Mixing Shifts of Finite Type with Non-Elementary Surjective Dimension Representations

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1 Mixing Shifts of Finite Type with Non-Elementary Surjective Dimension Representations Nicholas Long June 9, 010 Abstract The dimension representation has been a useful tool in studying the mysterious automorphism group of a shift of finite type, the classification of shifts of finite type, and surrounding problems However, there has been little progress in understanding the image of the dimension representation We discuss the importance of understanding the image of the dimension representation and discuss the candidate range for the fundamental case of mixing shifts of finite type We present the first class of examples of mixing shifts of finite type for which the dimension representation is surjective necessarily using non-elementary conjugacies 1 Statement of Results The group of automorphisms of a shift of finite type (X,σ X ), denoted Aut(σ X ), has been a vital tool in attempts to classify shifts of finite type, but in general, Aut(σ X ) is large and mysterious For example, the automorphism group of the (-sided) -shift is countably infinite, is residually finite, is not finitely generated, and contains a copy of many groups including every finite group, the free group on infinitely many generators, but not any group with unsolvable word problem [] The dimension group of a shift of finite type is a more tractable group to study In fact, the dimension group is the direct limit group of the action on Z n by a n by n non-negative integral matrix presentation Let A be a square non-negative integral matrix presenting the shift of finite type (X A,σ A ), and let Aut(Â) denote 1

2 the group of isomorphisms of the dimension group that commute with the action of  For many important cases, both the dimension group and Aut(Â) are easily computed and are finitely presented The dimension representation, ρ A : Aut(σ XA ) Aut(Â), has served as an important probe into the structure of the automorphism group of a shift of finite type Automorphisms in the kernel of the dimension representation are called inert Kim and Roush demonstrated a complete characterization of the actions of inert automorphisms on finite subsystems of shifts of finite type [9, 10] In stark contrast, there has been little progress in describing how non-inert automorphisms can act on finite subsystems An essential (and to a large extent sufficient) part of this description would be simply to know the image of the dimension representation Consequently, the fundamental question we consider is: Question 11 Given A, a primitive matrix, what is the image of the dimension representation, ρ A : Aut(σ XA ) Aut(Â)? Additionally, given a classification of irreducible shifts of finite type, Kim and Roush describe how the classification of (reducible) shifts of finite type (SFTs) can be found if Question 11 is answered [6] Further, insight from answering Question 11 may be relevant to classifying mixing SFTs and would provide a better understanding of Aut(σ XA ) Other surrounding problems are discussed in Boyle s open problems paper [1] Let A be a primitive matrix and Aut + (Â) be the positive automorphisms of the dimension group, ie automorphisms of the dimension group which multiply the Perron eigenvector by a positive constant If A is primitive, then the image of ρ A is in Aut + (Â) Now we regard ρ A as a map from Aut(σ XA ) Aut + (Â) and say that ρ A is surjective if its image is Aut + (Â) Question 1 Under what conditions does ρ A map Aut(σ XA ) onto Aut + (Â)? There are just a few sophisticated examples of SFTs for which the dimension representation is shown to be non-surjective [8, 15] In some easy cases (eg for full shifts) the dimension representation is known to be surjective There is just one general positive result known for showing elements of Aut(Â) lie in the image of the dimension representation Theorem 11 [] Suppose Φ Aut(Â), then for all sufficiently large n, there is a φ Aut(σ XA n) with ρ A (φ) = Φ, and moreover φ is presented as an elementary conjugacy of (X A n,σ XA n), ie φ arises from some Elementary Strong Shift Equivalence (R,S) from A n to A n

3 The first result of this paper, Proposition 34, presents an obstruction to generalizing the Elementary Strong Shift Equivalence result of Theorem 11 to the case n = 1 for a particular type of matrix A (even after considering the conjugacy class of (X A,σ XA )) The main result of this paper (Theorem 41) is the presentation of a non-trivial class of examples in which ρ A is surjective even though the Elementary Strong Shift Equivalence obstruction of Proposition 34 holds While these examples are not general, it is our hope that these examples will lead to some insight for more general constructions We refer the reader to Section for precise definitions In Section 3, we describe Aut + (Â), the candidate range of the dimension representation and compute several relevant examples In Section 4, we give examples of mixing shifts of finite type with surjective dimension representation Section 5 details the construction of the Strong Shift Equivalence used in Section 4, which relies on state splittings and amalgamations as described in Appendix A I am thrilled and honored to be able to contribute to this volume in memory of K H Kim not only because of the effect that his contributions have had on my work, but on mathematics in general Virtually all of the motivation for this paper is given by the referenced work of Kim and Roush, including the reduction of classifying reducible shifts of finite type to understanding the image of the dimension representation and the classification of irreducible shifts of finite type Much of my other work, including the main results of my PhD thesis, have also relied on the theorems and results of Kim and Roush I would also like to thank the referee and editors for their helpful comments Definitions and Motivations Let Z + = {0,1,, } and N = {1,,3, } Let G A be a finite directed graph with n ordered vertices and a finite edge set E A The graph G A is defined by its adjacency matrix, A, which is a n n non-negative integral matrix with A i j = the number of edges in G A from vertex i to vertex j We say that A or G A presents the shift of finite type (X A,σ XA ) defined as follows Let t(e) and i(e) denote the terminal and initial vertices of the edge e E A The set X A is the subset of E Z A given by {x = (x i ) i Z E Z A : t(e i) = i(e i+1 )for all i Z} The shift map, σ XA, slides the symbolic sequence one space to the left, (σ XA (x)) i = (x) i+1 The set X A can be thought of as the set of bi-infinite walks in the graph G A Further, X A is a compact metric space with discrete topology and σ XA is an expansive homeomorphism of X A The SFT (X [n],σ X[n] ) is called the n-shift and is the set of all allowed sequences 3

