TOPOLOGICAL PROPERTIES OF SELF-SIMILAR FRACTALS WITH ONE PARAMETER JUN JASON LUO AND LIAN WANG Abstract. In this paper, we study two classes of planar self-similar fractals T 7 ε 1 with a shifting parameter ε. The first one is a class of self-similar tiles by shifting 0 x-coordinates of some digits. We give a detailed discussion on the disk-likeness 2 (i.e., the property of being a topological disk) in terms of ε. We also prove that n T determines a quasi-periodic tiling if and only if ε is rational. The second one a ε J is a class of self-similar sets by shifting diagonal digits. We give a necessary and 5 sufficient condition for Tε to be connected. ] N G 1. Introduction . h t Let A be a d×d integer expanding matrix (i.e., all of its eigenvalues are strictly a m larger than one in modulus), let D = {d ,...,d } ⊂ Rd be a digit set with N = 1 N [ |det(A)|. Then we can define an iterated function system (IFS) {S }N where S j j=1 j 1 are affine maps v S (x) = A−1(x+d ), x ∈ Rd. 7 j j 0 Since A is expanding, each S is a contractive map under a suitable norm [13] of Rd, 3 j 1 there is a unique nonempty compact subset T := T(A,D) ⊂ Rd [10] such that 0 . N 1 (cid:91) 0 T = Sj(T) = A−1(T +D). 7 j=1 1 : The set T also has the radix expansion v i (cid:40) (cid:41) X ∞ (cid:88) T = A−kd : d ∈ D . (1.1) r j j a k k k=1 We call T a self-affine set generated by the pair (A,D) (or the IFS {S }N ). More- j j=1 over, if T has non-void interior (i.e., T◦ (cid:54)= ∅), then there exits a discrete set J ⊂ Rd satisfying T +J = Rd and (T◦ +t)∩(T◦ +t(cid:48)) = ∅ with t (cid:54)= t(cid:48),t,t(cid:48) ∈ J. Date: January 6, 2017. 2010 Mathematics Subject Classification. Primary 28A80; Secondary 52C20, 54D05. Key words and phrases. self-similar tile, connectedness, disk-likeness, quasi-periodic tiling. TheresearchissupportedbytheNNSFofChina(No.11301322),theFundamentalandFrontier Research Project of Chongqing (No.cstc2015jcyjA00035). 1 We call such T a self-affine tile and T + J a tiling of Rd. In particular, if A is a similarity, then T is called a self-similar set/tile. Since the fundamental theory of self-affine tiles was established by Lagarias and Wang ([13],[14],[15]), there have been considerable interests in the topological struc- ture of self-affine tiles T, including but not limited to the connectedness of T ([7],[8],[12],[1],[6]), the boundary ∂T ([2],[19],[22]), or the interior T◦ of a connected tile T ([24],[25]). Especially in R2, the study on the disk-likeness of T (i.e., the property of being a topological disk) has attracted a lot of attentions ([5],[16],[23],[11],[6]). For other related works, we refer to [20],[17],[18],[21] and a survey paper [3]. Any change on the matrix A and the digit set D may lead to some change on the topology of T(A,D). To simplify the analysis on the relations between those two types of “changes”, one may fix an expanding matrix A and focus on particular choicesofthedigitsetD. RecentlyDengandLau[6]consideredaclassofplanarself- affinetilesT