Herman Rings and Arnold Disks

For (λ,a)∈ C* × C, let fλ,a be the rational map defined by fλ,a(z) = λ z2 (az+1)/(z+a). If α∈ R/Z is a Brjuno number, we let Dα be the set of parameters (λ,a) such that fλ,a has a fixed Herman ring with rotation number α (we consider that (e2iπα,0)∈ Dα). Results obtained by McMullen and Sullivan imply that, for any g∈ Dα, the connected component of Dα(C* × (C/{0,1})) that contains g is isomorphic to a punctured disk.


Introduction
In this paper, we are mainly concerned with the dynamics of rational maps of the form f λ,a (z) = λz 2 az + 1 z + a , λ∈ C * , a ∈ C.
Note that f λ,a is conjugate to f λ,−a via the conjugacy z −→ −z. If λ ∈ S 1 and a = 0, then the map f λ,a is the rotation z −→ λz. Observe that when a is real and |λ| = 1, the map f λ,a is a Blaschke fraction z −→ λz 2 (z + b)/(1 +bz) with b = 1/a. However, as opposed to families of Blaschke fractions which only depend R-analytically on parameters, our family depends C-analytically on the parameters λ and a and is, in some sense, the simplest one that exhibits families of Herman rings. For all α ∈ R/Z, we denote by R α the rigid rotation of the complex plane R α (z) = e 2iπ α z. When λ ∈ S 1 and a ∈ (−1/3, 1/3), the map f λ,a restricts to a diffeomorphism of S 1 which has a rotation number ρ(λ, a) ∈ R/Z. Given a ∈ (−1/3, 1/3), the function t −→ ρ(e 2iπ t , a) is continuous and weakly increasing [15]. Moreover, for each fixed a ∈ (−1/3, 1/3) and for each irrational number α, there is a unique angle t ∈ R/Z such that ρ(e 2iπ t , a) = α (see for example [6]). By a theorem of Denjoy [7], when α = ρ(λ, a) is irrational, f λ,a : S 1 −→ S 1 is topologically conjugate to the rotation R α : S 1 −→ S 1 .    These sets are called Arnold tongues, and they intersect the line {a = 0} at the point t = α. If α is a rational number, the Arnold tongue T α has interior, and if α is irrational, the Arnold tongue T α is a Lipschitz curve [1].
If α ∈ R/Z is a Brjuno number and if (t, a) belongs to T α with a sufficiently close to 0, then the restriction of f e 2i π t ,a to S 1 is R-analytically conjugate to the rotation of angle α [3]. This conjugacy extends to a conjugacy in a neighborhood of S 1 , and so f e 2i π t ,a has a fixed Herman ring. Figure 2 shows an example of such a Herman ring for a = 1/4 and t = 0.615 173 215 88 . . . . Even in the case when λ / ∈ S 1 and a / ∈ (0, 1/3), the rational map f λ,a may have a fixed Herman ring. However, the unit circle is no longer invariant, and it is more difficult to locate parameters for which one can find a Herman ring.
It is known that in this particular family f λ,a , there is a Herman ring if and only if the rotation number is a Brjuno number (see for example [18] or [12]). This result is proved using a surgery construction due to Shishikura, and the optimality of the Brjuno condition for the existence of Siegel disks in the family of quadratic polynomials. More precisely, it uses the fact that the quadratic polynomial P α : z −→ e 2iπ α z(1 + z) is linearizable at 0 if and only if α is a Brjuno number. The sufficiency of this condition was proven by Brjuno [3] and Rüssmann [17], whereas the optimality was proven by Yoccoz [20]. Figure 3 shows the Siegel disk of the quadratic polynomial P α for α = ( √ 5 − 1)/2. This paper studies the set of complex parameters (λ, a) for which f λ,a has a fixed Herman ring with a given rotation number α. (If λ = e 2iπ α and a = 0, we consider that there is a Herman ring which is equal to C * .) Definition 1.1. Given a Brjuno number α ∈ R/Z, we let ∆ α be the Siegel disk of the quadratic polynomial P α and we let D α be the set of parameters (λ, a) ∈ C * × C such that f λ,a has a fixed Herman ring with rotation number α. We call D α the Arnold disk of rotation number α.
In some sense, this set is the complexification of the Arnold tongue T α , and it was studied for general families from a local point of view in [16]. The term Arnold disk is justified by the following theorem, which is proved in Section 4. We first recall the definitions of conformal radius and modulus.
If A is a round annulus A = {z ∈ C | r < |z| < R}, then the modulus of A is mod(A) = (1/2π) log(R/r). If H is any annulus conformally equivalent to A, then we define its modulus to be equal to mod(A).
