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What is an example of an entire non constant holomorphic function $\gamma: \mathbb{C} \to \mathbb{C}P^2$ such that the image of $\gamma$ is a leaf of a singular holomorphic foliation of $\mathbb{C}P^2$ arising from a non linear polynomial vector field on $\mathbb{R}^2$ or $\mathbb{C}^2$?

Moreover, is it true to say that every leaf of a singular holomorphic foliation of $\mathbb{C}P^2$ is the image of an entire function defined on whole $\mathbb{C}$.

The above question is included in the following post but it did not get an answer. So I ask it as an independent question

The error in Petrovski and Landis' proof of the 16th Hilbert problem

Ali Taghavi
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For your first question: $dy=y^2$ can be integrated by quadratures. The solutions are homographies.

For your second question: no it is not true. By the uniformization theorem, the universal covering of a leaf $L$ is either the Riemann sphere $\bar {\mathbb{C}}$, the complex line $\mathbb C$ or the unit disc $\mathbb D$. The latter is the generic situation for generic polynomial foliations. Now, any entire, surjective map $f : \mathbb C\to L$ would realize a holomorphic cover. Because $\mathbb C$ is simply connected, it means that $L$ has $\mathbb C$ for universal cover.

Loïc Teyssier
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  • 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)? – Ali Taghavi Sep 10 '17 at 10:03
  • 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. – Ali Taghavi Sep 10 '17 at 10:07
  • 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. – Loïc Teyssier Sep 10 '17 at 10:13
  • So you consider $\gamma(t)=-1/t$ as a an entire holomorphic map from C to CP^1. yes? – Ali Taghavi Sep 10 '17 at 10:16
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    But of course ! Why wouldn't it be ? – Loïc Teyssier Sep 10 '17 at 10:17
  • 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. – Loïc Teyssier Sep 10 '17 at 10:19
  • 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? – Ali Taghavi Sep 10 '17 at 10:19
  • 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. – Ali Taghavi Sep 10 '17 at 10:22
  • I think you are reffering to this already question: https://mathoverflow.net/questions/237877/is-there-an-entire-solution-for-the-van-der-pol-equation – Ali Taghavi Sep 10 '17 at 10:24
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    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. – Loïc Teyssier Sep 10 '17 at 10:25
  • 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? – Ali Taghavi Sep 10 '17 at 10:28
  • 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. – Loïc Teyssier Sep 10 '17 at 10:30
  • Thank you very much for your help and interesting answer. – Ali Taghavi Sep 10 '17 at 10:31