# Section of Hirzebruch Surface Semi Ample

We consider the Hirzebruch surface $$S = \mathbb{P}(\mathcal{E})$$ with locally free sheaf $$\mathcal{E} = \mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(2)$$. By definition of projectivation every section $$i:C \subset S$$ coresponds bijectively to a surjective bundle morphism $$\mathcal{E} \to \mathcal{L}$$ by taking into mind that $$\mathbb{P}(\mathcal{E})$$ represents a functor. Under this condition we have for this section $$C= \mathbb{P}(\mathcal{L})$$.

Futhermore it is known that $$C^2 = deg(\mathcal{L}) - deg(K_{\mathcal{L}})$$ where $$K_{\mathcal{L}} = Ker(\mathcal{E} \to \mathcal{L})$$.

Now back to our situation:

The considerations above provides us two obviously sections:

$$C:= \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}(2))$$ coming from the sequence

$$0 \to \mathcal{O}_{\mathbb{P}^1} \to \mathcal{E} \to \mathcal{O}_{\mathbb{P}^1}(2) \to 0$$

and $$D:= \mathbb{P}(\mathcal{O}_{\mathbb{P}^1})$$ coming from the sequence

$$0 \to \mathcal{O}_{\mathbb{P}^1}(2) \to \mathcal{E} \to \mathcal{O}_{\mathbb{P}^1} \to 0$$

One can show that $$(C \cdot D)=0$$ and $$C^2 = 2$$.

My question is why and how to show that $$C$$ is a semi ample divisor?

Two remarks:

1. by semi ample divisor I mean that there exist a semi ample invertible sheaf $$\mathcal{L}$$ (so for exery $$x \in X$$ there exist $$n \in \mathbb{N}$$ and a global section $$s \in \Gamma(X, \mathcal{L}^{\otimes n}$$) with property $$s(x) \neq 0$$)

2. Here: Divisor with positive Selfintersection Number Semi Ample I asked about a more general case: If we consider a general surface with a nef divisor $$N$$ with $$N^2 >0$$ then - by considering @Asal Beag Dubh's answer - $$D$$ is in general not semi ample.

So my question here is if concretely in the setting above with a Hirzebruch surface the section $$C$$ is semi ample? If yes, why?

• Are you sure you mean $\mathcal{O}_{\mathbb{P}^2}$ etc. and not $\mathcal{O}_{\mathbb{P}^1}$? In the previous case you get a 3-fold and not a surface. – Mohan Oct 4 '18 at 12:05
• @Mohan: Yes, you are right. Thanks. – KarlPeter Oct 4 '18 at 12:18

## 1 Answer

Yes, in this more specific case $$C$$ is semi-ample, in fact globally generated. To show it:

1. Note that it suffices to show that $$O_S(C)$$ has no base points along $$C$$.

2. Use the twisted ideal sheaf seqeuence \begin{align*} 0 \rightarrow O_S \rightarrow O_S(C) \rightarrow O_C(C) \rightarrow 0 \end{align*} to show that every global section of $$O_C(C)$$ comes from a global section of $$O_S(C)$$.

3. Note that $$C$$ is isomorphic to $$\mathbf P^1$$, and hence $$O_C(C)$$ is just $$O(2)$$, which is clearly globally generated.