The spaces of rational curves
on del Pezzo threefolds of degree one

Assume that $\kappa(2\beta^{*}H+K_{\widetilde{Y}})=0$. By [H1̈0] (see also [BLRT21, Theorem 8.10]), $(\widetilde{Y},\beta^{*}H)$ is birationally equivalent to a quadric surface $(Q,{\mathcal{O}}(1))$ as a polarized surface. The resolution $\beta$ factors as $\widetilde{Y}\rightarrow Y^{\prime}\rightarrow Y$ where $Y^{\prime}\rightarrow Y$ is the normalization. We denote it $\beta^{\prime}:Y^{\prime}\rightarrow Y$. Since $\beta^{\prime*}H$ is ample, we conclude that

Since $Y$ is a divisor in a smooth variety, $K_{Y}$ still makes sense. Hence we must have

Indeed, since $(Y^{\prime},\beta^{\prime*}H)$ is isomorphic to $(Q,{\mathcal{O}}(1))$, we have $2\beta^{\prime*}H+K_{Y^{\prime}}\sim 0$. By taking the pushforward, our assertion follows. Let $e\in{\mathbb{Z}}_{>0}$ be a positive integer such that $Y\sim eH$. Thus by adjunction we have $K_{Y}\sim(e-2)H$. But these relations imply that $eH\sim 0$, a contradiction. ∎

The proof below is taken from [LT21b, Lemma 8.1]. We include it for completeness of the paper. It is enough to show that the evaluation map

is dominant where $M^{(2)}$ is the component of $\overline{M}_{0,2}(X,d)$ above $M$. Suppose that it is not dominant. Since there is at least a one-parameter family of curves through a general point, the image must be an irreducible divisor $D$ in $X\times X$.

From the assumption we have $-K_{X}\cdot C=2H\cdot C=2d\geq 4$. For the fiber of the evaluation map $\mathrm{ev}:M\rightarrow X$ at a general point $x_{1}\in X$ , we have

Take a component $N\subset\mathrm{ev}^{-1}(x_{1})$ and let $S$ be the surface swept out by $N$. Then there exists a smooth resolution $\beta:\widetilde{S}\rightarrow S$ such that $\beta^{-1}(x_{1})$ is divisorial in $\widetilde{S}$. Let $\widetilde{C}$ be the strict transform of $C$. Then we must have

hence we have

Since $a(S,2H)2\beta^{*}H+K_{\widetilde{S}}$ is pseudo-effective, it follows that

Thus we obtain

Now we can write $a(S,2H)=\frac{2}{n}$ or $\frac{3}{n}$ by [LT21b, Lemma 4.5] where $n$ is a positive integer. When $d\geq 3$, we must have $a(S,2H)=1$. When $d=2$, we have $a(S,2H)=1$ or $\frac{3}{4}$. We will argue for each case.

When $a(S,2H)=1$, we have $\kappa(2\beta^{*}H+K_{\widetilde{S}})=1$ by Lemma 3.1. In this case the canonical map $\pi:\widetilde{S}\rightarrow B$ associated to $2\beta^{*}H+K_{\widetilde{S}}$ exists and we have the Zariski decomposition

with a general fiber $F\subset\widetilde{S}$ of $\pi$, an effective divisor $E$ and a positive integer $e$. Note that $F$ satisfies $\beta^{*}H\cdot F=1$ so its image on $X$ is an $H$-line. Then

implies that $(eF+E)\cdot\widetilde{C}>0.$ This means that $\widetilde{C}$ maps to $B$ dominantly and this implies that $B$ is rational. Thus we conclude that the variety of $H$-lines $\mathcal{R}_{1}$ is covered by rational curves $B$ as $x_{1}$ varies. Now let us consider the Abel-Jacobi mapping

where IJ$(X)$ is the intermediate Jacobian of $X$. By Theorem 2.9 $\mathrm{AJ}$ is generically finite to the image. Since $\mathrm{IJ}(X)$ has no rational curves, all such curves contained in $\mathcal{R}_{1}$ are contracted by this mapping. Thus there are only finitely many rational curves on $\mathcal{R}_{1}$ which contradicts with the fact that $B$ varies as $x_{1}$ varies.