Let us give an intuitive idea of how the parameterization F α (δ) = (λ(δ), a(δ)) is chosen. For any δ ∈ D, the map f Fα (δ ) possesses a Herman ring whose modulus is determined by |δ|; the argument of δ corresponds to the twist parameter, which roughly indicates how much one boundary of the ring is rotated with respect to the other (see Section 2 and Figure 7). Maps in the Arnold tongue T α correspond to δ ∈ (0, 1). (This parameterization was given in [12] for the rational family and [9] for the complex standard family.) Theorem B. Assume that α is a Brjuno number. Then the following hold. (a) The Arnold disk can be locally parameterized by a in a neighborhood of (e 2iπ α , 0) (that is, it is locally the graph of a holomorphic map a −→ λ(a)).
(b) As |a| → 0, the modulus m a of the Herman ring of f λ(a),a satisfies e π m a = r α |a| + O(a).
Assume that α is a Brjuno number. For a ∈ (0, 1/3), let t a ∈ R/Z be the unique parameter such that (t a , a) ∈ T α , and let m a be the modulus of the Herman ring of f e 2i π t a ,a (m a = 0 if there is no Herman ring). Then, as a → 0, we have This improves the estimate one would get by using the techniques developed by Fagella, Seara and Villanueva [10] in the case of the complex standard family.
In the last section, we explain how to adapt the arguments to the case of the complex standard family f λ,a (z) = λze (a/2)(z −1/z ) .
The following is work in progress. Theorem A describes the topology of any given Arnold disk and hence a natural question to ask is how these disks coexist in C 2 when we move the rotation number. We think that there is a nice lamination: if (α n ) n 1 is a sequence of Brjuno numbers converging to a Brjuno number α 0 such that lim n →∞ r α n = r, then the sequence of maps F α n converges uniformly on every compact subset of D to the map is well defined because the conformal radius r α depends upper semi-continuously on α, and so r r α 0 ). We think we can use this result to prove the existence of Brjuno numbers α ∈ R/Z such that the boundary of D α is a C ∞ Jordan curve and such that for all (λ, a) ∈ ∂D α , the map f λ,a restricts to a diffeomorphism of a C ∞ Jordan curve and is C ∞ (but not R-analytically) conjugate to the rotation of angle α on this curve. These results will appear in a forthcoming paper.

Preliminaries
In the whole of this section, we assume that α is a Brjuno number and that (λ, a) is in D α . For simplicity, we set f = f λ,a . By the definition of D α , the map f has a fixed Herman ring H on which it is conjugate to the rotation R α . More precisely, let m be the modulus of the Herman ring H and set r = e −2π m . Moreover, denote by A r the round annulus Then there exists an isomorphism φ : A r −→ H which conjugates R α : A r −→ A r to f : H −→ H and preserves the orientation in C of the invariant curves (that is, it sends S 1 to the outer boundary of H). This isomorphism is unique up to precomposition with a rotation centered at 0.

Basic properties
The rational map f λ,a has four critical points. One is fixed at 0, and one is fixed at ∞. Therefore there are only two free critical points. If f λ,a has a Herman ring, then the orbits of those critical points are trapped: the closure of those orbits must contain the boundary components of the Herman ring. In particular, the immediate basins of the superattracting fixed points at 0 and ∞ are simply connected. We claim that the Herman ring separates those two basins.
We first claim that d ∞ = 1 and d 0 = 2. Indeed, the image of γ (which is γ itself) turns exactly once around every point in V 0 . Therefore, by the argument principle, d 0 − d ∞ = 1. Since ∞ has at most one preimage in V 0 , we see that the only possibilities for (d ∞ , d 0 ) are (0, 1) or (1,2). The case (0, 1) is not possible, since otherwise, f : V 0 −→ V 0 would be an isomorphism, V 0 would be contained in the Fatou set of f , and this would contradict the fact that it contains a boundary component of the Herman ring H.
It follows that U ∞ is a topological disk compactly contained in V 0 and U 0 is a topological annulus. By the Riemannin-Hurwitz formula, U 0 contains two critical points of f . The two remaining critical points of f are contained in an annulus Figure 4).
We must finally show that 0 ∈ U 0 , which will complete the proof. This last result is more subtle, since we must use a dynamical characterization of 0: it is a superattracting fixed point. There are several possible arguments. It is possible to show that each component ofĈ \ γ must contain a critical point whose orbit accumulates a boundary component of H. Alternatively, we can argue that every connected component ofĈ\γ must contain two fixed points, one of which is repelling or has multiplier 1 (see for example [19] or [4]).
As mentioned previously, it is quite difficult to locate parameters in D α when a is not real. However, the following proposition asserts that all the sets D α are contained in some common compact subset of C * × C. Proof. As mentioned previously, the immediate basins of the superattracting fixed points at 0 and ∞ are simply connected, and the Herman ring separates those two basins. Let W 0 (respectively W ∞ ) be the immediate basin of 0 (respectively ∞). Let φ 0 : W 0 −→ B(0, r 0 ) (respectively φ ∞ : W ∞ −→Ĉ \ B(0, r ∞ )) be the conformal representation which is tangent to the identity at 0 (respectively ∞).
