Symbolic Dynamics on Free Groups

We study nearest-neighbor shifts of finite type (NNSOFT) on a free group G. We determine when a NNSOFT on G admits a periodic coloring and give an example of a NNSOFT that does not allow a periodic coloring. Then, we find an expression for the entropy of the golden mean shift on G. In doing so, we study a new generalization of Fibonacci numbers and analyze their asymptotics with a one-dimensional iterated map that is related to generalized continued fractions. 1. Introduction. Let G be a free group that is freely generated by σ = {σ 1 ,. .. , σ q } and fix a finite alphabet of colors A. A coloring of G is a function φ : G → A. We will consider colorings to be points in the dynamical system (A G , G), where the action is given by (wφ)(x) = φ(xw)


1.
Introduction.Let G be a free group that is freely generated by σ = {σ 1 , . . ., σ q } and fix a finite alphabet of colors A. A coloring of G is a function φ : G → A. We will consider colorings to be points in the dynamical system (A G , G), where the action is given by (wφ)(x) = φ(xw) for w, x ∈ G. Thus, wφ is a coloring that corresponds to "shifting" the coloring φ by the group element w.A coloring φ is periodic if the orbit {T φ : T ∈ G} is finite.A set F ⊂ A × σ × A is called a set of forbidden blocks.A coloring φ contains (a, σ i , b) ∈ F if there exists w ∈ G such that φ(w) = a and φ(wσ i ) = b.Thus F only restricts what colors can be adjacent over each of the σ i .Given a set F of forbidden blocks, define X F to be the set of colorings which do not contain any blocks in F .Since the cardinality of F is finite, we call X F a nearest-neighbor shift of finite type (NNSOFT).
We denote the length of a group element w as a reduced word on the generators of G by |w|.
Definition 1.1.The entropy h(X F ) of a NNSOFT X F is defined as where C n = {w ∈ G : |w| ≤ n} and B n is the number of allowed colorings of C n in X F .
Here, we first study NNSOFT on G and show that their behavior is considerably simpler than Z d for d > 1.We show how to determine if a NNSOFT on G is nonempty (Proposition 1).This is in contrast to, for example, Z×Z where nonemptyness is not necessarily computable for NNSOFT [1].We also determine when a NNSOFT on G admits a periodic coloring (Theorem 3.4), which can be reduced to finding a nontrivial nonnegative solution to a system of linear equations (Theorem 3.6).As in Z d for d > 1, there exist NNSOFT on G that do not admit periodic colorings [1,6], and an example is given in Section 4. In Z × Z, the existence of NNSOFT that do not admit periodic coloring is closely related to the undecidability of Z × Z nonemptyness.On G, we find no such relationship since there exist NNSOFT that do not allow periodic points, but nonemptyness is relatively easy to determine.
We also study the entropy of the golden mean shift on G.In one dimension (q = 1), the entropy of the golden mean shift is easy to determine [8].However, in Z × Z even determining if the entropy of the golden mean shift is algebraic is still an open problem.Much effort has been put towards achieving bounds on its value, which is important for understanding several physical systems and for many coding problems that arise in the study of two-dimensional run-length constrained channels [7,9,11].The golden mean shifts on Z d and G are instances of the hard core model from statistical mechanics, which corresponds to colorings of an arbitrary graph with the alphabet A = {0, 1} such that no two 1s are adjacent.Here we develop a new generalization of Fibonacci numbers to study the golden mean shift on G and find an expression for the entropy of the golden mean shift on G (Theorem 5.9).The results also hold for any k-regular trees, and thus extends to the hard core model on the Bethe lattice [2].The unique simple invariant Gibbs measure on the Bethe lattice has been studied previously [10,3].
2. Coloring G. Fix a set of forbidden blocks F .
∈ F } be the directed graph of allowed blocks of length 2 in the σ i direction.
Every finite set of forbidden blocks F defines q one-dimensional shifts of finite type: Definition 2.2.Let X σ i be the one-dimensional NNSOFT whose elements are given by bi-infinite walks on Γ σ i .
