Quintic del Pezzo sections

Proof. Both plane embeddings are defined by their complete linear systems. If $L_1\simeq L_2$, a projective change of coordinates identifies these embeddings. The bilinear equation of $V$ then restricts on $C$ to a quadratic form in the coordinates of one plane. A nonzero quadratic form cannot vanish on a smooth plane cubic. Therefore this quadratic form vanishes identically, so the matrix of the bilinear equation is skew-symmetric. A skew-symmetric matrix of size three is singular. This contradicts the smoothness of $V$. ◻

Proof. Let $q$ be the first plane image of $p$. Write $C_i$ for the two smooth plane cubic images of $C$. Choose a general line $L\subset\mathbb P^2_x$ through $q$. It meets $C_1$ transversely in three distinct points $x_1,x_2,x_3$. We may also require that its point $[L]\in\mathbb P^2_y$ does not belong to $C_2$: the pencil of lines through $q$ is a line in $\mathbb P^2_y$, and cannot be contained in the smooth cubic $C_2$.

The surface $V_L=V\cap(L\times\mathbb P^2_y)$ is the incidence surface of points of $L$ and lines of $\mathbb P^2_x$. Its second projection is the blow-up of $\mathbb P^2_y$ at $p_0=[L]$. Indeed, for $y\ne[L]$ the unique point above $y$ is $L\cap y$, whereas the fibre over $[L]$ is $L$. Let $c_i=(x_i,y_i)$ be the three points of $C\cap V_L$. They are distinct and avoid this exceptional curve, by the choice $[L]\notin C_2$. Since $V_L$ meets $C$ transversely, its inverse image $T$ in the blow-up $X$ is the blow-up of $V_L$ at $c_1,c_2,c_3$. In particular it is smooth and is the blow-up of $\mathbb P^2_y$ at $p_0,p_1=y_1,p_2=y_2,p_3=y_3$.

These four points are in general position. They are distinct because $C\to C_2$ is an embedding and $p_0\notin C_2$. If $p_0,p_i,p_j$ were collinear, the three corresponding lines $L,y_i,y_j$ in $\mathbb P^2_x$ would be concurrent. As $x_i=L\cap y_i$ and $x_j=L\cap y_j$, this would give $x_i=x_j$, a contradiction. If $p_1,p_2,p_3$ were collinear, the divisor $c_1+c_2+c_3$ on $C$ would belong to both $|L_1|$ and $|L_2|$. This contradicts Lemma 111.1. Thus $T$ is a smooth del Pezzo surface of degree five. The displayed divisor classes follow from the two projections and $F=H_1+H_2-E$.

This construction includes points on either component of every singular conic fibre of $X\to\mathbb P^2_x$, as well as their intersection: the Cartier inverse image of every line through $q$ contains the entire fibre over $q$. In particular there is no exception for $p\in E$. Transversality of $L$ to $C_1$ at $q$, when $q\in C_1$, is compatible with all the preceding choices. ◻

Proof. The inverse image of $C_2$ in $V$ under the second plane projection is a $\mathbb P^1$-bundle containing $C$ as a section. Its strict transform $D_2$ in $X$ is isomorphic to it, since blowing up a Cartier divisor on a smooth surface does not change the surface. For a point $y\in C_2$, the corresponding ruling fibre is the line represented by $y$ in the first plane. The line $L$ defining $T$ is different from each of these lines, because $[L]\notin C_2$. It therefore meets each ruling fibre once. Thus $T\cap D_2$ is a section of this bundle, isomorphic to $C_2$. Its class was computed in Remark 111.3. ◻

Proof. The divisors $A,B,F$ are nef. The divisor $J=F+2B-A$ is the exceptional divisor of the contraction to $\mathbb P^1\times\mathbb P^2$ given by $F,B$. A fibre of $J$ has intersection one with $A$ and zero with $B,F$, so its intersection with $J$ is $-1$. Thus $u>1$ forces the indicated fixed multiplicity along $J$. Subtraction leaves the stated nef divisor until $u=3/2$. For $u>3/2$ one has $(L-uA)BF=3-2u<0$, excluding pseudoeffectivity. The volumes are $6u^2-27u+24$ on $[0,1]$ and $-12u^3+60u^2-99u+54$ on $[1,3/2]$. Their integral, divided by $24$, is $69/128$. ◻

