2_4}-H))=0$ and surjective by the Castelnuovo genus bound $\pi(15,6)=13 \lt g$ by the same routine as we did in (i) in the proof of Claim 3. Therefore the residual series of $|\varphi^*{\mathcal{O}}_S(1)|_{|\widetilde{C}}$ is completely cut out by the linear system $|(4;3,1^3)|$ on ${\mathbb{P}}^2_4$. Note that $|(4;3,1^3)|$ is not very ample by [Reference Di Rocco14]. Furthermore, we have $(9;5,2^3)\cdot (2,1,1,0^2)=2$ whereas $(4;3,1^3)\cdot (2;1,1,0^2)=0$ contracting the $(-1)$ curve $l-e_1-e_2$ to a point. Hence, the image curve under the morphism induced by $|K_{\widetilde{C}}-\mathcal{E}|$ has a singularity. Therefore, the family of linear series arising from $\mathcal{L}=|4H+L|$ does not contribute to a component of $\mathcal{H}_{15,15,5}$.

(A3) $C_{\mathcal{E}}\in |3H+4L|$, $p_a(C_{\mathcal{E}})=15$: In this case, $C_{\mathcal{E}}$ is trigonal and by Claim 2, this case does not occur.

(A4) $C_{\mathcal{E}}\in |6H-5L|$, $p_a(C_{\mathcal{E}})=15$: In this case, $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ is smooth. For $C_\mathcal{E}\in |6H-5L|$, we have $|K_S+C_{\mathcal{E}}-H|=|3H-4L|.$ We may argue as in the two previous cases (A1) and (A2) to see that the restriction map

\begin{equation*}H^0(S,\mathcal{O}(3H-4L))\rightarrow H^0(C_{\mathcal{E}},\mathcal{O}(3H-4L)\otimes\mathcal{O}_{C_{\mathcal{E}}})\end{equation*}

is an isomorphism; by $H^0(S,\mathcal{O}(3H-4L-C_{\mathcal{E}}))=H^0(S,\mathcal{O}(-3H+L))=0$ and by Castelnuovo genus bound.

We also remark that the linear system $|3H-4L|=|2(H-L)+(H-2L)|$ has the fixed part $|H-2L|$;

\begin{equation*}H^0(S,\mathcal{O}(H-2L))=1 \mathrm{~and~} h^0(S,\mathcal{O}(3H-4L))=h^0(S,\mathcal{O}(2H-2L))=6.\end{equation*}

Hence the linear series $|K_{C_{\mathcal{E}}}-\mathcal{E}|$ cut out by the linear system $|3H-4L|$ has non-empty base locus B, $\deg B= (H-2L)\cdot C_{\mathcal{E}}=(H-2L)\cdot(6H-5L)=1$ and hence $C_{\mathcal{E}}\cdot 2(H-L)=14\neq 15$. This shows that curves in $|6H-5L|$ does not contribute to a component of $\mathcal{H}_{15,15,5}$.

(B) $\deg S=3$ and S is a cone over a twisted cubic in ${\mathbb{P}}^3$; $C_{\mathcal{E}}\subset S\subset{\mathbb{P}}^4$.

Let $\mathbb{F}_3\stackrel{|h+3f|}{\longrightarrow} S\subset{\mathbb{P}}^4$ be the minimal desingularization, which contracts h to the vertex $q\in S$; $h^2=-3, h\cdot f=0, f^2=0$. Let $\widetilde{C_{{\mathcal{E}}}}\subset \mathbb{F}_3$ be the strict transformation of $C_{{\mathcal{E}}}\subset{\mathbb{P}}^4$ and set $\widetilde{C_{{\mathcal{E}}}}\in |ah+bf|$. We have

\begin{equation*}\widetilde{C_{{\mathcal{E}}}}\cdot (h+3f)=(ah+bf)\cdot (h+3f)=b=13.\end{equation*}

By (1), $m:=h\cdot\widetilde{C_{\mathcal{E}}}=h\cdot (ah+13f)=-3a+13\ge 0,$ thus $a\le 4$ and hence a = 4 by Claim 2. Thus $\widetilde{C_{\mathcal{E}}}\in|4h+13f|$, m = 1, $p_a(\widetilde{C_{\mathcal{E}}})=18$ by adjunction formula, $C_{\mathcal{E}}$ passes through the vertex q of S and $C_{\mathcal{E}}$ is smooth at q.

Set $\mathcal{N}=|4h+13f|$ and let $\Sigma_{15,\mathcal{N}}$ be the Severi variety $\Sigma_{15,\mathcal{N}}$ consisting of curves of genus g = 15 in the linear system $\mathcal{N}$ on $\mathbb{F}_3$. Since a general element of $\Sigma_{15,\mathcal{N}}$ is a nodal curve, we may assume that $\widetilde{C_{\mathcal{E}}}$ has three nodes. Let $\mathbb{F}_{3,3}$ be the blow up of $\mathbb{F}_3$ at three nodes. Let $e_i~ (i=1,\ldots, 3)$ be exceptional divisors of the blow up and let $f_i~ (i=1,\ldots, 3)$ be three typical fibres containing the three nodal points of $\widetilde{C_{\mathcal{E}}}$. After resolving the three nodal singularities of $\widetilde{C_{\mathcal{E}}}$, we get a smooth curve ${\widehat{C_{\mathcal{E}}}}\subset\mathbb{F}_{3,3}$. We have

\begin{equation*}\widehat{C_{\mathcal{E}}}\in|4h+13f-\sum2e_i|, \mathrm{~~ and ~set}\end{equation*}

\begin{align*} \mathcal{M}:&=|\widehat{C_{\mathcal{E}}}+K_{\mathbb{F}_{3,3}}-(h+3f)|\\ &=|(4h+13f-\sum2e_i)+(-2h-5f+\sum e_i)-(h+3f)|\\ &=|h+5f-\sum e_i|. \end{align*}

