How to tinker with a homeo- morphism and get away with How to tinker with a homeomorphism and get it Judy Kennedy away with it Joint work with Jan Boronski, Xiaochuan Liu and Piotr Oprocha Judy Kennedy Joint work with Jan Boronski, Xiaochuan Liu and Piotr Oprocha Lamar University Beaumont, Texas, USA May 18, 2018
Mary Rees and BCL How to tinker with a homeo- morphism and get away with it Mary Rees published a paper “A minimal positive entropy Judy Kennedy homeomorphism of the 2-torus” in 1980. Joint work with Jan Boronski, Xiaochuan Liu In that paper she gave a construction that allowed the and Piotr Oprocha modification of a minimal homeomorphism to suit her purposes. That construction was intricate and hard to understand, so in 2011, F. B´ eguin, S. Crovisier, and F. Le Roux wrote a 66-page paper, “Construction of curious minimal uniquely ergodic homeomorphisms on manifolds”, one of whose goals was to make the Rees construction more accessible.
How to tinker with a homeo- morphism and Suppose we have a homeomorphism H : T 2 → T 2 . get away with it Judy Kennedy By a rectangle we mean any subset of T 2 homeomorphic to the Joint work with Jan unit disc in R 2 . Boronski, Xiaochuan Liu and Piotr Oprocha Let E , F be a collection of rectangles. We say that F refines E if: (a) every element of E contains at least one element of F ; (b) for elements X ∈ E , Y ∈ F either X ∩ Y = ∅ or Y ⊂ int X . We define mesh E = max { diam X : X ∈ E} .
p-times iterable How to tinker with a homeo- morphism and Let p ∈ N . A collection of rectangles E is p-times iterable if for get away with it rectangles X , Y ∈ E and integers − p ≤ k , s ≤ p , ether Judy Kennedy H k ( X ) = H s ( Y ) or H k ( X ) ∩ H s ( Y ) = ∅ . For any p -times Joint work with Jan iterable family of rectangles E and any 0 ≤ n ≤ p , Boronski, Xiaochuan Liu and Piotr E n = � H k ( E ) , Oprocha | k |≤ n where as usual H ( E ) = { H ( X ) : X ∈ E} . In particular, E 0 = E . Given an integer 0 ≤ n ≤ p we define an oriented graph G = G ( E n ), where the vertices are elements of E n and there is an edge from X to Y provided that H ( X ) = Y . For n < p we say that E n has no cycle if the graph G ( E n ) has no cycle.
F is compatible with E for p iterates How to tinker with a homeo- morphism and get away with it Judy Kennedy Joint work For a collection of rectangles E , let us denote by s ( E ) the union with Jan Boronski, of all rectangles in E . Fix an integer p ≥ 0 and let E , F be Xiaochuan Liu and Piotr collections of rectangles such that E is p -times iterable and F Oprocha is ( p + 1)-times iterable. Assume additionally that F p +1 refines E p . If for every k such that | k | ≤ 2 p + 1, we have H k ( s ( F )) ∩ s ( E ) ⊂ s ( F ), then we say that F is compatible with E for p iterates .
main axioms from B´ eguin, Crovisier, Le Roux How to tinker with a homeo- morphism and get away with Now we are ready to state the main axioms which are the it building blocks in the construction. Let ( E n ) n ∈ N 0 be a sequence Judy Kennedy Joint work of collections of rectangles. We introduce the following with Jan Boronski, hypotheses: Xiaochuan Liu and Piotr A 1 : For every n ∈ N 0 Oprocha a n : the collection E n is ( n + 1)-times iterable and E n n has no cycle; b n : the collection E n +1 refines the collection E m +1 for every n m 0 ≤ m < n ; c n : the collection E n +1 is compatible with E n for n + 1 iterates. A 3 : lim n →∞ mesh E n n = 0.
the homeomorphisms M i How to tinker with a homeo- morphism and get away with it Judy Kennedy Assume that ( M n ) n ∈ N is a sequence of homeomorphisms Joint work M n : T 2 → T 2 and that for every n the homeomorphisms Ψ n , g n with Jan Boronski, Xiaochuan Liu are defined by: and Piotr Oprocha Ψ n = M n ◦ . . . ◦ M 2 ◦ M 1 , Ψ − 1 = ◦ H ◦ Ψ n . g n n Finally we set Ψ 0 = id, g 0 = H .
More conditions on the M n In addition, assume that the homeomorphisms M n satisfy the How to tinker with a homeo- conditions specified below: morphism and get away with B 1 : For every n ∈ N 0 : it Judy Kennedy B 1 , n : The support of the homeomorphism M n is contained in Joint work the set E n − 1 n − 1 , where as usual the support of the with Jan Boronski, homeomorphism M n is defined by Xiaochuan Liu and Piotr supp M n = { x : M n ( x ) � = x } . Oprocha B 2 : For every n ∈ N 0 : B 2 , n : The homeomorphisms M n and H commute along edges of the graph G ( E n − 1 n − 1 ). B 3 : Denote A n = E n +1 \ E n − 1 for every n ∈ N 0 . n n B 3 , n : The mesh { Ψ − 1 n − 1 ( X ) : X ∈ A n } < 1 / n ; and, in particular, n →∞ mesh { Ψ − 1 lim n − 1 ( X ) : X ∈ A n } = 0 .