Then φ 0 (respectively φ ∞ ) conjugates f λ,a to a proper mapping of B(0, r 0 ) (respectivelyĈ\B(0, r ∞ )) of degree 2 which has a superattracting fixed point at 0 (respectively ∞). This proper mapping is z −→ z 2 /c 0 (respectively z −→ z 2 /c ∞ ). Since this map must have a fixed point on the boundary of B(0, r 0 ) (respectively B(0, r ∞ )), it follows that |c 0 | = r 0 (respectively |c ∞ | = r ∞ ). Now note that Therefore we deduce that Now W 0 does not contain −a, which is a preimage of ∞, and W ∞ does not contain −1/a, which is a preimage of 0. Therefore, by the Koebe one-quarter theorem, we have This, in turn, yields 1 4 |λ| 4.
Finally, the basins W 0 and W ∞ are disjoint. It follows that r 0 r ∞ (see for example [5,Section 3]). As a consequence, we have |a| |λ| which gives |a| 1.

Shishikura's surgery
We explain here a surgery construction originally due to Shishikura [18] that will be used in several instances in this paper. The general idea of the construction is, starting from a rational map with a Herman ring, to obtain a polynomial with a Siegel disk by means of gluing a rigid rotation to 'fill in the hole' of the Herman ring. In the particular case of our family of rational maps f λ,a , the polynomial that we obtain is precisely the quadratic polynomial P α . The result is summarized in the following proposition. Proposition 2.3. Suppose that f = f λ,a has a fixed Herman ring H with rotation number α, and let γ ⊂ H denote an invariant curve. Let U 0 be the bounded part ofĈ \ γ and let U ∞ be the unbounded one. Then the polynomial P α has a Siegel disk ∆ α and there exists a quasiconformal homeomorphism ψ :Ĉ −→Ĉ and a P α -invariant curve Γ in ∆ α such that the following hold.
Proof. Shishikura proved this using a surgery construction that has now become classic. We will give an outline of the procedure. Let φ : A r −→ H be a conformal map that conjugates the rigid rotation R α : Notice that φ −1 (γ) is a circle centered at 0 with radius ρ, r < ρ < 1. Denote bŷ φ : D −→ H ∪ U 0 a quasiconformal mapping that agrees with φ on A ρ , maps D ρ onto U 0 , and fixes 0. Define a mapf : The mapf :Ĉ −→Ĉ is a proper and quasiregular mapping which maps infinity to itself with local degree 2. By Proposition 2.1, f has a pole in U 0 , so by counting the preimages of infinity, we see thatf has degree 2, it fixes 0 and is conjugate to the rotation R α in a neighborhood of 0. Finally, since it is of degree 2, it has a unique critical point ω in C.
The mapf is not holomorphic on U 0 , but there it preserves the complex structure defined by the Beltrami from where ∂ denotes ∂/∂z and∂ denotes ∂/∂z. Pulling back this Beltrami form viaf , we see that there exists a Beltrami formμ which coincides with µ on U 0 , which vanishes onĈ \ n 0f −n (U 0 ), and which is invariant byf : By the measurable Riemann mapping theorem, there exists a quasiconformal homeomorphism ψ :Ĉ −→Ĉ which fixes 0 and ∞, sends ω to −1/2, and satisfieŝ Then the map ψ •f • ψ −1 :Ĉ −→Ĉ is proper and holomorphic of degree 2, it fixes 0 and is conjugate to the rotation R α in a neighborhood of 0, and it has a superattracting fixed point at infinity and a critical point at −1/2. The only such map is the quadratic polynomial P α , and thus The map ψ is the required conjugacy.
Remark 2.4. The modulus of the annulus bounded by γ and the outer boundary of H is equal to the modulus of the annulus bounded by Γ and the boundary of the Siegel disk ∆ α (since ψ gives a conformal isomorphism between these two annuli).

Case when a is real
Proof. Suppose that a ∈ (0, 1/3) and |λ| < 1. Then f maps the unit circle onto the circle of radius |λ| < 1 as an analytic diffeomorphism. We distinguish between the following two cases.
(a) We can cut out the dynamics in the unit disk and replace it by the attracting dynamics z −→ λz. Indeed, define φ : where Λ denotes multiplication by λ : Λ(z) = λz. Note that we have constructed φ so that the two definitions agree on S 1 . Nowf is a proper quasiregular mapping of degree 2 (we removed the pole in D). It is not holomorphic, but there exists an invariant Beltrami form µ that vanishes onĈ\ n 0 f −n (D) and satisfies It follows thatf is holomorphic with respect to the complex structure induced by µ, and one can verify thatf is quasiconformally conjugate to a quadratic polynomial. However, H is a Herman ring forf, which is impossible.
(b) Since D intersects H, we can choose an invariant curve γ ⊂ H so that D intersects the unbounded component ofĈ\γ. Let ψ and U 0 be defined by Proposition 2.3. Set W = ψ(D ∪ U 0 ) and W = ψ(λD ∪ U 0 ). Then P α (W ) ⊂ W , and W is strictly contained in W . Therefore P α : W −→ W is a strong contraction with respect to the hyperbolic metric of W which contradicts zero is an indifferent fixed point.
