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Rotation number：圆周同胚的提升、周期点与半共轭

旧博客原文

原题：Rotation number

Consider compact 1 dimension dynamic system.

We focus on $S_1$ , it does not mean $S_1$ is the only compact 1 dimensional system , but it is a typical example.

$T: S_1\to S_1$ .

If $T$ is a homomorphism then $T$ stay the order of $S_1$ (by continuous and the zero point theorem). That is just mean:

(may be do a reflexion $e^{2\pi i\theta}\to e^{-2\pi i\theta}$ ).

In the homomorphism case. We try to define the rotation number to describe the expending rate of the dynamic system.

$T: \mathbb S_1\to \mathbb S_1$ i.e. $T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ . Lifting to,

$\hat T:\mathbb R\to \mathbb R$ .

How to realize the lifting?

Step1: Periodic extend $T:\mathbb S_1\to \mathbb S_1$ to $T': \mathbb R\to \mathbb S_1$ . (Regard $\mathbb S_1$ as $[0,2\pi]$ ).

Step2: Consider the “flow” of $T':\mathbb R\to \mathbb S_1$ . we get $\hat T:\mathbb R\to \mathbb R$ .

The rotation number is defined as:

$\rho (T)=\limsup_{n\to \infty}\frac{\hat T^n(x)}{n}$ .

Following we will shall it is independent of the choice of $x$ and $\rho (T)=\lim_{n\to \infty}\frac{\hat T^n(x)}{n}$ in fact.

It is not difficult to proved the following property:

Property:

1.If $T'$ is conjugate (in fact semi-conjugate is enough ). Then 1.If $T'$ is conjugate (in fact semi-conjugate is enough ). Thenrotation number of $T'$ equal to rotation number of $T$ .

2.If $T'$ is conjugate to $T$ . Then rotation number of $T'$ equal to
rotation number of $T$ .

$T'(y)=\Psi\circ T\circ \Psi^{-1}(y)=\Psi(\Psi^{-1}(y)+t(\Psi^{-1}(y)))$ )

Example: $T: x\to x+\alpha$ , $\alpha\in \mathbb R$ . It is not difficult to prove the rotation number of $T$ is $\alpha$ .

Propersion:

1)For $n\geq 1$ we have that $\rho(T^n)=\rho(T) (mod 1)$ .

2).If $T$ has a periodic point, i.e. $x\in S_1,\exists n\in \mathbb N^*, T^{n}(x)=x$ . Then $latex\rho (T)$ is rational.

3) $T:\mathbb R/\mathbb Z \to \mathbb R/\mathbb Z$ has no periodic point then $\rho(T)$ is irrational.

4) The limit actually exists and we have: $\rho(T)=\lim_{n\to \infty}\frac{\hat T^n(x)}{n} (mod 1)$ .

pf of 1):

$\rho(T^n)=\lim_{k\to \infty}\frac{(\hat T^n)^{k}(x)}{k}$

$=\lim_{k\to \infty}n\frac{(\hat T)^{nk}(x)}{nk}$

$=n\rho (T)$ .

Used the property $|x-y|<k \leftrightarrow |T^{\omega}(x)-T^{\omega}(y)|<k+1, \forall \omega\in N^*, \forall k\in \mathbb Z^{+}$ .

pf of 2):

It is not difficult to prove $\rho(T)$ is independent with the choice of $x$ . So choose $x$ to be the periodic point.

Remark: but the inverse of 2) is not true. For example:

$x\to x+\frac{1}{2}+\frac{1}{100}sin(4\pi x)$ .

This dynamic system has both periodic points( $\{0,\frac{1}{2}\},\{\frac{1}{4},\frac{3}{4}\}$ ) and non-periodic pint (maybe orbits generated by $\{\frac{1}{\sqrt{2}}\})$ .

pf of 3):

If not. Assume $\rho(T)$ is rational number $\frac{q}{p}$ . Take any point $x\in \mathbb S_1$ , then:

$\lim_{n\to \infty}\frac{\hat T^n(x)}{n}=\frac{q}{p}$ .

$\Longrightarrow \lim_{n\to \infty}\frac{(\hat T^p)(x)}{n}=q$ .

$\Longrightarrow \lim_{n\to \infty}\frac{(\hat T^p-q)^n(x)}{n}=0$ .

Now assume $\hat T^p-q=\widetilde T$ .

Then $\widetilde x>x$ . $\forall x\in \mathbb S_1$ (if $\widetilde x<x, \forall x\in \mathbb S_1$ , take reflection $x\to -x$ ).

And there do not exists $n\in \mathbb N^*$ such that $\widetilde T^nx>x+1$ . If not, we could prove rotation number is large than $\frac{1}{n}$ lead a contradiction.

So $\{\widetilde T^nx\}_{n=1}^{\infty}$ is a bounded monotonically increasing sequences in $\mathbb S_1$ , it limits point $z\in \mathbb S_1$ must satisfied $\widetilde T^n (z)=z$ .

pf of 4):

Using the point wise approximation inequality induced from the monotonically and stay ordering property of $\mathbb S_1$ by $T$ .

Corollary:

Assume $\rho(T)$ is irrational.

1. Let $n_1,n_2,m_1,m_2\in \mathbb Z$ , and $x,y\in \mathbb R$ . If $\hat T^{n_1}(x)+m_1<\hat T^{n_2}(x)+m_2$ , then $hat T^{n_1}(y)+m_1<\hat T^{n_2}(y)+m_2$ .

2. The bijection $n\rho (T)+m\to \hat T^n(0)+m$ between the set $\Omega=\{n\rho(T)+m| n,m\in \mathbb Z\}$ and $\Gamma=\{\hat T^{n}(0)+m,n,m\in \mathbb Z\}$ precise the natural ordering on $\mathbb R$ .

This corollary is not difficult to prove use the established property.

Denjoy’s theorem

Proposition:

If $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is a minimal orientation presenving homomorphism with irrational rotation number $\rho$ then $T$ is topologically conjugate to the standard rotation $R_{\rho}: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ .

leave as a ex.

For $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ , $T': \mathbb R/\mathbb Z\to \mathbb R$ . We define the variation of $log|T'|: \mathbb R/\mathbb Z\to \mathbb R$ by:

$Var(log(|T'|))=$

$sup\{\sum_{i=0}^{n-1}|log|T'|(x_{i+1})-log|T'|(x_i)|: 0=x_0<x_1<...<x_n=1\}$

We say that the logarithm of $|T'|$ has bounded variation if this value $Var(log|T'|)$ is finite.

