Timeline for Connectedness of a section of an algebraic bundle
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2014 at 20:30 | vote | accept | aglearner | ||
| Feb 6, 2014 at 18:04 | comment | added | Sasha | If $E = L_1 \oplus \dots \oplus L_n$ is a sum of line bundles and so $s = (s_1,\dots,s_n)$, you can try to argue inductively --- just consider the sequence of subschemes $X = X_0 \supset X_1 \supset \dots \supset X_n = Y$ where $X_i$ is the zero locus of $s_i$ on $X_{i-1}$. Then each time you will have only a long exact sequence of cohomology. This is a way to avoid considering the spectral sequence. | |
| Feb 6, 2014 at 17:50 | comment | added | aglearner | Sasha, thanks again. I realised that in the case that I consider $E$ is just a sum of line bundles. But I suspect indeed that some higher cohomology will not vanish. Since I am not very familiar with the spectral sequence you are talking about, I would like to ask you if in the case when $E$ is a sum of line bundles, the calculation simplifies. Would you advise me some pedagogical reference where I could read about this? Or maybe there is an instructive example worked out somewhere? | |
| Feb 6, 2014 at 15:51 | history | edited | Sasha | CC BY-SA 3.0 |
added 2 characters in body
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| Feb 6, 2014 at 15:50 | comment | added | Sasha | Yes, thank you, I edited the answer. Note also that even if there is a nonzero cohomology, still the zero locus can be connected, if the cohomology is killed in the spectral sequence. | |
| Feb 6, 2014 at 15:02 | comment | added | aglearner | Sasha, many thanks! Maybe this is what will work for me. Do I understand correctly that one should check $H^i(X,\Lambda^iE^*)=0$ (i.e. you forgot to put "*")? | |
| Feb 6, 2014 at 10:41 | history | answered | Sasha | CC BY-SA 3.0 |