Fractal n -gons and their Mandelbrot sets

FRACTAL n-GONS AND THEIR MANDELBROT SETS CHRISTOPH BANDT AND NGUYEN VIET HUNG Abstract. We consider self-similar sets in the plane for which a cyclic group acts transitively on the pieces. Examples like n-gon Sierpiński gaskets, Gosper snowflake and terdragon are well-known, but we study the whole family. For each n our family is parametrized by the points in the unit disk. Due to a connectedness criterion, there are corresponding Mandelbrot sets which are used to find various new examples with interesting properties. The Mandelbrot sets for n > 2 are regular-closed, and the open set condition holds for all parameters on their boundary, which is not known for the case n = 2. MSC classification: Primary 28A80, Secondary 52B15, 34B45 1. Introduction Most of the classical fractals, like Cantor’s middle-third set, von Koch’s curve, Sierpiński’s gasket and carpet, Menger’s sponge etc. [20, 8, 12] are self-similar sets with certain symmetries. That is, they are compact sets which fulfil an equation A = f1 (A) ∪ ... ∪ fn (A) (1) where f1 , ..., fn are contracting similarity maps on Rd . And there is at least one symmetry map s with s(A) = A which permutes the pieces Ak = fk (A), so that s(Ak ) = Aj where j depends on k and s. Self-similarity and symmetry make the shapes look attractive, but what is more, they simplify the description and mathematical treatment of such fractals. An analytic theory including Brownian motion [19], Dirichlet forms, spectrum of the Laplacian [16] and geodesics [25] has been developed only on symmetric fractals. Recently, Falconer and O’Connor [13] have classified and counted certain symmetric fractals. This paper studies a large family of fractals with rich symmetry. A compact subset A of the complex plane is called a fractal n-gon if A fulfils equation (1) for similarity mappings fk (z) = λk z + ck , k = 1, ..., n, and there is a rotation in the plane which acts transitively on the pieces, permuting them in an n-cycle. Some fractal n-gons which generalize the Sierpiński gasket for arbitrary n have been considered by several authors [21, 25]. Some others like twindragon, terdragon, Gosper’s snowflake and the fractal cross [20, Ch. 6] are known as tiles. Figures 1 and 2 show some examples which seem to be new. For each n, all fractal n-gons are parametrized by one complex parameter λ running through the unit disk (Section 2). Since connectedness of A means that neighboring pieces intersect (Section 3), we can define a Mandelbrot set Mn for fractal 1 2 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 1. 3-gon fractals where pieces intersect in one point. Upper row: The point has addresses 01̄ ∼ 102̄ , 01̄ ∼ 10202̄ (Example 1). Lower row: Corresponding reverse fractals, see Section 5. n-gons in Section 4. M2 was introduced by Barnsley and Harrington in 1985 [9, 8] and studied by several authors [10, 11, 2, 22, 23, 24]. We shall see that Mn has similar properties. In section 5, we describe a fast algorithm to generate Mn , and discuss the overlap set of n-gons which determines their geometry. Finally, we consider the simplest n-gons with one-point intersection set, classify them and derive the basic analysis for this infinite family. 2. Parametrization We want to derive a very simple form for the mappings fk which generate a fractal n-gon. We start with the simple observation that a similarity map, that is, a change of the Cartesian coordinate system, will not change the structure of a self-similar set. S Remark 1. Let A = fk (A) be a self-similar set in Rd and h : Rd → Rd a similarity map. Then B = h(A) is the self-similar set generated by the mappings gk = hfk h−1 . S S Proof. B = h(A) = hfk (A) = hfk h−1 (B) ¤ In the complex plane, when h consists of translation, scaling and rotation, i.e. h(z) = µz + c, then gk will have the same factor λk as fk . If h includes a reflection and thus reverses orientation, h(z) = µz + c, then the gk will have the conjugate factors λk . In the present paper we consider all similitudes h as “isomorphisms” and do not distinguish between a self-similar set and its mirror image. FRACTAL n-GONS AND THEIR MANDELBROT SETS Proposition 1. Any fractal n-gon is similar to a self-similar set A = where fk (z) = λz + bk for k = 1, ..., n where b = cos 3 Sn k=1 fk (A) 2π 2π + i sin and |λ| < 1. n n Thus for each n, the points λ in the open unit disk parametrize fractal n-gons. Proof. By definition, a fractal n-gon A has n pieces Ak , k = 1, .., n, and a rotation s transforms the pieces cyclically into each other. We choose the numbering so that s(Ak ) = Ak+1 where + is taken modulo n. In the sequel we shall often write A0 and f0 instead of An , fn . By applying a translation h(z) = z +v we push the origin of our coordinate system to the fixed point of s so that s(z) = bz where bn = 1 since sn preserves all pieces of A and so must be the identity map. Since bk runs through all roots of unity, it is no loss of generality (but requires another renumbering of the Ak ) to assume that b = cos 2π + i sin 2π . n n We took the fk to be orientation-preserving, so f0 (z) = λz + u for some λ, u ∈ C with |λ| < 1. Applying the rotation h(z) = z/u we