PomeLemma 11.13.3.7055476D483C Both $E_0$ and $E_{\mathrm{amp}}$ are prioritary for the ruling.
Lemma 11.13.37055476D483C
Proof. Write $D=K_ X+f=-2S-(e+1)f$. The summands of $\operatorname {End}(E_0)(D)$ have classes
\[ D,\quad D,\quad 4S-(13e+7)f,\quad -8S+(11e+5)f. \]
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
\[ \operatorname {Hom}(E_{\mathrm{amp}},E_{\mathrm{amp}}(D)) \hookrightarrow H^0(X,E_{\mathrm{amp}}(D))^4. \]
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
\[ \begin{aligned} H^1(X,\mathcal O_ X(D-jH))& =H^1(X,\mathcal O_ X(jH-f))^*,\\ \pi _*\mathcal O_ X(jH-f)& =\bigoplus _{i=0}^ j\mathcal O_{\mathbb P^1}(j(3e+1)-1-ie). \end{aligned} \]
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 $E_{\mathrm{amp}}$. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.