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Möbius disjointness for C1+\epsC^{1+\eps} skew products

Alexandre de Faveri Department of Mathematics, Caltech, 1200 E. California Blvd., Pasadena, CA 91125, USA \hrefmailto:[email protected]@caltech.edu
Abstract.

We show that for ε>0\varepsilon>0, every C1+εC^{1+\varepsilon} skew product on 𝕋2\mathbb{T}^{2} over a rotation of 𝕋1\mathbb{T}^{1} satisfies Sarnak’s conjecture. This is an improvement of earlier results of Kułaga-Przymus-Lemańczyk, Huang-Wang-Ye, and Kanigowski-Lemańczyk-Radziwiłł.

1. Introduction

Let (X,d)(X,d) be a compact metric space and T:XXT:X\to X be a homeomorphism. If the topological dynamical system (X,T)(X,T) has (topological) entropy zero, then Sarnak’s conjecture [27, 26] predicts that

limN1NnNf(Tnx)μ(n)=0\lim_{N\to\infty}\frac{1}{N}\sum_{n\leq N}f(T^{n}x)\mu(n)=0

for any continuous f:Xf:X\to\mathbb{R} and every xXx\in X. When this holds, we say that the system (X,T)(X,T) is Möbius disjoint.

Sarnak’s conjecture has been proved for a variety of dynamical systems: see for instance [2, 3, 14, 7, 10, 8, 15, 24, 22, 25, 4]. A common feature of all the results listed is that the underlying system is regular, in the sense that for every xXx\in X the sequence 1NnNδTn(x)\frac{1}{N}\sum_{n\leq N}\delta_{T^{n}(x)} converges in the weak-* topology to some TT-invariant Borel probability measure on XX.

Let 𝕋:=/\mathbb{T}:=\mathbb{R}/\mathbb{Z} denote the circle. In this paper we will deal with the so-called Anzai skew products (𝕋2,Tα,ϕ)(\mathbb{T}^{2},T_{\alpha,\phi}), where α\alpha\in\mathbb{R}, ϕ:𝕋𝕋\phi:\mathbb{T}\to\mathbb{T} is a continuous map and the transformation is given by

Tα,ϕ(x,y):=(x+α,y+ϕ(x))T_{\alpha,\phi}(x,y):=(x+\alpha,y+\phi(x))

for all (x,y)𝕋2(x,y)\in\mathbb{T}^{2}. We often denote the system simply by Tα,ϕT_{\alpha,\phi}.

Observe that Tα,ϕT_{\alpha,\phi} is distal, so it has zero topological entropy and therefore we expect it to be Möbius disjoint. In fact, these skew products are the basic building blocks in Furstenberg’s classification of minimal distal flows [12], so understanding them is the first step towards establishing Sarnak’s conjecture for this important general case. The main novel dynamical challenge that arises when one deals with skew products is that they provide some of the simplest examples of irregular dynamics. Indeed, Furstenberg [11] showed that Tα,ϕT_{\alpha,\phi} is not regular for some α\alpha and some analytic ϕ\phi.

Lifting ϕ:𝕋𝕋\phi:\mathbb{T}\to\mathbb{T} to the real line, we can write ϕ(x)=cx+ϕ~(x)\phi(x)=cx+\tilde{\phi}(x) for all x𝕋x\in\mathbb{T}, where cc\in\mathbb{Z} is the topological degree of ϕ\phi and ϕ~:𝕋\tilde{\phi}:\mathbb{T}\to\mathbb{R} is a continuous 11-periodic function, unique up to shifts by \mathbb{Z} (we fix an arbitrary choice). Kułaga-Przymus and Lemańczyk [19] have shown that if ϕC1+\eps\phi\in C^{1+\eps} for some \eps>0\eps>0 then Tα,ϕT_{\alpha,\phi} is Möbius disjoint for a topologically generic set of α\alpha. Furthermore, they proved Möbius disjointness of Tα,ϕT_{\alpha,\phi} when α\alpha\in\mathbb{Q}, assuming only continuity of ϕ\phi [19, Proposition 2.3.1], so from now on we assume α\alpha\in\mathbb{R}\setminus\mathbb{Q}. A further consequence of their work [19, Remark 2.5.7] (see also [28, Corollary 2.6]) is that if ϕ\phi is assumed to be Lipschitz continuous, then Sarnak’s conjecture holds for Tα,ϕT_{\alpha,\phi} whenever c0c\not=0. Therefore, with the underlying assumption on ϕ\phi in mind, we can deal only with topological degree zero from now on, and with an abuse of notation we identify ϕ\phi with ϕ~\tilde{\phi}.

The first Möbius disjointness result for all α\alpha was established by Liu and Sarnak [21], who proved it for ϕ\phi analytic and satisfying the technical condition ϕ^(m)eτ|m|\widehat{\phi}(m)\gg e^{-\tau|m|} for some τ>0\tau>0. This was the first time Sarnak’s conjecture was proved for a system that is not regular (since Furstenberg’s example satisfies the condition). A refinement of this result was recently obtained by Wang [28], who removed the need for a lower bound on Fourier coefficients, obtaining Möbius disjointness of Tα,ϕT_{\alpha,\phi} for all analytic ϕ\phi. Huang, Wang, and Ye [17] later improved this to cover all ϕC\phi\in C^{\infty}. Finally, using the work of Matomäki and Radziwiłł [23] on the behavior of μ\mu in short intervals, Kanigowski, Lemańczyk, and Radziwiłł [18] established Möbius disjointness of Tα,ϕT_{\alpha,\phi} for all ϕC2+\eps\phi\in C^{2+\eps} subject to the condition ϕ^(0)=0\widehat{\phi}(0)=0, where \eps>0\eps>0 is arbitrary.

Our main result is a simultaneous improvement of the works of Kułaga-Przymus-Lemańczyk [19], Huang-Wang-Ye [17], and Kanigowski-Lemańczyk-Radziwiłł [18]:

Theorem 1.

Let \eps>0\eps>0. For any α\alpha\in\mathbb{R} and ϕ:𝕋𝕋\phi:\mathbb{T}\to\mathbb{T} of class C1+\epsC^{1+\eps}, the skew product Tα,ϕT_{\alpha,\phi} is Möbius disjoint.

The proof follows the ideas laid out by Kanigowski-Lemańczyk-Radziwiłł in [18], but instead of aiming for a polynomial rate of convergence for Tα,ϕrnIdT_{\alpha,\phi}^{r_{n}}\to\operatorname{Id} in the uniform norm (along some unbounded sequence {rn}n1\{r_{n}\}_{n\geq 1}), we establish a polynomial rate of convergence for Tα,ϕrnIdT_{\alpha,\phi}^{r_{n}}\to\operatorname{Id} in the L2(ν)L^{2}(\nu) norm, for each Tα,ϕT_{\alpha,\phi}-invariant Borel probability measure ν\nu. The difficulties in dealing with every such ν\nu are overcome because they all project to the Lebesgue measure in the first coordinate. We also remove the condition ϕ^(0)=0\widehat{\phi}(0)=0 present in [18] by slightly modifying their choice of the sequence {rn}n1\{r_{n}\}_{n\geq 1}.

Another important ingredient is better control of some sums related to the Fourier coefficients of ϕ\phi, where the Diophantine properties of α\alpha play an important role. The idea here is that not many qq’s at a given scale can make qα\|q\alpha\| small (i.e. be denominators of good rational approximations of α\alpha). Furthermore, the qq’s at a given scale that give rise to rational approximations of similar quality must be somewhat well-spaced. We apply the Denjoy-Koksma inequality to appropriately chosen functions in order to extract that information (see Section 3).

