Using Spectral Analysis in the Study of Sarnak s Conjecture

Transcription

1 Using Spectral Analysis in the Study of Sarnak s Conjecture Thierry de la Rue based on joint works with E.H. El Abdalaoui (Rouen) and Mariusz Lemańczyk (Toruń) Laboratoire de Mathématiques Raphaël Salem

2 The Möbius function ( 1) k if n is the product of k distinct µ(n) := primes (k 0), 0 otherwise. The Möbius function µ is multiplicative: µ(n 1 n 2 ) = µ(n 1 )µ(n 2 ) whenever gcd(n 1, n 2 ) = 1.

3 Sarnak s conjecture (2010) For any topological dynamical system (X, T ) with h top (X, T ) = 0, for any f : X C continuous, for any x X, 1 N 1 n N f(t n x) µ(n) N 0.

4 Measurable dynamics point of view Assume that (X, T ) is uniquely ergodic, with a unique invariant probability measure m.

5 Measurable dynamics point of view Assume that (X, T ) is uniquely ergodic, with a unique invariant probability measure m. Which properties of the measure preserving system (X, m, T ) imply the validity of Sarnak s conjecture for (X, T )?

6 Measurable dynamics point of view Assume that (X, T ) is uniquely ergodic, with a unique invariant probability measure m. Which properties of the measure preserving system (X, m, T ) imply the validity of Sarnak s conjecture for (X, T )? Spectral properties

7 Measurable dynamics point of view Assume that (X, T ) is uniquely ergodic, with a unique invariant probability measure m. Which properties of the measure preserving system (X, m, T ) imply the validity of Sarnak s conjecture for (X, T )? Spectral properties Which properties of the Koopman operator U T : f f T on L 2 (m) imply the validity of Sarnak s conjecture for (X, T )?

8 Main tool: KBSZ criterion Lemma (Katai, Bourgain-Sarnak-Ziegler) Assume that (a n ) is a bounded sequence of complex numbers, such that ( ) 1 lim sup lim sup a N pn a qn = 0. N p,q different primes n N Then, for any bounded multiplicative function ν, we have 1 a n ν(n) N 0. N n N

9 Application to Sarnak s conjecture For a n = f(t n x): find sufficient conditions to have ( 1 lim sup lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. p,q different primes n N

10 Application to Sarnak s conjecture For a n = f(t n x): find sufficient conditions to have ( 1 lim sup lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. p,q different primes n N For µ, we can assume that X f dm = 0

11 Application to Sarnak s conjecture For a n = f(t n x): find sufficient conditions to have ( 1 lim sup lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. p,q different primes n N For µ, we can assume that X f dm = 0 1 Indeed, N n N µ(n) 0. N

12 Application to Sarnak s conjecture For a n = f(t n x): find sufficient conditions to have ( 1 lim sup lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. p,q different primes n N For µ, we can assume that X f dm = 0 1 Indeed, N n N µ(n) 0. N Correlation of orbits of T p with orbits of T q, for p, q different large primes.

13 Joinings and disjointness 1 Any limit of N k n N k f ( (T p ) n x ) f ( (T q ) n x ) is of the form f(x 1 )f(x 2 ) dκ(x 1, x 2 ), X X where κ is a joining of T p and T q,

14 Joinings and disjointness 1 Any limit of N k n N k f ( (T p ) n x ) f ( (T q ) n x ) is of the form f(x 1 )f(x 2 ) dκ(x 1, x 2 ), X X where κ is a joining of T p and T q, i.e. a (T p T q )-invariant probability measure on X X with marginals m.

15 Joinings and disjointness 1 Any limit of N k n N k f ( (T p ) n x ) f ( (T q ) n x ) is of the form f(x 1 )f(x 2 ) dκ(x 1, x 2 ), X X where κ is a joining of T p and T q, i.e. a (T p T q )-invariant probability measure on X X with marginals m. J(T p, T q ) := {joinings of T p and T q } J e (T p, T q ) := {ergodic joinings of T p and T q }

16 Joinings and disjointness 1 Any limit of N k n N k f ( (T p ) n x ) f ( (T q ) n x ) is of the form f(x 1 )f(x 2 ) dκ(x 1, x 2 ), X X where κ is a joining of T p and T q, i.e. a (T p T q )-invariant probability measure on X X with marginals m. J(T p, T q ) := {joinings of T p and T q } J e (T p, T q ) := {ergodic joinings of T p and T q } T p and T q are disjoint if J(T p, T q ) = {m m}

17 Disjointness of prime powers Theorem (Bourgain, Sarnak, Ziegler) If for p, q different primes T p and T q are disjoint, then Sarnak s conjecture holds for (X, T ).

18 Spectral disjointness f L 2 0(m). The spectral measure of f associated to the transformation T p is the finite measure on the circle defined by σ f,t p(j) := f, U j T p f L2 (m) T p and T q are spectrally disjoint if for each f, g L 2 0(m), σ f,t p σ g,t q. Lemma Spectral disjointness implies disjointness.