4 on n symbols (typically {0,1,,n 1}) A block is a finite sequence of edges, b 1 b b n where b 1 E A and a block is allowed if t(b i ) = i(b i+1 ) for all i A SFT is mixing if there exists a N N such that for each pair of allowed blocks, u and v, and for each n N, there is a block w of length n such that uwv is an allowed block A k by k matrix, B, is primitive if its entries are nonnegative integers and there is some n N such that (B n ) i j > 0 for all 1 i, j k If all rows and columns of a square matrix A over Z + are nonzero, then A is primitive iff (X A,σ XA ) is a mixing shift of finite type The class of mixing shifts of finite type (MSFTs) are the fundamental class of SFTs and many problems involving SFTs can be reduced to the case of MSFTs Dynamical systems f : X X and g : Y Y are topologically conjugate if there exists a homeomorphism φ : X Y such that g φ = φ f Let (X A,σ XA ) or simply X A denote the shift of finite type defined by the non-negative integral matrix A For A and B square matrices over Z +, it is natural to ask under what conditions do A and B present topologically conjugate shifts of finite type Any two conjugate SFTs will have the same number of periodic points of each period and the same exponential growth rate of allowed blocks These properties dictate that if A and B present conjugate shifts of finite type, then A and B have the same non-zero spectrum But the non-zero spectrum is not enough to guarantee conjugacy, and in 1973 R Williams gave an algebraic framework with which to study conjugacy classes of shifts of finite type [16] 1 Strong Shift Equivalence Given matrices A and B over a unital semiring S, A is elementary strong shift equivalent (ESSE) to B (over S) if there exist matrices R and S over S with A = RS and B = SR An ESSE, (R,S), has direction from A to B for A = RS and B = SR, whereas the ESSE (S,R) has direction from B to A Because ESSE is not transitive, ESSE in not an equivalence relation For matrices A and B over S, A is strong shift equivalent (SSE) to B over S if there is a sequence of ESSEs (over S) beginning with A and ending with B Since SSE is the transitive closure of ESSE, SSE is an equivalence relation Theorem 1 [16] For A and B square matrices over Z +, (X A,σ XA ) is conjugate to (X B,σ XB ) iff A is SSE to B over Z + An elementary conjugacy is one that arises from an elementary strong shift equivalence By Theorem 1, SSE over Z + is an equivalence relation whose 4

5 equivalence classes correspond to conjugacy classes of shifts of finite type This algebraic characterization of conjugacy does not completely resolve the question of whether arbitrary square non-negative integral matrices present conjugate SFTs because there is no known finite procedure for deciding when two non-negative integral matrices are SSE over Z + Shift Equivalence Williams also defined the more tractable equivalence relation of shift equivalence For matrices A and B over a unital semiring S, A is shift equivalent (SE) to B over S if there exist matrices R and S over S and l N such that RA = BR AS = SB A l = RS B l = SR The integer l is referred to as the lag of the shift equivalence given by (R,S,l) The advantage of using SE rather than SSE is that SE over Z and Z + are well understood For example, matrices over Z are SE (over Z) to a non-singular matrix Further, two integral matrices are SE over Z iff they are SSE over Z Most importantly, SE over Z + is decidable [3, 4] In various important special cases, SE over Z + is classified by well understood invariants For example, all matrices over Z + with the same single non-zero eigenvalue, λ > 0, are SE over Z + It is not known whether they must also be SSE over Z + (Conjecture 31 of [1]) The relation of shift equivalence can be given more concretely, as we present now If A is an n n matrix over Z +, then the eventual range of A, R A, is given by Q n A k, for large enough k such that A is an isomorphism from Q n A k to Q n A k+1 By convention, the action of A is on row vectors The dimension group of A, D A Q n, is defined as D A = {v R A : va k Z n for some k 0} Let D + A be the set of positive vectors in D A, and let  be the automorphism of D A induced by A (Â(v) = va for v D A ) The ordered triple (D A,D + A,Â) is called the dimension module or dimension triple Dimension modules (D A,D + A,Â) and (D B,D + B, ˆB) are isomorphic if there exists an isomorphism, ψ : D A D B that takes the positive set D + A to D+ B and ψ  = ˆB ψ Theorem [11] Let A and B be square matrices over Z +, then A is SE to B over Z iff (D A,Â) = (D B, ˆB), and A is SE to B over Z + iff (D A,D + A,Â) = (D B,D + B, ˆB) 5