The smoothness exponent 1+\eps1+\eps seems to be the limit of this argument. Indeed, we prove in Section 5 that if one only assumes that ϕC1\phi\in C^{1} then, at least along the sequence of best rational approximations of the irrational α\alpha, the rate of rigidity of Tα,ϕT_{\alpha,\phi} can be logarithmic even when ϕ^(0)=0\widehat{\phi}(0)=0.

In Section 6 we show that our ideas can be used to extend some general rigidity results so far only known for functions of mean zero to the general case. A modification of Lemma 1 to obtain uniform polynomial rates of rigidity in the case ϕC1+\eps\phi\in C^{1+\eps} is also discussed.

Finally, in Section 7 we use our argument to deduce new Möbius disjointness results for flows in 𝕋2\mathbb{T}^{2} and Rokhlin extensions.

Notations

For a topological dynamical system (X,T)(X,T), let M(X,T)M(X,T) be the set of TT-invariant Borel probability measures on XX. Write \|\cdot\| for the distance to the nearest integer (which we use as the metric in 𝕋\mathbb{T}), d(,)d(\cdot,\cdot) for the product metric in 𝕋×𝕋\mathbb{T}\times\mathbb{T}, corresponding to \|\cdot\| in each coordinate, and L2(ν)\|\cdot\|_{L^{2}(\nu)} for the usual L2L^{2} norm with respect to a measure ν\nu. We also abbreviate e(x):=e2πixe(x):=e^{2\pi ix} and use the asymptotic notation f(x)g(x)f(x)\ll g(x) (respectively f(x)pg(x)f(x)\ll_{p}g(x)) to mean that there exists C>0C>0 absolute (respectively depending only on the parameter pp) such that |f(x)|C|g(x)||f(x)|\leq C|g(x)| for all xx in the relevant range. Furthermore, f(x)g(x)f(x)\asymp g(x) means f(x)g(x)f(x)f(x)\ll g(x)\ll f(x).

Acknowledgments

I would like to thank my PhD advisor, Maksym Radziwiłł, for introducing me to this problem and for general advice and encouragement. Thanks also to Adam Kanigowski and Mariusz Lemańczyk for pointing out a nice simplification to my initial proofs of Lemmas 2 and 3, and for providing valuable comments and references. I’m grateful to the American Institute of Mathematics (AIM) for their 2018 workshop on “Sarnak’s Conjecture”, which played a role in motivating this work.

2. Reduction of Theorem 1 to a rigidity result

As previously outlined, we can assume that α\alpha\in\mathbb{R}\setminus\mathbb{Q} and deg(ϕ)=0\deg(\phi)=0, so ϕ\phi can be realized as a function from 𝕋\mathbb{T} to \mathbb{R} of class C1+\epsC^{1+\eps}, which by an abuse of notation we still denote by ϕ\phi. Observe that ϕ\phi is in particular Lipschitz continuous, so we have pointwise convergence of its Fourier series, and the smoothness condition gives

(1) ϕ(x)=qcqe(qx)withcqϕ1|q|1+\epsforq0.\phi(x)=\sum_{q\in\mathbb{Z}}c_{q}e(qx)\quad\text{with}\quad c_{q}\ll_{\phi}\frac{1}{|q|^{1+\eps}}\ \text{for}\ q\not=0.

The key to the proof of Theorem 1 is the result below, which is motivated by [18].

Lemma 1.

Let 0<\eps<11000<\eps<\frac{1}{100} and α\alpha\in\mathbb{R}\setminus\mathbb{Q}. If ϕ:𝕋\phi:\mathbb{T}\to\mathbb{R} is of class C1+\epsC^{1+\eps}, then there exists an unbounded sequence of positive integers {rn}n1\{r_{n}\}_{n\geq 1} such that

𝕋×𝕋d(Tα,ϕrn(x,y),(x,y))2𝑑ν(x,y)ϕrn\eps/100\int_{\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))^{2}\,d\nu(x,y)\ll_{\phi}r_{n}^{-\eps/100}

for any νM(𝕋2,Tα,ϕ)\nu\in M(\mathbb{T}^{2},T_{\alpha,\phi}), where the implied constant does not depend on ν\nu.

Assuming Lemma 1, we can easily prove Theorem 1:

Proof of Theorem 1:.

Let {rn}n1\{r_{n}\}_{n\geq 1} be the sequence from Lemma 1. For any νM(𝕋2,Tα,ϕ)\nu\in M(\mathbb{T}^{2},T_{\alpha,\phi}), continuous f:𝕋2f:\mathbb{T}^{2}\to\mathbb{R}, and kk\in\mathbb{Z}, the triangle inequality and the Tα,ϕT_{\alpha,\phi}-invariance of ν\nu imply

(2) fTα,ϕkrnfL2(ν)2|k|j=1|k|fTα,ϕjrnfTα,ϕ(j1)rnL2(ν)2=k2fTα,ϕrnfL2(ν)2.\|f\circ T_{\alpha,\phi}^{kr_{n}}-f\|_{L^{2}(\nu)}^{2}\leq|k|\sum_{j=1}^{|k|}\|f\circ T_{\alpha,\phi}^{jr_{n}}-f\circ T_{\alpha,\phi}^{(j-1)r_{n}}\|_{L^{2}(\nu)}^{2}=k^{2}\cdot\|f\circ T_{\alpha,\phi}^{r_{n}}-f\|_{L^{2}(\nu)}^{2}.

If ff is also Lipschitz continuous, then using Lemma 1 we get

(3) fTα,ϕrnfL2(ν)2f𝕋×𝕋d(Tα,ϕrn(x,y),(x,y))2𝑑ν(x,y)ϕrn\eps/100.\|f\circ T_{\alpha,\phi}^{r_{n}}-f\|_{L^{2}(\nu)}^{2}\ll_{f}\int_{\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))^{2}\,d\nu(x,y)\ll_{\phi}r_{n}^{-\eps/100}.

Therefore, (2) and (3) together give

limn|k|rn\eps/400fTα,ϕkrnfL2(ν)2=0\lim_{n\to\infty}\sum_{|k|\leq r_{n}^{\eps/400}}\|f\circ T_{\alpha,\phi}^{kr_{n}}-f\|_{L^{2}(\nu)}^{2}=0

for every νM(𝕋2,Tα,ϕ)\nu\in M(\mathbb{T}^{2},T_{\alpha,\phi}), which is precisely the PR rigidity condition of [18] (using the linearly dense family \mathcal{F} of Lipschitz continuous functions) for the system (𝕋2,Tα,ϕ)(\mathbb{T}^{2},T_{\alpha,\phi}), so [18, Theorem 1.1] implies Möbius disjointness for this skew product, and Theorem 1 is proved.

3. Continued fractions and some arithmetic estimates

Before proceeding to the proof of Lemma 1, we recall some properties of continued fractions. Let pnqn\frac{p_{n}}{q_{n}}, with qn>0q_{n}>0 and (pn,qn)=1(p_{n},q_{n})=1, be the nn-th convergent of the continued fraction expansion [a0;a1,a2,][a_{0};a_{1},a_{2},\dots] of the irrational α\alpha, so that ai1a_{i}\geq 1 for i0i\not=0. Then:

  1. (P1)

    q0=1q_{0}=1, q1=a1q_{1}=a_{1} and qn+1=an+1qn+qn1q_{n+1}=a_{n+1}q_{n}+q_{n-1} for n1n\geq 1;

  2. (P2)

    1qn+1+qn<qnα<1qn+1\frac{1}{q_{n+1}+q_{n}}<\|q_{n}\alpha\|<\frac{1}{q_{n+1}};

  3. (P3)

    If 0<q<qn+10<q<q_{n+1}, then qnαqα\|q_{n}\alpha\|\leq\|q\alpha\|.

