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Sep 10, 2017 at 10:31 comment added Ali Taghavi Thank you very much for your help and interesting answer.
Sep 10, 2017 at 10:30 vote accept Ali Taghavi
Sep 10, 2017 at 10:30 comment added Loïc Teyssier I don't know offhand, but I think it's folklore. Try Il'Yashenko paper on topological rigidity ? I'll try to find a renerence.
Sep 10, 2017 at 10:28 comment added Ali Taghavi From what reference you read the proof of genericity of hyperbolic leaves of SHFC, as you indicated in your answer?could you please mention that reference?
Sep 10, 2017 at 10:25 comment added Loïc Teyssier The argument of Alexandre Eremenko works also for $\mathbb P_2(\mathbb C)$ since the argument only regards a bounded disk : if the function were to assume an infinite value on that disc, you could change the target coordinates so that the new function wouldn't.
Sep 10, 2017 at 10:24 comment added Ali Taghavi I think you are reffering to this already question: mathoverflow.net/questions/237877/…
Sep 10, 2017 at 10:22 comment added Ali Taghavi I had asked a question for existence of an entire solution for vander pol but in that question I was considering C to C^2 not CP^2.
Sep 10, 2017 at 10:19 comment added Ali Taghavi To what extent the entire assumption in the paper of Petrovski Landis paper was essential in their proof? Moreover do you have a downloadable version of the Ilyashenko video?
Sep 10, 2017 at 10:19 comment added Loïc Teyssier In your question you already obtained an example of polynomial vector field with no entire leaf, I don't really know what can be added. Since I don't have Petrovski and Landis paper, I don't know what they claim.
Sep 10, 2017 at 10:17 comment added Loïc Teyssier But of course ! Why wouldn't it be ?
Sep 10, 2017 at 10:16 comment added Ali Taghavi So you consider $\gamma(t)=-1/t$ as a an entire holomorphic map from C to CP^1. yes?
Sep 10, 2017 at 10:13 history edited Loïc Teyssier CC BY-SA 3.0
deleted 230 characters in body
Sep 10, 2017 at 10:13 comment added Loïc Teyssier Yes it's the constant vector field to start with but you twist the $y$-component by a homography so that the foliation is no longer linear in the original coordinates. Ok, that's cheating. Please see my edited answer.
Sep 10, 2017 at 10:07 comment added Ali Taghavi May I ask you to give comment to the linked question, too? Do you have a downladed version of the talk of Ilyashenko? I have difficulity for multiple watching of this video.
Sep 10, 2017 at 10:03 comment added Ali Taghavi Thank you. The first part of your answer, I think, correspond to the vector field $x'=1, y'=0$ but we require a non linear term.(As I indicated in my question). The second part of your answer is very helpful for me. But am I mistaken to think in the paper of Petrovski landis they wote in the first parts of their paper, "every solution is an entire holomorphic function from C to CP^2.(In their paper AMS translation)?
Sep 10, 2017 at 9:40 history answered Loïc Teyssier CC BY-SA 3.0