Infinitesimal twists along orbits. Hiromichi Nakayama (Hiroshima University)

Transcription

1 Infinitesimal twists along orbits Hiromichi Nakayama (Hiroshima University)

2 Contents I. Invariant foliations II. III. IV. Projectivized bundle Model case Properties of PSL(2,R) V. Twist along orbits VI. VII. Ruelle invariant Eyes of typhoons

3 1, Invariant foliations In Le Calvez s talk In this talk foliations dynamically transverse to the isotopy foliations invariant under diffeomorphisms f maps each leaf onto a leaf

4 Examples of invariant foliations f : T 2 T 2 ; a diffeomorphism Anosov Diffeom. Irrational transf. f x = 2 1 y 1 1 x y f (x, y) = (x +, y + ) / Q,, Q

5 Assumption In this talk, we will respect our attention to diffeomorphisms of the torus T 2 Remark (my original interest) When can a homeomorphism of R 2 without a fixed point be embedded in a flow? (flowability) leaf preserving homeoms foliation preserving diffeoms

6 The other examples (N-, F. Le Roux)

7 2, Projectivized bundles TT 2 : the tangent bundle of T 2 PT 2 = {(z,v) TT 2 ; v 0}/v ~ kv (k 0) projectivized bundle f : T 2 T 2 : a diffeomorphism Df : TT 2 TT 2 : the derivative of f Pf : PT 2 PT 2 : the diffeom induced from Df i.e. (z,[v]) PT 2, Pf (z,[v]) = (z,[df (v)])

8 Lemma. f is tangent to a C foliation There is a C section : T 2 PT 2 such that Pf ((z)) = ( f (z)) How to find such a section?

9 One of candidates is minimal sets of Pf (i.e. closed Pf - invariant sets whih is minimal w.r.t. the inclusion)

10 3, Model case Def. f is tangentially distal iff inf{ Df n (v) ; n Z} 0 for any v 0 Theorem. (Shigenori Matsumoto, N-, 1997) If f : T 2 T 2 is tangentially distal and minimal (i.e. all orbits dense), then there is a C 0 1- dim foliation tangent to f.

11 Sketch of proof 1) To find an invariant C 0 section for Pf. routine work 2) To find a tangent C 0 foliation. (not always uniquely integrable)

12 4, Properties of PSL(2,R) Cross ratio for straight lines (P 1,P 2,P 3,P 4 ) = P 1P 3 P 3 P 2 P 1 P 4 P 4 P 2 The cross ratio is invariant under SL(2,R) because SL(2,R) preserves the area of the triangles OP 1 P 3 OP 3 P 2 OP 1 P 4 OP 4 P 2

13 How to use the cross ratio. Here we consider the case when Pf has two minimal sets M 1, M 2. i.e. there are closed invariant sets M i (i =1,2) in PT 2 which are minimal among such closed invariant sets. Lemma. For any fiber {z} P 1, either M 1 ({z} P 1 ) or M 2 ({z} P 1 ) consists of a single point.

14 Proof of Lemma (by absurdity) small bounded below cross ratio P 1 P 3 P 3 P 2 P 1 P 4 P 4 P 2 = Q 1 Q 3 Q 3 Q 2 Q 1 Q 4 Q 4 Q 2 0 contradiction By the minimality of M 2, Q 1 Q 3 approaches to 0. On the other hand, Q 2 Q 3 and Q 1 Q 4 are bounded below.

15 Then we can 1) find a continuous section between M 1 and M 2. (N - and Noda, 2005) 2) control all the orbits between M 1 and M 2 by a single orbit between M 1 and M 2. (N-,2007) 1) 2)

16 5, Infinitesimal twist along orbits f : T 2 T 2 ; a C 2 diffeomorphism isotopic to id f t is its isotopy ( f 0 = id, f 1 = f ) T 1 T 2 : the unit tangent bundle of T 2 f : T 1 T 2 T 1 T 2 ; a diffeomorphism Df (v) defined by f (z,v) = f (z), z Df (v) z f (z,v)