The main technical tool that allows us to quickly explore the Diophantine properties of α\alpha through its continued fraction is the following inequality:

Proposition 1 (Denjoy-Koksma inequality).

Let α\alpha\in\mathbb{R}\setminus\mathbb{Q}. If f:𝕋f:\mathbb{T}\to\mathbb{R} is of bounded variation, which we denote by Var(f)\operatorname{Var}(f), then for any n0n\geq 0 and x𝕋x\in\mathbb{T} we have

|j=0qn1f(x+jα)qn𝕋f(z)𝑑z|Var(f).\left|\sum_{j=0}^{q_{n}-1}f(x+j\alpha)-q_{n}\int_{\mathbb{T}}f(z)\,dz\right|\leq\operatorname{Var}(f).
Proof of Proposition 1:.

See [16, VI.3.1].

The next two lemmas encapsulate estimates related to continued fractions that will be necessary to prove Lemma 1.

Lemma 2.

For any α\alpha\in\mathbb{R}\setminus\mathbb{Q} and k2k\geq 2,

0<|q|<qk1qα2qk2.\sum_{0<|q|<q_{k}}\frac{1}{\|q\alpha\|^{2}}\asymp q_{k}^{2}.
Proof of Lemma 2:.

The lower bound comes from positivity and the single term q=qk1q=q_{k-1}, by (P2). The upper bound follows from [1, Lemma 2.5] (see also [20, Lemma 1] for a partial result). We give a quick proof for completeness.

Assume 0<q<qk0<q<q_{k}, as the sum over negative qq is the same. Consider f:𝕋f:\mathbb{T}\to\mathbb{R} given by

f(z)={(2qk)2,ifz12qk1z2,ifz>12qk.f(z)=\begin{cases}(2q_{k})^{2},&\text{if}\ \|z\|\leq\frac{1}{2q_{k}}\\ \frac{1}{\|z\|^{2}},&\text{if}\ \|z\|>\frac{1}{2q_{k}}\end{cases}.

Observe that by (P2) and (P3), qα>12qk\|q\alpha\|>\frac{1}{2q_{k}} for all 0<q<qk0<q<q_{k}, so by the Denjoy-Koksma inequality we conclude that

0<q<qk1qα2=q=1qk1f(qα)|f(0)|+qk|𝕋f(z)𝑑z|+Var(f)=4qk2+qk(8qk4)+(8qk28)qk2,\begin{split}\sum_{0<q<q_{k}}\frac{1}{\|q\alpha\|^{2}}=\sum_{q=1}^{q_{k}-1}f(q\alpha)&\leq|f(0)|+q_{k}\left|\int_{\mathbb{T}}f(z)\,dz\right|+\operatorname{Var}(f)\\ &=4q_{k}^{2}+q_{k}(8q_{k}-4)+(8q_{k}^{2}-8)\ll q_{k}^{2},\end{split}

as desired.

Lemma 3.

For any α\alpha\in\mathbb{R}\setminus\mathbb{Q}, k1k\geq 1, and 1cqk1\leq c\leq q_{k},

qk|q|<qk+11q2min{1qα2,c2}cqk.\sum_{q_{k}\leq|q|<q_{k+1}}\frac{1}{q^{2}}\min\left\{\frac{1}{\|q\alpha\|^{2}},c^{2}\right\}\ll\frac{c}{q_{k}}.
Proof of Lemma 3:.

We can assume qkq<qk+1q_{k}\leq q<q_{k+1} since the sum over negative qq is the same.

Consider f:𝕋f:\mathbb{T}\to\mathbb{R} given by

f(z)={c2,ifz1c1z2,ifz>1c.f(z)=\begin{cases}c^{2},&\text{if}\ \|z\|\leq\frac{1}{c}\\ \frac{1}{\|z\|^{2}},&\text{if}\ \|z\|>\frac{1}{c}\end{cases}.

Observe that f(qα)=min{1qα2,c2}f(q\alpha)=\min\left\{\frac{1}{\|q\alpha\|^{2}},c^{2}\right\}, so (P1) gives

qkq<qk+11q2min{1qα2,c2}=j=1ak+11jqkq<(j+1)qkf(qα)q2+ak+1qkq<qk+1f(qα)q2j=1ak+111(jqk)2r=0qk1f(jqkα+rα)+4qk+12r=0qk11f(ak+1qkα+rα).\begin{split}&\sum_{q_{k}\leq q<q_{k+1}}\frac{1}{q^{2}}\min\left\{\frac{1}{\|q\alpha\|^{2}},c^{2}\right\}=\sum_{j=1}^{a_{k+1}-1}\sum_{jq_{k}\leq q<(j+1)q_{k}}\frac{f(q\alpha)}{q^{2}}+\sum_{a_{k+1}q_{k}\leq q<q_{k+1}}\frac{f(q\alpha)}{q^{2}}\\ &\leq\sum_{j=1}^{a_{k+1}-1}\frac{1}{(jq_{k})^{2}}\sum_{r=0}^{q_{k}-1}f(jq_{k}\alpha+r\alpha)+\frac{4}{q_{k+1}^{2}}\sum_{r=0}^{q_{k-1}-1}f(a_{k+1}q_{k}\alpha+r\alpha).\end{split}

Using the Denjoy-Koksma inequality for the sums over rr as in the proof of Lemma 2, since 𝕋f(z)𝑑zc\int_{\mathbb{T}}f(z)\,dz\asymp c and Var(f)c2\operatorname{Var}(f)\ll c^{2} by direct computation, we conclude that the remaining expression is

j=1qkc+c2(jqk)2+qk1c+c2qk+12cqk,\ll\sum_{j=1}^{\infty}\frac{q_{k}c+c^{2}}{(jq_{k})^{2}}+\frac{q_{k-1}c+c^{2}}{q_{k+1}^{2}}\ll\frac{c}{q_{k}},

so we are done.

4. Polynomial rate of rigidity in C1+\epsC^{1+\eps}: the proof of Lemma 1

At last, we are ready to prove our main lemma.

Proof of Lemma 1:.

Denoting Sr(g)(x):=g(x)+g(x+α)++g(x+(r1)α)S_{r}(g)(x):=g(x)+g(x+\alpha)+\dots+g(x+(r-1)\alpha), we have

Tα,ϕrn(x,y)=(x+rnα,y+Srn(ϕ)(x)),T_{\alpha,\phi}^{r_{n}}(x,y)=(x+r_{n}\alpha,y+S_{r_{n}}(\phi)(x)),

so that

d(Tα,ϕrn(x,y),(x,y))2rnα2+Srn(ϕ)(x)2.d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))^{2}\asymp\|r_{n}\alpha\|^{2}+\|S_{r_{n}}(\phi)(x)\|^{2}.

Therefore,

(4) Dn:=𝕋×𝕋d(Tα,ϕrn(x,y),(x,y))2𝑑ν(x,y)rnα2+𝕋×𝕋Srn(ϕ)(x)2𝑑ν(x,y).D_{n}:=\int_{\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))^{2}\,d\nu(x,y)\asymp\|r_{n}\alpha\|^{2}+\int_{\mathbb{T}\times\mathbb{T}}\|S_{r_{n}}(\phi)(x)\|^{2}\,d\nu(x,y).

