11.14 Stable extensions approaching the split comparison bundleD9D859664CFB

Proof. Extensions of line bundles are locally free: the quotient is free on a small trivializing neighborhood, so the sequence splits there. Their classes form

By lemma 11.13.1, the higher direct image of this line bundle vanishes and its direct image is the sum of the seven line bundles $\mathcal O_{\mathbb P^1}(-12e-6-ie)$, $0\leq i\leq 6$. Their first cohomology groups have total dimension

The Chern classes follow from the extension sequence. Direct computations of intersection numbers give

It suffices to test line-bundle subsheaves $M\hookrightarrow V$. Indeed, a rank-one torsion-free subsheaf on a smooth surface has a line bundle as its double dual, and its inclusion extends across the finitely many missing points by Hartogs extension in local bundle frames. The extended map remains injective and has the same first Chern class. This argument does not require the quotient to be locally free.

If the composite $M\to L_2$ is zero, then $M$ maps nontrivially into $L_1$, so $\deg _ P M\leq \mu _ P(L_1)<\mu _ P(V)$. Otherwise its zero divisor $D$ is effective and $M=L_2(-D)$. If $D=0$, the composite is an isomorphism and its inverse splits $q$, contrary to the hypothesis. Hence $D\neq 0$, and

Every effective class is $aS+bf$ with $a,b\geq 0$, and

For $e\geq 1$, every nonzero effective divisor has $P$-degree at least four, which proves the strict slope inequality.

For $e=0$, the only nonzero effective class of $P$-degree at most three is $S$. A lift of multiplication by a nonzero section $s\in H^0(X,\mathcal O_ X(S))$ from $L_2(-S)$ to $V$ would make the pullback of the nonzero extension class vanish. That pullback is the image of the class under the map

On $X=\mathbb P^1\times \mathbb P^1$, order the factors so that $\mathcal O_ X(S)=\mathcal O(1,0)$ and $\mathcal O_ X(f)=\mathcal O(0,1)$. The Künneth formula identifies this map with

Multiplication by a nonzero section is injective on global sections, and tensoring with a vector space preserves injectivity. Thus the nonzero extension class cannot pull back to zero. This excludes the possible equality of slopes and proves stability also for $e=0$. $\square$

Proof. Put $D=K_ X+f=-2S-(e+1)f$. The four line bundles $L_ i^*\otimes L_ j(D)$ have classes

Each has a negative coefficient in the basis $S,f$, and hence no nonzero section. Applying $\operatorname {Hom}_ X(L_ i,\cdot )$ to the extension twisted by $D$ gives $\operatorname {Hom}_ X(L_ i,V(D))=0$ for $i=1,2$. Applying $\operatorname {Hom}_ X(\cdot ,V(D))$ to the original extension then gives $\operatorname {Hom}_ X(V,V(D))=0$. By Serre duality this is the prioritary condition $\operatorname {Ext}^2_ X(V,V(-f))=0$.

Restriction to a ruling fiber is exact, because the quotient is locally free. The restricted line bundles have degrees five and minus one, and the restricted extension splits, since $\operatorname {Ext}^1_{\mathbb P^1}(\mathcal O(-1),\mathcal O(5)) =H^1(\mathbb P^1,\mathcal O(6))=0$. The quotient of negative degree on this curve excludes ampleness. $\square$

Proof. Choose a nonzero extension class and a smooth Dolbeault representative $\eta \in \Omega ^{0,1}(X,L_2^*\otimes L_1)$ of it, so that $\bar\partial \eta =0$. On the smooth direct sum, set

The covariant Dolbeault derivative of $A$ is zero and $A\wedge A=0$, so $\bar\partial _ t^2=0$. Adding the Dolbeault operator of the base $\mathbb C$ of the parameter $t$ also gives an integrable operator on $X\times \mathbb C$, because the family depends holomorphically on $t$. This defines a holomorphic family of bundles whose extension class is $t[\eta ]$. It is nonsplit for $t\neq 0$, so the two preceding propositions apply. For a fixed smooth background connection and fixed metrics,

and the norm $\| A\| _{C^ m}$ is finite for every $m$ because $X$ is compact. $\square$