分类： Sarnak conjecture

对数平均 Sarnak 猜想：从 BSZ 准则到熵下降
旧博客原文

原题：Log average sarnak conjecture

This is a note concentrate on the log average Sarnak conjecture, after the work of Matomaki and Raziwill on the estimate of multiplication function of short interval. Given a overview of the presented tools and method dealing with this conjectue.

1. Introduction

Sarnak conjecture \cite{Sarnak} assert that for any obersevable ${\{f(T^n(x_0))\}_{n=1}^{\infty}}$ come from a determination systems ${(T,X),T:X\rightarrow X}$ , where ${h(T)=0}$ , ${x_0\in X, f\in C(X)}$ . The correlation of it and the Liuvillou function is 0, i.e. they are orthongonal to each other, more preseicesly it is just to say,

$\displaystyle \sum_{n<x}\mu(x)f(T^n(x_0))=o(x) \ \ \ \ \ (1)$

This is a very natural raised conjecture, Liuville function is the presentation of primes, due to we always believe the distribution of primes in ${\mathbb N}$ should be randomness.

It has been known as observed by Landau \cite{Laudau} that the simplest case,

$\displaystyle \sum_{n<x}\mu(n)=o(x)$

already equivalent to the prime number theorem. It is not difficult to deduce the spetial case of Sarnak conjecture when with the obersevation in $latex {(1)}&fg=000000$ come from finite dynamic system is equivalent to the prime number theorem in athremetic progress by the similar argument. Besides this two classical result, may be the first new result was established by Davenport,

Theorem 1 Let ${T:S_1\rightarrow S_1, T(x)=x+\alpha}$ , ${\alpha}$ is a inrational, then the obersevation come from ${(T,S_1)}$ is orthogonal to Mobius function. due to ${\{e^{2\pi ikx}\}_{k\in \mathbb Z}}$ is a basis of ${C(S_1)}$ , suffice to proof,

$\displaystyle \sum_{n<x}e^{2\pi ikn\alpha}\mu(n)=o(x), \forall k\in \mathbb N$

There is a lots of spetial situations of Sarnak’s conjecture have been established, The parts I mainly cared is the following:
1. Interval exchange map.
2. Skew product flow.
3. Obersevable come from One dimensional zero entropy flow.
4. Nilsequences.
But in this note, I do not want to explain the tecnical and tools to establish this result, but considering an equivalent conjecture of Sarnak conjecture, named Chowla conjecture, and explain the underlying insight of the suitable weak statement, i.e. the log average Chowla conjecture and the underlying insight of it.

The note is organized as following way, in the next section $latex {(2)}&fg=000000$, we give a self-contained introduction on the tools called Bourgain-Sarnak-Ziegler critation, explain the relationship of this critation and the sum-product phenomenon, also given some more general critation along the philosephy use in establish the Bourgain-Sarnak-Ziegler critation, which maybe useful in following development combine with some other tools. The key point is transform the sum from linear sum to bilinear sum and decomposition the bilinear sum into diagonal part and off-diagonal part, use the assume in the critation to argue the off-diagonal part is small and on the orther hand the diagonal part is also small by the trivial estimate and the volume of diogonal is small, this is very similar to a suitable Caderon-Zugmund decomposition.

In section $latex {(4)}&fg=000000$, I try to give a proof sketch of the result of Matomaki and Raziwill, which is also a key tools to understanding the Sarnak conjecture, or equivalent the Chowla conjecture. The key points of the proof contains following:
1. Find a suitable fourier indentity
2. Construct a multiplication-addition dense subset ${S}$ , and proof that the theorem MR hold we need only to proof it hold for ${S\cap [1,2,...,n]}$ instead of ${[1,2,...,n]}$
3. Involve the power of euler product formula. divide the whole interval into a lot of small interval with smaller and smaller scale and a residue part. We look the part come from every small scale as a major term and look the residue part as minor term.
4. Deal with the major term at every scale, by a combitorios identity and second moments method.
5. find a enough decay estimate from a scale to the next smaller scale.
6. Deal with the minor term by the H… lemma.
Due to the theorem of MR do not exausted the method they developed, we trying to make some more result with their method, Tao and Matomaki attain the average version of Chowla conjecture is true by this way, and combine this argument and the entropy decresment argument they established the 2 partten of the log average Chowla conjecture is true. Very recently Tao and his coperator proved the odd partten case of log average chowla conjecture is true, combine an argument of frustenberg crresponding principle and entopy decresment argument. But it seems the even and large than 2 case is much difficult and seems need something new to combine with the method of MR and entropy decresment and frunstenberg corresponfing principle to make some progress.

So, in section $latex {(5)}&fg=000000$, we give a self-contain introduction to the entropy decresment argument of Tao, and combine with the frustenberg corresponding principle.

In the last section $latex {(6)}&fg=000000$, I state some result and method and phylosphy of them I get on nilsequences and wish to combine them with the previous method to make some progress on log average Chowla conjecture on the even partten case.

\newpage

2. Bourgain-Sarnak-Zieglar creation

We begin with the easiest one, this is the main result established in \cite{BSZ}, I try to give the main ideal under the proof, but with a no quantitative version is the following,

Theorem 2 (Bourgain-Sarnak-Zieglar creation, not quantitative version) if for all primes ${p,q>>1}$ we have:

$\displaystyle \sum_{n=1}^Nf(T^{pn}(x))\overline{f(T^{qn}(x))}=o(N) \ \ \ \ \ (2)$

Then for multiplication function ${g(n)}$ we have

$\displaystyle \sum_{n=1}^Ng(n)\overline{ f(T^n(x))}=o(N) \ \ \ \ \ (3)$

Remark 1 For simplify we identify ${f(T^n(x)):=F(n)}$ .

