The Moebius function and continuous extensions of rotations

Let $f\colon \mathbb{T}\to \mathbb{R}$ be of class $C^{1+\delta}$ for some $\delta>0$ and let $c\in\mathbb{Z}$. We show that for a generic $\alpha\in\mathbb{R}$, the extension $T_{c,f}\colon \mathbb{T}^2\to\mathbb{T}^2$ of the irrational rotation $Tx=x+\alpha$, given by $T_{c,f}(x,u)=(x+\alpha, u+cx+f(x))$ ($\bmod\ 1$) satisfies Sarnak's conjecture.


Introduction
Recall that the Möbius function µ : N → {−1, 0, 1} is a multiplicative function 1 defined as µ(p 1 · . . . · p k ) = (−1) k for distinct prime numbers p j , µ(1) = 1 and 0 otherwise. Its importance is reflected in the fact that the prime number theorem 2 is equivalent to the condition k≤x µ(k) = o(x) and the Riemann hypothesis is equivalent to the condition k≤x µ(k) = O ε (x 1 2 +ε ) for any ε > 0 * Research supported by Narodowe Centrum Nauki grant DEC-2011/03/B/ST1/00407. 1 Recall that ν : N → Z is said to be multiplicative if ν(mn) = ν(m)ν(n) for n, m relatively prime. 2 Recall that the prime number theorem states that π(x) = x ln x + o( x ln x ), where π(x) is the number of primes less than x.
(when x → ∞). The Möbius function appears to behave rather randomly and this statement was formalized in the following conjecture of Sarnak: Conjecture 1 ( [26]). Let X be a compact metric space and let T : X → X be a homeomorphism of zero topological entropy. Let x ∈ X, g ∈ C(X). Then n≤N g(T n x)µ(n) = o(N ). (1) Whenever condition (1) is true for some T for each x ∈ X and each g ∈ C(X), we say that Sarnak's conjecture holds for T or that T is disjoint from the Möbius function. 3 Sarnak's conjecture is known to hold in several situations, including rotations [9], nilsystems [15], horocycle flows [7], large class of rank one maps [6,2] and certain subclasses of dynamical systems generated by generalized Morse sequences [19], including the classical Thue-Morse system: [1,5,8,14,16,24] and the dynamical system generated by the Rudin-Shapiro sequence [25].
One of the most fruitful tools used for proving disjointness with the Möbius function turns out to be the following orthogonality criterion of Katai-Bourgain-Sarnak-Ziegler (we will refer to it as KBSZ criterion).
In order to use this theorem for proving Conjecture 1 for a given homeomorphism T : X → X, one takes F (n) := g(T n x) for n ∈ Z, x ∈ X and g ∈ C(X). (4) Notice that the expression (2) (for F (n) given by (4)) takes the form where x ∈ X. It follows that we are interested in the limit measures ρ = lim k→∞ 1 N k n≤N k δ (T r ×T s ) n (x,x) which are clearly T r × T s -invariant. Therefore, to prove disjointness of T with the Möbius function, it suffices to show that the following holds for T , for r, s relatively prime: (a) The ergodic components of T r × T s are pairwise disjoint closed sets filling up the whole space. 4 (b) The ergodic components are uniquely ergodic (this implies that all points are generic for T r × T s for relevant invariant measures).
(c) There exists a linearly dense set 5 of continuous functions F ⊂ C(X) such that for each g ∈ F, g • T = g, we have X 2 g ⊗ g dρ = 0 for any T r × T sergodic measure ρ, whenever r, s are sufficiently large.
We will use the strategy (a), (b), (c) to study disjointness with the Möbius function in the following setting. Denote by T = R/Z = [0, 1) the additive circle and consider where c ∈ Z and cx + f (x) is a lift of a continuous circle map, i.e. f : R → R is continuous, periodic of period 1, and c is the degree of the map in question.
In other words, we consider continuous case of the classical Anzai skew product extensions of a rotation on the circle [3]. Liu and Sarnak in their recent paper [23] proved the following. 23]). Assume that in (5), f : R → R is an analytic periodic function of period 1. Assume additionally that | f (m)| e −τ |m| for some τ > 0. Then T c,f satisfies Conjecture 1.