19 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X

20 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X in (X X, T p T q, κ), set F (x 1, x 2 ) := 1 A (x 1 ) m(a), G(x 1, x 2 ) := 1 B (x 2 ) m(b).

21 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X in (X X, T p T q, κ), set Then F (x 1, x 2 ) := 1 A (x 1 ) m(a), G(x 1, x 2 ) := 1 B (x 2 ) m(b). σ F,(T p T q ) = σ 1A m(a),t p

22 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X in (X X, T p T q, κ), set Then F (x 1, x 2 ) := 1 A (x 1 ) m(a), G(x 1, x 2 ) := 1 B (x 2 ) m(b). σ F,(T p T q ) = σ 1A m(a),t p σ G,(T p T q ) = σ 1B m(b),t q.

23 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X in (X X, T p T q, κ), set Then F (x 1, x 2 ) := 1 A (x 1 ) m(a), G(x 1, x 2 ) := 1 B (x 2 ) m(b). σ F,(T p T q ) = σ 1A m(a),t p σ G,(T p T q ) = σ 1B m(b),t q.

24 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X in (X X, T p T q, κ), set Then F (x 1, x 2 ) := 1 A (x 1 ) m(a), G(x 1, x 2 ) := 1 B (x 2 ) m(b). Hence F G in L 2 (κ), σ F,(T p T q ) = σ 1A m(a),t p σ G,(T p T q ) = σ 1B m(b),t q.

25 Spectral disjointness implies disjointness let κ be a joining of T p and T q, and A, B X in (X X, T p T q, κ), set Then F (x 1, x 2 ) := 1 A (x 1 ) m(a), G(x 1, x 2 ) := 1 B (x 2 ) m(b). σ F,(T p T q ) = σ 1A m(a),t p σ G,(T p T q ) = σ 1B m(b),t q. Hence F G in L 2 (κ), and κ(a B) = m(a)m(b).

26 Corollary If for p, q different primes T p and T q are spectrally disjoint, then Sarnak s conjecture holds for (X, T ). conditions for spectral disjointness of different prime powers?

27 Weak limits of powers (Ex. of Chacon)

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37 1/2 of T h n B is in B, 1/2 of T h nb is in T B

38 1/2 of T h n B is in B, 1/2 of T h nb is in T B U h n T w n 1 2 (Id +U T )

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40 1/6 of T 2h n B is in B, 1/6 is in T 2 B, 4/6 is in T B.

41 1/6 of T 2h n B is in B, 1/6 is in T 2 B, 4/6 is in T B. U 2h n T w n 1 6 (Id +4U T + U 2 T )

42 1/6 of T 2h n B is in B, 1/6 is in T 2 B, 4/6 is in T B. In general, U 2h n T w n 1 6 (Id +4U T + U 2 T ) U kh n T = U h n T k w n P k(u T )

43 Spectral disjointness of T and T 2 If T and T 2 were not spectrally disjoint, we could find H 1 L 2 (X, m), stable by U T H 2 L 2 (X, m), stable by U T 2 a continuous measure σ on S 1 such that (H 1, U T ) (H 2, U T 2) ( L 2 (S 1, σ), z ).

44 Spectral disjointness of T and T 2 If T and T 2 were not spectrally disjoint, we could find H 1 L 2 (X, m), stable by U T H 2 L 2 (X, m), stable by U T 2 and by U T a continuous measure σ on S 1 such that (H 1, U T ) (H 2, U T 2) ( L 2 (S 1, σ), z ).

45 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z

46 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z U T φ(z), where φ 2 (z) = z

47 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z U T φ(z), where φ 2 (z) = z wlim U h n T wlim U h n T 2 wlim z h n

48 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z wlim U h n T (Id +U T )/2 U T wlim U h n T 2 φ(z), where φ 2 (z) = z wlim z h n (1 + z)/2

49 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z wlim U h n T (Id +U T )/2 U T wlim U h n T 2 φ(z), where φ 2 (z) = z wlim z h n (1 + z)/2 Id +4U T +UT φ(z)+φ2 (z) 6

50 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z wlim U h n T (Id +U T )/2 Hence 1 + φ 2 (z) 2 U T wlim U h n T 2 φ(z), where φ 2 (z) = z wlim z h n (1 + z)/2 Id +4U T +UT φ(z)+φ2 (z) 6 = 1 + 4φ(z) + φ2 (z) 6 (σ-a.e.)

51 Spectral disjointness of T and T 2 H 1 H 2 L 2 (S 1, σ) U T U T 2 z wlim U h n T (Id +U T )/2 Hence 1 + φ 2 (z) 2 U T wlim U h n T 2 φ(z), where φ 2 (z) = z wlim z h n (1 + z)/2 Id +4U T +UT φ(z)+φ2 (z) 6 = 1 + 4φ(z) + φ2 (z) 6 Impossible! (σ-a.e.)

52 Spectral disjointness of powers For Chacon transformation, T p and T q are spectrally disjoint whenever 1 p < q.

53 Spectral disjointness of powers For Chacon transformation, T p and T q are spectrally disjoint whenever 1 p < q. This result extends to a large class of rank-one transormations, including all weakly mixing constructions with bounded parameters and non-flat towers.

