Timeline for Algebraicity and non-algebraicity of leaves of the characteristic foliation
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 20, 2016 at 13:17 | vote | accept | lks8271 | ||
| May 18, 2016 at 22:44 | comment | added | Jorge Vitório Pereira | Concerning your first question not much comes to mind. For the second question the answer is no (one algebraic leaf implies every leaf is algebraic ) if you assume that the $1$-form has no zeros. You have just to observe that you can find a primitive for $1$-form at an analytic neighbourhood of any algebraic leaf and one can choose these neighbourhoods foliated by algebraic leaves. If you allow singularities then you can easily produce examples with algebraic and non-algebraic leaves by blowing-up. | |
| May 18, 2016 at 16:32 | comment | added | lks8271 | @Jorge Vitório Pereira Thank you for the answer! And can something specific be said about the algebraic leaves if it is known that some of them are algebraic and some are not(not in your example but in general)? And, sorry for the off-topic, but can this situation(algebraic and non-algebraic leaves) occur when the dimension $n$ is $1$ and the foliation is genereated by a global holomorphic 1-form on a surface? | |
| May 18, 2016 at 11:05 | comment | added | Jason Starr | Just to mention one more thing about this question: the uniruledness conjecture (and thus the abundance conjecture) predicts that whenever $\mathcal{O}_X(D)|_D$ is not (eventually) effective, then the characteristic foliation is algebraically integrable with rational curves as leaves. For small dimensions this is known, but I believe this is wide open starting at dimension $6$. | |
| May 18, 2016 at 3:25 | history | edited | Jorge Vitório Pereira | CC BY-SA 3.0 |
The previous version contained an unnecessary argument to deal with divisors on $4$-folds.
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| May 18, 2016 at 2:59 | history | answered | Jorge Vitório Pereira | CC BY-SA 3.0 |