Timeline for FIltrations on a vector bundle on a curve
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2014 at 9:56 | vote | accept | Alexander Braverman | ||
| Aug 17, 2014 at 9:12 | answer | added | Vivek Shende | timeline score: 4 | |
| Aug 17, 2014 at 6:40 | comment | added | Edgardo | I guess this might be the motivation of your question, but for a curve over a finite field this is a consequence of reduction theory, i.e. I think it is precisely the translation of the fact that a "Siegel set" contains a fundamental domain. For that, there is an article of Springer "Reduction theory over global fields." | |
| Aug 17, 2014 at 5:55 | comment | added | Alexander Braverman | No, the point is that if $E$ has rank 2 then the claim is that we can find a line subbundle $E_1$ of $E$ such that $deg(E_1)-deg(E/E_1)$ is not too small (I think you can always make it smaller than $g$ in this case) -- this is true even when the degree of $E$ is very negative. | |
| Aug 17, 2014 at 5:52 | history | edited | Alexander Braverman | CC BY-SA 3.0 |
added 19 characters in body
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| Aug 17, 2014 at 1:33 | comment | added | Tony Pantev | I must be missing something, we have rank two (semi-)stable bundles of arbitrary negative degree. So you can not bound the degree of the line sub-bundles from below universally. Do you want a bound that depends on $g$, $n$, and the degree? | |
| Aug 16, 2014 at 15:10 | comment | added | Alexander Braverman | Yes, if you wish (otherwise the question is empty) | |
| Aug 16, 2014 at 14:26 | comment | added | user54268 | Do you mean to assume $n > 1$? | |
| Aug 16, 2014 at 11:38 | comment | added | Alexander Braverman | Well, it has something to do with Harder-Narasimhan, but it is not quite that. For example, if $n=2$ then the statement is that any rank 2 bundle has a line sub-bundle whose degree is not too small. | |
| Aug 16, 2014 at 11:32 | comment | added | Francesco Polizzi | This reminds me the Harder-Narashiman filtration, where $$\mu(E_i/E_{i-1}) > \mu(E_{i+1}/E_i)$$ and $\mu(F) :=\deg(F) / \textrm{rank}(F)$ is the slope. | |
| Aug 16, 2014 at 9:25 | history | asked | Alexander Braverman | CC BY-SA 3.0 |