Topological properties of self-similar fractals with one parameter

6.4 MB·

by  Jun Jason Luo

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 prop-
erty 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 prop-
erties 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 prop-
erties 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