Mixing Property of Z
2
Subshift of Finite
Type
Jung-Chao Ban
The National Center for Theoretical Sciences
Hsinchu 300, Taiwan
Hui-Ping Chen
Department of Applied Mathematics
National Chiao Tung University
Hsinchu 300 , Taiwan
Yin-Heng Lin
Department of Applied Mathematics
National Chiao Tung University
Hsinchu 300 , Taiwan
July 20, 2006
Abstract
In this paper, the primitivity of n-th order transition matrices An
de…ned on Z2 n are studied, this topic related to the mixing property
of 2-dimensional shift of …nite type (SFT). We provide a series of checkable conditions for ensuring a SFT is mixing.
The authors would like to thank the National Center for Theoretical Sciences for partially supporting this research
1
Introduction
Many systems have been studied as model for spatial pattern formation in biology, chemistry, engineering and physics. Lattices play important roles in modeling underlying spatial structure. We mention some works arising in biology (cf. [8, 9, 22, 23, 24, 25, 33, 34, 35]), chemical reaction and phase transitions ( cf. [7, 13, 14, 15, 19, 25]), image processing and pattern recognition ( cf. [13, 14, 15, 16, 17, 18, 20, 21, 26] )as well as material sciences( cf. [11, 12. 27]). In Lattice Dynamical Systems (LDS), especially Cellular Neural Networks (CNN), the complexity of the set of all global patterns has received considerable attention in recent years (cf. [1, 2, 3, 4, 5, 6, 28, 29, 30, 31, 32]). One of the interesting problem comes from the statistic mechanism, d-dimensional subshift of …nite type, states as follows: Let A be a …nite set with jAj > 1, and for a …nite set F Zd. Let F : AZ
d
! AF denote the restriction map, i.e., F (x) = xjF, where AZ
d
is viewed as the space as map Zd
! A with the product topology. The group Zd
acts on AZd
via the shift ; n(x)m = xn+m: A closed non-empty invariant subset
X AZd
is a d-dimensional subshift of …nite type ( d-dimensional SFT, brie‡y) if there is a …nite set F Zd and a collection of P
AF for which X = XP = n x2 AZd j F +n(x)2 P for every n 2 Zd o :
where F +n(x)is the restriction map of x 2 X on F and coordinate n 2 Z2:
We write Bn for the box f n; : : : ; ng d
and n(x) = Bn(x) :The collection
of all patterns f n(x) j x 2 Xg is denoted by Bn(X) : On the other hand,
the analogue of vertex shift in higher dimensional is de…ned by a collection of d transition matrices A1; : : : ; Adall indexed by the same set of symbols A;
We put
X (A1; : : : ; Ad) =
n
x2 AZd j Ai(xn; xn+ei) = 1 for all n;i 2 Z
do;
where ei is as usual the i th standard basis vector and Ai(a; b) denotes the
(a; b) entry of Ai:Such a space is called a d-dimensional matrix subshift:
In , every d-dimensional matrix shift is a subshift of …nite type. If X is a shift space with is a Zd-action on it, one of the most important invariant
measures the complexity is the topological entropy : h (X) is de…ned by h (X) = lim
n!1
logjBn(X)j
Given any P AF; it’s more is extremely di¢ cult to compute the exact
number for h (XP) for d 2 or for matrix shift X (A1; ; Ad) for
arbi-trary A1; ; Ad: For example, if d = 2; the exact solution of entropy for
2-dimensional matrix shift (or Golden-Mean Shift) with A1 = A2 =
1 1 1 0
is still open. However, in [4], for 2 dimensional SFT, the authors introduced the concept of n-th order transition matrices An and Bn de…ned on Z2 n and
Zn 2 respectively, i.e., the An (Bn) is a transition matrix indexed by 1 n
(n 1)patterns in XP. Once the A2 and B2 corresponding to P are de…ned,
the generation rule from A2 to An and B2 to Bn can also be found therein.