Remark 2

The idea is following, break the sum into a bilinear one, so, of course, we multiplication it with itself. i.e. we consider to control,

$\displaystyle |\sum_{i=1}^Ng(n)\overline{ F(n)}|^2=\sum_{n=1}^N\sum_{m=1}^Ng(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (4)$

To control 4, we need exhausted the mutiplication property of ${g(n)}$ , we have ${g(mn)=g(n)g(m),\forall\ m,n\in {\mathbb N}}$ . We can not get good estimate for all term,

$\displaystyle g(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (5)$

The condition in our hand if following,

$\displaystyle \sum_{n=1}^NF(pn)\overline{F(qn)}=o(N), \forall \ p,q\in \mathop{\mathbb P} \ \ \ \ \ (6)$

So, just like the situation of Cotlar-Stein lemma \cite{Cotlar-Stein lemma}, we wish to estimate like following:

$\displaystyle \begin{array}{rcl} |\sum_{p\in W}\sum_{n\in V}F(pn)g(pn)| & \leq & \sum_{n\in V}|g(n)|\cdot |\sum_{p\in W}F(pn)g(p)| \\ & \leq &\sum_{n\in V}|\sum_{p \in W}F(pn)g(p)|\\ & \overset{Cauchy-Schwarz}\leq & |V|^{\frac{1}{2}}[\sum_{n\in V}|\sum_{p\in W}F(pn)g(p)|^2]^{\frac{1}{2}}\\ & = & |V|^{\frac{1}{2}}[\sum_{p_1,p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}\\ \end{array}$

Then we consider divide the sum into diagonal part and non-diagonal part, as following,

$\displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}+|V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}} \ \ \ \ \ (7)$

But the first part is small, i.e.

$\displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}} =o(|W||V|) \ \ \ \ \ (8)$

Because of

$\displaystyle \sum_{n\in V}F(p_1n)\overline{F(p_2n)}=o(V), \forall p_1\neq p_2\in W \ \ \ \ \ (9)$

and the second part is small, i.e.

$\displaystyle |V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}}=o(|W||V|) \ \ \ \ \ (10)$

Because diagonal part is small in ${W\times W}$ and trivial inequality

$\displaystyle \sqrt{\sum_{n\in V}|F(pn)|^2}\leq |V|^{\frac{1}{2}} \ \ \ \ \ (11)$

But the method in remark 2 is not always make sense in any situation, we need to construct two suitable sets ${W,V}$ and then break up ${\{1,2,...,n{\mathbb N}\}}$ into ${W\times V}$ , this mean,

$\displaystyle \{1,2,...,N\}\sim W\times V+o(N) \ \ \ \ \ (12)$

But this ${W,V}$ could be construct in this situation, thanks to the prime number theorem,

Theorem 3 (Prime number theorem)

$\displaystyle \pi(n)\sim \frac{n}{ln(n)} \ \ \ \ \ (13)$

Morally speaking, this is the statement that the primes, which is the generator of multiplication function, is not very sparse.

3. Van der curpurt trick

There is the statement of Van der carport theorem:

Theorem 4 (Van der curpurt trick) Given a sequences ${ \{x_n\}_{n=1}^{\infty}}$ in ${ S_1}$ , if ${ \forall k\in N^*}$ , ${ \{x_{n+k}-x_n\}}$ is uniformly distributed, then ${ \{x_n\}_{n=1}^{\infty}}$ is uniformly distributed.

I do not know how to establish this theorem with no extra condition, but this result is true at least for polynomial flow. \newpage Proof:

$\displaystyle \begin{array}{rcl} |\sum_{n=1}^Ne^{2\pi imQ(n)}|& = &\sqrt{(\sum_{n=1}^Ne^{2\pi imQ(n)})(\overline{\sum_{n=1}^Ne^{2\pi imQ(n)}})}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi imQ(n+h_1)-Q(n)}}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi im \partial^1_{h_1}Q(n)}}\\ & \leq & \sqrt{\sum_{h_1=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)}|}\\ & = &\sqrt{\sum_{h_1=1}^N\sqrt{ (\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)} )(\overline{\sum_{n=1}^{N-h}e^{2\pi \partial^1_{h_1}Q(n)})}}}\leq\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi\partial^1_{h_2} \partial^1_hQ(n)} |}}\\ & \leq ....\leq & \\ & = & \sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N \sqrt{....\sqrt{\sum_{h_{k-1}=1}^{N-h_{k-2}}|\sum_{n=1}^{N-h_{k-1}}e^{2\pi\partial_{h_1h_2...h_{k-1}Q(n)}}|}}}} =o(1) \end{array}$

$\Box$

This type of trick could also establish the following result, which could be understand as a discretization of the Vinegradov lemma.

Remark 3

Uniformly distribution result of ${ F_p}$ : Given ${Q(n)=a_kn^k+...+a_1n+a_0}$ , ${\{Q(0),Q(1),...,Q(p-1)\}}$ coverages to a uniformly distribution in ${\{0,1,...,p-1\}}$ as ${p \rightarrow \infty}$ .

Remark 4 But I definitely do not know how to establish the similar result when ${Q(n)=n^{-1}}$ .

Remark 5

This trick could also help to establish estimate of correlation of low complexity sequences and multiplicative function, such as result:

$\displaystyle S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)=o(n)$

Maybe with the help of B-Z-S theorem.

\newpage

4. Matomaki and Raziwill’s work

In this section we explain the main idea underlying the paper \cite{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL}. But play with a toy model, i.e. the corresponding corollary of the original result on Liouville鈥檚 function.

Definition 5 (Lioville’s function)

$\displaystyle \lambda(n)=(-1)^{\alpha_1+\alpha_2+...+\alpha_k}, \forall \ n=p_1^{\alpha_1}...p_k^{\alpha_k}. \ \ \ \ \ (14)$

Remark 6

$\displaystyle |\int_{X}^{2X}\lambda(n)dx|=o(x) \ \ \ \ \ (15)$

is equivalent to the prime number theorem 3.

The most important beakgrouth of analytic number theory is the new understanding of multiplication function on share interval, this result is established by Kaisa Matom盲ki and Maksym Radziwill. Two very young and intelligent superstars.

