Christoph Bandt and Nguyen Viet Hung- Fractal n-Gons and their Mandelbrot Sets

8/3/2019 Christoph Bandt and Nguyen Viet Hung- Fractal n-Gons and their Mandelbrot Sets

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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 groupacts transitively on the pieces. Examples like n -gon Sierpinski gaskets, Gospersnowake and terdragon are well-known, but we study the whole family. For eachn our family is parametrized by the points in the unit disk. Due to a connect-edness criterion, there are corresponding Mandelbrot sets which are used to ndvarious new examples with interesting properties. The Mandelbrot sets for n > 2are regular-closed, and the open set condition holds for all parameters on theirboundary, which is not known for the case n = 2 .

MSC classication: Primary 28A80, Secondary 52B15, 34B45

1. Introduction

Most of the classical fractals, like Cantor’s middle-third set, von Koch’s curve,Sierpinski’s gasket and carpet, Menger’s sponge etc. [20, 8, 12] are self-similar setswith certain symmetries. That is, they are compact sets which full 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 onesymmetry map s with s(A) = A which permutes the pieces Ak = f k(A), so thats(Ak) = A j where j depends on k and s. Self-similarity and symmetry make theshapes look attractive, but what is more, they simplify the description and mathe-matical treatment of such fractals. An analytic theory including Brownian motion[19], Dirichlet forms, spectrum of the Laplacian [16] and geodesics [25] has beendeveloped only on symmetric fractals. Recently, Falconer and O’Connor [13] haveclassied 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 fulls equation (1) for similarity mappings f k(z) = λkz + 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 Sierpinski gasket for arbitrary n havebeen considered by several authors [21, 25]. Some others like twindragon, terdragon,Gosper’s snowake and the fractal cross [20, Ch. 6] are known as tiles. Figures 1and 2 show some examples which seem to be new.

For each n, all fractal n-gons are parametrized by one complex parameter λ run-ning through the unit disk (Section 2). Since connectedness of A means that neigh-

boring pieces intersect (Section 3), we can dene a Mandelbrot set Mn for fractal1



2 CHRISTOPH BANDT AND NGUYEN VIET HUNG

Figure 1. 3-gon fractals where pieces intersect in one point. Upperrow: The point has addresses 0 1 102 , 01 10202 (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 hassimilar properties. In section 5, we describe a fast algorithm to generate Mn , anddiscuss the overlap set of n-gons which determines their geometry. Finally, we con-sider the simplest n-gons with one-point intersection set, classify them and derivethe basic analysis for this innite family.

2. Parametrization

We want to derive a very simple form for the mappings f k which generate a fractaln-gon. We start with the simple observation that a similarity map, that is, a changeof the Cartesian coordinate system, will not change the structure of a self-similarset.

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 gk = hf kh−1.

Proof. B = h(A) = hf k(A) = hf kh−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 f k . If h includes a reectionand thus reverses orientation, h(z) = µz + c, then the gk will have the conjugatefactors λk . In the present paper we consider all similitudes h as “isomorphisms” anddo not distinguish between a self-similar set and its mirror image.



FRACTAL n -GONS AND THEIR MANDELBROT SETS 3

Proposition 1. Any fractal n-gon is similar to a self-similar set A = nk=1 f k(A)

where

f k(z) = λz + bk for k = 1 ,...,n where b = cos2πn

+ i sin2πn

and |λ| < 1.

Thus for each n, the points λ in the open unit disk parametrize fractal n-gons.

Proof. By denition, a fractal n-gon A has n pieces Ak , k = 1 ,..,n, and a rotations transforms the pieces cyclically into each other. We choose the numbering so thats(Ak) = Ak+1 where + is taken modulo n. In the sequel we shall often write A0 andf 0 instead of An , f n .

By applying a translation h(z) = z+ v we push the origin of our coordinate system

to the xed point of s so that s(z) = bz where bn

= 1 since sn

preserves all piecesof A and so must be the identity map. Since bk runs through all roots of unity, it isno loss of generality (but requires another renumbering of the Ak) to assume thatb = cos 2π

n + i sin 2πn .

We took the f k to be orientation-preserving, so f 0(z) = λz + u for some λ, u Cwith |λ| < 1. Applying the rotation h(z) = z/u we transform A into a positionwhere f 0(z) = λz + 1 .