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55 Applying KBSZ without disjointness? No hope to apply the method to non weakly-mixing systems:

56 Applying KBSZ without disjointness? No hope to apply the method to non weakly-mixing systems: If U T has an eigenvalue α 1,

57 Applying KBSZ without disjointness? No hope to apply the method to non weakly-mixing systems: If U T has an eigenvalue α 1, T p and T q share a common eigenvalue α pq,

58 Applying KBSZ without disjointness? No hope to apply the method to non weakly-mixing systems: If U T has an eigenvalue α 1, T p and T q share a common eigenvalue α pq, T p and T q are never disjoint.

59 Applying KBSZ without disjointness? In the case of disjoint powers, we have for f L 2 0(m) ( 1 lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. n N

60 Applying KBSZ without disjointness? In the case of disjoint powers, we have for f L 2 0(m) ( 1 lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. n N What we only need is ( 1 lim sup lim sup f ( (T p ) n x ) f ( (T N N q ) n x ) ) = 0. p,q different primes n N

61 AOP Property

62 AOP Property Definition (X, m, T ) has Asymptotic Orthogonal Powers (AOP) if f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0.

63 AOP Property Definition (X, m, T ) has Asymptotic Orthogonal Powers (AOP) if f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0. By KBSZ criterion, any uniquely ergodic model of a system with AOP satisfies Sarnak s conjecture.

64 AOP for (quasi-)discrete spectrum Theorem If (X, m, T ) has discrete spectrum and is totally ergodic, then it has AOP.

65 AOP for (quasi-)discrete spectrum Theorem If (X, m, T ) has discrete spectrum and is totally ergodic, then it has AOP. includes examples where all powers are isomorphic.

66 AOP for (quasi-)discrete spectrum Theorem If (X, m, T ) has discrete spectrum and is totally ergodic, then it has AOP. includes examples where all powers are isomorphic. extends to quasi-discrete spectrum systems, e.g. T : (x 1,..., x d ) T d (x 1 +α, x 2 +x 1,..., x d +x d 1 ).

67 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0?

68 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0? Enough to consider f and g eigenfunctions associated to irrational eigenvalues α and β S 1.

69 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0? Enough to consider f and g eigenfunctions associated to irrational eigenvalues α and β S 1. For κ J e (T p, T q ), in (X X, T p T q, κ)

70 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0? Enough to consider f and g eigenfunctions associated to irrational eigenvalues α and β S 1. For κ J e (T p, T q ), in (X X, T p T q, κ) f 1 is an eigenfunction associated to α p

71 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0? Enough to consider f and g eigenfunctions associated to irrational eigenvalues α and β S 1. For κ J e (T p, T q ), in (X X, T p T q, κ) f 1 is an eigenfunction associated to α p 1 g is an eigenfunction associated to β q

72 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0? Enough to consider f and g eigenfunctions associated to irrational eigenvalues α and β S 1. For κ J e (T p, T q ), in (X X, T p T q, κ) f 1 is an eigenfunction associated to α p 1 g is an eigenfunction associated to β q if α p β q, then f 1 1 g

73 Proof of AOP for discrete spectrum f, g L 2 0(m), lim p,q, p,q different primes sup κ J e (T p,t q ) X X f g dκ = 0? Enough to consider f and g eigenfunctions associated to irrational eigenvalues α and β S 1. For κ J e (T p, T q ), in (X X, T p T q, κ) f 1 is an eigenfunction associated to α p 1 g is an eigenfunction associated to β q if α p β q, then f 1 1 g But for α and β irrational eigenvalues, there exists at most one pair (p, q) such that α p = β q.

74 The case of rational eigenvalues

75 The case of rational eigenvalues KBSZ criterion ensures orthogonality to any bounded multiplicative function

76 The case of rational eigenvalues KBSZ criterion ensures orthogonality to any bounded multiplicative function Existence of periodic multiplicative functions (Dirichlet characters)

77 The case of rational eigenvalues KBSZ criterion ensures orthogonality to any bounded multiplicative function Existence of periodic multiplicative functions (Dirichlet characters) Those functions are output by rotations on a finite number of points

78 The case of rational eigenvalues KBSZ criterion ensures orthogonality to any bounded multiplicative function Existence of periodic multiplicative functions (Dirichlet characters) Those functions are output by rotations on a finite number of points KBSZ criterion cannot be used when there exist rational eigenvalues.

79 Sarnak for discrete spectrum systems Theorem (Huang, Wang, Zhang (2016)) Let (X, T ) be a uniquely ergodic system with unique invariant measure m. If (X, m, T ) has discrete spectrum, then Sarnak s conjecture holds for (X, T ). (even when there exist rational eigenvalues)

80 Sarnak for discrete spectrum systems An essential argument in the proof: an estimation by Matomäki, Radziwill and Tao 1 sup α S N 1 0 n<n 1 L 0 l<l µ(n + l)α n+l 0 as N, L, L N.