The main theorem in them article is :

Theorem 6 (Matomaki,Radziwill) As soon as ${H\rightarrow \infty}$ when ${x\rightarrow \infty}$ , one has:

$\displaystyle \sum_{x\leq n\leq x+H}\lambda(n)= o(H) \ \ \ \ \ (16)$

for almost all ${1\leq x\leq X}$ .

In my understanding of the result, the main strategy is:
1. Parseval indetity, transform to Dirchelet polynomial.
2. Involved by multiplication property, spectral decomposition.
3. From linear to multilinear , Cauchy schwarz inequality.
4. major term estimate.
5. Estimate the contribution of area which is not filled.
4.1. Parseval indetity, transform to Dirchelet polynomial

We wish to establish the equality,

$\displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|dx=o(H) \ \ \ \ \ (17)$

This is the ${L^1}$ norm, by Chebyschev inequality, this could be control by ${L^2}$ norm, so we only need to establish the following,

$\displaystyle \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx=o(H^2) \ \ \ \ \ (18)$

We wish to transform from the discretization sum to a continue sum, that is,

$\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x} \ \ \ \ \ (19)$
Remark 7 There are two points to understand why 19 and 18 are the same.
1. ${[xe^{-\frac{1}{T}},xe^{\frac{1}{T}}]\sim [x-H,x+H]}$ .
2. ${1_{x\leq n\leq 2x}}$ and ${\frac{1}{x}}$ is to make that ${x=O(X)}$ .
So the Magnitude of 18 and 19 are the same. i.e.

$\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}\sim \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \ \ \ \ \ (20)$
Now we try to transform 19 by Parseval indetity, this is something about the ${L^2}$ norms of the quality we wish to charge. It is just trying to understanding 19 as a quantity in physical space by a more chargeable quality in frequency space. Image,

$\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}:=\int_{{\mathbb R}}|f_X(x)|^2dx \ \ \ \ \ (21)$

Then ${f_X(x)=\int_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(x)1_{X\leq n\leq 2X}}$ . Note that,

$\displaystyle \begin{array}{rcl} \widehat{f_X(\xi)} & = & \int_{{\mathbb R}}f_X(x)e^{2\pi ix\xi}dx\\ & = & \sum_{x\leq n\leq 2x}\lambda(x)\int_{logn-\frac{1}{T}}^{logn+\frac{1}{T}}e^{2\pi ix\xi}dx, \ T=\frac{X}{H}\\ & = & \sum_{X\leq n\leq 2X}\lambda(x)e^{2\pi ilog(n)\cdot \xi}\cdot\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i-\frac{\xi}{T}}}{2\pi i\xi}\\ \end{array}$

So by Parseval identity, we have,

$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}}|f_X(x)|^2dx & = & \int_{{\mathbb R}}|\widehat{f_X(\xi)}|^2d\xi \\ & = & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2(\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2d\xi\\ & \sim & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2\frac{1}{T^2}1_{|\xi|^2\leq T}\\ \end{array}$

Remark 8 We know the Fejer kernel satisfied,

$\displaystyle (\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2\sim \frac{1}{T^2}1_{|\xi|\leq T} \ \ \ \ \ (22)$

So morally speaking, we get the following identity.

$\displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx\sim \frac{1}{(x/H)^2}\int_{0}^{\frac{X}{H}}|\sum_{x\leq n\leq 2x}\lambda(x)x^{2\pi i\xi}|^2d\xi \ \ \ \ \ (23)$

In fact we do a cutoff, the quality we really consider is just:

$\displaystyle \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (24)$

established the monotonically inequality:

Theorem 7 (Paserval type identity)

$\displaystyle \frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \sim聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (25)$

Remark 9

In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain ${ \mathbb N^*}$ with additive structure, it could be looked as a lots of wave with the periodic given by primes, so we could do a orthogonal decomposition in the fractional space, try to prove the cutoff is a error term and we get such a monotonically inequality.

But at once we get the monotonically inequality, we could look it as a聽compactification process and this process still carry most of the information so lead to the inequality.

It seems something similar occur in the attack of the moments estimate of zeta function by the second author. And it is also could be looked as something similar to the 聽spectral decomposition with some basis come from multiplication generators, i.e. primes.

4.2. Involved by multiplication property, spectral decomposition

I called it is “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for multiplication function ${\lambda(n)}$ , we have Euler-product formula:

$\displaystyle \Pi_{p,prime}(\frac{1}{1-\frac{\lambda(p)}{p^s}})=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s} \ \ \ \ \ (26)$

But anyway, we do not use the whole power of multiplication just use it on primes, i.e. ${\lambda(pn)=\lambda(p)\lambda(n)}$ leads to following result:

$\displaystyle \lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{\# \{p|m, p\in I\}+1}+\lambda(n)1_{p|n;p\notin I} \ \ \ \ \ (27)$

This is a identity about the function ${\lambda(n)}$ , the point is it is not just use the multiplication at a point,i.e. ${\lambda(mn)=\lambda(m)\lambda(n)}$ , but take average at a area which is natural generated and compatible with multiplication, this identity carry a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.

4.3. From linear to multilinear , Cauchy schwarz

Now, we do not use one sets ${I}$ , but use several sets ${I_1,...,I_n }$ which is carefully chosen. And we do not consider [X,2X] with linear structure anymore , instead reconsider the decomposition:

${[X,2X]=\amalg_{i=1}^n (I_i\times J_i) \amalg U}$

On every ${I_i\times J_i}$ it equipped with a bilinear structure. And ${U}$ is a very small set, ${|U|=o(X)}$ which is in fact have much better estimate.

${\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt =\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

Now we just use a Cauchy-Schwarz:

${\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

4.4. major term estimate

${=\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

${\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}$

4.5. estimate the contribution of area which is not filled

\newpage

5. Entropy dcrement argument

\newpage

6. Correlation with nilsequences

I wish to establish the following estimate: ${\lambda(n)}$ is the liouville function we wish the following estimate is true.

