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Let $X$ be a $n$ (complex) dimensional variety and let $\mathcal{F}$ be a coherent sheaf on $X$. Using the sheaf hom which I denote $\mathcal{H}om$, we can define the dual coherent sheaf as

$$\mathcal{F}^{\vee} = \mathcal{H}om(\mathcal{F}, \mathcal{O}_{X}).$$

With this definition, I'm hoping someone can explain how to compute the Chern character $\text{ch}(\mathcal{F}^{\vee})$ of the dual sheaf.

Given a class $v = \oplus_{i} v_{i} \in H^{*}(X, \mathbb{Q})$, I believe it is standard to define the dual class to be $v^{\vee} = \oplus_{i}(-1)^{i} v_{i} \in H^{*}(X, \mathbb{Q})$. This is, for example, done in Huybrechts and Lehn. The motivation behind this definition is supposed to be such that

$$\text{ch}^{\vee}(\mathcal{F}) = \text{ch}(\mathcal{F}^{\vee}).$$

Assuming someone can assist me in computing the Chern character of the dual sheaf, will it be consistent with this definition?

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    $\begingroup$ For any torsion coherent sheaf, the dual is zero. So, you can not expect a simple formula as you describe. $\endgroup$
    – Mohan
    Jul 26, 2017 at 22:29
  • $\begingroup$ @Mohan Interesting, thanks a lot. However, in general, a torsion coherent sheaf has a non-zero Chern character and therefore, a non-zero dual Chern character by my definition above. For example, you can explicitly compute the Chern character of a pushforward of a vector bundle. So perhaps $\text{ch}^{\vee}(\mathcal{F}) = \text{ch}(\mathcal{F}^{\vee})$ is only meant to hold for torsion-free coherent sheaves? $\endgroup$
    – Benighted
    Jul 26, 2017 at 22:39
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    $\begingroup$ Still, you will run into trouble. For simplicity, let me assume $X$ is smooth. Then the ideal sheaf defining a subvariety of codimension at least 2 has its dual just $\mathcal{O}_X$, independent of the subvariety. $\endgroup$
    – Mohan
    Jul 26, 2017 at 22:45
  • $\begingroup$ @Mohan Thank you, I suppose it will only work for locally-free sheaves. It is strange though that $\text{ch}^{\vee}$ appears in, for example, $\chi(\mathcal{F}, \mathcal{E})$ for general coherent sheaves. Because of this, I had assumed that the Chern character would commute with the dual for any coherent sheaf. $\endgroup$
    – Benighted
    Jul 26, 2017 at 23:05
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    $\begingroup$ In fact, $ch^\vee(F)$ is the Chern character of the DERIVED dual of $F$. $\endgroup$
    – Sasha
    Jul 28, 2017 at 19:56

1 Answer 1

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Let me convert Sasha's comment to an answer, since that comment is folded.

The dual defined in Huybrechts and Lehn is not $$\mathcal{H}om(\mathcal{F}, \mathcal{O}_{X}).$$ It's actually the derived dual.

Let's say, there exists an exact sequence $$0 \to E_1 \to E_0 \to \mathcal{F} \to 0$$ such that $E_1$ and $E_0$ are locally free. Then we have $$ 0 \to \mathcal{H}om(\mathcal{F}, \mathcal{O}_{X}) \to E_0^\vee \to E_1^\vee \to \mathcal{Ext}^1(\mathcal{F},\mathcal{O}_X) \to 0. $$ Obviously $$ch(\mathcal{H}om(\mathcal{F}, \mathcal{O}_{X})) \neq ch^\vee({\mathcal{F}}).$$ But the derived dual is
$$ {\mathcal{F}^\bullet}^\vee = [E_0^\vee \to E_1^\vee ]. $$ Hence $$ch({\mathcal{F}^\bullet}^\vee)=ch^\vee({\mathcal{F}}).$$ In general, it can be defined by the resolution.

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  • $\begingroup$ When you write a locally-free resolution with two terms, are you assuming we're on a surface? I'm guessing the derived dual is just the dual complex of the resolution. Whose Chern character then makes sense. $\endgroup$
    – Benighted
    May 3, 2019 at 2:47
  • $\begingroup$ @Benighted I just use two terms as an example. I don't think it's true for a surface if $\mathcal{F}$ has torsion part. Nevertheless, we can still define $ch$ through the alternating sum. $\endgroup$
    – WWK
    May 3, 2019 at 18:03

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