# A singular holomorphic foliation of $\mathbb{C}^2$ with a bounded leaf

Is there a polynomial vector field on $$\mathbb{C}^2$$ which possesses a bounded regular leaf? By bounded, I mean a bounded subset of $$\mathbb{C}^2$$.

## 1 Answer

Such a bounded leaf can not exist even if we assume that the vector field is analytic and the leaf is merely non-constant.

Indeed, suppose by contradiction that $$i:\cal F\to \mathbb C^2$$ is a leaf of an analytic vector field $$v$$ that is bounded subset of $$\mathbb C^2$$. Since $$i(\cal F)$$ is bounded, the vector field induces a flow on $$\cal F$$ that is defined for all time. So $$\cal F$$ has an action of $$\mathbb C$$ on it (without fixed points). Hence $$\cal F$$ is either $$\mathbb C$$ or $$\mathbb C^*$$ or an elliptic curve. But neither of these Riemman surfaces admits a bounded holomorphic function. I.e. $$i$$ is constant, contradiction.

PS. Maybe it's better to phrase the above in terms of the following claim.

Statement. Suppose by contradiction that there is a bounded leaf and $$x$$ is a point on it. Then there exists a holomorphic map $$i:\mathbb C\to \mathbb C^2$$ such that $$i(\mathbb C)$$ is tangent to the vector field, $$i(0)=x$$ and $$i(\frac{\partial}{\partial z})=v$$.

This statement holds just because the flow of $$v$$ is defined on the leaf for all times. And it leads to a contradiction.

• thanks and +1for your answer. It is possible that the leaf would be a non Haussdorf Riemann surface. Does your argument work, in this case? Moreover may you more explain about the reason you assume the action is free?Finaly why those three Riemann surface you mentioned are the only Riemann surface admitting free action of C? – Ali Taghavi Apr 14 at 10:06
• Ali, I added a slightly modified version of the argument that should be convincing, here the question of Hausdorfness doesn't appear. – Dmitri Panov Apr 14 at 10:36
• thank you for this new version. So i understand from your answer that the Liouville theorem implies such a leaf does not exist. To be honest, befor your answer, i had no idea of maximal domain of solution in VOMPLEX case. – Ali Taghavi Apr 14 at 11:24
• Yes, you are right. Of course, a solution to an ODE can run of to infinity in finite time and this is what happen for "majority" of polynomial vector fields. But if one assumes that it stays in a bounded region (which as we see is impossible), it would exist for all times. This should follow quickly form the fact that a solution always exists for finite time. – Dmitri Panov Apr 14 at 11:46
• Yes I was aware of this situation for the real case but I was not aware of the complex case. after your answer, I realized that the same real argument works for the complex case. BTW, you pointed out that the situation of " skip to infinity" occur for "majority" of algebraic vector field. This remind me of a theorem by Chicone and Sotomayor, an old paper in JDE, who characterize all complet algebraic vector field in terms of "existence or non existence of hyperbolic sink, source, or hyperbolic saddle. So the generic situation. – Ali Taghavi Apr 14 at 12:51