Notes on Rauzy Fractals

Transcription

1 Notes on Rauzy Fractals It is a tradition to associate symbolic codings to dynamical systems arising from geometry or mechanics. Here we introduce a classical result to give geometric representations to infinite words generated in algebraic or combinatorial way. 1 Backgrounds on Symbolic Dynamics An alphabet A is a finite set of letters. A word is a finite sequence of letters. Denote the set of words by A +. For a word u = u 1 u 2 u n A +, u is called the length of u. For each letter a, we denote by u a the number of occurrence of a in u. Full Shifts. Let A Z be the set of all biinfinite sequences of alphabets from A, i.e. A Z = {x = (x i ) i Z : x i A, i Z}. Consider a shifting operator σ : A Z A Z given by σ(x) n = x n+1. We call the dynamical system (A Z, σ) a full shift. The full shift endowed with the product topology on A Z. Two elements x, y A Z are regarded as close if they agree on a large block: x [ n,n] = y [ n,n] for some large n. Substitutions. Let A be an alphabet consisting of at least 2 letters. A substitution on A is a map σ : A A +. The map σ can be iterated, and σ k : A A + is again a substitution. 2 Geometric Construction of the Fibonacci Word Let A = {a, b} and a substitution σ : a ab, b a. Starting with the word a, we obtain a sequence of words {w n } n 0 w 0 = a, w 1 = ab, w 2 = aba, w 3 = abaab,. The length of the words are Fibonacci numbers (this is because w n = w n a + w n b. Since w n a = w n 1 and w n b = w n 1 a = w n 2 ). w n a φ = w n b 2 Moreover, the word w n is a prefix of the next one. 1

2 Definition 2.1. A point is periodic if σ d (a) = a.... A Fibonacci word is the infinite word abaab... that is fixed under σ. A Geometric Representation. We want to embed the Fibonacci word in a subspace of R n. Every a (resp. b) gives a step to the right (1, 0) (resp. a step up (0, 1)) (See Figure 1). The path has asymptotic slope 1/φ. Figure 1: Geometric construction of the Fibonacci word Let M be the associated matrix given by the substitution ( ) 1 1 M =. 1 0 The matrix M has one expanding direction (φ, 1) associated with eigenvalue φ and one contracting direction (1, 1 1φ ) with eigenvalue 1/φ = Therefore, the points associated with the finite Fibonacci words σ n (a) tend to the expanding space. More generally, any initial word U of the infinite Fibonacci word is of the format U = σ k (i k ) σ(i 1 )i 0, for i n {, a}. e.g. σ 4 (a) = abaababa = σ 3 (a)σ(a)a. The corresponding coordinates of U are finite sums and they are bounded by a geometric series whose ratio are eigenvalues of M. Therefore, the points U of the Fibonacci word are convergent in the contracting direction. The vertices of the path must lie in a bounded distance of the expanding direction. We can draw a corridor by sliding the unit square with upper right corner at the origin along the expanding direction, the path lies within this corridor and is determined by it. Therefore, we can recover the Fibonacci sequence by this corridor: its nth letter is an a (resp. b) if [(n + 1)/φ] [n/φ] is equal to 1 (resp. 0). Dynamical Interpretation. We can shift the Fibonacci word by erasing its first letter. If we consider the closure of the shift-orbit for the natural topology on infinite words, we get a symbolic dynamical system. This set projects continuously to the interval, and the projection conjugates the shift to the rotation by φ. 2

3 This can be seen on the window, which is divided into two intervals corresponding to the two letters; the point corresponding to the next vertex of the path is obtained by exchanging two intervals. This translation of intervals corresponds to the step to the right or up associated with each letter. 3 Rauzy Fractals Let σ be the substitution σ : 1 12, 2 13, 3 1. Consider the iteration of σ on the letter a. We have 1, 12, 1213, ,. The limit σ (a) is a fixed point of the substitution. 3.1 Primitive Subsutitution. Definition 3.1. A substitution σ is primitive if there exists k such that all letters appears in the image of all letters through σ k. For example, σ(1) = 12, σ(2) = 32, σ(3) = 23 is not primitive. There arises a natural question: how to know a substitution is primitive? Proposition 3.2. Let M be an incidence matrix. A substitution σ is primitive iff there exists a k > 0 such that M k has only positive coefficients. Example 3.3. Let σ(1) = 12, σ(2) = 13, σ(3) = 1 and M be the incidence matrix Since M 3 > 0, the substitution σ is primitive. Example 3.4. The substitution σ(1) = 12, σ2 = 32 and σ(3) = 23 is not primitive, since M n always has zeros in the first line where M the incidence matrix M = Geometric Interpretation of a Fixed Point Let P be a linear map taking letters to n independent vectors e 1,, e n. P : w 1 w k A + e w1 + + e wk R n. Theorem 3.5 (Commutation Formula). P (σ(w )) = MP (W ). 3