6 Clearly if A is SSE over Z + to B, then A is SE over Z + to B, but when does A being SE to B over Z + imply A is SSE to B over Z +? Williams conjectured in 1974 that for matrices over Z +, SE over Z + implies SSE over Z + [16] This conjecture was refuted by Kim and Roush for the reducible case in 199 [5] and for the irreducible and mixing cases in 1999 [7] but there remains much to be understood about the relation of SSE to SE Essential to the counterexamples was a deeper understanding of the dimension representation of the automorphism group of a shift of finite type Standing Convention 1 For the rest of this paper, ESSE, SE, and SSE refer to ESSE over Z +, SE over Z +, and SSE over Z + unless otherwise stated 3 The Dimension Representation An automorphism of a shift space X is a shift commuting homeomorphism of X to itself Let Aut(σ X ) denote the group of automorphisms on a shift space X In general, Aut(σ X ) is complicated and poorly understood Let Aut(Â) be the group of automorphisms of D A that commute with  The group Aut(Â) is more tractable to study and is typically finitely generated If A GL n (Z), then D A = Z n and  = A is the isomorphism given by multiplication by A, so Aut(Â) consists of invertible integral matrices that commute with A By Theorem 1, any φ Aut(σ XA ) can be realized as an automorphism induced by a sequence of ESSEs over Z + from A to A, A (R 1,S 1 ) A 1 (R,S ) A (R 3,S 3 ) (R k,s k ) A If (R,S) is an ESSE from A to B, then R induces an isomorphism from (D A,D + A,Â) to (D B,D + B, ˆB) For an automorphism φ and a corresponding SSE from A to A, (R 1,S 1 )(R,S )(R k,s k ), let ˆφ be the induced automorphism on (D A,D + A,Â), where ˆφ = ( ˆR i ) ε i and ε i is ±1 according to the direction that the i-th ESSE is traversed Since ˆφ does not depend on the choice of SSE representing φ, this gives a well defined map ρ A : Aut(σ XA ) Aut(Â) where ρ A (φ) = ˆφ The map ρ A is called the dimension representation and elements in its kernel are called inert automorphisms 3 The Group Aut + (Â) Note that D A = D A n, Aut(Â) Aut( Aˆ n ), and typically (eg if all eigenvalues of A n are simple roots of the characteristic polynomial of A n ) Aut(Â) = Aut( Aˆ n ) 6

7 Every element ˆφ Aut(Â) is the restriction of a unique invertible real linear transformation φ : R A R R A R The use of ˆφ and φ is an abuse of notation since we do not in general have an associated automorphism of the shift, φ, but we use the hat and tilde notations simply to refer to an element of Aut(Â) or its corresponding linear transformation Assume A is a primitive matrix with spectral radius λ A Let v A be a positive row eigenvector of λ A (a Perron eigenvector of A) Because ˆφ must commute with Â, ˆφ must preserve the eigendirections of A Thus φ(v A ) = αv A, where α depends only on φ We define Aut + (Â) = {ˆφ Aut(Â) : α > 0} It is well known that when A is primitive, ρ A (Aut(σ XA )) Aut + (Â) We say that the dimension representation ρ A is surjective if ρ A (Aut(σ XA )) = Aut + (Â) 31 Examples of Aut(Â) and Aut + (Â) Example 31 Full n-shifts Let A = [n], so X A is the full n-shift The dimension group, D A, is the ring Z[1/n] since Z[1/n] are the elements of Q that will be eventually mapped into Z by multiplication by n The group D + A is Z+ [1/n] and  is the isomorphism of Z[1/n] given by multiplication by n If n = p r 1 1 pr k k with each of the p i distinct primes, then Aut(Â) consists of elements of the form ˆφ(x) = ±p t 1 1 p t k k x for t i Z and Aut + (Â) are the automorphisms of D A of the form ˆφ(x) = p t 1 1 p t k k x for t i Z, Here Aut + (Â) is isomorphic to the finitely generated abelian group Z k Example 3 Invertible Integral Matrices Suppose A is a n n matrix over Z and det(a) = ±1 Then D A = Z n and  = A, since A is invertible over Z The group [ Aut(Â) ] consists of the elements 1 1 of GL(n,Z) that commute with A For A =, we have D 1 0 A = Z, Aut(Â) = {±A m : m Z}, and Aut + (Â) = {A m : m Z} = Z [ ] 8 5 Example 33 The matrix A = 5 8 [ ] 8 5 Let A = The matrix A has eigenvalues 13 and 3 with eigenvectors 5 8 u = [1,1] and v = [1, 1] If ˆφ Aut(Â), then φ sends u to α φ u and v to β φ v, where 7