Then Bn(XP) are stored in An 1n or Bn 1n , and
jBn(X)j = An 1n = B n 1 n
Thus computing h (XP) equal to computing the growth rate of maximum
eigenvalue of An;say (An), by Perron-Frobineous Theorem, i.e.,
h (XP) = lim n!1 log (An) n = limn!1 log (Bn) n
We refer reader to [4] for detail. Thus from An and Bn; some problems of
SFT can by transform to the problem of matrix by means of above discussion. For topological entropy, if A2 or B2 has special structure, the exact number
of entropy for h (XP) = h (A2) = h (B2) can be found. For those systems
which entropy are not known, trace and connecting operator are introduced for estimating the upper and lower bound respectively for d = 2 [6], the analogue n th order transition matrices, trace and connecting operators can be also found for d = 3 in [5]: Beside computing the topological entropy, it’s nature to consider the ergodic property for XP; the interesting property we
want to consider is the mixing property (de…ne later), in this paper, we check the mixing property of XP in terms of the An and Bn and by the following
theorem.
Theorem 1 Given P AF and XP is the 2-dimensional SFT de…ned on P
and F Z2;
if A2 and B2 are associate 2-nd order transition matrices, then
In Theorem 2, once the 2-nd order transition matrices is constructed, then the generation rule of Anand Bnis also de…ned, however, the size of the
An and Bn grow exponentially, i.e., An; Bn 2 M2n(Z) : Thus, checking the
primitivity of An and Bn for n 2 is not always very easy. In Theorem 24,
we search for some checkable conditions of Ak and Bk for k 3to guarantee
that An and Bn are primitive for n 2;thus XP is mixing in d = 2:
This paper is organized as follows, section2 review some notions and con-cepts in [4] for self-contained of this paper and de…ne some notations which will be used in the statement of the main theorem, a selection scheme for construct a closed graph (de…ned later) will be presented in section3 which will help us to proof the main theorem, and the main theorem and proof will be presented in section3.
2
Preliminary
In this section, we …rst recall some results in [4] which are crucial in this study. For simplicity, we only consider two symbols which are given on 2 2 lattice Z2 2 …rst; i.e., F is a 2 2 lattice and jAj = 2. Given P AF; we
begin with a consideration of given horizontal transition matrix
H2 = 0 B B @ h11 h12 h13 h14 h21 h22 h23 h24 h31 h32 h33 h34 h41 h42 h43 h44 1 C C A ;
which is related to a set of admissible local patterns on Z2 2as the ordering
matrices X2 in Figure1 ( X2 and Y2 are matrix with entries for which are all
possible patterns and indexed by 1 2and 2 1patterns respectively), that is, hij = 8 < : 1 if xij 2 P 0 otherwise ; thus hij 2 f0; 1g for 1 i; j 4:
The associated vertical transition matrix V2 corresponding to Y2 is also de-…ned by V2 = 0 B B @ v11 v12 v13 v14 v21 v22 v23 v24 v31 v32 v33 v34 v41 v42 v43 v44 1 C C A :
In [4], it has shown that H2 and V2 are related to each other as follows.
H2 = 0 B B @ v11 v12 v21 v22 v13 v14 v23 v24 v31 v32 v41 v42 v33 v34 v43 v44 1 C C A = H2;1 H2;2 H2;3 H2;4 ; (1) and V2 = 0 B B @ h11 h12 h21 h22 h13 h14 h23 h24 h31 h32 h41 h42 h33 h34 h43 h44 1 C C A = VV2;12;3 VV2;22;4 : (2)
The recursive formulas for n-th order horizontal transition matrices Hn
de…ned on Z2 n has been obtained [4] by the following procedure:
Hk+1 = 0 B B @ v11Hk;1 v12Hk;2 v21Hk;1 v22Hk;2 v13Hk;3 v14Hk;4 v23Hk;3 v24Hk;4 v31Hk;1 v32Hk;2 v41Hk;1 v42Hk;2 v33Hk;3 v34Hk;4 v43Hk;3 v44Hk;4 1 C C A ; whenever Hk= Hk;1 Hk;2 Hk;3 Hk;4
is de…ned. The number of all admissible patterns de…ned on Zm nwhich can
be generated from H2 is now de…ned by
m n(H2) =j Hm 1n j = the summation of all entries in H m 1 n
2
X
=
Y
2=
Figure1. Ordering matrices X2 and Y2
Notably since the relation(1) and (2), we will say if A2 = H2 represents
the horizontal (or vertical) transition matrix then B2 = V2 represents the
vertical (or horizontal) transition matrix. Results that hold for A2 are also
valid for B2: Therefore, for simplicity, only A2 is presented herein. We state
the generation rule for An as follows and omit the proof here.