Consider the projection map π(x,y)=x\pi(x,y)=x. Observe that the integrand in (4) is independent of the second coordinate, so we can rewrite the integral as

(5) 𝕋×𝕋Srn(ϕ)(π(x,y))2𝑑ν(x,y)=𝕋Srn(ϕ)(x)2d(πν)(x).\begin{split}\int_{\mathbb{T}\times\mathbb{T}}\|S_{r_{n}}(\phi)(\pi(x,y))\|^{2}\,d\nu(x,y)=\int_{\mathbb{T}}\|S_{r_{n}}(\phi)(x)\|^{2}\,d(\pi_{*}\nu)(x).\end{split}

Since π:(𝕋2,Tα,ϕ)(𝕋,Rα)\pi:(\mathbb{T}^{2},T_{\alpha,\phi})\to(\mathbb{T},R_{\alpha}) is a map of topological dynamical systems (where in the image the transformation is Rα(x):=x+αR_{\alpha}(x):=x+\alpha) and ν\nu is Tα,ϕT_{\alpha,\phi}-invariant, the Borel probability measure πν\pi_{*}\nu is RαR_{\alpha}-invariant. But α\alpha is irrational, so (𝕋,Rα)(\mathbb{T},R_{\alpha}) is uniquely ergodic and we conclude that πν\pi_{*}\nu is the Lebesgue measure on 𝕋\mathbb{T}.

Using the Fourier expansion of ϕ\phi we get Srn(ϕ)(x)=qcqSrn(eq)(x)S_{r_{n}}(\phi)(x)=\sum_{q\in\mathbb{Z}}c_{q}S_{r_{n}}(e_{q})(x), where eq(x):=e(qx)e_{q}(x):=e(qx). A computation shows that

Srn(eq)(x)=e(qx)1e(qrnα)1e(qα)S_{r_{n}}(e_{q})(x)=e(qx)\frac{1-e(qr_{n}\alpha)}{1-e(q\alpha)}

for q0q\not=0 and Srn(e0)(x)=rnS_{r_{n}}(e_{0})(x)=r_{n}, so we can plug this into (5) and conclude, using the triangle inequality and replacing \|\cdot\| by absolute values, that the integral there is bounded by a constant multiple of

(6) c0rn2+𝕋|q0cq1e(qrnα)1e(qα)e(qx)|2𝑑x=c0rn2+q0|cq|2|1e(qrnα)1e(qα)|2,\|c_{0}r_{n}\|^{2}+\int_{\mathbb{T}}\left|\sum_{q\not=0}c_{q}\frac{1-e(qr_{n}\alpha)}{1-e(q\alpha)}e(qx)\right|^{2}dx=\|c_{0}r_{n}\|^{2}+\sum_{q\not=0}|c_{q}|^{2}\left|\frac{1-e(qr_{n}\alpha)}{1-e(q\alpha)}\right|^{2},

where we have used Parseval for Srn(ϕ)c0rnL2(𝕋)S_{r_{n}}(\phi)-c_{0}r_{n}\in L^{2}(\mathbb{T}).

Now, we make a preliminary choice of the sequence {rn}n1\{r_{n}\}_{n\geq 1} by letting rn:=nqnr_{n}:=\ell_{n}q_{n}, where qnq_{n} is the denominator of the nn-th convergent of the continued fraction expansion of α\alpha, as before, and n\ell_{n}\in\mathbb{Z} is chosen so that

0<nqnδandnqnc0<qnδ,0<\ell_{n}\leq q_{n}^{\delta}\quad\text{and}\quad\|\ell_{n}q_{n}c_{0}\|<q_{n}^{-\delta},

where δ:=\eps/10\delta:=\eps/10. Such n\ell_{n} exist for all nn, by the Dirichlet approximation theorem.

Let λ:=\eps/100\lambda:=\eps/100. In what follows it is worth keeping in mind the rough hierarchy “λδ\eps\lambda\lll\delta\lll\eps” behind our choice of parameters. We wish to show that DnϕrnλD_{n}\ll_{\phi}r_{n}^{-\lambda}. With our choice of {rn}n1\{r_{n}\}_{n\geq 1} the first term in the RHS of (4) contributes at most

(7) n2qnα2<qn2δqn+12<qn2δ2<qnλ(1+δ)rnλ,\ell_{n}^{2}\cdot\|q_{n}\alpha\|^{2}<q_{n}^{2\delta}q_{n+1}^{-2}<q_{n}^{2\delta-2}<q_{n}^{-\lambda(1+\delta)}\leq r_{n}^{-\lambda},

so it is harmless. The first term of (6) contributes

(8) c0nqn2<qn2δ<qnλ(1+δ)rnλ,\|c_{0}\ell_{n}q_{n}\|^{2}<q_{n}^{-2\delta}<q_{n}^{-\lambda(1+\delta)}\leq r_{n}^{-\lambda},

and it is also harmless.

We break the remaining terms into two parts, corresponding to 0<|q|<qn0<|q|<q_{n} and |q|qn|q|\geq q_{n}. Observe that |1e(qα)|qα|1-e(q\alpha)|\asymp\|q\alpha\| and |1e(qrnα)|2|1-e(qr_{n}\alpha)|\leq 2. Furthermore, |Srn(eq)(x)|rn|S_{r_{n}}(e_{q})(x)|\leq r_{n} by a trivial bound, so

(9) |q|qn|cq|2|1e(qrnα)1e(qα)|2ϕ|q|qn1|q|2+2\epsmin{1qα2,rn2}<qn2\epsn2k=nqk|q|<qk+11q2min{1qα2,qn2}qn2\eps+2δk=nqnqkqn2\eps+2δ<qnλ(1+δ)rnλ,\begin{split}\sum_{|q|\geq q_{n}}|c_{q}|^{2}\left|\frac{1-e(qr_{n}\alpha)}{1-e(q\alpha)}\right|^{2}&\ll_{\phi}\sum_{|q|\geq q_{n}}\frac{1}{|q|^{2+2\eps}}\min\left\{\frac{1}{\|q\alpha\|^{2}},r_{n}^{2}\right\}\\ &<q_{n}^{-2\eps}\ell_{n}^{2}\sum_{k=n}^{\infty}\sum_{q_{k}\leq|q|<q_{k+1}}\frac{1}{q^{2}}\min\left\{\frac{1}{\|q\alpha\|^{2}},q_{n}^{2}\right\}\\ &\ll q_{n}^{-2\eps+2\delta}\sum_{k=n}^{\infty}\frac{q_{n}}{q_{k}}\ll q_{n}^{-2\eps+2\delta}<q_{n}^{-\lambda(1+\delta)}\leq r_{n}^{-\lambda},\end{split}

where we have used (1), Lemma 3 (for c=qnqkc=q_{n}\leq q_{k}) and the fact that qk+2>2qkq_{k+2}>2q_{k} by (P1), so the qkq_{k} grow exponentially.

It remains to deal with 0<|q|<qn0<|q|<q_{n}. In this case, we use |1e(qα)|qα|1-e(q\alpha)|\asymp\|q\alpha\| and |1e(qrnα)|2qnqnα2q2n2qnα2<q2qn2δqn+12|1-e(qr_{n}\alpha)|^{2}\asymp\|q\ell_{n}q_{n}\alpha\|^{2}\leq q^{2}\ell_{n}^{2}\cdot\|q_{n}\alpha\|^{2}<q^{2}q_{n}^{2\delta}q_{n+1}^{-2}, so those terms contribute

(10) 0<|q|<qn|cq|2|1e(qrnα)1e(qα)|2ϕqn2δqn+120<|q|<qn1|q|2\eps1qα2.\begin{split}\sum_{0<|q|<q_{n}}|c_{q}|^{2}\left|\frac{1-e(qr_{n}\alpha)}{1-e(q\alpha)}\right|^{2}&\ll_{\phi}\frac{q_{n}^{2\delta}}{q_{n+1}^{2}}\sum_{0<|q|<q_{n}}\frac{1}{|q|^{2\eps}}\frac{1}{\|q\alpha\|^{2}}.\end{split}

To deal with the sum over qq we consider two cases:

  • Case 1:

    There is a subsequence {qbn}n1\{q_{b_{n}}\}_{n\geq 1} of {qn}n1\{q_{n}\}_{n\geq 1} such that qbn+1qbn2q_{b_{n}+1}\geq q_{b_{n}}^{2} for all n1n\geq 1.

