MATH642. COMPLEMENTS TO INTRODUCTION TO DYNAMICAL SYSTEMS BY M. BRIN AND G. STUCK

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1 MATH642 COMPLEMENTS TO INTRODUCTION TO DYNAMICAL SYSTEMS BY M BRIN AND G STUCK DMITRY DOLGOPYAT 13 Expanding Endomorphisms of the circle Let E 10 : S 1 S 1 be given by E 10 (x) = 10x mod 1 Exercise 1 Show that there exists a point x such that E 10 -orbit of x is neither eventually periodic nor dense 15 Quadratic maps Let q µ (x) = µx(1 x) It has fixed points 0 and 1 1 Observe that q 1 µ µ (0) = {0, 1}, q 1 µ (1 1 ) = {1 1, 1 } µ µ µ Lemma 1 Consider the map f : x ax 2 + bx +c, a 0 Then either for all x we have x n as n or f is conjugated to some q µ, µ > 0 Proof By changing coordinates x x if necessary we can assume that a < 0 Then f(x) < x for large x Consider two cases (1) f(x) < x for all x Then x n is decreasing so it either has a finite limit or goes to Since f has no fixed points the second alternative holds (2) f(x) = x has two (maybe coinciding solutions) x 1 and x 2 let x 3 be the solution of f(x 3 ) = f(x 1 ) and x 4 be the solution of f(x 4 ) = f(x 2 ) We have x 1 + x 2 = b 1 a, x 1 + x 3 = b a, x 2 + x 4 = b a Hence (x 3 x 1 )+(x 4 x 2 ) = 2 > 0 So either x a 3 > x 1 or x 4 > x 2 In the first case make a change of coordinates y = x x 1 x 3 x 1 In this coordintes f takes the form y g(y) where g is quadratic with negative leading term Also g(0) = 0, g(1) = g(0) = 0 Thus g = q µ for some µ In the second case make a change of coordinates y = x x 2 x 4 x 2 Exercise 2 Let µ = 4 Then I = [0, 1] is invariant Show that (a) If x I then q4 n (x) (b) Show that the changes of variables y = 2x 1, y = cos z conjugate q 4 I to a piecewise linear map 1

2 2 DMITRY DOLGOPYAT Lemma 2 If 0 < µ < 1 x R then either (1) qµ n (x) or (2) n (x) 0 or (3) qµ(x) n = 1 1, n 2 µ Proof There are several cases to consider (1) x < 1 1, (2) 1 1 < µ µ x < 0, (3) 0 < x < 1, (4) 1 < x < 1, x > 1 We consider case µ µ (2), others are similar In this case by induction x n < x n+1 < 0 Let y = lim n x n Then q µ (y) = lim n =infty x n+1 = y So y is fixed Also y > x 0 > 1 1 since we are in case (2) It follows that y = 0 µ Exercise 3 Complete the proof of Lemma 2 16 Gauss map Let A be 2 2 matrix Since A(0) = 0 and A moves lines to lines, it acts on the projective line Let P A denote this action Coordinatizing projective space, by making the coordinate of a line its intersection with {y = 1} we get P A (x) = ax + b cx + d, x R { } if A = Note that P A P B = P AB ( a b c d Exercise 4 Describe the dynamics of P A When is P A conjugated to a rotation? Let f(x) = {1/x} Hence x 1 = (1/x) a 1 for some a 1 N Thus 1 x = = P a 1 + x x 1 A 1 a 1 Continuing we get where x = P a 1 1 Ax 1 = P a 1 Denoting the elements of M n 1 AP a 2 1 A ( ) 0 1 M n+1 = M n 1 a n+1 M n = we get A n = p n 1, B n = q n 1 and ( ) An p n B n q n ) x 2 = = P Mn x n p n+1 = p n 1 + a n+1 p n, q n+1 = q n 1 + a n+1 q n Exercise 5 q n f n where f n is the n-th Fibonacci number

