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Assume we are working over complex numbers. Let $\pi:\mathbb{P}^n\times\mathbb{P}^m\rightarrow \mathbb{P}^n (m,n>0)$ be the first projection. Suppose we are given a vector bundle $E$ over $\mathbb{P}^n$ such that $\mathbb{P}(E)=\mathbb{P}^n\times\mathbb{P}^m$ as bundles over $\mathbb{P}^n$ (for example, $E$ might be $\mathbb{P}^n\times\mathbb{C}^{m+1}$).

Let $\mathcal{Q}$ be the universal quotient bundle on $\mathbb{P}^n\times\mathbb{P}^m$ of rank $m$, coming from the short exact sequence $$ \begin{align} 0\to \mathcal{O}_{\mathbb{P}(E)}(-1)\to \pi^*(E)\to \mathcal{Q}\to 0. \end{align} $$

Let $l$ be a line in $\mathbb{P}^n$. Then, in Chow groups, $\pi ^*([l]) = [l\times \mathbb{P}^m]\in CH_{m+1}(\mathbb{P}^n\times\mathbb{P}^m)$.

Question: In $CH_1(\mathbb{P}^n\times\mathbb{P}^m)$, do we have $c_m(\mathcal{Q})\cap[l\times \mathbb{P}^m] = [l\times\{*\}]$?,

where $c_m(\_)$ denotes the $m$-th chern class, and $*$ denotes any point in $\mathbb{P}^m$?

I believe that they should not be equal (e.g. taking $m=n=1$, I feel they should not be equal) , but I can't prove it concretely. Any help/reference would be highly appreciated.

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  • $\begingroup$ What do you mean by $\mathbf P(E) = \mathbf P^n \times \mathbf P^m$? There exists an isomorphism of varieties? Compatible with the projection to $\mathbf P^n$? $\endgroup$ – R. van Dobben de Bruyn Jul 20 '19 at 11:53
  • $\begingroup$ @R.vanDobbendeBruyn : I mean they are isomorphic as $\mathbb{P}^m$-bundles over $\mathbb{P}^n$, as you said. Sorry I forgot to add it. $\endgroup$ – yojusmath Jul 20 '19 at 11:57
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The proposed equality is true, at least when $E$ is the trivial rank $m+1$ bundle on $\mathbb{P}^n$. Otherwise, your requirement that $P(E) \cong \mathbb{P}^n\times \mathbb{P}^m$ forces $E$ to be isomorphic to the $(m+1)$st power of a line bundle on $\mathbb{P}^n$. A reference for this is Corollary 9.5 of Eisenbud and Harris' book 3264.

The following shows how one can compute the class in question assuming $E$ is trivial. By the sum formula for Chern classes and your exact sequence, the total Chern class of $Q$ is $$c(Q) = \frac{1}{c(O_{P(E)}(-1))} = \frac{1}{1 + c_1(O_{P(E)}(-1))}.$$ That is, in the language of Fulton's intersection theory, $c(Q)$ is the Segre class of the tautological line bundle $O_{P(E)}(-1)$. This ratio is interpreted by expanding it out as a power series which is legal since all the Chern classes of index $>0$ are nilpotent in the ring of endomorphisms of ${CH}(\mathbb{P}^n\times\mathbb{P}^m)$ (multiplication corresponds to composition). We obtain:

$$c(Q) = 1 - c_1(O_{P(E)}(-1)) + \ldots + (-1)^m c_1(O_{P(E)}(-1))^m + \ldots,$$

and thus $c_m(Q) = (-1)^m c_1(O_{P(E)}(-1))^m$. So $$c_m(Q)\cap [\mathbb{P}^n\times\mathbb{P}^m] = ([\mathbb{P}^n\times\mathbb{P}^{m-1}])^m = [\mathbb{P}^n\times \{*\}].$$

The intersection product on $\mathbb{P}^n\times\mathbb{P}^m$ is defined so that $$c_m(Q)\cap [l\times \mathbb{P}^m] = (c_m(Q)\cap [l \times \mathbb{P}^m])\cdot [\mathbb{P}^n\times\mathbb{P}^m].$$

Then by Fulton example 8.1.6, this is $$[l\times \mathbb{P}^m]\cdot [\mathbb{P}^n\times \{*\}],$$ giving your result.

Fulton's excellent book is the end-all reference for this technology, though many results are stated in greater generality than what the situations one may most often encounter actually call for; in my experience at least, it can be a somewhat daunting task to specialize them. Chapter 8 defines the intersection product and the rigorous footing for Chern classes and their operations is developed in chapter 3.

If $E$ is nontrivial, then $$c(Q) = \frac{(1 + c_1(L))^{m+1}}{1 + c_1(O_{P(E)}(-1))},$$ for the pullback $L$ of some nontrivial line bundle on $\mathbb{P}^n$ to $P(E)$. It is messier, and I have not worked out the details, but an analogous computation as above should find $c_m(Q)\cap [l\times \mathbb{P}^m]$ in this case. Really, when we write $c(Q)\cap [l\times \mathbb{P}^m]$, this only makes sense if we are treating $Q$ as a bundle on $\mathbb{P}^n\times\mathbb{P}^m$ by applying the pullback map to it induced by the supposed isomorphism $P(E)\cong \mathbb{P}^n\times\mathbb{P}^m$. Then $$c(Q) = \frac{(1 + c_1(L^\prime))^{m+1}}{1 + c_1(O_{\mathbb{P}^n\times\mathbb{P}^m}(-1)) + c_1(L^\prime)},$$ where $L^\prime$ is the pullback of the line bundle on $\mathbb{P}^n$ to $\mathbb{P}^n\times\mathbb{P}^m$.

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