    In this case we take the subsequence {rbn}n1\{r_{b_{n}}\}_{n\geq 1} instead of the original sequence {rn}n1\{r_{n}\}_{n\geq 1}. Observe that (7), (8) and (9) still hold along any subsequence. In (10) we can use the given condition and Lemma 2 to get the upper bound

    qbn2δqbn40<|q|<qbn1qα2qbn2δ2<qbnλ(1+δ)rbnλ,\frac{q_{b_{n}}^{2\delta}}{q_{b_{n}}^{4}}\sum_{0<|q|<q_{b_{n}}}\frac{1}{\|q\alpha\|^{2}}\ll q_{b_{n}}^{2\delta-2}<q_{b_{n}}^{-\lambda(1+\delta)}\leq r_{b_{n}}^{-\lambda},

    and this finishes the proof.

  • Case 2:

    For all sufficiently large nn, we have qn+1<qn2q_{n+1}<q_{n}^{2}.

    In this case we stick with the original sequence {rn}n1\{r_{n}\}_{n\geq 1} and observe that for any 0<k<n0<k<n we can rewrite the sum in the RHS of (10) as

    (11) [0<|q|<qk+qk|q|<qn]1|q|2\eps1qα2<0<|q|<qk1qα2+qk2\epsqk|q|<qn1qα2qk2+qk2\epsqn2,\begin{split}\left[\sum_{0<|q|<q_{k}}+\sum_{q_{k}\leq|q|<q_{n}}\right]\frac{1}{|q|^{2\eps}}\frac{1}{\|q\alpha\|^{2}}&<\sum_{0<|q|<q_{k}}\frac{1}{\|q\alpha\|^{2}}+q_{k}^{-2\eps}\sum_{q_{k}\leq|q|<q_{n}}\frac{1}{\|q\alpha\|^{2}}\\ &\ll q_{k}^{2}+q_{k}^{-2\eps}q_{n}^{2},\end{split}

    where once again we have used Lemma 2.

    Take 0<k<n0<k<n such that qk[qn1/4,qn1/2]q_{k}\in\left[q_{n}^{1/4},q_{n}^{1/2}\right], which exists for all nn sufficiently large since we can find such terms in any interval of the form [a,a2][a,a^{2}] for aa sufficiently large, because of the given condition. Then the corresponding upper bound when we plug (11) into (10) is

    qn2δqn+12(qn+qn2\eps/2)qn2δ\eps/2<qnλ(1+δ)rnλ,\frac{q_{n}^{2\delta}}{q_{n+1}^{2}}\left(q_{n}+q_{n}^{2-\eps/2}\right)\ll q_{n}^{2\delta-\eps/2}<q_{n}^{-\lambda(1+\delta)}\leq r_{n}^{-\lambda},

    which establishes the result of Lemma 1.

5. Counterexample to polynomial rate of rigidity in C1C^{1}

Lemma 1 raises the question of how low one can push the smoothness of ϕ\phi and still have a polynomial rate of rigidity for Tα,ϕT_{\alpha,\phi}. We show that, at least along the sequence {qn}n1\{q_{n}\}_{n\geq 1} of denominators of best rational approximations for an irrational α\alpha, there is ϕC1\phi\in C^{1} with ϕ^(0)=0\widehat{\phi}(0)=0 such that

𝕋×𝕋d(Tα,ϕqn(x,y),(x,y))2𝑑ν(x,y)δqnδ\int_{\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{q_{n}}(x,y),(x,y))^{2}\,d\nu(x,y)\gg_{\delta}q_{n}^{-\delta}

for every δ>0\delta>0, unlike what happens for ϕC1+\eps\phi\in C^{1+\eps} with ϕ^(0)=0\widehat{\phi}(0)=0 (observe that in that case n=1\ell_{n}=1 works in Lemma 1).

Indeed, let α\alpha\in\mathbb{R}\setminus\mathbb{Q} and choose ϕ:𝕋\phi:\mathbb{T}\to\mathbb{R} given by

ϕ(x):=1Ck2e(qkx)+e(qkx)qk(logqk)2,\phi(x):=\frac{1}{C}\sum_{k\geq 2}\frac{e(q_{k}x)+e(-q_{k}x)}{q_{k}(\log{q_{k}})^{2}},

where C>0C>0 will be chosen to be sufficiently large. Since qk2(k1)/2q_{k}\geq 2^{(k-1)/2} by (P1), k2(logqk)2\sum_{k\geq 2}(\log q_{k})^{-2} is absolutely convergent and therefore ϕC1\phi\in C^{1}. Take C>0C>0 large enough so that Var(ϕ)<1/2\operatorname{Var}(\phi)<1/2. By the Denjoy-Koksma inequality, we have

|j=0qn1ϕ(x+jα)qn𝕋ϕ(z)𝑑z|Var(ϕ)<12\left|\sum_{j=0}^{q_{n}-1}\phi(x+j\alpha)-q_{n}\int_{\mathbb{T}}\phi(z)\,dz\ \right|\leq\operatorname{Var}(\phi)<\frac{1}{2}

for every x𝕋x\in\mathbb{T}. Since ϕ^(0)=0\widehat{\phi}(0)=0 we conclude that |Sqn(ϕ)(x)|<1/2|S_{q_{n}}(\phi)(x)|<1/2 , so that Sqn(ϕ)(x)=|Sqn(ϕ)(x)|\|S_{q_{n}}(\phi)(x)\|=|S_{q_{n}}(\phi)(x)| for all x𝕋x\in\mathbb{T}. Therefore, the beginning of the proof of Lemma 1 shows that

(12) 𝕋×𝕋d(Tα,ϕqn(x,y),(x,y))2𝑑ν(x,y)qnα2+k21qk2(logqk)4|1e(qnqkα)1e(qkα)|2.\int_{\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{q_{n}}(x,y),(x,y))^{2}\,d\nu(x,y)\asymp\|q_{n}\alpha\|^{2}+\sum_{k\geq 2}\frac{1}{q_{k}^{2}(\log{q_{k}})^{4}}\left|\frac{1-e(q_{n}q_{k}\alpha)}{1-e(q_{k}\alpha)}\right|^{2}.

If qnqnα<1/2q_{n}\cdot\|q_{n}\alpha\|<1/2 then qn2α=qnqnα\|q_{n}^{2}\alpha\|=q_{n}\cdot\|q_{n}\alpha\|, so

1qn2(logqn)4|1e(qn2α)1e(qnα)|21qn2(logqn)4qn2α2qnα2=1(logqn)4.\frac{1}{q_{n}^{2}(\log{q_{n}})^{4}}\left|\frac{1-e(q_{n}^{2}\alpha)}{1-e(q_{n}\alpha)}\right|^{2}\asymp\frac{1}{q_{n}^{2}(\log{q_{n}})^{4}}\frac{\|q_{n}^{2}\alpha\|^{2}}{\|q_{n}\alpha\|^{2}}=\frac{1}{(\log{q_{n}})^{4}}.

