FRACTAL n-gons AND THEIR MANDELBROT SETS. MSC classification: Primary 28A80, Secondary 52B15, 34B45

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1 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 = f 1 (A)... f n (A) (1) where f 1,..., f n are contracting similarity maps on R d. And there is at least one symmetry map s with s(a) = A which permutes the pieces A k = f k (A), so that s(a k ) = A j 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 f k (z) = λ k z + c k, 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 M n for fractal 1

2 2 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 1. 3-gon fractals where pieces intersect in one point. Upper row: The point has addresses , (Example 1). Lower row: Corresponding reverse fractals, see Section 5. n-gons in Section 4. M 2 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 M n has similar properties. In section 5, we describe a fast algorithm to generate M n, 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 f k 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. Remark 1. Let A = f k (A) be a self-similar set in R d and h : R d R d a similarity map. Then B = h(a) is the self-similar set generated by the mappings g k = hf k h 1. Proof. B = h(a) = hf k (A) = hf k h 1 (B) In the complex plane, when h consists of translation, scaling and rotation, i.e. h(z) = µz + c, then g k will have the same factor λ k as f k. If h includes a reflection and thus reverses orientation, h(z) = µz + c, then the g k will have the conjugate factors λ k. In the present paper we consider all similitudes h as isomorphisms and do not distinguish between a self-similar set and its mirror image.

3 FRACTAL n-gons AND THEIR MANDELBROT SETS 3 Proposition 1. Any fractal n-gon is similar to a self-similar set A = n k=1 f k(a) where f k (z) = λz + b k for k = 1,..., n where b = cos 2π n + i sin 2π n and λ < 1. 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 A k, k = 1,.., n, and a rotation s transforms the pieces cyclically into each other. We choose the numbering so that s(a k ) = A k+1 where + is taken modulo n. In the sequel we shall often write A 0 and f 0 instead of A n, f n. 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 b n = 1 since s n preserves all pieces of A and so must be the identity map. Since b k runs through all roots of unity, it is no loss of generality (but requires another renumbering of the A k ) to assume that b = cos 2π + i sin 2π. n n We took the f k to be orientation-preserving, so f 0 (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 f 0 (z) = λz + 1. Now we note that s k f 0 s k (A) = s k f 0 (A) = f k (A) for k = 1,..., n 1. Thus A = n 1 k=0 sk f 0 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 s k f 0 s k (z) = λz+b k, k = 0,..., n 1. These mappings will now be called f k. The points z in a self-similar set are often described by their addresses u 1 u 2..., sequences of symbols u m {0,..., n 1} which denote pieces and subpieces of A to which z belongs [8, 12, 5]. We have z = p(u 1 u 2...) = lim m f u 1 f um (0) (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 u 1 u 2..., u m {0,..., n 1} in the fractal n- gon A(λ) has the representation z = m=1 bum λ m 1. Thus A(λ) is the support of a random series of powers of λ with coefficients chosen from the n-th roots of unity. Proof. f u1 f um (0) = b u 1 + λb u λ m 1 b u m. Example 1. We explain how to determine λ for the first fractal in Figure 1. The intersection point z should have the addresses 0 1 and 10 2 where 1 = This is indicated by the relation which by the above remark turns into an equality for λ. (In Section 5 we shall see that this is indeed sufficient.) p(0 1) = 1 + b(λ + λ 2 λ +...) = 1 + b 1 λ = p(10 2) = b + λ + b 2 λ 2 1 λ

4 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 we have to consider p(10202 ) = 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 i. 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