We have shown that we cannot have |λ| < 1. The proof that we cannot have |λ| > 1 is analogous. Now suppose that (λ , a) ∈ D α . By what we have just shown, |λ | = 1, so f λ ,α is equal to the composition R β •f λ,α for some rigid rotation R β . For an arbitrary circle homeomorphism h, the rotation number of R β • h is a weakly increasing function of β, and there is only one value of β (modulo 1) for which the rotation number is equal to some given irrational number (see for example [6]). Since R β • f λ,α | S 1 and f λ,α | S 1 have the same irrational rotation number, we have β = 0 (modulo 1) and λ = λ .

The twist coordinate
As above, let f have a fixed Herman ring H of rotation number α and let φ : Our goal in this section is to define a twist coordinate for the map f λ,a . To fix ideas, we will first define it for the easiest case, that is, the case when the boundaries of the Herman ring H are Jordan curves and contain the two critical points of the map, say ω 1 on the outer boundary and ω 2 on the inner boundary.
We define the twist coordinate of f λ,a as the difference between these two arguments (see Figure 5), that is, Note that even though φ is uniquely defined only up to rotation (and hence so arẽ ω 1 andω 2 ), the twist coordinate is independent of this choice. Intuitively, the twist parameter measures the rotation of the boundaries with respect to each other.
We now turn to the general case, when the critical points are not on the boundary of H or φ cannot be extended as a homeomorphism ∂A r −→ ∂H. For this reason, the definition of Θ in this situation will be different. We then show that the two definitions coincide.
We first recall the definition of the equator of an annulus.
Definition 3.1. Let H be an annulus conformally equivalent to A r by a conformal map φ : A r −→ H. We define the equator of H (or core geodesic) as the simple closed curve Let γ be the equator of H let and ψ :Ĉ −→Ĉ, U 0 , U ∞ , Γ and V be as in . Moreover, ψ fixes 0 and ∞ and sends the critical point in We now proceed to create a parallel construction with the mapf (u) = 1/f (1/ū), that is, the conjugate of f under the change of coordinates u = τ (z) = 1/z. Observe thatf has a Herman ring H = τ (H) of rotation number α. More precisely, let γ = τ (γ) be the equator of H and letψ, U 0 , U ∞ , Γ and V be given by Proposition 2.3: We claim that Γ = Γ and V = V . Indeed, the modulus of H is the same as the modulus of H. Since γ andγ are equators of those annuli, the moduli of the annuli bounded by γ andγ on the one hand and by the outer boundaries of H and H on the other hand are equal. Therefore, the moduli of the annuli bounded by the curves Γ and Γ on the one hand, and by the boundary of the Siegel disk on the other hand, are equal. Therefore Γ = Γ and V = V .
We now denote by ϕ : D −→ ∆ α a linearizing map of the Siegel disk. The map ϕ conjugates the rigid rotation R α : D −→ D to P α : ∆ α −→ ∆ α and is unique up to precomposition with other rigid rotations. See Figure 6.
We define a map C : Observe that, by construction, C is a conjugacy between the rotation R α and itself on the Euclidean circle ϕ −1 (Γ). However, any such map can only be a rigid rotation itself, that is, C(z) = e 2π iΘ z for some Θ ∈ R/Z. We then define Θ to be the twist coordinate of f λ,a . It only remains to prove the following.  Proof. We first observe that if φ : Hence, observing the commutative diagram in Figure 6, we see that the maps are conformal maps from the annulus A √ r to the Euclidean annulus bounded by ϕ −1 (Γ) and S 1 , and they conjugate the rigid rotation R α to itself (this in fact implies that ϕ −1 (Γ) is the Euclidean circle of radius √ r). Such maps can only be rigid rotations themselves, and we determine their angles now.
We now do the same for R. First observe that the map rτ sends all points of modulus r to points of modulus 1 and vice versa, but preserves their arguments. Therefore the critical points are interchanged, that is, the pointω 2 is sent to a point of modulus 1 and argument Θ 2 . This point is in turn mapped to the critical point on the outer boundary of H which is later sent to ω = −1/2 byψ. Finally, its image is, as before, e 2π iβ . We deduce then that Now, the outmost diagram says that We now want to illustrate the twist coordinate with a picture. If λ ∈ S 1 and a ∈ (0, 1/3), the twist coordinate is always equal to 0, because of the symmetry z −→ 1/z with respect to S 1 . Now, if λ ∈ S 1 and a ∈ iR, then there is a symmetry z −→ −1/z, and if f λ,a has a Herman ring, the twist parameter is equal to 1/2. Even if S 1 is not invariant, it is still possible to locate parameters (λ, a) such that f λ,a has a Herman ring. For example, for a = i/4 and t = 0.624 098 187 . . . , the rational map f e 2i π t ,a has a fixed Herman ring with rotation number ( √ 5 − 1)/2. Now the twist coordinate is better seen by working in the universal covering of C * given by Z −→ z = e 2iπ Z . Figure 7 shows the Julia set of two rational maps having Herman rings with rotation numbers ( √ 5 − 1)/2 and with twist coordinates 0 on the top and 1/2 on the bottom. In both cases, we have marked the critical points with arrows. The real part of Z ranges from −1 to 1.