thataregeneratedbyalowertriangularexpandingmatrixandproduct- form digit sets. They gave a complete characterization on both connectedness and disk-likeness of T. Motivated by the above results, in this paper, we investigate the topological properties of the following two classes of self-similar fractals in R2. Assume that A is a diagonal matrix with equal nonzero entries, hence A is a similarity. In the first class, we consider a kind of digit sets D with a shift ε on the x-coordinates of some ε digits. We obtain an analogous result to [6]. Theorem 1.1. Let p be an integer with |p| = 2m + 1 where m ∈ N, let ε ∈ R. (cid:20) (cid:21) p 0 Suppose T is the self-similar set generated by A = and ε 0 p (cid:26)(cid:20) i+b (cid:21) 1−(−1)j (cid:27) D = j : b = ε, i,j ∈ {0,±1,...,±m} . ε j j 2 Then T is a self-similar tile. Moreover, ε (i) if |ε| < |p|, then T is disk-like; ε (ii) if |p|n ≤ |ε| < |p|n+1 for n ≥ 1, then T◦ has |p|n components and every closure ε of the component is disk-like. (see Figure 1) (cid:20) (cid:21) p 0 In fact, Theorem 1.1 can be proved in a more general setting where A = 0 q without causing much difficulty. We omit this for the completeness of the paper. Moreover, we further consider the quasi-periodic tiling property of T (the def- ε inition will be recalled in Section 3). Let D = D + AD + ··· + Ak−1D and ε,k ε ε ε D = (cid:83)∞ D . We prove that ε,∞ k=1 ε,k Theorem 1.2. With the same (A,D ) as in Theorem 1.1, T + D is a quasi- ε ε ε,∞ periodic tiling if and only if ε is a rational number. 2 The second one is a class of self-similar sets T with a shift ε on the diagonal digits ε along the diagonal line. In contrast with the first one, in this class, T might not be ε a tile, as the open set condition will not always hold for any ε (see the remark at the end of Section 4). Let δ = 1 if i = j; δ = 0 if i (cid:54)= j. Then we have ij ij Theorem 1.3. Let p be an integer with |p| > 2, ε ∈ R. Suppose T is the self-similar ε (cid:20) (cid:21) p 0 set generated by A = and 0 p (cid:26)(cid:20) (cid:21) (cid:27) i+a D = ij : a = δ ε, i,j ∈ {0,1,...,|p|−1} . ε j +a ij ij ij Then T is connected if and only if |ε| ≤ (|p|−1)2. (see Figure 3) ε |p|−2 For the organization of the paper, we prove Theorem 1.1 in Section 2, Theorem 1.2 in Section 3 and Theorem 1.3 in Section 4 respectively. 2. Self-similar tiles Let(A,D )bethepairasinTheorem1.1, andletT := T(A,D )betheassociated ε ε ε self-similar set. It suffices to prove the theorem for p > 0, otherwise we can replace A by A2 according to the fact T = A−1(T +D ) = A−2(T +D +AD ) ε ε ε ε ε ε where the digit set D +AD can be written as: ε ε (cid:26)(cid:20) (cid:21) (cid:27) pr+l+(pb +b ) D +AD = k t : r,l,k,t ∈ {0,±1,...,±m} ε ε pk +t (cid:26)(cid:20) (cid:21) (cid:27) i+b(cid:48) = j : i,j ∈ {0,±1,...,±(2m2 +2m)} , j where b(cid:48) = pb +b with j = pk +t. j k t We denote by I the set of i = i i ··· with i ∈ {0,±1,...,±m}. In view of (1.1), 1 2 n (cid:26)(cid:20) (cid:21) (cid:27) p(i)+b(j) T = : i = i i ··· ,j = j j ··· ∈ I (2.1) ε p(j) 1 2 1 2 where (cid:88) i (cid:88) b (cid:88) j p(i) = n, b(j) = jn and p(j) = n. pn pn pn n n n It follows from the above that the range of the y-coordinate of T is the interval ε [−1, 1]. For each fixed y = p(j) such that the radix expansion is unique, then the 2 2 horizontal cross section of T is an interval of length 1 with endpoints at −1 +b(j) ε 2 and 1+b(j); for the other y-coordinate that has two radix expansions, the horizontal 2 cross section of T is the union of two intervals with length 1. ε The following lemma is essentially the same as Proposition 2.2 in [6]. 3 Lemma 2.1. T is a self-similar tile. Moreover, for any sequence {(cid:96) } in R, let ε s s∈Z J = {(n+(cid:96) ,s)t : n,s ∈ Z}. Then T +J is a tiling of R2. s ε Proof. Let D = {0,±1,...,±m}. For any (x,y)t ∈ R2, since T(p,D) = [−1, 1], we 2 2 can find s ∈ Z such that y −s ∈ [−1, 1]. Let j ∈ I such that y −s = p(j). On the 2 2 other hand, there is n ∈ Z such that x−b(j)−(cid:96) −n ∈ [−1, 1]. This implies that s 2 2 x − b(j) − (cid:96) − n = p(i) for some i ∈ I. It follows that (x,y)t ∈ T + (n + (cid:96) ,s)t. s ε s Hence T +J = R2. ε Note that for almost all y ∈ R, the above s,j are unique. If we fix such y, then for almost all x ∈ R, the above n is also unique. Therefore, for almost all (x,y)t ∈ R2, the above n,s are unique. Hence {T +t : t ∈ J} are measure disjoint sets. That ε means T +J tile R2. (cid:3) ε Geometrically, the tile T has two sides on the horizontal line y = −1 and y = 1 ε 2 2 with length one. Lemma 2.1 implies that the tiling can be moved horizontally. The following is an elementary criterion for connectedness. Lemma 2.2 ([9],[12]). Let {S }N be an IFS of contractions on Rd and let K be its j j=1 attractor. Then K is connected if and only if, for any i (cid:54)= j ∈ {1,2,...,N}, there exits a sequence i = j ,j ,...,j = j of indices in {1,2,...,N} so that S (K) ∩ 1 2 n j k S (K) (cid:54)= ∅ for all 1 ≤ k < n. j k+1 Let (cid:18)(cid:20) (cid:21)(cid:19) (cid:18)(cid:20) (cid:21) (cid:20) (cid:21)(cid:19) x x i+b f = A−1 + j , i,j y y j where i,j ∈ {0,±1,··· ,±m}. Then f ’s form an IFS which generates T . By (2.1), i,j ε the elements of f (T ) are of the form i,j ε (cid:20) (cid:21) p(ii)+b(jj) (2.2) p(jj) where i = i i ··· , j = j j ··· ∈ I. For j j ···j with j ∈ {0,±1,...,±m}, if we 1 2 1 2 1 2 n t denote (cid:91) (cid:91) G = ··· f ◦f ◦···◦f (T ), (2.3) j1···jn i1,j1 i2,j2 in,jn ε i1 in then (cid:91) T = G . ε j1j2···jn j1,...,jn For simplicity of our statements, we write i = i ···i , j = j ···j and 0 = 0 1 n 0 1 n 0···0 where n ≥ 1. (cid:124) (cid:123)(cid:122) (cid:125) n Proposition2.3. Forj = j ···j ∈ {0,±1,...,±m}n andk,(cid:96) ∈ {0,±1,...,±m}, 0 1 n (i) if |k −(cid:96)| ≥ 2, then G ∩G = ∅; j0k j0(cid:96) 4 (ii) if |k −(cid:96)| = 1, then G ∩G is a line segment if and only if j0k j0(cid:96) |ε| < pn+1, and is a single point if and only if |ε| = pn+1. (see Figure 1) Proof. In view of (2.2) and (2.3), we obtain that (cid:26)(cid:20) (cid:21) (cid:27) p(i)+b(j kj) G = 0 : i,j ∈ I (2.4) j0k p(j kj) 0 From the expression of the y-coordinate, G is a part of T between the horizontal j0k ε lines y = (cid:80)n jt + k − 1 and y = (cid:80)n jt + k+1 − 1 . Hence the part (i) t=1 pt pn+1 2pn+1 t=1 pt pn+1 2pn+1 follows. From (2.3), we have (cid:18) (cid:20) (cid:21)(cid:19) (cid:91) p(i )+b(j ) G = A−nT + 0 0 j0 ε p(j ) 0 i1,...,in (cid:18) (cid:20) (cid:21)(cid:19) (cid:20) (cid:21) (cid:91) p(i ) b(j ) = A−nT + 0 + 0 ε 0 p(j ) 0 i1,...,in (cid:20) (cid:21) b(j ) = G + 0 . (2.5) 0 p(j ) 0 That is, every G is a translation of G . Hence, to prove the part (ii), we only need j0 0 to show the cases that G ∩ G and G ∩ G , as other situations are their 00 01 00 0(−1) translations. Since G and G are symmetric with respect to x-axis, it suffices 01 0(−1) to consider G ∩G . By making use of (2.4), 00 01 (cid:26)(cid:20) (cid:21) (cid:27) p(i)+b(00j) G = : i,j ∈ I 00 p(00j) (cid:26)(cid:20) (cid:21) (cid:27) p(i)+b(01j) G = : i,j ∈ I 01 p(01j) From the proof of part (i), we know that the intersection of G ∩G has a unique 00 01 y-coordinate 1 . By the expression of G , all digits in j are m. Since b = 0 if 2pn+1 00 i i is even; b = ε if i is odd. The x-coordinate of the element with y-coordinate of i 1 in G is as follows: 2pn+1 00 If m is even, b = 0, then b(00j) = 0. Hence the x-coordinate is x = p(i). Since m {p(i) : i ∈ I} = [−1, 1], the x’s form a unit interval 2 2 1 1 I = [− , ]. 1 2 2 If m is odd, b = ε, then b(00j) = (cid:80)∞ ε = ε . Hence the x-coordinate m t=n+2 pt pn+1(p−1) is x = p(i)+ ε . Such x’s form a unit interval pn+1(p−1) ε 1 ε 1 I = [ − , + ]. 1 pn+1(p−1) 2 pn+1(p−1) 2 5 In both two cases, we denote I = [α,α+1]. 1 Similarly, by the expression of G , all digits in j are −m. Then the x-coordinate 01 of the element with y-coordinate of 1 in G is of the following form: 2pn+1 01 If m is even, x(cid:48) = ε +p(i), which determines a unit interval pn+1 ε 1 ε 1 I = [ − , + ]. 2 pn+1 2 pn+1 2 If m is odd, x(cid:48) = ε + ε +p(i), which determines a unit interval pn+1(p−1) pn+1 ε ε 1 ε ε 1 I = [ + − , + + ]. 2 pn+1(p−1) pn+1 2 pn+1(p−1) pn+1 2 In both two cases, we denote I = [β,β +1]. 2 It follows that if G ∩G (cid:54)= ∅, then the y-coordinate of the intersection is 1 00 01 2pn+1 and the x-coordinate of the intersection is I ∩I . Note that I ∩I is an empty set 1 2 1 2 when |α −β| = ε > 1; a single point when |α −β| = ε = 1; and an interval pn+1 pn+1 when |α−β| = ε < 1. Therefore the part (ii) is proved. (cid:3) pn+1 Proposition 2.4. Let j = j j ···j ∈ {0,±1,...,±m}n for some n ≥ 1, then G 0 1 2 n j0 is a self-similar tile. Moreover, G is disk-like if and only if |ε| < pn+1. j0 Proof. Since G is a translation of G by (2.5), we only show the case for G . By j0 0 0 (2.3), we see that (cid:18) (cid:20) (cid:21)(cid:19) (cid:91) p(i ···i )+b(0j) G = A−n−1T + 1 n+1 0j ε p(0j) i1,...,in+1 (cid:18) (cid:20) (cid:21)(cid:19) (cid:91) p(i ···i )+b(0j) = A−1 A−nT +A 1 n+1 ε p(0j) i1,...,in+1 (cid:32) (cid:34) (cid:35)(cid:33) (cid:91) p(i ···i )+i + bj = A−1 A−nT + 2 n+1 1 pn ε j i1,...,in+1 pn   (cid:34) (cid:35) (cid:91) (cid:91) (cid:18) (cid:20) p(i ···i ) (cid:21)(cid:19) i + bj = A−1 A−nTε + 2 0 n+1 + 1 j pn  i1 i2,...,in+1 pn (cid:32) (cid:34) (cid:35)(cid:33) (cid:91) i + bj = A−1 G + 1 pn 0 j i1 pn From G = (cid:83)m G , it follows that 0 j=−m 0j G = A−1(G +D(cid:48)) (2.6) 0 0 where (cid:40)(cid:34) (cid:35) (cid:41) i+ bj D(cid:48) = pn : i,j ∈ {0,±1,...,±m} . j pn 6 (cid:20) (cid:21) 1 0 Let U = . Then UG = A−1(UG +UD(cid:48)), where 0 pn 0 0 (cid:40)(cid:34) (cid:35) (cid:41) i+ bj UD(cid:48) = pn : i,j ∈ {0,±1,...,±m} . j By Lemma 2.1, UG is a self-similar tile, so is G . Moreover, J(cid:48) = {(r +(cid:96) , s )t : 0 0 s pn r,s ∈ Z} is a tiling set of G for any sequence {(cid:96) } ⊂ R. 0 s s∈Z According to (2.6), we can write the corresponding IFS of G as 0 (cid:32) (cid:34) (cid:35)(cid:33) (cid:18)(cid:20) x (cid:21)(cid:19) (cid:20) x (cid:21) i+ bj f(cid:48) = A−1 + pn , i,j ∈ {0,±1,...,±m}. i,j y y j pn We will verify the disk-likeness of G through the following claims. 0 (i) We claim that f(cid:48) (G )∩f(cid:48) (G ) contains an interior point of G for any i,j. i,j 0 i+1,j 0 0 ItsufficestoshowthatG ∩(G +(1,0)t)containsapointof(G ∪(G +(1,0)t))◦. For 0 0 0 0 i,j ∈ I, we have p(i) ∈ [−1, 1] and p(j) ∈ [− 1 , 1 ]. Fix a point y ∈ (− 1 , 1 ) 2 2 pn 2pn 2pn 0 2pn 2pn so that there exists a unique j ∈ I such that p(j) = y . Let pn 0 b(j) 1 x = + , 0 pn 2 then (x ,y )t ∈ G ∩(G +(1,0)t). 0 0 0 0 On the other hand, since y is an interior point of [− 1 , 1 ] and the set J(cid:48) = 0 2pn 2pn {(r, s )t : r,s ∈ Z} is a tiling set of G (taking (cid:96) ≡ 0), then (x ,y )t ∈ G +(r, s )t pn 0 s 0 0 0 pn only if s = 0, i.e., (x ,y )t ∈ G +(r,0)t. Thus, x must be the form 0 0 0 0 b(j) x = +p(i)+r 0 pn for some i ∈ I. That implies p(i) + r = 1, hence r = 0 or 1 as p(i) ∈ [−1, 1]. 2 2 2 Then (x ,y )t ∈/ G + (r,0)t for any r ∈ Z \ {0,1}. Therefore (x ,y )t must be in 0 0 0 0 0 (G ∪(G +(1,0)t))◦. 0 0 (ii) We claim that G is connected if and only if |ε| ≤ pn+1. Indeed, if G is 0 0 connected, by Proposition 2.3, then G ∩G (cid:54)= ∅ for −m ≤ k ≤ m−1, which 0k 0(k+1) implies |ε| ≤ pn+1. On the contrary, if |ε| ≤ pn+1, then G ∩ G (cid:54)= ∅ for 0k 0(k+1) −m ≤ k ≤ m − 1, by Proposition 2.3. Thus, there exist i ,t ∈ {0,±1,...,±m} k k such that f(cid:48) (G )∩f(cid:48) (G ) (cid:54)= ∅. That together with (i), can help us select a i ,k 0 t ,k+1 0 k k finite sequence {S }N from {f(cid:48) } in the following zigzag order: j j=1 i,j i,j f(cid:48) ,f(cid:48) ,...,f(cid:48) ,f(cid:48) ,...,f(cid:48) ,f(cid:48) ,f(cid:48) ,..., −m,−m −m+1,−m m,−m m−1,−m i−m,−m t−m,−m+1 t−m+1,−m+1 f(cid:48) ,f(cid:48) ,...,f(cid:48) ,f(cid:48) ,...,f(cid:48) ,f(cid:48) ,..., m,−m+1 m−1,−m+1 −m,−m+1 −m+1,−m+1 i−m+1,−m+1 t−m+1,−m+2 f(cid:48) ,f(cid:48) ,...,f(cid:48) ,f(cid:48) ,...,f(cid:48) . im−1,m−1 tm−1,m m,m m−1,m −m,m Then each f(cid:48) appears at least once in the sequence {S }N and i,j j j=1 S (G )∩S (G ) (cid:54)= ∅, ∀ 1 ≤ j ≤ N −1. j 0 j+1 0 7 (a) ε=2 (b) ε=3 (c) ε=4 (d) ε=8 (e) ε=9 (f) ε=10 Figure 1. An illustration of Theorem 1.1 by taking p = 3. Hence G is connected by Lemma 2.2. 