Now we note that skf 0s−k(A) = skf 0(A) = f k(A) for k = 1 ,...,n −1. ThusA = n−1

k=0 skf 0s−k(A) so due the uniqueness of self-similar sets [12, 8] the set A isthe self similar set with respect to the mappings skf 0s−k(z) = λz + bk , 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 u1u2...,sequences of symbols um {0,...,n −1}which denote pieces and subpieces of A towhich z belongs [8, 12, 5]. We have

z = p(u1u2...) = limm→∞

f u 1 · · ·f u m (0) (2)

where p : {0,...,n −1}∞ →A is the so-called address map [8]. In our case, theaddress map has a particularly simple form.

Remark 2. The point z with address u1u2..., u m {0,...,n −1}in the fractal n-

gon A(λ

)has the representation z

= ∞m =1 b

u m

λm

−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 u 1 · · ·f u m (0) = bu 1 + λbu 2 + · · ·+ λm−1bu m .

Example 1. We explain how to determine λ for the rst fractal in Figure 1.The intersection point z should have the addresses 0 1 and 102 where 1 = 111...This is indicated by the relation 0 1 102 which by the above remark turns into anequality for λ. (In Section 5 we shall see that this is indeed sufficient.)

p(01) = 1 + b(λ + λ2 + ...) = 1 + bλ

1 −λ= p(102) = b + λ + b2 λ2

1 −λ




Figure 2. Some 4-gon fractals. In the upper row, pieces intersectin 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

λ =12

+ (1 −√32

)i .

For the upper-right part of Figure 1 with 0 1 10202 we have to consider p(10202) =b+ λ + b2λ2 + λ3 + b4λ4/ (1 −λ) which leads to the equation z4 −z3 + z2 + 2 bz = 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 lowerrow 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 fromthe dynamics of complex quadratic maps. For self-similar sets, it implies localconnectedness and arcwise connectedness [14, 6].

A self-similar set with n pieces Ak is connected if and only if the graph withvertices k = 1 ,...,n and with edges { j,k}for intersecting pieces A j ∩Ak = isconnected [14, 6]. Thus a self-similar set A with two pieces A0, A1 is connected if and only if A0 ∩A1 = , otherwise it is a Cantor set [8, Ch. 8]. It is not knownwhether a similar property holds for self-affine tiles A with respect to the mappingsf k(x) = M −1x + kv, k = 0 , 1,...,n −1 where M is an expanding integer d×d matrixwith determinant

nand

vR d

.Using algebraic methods, Kirat, Lau and Rao




[17, 18], Akiyama and Gjini [1] veried connectedness in this case for dimensionsd = 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 =

implies Ak ∩Ak+1 = for each k which implies connectedness of A due to theabove criterion. The necessity of A0 ∩A1 = is more difficult to prove.

Theorem 2. A fractal n-gon A is connected if and only if A0 ∩A1 = . If A isdisconnected, 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 smallestpositive number such that A B(0, γ ) = B0. For any word u = u1...u m {0,...,n −1

}m of length

|u

|= m let f u = f u 1

· · ·f u m . For m = 1 , 2,... let

Bm =|u |= m

f u (B(0, γ )) =|u |= m

B(f u (0), |λ|m γ )

be the m-th level ball approximation of A. We shall use the following fact.(*) If Bm is connected and f 0(Bm ) ∩f 1(Bm ) = , it follows that f j (Bm ) intersects

f j +1 (Bm ) for j = 1 ,...,n −1 and Bm +1 = n−1 j =0 f j (Bm ) is connected.

Now we distinguish two cases.Case one. A0 ∩A1 = . In this case (*) implies by induction that all Bm are

connected. Then A must be connected, too: if A would split into two disjoint non-empty closed sets F 0, F 1, then the distance of F 0 and F 1 is a positive and for

|λ|m

γ < the set Bm could not be connected.Case two. A0 ∩A1 = . Then A0 and A1 have some positive distance , and bythe argument just given there is a smallest m with f 0(Bm ) ∩f 1(Bm ) = . By (*),B l is connected for all l = 0 , 1,...,m. We show that all A j are disjoint and hence Ais totally disconnected.

Since Bm is a connected nite union of closed balls of radius |λ|m γ , the outerboundary of Bm (those points which can be connected with innity in the comple-ment of Bm ) forms a closed curve Γ consisting of nitely many arcs of circles of thisradius. Γ may have nitely many double points where balls touch each other, but noother multiple points. Thus we can consider Γ as a curve oriented in clockwise direc-tion, without self crossings, only touching points. Let Γ j = f j (Γ) for j = 0 ,...,n

−1.

