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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
Baragar-Bourgain-Gamburd-Sarnak 猜想：Markov triples 的模 p 连通性

旧博客原文

原题：Baragar-Bourgain-Gamburd-Sarnak conjecture

$M$ is the markov triple $(x,y,z)$ :

$x^2+y^2+z^2=xyz$ and $(x,y,x)\in \mathbb Z^3 \ \ \ \ (*)$ .

It is easy to see:

$R_1: (x,y,z)\to (3yz-x,y,z)$ .

map markov triple to markov triple.

This is also true for $R_2,R_3$ . and the transform $R_1,R_2,R_3$ and permutation a classical result of markov claim that all solution of (*) could be generated from $(1,1,1)$ . I get a similar result for a similar algebraic equation 1 half years ago when consider a $Q$ version of problem about 1-form given by Xu Bin.

Now we know the graph with root $(1,1,1)$ and with node generate by transform $R_1\cup R_2 \cup R_3 \cup S_3$ is connected.

The B-B-G-S conjecture is is the connected property still true for prime $p$ surfficed large?

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。

Markov triples 是方程

$$x^2+y^2+z^2=3xyz$$

的正整数解。经典 Markov 定理说，从根解 $(1,1,1)$ 出发，反复使用 Vieta involution

$$(x,y,z)\mapsto (x,y,3xy-z)$$

以及坐标置换，可以生成所有正整数解。

Markov triples 的 Vieta involution 生成解图；模 p 后的问题变成有限域曲面上的连通性问题。

1. 解图

把每个解看作图的一个顶点，若两个解由一次 Vieta involution 或置换相连，就连一条边。整数正解形成一棵以 $(1,1,1)$ 为根的巨大图。

2. 模 p 的问题

把方程放到有限域 $\mathbb F_p$ 上，得到有限集合

$$X_p=\{(x,y,z)\in\mathbb F_p^3:x^2+y^2+z^2=3xyz\}.$$

同样的 involution 仍然作用在 $X_p$ 上。Baragar-Bourgain-Gamburd-Sarnak 方向的问题是：当 $p$ 足够大时，这个作用图是否在主要部分上连通，甚至是否具有 expansion 性质？

3. 为什么这不是普通图论

图的边来自代数自同构，因此顶点集合有强烈的代数几何结构。连通性问题本质上是在问：这些简单变换能否在有限域曲面上产生足够大的轨道。

4. 可能工具

这类问题常混合使用代数几何、有限群展开、谱间隙和 sum-product 现象。若能证明没有大的不变子集，就能排除图分裂成多个大块的可能。

2017年11月22日
Metric entropy（二）：entropy map 的上半连续性与 infinity entropy
旧博客原文

原题：Metric entropy 2

I am reading the article “ENTROPY THEORY OF GEODESIC FLOWS”.

Now we focus on the upper semi-continuouty of the metric entropy map. The object we investigate is $(X,T,\mu)$ , where $\mu$ is a $T-$ invariant measure.

The insight to make us interested to this kind of problem is a part of variational problem, something about the existence of certain object which combine a certain moduli space to make some quantity attain critical value(maximum or minimum). The most simple example maybe Isoperimetric inequality and Dirichlet principle of Laplace. Any way, to establish such a existence result a classical approach is to proof the upper semi-continuouty and bounded for associate energy of the problem. In our case the semi-continuouty will be some thin about the regularity of the entropy map:

$E:M(X,T)\to h_{\mu}.$

We define the entropy at infinity:

$sup_{(\mu_n)}limsup_{\mu_n\to 0}h_{\mu_n}(T)$

Where $(u_n)_{n=1}^{\infty}$ varies in all sequences of measure coverage to $0$ in the sense for all $A\subset M$ , $A$ measurable then $\lim_{n\to \infty} \mu_{n}(A)=0$ .

Compact case

we say some thing about the compact case, In this case we have finite partition with smaller and smaller cubes, this could be understand as a sequences of smaller and smaller scales. A example to explain the differences is $\mathbb N^{\mathbb N},\sigma$ , shift map on countable alphabet.

Because of this thing, there is a good sympolotic model, i.e. h-expension, and it generalization asymptotically h-expension equipped on a compact metric space $X$ have been proved to be that the corresponding entropy map is upper semi-continous.

