Lecture 25 Markov Partitions

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1 Lecture 25 Markov Partitions March 24, 2008 Let Λ be a locally maximal hyperbolic set and ɛ, δ > 0 constants for the local product structure where δ < ɛ/2 and ɛ is also a constant for expansion on Λ. For A Λ denote int(a) as the interior of A relative to Λ and A as the boundary in Λ. Let W s (x, A) = W s (x) A and W u (x, A) = W u (x) A. For 0 < η ɛ we set W s η (x) = {y W s ɛ (x) d(x, y) η} and int(w s η (x)) = {y W s ɛ (x) d(x, y) < η}. Proposition 0.1 There exists a constant ρ (0, δ/2) such that for all x A the bracket is a homeomorphism of onto an open neighborhood of x in Λ. (int(w u ρ (x)) Λ) (int(w s ρ (x)) Λ) Proof. Let V δ (x) = {y Λ d(x, y) < δ}. Define Π s : V δ (x) W s ɛ (x, Λ) by y [x, y], and Π u : V δ (x) W u ɛ (x, Λ) by y [y, x]. The bracket i uniformly continuous on compact neighborhoods of the diagonal in Λ Λ so there exists a ρ (0, δ/2) such that for all x, y, z Λ, d(x, y) < ρ, and d(x, z) < ρ implies that d(x, [y, z]) < δ. So if (y, z) (int(w u ρ (x)) Λ) (int(w s ρ (x)) Λ), 1

2 then Π s and Π u are defined and (Π u, Π s ) is the inverse for [, ]. So the bracket is a homeomorphism onto the open set Π 1 s (int(wρ s (x)) Λ) Π 1 u (int(wρ u (x)) Λ). Definition 0.2 A subset R of Λ is a rectangle of diam(r) < δ provided that for all x, y R the point [x, y] is defined and contained in R. A rectangle is proper if int(r) = R. From the previous proposition we have the following result Proposition 0.3 If R is a rectangle and diam(r) < ρ with x R, then int(r) R = [int(w u (x, R)), int(w s (x, R)], = [ W u (x, R), W s (x, R)] [W u (x, R), W s (x, R)]. Corollary 0.4 Let R be a rectangle and diam(r) < ρ. Then 1. x int(r) if and only if x int(w u (x, R)) int(w s (x, R)), 2. if x int(r), then int(w s (x, R)) = W s (x, int(r)) and int(w u (x, R)) = W u (x, int(r)), and 3. if R is closed, then R = s R u R, where s R = {x R x / int(w u (x, R))} = {x R W s (x, R) int(r) = } and u R = {x R x / int(w s (x, R))} = {x R W u (x, R) int(r) = }. Proof. Parts (1) and (2) are clear from the previous arguments. For part (3) we see that for all z R we have [z, y] = y whenever y W s (z, R) and [y, z] = y whenever y W u (z, R). Hence, [{z} int(w u (z, R)), int(w s (z, R))] [{z} int(w u (z, R)), int(w s (z, R))] 2 = int(r) int(w s (z, R)) = int(r) W s (z, R), and = int(r) (int(w u (z, R))) = int(r) W u (z, R).

3 Definition 0.5 A Markov Partition of Λ for f is a finite collection of rectangles {R 1,..., R n } with the diameter of each rectangle less than ρ such that the following are satisfied: 1. if x int(r i ), f(x) int(r j ), then f(w s (x, R i )) W s (f(x), R j ), and 2. if x int(r i ), f(x) int(r j ), then f 1 (W u (x, R i )) W u (f 1 (x), R j ). Let A be the matrix of 0 s and 1 s such that a = 1 if f(int(r i )) int(r j ) and 0 else. Let Σ A be the SFT associated with A. Define h : Σ A Λ to be the itinerary map. This will be finite-to-one and hence have the same entropy as f Λ. Theorem 0.6 Every locally maximal hyperbolic set has a Markov Partition. Before proving the above the theorem we list another related result Theorem 0.7 (F.) For every hyperbolic set Λ and a neighborhood V of Λ there exists a hyperbolic set Λ n Z f n (V ) such that Λ Λ and Λ has a Markov partition. Proof. Let β (0, ρ/2 where ρ < δ/2 < ɛ. From the Shadowing Lemma there exists some α (0, β) such that every bi-infinite α-pseudo orbit in Λ is β-shadowed by a unique point in Λ. Since Λ is compact there exists some γ (0, α/2) such that for all x, y Λ, d(x, y) < γ we have d(f(x), f(y)) < α/2. Let p 1,..., p k be a set of γ-dense points (so k i=1 B γ(p i ) Λ). Define A a k k matrix by a = 1 if f(p i ) B α (p j ) and a = 0 else. Let Σ A be the SFT associated with A. For all s Σ A the sequence {p si } i Z is an α-pseudo orbit and β-shadowed by a unique x = θ(s). This defines a map θ : Σ A Λ. Proposition 0.8 The map θ is a semi-conjugacy of σ A and f. Proof. If x β-shadows the pseudo orbit {p si }, then f(x), β-shadows {p (σs)i } so fθ = θσ. Continuity follows from the Shadowing Lemma and expansiveness of Λ. Indeed, suppose that s n l, t n l, and θ(s n ) and θ(t n ) have different limits, denoted s and t. We know for all i Z and all n N that d(f i (θ(s n )), p s n i ) < β. So passing to the limit we have for all i Z that d(f i (s), p li ) β, and d(f i (t), p li ) β. 3

