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It is well known that a holomorphic vector field $z'=f(z), z\in \mathbb{C}$ does not have any limit cycle.See the last paragraph of this post

Orbits space of real-analytic planar foliations

One can imagine several reason for this fact. For example: The flow map is a holomorphic map. So if the limit cycle is of period $T$ then $\phi_T$ admits an infinite number of fixed point hence it is identically equal to the identity map.

Now we ask the following question

Is there a holomorphic vector field $z'=f(z)$ on an open region of $\mathbb{C}$ which admit an attractor homoclinic loop?

Ali Taghavi
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2 Answers2

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The answer is 'no' for much the same reason that the OP indicates: the existence of a homoclinic or heteroclinic connection implies that neighboring trajectories are periodic.

First, one needs to have poles in $f$ if one wishes to have heteroclinic connections between (real) saddle singularities. Indeed if $f$ is holomorphic, then the stationnary points of the underlying real vector field are either centers, foci or multiple singularities locally conformally conjugate to $z'=\frac{z^{k+1}}{1+\mu z^k}$. None of these singularities can be part of a candidate limit (poly)cycle.

Next, consider the time-coordinate $$T(z)=\int_{z_*}^z \frac{dz}{f(z)}.$$ Real-time trajectories of the vector field correspond to level sets $\mathrm{Im}(T)=\textrm{cst}$. The presence of a (poly)cycle $\Gamma$ connecting poles of $f$ implies that $\int_\Gamma \frac{dz}{f(z)}=:\tau\in \mathbb{R}$ is the period of $\Gamma$.

Let $\gamma$ be a neighboring trajectory connecting a transverse section $\Sigma$ to itself "after making one turn". Without loss of generality one may choose $\Sigma$ as a small piece of $\mathrm{Re}(T)=\textrm{cst}$. Then there exists a small curve $\sigma\subset\Sigma$ such that the concatenation $\sigma\gamma$ is a closed loop. Cauchy's formula implies that $\int_{\sigma\gamma} \frac{dz}{f(z)}=\tau$. But only $\sigma$ contributes to the imaginary part of the integral, therefore $\sigma$ is just one point and $\gamma$ is closed.

Loïc Teyssier
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  • Thank you very much for your answer to my question. – Ali Taghavi Apr 18 '22 at 10:26
  • I have some comment.: What is a reference for "conformally equivanent to $z'=z^{k+1}/(1+\mu z^k)$ – Ali Taghavi Apr 18 '22 at 10:27
  • Why the latter can not be the vertex of a homoclinic loop? or a part of polycycle? is there a complet classification of these singulaties? – Ali Taghavi Apr 18 '22 at 10:28
  • How do you compute $\int_{\Gamma} dz/f(z)$ but the denominator vanish at vertex? – Ali Taghavi Apr 18 '22 at 10:30
  • Observe that the integral I proposed is performed on a path linking poles of $f$, not zeroes. Hence it is (absolutely) convergent. – Loïc Teyssier Apr 19 '22 at 08:51
  • What about if we consider a zero of $f$? say a degenerate zero $f(z)=a_nz^n+\cdots$ – Ali Taghavi Apr 23 '22 at 10:51
  • I would appreciate if you answer to all my previous comments in particular the reason for "locally conformally equivalent to $z'=z^{k+1}/(1+\mu z^k)$. I guess that you already did some research on these area. could you plz give references? – Ali Taghavi Apr 23 '22 at 11:10
  • The case of a multiple zero is classified by the normal form I gave in the present answer. This is fairly classical (Kostov theorem), see e.g. https://link.springer.com/article/10.1007/s12346-020-00416-y – Loïc Teyssier Apr 25 '22 at 10:21
  • Such multiple points cannot be part of a limit (poly)cyle because of their local phase portraits organized in petals, whose $\alpha$- or $\omega$-limit is the singularity. – Loïc Teyssier Apr 25 '22 at 10:26
  • All 1-dimensional meromorphic singularities of holomorphic vector fields are locally conformally classified, see e.g. https://eudml.org/doc/125297 – Loïc Teyssier Apr 25 '22 at 10:28
  • I enjoyed the information you provided and materials of your answer. thanks for your contribution. – Ali Taghavi Apr 25 '22 at 12:38
  • However I wish to have more concentration on this question. As I told you the initial question mentioned in the post has a very short alternative proof. I am thinking to some possible generalizations(with several view points) – Ali Taghavi Apr 25 '22 at 12:46
  • Thank you again for your very interesting answer and your attention to my question – Ali Taghavi Apr 29 '22 at 16:14
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There is no any kind of, generally speaking, "Charactristic curve" for the vector field $z'=f(z)$ when $f$ is a holomorphic function. By charactristic curve I mean any kind of particular curve which is really rare!. Mathematically speaking: it is isolated in its neighborhood: there is no any thing similar to it, in its small neighborhood. For example a limit cycle, an attracor loop, a UINIQE hetroclinic or homoclinic connection, etc:

