Pome

Proposition 11.14.1A195B6162A49

Proposition 11.14.1.A195B6162A49 The extension space has dimension

\[ \dim \operatorname {Ext}^1_ X(L_2,L_1)=105e+35. \]

Every nonsplit extension

\[ 0\longrightarrow L_1\longrightarrow V\xrightarrow {q}L_2 \longrightarrow 0 \]

is a locally free, $\mu _ P$-stable bundle with $c_1(V)=4H$ and $c_2(V)=13H^2$.

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

\[ \operatorname {Ext}^1_ X(L_2,L_1) =H^1(X,\mathcal O_ X(6S-(12e+6)f)). \]

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

\[ \sum _{i=0}^6(12e+5+ie)=105e+35>0. \]

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

\[ \mu _ P(L_1)=40e+11,\qquad \mu _ P(V)=40e+14, \qquad \mu _ P(L_2)=40e+17. \]

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

\[ \deg _ P M=\mu _ P(V)+3-D\cdot P. \]

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

\[ S\cdot P=8e+3,\qquad f\cdot P=4. \]

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

\[ H^1(X,\mathcal O_ X(6S-6f)) \xrightarrow {\ \cdot s\ }H^1(X,\mathcal O_ X(7S-6f)). \]

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

\[ \bigl [H^0(\mathbb P^1,\mathcal O(6)) \xrightarrow {\ \cdot s\ }H^0(\mathbb P^1,\mathcal O(7))\bigr ] \otimes \operatorname {Id}_{H^1(\mathbb P^1,\mathcal O(-6))}. \]

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 $e=0$. $\square$

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.

Read this page’s discussion on GitHubPost a comment on GitHub