Theorem 2 (Theorem 3.1 of [4]) Let A2be a 2-nd transition matrix. Then,
for higher order transition matrices An; n 3; we have the following three
equivalent expressions
(I) An can be decomposed into n successive 2 2 matrices (or n-successive
Z-maps) as follows: An= An;1 An;2 An;3 An;4 ; An;j1 jk = An;j1 jk1 An;j1 jk2 An;j1 jk3 An;j1 jk4 ; for 1 k n 2 and An;j1 jn 1 = vj1 jn 11 vj1 jn 12 vj1 jn 13 vj1 jn 14 : Furthermore, An;k = vk1An 1;1 vk2An 1;2 vk3An 1;3 vk4An 1;4 : (3)
(II) Starting from A2 = A1 A2 A3 A4 with Ak = vk1 vk2 vk3 vk4 :
An can be obtained from An 1 by replacing Ak by Ak A2 according
the generating rule:
Ak 7 ! Ak A2 = vk1A1 vk2A2 vk3A3 vk4A4 : (III) An= (An 1)2n 1 2n 1 E2n 2 A2;1 A2;2 A2;3 A2;4 :
where denote the Hardamard product, i.e., if A, B 2 Mn(Z), then
A B = (aijbij)ni;j=1;
if A 2 Mn(R) and B = (Bij)ni;j=1 2 Mnk(Z), where Bij 2 Mk(R) ; then
we generalized the Hardamard product to di¤erent size of A and B by A B = (aijBij)
n i;j=1:
and denotes the Kronecker product, i.e., A B = (aijB)ni;j=1:
For reader’s convenience, we give an example to demonstrate the gener-ation rule for 2-dimensional Golden-Mean Shift:
Example 3 Consider the Golden-Mean shift with A1 = A2 =
1 1 1 0 ;
this means that consecutive 11 is not allowed in B2(X (A1; A2)) : It’s easy to
construct A2 and B2 From Figure1 as follows.
A2 = B2 = 0 B B @ 1 1 1 0 1 0 1 0 1 1 0 0 0 0 0 0 1 C C A :
From de…nition, we let A2 = A2;1 A2;2 A2;3 A2;4 ; where A2;1 = 1 1 1 0 ; A2;2 = 1 0 1 0 ; A2;3 = 1 1 0 0 and A2;1 = 0 0 0 0 : Then A3 = 0 B B @ 1 1 1 0 1 0 1 0 1 1 0 0 0 0 0 0 1 C C A 0 B B @ A2;1 A2;2 A2;1 A2;2 A2;3 A2;4 A2;3 A2;4 A2;1 A2;2 A2;1 A2;2 A2;3 A2;4 A2;3 A2;4 1 C C A = 0 B B @ A2;1 A2;2 A2;1 0 A2;3 0 A2;3 0 A2;1 A2;2 0 0 0 0 0 0 1 C C A = 0 B B B B B B B B B B @ 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 C C C C C C C C C C A : Thus An = 0 B B @ An 1;1 An 1;2 An 1;1 0 An 1;3 0 An 1;3 0 An 1;1 An 1;2 0 0 0 0 0 0 1 C C A whenever An 1 = An 1;1 An 1;2 An 1;3 An 1;4 :
Remark 4 (1) Sometimes we will deal with the product of Am
n; i.e., the m
power of n-th order transition matrix, and write Amn =
Am;n;1 Am;n;2
Am;n;3 Am;n;4
For n = 1; we write
An=
An;1 An;2
An;3 An;4
;
as usual for extension. On the other hand, for matrix multiplication, the indices An; 2 f1; 2; 3; 4g are conveniently express as
An =
An;11 An;12
An;21 An;22
: (4)
(2) All the discussion above can generalized to arbitrary jAj = p 2; for this case, the analogue X2 and Y2 are also de…ned by indexing the
lexicographic order of 1 2 and 2 1 patterns of p symbols, then A2;
B2 2 Mp2 p2(Z) can be de…ned similarly; the generation rule as in
Theorem 2 can generate An and Bn in the same fashion for n 2:
We de…ne the indicator matrix of A2; say A1 as follows for the main
Theorem.
