GGENERALIZATIONS OF DOUADY’S MAGIC FORMULA
ADAM EPSTEIN AND GIULIO TIOZZO
Abstract.
We generalize a combinatorial formula of Douady from the main car-dioid to other hyperbolic components H of the Mandelbrot set, constructing anexplicit piecewise linear map which sends the set of angles of external rays landingon H to the set of angles of external rays landing on the real axis. In the study of the dynamics of the quadratic family f c ( z ) := z + c , c ∈ C , acentral object of interest is the Mandelbrot set M := { c ∈ C : ( f ( n ) c (0)) n ≥ is bounded } . The Mandelbrot set contains two notable smooth curves: namely, the intersectionwith the real axis
M ∩ R = [ − , / main cardioid of M , which is the setof parameters c for which f c has an attracting or indifferent fixed point (of course,this is smooth except at the cusp c = 1 / R denote the set of angles of external rays landing on the real axis, and Ξ theset of angles of external rays landing on the main cardioid. The following “magicformula” is due to Douady (for a proof see [Bl]): Theorem 1 (Douady) . The map T ( θ ) := (cid:26) + θ if 0 ≤ θ < + θ if < θ ≤ sends Ξ into R . In this note, we prove the following generalization of Theorem 1 to other hyperboliccomponents. Let D ( θ ) := 2 θ mod 1 denote the doubling map; moreover, given afinite binary word S and a real number θ , by S · θ we denote the real number whosebinary expansion is the concatenation of S and the expansion of θ . Theorem 2.
Let H be a hyperbolic component in the upper half plane which belongsto a vein V of the Mandelbrot set, not in the / -limb. Let A H , B H be the binaryexpansions of the root of H , with A H < B H , and let δ V be the complexity of V . Thenthe map Φ H ( θ ) := D δ V ( B H A H · θ ) sends the set of external angles of rays landing on the upper part of H into the set R of angles of rays landing on the real axis. University of Warwick, UK,
[email protected] .University of Toronto, Canada, [email protected] . a r X i v : . [ m a t h . D S ] N ov ADAM EPSTEIN AND GIULIO TIOZZO
Figure 1.
The magic formula for the kokopelli component. The angle θ = 1 / H ( θ ) =21 /
40 which lands on the real axis.To illustrate such a formula, let us consider the “kokopelli” component of period4, which has root of angles θ = 3 /
15 = . θ = 4 /
15 = . A H = 0011, B H = 0100. This component lies on the principal vein with tip θ = 1 /
4, hence itscomplexity is δ V = 1. Thus,Φ H ( θ ) := 1000011 · θ = 67128 + θ . Let us note that if H is the main cardioid, then δ V = 0, A H = 0, B H = 1, hencewe recover Douady’s original formula from Theorem 1. By symmetry, an analogousformula to Theorem 2 holds for hyperbolic components in the lower half plane.Note that the map Φ H is very far from being surjective: indeed, the set of anglesof rays landing on a hyperbolic component has Hausdorff dimension zero, while theset R has dimension 1.We will prove our formula (Theorem 2) in two parts: first (Section 1.2), we willconstruct a map which sends the hyperbolic component into the real slice of thecorresponding small Mandelbrot set; then (Section 1.3), we will map the vein to whichthis small Mandelbrot set belongs to the real vein. This will be done by developing ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 3
Figure 2.
The graph of the function Ψ( x ) discussed in Remark 4.The red dots correspond to points on the graph of Douady’s magicformula function T = ϕ H for the main cardioid, the (thick) blue dotsto the image of ϕ H for the rabbit component, and the green dots to thekokopelli component.some notion of combinatorial veins and combinatorial Hubbard trees, which may be ofindependent interest. The final formula will be the composition of the two formulas.Then, in Section 2 we will prove that the rays given by the image of the formulaactually land, by showing they are not renormalizable. Finally, in Section 3 we willprove an alternate formula, which, however does not depend on the vein structure of M .During the preparation of this paper we have been informed of the generalizationof Douady’s formula by Bl´e and Cabrera [BC]. Let us remark that our formula isquite different, as the one in [BC] does not map into the real axis but rather into thetuned copies of the real axis inside small Mandelbrot sets. In fact, we will describetheir formula and how it relates to ours in Section 1.2. Acknowledgements.
We thank S. Koch and C. T. McMullen for useful conversa-tions. G.T. is partially supported by NSERC and the Alfred P. Sloan Foundation.1.
Combinatorics of external rays
Real rays and the original formula.