Since $\mathcal{O}_{\mathbb{F}_{3,3}}(\mathcal{M}-\widehat{C_{\mathcal{E}}})=\mathcal{O}_{\mathbb{F}_{3,3}}(-(3h+8f-\sum e_i))$ and $f\cdot (\mathcal{M}-\widetilde{C_{\mathcal{E}}}) \lt 0$, we have $h^0({\mathbb{F}_{3,3}},\mathcal{O}(\mathcal{M}-\widehat{C_{\mathcal{E}}}))=0$ implying that the restriction map

\begin{equation*}\rho: H^0({\mathbb{F}_{3,3}},\mathcal{O}(\mathcal{M}))\longrightarrow H^0(\widehat{C_{\mathcal{E}}},\mathcal{O}(\mathcal{M})\otimes\mathcal{O}_{\widehat{C_{\mathcal{E}}}})\end{equation*}

is injective. Note that $h^0({\mathbb{F}_{3,3}},\mathcal{O}(\mathcal{M}))= h^0(\mathbb{F}_3,\mathcal{O}(h+5f))-3=6$. Set $\mathcal{D}:={\mathbb{P}}(H^0(\widehat{C_{\mathcal{E}}},\mathcal{O}(\mathcal{M})\otimes\mathcal{O}_{\widehat{C_{\mathcal{E}}}}))$. Since $X\stackrel{iso}{\cong}\widehat{C_{\mathcal{E}}}\stackrel{bir}{\cong}{\widetilde{C_{\mathcal{E}}}}\stackrel{bir}{\cong}{C_{\mathcal{E}}}$, ${\mathcal{D}}=|K_{\widehat{C_{\mathcal{E}}}}-{\mathcal{E}}|$ is birationally very ample. If ρ is not surjective, we have $\dim{\mathcal{D}}\ge 6$ and $\pi(15,6)=13 \lt g=15$, which is a contradiction. Therefore, we have

\begin{equation*}\mathrm{Im}(\rho)= H^0(\widehat{C_{\mathcal{E}}},\mathcal{O}(\mathcal{M})\otimes\mathcal{O}_{\widehat{C_{\mathcal{E}}}})\end{equation*}

and the restriction map ρ is surjective. Denoting by $\widetilde{f}_i$ the proper transformation of three typical fibres of f_i under the blow up ${\mathbb{F}}_{3,3}\rightarrow{\mathbb{F}}_3$, we have

\begin{equation*}\tilde{f}_i^2=-1, \tilde{f}_i\cdot e_i=1, h\cdot\tilde{f}_i=1, \widehat{C_{\mathcal{E}}}\cdot \tilde{f}_i=2, (h+5f-\Sigma e_i)\cdot \tilde{f}_i=0.\end{equation*}

Hence the morphism ψ induced by $\mathcal{M}=|\widehat{C_{\mathcal{E}}}+K_{\mathbb{F}_{3,3}}-(h+3f)|$ contracts $(-1)$ curves $\tilde{f}_i$ and the image curve $\psi (\widehat{C_{\mathcal{E}}})\subset{\mathbb{P}}^5$ acquires singularities. It then follows that $\mathcal{M}\otimes\mathcal{O}_{\widehat{C_{\mathcal{E}}}}=|K_{\widehat{C_{\mathcal{E}}}}(-1)|={\mathcal{D}}$ is not very ample.

(C) $\deg S=4$ and S is smooth.

Note that $S\cong{\mathbb{P}}^2_5$ and $C_{\mathcal{E}}$ is smooth; if not, we have $15=g=\pi_1(13,4)\lneq p_a(C_{\mathcal{E}})\le \pi(13,4)=18$ and hence $C_{\mathcal{E}}$ is a nearly extremal curve lying on a cubic surface, which has been treated already in the steps (A) and (B). Setting $C_{\mathcal{E}}\in |(a;b_1,\ldots ,b_5)|$, we have

\begin{equation*}\deg C_{\mathcal{E}} =3a-\sum b_i=13, C_{\mathcal{E}}^2=a^2-\sum b_i^2=2g-2-K_S\cdot C=41.\end{equation*}

By Schwartz’s inequality, we have

\begin{equation*}(\sum b_i)^2\le 5(\sum b_i^2)\end{equation*}

and substituting $\sum b_i=3a-13$ and $\sum b_i^2=a^2-41$ we obtain

\begin{equation*}5\,({a}^{2}-41)- \left( 13-3\,a \right) ^{2}\ge 0 \Leftrightarrow a=9, 10, 11\end{equation*}

and therefore we have the following three cases:

\begin{equation*} (a;b_1,\cdots, b_5)= \begin{cases} (9;3^4,2)\\ (10;4^2,3^3)\\ (11;4^5). \end{cases} \end{equation*}

We need to check if $|K_{C_{\mathcal{E}}}-\mathcal{E}|=|K_S+C_{\mathcal{E}}-H|_{|C_{\mathcal{E}}}=|C_{\mathcal{E}}+2K_S|_{|C_{\mathcal{E}}}$ is very ample. The restriction map $\rho: H^0(S,\mathcal{O}(C_{\mathcal{E}}+2K_S)\rightarrow H^0(C_{\mathcal{E}},\mathcal{O}(C_{\mathcal{E}}+2K_S)\otimes\mathcal{O}_{C_{\mathcal{E}}})$ is an isomorphism; $\ker(\rho)=H^0(S, 2K_S)=0$, $h^0(S,\mathcal{O}(C_{\mathcal{E}}+2K_S))=6$ and by the Castelnuovo genus bound $\pi (15,6)=13 \lt g$ if ρ is not surjective. Assume the last case among three in the above list; $C_{\mathcal{E}}\in |(11;4^5)|$. We have

\begin{equation*}|K_S+C_{\mathcal{E}}-H|=|(11;4^5)-2(3;1^5)|=|(5;2^5)|,\end{equation*}

whose restriction on $C_{\mathcal{E}}$ does not induce an isomorphism onto its image. To see this, the $(-1)$ curve $(2;1^5)$ on S is contracted to a point; $(2;1^5)\cdot (5;2^5)=0$ whereas $(11;4^5)\cdot (2;1^5)=2$ and hence the image curve in ${\mathbb{P}}^5$ is singular. The verification for the other two cases $(9;3^4,2), (10;4^2,3^3)$ are similar which we omit. Hence we conclude that $|K_{C_{\mathcal{E}}}-\mathcal{E}|$ is not very ample.

(D) $\deg S=4$ and S is a cone over an elliptic curve $E\subset{\mathbb{P}}^3$:

Recall that a cone $S\subset{\mathbb{P}}^r$ over an elliptic curve $E\subset H\cong{\mathbb{P}}^{r-1}$ with vertex outside H is the image of the birational morphism $\mathbb{E}_{r}:=\mathbb{P}(\mathcal{O}_{E}\oplus\mathcal{O}_{E}(r))\rightarrow S\subset{\mathbb{P}}^r$ induced by $|\overline{h}|:=|h+rf|$, where $h^2=-r$ and f is the fibre of $\mathbb{E}_{r}\stackrel{\eta}{\rightarrow} E$. Let $C\subset S$ be an integral curve of degree d with the strict transformation $\widetilde{C}$ under $\mathbb{E}_{r}\rightarrow S$.  Setting $k=\widetilde{C}\cdot f$, we have $\widetilde{C}\equiv kh+df$, $\deg\eta_{|\widetilde{C}}= \widetilde{C}\cdot f=k$ and

(8)\begin{equation} p_a(\widetilde{C})= (k-1)(d-\frac{kr}{2})+1,~~ 0\le \widetilde{C}\cdot h=d-rk=m \end{equation}

where m is the multiplicity of C at the vertex.