The technical condition on the Fourier transform, namely | f (m)| e −τ |m| , seems to be necessary for the methods of [23] to work. On the other hand, there is no condition on α. Moreover, for some αs, the result is obtained using Theorem 1.0.1. Under some additional assumptions, also a quantitative version (i.e. concerning the speed of convergence to zero in (1)) of Theorem 1.0.2 is proved in [23]. This is achieved by treating the problem in a more direct way than applying Theorem 1.0.1.
A natural question arises whether the strong assumptions on f in Theorem 1.0.2 can be relaxed. We do so in the main result of the paper (Theorem 1.0.3 below) to obtain Sarnak's conjecture for each sufficiently smooth f at the cost of reducing "for each α" in Theorem 1.0.2 to "for a generic α". Hence, our result can be viewed as a complementary to Liu-Sarnak's result. Theorem 1.0.3. Let f : R → R be a function of class C 1+δ for some δ > 0, periodic of period 1. Let c ∈ Z. Then for a generic set 6 of α, the automorphism T c,f of T 2 given by (5) satisfies Conjecture 1. 4 The proof of the main result of the paper (Theorem 1.0.3 below) says also that the ergodic decomposition will be the same as the decomposition into minimal components which seems to be a fact of independent interest. In case of continuous compact group extensions the existence of the decomposition into minimal components is guaranteed by a result of Auslander [4] and Ellis [10] on distal systems. 5 Notice that we do not aim to prove (2) (for F (n) given by (4)) for each g ∈ C(X) (and each x ∈ X) -as a matter of fact (2) (for F (n) given by (4)) is not satisfied for some continuous functions already for an irrational rotation; we provide examples in Appendix, see Proposition 3.3.1. Our aim is to prove (2) for a linearly dense set of g ∈ C(X) (and each x ∈ X), as it implies that (3) holds for each g ∈ C(X). 6 The question of whether an analogous result to Theorem 1.0.3 is true for f which is only assumed to be continuous, remains open. We recall that under the continuity assumptions on f , even, it is open whether f is not a quasi-coboundary for a generic set of α.
Before we give the proof of Theorem 1.0.3, we will first show that Conjecture 1 holds in the following two natural cases: the case of an arbitrary continuous extension of a rational rotation (see Proposition 2.3.3 below) 7 and to get an independent proof of the purely affine case (i.e. when f = 0) for each α, first proved in a larger context in [23]. 23]). For any α, γ ∈ R and for any c ∈ Z, the automorphism (x, y) → (x + α, cx + y + γ) satisfies Conjecture 1.
Let us now describe how we check conditions (a), (b) and (c). Let α ∈ Q and let T x = x+α. In either setting (purely affine or with a non-trivial perturbation) the base rotation T r × T s is the same. Its ergodic components are obtained by taking the partition of T 2 into closed invariant sets A c1 = A r,s c1 := {(x, y + c 1 ) ∈ T 2 : sx = ry}, c 1 ∈ [0, 1 r ). In fact, these sets are at the same time the minimal components of T r × T s and they are uniquely ergodic. Thus, we are interested in the action of (T c,f ) r × (T c,f ) s on the sets I c1 = I r,s c1 := A c1 × T 2 . It turns out that this is equivalent to dealing with extensions of T by the following T 2 -valued cocycles: . This is the situation we aim for in course of the proof of Theorem 1.0.3. For a generic α we indeed obtain ergodicity of ψ c1 for all c 1 -for the details see Corollary 2.5.8. Statement (b) follows from the fact that we deal with compact group extensions of rotations. Finally, we show that also (c) holds: given a non-trivial character χ ∈ T 2 , we prove that for r and s relatively prime and large enough, the corresponding integrals of χ ⊗ χ are zero. In case of Theorem 1.0.4, the problem is in a sense more delicate -some of the sets I c1 are too large to be the ergodic components and they need to be partitioned further into smaller subsets. This refined partition will be however satisfying (a). Condition (b) is proved in the same way as in Theorem 1.0.3. To prove that also (c) holds, we take again F = T 2 . For reader's convenience, we added Appendix collecting some necessary facts concerning cocycles.

