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Almost orthogonality：Cotlar-Stein lemma、Schur test 与奇异积分

旧博客原文

原题：Almost orthogonality

Motivation and Cotlar’s lemma

We always need to consider a transform $T$ on Hilbert space $l^2(\mathbb Z)$ (this is a discrete model), or a finite dimensional space $V$ . If under a basis $T$ is given by a diagonal matrix this story is easy,

$\displaystyle A = \begin{pmatrix} \Lambda_1 & 0 & \ldots & 0 \\ 0 & \Lambda_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \Lambda_n \end{pmatrix} \ \ \ \ \ (5)$

Then $||T||=\max_{i}\lambda_i$ .

In fact, for $T$ is a transform of a finite dimensional space, $T$ is given by $(a_{ij})_{n\times n}$ by duality we have $||T||=||TT^*||$ , so we have,

$||T||=||TT^*||=|(\sum a_{ij}x_j)y_i|\leq |\sum_{i,j}\frac{1}{2}(|a_{ij}(|x_i|^2+|y_j|^2)|\leq M$

If we have given $\sum_{i}|a_{ij}|\leq M$ and $\sum_{j}|a_{ij}|\leq M$ $\forall i,j\in \{1,2,...,n\}$ .

But in application of this idea, the orthogonal condition always seems to be too restricted and due too this we have the following lemma which is follow the idea but change the orthogonal condition by almost orthogonal.

Lemma(Catlar-Stein)

Let $\{T_j\}_{j=1}^N$ be finitely many operators on some Hilbert space $H$ . Such that for some function $\gamma : \mathbb Z\to R^+$ one has,

$||T_j^*T_k||\leq \gamma^2(j-k),||T_jT_k^*||\leq \gamma^2(j-k)$

for any $1\leq j,k\leq N$ . Let $\sum_{l=-\infty}^{\infty}\gamma(l)=A<\infty$ . then ,

$||\sum_{j=1}^NT_j||\leq A$

Pf:

tensor power trick + duality $||T||=||TT^*||^{\frac{1}{2}}$ .

Singular integrals on $L^2$

Lemma(Schur)

Define $T$ is a operator on measure space $X\times Y$ equipped positive product measure $\mu\wedge \nu$ , via,

$(Tf)(x)=\int_YK(x,y)f(y)\nu(dy)$

$K$ is a measurable kernel, then,

1). $||T||_{1\to 1}\leq \sup_{y\in Y}\int_{X}|K(x,y)|\mu(dx)=:A$ .

2). $||T||_{\infty\to \infty}\leq \sup_{x\in X}\int_{Y}|K(x,y)|\nu(dy)=:B$ .

3). $||T||_{p\to p}\leq A^{\frac{1}{p}}B^{\frac{1}{p'}}$ , $\forall 1\leq p\leq \infty$ .

4). $||T||_{1\to \infty}\leq ||K||_{L^{\infty}(X\times Y)}$ .

Pf:

1),2),4) merely due to Fubini theorem and Bath lemma.

3) proof by the interpolation and combine 1) and 2).

Theorem

Let K be a Calderon-Zegmund operator, with the additional assumption

that $|\nabla K(x)|\leq B|x|^{-d-1}$ . Then

$||T||_{2\to 2} \leq CB$

with $C = C(d)$ .

Caldero ́n–Vaillancourt theorem

Hardy’s inequality

Theorem(Hardy inequality)

For any $0 \leq s < \frac{d}{2}$ there is a constant $C(s, d)$ with the prop-

arty that,

$|||x|^{-s} f||_2 \leq C(s,d)||f||_{H^s(R^d)}$

for all $f \in H^s(R^d)$ .