Informally, X σ i describes the space of possible colorings in the σ i direction at each element of G.It should be clear that the set of q one-dimensional shifts of finite type Then X σ1 is the golden mean shift and X σ 2 is the full shift on {0, 1}.Thus if φ ∈ X F and φ(w) = 1, then φ(σ 1 w) = 0, but it is not forbidden to have φ(σ 2 w) = 1.Definition 2.3.For any subset A ∈ A, a set of 2q functions,

Proposition 1. X F is nonempty if and only if for some A ⊆ A, a set of coloring functions exist that obey the forbidden blocks, meaning that we always have
Proof.It should be clear that if X F is nonempty, then there must exist a set of coloring functions that obey the forbidden blocks since each element of A can be adjacent to some other other element of A by each σ i , where A is the set of colors used in some φ ∈ X F .For the converse direction, note that coloring functions can specify colorings in the following way.We begin by defining φ(e) = a for some a ∈ A .We then repeat the following process: if φ(w) has been defined but φ(wx) has not yet been defined for x ∈ σ or x −1 ∈ σ, define φ(wx) = h x (φ(w)).Therefore, once we have colored one vertex, the h x inductively define the rest of the coloring.
Note that since the domain and range of coloring functions are finite, one can trivially check if a NNSOFT is nonempty by simply enumerating all possible sets of coloring functions and determining if any are consistent with the forbidden blocks.It should be clear that if there exists an alphabet A ⊆ A such that for each c ∈ A every X σi contains a point that uses c and only the colors in A , then it is possible to construct a set of coloring functions and the NNSOFT will be nonempty.
In the proof of Theorem 3.4, we require the following result which states that the h x are invertible, then the coloring specified by them will be periodic: Proposition 2. Let φ be a coloring produced by a set of coloring functions S. If Proof.Note that a coloring φ specified by S is color-isotropic, meaning that if φ(g 1 ) = φ(g 2 ) then φ(g 1 w) = φ(g 2 w) for all w ∈ G, as is evident from the fact that each h x is a bijection.But every color-isotropic coloring is necessarily periodic, since any shift T φ of φ will color e one of a finite number of colors, and each color of e will uniquely determine a coloring of G.
3. Periodic coloring of G.We now determine when a NNSOFT on G admits a periodic coloring.This question turns out to be much more subtle than it may initially seem.First, we represent a periodic point in a one-dimensional NNSOFT with a cycle: Definition 3.1.A cycle w is a closed path on the graph Γ σi .We represent a cycle through the vertices x 1 , x 2 , . . .x n , x 1 with x i ∈ A by the expression x 1 x 2 . . .x n .
A cycle w = x 1 x 2 . . .x n has length |w| = n and we will sometimes write (w Thus, a simple cycle is a closed path on Γ σi that does not contain any shorter closed paths.Definition 3.3.For a cycle w and a ∈ A, define If S is a set of cycles, define η a (S) = w∈S η a (w).
Theorem 3.4.A NNSOFT on G on q generators contains a periodic coloring if and only if there exist finite sets S 1 , S 2 , . . ., S q such that elements of S i are cycles that represent periodic points in X σi , and for all a ∈ A and 1 ≤ i < j ≤ q we have η a (S i ) = η a (S j ).
Proof.We first prove the ⇐ direction.Suppose such S i exist, and suppose . Each element x i j,s in B can be thought of as specifying a cycle w i j in S i , and a position s in that cycle.Next, we define χ : B → A by χ(x i j,s ) = (w i j ) s so that ).That is, χ maps the symbol x i j,s to the color at the location (in a cycle) specified by x i j,s .Since we assume η a (S i ) = η a (S j ) for all i, j and for all a ∈ A, we define η a to be the common value of all η a (S i ).Define N = a∈A η a and note we also have N = w∈S i |w| for every S i .Lemma 3.5.Given S 1 , S 2 , . . ., S q that satisfy the hypotheses of the theorem, there exist sets E 1 , E 2 , . . .E N that partition B and meet the following conditions: ). (ii) For i = 1, 2, . . ., q, each E r contains exactly one element of the form x i j,s (where j and s may depend on i).
Proof.In this proof we temporarily change notation.Suppose A = {1, 2, 3, . . ., m} and some values of the p r , where b i x = x i j,s for some j, s.For each fixed i ≤ q, we can "sort" the b i x by color so that , for appropriate choices of the y i z .The conditions on the S i assure that for all a ∈ A and 1 ≤ i, j ≤ q.Thus, when we sort by color, each row of b i x has the same number of elements (η a ) that χ maps to a ∈ A. That is, we can write for appropriate choices of y i a .To get the E r , simply read down the column of this array.That is, let Then the fact that the rows are sorted by color and each row contains the same number of each color implies (i).The fact that each E y contains exactly one element from each row implies (ii).
