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For globally generated line bundles it is quite clear to see that they are nef: Let $L$ be globally generated, then the global sections in $H^0(X,L)$ define a morphism $\Phi \colon X \to \mathbb{P}_N$, such that $L = \Phi^*(\mathcal{O}_{\mathbb{P}_N}(1))$. The line bundle $\mathcal{O}_{\mathbb{P}_N}(1)$ is ample which imples nef, and the pullback is nef as well, hence $L$ is nef.

How do we obtain this for vector bundles $E$ as well, such that if $E$ is globally generated it is nef?

Thanks in advance.

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  • $\begingroup$ For a vector bundle, what definition of nef are you using? $\endgroup$ – Mohan Sep 8 '16 at 1:56
  • $\begingroup$ I am working with the definition $E$ is nef if and olny if $\mathcal{O}_{\mathbb{P}(E)}(1)$ is nef on $\mathbb{P}(E)$. $\endgroup$ – L_K666 Sep 8 '16 at 7:00
  • $\begingroup$ $O(1)$ is the tautological quotient of $\pi^*E$. If $E$ is globally generated, then so is $\pi^* E$, hence so too is $O(1)$. Since it's a line bundle, that implies it's nef. $\endgroup$ – Nefertiti Sep 8 '16 at 10:16
  • $\begingroup$ @Nefertiti: I think I get it now. Global generation of $E$ fulfills spannendness of $\pi^*E$ and for any quotient bundle of $\pi^*E$ as well. $\mathcal{O}(1)$ is a quotient of $\pi^*E$, therefore globally generated and nef by the proof I gave above. Thanks for the hint! $\endgroup$ – L_K666 Sep 8 '16 at 10:36
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Let me make my comment an answer for completeness.

$O(1)$ is the tautological quotient of $\pi^* E$. If $E$ is globally generated, then so is $\pi^* E$, hence so too is $O(1)$. Since it's a line bundle, that implies it's nef.

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