Timeline for Subbundle generated by linearly dependent sections
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 24, 2019 at 16:41 | vote | accept | CommunityBot | ||
| Jan 24, 2019 at 16:29 | comment | added | Sasha | 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 $b$. | |
| Jan 24, 2019 at 16:15 | comment | added | user68440 | 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. | |
| Jan 24, 2019 at 16:12 | comment | added | Sasha | 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 $\mathcal{O}(-2) \oplus \mathcal{O}$. | |
| Jan 24, 2019 at 16:03 | comment | added | user68440 | 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 $\mathcal{O}(-2)\oplus\mathcal{O}$? | |
| Jan 24, 2019 at 15:51 | history | answered | Sasha | CC BY-SA 4.0 |