Lemma 42.40.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{E}$, $\mathcal{F}$ be finite locally free sheaves on $X$ of ranks $r$, $r - 1$ which fit into a short exact sequence
Then we have
in $A^*(X)$.
Proof. By Lemma 42.35.3 it suffices to show that if $X$ is integral then $c_ j(\mathcal{E}) \cap [X] = c_ j(\mathcal{F}) \cap [X]$. Let $(\pi : P \to X, \mathcal{O}_ P(1))$, resp. $(\pi ' : P' \to X, \mathcal{O}_{P'}(1))$ denote the projective space bundle associated to $\mathcal{E}$, resp. $\mathcal{F}$. The surjection $\mathcal{E} \to \mathcal{F}$ gives rise to a closed immersion
over $X$. Moreover, the element $1 \in \Gamma (X, \mathcal{O}_ X) \subset \Gamma (X, \mathcal{E})$ gives rise to a global section $s \in \Gamma (P, \mathcal{O}_ P(1))$ whose zero set is exactly $P'$. Hence $P'$ is an effective Cartier divisor on $P$ such that $\mathcal{O}_ P(P') \cong \mathcal{O}_ P(1)$. Hence we see that
for any cycle class $\alpha $ on $X$ by Lemma 42.31.1. By Lemma 42.38.2 we see that $\alpha _ j = c_ j(\mathcal{F}) \cap [X]$, $j = 0, \ldots , r - 1$ satisfy
Pushing this to $P$ and using the remark above as well as Lemma 42.26.4 we get
By the uniqueness of Lemma 42.38.2 we conclude that $c_ r(\mathcal{E}) \cap [X] = 0$ and $c_ j(\mathcal{E}) \cap [X] = \alpha _ j = c_ j(\mathcal{F}) \cap [X]$ for $j = 0, \ldots , r - 1$. Hence the lemma holds. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: