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Let $E$ be a globally generated vector bundle on a surface $S$ of rank $r\geq 2$. By standard facts about degeneracy loci, for a general $V\in G(r,H^0(E))$ one has:

(*)the evaluation map $ev: V\otimes \mathcal{O}_S\to E$ is injective and the cokernel is a line bundle supported on a smooth curve.

Now, let $E_1$ be a subvector bundle of $E$ and assume $E_1$ is globally generated as well. Is it possible to find $V\in G(r,H^0(E))$ which satisfies (*) and such that $V\cap H^0(E_1)$ has dimension equal to the rank of $E_1$?

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  • $\begingroup$ The "standard facts" you cite are roughly Bertini type theorems. These do work over algebraically closed fields. However, I expect there are counterexamples if $S$ is a surface over a finite field (just as there are counterexamples to other Bertini theorems). $\endgroup$
    – Jason Starr
    Jan 15, 2013 at 15:12
  • $\begingroup$ Sorry, I forgot to mention that I am over $\mathbb{C}$ $\endgroup$
    – ginevra86
    Jan 15, 2013 at 15:21
  • $\begingroup$ @ginevra86, why is the evaluation map ev injective? $\endgroup$
    – gradstudent
    May 11, 2015 at 11:34

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No.

Suppose that $E$ has rank $2$, and write $V_1=V\cap H^0(E_1)$. Then you have an exact sequence $$ 0\to coker(V_1\otimes \mathcal O_S\to E_1)\to coker(V\otimes \mathcal O_S\to E)\to coker((V/V_1)\otimes \mathcal O_S\to E/E_1)\to 0. $$ If the line bundles $E_1$ and $E/E_1$ are such that every divisor corresponding to $E_1$ has nonempty intersection with every divisor corresponding to $E/E_1$ (for example when $S=\mathbb P^2$ and both $E_1$ and $E/E_1$ are the hyperplane bundle), then there must be a point in $S$ where $coker(V\otimes \mathcal O_S\to E)$ is too big.

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  • $\begingroup$ Are you saying that the support of the cokernel of $V\otimes\mathcal{O}_S\to E$ is the union of the cokernel of $V_1\otimes\mathcal{O}_S\to E_1$ and $(V/V_1)\otimes\mathcal{O}_S\to E/E_1$? $\endgroup$
    – ginevra86
    Jan 15, 2013 at 23:58
  • $\begingroup$ Yes, if all the kernels are zero. Certainly in the case when $E_1$ is a direct summand of $E$. So that gives counterexamples. $\endgroup$
    – Tom Goodwillie
    Jan 16, 2013 at 3:21
  • $\begingroup$ There is something that I do not understand. Let $E$ be the direct sum of two hyperplane line bundles on the projective plane (let's say $H_1$ and $H_2$). Then, a general $V\in G(2,H^0(E))$ must satisfy (*). However, it seems to me that $V\cap H^0(H_i)$ is $1$-dimensional for $i=1,2$ because $V$ generically generates both the direct summands. Is this not correct? $\endgroup$
    – ginevra86
    Jan 16, 2013 at 9:39
  • $\begingroup$ In that case $H^0(E_1)$ is $3$-dimensional, $H^0(E)$ is $6$-dimensional, a typical $2$-dimensional vector subspace $V$ of the latter will intersect the former in zero alone. $\endgroup$
    – Tom Goodwillie
    Jan 16, 2013 at 13:54

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