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.

  • $\begingroup$ 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

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 $\mathbb{P}^n\times\mathbb{P}^m$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.