We assumed Γ 0 ∩Γ1 = 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 c0 and Γ j in the pointc j = b j c0.

The origin 0 is not inside Γ0, because then it would also be inside Γ1, whichimplies that Γ 0 and Γ1 intersect. Now there is a largest γ 2 > 0 such that the openball 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 d0 and each Γ j in the point d j = b j d0.

In the ring formed by D and C , the Γ j sit like spokes in a wheel, connecting d jwith c j . Of course Γ0 may touch D in other points than d0, but they all must liebetween

dn−1and

d1because otherwise the Jordan curves Γ

0and Γ

1(or Γ

n−1) would




Figure 3. 4-gon fractals where pieces intersect in a Cantor set

intersect. But this means that Γ 0 is enclosed by Γ1, Γn−1 and the arcs dn−1d1 on Dand cn−1c1 on C . Since Γ0 does not intersect the neighbor curves and does not crossthe circles D and C , it can not intersect the curves Γ 2, Γ3, ..., Γn−2. Thus all Γ j , andhence 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 thereis 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 Ak are disjoint, with pairwise distance ≥ > 0, then such U exists: we cantake U =

x AB(x,

2). For connected sets A in the plane, Bandt and Rao [7] proved

that OSC holds when pieces intersect in a nite set. Thus all examples in Figures 1and 2 full 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) = −log n

log |λ | [12]. On the other hand, dim H (A) ≤2 since Ais a subset of the plane. It follows that |λ| ≤ 1√n . This bound is sharp for n = 2 , 3, 4where we have fractal tiles (twindragon or rectangle, terdragon, square).

Proposition 4. If n < 25 and a fractal n-gon A fulls OSC then only consecutivepieces Ak , Ak+1 can intersect.

Proof. For n ≤3 all pieces are neighbors, so let n ≥4. Assume OSC holds andA0 ∩A j = for some j {2,...,n −2}. Since A0 B(1, r

1−r ) and A j B(b j , r1−r )

this implies |1−b j | ≤ 2r1−r . Since j {2,...,n −2}, it follows that |1−b2| ≤ 2r

1−r whichcan be reformulated as r ≥ |1−b2 |2+ |1−b2 |. However, calculation shows that |1−b2 |2+ |1−b2 | > 1√nfor 5 ≤ n ≤ 24 which contradicts Proposition 3. For n = 4 we just get equalitywhen r = |λ| = 1

2 , and the balls with centers 1 and b2 = −1 meet in zero. Now1 + ∞m =1 bkm λm = 0 is only possible for λ = ±1

2 which gives the square.We conjecture that the condition

n <25 is not needed here.




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 dene 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 discussedin Barnsley’s book [8] and investigated by Bousch [10, 11], Bandt [2], Solomyak andXu [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 fullled for all λin the boundary ∂ M2. We shall solve these questions for Mn with n > 2, n = 4 .

So let us consider Mn for n > 2. Figure 4 shows Mn for n = 3 , 4 as black subsetof the region 0 ≤arg( λ) ≤ π

2 , 0 ≤ |λ| < 1√n .

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 metriccannot be disconnected.

(ii). Since A(λ) is a mirror image of A(λ) (cf. Section 2), all sets Mn aresymmetric under reection at the real axis, λ →λ.

Moreover, A(λ) = A(b ·λ) for b = cos 2πn + i sin 2π

n , by an argument in the proof of Proposition 1. Thus Mn is also invariant under multiplication with b, that is,rotation around 2 π/n, as can be seen in Figure 4.

For odd n, however, Mn is also invariant under the map λ → −λ and hence underrotation around π/n. This is proved in Proposition 10,(i).




Figure 5. Magnication of

M4 in a square of side length 0.003.

For λ = 0 .3021 + 0.3375i taken from the upper right hole, the fractaln-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) willbe an interval but for nearby λ off the real axis it will be a Cantor set. For n > 2we cannot get intervals A(λ) and we have no antenna on Mn . On the whole, Mnseems to become “smoother” for larger n. Nevertheless, there are apparent holes in

M3 and M4 as were veried 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| ≥ 1√n belong to Mn . This bound can beimproved for n ≥5, and a lower bound for points in Mn can be given which showsthat for large n the boundary of Mn lies in a very thin ring.