Denjoy’s theorem:

If $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is a $C^1$ orientation preserving homomorphism of the circle with derivative of standard variation and irrational rotation number $\rho=\rho(T)$ then $T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is topologically conjugate to the standard rotation :

$R_{\rho}:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ .

Due to the upper proposition we only need show $T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is minimal. Proof pf minimal is splitting to following two sub lemmas.

Sublemma1:

If $T$ has irrational rotation number and there are a constant $C>0$ and a sequences of integers $q_n\to \infty$ such that the map: $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ Satisfy : $|(T^{q_n})'(x)||(T^{-q_n})'(x)|\geq C$ Then $T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is minimal.

Sublemma2:

Fix $x\in \mathbb R/\mathbb Z$ and write $x_n=T^n(x)$ , for $x\in \mathbb Z$ There exists an increasing sequences $q_n\to \infty$ of natural number such that the intervals $(x_0,x_{q_n}),(x_1,x_{q_n+1}),...,(x_i,x_{q_n+i}),...,(x_{q_n},x_{2q_n})$ are all disjoint.

Paradox and problem

Graph:

$\rho(T)>0$ because of existence of fix point.

$T_{\alpha}=T+\alpha$ for $\alpha< sup_x|T_x-x|$ .

Is $\rho(T_{\alpha})=0$ always true for $\alpha \in R$ ?

If it is right, then there is a contradiction with argument $(*)$ , but for what type of dynamic system $T$ ?

$T_{\alpha}=T+\alpha$ satisfied $\rho(T_{\alpha})=\rho(T)+\alpha$ . for all $\alpha\in \mathbb R$ ?

Problem:

If $T$ is not homomorphism but $T:x\to x+g(x)$ induced $g(x)=x-f(x)$ , $f(x)$ is striating increasing, Is the limit of $\lim_{x\to \infty}\frac{\hat T(x)}{n}$ always exists? it could not be increase with $x$ .

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。

rotation number 是一维动力系统中最基本的不变量之一。它衡量圆周同胚平均每次迭代旋转多少。

圆周同胚提升到实线后，rotation number 是迭代平均位移的极限。

1. 提升到实线

把圆周写成 $S^1=\mathbb R/\mathbb Z$。若 $f:S^1\to S^1$ 是保向同胚，可以取一个 lift $F:\mathbb R\to\mathbb R$，满足

$$F(x+1)=F(x)+1.$$

rotation number 定义为

$$\rho(F)=\lim_{n\to\infty}\frac{F^n(x)-x}{n}\pmod1.$$

这个极限存在，并且与 $x$ 的选择无关。

2. 基本性质

若 $f$ 与 $g$ 共轭，则它们有相同 rotation number。更弱的半共轭在很多情形下也保留 rotation number。标准旋转

$$R_\alpha(x)=x+\alpha$$

的 rotation number 就是 $\alpha$。

3. 周期点与有理数

如果 $f$ 有周期点，即 $f^q(x)=x$，那么

$$\rho(f)=\frac pq\in\mathbb Q.$$

反过来，对保向圆周同胚，若 rotation number 是有理数，则存在周期轨道。无理 rotation number 则排除周期点。

4. Denjoy 图像

若 rotation number 无理，系统常与无理旋转相关。足够光滑且导数变差有限时，Denjoy theorem 给出与刚性旋转的半共轭，甚至在更强条件下共轭。

5. 为什么它重要

rotation number 把一个非线性圆周动力系统压缩成一个算术量。这个量同时控制周期轨道、轨道排序和与刚性旋转的关系，是一维动力系统从拓扑进入数论的入口。

2017年12月4日
Floquet theory：周期系数线性系统与 monodromy matrix

旧博客原文

原题：Eloquent theory

Consider matrix ODE:

$\dot{\phi}(t)=A(t)\phi(t)$

Where $A(t)$ is a given periodic matrix with period $T$ , i.e. $A(x)=A(x+T), \forall x\in R$ .

Then the solution $\phi(t)$ satisfied identity:

$\phi(t+T)=\phi(t)\phi^{-1}(0)\phi(T)$ .

This could be explained as $\phi^{-1}\phi(T)=\int_{0}^T\phi(t)$ .

Now we consider to solve the equation: $e^{TB}=\phi^{-1}(0)\phi(T)$ . At least formally it could be solved:

$B=\frac{1}{T}log(\frac{\phi(T)}{\phi(0)})$ .

(Unfortunately $log$ is a multi-value function so $B=B_0+2\pi ik I$ , where $I$ is the identity matrix and $B$ is a solution of $e^{TB}=\phi^{-1}(0)\phi(T)$ .) This argument is false.

In fact matrix is not like numbers, the $log$ function is much more complicated. we have,

$log(A)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{A^n}{n} ...(*)$

So to solve $e^{TB}=\frac{1}{T}(\frac{\phi(T)}{\phi(I)})$ , it is equivalent to :

$B=\frac{1}{T}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(\frac{\phi(T)}{\phi(I)})^n}{n}$

But this type of identity only meaningful when $||\frac{\phi(T)}{\phi(I)}||<1$ , so is it true that for $||\frac{\phi(T)}{\phi(I)}||<1$ the equation is solved by $(*)$ , and for $||\frac{\phi(T)}{\phi(I)}||\geq 1$ it do not have solution?

The naive inspirit is wrong, the situation is similar to the $\mathbb Q_p$ case while $log_p$ could extend to $D(p^{\frac{-1}{p-1}-})$ and the identity

$exp_p(log_p(1+x))=1+x$

always holds for $x\in D(p^{\frac{-1}{p-1}-})$ . The key observation is $log[(1+Y)(1+Y)]=log(1+X)+log(1+Y)$ always holds when $||X||,||Y||<1$ , this will lead to a reasonable value of

$log[(1+X)(1+Y)]=log(1+X+Y+XY)$

even when $||X+Y+XY||\geq 1$ and this process could be continue to the whole matrix space and the identity enjoy the accosted principle so $log(X)$ is well-defined for all $X\in M_{2\times 2}$ .