De…nition 5 (A1) Given A2, the 2-nd transition matrix with P AF where
F is 2 2 lattice and jAj = p; the associated A1 = (aij)pij=1;where aij 2 f0; 1g
can be de…ned, aij = 8 < : 1 if A2;ij 6= 0 0 otherwise :
Note: A2;ij is the usual index in (4) for matrix multiplication.
Remark 6 As in (2) of Remark 4, we also de…ne A1 = (b )p
2 =1, where b 2 f0; 1g as b = 8 < : 1 if A2; 6= 0 0 otherwise ;
and 1 can be de…ned from the -indices to (i; j)-indices similarly, e.g.,
1(2) = 12:
From now on, we assume the number of symbol is p throughout this paper; and then continue the main study of mixing property.
De…nition 7 Let A 2 Mn(Z) ; we say A 2 Cn if A has no zero column,
A 2 Rn if A has no zero row, A 2 Pn if A is primitive A 2 Irn if A is
irreducible.
The following simple Proposition is crucial in our study, and in the rest of this research, we need to …nd conditions for guarantee An; 2 Cpn 1 (or
An; 2 Rpn 1) for essential 2 f1; ; p2g :
Proposition 8 Let A 2 Cn (or B 2 Rn) and denote the full matrix of size
n by En; then EnA En (or BEn En)
Proof. We just prove EnA Enfor A 2 Cn;and the other is similar, indeed,
(EnA)ij = n X i=1 eikakj n X i=1 akj 1:
The third inequality holds since A 2 Cn. Thus EnA En: This complete
the proof.
3
Closed graph
In this section, we provide a constructive scheme to select those 2 f1; ; p2g to guarantee that An; 2 Cpn 1 (or Rpn 1). First, we need some de…nitions.
De…nition 9 (CA; RA) Let A 2 Mn n(Z) ; de…ne s (A) = faij j aij 6= 0g ;
and let I s (A) be a collection of indices, we de…ne A jI = a0ij by
a0ij = 8 < : aij if aij 2 I 0 otherwise :
and we also de…ne CA=
n
(ai1j1; ; ainjn) j A j(ai1j1; ;ainjn) has no zero column
o ; similarly, we de…ne RA= n (ai1j1; ; ainjn) j A j(a
i1j1; ;ainjn) has no zero row
o :
Example 10 If A = 0 @ 1 0 1 0 1 1 0 0 1 1 A ; then CA=f(a11; a22; a13) ; (a11; a22; a23) ; (a11; a22; a33)g ; and RA=f(a11; a22; a33) ; (a13; a22; a33) ; (a13; a23; a33)g :
Remark 11 If A = An; in the De…nition 9, write An; = (b i)p
2
i=1: For
simplifying the notation, we de…ne CAn; =
n
(i1; ; ip) j An; j(b
i1; ;b ip) has no zero column
o ;
and similarly de…ne RAn; =
n
(i1; ; ip) j An; j(b
i1; ;b ip) has no zero row
o :
Example 12 For p = 3 and assume A2 is such that
A2;1= 0 @ b11 b12 b13 b14 b15 b16 b17 b18 b19 1 A = 0 @ 1 0 1 0 1 1 0 0 1 1 A : Then from Example 10, we have
CA2;1 =f(1; 5; 3) ; (1; 5; 6) ; (1; 5; 9)g and RA2;1 =f(1; 5; 9) ; (3; 5; 9) ; (3; 6; 9)g :
3.1
Selection Scheme
In this subsection, we will construct a selection scheme to choose those for which An; 2 C2n 1 (or R2n 1).Write A1 = (b ) as in Remark 6; and we start
from some i0 and de…ne IC;0=f i0g ; construct CA2; i0 as Remark 11; then
we have the following cases:
(ii) If CA2; i0 6= ?; i.e., A2; i0 2 Cp; then we pick some (j1; jp) 2
CA2; i0:We de…ne
(j1; jp) = ( i01; ; i0p) ;
then we have following subcases:
(A) If there exist some 1 i1 < p such that CA2; i0i1 = ?; then we stop;
(B) If CA2; i0i1 6= ? for all 1 i1 < p; then we process the following step:
(a) If i0i1 2 IC;0 =f i0g for some 1 i1 p;then we stop this procedure,
and say ( i0 i0i1) is a branch and call such i0i1 is a tail.
