Abstract
I. N. Baker established the existence of Fatou component with any given finite connectivity by the method of quasi-conformal surgery. M. Shishikura suggested giving an explicit rational map which has a Fatou component with finite connectivity greater than 2. In this paper, considering a family of rational maps that A. F. Beardon proposed, we prove that has Fatou components with connectivities 3 and 5 for any . Furthermore, there exists such that has Fatou components with connectivity nine.
1. Introduction and Main Results
By Sullivan’s theorem [1], each Fatou component of a rational map is eventually periodic. Moreover, for periodic Fatou components, there are only four possibilities: attracting basin, parabolic basin, Siegel disk, and Herman ring. Attracting basins and parabolic basins are either simply connected or infinitely connected, a Siegel disk is simply connected, and a Herman ring is doubly connected. However, for nonperiodic Fatou components, the corresponding connectivity may be bigger than two.
For any given , Baker et al. [2] proved that there exists a rational map which has a Fatou component with connectivity by the method of quasiconformal surgery. M. Shishikura suggested giving an explicit example such that it has a Fatou component with finite connectivity greater than two. Beardon [3] investigated the family of rational maps as follows:
He proved the following result.
Theorem A. For sufficiently small , there exists a Fatou component of with connectivity three or four.
At the same time, he claimed that one may be able to compute the connectivity of by further discussion. Qiao and Gao [4] verified that has connectivity three for . Moreover, for any given positive integer , two different families of rational maps were constructed such that one of them has a Fatou component with connectivity (see [4, 5]). However, the degree of rational maps satisfies with those conditions are increased as the number increases. As the first step to study the problem of connectivity number of Fatou components in rational maps space with fixed degree, we just investigate the connectivity of any other Fatou component of as the real parameter varies. In fact, we have the following results.
Theorem 1. Suppose that is defined as in (1); then we have the following.(1)For any , there exist two Fatou components and of with connectivities three and five, respectively. Moreover, is an unbranched covering.(2)There exists such that has one Fatou component with connectivity nine.
Remark 2. In order to draw the graphs of Julia sets and Fatou components of such rational maps in complex plane, we consider its conformal conjugate. Let , and put . It is easy to see that is a superattracting fixed point of for any . By Figure 1, has Fatou components with connectivities three, five, and nine for some since and have the same dynamical properties. Furthermore, by Figure 2, we know that has Fatou components with connectivities eight and fourteen, and we conjecture that for any large integral , there exists such that has a Fatou component with its connectivity bigger than .
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2. Preliminary Lemmas
For the fundamental concepts and classical results of iteration theory of rational maps, see [2, 3, 6, 7]. In order to prove Theorem 1, we need the following four lemmas. Except for Lemma 3, the others are certain modifications of results which have been verified in [3].
Lemma 3 (see [8, Proposition 2.5]). Let be a rational map of a degree larger than one, and let be its one (super)attractive or parabolic cycle of periodic Fatou components. If one of is not simply connected, then contains at least two different critical values for critical points in itself.
In what follows, in . It is easy to see that (resp., ) is a superattracting (resp., repelling) fixed point of . Let be the Fatou component that contains .
Lemma 4. The nonzero critical points of lie outside the circle .
Proof. has exactly eight critical points, that is, and , which are the solutions of the equation . Obviously, lies outside the circle . By a calculation, each satisfies the following equation: where . Putting we can deduce that By (2), the nonzero critical points of lie outside the circle .
Lemma 5. is simply connected and .
Proof. For , we have
If , then ; we can deduce that .
Suppose that meets the circle ; take a point and join to the origin by a curve . It is easy to see that uniformly on ; then there exists a unique positive integer , such that meets the circle , but does not meet for . Let be a point where meets ; we have
It is a contradiction and thus . Obviously, contains only one critical point by Lemma 4; then is simply connected by Lemma 3.
Lemma 6. only consists of two Fatou components, that is, and , which contains a triply connected domain . Here, .
Proof. Take ; by a simple calculation, we have
It is easy to see that ; we have
Moreover, . By Lemma 5, we have and , and there exists at least one Fatou component of . Noting that there are three (resp., two) zeros of in (resp., ), this implies that and .
