Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a rational map $f$ on the Riemann sphere, for their Fatou components, we can calculate the relations between the degree $k=\deg(f|_F)$, connectivity number $n=\mathrm{conn}(F)$ and number of critical points in the component $F$, using the Riemann-Hurwitz formula.

At least this is easy with simply connected components.

For an infinitely connected component things are more complicated. What are the general methods, if any, for knowing the degree of a rational map on an infinitely connected Fatou component?

In an example that I am now dealing with, I have a family of rational maps of degree 3, one superattracting, one repelling and one parabolic fixed point, multiplicities 0, 2 and 1 respectively.

The parabolic immediate basin contains always 2 or three critical points. So there is, all in all, the parabolic basin $A(\infty)$, the superattractive basin $A(1)$ and only possibly one more cycle of Fatou components (since only one free critical point left). I could show that all components outside the parabolic basin $A(\infty)$ are always simply connected.

Now, for the parabolic basin I could show that, if it contains three critical points, then it is infinitely connected and completely invariant, combining the Riemann-Hurwitz formula with some additional topological-combinatorial argument.

The argument is as follows:

Take some domain $V$ in $A^*(\infty)$ containing all three critical values, and a preimage $U$ of $V$ which contains some of the critical points. Since the degree on such $V$ must be at least two (containing a critical point) and the overall degree of $f$ is 3, there can be only one such component $V$, which must contain all three critical points. Let $m=\mathrm{conn}(U)$ and $n=\mathrm{conn}(V)$. The connectivity numbers $m$ and $n$ are the numbers of boundary componens in $U$ and $V$ respectively. Since each component of $\partial U$ is mapped onto a component of $\partial V$, and each component of $\partial V$ has at most $k$ preimage components in $\partial U$ we must have $m\leq kn$. Now inserting the Riemann-Hurwitz formula $m-2=k(n-2)+r \Leftrightarrow m=kn-2k+r+2$ we get: $kn-2k+r+2\leq kn \Leftrightarrow r+2 \leq 2k \Leftrightarrow 2k \geq 5$, so $k\geq 3$.

In another case, $A^*(\infty)$ containing two critical points, I could show that it is simply connected, and again completely invariant.

However, there is also the case, where it contains two critical points but is infinitely connected. I would like to know whether it is completely invariant in this case. Then I would know the most important things about the composition of the Fatou set and the connectivity of the Julia set.

Maybe my methods and patchwork understanding is a bit clumsy. Maybe there are some general useful ways on how to solve such problems?

So, in this case, I have the following idea:

In this paper by Krysztof Baranski (http://www.ams.org/journals/proc/1998-126-06/S0002-9939-98-04184-7/S0002-9939-98-04184-7.pdf) it is shown that for the immediate basin of an attractive fixed point, if it contains $2\deg(f)-4$ or more critical points, then it is completely invariant. So in my case, with $deg(f)=3$, an attractive immediate basin with at least 2 critical points would be completely invariant. Now my simple-minded idea would be: Can we not just perturb the map a little bit, bifurcating the parabolic fixed in one repulsive and one attractive while retaining the topology of the Julia set? Then we could apply Baranski's result to this and show that $A*(\infty)$ is completely invariant.

Are there results on such things? I guess nothing general. But maybe it is applicable to my case. I am not clear on where to look for these things. The term "parabolic implosion" comes up in relation to that. That looks very complicated. So I was glad when I came about this paper by Tomoki Kawahira: http://www.math.nagoya-u.ac.jp/~kawahira/works/rims0402.pdf

To quote:

In the general case, changes from “parabolic” to “hyperbolic” (=“attracting or repelling”), or opposite directions, are not easy as above. The difficulty comes from well-known parabolic implosion, but here we omit > to deal with it. Our main question is:

For a rational map with a parabolic cycle, can we give a way to perturb its parabolic cycle into another kind of cycle without changing most part of the dynamics?

...

Perturbation. A perturbation of a rational map (resp. polynomial) $f$ is a family of rational maps (resp. polynomials) ${f_\epsilon:\epsilon\in[0,\epsilon_0]}$ with some $\epsilon_0>0$ satisfying $f_0=f$; $\deg f_\epsilon = d$ and $d_\hat{\mathbb{C}}(f_\epsilon,f)\to 0$ as $\epsilon\to 0$. For simplicity we represent such a family in convergence form, $f_\epsilon\to f$.

...

Theorem 3.1 (Cui) Let $g$ be a geometrically finite rational map with parabolic cycles. Then there exists a subhyperbolic rational map $f $ satisfying the following:

  • There exist quasiconformal deformations ${f_\epsilon: 0<\epsilon\leq 1}$ of $f$($=f_1$) such that $f_\epsilon \to g$ is a perturbation.
  • Let $H_\epsilon$ denote the quasiconformal conjugacy from $f$ to $f_\epsilon$. Then $H_\epsilon$ converges uniformly as $\epsilon \to 0$ to the limit $h$ which semiconjugates $f$ to $g$.
  • For $y \in \hat{\mathbb{C}}$, $\mathrm{card}(h^{-1}(y))\geq 2$ iff y eventually lands on a parabolic cycle. In particular, such an $h^{-1}(y)$ is either a repelling periodic star like graph or a connected component of its preimages.

