Asymptotic composants of periodic unimodal inverse limit spaces. Henk Bruin Department of Mathematics, University of Surrey, Guildford, Surrey GU2 7XH, UK H.Bruin@surrey.ac.uk 1
Unimodal maps, e.g. f a (x) = 1 ax 2 on I = [1 a, 1] The critical point is c = 0. c n = f n (c). The unimodal map is periodic if f N (c) = c for some N. (To obtain interesting inverse limit spaces, we need N 3.) 2
The inverse limit space of f is X = X f X = {(x i ) i 0 : f(x i 1 ) = x i for all i 0} equipped with product topology and metric d(x, x) = i 0 2 i x i x i. The induced homeomorphism is ˆf(..., x 1, x 0 ) =..., x 1, x 0, f(x 0 ) and the projection maps are π i (x) = x i for i 0 3
Why study them? Hope to understand topology of Hénon and similar attractors, which are the global attractors of H a,b (x, y) = (1 ax 2 + by, x) = (f a (x) + by, x) Local structure of Cantor set of arcs is too naive. If a is such that f N a (0) = 0, then for b 0 small X is homeomorphic to the Hénon attractor (Barge & Holte 95). Main question: If f and g are nonconjugate, are X f and X g non-homeomorphic? 4
If the critical point is periodic (f N (c) = c), then X is locally a Cantor set of arcs, except at N endpoints:... c, f(c),..., f N 1 (c), c and its shifts. This alone is not enough to classify the periodic unimodal inverse limits. E.g. there are 3 different parameters such that f 5 a(0) = 0. 5
The inverse limit X f if f 3 (c) = c looks as follows: p This representation has a single infinite Wada channel. p is fixed under ˆf. 6
All composants of X f are all continuous copies of R or [0, ). Complete classification based on the way the composants with endpoints fold on themselves. Bruin 00, Kailhofer 03, Stima c 04 (preperiodic critical points), Block et al. 05. Complete proof of classification uncomfortably complicated. 7
There are more inhomogeneities! Barge & Diamond 95 discovered that some composants are asymptotic: There are parametrizations ϕ : R C and ϕ : R C such that d(ϕ(t), ϕ(t)) 0 as t This discovery relies on substitution tilings which are 2-to-1 coverings of inverse limit spaces. Every periodic unimodal inverse limit has them, and a finite number of them: Their number 2(N 2). How to find them? How to classify them? (Can they be homeomorphic to other composants?) 8
Patterns of asymptotic composants: Asymptotic composants can be arranged in many configurations, e.g. 5-fan 3-cycle two linked 2-fans This is only schematic. In reality the composant will fold backwards on themselves; their closure is X. 9
Symbolic dynamics: The kneading sequence K f is the right-infinite sequence where ν i = ν 1 ν 2 ν 3... { 0 if f i (0) < 0, 1 if f i (0) > 0, and ν kn = 0 or 1 to make Write #{0 < i kn : ν i = 1} is even. ϑ(n) = #{i n : ν i = 1} so ϑ(n) is even. 10
Backward symbolic itineraries: Let and I 0 = [1 a, 0) ( {0} if ν N = 0) I 1 = (0, 1] ( {0} if ν N = 1) The backward itinerary of x X is the left-infinite sequence e i (x), i 0, where e i (x) = k if π i (x) I k. Lemma: Two points x and y belong to the same composant of X if and only if their backward itineraries have the same tail. 11
A x := {y X : e(y) i = e(x) i i < 0}. The closure A x is an arc which can be parametrized by its zero-th component. Let e = e(x) and define τ R,L (e) = sup{n 0 : e n... e 1 = ν 1... ν n and ϑ(n) is even resp. odd} + 1 Lemma: If τ L,R (e(x)) are finite, then π 0 (A(x)) = [c τl (e), c τ R (e) ]. If τ R (e) or τ L (e) =, then A(x) contains an endpoint of the inverse limit. Lemma: A x and A y have an endpoint in common if e(x) i = e(y) i for all i < 0 except for one entry k = τ L (x) = τ L (y) ór k = τ R (x) = τ R (y). 12
