Lemma 42.55.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of rank $r$. Then

$\prod \nolimits _{n = 0, \ldots , r} c(\wedge ^ n \mathcal{E})^{(-1)^ n} = 1 - (r - 1)! c_ r(\mathcal{E}) + \ldots$

Proof. By the splitting principle we can turn this into a calculation in the polynomial ring on the Chern roots $x_1, \ldots , x_ r$ of $\mathcal{E}$. See Section 42.43. Observe that

$c(\wedge ^ n \mathcal{E}) = \prod \nolimits _{1 \leq i_1 < \ldots < i_ n \leq r} (1 + x_{i_1} + \ldots + x_{i_ n})$

Thus the logarithm of the left hand side of the equation in the lemma is

$- \sum \nolimits _{p \geq 1} \sum \nolimits _{n = 0}^ r \sum \nolimits _{1 \leq i_1 < \ldots < i_ n \leq r} \frac{(-1)^{p + n}}{p}(x_{i_1} + \ldots + x_{i_ n})^ p$

Please notice the minus sign in front. However, we have

$\sum \nolimits _{p \geq 0} \sum \nolimits _{n = 0}^ r \sum \nolimits _{1 \leq i_1 < \ldots < i_ n \leq r} \frac{(-1)^{p + n}}{p!}(x_{i_1} + \ldots + x_{i_ n})^ p = \prod (1 - e^{-x_ i})$

Hence we see that the first nonzero term in our Chern class is in degree $r$ and equal to the predicted value. $\square$

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