8 α φ = ±13 n for n Z and β φ = ±3 m for m Z, and the pair (α φ,β φ ) determines ˆφ The group Aut + (Â) consists of the automorphisms ˆφ such that α φ > 0 Clearly for ˆφ Aut + (Â) we have (α φ,β φ ) {(13 n,( 1) l 3 m ) : l,m,n Z} Thus L A : ˆφ (l,m,n) defines [ an ] embedding [ ] of [ the group] Aut + (Â) into Z/Z Z Z The integral matrices,, and commute with A and thus define elements of Aut + (Â) with (α,β) respectively being (13,1), (1, 1), and (1,3) The associated images under L A are respectively (0,0,1), (1,0,0), and (0,1,0) Because the L A images of (0,0,1), (1,0,0), and (0,1,0) generate all of Z/ Z Z, the embedding L A is an isomorphism from Aut + (Â) onto Z/ Z Z ([ There ] [ is ][ an automorphism ]) ψ induced by the ESSE (R, S) = , such that (α ψ,β ψ ) = (1, 1) and L A ( ˆψ) = (1,0,0) As an automorphism, the shift map σ XA corresponds to an ESSE of (A,Id) So, (α σxa,β σxa ) = (13,3) and L A ( σˆ XA ) = (0,1,1) However it is not obvious whether ρ A maps Aut(σ XA ) onto Aut + (Â) Let us consider if it is possible to create a generating set of Aut + (Â) using the image of ESSEs under the dimension representation If (R,S) is an ESSE from A to A, then R (and S) commute with A and thus R (and S) have eigenvectors [1,1] and [1, 1] This means that R (and S) will have fixed column sum of either 13 or 1 and column difference of either [ 1 or ] 3 If R has column sum of 13 and column 5 8 difference of 3, then R = A or Note that these two matrixes are used to 8 5 make the ESSE which induce automorphisms ψ and [ σ XA ] If R has column sum of and column difference of 1, then R = Id or R = These matrices are the 1 0 complementary ESSE matrices for the previous case The only other possibility is that either R or S has column sum of 1 and column difference of 3, which would imply that either R or S contains negative entries, which is a contradiction of the assumption that (R,S) is an ESSE over Z + So (1,0,0), (1,1,1), and (0,1,1) are the only possible coordinates in L A (Aut + (Â)) that can be the image of an ESSE Note that the subgroup generated by (1,0,0), (1,1,1), and (0,1,1) is not all of Z/ Z Z Using our construction from Section 5, Appendix A of [13] gives γ, an automorphism induced by a sequence of 4 ESSEs from A to A with (α γ,β γ ) = (13,1) 8

9 and L(ˆγ) = (0,1,0) The three automorphisms of X A given by ψ, γ, and σ XA will map to a generating set of Aut + (Â) given by their L A coordinates of (1,0,0), (0,1,0), and (0,1,1), and thus ρ A will be surjective The construction of the embedding L A from Aut + (Â) to (Z/Z) n 1 Z m is not particular to the preceding example Let A be a primitive matrix with simple integer eigenvalues λ 1,,λ n where λ 1 has the largest modulus If λ i is divisible by m i primes, then the map L A is an embedding of Aut + (Â) into (Z/Z) n 1 Z m 1 Z m n given by (p i 1 1 p i m 1 m 1,( 1) l q j 1 1 qj m m,,( 1) l n r k 1 1 rk mn m n ) (l,,l n,i 1,,i m1,,k 1,,k mn ) In general, one could hope that Aut + (Â) would be generated by the ρ A images of ESSEs The following proposition shows that under some simple conditions, the induced automorphisms of Aut + (Â) given by ESSEs are either finite order or generated by the shift map Proposition 34 Suppose C = RS = SR with C a primitive matrix such that its eigenvalue of largest modulus is a prime integer p Let φ Aut(σ XA ) be the conjugacy associated to the ESSE (R,S) Then there is a ˆψ Aut(Ĉ) and k Z + such that ˆψ k = Id and either ˆφ ˆψ = Ĉ or ˆφ = ˆψ Proof: The matrices R and S commute with C and thus R and S have the same eigenspaces as C Because C is primitive, then let λ C be the Perron eigenvalue with v C the positive eigenvector of λ C Because λ C is a simple eigenvalue of C, there are constants α, β > 0 such that v C R = αv C, v C S = βv C Now αβ = p because RSv = Av = pv Since p is prime, either α = 1 or β = 1 Suppose β = 1 Because v C > 0 and S i j 0 and v C β = v C S, we have that β is the spectral radius of S by the Spectral Radius Theorem If ˆψ = S, then for some k Z +, ˆψ k = Id since S will have eigenvalues of largest modulus that are k-th roots of unity This would imply that ˆφ ˆψ = ˆRŜ = Ĉ Suppose β = p and α = 1 The same argument as above shows that for ˆψ = R, there is some k Z + such that ˆψ k = Id Recall that the ESSE (A,Id) corresponds to the automorphism of X A given by the shift map In general, the subgroup of Aut + (Â) generated by the ρ A image of ESSEs is at least rank one If A has a single prime eigenvalue (as in Example 31 for n prime), then the image of the ESSE (A,Id) generates Aut + (Â) If A has a prime spectral radius and another simple eigenvalue (as in Example 33), all the automorphisms induced by ESSEs are either finite order or the shift map by 9

