Proc. Indian Acad. Sci. (Math. Sci.) Vol. 120, No. 4, September 2010, pp. 403–407.
© Indian Academy of Sciences
The rational maps Fλ (z) = zm + λ/zd have no Herman rings
YINGQING XIAO∗ and WEIYUAN QIU†
∗
College of Mathematics and Economics, Hunan University, Changsha 410082, China
School of Mathematical Sciences, Fudan University, Shanghai 200433, China
E-mail: [email protected]; [email protected]
†
MS received 30 January 2010; revised 26 February 2010
Abstract. It is proved that the rational maps in the family {z → zm + λ/zd : λ ∈ C\{0}}
for integers m, d ≥ 2 have no Herman rings.
Keywords.
Julia set; Herman ring.
1. Introduction
According to Sullivan’s classification theorem for dynamics of rational maps on the
Riemann sphere (see [10]), there are only five kinds of periodic Fatou components which
are attracting domains, super-attracting domains, parabolic domains, the Siegel disks and
the Herman rings. It is well-known that polynomials in the complex plane have no Herman
rings. The problem of the existence of Herman rings of a rational map has been studied by
Lyubich in [9], who considered the family F2 = {z → 1 + 1/ωz2 : ω ∈ C\{0}}. He asked
whether the maps in this family have Herman rings. Shishikura [13], using quasi-conformal
surgery techniques, proved that the quadratic rational maps have no Herman rings, thus
solving Lyubich’s question as a particular case. In [1], Bamon and Bobenrieth studied this
question for the family Fd = {z → 1 + 1/ωzd : ω ∈ C\{0}} where d is an integer not less
than 2. Using the purely topological methods, they proved that the maps in the family Fd
also have no Herman rings.
In this paper we consider the existence of Herman rings for the family
λ
Fm,d = z → zm + d : λ ∈ C\{0} ,
z
where m, d are integers such that m, d ≥ 2. This family has recently been the object of a
series of papers [3–6,12,14]. Our main result is as follows:
Theorem 1. The rational maps in the family Fm,d have no Herman rings.
2. Preliminaries
Let C and C̄ denote the complex plane and the Riemann sphere, respectively. Let R: C̄ → C̄
be a rational map of degree d ≥ 2. Let R n be the n-th iterate of the rational map R, namely
R n (z) = R(R n−1 (z)) for n ∈ {1, 2, 3, . . . } and
R 0 (z) = z.
403
404
Yingqing Xiao and Weiyuan Qiu
The Julia set J (R) of R is defined to be the set of points at which the family of iterates of
R fails to be a normal family in the sense of Montel [11]. Equivalently, the Julia set is the
closure of the union of repulsive fixed points of R n for all n ∈ {1, 2, 3, . . . } (see [7,8]).
F (R) = C\J (R) is called the Fatou set of R. It is known that J (R) is a closed set and
F (R) is an open set. A connected component of F (R) is called a Fatou component. The
critical point of R is defined to be the point at which R is not univalent in any neighborhood.
For properties of Julia sets and general background in holomorphic dynamics, we refer to
[2,9,10].
DEFINITION 1.
A Fatou component D is called a Herman ring if D is a topological annulus which is
conformally isomorphic to some annulus A(r, 1) = {z: r < |z| < 1} such that for some
integer p > 0, R p restricted on D is conjugate to an irrational rotation of A(r, 1).
For given integers m, d ≥ 2, we consider the rational maps
Fλ (z) = zm +
λ
∈ Fm,d .
zd
We introduce some basic properties of Fλ which can be found in [5]. Fλ is a rational
map of degree m + d. Points ∞ and 0 are critical points of Fλ of multiples m − 1 and
d − 1, respectively. ∞ is also a super-attracting fixed point and 0 is a pole. So their orbits
are very simple. There are also m + d simple critical points of Fλ which have the form
(dλ/m)1/(m+d) .