5 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, A 0 A 1 implies A k A k+1 for each k which implies connectedness of A due to the above criterion. The necessity of A 0 A 1 is more difficult to prove. Theorem 2. A fractal n-gon A is connected if and only if A 0 A 1. If A is disconnected, then it is totally disconnected, and all pieces A k 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) = B 0. For any word u = u 1...u m {0,..., n 1} m of length u = m let f u = f u1 f um. For m = 1, 2,... let B m = f u (B(0, γ)) = B(f u (0), λ m γ) u =m u =m be the m-th level ball approximation of A. We shall use the following fact. (*) If B m is connected and f 0 (B m ) f 1 (B m ), it follows that f j (B m ) intersects f j+1 (B m ) for j = 1,..., n 1 and B m+1 = n 1 j=0 f j(b m ) is connected. Now we distinguish two cases. Case one. A 0 A 1. In this case (*) implies by induction that all B m are connected. Then A must be connected, too: if A would split into two disjoint nonempty closed sets F 0, F 1, then the distance of F 0 and F 1 is a positive ɛ and for λ m γ < ɛ the set B m could not be connected. Case two. A 0 A 1 =. Then A 0 and A 1 have some positive distance ɛ, and by the argument just given there is a smallest m with f 0 (B m ) f 1 (B m ) =. By (*), B l is connected for all l = 0, 1,..., m. We show that all A j are disjoint and hence A is totally disconnected. Since B m is a connected finite union of closed balls of radius λ m γ, the outer boundary of B m (those points which can be connected with infinity in the complement of B m ) 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 Γ j = f j (Γ) for j = 0,..., n 1. We assumed Γ 0 Γ 1 = and want to show that all Γ j are disjoint. Now let γ 1 be the smallest positive number with Γ 0 B(0, γ 1 ), and C the circle around 0 with radius γ 1. Then C intersects Γ 0 in a point c 0 and Γ j in the point c j = b j c 0. The origin 0 is not inside Γ 0, because then it would also be inside Γ 1, which implies that Γ 0 and Γ 1 intersect. Now there is a largest γ 2 > 0 such that the open ball B(0, γ 2 ) does not intersect Γ 0. Let D be the circle around 0 with radius γ 2. Then D intersects Γ 0 in a point d 0 and each Γ j in the point d j = b j d 0. In the ring formed by D and C, the Γ j sit like spokes in a wheel, connecting d j with c j. Of course Γ 0 may touch D in other points than d 0, but they all must lie between d n 1 and d 1 because otherwise the Jordan curves Γ 0 and Γ 1 (or Γ n 1 ) would

6 6 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 3. 4-gon fractals where pieces intersect in a Cantor set intersect. But this means that Γ 0 is enclosed by Γ 1, Γ n 1 and the arcs d n 1 d 1 on D and c n 1 c 1 on C. Since Γ 0 does not intersect the neighbor curves and does not cross the circles D and C, it can not intersect the curves Γ 2, Γ 3,..., Γ n 2. Thus all Γ j, and hence all A j 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 f 1,..., f n if there is a nonempty open set U with f k (U) U and f j (U) f k (U) = for j, k {1,..., n}, j k. (3) If all A k are disjoint, with pairwise distance ɛ > 0, then such U exists: we can take U = x A B(x, ɛ ). For connected sets A in the plane, Bandt and Rao [7] proved 2 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 dim H (A) = log n [12]. On the other hand, dim log λ H (A) 2 since A is a subset of the plane. It follows that λ 1 n. 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 A k, A k+1 can intersect. Proof. For n 3 all pieces are neighbors, so let n 4. Assume OSC holds and r A 0 A j for some j {2,..., n 2}. Since A 0 B(1, ) and A 1 r j B(b j r, ) 1 r this implies 1 b j 2r. Since j {2,..., n 2}, it follows that 1 r 1 b2 2r which 1 r can be reformulated as r 1 b2. However, calculation shows that 1 b 2 > 2+ 1 b b 2 1 n for 5 n 24 which contradicts Proposition 3. For n = 4 we just get equality when r = λ = 1, and the balls with centers 1 and 2 b2 = 1 meet in zero. Now 1 + m=1 bkm λ m = 0 is only possible for λ = ± 1 which gives the square. 2 We conjecture that the condition n < 25 is not needed here.

7 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 M n = {λ λ < 1, A(λ) is connected }. The set M 2 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 M 2 is regular-closed. Moreover, it is not known whether OSC is fulfilled for all λ in the boundary M 2. We shall solve these questions for M n with n > 2, n 4. So let us consider M n for n > 2. Figure 4 shows M n for n = 3, 4 as black subset of the region 0 arg(λ) π, 0 λ < 1 2 n. Proposition 5. Elementary properties of M n for n 2. (i) M n is a closed subset of the open unit disk. (ii) For even n, it has the dihedral group D n as symmetry group. For odd n, D 2n 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 M n 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 M n is also invariant under multiplication with b, that is, rotation around 2π/n, as can be seen in Figure 4. For odd n, however, M n is also invariant under the map λ λ and hence under rotation around π/n. This is proved in Proposition 10,(i).