On the strategy of the proofs
Our approach to proving disjointness with the Möbius function was described in (a), (b) and (c). We will now make some more comments on this method. Recall that in view of Theorem 1.0.1, for a linearly dense set F of g ∈ C(X) and each x ∈ X, what we want to prove is for distinct, sufficiently large prime numbers r, s. When this is realized through (a), (b) and (c), we prove more. Namely, for each g ∈ F and for sufficiently large primes r = s, the following holds for all x, y ∈ X: (the condition on r, s is independent of the choice of x and y). (i) In view of the above discussion, in order to prove that Sarnak conjecture holds for T , it suffices to check conditions (a), (b) and (c) for T k for some k ≥ 1. Indeed, by applying (7) to T k , we obtain which implies that (6) holds for T .
(ii) Notice that whenever the conditions (a), (b) and (c) are satisfied for some homeomorphism T then they are also satisfied for T −1 . However, (iii) It is unclear how to prove directly that if T k for some k ∈ Z \ {0} is disjoint from µ then also T is disjoint from µ, or even to show that if the assumptions of the KBSZ criterion are satisfied for T k for some k ∈ Z \ {0} then they are satisfied for T .
Let now T ϕ : X × T → X × T be a continuous circle group extension of a homeomorphism T : X → X by ϕ : X → T.
Remark 2.1.2. Suppose that T ϕ satisfies Sarnak's conjecture. Then for any k ≥ 1 also T kϕ satisfies Sarnak's conjecture as it is a topological factor of T ϕ .
In the case of affine extensions of rotations, Remark 2.1.1 (i) and Remark 2.1.2 are complementary in the following sense. Let T (α) stand for the rotation

General remarks
From now on, our assumption will be that r, s ≥ 3 are odd and relatively prime.
Let α ∈ Q and denote by T : . It is the same as the ergodic decomposition. Moreover, (T r × T s )| Ac 1 is topologically isomorphic to T . The isomorphism is given by where a, b ∈ Z are such that ar + bs = 1.
Proof. Notice first that the sets A c1 are closed and invariant under T r × T s and ∈ Z be such that ar + bs = 1 and let W be given by (9).
have r(ax + by) = x and s(ax + by) = y. Therefore, W is bijective. Moreover, W preserves the measure, as rotations are uniquely ergodic.
For a measurable function ψ : T → T, let Ψ : T 2 → T 2 be given by Then clearly the automorphism . The isomorphism is given by with a, b ∈ Z such that ar + bs = 1.
Proof. Fix c 1 ∈ [0, 1 r ), let a, b ∈ Z be such that ar + bs = 1 and let V be given by (11). Then Moreover, r(ax + by) = x and s(ax + by) = y, which completes the proof.
Moreover, art + bst = t for a, b ∈ Z such that ar + bs = 1.

Continuous extensions of rational rotations
We will now show that all continuous extensions of rational rotations satisfy Sarnak's conjecture.
for sufficiently large prime numbers r = s, whenever χ = η ⊗ 1 T (η ∈ T) and Proof. Note that for each m ≥ 1, Let χ(x, y) = e 2πi(ax+by) for some a, b ∈ Z, with b = 0 by assumption. Hence If exactly one of the numbers f (x 1 ) and f (x 2 ) is irrational then the result follows from Weyl's criterion. If both f (x 1 ) and f (x 2 ) are irrational then there is at most one pair (r, s) of relatively prime numbers such that rf (x 1 ) − sf (x 2 ) ∈ Q and we can again make use of Weyl's criterion, this time for r, s sufficiently large.
Remark 2.3.2. Notice that the above proof says more. Namely, the convergence in (12) does not depend on y 1 , y 2 .
Given n ≥ 1, we take n such that n = qn + j with 0 ≤ j < q. Then, for each χ ∈ T 2 , we have where the first coordinates of the points R rj (x, y), R rj (x, y) belong to the finite for sufficiently large prime numbers r = s, we need to show that  (13), provided that By seting n = dn + k with 0 ≤ k < d, we obtain qn + j = qdn + (qk + j) and rewriting the sum to the form the result again follows from Sarnak's conjecture for finite systems.
Recall that given r, s ∈ N (odd and relatively prime) and c 1 ∈ [0, 1 r ), we have For where l 0 = l r,s,c1 0 is the smallest positive integer such that and k 0 = k r,s,c1 0 ∈ Z is such that where ϕ c1 : T → T 2 is given by ϕ c1 (x) = (r 2 x + rγ, s 2 x + sc 1 + sγ). The isomorphism is given by it follows that U given by (17) is indeed the required isomorphism. Notice that θ is continuous, whence U is a homeomorphism.