Folding patterns: Given t = e(x) you can follow the itinerary of the points in the composant of x by its symbolic folding pattern: α 1 = τ R (t) and switch the τ R -th symbol in t to obtain S(t). Next α 2 = τ L (S(t)) and switch the τ L -th etc. symbol in S(t) to obtain S 2 (t). The same can be done in the other direction: α 1 = τ L (t) and switch the τ L -th etc. symbol in t to obtain S 1 (t). 13
Example: If ν = 101101101101101101... we get t =... 11110111 α 1 = τ R (t) = 1 S(t) =... 11110110 α 2 = τ L (S(t)) = 6 S 2 (t) =... 11010110 α 3 = τ R (S 2 (t)) = 1 S 3 (t) =... 11010111 α 4 = τ L (S 3 (t)) = 2 S 4 (t) =... 11010101 α 5 = τ R (S 4 (t)) = 4 S 5 (t) =... 11011101 α 6 = τ L (S 5 (t)) = 2 S 6 (t) =... 11011111....... 14
Proposition: C and C are asymptotic composants if and only if there are x C, x C such that The discrepancies dis(s n (t), S n ( t)) as n where dis(y, ỹ) = min{k : y k ỹ k }. The folding patterns of x and x satisfy c αn c αn 0 as n. If f N (c) = c, the latter condition reduces to N divides α n α n for n sufficiently large. 15
Conclusions: Analysing the folding patterns of periodic inverse limits leads to: There are three groups of kneading sequences leading to different patterns of asymptotic composants: I: N 1-fans (when ν = 1000... 1.) II: k-cycles III: multiple 2-fans. These types are mutually exclusive. Additional linking is possible when there are additional symmetries in the kneading sequence. Some (although very few) are asymptotic on themselves: d(ϕ(t), ϕ( t)) 0. 16
All asymptotic composants have periodic symbolic tails. The bound of their number is sharp. 2(N 2) No composant with endpoint can be asymptotic. If c is strictly preperiodic, then there are no asymptotic composants. Unfortunately, we cannot completely classify the inverse limits by pattern of asymptotic composants. Conjecture: If C and C are asymptotic composant (not necessarily to each other), then there is k such that ˆf k (C) coincides or is asymptotic to C. 17
ν type periodic tail(s) k Case 1 101 1-cycle 1 2 I 2 1001 3-fan 101 3 I 3 10001 4-fan 1001 4 I 4 10010 3-cycle 101 3 II 5 10111 three 2-fans 101110 3 III 6 100001 5-fan 10001 5 I 7 100010 4-cycle 1001 4 II 8 100111 four 2-fans 10010011 4 III 9 101110 two linked 3-fans 1 10, 1 4 II, IV 10 1000001 6-fan 100001 6 I 11 1000010 5-cycle 10001 5 II 12 1000111 five 2-fans 1000100011 5 III 13 1000100 four 2-fans (l.i.p.) 10, 1001 4 II 14 1001101 four 2-fans 10011010 4 III 15 1001110 five 2-fans 10010, 10111 5 II 16 1001011 five 2-fans 1001011011 5 III 17 1011010 5-cycle 10111 5 II 18 1011111 five 2-fans 1011111110 5 III 19 10000001 7-fan 1000001 7 I 20 10000010 6-cycle 100001 6 II 21 10000111 six 2-fans 100001110000 6 III 22 10000100 five 2-fans 10001, 10010 5 II 23 10001101 five 2-fans 1000110100 5 III 24 10001110 six 2-fans 100010, 100111 6 II 25 10001011 six 2-fans 100010110011 6 III 26 10011010 six 2-fans (l.i.p.) 101, 100111 6 II 27 10011111 six 2-fans 100111110110 6 III 28 10011100 five 2-fans 10010, 10111 5 II 29 10010101 five 2-fans 1001010111 5 III 30 10010110 six 2-fans (l.i.p.) 100, 101110 6 II 31 10110111 three 3-cycles 101101110 3 III 32 10111110 two linked 4-fans 101110, 1 5 II, IV 18