Theorem 6. Bounds for Mn .

(i) All λ with |λ| < sin πn

1+sin πn

do not belong to Mn .

For odd n, a better estimate is |λ| < sin πn

cos π

2 n +sin πn

which equals 12 for n = 3 .

(ii) Mn includes all λ with

|λ| ≥ sin πn

cos2 π2n + √3

2 sin πn

for odd n,

and

|λ| ≥sin π

n

1 + √32 sin 2π

n

for even n.

(i) is proved after Remark 4 below, (ii) after Lemma 1 and 2 in Section 6. InSection 9 and 10 we show that the estimate (i) is sharp for all n which are notmultiples of 4.




5. How to generate Mn

To draw Mn , we must know when the pieces A0 and A1 of A(λ) intersect. ByRemark 2, a point z A0 ∩A1 can be written as z = 1 + ∞m =1 bkm λm = b +∞m =1 b j m λm where km , j m {0,...,n −1}. Thus

Remark 3. λ belongs to Mn if and only if λ is the solution of an equation

g(λ) = 1 −b +∞

m =1

dm λm = 0 (4)

where all dm are in ∆ n = bk −b j | j,k {0, 1,...,n −1} .

For xed λ, every choice of km , j m , m = 1 , 2,... such that the dm = bkm

−b j m 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 remainingsum ∞m = l+1 dm λm is at most δ|λ |

l +1

1−|λ | where δ = max {|d| |d ∆ n} ≤2. If the absolutevalue of the sum up to degree l is larger than this value, then there is no chance toextend the sum to obtain a series with g(λ) = 0 .

This argument, with division by λ at each step, provides an algorithm for generat-ing Mn . We search through a rooted tree with |∆ n | branches at every vertex. Eachvertex is assigned a complex number v. If |v| > δ|λ |1−|λ | then the vertex will be consid-ered 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 Mn , or if we reach a prescribed levellmax 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 = d1d2d3...d l ∆ l

n with l ≤ lmax and d1 = 1 −b, and corresponding points vu C according to the rule

vd1 = d1 = 1 −b , vud =vu

λ+ d , d ∆ n . (5)

When |vu | > δ|λ |1−|λ | then u and its successor words are removed. If no word of length 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 .

In particular, if |vd1 | > δ|λ |1−|λ | then λ is not in Mn . Since |vd1 | = 2 sin πn and δ = 2

for even n, while δ = 2 cos π2n for odd n, this proves Theorem 6,(i).

Remark 5. If the algorithm would be calculated up to innity, it would also decideabout OSC for A(λ) : The mappings h = f −1

j f k with j = j 1...j l , k = k1...k l aretranslations h(z) = z + w jk where

w01 =1 −b

λfor f −1

1 f 0, and w j j k k =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.




In particular, when vu = 0 is obtained after nitely many steps, that is, (4) holdswith a nite sum, then OSC cannot be true. In this case, certain pieces A j and Ak

with k1 = 0 and j 1 = 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 thatthey 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 di-rection, 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.∆ 3 consists of 0 and √3ei( π

6 + k π

3 ) , k = 0 , 1, ..., 5, vertices of a regular hexagon, allwith modulus √3. Thus we recover Theorem 6,(i): each point λ M3 fulls |λ| ≥ 1

2

because otherwise |v1| > δ|λ |1−|λ |. Moreover, |λ| = 12 can only lead to some v1d with

|v1d| ≤ |v1| if λ = ±12 , or λ = 1

2 bk . This is just the Sierpi nski gasket and its reverse,see section 8.

There is a signicant difference in the structure of ∆ n for odd and for even n.Let us start with odd n = 2 q + 1 . Sides and diagonals of all lengths of the reg-ular 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

2sin 2π

n,..., 2sin qπ

n= 2 cos π

2n. Thus

∆ n = {2ibl/ 2 sintπn | t = 0 , 1,...,q, l = 0 , 1,..., 2n −1}.

From a geometric viewpoint, ∆ n contains q points on 2n equally spaced rays around0. Let C 1,...,C q and D1,...,D q denote the points on two neighboring rays, andlet P Q denote the distance between points P and Q. Then 0C 1 = 0 D1 = s andC kDk+1 = DkC k+1 = s for k = 1 ,...,q −1 since 0C kDk+1 has a congruent trianglebuild from three vertices of the regular n-gon. We shall need an estimate for thesize of circular holes in ∆n .