In particular $C^{\infty}$ diffeomorphisms on compact manifold is asymptotically h-expensive.

Natural problem but I do not understand very well:

Why it is natural to assume the measure to be probability measure in the non-compact space?

Non-compact case

$(X,d)$ metric space

$T:X\longrightarrow X$ is a continuous map.

$d_n(x,y)=\sup_{0\leq k\leq n-1}d(T^kx,T^ky)$ , then $d_{n}$ is still a metric.

Easy to see $\frac{1}{n}h_{\mu}(T^n)=h_{\mu}(T)$ . This identity could be proved by the cretition of entropy by $\delta$ -seperate set and $\delta$ -cover set.

Kapok theorem:

$X$ compact, for every ergodic measure $\mu$ the following formula hold:

$h_{\mu}(T)=\lim_{\epsilon \to 0}limsup_{n\to \infty}\frac{1}{n}logN_{\mu}(n,\epsilon,\delta)$ .

Where $h_{\mu}(T)$ is the measure theoretic entropy of $\mu$ .

Riquelme proved the same formula hold for Lipchitz maps on topological manifold.

Let $M_e(X,T)$ defined the moduli space of $T$ -invariant portability measure.

Let $M_(X,T)$ defined the moduli space of ergodic $T$ -invariant probability measure.

Simplified entropy formula:

$(X,d,T)$ satisfied simplified entropy formula if $\forall \epsilon >0$ surfaced small and $\forall \delta\in (0,1)$ , $\mu\in M _e(X,T)$ .

$h_{\mu}(T)=\limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta))$ .

Simplified entropy inequality:

If $\epsilon>0$ suffciently small, $\mu \in M_{e}(X,T)$ , $\delta\in (0,1)$ .

$h_{\mu}(T)\leq \limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta))$ .

Weak entropy dense:

$M_e(X,T)$ is weak entropy dense in $M(X,T)$ . $\forall \lambda>0$ , $\forall \mu\in M(X,T)$ , $\exists \mu_n\in M_e(X,T)$ , satisfied:
1. $\mu_n\to \mu$ weakly.
2. $h_{\mu_n}(T)>h_{\mu}(T)-\lambda$ , $\forall \lambda>0$ .
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。

entropy map 的上半连续性是变分问题中的关键正则性。若想证明某个 invariant measure 使 entropy 或 pressure 达到最大，通常需要紧性和上半连续性。

entropy map 的上半连续性关系到最大熵测度存在性；非紧情形还要控制 entropy at infinity。

1. Entropy map

给定动力系统 $f:X\to X$，考虑 invariant measures 空间 $\mathcal M_f(X)$ 上的函数

$$\mu\mapsto h_\mu(f).$$

若 $\mu_j\to\mu$ 弱收敛，希望有

$$\limsup_{j\to\infty}h_{\mu_j}(f)\le h_\mu(f).$$

这就是上半连续性。

2. 紧空间情形

在紧空间上，可以用越来越细的有限 partition 近似 entropy。若系统具有 expansiveness 或 asymptotic h-expansiveness，entropy map 常有较好的上半连续性。

3. 非紧空间的困难

非紧空间中，测度可能逃向无穷远。即使 $\mu_j$ 弱收敛到某个极限，entropy 也可能在逃逸部分携带额外信息。这个额外损失可用 entropy at infinity 衡量。

4. Entropy at infinity

粗略地说，entropy at infinity 记录所有逃向无穷远的测度序列可能保留的 entropy：

$$h_\infty=\sup_{\mu_j\to0}\limsup h_{\mu_j}(f).$$

若 $h_\infty$ 小于系统的 topological entropy，就有机会证明最大熵测度存在。

5. 几何动力系统中的意义

在 geodesic flow 中，entropy 与轨道增长、曲率和测地线逃逸相关。上半连续性问题本质上是在问：复杂轨道是否可能全部跑到 cusp 或无穷远处。若不能，就能在内部找到达到最大 entropy 的测度。
2017年11月21日
Metric entropy（一）：Ruelle 不等式、Pesin 公式与 Lyapunov 指数

旧博客原文

原题：Metric entropy 1

Some basic thing, include the definition of metric entropy is introduced in my early blog.

Among the other thing, there is something we need to focus on:

1.Definition of metric entropy, and more general, topological entropy.

2.Spanning set and separating set describe of entropy.

