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Let $E$ be an ample rank $r\geq2$ vector bundle over a smooth projective surface $X$ defined on an algebraically closed field $\mathbb{K}$ of characteristic $0$.

In Kleiman S. L. - Ample Vector Bundles on Algebraic Surfaces, Proc. Amer. Math. Soc. 21 (1969) 673--676, it has been proved that $$ \int_Xs_2(E)=\int_Xc_1(E)^2-c_2(E)>0; $$ $s_2(E)$ is the second Segre class of $E$.

The author proves that in $\mathbb{P}(E)$ there exists a smooth surface $H$ which is finite over $X$ via the canonical projection. As usual, let $\xi=c_1\left(\mathcal{O}_{\mathbb{P}(E)}(1)\right)\in CH^1(\mathbb{P}(E))$, by hypothesis $\xi$ is ample and satisfies the Grothendieck equation $$ \xi^r-c_1(E)\xi^{r-1}+c_2(E)\xi^{r-2}=0 $$ where I skip the obvious pull-backs. He proves the following statements: $$ \exists n\in\mathbb{N}_{\geq1}\mid[H]=(n\xi)^{r-1}\in CH^{r-1}(\mathbb{P}(E)),\,\xi-c_1(E)\neq0\in CH^1(\mathbb{P}(E));\,\forall a\in CH^1(X),(\xi-c_1(E))\cdot a=0. $$ Let $i\colon H\hookrightarrow\mathbb{P}(E)$ be the inclusion, from the inequality $\displaystyle\int_Hi^{*}\xi^2>0$ he deduces the claim.

Why? What is the trick?

If $r=2$ I understood everything, but for $r=3$ I do not.

Thank you to whoever helps me.

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    $\begingroup$ The steps are explained quite clearly in Kleiman's paper. What is it you don't understand? $\endgroup$
    – abx
    Commented Feb 11, 2023 at 7:13
  • $\begingroup$ @abx Yes, it's true. My trouble is the last step: I don't understand how to prove $\displaystyle\int_Hi^{*}\xi^2>0\Rightarrow\int_Hs_2(E)>0$! $\endgroup$
    – Armando j18eos
    Commented Feb 11, 2023 at 7:19
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    $\begingroup$ Kleiman: "by (3.1) and (3.2) with $a=c_1$". $\endgroup$
    – abx
    Commented Feb 11, 2023 at 8:21
  • $\begingroup$ @abx My attempt: $\displaystyle0<\int_Hi^{*}\xi^2=\int_Hi^{*}(\xi-c_1(E)+c_1(E))^2=\int_Hi^{*}(\xi-c_1(E))^2+i^{*}c_1(E)^2=\int_Hi^{*}(\xi-c_1(E))i^{*}\xi+i^{*}c_1(E)^2$ if $r=2$ I find $\displaystyle0<\int_Hi^{*}c_1(E)^2-i^{*}c_2(E)=\int_Hi^{*}s_2(E)$. As I have wrote: I don't understand the Kleiman's reasoning. $\endgroup$
    – Armando j18eos
    Commented Feb 11, 2023 at 11:14
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    $\begingroup$ $(c_1^2-c_2)\cdot \xi ^{r-1}= c_1^2\cdot \xi ^{r-1}-c_1\cdot \xi ^r+\xi ^{r+1}=$ $c_1\cdot \xi ^{r-1}\cdot (c_1-\xi )+\xi ^{r+1}$. The first term is zero by (3.2), and $\xi ^{r+1}>0$. $\endgroup$
    – abx
    Commented Feb 12, 2023 at 7:39

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