11.13 Ample comparison quotients on Hirzebruch surfaces9C2F19628D6F

Proof. Over a trivializing chart of the base, sections of $\mathcal O_ X(aS)$ are homogeneous polynomials of degree $a$ in the fiber coordinates. Their transition law gives

and twisting by the base gives the formula. The first cohomology of $\mathcal O_{\mathbb P^1}(a)$ on each fiber vanishes, which gives the assertion on the higher direct image. If $a<0$, restriction to a general ruling fiber excludes a nonzero section. For $a\geq 0$, the direct-image formula gives a nonzero section exactly when $b\geq 0$.

Suppose that $a\geq 1$ and $b>ae$. All line bundles in the direct image then have degree at least one. Fiber polynomials of degree $a$ separate points and tangent directions within a fiber. To separate points over different base points, choose a fiber monomial that is nonzero at the first point and a coefficient section vanishing at the second base point but not at the first. At any point, choose such a nonzero monomial and a coefficient section with a simple zero at its base point; the resulting global section vanishes there, has nonzero derivative in a prescribed base direction and has zero vertical derivative. The complete linear system therefore separates points and first jets, so it defines an embedding. The formulas for $H^2$ and $h^0(H)$ follow from the intersection numbers and from the direct image $\mathcal O(3e+1)\oplus \mathcal O(2e+1)$. Finally, the relative Euler sequence gives $K_{X/\mathbb P^1}=-2S-ef$, and adding the pullback of the canonical class of $\mathbb P^1$ proves the last identity. $\square$

Proof. Step 1: the quadruple $a$. Since $\mathcal O_ X(H)$ is globally generated, a common zero at a fixed point imposes four independent linear conditions on a quadruple $a$. The incidence variety of quadruples with a common zero at some point of the surface therefore has codimension at least two in the space of quadruples, and its image there is closed by properness. Linear independence is also a nonempty open condition, because $h^0(H)\geq 4$, and the two open conditions intersect. Fix such an $a$ and write

Step 2: local freeness. At a fixed point $x$, the map $(a,b)$ drops rank exactly when $b(x)$ is proportional to the nonzero vector $a(x)$. Since evaluation on $V$ is surjective, this is a condition of codimension three. Its incidence over $X$ has dimension at most $N-1$, and its image in $V$ is closed. Outside that image the cokernel is locally free of rank two.

Step 3: the map to projective space. Dualizing the sequence identifies $E_{\mathrm{amp}}^*$ with a subbundle of $(\mathbb C^4)^*\otimes \mathcal O_ X$. This inclusion defines a proper morphism

For $w=[w_0:\cdots :w_3]$, its fiber is the common zero scheme of

Linear independence of $a$ gives $s_ w\neq 0$ for every $w$.

Step 4: no common curves. A nonzero effective subdivisor $D$ of $\operatorname {div}(s_ w)$ has class $uS+vf$, where

Write $|D|=\mathbb P H^0(X,\mathcal O_ X(D))$ for the complete linear system. For each of these finitely many classes $D_0$, let $T_{D_0}$ parametrize the triples $(w,[\sigma ],[\tau ])$ with

Since products of nonzero sections are nonzero, this is a closed projective parameter scheme. Its fiber over $w$ consists of the finitely many effective subdivisors of the fixed divisor $\operatorname {div}(s_ w)$, counted with their possible multiplicities. Thus $\dim T_{D_0}\leq 3$.

Vanishing on a nonzero effective divisor $D$ imposes at least four independent conditions on $H^0(X,\mathcal O_ X(3H))$. Indeed, choose an integral curve $C$ in its support. Very ampleness gives sections $r_0,r_1$ of $\mathcal O_ X(H)$ with nonconstant ratio on $C$. The four cubic monomials in their restrictions are linearly independent, since a linear relation would make the nonconstant ratio algebraic over $\mathbb C$, hence constant.

For fixed $(w,[\sigma ],[\tau ])$, the map $V\to H^0(X,\mathcal O_ X(3H))$, $b\mapsto t_ w$, is surjective. The condition that $t_ w$ vanish on $D=\operatorname {div}(\sigma )$ therefore has codimension at least four in $V$. This condition is closed in families: the universal multiplication map

is injective on fibers, so its image is a subbundle, and the condition is the vanishing of the image of $t_ w$ in the quotient bundle. Its incidence over $T_{D_0}$ has dimension at most $N-1$, and its image in $V$ is closed.

Step 5: conclusion. Choose $b$ outside these finitely many proper closed images and outside the image of Step 2. Then $s_ w$ and $t_ w$ have no common curve for any $w$, so every fiber of $q_ b$ is finite. A proper morphism with finite fibers is finite, and the pullback of $\mathcal O_{\mathbb P^3}(1)$ by a finite morphism is ample; hence $E_{\mathrm{amp}}$ is ample. Finally, the Whitney formula gives

Proof. Write $D=K_ X+f=-2S-(e+1)f$. The summands of $\operatorname {End}(E_0)(D)$ have classes

Each has a negative coefficient in the basis $S,f$, so by lemma 11.13.1 none of them has a nonzero section. This proves the required vanishing of $\operatorname {Hom}$ for $E_0$.

Precomposition with the surjection $\mathcal O_ X^4\to E_{\mathrm{amp}}$ gives an injection

Twist the defining sequence by $D$. Its middle term has no sections. For $j=1,3$, Serre duality and the formula of lemma 11.13.1 give

The higher direct image vanishes, and every degree in this sum is at least $j(2e+1)-1\geq 0$. Hence all these first cohomology groups vanish, and the long exact sequence of the twisted sequence gives $H^0(X,E_{\mathrm{amp}}(D))=0$. This proves the vanishing of $\operatorname {Hom}$ for $E_{\mathrm{amp}}$. $\square$

Proof. The integral classes

are very ample by lemma 11.13.1. Pullbacks of normalized Fubini–Study forms under the corresponding embeddings give Kähler forms $\eta _ s$ representing $c_1(\mathcal O_ X(A_ s))$. A direct computation of intersection numbers gives

so the degrees of the duals are equal to one. Since $n=2$, the closed positive $(1,1)$-forms $\eta _ s$ are admissible test forms $\Theta _ s$ in theorem 11.9.4. $\square$