There are m+d zeros of Fλ at points (−λ)1/(m+d) , which we call prepoles. Let q0 denote
one of the prepoles and set ω = e2πi/(m+d) . Then the m + d prepoles can be expressed
as qk = ωk q0 , k = 0, 1, . . . , m + d − 1. They are arranged symmetrically on the circle
|z| = |λ|1/(m+d) .
It is easy to verify the symmetry of Fλ :
Fλ (ωz) = ωm Fλ (z).
(1)
This follows that the orbit of ωk z for every integer k behaves symmetrically under iteration
of Fλ . So the orbits Fλn (ωk z) are bounded or tend to ∞ as n → ∞ for k = 0, 1, . . . , m +
d − 1 simultaneously.
The point at ∞ is a super-attracting fixed point of order m for Fλ . By Böttcher’s theorem,
Fλ is conjugate to z → zm in a neighborhood of ∞. There is an immediately attracting
basin B at ∞ which is a Fatou component containing ∞ such that Fλ : B → B and for
every z ∈ B the orbit Fλn (z) tends to infinity as n → ∞. Since Fλ has a pole of order d at
0, there is an open neighborhood of 0 that is mapped d to 1 onto a neighborhood of ∞ in
B. If B contains this neighborhood of 0, then B is completely invariant under Fλ and B
is the only Fatou component. If B does not contain this neighborhood of 0, then there is a
disjoint Fatou component T containing 0 which is mapped d to 1 onto B. In this case, Fλ
is a m to 1 mapping from B onto itself. Since the degree of Fλ is m + d, all points in the
preimages of B lie either in B or in T .
By symmetry (1), it is shown in [5] as follows:
Lemma 1. Both B and T have (m + d)-fold symmetry, i.e., if z ∈ B, then ωk z ∈ B for
every integer k.
The rational maps Fλ (z) = zm + λ/zd have no Herman rings
405
Consequently, we have as follows:
Lemma 2. If D is a Fatou component of Fλ , then ωj D is also a Fatou component for every
integer j , where ωj D = {ωj z: z ∈ D}.
Proof. Suppose that D is a Fatou component of Fλ . Then either Fλn (D) = B for some
integer n ≥ 1 or the orbit Fλn (D) is disjoint from B. For the first case, by symmetry (1),
Fλn (ωj D) = ωk B for some integer k. Then Lemma 1 implies that Fλn (ωj D) = B. For the
second case, again by (1), the orbit Fλn (ωj D) is also disjoint from B. So {Fλn }n>0 forms a
normal family on ωj D. In both cases, ωj D is a Fatou component of Fλ .
2
Lemma 3. Suppose γ is a closed curve and there exists a bounded component of C̄\γ
containing 0, then σ γ ∩ γ = ∅ for all σ = e2πiθ , θ ∈ [0, 1).
Proof. We take two points z1 and z2 on γ such that
d(z1 , 0) = inf d(z, 0)
z∈γ
and d(z2 , 0) = sup d(z, 0),
z∈γ
where d(·, ·) denotes the Euclidean distance in C. Then γ is contained in A(r1 , r2 ) =
{z, r1 ≤ |z| ≤ r2 } where r1 = d(z1 , 0), r2 = d(z2 , 0) and γ connects two boundary
circles of A(r1 , r2 ). For any σ = e2πiθ with θ ∈ [0, 1), we have σ z1 ∈ {|z| = r1 } and
σ z2 ∈ {|z| = r2 }. If none of them lie on the curve γ , then they must lie in different
components of C̄\γ . Since γ is a closed curve in A(r1 , r2 ) surrounding 0, any curve that
connects the boundary circles {|z| = r1 } and {|z| = r2 } must intersect γ . Hence, the arc
2
α ⊂ σ γ that connects σ z1 and σ z2 must intersect γ . This implies σ γ ∩ γ = ∅.
Lemma 4. Suppose D is a Fatou component of Fλ . Suppose also that both z0 and ωj z0
belong to D, where ωj = 1. Then ωk z0 belongs to D for every integer k and, as a
consequence, D has (m + d)-fold symmetry and surrounds the origin.
