Determine the splitting type of a bundle with cokernel as $\mathcal{O}_2p$ or $\mathcal{O}_p\oplus\mathcal{O}_q$

Question

In my situation, I want to determine the splitting type of a rank 2 vector bundle $\mathcal{F}$ fitted in an exact sequence $$0\rightarrow\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(-2)\rightarrow\mathcal{F}|_{\mathbb{P^1}}\rightarrow\mathcal{O}_{2p}\rightarrow 0,$$ and at the same time when $p\ne q$ are the points of degree 1 in $\mathbb{P^1}$, $$0\rightarrow\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(-2)\rightarrow\mathcal{G}|_{\mathbb{P^1}}\rightarrow\mathcal{O}_{p}\oplus\mathcal{O}_{q}\rightarrow 0.$$

I think that the candidates of splitting type of each case are (2, -2), (1, -1), or (0, 0).

My question is that :

Q1. How to determine which is the type of bundle in each case?

Q2. If cannot in just this situation, what are the additional conditions to achieve?

Q3. Those cases are actually the different situation? Means that could I have $\mathcal{F}|_{\mathbb{P^1}}\ncong\mathcal{G}|_{\mathbb{P^1}}$?

Thank you for your interest.

Answer 1

Consider the dual sequence twisted by $\mathcal{O}(-1)$: $$ 0 \to F^\vee(-1) \to \mathcal{O}(-1) \oplus \mathcal{O}(1) \to \mathcal{O}_{2p} \to 0. $$ From the degree reasons it follows that $F^\vee(-1) \cong \mathcal{O}(-1-a) \oplus \mathcal{O}(-1+a)$ for some $a \ge 0$, and clearly $$ a = \dim H^0(\mathbb{P}^1, F^\vee(-1)) = H^1(\mathbb{P}^1, F^\vee(-1)). $$ On the other hand, from the exact sequence we see \begin{align*} H^0(\mathbb{P}^1, F^\vee(-1)) &= \mathrm{Ker}( H^0(\mathbb{P}^1, \mathcal{O}(-1) \oplus \mathcal{O}(1)) \to H^0(\mathbb{P}^1, \mathcal{O}_{2p}) ) \\ &= \mathrm{Ker}( H^0(\mathbb{P}^1, \mathcal{O}(1)) \to H^0(\mathbb{P}^1, \mathcal{O}_{2p}) ). \end{align*} Now, there are three options for the morphism $\mathcal{O}(1) \to \mathcal{O}_{2p}$:

• an epimorphism with kernel $\mathcal{O}(-1)$;
• a composition $\mathcal{O}(1) \to \mathcal{O}_{p} \to \mathcal{O}_{2p}$ with kernel $\mathcal{O}$;
• the zero map with kernel $\mathcal{O}(1)$.

These correspond to cases $a = 0$, $a = 1$, and $a = 2$, respectively.

A similar analysis applies to the case of vector bundle $G$ and gives exactly the same three possibilities.