transform A into a position where f0 (z) = λz + 1. Now we note that sk f0 s−k (A) = sk f0 (A) = fk (A) for k = 1, ..., n − 1. Thus Sn−1 A = k=0 sk f0 s−k (A) so due the uniqueness of self-similar sets [12, 8] the set A is the self similar set with respect to the mappings sk f0 s−k (z) = λz+bk , k = 0, ..., n−1. These mappings will now be called fk . ¤ The points z in a self-similar set are often described by their addresses u1 u2 ..., sequences of symbols um ∈ {0, ..., n − 1} which denote pieces and subpieces of A to which z belongs [8, 12, 5]. We have z = p(u1 u2 ...) = lim fu1 · · · fum (0) m→∞ (2) where p : {0, ..., n − 1}∞ → A is the so-called address map [8]. In our case, the address map has a particularly simple form. Remark 2. The point z with address um ∈ {0, ..., n − 1} in the fractal nPu1 u2 ..., um m−1 λ . Thus A(λ) is the support of a gon A(λ) has the representation z = ∞ b m=1 random series of powers of λ with coefficients chosen from the n-th roots of unity. Proof. fu1 · · · fum (0) = bu1 + λbu2 + · · · + λm−1 bum . ¤ Example 1. We explain how to determine λ for the first fractal in Figure 1. The intersection point z should have the addresses 01̄ and 102̄ where 1̄ = 111... This is indicated by the relation 01̄ ∼ 102̄ which by the above remark turns into an equality for λ. (In Section 5 we shall see that this is indeed sufficient.) p(01̄) = 1 + b(λ + λ2 + ...) = 1 + b λ λ2 = p(102̄) = b + λ + b2 1−λ 1−λ 4 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 2. Some 4-gon fractals. In the upper row, pieces intersect in two points, in the lower row in four points. See Example 2. which leads to the quadratic equation λ2 (1 + b) − 2λ + 1 = 0. The solution with |λ| ≤ 1 is √ 3 1 λ = + (1 − )i . 2 2 For the upper-right part of Figure 1 with 01̄ ∼ 10202̄ we have to consider p(10202̄) = b + λ + b2 λ2 + λ3 + b4 λ4 /(1 − λ) which leads to the equation z 4 − z 3 + z 2 + 2bz = b. Again there is only one solution with |λ| ≤ 1 which is determined numerically: λ ≈ 0.5135 + 0.1004i. In both cases 1/λ is a Pisot number over Q(b). In the lower row of Figure 1, λ was replaced with −λ (see Section 10). 3. Connectedness Connectedness of fractals is a very important property, which is well-known from the dynamics of complex quadratic maps. For self-similar sets, it implies local connectedness and arcwise connectedness [14, 6]. A self-similar set with n pieces Ak is connected if and only if the graph with vertices k = 1, ..., n and with edges {j, k} for intersecting pieces Aj ∩ Ak 6= ∅ is connected [14, 6]. Thus a self-similar set A with two pieces A0 , A1 is connected if and only if A0 ∩ A1 6= ∅, otherwise it is a Cantor set [8, Ch. 8]. It is not known whether a similar property holds for self-affine tiles A with respect to the mappings fk (x) = M −1 x + kv, k = 0, 1, ..., n − 1 where M is an expanding integer d × d matrix with determinant n and v ∈ Rd . Using algebraic methods, Kirat, Lau and Rao FRACTAL n-GONS AND THEIR MANDELBROT SETS 5 [17, 18], Akiyama and Gjini [1] verified connectedness in this case for dimensions d = 2, 3 and 4. With a topological argument we prove now that the alternative “connected or Cantor set” also holds for fractal n-gons. Due to the rotational symmetry, A0 ∩A1 6= ∅ implies Ak ∩ Ak+1 6= ∅ for each k which implies connectedness of A due to the above criterion. The necessity of A0 ∩ A1 6= ∅ is more difficult to prove. Theorem 2. A fractal n-gon A is connected if and only if A0 ∩ A1 6= ∅. If A is disconnected, then it is totally disconnected, and all pieces Ak are disjoint. Proof. Let B(x, r) denote the ball around x with radius r. Let γ be the smallest positive number such that A ⊂ B(0, γ) = B0 . For any word u = u1 ...um ∈ {0, ..., n− 1}m of length |u| = m let fu = fu1 · · · fum . For m = 1, 2, ... let [ [ B(fu (0), |λ|m γ) Bm = fu (B(0, γ)) = |u|=m |u|=m be the m-th level ball approximation of A. We shall use the following fact. (*) If Bm is connected and f0 (Bm ) ∩ f1 (Bm ) 6= ∅, it follows that fj (Bm ) intersects S fj+1 (Bm ) for j = 1, ..., n − 1 and Bm+1 = n−1 j=0 fj (Bm ) is connected. Now we distinguish two cases. Case one. A0 ∩ A1 6= ∅. In this case (*) implies by induction that all Bm are connected. Then A must be connected, too: if A would split into two disjoint nonempty closed sets F0 , F1 , then the distance of F0 and F1 is a positive ǫ and for |λ|m γ < ǫ the set Bm could not be connected. Case two. A0 ∩ A1 = ∅. Then A0 and A1 have some positive distance ǫ, and by the argument just given there is a smallest m with f0 (Bm ) ∩ f1 (Bm ) = ∅. By (*), Bl is connected for all l = 0, 1, ..., m. We show that all Aj are disjoint and hence A is totally disconnected. Since Bm is a connected finite union of closed balls of radius |λ|m γ, the outer boundary of Bm (those points which can be connected with infinity in the complement of Bm ) forms a closed curve Γ consisting of finitely many arcs of circles of this radius. Γ may have finitely many double points where balls touch each other, but no other multiple points. Thus we can consider Γ as a curve oriented in clockwise direction, without self crossings, only touching points. Let Γj = fj (Γ) for j = 0, ..., n − 1. We assumed Γ0 ∩ Γ1 = ∅ and want