0 (iii) We claim that (G )◦ is connected if |ε| < pn+1. For −m ≤ k ≤ m − 1, if 0 |ε| < pn+1, then there exist i ,t such that f(cid:48) (G ) ∩ f(cid:48) (G ) is a horizontal k k i ,k 0 t ,k+1 0 k k line segment by Proposition 2.3. Suppose z = f(cid:48) (z˜) be the mid-point of the line i ,k k segment. Since (cid:32) (cid:34) (cid:35)(cid:33) i + bk f(cid:48) (G ) = A−1 G + k pn , ik,k 0 0 k pn (cid:32) (cid:34) (cid:35) (cid:33) i + bk (cid:20) c (cid:21) f(cid:48) (G ) = A−1 G + k pn + , tk,k+1 0 0 k 1 pn pn where c = t −i + bk+1−bk. Then G ∩(G +(c, 1 )t) is a horizontal line segment k k pn 0 0 pn with positive length and z˜ is its mid-point. Note that the top and bottom sides of G are horizontal line segments with length one and the height of G is 1 , it is easy 0 0 pn to verify that s s 1 z˜∈ G +(r+(cid:96) , )t if and only if (r+(cid:96) , )t = (0,0)t or (c, )t. 0 s pn s pn pn Then z˜ is an interior point of G ∪(G +(c, 1 )t), this means z is an interior point 0 0 pn of f(cid:48) (G )∪f(cid:48) (G ). Thus, f(cid:48) (G )∩f(cid:48) (G ) contains an interior point of i ,k 0 t ,k+1 0 i ,k 0 t ,k+1 0 k k k k G . Hence (G )◦ is connected by Lemma 2.2. 0 0 Now we prove the second part of the proposition. Suppose |ε| < pn+1, then claim (iii) implies (G )◦ is connected, which yields the disk-likeness of G by a theorem 0 0 of Luo et al. [23]. Suppose G is disk-like, then G and (G )◦ are both connected, 0 0 0 hence we have |ε| ≤ pn+1 by claim (ii). If the equality holds, then G ∩ G 0k 0(k+1) 8 contains only one point for any k, and G ∩G = ∅ for |k−(cid:96)| ≥ 2 by Proposition 0k 0(cid:96) 2.3. Since G = (cid:83)m G , G can be divided into p parts, and every two adjacent 0 k=−m 0k 0 parts intersect at one common point. Hence the intersection point must be at the boundaryofG . Thatis,(G )◦ isnotconnected,yieldingacontradiction. Therefore, 0 0 G is disk-like if and only if |ε| < pn+1. (cid:3) 0 Proof of Theorem 1.1: (i) If |ε| < |p|, by using the same argument as in the proof of Proposition 2.4, we can prove that T is disk-like (see also Theorem 3.1 of ε [6]); If |ε| < |p|n+1 for n ≥ 1, by Proposition 2.4, then G are disk-like tiles. j1···jn Moreover, from Proposition 2.3 and the assumption |ε| ≥ |p|n, it follows that the tiles G are either disjoint or meet each other at a single point (see Figure 1). j1···jn Therefore, we conclude with proving (ii). (cid:50) 3. Quasi-periodic tiling A tiling T +J of Rn with a tile T is called a self-replicating tiling with a matrix B if for each t ∈ J there exists a finite set J(t) ⊂ J such that (cid:91) B(T +t) = (T +t(cid:48)). t(cid:48)∈J(t) Moreover, we say a tiling T +J is a quasi-periodic tiling if the following two properties hold: (1) Local isomorphism property: For any finite set Σ ⊂ J, there exists a con- stant R > 0 such that