Parameterization of Arnold disks: proof of Theorem A
We will work with a slightly different family of rational maps. The purpose of introducing a new family of rational maps is to get rid of the conjugacy between f λ,a and f λ,−a . Definition 4.1. For λ ∈ C * and b ∈ C, we define The map f λ,a is conjugate to the map g λ,a 2 via z −→ w = az, that is, Now, if g λ1,b1 is conjugate to g λ2,b2 by a scaling map, then (λ 1 , b 1 ) = (λ 2 , b 2 ). Moreover, the maps g λ,b are exactly the cubic rational maps which have superattracting fixed points at 0 and ∞ and which map −1 to 0. Observe that g λ,0 (z) = λz(1 + z), a quadratic polynomial, and finally note that if λ ∈ S 1 and b > 0, then the invariant circle now has radius √ b.
Definition 4.2. Given a Brjuno number α ∈ R/Z, let D α be the set of parameters (λ, b) such that g λ,b has a fixed Herman ring with rotation number α. By convention, we consider that for λ = e 2iπ α and b = 0, the quadratic polynomial g λ,b = P α has a Herman ring ∆ α \{0} of infinite modulus. We now state a weak version of Theorem A for the family g λ,b . 2iπ α , 0). Moreover, for any δ ∈ D, the modulus of the Herman ring of g Gα (δ ) is equal to (1/2π) log(1/|δ|).

Proof of Proposition 4.3
We will prove this result in several steps and later see how it implies Theorem A. In the previous section, we defined a twist coordinate from D α to R/Z. Similarly, we can define a twist coordinate Θ : D α −→ R/Z. Let us also consider the modulus coordinate m : D α −→ (0, +∞] which maps (λ, b) ∈ D α to the modulus of the Herman ring of g λ,b . We can define a map Π : We will show that this map provides an isomorphism between D (α) and D. See Figure 8.
For this purpose, given (λ 1 , b 1 ) ∈ D α , we will construct an analytic map G : D −→ C * × C whose image is contained in D α and such that 2iπ α , 0) and Π • G = Id, which shows that Π is injective. We will then show that the map G does not depend on the choice of (λ 1 , b 1 ), which proves that G is an isomorphism between D and D α , and Π is its inverse.
Note that L η is a homeomorphism between the vertical strips which commutes with the vertical translation by 2πi. See Figure 9. Thus it projects to a K-quasiconformal homeomorphism h η : A r1 −→ A rη . Moreover, h η (e τ1 ) = e τη and h η = Id on S 1 .
Finally, for any angle α ∈ R/Z, the homeomorphism h η : A r1 −→ A rη conjugates the rotation R α : A r1 −→ A r1 to the rotation R α : A rη −→ A rη . As a consequence, the rotation R α : A r1 −→ A r1 preserves the complex structure defined by the Beltrami form∂h η /∂h η .
Step 2 (a map S g1 : H + −→ D α ): For simplicity, we identify (λ, b) ∈ D α with the rational map g λ,b . Let us consider a rational map g 1 := g λ1,b1 ∈ D α . Let H 1 be the Herman ring of g 1 and let (m 1 , Θ 1 ) be the modulus and twist coordinate of H 1 . Set τ 1 = 2π(−m 1 + iΘ 1 ) and r 1 = e −2π m 1 , and let φ 1 : A r1 −→ H 1 be an isomorphism which conjugates R α : A r1 −→ A r1 to g 1 : Now, let η ∈ H + be an arbitrary complex number and let r η and h η : A r1 −→ A rη be as in step 1. Since R α : A r1 −→ A r1 preserves the complex structure defined by the Beltrami form∂h η /∂h η , and since φ 1 : Remark 4.4. Note that µ η does not depend on the choice of isomorphism There exists a unique extension of µ η to C which is g 1 -invariant and is 0 outside the preimages of H 1 . This Beltrami form µ η depends holomorphically on η ∈ H + .
Next, let χ η : C −→ C be the unique quasiconformal homeomorphism which fixes 0 and −1 and integrates µ η : Since g 1 preserves the complex structure defined by µ η , the map is holomorphic. Observe that g η must be a rational map of degree 3 which has superattracting fixed points at 0 and ∞ and maps −1 to 0. Moreover, by construction, H η := χ η (H 1 ) is a Herman ring for g η with rotation number α. Thus g η ∈ D α and the above construction defines a map S g1 : H + −→ D α which maps η to g η . The construction above is summarized in the commutative diagram shown in Figure 10.
As a consequence, the map S g1 : H + −→ D α is analytic.
Step 3 (modulus and twist of H η ): Our goal is to compute the modulus and the twist coordinate of the Herman ring H η of the new map g η = S g1 (η), in terms of η, m 1 and Θ 1 . We will prove the following.
