Let $\mathcal{E}$ be a rank two vector bundle on $\mathbb{P}^2$ fitting in the following exact sequence

$$0\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow \mathcal{E}\rightarrow \mathcal{I}_p(-1)\rightarrow 0$$

where $\mathcal{I}_p$ is the ideal sheaf of a point $p\in\mathbb{P}^2$. Then $c_1(\mathcal{E})=-1$ and $c_2(\mathcal{E})=1$.

Let $\pi:X = \mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$, and $-K_X$ the anti-canonical divisor of $X$. Finally, let $C\subset X$ be an irreducible curve such that $\pi(C) = L_p$ is a line through $p\in\mathbb{P}^2$.

How can we compute the intersection number $-K_X\cdot C$ ?


Here are the details.

Let $L_p\cong\mathbb{P}^1$ be a line in $\mathbb{P}^2$ through $p$. Your exact sequence restricted to $L_p$ yields the following exact sequence $$0\rightarrow\mathcal{O}_{\mathbb{P}^1}(1)\rightarrow\mathcal{E}_{|L_p}\rightarrow\mathcal{O}_{\mathbb{P}^1}(-2)\rightarrow 0 $$ and hence $\mathcal{E}_{|L_p}\cong\mathcal{O}_{\mathbb{P}^1}(1)\oplus\mathcal{O}_{\mathbb{P}^1}(-2)$. Therefore, $\pi^{-1}(L_p) = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}(1)\oplus\mathcal{O}_{\mathbb{P}^1}(-2))$ is the Hirzebruch surface $\mathbb{F}_3$. Let $\overline{\xi}$ be the class of the only curve with self-intersection $-3$ on $\mathbb{F}_3$, and let $f$ be the class of a fiber of $\pi_{|\mathbb{F}_3}:\mathbb{F}_3\rightarrow\mathbb{P}^1$. Then if $\widetilde{H} = \pi^{*}\mathcal{O}_{\mathbb{P}^2}(1)$ and $\xi = c_1(\mathcal{O}_X(1))$ we have $\widetilde{H}_{|\mathbb{F}_3} = f$ and $\xi_{|\mathbb{F}_3} = f+\overline{\xi}$.

Since $c_1(\mathcal{E})=-1$ and $c_2(\mathcal{E}) = 1$ the canonical divisor of $X$ is given by $-K_X = 4\widetilde{H}+2\xi$ and we may compute the intersection $-K_W\cdot\overline{\xi}$ inside $\mathbb{F}_3$, that is $$-K_X\cdot\overline{\xi} = (4\widetilde{H}+2\xi)_{|\mathbb{F}_3}\cdot\overline{\xi} = (4f+2(f+\overline{\xi}))\cdot\overline{\xi} = (6f+2\xi) = 6+2(-3) = 0.$$


It seems $C$ is not well-defined. Let $\ell$ be the line such that $\pi(C) = \ell$. Then we have the $\mathbb{P}^1$-bundle $$ \pi'\colon\mathbb{P}(\mathcal{E}|_\ell) \rightarrow \ell. $$ Any section of this bundle will give rise to a curve $C$ as you defined (and indeed, such $\mathbb{P}^1$ bundles admit infinitely many sections). The intersection number $-K_X \cdot C$ will depend on which section you choose.

  • 3
    $\begingroup$ The unique cross-section over $\ell$ with negative self-intersection inside the ruled surface $\mathbb{P}(\mathcal{E}_\ell)$ has self-intersection equal to $-3$. Thus, the relative dualizing sheaf of $\pi$ has degree $+3$ on this curve. Also, the dualizing sheaf of $\mathbb{P}^2$ has degree $-3$ on $\ell$. Altogether, the dualizing sheaf of $X$ has degree $0$ on this cross-section. $\endgroup$ – Jason Starr Oct 10 '17 at 0:46
  • $\begingroup$ What Jason said. $\endgroup$ – Sándor Kovács Oct 10 '17 at 4:47

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