Proof. Let $\rho_2:X\longrightarrow Y=\mathbb P^1\times\mathbb P^2$ be the birational morphism defined by $F$ and $B$, with exceptional divisor $D_2$. Recall that $A=F+2B-D_2$, so the image of a general $D\in|A|$ is an integral divisor of bidegree $(1,2)$ containing the centre $Z_2$ of $\rho_2$. Moreover, $$(F+B)|_D=3l-e_0-e_1-e_2-e_3=-K_D.$$ The six sections of $\mathcal O_Y(1,1)$ restrict independently to $D$: a relation would give a divisor of bidegree $(1,1)$ containing its image of bidegree $(1,2)$, which is impossible. Since $h^0(D,-K_D)=6$ and $-K_D$ is very ample, these sections give the complete anticanonical embedding. Thus $\rho_2|_D$ is an isomorphism onto a smooth divisor of bidegree $(1,2)$, and $D_2|_D$ is its smooth elliptic curve $Z_2$.

For rational $0\leqslant u<1$, the divisor $P_X(u)=(1-u)A+B+F$ on $X$ is ample. After clearing denominators, Serre vanishing applied to $$0\longrightarrow\mathcal O_X(mP_X(u)-D) \longrightarrow\mathcal O_X(mP_X(u)) \longrightarrow\mathcal O_D(mP(u))\longrightarrow0$$ gives surjectivity of restriction for all sufficiently large divisible $m$.

For $1\leqslant u\leqslant3/2$, removing the fixed divisor $m(u-1)D_2$ identifies the global moving sections with the sections of $\mathcal O_Y(a,b)$, where $$a=m(2-u),\qquad b=m(3-2u).$$ Both integers are non-negative. As $D\subset Y$ has bidegree $(1,2)$, the restriction sequence is $$0\longrightarrow\mathcal O_Y(a-1,b-2) \longrightarrow\mathcal O_Y(a,b) \longrightarrow\mathcal O_D(mP(u))\longrightarrow0.$$ Its first cohomology vanishes: $H^1(\mathbb P^1,\mathcal O(a-1))=0$ for $a\geqslant0$, and $H^1(\mathbb P^2,\mathcal O(b-2))=0$. Thus the moving restrictions are complete also on this interval.

Refining on $D$ by a flag curve gives the usual surface Zariski-decomposition bounds. One may enlarge the resulting spaces on the curve to complete series, since a basis compatible with the point filtration extends to a compatible basis of the larger space and all added vanishing orders are non-negative. This only increases the moments. Rounding errors vanish in the normalized limit. Finally, the fixed divisor removed on the second interval restricts to $m(u-1)C_2$; it is retained in the refined divisor and contributes $7(C_2\cdot T)_p/384$ to the point moment. Omitting this contribution would give an incorrect calculation at points of $C_2$. ◻

Proof. The curve $Q$ is a smooth rational curve, since the reducible fibres of $|l-e_0|$ are the unions $e_i+(l-e_0-e_i)$, for $1\leqslant i\leqslant3$. Put $$\begin{gathered} L_{ij}=l-e_i-e_j,\qquad R=L_{12}+L_{13}+L_{23},\\ a=3-2u,\quad b=2-u,\quad k=8-5u-2v,\quad r=2-u-v. \end{gathered}$$ The four curves $e_0,L_{12},L_{13},L_{23}$ are pairwise disjoint. The positive and negative parts $M(u,v),N(u,v)$ of $P(u)-vQ$ are as follows. The last column records $M(u,v)\cdot Q$.

These are Zariski decompositions: the listed negative curves are disjoint, the positive part is orthogonal to each component of the negative part, and its intersection with every $(-1)$-curve is non-negative. Since the cone of curves is generated by the ten $(-1)$-curves, the positive part is nef. The upper endpoints are the pseudoeffective thresholds. At these endpoints the positive part is zero, except when $4/3\leqslant u\leqslant3/2$, where it is a multiple of the nef class $2l-e_0-e_1-e_2-e_3$ of square zero. Increasing $v$ then gives negative intersection with this nef class, or with $T$, as appropriate.

