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On $\mathbb{P}^1$ consider the trivial bundle $\mathcal{O}\oplus \mathcal{O}$, and the subbundle $\mathcal{L}_{a,b}\subset\mathcal{O}\oplus \mathcal{O}$ that on an open subset $U$ of $\mathbb{P}^1$ is generated by the local sections $(f,g)$ of $\mathcal{O}\oplus \mathcal{O}$ such that $af(p)+bg(p)=0$, where $p\in\mathbb{P}^1$ is a point and $a,b$ are scalars. How can we write $\mathcal{L}_{a,b}$?

For instance if $a =1, b=0$ then the condition is just $f(p)=0$ and hence $\mathcal{L}_{1,0} = \mathcal{O}(-1)\oplus\mathcal{O}$. If $a=0, b=1$ the condition is $g(p)=0$ and hence $\mathcal{L}_{0,1} = \mathcal{O}\oplus\mathcal{O}(-1)$. But what is for instance $\mathcal{L}_{1,1}$?

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$\mathcal{L}_{1,1}$ is still $\mathcal{O}(-1) \oplus \mathcal{O}$. Actually, this is the only subsheaf in $\mathcal{O} \oplus \mathcal{O}$ of rank 2 and degree $-1$.

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  • $\begingroup$ Now, let's say we do the same thing starting with $\mathcal{O}(-1)\oplus\mathcal{O}$ instead of $\mathcal{O}\oplus\mathcal{O}$. Do we get $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$ or $\mathcal{O}(-2)\oplus\mathcal{O}$? $\endgroup$
    – user68440
    Commented Jan 24, 2019 at 16:03
  • $\begingroup$ That now depends on the scalars $a$ and $b$. Typically you will get $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$, but sometimes (in fact, when $b = 0$), you will get $\mathcal{O}(-2) \oplus \mathcal{O}$. $\endgroup$
    – Sasha
    Commented Jan 24, 2019 at 16:12
  • $\begingroup$ If I understand correctly you are saying that for $a,b$ general we get $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$. Could you please give me an intuitive idea of why this is the case? Thank you. $\endgroup$
    – user68440
    Commented Jan 24, 2019 at 16:15
  • $\begingroup$ Any sheaf on $\mathbb{P}^1$ with no cohomology is a sum of $\mathcal{O}(-1)$. Now your vector bundle (say $E$) comes in an exact sequence $0 \to E \to \mathcal{O}(-1) \oplus \mathcal{O} \to \mathcal{O}_p \to 0$ (the last term is the structure sheaf of the point $p$), so the only question is to understand the map $H^0(\mathcal{O}(-1) \oplus \mathcal{O}) \to H^0(\mathcal{O}_p)$. This turns out to be given by $b$. $\endgroup$
    – Sasha
    Commented Jan 24, 2019 at 16:29

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