3. Proofs
Let be a bounded Fatou component of . We denote the connectivity of by and the unbounded component of by . Let and . Moreover, we say that a component of is a component such that . We say that surrounds a point (or a domain ) if it satisfies (or ) and denote by (or ). Denote the number of zeros and poles of in the interior of Jordan curve by and . In order to prove Theorem 1, we need the following propositions. Considering the connectivity of in Lemma 6, we have the following result.
Proposition 7. is a triply-connected domain.
Proof. By a simple calculation, we have Here Note that we can deduce that there exists a point such that ( is a critical point of ). Note that and and for any , we have ; thus . By Lemma 5, it follows that . Moreover, three critical points 0, (twice) lie outside of . Hence, contains at most four critical points. By Lemma 6, since the triply-connected domain and , then . Applying the Riemann-Hurwitz formula to the threefold covering map , we have so . It follows that and , and thus .
By Lemma 5, is bounded. Since is a repelling fixed point, then each component in the preimage of is bounded. In fact, we have the following result.
Proposition 8. Each Fatou component of is bounded.
Proof. By Proposition 7, contains four critical points and they tend to under . Note that ; then the dynamics of are decided by the forward orbit of the critical point in Proposition 7. If , there exists at most one cycle of periodic components which is distinct from by Sullivan’s theorem. Assume that this cycle exists, denoted by ; then
Otherwise, .
Below we will prove that each component of is bounded. From the above analysis, let be any component of ; there exists such that or (if it exists). In order to show that is bounded, we need to prove that is bounded. Suppose that exist, and note that ; we have which are (super)attracting or parabolic components by the Sullivan theorem, and thus . Without loss of generality, . By Lemma 3, is simply connected; then is contained in the bounded component of .
Remark 9. By Proposition 8, and its preimages are buried points; here a point in Julia set is called buried point if it is not on the boundary of any Fatou component.
By Lemma 6 and Propositions 7 and 8, let be the component of which contains . Clearly, , and in the following, we denote ; then . Below we consider the connectivity of .
Proposition 10. consists of three Fatou components with connectivities 5, 3, and 3, respectively.
Proof. We claim that consists of three Fatou components. On one hand, since and , there exists at least one component of in and , respectively. On the other hand, since is monotone increasing from 0 to for and by Lemma 6, there exists a unique component of with and . We claim that
Assume that (15) is true (in what follows, we will return to the proof of this fact later in the proof), and by the definition of interior at the beginning of this section, . Since is a critical point with multiplities and , then is a 3-fold map from to some neighborhood of . Furthermore, we can easily deduce that is the unique component of in (otherwise, is at least a 4-fold map; it is a contradiction). Hence, owing to which is a proper map. Obviously, we have , so the number of components of in and is exactly one, respectively, denoted by and . Clearly, , , and are mutually disjoint preimage components of and .
Below we prove (15). In fact, if does not surround , we distinguish two cases to discuss and get a contradiction.(i)If does not surround , it is easy to see that and , and since has no pole, then , but it is a contradiction to Lemma 6.(ii)If , note that for and , and we can deduce that there exists at least one component (denoted by ) of with . Obviously, and since does not surround . If , then , and by a similar discussion as used in the case of , it is easy to see that , but it is a contradiction to the fact that both and contain some connected components of . If does not surround , we also get a contradiction by a similar discussion of case (i). Hence, we get .
Next we will acquire the connectivity of . Obviously, , and by the Riemann-Hurwitz formula, . Furthermore, we claim that the “free” critical point in Proposition 7 is not contained in , and thus . In order to prove that , we turn to show the stronger result as follows:
Otherwise, assume that . Note that is a real parameter and ; then is symmetric with respect to real axis . We choose a Jordan curve in the interior of such that is very close to and symmetric with respect to , and we take a point in such that is one of the nearest points from ; denote the arc of from to in counterclockwise direction by . Moreover, we can also choose a Jordan arc between and in such that . Set and take ; then is a Jordan curve in and . Since , , and changes by as goes around by argument principle, but changes by since ; it is a contradiction. Hence, we get (16).
By a similar argument as the one used in (15), we can deduce that and . In fact, it is decided by the “similarity” of the Julia sets . By the definition of and Lemma 6, for any , . Let be the largest point of and let be the largest point of ; by Lemma 5, . Furthermore, by Lemma 6 and (16), the unique “free” critical point in Proposition 7 satisfies . Note that is monotone increasing in and monotone decreasing in ; by a calculation, we can easily deduce that for any . It is easy to see that is monotone increasing in from to ; then the equation has only one real root in since for . Since and , we can deduce that the number of bounded components of which intersects with is two (see Figure 3).