4 Figure 2: Geometric interpretation for Piscot Case When the incidence matrix M has one expanding eigenvalue β and all other eigenvalues β (i) are contracting. We say the substitution σ of Piscot type. Let σ be a substitution of Pisot type on a n letter-alphabet. Suppose H e is the expanding line of M and H c is the contracting plane of M. Let π : R n H c be the projection along H e. Then h = πm is a contraction on the hyperplane H c. It follows that the projection of P (σ(w )) is smaller than P (W ). Proposition 3.6. If σ is of Piscot type, then the stair of any periodic point remains at a finite distance from the expanding line. Proof. 1. Let the word u 0 u k be a periodic point fixed by σ d. σ d (u 0 u l ) u 0 u k < σ d (u 0 u l+1 ) 2. There exists a prefix p 0 of σ d (u l ) such that u 0 u k = σ d (u 0 u l )p 0, σ d (u l ) = p 0 u k+1 s 3. Iterate this decomposition u 0 u k = σ d (u 0 u l )p 0 = σ d (σ d (p 2 )p 1 )p 0 = σ dn p N σ N 1 (p N 1 ) σ d (p 1 )p Linearize: πp (u 0 u k ) = h dn P (p N ) + h d(n 1) P (p n 1 ) + h d P (p 1 ) + P (p 0 ) 5. Since h is a contraction and p k s are in finite number, the projections of the stair along the expanding line are bounded. 4

5 Figure 3: The projection of stairs 3.4 Rauzy Fractal A Rauzy fractal (central tile) is the closure of the image under the projection of the nodes in the stair to the contraction plane, i.e. We set for each i A T = {π P (u 0 u N 1 ) N N}. T (i) = {π P (u 0 u N 1 ) N N, u N = i}. Figure 4: The central tile and its subtitles for the Tribonacci substitution Theorem 3.7. Let σ be a unit Pisot irreducible substitution. The central tile T and the subtiles T (i) are compact sets. Proof. The boundness of the subtiles T (i) is a direct consequence of the compactness of T, since they are closed subsets of T σ. To prove the compactness of T, it is sufficient to show that the points π P (u 0 u N 1 ) remain at a uniformly bounded distance of the origin in H c. We have shown this by the previous theorem. 3.5 Decomposition Definition 3.8. Given a collection of finitely many compact sets {K 1,, K q }. Each set K i can be decomposed as a union of contracted copies of itself and the other sets K j. We associated these relations with a natural graph G: 1. the vertices is given by {K i : 1 i q} 5

6 2. there is an edge e from K i to K j if the there is a contraction τ e such that τ e (K j ) K i. We call (G, {τ e } e E ) a graph-directed iterated function system (GIFS). By a fixed point argument that given a GIFS (G, {τ e } e E ) there exists a unique collection of non-empty compact sets K 1,, K q R n K i = τ e (K j ) i j where the union runs over all edges in G (see [1]). We apply this GIFS to the the subtiles T (i) which we call prefix-suffix graph which is given by the formula: i j iff σ(j) = pis. An edge from the vertex i to j is labeled by (p, i, s) A + A A + such that σ(j) = pis. Figure 5: The prefix-suffix graph for σ(1) = 12, σ(2) = 13, σ(3) = 1 Consider the substituion σ(1) = 12, σ(2) = 13 and σ(3) = 1. The map h = πm is a contraction on the hyperplane H c of angle 2/3. We have a decomposition of each tiles: T (1) = ht (1) ht (2) ht (3). T (2) = ht (1) + πp (1), T (3) = ht (2) + πp (1). Theorem 3.9. T (i) = ht (j) + πp (p). j A, i j Proof. Fix i A and assume that u k = i. By definition π P (u 0 u k 1 ) T (i). By previous argument, there exist L and a decomposition of σ(σ l ) as σ(u l ) = pu k s = pis such that P (u 0 u k 1 ) = h π P (u 0 u l 1 ) + π P (p), thus P (u 0 u k 1 ) h(t (u l )) + π P (p). This is true for each N with u N = i, T (i) = h(t (j)) + π P (p). (p,i,s),σ(j)=pis 6

7 To show the disjointness, it is sufficient to show that βµ(t (i)) = j A m ij µ(t (j)) where β is the dominant eigenvalue of the incident matrix M. By measure theory, we know that βµ(t (i)) j A m ij µ(t (j)). The reverse sign is given by the Perron-Frobenius theorem: Assume that M is a matrix with positive entries. Let β be its dominant eigenvalue. Let v be a vector with positive entries. Then we have Mv βv. The equality holds only when v is a dominant eigenvector of M. It follows that the pieces are disjoint in the decomposition of each subtile. Now, we want to know that if the subtitles are disjoint. Definition A substitution σ over the alphabet A satisfies strong coincidence condition if for every (j 1, j 2 ) A 2, we have with P (p 1 ) = P (p 2 ) or P (s 1 ) = P (s 2 ). σ k (j 1 ) = p 1 is 1 σ k (j 2 ) = p s is 2 Theorem If σ satisfies the strong coincidence condition, then the subtile T (i) of the central tile T have disjoint interiors. Proof. The pieces ht (j 1 )+P (p 1 ) and ht (j 2 )+P (p 2 ) both appear in the GIFS decomposition of T (i). References [1] D. Mauldin, and S. Williams. Hausdorff dimension in graph directed constructions. Transactions of the AMS, Vol. 309, No.2,