Now it is time to consider the rotation number, which is defined by $\lim_{n\to \infty}\frac{f^{n}(x)-x}{n}$ for $f:R\to R$ is a continuous increasing function.

And I do not know how to associated a dynamic system for the matrix $B$ given here, but in any case it seems iff it is given by a hemoermorphifm then the rotation number is zero due to the following reason:

Consider $\mathbb S^1$ as the quotient $\mathbb R/\mathbb Z$ . Your homeomorphism $f$ lifts to a homeomorphism

$\phi : \mathbb R \to \mathbb R$ such that $\phi(x+1)=\phi(x)+1$ .

Form the map $h:=\frac{1}{q} \sum _{n=1} ^q (\phi^{\circ n}-pn)$ , where $\phi ^{\circ n}$ is the composition $n$ times of $\phi$ with itself. By construction $h\circ \phi = h+\frac{p}{q}$ and $h(x+1)=1+h(x)$ , so that $h$ factors as a homeomorphism of the circle conjugating $f$ to the rotation.
By the way this approach wors in $\mathbb R^n$ too.

Maslov index of a holomorphic disk

A natural way to understand the rotation number here is according the way of maslov index, we have the following formula:

$f(A)=\int_{\Gamma}\frac{1}{2\pi i}\frac{f(\lambda)}{\lambda I-A}f(\lambda)d\lambda$

$TB=log(\frac{\phi(T)}{\phi(0)})=log(\int_0^T \phi'(\lambda)d\lambda)=\int_0^T log(\phi'(\lambda))d\lambda=\int_0^T log(A+F(t))d\lambda$

Proof sketch:

1. $B=\frac{1}{T}log(e^{\int_0^T A+f(t)dt})= \frac{1}{T}(\int_{0}^T A+f(t)dt)$ .

2. The dynamic system is defined by : $W: R^2-\{0\} \to R^2-\{0\}$ , $W( x)=B x$ .

3. this dynamic system $(R^2-\{0\},W)$ is conjugate to the dynamic system $T:S_1\to S_1$ , Not difficult to proof it is a homomorphism on $S_1$ and it is zero entropy by Pesin’s formula

If $T: S_1\to S_1$ could lifting to $\hat T:R \to R$ the rotation number is defined as :

$\lim_{n\to \infty}\frac{\hat T^n(x)}{n}$

This problem is not a good problem due to the philosophy, i.e. use rotation number to describe the information of a hamiltonian flow is not satisfied, in fact it is difficult to establish a suitable definition of “rotation number”! But this is the first crucial thing to establish a theorem!

Hamiltonian flow

In mathematics and physics, a Hamiltonian vector field on a symplectic manifold is a vector field, defined for any energy function or Hamiltonian. A Hamiltonian vector field is a geometric manifestation of Hamilton’s equations in classical mechanics. The integral curves of a Hamiltonian vector field represent solutions to the equations of motion in the Hamiltonian form. The diffeomorphisms of a symplectic manifold arising from the flow of a Hamiltonian vector field are known as canonical transformations in physics and (Hamiltonian) symplectomorphisms in mathematics.[1]

Hamiltonian vector fields can be defined more generally on an arbitrary Poisson manifold. The Lie bracket of two Hamiltonian vector fields corresponding to functions f and g on the manifold is itself a Hamiltonian vector field, with the Hamiltonian given by the Poisson bracket of f and g.

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。

Floquet theory 研究周期系数线性微分方程

$$\dot x=A(t)x,\qquad A(t+T)=A(t).$$

它告诉我们，周期系统的长期行为由一个周期部分和一个指数部分共同决定。

Floquet theory 把周期系数线性系统分解成周期部分和指数部分，monodromy matrix 控制稳定性。

1. 基本解矩阵

令 $X(t)$ 是基本解矩阵，$X(0)=I$。周期性给出

$$X(t+T)=X(t)X(T).$$

矩阵 $X(T)$ 称为 monodromy matrix。它记录系统经过一个周期后的净变化。

2. Floquet 分解

Floquet theorem 说，在复数域上可以写成

$$X(t)=P(t)e^{tB},$$

其中 $P(t+T)=P(t)$，$B$ 是常矩阵。也就是说，周期系统可以拆成周期振荡和指数增长/衰减。

3. 矩阵对数的细节

形式上想令

$$B=\frac1T\log X(T).$$

但矩阵对数是多值的，而且实矩阵上未必能选到实对数。正确表述通常在复数域成立；若要实形式，需要加入额外周期或 Jordan 分解的讨论。

4. 稳定性

monodromy matrix 的特征值称为 Floquet multipliers。若所有 multiplier 的模都小于 $1$，零解渐近稳定；若有模大于 $1$ 的 multiplier，则出现不稳定方向。

5. 与 rotation number 的关系

二维或辛系统中，monodromy 的作用可能诱导圆周或射影线上的动力系统，此时 rotation number 可以描述方向的平均旋转。这把 Floquet theory 和一维动力系统联系起来。

2017年12月3日
Van der Corput trick：从差分到均匀分布
旧博客原文

原题：Van der curpurt trick

There is the statement of Van der carport theorem:

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.

$|\sum_{n=1}^Ne^{2\pi imQ(n)}|= \sqrt{(\sum_{n=1}^Ne^{2\pi imQ(n)})(\overline{\sum_{n=1}^Ne^{2\pi imQ(n)}})}$

$= \sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi imQ(n+h_1)-Q(n)}}=\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi im \partial^1_{h_1}Q(n)}} \leq \sqrt{\sum_{h_1=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)}|}$

$= \sqrt{\sum_{h_1=1}^N\sqrt{ (\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)} )(\overline{\sum_{n=1}^{N-h}e^{2\pi \partial^1_{h_1}Q(n)})}}}\leq\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi\partial^1_{h_2} \partial^1_hQ(n)} |}}$

$\leq ....\leq$

$\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N \sqrt{....\sqrt{\sum_{h_{k-1}=1}^{N-h_{k-2}}|\sum_{n=1}^{N-h_{k-1}}e^{2\pi\partial_{h_1h_2...h_{k-1}Q(n)}}|}}}} =o(1)$

This type of trick could also establish the following result, which could be understand as a discretization of the Vinegradov lemma.
```
Uniformly distribution result of :
Given ,  coverages
 to a uniformly distribution in 
 as .
```
This trick could also help to establish estimate of correlation of low complexity sequences and multiplicative function, such as result:

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

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

The standard estimate of Mobius function is:

$\sum_{n\leq X:n\equiv a~(mod~q)} \mu^2(n)=\frac{6}{\pi^2} \prod_{p|q} \left(1-\frac{1}{p^2} \right)\frac{X}{q}+E(X,q,a)$

The error term $O_{\varepsilon}\left(\sqrt{X/q} +q^{\frac{1}{2}+\varepsilon}\right)$ is true for $q\leq X^{\frac{2}{3}-\varepsilon}$ .