10 Proposition 34 Therefore, the image of ESSEs are not enough to generate all of Aut + (Â) because Aut + (Â) has rank at least two and the subgroup generated by the images of ESSEs is only rank one While the conditions[ of Proposition ] 34 are 8 5 not general, the consequence is that for matrices like A =, we necessarily 5 8 to look at more than just the image of ESSE to generate all of Aut + (Â) Lastly, as an obstruction to another proof strategy, we give a cautionary example where the image of Aut + (Â) under the embedding L A constructed above need not be all of the lattice Z/ Z Z This does not preclude A from having a surjective dimension representation, but it shows that one cannot find a general proof which simply realizes automorphisms whose L A images are arbitrary elements of the lattice [ ] 4 1 Example 35 If A =, then L 3 A Z/ Z Z [ ] 4 1 Let A = The matrix A has eigenvalues of 5 and with eigenvectors 3 u = [,1] and v = [1, 1] In order to compute the image of L A, we need to examine matrices that commute with A and have non-zero spectrum 5 p 1 and ± p with corresponding eigenvectors u and v The[ unique ] matrix that has eigenvalues 5 and 1 with eigenvectors u and v is C = The matrix C corresponds to 8 7 the elementary vector (0,0,1) Z/ Z Z, but C Aut(Â), because for all n N, [1,0]CA n Z Therefore (0,0,1) L A (Aut + (Â)) In fact, (0,n,m) L A (Aut + (Â)) if n + m is odd 4 Examples of Surjective Dimension Representations Example 41 For n N, the dimension representation of the full n-shift is surjective Let A = [n], so X A is the full n-shift The ring D A is Z[1/n], D + A is Z+ [1/n], and  is the isomorphism of Z[1/n] given by multiplication by n If n = p r 1 1 pr k k for primes p 1,, p k, then Aut(Â) consists of elements of the form ˆφ(x) = ±p t 1 1 p t k k x for t i Z and Aut + (Â) = {A m : m Z} Clearly, Aut(Â) = Z/ Z k and Aut + (Â) = Z k Consider γ i, the automorphism induced by the ESSE from A to A ([p i ],[n/p i ]) So ρ A (γ i ) = [p i ] and L A ( ˆγ i ) = e i, where e i is the i-th elementary row vector Thus 10

11 γ 1,,γ k get mapped by ρ A to a generating set of Aut + (Â), and the dimension representation of A is surjective Example 4 Let B = na where A is a primitive symmetric matrix with eigenvalues n and 1, both of multiplicity 1 If n is prime, then the dimension representation of B is surjective If A has integer eigenvectors u and v for eigenvalues n and 1, then B has eigenvectors u and v for eigenvalues n and n (by Perron-Frobenius theory, we assume u is positive) The automorphism ˆB is multiplication by B on D B The group Aut( ˆB) consists of matrices over Q that are automorphisms of D A and commute with B (thus must have the same eigenspaces) So Aut + ( ˆB) will consist of the matrices that have eigenvalue n j on u and ±n k on v for j,k Z We will show that L B will map Aut + ( ˆB) isomorphically onto Z/ Z Z by giving elements of Aut + ( ˆB) whose images under L B generate all of Z/ Z Z Since A is symmetric, B will be symmetric Let ψ be the automorphism induced by the ESSE [D,BD] from B to B where D is the permutation matrix such that conjugation by D gives the transpose of a matrix In this case, (α ψ,β ψ ) = (1, 1) and L B ( ˆψ) = (1,0,0) Also note that ρ B (σ XB ) = ˆB and L B ( ˆB) = (0,,1) If γ is the ESSE from B to B given by (A,n Id), then L B (ˆγ) = (0,1,0) Since (1,0,0), (0,,1), and (0,1,0) will generate all of Z/ Z Z = Aut + ( ˆB) and each of the generators of Aut + ( ˆB) is a ρ B image of an ESSE, the dimension representation of B is surjective Alternatively, it is possible to view X B as a product shift of X [n] X A A point in X B is a point in the full n-shift cross a point in X A and γ corresponds to the automorphism of the product shift given by id [n] σ XA Theorem 41 Let n] and k be prime odd integers such that n > 1 and 0 < k < n For A = [ n+k n k n k n+k, the dimension representation of A is surjective Note that the previous example shows for the case n = k, A will have surjective dimension representation Also note that because of Proposition 34, we will necessarily need to look at more than the image of ESSEs Proof: The matrix A has simple spectrum of n and k, with eigenvectors of u = [1,1] and v = [1, 1] respectively For ˆφ Aut + (Â) with (α φ,β φ ) = (n t,( 1) l k s )), L A maps ˆφ to (l,s,t) Z/ Z Z Further, L A will map Aut + (Â) onto Z/ 11

