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:) Feb 18 '18 at 5:32
• @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. Jan 20 '19 at 11:29

A revision: Novembre 2020

I am realy indebted to Loic Teyssier for his $$2$$ very valuable comments and suggestions. I summarize his comments as follows:

1. To have a hyperbolic complex limit cycle it is not sufficient you check $$\int_{\gamma} \alpha \neq 0$$ but you should also check that this integral is different from $$2k\pi i$$

2. If your example realy work, you can obtain a similar example with real coefficient if you replace $$z^2+w^2+1=0$$ with $$(z^2+w^2-4i=0)$$.

Now after more than one year of his suggestions, I look at my answer again.

His first comment leads me to compute the corresponding integral again. Then I just realiz that this integral is equal to $$0$$!. More over I realize that not only this example is not appropriate for the purpose of this question but also every possible reform of this example is not appropriate. For example consideration of $$\begin{cases} z'=w+z(z^2+w^2-4i)\\ w'=-z+w(z^2+w^2-4i) \end{cases}$$ does not work. For all these examples the holonomy would be tangent to the identity maps. Hence a relevant question would be that: Are the corresponding leaf $$z^2+w^2=4i$$ is a leaf with non trivial holonomy?

His second comment help me to realize that the following system has a complex limit cycle $$z^2+w^2+1=0$$ which obviously does not intersect the real plane $$\mathbb{R}^2$$. Here is the true example as required as an answer to this post:

$$\begin{cases} z'=w+z(z^2+w^2+1)\\ w'=-z+w(z^2+w^2+1) \end{cases}$$

Finally we include the following question in our answer:

Can a real polynomial vector field possess a hyperbolic complex limit cycle $$\gamma$$ which is not algebraic and does not intersect the real plane?

The previous version of my answer:

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. $$\square$$