8 8 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 5. Magnification of M 4 in a square of side length For λ = i taken from the upper right hole, the fractal n-gon must be a Cantor set, for nearby λ it will be connected. M 2 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 M n. On the whole, M n seems to become smoother for larger n. Nevertheless, there are apparent holes in M 3 and M 4 as were verified for M 2 [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 1 n belong to M n. This bound can be improved for n 5, and a lower bound for points in M n can be given which shows that for large n the boundary of M n lies in a very thin ring. Theorem 6. Bounds for M n. (i) All λ with λ < 1+ do not belong to M n. For odd n, a better estimate is λ < (ii) M n includes all λ with cos π 2n + which equals 1 2 for n = 3. λ cos 2 π + 3 sin π 2n 2 n for odd n, and λ sin 2π 2 n for even 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.

9 FRACTAL n-gons AND THEIR MANDELBROT SETS 9 5. How to generate M n To draw M n, we must know when the pieces A 0 and A 1 of A(λ) intersect. By Remark 2, a point z A 0 A 1 can be written as z = 1 + m=1 bk m λ m = b + m=1 bjm λ m where k m, j m {0,..., n 1}. Thus Remark 3. λ belongs to M n if and only if λ is the solution of an equation g(λ) = 1 b + d m λ m = 0 (4) where all d m are in m=1 n = { b k b j j, k {0, 1,..., n 1} }. For fixed λ, every choice of k m, j m, m = 1, 2,... such that the d m = b km b jm solve the above equation, corresponds to one point in the intersection set A 0 A 1. Of course the choice of d l is restricted since the absolute value of the remaining sum m=l+1 d mλ m is at most δ λ l+1 where δ = max{ d d 1 λ 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 M n. We search through a rooted tree with n branches at every vertex. Each vertex is assigned a complex number v. If v > δ λ 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 M n, or if we reach a prescribed level l max in which case λ belongs to our approximation of M n. Remark 4. (Algorithm to generate M n ) For given λ and l max we consider words u = d 1 d 2 d 3...d l l n with l l max and d 1 = 1 b, and corresponding points v u C according to the rule v d1 = d 1 = 1 b, v ud = v u λ + d, d n. (5) When v u > δ λ 1 λ then u and its successor words are removed. If no word of length l max remains, λ does not belong to M n. If at least one word of length l max remains, λ is considered as an element of M n. In particular, if v d1 > δ λ then λ is not in M 1 λ n. Since v d1 = 2 sin π and δ = 2 n for even n, while δ = 2 cos π for odd n, this proves Theorem 6,(i). 2n Remark 5. If the algorithm would be calculated up to infinity, it would also decide about OSC for A(λ) : The mappings h = f 1 j f k with j = j 1...j l, k = k 1...k l are translations h(z) = z + w jk where w 01 = 1 b for f1 1 f 0, and w jj λ kk = w jk λ + bk b j. 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 10 CHRISTOPH BANDT AND NGUYEN VIET HUNG In particular, when v u = 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 A j and A k with k 1 = 0 and j 1 = 1 coincide, so that the overlap set A 0 A 1 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 b k, 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. 3 consists of 0 and 3e i( π 6 +k π 3 ), k = 0, 1,..., 5, vertices of a regular hexagon, all with modulus 3. Thus we recover Theorem 6,(i): each point λ M 3 fulfils λ 1 2 because otherwise v 1 > δ λ 1 λ. Moreover, λ = 1 2 can only lead to some v 1d with v 1d v 1 if λ = ± 1, or λ = 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, ±ib 1/2,..., ±ib (2n 1)/2 exactly once. The length of a side is s = 2 sin π, and the diagonals have lengths n 2 sin 2π,..., 2 sin qπ = 2 cos π. Thus n n 2n n = {2ib l/2 sin tπ n t = 0, 1,..., q, l = 0, 1,..., 2n 1}. From a geometric viewpoint, n contains q points on 2n equally spaced rays around 0. Let C 1,..., C q and D 1,..., D q denote the points on two neighboring rays, and let P Q denote the distance between points P and Q. Then 0C 1 = 0D 1 = s and C k D k+1 = D k C k+1 = s for k = 1,..., q 1 since 0C k D k+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. Lemma 1. For odd n, the set n intersects each ball B(x, r) with x 2 and r 2 sin π. The estimate of r is sharp. 2n 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 circumcircle of each trapezium C k C k+1 D k+1 D k equals 2 sin π. (The points x with 2n x 2 outside the segment C q D q between the two rays are covered by the balls of radius 2 sin π around C 2n q and D q.) Let M denote the center of this circle, and let α = 0D k+1 C k. In 0D k+1 C k we have 0C k / sin α = s/ sin π = 2. Since C n kmd k = 2α, we have in 0MC k a similar relation: 0C k / sin α = r/ sin π. Now r is obtained from the two equations. 2n