12 Z Z because [ n t +( 1) l k s n t ( 1) l k s n t ( 1) l k s n t +( 1) l k s ] will be an integral matrix that commutes with A for any (l,s,t) Z/ Z Z Since [ A] is symmetric, there exists ψ, an ESSE [D,AD] from A to A with 0 1 D =, and L 1 0 B ( ˆψ) = (1,0,0) Another generator of Aut + (Â) is given by ρ A (σ XA ) = Â with L A (Â) = (0,1,1) In the construction of Section 5, we produce a SSE with induced automorphism γ, such that (α γ,β γ ) = (n,1) and ρ A composed with L A maps γ to the (0,1,0) element of Z/ Z = Aut + (Â) The three automorphisms of X A given by ψ, γ, and σ XA will map to a generating set of Aut + (Â) given by their L A coordinates of (1,0,0), (0,1,0), and (0,1,1), and thus ρ A will be surjective 5 Construction for Proposition 41 The automorphism γ is induced by a sequence of ESSEs A (D 1,S 1 ) (D,S ) A 1 (D 3,S 3 ) (D 4,S 4 ) A A 3 A where D 1 and D are subdivision matrices for row splittings and D 3 and D 4 are amalgamation matrices for row amalgamations Further, we will show that [1,1]D 1 D D 3 D 4 = [n,n] and [1, 1]D 1 D D 3 D 4 = [1, 1], which implies (α γ,β γ ) = (n,1) and ρ A composed with L A maps γ to the (0,1,0) element of Z/ Z = Aut + (Â) The following construction is based on state splitting and amalgamations, described in Appendix A We will now briefly describe the general procedure for the row splittings (D 1,S 1 ) and (D,S ), and the row amalgamations (D 3,S 3 ), and (D 4,S 4 ) The splitting (D 1,S 1 ): The ESSE (D 1,S 1 ) is a row splitting of the two rows of A The first row, k+1 ] is split into rows of the form [k,0], k 1 rows of the form [0,k], [ n+k, n k and one row of the form [ n k, n k ] This is a valid splitting because n > k and k+1 k 1 n k [k,0] + [0,k] + [, n k ] = [ n+k, n k n k ] The second row, [, n+k ], is split into k 1 rows of the form [k,0], k+1 rows of the form [0,k], and one row of the form [ n k, n k ] This is a valid splitting because n > k and k 1 [k,0] + k+1 n k [0,k] + [, n k ] = [ n k, n+k ] The matrix S 1 will have columns and k + rows because both rows of A 1

13 are split k + 1 times S 1 = n k k 0 k 0 0 k 0 k k 0 k 0 0 k k+1 rows k 1 rows n k 1 row k 1 rows k+1 rows 0 k n k n k 1 row Let A 1 = S 1 D 1 The matrix A 1 is k + 1 copies of the first column of S 1 and k + 1 copies of the second column of S 1 because the first row of A was split k + 1 times and the second row of A was split k + 1 times 13

14 # of cols k + 1 k + 1 k k 0 0 k k k k A 1 = 0 0 k k n k n k k k 0 0 n k k+1 rows k 1 rows n k 1 row {}}{{}}{ k k k k 0 0 k k n k n k n k n k 1 row k 1 rows k+1 rows The splitting (D,S ): The matrix A 1 has 3 different rows, [k k 0 0], [0 0 k k], and [ n k n k ] Each of the k rows of A 1 with the form # of cols = {}} k + 1 { k {}} + 1 { [k k 0 0] should be split into k rows of the form # of cols = k k + 1 {}}{{}}{ [k ] [0 k ] [0 0 k 1 0 0] For 1 i k, we will call the i-th row above a type (1,i) row 14

15 Each of the k rows of A 1 with the form should be split into k rows # of cols = {}} k + 1 { k {}} + 1 { [0 0 k k] # of cols = {}} k + 1 { k {}}{ [0 0 k 0 0 1] [0 0 0 k 0 1] [ k 1] For 1 i k, we will call the i-th row above a type (,i) The two rows of A 1 of the form [ n k n k n k ] should be split into pairs of rows with each pair summing to [1 1] and such that the first row of the pair has ones in the first k+1 entries and from the k + 1 entry to the 3k+1 entry, and zeros otherwise This pair is chosen such that the transpose will match the resulting columns that show up in A Each pair of rows will look like k+1 k 1 1 k 1 k+1 1 {}}{{}}{{}}{{}}{ We will refer to this pair of rows as complementary rows So R will have the form of (k + 1)/ blocks of type 1 rows (k 1)/ blocks of type rows (n k )/ pairs of complementary rows (k 1)/ blocks of type 1 rows (k + 1)/ blocks of type rows (n k )/ pairs of complementary rows The matrix A will have k copies of the first (k + 1)/ columns of R because the first (k + 1)/ rows of A 1 are split k times Then A will have k copies of the 15

16 (k + 1)/ + 1 to (k + 1)/ + (k 1)/ columns of R because the (k + 1)/ + 1 to (k + 1)/ + (k 1)/ rows of A 1 are split k times, and so on # of cols = k n k k n k DK k(k + 1)/ rows DK DK 1 k(k 1)/ rows 0 0 DK 1 A = P P P P n k rows DK k(k 1)/ rows DK DK 1 k(k + 1)/ rows 0 0 DK 1 P P P P n k rows where 0 and 1 represent matrices filled with zeros and ones respectively, DK is the k by k matrix # of cols = k k k {}}{{}}{{}} { k k DK = 0 0 k k 0 0 and the rows of P are given by pairs of the form k k k(k+1) k(k 1) (n k k(k 1) k(k+1) ) (n k )