$\displaystyle \int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH). \ \ \ \ \ (28)$

Where we have ${ H\rightarrow \infty}$ as ${ x\rightarrow \infty}$ ,

$\displaystyle \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\}$

is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem 8 (multiplication function in short interval)

${f(n): \mathbb N\rightarrow \mathbb C}$ is a multiplicative function, i.e. ${ f(mn)=f(n)f(m), \forall m,n\in \mathbb N}$ . ${H\rightarrow \infty}$ as ${x\rightarrow \infty}$ , then we have the following result,

$\displaystyle \int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH). \ \ \ \ \ (29)$

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem 9 (correlation of multiplication function and nil-sequences in long interval)

${f(n): \mathbb N\rightarrow \mathbb C}$ is a multiplicative function, i.e. ${ f(mn)=f(n)f(m), \forall m,n\in \mathbb N}$ . ${g(n)=a_n^m+...+a_1n+a_0}$ is a polynomial function then we have the following result,

$\displaystyle \int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X) \ \ \ \ \ (30)$

\newpage {9} \bibitem{Sarnak} Peter Sarnak, Mobius Randomness and Dynamics.

\texttt{https://publications.ias.edu/sites/default/files/Mahler }. \bibitem{Laudau} JA 虂NOS PINTZ (BUDAPEST). LANDAU鈥橲 PROBLEMS ON PRIMES.

\texttt{https://users.renyi.hu/~pintz/pjapr.pdf} \bibitem{BSZ} Knuth: Computers and Typesetting,

\texttt{http://www-cs-faculty.stanford.edu/\~{}uno/abcde.html}

\bibitem{Cotlar-Stein lemma} Almost orthogonality

\texttt{https://hxypqr.wordpress.com/2017/12/18/almost-orthogonality/}

\bibitem{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL} KAISA MATOMA 虉KI AND MAKSYM RADZIWIL, MULTIPLICATIVE FUNCTIONS IN SHORT INTERVALS.

\texttt{https://arxiv.org/abs/1501.04585v4/}.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

这篇笔记想整理的是 Sarnak 猜想的一条现代路线：从莫比乌斯函数和零熵动力系统的正交性出发，经过 Bourgain-Sarnak-Ziegler 准则，把线性相关和转成双线性相关；再借助 Matomaki-Radziwill 的短区间乘法函数估计，以及 Tao 的 entropy decrement，把问题推向对数平均 Chowla 猜想。

Sarnak/Chowla 的对数平均路线：线性相关先转成双线性相关，再用短区间估计和熵下降选择合适尺度。

1. Sarnak 猜想的基本形状

Sarnak 猜想说，如果 $(X,T)$ 是零拓扑熵动力系统，$f\in C(X)$，那么对任意 $x\in X$，莫比乌斯函数与观测序列 $f(T^n x)$ 应该正交：

$$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to 0.$$

这里的哲学是：$\mu(n)$ 携带素数分布中的振荡，而零熵系统产生的是低复杂度的确定序列；二者不应该长期对齐。

最简单的有限动力系统情形已经包含素数定理和算术级数中的素数定理。圆周无理旋转情形则接近 Davenport 型估计：对任意无理数 $\alpha$，指数序列 $e(n\alpha)$ 与 $\mu(n)$ 的相关和具有消失。

2. 从 Sarnak 到 Chowla

Chowla 猜想更像是对 $\mu$ 或 Liouville 函数自身随机性的表述。一个典型的 $k$ 点相关形式是

$$\frac1N\sum_{n\le N}\lambda(n+h_1)\cdots \lambda(n+h_k)\to 0,$$

其中 $h_i$ 两两不同。Sarnak 关心的是乘法函数与低复杂度确定序列的相关，Chowla 关心的是乘法函数自身不同平移之间的相关。二者之间可以用 Furstenberg correspondence principle 和动力系统模型联系起来。

对数平均版本把普通平均换成

$$\frac1{\log N}\sum_{n\le N}\frac{a(n)}n.$$

这个权重让尺度选择更加稳定，也更适合短区间分析。很多情况下，对数平均结论比普通平均结论先被证明，因为它容许把不同尺度上的误差以更柔和的方式叠加。

3. Bourgain-Sarnak-Ziegler 准则

BSZ 准则的核心是一个线性到双线性的转换。设 $a_n$ 是有界序列，如果对很多不同素数 $p\ne q$，都有

$$\frac1N\sum_{n\le N}a_{pn}\overline{a_{qn}}\to 0,$$

那么可以推出

$$\frac1N\sum_{n\le N}\mu(n)a_n\to 0.$$

直观上，$\mu$ 的乘法结构允许我们把原来的线性相关拆成不同素数伸缩后的相关。对角部分 $p=q$ 用平凡估计和体积小来处理；非对角部分则由上面的假设控制。这一点和 Calderon-Zygmund 分解的精神很像：把难对象拆成一个小的对角坏集和一个可估计的非对角主体。

4. Matomaki-Radziwill 的短区间输入

Matomaki-Radziwill 的工作说明，乘法函数在大多数短区间里的平均行为可以被控制。粗略地说，对许多短区间 $[x,x+H]$，有

$$\frac1H\sum_{x
接近它在长区间中的平均。这使得我们可以把一个全局相关和切成许多短尺度，再用欧拉乘积、二阶矩和组合恒等式逐层比较。

短区间估计真正有用的地方在于：它让乘法函数的局部随机性可以被拿来攻击动力系统中的相关问题，而不是只停留在平均阶的数论命题。

5. Entropy decrement 的作用

entropy decrement 的思想是寻找一个尺度，使得随机变量 $n$ 与它的素数倍 $pn$ 之间的条件信息变少。换句话说，在合适尺度上，系统看到的结构不会因为乘一个小素数而增加太多复杂度。

这一步的意义是把“乘法平移”转成“动力系统里可比较的两个观测”。当这个信息损失足够小的时候，短区间估计、BSZ 双线性结构和 correspondence principle 就可以接上。

6. Nilsystem 方向

nilsequence 是零熵系统中非常重要的一类模型。它既有足够丰富的几何结构，又保留了可计算的 Fourier 分析。把 Sarnak、Chowla、短区间乘法函数和 nilsequence 放在一起看，真正的问题是：乘法函数的随机性如何穿过 nilmanifold 上的低复杂度轨道。

这条路线目前最有价值的地方，不是把所有情形一次性解决，而是提供了一张方法地图：线性相关转双线性相关；短区间估计提供局部随机性；entropy decrement 选择尺度；Furstenberg 原理把数论相关放回动力系统。
2018年1月8日
短区间中 Mobius 函数与 nil-sequence 的相关估计

旧博客原文

原题：The correlation of Mobius function and nil-sequences in short interval

I wish to establish the following estimate:

Conjecture :(correlation of Mobius function and nil-sequences in short interval)

$\lambda(n)$ is the liouville function we wish the following estimate is true.

