Let $S$ be a K3 surface with $Pic(S) = \mathbb{Z}.[C]$, where $C$ is a smooth curve. Let $A$ be a complete, base point free $g_d^r$ on $C$. How to show that the Lazarsfeld-Mukai bundle $E_{C,A}$ associated to $(C,A)$ is $C$-stable on $S$? (The bundle $E_{C,A}$ is as defined in Lazarsfeld's paper 'Brill-Noether-Petri without degenerations').
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2$\begingroup$ What is $g^r_d$? $\endgroup$– Atsushi KanazawaCommented Jun 2, 2013 at 20:42
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2$\begingroup$ A good start could be defining the main objects that you ask about, without giving just a reference ;) $\endgroup$– IMeasyCommented Jun 2, 2013 at 20:44
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2 Answers
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Since the Lazarsfeld-Mukai bundle $E = E_{C,A}$ fits into the exact sequence
$$0 \to H^0(C,A)^\vee \otimes \mathcal{O}_S \to E \xrightarrow{\phi} K_C(-A) \to 0,$$
the first Chern class of $E$ is $[C]$. Let $F$ be a subsheaf of $E$ and let $K = \ker(\phi_{|F})$ and $L= \mathrm{im}(\phi_{|F})$. Since $K$ and $L$ are subsheaves of $V :=H^0(C,A)^\vee \otimes \mathcal{O}_S$ and $K_C(-A)$ respectively, one has
$$c_1(F) \cdot [C] = (c_1(K) +c_1(L))\cdot [C] \le [C]\cdot [C] = c_1(E) \cdot [C].$$
The equality holds when $L = K_C(-A)$ and $K$ is a direct sum of copies of $\mathcal{O}_S$. In this case, the quotient $E/F \simeq V/K$ is also a direct sum of copies of $\mathcal{O}_S$. As $H^0(E^\vee) = 0$, we conclude that $E = F$.
Therefore if $F$ is a proper saturated subsheaf of $E$ such that $\mu_C(F) \ge \mu_C(E)$, we would have $$[C]\cdot [C] > c_1(F)\cdot [C] \ge \frac{rkF}{rkE}[C]\cdot [C] > 0,$$ which is in contradiction to the assumption that $Pic(S) = \mathbf{Z} \cdot[C]$.
(EDIT: Last inequality corrected following abx's comment.)
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$\begingroup$ I do not understand the last line of inequalities. From $\mu _C(F)\geq \mu _C(E)$ you get $c_1(F)\cdot [C] \ge \left(\frac{\mathrm{rk}\,F}{\mathrm{rk}\,E}\right)[C]\cdot [C] $. Where does your $1-\cdots $ come from? $\endgroup$– abxCommented Feb 9, 2015 at 8:02
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$\begingroup$ Thanks, there's no "$1-$...". The contradiction still holds. $\endgroup$– HYLCommented Feb 11, 2015 at 23:41
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$\begingroup$ I still don't understand your proof. Why does $c_1(F)=[C]$ (the case where equality holds) imply $L=K_C(-A)$? It could be any sub-line bundle of $K_C(-A)$. $\endgroup$– abxCommented Feb 12, 2015 at 8:55
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$\begingroup$ Oh sorry! I wanted to assume that $F$ is a "saturated" subsheaf of $E$. $\endgroup$– HYLCommented Feb 14, 2015 at 10:33
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You can argue as follows. Put $E:=E_{C,A}$. You need only to know : $c_1(E)$ is the positive generator of $\mathrm{Pic}(S)$, there is a homomorphism $\mathscr{O}_S^r\rightarrow E$ which is generically surjective, and $H^0(S,E^*)=0$.
If $E$ is not stable, it admits a torsion-free quotient $\mathscr{F}$ with $c_1(\mathscr{F})\leq 0$. The bidual $F$ of $\mathscr{F}$ is a vector bundle with the same $c_1$, and a generically surjective homomorphism $\mathscr{O}_S^r\rightarrow F$. Let $s:=\rm{rk}( F)$; we get a generically surjective homomorphism $\wedge^s\mathscr{O}_S^r\rightarrow \det(F)$. Therefore $\det(F)=\mathscr{O}_S$, and there is a direct factor $\mathscr{O}_S^s$ of $\mathscr{O}_S^r$ such that the induced homomorphism $\mathscr{O}_S^s\rightarrow F$ has nonzero determinant, hence is an isomorphism. But this contradicts $H^0(S,E^*)=0$.
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$\begingroup$ how is that you get a direct factor $O_S^s$ of $O_S^r$ such that the induced morphism has non-zero determinant? $\endgroup$ Commented Jun 20, 2016 at 8:59
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1$\begingroup$ The homomorphism $\wedge^s\mathcal{O}_S^r\rightarrow F$ is given by $r$ global sections $f_1,\ldots ,f_r$ of $F$, which generate $F$ over the function field $K$ of $S$. Thus there are $s$ among them, say $f_1,\ldots ,f_s$, which form a basis of $F_K$. This means that $f_1\wedge\ldots \wedge f_s$ is a nonzero section of $\det(F)\cong \mathcal{O}_S$, hence the corresponding map $\mathcal{O}_S^s\rightarrow F$ has nonzero determinant. $\endgroup$– abxCommented Jun 20, 2016 at 9:22