to show that all Γj are disjoint. Now let γ1 be the smallest positive number with Γ0 ⊂ B(0, γ1 ), and C the circle around 0 with radius γ1 . Then C intersects Γ0 in a point c0 and Γj in the point cj = bj c0 . The origin 0 is not inside Γ0 , because then it would also be inside Γ1 , which implies that Γ0 and Γ1 intersect. Now there is a largest γ2 > 0 such that the open ball B(0, γ2 ) does not intersect Γ0 . Let D be the circle around 0 with radius γ2 . Then D intersects Γ0 in a point d0 and each Γj in the point dj = bj d0 . In the ring formed by D and C, the Γj sit like spokes in a wheel, connecting dj with cj . Of course Γ0 may touch D in other points than d0 , but they all must lie between dn−1 and d1 because otherwise the Jordan curves Γ0 and Γ1 (or Γn−1 ) would 6 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 3. 4-gon fractals where pieces intersect in a Cantor set intersect. But this means that Γ0 is enclosed by Γ1 , Γn−1 and the arcs dn−1 d1 on D and cn−1 c1 on C. Since Γ0 does not intersect the neighbor curves and does not cross the circles D and C, it can not intersect the curves Γ2 , Γ3 , ..., Γn−2 . Thus all Γj , and hence all Aj are pairwise disjoint. ¤ When dealing with self-similar sets, it is very natural to require the open set condition, OSC [12, 8, 7]. We say that OSC holds for the mappings f1 , ..., fn if there is a nonempty open set U with fk (U ) ⊂ U and fj (U ) ∩ fk (U ) = ∅ for j, k ∈ {1, ..., n}, j 6= k. (3) If all Ak are S disjoint,ǫ with pairwise distance ≥ ǫ > 0, then such U exists: we can take U = x∈A B(x, 2 ). For connected sets A in the plane, Bandt and Rao [7] proved that OSC holds when pieces intersect in a finite set. Thus all examples in Figures 1 and 2 fulfil OSC. Proposition 3. If |λ| > √1 n for n ≥ 2, then OSC fails, and A is connected. Proof. Assume OSC holds (which is true if A is disconnected). Then the Hausdorff dimension of A is dimH (A) = −loglog|λ|n [12]. On the other hand, dimH (A) ≤ 2 since A is a subset of the plane. It follows that |λ| ≤ √1n . This bound is sharp for n = 2, 3, 4 where we have fractal tiles (twindragon or rectangle, terdragon, square). ¤ Proposition 4. If n < 25 and a fractal n-gon A fulfils OSC then only consecutive pieces Ak , Ak+1 can intersect. Proof. For n ≤ 3 all pieces are neighbors, so let n ≥ 4. Assume OSC holds and r r ) and Aj ⊂ B(bj , 1−r ) A0 ∩ Aj 6= ∅ for some j ∈ {2, ..., n − 2}. Since A0 ⊂ B(1, 1−r 2r 2r 2 j this implies |1−b | ≤ 1−r . Since j ∈ {2, ..., n−2}, it follows that |1−b | ≤ 1−r which 2 2 |1−b | |1−b | √1 can be reformulated as r ≥ 2+|1−b 2 | . However, calculation shows that 2+|1−b2 | > n for 5 ≤ n ≤ 24 which contradicts Proposition 3. For n = 4 we just get equality when r = |λ| = 21 , and the balls with centers 1 and b2 = −1 meet in zero. Now P 1 km m ¤ 1+ ∞ m=1 b λ = 0 is only possible for λ = ± 2 which gives the square. We conjecture that the condition n < 25 is not needed here. FRACTAL n-GONS AND THEIR MANDELBROT SETS 7 Figure 4. Mandelbrot sets for 3- and 4-gons with symmetry lines 4. Mandelbrot set for n-gons To get an overview over all fractal n-gons, we work in the parameter space. Let us define the Mandelbrot set for fractal n-gons as Mn = {λ | |λ| < 1 , A(λ) is connected }. The set M2 was introduced 1985 by Barnsley and Harrington [9]. It was discussed in Barnsley’s book [8] and investigated by Bousch [10, 11], Bandt [2], Solomyak and Xu [24] and Solomyak [22, 23]. Nevertheless, it is still an open question whether M2 is regular-closed. Moreover, it is not known whether OSC is fulfilled for all λ in the boundary ∂M2 . We shall solve these questions for Mn with n > 2, n 6= 4. So let us consider Mn for n > 2. Figure 4 shows Mn for n = 3, 4 as black subset of the region 0 ≤ arg(λ) ≤ π2 , 0 ≤ |λ| < √1n . Proposition 5. Elementary properties of Mn for n ≥ 2. (i) Mn is a closed subset of the open unit disk. (ii) For even n, it has the dihedral group Dn as symmetry group. For odd n, D2n acts as symmetry group. Proof. (i). A(λ) depends continuously on λ with respect to the Hausdorff metric [8], and the limit of connected closed sets with respect to the Hausdorff metric cannot be disconnected. (ii). Since A(λ) is a mirror image of A(λ) (cf. Section 2), all sets Mn are symmetric under reflection at the real axis, λ → λ. Moreover, A(λ) = A(b · λ) for b = cos 2π + i sin 2π , by an argument in the proof n n of Proposition 1. Thus Mn is also invariant under multiplication with b, that is, rotation around 2π/n, as can be seen in Figure 4. For odd n, however, Mn is also invariant under the map λ → −λ and hence under rotation around π/n. This is proved in Proposition 10,(i). ¤ 8 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 5. Magnification of M4 in a square of side length 0.003. For λ = 0.3021 + 0.3375i taken from the upper right hole, the fractal n-gon must be a Cantor set, for nearby λ it will be connected. M2 has an antenna on the real axis since for real λ ∈ [0.5, 0.6] the set A(λ) will be an interval but for nearby λ off the real axis it will be a Cantor set. For n > 2 we cannot get intervals A(λ) and we have no antenna on Mn . On the whole, Mn seems to become “smoother” for larger n. Nevertheless, there are apparent