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It is easy to see that for any entire function $f : \mathbb C \to \mathbb C$, its graph $G(f) = \{(z,f(z)) \in \mathbb C^2 \mid z \in \mathbb C\}$ can be translated by $(0,c)$ for any $c \in \mathbb C$, so that all the translated graphs $\{G(f) + (0,c) \mid c \in \mathbb C\}$ form the leaves of a holomorphic foliation of $\mathbb C^2$.

If $V : \mathbb C^2 \to \mathbb C^2$ defines a holomorphic and nowhere zero vector field on $\mathbb C^2$, then there is a unique non-singular holomorphic foliation of $\mathbb C^2$ by holomorphic curves that are everywhere tangent to $V$. These are the orbits of the local $\mathbb C$-action that $V$ induces on $\mathbb C^2$. It is known that the leaves of such a foliation must be topologically either planes or cylinders.

Is it known exactly which holomorphic curves $X \subset \mathbb C^2$ can be a leaf of a non-singular holomorphic foliation of $\mathbb C^2$?

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    $\begingroup$ How do you know that $V$ induces an action of the additive group $\mathbb C$? Don't you just get a (holomorphic) local flow in general, rather than a flow? $\endgroup$ Commented Oct 25, 2023 at 20:39
  • $\begingroup$ Good point; it may not be all of ℂ, just a local ℂ-action. (Though there is always a holomorphic foliation induced.) I will modify the question. $\endgroup$ Commented Oct 25, 2023 at 22:14
  • $\begingroup$ It was pointed out to me, I think by @TheAmplitwist, that using Unicode like ℂ instead of TeX like $\mathbb C$ makes questions unsearchable—I guess the search engine only sees ASCII. So I think that it is friendlier to use TeX at the very least in the body. $\endgroup$
    – LSpice
    Commented Oct 25, 2023 at 22:22
  • $\begingroup$ I would if I knew how, but I don't. $\endgroup$ Commented Oct 25, 2023 at 23:01
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    $\begingroup$ Thanks, @C7X!.. $\endgroup$
    – LSpice
    Commented Oct 27, 2023 at 14:19

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