4 Hence, for all i Z we have d(f i (s), f i (t)) 2β < δ/2 < ɛ, but ɛ is an expansive constant for Λ, a contradiction. Hence, θ is continuous. To continue to proof of the theorem we will see that θ(c 0 i ) is a rectangle and the image of the rectangles cross correctly. However, they may overlap. Proposition 0.9 The map θ is a morphism of the local product structure. More precisely, for all t Σ A, θ(w s 1/3 (t)) W s ɛ (θ(t)), θ(w u 1/3 (t)) W u ɛ (θ(t)), if d(t, t ) < 1/2, then d(θ(t), θ(t )) < ρ δ, and θ([t, t ]) = [θ(t), θ(t )]. Proof. Note that d(t, t ) < frm[o] /2 implies that t 0 = t 0. Also we have d(θ(t), θ(t )) d(θ(t), p t0 ) + d(p t0, θ(t )) 2β < ρ. So [θ(t), θ(t )] is defined. Let t Σ A and t W u 1/3 (t). Then t i = t i for all i 0. Hence, for all i 0 we have d(f i (θ(t)), f i (θ(t ))) d(f i (θ(t))p ti ) + d(p ti, f i (θ(t ))) 2β < ɛ. So θ(t ) Wɛ s (θ(t)). Similarly, we can show that θ(w1/3 u (t)) Su ɛ (θ(t)). From the definition of the bracket we then have θ([t, t ]) = θ(w s frm[o] /3 (t) W u 1/3 (t )) W s ɛ (θ(t)) W u ɛ (θ(t )) = [θ(t), θ(t )]. Lemma 0.10 Let 1 i k and t C i = {t Σ A t 0 = i}. Then T i = θ(c i ) is a closed rectangle, diam(t i ) < ρ, θ(w s (t, C i )) = W s (θ(t), T i ), and θ(w u (t, C i )) = W u (θ(t), T i ). 4

5 Proof. We know that C i is a rectangle of diameter less than or equal to 1/2 in Σ A. Then the previous proposition says that T i is a rectangle and diam(t i ) < ρ. The set T i is closed since C i is compact and θ is continuous. Furthermore, W s (t, C i ) = W1/3 s (t) so θ(w s (t, C i )) W s ɛ (θ(t)) T i = W s (θ(t), T i ) and θ(w u (t, C i )) W u ɛ (θ(t)) T i = W u (θ(t), T i ). Let x W s (θ(t), T i ) and s θ 1 (x) C i. Then θ([t, s]) = [θ(t), θ(s)] = [θ(t), x] = x and W s (θ(t), T i ) θ(w s (t, C i )). Likewise, W u (θ(t), T i ) θ(w u (t, C i )). Proposition 0.11 Is s C i, σ(s) C j, and x = θ(s), then f(w s (x, T i )) W s (f(x), T j ), and f 1 (W u (f(x), T j )) W u (x, T i ). Proof. From the previous lemma we have f(w s (x, T i )) = f(w s (θ(s), T i )) = f(θ(w s (s, C i ))) = θσ(w s (s, C i )) = W s (θ(σs), T j ) = W s (f(x0, T j ). Similarly, we get the result for the unstable direction. Corollary 0.12 Let x T i. Then the following are satisfied: 1. There exists 1 j k such that f(x) T j, f(w s (x, T i )) W s (f(x), T j ), and f 1 (W u (f(x), T j )) W u (x, T i ). 2. There exists 1 m k such that f 1 (x) T m, f(w s (f 1 (x), T m )) W s (x, T i ), and f 1 (W u (x, T i )) W u (f 1 (x), T m ). Proof. Let s C i such that θ(s) = x, s 1 = j, and s 1 = m. Then the result follows from the previous proposition. 5