3 Exercise 6 Find all x such that q n = f n Lemma 3 p n 1 q n q n 1 p n = ( 1) n Proof det(m n ) = COMPLEMENTS 3 n j=1 ( ) 0 1 det 1 a j Lemma 4 pn q n x, n Moreover x p n q n 1 qn 2 Proof We have x pn q n = p n 1 x n+p n q n 1 x n+q n pn q n = x p n 1 q n q n 1 p n q n(q n 1 x n+q n) x = q n(q n 1 x n+q n) (by Lemma 3) (since 0 x n 1) 1 q 2 n Thus for every number x p/q 1/q 2 has infinitely many solutions Lemma 5 Suppose that x is an irrational number satisfying the quadratic equation F(x) = ax 2 + bx + c = 0 with integer coefficients When there is a constant C such that x p/q C q 2 Proof Consider two cases (1) x p/q > 1 Then x p/q > 1/q 2 since q > 1 (2) x p/q 1 Decompose F(z) = a(z x)(z y) where y is the second root of F Then ( ) p (1) F = ap2 + bpq + cq 2 1 q q 2 q 2 since the denominator is a non-zero integer On the other hand y p q y + p q y + x + 1 we have ( (2) p q) F a x p q y p q a x p q ( y + x + 1) (1) and (2) imply the result

4 4 DMITRY DOLGOPYAT Exercise 7 Let x be an irational root of an equation F(z) = a m z m + a m 1 z m 1 + a 1 z + a 0 = 0 Prove that there exists a constant C such that x p/q C/q m so Lemma 3 implies p n p n+1 = ( 1)n+1 p n+1 p n+2 = ( 1)n q n q n+1 q n q n+1 q n+1 q n+2 q n+1 q n+1 p n p ( n+1 = ( 1)n ) q n q n+1 q n+1 q n q n+2 Corollary 6 p 2k /q 2k is increasing, p 2k+1 /q 2k+1 is decreasing In particular p 2k < x < p 2k+1 q 2k q 2k+1 Observe that f(0) is not defined Lemma 7 The orbit of x contains 0 if and only if x is rational Proof If x n = 0 then x = p n /q n Conversely if x is rational then f(x) is rational with smaller denominator Theorem 1 x is eventually periodic if and only if it is a quadratic irrational Proof (1) Observe that if x satisfies a quadratic equation then ax 2 + bx + c then x n satisfies the equation A n x 2 n + B nx n + C n = 0 where A n = ap 2 n 1 + bp n 1q n 1 + c B n = 2ap n 1 p n + b(p n 1 q n + q n 1 p n ) + 2cq n 1 q n C n = ap 2 n + bp nq n + c We claim that A n, B n and C n are uniformly bounded so they must eventually repeat giving eventual periodicity of x Indeed using the notation of Lemma 5 we get C n = qn F(p/q) 2 = qn a x 2 p/q y p/q qn 2 a ( y + 1) x p/q ( x < 1) a ( y + 1) (Lemma 5) Likewise A n a ( y + 1) Exercise 8 Show (eg by induction) that b 2 4ac = B 2 n 4A nc n Hence Thus x is eventually periodic B n b 2 4ac + 4a 2 ( y + 1) 2

5 (2) If x is periodic then for some n Hence x satisfies the equation Next if COMPLEMENTS 5 x = p n 1x + p n q n 1 x + q n q n 1 x 2 + (q n p n 1 )x p n = 0 x = p m 1y + p m q m 1 y + q m with y periodic then y satisfies a quadratic equation an a computation similar to the one done part (1) shows that x satisfies a quadratic equation as well 512 Markov partitions Exercise 9 Show that any linear hyperbolic automorhism of T 2 has a Markov partition Exercise 10 Show that no Markov partition of T 2 gives a full shift Hint Compare periodic points 24 Expansive transformations Exercise 11 Show that no isometry of infinite compact metric space is expansive 28 Applications of topological dynamics Exercise 12 Let T 1, T 2 T N be commuting homeomorphisms of a compact metrix space X Prove that there exist x X and a sequence n k such that d(x, T n k j x) 0 for all j Hint Let F(x) = inf n 1 max j d(x, T j x) Let A ε = {F < ε} Show that A ε is open and dense 33 The Perron-Frobenius Theorem A subset K reals d is called a cone if for any v K, λ > 0 λv K Let K be a convex closed cone satisfying (K1) K ( K) = {0} and (K2) Any vector u in R d can be represented as u = v 1 v 2 with v j K Lemma 8 Any line l containing points inside K intersects the boundary of K