$\int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH)$ .

Where we have $H\to \infty$ as $x\to \infty$ , $\Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\}$ is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem (multiplication function in short interval)

$f(n): \mathbb N\to \mathbb C$ is a multiplicative function, i.e. $f(mn)=f(n)f(m), \forall m,n\in \mathbb N$ . $H\to \infty$ as $x\to infty$ , then we have the following result,

$\int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH)$ .

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem(correlation of multiplication function and nil-sequences in long interval)

$f(n): \mathbb N\to \mathbb C$ is a multiplicative function, i.e. $f(mn)=f(n)f(m), \forall m,n\in \mathbb N$ . $g(n)=a_n^m+...+a_1n+a_0$ is a polynomial function then we have the following result,

$\int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X)$ .

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

希望建立的估计可以粗略写成：对 Liouville 函数或 Mobius 函数 $\lambda(n)$，以及复杂度受控的 nil-sequence $F(g^n x)$，在短区间 $I=[X,X+H]$ 上有

$$\frac1H\sum_{n\in I}\lambda(n)F(g^n x)=o(1).$$

这里 $H=H(X)\to\infty$，但 $H$ 可以远小于 $X$。这类估计如果成立，会把短区间乘法函数理论和 Sarnak/Chowla 型问题连接起来。

短区间中的 Mobius-nilsequence 相关估计试图在局部窗口内捕捉乘法函数的随机性。

1. 为什么是 nil-sequence

nil-sequence 是低复杂度动力系统轨道的模型。多项式相位 $e(P(n))$ 是最基本例子，更高阶 nilmanifold 上的轨道则对应高阶 Fourier 分析中的结构部分。若 Mobius 与所有这类低复杂度序列正交，就说明它在动力系统意义下表现得像随机噪声。

2. 短区间困难

长区间中可以使用 Bourgain-Sarnak-Ziegler criterion、Vinogradov 型估计和 nilsequence equidistribution。短区间的问题更硬，因为平均长度不够，许多全局消去无法直接使用。

Matomaki-Radziwill 的定理说明，乘法函数在几乎所有短区间中仍有平均消去。这给出一个强烈信号：如果 nil-sequence 的结构在短窗口上足够规则，那么相关和也应当消失。

3. 与 Chowla 的关系

对数平均 Chowla 猜想可以通过 Fourier 展开和结构分解，转化为乘法函数与低复杂度序列的相关估计。这里的短区间版本相当于把“全局随机性”压缩到局部窗口中观察。

4. 可能路线

一个可行框架是：先用短区间乘法函数定理处理非结构部分，再对 nil-orbit 做定量 equidistribution 分解，最后用 BSZ 型准则控制剩余相关。核心瓶颈是所有常数都必须对短区间长度 $H$ 有足够好的依赖。

2017年12月18日
Sarnak 猜想的标准模型：skew product 与 interval exchange

旧博客原文

原题：Sarnak conjecture, understand with standard model

Sarnak conjecture is a conjecture lie in the overlap of dynamic system and number theory. It is mainly focus on understanding the behavior of entropy zero dynamic system by look at the correlation of an observable and the Mobius function .

We state it in a rigorous way:

let $(X,T)$ be a entropy zero topological dynamic system. Let Mobius function be defined as $\mu(n)=(-1)^t$ , where $latex$ is the number of different primes occur in the decomposition of $n$ .

Then for any continuous function $f:X\to R$ and $x\in X$ , observable $\xi(n)=f(T^n(x))$ is orthogonal to the Mobius function; i.e. ,

$\lim_{N\to \infty}\frac{1}{N}\sum_{n=0}^{N-1}\mu(n)\xi(n)=o(N).$

I mainly focus on the special cases when dynamic system $X$ is the skew product on $T^2$ and when the dynamic system which is a interval exchange in $[0,1]$ .

Skew product

For the first one, $\Theta=(T,T^2)$ , $T:T^2\longrightarrow T^2$ :
$T(x)=x+\alpha,T(y)=cx+y+h(x)$
$y_1(n)=T^{n}(x)=x+n\alpha,y_2(n)=T^n(y)=nx+\frac{n(n-1)}{2}\alpha+y+\sum_{n=1}^{N-1}h(x+i\alpha)$ , where $c=1,-1$ .

by Bourgain-Ziegelar-Sarnak theorem we know the difficulties is focus on deal with the exponent

$S_{p,q}(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)(\frac{e(npm\alpha)-1}{e(m\alpha)-1}- \frac{e(nqm\alpha)-1}{e(m\alpha)-1})}$

for all $p,q$ is suffice large primes pair.

and a much simper case is the affine map: $T:(x,y)\to (x+\alpha,cx+y+\beta)$ on $\mathbb T^2$ and the general case $T:(x_1,...,x_n)\to A(x_1,...,x_n)$ where A is a upper-triangle matrix with diagonal 1; i.e. $A=I+B$ , B is nilpotent. So the sarnak conjecture in this case is reduce to the Davenport estimate on exponent by B-Z-S theorem:

$|\sum_{n=0}^{N}e^{2\pi if(n)}|\leq c_A\frac{N}{(log N)^A}$ , $\forall A>0$ .

