Lemma 81.31.2. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module with dual $\mathcal{E}^\vee$. Then

$c_ i(\mathcal{E}^\vee ) = (-1)^ i c_ i(\mathcal{E})$

in $A^ i(X)$.

Proof. Choose a morphism $\pi : P \to X$ as in Lemma 81.31.1. By the injectivity of $\pi ^*$ (after any base change) it suffices to prove the relation between the Chern classes of $\mathcal{E}$ and $\mathcal{E}^\vee$ after pulling back to $P$. Thus we may assume there exist invertible $\mathcal{O}_ X$-modules ${\mathcal L}_ i$, $i = 1, \ldots , r$ and a filtration

$0 = \mathcal{E}_0 \subset \mathcal{E}_1 \subset \mathcal{E}_2 \subset \ldots \subset \mathcal{E}_ r = \mathcal{E}$

such that $\mathcal{E}_ i/\mathcal{E}_{i - 1} \cong \mathcal{L}_ i$. Then we obtain the dual filtration

$0 = \mathcal{E}_ r^\perp \subset \mathcal{E}_1^\perp \subset \mathcal{E}_2^\perp \subset \ldots \subset \mathcal{E}_0^\perp = \mathcal{E}^\vee$

such that $\mathcal{E}_{i - 1}^\perp /\mathcal{E}_ i^\perp \cong \mathcal{L}_ i^{\otimes -1}$. Set $x_ i = c_1(\mathcal{L}_ i)$. Then $c_1(\mathcal{L}_ i^{\otimes -1}) = - x_ i$ by Lemma 81.18.2. By Lemma 81.30.4 we have

$c(\mathcal{E}) = \prod \nolimits _{i = 1}^ r (1 + x_ i) \quad \text{and}\quad c(\mathcal{E}^\vee ) = \prod \nolimits _{i = 1}^ r (1 - x_ i)$

in $A^*(X)$. The result follows from a formal computation which we omit. $\square$

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