Timeline for A complex limit cycle not intersecting the real plane(2)
Current License: CC BY-SA 4.0
16 events
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| Jan 2, 2023 at 19:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 4, 2022 at 19:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 7, 2022 at 18:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 7, 2022 at 17:28 | answer | added | Ali Taghavi | timeline score: 1 | |
| May 2, 2019 at 10:34 | comment | added | Ali Taghavi | @LoïcTeyssier No but your comment was interesting. May be it is an indirect motivation to consider the following example: $\begin{cases}x'=y-x^2\\y'=-x \end{cases} $ Now what about the curve $e^{-2y}(y-x^2+1/2)=i$? What can be said about its holonomy?Is it nontrivial? I appreciate your comments to this question. | |
| Apr 28, 2019 at 19:14 | comment | added | Loïc Teyssier | I'm not convinced that the algebraic nature of the leaf is relevant here. But I already was wrong on this question ;) | |
| Apr 26, 2019 at 8:40 | comment | added | Ali Taghavi | @LoïcTeyssier Now I think that the example i provided in the linked question is somewhat a fake example, in the sense that the leaf is an algebraic leaf. now I am thinking to find an algebraic vector field with a non algebraic complex limit cycle not intersecting the real plane(both real or complex coefficents). As we know a generic algebraic vector field does not have an algebraic leaf.So in this new formulation is the question still an obvious question? | |
| Apr 26, 2019 at 7:25 | comment | added | Ali Taghavi | @LoïcTeyssier Yes. Many thanks for this comment. | |
| Apr 25, 2019 at 14:10 | comment | added | Loïc Teyssier | This case is the same as your example in the complex version of the question. Just consider $$z′=w+(z^2+w^2+4) \\ w′=−z+(z^2+w^2+4)$$. | |
| S Apr 22, 2019 at 13:02 | history | bounty ended | CommunityBot | ||
| S Apr 22, 2019 at 13:02 | history | notice removed | CommunityBot | ||
| Apr 14, 2019 at 11:34 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 1 character in body
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| S Apr 14, 2019 at 11:27 | history | bounty started | Ali Taghavi | ||
| S Apr 14, 2019 at 11:27 | history | notice added | Ali Taghavi | Draw attention | |
| Apr 10, 2019 at 22:06 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
edited body
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| Apr 10, 2019 at 21:53 | history | asked | Ali Taghavi | CC BY-SA 4.0 |