3.amernov theorem:

$h_{\mu}(T)=\frac{1}{n}h_{\mu}(T^n)$ .

Now we state the result of Margulis and Ruelle:

Let $M$ be a compact riemannian manifold, $f:M\to M$ is a diffeomorphism and $\mu$ is a $f$ -invariant measure.

Entropy is always bounded above by the sum of positive exponents;i.e.,

$h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).$

Where $dimE_i(x)$ is the multiplicity of $\lambda_i(x)$ and $a^{+}=max(a,0)$ .

Pesin show the inequality is in fact an equality if $f\in C^2$ and $m$ is equivalent to the Riemannian measure on $M$ . So this is also sometime known as Pesin’s formula.

F.Ledrappier and L.S.Young generate the result of Pesin.

One of their main result is:

$f:M\to M$ is a $C^2$ diffemoephism, where $M$ is a compact riemanian manifold, f is compatible with the Lesbegue measure on $M$ , and

$h_m({f,\mu})=\int_{M}\lambda_idim(V_i)dm$

If and only if on the canonical defined quation manifold $M/W_{\mu}$, i.e. the manifold mod unstable manifold $W_{\mu}$, the induced conditional measure $m_{\xi}$ is absolute continuous.

Remark: according to my understanding, the equality just mean in some sense we have the inverse estimate:

$h_{m}(f,\mu)\geq \int_{M}\lambda_idim(V_i)dm.$

This result maybe just mean near the fix point of $f$ ,i.e. the place charge the topology of the foliation, we have the inverse estimate. Such a inverse estimate will lead a control of the singularity of the push forward measure $m_{\xi}$ on the quation manifold. So $m_{\xi}$ have good regularity. But this idea is not complete to solve the problem.

Now we begin to get a geometric explain and which will lead a rigorous proof of the inequality:

$h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).$

At first we could observe that the long time average $\lim_{n\to \infty}\frac{1}{n}log||Df^n||$ of $Df$ could be diagonal. Assume after diagonal the eigenvalue is

$\lambda_1\leq \lambda_2\leq \lambda_3\leq...\leq \lambda_{n-1}\leq \lambda_n$ .

This eigenvalue could divide into 3 parts: <0,=0,>0.

This will lead to a direct sum decomposition of the tangent bundle $TM$ :

$TM\simeq E_{u}\otimes E_s\otimes E_c.$

Where $E_u$ is the part corresponding to the eigenvalue>0, For this part we consider the more refinement decomposition:

$E_u=\otimes_{k=1}^rV_k$ , $V_k$ is the eigenvector space of $\lambda_k$ . The dimension of $V_k$ is $dim V_k$.

On the other hand, we have a equality of metric entropy:

$h_{m}(f)=\frac{1}{n}h_{m}(f^n)=\sup_{\alpha\in partition \ set}\frac{1}{n}h_m(f^n,\alpha)$ .

For the later one, $\alpha$ is a measurable partition of $M$ , then $\alpha$ could always be refine to a smaller partition $\beta$ , and we have:

$h_{m}(f,\alpha)\leq h_{m}(f,\beta)$ .

Now we arrive the central place of the proof:

every partition could be refine by a partition with boundary of almost all cubes is parallel to the foliation. So we focus ourselves on the portion $\beta$ and all boundary of cubes in $\beta$ is parallel to the eigenvector.

Under this situation, we need only estimate the numbers of $\vee_{i=1}^nT^i\beta$ . Estimate it is not very difficult. we need only observe the following two thing:

1.

$\lim{n\to \infty}$ exists a.e. in $M$ . So this lead to the definition of foliation almost everywhere, and except a measurable zero set. In fact this set is the set of fix point of $M$ under $f$ .

2.

After a rescaling, every point which is not a fix point of $f$ could be understand as it is far away from fix points. Then the foliation could be understand as a product space locally. The flow with the direction which the eigenvalue is less than 1 cold not change $\vee_{i=1}^nT^i\beta$ . The direction with eigenvalue equal to 1 is just transition and just change the number of $\vee_{i=1}^nT^i\beta$ with polynomial growth. But the central thing is the direction with eigenvalue large than one and will make $\vee_{i=1}^nT^i\beta$ change with viscosity $e^{\lambda_i}$ . and we product it and get :

$h_{m}(f,\mu)\leq \int_{M}\lambda_i dim(V_i)dm$ .