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。

Van der Corput trick 的核心是：不要直接估计一个振荡平均，而是估计它与平移后的相关。若所有非零差分序列都足够均匀，那么原序列本身也应当均匀。

Van der Corput trick 用差分相关替代原始平均，在多项式相位问题中会降低次数。

1. 均匀分布版本

设 $(a_n)$ 是 $\mathbb T^d$ 中的序列。一个典型命题是：若对每个 $h\ne0$，差分序列

$$a_{n+h}-a_n$$

都在 $\mathbb T^d$ 中均匀分布，那么 $a_n$ 也均匀分布。证明通常通过 Weyl criterion，把问题化成指数和估计。

2. 指数和不等式

对复数序列 $u_n$，Van der Corput 不等式给出

$$\left|\frac1N\sum_{n\le N}u_n\right|^2
\lesssim \frac1H+\frac1H\sum_{1\le h\le H}\left|\frac1N\sum_{n\le N-h}u_{n+h}\overline{u_n}\right|.$$

右侧出现的就是相关项。若相关项都小，则原平均小。

3. 多项式相位

当 $u_n=e(P(n))$ 且 $P$ 是次数 $k$ 的多项式时，差分 $P(n+h)-P(n)$ 的次数降为 $k-1$。因此可以用归纳证明 Weyl 型均匀分布结论。

4. 与乘法函数相关

在 Mobius 或 Liouville 与低复杂度序列的相关估计中，Van der Corput trick 常用于把一个序列的复杂度下降一层，再配合 Bourgain-Sarnak-Ziegler criterion。它的角色不是给出最终消去，而是把问题改写成更适合结构分析的形式。
2017年12月1日
短区间上的乘法函数：Matomaki-Radziwill 定理的分析图像
旧博客原文

原题：Multiplication function on short interval

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

The main theorem in them article is :
```
Theorem(Matomaki,Radziwill)
As soon as  when , one has:

                    

for almost all  .
```
In my understanding of the result, the main strategy is:

Step 1:Parseval indetity, monotonically inequality

Parseval indetity, monotonically inequality, this is something about the $L^2$ norms of the quality we wish to charge. It is just trying to understanding

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

as a fuzzy thing by a more chargeable quality:

$\frac{1}{X^2}\int_{0}^{\infty}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt$

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

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

established the monotonically inequality:

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

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 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 spectral decomposition with some basis come from multiplication unclear, i.e. primes.

Step 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:
```
Euler-product formula:
                      
```
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:

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

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.

Step 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.

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

Now we just use a Cauchy-Schwarz:

\sum_{i=1}^n\int_{I_i\times J_i} \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$

Step 4: major term estimate

step 5:minor term estimate

step 6: estimate the contribution of area which is not filled

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。

Matomaki-Radziwill 的工作改变了我们对乘法函数短区间平均的理解。它说明，有界乘法函数在几乎所有短区间中的平均，通常接近其长区间平均。

短区间乘法函数问题把局部平均转化为 Dirichlet polynomial 的频率估计。

1. 基本问题

设 $f$ 是有界乘法函数。我们关心

$$\frac1H\sum_{x
对多数 $x$ 的行为。传统解析数论更擅长长区间平均，而短区间要求理解局部波动。

2. Matomaki-Radziwill 定理

粗略地说，只要 $H\to\infty$ 不太慢，对几乎所有 $x\le X$，短区间平均可以由长区间信息控制。这一结果为 Chowla、Sarnak 和 pretentious multiplicative functions 提供了关键输入。

3. Dirichlet polynomial 视角

把乘法函数平均转化成 Dirichlet polynomial：

$$\sum_{n\le X}\frac{f(n)}{n^{1+it}}.$$

Parseval 型恒等式把短区间均方问题变成 $t$-空间上的积分估计。这是从 additive intervals 进入 multiplicative Fourier analysis 的桥。

4. Cut-off 与单调性

证明中需要去掉某些坏尺度，并建立类似单调性的控制：截断后的对象仍然保留主要信息。这个过程有点像 compactification，把原来粗糙的短区间平均换成更可估的频率对象。

5. 为什么它重要

短区间乘法函数估计让“乘法随机性”可以在局部尺度上使用。Sarnak 和 Chowla 的许多对数平均进展，都依赖这种把局部平均、Dirichlet polynomial 和 entropy decrement 结合起来的能力。
2017年11月30日
Transverse intersections：从 Sard 定理到 generic position

旧博客原文

原题：transverse intersections

https://en.wikipedia.org/wiki/Transversality_(mathematics)

This problem may be a embarrassed one, but I even could not prove it for the 1 dimensional case.

Here is the problem:

>**Question 1** $M$ is a compact $n$ -dimensional smooth manifold in $R^{n+1}$ , take a point $p\notin M$. prove there is always a line $l_p$ pass $p$ and $l_p\cap M\neq \emptyset$ , and $l_p$ intersect transversally with $M$ .

You can naturally generated it to:
>**Qusetion 2** $M$ is a compact $n$-dimensional smooth manifold in $R^{n+m}$ , take a point $p\notin M$ . Prove $\forall 1\leq k\leq m$, there is always a hyperplane $P_p, dim(P_p)=k$ pass $p$ and $P_p\cap M\neq \emptyset$ , and $P_p$ intersect transversally with $M$ .

Thanking for Piotr pointed out, assuming “transverse” means “the tangent spaces intersect only at 0”.

We focus on question 1 for simplified.

Even in 1 dimension it is not easy at least for me, **warning**: a line $l$ pass $p$ may be intersect $M$ at several points combine a set $A_l$ , $A_l$ could be finite, countable or even it is not countable (consider $M$ is induced by a smooth function for which the zeros set is Cantor set.)… And if there is one point $a\in A_l$ , $l$ is tangent with the tangent line of $M$ at $a$ , then $l$ is not intersect transversally with $M$ .

