github-backend

分类： Dynamic system

  • Sturm-Liouville theory 与周期轨道：谱分解、边界条件和 monodromy

    旧博客原文

    原题：Periodic orbits and Sturm–Liouville theory

    I thinks there is some problem related to the solution of a 2 order differential equation given by Sturm-Liouville system which is nontrivial.

    It is well-know that the power of Sturm-Liouville theory see  wiki, is due to it is some kind of “spectral decomposition” in the solution space.

    Two kind of problem is interesting, one is the eigenvalue estimate, both upper bound and lower bound, this already investigated in ESTIMATING THE EIGENVALUES OF STURM-LIOUVILLE. PROBLEMS BY APPROXIMATING THE DIFFERENTIAL EQUATION.

    I post two problem here, this is a product due to a random walk along the boundary of topology and the analysis,

    Problem 1.

    Fix a set A=\{k_1<k_2<...<k_l\}, is there a 2 order ordinary differential equation given by Sturm–Liouville theory  such that the eigenfunction f_{k} is periodic if and only if k\in A?

    There is also some weak version of this and a infinity version of this.

    Of course we have the following map, from the high order ordinary differential equation to the 1 order differential equation in high dimension. But the key point is that it is not a bijection! The Frobenius condition play a crucial role.

    Problem 2.

    There is a homotopy in the moduli space of differential equation, and we could define a direct product operator in this space, and we consider the topology defamation of the eigenfunction, could there be some equality, one side of it explain the topology information, the other side explain the spectral (or analysis) information?

    There is another interesting problem.

    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。

    Sturm-Liouville theory 把二阶线性微分方程变成谱问题。若再加入周期边界条件，就会自然出现 monodromy matrix、周期解和一维动力系统的交界。

    Sturm-Liouville theory 与周期轨道：谱分解、边界条件和 monodromy
    Sturm-Liouville 周期谱可以通过一阶系统的 monodromy matrix 来刻画。

    1. Sturm-Liouville 系统

    标准形式为

    $$-(p(x)y’)’+q(x)y=\lambda w(x)y.$$

    在合适边界条件下，这是自伴特征值问题，特征函数构成正交基。

    2. 周期边界条件

    若区间为 $[0,T]$，周期解满足

    $$y(0)=y(T),\qquad y'(0)=y'(T).$$

    这等价于一阶系统的 monodromy matrix 有特征值 $1$。

    3. 从二阶到一阶系统

    令 $Y=(y,py’)$，二阶方程可写为

    $$Y’=A_\lambda(x)Y.$$

    基本解矩阵 $M_\lambda(T)$ 描述一个周期后的变化。周期谱由

    $$\det(M_\lambda(T)-I)=0$$

    刻画。

    4. 反问题

    一个自然问题是：能否指定某个集合 $S$，使得周期特征值恰好落在 $S$ 中？这类问题接近 inverse spectral theory，需要理解势函数 $q(x)$ 如何控制 monodromy。

    5. 拓扑与分析

    周期轨道是动力系统对象，Sturm-Liouville 是谱分析对象。monodromy matrix 把二者联系起来：谱参数变化时，monodromy 在矩阵群中运动，周期解对应它穿过特定子集。

  • 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:

    img_0510.jpg

    (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.

     

    img_0511

    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:img_0513.jpg

    \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。

    Rotation number：圆周同胚的提升、周期点与半共轭
    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。

  • 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：周期系数线性系统与 monodromy matrix
    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。

  • 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.

    img_0069.jpg
    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：skew product 与 interval exchange
    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。

  • 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。

    Metric entropy（二）：entropy map 的上半连续性与 infinity entropy
    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。

  • 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。

    Metric entropy（一）：Ruelle 不等式、Pesin 公式与 Lyapunov 指数
    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。

  • 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:

     

    06458F94-5D06-4578-8588-3962E07291FB.png

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

    37D51278-96BF-4A32-B8D5-6F900BAB10D6.png27E8B04D-FD07-449C-95E3-8D1D4D2D92B8.png1F966065-6A6A-42E0-AA2C-096B1DD73FD8.png