Proposition 4.5. Let H 1 be the Herman ring of g 1 and let m 1 and Θ 1 be its modulus and twist coordinate. Given η ∈ H + , let H η be the fixed Herman ring of g η = S g1 (η). Denote its modulus by m η and its twist coordinate by Θ η . Then the following hold.
Proof. For the proof we shall use the notation in step 2.
We proceed now to compute Θ η . To this end, we need to create the general construction of Section 3 with the map g η and its Herman ring H η . Using the notation in that section, observe that we have the diagram shown in Figure 11 (compare with Figure 6). In Figure 11 we introduce three new composition maps, ψ, C and C, which are defined so as to make the diagram commute.
We first comment on the bottom row, which is the most important for our argument. Observe that C 1 and C η are the maps that define the twist coordinates for g 1 and g η . Indeed, let γ 1 , γ η , Γ 1 , Γ η be as in Section 3. Then C 1 (respectively C η ) restricted to the circle ϕ −1 (Γ 1 ) (respectively ϕ −1 (Γ η )) is a rigid rotation of angle Θ 1 (respectively Θ η ).
Next observe that both maps C, C : ϕ −1 (Γ 1 ) −→ ϕ −1 (Γ η ) are also analytic maps that conjugate the rigid rotation R α to itself. Hence they must be rigid rotations composed with scaling maps. That is, where s ∈ R and Θ, Θ ∈ R/Z. Our goal is then to compute the angles Θ and Θ, since we have A key observation that we shall use later is the following.
Claim 4.6. The map C and C extend as the identity to the unit circle.
Proof. Observe that the map χ := ψ η • χ η • ψ −1 1 is a globally defined quasiconformal map which conjugates the polynomial P α to itself. It follows that χ can be extended as the identity to the boundary of the Siegel disk (for example, one can argue that χ must fix the orbit of the critical point which accumulates on the boundary of the Siegel disk).
Moreover, the restriction of χ to the annulus bounded by Γ and the boundary of the Siegel disk is quasiconformal. It follows that χ can be extended as the identity to the ideal boundary of the Siegel disk, and hence C extends as the identity to the unit circle. The same argument applied toχ : We proceed by computing some rotation angles of the maps in the topmost row. It is easy to check from the expressions of L η and h η that Figure 12.
To transfer this information to the bottom maps, we need to consider the vertical compositions. Let us then consider all the maps of the diagram restricted to the annuli bounded by the equator curves and the outer boundaries. Then all the 'vertical' compositions are conformal. It follows that the circle ϕ −1 (Γ 1 ) (respectively ϕ −1 (Γ η )) has radius √ r 1 (respectively √ r η ), and hence we have the diagram shown in Figure 12, where R 1 , R η , R 1 and R η are rigid rotations.
We first analyze the innermost diagram to observe that, given that h η commutes with rigid rotations, the map C is equal to h η up to rotation. However, we observed in Claim 4.6 that C extends as the identity to the unit circle as h η does. It follows that C = h η on A √ r1 , and hence Θ = − 1 2 m 1 Im(η). Studying the outermost diagram and arguing as above, we have C is equal toĥ η up to composition with a rigid rotation. However, again C is the identity on S 1 . It then follows that on A √ r1 , C = e 2π im 1Im(η ) ·ĥ η , and hence, by knowingĥ η on the equator, we conclude that Θ = 1 2 m 1 Im(η). Finally, from equation (1), we obtain Step 4 (a group property): Let S : D α × H + −→ D α be the map S : (g, η) −→ S g (η). Assume that η 1 and η 2 are two complex numbers in H + , and g 1 ∈ D α . Our goal is to prove the following. S(S(g 1 , η 1 ), η 2 ) = S(g 1 , L η2 (η 1 )).
Then we want to show that Indeed, for i = 1, 2 or 3, let H i be the Herman ring of g i and let m i be the modulus of H i , and set r i = e −2π m i . Observe that L η3 = L η2 • L η1 . These homeomorphisms project to homeomorphisms h 1 : Now, let µ 1 be the Beltrami form which coincides with 0 outside n 0 g −n on H 1 , and which satisfies g * 1 (µ 1 ) = µ 1 everywhere. Let χ 1 : C −→ C be the integrating map that fixes 0 and −1. Observe that the map a homeomorphism that preserves the standard complex structure, and is thus an isomorphism (and a linearizing map). Therefore we can define µ 2 on H 2 and, χ 2 : C −→ C and, as above, On the other hand, letχ 3 ,μ 3 be the map and complex structure used in the construction ofg 3 := S(g 1 , η 3 ). We will show that the complex structure µ 3 induced by φ 3 equalsμ 3 . Since χ 3 andχ 3 are normalized in the same way, these two integrating maps must be the same. Hence we can conclude that g 3 =g 3 .
Thanks to the previous step, we can assume, without loss of generality, that η = 1. Indeed, if we know that S h (1+i/m) = S h (1) for all h ∈ D α , then, taking h = S g1 (η), the general equality follows.