Lemma 1. For odd n, the set ∆ n intersects each ball B (x, r ) with

|x

| ≤2 and

r ≥2sin π2n . The estimate of r is sharp.

Proof. If the center of a circle lies in a triangle (or a trapezium), and the verticesare outside the circle, then the circle must be smaller than the circumscribed circleof the triangle (or trapezium). Thus it suffices to show that the radius r of thecircumcircle of each trapezium C kC k+1 Dk+1 Dk equals 2sin π

2n . (The points x with

|x| ≤2 outside the segment C qDq between the two rays are covered by the balls of radius 2 sin π

2n around C q and Dq.)Let M denote the center of this circle, and let α = 0Dk+1 C k . In 0Dk+1 C k we

have 0C k / sin α = s/ sin πn = 2 . Since C kMD k = 2 α, we have in 0MC k a similar

relation: 0C k/ sin α = r/ sin π

2n. Now r is obtained from the two equations.




The proof of Theorem 6,(ii) for odd n is quite similar. We show that when λ fullsthe condition, the algorithm of Remark 4 will produce points vu B = B(0, 2sin π

n )for words u of arbitrary length. Since v1 = 1 −b B, it is enough to prove thatv B implies ( v

λ +∆ n )∩B = . The lemma takes care of |vλ | ≤2. So we assume that

the circle C of radius 2 cos π2n does not contain 0. We check under which condition

C intersects B in an arc subtending an angle ≥ πn at M = v

λ , because at least onepoint of such an arc belongs to v

λ + ∆ n .Let D be an intersection point of C with the boundary of B. Since DM 0 in-

creases when the distance y = 0 M decreases, it suffices to determine y for the caseDM 0 = π

2n . The cosine formula gives y2 −4y cos2 π2n +4(cos 2 π

2n −sin2 πn ) = 0 . Since

y > 2cos π2n , calculation gives y = 2 cos2 π

2n + √3sin πn . The condition |v

λ | ≤2sin π

n

|λ | ≤yprovides the estimate for

|λ

|.

For even n = 2 q, 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, thelength of a side is s = 2 sin π

n , and the diagonals have lengths 2 sin 2πn , ..., 2sin qπ

n = 2 .However, now there are n directions for the sides and diagonals with “odd” length2sin 3π

n , 2sin 5πn ,... alternating with n other directions for the “even” diagonals. For

n = 4 p we get

∆ n = {2bl sin tπn | l = 0 ,...,n −1, t = 0 , 2, ..., 2 p}

{2bl+ 12 sin tπ

n | l = 0 ,...,n −1, t = 1 , 3, ..., 2 p−1},

and for n = 4 p + 2 we have t = 1 , 3, ..., 2 p + 1 in the rst part and t = 0 , 2, ..., 2 pin 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 representthe diagonals, we have C kC k+1 = s and C kC k−1C k+1 = C kC k+1 C k−1 = kπ

n fork = 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| ≤2and r ≥√2sin π

n = s/ √2. For n = 4 p the estimate of r is sharp.

Proof. We consider the triangulation of ∆ n into isosceles triangles with sidelength s described above. We can modify it so that the triangles contain no angles> π

2 . (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 twosides 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 = 4 p where C kC k+1 C k−1 = kπ

n = π4

happens for k = p. For n = 4 p + 2 the largest possible angle between the two equalsides is π −2 ( p+1) π

n = 2 p2 p+1

π2 which leads to the sharp estimate. In particular, for

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, 2sin π

n) implies ( v

λ+ ∆

n)

∩B = so that the algorithm of Remark 4 gives




vu B for arbitrary long words u. Thus λ has to be taken so that

B(0,2sin π

n

|λ| ) ∆ n + B.

We consider F = B(0, 2) 2nk=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 onthe boundary ∂F are intersection points of two neighboring outer circles. Let w bethe outer intersection point of z1 + ∂B and z2 + ∂B, and assume |z1| = 2 , |z2| =2cos π

n . Then |z1 −z2| = 2 sin πn since 0z2z1 is a right angle. Thus z1z2w is an

equilateral triangle. Choose y on the ray 0z1 such that 0yw is a right angle. Sinceα := wz1y = π

6 + πn , we have by Pythagoras

|w|2

= (2 + 2 sinπn cos α)

2

+ (2sinπn sin α)

2

= 4(1 +√32 sin

2πn ).