**My attempt**:
I could use a dimensional argument and Sard’s theorem to establish a similar result but instead of a fix point $p$ , we proof for generic point in $R^{n+1}$ which is not in $M$ we can choose such a line.

So it seems reasonable to develop the dimensional technique to attach the question 1, in 1 dimensional, it will relate to investigate the ordinary differential equation:

$\frac{f(x)-b}{x-a}=f'(x)$

Where $p=(a,b)$ , $M$ have a parameterization $M=\{x,f(x)\}$ . If there is a counterexample for the question 1, then there is another solution which satisfied the ODE in the sense:

at least for every line $l$ there is a intersection point $a_l\in l\cap M$ , $f$ satisfied ODE at $a_l$ .

This is just like the uniqueness of the solution of such a ODE is destroyed at some subspace of a line which have some special linear structure, I do not know if this point of view with be helpful.

I will appreciate for any useful answers and comments.

Proof 1(provided by fedja)

Area trick.(weakness:it seems we could not proof the transtivasally intersection point have positive measure by this way).

Proof 2(provided by Piotr)

#For the codimension 1 case.#

###Using Thom transversality theorem.###
Consider the maps $f_s:\mathbb{R} \to \mathbb{R}^n$ parametrized by $s \in S^{n-1}$ and given by $f_s(t) = p + t \cdot s$ . The map $F(s,t) = f_s(t)$ , $F:S^{n-1} \times \mathbb{R} \to \mathbb{R}^n$ is clearly transverse to $M$ , thus Thom’s transversality says that $f_s$ is transverse to $M$ for almost all $s$ . Now it suffices to prove that for an open set in $S^{n-1}$ , the line given by $f_s$ intersects $M$ . Proven below.

###Using Sard’s theorem directly.###
Thom’s transversality is usually proven using Sard’s theorem. Here is the idea.

Consider the projection $\Pi:\mathbb{R}^n \setminus \{p\}\to S^{n-1}_p$ onto a sphere centered at $p$ . A line $l_p$ through $p$ intersects $M$ transversally if the two points $l_p \cap S^{n-1}_p$ are regular values of $\Pi$ (indeed, the critical points of $\Pi$ are exactly the points $x \in M$ at which the normal $\vec n_x$ is perpendicular to the radial direction (with respect to $p$)). By Sard’s theorem, the set of regular values is dense in $S^{n-1}_p$ .

We need to choose any point $s$ on the sphere for which both $s$ and $-s$ are regular values, and the line $f_s$ through $p$ and $s$ actually intersects $M$ . It suffices to prove that the set of points $s$ for which this line intersects $M$ contains an open set. We could now use the Jordan-Brouwer Separation Theorem and we would be done, but we can do it more directly (and in a way that seems to generalize).

###The set of points $s \in S$ for which $f_s$ intersects $s$ has nonempty interior.###
For each point $q \notin M$ the projection $\Pi:M \to S_{q,\varepsilon_q}^{n-1}$ onto the sphere centered at $q$ , of radius $\varepsilon_q$ small enough so that the sphere does not intersect $M$ , has some (topological) degree $d_q$ . It is easy to check that if one takes any point $x \in M$ and considers the points $x \pm \delta \vec n_x$ for small $\delta$ , the degrees of the corresponding maps differ by $1$ . It follows that we can find a point $q$ for which $d_q \neq d_p$ , which guarantees that for every point $q'$ in a small open ball $B$ around $q$ (all these points have same degree $d_q$ ), the line joining $p$ and $q'$ intersects $M$ . Projection of $B$ on $S_p^{n-1}$ is an open set which we sought.

#For the general case (partial solution).

I think a similar reasoning should work, however, notice that for $k < m$ we cannot make $P_p$ intersect transversally with $M$ because of dimensional reasons: the dimensions of $M$ and $P_p$ don’t add up to at least $n+m$ . Recall that transversality implies Thus, either (1) you want to consider $k \geq m$ , or (2) define “transversal intersection” for such manifolds saying that the tangent spaces have to intersect at an empty set.

Also, for $k>n$ we can just take any plane $P_p$ which works for $k=m$ and just extend it to a $k$ -dimensional plane.

###Assuming $k = m$ .###
A similar reasoning should work for $f_s:\mathbb{R}^m \to \mathbb{R}^{n+m}$ with $s = (s_1, \ldots, s_m)$ going over all families of pairwise perpendicular unit vectors, and $f_s(t_1,\ldots,t_m) = p+\sum_{j=1}^m t_i \cdot s_i$ . Thom’s transversality says that for almost all choices of $s$ , the plane $f_s$ is transverse to $M$ .

### The nonempty interior issue. ###
The only thing left is to prove that the set of $s$ for which the intersection is nonempty has nonempty interior. Last time we proved that there is a zero-dimensional sphere containing $p$ , namely $\{p,q\}$ , which has nonzero linking number with $M$ , and by deforming if to spheres $\{p,q'\}$ and taking lines through pairs $p,q'$ , we got an open set of parameters for which the line intersects $M$ .

Here should be able to do a similar trick by finding a $m-1$ -dimensional sphere with nonzero linking number with $M$ . The ball that bounds that sphere has to intersect $M$ , thus the plane $P$ containing the sphere has to intersect $M$ . By perturbing the sphere we get spheres with the same linking numbers, and get all the planes that lie in a neighbourhood of $P$ ; in particular, we get an open set of parameters $s$ for which $f_s$ intersects $M$ .

Well, we don’t actually need a *round* sphere, but we do need a *smooth* sphere that lies in a $m$ -dimensional plane. There’s some trickery needed to do this, but I am sure something like this can be done.

Maybe somebody else can do it better?

### For $k<m$ ###

I don’t really know how to attack this case, assuming “transverse” means “the tangent spaces intersect only at $0$ “.