For $C=C_{\mathcal{E}}$, d = 13, and r = 4, we get $(k,m,p_a(\widetilde{C_{\mathcal{E}}}))\in\{(3,1,15), (2,5,10)\}$ by (8). In the first case, since $g=15=p_a(\widetilde{C_{\mathcal{E}}})$, $\widetilde{C_{\mathcal{E}}}\in |3h+13f|=|3\overline{h}+f|$ is smooth and is a triple covering of an elliptic curve. The second case is not possible since $p_a(\widetilde{C_{\mathcal{E}}}) \lt g$. We now check if

\begin{equation*}|K_{\mathbb{E}_4}+\widetilde{C_{\mathcal{E}}}-\overline{h}|_{|\widetilde{C_{\mathcal{E}}}}=|(-2\overline{h}+4f+(k\overline{h}+(d-4k)f)-\overline{h}|_{|\widetilde{C_{\mathcal{E}}}}=|5f|_{|\widetilde{C_{\mathcal{E}}}}\end{equation*}

is very ample.

Claim: The restriction map

\begin{equation*}\rho: H^0(\mathbb{E}_4,|K_{\mathbb{E}_4}+\widetilde{C_{\mathcal{E}}}-\overline{h}|)\rightarrow H^0(\widetilde{C_{\mathcal{E}}},K_{\widetilde{C_{\mathcal{E}}}}((-1)))\end{equation*}

fails to be surjective: We consider the exact sequence

\begin{equation*}0\rightarrow\mathcal{I}_{\widetilde{C_{\mathcal{E}}}}(5f)=\mathcal{O}_{{\mathbb{E}}_4}(-3\overline{h}+4f)\rightarrow\mathcal{O}_{{\mathbb{E}}_4}(5f)\rightarrow\mathcal{O}(5f)_{|\widetilde{C_{\mathcal{E}}}}\rightarrow 0.\end{equation*}

We have $h^0({\mathbb{E}}_4,\mathcal{O}_{{\mathbb{E}}_4}(-3\overline{h}+4f))=0$, $h^0({{\mathbb{E}}_4},\mathcal{O}_{{\mathbb{E}}_4}(5f))=5$, $h^1({{\mathbb{E}}_4},\mathcal{O}_{{\mathbb{E}}_4}(5f))=0$ and by Serre’s duality

\begin{equation*}h^1({\mathbb{E}}_4,\mathcal{O}_{{\mathbb{E}}_4}(-3\overline{h}+4f))=h^1(\mathcal{O}_{{\mathbb{E}}_4}(\overline{h}))=1.\end{equation*}

Thus ρ is not surjective and $h^0(\widetilde{C_{\mathcal{E}}}, \mathcal{O}(5f)_{|\widetilde{C_{\mathcal{E}}}})=6$.

Claim: $|K_{\widetilde{C_{\mathcal{E}}}}(-1)|=|K_{\mathbb{E}_4}+\widetilde{C_{\mathcal{E}}}-\overline{h}|_{|\widetilde{C_{\mathcal{E}}}}$ is not very ample.

Let $V:=\mathrm{Im}(\rho)\subset H^0(\widetilde{C_{\mathcal{E}}},\mathcal{O}(5f)_{|\widetilde{C_{\mathcal{E}}}})$ and we assume $|K_{\widetilde{C_{\mathcal{E}}}}(-1)|$ is very ample inducing an isomorphism φ onto X ₁. Since ${\mathbb{P}}(V)\subsetneq |K_{\widetilde{C_{\mathcal{E}}}}(-1)|$ and $\dim V=5$, we have the following commutative diagram:

(a) ψ is the morphism on ${\mathbb{E}}_4$ induced by the base-point-free ${\mathbb{P}}(V)=|5f|$, $\deg E_1=\overline{h}\cdot 5f=5$, $\deg\psi_{|\widetilde C_{\mathcal{E}}}=\widetilde{C_{\mathcal{E}}}\cdot f=(3\overline{h}+f)\cdot f =3$. E ₁ is elliptic by Claim 2.

(b) τ is the projection map with centre of projection $p_1\in{\mathbb{P}}^5$ corresponding to ${\mathbb{P}}(V)$, i.e. the intersection of all hyperplanes corresponding to divisors in ${\mathbb{P}}(V)$, inducing a morphism $X_1\stackrel{\tau}{\rightarrow{E}}_1$, $\deg\psi=\deg\tau =3$, $\tau\circ\varphi=\psi$, and $p_1\notin X_1$ since ${\mathbb{P}}(V)$ is base-point-free.

Let $T_1\subset{\mathbb{P}}^5$ be a cone over E ₁ with vertex p ₁. T ₁ is the image of the morphism on $\mathbb{E}_5:=\mathbb{P}(\mathcal{O}_{E_1}\oplus\mathcal{O}_{E_1}(5))$ induced by $|\widetilde{h}|:=|h+5f|$ and we have $X_1\subset T_1$. Let $\widetilde{X}_1$ be the strict transformation of X ₁ via ${\mathbb{E}}_5\rightarrow T_1$. Setting $k=\widetilde{X}_1\cdot f$, we have $\widetilde{X}_1\equiv k\widetilde{h}+(d-5k)f$. By (8) (for d = 15, r = 5), we get $(k,m,p_a({\widetilde{X}_1}))\in\{(3,0,16), (2,5,11)\}$. In the first case $p_a(\widetilde{X})=p_a(X)=16 \gt g=15$, a contradiction since X ₁ and $\widetilde{X}_1$ are smooth. The second case is not possible since $p_a(\widetilde{X}) \lt g$. Thus we deduce that $|K_{\widetilde{C_{\mathcal{E}}}}(-1)|$ is not very ample.