Proof. Let γ0 be a curve in D which connects z0 and ωj z0 and let γi = ωj i γ0 for every
integer i. Then γi ∈ ωj i D and, by Lemma 2, ωj i D is a Fatou component
for every i.
Since ωj i D ∩ ωj (i−1) D ωj i z0 , we get ωj i D = D for all i. So γ = m+d
i=0 γi forms a
closed curve in D and surrounds the origin 0. Now for every integer k, taking σ = ωk in
2
Lemma 3 gives ωk γ ∩ γ = ∅. We then get ωk D = D.
3. Proof of the theorem
Proof of Theorem 1. Let Fλ (z) = zm + λ/zd , λ ∈ C\{0}. If B contains a neighborhood
of 0, then B is the only Fatou component of Fλ . So it has no Herman rings. Suppose
that the Fatou component T exists and assume that Fλ has a cycle of Herman rings
p
{D0 , D1 , . . . , Dp = D0 } with period p ≥ 1. Then Fλ : D0 → D0 is conjugate to an irrational rotation z → σ z on the annulus A(r, 1) = {z, r < |z| < 1} where σ = e2πiθ , θ is an
irrational number and 0 < r < 1, and Fλ : Dk → Dk+1 is a conformal mapping for every
k = 0, 1, . . . , p − 1. Furthermore, Dk is disjoint from the domain B and all of its preimp
ages under iteration. There exists an Fλ -invariant analytic Jordan curve γ0 in D0 which
corresponds to the circle {|z| = ρ} in A(r, 1) for some 0 < ρ < r. For a given integer
0 ≤ k ≤ p − 1, γk = Fλk (γ0 ) is an analytic Jordan curve in Dk and Fλ maps γk to γk+1
406
Yingqing Xiao and Weiyuan Qiu
in a one-to-one manner, where γp = γ0 . Let Uk denote the bounded component of C\γk ,
k = 0, 1, . . . , p − 1, and Up = U0 . Then Uk are all disjoint from B but Uk ∩ J (Fλ ) = ∅.
We claim that Fλ restricted to Uk is a holomorphic mapping for k = 0, 1, . . . , p − 1.
It is sufficient to prove that every Uk does not contain the pole 0.
In fact, assume that Uk contains 0 for some k. Then Fλ has a pole 0 of order d and 0 is
the only pole in Uk . Now Fλ maps the boundary curve γk of Uk one-to-one to the curve
γk+1 . Since γk+1 is a Jordan curve, the possible winding number of γk+1 with respect to
0 is 0, 1 or −1. By the argument principal, the number of zeros of Fλ in Uk is equal to d,
d + 1 or d − 1. Since d ≥ 2, there exists at least a zero (prepole) in Uk , say q0 ∈ Uk .
Let Vk = Uk \Dk . Since Fλ maps Dk to Dk+1 and Dk+1 is disjoint from T , we have
the prepole q0 ∈ Vk . Note that γk is a Jordan curve and surrounding 0. By Lemma 3,
ωj γk ∩ γk = ∅ for every integer j = 1, 2, . . . , m + d − 1 which implies that Dk is
(m + d)-fold symmetry. Since the boundary of Vk is the inner boundary of Dk which is
also (m + d)-fold symmetry, we get that Vk is (m + d)-fold symmetry. Thus, every prepole
qj = ωj q0 ∈ Vk , j = 0, 1, . . . , m + d − 1, which implies that Fλ has m + d > d + 1
zeros in Uk . This is a contradiction.
Now, we have proved that Fλ is holomorphic in Uk for every integer k = 0, 1, . . . , m +
d − 1. Since Fλ maps γk to γk+1 , by the maximal modulus principal, Fλ maps Uk to Uk+1 .
In particular, Fλn (U0 ) are disjoint from B for all positive integers n. Hence, by Montel’s
theorem, {Fλn } forms a normal family on U0 , which contradicts J (Fλ ) ∩ U0 = ∅.
2
Acknowledgement
This work is supported by NNSF No. 10831004, 10871047.
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