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。

Transversality 是微分拓扑中最基础也最有用的原则：如果两个几何对象不是被特殊关系强迫相切，那么经过任意小扰动后，它们通常会横截相交。许多“存在一个好方向”的问题，本质上都是 transversality 和 Sard 定理的影子。

Transversality 把几何相交问题转化为参数空间中的 regular value 问题。

1. 定义

若 $A,B\subset X$ 是光滑子流形，在交点 $x\in A\cap B$ 处称它们 transverse，如果

$$T_xA+T_xB=T_xX.$$

对一条曲线和一条直线来说，这表示交点处切线方向不同；对高维子流形来说，它表示两者的切空间张成了整个环境空间。

2. 固定点与过点直线的问题

一个自然问题是：给定 compact smooth submanifold $M\subset\mathbb R^m$ 和一点 $p\notin M$，是否存在过 $p$ 的直线与 $M$ 横截相交？

困难在于，一条直线可能与 $M$ 有多个交点，甚至交点集合不一定有限。如果某个交点处直线方向落在 $T_xM$ 中，就会失去横截性。因此不能只检查一个交点，而要同时控制所有交点。

3. Sard 定理的用法

把方向空间看成 projective space。对每个 $x\in M$，从 $p$ 指向 $x$ 的方向给出映射

$$\Phi:M\to \mathbb RP^{m-1},\qquad \Phi(x)=[x-p].$$

若选择的方向是 $\Phi$ 的 regular value，那么对应射线与 $M$ 的交点满足横截条件。Sard 定理告诉我们，critical values 的测度为零，所以 generic direction 是好的。

4. 为什么“generic”比“显式构造”容易

显式找一条好直线可能很难，因为坏方向集合由所有相切条件组成。Sard 定理的强处在于，它不需要列出坏方向，只需证明坏方向是某个光滑映射的 critical values。

这就是微分拓扑常见的思想：把几何条件变成参数空间中的 regular value 问题。

5. 高维推广

对过 $p$ 的 $k$-平面，也可以考虑相应的 Grassmannian 参数空间。横截性条件仍然可以写成 evaluation map 的 regular value 条件。Thom transversality theorem 进一步说明，在函数空间中 transverse maps 构成 residual set。

所以这类问题的正确答案通常不是找一条神奇的线，而是证明坏参数集合很小，因而几乎所有选择都好。

2017年11月29日
Vinogradov 估计的一条思路：环面均匀分布与连分数尺度

旧博客原文

原题：An approach to Vinogradov estimate

Vinogradov estimate is:

$|\sum_{n=1}^{N}e^{2\pi i\alpha P(n)}|\leq c_A\frac{N}{log^A N}$

For fix $\alpha$ is irrational and $\forall A>0 ... (*)$ .

Assume $deg(P)=n$ , this could view as a effective uniformly distribute result of dynamic system: $([0,1]^n,T)$ , where $T: x\to (A+B)x$ , $b$ is a nilpotent matrix, matrix $A$ is identity but with a irrational number $\alpha$ in the $(n, n)$ elements.

First approach

we could easily to get a “uniform distribute on fiber” result without very much tough estimate to attach the theorem. That is just a application by my “rigid trick” that is describe in my early note. But this approach is according to the understanding of the result as a uniformly distribute result on Torus $T^n$ , we could do this approach with the last $S^1$ , which will corresponding to $\partial^{n-1}x_k$ , i.e. we could apply the “rigid trick” to prove sequences $(x_k,\partial^1 x_k,..., \partial^{n-1} x_k)$ is uniformly distribute according to $\partial^{n-1} x_k\in S^1$ .

Graph

But this approach seems difficult to generate. The difficulty is come from both there is no similar uniformly distribute of the other perimeter use the rigid trick (At least as I know, I try to prove there could be one but I failed) and if in the best case we have the similar uniformly distribute result for other perimeter there is still some thing more need to be established. See this graph for a counterexample that the uniformly distribute for all fiberation could not derive a uniformly distribute for the original space.

Second approach

In this approach we need use the information of continue fractional to get some information (Which is of course critical to get some information about the estimate). But I do not know if it is necessary, maybe this could be a interesting question weather the information come from continue fractional must involve to get such a estimate in the future, but not today.

Any way, there is two different type of continue fractional:

1. $\alpha=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+...}}}$ .

2. $\alpha=q_0+\frac{1}{q_1}+\frac{1}{q_1q_2}+\frac{1}{q_1q_2q_3}+\frac{1}{q_1q_2q_3q_4}+...$ .

Anyway, these could be understand as a same thing more or less (if fact we can calculate some quantitive with $a_i,q_i$ which is roughly the same). That is just the orbits $\{e^{2\pi i\alpha}\}$ have quasi-period property, that is to say, under certain norms, it could be understand as the limits of periodic sequences. So it is natural to approximation $\{e^{2\pi i\alpha}\}$ by periodic sequences and will lead to a very good point-wise coverage result:

$T_k^{n}(x) \longrightarrow T^{n}(x)$

Where $T_k^{n}(x)=e^{2\pi i\sum_{i=1}^k\frac{1}{p_1...p_i}}$ is just the periodic approximation sequence which come from the best approximation (critical point of $||\frac{q}{p}-\alpha||$ ), which natural occur in continue fractional. And by this we already arrive a non qualitative form result of $(*)$ with $deg(P)=1$ .

But unfortunately this approximation is too good to be true for $deg(P)\geq 2$ case. The reason of this result could be true is just because the natural estimate for the best approximation of $\alpha$ ; i.e. Dirichlet approximation theorem.

But for higher degree case, although we could not expect this thing to be true, we still could image a weaker but enough result to be true:

$\{T_k^{n}(x)\} \longrightarrow \{T^{n}(x)\}$

in the Gromov Hausdorff metric sense, and the $T_k^n(x)$ is carefully chose, which have a finite torsion structure(which could be view as a multilinear structure which will play a central role in the estimate). Here is a graph for $deg(P)=2$ :

Roughly speaking, in general $deg(P)=n$ case, there is a cube structure in the orbits $e^{2\pi iP(n\alpha)}$ and is critical to observe that the progression of difference structure in it. The goal of this approach is to establish some result from the finite torsion structure(multilinear structure). That is to say, the boundary is high order thing in all direction but there is only one direction attend to infinity the other is just a finite torsion, and we wish to get more information from the extra structure.