Continuing the proof of the main theorem, we can apply Lemma 3.5 to choose E 1 , E 2 , . . .E N that satisfy (i) and (ii).Note that (i) assures that elements of a given E r all specify locations in cycles that are colored the same color, while (ii) assures that colors from cycles in each S i are represented exactly once in each E r , so that |E r | = |σ| = q for all r.
We will now use the E r to construct a coloring of G with the alphabet {1, 2, . . .N }.Later, this will be projected down to a periodic coloring by A. Given an r ∈ {1, 2 . . ., N } and i with 1 ≤ i ≤ q, find the unique j, s ∈ Z such that x i j,s ∈ E r .Define h σ i (r) = r , where r is such that That is, given r, we find the element of the form x σ i j,s ∈ E r , and set h σ i (x σ i j,s ) = r such that E r contains the symbol adjacent to x σ i j,s in the cycle w σ i j .Each h σ i is invertible, so we can define h σ −1 i = h −1 σ i .Note that each h σ i is necessarily a bijection, and therefore the h σ i define a periodic coloring by Proposition 2. Suppose this periodic coloring is Φ : G → {1, 2, . . ., N }.We can project Φ down to a coloring of G by using χ.Define φ : G → A by φ(w) = χ(x), where x is any arbitrary element of E Φ(w) .Note the specific choice of x does not matter because of condition (i) above.The orbit {T Φ : T ∈ G} is finite since Φ is periodic, and χ maps each of these colorings to at most one distinct coloring, so the orbit must be finite.This shows that φ is periodic.
To prove the ⇒ direction, we show that any periodic coloring defines a set of cycles that meet the conditions of the theorem.Let φ be a periodic coloring that satisfies the NNSOFT constraints.We can consider the stabilizer of φ, H = {T ∈ G : T φ = φ}.H is a finite index subgroup since φ is periodic, and can be thought of as a fundamental region of the periodic coloring.
H is also a normal subgroup, so consider the group of right cosets G \ H. Since H is a stabilizer of φ, hφ(r) = φ(r) for all h ∈ H and r ∈ G. Thus, elements of the coset (Ha) are all colored the color φ(a), but these colors may not be distinct for different cosets.Now fix σ i ∈ σ, and consider any coset (Hs).We define the coset orbit Since G \ H is finite, each coset orbit O σ i (Hs) must be finite.Note that two coset orbits O σ i (Hs 1 ) and O σ i (Hs 2 ) are either equal or disjoint.Therefore, there are some finite number of distinct coset orbits, which for each σ i ∈ σ form a disjoint partition of G \ H.We can regard each coset orbit O σi (Hs j ) as specifying a cycle C σi (Hs j ) given by where n is the cardinality of O σ i (Hs j ).Since H is the stabilizer of a coloring in X F , each C σ i (Hs j ) represents a point in X σ i .
We then define the set where m i is the number of distinct coset orbits.Note that is independent of σ i , which shows that η a (S i ) = η a (S j ) for all σ i , σ j ∈ σ.This shows that the S i meet the hypotheses of the theorem.

Corollary 1.
If there is a one-dimensional periodic point x which is an element of X σ i for all σ i ∈ σ, then X F contains a periodic coloring.
Proof.The hypotheses of the theorem are satisfied by choosing S i = {y} for all i ≤ q, where y is some cycle representing x.
Theorem 3.6.The existence of sets S 1 , S 2 , . . ., S q satisfying the conditions of Theorem 3.4 is equivalent to the existence of a nontrivial nonnegative solution to a system of linear equations.
Proof.Note that there are a finite number of simple cycles in each Γ σ i and let c i,1 , c i,2 , . . ., c i,l i be the set of simple cycles in Γ σ i .The fact that the c i,r are simple implies that η a (c i,r ) ∈ {0, 1} for every a ∈ A. Define x i,r to be the number of times that c i,r appears in all of the cycles of S i .Then we have Thus, the requirement of Theorem 3.4 that η a (S i ) = η a (S j ) for 1 ≤ i < j ≤ q gives q(q − 1)/2 homogeneous equations in the l i unknowns, x i,r , for each a ∈ A. Since we must solve the system for all a ∈ A, determining if a periodic coloring is allowed corresponds to solving |A|q(q − 1)/2 equations in l 1 + l 2 + . . .+ l q unknowns.The entire system is given by a matrix with rational entries, so the existence of a nonnegative solution implies the existence of a nonnegative integer solution.
