In his "Intersection theory" book Fulton proves in lemma 3.2 the following fact:
if a $E$ is a filtered vector bundle of rank $r$ over $X$ with quotients line bundles $L_i$, $s$ is a section of $E$ and $Z$ is the zero-set of $s$, then for any $k$ cycle $\alpha$ on $X$, there exists a $(k-r)$ cycle on $\beta$ on $Z$ with $$c_r(E) \cap \alpha = \prod_i c_1(L_i) \cap \alpha = \beta \ \ \text{in} \ A_{k-r}(X).$$

Now in example 3.2.16 it is claimed that it holds without the assumption that $E$ is filtered as a consequence of the splitting principle.

I don't quite see how to proceed.

All previous applications of splitting principle first took the flat pullback along $p: \mathbb{P}(E) \to X$. So I first tried considering the following diagram:

$\hspace{6cm}$enter image description here

Then using flat pullback I get $$p^{*} (c_r (E) \cap \alpha) = c_r(p^*E) \cap (p^* \alpha) = \beta$$ where $\beta$ now is an element of $A_{k-1} (p^{-1}(Z))$, since $s$ induces $p^*s$ with zero set $p^{-1}Z$.

I am not quite sure what to do next, how to descend $\beta$ to an element of $A_{k-r} (Z)$ is not clear at all.

One idea I had is to actually use later theorem 3.3, but I had no success with it too.

Any help would be appreciated!

  • $\begingroup$ $p^{-1}Z\to Z$ is just the projective bundle $\mathbb{P}(E|Z)\to Z$ and thus $p^*$ is injective on Chow groups. Notice that $c_r(E)\cap \alpha\in A(Z)$ and that says $\beta$ descends. $\endgroup$ – Mohan Sep 3 '18 at 21:26
  • $\begingroup$ @Mohan, $c_r(E) \cap \alpha$ belonging to $A(Z)$ is exactly what we are trying to show, so how do you assume we know it? $\endgroup$ – Bananeen Sep 7 '18 at 20:14

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