Now I am not sure whether this would be enough for me? It is hard for me to decipher, but it gives me the vague impression that it might provide the solution for my case. I guess what is important is the semiconjugacy between the map $g$ with parabolic fixed points and $f$ without parabolic fixed points. Would such a semiconjugacy "preserve most of the dynamics" in the sense that, say, the Julia set stays topologically the same?

The paper talks about star-like graphs and Thurston obstructions and things I don't know of.

Here, as well as in Baranski's paper use is made of "rational-like maps" and quasiconformal methods, of which I don't understand anything. I understand that "rational-like map" refers to some kind of abstract model simplification of a rational map with one designated attractive fixed point.

So I have only very vague ideas. Now, before I get lost in this I would like to ask whether this (Theorem by Cui) would seem to amount to a solution to my problem.

If that is the case I would have to learn about what is necessary to understand it.

Thank you.

*Edit (adding some specific questions about the result by Cui, from the comments to Lasse Rempe's answer): * I don't really understand all the technical language involved here. So some specific questions for me would be for example: 1) If $f$ and $f_\epsilon$ are quasiconformally conjugate, does this mean their Julia sets and Fatou components are homeomorphic and dynamics on each of them topologically conjugate? 2) As the limit $f_\epsilon\to g$ obtained here is only semiconjugate to $f$ by the limit $H_\epsilon \to h$ of quasiconformal conjugacies: Is there anything important that gets lost in this process?

What I would foremostly like to know, really, is whether this result is applicaple "as is".

As my understanding is very vague, I am aware that this might seem difficult. And if the only possible answer should be "learn about these things and figure them out for yourself" then I shall also be satisfied with that.

Thank you very much for your help given already, and I will be grateful for any further hints.

share|improve this question
add comment

1 Answer 1

Your question is rather long and a little bit difficult to parse (for example, in the fifth paragraph, you appear to mean 'critical' rather than 'fixed' points), so I apologize for not having all of it very carefully.

Indeed, there exist general results regarding the perturbation of parabolic to attracting cycles. The result of Cui that you cite here is of this type. I also recall that Peter Haissinsky talked about this kind of construction some time ago, using a technique sometimes known as "trans-quasiconformal surgery". This method often allows you to reduce questions about parabolic basins to those about attracting ones.

Tomoki's article gives some references, which may be a good place to start if you wish to learn about these techniques. I suspect that you will find results that you can apply there, provided that the maps you are interested in are sufficiently nice (e.g. 'weakly hyperbolic' in the sense of Tan Lei and Peter Haissinsky).

EDIT: I think actually that Cui's result as cited MAY give you precisely what you are looking for. If I understand Tomoki's article correctly, a "repelling periodic star-like graph" intersects the Julia set in precisely one point (a repelling periodic point). But it is not 100% clear in the definition. If my interpretation is correct, then the map will actually be a homeomorphism on the Julia set. But you should check this carefully, including that the citation to the original article is accurate.

On the other hand, often the results regarding attracting basins can be carried over, by similar techniques but with some more effort, to parabolic ones. (An example is the original proof by Douady and Hubbard of local connectivity of quadratic polynomial Julia sets in the parabolic case. For maps with an attracting cycle, this is proved using uniform expansion in a suitable hyperbolic metric. For parabolic points, one has to work somewhat harder, but can still find a suitable expansion argument with respect to the right metrics.)

I haven't read Baranski's paper, but would be surprised if his methods did not extend to the parabolic case.

(EDIT: I meant "I WOULD be surprised", not "I wouldn't" as I wrote in the original post.)

"Parabolic implosion" is an important phenomenon, but it shouldn't really apply here, as you are going to perturb in the "good" direction, where the parabolic becomes attracting.

I hope this is of some help!

share|improve this answer
    
Thank you very much, Mr. Rempe! Yes, my question was a bit messy and complex. So thank you especially for affording the patience to respond to it. I corrected "fixed point -> critical point" in the fifth paragraph. Thanks for pointing it out. So it is as probably as I thought: "This seems to be the approach I need". I also don't think Baranski's result would carry over to the parabolic case. So I would have to use some results like that of Cui mentioned here. –  idiot_1337 Oct 17 '13 at 15:07
    
Now, as my situation is somewhat urgent in that I have to finish my diploma thesis within 11 days and I would have to learn about what exactly I have to learn and apply here very fast (yes yes, this seems quite presumptuous...), let me ask again very specifically if the result by Cui mentioned here would seem to amount to a solution already, given that the map that I have is geometrically finite. I don't really understand all the technical language involved here. –  idiot_1337 Oct 17 '13 at 15:08
    
So some specific questions for me would be for example: 1) If $f$ and $f_\epsilon$ are quasiconformally conjugate, does this mean their Julia sets and Fatou components are homeomorphic and dynamics on each of them topologically conjugate? 2) As the limit $f_\epsilon\to g$ obtained here is only semiconjugate to $f$ by the limit $H_\epsilon \to h$ of quasiconformal conjugacies: Is there anything important that gets lost in this process? –  idiot_1337 Oct 17 '13 at 15:08
    
What I would foremostly like to know, really, is whether this result is applicaple "as is". As my understanding is very vague, I am aware that this might seem difficult. And if the only possible answer should be "learn about these things and figure them out for yourself" then I shall also be satisfied with that. Thank you very much for your help given already, and I will be grateful for any further hints. –  idiot_1337 Oct 17 '13 at 15:09
    
See my edits - but I suggest that you should speak with your diploma thesis supervisor about how to proceed. –  Lasse Rempe-Gillen Oct 17 '13 at 21:43
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.