     

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

    452086C1-4237-45AC-AEFF-C0B8CB68496E.png

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

    2A34FA37-4C47-4D87-8772-701FD2B05324.png

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

    F107D058-486E-4F62-92FB-7FC03C5668A2.png

     

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

    7E54B6CB-B085-43F6-809C-918EAA380E9D

    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

    0B2AEAFD-9FD7-491C-A28A-BB969E36AAB92709DFBC-AD6C-4645-90D3-6F602701ADBD.png8251B027-35DF-4576-9695-C939BD84BC20.png963F63F4-FF84-4E93-9F7B-90C4027DE72C.png

     

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

    29D2E486-CAA7-4759-BBA5-76B0AAFAC32A.pngBBC9D6CA-8E5D-4FF9-AFCC-FCCA1F9DDFDB.png

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

    BA20336A-77C8-4F28-959B-45FF66E50259.png

    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)

    9469795B-03E4-4F80-BB52-EF3B58FF29BB.png

    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.

    img_0508Every 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.

    img_0509.jpg

    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 第十六问题笔记：limit cycle、奇点与拓扑图像
    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。

  • 动力系统笔记（一）：transitivity, minimality 与 Birkhoff 回复

    旧博客原文

    原题：动力系统笔记

     

    \section{基本性质，例子}
    \subsection{例子和基本性质}
    在这一章的第一节引入了我们的研究对象，一般是一个紧的度量空间$X$装备上了一个同胚
    T:X\longrightarrow X
    介绍了三个简单例子，包括S_1上的加倍映射，旋转映射以及X_k=\Pi_{n\in Z}\{1,2,...,k\} 上的平移映射。
    加倍映射会出现在微分流形中一些函数f的singular point，也就是hess f=0的地方附近的环绕数计算，还有一些scalling变换或者是一些多尺度的问题里。\\
    旋转映射会和旋转数是有理数还是无理数有关,相关的wely准则告诉我们如果是无理数的话会是每个点的轨道均匀分布的，稠密性在动力系统里面说就是这个动力系统是minimal的。相关的问题有sarnack猜想在Torus上的特殊情形，目前半解析的$T^2$情形已经解决，这是最近的工作，后续很多工作在进行，本质困难来自解析数论。\\
    平移映射我不是很懂，第二章中讲的Van der warden定理的证明是一个好例子，动力系统中的回复定理主要是用来刻画这些动力系统内蕴的算术性质的，basic ideal是如下事实：\\
    将一个大的集合分类，同一类有序的出现的存在性。

    \subsection{Transitivity}
    这个性质是指一个动力系统中存在轨道在动力系统中稠密。\\
    动力系统往往具有transitivity的性质，加倍映射的例子用二进制分解构造，平移映射构造transitivity point的方法与之雷同，旋转映射情形这是初等的。\\
    transitivity会有很多等价的刻画，包括四种：

    \begin{thm}(transitivity的等价定义)\\
    1.transitivity\\
    2.U open,TU=U \Longrightarrow U=\emptyset或者U是一个稠密集\\
    3.U,V开集,\exists N\in N^*,T^n U \cap V\neq \emptyset \\
    4.\{x\in X|\{T^nx\}_{n\in Z}dense\}是一个G_{\delta}集合
    \end{thm}
    证明都是标准的，提两个关键点，第一点是注意到\cup_{n\in Z}T^n U这个集合是$T$不变的，第二点是注意到transitive point可以通过选取一组开集集进行描述从而有集合等式：
    \{x\in X|\{T^nx\}_{n\in Z}dense\}=\cap_{n\in N^*}\cap_{k\in N^*}\cup_{m\in Z} T^mB_{\frac{1}{k}}(x_n)
    \subsection{一个和矩阵有关的例子}
    定义了一个和矩阵有关的动力系统，并且说明了这个动力系统是transitive的当且仅当底层的矩阵是不可约的，对于矩阵不可约这个概念不熟所以这个例子没有仔细看。
    \subsection{minimality和Birkhoff回复定理}
    我认为这部分内容是Pollicott书第一章最有趣的部分。\\
    minimality定义是动力系统T:X\rightarrow X所有的点都是transitivity point。\\
    也有三个等价定义，其他两个是：\\
    T不变集只有X和空集\\
    任何开集通过T作用生成的集合是全空间\\

