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Let $X$ be a projective manifold. Then we can define ample sheaves on $X$, and many results of ample vector bundles still hold in this more general case (See K. Kubota, Ample sheaves).

Now I was wondering if the double dual of an ample torsion-free sheaf $\mathcal F$ is again ample? This is trivial for reflexive sheaves, since the double dual of a reflexive sheaf is itself. However, if $\mathcal F$ is not reflexive, is this still true?

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    $\begingroup$ Finally, I find that this may be not true in general. If $\mathcal F$ is of rank $1$, the double dual of $\mathcal F$ is just the determinant bundle $\det(\mathcal F)$. Moreover, there is a closed subscheme $Z\subset X$ of codimension $\geq 2$ such that $\mathcal F= \mathbb P(\det(\mathcal F))\otimes \mathcal I_Z$. Since $\mathbb P(\mathcal F)$ is just the blowing-up of $\pi\colon Bl_Z X\to X$, if $Z$ is smooth, ampleness of $\mathcal F$ just means $\pi^*\det(\mathcal F)-E$ is ample, where $E$ is the exceptional divisor. In general, this doesn't imply that $\det(\mathcal F)$ is ample. $\endgroup$
    – Chieh LIU
    Commented Jul 1, 2016 at 7:43

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