This also have a physics explaining, for which see the graph:

Third approach

For $P(n)=an^2+bn+c$ case:

$P(k+\Delta)=P(k)+(ka+b)\Delta+\Delta^2$

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。

Vinogradov 型估计可以看成带有有效误差的均匀分布定理。若相位含有无理参数，问题往往在两个语言之间切换：动力系统上是环面轨道的均匀分布，解析数论上是指数和的消失。

Vinogradov 型估计可以看成环面轨道均匀分布的定量版本，核心是控制指数和。

1. 动力系统图像

考虑环面上的序列

$$n\mapsto (n\alpha,n^2\alpha,\ldots)\pmod1.$$

当 $\alpha$ 无理时，这类序列常常均匀分布。Vinogradov 估计要求更强：不仅要知道平均极限，还要给出可用的误差项。

2. 指数和形式

Weyl criterion 把均匀分布转成指数和：

$$\frac1N\sum_{n\le N}e(P(n))\to0.$$

有效估计则要控制

$$\left|\sum_{n\le N}e(P(n))\right|.$$

当 $P$ 的系数包含无理数时，连分数近似会决定哪些尺度上相位最接近有理、哪些尺度上可以得到抵消。

3. Fiber 均匀分布的限制

一种诱人的想法是先证明每个 fiber 上的均匀分布，再合成整个空间的均匀分布。但这并不总是成立：所有纤维方向看起来平均，不代表整体分布没有隐藏相关性。这个失败提示我们必须直接控制整体指数和。

4. 连分数尺度

设 $\alpha$ 的 convergents 是 $p_k/q_k$。在长度接近 $q_k$ 的区间上，$n\alpha$ 的分布有特别好的结构。把 $[1,N]$ 拆成这些标准尺度，可以把任意长度的问题化成一族可估的块。

5. 证明路线

一个合理策略是：先用 Weyl differencing 降低多项式相位阶数；再用连分数控制主要尺度；最后把误差在不同块上求和。动力系统语言提供几何直觉，真正的定量估计则来自指数和技术。

2017年11月27日
From periodic to quasi-periodic：周期轨道、无理旋转与逼近

旧博客原文

原题：From periodic to quasi periodic

旧站归档中的这篇正文原本为空。

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。

从 periodic 到 quasi-periodic，最简单的模型是圆周旋转

$$R_\alpha:x\mapsto x+\alpha\pmod 1.$$

当 $\alpha=p/q$ 为有理数时，每条轨道都是周期的；当 $\alpha$ 为无理数时，轨道不闭合，而是在圆周上稠密。这两种现象之间由有理逼近连接。

周期旋转与拟周期旋转之间由有理逼近连接，时间尺度决定了两者看起来有多接近。

1. 周期轨道

若 $\alpha=p/q$ 且 $(p,q)=1$，则

$$R_\alpha^q x=x.$$

因此动力系统分解成长度为 $q$ 的有限轨道。许多谱量、平均量和相关函数都可以化成有限和。

2. 拟周期轨道

若 $\alpha$ 无理，则 $n\alpha$ 模 $1$ 稠密。更强地，Weyl criterion 给出均匀分布：

$$\frac1N\sum_{n\le N}e^{2\pi i k n\alpha}\to0,\qquad k\ne0.$$

这说明拟周期不是“没有规律”，而是有一种全局平均意义下的均匀规律。

3. 有理逼近的两面性

连分数给出 $\alpha$ 的最佳有理逼近 $p_j/q_j$。在时间尺度 $q_j$ 以下，无理旋转很像周期 $q_j$ 的旋转；但超过这个尺度后，误差积累会显露真正的拟周期行为。

4. 推广到高维环面

在 $\mathbb T^d$ 上，平移 $x\mapsto x+\omega$ 的轨道性质由整数关系 $k\cdot\omega$ 控制。没有非平凡整数关系时，轨道稠密；若还满足 Diophantine 条件，则许多小除数估计也会变得可控。

2017年11月26日
Dirichlet 原理：从离散调和函数到连续调和函数

旧博客原文

原题：A discret to continuous approach to the Dirichlet principle.

Direchlet principle:
$\Omega \subset R^n$ is a compact set with $C^1$ boundary. then there exists unique solution $f$ satisfied $\Delta f=0$ in $\Omega$, $f=g$ on $\partial \Omega$ .

Perron lifting and barrier function

We know the standard approach of the Dirichlet principle is perron lifting and construction of barrier function on the boundary.

The key point is if we define the variation energy $E(u)=\int_{\Omega}|\nabla u|^2$ , then it is easy to see for $u_1,u_2$ is in perron set, $E(sup (u_1,u_2))\geq \max\{E(u_1),E(u_2)\}$ . So we can begin from a maximization sequence to construct a Cauchy sequence by perron lifting and by the involve of barrier function to make the solution compatible with the boundary condition then arrive a proof.

But when I was a freshman in undergraduate school and I do not know the method of perron lifting I try something I name it from discret to continuous approach to try to solved the problem. It is always a puzzle in my mind iff we can solve the Dirichlet principle in this way, roughly speaking, it is divid into two part:

1. Investigate the discretization of harmonic function in smaller and smaller scale. The discretization I consider is just $\Omega\cap \epsilon \mathbb Z^2$ i.e. the $\epsilon$ -latties in $\Omega$,and discretization Laplace operator $\Delta_{\epsilon}u(x_1,...,x_n)=\sum_{i_1,...,i_n\in\{-1,1\}}\frac{u(x_1+i_1,...,x_n+i_n)}{2^n}$ . Some result is much easier to arrive with the discretization thing, you know ,such as the existence of solution is just come from simple linear algebra. and we can deduce harneck inequality, gradient estimate, even green function. So we get a solution $\hat f_{\epsilon}$ of $\epsilon$ discretization and we do a extension $\Omega\cap \epsilon \mathbb Z^2$ to $\Omega$ by take value of a small tube by the center of the tube, where the value have a definition by $\hat f_{\epsilon}$ , and now we get $f_{\epsilon}$ .

2. The second step is to proof the solution $f_{\epsilon}$ with $\epsilon$ -discretization problem will coverage to the solution of original problem;i.e. we want to proof a $L^{\infty}$ estimate;i.e. $\forall \delta>0$ , $\exists \epsilon>0, \forall 0<\epsilon_1,\epsilon_2<\epsilon$ we have $\forall x\in \Omega$ , $|f_{\epsilon_1}(x)-f_{\epsilon_2}(x)|<\delta$ . and by Albano-Ascoli theorem to construct $f$ . Then we need to proof $f$ is the harmonic function we find, to verify this information we use the mean-value property. So we need to prove $f$ satisfied mean-value property for every ball in $\Omega$ .