     

    可以看出来这三个定义都是transitivity情形对应定义的加强版。这些证明也是标准的，接下来一个定理表明X这个空间可以在T不变的意义下分解成很多小的空间，每个都是不能再分解的，这个定理的证明的两个关键点是：
    1.zorn lemma，2.minimal性质的第二点\\

     

    那么马上我们就可以得到minimality的定理系统满足birkhoff回复定理：\\
    \exists x\in X,\exists \{n_i\},\lim_{i\to \infty}T^{n_i}x=x
    Birkhoff回复定理在高维情形也会很有趣，我们这时就需要多个可交换的动力系统（为什么一定要可交换？一种解释是可交换大幅降低复杂度）一旦这些动力系统被证明是minimal的，我们用类似的路线建立起以上定理是没有本质困难的。\\
    步骤一：建立起transitive的相关定理\\
    步骤二：建立起minimal的相关定理\\
    步骤三：说明T^i不变的集合满足zorn lemma，所以有最小元\\
    整个过程在乘积空间中进行
    \newpage

    \section{Birkhoff回复定理蕴含Van Der Warden定理}
    \subsection{Van Der Warden定理与它的动力系统解释}
    这是一个组合定理，原始证明是很trick的，单遵老先生有一个证明，很trick，高中的时候尝试过证明，自己证了一个星期证明不出来就看掉了，现在回想起来应该跟当时的工具太原始了有关系。我想强调的是并不是数学思想的飞跃，而是数学工具的升级使得这个问题变简单了。\\

     

    原始问题是将N^*分成若干个类，一定存在一个类存在任意长等差数列。\\

     

    怎么转化成一个组合问题呢，其实用第一章中的X_k装备上平移这个同胚构成的动力系统就够了，等差数列的存在性等价于若干个可以交换的映射，其实就是平移的步长不一样下的都回到原始值附近，这是birkhoff回复定理能够告诉我们的。\\

    关于细节的建立
    第一步是简单的，问题出在第二步，也就是证明整个动力系统是minimal的这一步上，我们知道这个动力系统是初始状态通过平移生成再取闭包得到的，所以天然是transitivity的，如果是minimality的，那么就可以用birkhoff定理得到结果了。这其实不难，因为这个距离空间是non-archimeadian的，用初始状态的平移去逼近就好了。\\

    上面这一段划去，实际上要想真正建立一个动力系统本身是minimal的性质，本质上需要比连续性更强的某种正则性，比如一个lipchitz连续性的动力系统就是minimal的。但是仅仅找到一个紧集是minimal的时简单是事情，用minimal等价定义第二条加zorn引理就可以做到。好了现在我们有了一个minimal的动力系统，我们只需要建立多重birkhoff回复定理就完成了证明。\\

    在思考多重birkhoff回复定理的过程中我发现了几种方式来构建整个框架，pollicott上标准的证明是利用乘积空间的对角线作为低空间加上归纳法，我尝试过将对角线作为低空间证明但是失败了，主要原因是对角线在乘积空间中是低维子集我不知道怎么将合适映射限制到这个空间上，事实也证明做归纳法的话我们可以转而对映射而不是空间做文章而规避这个困难。\\

    但是在这个过程中我发现了另外一个有意思的现象，就是我们可以归纳的构造出一个集合，至少有限步的构造在逻辑上式对的，利用算子T^i之间的交换性得到一个很好地X的子空间，T^i在上面的作用也有很好的性质，但是还不够好。具体的说，是一种纤维结构的空间，T^1在底空间上作用是transitive的，T^2在第一层纤维上的作用是transitive的，依次类推。由于交换性可以导致在每一个section上T^i的作用都是trsnsitive的。但是不好的地方在于每个T^i想要在全空间中transitive都必须借助别的T^i，换而言之每个T^i都只管一层。所以这并不是我们想要的空间。\\