(E) $\deg S=4$ and S is singular with isolated singularities; $C_{\mathcal{E}}\subset S\subset{\mathbb{P}}^4$. We see that $C_{\mathcal{E}}$ is smooth; otherwise, we have $15=g=\pi_1(13,4)\lneq p_a(C_{\mathcal{E}})\le \pi(13,4)=18$, hence $C_{\mathcal{E}}$ is a nearly extremal curve lying on a cubic surface which has been treated already in the steps (A) and (B). We assume that there is a smooth curve C—which we may take as $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ under our current situation—of degree d = 13 and genus g = 15 on a singular del-Pezzo surface S with isolated singularities. We also assume that the dual curve of $C_{\mathcal{E}}$ is a smooth curve $X\subset{\mathbb{P}}^5$. By proposition 2.1, we let $\{C_t\}$ be a one parameter flat family of (smooth) curves with $C_0=C=C_{\mathcal{E}}$ lying on a singular del Pezzo $S\subset{\mathbb{P}}^4$ and $\{C_t; t\neq 0 \},$ lying on smooth del Pezzo surfaces. If the dual curve $X=C_0^\vee\subset{\mathbb{P}}^5$ is smooth, dual curves $\{C_t^\vee; t\neq 0\}$ are also smooth since singular curves cannot specialize to a smooth curve $X=C_0^\vee$. However, this is contradictory to what we have verified in (C), i.e. every smooth curve $C_t\subset {\mathbb{P}}^4$ with $(d,g)=(13,15)$ lying on a smooth del Pezzo has its dual curve in ${\mathbb{P}}^5$ which is always singular.

Conclusion: We have exhausted all the possibilities for the surfaces $S\subset {\mathbb{P}}^4$ on which dual curves $C_{\mathcal{E}}$ of $X\in\mathcal{H}_{15,15,5}$ may sit. Our lengthy discussion in parts (A)–(E) shows that the only case such that the residual series of the hyperplane series of $C_{\mathcal{E}}\subset{\mathbb{P}}^4$ is very ample—among all the possibilities for the surface S—is the case (A1); $C_{\mathcal{E}}\subset S\subset{\mathbb{P}}^4$ lies a smooth cubic surface S, $p_a(C_{\mathcal{E}})=18$, $C_{\mathcal{E}}\in \Sigma_{15,\mathcal{M}}$ where $\mathcal{M}=|5H-2L|$. Part (A1) also shows that the curve X corresponding to a general element in the Severi variety $\Sigma_{15,\mathcal{M}}$ lies on a smooth del Pezzo surface in ${\mathbb{P}}^5$; cf. (7). From (6), we see that a general $X\in\mathcal{H}_{15,15,5}$ has a plane model of degree 8 with an ordinary triple point and lines through the triple point cut out a base-point-free $g^1_5$. X does not have $g^1_4$ by Castelnuvo–Severi inequality.

Proof. X has genus g = 15 by the assumption that Y has three ordinary nodes or ordinary cusps at points in A and an ordinary triple point at p. Let 2p the closed subscheme of ${\mathbb{P}}^2$ with $({\mathcal{I}}_p)^2$ as its ideal sheaf. We have $\deg (2p)=3$ and $(2p)_{\mathrm{red}} =\{p\}$. Set and fix $Z:=A \cup 2p$ once and for all; $\deg (Z)=6$. We note that Z is the conductor of the normalization map, i.e. the complete linear system $|K_X|$ is induced by $|{\mathcal{I}}_Z(5)|$. Thus to prove that $h^0(X,{\mathcal{L}})=3$, it is sufficient to show that Z imposes six independent conditions on $|{\mathcal{O}}_{{\mathbb{P}}^2}(4)|$, i.e. $h^1({\mathbb{P}}^2,{\mathcal{I}}_Z(4)) =0$. Recall that for a degree 6 zero-dimensional scheme $F\subset {\mathbb{P}}^2$, $h^1({\mathbb{P}}^2,{\mathcal{I}}_F(4))=0$ if and only if F is not contained in a line; cf. [Reference Bernardi, Gimigliano and Idà8, lemma 34]. Therefore, we have $h^1({\mathbb{P}}^2,{\mathcal{I}}_Z(4)) =0$ by the assumption on Z and hence $h^0(X, {\mathcal{L}})=3$. We can also deduce easily that the line bundle ${\mathcal{R}}$ described in (a) and the three line bundles ${\mathcal{L}}_1,{\mathcal{L}}_2,{\mathcal{L}}_3$ described in (b) are complete pencils. After the characterization of the (unique) $g^1_5$ and the (only) three $g^1_6$’s in the following part (a) and (b), the uniqueness of the $g^2_8$ is shown in (c).

For a line $L\subset {\mathbb{P}}^2$, $\deg (L\cap Z)\le 3$ by our assumptions on $A\cup \{p\}$. Note that

1. (i) $\deg (L\cap Z)=3$ if and only if L is one of the three lines, say $R_1,R_2,R_3$, containing p and one of the points of A.

2. (ii) $\deg (L\cap Z)=2$ if and only if either L is one of the lines, $L_1,L_2,L_3$, containing two of the points of A or $p\in L$ and $L\notin \{R_1,R_2,R_3\}$.

Recall that a conic $D\subset {\mathbb{P}}^2$ contains 2p if and only D is singular at p if and only if D is a union of two lines intersecting at p or a double line through p. Thus

(11)\begin{equation} \begin{cases} h^0({\mathbb{P}}^2,{\mathcal{I}}_Z(2)) =0,\\ h^1({\mathbb{P}}^2,{\mathcal{I}}_Z(2)) =0 \mathrm{~since ~} \deg (Z)=6 \mathrm{~and ~therefore }\\ h^1({\mathbb{P}}^2,{\mathcal{I}}_Z(t)) =0 \,\textrm{for all } t\ge 3. \end{cases} \end{equation}

(a) X has no pencil of degree four or less by the Castelnuovo–Severi inequality [Reference Accola1, theorem 3.5] and hence X is 5-gonal with a $g^1_5$ cut out by lines through p. Take a complete base-point-free pencil ${\mathcal{R}}$ on X such that $\deg ({\mathcal{R}}) = 5$. Fix a general $E\in {\mathcal{R}}$ and set $B:= v(E)$. We note that

1. (ai) $B\cap Z=\emptyset$ since ${\mathcal{R}}$ is base-point-free and E is general,

2. (aii) $h^1({\mathbb{P}}^2,{\mathcal{I}}_{Z\cup B}(5)) \gt 0$ since $|K_X|$ is induced by $|{\mathcal{I}}_Z(5)|$ and ${\mathcal{R}}$ is a pencil,

3. (aiii) $h^1({\mathbb{P}}^2,{\mathcal{I}}_{Z\cup B^{\prime}}(5)) =0$ for all $B^{\prime}\subsetneq B$ since ${\mathcal{R}}$ is complete and base-point-free,

4. (aiv) $|{\mathcal{I}}_B(2)|\ne \emptyset$ since $\#B=5$.