Here is my first question,
> **Question 1:** How to prove the $L^{\infty}$ estimate and the MVP of $\epsilon$ -discretization will coverage to the MVP in $R^n$ case occor in second step?

My attempt to the $L^{\infty}$ estimate is by renomelazation which seems could work, but the annoying thing is to proof the mean-value property will coverage to the real one, I try to use some result of random walk, but it seem not works…

My second question is:
> **Question 2:** Are this approach a universal phenomenon? At least could we use this approach to establish the existence of solution for linear elliptic and parabolic equation?

The Third question is:
> **Question 3:** If we consider some inverse problem, that is to say, form a MVP instead of a PDE to derive a solution, could this always be possible? some example is, if we change the mean value property for harmonic function from the average of ball to cube or triangle or elliptic or something else, what happen? Is there always a solution satisfied the news MVP point-wise? If not, Is there some counterexample? on another hand, if yes, are them came from some PDE?

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。

Dirichlet 原理说：给定边界值，调和函数是 Dirichlet energy 的极小者。除了 Perron 方法和 barrier function，也可以从离散问题逼近连续问题。

离散 Dirichlet 问题可由线性代数求解，再通过统一估计和紧性逼近连续调和函数。

1. 连续问题

设 $\Omega$ 有光滑边界，给定 $g$，要求

$$\Delta u=0\quad\text{in }\Omega,\qquad u|_{\partial\Omega}=g.$$

变分形式是极小化

$$E(u)=\int_\Omega |\nabla u|^2\,dx.$$

Euler-Lagrange 方程正是 Laplace 方程。

2. 离散化

取 $\varepsilon$-lattice 上的区域 $\Omega_\varepsilon$，定义离散 Laplacian

$$\Delta_\varepsilon u(x)=\frac1{\varepsilon^2}\sum_{y\sim x}(u(y)-u(x)).$$

离散 Dirichlet 问题是一个有限维线性代数问题，因此存在唯一解。

3. 离散估计

离散调和函数仍有最大值原理、Harnack inequality、Green function 和能量估计。这些估计若能与 $\varepsilon$ 无关，就可以取极限。

4. 延拓与紧性

把格点函数延拓为分片常数或分片线性函数。若能得到统一的 Holder 或 Sobolev 控制，就能通过 compactness 取出收敛子列。极限函数满足弱形式

$$\int_\Omega \nabla u\cdot\nabla\varphi=0.$$

5. 边界条件

最细的是边界兼容性。Perron 方法中 barrier function 正是为处理边界极限；离散方法也需要离散 barrier，保证延拓后的极限真正取到给定边界值。

2017年11月25日
Weyl law：特征值计数、相空间体积与热核直觉

旧博客原文

原题：Weyl law

In 1911 year, when Weyl is a young mathematician specticlizing in integrable system and PDE, He proved the important result about the asystomztion of eigenvalues of Dirichelet problem in $\Omega\subset R^n$ is a compact domain;i.e.

$N(\lambda)=(2\pi)^d Vol(\Omega)\lambda^{\frac{d}{2}}(1+o(1))$

Which in fact is a conjecture of *** in *** in published in 1910.

This is definitely a very amazing achievement of mathematician, The realist meaning we can actually charge with the spectrum asyspesion.

In fact, we know, the only thing we know is that the eigenfunction with different spectrum is orthogonal and we have are a cretition named maximum-minmum principle for the k eigenvalue. but how could we charge with the asymotum of them? It seems not to be chargeable, though we have a $L^2$ isometry, the spectrum expansion, ut it still not seems to be chargeable the main difficult come from the compacness this just mean a divide of the whole space, and we consider the X-ray tansigation from every point to the whole space, it need to be passion kernel, or we change it to be a pare matrix.

Yes, We can just look this phenomenon as there are two different world, one is the real world in the Ecliud space, there other is the a wave function world, in the second world it is composted by the unique of the solution $u$ for $\Delta u=\lambda u$ for some eigenvalue $\lambda$ and all units in this world is the translation and rescaling of u. Then thing become interesting, now how to understand the other guy, i.e. the other eigenvalue and eigenfunctions? They must be the u after some translation combine with rescaling and trslation and rotation!!! so there is a dynamic system action on it! and if we only let it to be affine map,i.e. combine only taslation and rotation, then we just get the all eigenfunctions with the same eigenvalue.

Now let us see what is it, it is just need to be compatible with the boundary condition, so it need to be moduli space that the boundary map is a measure that is arrive able by only affine translation of the function (we look it as a obsevalbel) So it is a restriction from a high dimensional space to the boundary of it. and very fortunately it could assume a

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。

Weyl law 描述 Laplace 算子特征值的渐近分布。它说明高频谱的主要项只看区域体积和相空间体积，而不看边界的细节。

Weyl law 把 Laplace 特征值计数的主项解释为 phase space volume。

1. Dirichlet 特征值问题

设 $\Omega\subset\mathbb R^n$ 有界，考虑

$$-\Delta u=\lambda u,\qquad u|_{\partial\Omega}=0.$$

特征值排成

$$0<\lambda_1\le\lambda_2\le\cdots\to\infty.$$

定义计数函数 $N(\lambda)=\#\{j:\lambda_j\le\lambda\}$。

2. Weyl 渐近

Weyl law 说

$$N(\lambda)\sim \frac{\omega_n}{(2\pi)^n}|\Omega|\lambda^{n/2}.$$

右边正是相空间中

$$\{(x,\xi):x\in\Omega,\ |\xi|^2\le\lambda\}$$

的体积除以 $(2\pi)^n$。

3. 为什么是相空间体积

高频 eigenfunctions 局部上像平面波 $e^{ix\cdot\xi}$。频率满足 $|\xi|^2\le\lambda$，位置在 $\Omega$ 中。于是特征态数量近似等于可用 phase space cells 的数量。

4. 热核证明直觉

热 trace 为

$$\operatorname{Tr}(e^{t\Delta})=\sum_j e^{-t\lambda_j}.$$

当 $t\to0$，热核对角线主项为

$$K_t(x,x)\sim (4\pi t)^{-n/2}.$$

积分得到 trace 主项，再由 Tauberian theorem 推出 Weyl law。

5. 边界和低阶项

边界会出现在下一阶项中。主项只看体积，边界面积、曲率和动力学信息会在更细的谱渐近或余项估计中出现。这正是谱几何的入口。

2017年11月24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