6 6 DMITRY DOLGOPYAT Proof Take two points v, u K K Then l = {z t = tu + (1 t)v} Rewrite z t = v + t(u v) Let z t K for all positive t Since K is a cone, (v/t) + u v K and since K is closed u v K Likewise if z t K for all negative t then v u K By (K2) both inclusions can not be true Let K be a subset of the RP d 1 consisting of directions having representatives in K Define a distance on K as follows If ũ 1, ũ 2 are rays in K choose b ũ 1, c u 2 and let l be the line through b and c Let a and d be the points where l crosses the boundary of K such that a, b, c, d is the correct order on this line and let t be an affine parameter on l Define ( ) (tc t a )(t d t b ) d K (u 1, u 2 ) = ln (t b t a )(t d t c ) To see that this distance correctly defined it is enough to consider the case of the plane since d K only depends on the section of K be the plane containing u 1 and u 2 Now let v 1 and v 2 are two vectors on the boundary of the cone and u is a vector on the line joining v 1 and v 2 Thus u = tv 1 + (1 t)v 2 Now if we consider another line say through v 1 and v 2 then this line crosses the ray through u at a point ū = sv 1 + (1 s)v 2 Denoting by the vector product we get Thus (v 2 + t(v 1 v 2 )) (v 2 + s(v 1 v 2)) = 0 s = v 2 (v 2 + t(v 1 v 2 )) (v 2 + t(v 1 v 2 )) (v 1 v 2) That is, the map between the affine parameters corresponding to different lines is fractional linear Since fractional linear maps preserve the cross ratio d K is correctly defined Exercise 13 Let K be the cone of vecors with non-negative components and K L = {u K : max u i L min u i } i i Show that K L has finite d K -diameter Theorem 2 Let A be a matrix with positive entries Then (a) A has a positive eigenvalue λ; (b) The corresponding eigenvector v is positive; (c) All other eigenvalues have absolute value less than λ; (d) There are no other positive eigenvectors

7 COMPLEMENTS 7 Lemma 9 Let P A be the projecive transformation defined by A Then there exists ṽ K and θ < 1 such that P A (ṽ) = ṽ and for all ũ K (3) dist(p n Aũ, ṽ) Constθ n Proof of Theorem 2 Let ṽ be as in Lemma 9 and v be a positive vector projecting to ṽ Then P A (ṽ) = (ṽ) means that A(v) = λ(v), so A has positive eigenvector Take i such that v i 0 Then λ = (Av) i /v i is positive Also it v is another positive eigenvector then P A (ṽ ) = (ṽ ) contradicting Lemma 9, so there are no other positive eigenvectors We can assume without the loss of generality that v = 1 Next we claim that for for all u R d there exist the limit A n (u) (4) l(u) = lim n λ n and moreover (5) A n u l(u)λ n v Constλ n θ n Indeed assume first that u K denote S(ũ) = Au (this definition u is clearly independent of the choice of the vector projecting to ũ We have n 1 = ( u λ n ) n 1 A n u = u j=0 S(P j A (u)) S(v) j=0 A j+1 u A j u = ( u λ n ) exp n 1 = u S(P j A (u)) [ n 1 j=0 j=0 Since S is Lipshitz we have (6) ln S(P j A (u)) ln S(v) Constθ j ( ln S(P j A (u)) ln S(v))] which proves (4) with l(u) = u j=0 Moreover the exponential convergence of (6) implies A n u = l(u)λ n (1+O(θ n )) Now by Lemma 9 we have A n u A n u = v + O(θn ) This proofs (5) for positive vectors Now (5) in general case follows by (K2) From the properties of limit it follows that l is a linear functional Let L = Ker(l) Then L is d 1 dimensional hypersurface and since l(a(u)) = λl(u), L is A-invariant Thus all other eigenvectors lie in L It follows from (5) that A n L Constλ n θ n so all eigenvalues are less than λθ in asbsolute value