every ball of radius R in Rn contains a translating copy Σ+t of Σ such that Σ+t ⊂ J. (2) Local finiteness property: For any k ≥ 1 and r > 0, there are finitely many translation-inequivalent arrangements of k points in J which lie in some ball of radius r. Under the assumption of Theorem 1.1, we let D = D + AD + ··· + Ak−1D ε,k ε ε ε and D = (cid:83)∞ D . For j ∈ {0,±1,...,±m}, denote by ε,∞ k=1 ε,k (cid:26) 1 j is odd (cid:98)j = 0 j is even. Then it can be easily seen that D = (cid:26)(cid:20) (cid:80)∞n=0inpn +ε(cid:80)∞n=0(cid:98)jnpn (cid:21) : i ,j ∈ {0,±1,...,±m}(cid:27) ε,∞ (cid:80)∞ j pn n n n=0 n (cid:40) (cid:41) (cid:20) (cid:21) ∞ ∞ m +m ε (cid:88) (cid:88) = 2m (cid:98)1 : m1,m2 ∈ Z,m(cid:98)1 = (cid:98)jnpn when m1 = jnpn . 1 n=0 n=0 Proposition 3.1. T +D is a self-replicating tiling. ε ε,∞ 9 Proof. By Lemma 2.1, obviously D is a tiling set of R2. For any t ∈ D , we let ε,∞ ε,∞ J(t) = At+D . Since AT = T +D , we have AT +At = T +D +At. Hence ε ε ε ε ε ε ε (cid:91) A(T +t) = (T +t(cid:48)). ε ε t(cid:48)∈J(t) Therefore T +D is a self-replicating tiling by letting B = A. (cid:3) ε ε,∞ Theorem 3.2. T + D is a quasi-periodic tiling if and only if ε is a rational ε ε,∞ number. Proof. By the definition, we first show the local isomorphism property holds. Let Σ be a finite subset of D , then there exists an integer (cid:96) such that Σ ⊂ D . Let ε,∞ ε,(cid:96) (cid:91) R = diam(A(cid:96)T ) = diam( (T +d)). ε ε d∈D ε,(cid:96) We claim that every ball B (3R) with center x ∈ R2 and radius 3R, contains a x translatingcopyofΣ. Indeed,forthex,thereexistsad ∈ D suchthatx ∈ T +d x ε,∞ ε x by the tiling property. If d ∈/ D , then there is a larger integer (cid:96)(cid:48) > (cid:96) such that x ε,(cid:96) (cid:96)−1 (cid:96)(cid:48) (cid:88) (cid:88) d ∈ Ajd + Ajd , where d ∈ D . x ij ij ij ε j=0 j=(cid:96) Letd(cid:48) = (cid:80)(cid:96)(cid:48) Ajd ,thend −d(cid:48) ∈ D . Byusing0 ∈ T andthetriangleinequality, x j=(cid:96) ij x x ε,(cid:96) ε for d ∈ D , we have, ε,(cid:96) (cid:107)d(cid:48) +d−x(cid:107) ≤ (cid:107)x−d (cid:107)+(cid:107)d −d(cid:48)(cid:107)+(cid:107)d(cid:107) ≤ 3R. x x x x Thus d(cid:48) +D ⊂ B (3R), yielding d(cid:48) +Σ ⊂ B (3R). If d ∈ D , the above is still x ε,(cid:96) x x x x ε,(cid:96) true as d(cid:48) = 0. x Now we show T +D has the local finiteness property if and only if ε is rational, ε ε,∞ then the desired result follows. Suppose ε is irrational. Consider m = m(1+p+···+pk−1) = 1(pk −1), n = 1 2 1 (−m)(1 + p + ··· + pk−1) + 1 · pk = 1(pk + 1). If m is an even integer, then 2 m = 0,n = pk. We have (cid:98)1 (cid:98)1 (cid:20) (cid:21) (cid:20) (cid:21) m n +εpk t := 2 , t(cid:48) := 2 ∈ D , ∀ m ,n ∈ Z. k m k n ε,∞ 2 2 1 1 Ifmisanoddinteger,thenm = 1+p+···+pk−1 = 1−pk, n = 1+p+···+pk−1+pk = (cid:98)1 1−p (cid:98)1 1−pk +pk. We have 1−p (cid:34) (cid:35) (cid:34) (cid:35) m + ε(1−pk) n +εpk + ε(1−pk) t := 2 1−p , t(cid:48) := 2 1−p ∈ D , ∀ m ,n ∈ Z. k k ε,∞ 2 2 m n 1 1 10

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