11 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 v u B = B(0, 2 sin π) n for words u of arbitrary length. Since v 1 = 1 b B, it is enough to prove that v B implies ( v + λ n) B. The lemma takes care of v 2. So we assume that λ the circle C of radius 2 cos π does not contain 0. We check under which condition 2n C intersects B in an arc subtending an angle π at M = v, because at least one n λ point of such an arc belongs to v + λ n. Let D be an intersection point of C with the boundary of B. Since DM0 increases when the distance y = 0M decreases, it suffices to determine y for the case DM0 = π. The cosine formula gives 2n y2 4y cos 2 π 2n +4(cos2 π 2n sin2 π ) = 0. Since n y > 2 cos π, calculation gives y = 2 2n cos2 π + 3 sin π. The condition v 2 y 2n n λ λ 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 length of a side is s = 2 sin π, and the diagonals have lengths 2 sin 2π,..., 2 sin qπ = 2. n 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 = {2b l sin tπ n l = 0,..., n 1, t = 0, 2,..., 2p} {2b l+ 1 2 sin tπ n l = 0,..., n 1, t = 1, 3,..., 2p 1}, 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 C 0 = 0, C 1,...C q. Since 0C k represent the diagonals, we have C k C k+1 = s and C k C k 1 C k+1 = C k C k+1 C k 1 = kπ for 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 = 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 > π. (Where such angle appears, there is an adjacent triangle so that the union of 2 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, and we see that the estimate is sharp for n = 4p where C k C k+1 C k 1 = kπ = π n 4 happens for k = p. For n = 4p + 2 the largest possible angle between the two equal sides is π 2 (p+1)π = 2p π which leads to the sharp estimate. In particular, for n 2p+1 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 ) implies ( v λ + n) B so that the algorithm of Remark 4 gives

12 12 CHRISTOPH BANDT AND NGUYEN VIET HUNG v u B for arbitrary long words u. Thus λ has to be taken so that B(0, 2 ) n + B. λ We consider F = B(0, 2) 2n k=1 z k + B where z k 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 z 1 + B and z 2 + B, and assume z 1 = 2, z 2 = 2 cos π. Then z n 1 z 2 = 2 sin π since 0z n 2z 1 is a right angle. Thus z 1 z 2 w is an equilateral triangle. Choose y on the ray 0z 1 such that 0yw is a right angle. Since α := wz 1 y = π + π, we have by Pythagoras 6 n w 2 = (2 + 2 cos α)2 + (2 sin α)2 = 4(1 + If the condition (ii) for λ is fulfilled, then 2 λ 3 2 sin 2π n ). w and B(0, 2 λ ) F. 7. OSC holds on the boundary of M n Now let us improve the algorithm of Remark 4. We give a condition for the v u which guarantees that λ belongs to M n, for all n 5 and n = 3. In our pictures of M n, 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 l max = 60 was sufficient for magnification up to factor Theorem 7. For n 2, 4 there is a critical radius r n with the following property. If v u r n for some u in the algorithm of Remark 4, where λ 1 for n = 3, and 2 λ for n 5, then λ is in M n. 1+ Proof. Let r n = 2 sin π for odd n 3, r 2n n = 2 sin π for even n 8, and r n 6 = 1 3. We assume v u r n and show that v ud r n for some d n. Then we can use induction to extend u to an infinite sequence d 1 d 2... which fulfils (4). Thus we want to show v ud = vu + d belongs to B(0, r λ n). If we verify x 2 for x = vu then Lemma 1 and 2 say that B(x, r λ n) intersects n. Thus there is b with b r n and d n with vu + b = λ d. In other words, v uλ d = b belongs to B(0, r n ). Since v u r n was assumed, it remains to show r n 2. For n = 3 we have λ r 3 = 1 and λ 1. For odd n > 3, 2 r n λ 2 sin π 2n (1 + ) = 1 + cos π 2n which is < 2 for n = 5 and, by monotonicity, for n > 5. For n = 6 we have r 6 = 1 3 and λ 1 3. For even n > 6, cancellation results in r n λ 2(1 + ) < 2.