If instead qnqnα>1/2q_{n}\cdot\|q_{n}\alpha\|>1/2 then from qnα<1/qn+1\|q_{n}\alpha\|<1/q_{n+1} (by (P2)) we get qn+1<2qnq_{n+1}<2q_{n}. Since qn+2>2qnq_{n+2}>2q_{n} we have qnqn+1α<qn/qn+2<1/2q_{n}\cdot\|q_{n+1}\alpha\|<q_{n}/q_{n+2}<1/2, so qnqn+1α=qnqn+1α\|q_{n}q_{n+1}\alpha\|=q_{n}\cdot\|q_{n+1}\alpha\|, and in conclusion

1qn+12(logqn+1)4|1e(qnqn+1α)1e(qn+1α)|21qn2(logqn)4qnqn+1α2qn+1α2=1(logqn)4.\frac{1}{q_{n+1}^{2}(\log{q_{n+1}})^{4}}\left|\frac{1-e(q_{n}q_{n+1}\alpha)}{1-e(q_{n+1}\alpha)}\right|^{2}\asymp\frac{1}{q_{n}^{2}(\log{q_{n}})^{4}}\frac{\|q_{n}q_{n+1}\alpha\|^{2}}{\|q_{n+1}\alpha\|^{2}}=\frac{1}{(\log{q_{n}})^{4}}.

Taking respectively the terms corresponding to k=nk=n and k=n+1k=n+1 in (12) and using positivity of the other terms we conclude that the whole expression is (logqn)4\gg(\log{q_{n}})^{-4}, so there is no polynomial rate of convergence to zero along any subsequence of {qn}n1\{q_{n}\}_{n\geq 1}. In fact, [18, Lemma 3.2] shows that a decay of the form exp((loglogqn)1+δ)\exp(-(\log{\log{q_{n}}})^{1+\delta}) for any δ>0\delta>0 would be enough for Möbius disjointness, but that too is false by our counterexample.

6. Extension of general rigidity results to ϕ\phi of non-zero mean

Recall that a topological dynamical system (X,T)(X,T) is called rigid if for each νM(X,T)\nu\in M(X,T) there exists a sequence {rn}n1\{r_{n}\}_{n\geq 1} of positive integers such that gTrngg\circ T^{r_{n}}\to g in L2(ν)L^{2}(\nu) for all gL2(ν)g\in L^{2}(\nu).

By theorems of Herman [16, XIII.4.8] and Gabriel, Lemańczyk, and Liardet [13, Théorème 1.1], if α\alpha\in\mathbb{R}\setminus\mathbb{Q} and ϕ\phi is absolutely continuous, has topological degree zero, and satisfies ϕ^(0)=0\widehat{\phi}(0)=0, then the skew product Tα,ϕT_{\alpha,\phi} is rigid, and in fact they show that Tα,ϕqnIdT_{\alpha,\phi}^{q_{n}}\to\operatorname{Id} uniformly by obtaining

limnsupx𝕋|Sqn(ϕ)(x)|=0.\lim_{n\to\infty}\sup_{x\in\mathbb{T}}|S_{q_{n}}(\phi)(x)|=0.

Lemańczyk and Mauduit [20, Theorem 1] (see also [1, Corollary 2.8]) generalized111If ϕ:𝕋\phi:\mathbb{T}\to\mathbb{R} is absolutely continuous then ϕL1(𝕋)\phi^{\prime}\in L^{1}(\mathbb{T}), so the Riemann-Lebesgue lemma gives ϕ^(m)=o(1/|m|)\widehat{\phi}(m)=o(1/|m|). these theorems to show rigidity (though not uniformly) of Tα,ϕT_{\alpha,\phi} for all α\alpha\in\mathbb{R}\setminus\mathbb{Q} and ϕL2(𝕋)\phi\in L^{2}(\mathbb{T}) (of topological degree zero) satisfying ϕ^(0)=0\widehat{\phi}(0)=0 and ϕ^(m)=o(1/|m|)\widehat{\phi}(m)=o(1/|m|).

The techniques of this paper may be employed to extend both results to cover the case ϕ^(0)0\widehat{\phi}(0)\not=0. Furthermore, in the case ϕC1+\eps\phi\in C^{1+\eps} we can modify Lemma 1 to recover a uniform polynomial rate of rigidity instead of just the result in L2(ν)L^{2}(\nu) presented previously.

6.1. Uniform rigidity for ϕ\phi absolutely continuous

Proposition 2.

If α\alpha\in\mathbb{R}\setminus\mathbb{Q} and ϕ\phi is absolutely continuous of topological degree zero, then the skew product Tα,ϕT_{\alpha,\phi} is uniformly rigid.

Proof of Proposition 2:.

We can simply use the original result for the zero mean case to conclude that there is λ(n)\lambda(n)\to\infty as nn\to\infty such that

supx𝕋|Sqn(ϕϕ^(0))(x)|λ(n)1,\sup_{x\in\mathbb{T}}|S_{q_{n}}(\phi-\widehat{\phi}(0))(x)|\leq\lambda(n)^{-1},

so choose n\ell_{n}\in\mathbb{Z} with

0<nλ(n)1/2andnqnϕ^(0)<λ(n)1/20<\ell_{n}\leq\lambda(n)^{1/2}\quad\text{and}\quad\|\ell_{n}q_{n}\widehat{\phi}(0)\|<\lambda(n)^{-1/2}

using Dirichlet’s approximation theorem to get

Snqn(ϕ)(x)nqnϕ^(0)+|Snqn(ϕϕ^(0))(x)|<λ(n)1/2+k=0n1|Sqn(ϕϕ^(0))(x+kqnα)|λ(n)1/20\begin{split}\|S_{\ell_{n}q_{n}}(\phi)(x)\|&\leq\|\ell_{n}q_{n}\widehat{\phi}(0)\|+|S_{\ell_{n}q_{n}}(\phi-\widehat{\phi}(0))(x)|\\ &<\lambda(n)^{-1/2}+\sum_{k=0}^{\ell_{n}-1}|S_{q_{n}}(\phi-\widehat{\phi}(0))(x+kq_{n}\alpha)|\ll\lambda(n)^{-1/2}\to 0\end{split}

uniformly in x𝕋x\in\mathbb{T}. Therefore, Tα,ϕT_{\alpha,\phi} is uniformly rigid along the sequence {nqn}n1\{\ell_{n}q_{n}\}_{n\geq 1}.

6.2. Rigidity for ϕ\phi with tamely decaying Fourier coefficients

Proposition 3.

If α\alpha\in\mathbb{R}\setminus\mathbb{Q} and ϕL1(𝕋)\phi\in L^{1}(\mathbb{T}) (of topological degree zero) satisfies ϕ^(m)=o(1/|m|)\widehat{\phi}(m)=o(1/|m|), then the skew product Tα,ϕT_{\alpha,\phi} is rigid.

Proof of Proposition 3 (Sketch):.

We have the Fourier expansion222It follows from the conditions that ϕL2(𝕋)\phi\in L^{2}(\mathbb{T}). (in L2(𝕋)L^{2}(\mathbb{T}))

ϕ(x)=qcqe(qx)with|cq|1|q|ψ(|q|)forq0,\phi(x)=\sum_{q\in\mathbb{Z}}c_{q}e(qx)\quad\text{with}\quad|c_{q}|\leq\frac{1}{|q|\cdot\psi(|q|)}\ \text{for}\ q\not=0,

where ψ:(0,)(0,)\psi:(0,\infty)\to(0,\infty) satisfies ψ(z)\psi(z)\to\infty as zz\to\infty and for technical reasons we can of course also assume that it is non-decreasing and does not grow too fast, say ψ(z)ϕz1/100\psi(z)\ll_{\phi}z^{1/100}.