If the condition (ii) for λ is fullled, then 2sin πn

|λ | ≤ |w| and B(0, 2sin πn

|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 vuwhich guarantees that λ belongs to Mn , for all n ≥ 5 and n = 3 . In our picturesof Mn , either this condition or the Cantor set condition of Remark 4 was fullledfor each single pixel. To obtain precise gures, we did not need to go far into therecursion: level lmax = 60 was sufficient for magnication up to factor 10 8.

Theorem 7. For n = 2 , 4 there is a critical radius r n with the following property.If |vu | ≤r n for some u in the algorithm of Remark 4, where |λ| ≥ 1

2 for n = 3 , and

|λ| ≥sin π

n

1+sin πn

for n ≥5, then λ is in Mn .

Proof. Let r n = 2 sin π2n for odd n ≥3, rn = √2sin π

n for even n ≥8, and r 6 = 1√3 .We assume |vu | ≤r n and show that |vud | ≤r n for some d ∆ n . Then we can useinduction to extend u to an innite sequence d1d2... which fulls (4). Thus we wantto show vud = vu

λ + d belongs to B(0, r n ).If we verify

|x

| ≤2 for x = vu

λ then Lemma 1 and 2 say that B(x, r n ) intersects∆ n . Thus there is b with |b| ≤r n and d ∆ n with vu

λ + b = d . In other words,vu

λ −d = −b belongs to B(0, r n ).Since |vu | ≤r n was assumed, it remains to show |r n |

|λ | ≤ 2. For n = 3 we haver 3 = 1 and |λ| ≥ 1

2 . For odd n > 3,

|r n ||λ| ≤

2sin π2n

sin πn

(1 + sinπn

) =1 + sin π

n

cos π2n

which is < 2 for n = 5 and, by monotonicity, for n > 5. For n = 6 we have r 6 = 1√3

and |λ| ≥ 13 . For even n > 6, cancellation results in |r n |

|λ | ≤√2(1 + sin πn ) < 2.




Theorem 8. For all n = 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 accumulationpoint of the vu by Remark 5, and so |vu | < r n for some u in the algorithm of Remark4.

Since vu depends continuously on λ there is a neighborhood B = B(λ, ε ) of λ suchthat for each λ B we also have |vu | < r n . By Theorem 6,(i), ε can be taken sothat all λ B full the lower bound in Theorem 7. (The only exceptions are n = 3and λ = 1

2 , which concerns the Sierpinski gasket and its reverse, see Section 10, andλ = sin π

n

1+sin πn

for even n = 4 p + 2 ≥ 6, which are discussed in Example 4. In bothcases, 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 thetwindragon and tame twindragon for n = 2 , the terdragon ( λ = 1

2 + √36 i) for n = 3 ,

and Gosper’s snowakes (see Section 9) for n = 6 .Question. Are there λ in the interior of Mn which full OSC and are not

related to tilings? In particular, do such parameters exist for n = 5 ? Is the number of such examples nite 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 = 2 , 4, the set Mn is the closure of its interior.

Proof. Let λ Mn , that is, g(λ) = ∞m =0 dm λm = 0 for certain dm ∆ n .Taking any ε > 0 we have to verify that there is µ int Mn with |µ −λ| < ε. ByProposition 3 this is true for n = 3 , λ ≥ 1√3 and for n ≥5, λ ≥ 1

2 .Otherwise, since |dm | ≤2 for all n and |dm | ≤√3 for n = 3 , the function g isanalytic in λ. So the zeros of g do not accumulate at λ, and we nd ε ≤ ε suchthat in the ball B (λ, ε ), only z = λ fulls g(z) = 0 , and moreover

|z

|< 1

2 (< 1√3in case n = 3). Let δ = min {|g(z)| | |z −λ| = ε }, and let M be so large thath(z) = ∞m = M dm zm < δ for |z −λ| ≤ε . Now the well-known theorem of Rouche(cf. [22, 23, 2]) implies that the polynomial g(z) −h(z) = M −1

m =0 dm zm has exactlyone zero µ in B(λ, ε ). By Remark 5, µ int Mn .

We do not study connectedness and local connectedness of Mn by the methodsof Bousch [10, 11]. See [2, Section 10] for a discussion of these arguments in casen = 2 . For odd n ≥3, the previous results and computer experiments like Figure 6indicate 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.




