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The Connecting Lemma(s) Following Hayashi, Wen&Xia, Arnaud.

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Presentation on theme: "The Connecting Lemma(s) Following Hayashi, Wen&Xia, Arnaud."— Presentation transcript:

1 The Connecting Lemma(s) Following Hayashi, Wen&Xia, Arnaud

2 Pughs Closing Lemma If an orbit comes back very close to itself

3 Pughs Closing Lemma If an orbit comes back very close to itself Is it possible to close it by a small pertubation of the system ?

4 Pughs Closing Lemma If an orbit comes back very close to itself Is it possible to close it by a small pertubation of the system ?

5 An orbit coming back very close

6 A C 0 -small perturbation

7 The orbit is closed!

8 A C 1 -small perturbation: No closed orbit!

9 For C 1 -perturbation less than, one need a safety distance, proportional to the jump:

10 Pughs closing lemma (1967) If x is a non-wandering point of a diffeomorphism f on a compact manifold, then there is g, arbitrarily C 1 -close to f, such that x is a periodic point of g. Also holds for vectorfields Conservative, symplectic systems (Pugh&Robinson)

11 What is the strategy of Pugh? 1) spread the perturbation on a long time interval, for making the constant very close to 1. For flows: very long flow boxes

12 For diffeos

13 2) Selecting points:

14 The connecting lemma If the unstable manifold of a fixed point comes back very close to the stable manifold Can one create homoclinic intersection by C 1 - small perturbations?

15 The connecting lemma (Hayashi 1997) By a C 1 -perturbation:

16 Variations on Hayashis lemma x non-periodic point Arnaud, Wen & Xia

17 Corollary 1: for C 1 -generic f, H(p) = cl(W s (p)) cl(W u (p))

18 Other variation x non-periodic in the closure of W u (p)

19 Corollary 2: for C 1 -generic f cl(W u (p)) is Lyapunov stable Carballo Morales & Pacifico Corollary 3: for C 1 -generic f H(p) is a chain recurrent class

20 30 years from Pugh to Hayashi : why ? Pughs strategy :

21 This strategy cannot work for connecting lemma: There is no more selecting lemmas Each time you select one red and one blue point, There are other points nearby.

22 Hayashi changes the strategy:

23 Hayashis strategy. Each time the orbit comes back very close to itself, a small perturbations allows us to shorter the orbit: one jumps directly to the last return nearby, forgiving the intermediar orbit segment.

24 What is the notion of « being nearby »? Back to Pughs argumentFor any C 1 -neighborhood of f and any >0 there is N>0 such that: For any point x there are local coordinate around x such that Any cube C with edges parallela to the axes and C f i (C)= Ø 0

25 Then the cube C verifies:

26 For any pair x,y

27 There are x=x 0, …,x N =y such that

28 The ball B( f i (x i ), d(f i (x i ),f i (x i+1 )) ) where is the safety distance is contained in f i ( (1+ )C )

29 Perturbation boxes 1) Tiled cube : the ratio between adjacent tiles is bounded

30 The tiled cube C is a N-perturbation box for (f, ) if: for any sequence (x 0,y 0 ), …, (x n,y n ), with x i & y i in the same tile

31 There is g -C 1 -close to f, perturbation in C f(C) … f N-1 (C)

32 There is g -C 1 -close to f, perturbation in C f(C) … f N-1 (C)

33 There is g -C 1 -close to f, perturbation in C f(C) … f N-1 (C)

34 The connecting lemma Theorem Any tiled cube C, whose tiles are Pughs tiles and verifying C f i (C)= Ø, 0

35 Why this statment implies the connecting lemmas ?

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42 x 0 =y 0 =f i(0) (p) x 1 =y 1 =f i(1) (p) … x n =f i(n) (p); y n =f –j(m) (p) x n+1 =y n+1 =f -j(m-1) (p) … x m+n =y m+n =f –j(0) (p) By construction, for any k, x k and y k belong to the same tile

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44 For definition of perturbation box, there is a g C 1 -close to f

45 Proof of the connecting lemma:

46 Consider (x i,y i ) in the same tile

47 Consider the last y i in the tile of x 0

48 And consider the next x i

49 Delete all the intermediary points

50 Consider the last y i in the tile

51 Delete all intermediary points

52 On get a new sequence (x i,y i ) with at most 1 pair in a tile x 0 and y n are the original x 0 and y n

53 Pugh gives sequences of points joining x i to y i

54 There may have conflict between the perturbations in adjacent tiles

55 Consider the first conflict zone

56 One jump directly to the last adjacent point

57 One delete all intermediary points

58 One does the same in the next conflict zone, etc, until y n

59 Why can one solve any conflict?

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61 There is no m other point nearby! the strategy is well defined

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