With the conditions above, we can show that there is a sequence of positive integers {rn}n1\{r_{n}\}_{n\geq 1} such that

𝕋×𝕋d(Tα,ϕrn(x,y),(x,y))2𝑑ν(x,y)ϕψ(qn1/4)1/1000asn\int_{\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))^{2}\,d\nu(x,y)\ll_{\phi}\psi(q_{n}^{1/4})^{-1/100}\to 0\quad\text{as}\quad n\to\infty

for any νM(𝕋2,Tα,ϕ)\nu\in M(\mathbb{T}^{2},T_{\alpha,\phi}).

The proof is a simple modification of the proof of Lemma 1, substituting qn\epsq_{n}^{\eps} with ψ(qn)\psi(q_{n}), so for instance n\ell_{n}\in\mathbb{Z} is chosen so that

0<nψ(qn)1/10andnqnc0<ψ(qn)1/10.0<\ell_{n}\leq\psi(q_{n})^{1/10}\quad\text{and}\quad\|\ell_{n}q_{n}c_{0}\|<\psi(q_{n})^{-1/10}.

Observe that we do not have multiplicativity of ψ\psi, which is why the bound is not of the form ψ(rn)1/100\psi(r_{n})^{-1/100}, but it is enough to prove that Tα,ϕT_{\alpha,\phi} is rigid333The bound implies rigidity for Tα,ϕT_{\alpha,\phi} since the Lipschitz continuous functions on 𝕋2\mathbb{T}^{2} are dense in L2(ν)L^{2}(\nu), for any νM(𝕋2,Tα,ϕ)\nu\in M(\mathbb{T}^{2},T_{\alpha,\phi}). This follows by the Stone-Weierstrass theorem and the fact that C(𝕋2)C(\mathbb{T}^{2}) is dense in L2(ν)L^{2}(\nu), since ν\nu is a Radon measure – see for instance [9, Proposition 7.9]. (the latter bound could be obtained if we imposed extra attainable conditions on ψ\psi).

6.3. Uniform polynomial rate of rigidity for ϕC1+\eps\phi\in C^{1+\eps}

Finally, we point out that the conclusion of Lemma 1 can actually be strengthened to a uniform polynomial rate of rigidity:

Proposition 4.

Let 0<\eps<11000<\eps<\frac{1}{100} and α\alpha\in\mathbb{R}\setminus\mathbb{Q}. If ϕ:𝕋\phi:\mathbb{T}\to\mathbb{R} is of class C1+\epsC^{1+\eps}, then there exists an unbounded sequence of positive integers {rn}n1\{r_{n}\}_{n\geq 1} such that

sup(x,y)𝕋×𝕋d(Tα,ϕrn(x,y),(x,y))ϕ,\epsrn\eps/200.\sup_{(x,y)\in\mathbb{T}\times\mathbb{T}}d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))\ll_{\phi,\eps}r_{n}^{-\eps/200}.
Proof of Proposition 4 (Sketch):.

We start by substantially modifying the results of Lemma 2 and Lemma 3. Namely one can show, using the same techniques as in the corresponding results of Section 3 but this time for the functions

f1(z)={2qk,ifz12qk1z,ifz>12qkandf2(z)={c,ifz1c1z,ifz>1c,f_{1}(z)=\begin{cases}2q_{k},&\text{if}\ \|z\|\leq\frac{1}{2q_{k}}\\ \frac{1}{\|z\|},&\text{if}\ \|z\|>\frac{1}{2q_{k}}\end{cases}\quad\text{and}\quad f_{2}(z)=\begin{cases}c,&\text{if}\ \|z\|\leq\frac{1}{c}\\ \frac{1}{\|z\|},&\text{if}\ \|z\|>\frac{1}{c},\end{cases}

respectively, that if \eps>0\eps>0, α\alpha\in\mathbb{R}\setminus\mathbb{Q}, k1k\geq 1, and 1cqk1\leq c\leq q_{k} then

(13) 0<|q|<qk1qαqklog(qk+1)\sum_{0<|q|<q_{k}}\frac{1}{\|q\alpha\|}\ll q_{k}\log(q_{k}+1)

and

(14) qk|q|<qk+11|q|1+\epsmin{1qα,c}\epslog(c+1)qk\eps.\sum_{q_{k}\leq|q|<q_{k+1}}\frac{1}{|q|^{1+\eps}}\min\left\{\frac{1}{\|q\alpha\|},c\right\}\ll_{\eps}\frac{\log(c+1)}{q_{k}^{\eps}}.

Then expanding Srn(ϕ)(x)S_{r_{n}}(\phi)(x) into a Fourier series and trivially bounding it we get

d(Tα,ϕrn(x,y),(x,y))rnα+c0rn+q0|cq||1e(qrnα)1e(qα)|,d(T_{\alpha,\phi}^{r_{n}}(x,y),(x,y))\leq\|r_{n}\alpha\|+\|c_{0}r_{n}\|+\sum_{q\not=0}|c_{q}|\left|\frac{1-e(qr_{n}\alpha)}{1-e(q\alpha)}\right|,

so we can proceed as in the proof of Lemma 1 with the expression above corresponding to (6) and the bounds of (13) and (14) corresponding to Lemma 2 and Lemma 3, respectively, to get the desired uniform polynomial decay.

Remark 1.

The proof actually shows that for every ϕ:𝕋\phi:\mathbb{T}\to\mathbb{R} of class C1+\epsC^{1+\eps} and of mean zero,

(15) supx𝕋|Sqn(ϕ)(x)|ϕ,\epsqn\eps/200,\sup_{x\in\mathbb{T}}|S_{q_{n}}(\phi)(x)|\ll_{\phi,\eps}q_{n}^{-\eps/200},

since in that case we can take n=1\ell_{n}=1 throughout the argument.

Remark 2.

Even though Proposition 4 gives a stronger result than Lemma 1, we chose to emphasize the latter in our presentation because the L2L^{2} methods employed there seem more suitable for generalization (and the proof is slightly more complicated). For instance, an approach to Proposition 3 using LL^{\infty} methods would already be frustrated by the presence of the extra logarithmic factor in (13), if the decay of the Fourier coefficients is sufficiently slow. Therefore, the use of L2L^{2} methods seems to allow us to go a bit further.

7. Smooth flows on 𝕋2\mathbb{T}^{2} and Rokhlin extensions

We can adapt the result of this paper, following [18], to give Möbius disjointness for new cases of smooth flows on the torus and Rokhlin extensions.

7.1. Smooth flows on 𝕋2\mathbb{T}^{2}

For α\alpha\in\mathbb{R}\setminus\mathbb{Q}, let f:𝕋f:\mathbb{T}\to\mathbb{R} be a strictly positive continuous function. Let

𝕋f:={(x,s)𝕋×:0sf(x)}/\mathbb{T}^{f}:=\{(x,s)\in\mathbb{T}\times\mathbb{R}:0\leq s\leq f(x)\}/\sim

where \sim denotes the equivalence relation (x,s+f(x))(Rα(x),s)(x,s+f(x))\sim(R_{\alpha}(x),s) in 𝕋×\mathbb{T}\times\mathbb{R} and Rα:𝕋𝕋R_{\alpha}:\mathbb{T}\to\mathbb{T} is the irrational rotation by α\alpha. We can define a special flow Tf={Ttf}tT^{f}=\{T^{f}_{t}\}_{t\in\mathbb{R}} over RαR_{\alpha} with roof function ff, which acts on 𝕋f\mathbb{T}^{f} by

Ttf(x,s):=(x,s+t)T^{f}_{t}(x,s):=(x,s+t)

for all (x,s)𝕋f(x,s)\in\mathbb{T}^{f}. More explicitly, if we extend a previous definition to

SN(f)(x):={0j<Nf(Rαj(x)),ifN>00,ifN=0Nj<0f(Rαj(x)),ifN<0S_{N}(f)(x):=\begin{cases}\sum_{0\leq j<N}f(R_{\alpha}^{j}(x)),&\text{if}\ N>0\\ 0,&\text{if}\ N=0\\ \sum_{N\leq j<0}f(R_{\alpha}^{j}(x)),&\text{if}\ N<0\end{cases}

then

Ttf(x,s)=(RαN(x),s+tSN(f)(x))T^{f}_{t}(x,s)=(R_{\alpha}^{N}(x),s+t-S_{N}(f)(x))

for all (x,s)𝕋f(x,s)\in\mathbb{T}^{f}, where N=N(x,s,t)N=N(x,s,t)\in\mathbb{Z} is such that

SN(f)(x)s+t<SN+1(f)(x),S_{N}(f)(x)\leq s+t<S_{N+1}(f)(x),

which exists and is unique as ff is continuous and strictly positive.