17 The amalgamation (D 3,S 3 ): The matrix A will have 3 types of rows patterns corresponding to the 3 different types of splittings of rows in the state splitting (D,S ) The amalgamation matrix D 3 will be determined by the total 1-step row amalgamation of A So the matrix S 3 consists of the k + distinct rows of A Specifically S 3 = # of cols k k n k k k n k k k k k k k S 3 = k k The matrix A 3 can be computed from S 3 as follows: for 1 i k, the i-th column of A 3 is the sum of the i + jk columns of S 3 for 0 j k+1 1 and the n + i + jk columns of S 3 for 0 j k 1 1 For 1 i k, the (k + i)-th column of A 3 is the sum of the k(k+1) + i + jk columns of S 3 for 0 j k 1 k(k 1) 1 and the n + +i+ jk columns of S 3 for 0 j k+1 1 The k + 1 column of A 3 will be the sum of the k + 1 to n columns of S 3 The k + column of A 3 will be the sum of the n + k + 1 to n columns of S 3 17

18 # of cols = k k 1 1 {}}{{}}{ k k 0 0 n k n k k k 0 0 n k 0 0 k k n k n k n k k+1 rows 0 0 k k n k A 3 = k k 0 0 n k n k n k k 1 rows k k 0 0 n k 0 0 k k n k n k n k k 1 rows 0 0 k k n k n k k+1 rows k k 0 0 n k n k 0 0 k n k k n k The amalgamation (D 4,S 4 ): As shown above, A 3 will have only different row patterns, n k [k k 0 0 n k ] and [0 0 k k n k The matrix A 4 is the total 1-step row amalgamation of A 3 So, n k ] 18

19 # of cols = k k 1 1 S 3 = k k 0 0 n k n k 0 0 k n k k n k We can compute A 4 with the following rules: The first column of A 4 will be the sum of columns 1 to (k + 1)/, k + 1 to k + (k 1)/, and the k + 1 column of S 3 The second column of A 4 is the sum of columns (k +1)/ +1 to k, k +(k 1)/ + 1 to k, and the k + column of S 3 The matrix A 4 is therefore [ n+k n k n k n+k All that remains to prove Theorem 41 is to show [1,1]D 1 D D 3 D 4 = [n,n] and [1, 1]D 1 D D 3 D 4 = [1, 1] The matrices D 1 and D will copy columns according to how the rows of A and A 1 are split and the matrices D 3 and D 4 will sum columns according to how the rows of A and A 3 are amalgamated Because the first n rows of A are split from the first row of A and the second n rows of A are split from the second row of A, n cols ] = A n cols [1,1]D 1 D = [ ] Because there are k copies of the first k rows of S 3 in A and n k copies of the each of the last two rows of S 3 in A, [1,1]D 1 D D 3 = [1 1]D 3 = [k k k k n k n k ] The first to (k + 1)/, k + 1 to k + (k 1)/, and k + 1 rows of A 3 are the same as the first row of S 4, so D 4 will sum these columns and k(k + 1)/ + k(k 1)/ + n k = n 19

20 The (k + 1)/ + 1 to k, k + (k 1)/ + 1 to k, and k + rows of A 3 are the same as the second row of S 4, so D 4 will sum these columns and k(k + 1)/ + k(k 1)/ + n k = n [1,1]D 1 D D 3 D 4 = [k k n k n k ] D 4 = [n,n] Note now that the first n rows of A are split from the first row of A and the second n rows of A are split from the second row of A, # of cols n n [1, 1]D 1 D = [ ] Let (S 3 ) i be the i-th row of the matrix S 3 The i-th coordinate of [1, 1]D 1 D D 3 = [ ]D 3 is the difference between the number of the first n rows of A that equal (S 3 ) i and the number of the second n rows of A that are equal to (S 3 ) i There are k+1 copies of (S 3 ) 1 in the first n rows of A 3 and k 1 copies of (S 3 ) 1 in the second n rows of A 3, which means that the first coordinate of [1, 1]D 1 D D 3 is 1 The same argument applies to the first k coordinates of [1, 1]D 1 D D 3 For k + 1 i k, there are k 1 copies of (S 3 ) i in the first n rows of A 3 and k+1 copies of (S 3 ) i in the second n rows of A 3, so the i-th coordinate of [1, 1]D 1 D D 3 is -1 For i = k + 1,k +, there are n k copies of (S 3 ) i in the first n rows of A 3 and n k copies of (S 3 ) i in the second n rows of A 3, so the i-th coordinate of [1, 1]D 1 D D 3 is 0 Therefore, # of cols k k [1, 1]D 1 D D 3 = [ ] In order to compute 0

