A complex limit cycle not intersecting the real plane

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 $$im (z)=im(w)=0,\;(z,w) \in \mathbb{C}^2$$.

Note:

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

• 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 :) – YCor Feb 18 '18 at 0:30
• @YCor Yes. As you have predicted, I could not resist myself from adding this tag:) – Ali Taghavi Feb 18 '18 at 5:32
• 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. – Loïc Teyssier Mar 18 '18 at 20:02
• @LoïcTeyssier Very interesting point. It is a very good idea. – Ali Taghavi Mar 23 '18 at 11:42
• @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. – 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. $$\square$$