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。

正交性是 Hilbert space 中最强的简化机制；almost orthogonality 则是在真实分析问题中更常见的替代品。频率块、空间块和算子族通常不完全正交，但交互足够小。

almost orthogonality 用交互衰减替代严格正交，是奇异积分和伪微分算子估计的基础。

1. 从正交到几乎正交

若 $T_j$ 的像彼此正交，则

$$\left\|\sum_j T_j f\right\|_2^2=\sum_j\|T_jf\|_2^2.$$

实际中通常只有 $T_i^\ast T_j$ 和 $T_iT_j^\ast$ 随 $|i-j|$ 衰减。

2. Cotlar-Stein lemma

若

$$\|T_i^\ast T_j\|+\|T_iT_j^\ast\|\le a(i-j)$$

且 $\sum_k a(k)^{1/2}<\infty$，则

$$\left\|\sum_jT_j\right\|_{2\to2}<\infty.$$

证明可用 tensor power trick 和对偶性。

3. Schur test

对 kernel operator

$$Tf(x)=\int K(x,y)f(y)\,dy,$$

若

$$\sup_x\int |K(x,y)|dy<\infty,\qquad \sup_y\int |K(x,y)|dx<\infty,$$

则 $T$ 在 $L^2$ 上有界。

4. 奇异积分中的用途

Calderon-Zygmund theory 中，常把算子分解成不同尺度的 pieces。尺度相隔很远时，kernel 的光滑性带来交互衰减；相近尺度则只需有限重叠。

5. Calderon-Vaillancourt 方向

伪微分算子的 $L^2$ 有界性也可以看成 almost orthogonality 的结果：把相空间切成小块后，不同块之间的交互由 symbol 的导数控制。

2017年12月18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
一个 Fourier 系数衰减估计：解析延拓、分歧提升与矩阵值函数

旧博客原文

原题：An Fourier coefficient decay estimate.

f is a matrix value analytic function on $\mathbb T$ , we know $h(f)>\alpha$ , this is just mean $\forall k\in \mathbb Z$ , assume $| \hat f(k)|\leq e^{-|k|\alpha}$ , $g=log(f)$ for which we assume $g$ is a lifting of f use the inverse of ramification map ,

$M_{n\times n} \to M_{n\times n} , A \to e^A$ .

Then exists $\beta=c(\alpha)>0$ such that ,

$\forall k \in \mathbb Z$ , $|\hat g(k)| \leq e^{-k\beta}$ .

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。

设 $f$ 是圆周或环面上的矩阵值解析函数。解析性最直接的 Fourier 后果是指数衰减：

$$\|\widehat f(k)\|\le C e^{-\rho |k|}.$$

这里 $\rho$ 不是一个装饰性常数，而是由 $f$ 能延拓到多宽的复邻域决定。若函数还经过分歧映射的提升，衰减率会被分歧阶数重新缩放。

解析函数的 Fourier 系数指数衰减，衰减率由复邻域宽度和可能的分歧提升共同决定。

1. 一维解析情形

如果 $f$ 在带状区域 $|\operatorname{Im} z|<\rho$ 内解析且有界，那么

$$\widehat f(k)=\int_0^1 f(x)e^{-2\pi i kx}\,dx.$$

当 $k>0$ 时把积分路径向上平移 $i\sigma$，得到因子 $e^{-2\pi k\sigma}$；当 $k<0$ 时向下平移。令 $\sigma<\rho$，就得到指数衰减。

2. 矩阵值函数没有本质困难

若 $f(x)$ 取值于矩阵空间，只需要把绝对值换成任意相容矩阵范数。Cauchy 积分和路径平移仍逐项成立，因此

$$\|\widehat f(k)\|\le \sup_{|\operatorname{Im}z|\le\sigma}\|f(z)\|\,e^{-2\pi\sigma |k|}.$$

3. 分歧提升的影响

若 $f$ 通过某个 ramification map 被提升，例如局部写成 $z=w^m$，那么 $w$ 平面中的解析带宽会按比例改变。直观上，绕分歧点一圈的角变量被拉伸，Fourier 频率也随之重标定。

4. 估计的用途

这类估计常出现在解析 cocycle、准周期系统和小除数问题中。指数衰减提供了“高频很小”的精确版本，使得截断 Fourier 级数时可以把尾项压到可控范围。

2017年12月12日
非线性椭圆方程平均值性质的几何直觉

旧博客原文

原题：Geometric intuition of mean value property of nonlinear elliptic equation

I wish to gain some understanding of the MVP of nonlinear elliptic equation by geometric intuition.

