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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

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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.

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  • $\begingroup$ 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)? $\endgroup$
    – Ali Taghavi
    Commented Sep 10, 2017 at 10:03
  • $\begingroup$ 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. $\endgroup$
    – Ali Taghavi
    Commented Sep 10, 2017 at 10:07
  • $\begingroup$ 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. $\endgroup$
    – Loïc Teyssier
    Commented Sep 10, 2017 at 10:13
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    $\begingroup$ But of course ! Why wouldn't it be ? $\endgroup$
    – Loïc Teyssier
    Commented Sep 10, 2017 at 10:17
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    $\begingroup$ 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. $\endgroup$
    – Loïc Teyssier
    Commented Sep 10, 2017 at 10:25

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