In fact the proof only need $f$ to be $C^1$

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。

metric entropy 衡量一个保测动力系统在可测意义下产生信息的速率。对光滑动力系统，它和 Lyapunov exponents 之间有深刻联系：正 Lyapunov 指数给出不稳定方向上的体积增长，而 entropy 记录可区分轨道的增长。

Ruelle 不等式和 Pesin 公式把 entropy 与正 Lyapunov 指数联系起来。

1. Entropy 的基本图像

给定有限 measurable partition $\mathcal P$，Shannon entropy 是

$$H_\mu(\mathcal P)=-\sum_{A\in\mathcal P}\mu(A)\log\mu(A).$$

迭代下的平均信息增长定义为

$$h_\mu(f,\mathcal P)=\lim_{n\to\infty}\frac1n H_\mu\left(\bigvee_{j=0}^{n-1}f^{-j}\mathcal P\right).$$

再对所有 partition 取上确界，得到 $h_\mu(f)$。

2. Spanning 与 separating

拓扑 entropy 可以用 $(n,\varepsilon)$-spanning set 或 separating set 描述。两个点若在前 $n$ 次迭代中始终很近，就被看作同一条轨道影子。entropy 记录为了覆盖所有轨道影子，需要多少个名字。

这个图像和 metric entropy 的 partition 定义相互对应：一个是拓扑尺度，一个是测度尺度。

3. Ruelle inequality

设 $f$ 是紧 Riemannian manifold 上的 $C^1$ diffeomorphism，$\mu$ 是 $f$-invariant measure。Ruelle 不等式说

$$h_\mu(f)\le \int \sum_{\lambda_i(x)>0}\lambda_i(x)m_i(x)\,d\mu(x).$$

右端是正 Lyapunov exponents 的总和。直观上，系统能产生的信息不可能超过不稳定方向的体积膨胀能力。

4. Pesin formula

在更光滑并且测度与 Riemannian volume 绝对连续的情形，Pesin 公式给出等号：

$$h_\mu(f)=\int \sum_{\lambda_i(x)>0}\lambda_i(x)m_i(x)\,d\mu(x).$$

这说明所有不稳定方向上的 expansion 都真正转化成了信息增长，没有被 singular conditional measures 损失掉。

5. Ledrappier-Young 的视角

Ledrappier-Young 理论进一步解释等号何时成立：关键在于 unstable foliation 上的 conditional measures 是否绝对连续。若条件测度太奇异，几何膨胀不一定变成可测 entropy；若条件测度足够连续，膨胀就能被 entropy 读出来。

因此 entropy、Lyapunov exponents 和 foliation 上的条件测度，实际上是同一件事的三个侧面。

2017年11月21日
Gromov 式尺度思想在抛物方程中的应用

旧博客原文

原题：Gromov’s idea applicate to parabolic equation

Gromov’s idea applicate to parabolic equation.

Key point:

1.rescaling+renormalization.

2.analysis it on every scale.

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。

抛物方程的一个基本特征是空间和时间的尺度不同：若空间尺度缩小为 $r$，时间尺度就应缩小为 $r^2$。Gromov 式思想强调：不要只在一个固定尺度看方程，而要不断 rescale、renormalize，并在每个尺度上寻找紧性和极限模型。

抛物方程的自然尺度是空间 $r$、时间 $r^2$；正则性分析通常要在所有尺度上比较。

1. 抛物缩放

以热方程为例，若 $u_t-\Delta u=0$，定义

$$u_r(x,t)=u(x_0+rx,t_0+r^2t).$$

则 $u_r$ 仍满足同类方程。这个不变性说明，局部正则性问题可以被搬到单位尺度上处理。

2. 重整化

如果在某点附近梯度或曲率变大，就按最大量归一化。得到的序列若有紧性，极限通常是一个定义在全空间或半空间上的古老解。然后利用 Liouville theorem 排除非平凡极限。

3. 每个尺度的分析

尺度方法的关键不是一次缩放，而是在 dyadic scales 上反复比较：

$$Q_r(z_0)=B_r(x_0)\times(t_0-r^2,t_0).$$

若某个量在小尺度上不能衰减，就会产生 blow-up 极限；若 blow-up 极限被排除，就得到衰减估计。

4. 与几何分析的联系

Ricci flow、mean curvature flow 和非线性扩散方程中都可以看到这个结构。Gromov 风格的贡献在于把局部估计转成“所有尺度上的紧性与极限分类”。

2017年11月20日
Hilbert 第十六问题笔记：limit cycle、奇点与拓扑图像
旧博客原文

原题：Hilbert 16th problem

Introduction

the statement of Hilbert’s 16th problem:

$H(n)<\infty?$

definition of $H(n)=max$

Limit cycle:

Try beginning with Bendixon-Poincaré theorem, which is classical stuff and belongs to a lot of textbooks on vector fields.