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 furthermagnication 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| ≤√5and 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 ∂ 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 havebk {1, −1}so that d = 0 has two representations bk −b j while d = 2 and d = −2admit a unique representation. If λ is the root of just one power series g(λ) andthis g has only coefficients ±2 then D is a single point. If N of the coefficients arezero, D has cardinality 2 N , and if there are innitely many zero coefficients, D is aCantor set [23, 5].

For even n > 2, the situation is very similar. We shall not consider zero coefficientssince 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 tworepresentations bk −b j .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 dmwhich do not correspond to the longest diagonals, we shall have 2N points in D, seeFigure 2. And when there are innitely 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 bk form a square. The 4-gonin the lower row of Figure 2 was obtained from the relation of addresses u1u2... =01 130303 = v1v2... Here um = vm + 2 mod 4, corresponding to a diagonal of thesquare, for all

m >1 except

m= 3 and

m= 5 where (

uk , vk) = (1

,0) which can




Figure 7. Fractal 6-gons with central piece. Note the differencebetween Gosper snowakes with λ = 5+ √3i

14 and 2+ √3i7 .

be replaced by (2, 3) since i −1 = i2 −i3. Thus there are 4 pairs of addresses whichgive the same equation

p(01) = 1 + iλ

1 −λ= p(130303) = i −iλ + λ2 −iλ3 + λ4 −i

λ5

1 −λ.

which results in λ5 −λ4 + λ3 −λ2 + λ(1 + 2 i) = 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 + i1 + √5

.

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 squarewhich fulls OSC.

It is possible to determine λ by identifying addresses u = u1u2... = 010033221 andv = 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 2z2 +(1+ i)z−i has λ as the root with modulus ≤1. Thus 1/λ is a quadraticPisot number over Q (i)

.




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 un-countable, and it is a linear Cantor set. In fact D is a self-similar set with respectto two homotheties with factor r 2 = |λ|2 : For E = f −1

0 (D), inspection of Figure 3shows that E = g1(E ) ∩g2(E ) with g1(z) = f 21 (−iz) and g2(z) = f 1f 0(−iz). So theHausdorff dimensions of A and D are log4

−log r and log2

−log r 2 , respectively [12], and thedimension of D is just one quarter of the dimension of A.

Clearly D fulls OSC. For A we can explicitly construct an open set - an octagonwith interior angles 3

4 π and with alternating side lengths s,rs for some constant s

(for our choice of mappings s = 4 r ). Drawing this octagon U and the f i(U ), we candetermine r in alternative way, by elementary geometry.Example 4: Sierpi´ nski n-gons for even n. We determine all λ Mn with

minimal modulus |λ| = sin πn

1+sin πn

(cf. Theorem 6,(i)). We use the algorithm of Remark

4, starting with v = vd1 = 1 −b which just has the critical modulus 2|λ |1−|λ |. Since

|vλ | − |v| = 2 , there is at most one choice of d2 = bk −b j for which |vd1 d2 | is not

larger than the critical modulus: d2 must be a diameter of the unit circle which is adiagonal of the regular n-gon and parallel to 1 −b. For n = 4 p such diagonal doesnot exist. For n = 4 p + 2 , the only choice is k = p + 1 , j = 3 p + 2 . This leads tovd1 d2 = vd1 , so that by induction dk = d2 for all k > 2. So the single λ with minimalmodulus is the real number λ = sin

πn

1+sin πn

. The formula in [21] is more complicated.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 f 6(z) = λz, and the extended system of mappings fulls OSC. This follows from Remark 5. Moreover,A(λ) is connected if and only if the extended self-similar set is, because A0 ∩A1 =A6 ∩A2. Figure 7 shows some examples. The Gosper tiles full |λ| = 1 / √7 which explains the bound in Theorem 6,(ii) for n = 6 .




