The Stacks project

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 ( 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 $\mathcal{E}$.

  1. 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
    \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}
  2. 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 $X$.

  3. 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 ( 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. $\square$

Remark 42.37.3. We could also rewrite equation as
\begin{equation} \label{chow-equation-signs} \sum \nolimits _{i = 0}^ r c_1(\mathcal{O}_ P(-1))^ i \cap \pi ^*c_{r - i} = 0. \end{equation}

but we find it easier to work with the tautological quotient sheaf $\mathcal{O}_ P(1)$ instead of its dual.

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