We will construct a quasiconformal homeomorphismχ that satisfies We will then show thatχ integrates µ 1+i/m and fixes 0, ∞ and −1. Therefore χ =χ and The quasiconformal homeomorphism L 1+i/m preserves vertical strips and, in particular, it preserves It coincides with the identity on iR and with the translation by −2iπ on log r 1 + iR. Therefore the quasiconformal homeomorphism h 1+i/m : A r1 −→ A r1 extends by the identity to the two boundary components of A r1 (it is a Dehn twist). Let on the Herman ring. Note that∂χ/∂χ = µ 1+i/m on H 1 (since it preserves the standard complex structure), andχ is equal to the identity on the two ideal boundary components of H 1 . In particular,χ extends by the identity to the two boundary components of H 1 in C (see [8]). Next, given a connected component U of n 0 g −n 1 (H 1 ), let n 0 be the smallest integer such that g n 1 (U ) = H 1 . The map g n 1 : U −→ H 1 is an isomorphism, and we set Note thatχ| U extends by the identity to ∂U . We can extendχ to a homeomorphism χ : C −→ C such that the restriction ofχ to C \ n 0 g −n 1 (H 1 ) is the identity. By a lemma of Rickman,χ : C −→ C is quasiconformal. By construction, Finally, observe that −1, 0 and ∞ are not eventually mapped to H 1 . Henceχ fixes these three points.
Proof. Just from the diagram we see that Recall that Π(g 1 ) = r 1 e 2π iΘ1 , and therefore we have shown that G g1 (Π(g 1 )) = g 1 . We now want to show that Π • G g1 = Id.
In order to do this, let us first determine the 'inverse' of p. It is easy to check that, for e −2π im = r < 1 and Θ ∈ R/Z, we have where g ∈ D α has a Herman ring H of modulus m 1 (m/m 1 ) = m and twist Θ 1 − m 1 (Θ 1 − Θ)/m 1 = Θ (see Proposition 4.5). It then follows that Π(g) = re 2π iΘ and hence Π • G g1 = Id.

Proof. By Proposition 2.2, we know that
Thus, by the removable singularity theorem, G g1 : D * −→ D α extends holomorphically at 0. We know that the modulus of the Herman ring H δ of the map G g1 (δ) is (1/2π) log(1/|δ|) for δ ∈ D * . Thus the modulus of the Herman ring tends to +∞, when δ tends to the origin. The Herman ring H δ separates −1, which is mapped to 0, and −b, which is mapped to ∞. Therefore we have b → 0 as δ → 0.
Step 7 (the map G g1 is surjective): In order to complete the proof of Proposition 4.3, it is now enough to show that G g1 is surjective. Thus we must prove that for all g 1 , g 2 ∈ D α , there exists δ ∈ D * such that g 2 = G g1 (δ). Thanks to step 3, it is enough to prove that there exist δ 1 and δ 2 such that G g1 (δ 1 ) = G g2 (δ 2 ).
We can then take G α = G g1 , and this concludes the proof of Proposition 4.3.
Moreover we claim that ∂a ∂δ This is a consequence of the following proposition. The analogue for the complex standard family has been proved by Fagella, Seara and Villanueva in [10]. Proof. Let us again work with the family g λ,b with b = a 2 . In the whole proof, we consider that (λ, b) varies in the Arnold disk D α . More precisely, we assume that Observe that when δ → 0, the maps g λ,b converge to P α uniformly on every compact subset of C * . In particular, one of the critical points of g λ,b , say ω(δ), converges to the critical point ω(0) = −1/2 of P α .
The maps δ −→ g n λ(δ ),b(δ ) (ω(δ)) are all well defined and holomorphic for δ ∈ D. Moreover they define a holomorphic motion of the orbit of the critical point ω(δ), parameterized by δ ∈ D. By the Mañe-Sad-Sullivan λ-lemma [13], this holomorphic motion extends to a holomorphic motion of the closure of this critical orbit.
Let B(δ) be the outer boundary component of the Herman ring of g λ,b for δ = 0, and the boundary of the Siegel disk for δ = 0. Moreover, let ∆(δ) be the bounded component of C \ B(δ). Since B(δ) is contained in the closure of the orbit of ω(δ), we see that B(δ) moves holomorphically with respect to δ ∈ D. As a consequence, (∆(δ), 0) depends continuously on δ ∈ D for the Carathéodory topology. Thus its conformal radius at 0 varies continuously with respect to δ ∈ D.
Let us now assume that δ ∈ (0, 1), and thus that b(δ) > 0 (see Corollary 4.10). Then the Herman ring is symmetric with respect to the circle centered at 0 with radius √ b. It follows that the modulus m(δ) of the Herman ring is equal to twice the modulus of the annulus bounded by this circle and B(δ). As δ → 0, we have b → 0, and the conformal radius of ∆(δ) tends to r α . It follows from Lemma 4.12 below that Using the relation b = a 2 and m a = m(δ), we get m a = 1 π log r α a +o(1).
By the Grötzsch inequality, we have We get the required inequality by letting ε tend towards 0.
By the inverse function theorem, it follows that In particular, as a → 0, we have e π m a = 1 |δ(a)| = r α |a| + O(a).