10. Overlap set and examples for odd n

A regular n-gon with odd n has no parallel sides, and no parallel diagonals of thesame length. So d = 0 is the only vector in ∆ which can be represented as d = bk −b j

in different ways - actually in n ways. When we have OSC, however, coefficients 0are unlikely to appear in the power series of g(λ), at least for n ≥5. So it seems wehave 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: Sierpi´ nski n-gons for odd n. We determine all λ Mn withminimal modulus |λ| = sin π

n

cos π

2 n +sin πn

(cf. Theorem 6,(i)). We use the algorithm of Remark 4, starting with v = vd1 = 1 −b which has the critical modulus 2 sin π

n =δ|λ |1−|λ |, where δ = 2 cos π

2n . Since |vλ |−|v| = δ, there is at most one choice of d2 = bk−b j

for which |vd1 d2 | is not larger than the critical modulus: d2 must be the longestdiagonal of the regular n-gon which is parallel to 1 −b. As for even n, we concludethat dk = d2 for all k > 2. So the single λ with minimal modulus is the real numberλ = sin π

n

cos π

2 n +sin πn

. 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 whichis root of several power series g(λ) and still admits nite intersection set D. Forn = 3 , Figure 8 shows an example, with address identications

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 onthe symmetry line for every odd n, and we do not understand the algebraic reason.For n = 3 , the power series derived from the given identications both lead topolynomials with the factor 2 z4 + ( 3 + √3i)z3 +2(1+ √3i)z2 + 1 −√3i, and λ is aroot of this polynomial.

Also on the symmetry line, but far inside M3, there is the parameter λ = 12 + √3

6 ifor the terdragon for which we have lots of power series with g(λ) = 0 . Instead of this well-known gure, we discuss a similar example introduced in [3].

Example 7. For the second picture in Figure 8, we have the identication012 100

and many others, since D is a Cantor set [3]. Actually, E = f −10 (D) is self-similar:

E = f 21 (E ) ∩σf 21 (E ) where σ denotes a clockwise 2π

3 rotation with appropriatecenter. In this case, λ ≈0.5222−0.0893i fulls the equation 1 −λ = (1 −b2)λ2.

It seems not possible to use this equation directly to replace nitely many symbolsin the above address identication. But a calculation shows that we can replace 12and 00 by 1122 and 0001, respectively, which leads to identications 0(12) k1121(00)k0001 and corresponding qk(λ) = 0 for k = 1 , 2,... Thus we have innitelymany power series.




The reverse n-gon. For odd n, we call A(−λ) the reverse of A(λ). TheSierpinski gasket is self-reverse, and for Example 4 and the terdragon, the reverse ismirror-symmetric to A(λ). If we are not on a symmetry line, however, the appearanceof A and its reverse can differ considerably, as Figure 1 shows. Nevertheless, theirstructure is similar. Since the reverse of the reverse is A itself, “if” in the followingstatements 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 satises OSC if the reverse does.(v) A is of nite type if the reverse is.

Proof. For (i) and (ii) we use Remark 3. D(λ) is described by the sequencesd1, d2,... from ∆ for which 1 −b+ ∞m =1 dm λm = 0 , and D(−λ) is described by thesequences d1, d2,... from ∆ for which 1 −b + ∞m =1 dm (−λ)m = 0 . Thus betweenboth sets there is a one-to-one correspondence, given by dm = dm for even m anddm = −dm for odd m. If one set is empty or nite, then the other is, too.Moreover, when a point of D(λ) has preperiodic addresses, then the correspondingpoint 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 f −1u 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) = λm z + mk=1 bvk λk−1 and

f −1

u (w) =w

λm

−m

k=1 bu k

λk

−m

−1

. Thus all neighbor maps are translations:

f −1u f v(z) = z +

m

k=1

(bvk −bu k )λk−m−1 .

Now dening u , v by uk = uk , vk = vk for even k and uk = vk , vk = uk we seethat an n-gon and its reverse have just the same neighbor maps. OSC means thatneighbor maps cannot approach the identity map [4], so (iii) is proved. Finite typesays that there are only nitely many neighbor maps f −1

u f v with Au ∩Av = [7].Au , Av are disjoint iff the dk = bvk −bu k can be extended to a sequence d1, d2,... forwhich 1 −b + dkλk = 0 . With the above construction, this implies (iv).

Acknowledgement. Most of this work was done when Nguyen Viet Hungvisited the University of Greifswald, supported by the Ministry of Education andTraining of Vietnam and by the DAAD.

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Christoph BandtInstitute for Mathematics and Informatics

Arndt University17487 Greifswald, [email protected]

Nguyen Viet HungDepartment of Mathematics

Hue UniversityHue, [email protected]

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Christoph Bandt and Nguyen Viet Hung- Fractal n-Gons and their Mandelbrot Sets

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