The decomposition of T 4 into minimal components for (T r ×T s ) Ψ is the same as the decomposition into ergodic components. It consists of sets of the form I c1 for c 1 ∈ αQ + Q and J c1,c2 for . Moreover, on each such component, (T r × T s ) Ψ is uniquely ergodic. In particular, each point in T 4 is generic (for a relevant invariant measure).
Then the solution to (18) is given by In particular, all measurable solutions to (18) are continuous. It follows by Remark 3.2.7 that the decomposition into minimal components for (T r × T s ) Ψ is the same as the decomposition into ergodic components.
We will now describe the ergodic (i.e. minimal) components of each T ϕc 1 and show that they are uniquely ergodic. Suppose that χ • ϕ c1 is a coboundary. It follows from (19) (recall that r = s are coprime) that r 2 |B and s 2 |A, and therefore A = A s 2 and B = −A r 2 for some A ∈ Z. Hence, the second part of condition (19) takes the form Having this in mind, we consider two cases: In case (i), it follows immediately from the first part of the proof that (T r × T s ) Ψ | Ic 1 is ergodic and minimal. Unique ergodicity follows from Proposition 3.2.8.
We consider now case (ii). We will describe characters χ, such that (18) has a (measurable and continuous) solution. In view of and (15) and (20), A = al 0 for some a ∈ Z. Therefore, a measurable solution ξ : T → T to (18) exists precisely for the characters χ of T 2 of the form Denote the set of such characters by Γ = Γ top . It is easy to see that ann Γ = (u, v) ∈ T 2 : l 0 s 2 u = l 0 r 2 v .
We claim that are minimal components of T ϕc 1 . Indeed: • each J r,s c1,c2 is closed, • by (16), each J r,s c1,c2 is T ϕc 1 -invariant, • c2 J r,s c1,c2 = T 3 (notice that the projection of J r,s c1,c2 onto the first two coordinates is equal to T 2 ), described above, we need to find the preimage of J r,s c1,c2 via U c1 • V c1 . We have whence, by Remark 2.2.3, Therefore (U c1 V c1 ) −1 J r,s c1,c2 = J r,s c1,c2 , which completes the proof.
Remark 2.4.4. The sets I r,s c1 are translates of I r,s 0 , which is a subgroup of T 4 . Therefore, the conditional measures given by the ergodic decomposition of (T r × T s ) Ψ , which are supported on the sets I c1 for c 1 such that (s − r)γ − rc 1 ∈ Qα + Q, are translates of Haar measure on I r,s 0 . For c 1 with (s − r)γ − rc 1 ∈ Qα + Q, the sets J r,s c1,c2 are translates of J r,s c1,0 , which is a subgroup of T 3 . As before, the conditional measures given by the ergodic decomposition of T ϕc 1 , are translates of Haar measure on J r,s c1,0 . Notice that V c1 carries a translate of Haar measure to Haar measure. Therefore, also the conditional measures given by the ergodic decomposition of (T r × T s ) Ψ , which are supported on the sets J r,s c1,c2 , are translates of the corresponding Haar measures. Proof. Let d stand for the metric on G. Since χ is uniformly continuous, there exists δ > 0 such that d(x, y) < δ implies |χ(x) − χ(y)| < 1 4 . Since {y 1 , . . . , y n } is a δ-net, it follows that |χ(G) − 1| < 1 2 . This is however possible only when χ ≡ 1, as χ(G) is a closed subgroup of S 1 . Lemma 2.4.8. Let χ ∈ T 4 be a non-trivial character. If r, s ∈ N are large enough then χ| Ic 1 is not constant for c 1 ∈ [0, 1 r ). Proof. Fix 1 ≡ χ ∈ T 4 and let δ > 0 be as in Lemma 2.4.7. In view of Lemma 2.4.7, we need to show that whenever r, s are large enough then there exists a δ-net of T 4 contained in I r,s c1 for c 1 ∈ [0, 1 r ). Since I r,s c1 is a translate of I r,s 0 and the third and fourth coordinate in I 0 is arbitrary, it suffices to prove that for r, s sufficiently large we can always find a δ-net of T 2 contained in the set A r,s 0 . To complete the proof it suffices to consider sets of the form i s , j r : 0 ≤ i < s, 0 ≤ j < r .