(b) If i0i1 2 I= C;0 = f i0g for some 1 i1 p; then construct CA2; i0i1;
(Note that i0i1 2 f1; ; pg is an integer) then we check i0i1i2 for
1 i2 p for the following subcases:
(1) If there exists an i2 2 f1; ; pg such that CA2; i0i1i2 = ? ; then we stop
this selection of ( i0i11; ; i0i1p) 2 CA2; i0i1 and go back to choose
another ( i01; ; i0p)2 CA2; i0 in process (ii), and this step will stop
when exhausted the element of CA2; i0; i.e., return to step (i);
(2) If CA2; i0i1i2 6= ? for all 1 i2 < p; then let
IC;1=f i01; ; i0pg ;
and continue step(ii) with CA2; i0 = CA2; i0i1i2.Finally, if
( i0; i0i1; i0i1i2; ; i0i1i2 in 1) is a branch in this collection, and IC;0;
; IC;n 2 is de…ned, if CA2; i0i1i2 in 1 = ?; then we follow step (ii)-(B)-(b)
to pick another choice of CA2; i0i1i2 in 2; if CA2; i0i1i2 in 1 6= ? and let IC;n 1
can be de…ned, then we follow step(ii), i.e., check all
( i0; i0i1; i0i1i2; ; i0i1i2 in 1; i0i1i2 in 1j) for1 j p;
if i0i1i2 in 1j 2
n 1[ i=1
IC;i; then we stop, and say
is a branch and i0i1i2 in 1j is a tail. If i0i1i2 in 1j 2=
n 1[ i=1
IC;i; then we
con-tinue the above process (B).Then from above inductive process with initial symbol i0;a tree graph GC; i0 can be de…ned and let B GC; i0 be the
collec-tion of all branch of this graph. For each branch ( i0; i0i1; i0i1i2; ; i0i1i2 in 1)
in this tree graph, we call i0i1i2 in 1 the tail of this branch. we denote by
GC; i0 the maximum length of each branches in GC; i0:We call a tree graph
GC; i0 is closed if for each branch
i0; i0i1; i0i1i2; ; i0i1i2 ik 1 2 B GC; i0 with k GC; i0 ;
we have i0i1i2 ik 1 2
k 2[ i=0
IC;k 2:Finally, we de…ne
A (GC; ) = f 0 j 0 is a symbol appear in some branch of GC; g :
i.e., if 0 2 A (G
C; ) ; then there is a branch ; ; 0; ; i0i1 ik 1 2
B (GC; ) : IR;n; GR; and jGR; j can be de…ned similarly.
Example 13 Let p = 2 and
A2 = 0 B B @ 1 1 1 1 1 1 0 0 1 0 0 1 1 1 0 1 1 C C A = A2;1 A2;2 A2;3 A2;4
and if we start from i0 = 1; it can be easily computed that
CA2;1 =f(1; 2) ; (1; 4) ; (3; 2) ; (3; 4)g ;
Since CA2;4 = ?; only graph (a) is closed. ( a ) 1 1 2 1 2 1 1 4 ( b ) ( c ) ( d ) 1 2 3 3 4 2 1 1 3 4
Figure2. Tree graph with initial symbol bi0 = 1
Lemma 14 Let GC; i0 is a closed graph with initial symbol i0;then for each
is not a tail of some branch of GC; i0; we have A2; 2 Cp:
Proof. This is obviously from the procedure of the selection of closed graph.