Consider the Riemann map Φ M : C \ D → C \M , which is unique if we normalize it so that Φ M ( ∞ ) = ∞ , Φ (cid:48) M ( ∞ ) = 1.For each θ ∈ R / Z , we define the external ray of angle θ as R M ( θ ) := { Φ M ( re πiθ ) : r > } . The ray R M ( θ ) lands if there exists γ ( θ ) := lim r → + Φ M ( re πiθ ). It is conjectured thatthe ray R M ( θ ) lands for any θ ∈ R / Z ; to circumvent this issue, we define as (cid:98) R M ( θ ) ADAM EPSTEIN AND GIULIO TIOZZO the impression of R M ( θ ), and the set of real angles as (cid:98) R := { θ ∈ R / Z : (cid:98) R M ( θ ) ∩ R (cid:54) = ∅} . Conjecturally, this is exactly the set of angles for which the corresponding ray landson the real axis. For further details, see e.g. [Za].The following lemma gives a characterization of (cid:98) R in terms of the dynamics of thedoubling map. Lemma 3 ([Do]) . The set of real angles equals (cid:98) R := { θ ∈ R / Z : | D n ( θ ) − / | ≥ | θ − / | ∀ n ∈ N } . Proof of Theorem 1.
By symmetry, we can assume 0 < θ < . Since the external rayof angle θ lands on the main cardioid, then the forward orbit P := ( D n ( θ )) n ∈ N ofthe angle θ does not intersect the half-circle I = ( θ + , θ ) (the one which contains0) (see [BS]). The preimage of I by the doubling map is the union of two intervals( − θ , θ ) ∪ ( − θ , + θ ). In particular, the forward orbit of θ does not intersect J = ( − θ , + θ ). Thus, if we look at the orbit of θ (cid:48) = T ( θ ), we have 2 θ (cid:48) = θ / ∈ J and D n ( θ (cid:48) ) ∈ P for any n ≥
2, hence the forward orbit of θ (cid:48) does not intersect J ,hence θ (cid:48) is a real angle by Lemma 3. (cid:3) Remark 4.
By Lemma 3, one recognizes that a way to map any angle θ ∈ R / Z tothe set of real angles is to consider the functionΨ( x ) := 12 + inf k ≥ (cid:12)(cid:12)(cid:12)(cid:12) D k ( x ) − (cid:12)(cid:12)(cid:12)(cid:12) , which indeed satisfies Ψ( R / Z ) ⊆ (cid:98) R . As you can see from Figure 2, this function isdiscontinuous, while its graph “contains” the graphs of all the magic formula functions ϕ H given by Theorem 2 (which are indeed continuous) for all components H .1.2. Tuning and the Bl´e-Cabrera magic formula.
Given any hyperbolic com-ponent H in the Mandelbrot set, let us recall that there is a tuning map which sendsthe main cardioid to the hyperbolic component H , and the Mandelbrot set to a smallcopy of itself which contains H .In order to define the map precisely, let a H = .a . . . a k < b H = .b . . . b k denotethe two external angles of rays landing at the root of H , and denote A H = a . . . a k and B H = b . . . b k the two corresponding finite binary words. Moreover, denote a (cid:48) H = .a . . . a k b . . . b k and b (cid:48) H = .b . . . b k a . . . a k .The tuning map T H on the set of external angles is now defined as follows. If θ = .(cid:15) (cid:15) . . . is the binary expansion of θ , then the angle T H ( θ ) has binary expansion T H ( θ ) = .A (cid:15) A (cid:15) . . . where A = A H , A = B H . Then, the set Ξ H := T H (Ξ) is the set of rays landing onthe boundary of H .If θ ∈ T = R / Z has infinite binary expansion (i.e. it is not a dyadic number) and S = s . . . s n is a finite word on the alphabet { , } , we denote by S · θ the element ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 5 of T S · θ := n (cid:88) k =1 s k − k + 2 − n θ i.e. the point whose binary expansion is the concatenation of S and the binaryexpansion of θ . Recall that tuning behaves well with respect to concatenation, namelyfor any S, H and θ we have T H ( S · θ ) = T H ( S ) · T H ( θ ) . We define the small real vein associated to H , and denote it as R H , to be the realvein of the small Mandelbrot set associated to H . Let us denote as R H the set ofexternal angles of rays landing on R H , and as (cid:98) R H the set of external angles of rayswhose impression intersects R H . Clearly, R H = R if H is the main cardioid, and R H = R . Proposition 5 (Bl´e-Cabrera [BC]) . The map T H := T H ◦ T ◦ T − H maps Ξ H into thesmall real vein (cid:98) R H . Moreover, this map (restricted to Ξ H ) is piecewise affine: in fact,it can be written in terms of binary expansions as T H ( θ ) = B H A H · θ if θ ∈ ( a H , a (cid:48) H ) , and T H ( θ ) = A H B H · θ if θ ∈ ( b (cid:48) H , b H ) .Proof. By construction, the tuning map T H : Ξ \ { } → Ξ H \ { a H , b H } is a bijection,so the first statement follows by looking at the diagram:Ξ T (cid:47) (cid:47) T H (cid:15) (cid:15) (cid:98) R T H (cid:15) (cid:15) Ξ H T H (cid:47) (cid:47) (cid:98) R H . Moreover, since T ( θ ) = 10 · θ for θ < / T H ( θ ) = T H (10 · T − H ( θ )) = T H (10) · θ which proves the second claim. (cid:3) Veins, pseudocenters and complexity.