holes in M3 and M4 as were verified for M2 [2]. That means there are Cantor sets A(λ) such that when we decrease λ continuously to reach λ = 0 we must get a connected A on the way. See Figure 5. Proposition 3 says that all λ with |λ| ≥ √1n belong to Mn . This bound can be improved for n ≥ 5, and a lower bound for points in Mn can be given which shows that for large n the boundary of Mn lies in a very thin ring. Theorem 6. Bounds for Mn . (i) All λ with |λ| < π sin n π 1+sin n do not belong to Mn . For odd n, a better estimate is |λ| < (ii) Mn includes all λ with |λ| ≥ π sin n π cos 2n +sin sin nπ cos2 π 2n + √ 3 2 sin πn π n which equals 1 2 for n = 3. for odd n, and |λ| ≥ q sin nπ 1+ √ 3 2 sin for even n. 2π n (i) is proved after Remark 4 below, (ii) after Lemma 1 and 2 in Section 6. In Section 9 and 10 we show that the estimate (i) is sharp for all n which are not multiples of 4. FRACTAL n-GONS AND THEIR MANDELBROT SETS 9 5. How to generate Mn To draw Mn , we must know when the pieces A0 and A1 ofP A(λ) intersect. By ∞ km m Remark P∞ jm2, ma point z ∈ A0 ∩ A1 can be written as z = 1 + m=1 b λ = b + m=1 b λ where km , jm ∈ {0, ..., n − 1}. Thus Remark 3. λ belongs to Mn if and only if λ is the solution of an equation ∞ X g(λ) = 1 − b + dm λm = 0 (4) m=1 where all dm are in © ª ∆n = bk − bj | j, k ∈ {0, 1, ..., n − 1} . For fixed λ, every choice of km , jm , m = 1, 2, ... such that the dm = bkm − bjm solve the above equation, corresponds to one point in the intersection set A0 ∩ A1 . Of course the choice of dl is restricted since the absolute value of the remaining P δ|λ|l+1 m sum ∞ m=l+1 dm λ is at most 1−|λ| where δ = max{|d| | d ∈ ∆n } ≤ 2. If the absolute value of the sum up to degree l is larger than this value, then there is no chance to extend the sum to obtain a series with g(λ) = 0. This argument, with division by λ at each step, provides an algorithm for generating Mn . We search through a rooted tree with |∆n | branches at every vertex. Each δ|λ| vertex is assigned a complex number v. If |v| > 1−|λ| then the vertex will be considered as dead, and the search will not go deeper at this place. We stop the search if either all vertices are dead, in which case λ 6∈ Mn , or if we reach a prescribed level lmax in which case λ belongs to our approximation of Mn . Remark 4. (Algorithm to generate Mn ) For given λ and lmax we consider words u = d1 d2 d3 ...dl ∈ ∆ln with l ≤ lmax and d1 = 1 − b, and corresponding points vu ∈ C according to the rule vu vd1 = d1 = 1 − b , vud = + d , d ∈ ∆n . (5) λ δ|λ| then u and its successor words are removed. If no word of length When |vu | > 1−|λ| lmax remains, λ does not belong to Mn . If at least one word of length lmax remains, λ is considered as an element of Mn . δ|λ| then λ is not in Mn . Since |vd1 | = 2 sin πn and δ = 2 In particular, if |vd1 | > 1−|λ| π for even n, while δ = 2 cos 2n for odd n, this proves Theorem 6,(i). Remark 5. If the algorithm would be calculated up to infinity, it would also decide about OSC for A(λ) : The mappings h = fj−1 fk with j = j1 ...jl , k = k1 ...kl are translations h(z) = z + wjk where ′ ′ 1−b wjk bk − bj w01 = for f1−1 f0 , and wjj ′ kk′ = + . λ λ λ So the w’s differ from the v’s above only by the factor λ. Now OSC holds if and only if 0 is an accumulation point of the w’s [4, 3], or equivalently, of the v’s. 10 CHRISTOPH BANDT AND NGUYEN VIET HUNG In particular, when vu = 0 is obtained after finitely many steps, that is, (4) holds with a finite sum, then OSC cannot be true. In this case, certain pieces Aj and Ak with k1 = 0 and j1 = 1 coincide, so that the overlap set A0 ∩ A1 is really big. For n = 2, the algorithm correctly states when some λ is not in M, and hence OSC holds, but only for the special values λ just mentioned it can guarantee that they belong to M [2]. We shall see that for n > 2 the algorithm works much better, due to the structure of ∆n . 6. The difference sets ∆n ∆n contains 0, and all sides and diagonals, considered as vectors in either direction, of the regular n-gon formed by the vertices bk , k = 0, 1, ..., n − 1. Thus ∆2 = {−2, 0, 2}, and the√8 non-zero vertices ±2, ±2i, ±1± i of ∆4 lie on a square. π π ∆3 consists √ of 0 and 3ei( 6 +k 3 ) , k = 0, 1, ..., 5, vertices of a regular hexagon, all with modulus 3. Thus we recover Theorem 6,(i): each point λ ∈ M3 fulfils |λ| ≥ 12 δ|λ| . Moreover, |λ| = 21 can only lead to some v1d with because otherwise |v1 | > 1−|λ| |v1d | ≤ |v1 | if λ = ± 12 , or λ = 21 bk . This is just the Sierpiński gasket and its reverse, see section 8. There is a significant difference in the structure of ∆n for odd and for even n. Let us start with odd n = 2q + 1. Sides and diagonals of all lengths of the regular n-gon appear in each of the possible 2n directions ±i, ±ib1/2 , ..., ±ib(2n−1)/2 exactly once. The length of a side is s = 2 sin πn , and the diagonals have lengths π 2 sin 2π , ..., 2 sin qπ = 2 cos 2n . Thus n n tπ | t = 0, 1, ..., q, l = 0, 1, ..., 2n − 1}. n From a geometric viewpoint, ∆n contains q points on 2n equally spaced rays around 0. Let C1 , ..., Cq and D1 , ..., Dq denote the points on two neighboring rays, and let P Q denote the distance between points P and Q. Then 0C1 = 