Lemma 2.4.9. Let χ ∈ T 4 be a non-trivial character. If r, s ∈ N are large enough then χ| J r,s c 1 ,c 2 is not constant for c 1 ∈ [0, 1 r ) such that c 1 ∈ αQ + Q and c 2 ∈ [0, 1 r 2 ). Proof. Fix 1 ≡ χ ∈ T 4 and let δ > 0 be as in Lemma 2.4.7. Since J r,s c1,c2 is a translation of J r,s c1,0 , by Lemma 2.4.7, it suffices to prove that for r, s sufficiently large there exists a δ-net of T 4 in J r,s c1,0 . Notice first that the projection of J r,s c1,0 onto the first two coordinates is equal to A r,s c1 . Indeed, for any z ∈ T the equation l 0 s 2 u = l 0 r 2 v + z has a solution (u, v) as T is an infinitely divisible group. By the proof of the previous lemma, we can find a δ-net of T 2 Therefore, whenever r, s are sufficiently large, the set 10 This completes the proof. 10 Notice that since l 0 ≥ 1, the condition on r, s is independent of c 1 .  They are at the same the ergodic components which are moreover uniquely ergodic, and we conclude as previously.

Generic case -compact group extensions
Let f : R → R (periodic of period 1) be in L 2 (T). Denote by the Fourier expansion of f . Recall that our goal is to prove disjointness of with the Möbius function µ for a generic set of α (under some additional assumptions on f ).
Recall the following result. . Suppose that f (x) = n∈Z f (n)e 2πinx is in C 1+δ (T) for some δ > 0, and has zero mean. Denote by T : T → T the irrational rotation x → x + α. Assume that for a sequence (p n /q n ) n∈N of rational numbers we have and Then for each λ ∈ S 1 the cocycle λe 2πif (·) is not a T -coboundary.
We will now prove a modified version of the above theorem. It will be our main tool in course of the proof of Theorem 1.0.3. Theorem 2.5.2. Suppose that f (x) = n∈Z f (n)e 2πinx is in C 1+δ (T) for some δ > 0, and has zero mean. Denote by T : T → T the irrational rotation x → x + α. Let r, s ∈ N (r > s) be relatively prime. Assume that (p n /q n ) n∈N is a subsequence of convergents of α in its continued fraction expansion such that and Then for each λ ∈ S 1 , h ∈ R and A, B ∈ R with A 2 + B 2 = 0 the cocycle λe 2πi(Af (r) (rx)+Bf (s) (sx+h)) is not a T -coboundary.
Proof. Fix A, B ∈ R with A 2 + B 2 = 0, relatively prime numbers r > s and For any k ∈ N we have Therefore, 1 − e 2πinα e 2πin(sx+h) . (27) Suppose first that A · B = 0. We may assume without loss of generality that A = 0 and B = 0. Then, for all n ∈ N, by (27), we have Therefore, since and for all n ∈ N, m ∈ Z. It follows immediately by (29) and (30) that for n sufficiently large Now, (31) and (32) and the assumptions (24) and (26) imply that for n sufficiently large In view of Theorem 2.5.1, this completes the proof in case A · B = 0. Suppose now that A · B = 0. Applying the fact that for any absolutely summable sequence (y n ) n∈Z to the second summand in formula (27), we obtain For m ∈ N, let B m := {n ∈ N : m q n }. We will consider the following cases: (i) B r or B s is infinite, (ii) both sets B r and B s are finite.
We will cover first case (i). Without loss of generality we may assume that B r is infinite. Suppose that λe 2πi(Af (r) (rx)+Bf (s) (sx+h)) is a coboundary.
Hence, dividing the expression from (34) by the one from (35), we conclude that is a coboundary as well.
We will show that this is impossible in view of Theorem 2.5.1. Indeed, we have We claim that for n ∈ B r , To prove (43), we will estimate |a rqn | from below and m≥1 |a mrqn | from above in an appropriate way. We begin by estimating |a rqn |. We have where the latter inequality follows from (28) and is valid for n sufficiently large. It follows by (25) that for n ∈ B s , We will now estimate |1 − e 2πiqn r s α |. Notice that Therefore and by (47), for such n, we have Since for all x, y ∈ R and all k ∈ N |e 2πikx − e 2πiky | ≤ k|e 2πix − e 2πiy |, it follows from (48) that Since q n is a denominator of α and s|q n , we have |c − qn s α| > |p n − q n α| for all 0 ≤ c ≤ qn s (see(52) in Section 3.1). Using this inequality and (47), we obtain By the first two lines of (48) and (49), we have Suppose that s|[q n α]. Then (50) implies 1 < π 2s · π 2 , i.e. s < (π/2) 2 . This is however impossible, as s ≥ 3. Therefore s [q n α], which implies s [q n α]r, i.e.