Interval exchange map

For the interval exchange map, we can explain it by a composition of rotation of some part of $S_1$ step by step and with a renormalization process to glue the neighbor rotations.

Now let us explain a little with this interesting dynamic system. We focus in the simplest nontrivial case, which is the 3-interval exchange map. In this case, just consider the permutation of intervals $I_1,I_2,I_3$ , and it is easy to see there is only one case is nontrivial that is permutation: $I_1\to I_3,I_2\to I_2,I_3\to I_1$ . We explain a little more with other trivial case:

When $I_1\to I_2,I_2\to I_3,I_3\to I_1$ , the interval exchange map is just a rotation and for which the sarnak conjecture is just come from:

$|\sum_{n=0}^{N}e^{2\pi in\alpha}\mu(n)|=o(N)$ , $\forall \alpha\in R$ .

Which is trivial because $\sum_{n=0}^{N}e^{2\pi in\alpha}\mu(n)=\frac{1-e^{2\pi iN\alpha}}{1-e^{2\pi i\alpha}}$ .

For the case $I_1\to I_2, I_2\to I_1, I_3\to i_3$ the map $T$ is a rotation on $I_1\cap I_2$ but it is a identity map on $I_3$ and the orbits of point only lying one of $I_1\cap I_2, I_3$, lying in which one depend on the original point $x$ we take is lying in which one.

Now we focus on the most difficult situation. It is annoying but it is the obstacle we must get over to go far. Fortunately it could be explained as in the following picture.

3-Interval exchange map as two rotation map glue with a renormalization map.

Now we explain what happen in the picture, it is mainly say one identity, which explain how to look 3-interval exchange map as a composition of rotation map with a renormalization map to glue them. Rotation is a kind of map we have good understanding but we do not understand very well with the renormalization map which is glue the two endpoints of $I_2,I_3$ which are not the common endpoint of them. Then you get two circle glue like a “8” , and $T_2$ is just rotate one of it and make the other one to be invariance.

Now we roughly could think about what is the thing we need to charge with, it is just:

$\sum_{n=0}^{N}f((T_1\circ R\circ T_1)^n(x))\mu(n)=o(N)$ .

Now we do some calculate with this geometric explain of interval exchange map.

Let $A=I_1, B=I_2\cap I_3$ , then $A\cap B=\emptyset, A\cup B=[0,1]$ . And $|A|=\alpha$ , $0<\beta<|B|$ . the rotation $T_1:x\to x-\alpha$ , $T_2:x\to x+\beta$ .

Standard model

Is there a standard model of entropy zero dynamic system?

This problem seems to be too ambitious. But it occur naturally when I an trying to have a global understand of the Sarnak conjecture.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Sarnak 猜想位于动力系统和解析数论的交界处。它说零熵动力系统产生的确定序列，应该和莫比乌斯函数这样的算术随机序列正交。

Sarnak 猜想的标准模型包括 skew product、unipotent affine maps 和 interval exchange maps。

1. 基本陈述

设 $(X,T)$ 是零拓扑熵系统，$f\in C(X)$。Sarnak 猜想断言

$$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to0.$$

这里 $\mu(n)$ 是 Mobius function。零熵表示轨道复杂度低，而 $\mu(n)$ 预期具有强随机性。

2. Skew product 模型

典型例子是

$$T(x,y)=(x+\alpha,y+h(x))\pmod1.$$

对 Fourier character 展开后，问题会变成

$$\sum_{n\le N}\mu(n)e(P(n))$$

或更一般的旋转 Birkhoff sum 相位。Bourgain-Sarnak-Ziegler 准则可以把莫比乌斯相关转为不同素数伸缩下的双线性相关。

3. Affine nilsystem 情形

若环面自同态由上三角 unipotent 矩阵给出，例如 $A=I+B$ 且 $B$ nilpotent，那么 $T^n$ 的坐标是 $n$ 的多项式。因此 Sarnak 猜想可归约到 Davenport 型多项式指数和估计。

4. Interval exchange maps

interval exchange map 可以看作把区间切成有限段后重排。它通常是零熵，但没有简单的光滑结构。它的 renormalization 来自 Rauzy induction，类似连续分数在旋转中的作用。

这里的困难是：相位不再是一个光滑多项式，而是经过多次 induction 拼接出来的低复杂度序列。

5. 标准模型的意义

skew product 展示了“低熵加光滑结构”如何导出指数和；interval exchange 展示了“低熵但不光滑”的困难。理解这两个模型，就能看清 Sarnak 猜想里动力系统复杂度与数论随机性之间的真正接口。

2017年11月23日
Bourgain-Sarnak-Ziegler criterion：Mobius 正交性的有限检验

旧博客原文

原题：Bourgain-Sarnak-Ziegler Criterion

Bourgain-Sarnak-Ziegler定理可以视为Vingrodov均值定理的有限版本。

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Bourgain-Sarnak-Ziegler criterion 可以看成 Vinogradov 均值思想的有限版本：若一个有界序列在不同素数倍采样下彼此近似正交，那么它与 Mobius 函数也应当正交。

BSZ criterion 用不同素数倍采样的相关消失来推出 Mobius 正交性。

1. 判别法的形式

设 $a_n$ 是有界序列。若对不同素数 $p\ne q$，有

$$\sum_{n\le N}a_{pn}\overline{a_{qn}}=o(N)$$

并且这个估计对一批素数足够一致，那么可以推出

$$\sum_{n\le N}\mu(n)a_n=o(N).$$

2. 为什么素数倍相关重要

Mobius 函数的困难在于它携带素因子结构。BSZ criterion 的想法是：不直接分析 $\mu(n)$，而是检查序列 $a_n$ 对不同素数尺度是否产生相关。如果所有这些相关都小，Mobius 就没有可利用的结构与之耦合。