Given a dyadic number θ ∈ S = R / Z , we define its complexity as (cid:107) θ (cid:107) := min { k ≥ D k ( θ ) = 0 mod 1 } . Of course, if θ = p q with p odd, then (cid:107) θ (cid:107) = q . Given an interval ( θ − , θ + ) with θ − < θ + , we define its pseudocenter θ to be the dyadic rational of lowest complexityinside the interval ( θ − , θ + ).A pair of elements ( θ − , θ + ) in T is a ray pair if the two external rays of angle θ − and θ + combinatorially land at the same parameter on the boundary of the Mandelbrot ADAM EPSTEIN AND GIULIO TIOZZO set. To be precise, one starts by defining a relation on Q / Z by setting that θ ∼ Q θ if R M ( θ ) and R M ( θ ) land at the same point. Then, one takes the transitive closureof this relation and finally its topological closure to define an equivalent relation ∼ on R / Z . As constructed by Thurston [Th], there is a lamination QM L on the unit disksuch that its induced equivalent relation is precisely ∼ . If MLC holds, then θ ∼ θ if and only it R M ( θ ) and R M ( θ ) land on the same point.Ray pairs are partially ordered: in fact, we say ( θ − , θ +1 ) ≺ ( θ − , θ +2 ) if the leaf( θ − , θ +1 ) separates ( θ − , θ +2 ) from 0. Let us denote as N ( θ ) the number of ends of theHubbard tree associated to the angle θ . Recall that if θ ≺ θ , then N ( θ ) ≤ N ( θ ).A dyadic number θ defines a combinatorial vein in the Mandelbrot set as follows. Definition 6.
Given a dyadic rational number θ , we define the combinatorial vein of θ as the set of ray pairs ( θ , θ ) such that: (1) θ is the pseudocenter of ( θ , θ ) ; (2) N ( θ ) = N ( θ ) . For instance, if θ = , then the vein extends all the way to ( , ), since N ( ) = 3,and one can check easily that N ( ) = 3 (the “rabbit”). On the other hand, if θ = ,then the vein extends up to ( , ). Indeed, N ( ) = 5, and N ( θ ) = 3 if θ ∈ ( , ).Note that condition (2) in the above definition is needed. For instance, if oneconsiders the ray pair ( θ , θ ) = ( , ), then its pseudocenter is θ = 39 / N ( ) = 6 < N ( ) (see Figure 1.3). Definition 7.
The complexity of a vein V with pseudocenter θ is given by δ V = (cid:107) θ (cid:107) − . The complexity δ V equals: • the smallest k such that D k ( θ ) = 1 / • the smallest k such that D k ( θ − ) and D k ( θ + ) lie on opposite connected com-ponents of the set T \ { , / } ; • the smallest k such that f kc ( c ) lies on the spine [ − β, β ] of the Julia set of f c for c which belongs to the vein.Clearly, δ V = 0 if and only if V is the real vein. Given a vein V in the upper halfplane (i.e. with θ < / lower side is the set of angles θ + for which there exists θ − < θ + such that ( θ − , θ + ) belongs to the vein. Proposition 8.
Let V be a vein in the upper half plane, and δ V be its complexity.Then for each ray pair ( θ − , θ + ) which belongs to V we have D δ V ( θ + ) ⊆ (cid:98) R . In order to prove the proposition, we need the following
Lemma 9.
Let ( θ − , θ + ) be a ray pair, and θ be its pseudocenter, with θ < / (i.e.the hyperbolic component lies in the upper half plane, and not in the / limb). Thenwe have θ + − θ < θ − θ − . ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 7
Figure 3.
The Julia set for the ray pair ( θ − , θ + ) = ( , ). One cansee that the Hubbard tree for θ − and θ + has 6 ends, while the Hubbardtree for the pseudocenter θ = has 8 ends. Here, (cid:107) θ (cid:107) = 7, so thevein V associated to θ has δ V = 6, but θ + does not belong to V . Thisshows that the formula of Proposition 8 cannot be extended beyondthe vein V : indeed, the point in the forward orbit of θ + which is closestto is ∈ (cid:98) R , while D δ V ( θ + ) = , so the formula would break down. Proof.
