分类： Sarnak conjecture

Log average sarnak conjecture

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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/}.

 

广告

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

hxypqr留下评论

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).

Sarnak猜想在skew product上的情形。

hxypqr留下评论

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