观点

Lemma 18 ([The determinant obstruction map]). Let 0→F→V_0→Q_0→0 be an exact sequence on a smooth projective surface, where V_0 is a vector bundle of rank r and Q_0 is torsion. If K is ample and Kc_1(Q_0)>rK^2 , the map ^1(F,Q_0)⟶^2(F,F) …

Lemma 18 ([The determinant obstruction map]). Let 0→F→V0→Q0→0 be an exact sequence on a smooth projective surface, where V0 is a vector bundle of rank r and Q0 is torsion. If K is ample and Kc1(Q0)>rK2 , the map1(F,Q0)⟶2(F,F) trH2( OX) is surjective.

证明 1证否 0作为观点被引用 2链条图发起推演

发起推演1

证明
Lemma 17 ([The open arrow enhancement]). The derived stack of arrows with…Theorem 5.1 ([Surface bound]). Suppose b≥5 and S<χ-1 . Let (w_α)_ α∈I be any…观点
By duality, a canonical form ω annihilating the image has the property that ω_F:F→F(K) extends to a map a:V_0→F(K) . If i:F→V_0 , the maps i(K)a and ω_ V_0 agree where F=V_0 , hence everywhere. It follows that ωQ_0=0 . For nonzero ω , write (ω)=∑a_jC_j . At the generic point of C_j a quotient of ( O/(ω))^r has length at most ra_j . These generic lengths are the multiplicities of c_1(Q_0) . Thus c_1(Q_0)≤r(ω) as divisorial cycles, and ampleness gives Kc_1(Q_0)≤rK^2 , a contradiction. The argument permits embedded points and nonreduced curve support.
更多详情 →