A lemma How to tinker with a homeo- morphism and The following fact is [BCL, Proposition 3.1]. It ensures proper get away with it convergence of the constructed functions. Judy Kennedy Joint work Lemma with Jan Boronski, Assume that hypotheses A 1 , 3 , B 1 , 2 , 3 are satisfied. Then: Xiaochuan Liu and Piotr Oprocha 1 The sequence of homeomorphisms (Ψ n ) n ∈ N converges uniformly to a continuous surjective map Ψ: T 2 → T 2 . 2 The sequence of homeomorphisms ( g n ) n ∈ N converges uniformly to a homeomorphism map g : T 2 → T 2 and ( g − 1 n ) n ∈ N converge uniformly to its inverse g − 1 . 3 The homeomorphism g is an extension of H by Ψ , that is, H ◦ Ψ = Ψ ◦ g.
another lemma How to tinker with a homeo- The following is Proposition 3.4 in BCL. morphism and get away with it Lemma Judy Kennedy n ∈ N s ( E n ) and assume that hypotheses A 1 , 3 , B 1 , 2 , 3 Let K = � Joint work with Jan are satisfied. Boronski, Xiaochuan Liu 1 Fix x ∈ T 2 and suppose that there is m ∈ Z such that and Piotr Oprocha x ∈ H m ( K ) . Let ( X n ) n ≥ m be the decreasing sequence of rectangles in E m n containing x. Then Ψ − 1 ( x ) = � Ψ − 1 n ( X n ) . n ≥ m 2 For every x which does not belong to the orbit of K the set Ψ − 1 ( x ) is a single point.
minimal noninvertible pseudocircle map How to tinker with a homeo- morphism and get away with it We wish to construct a minimal map on the pseudocircle which Judy Kennedy Joint work is not a homeomorphism. The main step of the construction is with Jan Boronski, the following theorem. We denote the annulus by A . Xiaochuan Liu and Piotr Oprocha Theorem There exists a homeomorphism g : A → A with an invariant pseudocircle P ⊂ A such that ( g , P ) is minimal and there exists a pseudoarc A ⊂ P such that lim | n |→∞ diam g n ( A ) = 0 .
main ideas of the proof of the theorem How to tinker Let H : A → A be the annulus homeomorphism defined by with a homeo- morphism and Handel. In particular, (1) H is a rotation on (both) circles get away with it that form the boundary of A , (2) there exists an essential Judy Kennedy pseudocircle P ⊂ A that is a minimal, invariant subset Joint work with Jan under H , and (3) every point from the interior of A is Boronski, Xiaochuan Liu attracted by P . and Piotr Oprocha We can “glue” the boundary of A to a single circle, call it S , which turns A into T 2 , and now (our modified) H : T 2 → T 2 . If we perturb H to a homeomorphism H ′ in such a way that H and H ′ coincide in a neighborhood of S , then we can again “cut back” T 2 to A obtaining a well defined homeomorphism H ′ : A → A . In particular, if, in a sufficiently small neighborhood of P there is an H ′ -invariant set P ′ , which is a hereditarily indecomposable circlelike continuum, then it must also be a pseudocircle.
Step 1. Definition of the maps M n . To start the How to tinker with a homeo- construction, fix a pseudoarc A ⊂ P and a point p ∈ A . There morphism and get away with exists a sequence of rectangles ( U n ) n ∈ N , U n +1 ⊂ int U n ⊂ T 2 it such that � n ∈ N U n = A . There also exists a decreasing Judy Kennedy Joint work sequence of rectangles ( V n ) n ∈ N such that (1) V n ⊂ U n for each with Jan Boronski, n , (2) p ∈ int V n for every n , and (3) ∩ V n = { p } . The Xiaochuan Liu and Piotr pseudocircle P is an invariant set of H without fixed points, the Oprocha pseudoarcs H i ( A ) belong to different composants of P for different i ∈ Z . In particular H i ( A ) ∩ H j ( A ) = ∅ for i � = j , hence we may assume that for | i | ≤ 3 n the sets H i ( U n ) are pairwise 1 disjoint. Furthermore, we may assume that diam H i ( V n ) < n +1 for | i | ≤ 3 n . Since the pseudoarc A can be chosen to be arbitrarily small, we may assume that V 0 = U 0 . Let E 0 = { V 0 } . Take any k 1 > 2 and let M 1 : T 2 → T 2 be a homeomorphism such that M 1 | U 1 is a homeomorphism between U 1 and V k 1 and M 1 | T 2 \ int U 0 = id. Require additionally that M 1 ( p ) = p .
Main Theorem How to tinker with a homeo- morphism and get away with it Judy Kennedy Joint work with Jan Boronski, Theorem Xiaochuan Liu and Piotr There exists a continuous surjection G : A → A with an Oprocha invariant pseudocircle P ⊂ A such that ( G , P ) is minimal but is not one-to-one.
How to tinker with a homeo- morphism and get away with it Judy Kennedy Joint work with Jan Boronski, Xiaochuan Liu and Piotr Thanks so much for listening! Oprocha
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