Recall that, for each rational pq with p, q coprime, there exists a unique set C p/q of q points on T such that the doubling map acts on C p/q with rotation number p/q , i.e. D ( x i ) = x i + p , where the indices are considered modulo q . We know that theset C p/q is precisely the set of external rays which land on the α fixed point for f c ,when c belongs to the p/q -limb. Let us call sectors the q connected components of T \ C p/q . In particular, we call critical sector S the smallest sector, and central sector S the largest sector. We say an interval I ⊆ T is embedded if it is entirely containedwithin one sector. Let us consider the interval I = [ θ − , θ + ], which is embedded byconstruction.Let k ≥ I k +1 := D k +1 ( I ) is not embedded(it must exist, since D doubles lengths, and the length of an embedded interval isbounded above by a constant c < ADAM EPSTEIN AND GIULIO TIOZZO between each non-central sector and its image, so I k must be contained in the centralsector.Let us now consider the set (cid:98) C p/q = C p/q + 1 /
2, which is a subset of the centralsector and equals the set of external rays which land on − α , the preimage of the α fixed point. Thus, let us call subsectors the connected components of S \ (cid:98) C p/q .In particular, exactly two subsectors S − and S + map to the critical sector S ; letus call them central subsectors . Since the rays D k ( θ − ) and D k ( θ + ) land together,then either they are both contained in the same subsector, or one is contained in onecentral subsector and the other one in the other. If they are contained in the samesubsector, then the whole I k is contained in the same subsector, hence its image I k +1 is still embedded, contradicting the definition of k . Thus, the two endpoints of I k are contained in two distinct central subsectors. Since I k is embedded, then it mustcontain 0; moreover, since 0 is the dyadic of lowest possible complexity, this meansthat D k ( θ ) = 0. Now, by construction the images D k +1 ( θ − ) and D k +1 ( θ + ) lie inthe critical sector S , which is contained inside the arc [0 , /
2] since the hyperboliccomponent lies in the upper half plane. This means that (cid:96) ([0 , D k +1 ( θ + )]) < / < (cid:96) ([ D k +1 ( θ − ) , (cid:96) denotes the length of the intervals) and, since D k +1 is a homeomorphismwhen restricted to [ θ , θ + ] and to [ θ − , θ ],2 k +1 (cid:96) ([ θ , θ + ]) = (cid:96) ( D k +1 [ θ , θ + ]) = (cid:96) ([0 , D k +1 ( θ + )]) < / k +1 (cid:96) ([ θ − , θ ]) = (cid:96) ( D k +1 [ θ − , θ ]) = (cid:96) ([ D k +1 ( θ − ) , > / (cid:96) ([ θ , θ + ]) < (cid:96) ([ θ − , θ ])which proves the claim. (cid:3) Combinatorial Hubbard trees.
Recall that every angle θ ∈ T has an associ-ated lamination on the disc which is invariant by the doubling map (see [Th]). The(two) longest leaves of the lamination are called major leaves , and their common im-age is called minor leaf and will be denoted by m . Moreover, we let β denote the leaf { } , which we will take as the root of the lamination (the notation is due to the factthat the ray at angle 0 lands at the β -fixed point). The dynamics on the laminationis induced by the dynamics of the doubling map on the boundary circle. In particu-lar, let us denote f : D → D to be a continuous function on the filled-in disc whichextends the doubling map on the boundary S . Let us denote by ∆ the diameter ofthe circle which connects the boundary points at angles θ/ θ + 1) /
2. Then weshall also choose f so that it maps homeomorphically each connected component of D \ ∆ onto D .Let L and L be two distinct leaves. Then we define the combinatorial segment [ L , L ] as the set of leaves L of the lamination which separate L and L . Somesimple properties of combinatorial segments are the following: ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 9 (a) if
L ∈ [ L , L ], then [ L , L ] ⊆ [ L , L ];(b) for any choice of leaves L , L , L we have [ L , L ] ⊆ [ L , L ] ∪ [ L , L ];(c) the image of [ L , L ] equals: f ([ L , L ]) = (cid:26) [ f ( L ) , f ( L )] if ∆ does not separate L and L , [ f ( L ) , m ] ∪ [ f ( L ) , m ] if ∆ separates L and L . (d) in any case, for any leaves L , L we have[ f ( L ) , f ( L )] ⊆ f ([ L , L ]) ⊆ [ f ( L ) , m ] ∪ [ f ( L ) , m ] . Finally, we shall say that a set S of leaves is combinatorially convex if whenever L and L belong to S , then the whole set [ L , L ] is contained in S .We now define H n := (cid:91) ≤ i ≤ n [ β, f i ( m )]and H := (cid:91) n ∈ N H n . We call the set H the combinatorial Hubbard tree of f , as it is a combinatorialversion of the (extended) Hubbard tree. Lemma 10.