For completeness, the squares in the six rows are, respectively, $$9-4u-4v,\quad(5-2u-2v)^2,\quad 33-40u+12u^2-2v(5-3u),\quad k^2-r^2,\quad k^2-r^2,\quad k^2.$$ Integrating these polynomials over their indicated ranges gives $$\int\!\!\int M(u,v)^2\,dv\,du=\frac{245}{36},\qquad \int\!\!\int (M(u,v)\cdot Q)^2\,dv\,du=\frac{133}{18},$$ where each integral means the sum over the six rows. Moreover, $$\int_1^{3/2}\int_0^{t(u)}(u-1)(M(u,v)\cdot Q)\,dv\,du =\frac{7}{96}.$$ Here $t(u)$ is the upper endpoint in the table. Since $p$ is outside the ten negative curves, $N(u,v)|_Q$ has zero multiplicity at $p$. The Abban–Zhuang formula, with factors $3/24$ and $6/24$, now gives the two asserted bounds. Finally, $C_2\cdot Q=3$, so $(C_2\cdot Q)_p\leqslant3$. Applying the flag estimate proves the assertion. ◻

Proof. Write $\pi(p)=(x_0,y)$ and let $c_0=(x_0,y_0)$ be the unique point of $C$ above $x_0$. The line $\ell$ is the strict transform of the first-projection fibre over $x_0$. The divisor $D_2$ is the strict transform of the inverse image of the plane cubic $C_2$ under the second projection. Thus $p\notin E\cup D_2$ implies $y\notin C_2$.

Choose a general line $M\subset\mathbb P^2_y$ through $y$. The proof of Lemma 111.2, with the projections interchanged, identifies its inverse image $T$ with the blow-up of $\mathbb P^2_x$ at four points $r=[M],x_1,x_2,x_3$, where $(x_i,y_i)=c_i\in C$ and $\{y_1,y_2,y_3\}=M\cap C_2$. We may take these four points in general position. We may also assume $r\ne x_0$ and $M\not\ni y_0$. Consequently $p$ is on none of the four exceptional curves.

The line through $r$ and $x_i$ in $\mathbb P^2_x$ is precisely the line represented by $y_i$: both points lie on it. Thus $p$ is on the strict transform of this line only if $y_i(x_0)=0$. There are only finitely many such points $y_i$ on $C_2$, and none equals $y$, since $y\notin C_2$. Avoiding the finitely many lines joining $y$ to these points ensures that $p$ lies on none of the three curves of classes $l-e_0-e_i$.

Finally, suppose that infinitely many choices of $M$ put $p$ on a line through two of $x_1,x_2,x_3$. Projection of $C_1$ from $x_0$ gives the degree-two morphism defined by $|L_1-c_0|$; let $\iota$ be its involution. Projection of $C_2$ from $y\notin C_2$ gives a degree-three morphism $g\colon C\to\mathbb P^1$. The supposed incidences give infinitely many points $c$ with $g(c)=g(\iota(c))$. Therefore these two morphisms are identical, and $g$ factors through the quotient by $\iota$. Its degree would be even, a contradiction. Only finitely many further choices of $M$ are excluded. This proves the first assertion.

For completeness, the exceptional points can be read directly from the Jordan forms. At an $A_1$ fibre take $$A=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}, \qquad \ell=\{x=[1:0:0],\ y=[0:s:t]\}.$$ If $B=(b_{ij})$ defines the other pencil member, the plane cubic $C_2$ restricted to this line has equation, up to nonzero scalar, $$st(b_{10}s+b_{20}t)=0.$$ The last factor cuts out $C\cap\ell$. Removing it in the blow-up leaves $st=0$. The point $[s:t]=[1:0]$ is the node, and $[s:t]=[0:1]$ is the smooth intersection with the good line contracted by the second projection. Smoothness of $C$ ensures $b_{10}\ne0$, so $C$ misses the node. The other point may coincide with $C\cap\ell$, but the strict-transform computation still gives one residual intersection there. For an $A_2$ fibre take $A$ to be the nilpotent block of size three. On the same line the equation is $$s^2(b_{10}s+b_{20}t)=0.$$ Here $b_{20}\ne0$, since $C$ misses the singular point $[s:t]=[0:1]$. After removing the base-curve factor, the intersection is twice this point. Transposing the matrices proves the symmetric assertions. ◻