Linear elliptic equation case

First of all, I have a very good geometric explain of the MVP of Laplace equation, i.e.

MVP of laplace equation

$\Delta u=0$ in $\Omega$ , $\forall B(x_0,r)\subset \Omega$ is a Ball, we have following identity:

$\frac{1}{\mu(\partial(B))}\int_{\partial B}u(x)dx=u(x_0)$

I need to point out first, this property is not difficult to proof by standard integral by part method, but the following method have more geometric intuition. And in some sense explain why this property holds.

The proof is not very difficult to explain by mathematic formula, but I wish to divide the proof into two part, explain one part by graph and literal interpretation.

part 1 of the proof:

we consider a 1-parameter group of foliation, and consider the integral identity with this foliation.

$\int_{v\in S_{n-1}}\int_{\gamma_v} \frac{\partial \partial_{v}u}{\partial t}dt=\int_{B(x_0,r)}\partial_{n}u-\int_{S_{n-1}}\partial_{v}udv=\int_{B(x_0,r)}\partial_{n}u$

Part 2 of the proof:

and we have:

$\int_{v\in S_{n-1}}\int_{\gamma_v} \frac{\partial \partial_{v}u}{\partial t}dt=0$

by the pointwise equation $\Delta u=0$ , one key point is $\partial_{-v-v}u=\partial_{vv}u, \forall v\in S_{n-1}$ .

This approach cloud easily to transform to the general elliptic equation case and it seems a little difficult to transform to Possion equation, the non-hemomorphism case.

Nonlinear elliptic equation case

A-B-P estimate for general nonlinear uniformly elliptic equation

ABP estimate is the most basic estimate in fully nonliear elliptic equation.
The ABP maximum principle states (roughly) that, if

$a^{ij} \partial _i \partial _j u \geq f, in \ \Omega \subset \mathbb{R}^n (a^{ij} \geq C Id >0),$

Then (assuming sufficient regularity of the coefficients),

\sup _{\Omega} u \leq \sup _{\partial \Omega} u + C (\int _{\Omega} \vert f \vert^n )^{1/n} ………. (*)

I will give an intuitive explanation of the proof of (*) .
Usually, in order to prove maximum principles, the key idea is to use that at a local max the second derivative is negative-definite, then choose a good basis and get some identity of 1-order drivative and inequality for 2-order’s. This process is used in such like the proof of the Hopf lemma, and some inter gradient estimate, consider some flexiable function like $e^{Au}$ or sometihng else anyway.

But in the proof of ABP we need more geometric intution and more trick.

First we do a rescaling:
if $a^{ij} \partial _i \partial _j u \geq 1, in B_1 \subset \mathbb{R}^n (a^{ij} \geq C Id >0),u|_{\partial B_1}\geq 0.$
then:

$|inf_{B_1}u| \leq C |A|^{1/n} ....(**)$

And then We explain what is the contact set. It is the subset $\Gamma^{+} of \Omega$ such that u agree with it convex envelop. i.e. $\Gamma^{+}=\{x|u=convex \ evolap \ of \ u\ at \ x\}$ . The geometric meaning is it has at least one lower support plane. So what is $\Gamma^{+}$ it is just the set that u is very low on it. Or in another way of view you consider $-u$ as a lot of mountains then $\Gamma^+$ is the place near the tops of which mountain can see every thing (locally).
Then we look at every point in $\Gamma^{+}$ , then determination of the hessian matrix $det(u_{ij})$ at this point have a control due to the PDE $a^{ij} \partial _i \partial _j u \geq 1$ , and the uniformly elliptic property.