6 We now want to subdivide the T i s so thy do not overlap. If int(t i ) int(t j ) we let τ 1 = {x T i W s (x, T i ) T j, W s (x, T i ) T j }, τ 2 = {x T i W s (x, T i ) T j, W s (x, T i ) T j = }, τ 3 = {x T i W s (x, T i ) T j =, W s (x, T i ) T j }, and τ 4 = {x T i W s (x, T i ) T j =, W s (x, T i ) T j = }. Lemma 0.13 The τ m form a partition of T i, τ 1 = T i T j, and τ, 1 τ 1 τ, 2 and τ 1 τ 3 are closed. Proof. The partition follows from the previous results. Since T i and T j are closed we see that τ 1 is closed. The other two conditions also follow from the closure of T i and T j. Now let T m = τ m. Lemma 0.14 The sets T m satisfy the following: 1. T 1 = T i T j, 2. int(t l ) T m = if l < m or (l, m) = (3, 2), 3. the T m have disjoint interiors, 4. T l τ l ( 4 m=1 T m ), and 5. T m is a rectangle and diam(t m ) < ρ. Proof. Numbers (1)-(3) are clear from the definitions and previous results. (For (2) a picture is sufficient.) Since the interiors are disjoint we know that int(t l ) ( 4 m=1 T m T i m l T m τ l. So (4) follows. For (5) we let the closure of the interior of the rectangles of diameter be less than ρ is a rectangle by continuity of the bracket. So we need to show that τ m is a rectangle. Since τ m is closed under the bracket and the diameter is less than ρ we know that τ m is a rectangle. 6

7 For Z = Λ 4 k m=1 i,j=1 T m the previous lemma shows that Z is open and dense in Λ and Let x Z and Z T m = Z int(t m ) = Z τ m. K(x) = {T i x T i }, K (x) = {T j there exists T i K(x) such that T i T j }, R(x) = {int(t m ) T i K(x), T j K (x), x T m }. By construction R(x) is an open rectangle and diam(r(x)) < ρ for all x Z. Lemma 0.15 If x, y Z, then R(x) = R(y) or R(x) R(y) =. Proof. Let z Z R(x). By the definition of R(x) we see that K(z) K(x). If T j K(z) and x T i, then T j is in K (x) and x int(t m ) for some 1 m 4. Since z int(t l we know from the previous lemma that l = m so l = m = 1. Therefore, x T 1 and T j K(x). Hence, K(z) = K(x) and K (x) = K (z) and R(z) = R(x) for z R(x) Z. For x, y Z and R(x) R(y) we know there exists some z R(x) R(y) Z. So R(x) = R(z) = R(y). Lemma 0.16 Let x, y Z f 1 (Z), R(x) = R(y), and y W s ɛ (x). Then R(f(x)) = R(f(y)). Proof. If f(x) T i, then there exists some s Σ A with s 0 = i such that f(x) = θ(s). Let j = s 1. Since x θσ 1 (s), and θ(w s (σ 1 (s), C j )) = W s (x, T j ) we see that there exists some t Σ A such that y = θ(t) and t W s (σ 1 (s), C j ). So t 1 = s 0 = i and f(y) = θσ(t) T i. Then K(f(x)) K(f(y)). Exchanging x and y we see that K(f(x)) = K(f(y)) and K (f(x)) = K (f(y)). Fix T i in K(f(x)) and T j in K (f(x)). Since W s (f(x), T i ) = W s (f(y), T i ) we have W s (F (x), T i ) T j. This implies that W s (f(y), T i ) T j. Similarly, for the unstable we see that if T m is in K(x) where f 1 (W u (f(x), T i )) W u (x, T m ) and W u (f(x), T i ) T j contains a point z, then there exists T p in K(f 1 (z)) such that f(w s (f 1 (z), T p )) W s (z, T j ). 7

8 Let t = [z, f(y)]. Then t T i since z and f(y) are in T i as well as Wɛ u (f(y)). From the choice of z we see that f 1 (z) is in T p W u (x, T m ). Then x τmp 1 τmp 3 and y τmp 1 τmp 3 so W u (y, T m ) T p. If y W u (y, T m ) T p we have [f 1 (z), u] = [f 1 (z), y] W s (f 1 (z), T p. So t = [z, f(y)] = f([f 1 (z), y]) W s (z, T j ) T j, and if W u (f(x), T i ) T j, then W u (f(y), T i ) T j. Exchanging x and y the converse holds. Then W s (f(x), T i ) T j if and only if W s (f(y), T i ) T j, and W u (f(x), T i ) T j if and only if W u (f(y), T i ) T j. Since f(x), f(y) Z we have R(f(x)) = R(f(y)). Let R = {R 1,..., R s } = {R(x) x Z}. Proposition 0.17 The collection R is a Markov partitions of Λ for f. Proof. We know be construction that s i=1 R i = Λ, the diam(r i ) < ρ for all 1 i s, the rectangles are proper, and the interior of the rectangles are disjoint. The crossing property follows since the T i rectangles crossed properly and the fact that the R(x) were open (i.e. if x int(r i ) and f(x) int(r j ), then f(w s (x, R i ))) W s (f(x), R j ), and if x int(r i ) and f 1 (x) int(r j ), then f 1 (W u (x, R i ))) W s (f 1 (x), R j )). 8