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For any locally free sheaf $\mathcal F$ on a scheme $(X,\mathcal O_X)$ of finite rank, its dual is defined as $\mathcal F^{\vee}:=\mathscr Hom(\mathcal F,\mathcal O_X)$. So,

if $\mathcal F,\mathcal G$ are locally free sheaves, what is the dual of $\mathcal F\oplus\mathcal G$?

I would like to think that $(\mathcal F\oplus\mathcal G)^{\vee}=\mathcal F^{\vee}\oplus\mathcal G^{\vee}$, since for every short exact sequence of locally free sheaves on $X$ the $\mathscr Hom(-,\mathcal O_X)$ functor is exact, but I think that this is not true because short exact sequences of locally free sheaves don't necessarily split, (see this answer: Short exact sequence of vector bundles vs locally free sheaves).

In the particular if $X$ is a projective variety, what is the dual of $\mathcal O(1)\oplus\mathcal O(2)$ ?

Thanks for your answers in advance.

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    $\begingroup$ The formula $(F \oplus G)^\vee \cong F^\vee \oplus G^\vee$ is correct. $\endgroup$
    – Sasha
    Jan 20, 2022 at 14:29

1 Answer 1

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Any additive functor between abelian categories sends split exact sequences to split exact sequences. For example you can see a proof here for abelian groups. Since $^{\vee}$ is an additive functor, it sends the split sequence

$0 \to \mathcal F \to \mathcal F \oplus \mathcal G \to \mathcal G \to 0$

to

$0 \to \mathcal G^{\vee} \to \mathcal (F \oplus \mathcal G)^{\vee} \to \mathcal F^{\vee} \to 0$.

and since this is again split, $(F \oplus \mathcal G)^{\vee} \cong F^\vee \oplus \mathcal G^{\vee}$. Note that this applieas to $\mathcal O_X$-modules in general, not only locally free sheaves, and to any other functor of $\mathcal O_X$-modules which is additive (almost all of them are)

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