Weak del Pezzo generation and the smooth degree-one pencil

Empty-boundary nef-big adjoint vanishing applied to $K_T+L=0$ gives $H^i(O_T)=0$ for $i>0$, and to $K_T+2L=L$ gives $H^i(L)=0$ for $i>0$. Write $d=L^2>0$. Surface RR gives h0(L)=d+1, χ(N)=1+(N²+LN)/2

for every line bundle N. In particular N²+LN is even. The pullback degree-zero curves form the negative-definite exceptional locus. The minimal-resolution operation is the direct projective (−1)-curve contraction described in the companion nef-K3 note.

Assume $d\ge 2$ and evaluation of $L$ fails at a point $z$. Apply $\mathrm{Hom}(-,O_T)$ to $0\to I_z(2L)\to O_T(2L)\to O_z(2L)\to 0$. By Serre duality and $H^1(L)=0$, ${\mathrm{Ext}}^1(O_T(2L),O_T)=0$; the map ${\mathrm{Ext}}^2(O_z(2L),O_T)\to {\mathrm{Ext}}^2(O_T(2L),O_T)$ is dual to evaluation $H^0(K_T+2L)=H^0(L)\to L{\mathrm{\mid }}_z$, hence is zero. Thus ${\mathrm{Ext}}^1(I_z(2L),O_T)\cong \mathbb{C}$. Choose a nonzero class. Its restriction to the regular local ring at $z$ is the nonzero class in ${\mathrm{Ext}}^1({\mathfrak{m}}_z,O_{T,z})\cong \mathbb{C}$, whose middle module is free of rank two; away from $z$ the quotient is a line bundle and the extension is locally free. We obtain $0\to O_T\to V\to I_z(2L)\to 0$, with $V$ a rank-two vector bundle, $c_1(V)=2L$ and $c_2(V)=1$.

Suppose $V$ is $L$-slope-stable, where ${\mu }_L(F)=c_1(F)\cdot L\mathrm{/}\mathrm{rk}(F)$ for a torsion-free sheaf $F$, and stability tests every rank-one subsheaf. Then ${\mu }_L(V)=d$, and twisting by $-L$ preserves stability while changing this slope to $0$. Any nonscalar endomorphism of $V$ has, after subtracting a constant eigenvalue, a nonzero rank-one map $V\to V$; its rank-one kernel and image would both have slope $<d$, contradicting additivity of $L$-degree. Hence $h^0(\mathrm{End}V)=1$. A nonzero map $V\to V(-L)$ cannot have generic rank two, since its determinant would be a nonzero section of $O_T(-2L)$, impossible because $L$ is nef and $L^2>0$. If it has rank one, its kernel has slope $<d$, so its image has slope $>d$, contradicting stability of $V(-L)$, whose rank-one subsheaves have slope $<0$. Thus $h^2(\mathrm{End}V)=\dim \mathrm{Hom}(V,V(K_T))=0$ by Serre duality. Riemann--Roch gives $\chi (\mathrm{End}V)=4\chi (O_T)+c_1(V{)}^2-4c_2(V)=4d$, whereas $h^0=1$ and $h^2=0$ imply $\chi \le 1$. This contradiction proves that $V$ is not $L$-slope-stable.

A saturated rank-one destabilizing subsheaf is a line bundle $M$ and satisfies $LM\ge d$. Its projection into $I_z(2L)$ is nonzero: otherwise a nonzero map $M\to O_T$ would give an effective divisor of $L$-degree $-LM<0$, contrary to nefness. Consequently $N=2L-M$ has a nonzero section vanishing at $z$, so $N$ is a nonzero effective divisor containing $z$. Put $a=LN$ and $q=N^2$. Then $0\le a\le d$; writing the torsion-free rank-one quotient by $M$ as $I_Z(N)$, Chern additivity gives $2a-q=M\cdot N\le c_2(V)=1$, hence $q\ge 2a-1$. Hodge index gives $q\le a^2\mathrm{/}d$, and surface Riemann--Roch gives $q+a$ even.

If $a=0$, then $N$ is supported on the contracted curves, so negative definiteness gives $q<0$; together with $q\ge -1$ this forces $q=-1$, contradicting the evenness of $q+a$. If $a>0$, then $2a-1\le q\le a^2\mathrm{/}d\le a$, so $a=1$. But then $q\ge 1$ and $q\le 1\mathrm{/}d<1$, impossible. There is no base point. Hence the anticanonical pullback $L$, and therefore $H$ on $S$, is globally generated for every $d\ge 2$.

Now $d=1$ and h0(L)=2. Write |L|=B+|P| with B fixed and |P| a nonzero movable pencil. Since L is nef, $0\le LP\le 1$. The value $LP=0$ is impossible: the class P would lie in the negative-definite exceptional subspace although $P^2\ge 0$. Thus $LP=1$ and $LB=0$; B is exceptional. The identity $LP=P^2+PB=1$ gives $P^2=0$ or1.

The value $P^2=0$ is impossible. To see this without assuming irreducibility of a possibly composed moving divisor, take the connected-fibre morphism associated to its basepoint-free movable part after blowing up any base points. Since $P^2=0$, two general members have intersection zero, so the original pencil already has no base point. Its connected generic fibre F is a smooth integral curve, $P=kF$, and LF is a positive integer. Since $LP=1,k=1$ and $LF=1$. Adjunction with $K_T=-L$ would give 2g(F)−2=F²−LF=−1, impossible. Hence $P^2=1$.

Then B²=(L−P)²=1−2+1=0. Negative definiteness forces $B=0$. Thus |L| is a pencil without a fixed component. Two independent sections have total intersection one, so their common zero scheme consists of a single point q of length one. In regular local parameters, length one means that their linear terms are independent. Every general member is therefore smooth at q, and ordinary Bertini makes it smooth elsewhere.

This base point cannot lie on an exceptional irreducible curve E: L|E is trivial (E is a rational (−2)-curve by $K_T=-L$ and adjunction), so vanishing of every section at q∈E would force E to be a fixed component. A general member contains no exceptional curve and has zero intersection with each, hence is disjoint from the exceptional locus. Every nonexceptional component has positive integral L-degree, while the total member has L-degree one; there is exactly one such component and no exceptional component. The smooth member is therefore connected and integral. Its image on S is smooth, and adjunction gives genus one. For every integer $k\ge 2$, the line bundle kL restricts to degree k on this smooth genus-one curve and is generated there: curve RR and Serre duality give h0(kL|C)−h0(kL|C(−x))=1 at every x, since both relevant positive degrees have H1 zero. The restriction map H0(kL)→H0(C,kL|C) is onto because H1((k−1)L)=0 by nef-big adjoint vanishing. Thus a kL section does not vanish at the single base point q of L. Away from q, powers of generated L sections suffice. Therefore every kL with $k\ge 2$ is globally generated.