Affine invariance

The number of limit cycle is invariant under affine map.

Classification of singular point

Bezout theorem

Example

1. $\frac{dx}{dt}=y,\frac{dy}{dt}=x$ .The graph is just like:

2. $\frac{dx}{dt}=x^2+y^2,\frac{dy}{dt}=x-y$ .The graph is just like:

3. $\frac{dx}{dt}=x^2-y^2,\frac{dy}{dt}=5-y$ .The graph is just like:

4. $\frac{dx}{dt}=x^2-y^2,\frac{dy}{dt}=-y$ .The graph is just like:

5. $\frac{dx}{dt}=x^3-y^3,\frac{dy}{dt}=5-y$ .The graph is just like:

6. $\frac{dx}{dt}=y^2-x^2+1,\frac{dy}{dt}=y$ . The graph of it is just like:

Now we try to explain the phenomenon we see. At first we can see there is no limit cycle in the picture. The bifurcation place is just the place $\frac{dx}{dt}=0\ or \frac{dy}{dt}=0$ and is just like 3 lines. And there are two singularity $(-1,0),(1,0)$ .

7. $\frac{dx}{dt}=y,\frac{dy}{dt}=y-x^2-x^2y$

8. $\frac{dx}{dt}=y,\frac{dy}{dt}=y-x-x^2y$ :

9. $\frac{dx}{dt}=y(y-3),\frac{dy}{dt}=(y-3-x-x^2y)(y-x-x^2y)$ :

exist 2 limits cycles.

10. $\frac{dx}{dt}=y(y-3)(y-5),\frac{dy}{dt}(y-5-x-x^2y)(y-3-x-x^2y)(y-x-x^2y)$

Tree structure

In general, the lower bound of $H(n)$ is established. first $H(n)=O(n^2)$ by Otrokov,and later proved to be $H(x)=O(ln(n)n^2)$ .

If right, this upper bound estimate is combine of two things:

1.every limit cycle do not intersect.

2.every unclear limit cycle contain a singularity.

Rough strategy attack Hilbert 16th problem

Step1:

The first step is to do some simplify, we know the number of limit cycles do not change under a affine map $(x,y)\to (\hat x,\hat y)=(x,y)A$ , where A is a nonsingular 2*2 matrix. So we could classify the topological graph of the dynamic system.

Step2:

Classification the singularity under affine map, and investigate the topological graph of the topological graph of the singularity by floer cohomology. There is only finite type and we could focus on them one by one.

Every color in the graph is an area $P,Q$ do not have same component in it. And the boundary of areas is just the same component of $P,Q$ if it exists.

Step3:

Bezout theorem tell us if two polynomials $P(x,y), Q(x,y)$ do not have same connected component then the intersection $I_{P,Q}\leq deg(P)deg(Q)$ . And we can divide the space $R^2$ into finite parts, $P(x,y),Q(x,y)$ restrict to every part do not have the same component. And we have a upper bound control on the number of part when $\max\{deg(P(x,y)),deg(Q(x,y))\} \leq n$ . A easy arrive bound could be $4^n$ .

Step4:

Now we focus ourself on 1 part where $P(x,y),Q(x,y)$ do not have same component on it. Now we begin to proof there will be a relationship between the limit cycle, very like a tree structure, it will combine with the following two thing:
1. Every limit cycle contains at least one singularity or one smaller limit cycle inside it.
2. Every pair of limit cycle $(A,B)$ , $A,B$ are limit cycle and $A$ is inside of $B$ , then there is at least one singularity or another limit cycle contain in $\Omega$ .

This kind of topological result will lead to a upper bound of number of limit cycle and end the proof of Hilbert 16th problem.

