# Question on top Chern class of ample vector bundle

In W. Fulton, R. K. Lazarsfeld - Positive polyonomials for vector bundles, Ann. of Math. 118 (1983) 35-60, they prove the positivities of the top Chern class of a rank $$r$$ ample vector bundle $$E$$ over a smooth complex projective variety $$X$$ of dimension $$n\leq r$$ (Theorem in Appendix B), that is $$\displaystyle\int_Xc_n(E)>0$$.

Let $$\mathbb{P}\equiv\mathbb{P}\left(E^{\vee}\right)$$, by hypothesis $$\xi=c_1\left(\mathcal{O}_{\mathbb{P}}(1)\right)$$ is the Chern class of an ample line bundle, and by construction $$\xi^r-\xi^{r-1}\pi^{*}c_1(E)+\dotsc+(-1)^n\xi^{r-n}\pi^{*}c_n(E)=0\in H^{2r}\left(\mathbb{P},\mathbb{Z}\right)$$ where $$\pi:\mathbb{P}(E)\to X$$ is the obvious projection.

They consider the class $$\alpha=\xi^{n-1}-\xi^{n-2}\pi^{*}c_1(E)+\dotsc+(-1)^{n-1}\pi^{*}c_{n-1}(E)\in H^{2(n-1)}(\mathbb{P},\mathbb{Z})$$ and state $$\pi_{*}\left(\left(\xi^{r-n}\smile\alpha\right)\frown[\mathbb{P}]\right)=[X]$$; where $$[\cdot]$$ is the fundamental class of $$\cdot$$ in homology, $$\smile$$ is the cup product in cohomology, and $$\frown$$ is the cap product between cohomology and homology.

I do not object on the rightness of this equality; the point is: I am not able to figure out the geometric meaning of this equality!

Roughly speaking: how can I compute this?

I am sorry for the poor clarity, but I am very confused by all this.

## 1 Answer

You can compute this by the projection formula. $$\pi_* (\xi^{r-1-k} . \pi^* c_k(E)) = (\pi_* \xi^{r-1-k}) . c_k(E)$$ Note $$\xi^{r-1-k} \in H^{2(r-1-k)}(\mathbb{P}) \cong H_{2(n+k)}(\mathbb{P})$$, so when $$k \geq 1$$, the pushforward class $$\pi_* \xi^{r-1-k} = 0$$. Therefore the equality claimed is just $$\pi_* \xi^{r-1} = [X]$$. Geometrically the class $$\xi^{r-1}$$ is the class of a point in each fibre $$\mathbb{P}^{r-1} = \mathbb{P}(E_x^{\vee})$$, so pushing this class to $$X$$ you get the fundamental class of $$X$$.

• Thank you for your nice and clear answer! Now the geometric meaning of these objects is less mysterious for me. Just another one stupid question (I am squabble with algebraic topology): why is $\pi_{*}\xi^{r-1-k}=0$ for $k\geq1$? Nov 25, 2022 at 14:58
• @Armandoj18eos Pushforward should preserve the dimensions of the classes. Since $\xi^{r-1-k}$ is a class of dimension $2(n+k)$ and $H_{i}(X) = 0$ for $i > 2n$, the pushforward must vanish when $k \geq 1$.
– SLX
Nov 25, 2022 at 18:39
• Oh yes: the push-forward is to cohomology of $X$. Thank you so much! Nov 25, 2022 at 22:00