A PLANAR INTEGRAL SELF-AFFINE TILE WITH CANTOR SET INTERSECTIONS WITH ITS NEIGHBORS

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1 #A INTEGERS 9 (9), 7-37 A PLANAR INTEGRAL SELF-AFFINE TILE WITH CANTOR SET INTERSECTIONS WITH ITS NEIGHBORS Shu-Juan Duan School of Math. and Computational Sciences, Xiangtan University, Hunan, China jjmmcn@6.com Dan Liu School of Math. and Computational Sciences, Xiangtan University, Hunan, China liudan8@6.com Tai-Man Tang School of Math. and Computational Sciences, Xiangtan University, Hunan, China tmtang xtu@yahoo.cn Received: 9/3/8, Accepted: //9 Abstract We give an example of a planar self-affine tile that intersects some of its neighbors in the tiling it generates in Cantor sets. The reasoning also shows that it has nontrivial fundamental group. The technique used is to obtain information from a good approximation of the tile, namely, an estimation of its convex hull.. Introduction A planar integral self-affine tile is a compact set T in R of positive Lebesgue measure for which there is an expanding integral matrix B and a digit set D = {d,..., d m } Z such that m B(T ) = (T + d i ), i= with (T +d i ) (T +d j ) of measure zero. In this case, we must have det B = D = m. See [8]. We consider the tile T = T (B, D), where B = ( 3 3 ), D = { d i = ( i ) }, i =,..., 7. () We know that T + Z is a tiling of R (see Lagarias Wang [8, Theorem.(ii)]). We will prove the following two theorems. Theorem Let T = T (B, D) be the self-affine tile with B, D given in Equation (). In the planar tiling generated by T, there are neighbors intersecting in Cantor sets, that is, totally disconnected perfect sets. More precisely, T intersects with at least four of its neighbors in Cantor sets. (See Figure.) Theorem T has nontrivial fundamental group. Hence its fundamental group is uncountably generated []. The research was supported by the National Natural Science Foundation of China, #778

2 INTEGERS: 9 (9) 8 Figure : The central tile T, painted red, intersects four of its neighbors in Cantor sets, the blue and the black ones. To put these into perspectives, we recall two theorems on tilings. Leung and Lau [9] give a necessary and sufficient condition for certain planar self-affine tiles to be disklike. Theorem A [9, Theorem.4] Let C M (Z) be an expanding matrix with characteristic polynomial f(x) = x + px + q. Let v Z {(, )} with v, Cv independent. Let D = {, v, v,..., ( q )v}. Then T := T (C, D) is disklike if and only if p q +. Akiyama and Thuswaldner [, ] gave a proof for the special case of tilings generated by canonical number systems, hence requiring f(x) to be irreducible, p q and q, that T is disklike if and only if p q +. We next state a theorem of Bandt and Gelbrich. Recall that two tiles T, T are tiling neighbors if T T. They are vertex neighbors if their intersection is a point. They are edge neighbors if their intersection contains a point inside (T T ). Theorem B [3, Lemma 5.] Let T be a topological disk which tiles R by the lattice L. Then in the tiling T + L, one of the followings is true: (i) T has no vertex neighbors and six edge neighbors T ± α, T ± β, T ± (α + β) for some α, β L, with Zα + Zβ = L. (ii) T has four edge neighbors T ± α, T ± β, and four vertex neighbors T ± α ± β for some α, β L, with Zα + Zβ = L. From Theorem A, the tile T = T (B, D) is not disklike, with B, D given in Equation (). In contrast to Theorem B for disklike tiles,theorem shows that Cantor set intersections among neighbors can occur for non disklike self-affine Z tiles. While

3 INTEGERS: 9 (9) 9 one can construct non self-affine tiles that have this property, the authors are not aware of any example that is self-affine. There are certainly self-affine tiles that intersect in finite but more than one point with some of its neighbors. For example, the fundamental domain of the canonical number system of base + i. There are different ways to be non disklike, even for fractal tiles. For example, the Heighway dragon is pinched, but doesn t have holes. Theorem says that the tile we study has holes, and that is not clear from its picture. The method used here is to get conclusions from a good enough approximation of the tile T, namely, an estimation of the convex hull of T from the outside. This estimation is given by Duda [4]. He gives an example where the estimation gives exactly the convex hull. For our tile, the method also works very well. Indeed, a little more work gives the exact convex hull in our case. The paper is organized as follows. In Section, we give an estimation of the convex hull of the tile we study. We prove Theorem in Section 3 and Theorem in Section 4.. Convex Hull Estimation We recall the theory of convex hull estimation [4] for the tiles we are dealing with. Let T be self-affine tiles satisfying the equation T = i=,...,m A(T + d i ), d i = (i, ) t, where A = B for an integral expanding matrix B. Let S R be the unit circle and d S be a direction. The width function centered at zero is the function h : S R given by h(d) := inf{t : T H(d, t)}, where H(d, t) is the half space {x R : x d t}. We have the following theorem. Theorem C [4, Observation 4] Let T be as above. Then for d S, h(d) = A j d h (Âj d). j= Here, Âj d = A j d/ A j d, and for any v S, h (v) = max i=,...,m v Ad i. Theorem D [4, Observation 5] Let x be the center of symmetry of T : T x = x T. Then x = (B I) d m.