13 FRACTAL n-gons AND THEIR MANDELBROT SETS 13 Theorem 8. For all n 2, 4, the open set condition holds for all λ on the boundary M n of M n. Proof. Take λ M n, and assume OSC does not hold. Then 0 is an accumulation point of the v u by Remark 5, and so v u < r n for some u in the algorithm of Remark 4. Since v u depends continuously on λ there is a neighborhood B = B(λ, ε) of λ such that for each λ B we also have v u < r n. By Theorem 6,(i), ε can be taken so that all λ B fulfil the lower bound in Theorem 7. (The only exceptions are n = 3 and λ = 1, which concerns the Sierpiński gasket and its reverse, see Section 10, and 2 λ = for even n = 4p + 2 6, which are discussed in Example 4. In both 1+ cases, OSC holds.) Theorem 6 now says that B M n, so λ cannot be a boundary point of M n. We proved that all λ for which OSC fails are in the interior of M n. The converse is not true: a few λ in the interior of M n are known for which OSC holds, like the twindragon and tame twindragon for n = 2, the terdragon (λ = 1 + 3i) for n = 3, 2 6 and Gosper s snowflakes (see Section 9) for n = 6. Question. Are there λ in the interior of M n 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. M n 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 2, 4, the set M n is the closure of its interior. Proof. Let λ M n, that is, g(λ) = m=0 d mλ m = 0 for certain d m n. Taking any ε > 0 we have to verify that there is µ int M n with µ λ < ε. By Proposition 3 this is true for n = 3, λ 1 3 and for n 5, λ 1. 2 Otherwise, since d m 2 for all n and d m 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 < 1 (< in case n = 3). Let δ = min{ g(z) z λ = ε }, and let M be so large that h(z) = m=m d mz m < δ for z λ ε. Now the well-known theorem of Rouché (cf. [22, 23, 2]) implies that the polynomial g(z) h(z) = M 1 m=0 d mz m has exactly one zero µ in B(λ, ε ). By Remark 5, µ int M n. We do not study connectedness and local connectedness of M n 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 M n is very simple. Conjecture. For odd n 3, the boundary of M n is a Jordan curve. For even n 4 it is a countable union of disjoint Jordan curves.

14 14 CHRISTOPH BANDT AND NGUYEN VIET HUNG Figure 6. Left: M 3 in a square of side length 10 5 with center ( , ). This is one of the suspicious places, but further magnification shows no holes. Right: a hole in M 6. Side length 10 4, center ( , ). 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 r 4 = 1, under the assumption λ 1/ 5. This proves Theorem 8 for all λ M 4 with modulus 1/ 5, that is, for the region x 0.4, y 0.2 in Figure 4. In this region, M 4 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 = A 0 A 1. For n = 2 we have b k {1, 1} so that d = 0 has two representations b k b j 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 2 N, 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 b k, 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 b k b j. Thus only those power series g(λ) where every d m has the form b k m b k m+n/2 can yield a single intersection point, see Example 4 below. If we have N coefficients d m which do not correspond to the longest diagonals, we shall have 2 N 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. Example 2. For n = 4 we have b = i and the b k form a square. The 4-gon in the lower row of Figure 2 was obtained from the relation of addresses u 1 u 2... = = v 1 v 2... Here u m = v m + 2 mod 4, corresponding to a diagonal of the square, for all m > 1 except m = 3 and m = 5 where (u k, v k ) = (1, 0) which can

15 FRACTAL n-gons AND THEIR MANDELBROT SETS 15 Figure 7. Fractal 6-gons with central piece. Note the difference between Gosper snowflakes with λ = 5+ 3i and 2+ 3i be replaced by (2, 3) since i 1 = i 2 i 3. Thus there are 4 pairs of addresses which give the same equation λ p(0 1) = 1 + i 1 λ = p( ) = i iλ + λ 2 iλ 3 + λ 4 λ 5 i 1 λ. which results in λ 5 λ 4 + λ 3 λ 2 + λ(1 + 2i) = i with numerical solution λ i. Example 3. As a landmark point in M 4, we consider the intersection of the symmetry line in Figure 4 with the boundary of M 4. It turns out that this is λ = 1 + i The corresponding fractal, shown on the left of Figure 3, is mirror-symmetric: A( λ) = 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 = u 1 u 2... = and v = 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 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 (u k, v k ) can be replaced by one other choice. For instance (0, 3) and (1, 2) give the same d = 1 + i. Thus D = A 0 A 1 is uncountable, and it is a linear Cantor set. In fact D is a self-similar set with respect to two homotheties with factor r 2 = λ 2 : For E = f0 1 (D), inspection of Figure 3 shows that E = g 1 (E) g 2 (E) with g 1 (z) = f1 2 ( iz) and g 2 (z) = f 1 f 0 ( iz). So the Hausdorff dimensions of A and D are log 4 log 2 and, 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 3 π and with alternating side lengths s, rs for some constant s 4 (for our choice of mappings s = 4r). Drawing this octagon U and the f i (U), we can determine r in alternative way, by elementary geometry. Example 4: Sierpiński n-gons for even n. We determine all λ M n with minimal modulus λ = (cf. Theorem 6,(i)). We use the algorithm of Remark 1+ 2 λ 4, starting with v = v d1 = 1 b which just has the critical modulus. Since 1 λ v v = 2, there is at most one choice of d λ 2 = b k b j for which v d1 d 2 is not larger than the critical modulus: d 2 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 v d1 d 2 = v d1, so that by induction d k = d 2 for all k > 2. So the single λ with minimal modulus is the real number λ =. The formula in [21] is more complicated. 1+ 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 b k, not only their differences. Thus whenever OSC holds for some λ, we can add an seventh map f 6 (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 A 1 = A 6 A 2. Figure 7 shows some examples. The Gosper tiles fulfil λ = 1/ 7 which explains the bound in Theorem 6,(ii) for n = 6.