Every sufficiently smooth area-preserving flow on 𝕋2\mathbb{T}^{2} with no fixed points or closed orbits can be represented by such a special flow for ff with corresponding smoothness properties (see [6]).

We have the following consequence of our work:

Corollary 1.

Let \eps>0\eps>0 and α\alpha\in\mathbb{R}\setminus\mathbb{Q}. If fC1+\epsf\in C^{1+\eps} then all the maps of the special flow Tf={Ttf}tT^{f}=\{T_{t}^{f}\}_{t\in\mathbb{R}} over the irrational rotation RαR_{\alpha} are Möbius disjoint.

Proof of Corollary 1:.

There is a natural quotient metric DD making 𝕋f\mathbb{T}^{f} a compact metric space (see [5, Appendix 9.1]), and it satisfies

(16) D(Ttf(x,s),(x,s))|t|for alltand(x,s)𝕋f.D(T^{f}_{t}(x,s),(x,s))\leq|t|\quad\text{for all}\quad t\in\mathbb{R}\quad\text{and}\quad(x,s)\in\mathbb{T}^{f}.

Denote β:=f^(0)>0\beta:=\widehat{f}(0)>0 and let qnq_{n} be the denominators of convergents of the continued fraction of α\alpha, as before. For a fixed tt\in\mathbb{R}, let vnv_{n}\in\mathbb{Z} be such that

0<vnqn1+γandvntqnβ<qn1γ,0<v_{n}\leq q_{n}^{1+\gamma}\quad\text{and}\quad\left\|v_{n}\frac{t}{q_{n}\beta}\ \right\|<q_{n}^{-1-\gamma},

where γ>0\gamma>0 will be chosen later to be sufficiently small (vnv_{n} exists by Dirichlet’s approximation theorem). Then there is jnj_{n}\in\mathbb{Z} such that

(17) |tvnjnqnβ|<qnβqn1+γfqnγ|tv_{n}-j_{n}q_{n}\beta|<\frac{q_{n}\beta}{q_{n}^{1+\gamma}}\ll_{f}q_{n}^{-\gamma}

and

(18) |jn|<|vntqnβ|+1f,tqnγ.|j_{n}|<\left|\frac{v_{n}t}{q_{n}\beta}\right|+1\ll_{f,t}q_{n}^{\gamma}.

For every (x,s)𝕋f(x,s)\in\mathbb{T}^{f} we have

D(Ttvnf(x,s),(x,s))=D(TtvnSjnqn(f)(x)fTSjnqn(f)(x)f(x,s),(x,s))D(TtvnSjnqn(f)(x)f(Rαjnqn(x),s),(Rαjnqn(x),s))+D((Rαjnqn(x),s),(x,s))|Sjnqn(fβ)(x)|+|tvnjnqnβ|+jnqnαf,tqnγsupz𝕋|Sqn(fβ)(z)|+qnγ+qnγqnα,\begin{split}D(T^{f}_{tv_{n}}(x,s),(x,s))&=D(T^{f}_{tv_{n}-S_{j_{n}q_{n}}(f)(x)}\circ T^{f}_{S_{j_{n}q_{n}}(f)(x)}(x,s),(x,s))\\ &\leq D(T^{f}_{tv_{n}-S_{j_{n}q_{n}}(f)(x)}(R_{\alpha}^{j_{n}q_{n}}(x),s),(R_{\alpha}^{j_{n}q_{n}}(x),s))+D((R_{\alpha}^{j_{n}q_{n}}(x),s),(x,s))\\ &\leq|S_{j_{n}q_{n}}(f-\beta)(x)|+|tv_{n}-j_{n}q_{n}\beta|+\|j_{n}q_{n}\alpha\|\\ &\ll_{f,t}q_{n}^{\gamma}\cdot\sup_{z\in\mathbb{T}}|S_{q_{n}}(f-\beta)(z)|+q_{n}^{-\gamma}+q_{n}^{\gamma}\cdot\|q_{n}\alpha\|,\end{split}

where we have used (16), (17) and (18). Choosing γ:=\eps/1000\gamma:=\eps/1000 and using (15) (we could also take the L2L^{2} norm and use the proof of Lemma 1) we get the bound f,t,\epsqn\eps/1000<vn\eps/2000\ll_{f,t,\eps}q_{n}^{-\eps/1000}<v_{n}^{-\eps/2000}, which gives a polynomial rate of rigidity for Ttf:𝕋f𝕋fT_{t}^{f}:\mathbb{T}^{f}\to\mathbb{T}^{f} along the (unbounded, unless t=0t=0) sequence {vn}n1\{v_{n}\}_{n\geq 1}, and this implies Möbius disjointness for TtfT_{t}^{f}.

7.2. Rokhlin extensions

As before, let α\alpha\in\mathbb{R}\setminus\mathbb{Q} and let Rα:𝕋𝕋R_{\alpha}:\mathbb{T}\to\mathbb{T} denote the irrational rotation by α\alpha. Given a continuous function f:𝕋f:\mathbb{T}\to\mathbb{R}, a compact metric space (Y,ρ)(Y,\rho) and a continuous flow L={Lt}tL=\{L_{t}\}_{t\in\mathbb{R}} acting on YY, we can define a Rokhlin extension Ef,LE_{f,L} of RαR_{\alpha}, acting on 𝕋×Y\mathbb{T}\times Y by

Ef,L(x,y):=(Rα(x),Lf(x)(y))E_{f,L}(x,y):=(R_{\alpha}(x),L_{f(x)}(y))

for all (x,y)𝕋×Y(x,y)\in\mathbb{T}\times Y (observe that if Y=𝕋Y=\mathbb{T} and LL is the linear flow we recover the Anzai skew product Tα,fT_{\alpha,f}). We have the following disjointness result in this case:

Corollary 2.

Let \eps>0\eps>0. If fC1+\epsf\in C^{1+\eps} has mean zero and the flow L={Lt}tL=\{L_{t}\}_{t\in\mathbb{R}} is uniformly Lipschitz continuous in tt, then Ef,LE_{f,L} is Möbius disjoint.

Proof of Corollary 2:.

If DD denotes the product metric in 𝕋×Y\mathbb{T}\times Y then

D(Ef,Lqn(x,y),(x,y))=qnα+ρ(LSqn(f)(x)(y),y)Lqnα+|Sqn(f)(x)|,D(E_{f,L}^{q_{n}}(x,y),(x,y))=\|q_{n}\alpha\|+\rho(L_{S_{q_{n}}(f)(x)}(y),y)\ll_{L}\|q_{n}\alpha\|+|S_{q_{n}}(f)(x)|,

where the implied constant does not depend on (x,y)(x,y). Using (15) we get a polynomial rate of rigidity for Ef,LE_{f,L} along {qn}n1\{q_{n}\}_{n\geq 1} (we could also take the L2L^{2} norm and use the proof of Lemma 1), so the corollary follows.

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