Fix a general $C\in |{\mathcal{I}}_B(2)|$. Set $M:= C\cap (Z\cup B)$. Since $B\subset C$, we have $\operatorname{Res}_C(Z\cup B) =\operatorname{Res}_C(Z)$. We consider the residual exact sequence

(12)\begin{equation} 0 \rightarrow {\mathcal{I}}_{\operatorname{Res}_C(Z)}(3)\rightarrow {\mathcal{I}}_{Z\cup B}(5) \rightarrow {\mathcal{I}}_{M,C}(5) \rightarrow 0 \end{equation}

of $Z\cup B$ with respect to C. Since $\operatorname{Res}_C(Z)\subseteq Z$ and $h^1({\mathbb{P}}^2,{\mathcal{I}}_Z(2)) =0$ by (11), we have $h^1({\mathbb{P}}^2,{\mathcal{I}}_{\operatorname{Res}_C(Z)}(3)) =0$. Hence by (aii), the long cohomology sequence of (12) yields

\begin{equation*}h^1(C,{\mathcal{I}}_{M,C}(5)) \gt 0.\end{equation*}

We first assume that C is smooth. Since $\deg M\le\deg (Z\cup B) =11$, $\deg ({\mathcal{O}}_C(5)) =10$ and $C\cong {\mathbb{P}}^1$, we have $h^1(C,{\mathcal{I}}_{M,C}(5))=0$, a contradiction.

We next assume that C is singular (a priori even a double line), say $C=C_1+C_2$ with $\deg (C_1\cap (Z\cup B)) \ge \deg (C_2\cap (Z\cup B))$ if $C_1\ne C_2$.

Since $M\subset C_1+C_2$, we have $\operatorname{Res}_{C_1+C_2}(M) =\emptyset$, $\operatorname{Res}_{C_2}(\operatorname{Res}_{C_1}(M)) =\emptyset$ and hence $\operatorname{Res}_{C_1}(M)\subset C_2$. By the basic property (9) on residual schemes in remark 4.3, $\deg (\mathrm{Res}_{C_1}(M)) =\deg (M) -\deg (M\cap C_1)$. Since $\deg (M)\le 11$ and $\deg (M\cap C_1)\ge \deg(M\cap C_2)$ by assumption, we have $\deg (M\cap C_2)\le 5$. Since $\operatorname{Res}_{C_1}(M)\subset C_2$ and $\operatorname{Res}_{C_1}(M)\subseteq M$, we have

\begin{equation*}\deg (\operatorname{Res}_{C_1}(M)) =\deg (\operatorname{Res}_{C_1}(M)\cap C_2) \le \deg (M\cap C_2) \le 5.\end{equation*}

Since $h^1(C,{\mathcal{I}}_{M,C}(5)) \gt 0$, we have $h^1({\mathbb{P}}^2,{\mathcal{I}}_M(5)) \gt 0$ by usual cohomology computation of the sequence $0\rightarrow {\mathcal{I}}_C(5)\rightarrow {\mathcal{I}}_M(5) \rightarrow{\mathcal{I}}_{M,C}(5)\rightarrow 0$. Consider the residual exact sequence of M with respect to the line C ₁;

(13)\begin{equation} 0 \rightarrow {\mathcal{I}}_{\mathrm{Res}_{C_1}(M)}(4)\rightarrow {\mathcal{I}}_M(5) \rightarrow {\mathcal{I}}_{C_1\cap M,C_1}(5)\rightarrow 0. \end{equation}

Since $\deg (\operatorname{Res}_{C_1}(M)) \le 5$, we have $h^1({\mathbb{P}}^2,{\mathcal{I}}_{\operatorname{Res}_{C_1}(M)}(4)) =0$ [Reference Bernardi, Gimigliano and Idà8, lemma 34]. Thus, the long cohomology exact sequence of (13) and $h^1({\mathbb{P}}^2,{\mathcal{I}}_M(5)) \gt 0$ yield

\begin{equation*}h^1(C_1,{\mathcal{I}}_{C_1\cap M}(5)) \gt 0 ~ \textrm{implying} ~\deg (C_1\cap M)\ge 7.\end{equation*}

From this, we have the following two cases:

\begin{equation*} \begin{cases} B\subset C_1 \mathrm{~and ~}\deg (Z\cap C_1)\ge 2 \mathrm{~ or ~} \\ \#(B\cap C_1)=4 \mathrm{~and~ } \deg (Z\cap C_1)=3. \end{cases} \end{equation*}

The latter possibility is excluded for a general $E\in |{\mathcal{R}}|$, because only three lines, $R_1,R_2,R_3$, intersect Z in a degree 3 schemes. Thus $B\subset C_1$ and $\deg (Z\cap C_1)=2$. Since there are only three lines, $L_1,L_2,L_3$, intersecting Z in a degree 2 scheme and not containing p, we get $p\in C_1$, concluding the proof of (a).

(b) Fix a base-point-free line bundle ${\mathcal{M}}$ on X such that $\deg ({\mathcal{M}})=6$. Since X has a unique $g^1_5$ by part (a), we have $h^0(X,{\mathcal{M}})=2$. To see this, X has no birationally very ample $g^2_6$ by the uniqueness of $g^1_5$ or by genus reason. X does carry a compounded $g^2_6$ either since C is neither trigonal nor bi-elliptic.

Fix a general $G\in |{\mathcal{M}}|$ and set $F:= v(G)$. Since ${\mathcal{M}}=g^1_6$ is base-point-free and complete, we have $h^1({\mathbb{P}}^2,{\mathcal{I}}_{Z\cup F}(5)) \gt 0$ and $h^1({\mathbb{P}}^2,{\mathcal{I}}_{Z\cup F^{\prime}}(5)) =0$ for all $F^{\prime}\subsetneq F$. Choose any $F^{\prime}\subset F$ formed by five points and take $D\in |{\mathcal{I}}_{F^{\prime}}(2)|\neq\emptyset$. Set $N:= D\cap (Z\cup F)$. Assume D is smooth thus no 3 among F ^ʹ is collinear. Consider the residual exact sequence of $Z\cup F$ with respect to D;

(14)\begin{equation} 0\rightarrow {\mathcal{I}}_{\mathrm{Res}_D(Z\cup F)}(3)\rightarrow {\mathcal{I}}_{Z\cup F}(5) \rightarrow {\mathcal{I}}_{N,D}(5)\rightarrow 0. \end{equation}