It follows that This and (45), (46) imply that for n sufficiently large We will estimate now m≥1 |a mrqn |. By (24) and (25) we have Using this estimate and (51), we obtain and (43) follows. Notice that (51) together with (26) implies that also (44) is true. By Theorem 2.5.1, we conclude that λe 2πiF (x) cannot be a coboundary, which completes the proof in case (ii).
Lemma 2.5.4 (cf. Lemma 4 in [21]). Let g : N → (0, ∞) be a non-increasing positive function such that g(mn) ≤ g(m)g(n) for all m, n ∈ N and m≥1 g(m) = C < ∞. Let (x n ) n∈N ⊂ [0, ∞) be a summable sequence such that x n = o(g(n)), n ∈ N. Let (x n k ) k∈N be a subsequence of (x n ) n∈N such that x n k > 0. Let b ≥ 1 and let Then ε k → 0.
Proof. We will choose a subsequence of (ε k ) k∈N recursively. Let k 1 ≥ 1 and δ 1 > 0 be such that x n k 1 g(n k1 ) > δ 1 and x n g(n) ≤ δ 1 for n > n k1 .
As before, we obtain m≥1 x m[bn k j+1 ] < (C + 2)x n k j+1 which completes the proof.
Remark 2.5.5. Given b 1 , . . . , b l ≥ 1, under the assumptions of the above lemma, by a diagonalizing procedure we can find an increasing sequence (n k ) ⊂ N such that ε n k (b i ) > c > 0 for 1 ≤ i ≤ l.
Corollary 2.5.6. Suppose that f ∈ C 1+δ (T) for some δ > 0 and it is not a trigonometric polynomial. Then for a generic α and any A, B ∈ R 2 with A 2 + B 2 = 0, any relatively prime numbers r and s, any h ∈ R and any λ ∈ T, the cocycle λe 2πi(Af (r) (rx)+Bf (s) (sx+h)) is not a T -coboundary for T x = x + α.
Proof. Let (q n ) n∈N be such that f (q n ) = 0. Then by (53), for a residual set of irrationals α, we have α − pn qn q n | f (q n )| → 0 along some subsequence of (q n ) n∈N which, for convenience, we will still denote by (q n ) n∈N . In other words, (26) holds. It follows from Remark 2.5.3, Lemma 2.5.4 and Remark 2.5.5 (for l = 2, b 1 = 1, b 2 = r s ) that (24) and (25) also hold. Therefore, we can apply Theorem 2.5.2 to complete the proof.
Proof. In view of Remark 3.2.2, the assertion follows immediately from Corollary 2.5.6 and Remark 2.5.7.
Proof of Theorem 1.0.3. Recall that T c,f (x, y) = (x + α, y + cx + f (x)). We can assume that f is not a trigonometric polynomial. Indeed, otherwise f is a coboundary with the transfer function also being a trigonometric polynomial and the problem is reduced to the affine case.
It follows from Lemma 2.2.2 and Corollary 2.5.8 that for a generic α the decomposition of T 4 into minimal components of (T c,f ) r × (T c,f ) s is the same as the decomposition into ergodic components: it consists of sets I c1 , c 1 ∈ [0, 1 r ) (see (8)  To complete the proof, it suffices to show that the set T 2 satisfies condition (c). This is however true by Lemma 2.4.8 and the result follows.
Recall also (see e.g. [21]) that given an infinite set {q n } n∈N ⊂ N and a positive real valued function R = R(q n ) the set A = α ∈ [0, 1) : for infinitely many n we have α − p n q n < R(q n ), where p n q n are convergents of α is residual in T. (53) Cocycles and group extensions Let T ∈ Aut(X, B, µ). For a locally compact second countable Abelian group G 13 with a Haar measure λ G , a measurable map ϕ : The generator ϕ(x) = ϕ (1) (x) determines ϕ (n) (x) for any (n, x) ∈ Z × X: Thus, we will call a cocycle any measurable function ϕ : X → G as well. Given ϕ, we consider a G-extension of T , acting on (X × G, µ ⊗ λ G ), defined by the formula T ϕ (x, g) = (T x, ϕ(x)g).