3. 与 Sarnak 猜想

在动力系统中，常取 $a_n=f(T^n x)$。于是 Mobius disjointness 变成

$$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to0.$$

BSZ 把这个问题转化为比较 $T^p$ 和 $T^q$ 产生的两个轨道序列。

4. 有限版本的意义

称它为 Vinogradov 均值定理的有限版本，是因为它同样通过“多重相关消失”来控制原始振荡和。它特别适合低复杂度系统，例如 nilsequence、skew product 或 substitution dynamics。

2017年9月25日
Sarnak 猜想在 skew product 上的情形

旧博客原文

原题：Sarnak猜想在skew product上的情形。

Cylinder map:
Cylender map:这是一个动力系统 $\Theta=(T,T^2)$ , $T:T^2\longrightarrow T^2$ 满足：\\
$T(x)=x+\alpha,T(y)=cx+y+h(x)$
因此
$y_1(n)=T^{n}(x)=x+n\alpha,y_2(n)=T^n(y)=nx+\frac{n(n-1)}{2}\alpha+y+\sum_{n=1}^{N-1}h(x+i\alpha)$
来自动力系统 $\Theta$ 中的可观测量是指 $\xi(n)=f(T^n(x))$ ,其中 $x\in T^2$ ,$f\in C(T^2)$.
由于Cylender map是零熵的，这个情形下Sarnak猜想成立等价于:
$S(N)=\sum_{n=1}^N\mu(n)\xi(n)=\sum{n=1}^N \mu(n)f(T^nx)$
满足 $S(N)=o(N)$ ,由于 $f_{\lambda_1\lambda_2}=e^{2\pi i(\lambda_1 x+\lambda_2 y)}$ 是 $C(T^2)$ 的一组基，只需对 $f_{\lambda_1\lambda_2}$ 证明 $S(N)=o(N)$ \\
展开 $S(N)$ ,我们有\\
$S(N)=\sum_{n=1}^N\mu(n)\xi(n)=\sum_{n=1}^N \mu(n)f(T^nx)$ \\

$=\sum_{n=1}^N\mu(n)e^{2\pi ik(\lambda_1(x+n\alpha)+\lambda_2(nx+\frac{n(n-1)}{2}+y\sum_{i=1}^{n-1}h(x+i\alpha)))}$ \\

$=\sum_{n=1}^N\mu(n)e^{2\pi i(\phi(n)+\sum_{i=1}^{n-1}h(x+i\alpha))}$ \\

$=\sum_{n=1}^N\mu(n)e^{2\pi i(\phi(n)+\sum_{i=1}^{n-1}\sum_{m\in Z}\hat h(m)e^{2\pi im(x+i\alpha)})}$ \\

$=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}$ \\
其中我们暂时假定h是解析的，实际上我们要求对h的fourior级数有下界控制，总的来说就是 $\exists \tau_1,\tau_2$ :
$e^{\tau_1 m}<<\hat h(m)<<e^{\tau_2 m}$

\begin{lemma}
$\forall A>0,\forall \phi(n)$ 为多项式函数,我们有指数和估计：
$|\sum_{n=1}^{N}\mu(n)e^{\phi(n)}|<<\frac{N}{(logN)^A}$
\end{lemma}

此引理来自解析数论指数和理论，那么 $\alpha \in Q$ 情形是引理的直接推论。接下来处理 $\alpha \in R-Q$ 情形，这种情形下，我们定义 $\alpha$ 的连分数展开为：
$\alpha=[q_1,q_2,q_3,....]$

\begin{lemma}
如果 $\alpha$ 的连分数展开有一致的上界，即存在 $C\in N^*$ , $\forall n\in N^*,1\leq q_n\leq C$ 那么：
$sup_{0\leq a<b\leq 1}|\sum_{k=0}^{N-1}\chi_{(a,b)}(\{k\alpha\})-N(b-a)|=O(log N)$

\end{lemma}

这个引理的证明由三部分组成，第一部分用一个初等的trick加上连分数表示得到一系列长度区间上的更好的估计，第二部分建立一个有效性估计，第三部分将任何区间拆分成第一种区间的并，并使得余项被有效性估计控制。

我们现在考察最后这个式子：
$S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}$

我们对这个式子建立有效的估计，指的是能够证明:
$S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)$
那么我们接下来建立这个估计,这个估计主要由三部分组成，我们分成三节处理这三部分，最后一节是总结。\\
1.带密度的指数和估计。\\
2.cut-off估计。\\
3.一致性均匀估计。\\

\newpage
\section{带密度的指数和估计}
$S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)$
令： $A_n=\mu(n)e(\phi(n)),B_n=e(\sum_{m\in Z}e(mx)\hat H(m))$
经典的指数和估计是：
theorem：
$\forall A>0,\forall \phi(n)$ 为多项式函数,我们有指数和估计：
$|\sum_{n=1}^{N}\mu(n)e^{\phi(n)}|<<\frac{N}{(logN)^A}$

theorem：
对于 $P$ 是一个质数，对于 $P<<N_1<<N$ :\\定义 $\chi_{p}(n)=e^{\frac{2\pi in}{p}}=e_p(n)$ , 定义 $f:N^*\to Im(\chi_p)$ 满足:\\
对于任何长度为 $N_1$ 的一段区间$I$，对任意 $k\in \{0,1,...,p-1\}$ ,
$\sharp\{n\in I|f(n)=e_p(k)\}=\frac{N_1}{p}+O(1)$

$|\sum_{n=1}^{N}\mu(n)f(n)e^{\phi(n)}|<<_{C}\frac{N}{(logN)^A}$
其中 $C\sim P,A$ \\
$\mu$ 是Mobius函数

cut-off 估计
在式子 $S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)$
中，我们希望对 $m\in Z$ 和 $1\leq n \leq N$ 做cut off来简化问题。\\
后者是简单的，我们待定一个常数c，有:
$S(N)=\hat S(n)+\sum_{n=1}^{cN}\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=\hat S(n)+O(cN)$
c可以待定，之后取得任意小，所以这一部分误差不影响我们最后的结果。\\
对m做cut off会稍微复杂一些，根据Fourior分析我们知道:\\
1如果 $h\in C^{\omega}(T)$ ,则
$\hat h(m)=O(e^-\tau m)$ .
2.若 $h\in C^{d}(T)$ ，则根据分部积分公式 $\hat h(m)=O(m^{-d})$ .\\
接下来的结果可能可以用调和分析中的几乎正交性改进到更好的结果，但是至少我们有:\\
$e(\sum_{|m|>\delta}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1})\sim \sum_{|m|>\delta}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}$
$=O(\sum_{|m|>\delta}m \cdot m^{-d})=O(\delta^{d-2})$
所以至少当 $d>2$ 时，我们可以找到 $\delta \to \infty$ 当$N\to \infty$，使得 $|m|>\delta$ 的部分可以被cut off.