Planar polynomial vector field for a harmonic pair of polynomials

In this case you can consider the heat equation $\partial_t u(z,t)=\Delta u(z,t)$ . If the number of the limit cycle change, it must be the time to pass a singularity. and take $t=\infty$ , the dynamic system coverage to a very simlpe one and in particular it do not have limit cycle. So we need only look at the moments passing singularity.

Has the system of ODEs:

$\frac{dx}{dt}=P(x,y)\\ \frac{dy}{dt}=Q(x,y)$

been studied for the special case of the polynomials $P$ and $Q$ being a harmonic pair, i.e. the real and imaginary part of a holomorphic polynomial $F=F(z)$ , $z=x+iy$ ?

I am looking to learn a bit about (complex) ODEs and their interplay with algebraic geometry by some examples, but I couldn’t find anything on this special case in Ilyashenko’s survey on Hilbert 16 (I guess this case is too special and/or not very interesting as far as Hilbert 16 is concerned).

Nontheless, it seems very natural. If we set $\gamma(t)=x(t)+iy(t)$ , this amounts to the equation $latex \int_{\gamma_t}\frac{dz}{F(z)}=t$
where $\gamma_t$ is the curve $\gamma$ “truncated” at $t$ and the RHS is in particular **real**. This can be taken further, for example by assuming $\gamma$ is closed and using the residue theorem to obtain constraints on (the coefficients of) $F$ .

First, this case is totally uninteresting regarding Hilbert XVI. Indeed, there are no limit cycles in such systems. The $\alpha / \omega$ -limit of a trajectory is either a point or a non-isolated cycle (center case).

A singularity at $a\in \mathbb C$ (*i.e.* a root of $F$) can only be of three types, according to the value of $F'(a)$ :

1. Source/focus: $F'(a)\notin i\mathbb R$ .
2. Center: $F'(a)\in i\mathbb R_{\neq 0}$ .
3. Flower with $2k$ petals: $F'(a)=0$ with multiplicity $k$ .

In addition there is a pole at infinity (if $\deg(F)>0$ ) with exactly $2\deg(F)$ separatrices, reaching the singularity in finite time. The bassins of attraction / center regions attached to the above singularities are delimited by the separatrices.

[![enter image description here][2]][2]

S. Smale began to get interested in the question in the early 80’s while laying the foundations for BSS computational model (*The fundamental theorem of algebra and complexity theory*, 1981). He proposed a numerical root solver for polynomials by following the flow of $\frac{F}{F'}$ . This started some works on the topic, for instance by Schub, Tischler, William (*The Newtonian graph of a complex polynomial*, 1988) or Benzinger (*Plane autonomous systems with rational vector fields*, 1991)…

In the case of these vector fields, the topological class is entirely encoded by their Newtonian graph (or the «dual» spinal graph) given by the incidence graph of the $\alpha / \omega$ -limits of trajectories (in red on the picture). The main result for polynomials is that it is a tree. See *e.g.* Sverdlove (*Inverse problems for dynamical systems*,1981) and Schecter, Singer (*A class of vectorfields on $\mathbb S^2$ that are topologically equivalent to polynomial vectorfields*,1985) and Jongen, Jonker, Twilt (*On the classification of plane graphs representing structurally stable rational Newton flows*,1991).

The conformal classification has been initiated by Douady, Estrada and Sentenac (unpublished monograph, 2005) for the generic case (only focus/source singularities) and completed by Branner and Dias (*Classification of complex polynomial vector fields in one complex variable*, 2010). In addition to the combinatorial (topological) invariant, a complex «time-shift» (related to the integrals $\int_\gamma\frac{1}{F(z)} dz$ ) is associated to the separatrices, providing a complete conformal invariant.

In that latter context, the function $\int\frac{1}{F(z)} dz$ is called a Fatou coordinates. It is a rectifying chart for the vector field, and has many interesting dynamical properties.

Notice also the deep and beautiful relationship between spinal graph and *Dessins d’enfants*, as established by Pilgrim (*Polynomial vector fields, dessins d’enfants, and circle packings*,2006), related to [this question](https://mathoverflow.net/questions/118527).

Reference

Classification of Singularities and Bifurcations of Critical Points of Even Functions

E.A.Kudryavtseva, E.Lakshtanov

Classification of the singularity in even degree case.