When $a(S,2H)=\frac{3}{4}$, let $S^{\prime}$ be the normalization of $S$. From [She12, Lemma 2.4 and Proposition 2.5], $S^{\prime}$ is smooth along the image of $\widetilde{C}$ which is the strict transform of $C$. Furthermore take the minimal resolution $\beta:\widetilde{S}^{\prime}\to S$. Then there is a birational map $f:\widetilde{S}^{\prime}\rightarrow{\mathbb{P}}^{2}$ by [LT21b, Theorem 5.5] and it maps the 2-dimensional family of $\widetilde{C}^{\prime}\subset\widetilde{S}^{\prime}$ which is the strict transform of $C$ to the lines passing through one point on $\mathbb{P}^{2}$. Indeed, it follows from [LT21b, Theorem 5.5] that we have $2\beta^{*}H=f^{*}\mathcal{O}(4)-K_{\widetilde{S}^{\prime}/\mathbb{P}^{2}}$ so that $f_{*}\widetilde{C}^{\prime}\cdot\mathcal{O}(1)=1$. Note that we have $d=2$ in our situation. Then such lines admit only 1-dimensional family on ${\mathbb{P}}^{2}$, a contradiction. ∎

Let $C$ be a general very free rational curve on $X$ such that $H.C=2$. Then there are three possibilities for the image $C^{\prime}$ of $C$ via $\Phi_{|-K_{X}|}$:

• •

  $C^{\prime}$ is a line and $\Phi_{|-K_{X}|}|_{C}:C\rightarrow C^{\prime}$ has degree 4;

• •

  $C^{\prime}$ is a conic and $\Phi_{|-K_{X}|}|_{C}:C\rightarrow C^{\prime}$ has degree 2;

• •

  $C^{\prime}$ is a quartic rational curve and $\Phi_{|-K_{X}|}|_{C}:C\rightarrow C^{\prime}$ is birational.

We will consider these situations. For the first case every line on $W$ passes through the singular point, but this contradicts with the assumption that $C$ is general so that $C$ avoids any codimension $2$ locus.

For the second case note that since $C$ and $C^{\prime}$ are smooth by [Kol96, 3.14 Theorem] , we have the Hurwitz formula

Let $Q$ be a cubic hypersurface in ${\mathbb{P}}^{6}$ such that $\Phi_{|-K_{X}|}:X\rightarrow W$ is ramified along $Q\cap W$. Since we have $S^{\prime}\cdot C^{\prime}=2$ where $S^{\prime}$ is a hyperplane class in $\mathbb{P}^{6}$, we have $Q\cdot C^{\prime}=6$ so that the degree of the ramification divisor of $\Phi_{|-K_{X}|}|_{C}:C\rightarrow C^{\prime}$ is less than or equal to 6. Since we have $-K_{W}\sim\frac{5}{2}S^{\prime}$, we conclude $-K_{W}\cdot C^{\prime}=5$. But then among points in $C^{\prime}\cap Q$ there must be two points with multiplicity 2 and the dimension decreases by 1 for having a multiplicity 2 point, so that one can show that the dimension is $5-1-1=3<4$ and this contradicts with $-K_{X}\cdot C=4$. Thus we conclude that for a general member $C$ we only have the third case. ∎

Let $C^{\prime}\subset W$ be a general quartic rational curve avoiding cone point. Then we have $-K_{W}\cdot C^{\prime}=10$ so that the parameter space of normal quartic rational curves on $W$ has dimension 10. Assume that we have a general hyperplane $S^{\prime}$ containing $C^{\prime}$ but not containing the cone point. There is a composition $S^{\prime}\cap W\cong{\mathbb{P}}^{2}\hookrightarrow S^{\prime}\hookrightarrow{\mathbb{P}}^{6}$ which is the Veronese embedding of $\mathbb{P}^{2}$ up to linear transformations. Considering such the embedding as

with $S^{\prime}\cong{\mathbb{P}}^{5}$. One can see that $C^{\prime}$ is realized as some conic in ${\mathbb{P}}^{2}$ and its image spans ${\mathbb{P}}^{4}$.