The combinatorial Hubbard tree H has the following properties. (1) The set H is the smallest combinatorially convex set of leaves which contains β , m and is forward invariant. (2) Let N ≥ be an integer such that f N +1 ( m ) ∈ H N . Then we have H = H N . Note that N + 2 coincides with the number of ends of the extended Hubbard tree. Proof. (1) Let us check that H is combinatorially convex. Suppose L , L belong to H , so that say L ∈ [ β, f i ( m )] and L ∈ [ β, f j ( m )]. Then[ L , L ] ⊆ [ L , β ] ∪ [ β, L ] ⊆ [ f i ( m ) , β ] ∪ [ β, f j ( m )] ⊆ H as required. In order to check that H is forward invariant, note that by point (d) andusing that f ( β ) = β yields the inclusion(1) f ([ β, f i ( m )]) ⊆ [ f ( β ) , m ] ∪ [ m, f i +1 ( m )] ⊆ [ β, m ] ∪ [ β, f i +1 ( m )]and the right-hand side is contained in H by construction, so f ( H ) ⊆ H .In order to check the minimality, let S be a combinatorially convex, forward in-variant set of leaves which contains β and m . Then by convexity S contains [ β, m ].Moreover, by forward invariance for any i ≥ S ⊇ f i ([ β, m ]) ⊇ [ f i ( β ) , f i ( m )] = [ β, f i ( m )]thus S ⊇ H .(2) Note that by construction H N ⊆ H , and the same proof as in (1) shows that H N is combinatorially convex. In order to prove the claim it is thus enough to check that H N is forward invariant, because then by minimality we get H N ⊇ H and theclaim is proven. To prove forward invariance, note that f ( H N ) = (cid:91) ≤ i ≤ N f ([ β, f i ( m )]) ⊆ H N ∪ [ β, f N +1 ( m )]and since f N +1 ( m ) ∈ H N then [ β, f N +1 ( m )] ⊆ H N , proving the claim. (cid:3) Lemma 11.
Let m < m , and H , H be the corresponding combinatorial Hubbardtrees. Then H ⊆ H . As a corollary,
Ends ( T ) ≤ Ends ( T ) . Proof.
By definition, m < m means that m ∈ [ β, m ]. Thus, m ∈ H and since H is forward invariant we have f i ( m ) ∈ H for any i ≥
0. Since also β ∈ H and H is combinatorially convex, then [ β, f i ( m )] ⊆ H for any i ≥
0, thus H ⊆ H as required. For the corollary, note that since the trees are dual to the laminations, T ⊆ T , and in general a connected subtree of a tree cannot have more ends thanthe ambient tree. (cid:3) Lemma 12.
Let ( θ − , θ + ) be a ray pair, and θ its pseudocenter, with (cid:107) θ (cid:107) = q . Then Ends ( T θ + ) = Ends ( T θ ) if and only if the arcs I k = ( D k ( θ − ) , D k ( θ + )) for k = 0 , . . . , q − are disjoint.Proof. Note that the forward orbit of θ has cardinality q + 1, and since its minorleaf is a point (and so are all its forward iterates), the number of ends of T θ is also q + 1. Now, consider the minor leaf m = ( θ − , θ + ) and its forward iterates. Note thatby Lemma 13 the arc I k for k ≤ q − I j and I k are disjoint if and only if neither f j ( m ) ∈ [ β, f k ( m )] nor f k ( m ) ∈ [ β, f j ( m )]. Now,if all the intervals are disjoint, then f k ( m ) / ∈ [ β, f i ( m )] for any 0 ≤ i < k ≤ q − T θ + is at least q + 1, which implies it is exactly q + 1since by the previous Lemma it cannot exceed the number of ends of T θ . Vice versa,if 0 ≤ i < k ≤ q − f k ( m ) / ∈ [ β, f i ( m )], then either I k is disjoint from I i , or f i ( m ) ∈ [ β, f k ( m )]. This however is impossible, as it implies I k ⊆ I i with i < k .Indeed, let us denote as x k and x i the pseudocenters of, respectively, I k and I i . Asthe complexity of the pseudocenter decreases precisely by 1 under iteration, since i < k one has (cid:107) x i (cid:107) > (cid:107) x k (cid:107) . However, since I k ⊆ I i , then x k ∈ I i , hence by definitionof pseudocenter (cid:107) x k (cid:107) ≥ (cid:107) x i (cid:107) , contradicting the previous statement. (cid:3) Lemma 13.