21 [1, 1]D 1 D D 3 D 4 = [ ]D 4 Note that (k + 1)/ of the first k rows and (k 1)/ of the second k rows of A 3 are equal to the first row of S 4, and (k 1)/ of the first k rows and (k + 1)/ of the second k rows of A 3 are equal to the second row of S 4 This means that [1, 1]D 1 D D 3 D 4 = [1, 1] This completes the proof of Theorem 41 A State splittings State splitting is a fundamental type of ESSE between matrices over Z + Any SSE between shifts of finite type can be decomposed into state splittings and the inverse operations of state amalgamations Theorem A1 (Theorem 71) [1] Let φ be a conjugacy from X A to X B Then φ is a composition of conjugacies given by ESSEs from splittings and amalgamations Furthermore, it is possible to decompose an automorphism of X A, φ, into the composition of k conjugacies arising from row splittings and k conjugacies arising from row amalgamations [14] In particular, state splittings are used to generate the SSEs used in Proposition 41 Let A be a n n matrix over Z + A row splitting of A is given by some splitting of the rows of A, ie the i-th row of A, a i, is split into k i rows over Z +, b 1,,b ki, such that k i j=1 b j = a i Let k = n i=1 k i The row splitting matrix of A is the k by n matrix, R, of the split rows of A, ie that the first k 1 rows of R are the rows split from a 1, the k 1 +1 to k 1 +k rows of R are the split rows of a, and so on The split matrix, B, is created by taking R and copying the i-th column of R k i times Let S be the n k matrix such that S i j = 1 if the j-th row of R is split from the i-th row of A and S i j = 0 otherwise The matrix S is a so called subdivision matrix in which every row has exactly one entry equal to 1 and every column has at least one entry equal to 1 So A = RS since S will sum the columns of R that are split from the same column of A Also B = SR since S will copy the the rows of R according to how the columns of R were split from the columns of A Thus A = SR, B = RS and (R,S) is an ESSE from B to A The matrix A is called a column amalgamation of B if B can be made from a finite sequence of row splittings of A 1

22 [ ] 3 1 Example A1 Let A = and let the first row, [3,1], be split into [1,1] and 4 [,0] and the second row, [,4] be split into [1,1], [1,], and [0,1] [ ] Then R = and S = 0 0 0, so B = There is an analogous procedure for the [ ] column[ splitting ] [ of ] a matrix A For example, if we split the first column of A,, into and, and the second [ ] [ ] [ ] [ ] column,, into,, and, then [ ] R =, S = , and B = We say that B is a column splitting of A A matrix B is a row amalgamation of A if B can be obtained by a finite sequence of column splittings of A A matrix B is a 1-step splitting of a matrix A if B can be obtained as a single splitting of A, ie if A and B are ESSE by some (R,S), given by a splitting The matrix R is called the row/column splitting matrix (or the column/row matrix of an amalgamation) for the row/column splitting of A to B The matrix S is called the subdivision matrix for the splitting of A to B (or the amalgamation matrix for the amalgamation of B to A) The total 1-step row amalgamation of A is defined as follows If A is n by n and A has k( n) distinct rows, then let R be the k by n matrix made up of the distinct rows of A The splitting matrix R is unique up to some permutation of its rows For a fixed choice of the rows of R, S is given by a unique subdivision matrix such that A = SR If B = RS, then B is called the total 1-step row amalgamation of A and is uniquely determined by A up to conjugation by a permutation matrix The total 1-step column amalgamation is defined similarly The total row/column amalgamation of a matrix A is the matrix arrived at by performing total 1-step row/column amalgamations until every row/column is distinct

23 1 0 1 [ ] Example A Let C = , B =, and A = [] The total 1-step row amalgamation of C is B and the total row amalgamation of C is A References [1] Boyle M, Open problems in Symbolic Dynamics Geometric and probabilistic structures in Dynamics Contemp Math, 469, Amer Math Soc, Providence, RI, (008), [] Boyle, M, Lind, D, and Rudolph, D, The automorphism group of a shift of finite type Trans Amer Math Soc 306, no 1 (1988), [3] Kim, K H and Roush, F W, Some results on the decidability of Shift Equivalence J Comb Inform System Dci 4, (1979), [4] Kim, K H and Roush, F W, Decidablilty of shift equivalence, Dynamical Systems, Lecture Notes in Mathematics vol 134, Springer Verlag, Heidelberg, 1988 [5] Kim, K H and Roush, F W, Williams Conjecture is false for reducible subshifts J Amer Math Soc 5 (199), [6] Kim, K H and Roush, F W, Topological classification of reducible subshifts Pure Math Appl Ser B 3, no -4, (1993), [7] Kim, K H and Roush, F W, The Williams conjecture is false for irreducible subshifts Annuls of Mathematics () 149, (1999), [8] Kim, KH, Roush, FW and Wagoner, JB, Automorphisms of the dimension group and gyration numbers JAMS 5 (199), [9] Kim, KH, Roush, FW and Wagoner, JB, Characterization of inert actions on periodic points I Forum Math 1 (000),

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