Lemma 15 Let GC; i0 is a closed graph with initial symbol i0;then for each
is a tail of some branch of B GC; i0 ; we have A2; 2 Cp:
Proof. If is a tail of some branch ( i0; ; k 2; ) 2 B GC; i0 with
length k, GC; i0 is closed, then by the collection scheme, 2 k 2[
i=0
IC;i;and we
may assume = 0 for which 0 2 I
C;k0 for some k0 k 2; if 0 is a tail,
then use the same argument to …nd next symbol until there is a ^ for which is not a tail, and closed graph guarantee that the scheme will stop. Then Lemma 14 applied to show
A2; = A2; 0 = = A2;^ 2 Cp:
De…nition 16 Let GC; i0 is a closed graph with initial symbol i0; if 2
A GC; i0 is a symbol of some branch ; we call 0 is the root of if there is
a branch of length m with 0; are two consecutive symbols in this branch, i.e.,
i0; ;
0; ; ;
im 1 2 B GC; i0 ;
we write r ( ) = 0 and in this situation we also call is one leave of 0;
write = l ( 0) ; more generally, we de…ned 0 is a n-root of ( is one
n-leave of 0) if
i0; ;
0 =
i0 ik; ; = i0 ik ik+n; ; im 1 2 B GC; i0
, and similarly for GR; i0:
Lemma 17 Let GC; i0 is a closed graph with initial symbol i0; if
i0; ; ; i0i1 im 1j for 0 j p
are branch in B GC; i0 ; then A3; 2 Cp2:
Proof. Since is not a tail, and Lemma 14 shows that A2; 2 Cp; and by
Lemma 15, we have A2; i0i1 im 1j 2 Cp for 1 j p:Since
(im;1; ; im;p)2 CA2; i0i1 im 1;
apply the generation rule (3) for jAj = p, we have A3; = 0 B @ v 1A2;1 v pA2;p .. . . .. ... v (p2 p+1)A2;(p2 p+1) v p2A2;p2 1 C A : Thus we have A3; 2 Cp2:
Lemma 18 Let GC; i0 is a closed graph with initial symbol i0; we have
An+1; 0 2 Cpn if Ak; 2 Cpk 1 for k n for all is a leave of 0; i.e.,
= l ( 0) :
Proof. Since 0 is not a tail of G
C; i0, Lemma 14 applied to show A2; 0 2 Cp:
Without loss of generality, we number the leaves of 0 as l1( 0) ; ; lp( 0),
then we have (l0( 0) ; ; lp 1( 0)) 2 CA2; 0: From the assumption, we
as-sume An;li( 0) 2 Cpn 1 for 1 i p; by the same fashion of proof in Lemma
Theorem 19 Given A2 and write A1 = (b ) as in Remark 6: If there exists
a closed graph with initial symbol i0, then for all 2 A GC; i0 ; we have
An; 2 Cpn 1 for all n 2: Similarly, for 0 2 A GR;
i0 we have An; 0 2
Rpn 1 for all n 2:
Proof. We just prove for 2 A GC; i0 ;we have An; 2 Cpn 1 for all n 2;
and the other is similar. We assume GC; i0 = m: We …rst claim that for all
2 A GC; i0 is a tail, we have An; 2 Cpn 1 for all n 2; Once this claim
holds, we consider two cases: 1. for some 0 is not a tail, but all its leaves
are tails, then Lemma 17 applied to show that An+1; 0 2 Cpn; 2. for some 0
is not a tail, and not all its leaves are tails, i.e., there are I1 and I2 such that
if i 2 I1; li( 0) is a tail and for i0 2 I2; li0( 0) is not a tail. In this case, we
have An;li( 0)2 Cpn 1 for i 2 I1; and An+1;li0( 0) 2 Cpn for i 2 I2; thus Lemma
18 applied to show that An+1; 0 2 Cpn: Thus we can show that for all 0 is a
root of a tail, we have An+1; 0 2 Cpn, since if 0 6= i
0, then i0 is also a leave
of some 00; then we can show that A
n+2; 0 2 Cpn+1 inductively. Generally,
we have
An+k; 0 2 Cpn+k 1
for 0 is a k-root of some tail, then A
n; 2 Cpn 1 for all n 2; for all
2 A GC;bi0 ;thus we can end the proof: For complete the claim, we prove
it by induction, Lemma 15 shows that n = 2 is OK, we assume for n = k holds; that is, Ak; 2 Cpk 1 for all is a tail, then for n = k + 1; we just
show that Ak+1; 2 Cpk for those 2 IC;m (recall that m = GC;
i0 ), and
the proof for other tail is similar, there are two cases: (a). If 2 IC;m; and
by the collection of GC; i0; if = 1 2 IC;m r1 is not a tail for some r1 2 Z,
then above discussion to show that Ak+1; 2 Cpk;(b). If = 2 = IC;m r 2 is
a tail, then it must equal to 3 2 IC;m r2 r3, if 3 is a symbol belong to case
(a), i.e., 3is not a tail, then we are done, otherwise, we …nd 4; ; l, since
GC; i0 is closed, then, the procedure will stop up to …nd l= i0;that means
we always can …nd 0 2 A G
C; i0 which is not a tail such that = 0;then
we have Ak+1; 0 = Ak+1; 2 Cpk, then induction shows that An; 2 Cpn 1 for
all n 2 and for all 2 A GC; i0 is a tail. This complete the proof.