4. An example that is not periodic.Now we present an example of a NNSOFT on G where each one-dimensional NNSOFT is irreducible, but G still cannot be colored periodically.This is perhaps a counterintuitive result because irreducibility is a strong condition and implies the existence of many kinds of periodic points.

Cycles representing points in X σ2 are always of the form
where each a j ∈ {G, B}.This shows that for any set S 1 of cycles in X σ1 , we must have In addition, for any set S 2 of cycles in X σ 2 we must have If η a (S σ 1 ) = η a (S σ 2 ) = η a for all a ∈ A, then (1) and (2) become and Clearly we cannot simultaneously solve (3) and ( 4) unless η G = 0 or η B = 0, which would imply that S 1 is empty, since every point in X σ 1 uses G and B. Therefore, even though both X σ 1 and X σ 2 are irreducible, X F does not allow a periodic coloring.

5.
Entropy of the golden mean shift on G. Let X F denote the golden mean shift on the free group G on q free generators, defined in Example 1 of the introduction.
For notational simplicity we define k = 2q − 1.However, the results of this section are general enough to apply to any integer k > 1.
We first analyze finite blocks in X F by using a combinatorial argument.Fix any σ i ∈ σ and define the set T n ⊂ G by T n is called a tree and can be represented by a subset of the Cayley graph of G.Such a graph is shown in Figure 2. In the graph of T n , we label the vertex corresponding to e by α and the vertex corresponding to σ i by β.We call α the "root" of T n and we say that T n has height n.Definition 5.1.Let c 0 (n) and c 1 (n) be the numbers of possible colorings of T n when α is colored 0 and 1, respectively.Theorem 5.2.c 0 (n) satisfies the following recursion relationship: Proof.It is easy to see that c 0 (1) = 2 and c 0 (2) = 2 k + 1.
When α is colored 1, then β must be colored 0. But we can regard β as the "root" of k different trees each of height n − 1.The colorings of each of the k trees for which β is the "root" can be chosen independently.Thus, Similarly, when α is colored 0, β can be colored either 0 or 1, so from which the result follows.
Define C n = {w ∈ G : |w| ≤ n} and denote the number of colorings of C n allowed in X F by B n .
Proof.Note that C n consists of k + 1 trees of height n that all share e as their root.
When e is colored 0, there are c 0 (n) possible colorings of each of the k + 1 trees of height n, each of which can be chosen independently, giving a total of [c 0 (n)] k+1 possible colorings.When e is colored 1, each of the k + 1 trees has a root colored 1, and each tree can be colored independently.Therefore, there are [c possible colorings of C n when e is colored 1.
For notational simplicity, define a n = c 0 (n) so that we have the recursion formula with a 1 = 2 and a 2 = 2 k +1.When k = 1, so that the free group under consideration is Z, the a n are Fibonacci numbers with the standard recursion formula.When k = 1, the nonlinear recursion sequence a n will be central for understanding the golden mean shift on the free group.We begin by studying the general growth properties of (a n ).Define q n = a n /a k n−1 .We first determine when lim exists.We will show that the limit exists for sufficiently small k and equals the solution of x k+1 = x k + 1 in the interval I = [1,2].When k is sufficiently large, the limit may not exist but the values of q n oscillate between two limits.In this case, the odd terms and even terms respectively converge to different limits.
Proof.For the first part, q 2 < q 3 only if a k+1 . This is the same as requiring that which is the same as ) k , which is obviously true.
Proposition 4. For each n = 2, 3, . . .we have Proof.We will prove only (i), as the proof for (ii) is analogous.Note that q n+1 < q n−1 if and only if . This is the same as q n+2 > q n .Let E n = q 2n and O n = q 2n+1 for n = 1, 2, . . .denote the subsequences of (q n ) which consist of the even-and odd-numbered terms of (q n ) respectively.Equation ( 6) implies q n+1 = 1 + 1/q k k .Therefore, we can understand the sequence (q n ) by studying the iterated map f : [1, 2] → [1,2] given by f (x) = 1 + 1/x k .Define α to be the unique fixed point of f in [1,2] and note that α satisfies α k+1 = α k + 1.
Proof.This follows from Proposition 3, Proposition 4, and the fact that the q n are bounded.
