42.37 The Chern classes of a vector bundle
We can use the projective space bundle formula to define the Chern classes of a rank r vector bundle in terms of the expansion of c_1(\mathcal{O}(1))^ r in terms of the lower powers, see formula (42.37.1.1). The reason for the signs will be explained later.
Definition 42.37.1. Let (S, \delta ) be as in Situation 42.7.1. Let X be locally of finite type over S. Assume X is integral and n = \dim _\delta (X). Let \mathcal{E} be a finite locally free sheaf of rank r on X. Let (\pi : P \to X, \mathcal{O}_ P(1)) be the projective space bundle associated to \mathcal{E}.
By Lemma 42.36.2 there are elements c_ i \in \mathop{\mathrm{CH}}\nolimits _{n - i}(X), i = 0, \ldots , r such that c_0 = [X], and
42.37.1.1
\begin{equation} \label{chow-equation-chern-classes} \sum \nolimits _{i = 0}^ r (-1)^ i c_1(\mathcal{O}_ P(1))^ i \cap \pi ^*c_{r - i} = 0. \end{equation}
With notation as above we set c_ i(\mathcal{E}) \cap [X] = c_ i as an element of \mathop{\mathrm{CH}}\nolimits _{n - i}(X). We call these the Chern classes of \mathcal{E} on X.
The total Chern class of \mathcal{E} on X is the combination
c({\mathcal E}) \cap [X] = c_0({\mathcal E}) \cap [X] + c_1({\mathcal E}) \cap [X] + \ldots + c_ r({\mathcal E}) \cap [X]
which is an element of \mathop{\mathrm{CH}}\nolimits _*(X) = \bigoplus _{k \in \mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ k(X).
Let us check that this does not give a new notion in case the vector bundle has rank 1.
Lemma 42.37.2. Let (S, \delta ) be as in Situation 42.7.1. Let X be locally of finite type over S. Assume X is integral and n = \dim _\delta (X). Let \mathcal{L} be an invertible \mathcal{O}_ X-module. The first Chern class of \mathcal{L} on X of Definition 42.37.1 is equal to the Weil divisor associated to \mathcal{L} by Definition 42.24.1.
Proof.
In this proof we use c_1(\mathcal{L}) \cap [X] to denote the construction of Definition 42.24.1. Since \mathcal{L} has rank 1 we have \mathbf{P}(\mathcal{L}) = X and \mathcal{O}_{\mathbf{P}(\mathcal{L})}(1) = \mathcal{L} by our normalizations. Hence (42.37.1.1) reads
(-1)^1 c_1(\mathcal{L}) \cap c_0 + (-1)^0 c_1 = 0
Since c_0 = [X], we conclude c_1 = c_1(\mathcal{L}) \cap [X] as desired.
\square
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