>Proof: observe that $[f,if]=0$ and $f$ and $if$ are independent vector fields \QED

Remark: The above 1-line proof can be accepted via the same argument as page 2 item 1 of my paper https://arxiv.org/abs/math/0507516 which says: an attracting limit cycle for a vector field $X$ is an invariant curve for any vector field $Y$ with $[X,Y]=0$. a common observation in investigation of limit cycles, isochronois centers, etc.

When I was asking this MO question I forgot the simple fact that $[f,if]=0$ when $f$ is a holomorphic function. I observed it imeddiate after my first meet with the concept "Lie bracket of vector fields" when I was a master student of the course "Differentiable manifold"(about 27 years ago). In that period I was not familliar with the concept limit cycle but it seems that "limit cycle" was in my inconscient.

Any way I lookforward to hear the answer to my commented questions to other existing answer to this question. a possible answer to my commented questions would help me to understand:

In the holomorphic setting, is it possible we have a familly of homoclinic loop $\gamma_t$ which homotopically vanish at the vertex of homoclinic loop? So this would be a contradiction to possible common belife that in the interior of a homoclinic loop we necessarilly have a singularity(of course different from the vertex).

Ali Taghavi
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  • As I said in my answer, a homoclinic connection cannot occur at a point where $f$ is holomorphic. If you allow poles of $f$, you cannot have such an accumulation, because through a pole only pass finitely many separatrices. If you consider essential singularities (say, $f:=\exp$), you can have such homoclinic loops that accumulate on the singularity (say, $\infty$). – Loïc Teyssier Apr 25 '22 at 12:07
  • Notice though that such isolated curves have «infinite period» (one end of the loop behaves like a saddle, needing a finite amount of time to reach, the other end behaves like a node, at infinite time). – Loïc Teyssier Apr 25 '22 at 12:08
  • See https://www.jstor.org/stable/2691391 – Loïc Teyssier Apr 25 '22 at 12:12
  • @LoïcTeyssier let $\gamma$ be periodic orbit according to Jordan curve theorem it has an interior so thats what I mean by interior – Ali Taghavi Apr 25 '22 at 12:28
  • @LoïcTeyssier i think I explained the proof in my linked note of mine. But let me mention here: Assume $[X,Y]=0$ let $\phi_t, \psi_s$ be the corresponding flow of X,Y. Then $\phi_t \circ \psi_s=\psi_s\circ \phi_t$, namely the twon flows commute. This means that every solution curve of X is maped to another solution curve of X via the flow $\psi_t$ of Y – Ali Taghavi Apr 25 '22 at 12:32
  • @LoïcTeyssier So a limit cucle or an isolated loop of f is maped to another nearby limit cycle of f via the flow of $if$ – Ali Taghavi Apr 25 '22 at 12:34
  • @LoïcTeyssier non triviality of the centralizer of a holomorphic vector field is a significance dynamical fact for consideration of holomorphic vector field. may be it would be interesting in line of your reserach – Ali Taghavi Apr 25 '22 at 12:42
  • @LoïcTeyssier Please look at my linked paper in the body of my answer: inspired by centralizer problem one can ask about the dimension of the centralizer of holomorphic vector fields – Ali Taghavi Apr 25 '22 at 12:50