0D1 = s and Ck Dk+1 = Dk Ck+1 = s for k = 1, ..., q − 1 since △0Ck Dk+1 has a congruent triangle build from three vertices of the regular n-gon. We shall need an estimate for the size of circular holes in ∆n . ∆n = {2ibl/2 sin Lemma 1. For odd n, the set ∆n intersects each ball B(x, r) with |x| ≤ 2 and π r ≥ 2 sin 2n . The estimate of r is sharp. Proof. If the center of a circle lies in a triangle (or a trapezium), and the vertices are outside the circle, then the circle must be smaller than the circumscribed circle of the triangle (or trapezium). Thus it suffices to show that the radius r of the π . (The points x with circumcircle of each trapezium Ck Ck+1 Dk+1 Dk equals 2 sin 2n |x| ≤ 2 outside the segment Cq Dq between the two rays are covered by the balls of π around Cq and Dq .) radius 2 sin 2n Let M denote the center of this circle, and let α = ∠0Dk+1 Ck . In △0Dk+1 Ck we have 0Ck / sin α = s/ sin πn = 2. Since ∠Ck M Dk = 2α, we have in △0M Ck a similar π . Now r is obtained from the two equations. ¤ relation: 0Ck / sin α = r/ sin 2n FRACTAL n-GONS AND THEIR MANDELBROT SETS 11 The proof of Theorem 6,(ii) for odd n is quite similar. We show that when λ fulfils the condition, the algorithm of Remark 4 will produce points vu ∈ B = B(0, 2 sin πn ) for words u of arbitrary length. Since v1 = 1 − b ∈ B, it is enough to prove that v ∈ B implies ( λv +∆n )∩B 6= ∅. The lemma takes care of | λv | ≤ 2. So we assume that π the circle C of radius 2 cos 2n does not contain 0. We check under which condition C intersects B in an arc subtending an angle ≥ πn at M = λv , because at least one point of such an arc belongs to λv + ∆n . Let D be an intersection point of C with the boundary of B. Since ∠DM 0 increases when the distance y = 0M decreases, it suffices to determine y for the case π π π ∠DM 0 = 2n . The cosine formula gives y 2 − 4y cos2 2n + 4(cos2 2n − sin2 πn ) = 0. Since √ 2 sin π π π y > 2 cos 2n , calculation gives y = 2 cos2 2n + 3 sin πn . The condition | λv | ≤ |λ| n ≤ y provides the estimate for |λ|. ¤ For even n = 2q, every side and diagonal of the regular n-gon has a parallel side, except for the longest diagonals which are diameters of the unit circle. Again, the , ..., 2 sin qπ = 2. length of a side is s = 2 sin nπ , and the diagonals have lengths 2 sin 2π n n However, now there are n directions for the sides and diagonals with “odd” length 2 sin 3π , 2 sin 5π , ... alternating with n other directions for the “even” diagonals. For n n n = 4p we get ∆n = {2bl sin tπ n ∪ {2b l+ 12 | l = 0, ..., n − 1, t = 0, 2, ..., 2p} sin tπ | l = 0, ..., n − 1, t = 1, 3, ..., 2p − 1} , n and for n = 4p + 2 we have t = 1, 3, ..., 2p + 1 in the first part and t = 0, 2, ..., 2p in the second. In both cases, we can label the points on any two neighboring rays, ordered with respect to their distance to 0, as C0 = 0, C1 , ...Cq . Since 0Ck represent for the diagonals, we have Ck Ck+1 = s and ∠Ck Ck−1 Ck+1 = ∠Ck Ck+1 Ck−1 = kπ n k = 1, ..., q − 1. Again, we can derive an estimate for the circular holes in ∆n . Lemma √ 2. For even √ n, the set ∆n intersects each ball B(x, r) with center |x| ≤ 2 and r ≥ 2 sin πn = s/ 2. For n = 4p the estimate of r is sharp. Proof. We consider the triangulation of ∆n into isosceles triangles with side length s described above. We can modify it so that the triangles contain no angles > π2 . (Where such angle appears, there is an adjacent triangle so that the union of both triangles is a rhombus, and then we change the diagonal of this rhombus.) Now √ it suffices to show that the radius r of the circumcircle of each triangle is ≤ s/ 2. However, this is just the value for a right-angled isosceles triangle with two sides s, and for an acute isosceles triangle the value is smaller. The proof is complete, = π4 and we see that the estimate is sharp for n = 4p where ∠Ck Ck+1 Ck−1 = kπ n happens for k = p. For n = 4p + 2 the largest possible angle between the two equal 2p π = 2p+1 which leads to the sharp estimate. In particular, for sides is π − 2 (p+1)π n 2 √ n = 6 we have r ≤ 1/ 3 which is also obvious from the structure of ∆6 . ¤ Proof of Theorem 6,(ii) for even n. As for odd n, we want that v ∈ B = B(0, 2 sin πn ) implies ( λv + ∆n ) ∩ B 6= ∅ so that the algorithm of Remark 4 gives 12 CHRISTOPH BANDT AND NGUYEN VIET HUNG vu ∈ B for arbitrary long words u. Thus λ has to be taken so that 2 sin πn ) ⊆ ∆n + B. B(0, |λ| S We consider F = B(0, 2) ∪ 2n k=1 zk + B where zk are the outmost points on the 2n rays of ∆n . By Lemma 2, F ⊆ ∆n + B. The points with minimal modulus on the boundary ∂F are intersection points of two neighboring outer circles. Let w be the outer intersection point of z1 + ∂B and z2 + ∂B, and assume |z1 | = 2, |z2 | = 2 cos πn . Then |z1 − z2 | = 2 sin πn since ∠0z2 z1 is a right angle. Thus △z1 z2 w is an equilateral triangle. Choose y on the ray 0z1 such that ∠0yw is a right angle. Since α := ∠wz1 y = π6 + πn , we have by Pythagoras √ π 2π 3 π 2 2 2 sin ). |w| = (2 + 2 sin cos α) + (2 sin sin α) = 4(1 + n n 2 n If the condition (ii) for λ is fulfilled, then π 2 sin n |λ| ≤ |w| and B(0, π 2 sin n ) |λ| ⊆ F. ¤ 7. OSC holds on the boundary of Mn Now let us improve the algorithm of Remark 4. We