For any component in the preimage of , it is easy to see that by the Riemann-Hurwitz formula. Furthermore, the connectivity is decided by the number of critical points in and local degree . It is easy to see that there exists at most one Fatou component which contains the free critical point ; the following Proposition 13 shows that even if there is no critical point in , its local degree may be larger than one. Therefore, we cannot give a complete description of connectivity of this family of rational maps. In order to get Proposition 13, we first consider the number of poles in .
Proposition 11. No preimage component of or surrounds 0, or .
Proof. We argue by contradiction and induction. Let be a preimage component of or ; we distinguish the following three cases to discuss.
(i) Suppose that ; it is easy to see that since and .
(ii) Suppose that ; then or . If , since for any and for any , by symmetry. If , then or there exist two points and . Since and , then surrounds by symmetry.
(iii) Suppose that ; then either or surrounds by the similar proof of (i) and (ii).
In all, if surrounds any of 0, or , then either or surrounds . Note that if surrounds , we have and since and . To get the conclusion in this proposition, it suffices to prove that no preimage component of or surrounds .
Below we prove that no preimage component of surrounds by induction.
Let be a component of . Obviously, we have . Suppose that ; we will get contradictions by discussion.
(iv) Suppose that , and since , we get which contradicts with .
(v) Suppose that , and choosing a Jordan curve in such that , it is easy to know that and or 3. Since and does not surround , the contradiction can be deduced by argument principle.
Hence, any component of cannot surround . Assume that any component of cannot surround . Again, let be a component of . Suppose that ; we still distinguish two cases to discuss.
(vi) Suppose that , and since for any and for any , then surrounds by symmetry which contradicts with the assumption.
(vii) Suppose that , and since is a component of and , then cannot surround by assumption. By a similar analysis as used in the case (v), we also deduce a contradiction.
Therefore, we get that no preimage component of surrounds . By a similar discussion as the above used in , any preimage component of cannot surround .
However, the conclusion in Proposition 11 cannot fit for . For simplicity, the symbol is and for .
Proposition 12. Given that , only one component of surrounds 0, , and . Moreover, only one component of surrounds but does not surround 0 and . In addition, only one component of surrounds but does not surround 0 and .
Proof. By induction and a similar discussion as the one used in Proposition 10, it is easy to get the following conclusion. For any integer , consists of three components: one is contained in , denoted by ; one is contained in , denoted by ; one is contained in , denoted by . Moreover, and . By a similar discussion as used in Proposition 11, we can deduce that any preimage component of or () cannot surround 0 or . Since , then , and are satisfied with conditions of this proposition in turn as follows:
By Proposition 10, we can deduce that , for any . However, for the other components in the preimage of , we have the following result.
Proposition 13. Given that , let be a component of preimage of or , and if , then or .
Proof. By Propositions 11 and 12, contains no pole of , and thus and is a proper map in . We distinguish three cases to discuss the local degree of as follows.(1)If , then by the Riemann-Hurwitz formula. Obviously, .(2)If , then . By the Riemann-Hurwitz formula, , so .(3)If , then by the Riemann-Hurwitz formula. Furthermore, we can deduce that . Otherwise, . Note that has no poles of , and for any bounded components of , the forward components of it in are corresponding to all bounded components of . Hence, ; it is a contradiction to .
Proof of Theorem 1. (1) It is an immediate result of Propositions 7 and 10.
(2) We show that has a Fatou component with connectivity 9.
Put , and by Proposition 7, the “free” critical point satisfies . Set , , , , (); then . Put , and for any , we have
Furthermore, we can deduce that
If necessary, we enlarge or reduce the length of corresponding interval, by a similar discussion as the one used in inequality 5, as follows:
For example,
By Lemmas 5 and 6, and . Note that ; then . Let be the Fatou component that contains , and since , then by Proposition 11, and thus . As has an approximate root , we have . By Proposition 12, we can deduce that cannot surround . By Proposition 13, and , and thus , by the Riemann-Hurwitz formula.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors would like to thank the referees for their valuable suggestions for improving this paper. This research was supported by the National Natural Science Foundation of China (nos. 11371363, 11231009, and 11261002), the Fundamental Research Funds for the Central Universities (no. 2009QS15), and the construct program of the key discipline in Hunan province.