To simplify notation, we also define a function g : [1, 2] → [1, 2] by g(x) = f 2 (x).Note that g(q n ) = q n+2 , so that g(O n ) = O n+1 and g(E n ) = E n+1 .This implies that the (possibly non-distinct) fixed points of g are L E , L O , and α.It is not difficult to show that x = k √ k − 1 is a maximum for g (x) on [1,2] and that g Note that for for k > Ω we have (k − 1) k+1 > k k and for k < Ω we have (k − 1) k+1 < k k .Thus, for k < Ω we know that sup { g (x) : x ∈ [1, 2]} = C < 1 for all x.Therefore, for k < Ω the Contraction Mapping Theorem implies that g has a unique fixed point in [1,2].This implies that the sequences (E n ) and (O n ) both converge to L E = L O = α.Thus, for k < Ω, we know that the limit Next, we show that for k > Ω there are two distinct fixed points besides α, one strictly less than α and one strictly greater than α.Therefore, L E = L O , and neither equals α.Lemma 5.5.k > Ω implies g (α) > 1. ) and g n (q 3 ) respectively.In addition, To see that there cannot be more than three fixed points, note that g (1) < 1 and g (2) < 1.Since g has only one zero in I, this implies that g (x) = 1 for at most two values in I. Therefore, g can cross the line y = x at most three times.
We now determine the asymptotics of (a n ), in which the limits L E and L O will arise naturally.We denote exponentiation with base 2 by exp 2 (x).By the definition of (q n ) we have, Define the sum so that a n = exp 2 (k n−1 A n ).Since the q j are bounded, (A n ) must converge to some limit, A, given by Note that A n , A, and q j all depend on k.It is reasonable to guess that a n is asymptotic to λ exp 2 (k n−1 A) for some λ, and this is indeed true.
Theorem 5.7.For k = 2, 3, . . .and n even, we have where A is as above and L O and L E are the limits of (O n ) and (E n ) respectively.
Proof.We will prove this by showing that We define For all r ≥ 3, we have 1 < q r < 2, so log 2 q r > 0. Then we can rearrange the sum in (10), splitting it into even and odd terms to give Fix > 0. We have already shown that Then we know that for n even, Therefore for n even, Theorem 5.8.For k = 2, 3, . . .and n odd, we have where A is as above and L E and L O are the limits of (O n ) and (E n ) respectively.
Proof.The proof is identical to the proof of the previous theorem, except that for odd n, L E and L O are switched in (12) and every step after.
Note that for n even we have, and for n odd we have, When k < Ω, we know that L E = L O = α, so these expressions both simplify to Theorem 5.9.The entropy of the golden mean shift on the free group G is given by h(X F ) = A(k − 1)/k.
Proof.To prove the theorem we show that if a n ∼ λ exp 2 (Ak n−1 ), then In the free group, Next by Theorem 5. Remark 1.Since λ does not appear in the expression for h(X F ), it is irrelevant to the entropy calculation that λ may depend on whether we look at the even or the odd terms of a n .
Remark 2. Iterations of the map f (x) = 1 + x −k produce an expression similar to the continued fraction for the golden mean, φ, but with k'th powers on each denominator.The above results show that for k < Ω, the limit f n (x) as n → ∞ is independent of x.Such continued fractions with exponents on the denominator are called generalized continued fractions and can be defined using a nonlinear version of the Gauss map given by φ k (x) = x −1/k mod 1.It is easy to show that for k > Ω generalized continued fractions are not well-defined.Note that φ k acts as the shift on generalized continued fraction expansions and is conjugate to the usual Gauss map where k = 1.Generalized continued fractions and their invariant measures have been investigated in [4,5].
Theorem 5.9 shows that the entropy of the golden mean shift on G is therefore given by The form of (13) is useful for numerical approximation because the q j are bounded, so an upper bound can easily be found.The below table shows upper and lower bounds on the entropy.Lower bounds were found by taking n terms of Equation (13) for each k.Upper bounds were found by taking n terms of the sum and then taking the q j = 2 for j = n + 1 to ∞. Finding a closed form for h(X F ) is still an open problem.We did find a closed form for λ when k < Ω, which implies that perhaps it will be easier in other problems, such as the hard core model on Z×Z, to find the constant in front of the exponential, rather than the actual entropy.

Example 3 .
Consider the shifts of finite type shown in Figure 1 on the alphabet A = {R, G, B}.Cycles representing points in X σ1 are always of the form RGBRGB . . .RGB.

Figure 2 .
Figure 2. The tree T 3 for k = 3.Here group elements are represented as circles.