## 42.36 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 (42.36.1.1). The reason for the signs will be explained later.

Definition 42.36.1. 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 42.35.2 there are elements $c_ i \in \mathop{\mathrm{CH}}\nolimits _{n - i}(X)$, $i = 0, \ldots , r$ such that $c_0 = [X]$, and

42.36.1.1
\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 42.36.2. 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 42.36.1 is equal to the Weil divisor associated to $\mathcal{L}$ by Definition 42.23.1.

**Proof.**
In this proof we use $c_1(\mathcal{L}) \cap [X]$ to denote the construction of Definition 42.23.1. 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 (42.36.1.1) 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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