Let [ α, β ] be an embedded arc in S which does not contain , and θ itspseudocenter, with q = (cid:107) θ (cid:107) . Then for all ≤ k ≤ q − , the arc [ D k ( α ) , D k ( β )] isembedded and does not contain .Proof. We claim that there exists a minimal k such that [ D k ( α ) , D k ( β )] contains 1 / D h ( α ) , D h ( β )] for 0 ≤ h ≤ k are embedded. Indeed, themap D doubles lengths of arcs of length less than 1 /
2, hence eventually the length
ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 11 of one of its images is more than 1 /
2. Let k be minimal such that the length of J = [ D k ( α ) , D k ( β )] is at least 1 /
2. Note that all previous iterates are embedded arcs,moreover J must contain either 0 or 1 /
2. However, by minimality of k , since thepreimages of 0 are 0 and 1 / J must contain1 /
2. Since J contains 1 / /
2, andsince J = D k ([ α, β ]) one has D k ( θ ) = 1 /
2, hence k = q −
1, which completes theproof of the claim. (cid:3)
Proof of Proposition 8.
Let k = δ V , and let θ − < θ + be the two endpoints of theleaf. We need to show that D k ( θ + ) belongs to (cid:98) R . Let us set I := ( θ − , θ + ) and I h := D h ( I ). Since the component belongs to V , we have by Lemma 12 that theintervals I h for 0 ≤ h ≤ k + 1 are pairwise disjoint. Moreover, by definition of δ V the interval I k contains 1 /
2. Now if one looks at the lamination it follows that f k +1 ( m ) ∈ H k , so we are in the hypothesis of Lemma 10. Thus, all higher iterates f i ( m ) for i ≥ k + 1 belong to H k . This means that the leaf f k ( m ) separates thepoint { / } from all postcritical leaves f i ( m ) with i ≥
0. Thus, if we consider theorbit { D i ( θ + ) , i ≥ } , we have that no iterate lies in the interior of I k , and pointsin complement of I k are by Lemma 9 at least at distance | D k ( θ + ) − / | from 1 / / D k ( θ + ), so D k ( θ + ) belongs to (cid:98) R , asrequired. (cid:3) Renormalization and landing
We proved that any element in the image of Φ H combinatorially lands on the realaxis. To complete the proof of the main theorem, we need to show it actually lands. Inorder to do so, we will prove that it is not renormalizable, hence it lands by Yoccoz’stheorem. Proposition 14.
Let H be a hyperbolic component which does not lie in the / -limb, and let θ ∈ R / Z be an irrational angle of an external ray which lands on theboundary of H . Then the external ray at angle Φ H ( θ ) is not renormalizable, hencethe corresponding ray lands. The proof uses the concept of maximally diverse sequence from [Sh], which wediscuss in the following section.2.1.
Maximally diverse sequences.Definition 15.
A sequence ( s n ) ∈ { , } N is maximally diverse if the subsequences ( s i + np ) n ∈ N with p ≥ , ≤ i ≤ p − are all distinct. Lemma 16.
A sequence ( s n ) is maximally diverse if and only if the subsequences ( s i + np ) n ∈ N with p ≥ , i ≥ are all distinct. Proof.
Suppose that the two subsequences ( s i + np ) n ∈ N and ( s j + nq ) n ∈ N with p, q ≥ i, j ≥ k ≥ s i + npk ) n ∈ N and ( s j + nqk ) n ∈ N are equal. Now, we can choose k large enough so that i < pk and j < qk ; then s isnot maximally diverse by definition. (cid:3) Corollary 17. If s is maximally diverse, then for any i ≥ and p ≥ the subsequence ( s i + np ) n ∈ N is maximally diverse. Lemma 18.
Let s ∈ { , } N and s (cid:48) = σ ( s ) its shift, i.e. ( s (cid:48) ) n = s n +1 for any n . Then s is maximally diverse if and only if s (cid:48) is maximally diverse.Proof. If s is not maximally diverse, then there exist i, j ≥ p, q ≥ s i + np = s j + nq for all n ≥
0. Hence, also s i + p + np = s j + q + nq for all n ≥
0. Since i + p ≥ j + q ≥
1, this also implies s (cid:48) i + p − np = s (cid:48) j + q − nq for all n ≥
0, hence s (cid:48) is notmaximally diverse. Vice versa, if s (cid:48) is not maximally diverse then there exist i, j ≥ p, q ≥ s (cid:48) i + np = s (cid:48) j + nq for all n ≥
0, which implies s i +1+ np = s j +1+ nq for all n ≥
0, hence s is not maximally diverse by the previous Lemma. (cid:3) Recall an infinite sequence ( (cid:15) n ) is Sturmian if there exists α ∈ (0 , \ Q , β ∈ R such that (cid:15) n = (cid:98) ( n + 1) α + β (cid:99) − (cid:98) nα + β (cid:99) − (cid:98) β (cid:99) for all n ≥ Theorem 19 (Shallit [Sh]) . Sturmian sequences are maximally diverse.