Since $\#(F\setminus F\cap D)\le 1$, $\deg{\mathrm{Res}_D(Z\cup F)}\le 7$. Since no 5 among $\mathrm{Res}_D(Z\cup F)$ is collinear, $h^1({\mathbb{P}}^2,{\mathcal{I}}_{\mathrm{Res}_D(Z\cup F)}(3))=0$ by [Reference Bernardi, Gimigliano and Idà8, lemma 34]. Thus, the long cohomology sequence from (14) gives $h^1(D,{\mathcal{I}}_{N,D}(5)) \gt 0$. Since $D\cong {\mathbb{P}}^1$, $\deg({\mathcal{O}}_D(5)) =10$, and $h^1(D,{\mathcal{I}}_{N,D}(5)) \gt 0$, we get

\begin{equation*}\deg(N)\ge 12.\end{equation*}

On the other hand, we have $\deg (D\cap Z)\le 5$. To see this, we assume $\deg (D\cap Z)\ge 6$ and hence D is smooth. From the exact sequence $0\rightarrow{\mathcal{I}}_{D\cap Z, D}(2)\rightarrow{\mathcal{O}}_D(2)\rightarrow{\mathcal{O}}_{D\cap Z}(2)\rightarrow 0$, we have $h^1(D,{\mathcal{I}}_{D\cap Z,D}(2)) \gt 0$ following from $h^0(D,{\mathcal{O}}_D(2))=5$ and $h^0({\mathcal{O}}_{D\cap Z})=\deg (D\cap Z)\ge 6$. From the exact sequence $0\rightarrow{\mathcal{I}}_{D}(2)\rightarrow {\mathcal{I}}_{D\cap Z}(2)\rightarrow{\mathcal{I}}_{D\cap Z,D}(2)\rightarrow 0$ and by $h^i({\mathbb{P}}^2,{\mathcal{I}}_D(2))= h^i({\mathbb{P}}^2,{\mathcal{O}}_{{\mathbb{P}}^2}) =0~ (i\ge 1)$, we get $h^1({\mathbb{P}}^2,{\mathcal{I}}_{D\cap Z}(2)) \gt 0$. Since $D\cap Z\subseteq Z$, we get $h^1({\mathbb{P}}^2,{\mathcal{I}}_Z(2)) \gt 0$ contrary to (11) concluding $\deg (D\cap Z)\le 5$, thus

\begin{equation*}\deg (N)\le 11.\end{equation*}

This contradiction shows that D is not smooth. Set $D=D_1+D_2$. Exactly as in step (a), we may prove the existence of a line D ₁ with $\deg (D_1\cap N)\ge 7$. Since ${\mathcal{M}}$ is not induced by the pencil of lines through p, $p\notin D_1$. For a general $G\in|{\mathcal{M}}|$, we have $D_1\notin \{L_1,L_2,L_3\}$. Thus $F\subset D_1$ and D ₁ contains one of the points of A, concluding the proof of (b).

(c) We only need to prove the uniqueness part. Take a line bundle ${\mathcal{N}}$ on X such that $h^0({\mathcal{N}})\ge 3$ and $\deg ({\mathcal{N}})\le 8$. Since X has only finitely many base-point-free $g^1_6$’s by part (b), $\deg({\mathcal{N}})=8, h^0(X,{\mathcal{N}})=3$ and ${\mathcal{N}}$ is base-point-free. Part (a) implies that $|{\mathcal{N}}|$ is not compounded; if it were then either X is 4-gonal or a double covering of a smooth plane curve of degree 3, which may be excluded by the Castelnuovo–Severi inequality. We want to apply remark 4.4(a) to the linear series $|{\mathcal{N}}|$.

Fix a general $V\in |{\mathcal{N}}|$ and set $U:= {v}(V)$. To conclude the proof we need to prove that U is formed by eight collinear points. For a general V, we have $Z\cap {v}(V) =\emptyset$. Since $h^0(X,{\mathcal{N}})=3$, we have

(15)\begin{equation} h^1({\mathbb{P}}^2,{\mathcal{I}} _{Z\cup U}(5))\ge 2. \end{equation}

Observation 1:

1. (0) Since V is general, remark 4.4(a) implies that if $\#(U\cap L)\ge 3$ for some line L, then $U\subset L$, concluding the proof.

2. (i) Thus from now on we may assume that no 3 points among U are collinear.

3. (ii) Suppose $\#(U\cap D)\ge 6$ for some conic D, i.e. $U\cap D$ fails to impose independent conditions on $|{\mathcal{O}}_{{\mathbb{P}}^2}(2)|$. Hence by remark 4.4(a), we have $U\subset D$.

Observation 2:

1. (0) The zero-dimensional scheme $Z =2p\cup A$ has degree 6 and is not contained in a conic; just because A is not formed by three collinear points and any conic containing Z is singular at p.

2. (i) We set ${\mathcal{Z}}:=\{D\in |{\mathcal{O}}_{{\mathbb{P}}^2}(2)| | \deg Z\cap D=5\}, {\mathcal{Z}}_1:=\{D\in {\mathcal{Z}} | 2p\subset D\},$ ${\mathcal{Z}}_2:= {\mathcal{Z}}\setminus {\mathcal{Z}}_1.$ Each $D\in {\mathcal{Z}}_1$ is singular at p and hence $\#{\mathcal{Z}}_1=3$; each $D\in {\mathcal{Z}}_1$ is the union of two lines through p containing one of the points of A. For $D\in {\mathcal{Z}}_2$, $D\cap Z=A\cup w$, where w is degree 2 connected zero-dimensional subscheme with p as its reduction. Since no conic contains Z, $A\cup w$ uniquely determines D. Thus ${\mathcal{Z}}_2$ is a one-dimensional family and ${\mathcal{Z}}$ is an algebraic family of dimension 1.

