Weak quadric and projective-plane polarization generation

Let T be a smooth projective surface, L nef and big, $K_T=-2L$ and L²=2. Adjoint nef-big vanishing gives $H1(O_T)=0$ and H1(L)=0. RR gives h0(L)=4 and χ(N)=1+(N²+2LN)/2. If L has a base point, its point-Serre bundle V has determinant L and c2=1. A stable V would have h0(End V)=1 and h2(End V)=0 (twist by $K_T$ lowers slope), whereas χ(End V)=L²=2. Hence it has a saturated destabilizing line M. As before $N=L-M$ is nonzero effective with $a=LN\ge 0$, $q=N^2$, $a\le 1$, $a-1\le q\le a^2\mathrm{/}2$, and q even. Negative definiteness excludes $a=0$; therefore $a=1$ and $q=0$.

Reflect N in an L-degree-zero exceptional irreducible curve E whenever $NE<0$. Negative definiteness and adjunction with $K_T$ $E=0$ give $E^2=-2$ and arithmetic genus zero, so E is smooth rational. Put $N_{prime}=N+(NE)E$. Then $N_{prime}^2=N^2=0$ and $L{N}_{prime}=LN=1$. RR gives $\chi (N_{prime})=1+(N_{prime}^2+2L{N}_{prime})\mathrm{/}2=2$. Also $h2(N_{prime})=h0(K_T-N_{prime})=0$ because that class has negative L-degree. Hence $h0(N_{prime})\ge 2$, so $N_{prime}$ is effective. For a fixed ample Cartier divisor H its degree drops by the positive integer -(NE)HE. Positive integral H-degree cannot decrease indefinitely; the process terminates with an effective D of square zero and L-degree one having nonnegative intersection with every exceptional curve. There is a unique nonexceptional component C, with coefficient one; write $D=C+B$, with B effective exceptional. Adjunction and Hodge give $C^2=0$: $C^2-2=2p_a(C)-2$, so $C^2=2p_a(C)\ge 0$, while $C^2\le$ $(LC{)}^2\mathrm{/}{L}^2=1\mathrm{/}2$. Therefore $DC=C^2+BC=BC\ge 0$. Every other component of D is exceptional and has already been treated; a curve outside its support intersects the effective D nonnegatively. Therefore D is nef.

Its complete system has no fixed part and no base point: the movable part is numerically proportional by Hodge, has positive integral L-degree one, and its fixed exceptional part has square zero, hence vanishes. Let f:T→P1 be the connected-fibre morphism. A general fibre F has F²=0, LF=1 and genus zero by adjunction. There is no singular fibre. For a multiple irreducible fibre, LF=1 forces multiplicity one and arithmetic genus zero makes it smooth. In a reducible fibre, every component has negative square, and $\sum m_i$ $K_T$ $C_i=K_T$ F=−2 gives a component with $K_T$ $C_i<0$; adjunction forces it to be a rational (−1)-curve. This is impossible since $K_T=-2L$ has even degree on every curve.

Thus f is a smooth P1 fibration. Since L restricts to O(1) on every fibre, the Cech cohomology of P1 gives a rank-two bundle $E=f_{\ast }L$, base change, and surjective evaluation f*E→L. The induced T→P(E) is an isomorphism on every fibre and hence an isomorphism. Higher direct images of L vanish, and χ(L)=4 gives deg $E=2$. A maximal-degree line subbundle O(a)⊂E has $a\ge 1$ because χ(E(−1))=2; its quotient is O(2−a). The extension splits, since H1(O(2a−2))=0. Nefness of the tautological L forces both quotient degrees a and2−a to be nonnegative. Thus E is a sum of generated line bundles and L is generated. This contradicts the assumed base point. Consequently the weak quadric polarization is globally generated.

The argument constructing the maximal line subbundle is elementary on P1: twist down until sections cease, saturate a nonzero section to a line, and choose the largest resulting degree; the degrees are bounded above by embedding E into a sum of sufficiently positive line bundles. No splitting theorem is imported.

Let T be smooth, L nef and big, $K_T=-3L$ and L²=1. Adjoint vanishing gives $H1(O_T)=H1(L)=0$ and RR gives h0(L)=3. Write |L|=B+|P| with B fixed. The moving part has $LP>0$, and $LP\le 1$, so LP=1, LB=0 and $P^2\le 1$. Its square cannot be zero: a square-zero moving pencil has no base point, and its connected primitive fibre F would have LF=1 and F²=0, giving 2g(F)−2=−3 by adjunction. Hence P²=1, and B²=(L−P)²=0 forces $B=0$ by negative definiteness on the L-null locus.

If |L| had a base point, two general members would have common zero scheme of length one. Their independent first-order local terms make a general member C smooth there; Bertini makes it smooth elsewhere. No exceptional component occurs in a general member, and the total L-degree one permits only one nonexceptional component, so C is connected and integral. Adjunction gives genus zero. But H0(L)→H0(C,L|C) is surjective because $H1(O_T)=0$, and $L\mathrm{\mid }C=O_{P1}(1)$ is generated. A common base point on C is impossible. Thus L is generated.