连分数与Ostrowoski表示
我们知道任何一个(0,1)中的数都有连分数表示，并且这个表示是唯一的。
$\alpha=(q_1,q_2,....,q_n,...)$
此时我们定义正整数集 $N^*$ 关于 $\alpha$ 的Ostrowoski表示（wangzhiren 2）：
定义：
每一个正整数n可以唯一的表示为：
$n=\sum_{i=0}^{\infty}(\Pi_{j=0}^{i-1}q_j)r_i$
其中 $r_i \in [0,q_{i}-1]$

很明显上面表示中只有有限个 $r_i$ 不为0,为什么要利用Ostrowoski表示，关键在于Ostrowoski表示中的标架 $\{q_1...q_k\}$ 是最佳逼近下最好的标架。\\
实际上我们归纳定义 $\alpha-$ 标准长度 $\{l_k\}_{k=1}^{\infty}$ 如下：

$l_1=\alpha,l_{k+1}=1-[\frac{1}{l_k}]l_k$

容易知道 $l_{k+1}<l_{k}$ ,做一些微小的计算会发现第k个 $\alpha$ -标准长度和连分数展开的前k项系数乘积之间能够相互控制。
lemma:
$\forall k\in N^*$
$\frac{1}{2q_1...q_k}<l_k<\frac{1}{q_1....q_k}$

直接将 $\alpha$ 的连分数展开代入计算即可证明。 $\alpha$ -标准长度的关键性质是 $\{n\alpha\}$ 在这个区间中的均匀分布性的余项可以得到很好地控制。
lemma:
对任意 $k\in N^*$ ,对任意长度为 $l_k=(a,b)$ 的区间 $I_k\subset (0,1)$ , $\forall N\in N^*$ 我们有：
$\sum_{n=1}^N\chi_{(a,b)}(\{n\alpha\})=N(b-a)+O(1)$

证明是对k归纳，实际上k等于1的时候将 $f(n)=\{n\alpha\}$ 提升为 $g=n\alpha$ ,因为实轴上长度为 $\alpha$ 的区间中一定会包含一个 $\{g(1),...,g(n)\}$ 中的元素，有由于长度为 $n\alpha$ 的区间中有 $[n\alpha]$ 个整数，所以：
$\sum_{n=1}^N\chi_{(a,b)}(\{n\alpha\})=[N\alpha]=N(b-a)+O(1)$
归纳过渡也是简单的。

\section{一致性均匀估计}
最后我们要建立一致性均匀估计，将对
$S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}$
进行多尺度分解，并且说明他和一个多重带密度的指数和的差是$o(N)$
\newpage

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

这篇笔记讨论一个很典型的零熵动力系统：圆环或二维环面上的 skew product。Sarnak 猜想在这里会变成一个指数和问题。动力系统给出相位，莫比乌斯函数给出算术权重，最后要证明两者没有长期相关。

Skew product 的迭代把 Fourier character 转成带有旋转 Birkhoff sum 的指数和。

1. Skew product 的形式

考虑二维环面上的映射

$$T(x,y)=(x+\alpha,\;y+h(x))\pmod 1.$$

这里 $\alpha$ 是旋转数，$h$ 是足够光滑或解析的函数。若 $f\in C(\mathbb T^2)$，Sarnak 猜想要求

$$\frac1N\sum_{n\le N}\mu(n)f(T^n(x,y))\to0.$$

由于 trigonometric polynomials 在 $C(\mathbb T^2)$ 中稠密，可以先检验 Fourier characters $f(x,y)=e(mx+ny)$。

2. 相位展开

迭代 $T$ 得到

$$T^k(x,y)=\left(x+k\alpha,\;y+\sum_{j=0}^{k-1}h(x+j\alpha)\right).$$

所以相关和变成

$$\sum_{k\le N}\mu(k)e\left(m(x+k\alpha)+n y+n\sum_{j
问题的核心是控制这个由旋转 Birkhoff sum 产生的相位。若 $h$ 是多项式或 Fourier 支持很简单，相位可以化成多项式相位，经典解析数论的指数和估计可以直接进入。

3. 有界型旋转数

当 $\alpha$ 的连分数展开系数有一致上界时，旋转轨道具有较好的均匀分布余项。Ostrowski 表示把任意长度拆成由分母 $q_k$ 控制的标准块：

$$N=\sum_k b_k q_k.$$

这些标准块是处理旋转和的自然尺度。每一块上相位的波动可控，块与块之间再通过 cut-off 和 Fourier 展开拼接。

4. Cut-off 与 Fourier 级数

若 $h$ 解析，它的 Fourier 系数指数衰减；若只要求有限光滑性，则系数只有多项式衰减。把高频部分 cut off 后，误差由

$$\sum_{|r|>R}|\widehat h(r)|$$

控制。低频部分则给出有限多个可估的指数和。这里调和分析中的 almost orthogonality 可以改进一些粗糙估计，但最重要的是把问题压缩到有限频率。

5. 证明图像

整个论证可以理解成三层：先把 observables 化成 Fourier characters；再把 skew product 的迭代化成旋转和；最后用连分数分块、cut-off 和带密度的指数和估计控制莫比乌斯相关。这个模型清楚展示了 Sarnak 猜想在零熵系统中常见的结构：动力系统低复杂度负责给出可分解相位，解析数论负责证明乘法函数无法跟随这些相位。

2017年9月25日