Proof of Proposition 14.
Let θ be an external angle of a ray landing on theboundary of the hyperbolic component H , of period p . If θ is irrational, then θ isthe tuning of an irrational angle η of a ray landing on the main cardioid. Thus, thebinary expansion of θ is θ := S (cid:15) S (cid:15) S (cid:15) . . . where ( S , S ) are the binary expansions of the two rays landing at the root of H ,and ( (cid:15) n ) is the binary expansion of the angle of a ray landing on the main cardioid.Thus, by [BS] the sequence ( (cid:15) n ) is a Sturmian sequence, hence by Theorem 19 it is amaximally diverse sequence.Hence, for any α = 0 , . . . , p − θ α + np ) n ∈ N is maximally diverse(if ( S ) α (cid:54) = ( S ) α ) or is constant (if ( S ) α = ( S ) α ). Thus, if we consider the image σ := Φ H ( θ ) we have by Lemma 18 that either ( σ α + np ) n ∈ N is maximally diverse or iseventually constant.Note that since H does not intersect the real axis, then either ( S ) = ( S ) = 0(if H lies in the upper half plane) or ( S ) = ( S ) = 1 (if H lies in the lower halfplane). In both cases, the subsequence ( θ np ) n ∈ N is eventually constant. Thus, thereexists j ≥ σ j + np ) n ∈ N is eventually constant. ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 13
Let us now suppose by contradiction that the angle σ is renormalizable. Thisimplies that there exist two words Z , Z with | Z | = | Z | = q such that σ ∈ { Z , Z } N .Since by construction the angle σ is real, the only possibility is that ( Z ) i = 1 − ( Z ) i for any i = 1 , . . . , q . Case 1.
Let us first assume that p is not a multiple of q .Then let l := lcm( p, q ), and consider all remainder classes modulo l . Definition 20.
Let us define two remainder classes α, α (cid:48) modulo p to be q -equivalentif there exists m ∈ { , . . . , lq − } and β, β (cid:48) ∈ { , . . . , q − } such that α ≡ qm + β mod p and α (cid:48) ≡ qm + β (cid:48) mod p . Lemma 21.
Let p, q ≥ be two integers, with p not a multiple of q . Now, supposethat a set A ⊆ { , . . . , p − } is not empty and has the following property: if α ∈ A and α (cid:48) ∈ A is q -equivalent to α , then α (cid:48) ∈ A . Then A = { , . . . , p − } .Proof. If p ≤ q , the claim is almost trivial: indeed, the set { , . . . , q − } projects toall possible classes modulo p . Let us suppose now p > q , and let k := (cid:98) pq (cid:99) . Theneach interval A m := [ qm, qm + q −
1] with 0 ≤ m ≤ k lies in some equivalence class.Moreover, if we choose j ∈ { , . . . , p − } such that j ≡ ( k + 1) q mod p , then alsoeach interval B m := [ qm + j, qm + j + q −
1] with 0 ≤ m ≤ k lies in some equivalenceclass. Then, note that B m intersects both A m and A m +1 , hence all remainder classesmust belong to the same equivalence class. (cid:3) Now, let us pick α ∈ { , . . . , p − } such that ( σ α + np ) is eventually constant. Find γ in { , . . . , l − } such that γ ≡ α mod p , and let β ∈ { , . . . , q − } such that β ≡ γ mod q . Then one can write γ = β + mq with 0 ≤ m ≤ lq − σ = Z (cid:15) Z (cid:15) . . . with (cid:15) i ∈ { , } .If for some α ∈ { , . . . , p − } the sequence ( σ α + np ) n ∈ N is eventually constant(e.c.), so is its subsequence ( σ γ + nl ) n ∈ N , which coincides with (( Z (cid:15) m + nl/q ) β ) n ∈ N . Since( Z ) β (cid:54) = ( Z ) β , this implies ( (cid:15) m + nl/q ) n ∈ N is also eventually constant. On the otherhand, if ( σ α + np ) n ∈ N is not eventually constant, then it is maximally diverse, hence byLemma 17 so is ( σ γ + nl ) n ∈ N , hence also ( (cid:15) m + nl/q ) n ∈ N is not eventually constant.Let us now consider another α (cid:48) ∈ { , . . . , p − } which is q -equivalent to α . Thenby the above discussion( σ α + np ) n ∈ N e.c. ⇔ ( (cid:15) m + nl/q ) n ∈ N e.c. ⇔ ( σ α (cid:48) + np ) n ∈ N e.c.By Lemma 21, since we know that there exists at least some α for which ( σ α + np ) n ∈ N is e.c., then each sequence ( σ α (cid:48) + np ) n ∈ N is e.c. for any α (cid:48) ∈ { , . . . , p − } , hence σ isalso eventually periodic, thus the angle θ cannot be irrational. Case 2.