Given $U={v}(V)$, let $D_U\in |{\mathcal{O}}_{{\mathbb{P}}^2}(2)|$ be such that $\#(D_U\cap U)$ is maximal. Note that $\#(U\cap D_U)\ge 5$ since $\dim |{\mathcal{O}}_{{\mathbb{P}}^2}(2)| =5$. By Observation 1(i), D_U is a smooth conic. Bezout gives $\#(D_U\cap v(X)) \le 16$. Thus D_U contains at most one other set U ^ʹ with $\#U^{\prime} =8$ with $U^{\prime}=\#v(V^{\prime}) =8$ for some $V^{\prime}\in |{\mathcal{N}}|$ and $U^{\prime}\cap U=\emptyset$

\begin{equation*} \begin{cases} \#(D_U\cap U) =5 \,\textrm{or } \\ U\subset D_U. \end{cases} \end{equation*}

Now we know that there are two-dimensional family of conics

\begin{equation*}{\mathcal{U}}:=\{D_U| U={v}(V), V\in{\mathcal{N}}, \#(D_U\cap U)\ge 5\}\subset|{\mathcal{O}}_{{\mathbb{P}}^2}(2)|.\end{equation*}

On the other hand, the family of conics

\begin{equation*}{\mathcal{Z}}=\{D|\deg(D\cap Z)=5\}\subset|{\mathcal{O}}_{{\mathbb{P}}^2}(2)|\end{equation*}

moves only in one-dimensional family by Observation 2(i). Hence for general $D_U\in{\mathcal{U}}$, we have

(16)\begin{equation}\deg (Z\cap D_U)\le 4. \end{equation}

(c1) Assume $\#(D_U\cap U)=5$ for general $D_U\in{\mathcal{U}}$. Recall that by Observation 1(i), D_U is a smooth conic and we set $W:= U\setminus D_U\cap U$ consisting of three non-collinear points. Note that

\begin{equation*}\deg (D_U\cap (Z\cup U))=\deg (D_U\cap Z)+\deg (D_U\cap U) \le 9\end{equation*}

and hence $\deg{\mathcal{I}}_{D_U\cap (Z\cup U),D_U}(5)\ge 1$ implying $h^1(D_U,{\mathcal{I}}_{D_U\cap (Z\cup U),D_U}(5)) =0$. Consider the residual exact sequence of $Z\cup U$ with respect to D:

(17)\begin{equation} 0\rightarrow {\mathcal{I}}_{\mathrm{Res}_D(Z\cup U)}(3)\rightarrow {\mathcal{I}}_{Z\cup U}(5) \rightarrow {\mathcal{I}}_{D\cap (Z\cup U),D}(5)\rightarrow 0. \end{equation}

From the long cohomology exact sequence of (17) and (15), we have

\begin{equation*}h^1({\mathbb{P}}^2,{\mathcal{I}}_{\mathrm{Res}_D(Z\cup U)}(3))\ge 2\end{equation*}

and hence

(18)\begin{equation} h^1({\mathbb{P}}^2,{\mathcal{I}}_{Z\cup W}(3))\ge h^1({\mathbb{P}}^2,{\mathcal{I}}_{\mathrm{Res}_D(Z\cup U)}(3))\ge 2. \end{equation}

Take a line L containing two points of W and set $\{o\}:= W\setminus W\cap L$. We consider the residual exact sequence of $Z\cup W$ with respect to L; note that $\operatorname{Res}_L(Z\cup W)=\operatorname{Res}_L(Z)\cup\{o\}$. Hence we have the exact sequence

(19)\begin{equation} 0 \rightarrow {\mathcal{I}}_{\operatorname{Res}_L(Z)\cup \{o\}}(2)\rightarrow {\mathcal{I}}_{Z\cup W}(3) \rightarrow {\mathcal{I}}_{(Z\cup W)\cap L,L}(3) \rightarrow 0. \end{equation}

Since $\operatorname{Res}_L(Z)\subseteq Z$ and $h^1({\mathbb{P}}^2, {\mathcal{I}} _Z(2)) =0$ (by (11)), we have

\begin{equation*}h^1({\mathcal{I}}_{\operatorname{Res}_L(Z)\cup \{o\}}(2))\le 1.\end{equation*}

From the long cohomology exact sequence of (19) together with (18), we have $h^1(L,{\mathcal{I}}_{(Z\cup W)\cap L,L}(3))\ge 1$. Thus $\deg ((Z\cup W)\cap L)) \ge 5$. Since $\deg (W\cap L) =2$, we obtain $\deg (Z\cap L)\ge 3$. Thus L is one of the three lines spanned by p and one of the points of A, say $L=R_1$; remember that the three lines $R_i, i=1,2,3$ do not depend on the choice of $V\in|{\mathcal{N}}|$ and $U=v(V)$. On the other hand, since $R_1\cap v(X)$ is a finite set, for a general $V\in |{\mathcal{N}}|$, we have $v(V)\cap R_1=U\cap R_1=\emptyset$. However, we took the line L containing two of the points of $W\subset U$, a contradiction.

(c2) Assume $U\subset D_U$. In this case, we have $\mathrm{Res}_{D_U}(Z\cup U)\subseteq Z$ and hence $h^1({\mathbb{P}}^2,{\mathcal{I}}_{\mathrm{Res}_{D_U}(Z\cup U)}(3)) =0$ since $h^1({\mathbb{P}}^2, {\mathcal{I}}_Z(3))=0$ by (11). The long cohomology exact sequence of (17) together with (15), i.e. $h^1({\mathbb{P}}^2,{\mathcal{I}} _{Z\cup U}(5))\ge 2$ yields

\begin{equation*}h^1({D_U},{\mathcal{I}}_{{D_U}\cap (Z\cup U),{D_U}}(5)) \ge 2.\end{equation*}

Recalling $\deg (Z\cap {D_U})\le 4$, on a smooth conic ${D_U}$, we have $\deg ({D_U}\cap (Z\cup U)) \le 12$, $\deg{\mathcal{I}}_{{D_U}\cap (Z\cup U),{D_U}}(5)\ge -2$ and hence $h^1({D_U},{\mathcal{I}}_{{D_U}\cap (Z\cup U), {D_U}}(5))\le 1,$ a contradiction.

Conclusion: (c1) and (c2) show that there is a subset $B\subset U$ with $\#B=3$ and B is collinear, i.e. $v^{-1}(B)$ fails to impose independent conditions on ${\mathcal{L}}=v^*({\mathcal{O}}_{{\mathbb{P}}^2}(1))$, hence any subset $B^{\prime}\subset U$ with $\#B^{\prime}=3$ is collinear by remark 4.4 and we are done.

Proof of theorem 4.2

Recall that a general element of $X\in\mathcal{H}_{15,15,5}$ has a plane model of degree 8 with one ordinary triple point and three nodes as its only singularities; cf. part (A1) in the proof of theorem 4.1. We also recall that a curve with a plane model C of degree 8 with such prescribed singularities is embedded into ${\mathbb{P}}^5$ as a smooth curve of degree 15 and genus g = 15 in the following way:

(i) Blowing up ${\mathbb{P}}^2$ at four (singular) points in general position in ${\mathbb{P}}^2$ and then take the strict transformation $\widetilde{C}$ of C in ${\mathbb{P}}^2_4$ under this blow up.