give a condition for the vu which guarantees that λ belongs to Mn , for all n ≥ 5 and n = 3. In our pictures of Mn , either this condition or the Cantor set condition of Remark 4 was fulfilled for each single pixel. To obtain precise figures, we did not need to go far into the recursion: level lmax = 60 was sufficient for magnification up to factor 108 . Theorem 7. For n 6= 2, 4 there is a critical radius rn with the following property. If |vu | ≤ rn for some u in the algorithm of Remark 4, where |λ| ≥ 12 for n = 3, and sin π |λ| ≥ 1+sinn π for n ≥ 5, then λ is in Mn . n √ π for odd n ≥ 3, rn = 2 sin πn for even n ≥ 8, and r6 = √13 . Proof. Let rn = 2 sin 2n We assume |vu | ≤ rn and show that |vud | ≤ rn for some d ∈ ∆n . Then we can use induction to extend u to an infinite sequence d1 d2 ... which fulfils (4). Thus we want to show vud = vλu + d belongs to B(0, rn ). If we verify |x| ≤ 2 for x = vλu then Lemma 1 and 2 say that B(x, rn ) intersects ∆n . Thus there is b with |b| ≤ rn and d′ ∈ ∆n with vλu + b = d′ . In other words, vu − d′ = −b belongs to B(0, rn ). λ Since |vu | ≤ rn was assumed, it remains to show |r|λ|n | ≤ 2. For n = 3 we have r3 = 1 and |λ| ≥ 21 . For odd n > 3, π 2 sin 2n 1 + sin nπ |rn | π ≤ ) = (1 + sin π |λ| sin πn n cos 2n which is < 2 for n = 5 and, by monotonicity, for n > 5. For n = 6 we have r6 = √13 √ ¤ and |λ| ≥ 13 . For even n > 6, cancellation results in |r|λ|n | ≤ 2(1 + sin πn ) < 2. FRACTAL n-GONS AND THEIR MANDELBROT SETS 13 Theorem 8. For all n 6= 2, 4, the open set condition holds for all λ on the boundary ∂Mn of Mn . Proof. Take λ ∈ ∂Mn , and assume OSC does not hold. Then 0 is an accumulation point of the vu by Remark 5, and so |vu | < rn for some u in the algorithm of Remark 4. Since vu depends continuously on λ there is a neighborhood B = B(λ, ε) of λ such that for each λ̄ ∈ B we also have |vu | < rn . By Theorem 6,(i), ε can be taken so that all λ̄ ∈ B fulfil the lower bound in Theorem 7. (The only exceptions are n = 3 and λ = 21 , which concerns the Sierpiński gasket and its reverse, see Section 10, and sin π λ = 1+sinn π for even n = 4p + 2 ≥ 6, which are discussed in Example 4. In both n cases, OSC holds.) Theorem 6 now says that B ⊂ Mn , so λ cannot be a boundary point of Mn . ¤ We proved that all λ for which OSC fails are in the interior of Mn . The converse is not true: a few λ in the interior of Mn are known for which OSC √holds, like the twindragon and tame twindragon for n = 2, the terdragon (λ = 12 + 63 i) for n = 3, and Gosper’s snowflakes (see Section 9) for n = 6. Question. Are there λ in the interior of Mn which fulfil OSC and are not related to tilings? In particular, do such parameters exist for n = 5 ? Is the number of such examples finite for each n ? 8. Mn is regular-closed Our next statement, as well as Theorem 8, remains an open problem for n = 2 [22, 23, 24]. Theorem 9. For all n 6= 2, 4, the set Mn is the closure of its interior. P∞ m Proof. Let λ ∈ Mn , that is, g(λ) = = 0 for certain dm ∈ ∆n . m=0 dm λ Taking any ε > 0 we have to verify that there is µ ∈ int Mn with |µ − λ| < ε. By Proposition 3 this is true for n = 3, λ ≥ √13 and for n ≥ 5, λ ≥ 12 . √ Otherwise, since |dm | ≤ 2 for all n and |dm | ≤ 3 for n = 3, the function g is analytic in λ. So the zeros of g do not accumulate at λ, and we find ε′ ≤ ε such that in the ball B(λ, ε′ ), only z = λ fulfils g(z) = 0, and moreover |z| < 21 (< √13 in case P n = 3). Let δ = min{|g(z)| | |z − λ| = ε′ }, and let M be so large that m h(z) = ∞ < δ for |z − λ| ≤ ε′ . Now the well-known theorem of Rouché m=M dm z P −1 m has exactly (cf. [22, 23, 2]) implies that the polynomial g(z) − h(z) = M m=0 dm z one zero µ in B(λ, ε′ ). By Remark 5, µ ∈ int Mn . ¤ We do not study connectedness and local connectedness of Mn by the methods of Bousch [10, 11]. See [2, Section 10] for a discussion of these arguments in case n = 2. For odd n ≥ 3, the previous results and computer experiments like Figure 6 indicate that the structure of Mn is very simple. Conjecture. For odd n ≥ 3, the boundary of Mn is a Jordan curve. For even n ≥ 4 it is a countable union of disjoint Jordan curves. 14 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 6. Left: M3 in a square of side length 10−5 with center (0.479227, 0.276787). This is one of the suspicious places, but further magnification shows no holes. Right: a hole in M6 . Side length 10−4 , center (0.33643, 0.15616). Remark 6. (The case n = 4) √ For n = 4, Lemma 2 reads as follows. B(x, r) intersects ∆4 whenever |x| ≤ 5 and r ≥√1. This allows to prove Proposition 7 with r4 = 1, under the √ assumption |λ| ≥ 1/ 5. This proves Theorem 8 for all λ ∈ ∂M4 with modulus ≥ 1/ 5, that is, for the region x ≥ 0.4, y ≤ 0.2 in Figure 4. In this region, M4 is also regular-closed. Thus although n = 4 was excluded from the theorems, there is at least a partial result, as was proved for n = 2 by Solomyak and Xu [24]. 