Next assume all $S^{\prime}$ containing $C^{\prime}$ contain the cone point. In this case we will prove that there exists $S^{\prime}$ containing $C^{\prime}$ such that $W\cap S^{\prime}$ is not integral. There exists 1-parameter family $S^{\prime}_{t}\supset C^{\prime}$ of hyperplanes where $t$ is another variable. By assumption $S^{\prime}_{t}$ contains the cone point. Now the intersection $W\cap S^{\prime}_{t}$ is a quadratic cone and it degenerates to the union of two planes. Indeed, $W\cap S^{\prime}_{t}$ is a cone over a quartic in the Veronese surface which is a conic in $\mathbb{P}^{2}$. Any $1$-parameter family of conics in $\mathbb{P}^{2}$ degenerates to the union of two lines. So the double cone breaks into the union of two irreducible surfaces, and the curve $C^{\prime}$ is contained one of these and such a surface must be general because $C^{\prime}$ is general.

We see that $C^{\prime}$ is contained in $H^{\prime}\subset W$ such that $\Phi_{|-K_{X}|}^{*}H^{\prime}\in|H|$. Then we have

and

because of $S^{\prime}\cdot C^{\prime}=4$. Since the ramification is 6 points with multiplicity 2, $C^{\prime}$ deforms in dimension $7-6+2=3<4$. This is a contradiction. ∎

Let $p\in X$ be a general point and $l$ be a general free $H$-line on $X$. Since a general $C$ is very free by Proposition 3.2, the locus $\{C\in M\mid p\in C,l\cap C\neq\phi\}$ of $M$ is $1$ dimensional. Pick a component $N$ of this locus. We set

Every fiber of the projection $p_{1}:{\mathcal{U}}\rightarrow N$ is a line in $|-K_{X}|$ and the image $p_{2}({\mathcal{U}})$ of the other projection $p_{2}:{\mathcal{U}}\rightarrow|-K_{X}|={\mathbb{P}}^{6}$ is a surface. The locus ${\mathcal{H}}=\{H_{1}+H_{2}\in|-K_{X}|\mid H_{i}\in|H|\}$ has dimension 4 so by looking at these dimensions, we see that we have $p_{2}({\mathcal{U}})\cap{\mathcal{H}}\neq\phi$. Then there exists a curve $C_{0}$ such that $p_{1}^{-1}(C_{0})\ni H_{1}+H_{2}$.

If $C_{0}$ is not integral, then it is the union of two $H$-lines so $C_{0}=l_{1}+l_{2}$ and we may assume that they satisfy that $p\in l_{1},l\cap l_{2}\neq\phi$ and $l_{1}\cap l_{2}\neq\phi$. Note that there are only finitely many such $l_{1}$ and $l_{2}$ satisfying these conditions. Indeed, there are only finitely many $l_{1}$ containing $p$. Any locus parametrizing $l_{2}$ such that $l\cap l_{2}\neq\emptyset$ is $1$-dimensional, and among them there are only finitely many $l_{2}$ meeting with $l_{1}$. Since $l_{1}$ contains a general point it is free. Then since $l$ and $l_{1}$ are general in its moduli, we conclude that $l_{2}$ is also general in its moduli so that it is free. Thus we conclude that $l_{1}+l_{2}$ are distinct free $H$-lines. Thus our assertion follows.

If $C_{0}$ is integral, we may assume it is contained in $H_{1}$. Then there are only finitely many $C_{0}$ such that $p\in C_{0},l\cap C_{0}\neq\emptyset$ and $C_{0}\subset H_{1}$. Indeed, any locus parametrizing $C_{0}$ containing $p$ is $2$-dimensional. In this $2$-dimensional locus, $C_{0}$’s meeting with $l$ are parametrized by $1$-dimensional family. A general member in this $1$-dimensional family is not contained in $H_{1}$. Thus we conclude that there are only finitely many $C_{0}$ such that $p\in C_{0},l\cap C_{0}\neq\emptyset$ and $C_{0}\subset H_{1}$. Then there are only finitely many $H_{1}$ containing $C_{0}$ so that $H_{1}$ is general in $|H|$ and smooth. Thus it is a smooth del Pezzo surface of degree $1$. The image $H^{\prime}_{1}=\Phi_{|-K_{X}|}(H_{1})$ is the quadric cone in ${\mathbb{P}}^{6}$, we have $-K_{H^{\prime}_{1}}=2S^{\prime}|_{H_{1}^{\prime}}$. Then we have the equivalence $C^{\prime}_{0}\sim 2S^{\prime}|_{H^{\prime}_{1}}$ on $H^{\prime}_{1}$ since $S^{\prime}|_{H^{\prime}_{1}}\cdot C^{\prime}_{0}=4$. Thus we must have