17 6, Ruelle invariant f : T 2 R T 2 R : the lift of f with respect to the isotopy f t f n (z,0) (z) = lim n if it exists n Def. μ :an f - invariant prob. measure of T 2 The Ruelle invariant R μ ( f ):= T 2 (z)dμ "Average of the twist along the orbits"

18 Another def of Ruelle invariant G = SL(2,R) G : the universal cover of G Let A be an element of G. i.e. A(t) G, A(0) = e (0 t 1) We will define the angle of A Polar decomposition of A(t) A(t) = O(t)S(t) where S(t) : (positive) symmetric matrix O(t) : orthogonal matrix (S(t) = T A(t)A(t),O(t) = A(t)S t 1 ) by Ruelle

19 (t):= (the angle of O(t)) S 1 O(t) = cos(t) sin(t) sin(t) cos(t) (A) R; the variation of (t) For n Z + A(z,n):= t a Df (z) nt det Df nt (z) G (0 t 1) Lemma. (z) = lim n (A(z,n)) n

20 The Ruelle invariant R μ ( f ) = T 2 lim n (A(z,n)) n dμ Remark. The Ruelle invariant can be defined for symplectic matrices (by Ruelle).

21 Another def of Ruelle invariant μ : an f - invariant measure of T 2 : T 1 T 2 T 2 ; the projection Then there is a measure on T 1 T 2 s.t. (f ) * = and * = μ : T 1 T 2 R defined by f (z,v) = ( f (z),v + f (z,v)) for (z,v) T 2 R Theorem (T. Inaba and N-, 2004) R μ ( f ) = T1 T 2 f d by Inaba, N- T 2 R T 1 T 2 f T 2 R f T 1 T 2 TT 2 T 2 Df TT 2 f T 2

22 Outline of proof By the disintegration theorem, there is a prob. measure z on each fiber {z} S 1 s.t. (z,v) d = dμ (z,v) d T 1 T 2 T 2 {z}s 1 z for a continuous function. lim 1 n n lim 1 n n T 1 T 2 T 1 T 2 T 1 T 2 f n (z,v) d = n1 lim 1 n n f ( f i (z,v)) d = i= 0 f (z,s)d = T 2 (z)dμ T 2 dμ lim 1 n n {z}s 1 T 2 f n (z,v) d z f n (z,0) dμ

23 Theorem (Shigenori Matsumoto and N-, 2002) f : T 2 T 2 ; a C - diffeomorphism isotopic to id there is an f - invariant prob. measure μ such that R μ ( f ) = 0 Key lemma. f : T 2 T 2 ; a C - diffeomorphism isotopic to id. f (z,v) = 0 for some point (z,v) T 1 T 2

24 Proof of (Key lemmaheorem) There is (z n,v n ) s.t. f n (z n,v n ) = 0 Thus n := 1 n n-1 f ( f i (z n, n )) = 0 i= 0 n1 i= 0 ( f i (z n,v n )) where is the Dirac measure at f i (z n,v n ) : accumulation of n Then R μ ( f ) = f (z,s) d = lim n f (z,s) d = lim n f (z,s) 1 n1 ( f i (z n n,v n )) = lim n 1 n n i= 0 n1 f ( f i (z n,v n )) = 0 i= 0

25 7, Eyes of typhoons Eye from When does an Eye of typhoon turn out? strong twist + slow move How to describe this situation?

26 For t R, f t = f t[t ] o f [t ] f t :R 2 R 2 ; the lift of f t with respect to f t For x R 2, K n (x) = max (Df ) (Df ) n y n x ; x y, y R 2 y x S n (x) : symmetric part of the polar decomposition for Df n (x) for x R2

27 Theorem. f : T 2 T 2 ; a C - diffeomorphism isotopic to id s.t. f has no periodic point and the Ruelle invariant R μ ( f ) > 0 3 N := +1 R μ ( f ) If S n (x) id <1/2 for any x and 1 n N, then there is a point x 0 s.t. 2 32K n (x 0 ) d(x 0, f n (x 0 ))