The complex standard family
We now explain how to adapt the results to the complex standard family f λ,a (z) = λze (a/2)(z −1/z ) .
In this section, we will no longer consider cubic rational maps and quadratic polynomials. We will therefore feel free to use the same notations as in the rest of the paper to denote similar but different maps and objects. The complex standard family has been extensively studied in the case λ = e 2iπ t ∈ S 1 and a ∈ R: it lifts via Z −→ z = e 2iπ Z to the Arnold family F α,a (Z) = Z + t + a 2π sin(2πZ).
For t ∈ R and a ∈ (−1, 1), these maps restrict to analytic diffeomorphisms of R/Z.
Let us now assume that λ and a are complex parameters, and that f λ,a has a fixed Herman ring H ⊂ C * with rotation number α. By the maximum modulus principle, H separates the essential singularities 0 and ∞. Following [11], we can perform Shishikura's surgery in order to obtain an entire mapping fixing 0 and having a Siegel disk with rotation number α around 0. By construction, this map does not vanish, except at 0, and it may be normalized to have a critical point at −1. In fact, Geyer has shown that the resulting map is given by E α (z) = e 2iπ α ze z , and, moreover, that E α has a Siegel disk around 0 if and only if α is a Brjuno number. Therefore a map f λ,a may have a fixed Herman ring only when α is a Brjuno number. Definition 6.1. Given a Brjuno number α ∈ R/Z, we let ∆ α be the Siegel disk of the entire mapping E α , and we let D α be the set of parameters (λ, a) ∈ C * × C such that f λ,a has a fixed Herman ring with rotation number α. We call D α the Arnold disk of rotation number α.
As in the case of cubic rational maps, to any map f ∈ D α we can associate two coordinates: the modulus of the Herman ring and a twist coordinate defined via the surgery construction (see Section 3).
Also, the change of coordinates z −→ −z conjugates f λ,a to f λ,−a , and it is thus useful to introduce a new family The map f λ,a is conjugate to the map g λ,a 2 via z −→ w = az/2, that is, f λ,a (z) = 2 a g λ,a 2 a 2 z .
Note that g λ,0 is the entire mapping w −→ λwe w .
Definition 6.2. Given a Brjuno number α ∈ R/Z, let D α be the set of parameters (λ, b) such that g λ,b has a fixed Herman ring with rotation number α.
By convention, we consider that for λ = e 2iπ α and b = 0, the entire mapping g λ,b = E α has a Herman ring ∆ α \ {0} of infinite modulus. As in the case of the cubic rational maps studied in this paper, one can show the following proposition. Proposition 6.3. For any Brjuno number α ∈ R/Z, there is a holomorphic bijection G α : D −→ D α with G α (0) = (e 2iπ α , 0). Moreover, for any δ ∈ D, the modulus of the Herman ring of g Gα(δ ) is equal to (1/2π) log (1/|δ|).
The only difference in the proof appears when one shows that a certain map G : D * −→ D α has a removable singularity at 0 (this corresponds to Claim 4.9). Indeed, we do not know whether the Arnold disks D α are compactly contained in C * × C. Remark 6.4. This raises the following questions. Are the Siegel disks ∆ α bounded in C? Are the Arnold disks D α and D α bounded in C * × C?
In order to overcome this difficulty, we set G(δ) = (λ(δ), b(δ)) for δ ∈ D * and argue as follows. First, as in the case of the cubic rational maps, the modulus of the Herman ring tends to ∞ as δ → 0. The Herman ring separates the critical points which are the roots of the equation 4z 2 + 4z + b(δ) = 0. It follows that as δ → 0, we have b(δ) → 0 (one of the critical points tends to 0 and the other tends to −1). To show that G has a removable singularity at δ = 0, it is sufficient to show that when δ → 0, λ(δ) remains uniformly bounded away from 0 and ∞. This is a consequence of the following lemma and the fact that b(δ) −→ 0 δ →0 . Proof. Suppose that (λ, a) ∈ D α , |a| < 1 and |λ| < e −|a| . Then the image Γ = f λ,a (S 1 ) of the unit circle is an R-analytic Jordan curve in D, and f λ,a : S 1 −→ Γ is a diffeomorphism (this is easily seen by working in the lifted F ω ,a (Z) = Z + ω + (a/2π) sin(2πZ) with Im(ω) > 0 and λ = e 2iπ ω ). We then get a contradiction as in the proof of Proposition 2.5. Similarly, we cannot have |λ| > e |a| .
With the help of Proposition 6.3, we can now proceed exactly as in the case of cubic rational maps, and we obtain the following results.
Theorem B . Assume that α is a Brjuno number. Then the following hold. (a) The Arnold disk can be locally parameterized by a in a neighborhood of (e 2iπ α , 0) (that is, it is locally the graph of a holomorphic map a −→ λ(a)).
(b) As |a| → 0, the modulus m a of the Herman ring of f λ(a),a satisfies e π m a = 2r α |a| + O(a).