Example 20 For p = 2 and assume that
A2 = 0 B B @ 1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 1 C C A ;
then there exist two closed graphs GC;1 and GR;1 with initial symbol 1 as in
Figure2: Theorem 19 applies to show that
An; 2 C2n 1 for all n 2; and 2 f1; 2g
and
An; 2 R2n 1 for all n 2; and 2 f1; 3g :
1
1
2
1
2
1
1
3
1
3
,1 CG
G
R,1Figure3. Closed tree graph GC;1 and GR;1
with initial symbol 1
Remark 21 (1) In Example, the authors call this system as Simpli…ed Golden Mean shift (SGM), the exact number of entropy still unknown for such system, however, a good numerical approximation can be found in [6] by means of trace and connecting operators.
(2) If one let A2;32= 1 in the SGM system, such system is call hard square
system comes from the statistic mechanism, the exact number of entropy for such system can be found in [] by mean of the knowledge of modular function in number theory.
4
Main Theorem
In this section, the main result of this paper will be presented and we give some de…nitions …rst.
De…nition 22 Let A2 2 Mp2(Z) and write A2 = (b ) in the -indices; we
call A2 is consistent if A2; 6= 0p p whenever b 6= 0:
De…nition 23 Given A 2 Mn(Z) ; we de…ned
t (A; q) =faiq j aiq = 1 for some i2 f1; ; ng n qg
and
i (A; q) =faqi j aqi = 1 for some i2 f1; ; ng n qg :
Given A2, let A1 be the associated indicator matrix in De…nition 5; we de…ne
tq0 = t0(A1; q) = 11(f(i; j) j aij 2 t (A1; q)g) ;
Recall that 1 is the bijection from b coordinate to aij of A1 in Remark 6,
and also de…ne
i0q = 11(f(i; j) j aij 2 i (A1; q)g)
similarly:
The following is the main result of this paper.
Theorem 24 Given A2 2 Mp(Z) and let A1 = (b ) = (aij) satis…es the
following conditions,
(1) A2 is consistent and A1 2 Pp;
(2) There exist an index 2 f1; ; pg with 1( ) = (qq) for some q such that b = 1 and A2; 2 Pp;
(3) For those in condition(2), there exist closed GC; and GR; such that
t0
q A (GC; ) and i0q A (GC; ) ; Then
An 2 Ppn for all n 2:
Proof. It su¢ ces to show that for any n there exist a number k (n) 2 Z such that for all 1 i; j p; Ak(n)n;ij E2n 1 for all n 2: Indeed, since A1 2 Pp;
then there exist two pathes from i to q with length r and from q to j with length s; write i = i0; i1 ; ir 1 = q and q = j0; ; js 1 = j;by Lemma 25
below, there exist k1 such that
Ak1(n)
n; 11(qq) = A k1(n)
Since t0
q A (GC;q) ;and i0 A (GR;q), then by Theorem 19, we have
An; 1
1 (ilil+1) 2 R2n 1 for all 0 l r 1 and for all n 2
and
An; 1
1 (jljl+1)2 C2n 1 for all 0 l s 1 and for all n 2:
Thus
An;ilil+1 2 R2n 1 for all 0 l r 1 and for all n 2
and
An;jljl+1 2 C2n 1 for all 0 l s 1 and for all n 2:
Then we consider the path
i0; ; ir 1 = q; q; ; q = j0
| {z }
k1+1 times
; j1; ; js 1
we can conclude that Ar+(k1+1)+s
n;ij An;i0i1An;i1i2 A
k1
n;ir 1qAn;qj1An;j1j2 An;js 1j E2n 1
The third inequality is by Proposition 8: Note that the power r + (k1+ 1) + s
may depend on the choice of i and j; however, since A1 2 Pp; there exists a
k2 such that
Ak2
1 Ep:
And follow the same argument above, we have A(k1+1)+2k2
n;ij E2n 1:
and k1 and k2 are independent of choice of i and j: This complete the proof.