(ii) We then embed ${\mathbb{P}}^2_4$ and $\widetilde{C}$ by the anticanonical system $|-K_{{\mathbb{P}}^2_4}|=|(3;1^4)|$ to get a smooth del Pezzo $S\subset{\mathbb{P}}^5$ and smooth $\widetilde{C}\cong X\subset{\mathbb{P}}^5$.

Now we fix four points $A\subset{\mathbb{P}}^2$ in general linear position. Let $C_i,~ 1\le i\le2$ be two plane curves of degree 8 with one triple point and three nodes with $\mathrm{Sing}(C_i)=A$. By theorem 4.5, we have the following equivalent conditions:

1. (a) Two curves $X_i\subset{\mathbb{P}}^5, i=1,2$ such that $X_i\cong\widetilde{C}_i$ are isomorphic.

2. (b) Two singular plane models C_i of X_i are projectively equivalent under a projective motion of ${\mathbb{P}}^2$ inducing a permutation on the set $A\subset{\mathbb{P}}^2$.

3. (c) X_i lies on a same smooth del Pezzo surface $S\subset{\mathbb{P}}^5$.

4. (d) There exists $\tau\in{\operatorname{Aut}}(S)$ such that $X_1=\tau(X_2)$.

Note that for a smooth del Pezzo $S\subset{\mathbb{P}}^5$ and $\tau\in{\operatorname{Aut}}(S)$, there is $\beta\in {\operatorname{Aut}}({\mathbb{P}}^5)$ such that $\beta_{|S} =\tau$ since S is anticanonically embedded in ${\mathbb{P}}^5$. Hence we have $\beta(X_2)=\tau(X_2)=X_1$.

Proof. Note that $\deg ({\mathcal{O}}_X(2)) =\deg (K_X)$ for $X\in \Gamma_1$. Suppose $S\cong{\mathbb{F}}_0$. By remark 5.3 (i), $X\in |{\mathcal{O}}_Q(3,9)|$. From the standard exact sequence

\begin{equation*}0\rightarrow{\mathcal{O}}_Q(-1,-5)\rightarrow {\mathcal{O}}_Q(2,4)\rightarrow {\mathcal{O}}_X(2,4)\rightarrow 0\end{equation*}

and by $h^0({\mathcal{O}}_Q(-1,-5))=h^1({\mathcal{O}}_Q(-1,-5))=0$, we have $h^0(X, {\mathcal{O}}_X(2))= h^0(X,{\mathcal{O}}_X(2,4)) =h^0( {\mathcal{O}}_Q(2,4))=15\neq g$. The case $S\cong{\mathbb{F}}_2$ is similar.

Proof. (i) If $S\cong{\mathbb{F}}_0$, $h^1({\mathcal{O}}_X(2))=0$ follows from the same computation as in lemma 5.4.

(ii) For the case $S\cong{\mathbb{F}}_2$, we have $X\in |{\mathcal{O}}_{{\mathbb{F}}_2}(5h+10f)|$; remark 5.3 (iii). The exact sequence $0\rightarrow{\mathcal{O}}_{{\mathbb{F}}_2}(-3h-4f)\rightarrow {\mathcal{O}}_{{\mathbb{F}}_2}(2h+6f)\rightarrow {\mathcal{O}}_X(2h+6f)\rightarrow 0$ and the cohomology of line bundles on ${\mathbb{F}}_2$ [Reference Laface27, proposition 2.3] yield

\begin{align*}h^0(X, {\mathcal{O}}_X(2))&=h^0({\mathbb{F}}_2, {\mathcal{O}}_{{\mathbb{F}}_2}(2h+6f))+h^1({\mathbb{F}}_2,{\mathcal{O}}_{{\mathbb{F}}_2}(-3h-4f))\\ &=h^0({\mathbb{F}}_2, {\mathcal{O}}_{{\mathbb{F}}_2}(2h+6f))+h^1({\mathbb{F}}_2,{\mathcal{O}}_{{\mathbb{F}}_2}(h))=15+1=16. \end{align*}

Therefore $h^1({\mathcal{O}}_X(2)) =1$ by Riemann–Roch.

Proof. Since S is ACM, $h^1({\mathbb{P}}^5,{\mathcal{I}}_X(t)) =h^1(S,{\mathcal{I}}_{X,S}(t))$ for all $t\in {\mathbb{N}}$.

(a) Take $X\in \Gamma_1$:

1. (a-1) Assume $S\cong Q$. Since $X\in |{\mathcal{O}}_Q(3,9)|$, $h^1(Q,{\mathcal{I}}_{X,Q}(t)) = h^1(Q,{\mathcal{O}}_Q(t-3,2t-9))$. We have $h^1(Q,{\mathcal{O}}_Q(0,-3)) =2$ and $h^1(Q,{\mathcal{O}}_Q(t-3,2t-9)) =0$ for all $t\ge 4$ by the Künneth formula.

2. (a-2) Assume $S\cong {\mathbb{F}}_2$. We have ${\mathcal{O}}_S(1) \cong {\mathcal{O}}_{{\mathbb{F}}_2}(h+3f)$, $X\in |{\mathcal{O}}_{{\mathbb{F}}_2}(3h+ 12f)|$ and ${\mathcal{I}}_{X,S}(t) \cong {\mathcal{O}}_{{\mathbb{F}}_2}((t-3)h+(3t-12)f)$. For t = 3, $h^1({\mathbb{P}}^5,{\mathcal{I}}_X(3)) =h^1({\mathbb{F}}_2,{\mathcal{O}}_{{\mathbb{F}}_2}(-3f)) = 2$. For $t\ge 4$, $h^1({\mathbb{P}}^5,{\mathcal{I}}_X(t))=h^1({\mathbb{F}}_2, {\mathcal{O}}_{{\mathbb{F}}_2}((t-3)h+(3t-12)f)) = 0$; cf. [Reference Laface27, proposition 2.3].

(b) Take $X\in \Gamma_2$:

1. (b-1) If $S\cong Q$, we have $X\in |{\mathcal{O}}_Q(5,5)|$ and ${\mathcal{I}}_{X,S}(t) \cong {\mathcal{O}}_Q(t-5,2t-5)$.

2. (b-2) If $S\cong {\mathbb{F}}_2$, we have $X\in |{\mathcal{O}}_{{\mathbb{F}}_2}(5h+ 10f)|$ and ${\mathcal{I}}_{X,S}(t)\cong {\mathcal{O}}_{{\mathbb{F}}_2}((t-5)h+(3t-10)f)$.

The verification for this case (b) is similar and we omit the routine.