4 INTEGERS: 9 (9) 3 We now estimate the convex hull of T (B, D), with B, D as given in Equation (). Proposition. Let B, D be given as in equation. Then T (B, D) lies in the parallelogram P with vertices (, ), (7/, 7/6), (4/3, 7/3), (7/6, 7/6). Moreover T contains (, ), (4/3, 7/3). Proof. Let A = B. The eigenvalues of A are λ = / and λ = /4, with eigenvectors v = (, ) t, v = (, ) t respectively. Notice that the vectors Ad i = (3i/8, i/8) t are nonnegative multiples of (3, ) t. Let d = (, 3) t. As the eigenvector v = (, ) correspond to the larger eigenvalue and v = (, ) corresponds to the smaller one, Âj d ( /, / ) t monotonically as j. Hence for i =,..., 7 and nonnegative integer j, Âj d Ad i. By definition, h (Âj d) = max i=,...,7 Â j d Ad i = for j =,,,..., implying that h(d) = from Theorem C. That is, T is below the line x + 3y =. See Fig.. The same reasoning applied to d = (, ) t shows that T is to the right of the line x + y =. From Theorem D, the center of symmetry of T is (7/3, 7/6). Obviously, (, ) T from the definition of T. Hence (4/3, 7/3) T. Therefore T and its convex hull lies in the region bounded by the lines x + 3y =, x + y =, x + y = 7/3, x + 3y = 7/3. See Figure for a picture for the convex hull approximation. Indeed, a little further work can give us the exact convex hull of the tile we are working with..5 x+y= x+3y= (7/,-7/6).5 (7/6,-7/6) x+3y=-7/3 x+y=7/3 (4/3,-7/3) Figure : An approximation of the convex hull of the tile T.

5 3 INTEGERS: 9 (9) T^(-4,).5 T^(-,) T.5 T Figure 3: (left) T ( 4,) T is a Cantor set. The same is true for T (,) T (right) 3. Cantor Set Intersections: Proof of Theorem In this section, we show that there are neighbors in the tiling generated by T (B, D) that intersects in Cantor sets. See Fig. 3. We first introduce some notations. They are general though we only state them for our tile. For i =,..., 7, let fi (x) := A(x + di ) for x R. Let fi ik := fi fik be their compositions, where k is a positive integer and for j =,..., k, ij {,..., 7}. (7 For E R, we also write Ei ik for fi ik (E). Let F k (E) = i,...,ik = fi ik (E). For (i, j) Z, define E (i,j) := E + (i, j)t to be the translate of E by (i, j)t. We (i,j) will write E for E (,). We call Ti ik a k-th level piece of T (i,j), with the tile T (i,j) itself being the -th level piece. For the polygon P given in Proposition., we (i,j) (i,j) will call (for convenience) Pi ik a k-th level polygon of P (i,j). Notice that Pi ik (i,j) contains the k-th level piece Ti ik of T (i,j). We now show that T T ( 4,) is a Cantor set. Then we can conclude that T T (,), T T (, ) and T T (4, ) are also Cantor sets as they are either a translate of T T ( 4,) or unions of translates of T T ( 4,) (see Fig. ). The following lemma shows that it suffices to figure out F k (P ) F k (P )( 4,) for each positive integer k. Lemma 4 For all positive integers k, T T ( 4,) = ) [F k (P ) F k (P )( 4,) ]. k= Proof. As F (P ) P, we have F k (P )(i,j) decreasing in k, and T = ) k= F k (P ), T ( 4,) = see, for example, [5]. The lemma follows. ) k= F k (P )( 4,) ;!

