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Let $X$ be a smooth projective variety, and let $E=\mathcal{O}\oplus \mathcal{O}(1)$ be a vector bundle of rank $2$. Then $L=\wedge ^{2} E $ is a line bundle on $X$. Is $L(-2)$ $\mathbb{Q}$-linear to an effective divisor?

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up vote 3 down vote accepted

The answer is no already for $X=\mathbb{P}^1$. Indeed $c_1(L)=c_1(\mathcal{O}(1))$ and so $c_1(L(-2))=c_1(\mathcal{O}(-1))$. Therefore $L(-2)$ is linearly equivalent to the tautological bundle, which is not $\mathbb{Q}$-effective.

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