The determination of hessian matrix could be view as a determination of Jacobe matrix of the map $(u_1,...,u_n)\to (e_1,...,e_n)$ .and by Area formula we have:

$\int_{\Phi(\Omega)} f( \Phi^{-1}(y)) dy =\int_{\Omega}f(x)|J({\Phi(x)})| dx,$

It is easy to see for a constant $c$ , $B_{c|sup_{\Omega}|u||}(0)\subset \Phi(\Omega)$ (Base on the PDE on every point, the geometric intution is just the function u could not be very narrow cone at every point). So we have , take $f=\chi_{\Gamma^{+}}$ ,

$|B_{c\sup_{\Omega}|u|}(0)|^{n}\leq \int_{\Gamma^{+}}\chi_{\Gamma_{+}}(x)|J_{\Phi}(x)|dx$

so we have:

$|B_{c\sup_{\Omega}|u|}(0)|\lesssim |{\Gamma_{+}}|^{\frac{1}{n}}...(***)$

and the classical matrix inequality for every positive definite matrix A we have
:

$det(AB)\leq (\frac{tr(AB)}{n})^n...(****) .$

combine (***),(****) ,we have:

$sup_{\Omega}|u|\lesssim ||\frac{a^{ij}u_{ij}}{D^*}||_{L^n({\Gamma^+})} .$

graph

General approach to get MVP for elliptic equation which is come from geometry

MVP for K-hessian equation, with geometric explanation

MVP for K-curvature equation, with geometric explanation

MVP for p-Laplace equation, with geometric explanation

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。

调和函数的平均值性质是椭圆方程最漂亮的几何现象之一。在线性情形，它来自球面对称和分部积分；在非线性椭圆方程中，真正的替代物往往不是严格平均值公式，而是 ABP estimate、contact set 和凸包几何。

线性调和函数有球面平均值性质，非线性椭圆方程则用 ABP estimate 和接触集合几何替代。

1. Laplace 方程的平均值性质

若 $\Delta u=0$，则对球 $B_r(x)$，

$$u(x)=\frac1{|\partial B_r|}\int_{\partial B_r(x)}u\,d\sigma.$$

证明可用 Green identity，也可以看成沿同心球 foliation 的通量守恒。

2. 几何解释

令

$$A(r)=\frac1{|\partial B_r|}\int_{\partial B_r(x)}u.$$

对 $r$ 求导后，$\Delta u=0$ 让通量项消失，于是 $A(r)$ 为常数。球面对称和散度结构共同产生平均值性质。

3. 非线性情形的问题

对一般 fully nonlinear elliptic equation

$$F(D^2u,x)=f,$$

没有线性叠加，也没有简单的球面平均公式。我们需要用比较原理和接触几何替代平均值。

4. ABP estimate

ABP 最大值原理粗略地说给出

$$\sup_\Omega u\le \sup_{\partial\Omega}u+C\|f\|_{L^n(\Omega)}.$$

其证明看 $u$ 的凸包和 contact set。梯度映射的面积公式把函数的最大值和右端 $f$ 的积分联系起来。

5. 一条直觉

线性平均值性质说：每个点由周围平均决定。非线性椭圆理论说：最大值由接触集合和 Monge-Ampere 型几何控制。二者背后的共同点是椭圆性阻止信息沿单一方向集中，迫使解受到周围区域的整体约束。

2017年12月11日
代数数的 Diophantine approximation：Liouville、Roth 与 Vandermonde 约束

旧博客原文

原题：Diophantine approximation of algebraic number

An important theorem in Diophantine approximation is the theorem of Liuoville:

**Liuoville Theorem** If x is a algebraic number of degree $n$ over the rational number then there exists a constant $c(x) > 0$ such that: $\left|x-{\frac {p}{q}}\right|>{\frac {c(x)}{q^{{n}}}}$

holds for every integer $p,q\in N^*$ where $q>0$ .

This theorem explain a phenomenon, the approximation of algebraic number by rational number could not be very well. Which was generated later to **Thue–Siegel–Roth theorem**, them could be used to proof a lots of constant is not algebraic, i.e. transcendentals .

My questions is in another direction, now let us not just consider one root $\alpha_1$ of a integer polynomial $P(x)=a_mx^m+...+a_1x+a_0$ but consider all roots of it, i.e. $\{\alpha_1,...,\alpha_m\}$ , which is based on a observation : If we define

$\sigma_k(P(x))=\sum_{1\leq \alpha_{i_1}<\alpha_{i_2}<...<\alpha_{i_k}\leq m}\alpha_{i_1}\alpha_{i_2}...\alpha_{i_k}$

By **Vieta theorem** we know $\sigma_k(n)\in \mathbb Q$ for all $k\in N^*$ , this will lead to some restriction and in fact destroy the uniformly distribution of $(\alpha_1,...,\alpha_m)\in [0,1]^m$ . In fact the most important one is the determination of Vandermon Determinant:
$V(P(x))=\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)$ .