6 INTEGERS: 9 (9) 3 Lemma 5 Let P be the parallelogram in Proposition.. Let k be a positive integer and for j =,..., k, i j {,..., 7}. Then (a) AP ( 4,) i i k = P ( 4,) 6i i k. (b) AP i i k = P i i k. (c) P ( 4,) 6i i k + (3/8, /8) t = P ( 4,) 7i i k. Notice that (3/8, /8) t = Ad. As usual, A stands for the map multiplication by A. Proof. By the definitions given at the beginning of this section, Hence On the other hand, P ( 4,) i i k = f i i k (P ) + ( 4, ) t = A k P + AP ( 4,) i i k = A k+ P + P ( 4,) 6i i k = A k+ P + k A j d ij + ( 4, ) t. j= k A j+ d ij + A( 4, ) t. j= k A j+ d ij + Ad 6 + ( 4, ) t. j= As (A I)( 4, ) t = Ad 6, (a) is proved. (b) can be proved similarly. (c) follows from Ad = (3/8, /8) t. Lemma 6 For k =,, 3,..., F k (P ) ( 4,) F k (P ) = [P ( 4,) i i k i,...,i k =6,7 P (i 6) (i k 6)]. Proof. We use induction. For k =, it can be verified directly using the equations of the boundary of P, or can be seen clearly from Figure 4 that F (P ) ( 4,) F (P ) = (P ( 4,) 6 P ) (P ( 4,) 7 P ). Suppose the statement is true for k. For k +, as F k+ (P ) ( 4,) F k+ (P ) F (P ) ( 4,) F (P ) = (P ( 4,) 6 P ) (P ( 4,) 7 P ), we just need to find out the portions of F k+ (P ) ( 4,) F k+ (P ) in P ( 4,) 6

7 INTEGERS: 9 (9) P^(-4,)_7.5 P^(-4,)_77 P^(-4,)_76 P_ P_ P_.5.5 P^(-4,)_6.5 P_ Figure 4: (left) The intersection F (P ) ( 4,) F (P ), consisting of intersections of first level polygons P ( 4,) 6 P and P ( 4,) 7 P. (right) The polygons drawn are second level polygons of P ( 4,) and P inside P ( 4,) 6, P ( 4,) 7 and P, P. The intersection shown is F (P ) ( 4,) F (P ). It consists of four pieces, two in P ( 4,) 6 P and two in P ( 4,) 7 P P and in P ( 4,) 7 P. Notice that for i = 6, 7, the only -st level piece of P intersecting P ( 4,) i is P i 6. Similarly, for i =,, the only -st level piece of P ( 4,) intersecting P i is P ( 4,) i+6. Hence the portion of F k+ (P ) ( 4,) F k+ (P ) in P ( 4,) 6 P is 7 P ( 4,) 6i i k+ i,...,i k+ = = = A 7 7 i,...,i k+ = AP ( 4,) i i k+ i,...,i k+ = 7 P i i k+ P ( 4,) i i k+ i,...,i k+ = = A[F k (P ) ( 4,) F k (P )], 7 i,...,i k+ = 7 i,...,i k+ = AP i i k+ P i i k+ where we have used Lemma 3.(a)(b) in the first equality. By the induction hypothesis, this is equal to A [P ( 4,) i i k+ P (i 6) (i k+ 6)] i,...,i k+ =6,7 = i,...,i k+ =6,7 [P ( 4,) 6i i k+ P (i 6) (i k+ 6)].

8 INTEGERS: 9 (9) 34 Next, the part of F k+ (P ) ( 4,) F k+ (P ) in P ( 4,) 7 P is, using Lemma 3. and the induction hypothesis, 7 7 P ( 4,) i,...,i k+ = = = A = A = = 7i i k+ 7 P ( 4,) 6i i k+ i,...,i k+ = 7 i,...,i k+ = P ( 4,) i i k+ i,...,i k+ = i,...,i k+ =6,7 i,...,i k+ =6,7 i,...,i k+ =6,7 [P ( 4,) [P ( 4,) P i i k+ 7 i,...,i k+ = 7 i,...,i k+ = P i i k+ P i i k+ + (3/8, /8) t i i k+ P (i 6) (i k+ 6)] + (3/8, /8) t 6i i k+ P (i 6) (i k+ 6)] + (3/8, /8) t [P ( 4,) 7i i k+ P (i 6) (i k+ 6)]. + (3/8, /8) t Hence F k+ (P ) ( 4,) F k+ (P ) = i,...,i k+ =6,7 [P ( 4,) i i k+ P (i 6) (i k+ 6)]. Lemma 7 The union on the right-hand side of Lemma 3.3 is a disjoint union. Proof. Use induction on k. For k =, P ( 4,) 6 P and P ( 4,) 7 P are disjoint, as can be verified directly, or seen clearly from Figure 4. Suppose the statement is true for k. Consider the pieces R := P ( 4,) i i k+ P (i 6) (i k+ 6), S := P ( 4,) l l k+ P (l 6) (l k+ 6) in F k+ (P ) ( 4,) F k+ (P ). That is, i j, l j {6, 7} for all j =,..., k +. If i j l j for some j {,..., k}, then R and S are disjoint as they are contained in different pieces of F k (P ) ( 4,) F k (P ), which are disjoint by the induction hypothesis. Suppose that i j = l j for j {,..., k}, and i k+ l k+, say, i k+ = 6 and l k+ = 7. Then R and S are respectively the images of the disjoint sets P ( 4,) 6 P and P ( 4,) 7 P under the same sequence of affine maps consisting of A( ) and A( ) + (3/8, /8) t, and hence are disjoint.