A cocycle ϕ is called a T -coboundary (or simply a coboundary) if it is of the form ϕ(x) = ξ(x)(ξ(T x)) −1 for some measurable function ξ : X → G (called a transfer function). Two cocycles φ, ψ : X → G are cohomologous if for some measurable function f : Analogous notions to the above ones are also present in topological dynamics. Let T : X → X be a minimal homeomorphism of a compact metric space. Let ϕ : X → G be a continuous function. We say that ϕ is a topological T -coboundary if it is a measurable T -coboundary with a continuous transfer function.

Compact group extensions: ergodicity and minimality
Let G be a compact Abelian metrizable group and assume that T ∈ Aut(X, B, µ) is ergodic. Let ϕ : X → G be a cocycle. Then For example, see [11], if T x = x + α, ϕ : T → S 1 is Lipschitz with non-zero degree then T ϕ is ergodic.
If T ϕ is ergodic, we will also say that ϕ is ergodic (for T ).
Recall that G acts on X × G via g → R g , where: Ergodic components Let ϕ : X → G be a cocycle. Let P(T ϕ , B, µ) stand for the set of T ϕ -invariant Borel measures whose projection on X is µ. Fix an ergodic measure λ ∈ P(T ϕ , B, µ). Denote by H be the stabilizer of λ in G, i.e.
The proof of the following result is analogous to the one in the measuretheoretical case. We include it here for the sake of completeness. Proof. Let g 0 ∈ H top and let χ ∈ Γ top , i.e.
Relation between the ergodic and the minimal components Let T be a minimal homeomorphism of a compact metric space X and let µ be a Tinvariant probability Borel measure, ergodic with respect to T . Let ϕ : X → G be continuous. Then T ϕ is a homeomorphism of X × G and T ϕ ∈ Aut(X × G, B ⊗ B(G), µ ⊗ λ G ). Let λ ∈ P(T ϕ , B, µ). There are two natural partitions associated to T ϕ : • P erg -the partition into the ergodic components of T ϕ , • P min -the partition into the minimal components of T ϕ .
Partition P erg is clearly measurable.
settles an equivariant Borel isomorphism of X × H (considered with T ϕ ) and M (considered with T ϕ ). Moreover, (58) can be naturally extended to a Borel isomorphism of X × H top × G/H top and X × G. Clearly the partition of X × H top × G/H top given by relevant translations of X × H top × {1} (indexed by G/H top ) is measurable for the product measure µ ⊗ λ Htop ⊗ λ G/Htop . Hence its image by the Borel extensions of (58) is also a measurable partition for the image of the measure µ ⊗ λ Htop ⊗ λ G/Htop . This image is equal to µ ⊗ λ G , so the partition into minimal components is indeed measurable.
Remark 3.2.6. Since the partition into the ergodic components can be defined as the finest measurable partition whose atoms are invariant under the action of the homeomorphism in question, P erg is finer than P min .
As a direct consequence of the above remark we obtain the following:  [12] and Proposition 3.10 in [13]). Let T : X → X be uniquely ergodic and let ϕ : X → G be a continuous cocycle with values in a compact Abelian group. If T ϕ is ergodic with respect to µ ⊗ λ G then it is uniquely ergodic.

A remark on the KBSZ criterion for T x = x + α
Let T x = x + α be an irrational rotation on T. (i) If A is such that rA ∩ sA = ∅ for sufficiently large prime numbers r = s, then (6) holds true for every f ∈ C(T) such that supp f := {n ∈ Z : f (n) = 0} = A.
(ii) Suppose that A contains infinitely many primes. Let f ∈ C(T) be such that supp f = A and all nonzero Fourier coefficients are positive. Then (6) fails for f . Proof of Proposition 3.3.1. We will consider the behavior of the sums in (6) at (0, 0). Given r, s, two different prime numbers, we set I r,s := {(x, y) : sx = ry}, which is a closed subgroup of T 2 (of course (0, 0) ∈ I r,s ), invariant under T r ×T s . It is not hard to see that and that W : I r,s → T, W (x, y) = ax + by (ar + bs = 1) is a continuous group isomorphism, in particular, it sends λ Ir,s to λ T . In view of (60),