Lemma 25 Under the same assumption of A2 in Theorem 24; then for any
n; there exists k1 = k1(n) such that
Ak1(n)
n;q E2n 1:
Proof. We prove it by induction, for simplicity, we assume q = 1; then for n = 2; from condition(2) of Theorem 24, b1 = 1, q = 1and A2;1 2 Pp; then
just choose the k1 = k1(2) such that Ak2;11 Ep; then we have Ak2;11 Ep; we
assume for n = m; there exists k1 = k1(m) such that
Ak1(m) m;1 E2m 1: (5) By the formula(3) of Am+1 Am+1;1 = 0 B @ b11Am;1 b1pAm;p .. . . .. ... b1(p2 p+1)Am;(p2 p+1) b1p2Am;p2 1 C A
Since A2 is consistent, A2;1 2 Pp and induction assumption for Am;1(5); the
same argument of proof in Theorem 24 can be applied to show that there exists a k1(m + 1) such that A
k1(m+1)
m+1;1 E2m: This complete the proof.
Example 26 In Example 20, A1 =
1 1
1 0 2 P2; choose = 1 and thus q = 1 such that A2;1 2 P2; therefore (1) and (2) of Theorem 24 hold. For
(3), consider = 1; the GC;1 and GR;1 are developed in Example 20 with
A (GC;1) = f1; 2g and A (GR;1) = f1; 3g : One can easily checked that
t01 = 22 A (GC;1) and i01 = 3 2 A (GR;1) ;
therefore (3) holds, thus Theorem 24 applied to show that An2 P2n;
for all n 2: Since from (1) and (2), we have A2 = B2;
then
Bn 2 P2n 1 for n 2:
Then Theorem 1 applied to show that SGM shift is mixing.
Remark 27 For Golden Mean Shift in Example 3, if one delete the all zero column and row in every An; then the same argument in Example 26 can be
shown that such system is mixing.
Theorem 28 Given A2 and write A1 = (b ) = (aij) satis…ed the (i) and (iii)
of Theorem 24, if there exists a sequence ij; where 0 j k with i0 = ik = q
as in statement (ii) of Theorem 24 and
(ii)1 (A2;i0i1)i0i1(A2;i1i2)i1i2 A2;ik 1ik ik 1ik = 1 where (A)ij is the ij coordinate
of A; and
(ii)2 A2;i0i1A2;i1i2 A2;ik 1ik 2 Pp; then
An 2 Ppn for all n 2:
Proof. It su¢ ce to show that
An;i0i1An;i1i2 An;ik 1ik 2 Ppn 1 for n 2;
then the rest is the same as the proof in Theorem 24. We prove it by induc-tion, for n = 2 is by (ii)00; we assume for n = m
Am;i0i1Am;i1i2 Am;ik 1ik 2 Ppm 1: (6)
Then since Am+1;i0i1Am+1;i1i2 Am+1;ik 1ik equal to
0 B B @ b 1 1 (ai0i1)1Am;1 b 11(ai0i1)pAm;p .. . . .. ... b 1 1 (ai0i1)(p2 p+1)Am;(p2 p+1) b 11(ai0i1)p2Am;p2 1 C C A (7) 0 B B @ b 1 1 (aik 1ik)1Am;1 b 11(aik 1ik)pAm;p .. . . .. ... b 1 1 (aik 1ik)(p2 p+1)Am;(p2 p+1) b 11(aik 1ik)p2Am;p2 1 C C A And also ai0i1ai1i2 aik 1ik = 1;then we always can extract a path from (7),
from consistency of A2; and (6), then the same argument as in Lemma 25
can be shown
Am+1;i0i1Am+1;i1i2 Am+1;ik 1ik 2 Ppm:
Thus induction shows that
An;i0i1An;i1i2 An;ik 1ik 2 Ppn 1 for n 2:
Example 29 Let A2 = 0 B B @ 1 0 0 1 0 1 1 1 1 0 0 1 1 1 1 0 1 C C A :
Then one can choose q = = 1; and there exist GC;1 = GR;1 as in Figure4.
Thus
An; 2 Cpn 1 \ Rpn 1 for n 2 and 2 f1; 2; 3; 4g :
Finally, one can choose A2;12 and A2;21 such that
(A2;12)12(A2;21)21 = 1; and A2;12A2;21= 1 1 2 1 2 P2: Thus An 2 P2n for all n 2:
1
1
1
4
1
4
4
2
3
Figure4. GC;1 = GR;1Acknowledgement 30 The authors would like to thank Prof. S. S. Lin for valuable discussion on construct the selection scheme.
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