Finally, let us consider the case when p is a multiple of q .Then recall one can write σ = P S (cid:15) S (cid:15) · · · = Z η Z η . . . where P is a finite word of some length k ≥ k is not a multiple of q . Note that since H is contained ineither the upper or the lower half plane, we have ( S ) = ( S ) ; moreover, we claimthat ( S ) p (cid:54) = ( S ) p : in fact, if one considers the external angles θ = .S and θ = .S ,the number of periodic external rays of period which divides p and lie in the interval( θ , θ ) is even, since on each landing point exactly two rays land. Hence, in theirbinary expansion the last digit of S must be opposite to the last digit of S .Now, for each i ≥ j such that Z η j overlaps with both S (cid:15) i and S (cid:15) i +1 . As the last part of Z i must coincide with the first part of S (cid:15) i +1 , and since S and S start with the same symbol, this forces Z η j to be either Z or Z , independentlyof i . However, since the last symbols of S and S are different, this means that S (cid:15) i is also fixed, hence the sequence ( (cid:15) i ) must be eventually constant, which contradictsthe irrationality of θ .Finally, if k is multiple of q , then S and S are finite concatenations of Z , Z ,which means that θ lies already in the small copy of the Mandelbrot set with roots { Z , Z } . Since { Z , Z } represent a real pair, such a small copy of the Mandelbrotset lies in the 1 / H is outside such limb andcompletes the proof of Proposition 14.2.3. Proof of the magic formula.
Proof of Theorem 2. If θ belongs to Ξ H , then by Proposition 5 the angle B H A H · θ belongs to the upper part of the combinatorial vein V on which H lies. Hence,by Proposition 8 the angle Φ H ( θ ) := D δ V ( B H A H · θ ) belongs to (cid:98) R . Finally, byProposition 14 the ray actually lands, hence Φ H ( θ ) belongs to R , as claimed. (cid:3) An alternate formula
We conclude with another possible generalization of Douady’s formula. This versiondoes not depend on the vein structure of M . Proposition 22.
Let H be any hyperbolic component, and Θ H be the set of externalangles of rays landing on the small Mandelbrot set with root H (in particular, this setcontains the set of external angles landing on the boundary of H ). Then there existsan affine map ϕ H such that ϕ H (Θ H ) ⊆ R . Lemma 23.
For each hyperbolic component H of period p > , one has (cid:12)(cid:12)(cid:12)(cid:12) D n ( θ ) − (cid:12)(cid:12)(cid:12)(cid:12) ≥ p for all θ ∈ Θ H , for all n ≥ .Proof. Let Σ = A H and Σ = B H be the binary words which give the binaryexpansion of the two angles landing at the root of H , and let p be the period of H ,which equals the length of both Σ and Σ . By the construction of tuning operators, ENERALIZATIONS OF DOUADY’S MAGIC FORMULA 15 any external angle θ landing on the small Mandelbrot copy associated to H has binaryexpansion of type θ = . Σ (cid:15) Σ (cid:15) . . . Σ (cid:15) n . . . where (cid:15) i ∈ { , } for all i . Since H is not the main cardioid, then both Σ andΣ contain both the symbol 0 and the symbol 1. As a consequence, any block ofconsecutive equal digits in the binary expansion of θ cannot have length larger than2 p −
2. However, all numbers in the interval U p = (cid:0) − p , + p (cid:1) have binaryexpansion of type either . . . . (cid:124) (cid:123)(cid:122) (cid:125) p − or . . . . (cid:124) (cid:123)(cid:122) (cid:125) p − , hence none of the iterates D n ( θ ) can lie in the interval U p . (cid:3) Proof of Proposition 22.
The map is given by ϕ H ( θ ) := 0 1 . . . (cid:124) (cid:123)(cid:122) (cid:125) p − · θ In fact, by the above observation, ϕ H ( θ ) ∈ U p = (cid:0) − p , + p (cid:1) . On the otherhand, consider the other iterates D n ( ϕ H ( θ )) for n ≥
1. If n < p −
1, then D n ( ϕ H ( θ ))begins with 11 so it does not lie in U p . For n = 2 p −
1, then D n ( ϕ H ( θ )) = 1 · θ alsodoes not belong to U p , as θ cannot begin with 2 p − n ≥ p , then D n ( ϕ H ( θ )) = D n − p ( θ ) / ∈ U p again by the Lemma. In conclusion, since ϕ H ( θ ) belongs to U p and none of its forwarditerates does, then ϕ H ( θ ) is closer to 1 / ϕ H ( θ ) belongs to R . (cid:3) References [Bl]
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