9. Overlap set and examples for even n Now we discuss the size of the overlap set D = A0 ∩ A1 . For n = 2 we have b ∈ {1, −1} so that d = 0 has two representations bk − bj while d = 2 and d = −2 admit a unique representation. If λ is the root of just one power series g(λ) and this g has only coefficients ±2 then D is a single point. If N of the coefficients are zero, D has cardinality 2N , and if there are infinitely many zero coefficients, D is a Cantor set [23, 5]. For even n > 2, the situation is very similar. We shall not consider zero coefficients since we conjecture that λ cannot be the root of only one g(λ) with coefficients in ∆ when there are zero coefficients in q. However, in a regular n-gon with even n ≥ 4, as it is formed by the roots bk , k = 0, ..., n − 1, every side and diagonal, except for the longest ones, has one parallel side or diagonal, and the corresponding d has two representations bk − bj . Thus only those power series g(λ) where every dm has the form bkm − bkm +n/2 can yield a single intersection point, see Example 4 below. If we have N coefficients dm which do not correspond to the longest diagonals, we shall have 2N points in D, see Figure 2. And when there are infinitely many such d then D will be a Cantor set, as in Figure 3. k Example 2. For n = 4 we have b = i and the bk form a square. The 4-gon in the lower row of Figure 2 was obtained from the relation of addresses u1 u2 ... = 01̄ ∼ 130303̄ = v1 v2 ... Here um = vm + 2 mod 4, corresponding to a diagonal of the square, for all m > 1 except m = 3 and m = 5 where (uk , vk ) = (1, 0) which can FRACTAL n-GONS AND THEIR MANDELBROT SETS 15 Figure 7. Fractal 6-gons with central piece. Note the difference √ √ 5+ 3i 2+ 3i between Gosper snowflakes with λ = 14 and 7 . be replaced by (2, 3) since i − 1 = i2 − i3 . Thus there are 4 pairs of addresses which give the same equation λ5 λ 2 3 4 = p(130303̄) = i − iλ + λ − iλ + λ − i . p(01̄) = 1 + i 1−λ 1−λ which results in λ5 − λ4 + λ3 − λ2 + λ(1 + 2i) = i with numerical solution λ ≈ 0.4587 + 0.1598i. Example 3. As a landmark point in M4 , we consider the intersection of the symmetry line in Figure 4 with the boundary of M4 . It turns out that this is λ= 1+i √ . 1+ 5 The corresponding fractal, shown on the left of Figure 3, is mirror-symmetric: A(λ̄) = A(−iλ) = A(λ). It seems the only mirror-symmetric 4-gon beside the square which fulfils OSC. It is possible to determine λ by identifying addresses u = u1 u2 ... = 010033221 and v = 133221100. The corresponding equation (4) results in a geometric series with −λ, and the resulting polynomial of degree 4 factors into two quadratic polynomials, of which 2z 2 +(1+i)z−i has λ as the root with modulus ≤ 1. Thus 1/λ is a quadratic Pisot number over Q(i). 16 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 8. n-gons for odd n with several series g(λ) : Examples 6,7. Note that every second combination (uk , vk ) can be replaced by one other choice. For instance (0, 3) and (1, 2) give the same d = 1 + i. Thus D = A0 ∩ A1 is uncountable, and it is a linear Cantor set. In fact D is a self-similar set with respect to two homotheties with factor r2 = |λ|2 : For E = f0−1 (D), inspection of Figure 3 shows that E = g1 (E) ∩ g2 (E) with g1 (z) = f12 (−iz) and g2 (z) = f1 f0 (−iz). So the 4 2 Hausdorff dimensions of A and D are −log and −log , respectively [12], and the log r log r 2 dimension of D is just one quarter of the dimension of A. Clearly D fulfils OSC. For A we can explicitly construct an open set - an octagon with interior angles 34 π and with alternating side lengths s, rs for some constant s (for our choice of mappings s = 4r). Drawing this octagon U and the fi (U ), we can determine r in alternative way, by elementary geometry. Example 4: Sierpińskiπ n-gons for even n. We determine all λ ∈ Mn with sin minimal modulus |λ| = 1+sinn π (cf. Theorem 6,(i)). We use the algorithm of Remark n 2|λ| . Since 4, starting with v = vd1 = 1 − b which just has the critical modulus 1−|λ| v k j | λ | − |v| = 2, there is at most one choice of d2 = b − b for which |vd1 d2 | is not larger than the critical modulus: d2 must be a diameter of the unit circle which is a diagonal of the regular n-gon and parallel to 1 − b. For n = 4p such diagonal does not exist. For n = 4p + 2, the only choice is k = p + 1, j = 3p + 2. This leads to vd1 d2 = vd1 , so that by induction dk = d2 for all k > 2. So the single λ with minimal sin π modulus is the real number λ = 1+sinn π . The formula in [21] is more complicated. n By Remark 5, OSC holds. Remark 7. (Adding a central piece for n = 6) The set ∆6 is special in the sense that it contains the bk , not only their differences. Thus whenever OSC holds for some λ, we can add an seventh map f6 (z) = λz, and the extended system of mappings fulfils OSC. This follows from Remark 5. Moreover, A(λ) is connected if and only if the extended self-similar set is, because A√0 ∩ A1 = A6 ∩ A2 . Figure 7 shows some examples. The Gosper tiles fulfil |λ| = 1/ 7 which explains the bound in Theorem 6,(ii) for n = 6. FRACTAL n-GONS AND THEIR MANDELBROT SETS 17 10. Overlap set and examples for odd n A regular n-gon with odd n has no parallel sides, and no parallel diagonals of the same length. So d = 0 is the only vector in ∆ which can be represented as d = bk −bj in different ways - actually in n ways. When we have OSC, however, coefficients 0 are unlikely to appear in the power series of g(λ), at least for n ≥ 5. So it seems we have the following