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The Stacks project

Lemma 42.44.2. Let (S, \delta ) be as in Situation 42.7.1. Let X be a scheme locally of finite type over S. Let

0 \to \mathcal{N}' \to \mathcal{N} \to \mathcal{E} \to 0

be a short exact sequence of finite locally free \mathcal{O}_ X-modules. Consider the closed embedding

i : N' = \underline{\mathop{\mathrm{Spec}}}_ X(\text{Sym}((\mathcal{N}')^\vee )) \longrightarrow N = \underline{\mathop{\mathrm{Spec}}}_ X(\text{Sym}(\mathcal{N}^\vee ))

For \alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X) we have

i_*(p')^*\alpha = p^*(c_{top}(\mathcal{E}) \cap \alpha )

where p' : N' \to X and p : N \to X are the structure morphisms.

Proof. Here c_{top}(\mathcal{E}) is the bivariant class defined in Remark 42.38.11. By its very definition, in order to verify the formula, we may assume that \mathcal{E} has constant rank. We may similarly assume \mathcal{N}' and \mathcal{N} have constant ranks, say r' and r, so \mathcal{E} has rank r - r' and c_{top}(\mathcal{E}) = c_{r - r'}(\mathcal{E}). Observe that p^*\mathcal{E} has a canonical section

s \in \Gamma (N, p^*\mathcal{E}) = \Gamma (X, p_*p^*\mathcal{E}) = \Gamma (X, \mathcal{E} \otimes _{\mathcal{O}_ X} \text{Sym}(\mathcal{N}^\vee ) \supset \Gamma (X, \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{N}, \mathcal{E}))

corresponding to the surjection \mathcal{N} \to \mathcal{E} given in the statement of the lemma. The vanishing scheme of this section is exactly N' \subset N. Let Y \subset X be an integral closed subscheme of \delta -dimension n. Then we have

  1. p^*[Y] = [p^{-1}(Y)] since p^{-1}(Y) is integral of \delta -dimension n + r,

  2. (p')^*[Y] = [(p')^{-1}(Y)] since (p')^{-1}(Y) is integral of \delta -dimension n + r',

  3. the restriction of s to p^{-1}Y has vanishing scheme (p')^{-1}Y and the closed immersion (p')^{-1}Y \to p^{-1}Y is a regular immersion (locally cut out by a regular sequence).

We conclude that

(p')^*[Y] = c_{r - r'}(p^*\mathcal{E}) \cap p^*[Y] \quad \text{in}\quad \mathop{\mathrm{CH}}\nolimits _*(N)

by Lemma 42.44.1. This proves the lemma. \square


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