17 FRACTAL n-gons AND THEIR MANDELBROT SETS 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 = b k b j 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 M n with sin minimal modulus λ = π n (cf. Theorem 6,(i)). We use the algorithm of cos π 2n + Remark 4, starting with v = v d1 = 1 b which has the critical modulus 2 sin π = n δ λ, where δ = 2 cos π. Since v v = δ, there is at most one choice of d 1 λ 2n λ 2 = b k b j for which v d1 d 2 is not larger than the critical modulus: d 2 must be the longest diagonal of the regular n-gon which is parallel to 1 b. As for even n, we conclude that d k = d 2 for all k > 2. So the single λ with minimal modulus is the real number λ = cos π 2n +. The formula in [21] is more complicated. OSC holds by Remark 5. 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 and This is obtained from a landmark point λ i : the intersection of the 30 degree symmetry line with the boundary of M 3. 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 polynomials with the factor 2z 4 + (3 + 3i)z 3 + 2(1 + 3i)z i, and λ is a root of this polynomial. Also on the symmetry line, but far inside M 3, there is the parameter λ = i 2 6 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 and many others, since D is a Cantor set [3]. Actually, E = f0 1 (D) is self-similar: E = f1 2 (E) σf1 2 (E) where σ denotes a clockwise 2π rotation with appropriate 3 center. In this case, λ i fulfils the equation 1 λ = (1 b 2 )λ 2. 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 q k (λ) = 0 for k = 1, 2,... Thus we have infinitely many power series.

18 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 Remark 3. D(λ) is described by the sequences d 1, d 2,... from for which 1 b + m=1 d mλ m = 0, and D( λ) is described by the sequences d 1, d 2,... from for which 1 b + m=1 d m( λ) m = 0. Thus between both sets there is a one-to-one correspondence, given by d m = d m for even m and d m = d m 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 f v for words u, v {0, 1,..., n 1} m, m = 1, 2,... [4, 7, 3]. For fractal n-gons we have w = f v (z) = λ m z + m k=1 bv k λ k 1 and fu 1 (w) = w m λ m k=1 bu k λ k m 1. Thus all neighbor maps are translations: m fu 1 f v (z) = z + (b v k b u k )λ k m 1. k=1 Now defining u, v by u k = u k, v k = v k for even k and u k = v k, v k = u k 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 f v with A u A v [7]. A u, A v are disjoint iff the d k = b v k b u k can be extended to a sequence d1, d 2,... for which 1 b + d k λ k = 0. With the above construction, this implies (iv). Acknowledgement. Most of this work was done when Nguyen Viet Hung visited the University of Greifswald, supported by the Ministry of Education and Training of Vietnam and by the DAAD. References [1] S. Akiyama and N. Gjini, Connectedness of number-theoretic tilings, Discrete Math. Theoret. Computer Science, to appear [2] C. Bandt, On the Mandelbrot set for pairs of linear maps, Nonlinearity 15 (2002), [3] C. Bandt, Self-similar measures, Ergodic Theory, Analysis, and Efficient Simulation of Dynamical Systems, B. Fiedler (ed.), Springer (2001), [4] C. Bandt and S. Graf, Self-similar sets VII. A characterization of self-similar fractals with positive Hausdorff measure. Proc. Amer. Math. Soc. 114 (1992),

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