8 8 DMITRY DOLGOPYAT Proof of Lemma 9 For any u in K we have min ij A ij max j u j (Au) i d max ij A ij max u j j so A(u) K L with L = d max ij A ij min ij A ij Now K L is compact in both dist and d K metrics so it is enough to establish (3) for d K Lemma 10 Given D there exists θ < 1 such that for and d for any linear map A : R d R d, such that A(K) K and P A ( K) has diameter less than D in K d(p A ṽ 1, P A ṽ 2 ) θd(ṽ 1, ṽ 2 ) Proof of Lemma 10 Since the definition of d K (ṽ 1, ṽ 2 ) deepends only on the section of K by the plane through v 1 and v 2 it is enough to establish the result for linear map from plane to plane Now on the plane we can use y/x as a coordinate for the point (x, y) In this case K = [0, ] and P A is a fractional linear transformation Also for z 1 < z 2 (7) d K (z 1, z 2 ) = ln ( z2 0 z 1 0 lim z z 1 z z 2 z ) = ln z 2 z 1 = z2 z 1 dz z So we have to prove that any fractional linear transformation s from [0, ] to itself such that s( )/s(0) < e D contracts distance (7) by a factor which depoends only on D Since dilations preserve (7) we can assume that s( ) = 1 thus we have w := s(z) = z + a z + b with a/b > e K We have so that But It follows that dw dz = b a (z + b) 2 dw w = z(b a) dz (z + a)(z + b) z z(b a) (z + a)(z + b) b a b d K (s(z 1 ), s(z 2 )) = z2 < (1 e D ) z 1 w 2 w 1 < 1 e D dw z2 w = z 1 z dw dz w dz z dz z < (1 e D )d(z 1, z 2 ) (3) follows from Lemma 10 and contraction mapping principle

9 COMPLEMENTS 9 Theorem 3 Let A be the matrix with non-negative entries such that for all i, j there exists n such that A n ij > 0 Then there exist lambda > 0 and c such that (a) λe 2πir/c are eigenvalues of A (b) A c has c linearly independent positive eigenvectors with eigenvalues λ c (c) All other eigenvalues have absolute value less than λ (d) There are no other positive eigenvectors Proof Let Z ij = {n : A n ij > 0 Observe that Z ij + Z jk Z ik in particular Z ii are semigroups Lemma 11 If a, b Z ii let q = gcd(a, b) Then for large N Nq Z ii Proof We have q = ma nb for some m, n N Write N = Lb + k then Nq = (Lq n)b + mka Corollary 12 Let c i be the greatest common divisor of all numbers in Z ii Then Z ii = c i N finitely many numbers Proof Let c (1) i be any number in Z ii Then c (1) i N is in Z ii and if there are no other numbers in Z ii then we are done Otherwise if b (1) i Z ii c (1) i N then let c (2) i = gcd(c (1) i, b (1) i ) Then c (2) i N a finite set is in Z ii and if there are no other numbers in Z ii then we are done Otherwise if b (1) 2 Z ii c (2) i N then let c (3) i = gcd(c (2) i, b (2) i ) etc Since c (k) i is decreasing it must stabilize Lemma 13 (a) Any two numbers in Z ij are comparable mod c i (b) Any two numbers in Z ji are comparable mod c i (c) c i = c do not depend on i Proof (a) Let n, n Z ij, n Z ji, m Z jj then n + n = n + n = n + n + m mod c i This prove (a) and (c) (b) is similar to (a) Let V m = {i : Z 1i = m mod c}, L m = span(e i, i V m } Then There exists N such that A N ij > 0 if and only if i and j belong to the same V m Thus by Theorem 2 A N has uniques eigenvector v 1 on L 1 with positive eigenvalue ν Let v i = A i 1 v 1 Since A commutes with A N v i are eigenvectors of A N and since A has non-negative entries v i are non-negative Also v i L i mod c In particular by uniqueness of the positve eigenvector A c v 1 = νv 1 Let λ = ν 1/c Then A c (v i ) = λ c (v i ) Let w r = c e 2πijr/c j=1 v λ j j Then Aw r = e 2πjr/c w r Finally by Theorem 2 all other eigenvalues of A N are less than ν = λ N in absolute value Theorem 3 is proven