Computing the intersection with $C_{0}$, we have

Moreover note that $2\iota(C_{0})\cdot C_{0}=C^{\prime}_{0}\cdot 3S=12$. So we conclude that $C_{0}^{2}=2$. Let $p_{a}(C_{0})$ be the arithmetic genus of $C_{0}$, then the adjunction formula

implies that $p_{a}(C_{0})=1$. This means that there exists a blow down $\beta:H_{1}\rightarrow\widetilde{H}_{1}$ between del Pezzo surfaces of degree 1 and degree 2 such that $C_{0}\sim-\beta^{*}K_{\widetilde{H}_{1}}$. (See, e.g., [BLRT21, The proof of Lemma 3.4] for this claim.) Then $|-K_{\widetilde{H}_{1}}|$ contains the union of two $(-1)$-curves $l_{1}+l_{2}$. Since $H_{1}$ is general, the $H$-lines $l_{1},\,l_{2}$ are general and must be free. ∎

Let $l_{1}+l_{2}$ be the union of general two free $H$-lines meeting each other at exactly two points. Then there exists a unique $H^{\prime}\in|H|$ such that $H^{\prime}$ contains both $l_{1}$ and $l_{2}$. Since $H$-lines are general, $H^{\prime}$ is also general proving that $H^{\prime}$ is a smooth del Pezzo surface of degree $1$. Then one can find a blow down $\beta:H^{\prime}\to\widetilde{H}^{\prime}$ to a degree $2$ del Pezzo surface such that

Since the monodromy on smooth members of $|H|$ is the maximum Weyl group of $\mathbb{E}_{8}$ type, we conclude that the locus $N^{\prime}$ of $\mathcal{R}_{1}\times\mathcal{R}_{1}$ parametrizing the union of general free $H$-lines meeting each other at two points is irreducible. Then a natural map $N\to N^{\prime}$ is a degree $2$ covering. This implies that $N$ has at most two components.

Now $N^{\prime}$ admits a finite cover $N^{\prime}\to|H|$. We consider a general pencil $\ell\subset|H|$ and the base change $N^{\prime}_{\ell}\to\ell$. Note that $N^{\prime}_{\ell}$ is irreducible by Lefschetz theorem of the monodromy. Then a point in $N^{\prime}_{\ell}$ corresponds to $H^{\prime}\in|H|$ and a birational morphism $\beta:H^{\prime}\to\widetilde{H}^{\prime}$ to a del Pezzo surface of degree $2$ with a pair of $(-1)$-curves $E_{1},E_{2}$ on $\widetilde{H}^{\prime}$ such that $E_{1}+E_{2}\sim-K_{\widetilde{H}^{\prime}}$. The anticanonical sytem $-K_{\widetilde{H}^{\prime}}$ defines a degree $2$ covering $\widetilde{H}^{\prime}\to\mathbb{P}^{2}$ ramified along a quartic curve $B_{\widetilde{H}^{\prime}}$ and the image of $E_{1}+E_{2}$ is a bitangent line to $B_{\widetilde{H}^{\prime}}$. For the quartic curve $B_{\widetilde{H}^{\prime}}$, having an inflection point of order $4$ is codimension $1$ condition, so we may assume that there exists $\widetilde{H}^{\prime}\in N^{\prime}_{\ell}$ such that $\widetilde{H}^{\prime}$ is smooth and $B_{H^{\prime}}$ admits an inflection point of order $4$ by generality of $X$ and $\ell$. This means that $N_{\ell}\to N^{\prime}_{\ell}$ is not étale. On the other hand, deformation theory tells us that $N_{\ell}$ is smooth at $(H^{\prime},\beta,E_{1}+E_{2},p)$ where $p$ is one of points in $E_{1}\cap E_{2}$. Indeed, this follows from the fact that $X$ and $\ell$ are general. Thus we conclude that $N_{\ell}$ is irreducible, proving our claim. ∎