alternative: either D is a singleton (Figure 1), or λ is the root of several power series g(λ) (see Figure 8). Example 5: Sierpiński n-gons for odd n. We determine all λ ∈ Mn with π sin n minimal modulus |λ| = cos π +sin π (cf. Theorem 6,(i)). We use the algorithm of 2n n Remark 4, starting with v = vd1 = 1 − b which has the critical modulus 2 sin nπ = δ|λ| π , where δ = 2 cos 2n . Since | λv |−|v| = δ, there is at most one choice of d2 = bk −bj 1−|λ| for which |vd1 d2 | is not larger than the critical modulus: d2 must be the longest diagonal of the regular n-gon which is parallel to 1 − b. As for even n, we conclude that dk = πd2 for all k > 2. So the single λ with minimal modulus is the real number sin n λ = cos π +sin π . The formula in [21] is more complicated. OSC holds by Remark 5. 2n n Example 6. Considering 2-gons, Solomyak [23] asked whether there is a λ which is root of several power series g(λ) and still admits finite intersection set D. For n = 3, Figure 8 shows an example, with address identifications 0110022221111000 ∼ 1022111000022221 and 0111110000222211 ∼ 1222222211110000 . This is obtained from a landmark point λ ≈ 0.4793 + 0.2767i : the intersection of the 30 degree symmetry line with the boundary of M3 . Supported by experiment, we conjecture that we have the same situation for the corresponding parameter on the symmetry line for every odd n, and we do not understand the algebraic reason. For n = 3, the power series derived from the given √ identifications√ both lead to √ 4 3 polynomials with the factor 2z + (3 + 3i)z + 2(1 + 3i)z 2 + 1 − 3i, and λ is a root of this polynomial. √ Also on the symmetry line, but far inside M3 , there is the parameter λ = 21 + 63 i for the terdragon for which we have lots of power series with g(λ) = 0. Instead of this well-known figure, we discuss a similar example introduced in [3]. Example 7. For the second picture in Figure 8, we have the identification 012 ∼ 100 and many others, since D is a Cantor set [3]. Actually, E = f0−1 (D) is self-similar: E = f12 (E) ∩ σf12 (E) where σ denotes a clockwise 2π rotation with appropriate 3 center. In this case, λ ≈ 0.5222 − 0.0893i fulfils the equation 1 − λ = (1 − b2 )λ2 . It seems not possible to use this equation directly to replace finitely many symbols in the above address identification. But a calculation shows that we can replace 12 and 00 by 1122 and 0001, respectively, which leads to identifications 0(12)k 112 ∼ 1(00)k 0001 and corresponding qk (λ) = 0 for k = 1, 2, ... Thus we have infinitely many power series. 18 CHRISTOPH BANDT AND NGUYEN VIET HUNG The reverse n-gon. For odd n, we call A(−λ) the reverse of A(λ). The Sierpiński gasket is self-reverse, and for Example 4 and the terdragon, the reverse is mirror-symmetric to A(λ). If we are not on a symmetry line, however, the appearance of A and its reverse can differ considerably, as Figure 1 shows. Nevertheless, their structure is similar. Since the reverse of the reverse is A itself, “if” in the following statements means “if and only if”. Proposition 10. Let n be odd and A = A(λ) a fractal n-gon. (i) A is connected if the reverse is connected. (ii) The cardinalities of the intersection sets D of A and of its reverse coincide. (iii) A is a p.c.f. fractal if the reverse is. (iv) A satisfies OSC if the reverse does. (v) A is of finite type if the reverse is. Proof. For (i) and (ii) we use P Remark 3. D(λ) is described by the sequences m d1 , d2 , ... from ∆ for which 1 − b + ∞ = 0, and D(−λ) is described by the m=1 dm λ P ′ m ′ ′ = 0. Thus between sequences d1 , d2 , ... from ∆ for which 1 − b + ∞ m=1 dm (−λ) ′ both sets there is a one-to-one correspondence, given by dm = dm for even m and d′m = −dm for odd m. If one set is empty or finite, then the other is, too. Moreover, when a point of D(λ) has preperiodic addresses, then the corresponding point of D(−λ) also has. So if A is p.c.f. [16], then the reverse is p.c.f. For (iv) and (v) we use neighbor maps fu−1 fv for words u, v ∈ {0, P 1, ..., n−1}m , m = vk k−1 and 1, 2, ... [4, 7, 3]. P For fractal n-gons we have w = fv (z) = λm z + m k=1 b λ m w uk k−m−1 −1 . Thus all neighbor maps are translations: fu (w) = λm − k=1 b λ fu−1 fv (z) =z+ m X k=1 ′ ′ u′k (bvk − buk )λk−m−1 . uk , vk′ Now defining u , v by = = vk for even k and u′k = vk , vk′ = uk we see that an n-gon and its reverse have just the same neighbor maps. OSC means that neighbor maps cannot approach the identity map [4], so (iii) is proved. Finite type says that there are only finitely many neighbor maps fu−1 fv with Au ∩ Av 6= ∅ [7]. vk uk Au , Av are disjoint P iffk the dk = b − b can be extended to a sequence d1 , d2 , ... for which 1 − b + dk λ = 0. 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Xu, On the ’Mandelbrot set’ for a pair of linear maps and complex Bernoulli convolutions, Nonlinearity 16 (2003), 1733–1749. [25] R.S. Strichartz, Isoperimetric estimates on Sierpiński gasket type fractals, Trans. Amer. Math. Soc. 351 (1999), 1705-1752. Christoph Bandt Institute for Mathematics and Informatics Arndt University 17487 Greifswald, Germany bandt@uni-greifswald.de Nguyen Viet Hung Department of Mathematics Hue University Hue, Vietnam nvh0@yahoo.com