Edit: This is a real coefficient version of the current post.

Is there a polynomial vector field $X$ with complex coefficients on $\mathbb{C}^2$ with the property quoted bellow?

There is a regular leaf $L$ whose holonomy, along at least one closed curve on it, is not trivial but $L$ does not intersect the real part $im (z)=im(w)=0,\;(z,w) \in \mathbb{C}^2$.


A leaf with non trivial holonomy is called a complex limit cycle, according to the terminology used in the video lecture by Ilyashenko described in the following answer:

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

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    $\begingroup$ When I saw your question I did the bet that you would eventually, at some point, edit its tags just to add "limitcycle". It's too predictable :) $\endgroup$ – YCor Feb 18 '18 at 0:30
  • $\begingroup$ @YCor Yes. As you have predicted, I could not resist myself from adding this tag:) $\endgroup$ – Ali Taghavi Feb 18 '18 at 5:32
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    $\begingroup$ I think your question stands even without asking about a non-trivial holonomy. What may come into play is the fact that a non-constant entire function omits at most 2 values. Of course solutions to polynomial ODE are not always entire, but they can be extended almost everywhere, hence my guess is that the answer is no. $\endgroup$ – Loïc Teyssier Mar 18 '18 at 20:02
  • $\begingroup$ @LoïcTeyssier Very interesting point. It is a very good idea. $\endgroup$ – Ali Taghavi Mar 23 '18 at 11:42
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    $\begingroup$ @YCor I will just add that (limitcycle) seems to be renamed to (limit-cycles). I suppose that was your suggestion on meta - although I no longer can see the post, since it was deleted. $\endgroup$ – Martin Sleziak Jan 20 at 11:29

The answer to this question is yes. There is a complex polynomial vector field on $\mathbb{C}^2$ with a complex limit cycle which does not intersect the real plane $im(z)=im(w)=0$.

Consider the differential equation $$\begin{cases}z'=w+(z^2+w^2-4i)\\ w'=-z+(z^2+w^2-4i) \end{cases}$$

The regular leaf $L: z^2+w^2=4i$ of this singular foliation does not intersect the real part of $\mathbb{C}^2$. This leaf, which is topologically a cylinder, has a non trivial holonomy. In fact we have more: there is a closed curve on this leaf whose corresponding holonomy map is a hyperbolic map: namely the holonomy is not tangent to the identity map. Here is the argument:

The hyperbolicity, hence non triviality, of the holonomy of this leaf is a consequence of Theorem 3.2 Page 333 of the paper: First Variation of Holomorphic forms and some applications.

Elaboration: The foliation is defined by $$\omega= (w+(z^2+w^2-4i))dw-(-z+(z^2+w^2-4i))dz=0$$

To apply the theorem 3.2 in the above paper we find a $1-$ form $\alpha$ which satisfies $d\omega=\alpha \wedge \omega$, locally around an appropriate closed curve $\gamma$ in $L$.

Represent the above $1$- form $\omega$ in the form $\omega=Pdw-Qdz$. Then for $$\alpha=(P_z+Q_w)/(P^2+Q^2)(Pdz+Qdw)$$ we have $d\omega=\alpha \wedge \omega$. Note that $P^2+Q^2$ does not vanish on $L$. Now we have to compute $\int_{\gamma} \alpha$, along an appropriate closed curve $\gamma \subset L$, and show that this integral is non zero.

To compute this integral we parametrize the cylinder $L$ with
$$ \phi(t)= \begin{cases} z(t)=t+i/t\\w(t)=t/i-1/t \end{cases}$$ where $$\phi:\mathbb{C}\setminus \{0\}\to \mathbb{C}^2$$ is the global parametrization of $L$. We will see that the desired appropriate curve $\gamma$ is $\phi(S^1)$.

We denote by $\phi^*(\alpha) $, the pull back of $\alpha$ under embedding $\phi$. Now a very simple computation shows that $\int_{S^1} \phi^* \alpha$ is non zero since we obtain a pole of order 1 at the origin. In fact the later integral is $\int_{S^1} 2(z(t)+w(t))(wdz-zdw)$. An straightforward and short computation shows that we have a non degenerate pole, namely a pole of order 1. so the integral does not vanish. So the multiplier $e^{\int _{S^1} \alpha} $ is different from $1$. Then the leaf $L$ is a hyperbolic leaf. $\square$


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