Note that for a free $H$-line $l$, its involution $\iota(l)$ is the only $H$-line meeting with $l$ at three points. Indeed, let $l^{\prime}$ be a line meeting with $l$ at three points. Let $C_{1}$ and $C_{2}$ be the images of $l$ and $l^{\prime}$ via $\Phi_{|-K_{X}|}:X\to W$ respectively, then $C_{i}$ is a conic. Since $W$ is the intersection of quadrics in $\mathbb{P}^{6}$, we conclude that for any plane $P$ in $\mathbb{P}^{6}$, $P\cap W$ can contain at most one conic. Thus when $C_{1}$ and $C_{2}$ are distinct, two planes spanned by $C_{1}$ and $C_{2}$ are different as well. As a result, $C_{1}$ can meet $C_{2}$ with at most two intersection points which contradicts with our assumption that $l$ and $l^{\prime}$ meet at three points. Hence we conclude that $C_{1}=C_{2}$ and $l^{\prime}=\iota(l)$. Thus if we denote the closure of the locus of $\mathcal{R}_{1}\times\mathcal{R}_{1}$ parametrizing a pair of free $H$-lines meeting at three points by $K^{\prime}$, then $K^{\prime}$ is isomorphic to $\mathcal{R}_{1}$. Then a natural map $K\to K^{\prime}$ is a degree $3$ covering. The remaining of the proof is similar to a proof of Lemma 3.6. ∎

Since $\mathcal{R}_{1}$ is smooth by Theorem 2.9, $\mathcal{R}_{1}^{(1)}$ is also smooth. Then

is a generically finite dominant cover and let $B\subset X$ be the branch divisor on $X$. Let $l\in\mathcal{R}_{1}$ be a general free $H$-line so that $l$ meets with $B$ transversally. Then $\mathrm{ev}^{-1}(l)$ is smooth and $1$-dimensional, and contains $p^{-1}(l)$ as a component where $p:\mathcal{R}_{1}^{(1)}\rightarrow\mathcal{R}_{1}$ is the family map.

Let $D_{l}$ be the union of components of $\mathrm{ev}^{-1}(l)$ other than $p^{-1}(l)$. Since $D_{l}$ is smooth, $D_{l}$ is the disjoint union of smooth irreducible curves as $D_{l}=D_{1}+\cdots+D_{r}$. Then $D_{l}$ is isomorphic to the fiber at $l$ for a morphism

.

For an irreducible curve $L\subset\mathcal{R}_{1}$, we show that $L\cdot p_{*}(D_{l})>0$. Take an another line $l^{\prime}$. Since $p_{*}(D_{l})$ is algebraically equivalent to $p_{*}(D_{l^{\prime}})$, we only have to show that $L\cdot p_{*}(D_{l^{\prime}})>0$. The curve $L$ gives us a 1-parameter family of $H$-lines, thus $H$-lines parametrized by $L$ sweep out a surface $T$. Since $X$ has Picard rank $1$, the intersection $T\cap l^{\prime}$ is not empty and so $L\cap p_{*}(D_{l^{\prime}})\neq\phi$.

A general $H$-line $l^{\prime}$ meeting with $l$ meets with it at one point. There are finitely many $H$-lines meeting with $l$ twice and $\iota(l)$ is the only $H$-line meeting with $l$ three times. Assume that there is $i$ such that any $H$-line parametrized by $D_{i}$ meets with $l$ at one point. Then $p_{*}D_{i}$ is disjoint from $\sum_{j\neq i}p_{*}D_{j}$. Thus we conclude that

as well as

These imply that

but this contradicts with Hodge index theorem.

Let $N$ and $K$ be the irreducible loci defined in Lemma 3.6 and 3.7. Since the union of free $H$-lines is a smooth point of $\mathcal{R}_{1}^{(1)}\times_{X}\mathcal{R}_{1}^{(1)}$, we conclude that there are at most two components of $R$, one containing $N$ and another containing $K$. Assume that there are exactly two components, denoted by $R_{N}$ and $R_{K}$ each containing $N$ and $K$ respectively. Let $D_{N}$ be the union of $D_{i}$’s contained in $R_{N}$ and $D_{K}$ be the union of $D_{i}$’s contained $R_{K}$. Then $p_{*}D_{N}$ and $p_{*}D_{K}$ are disjoint. Thus we conclude that

This contradicts with Hodge index theorem again. Thus we conclude that $R$ is irreducible. ∎