10 10 DMITRY DOLGOPYAT 410 Weak Mixing Let U be a unitary operator and φ be a unit vector Let R n =< U n φ, φ > Lemma 14 There is a probability measure σ φ on [0, 1] such that R n = e 2πiu dσ(u) Proof For 0 < ρ < 1 let f(u, ρ) = m=0 n=0 R n m e 2πi(m n) ρ n+m Since R n m =< U n φ, U m φ > it follows that f(u, ρ) = n ρ n e 2πiu U n φ 2 is a real positive number Estimateing terms in () by their absolute values we get f (1 ρ) 2 Thus f L (du) L 2 (du) Let us examine its Fourier series We have f = k R k e 2πiu m ρ k+2m = k e 2πk R k ρ k (1 ρ 2 ) 1 Consider measures We have Hence as ρ 1 dσ ρ = (1 ρ 2 )f(u, ρ)du e 2πiu dσ ρ (u) = R k ρ k e 2πiu dσ ρ (u) R k Since linear combinattions of e 2πiu are dense in C(S 1 ) it follows that for any continuous function A A(u)dσ ρ (u) σ(a) Exercise 14 Consider a full two shift with Bernoulli measure (that is the measure of each cylinder of size n is (1/2) n ) and let φ(x) = 2(I x0 =1,x 1 =1 I x0 =0,x 1 =0) Find the spectral measure of φ with respect to U(φ) = φ(σx)

11 COMPLEMENTS 11 Exercise 15 Let Φ n (u) = 1 n 2 n j,k=1 e2πi(j k)u Show that Φ n (u) 0, if u 0 (and Φ n (0) = 1) 51 Expanding endomorphisms Let f : S 1 S 1 be a map such that f θ 1 for some θ < 1 We call such map expanding Given any diffeomorphism of S 1 let ḡ : R R be its lift, that is π ḡ = g π, where π : R S 1 is the natural projection Define deg(g) = ḡ(x + 1) ḡ(x) (this number is easily seen to be independent of x and the lift ḡ) Let L be the space of maps τ : R 1 R 1 which are lifts of degree 1 maps That is τ x is periodic We endow L with the distance d( τ 1, τ 2 ) = sup τ 1 (x) τ 2 (x) = max τ 1 (x) τ 2 (x) x R x [0,1] Lemma 15 Let f be an expanding map and g be a map of the same degree Then given any two lifts f and ḡ there is unique τ L such that f τ = τ ḡ Proof τ must satisfy τ(x) = f 1 ( τ(ḡ(x))) Define K : L L by K( τ)(x) = f 1 ( τ(ḡ(x))) Since f is expanding it follows from the Intermidiate Value Theorem that f 1 (x 1 ) f 1 (x 2 ) θ x 1 x 2 Hence d(k( τ 1 ), K( τ 1 )) θd( τ 1, τ 2 ) Now the result follows from the contraction mapping principle Theorem 4 Any two expanding maps of the same degree are topologically conjugated Proof Let f 1 and f 2 be lifts of two expanding maps By Lemma 15 there are maps τ 1, τ 2 such that τ 1 f 1 = f 2 τ 1 and τ 2 f 2 = f 1 τ 2 Let τ = τ 2 τ 1 Then τ f 1 = f 1 τ By uniquesness part of Lemma 15 τ 2 τ 1 = id Likewise τ 1 τ 2 = id Exercise 16 Show that this conjugacy is typically NOT C 1