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Consider the Hirzebruch surface $\mathbb{F}_n = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}(n))\rightarrow\mathbb{P}^1$. The Picard group of $\mathbb{F}_n$ is generated by the class of the negative section $\Gamma$, such that $\Gamma^2 = -n$, and the class of a fiber $F$ of the morphism onto $\mathbb{P}^1$. Furthermore, $\Gamma$ and $F$ generate the effective cone of $\mathbb{F}_n$ so that all divisors of the form $D_{a,b} = a\Gamma + bF$ with $a,b\in \mathbb{N}$ have sections.

Does there exist a closed formula for the dimension of the vector space $H^0(\mathbb{F}_n,D_{a,b})$?

Thank you very much.

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Sure. To compute $H^0$ one can first pushforward to the base $\mathbb{P}^1$. If $a \ge 0$ one obtains $$ p_*\mathcal{O}(a\Gamma + bF) \cong p_*\mathcal{O}(a\Gamma) \otimes \mathcal{O}(b) \cong S^a(\mathcal{O} \oplus \mathcal{O}(-n)) \otimes \mathcal{O}(b) \cong \bigoplus_{i=0}^a \mathcal{O}(b-in), $$ and if $a < 0$ the pushforward is zero. Therefore, $$ \dim H^0(\mathbb{F}_n, \mathcal{O}(D_{a,b})) = \begin{cases} \sum_{i = 0}^{\min(a,\lfloor b/n \rfloor)} (b - in + 1), & \text{if $a \ge 0$},\\ 0, & \text{if $a < 0$}. \end{cases} $$

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  • $\begingroup$ Thank you. Is there also a formula for the case $n = 0$? In this case we could take $C_0$ to be a fiber of the other ruling of $\mathbb{F}_0$. $\endgroup$
    – user114666
    Commented Sep 9, 2021 at 13:32
  • $\begingroup$ For $n=0$, use the Künneth formula: $h^0(D_{a,b})=(a+1)(b+1)$ if $a,b\geq 0$, $=0$ otherwise. $\endgroup$
    – abx
    Commented Sep 9, 2021 at 13:52
  • $\begingroup$ For $n = 0$ the same formula works, and it gives precisely $(a+1)(b+1)$ (for nonnegative $a$ and $b$), as @abx mentioned. $\endgroup$
    – Sasha
    Commented Sep 9, 2021 at 14:40

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