9 INTEGERS: 9 (9) 35 P_.5.5 P^(-,)_ P^(-,)_.5 the `hole' R P^(-,) P.5 P_4.5 P_ P_ Figure 5: (left) Second level polygons of P in P, P are drawn. P and P 4 are drawn non-filled, bounded by solid and dotted lines respectively. Part of the intersection P P 4 is not covered by other second level polygons of P. (right) The affine map F, defined in step in section 4, maps P (,) P onto P P 4. Proof of Theorem. The maps A( ) and A( ) + (3/8, /8) t are contractions. As the eigenvalues of A are not bigger than /, for any bounded set E R with diameter diam E, diam A(E) /(diam E). The same is true for the map A( ) + (3/8, /8) t. Hence from Lemmas 5, 6, and 7, each piece P ( 4,) i i k P (i 6) (i k 6) of F k (P ) ( 4,) F k (P ) contains two disjoint pieces of F k+ (P ) ( 4,) F k+ (P ), each of diameter less than / of its own. This is true for every k =,,.... It is standard to prove that k= [F k (P ) F k (P ) ( 4,) ] is a totally disconnected perfect set. Theorem follows from Lemma Nontrivial Fundamental Group: Proof of Theorem In this section, we prove that π (T ) is nontrivial. In short, the reason is that there are affine images of T ( 4,) 6 T ( 4,) 7 T T (an area studied in section 3) in T, with the hole it bounds mapped to points not in T. Proof of Theorem. Step. Look at the second level polygons P 4 and P of P (see Figure 5). They have nonempty intersection. Part of this intersection is not covered by other second level polygons of P. Call this set W := (P 4 P ) { Pii : (i, i ) (, 4), (, ) }. We will show that there is a hole of T in W. Step. The affine map F (x) = A (x+d 4 ) for x R maps P to P 4, and P (,) to P. For example, the reason for the latter is F (P (,) ) = A (P + (, ) t + d 4 ) = A P + Ad = P.

10 INTEGERS: 9 (9) 36 So F maps P (,) P (,) P P 3 homeomorphically onto P P P 4 P 43, also mapping the hole R bounded by the former (see Fig. 5) to a hole called S bounded by the latter. Step 3. The set S (the hole) does not intersect the second level polygons in P other than P 4 and P. To see this, we can check that the images under F ( ) = B ( ) d 4 of these other second level polygons of P do not intersect R. For example, we check whether P intersects S as follows: F (P ) = F (A P + A d + Ad ) = B (A P + A d + Ad ) d 4 = P + (, ) t, and it doesn t intersect the hole R. Others are checked similarly. Therefore, S is not contained in F (P ) and hence not in T. Step 4. Notice that the intersection (P (,) P (,) ) (P P 3 ) is congruent to (P ( 4,) 6 P ( 4,) 7 ) (P P ), the set we have studied in detail in Section 3. They differ by a translation of (3/8, /8) t. Similarly, the intersection T (,) T is congruent to three copies of T ( 4,) T, differing by translations by multiples of (3/8, /8) t (see Figure 3). Now pick any point p T T (,) in P (,) P, and any point q T T (,) in P (,) P 3. Step 5. Notice that T is connected [7] and hence path connected [6]. The path connectedness of T (,) guarantees that there is a path γ T (,) 7 i= P (,) i joining p, q. The same reason gives a path γ T 7 i= P i joining p, q. Then γ γ is a nontrivial loop in T T (,), surrounding the hole R. Hence F (γ γ ) is a nontrivial loop in T because of the hole S. Step 6. By Luo and Thuswaldner [], π (T ) nontrivial implies that it is uncountably generated. Acknowledgments. manuscript. We thank the referee for the detailed comments on the References [] S. Akiyama and J. Thuswaldner, A survey on the topological tiles related to number systems, Geom. Dedicata 9 (4), [] S. Akiyama and J. Thuswaldner, On the topological properties of fractal tilings generated by quadratic number systems, Comput. Math. Appl. 49(7-) (5), [3] C. Bandt and G. Gelbrich, Classification of self-affine lattice tilings, J. London Math. Soc.() 5 (994), [4] J. Duda, Analysis of the convex hull of the attractor of an IFS, arxiv:7.3863v, math.ca, Oct 7.

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