Adjoint harmonic case have been studied. Look into the recent paper
Langley, J. K. Trajectories escaping to infinity in finite time. Proc. Amer. Math. Soc. 145 (2017), no. 5, 2107–2117, and the reference list in this paper.

They were also studied by physicists:

Bender, Carl M.; Hook, Daniel W.
Complex classical motion in potentials with poles and turning points.
Stud. Appl. Math. 133 (2014), no. 3, 318–336.

EDIT. I forgot to mention this:

B. Branner, K. Dias, Classification of complex polynomial vector fields in
one complex variable, Journal
Journal of Difference Equations and Applications
Volume 16, 2010 – Issue 5-6:

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。

Hilbert 第十六问题的第二部分问：平面多项式向量场的 limit cycles 数量能否只用次数控制？这看起来像一个拓扑问题，因为 limit cycles 是平面轨道的闭曲线；但真正困难在于解析和代数几何结构如何控制这些闭轨道的产生与消失。

Hilbert 第十六问题把 limit cycles、奇点、separatrix graph 和 return map 零点数放到同一个问题里。

1. 基本对象

考虑平面多项式系统

$$\dot x=P(x,y),\qquad \dot y=Q(x,y),$$

其中 $P,Q$ 是次数不超过 $d$ 的多项式。limit cycle 是孤立的周期轨道。Hilbert 第十六问题想问是否存在一个只依赖 $d$ 的上界 $H(d)$，控制所有这类系统的 limit cycles 数量。

对 $d=1$，线性系统没有孤立 limit cycles。对 $d\ge2$，问题迅速变得非常困难；一般情形至今仍未完全解决。

2. Poincare-Bendixson 的图像

Poincare-Bendixson theorem 告诉我们，平面系统的紧 $\omega$-limit set 若没有奇点，通常会落到周期轨道上。这使得二维动力系统比高维系统更有拓扑可视化：轨道、奇点、separatrices 和 limit cycles 共同组成一张平面图。

但可视化不等于容易。limit cycle 可以嵌套，可以从 polycycle bifurcation 中产生，也可能在参数变化时通过奇点附近的精细结构出现。

3. Affine invariance 与奇点分类

非退化 affine 变换不会改变 limit cycles 的数量。因此可以尝试先把系统化到较简单的坐标，再分类奇点局部模型。奇点由

$$P(x,y)=Q(x,y)=0$$

给出。若 $P,Q$ 没有公共因子，Bezout theorem 给出复射影意义下交点数不超过 $d^2$。这给了奇点数量的粗上界。

但是 limit cycles 不只由奇点数量决定。一个奇点周围可能有复杂的 separatrix structure；多个奇点之间的连接也可能产生 bifurcation。

4. 树结构的想法与局限

一个自然设想是把嵌套的 limit cycles 看成树：外层 cycle 包含内层 cycle 或奇点；两层之间的 annulus 里如果没有奇点，也许能排除新的孤立周期轨道。这样的拓扑图像确实有启发性。

困难在于，排除一个 annulus 中的周期轨道需要 Dulac function、Abelian integral、return map 或更精细的解析控制。拓扑结构给出框架，真正的上界需要估计 Poincare return map 的零点数。

5. Harmonic pair 的特殊情形

若 $P$ 和 $Q$ 是某个 holomorphic polynomial $F(z)$ 的实部和虚部，即

$$P+iQ=F(z),$$

则系统具有复分析结构。此时很多 Hilbert 第十六问题中的困难现象会消失：轨道可以通过

$$\int \frac{dz}{F(z)}=t$$

来理解，奇点由 $F$ 的零点控制。这样的系统通常不会给出 Hilbert XVI 中真正困难的孤立 limit cycles；它更像是一个用来学习复 ODE、Newtonian graph 和代数拓扑图像的模型。

6. 这条路线的意义

一个可能的攻击路线是：先用 affine 变换和 Bezout 控制奇点类型；再用 separatrix graph 把平面分解成有限区域；最后在每个区域中估计 return map 或 Dulac integral 的零点。这个策略很自然，但每一步都需要强解析输入。

Hilbert 第十六问题之所以难，正是因为它站在三件事的交界处：平面拓扑告诉我们轨道如何嵌套，代数几何控制多项式奇点，分析估计决定 limit cycles 能否真正出现。
2017年11月18日