We know $\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\in \mathbb Q$ so when $\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\neq 0$ we could use this to proof a nontrivial estimate for $\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}$ .
$\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).$

by combine the A-G inequality and $\Pi_{1\leq \alpha_i<\alpha_j\leq n}(\alpha_i-\alpha_j)=\lambda\neq 0$ .While by continue fractional expansion we only know a trivial estimate of type $\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n})$ .

my question is the following:
Is there still have a nontrivial estimate for $\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}$ (which could be slight weaker), if we don’t have the whole power of **Vieta theorem**? more precisely:

**problem 1**

if we have $\sigma_k((\alpha_1,...,\alpha_m))=\lambda_k\in \mathbb Q$ for all $k\in \{1,2,...,m'\}$ where $m'<m$ , is there still some nontrivial estimate of,

$\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}$

hold for all $n\in N^*$ ?

One reason to consider this could be true is that although $\{\alpha_1,...,\alpha_m\}$ is not roots of a integer polynomial but we could image in some suitable metric space $X$ the gromov-hausdorff distance of tuple $(\alpha_1,...,\alpha_m)$ and a tuple come form roots of integer polynomial is small . And it seems reasonable to image this type of asymptotic quality is continue with the G-H distance on $X$ .

Another problem is what happen when $V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0$ . More precisely,

**problem 2**

What happen when $V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0$ , is this result,

$\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).$

still true?

Let us go a litter further, if these two problem both have a satisfied answer, what is the higher dimensional case?

**problem 3**

Given $m\in \mathbb N^*$ . If tuple $(y_1,...,y_k)$ is very closed to the zero set of a variety in $\mathbb Z[x_1,...,x_m]$ in $\mathbb (Z^{m})^k$ in the sense a lots of symmetric sum of $y_1,...,y_k$ belong to $\mathbb Q^m$ , will this lead to some good estimate for

$\sum_{1\leq s\leq k}||y_sn||_{\mathbb R^m/\mathbb Z^m}?$

I think these type of result should be investigated very well, Iappreciate to any pointer with useful comments and answer, both on given some strategy to solve these problems or given some reference about these problems.

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。

代数数不能被有理数“过分好”地逼近。Liouville theorem 是这件事的第一层形式，Thue-Siegel-Roth theorem 则给出几乎最优的结论。

Liouville 和 Roth 定理说明代数数不能被有理数过分好地逼近。

1. Liouville theorem

若 $\alpha$ 是次数 $d$ 的代数数，则存在 $C(\alpha)>0$，使对所有有理数 $p/q$，

$$\left|\alpha-\frac pq\right|\ge \frac{C(\alpha)}{q^d}.$$

证明的核心是把 $\alpha$ 代入整数多项式 $P$，再估计 $P(p/q)$ 不可能是太小的非零有理数。

2. Roth theorem

Roth theorem 大幅加强 Liouville：若 $\alpha$ 是无理代数数，则对任意 $\varepsilon>0$，

$$\left|\alpha-\frac pq\right|<\frac1{q^{2+\varepsilon}}$$

只有有限多个有理解。换句话说，代数数的有理逼近指数不能超过 $2$ 太多。

3. 多个根的约束

若考虑一个整数多项式的所有根 $\alpha_1,\ldots,\alpha_d$，Vieta 定理给出系数与根的对称函数之间的整数关系。Vandermonde determinant

$$\prod_{i
又控制根之间不能全部过分靠近。

4. 从单点逼近到整体结构

单个根的有理逼近只看一个 $\alpha$；所有根一起看时，还会出现判别式、对称多项式和高度的约束。整体代数结构比单个连分数展开更刚性。

5. 一个自然问题

如果没有完整 Vieta 结构，只知道某些弱的对称约束，是否仍能推出非平凡逼近下界？这类问题位于 Diophantine approximation、代数高度和几何不等式之间。

2017年12月5日
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