    这样构造出来的空间在这里可能没有用，但是这个空间本身具备很好的性质，而就算我们知道多重回复定理这样的空间也是构造不出来的，注意我们并不是因为构造了一个T^i在上面”一致的”transitivity的空间而把回复定理证明出来了，而是用了一些更弱的argument达到目的。这个空间可能在计算全空间上某些可交换的映射的特征时有用，尤其在可以证明这个空间和全空间只差一个零测集的情况下。\\

    猜想：存在一个T^i:X\longrightarrow X ,$T^iT^j=T^jT^i$，并不存在满足某种”一致”minimal的子空间。但是我们知道多重回复定理是对的。\\

    总之用归纳的方法加上一些拓扑的标准的方法我们可以得到多重回复定理从而完成证明。
    \newpage

    \section{拓扑熵}
    拓扑熵的定义可复杂了，拓扑熵是一个描述拓扑动力系统复杂程度的量。顺序是先引入标准定义和基本性质，然后给出一个计算方法，再然后引入spanning set和separeting set，利用这两种集合引入等价的定义方法，再证明amernov定理：h(T^m)=mh(T),最后证明动力系统之间的半共轭会导致熵之间的不等式。

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

    \newpage
    $f(x),g(x)\in C_{c}^{\infty}(R^n)$
    \[f*g(x)=\int_{R^n}f(\xi)g(x-\xi)d\xi \]
    \[x=(x_1,x_2,…,x_n)\in R^n\]
    \[g(x)=\frac{1}{x_1^2+x_2^2…+x_n^2+1}\]
    \[f*g(x)=\int_{R^n}f(\xi)g(x-\xi)d\xi\sim\sum_{k_1=-\infty}^{\infty}…\sum_{k_n=-\infty}^{\infty}\frac{f(x_1-k_1,x_2-k_2,…,x_n-k_n)}{k_1^2+k_2^2+…+k_n^2+1} \]
    \[\sum_{k_1=-\infty}^{\infty}…\sum_{k_n=-\infty}^{\infty}\frac{f(x_1-k_1,x_2-k_2,…,x_n-k_n)}{k_1^2+k_2^2+…+k_n^2+1}=\sum_{\xi\in Z^n}f(x-\xi)g(\xi)=\int_{\xi\in R^n}f(x-\xi)\delta g(\xi) d\xi\]\\
    (Young inequality)
    $f\in L_1(R^n)$,$g\in L^p(R^n)$:
    \[ ||f*g||_{p}\leq ||f||_1||g||_{p} \]

    (hardy-litterwood-soblev inequality)
    $p,r>1$,$0<\lambda<n$,$\frac{1}{p}+\frac{\lambda}{n}+\frac{1}{r}=2$,$f\in L^p(R^n),h\in L^r(R^n)$.exists a constant C,$C\sim n,\lambda,p$.
    \[|\int_{R^n}\int_{R^n} f(x)|x-y|^{\lambda}g(y)dxdy|\leq C(n,\lambda,p)||f||_p||h||_r\]

    \section{热核正则性}
    t \to 0^+情况的技巧
    gap太多了，主要集中在两条，第一条是需要研究billiard上热核的正则性，这需要建立大量的耗散性先验估计。连续是显然的，我目前连C1都证明不出来，因为其中需要处理一个级数和。如果这一条对了，那么我们集中看t趋于0正时的热核。

     

    \section{Caldron Zygmund算子的谱}
    我们需要刻画Caldron Zygmund算子作用在某个区域上之后产生的谱会携带多少区域的形状的信息。通过在热核中令t\longrightarrow 0这个会化简为简单的情况，再加上凸性。
    第二条是热核对t求任意次导以后是Caldero ́ n Zygmund算子，对这个算子卷积上一个具备C^1Boudary正则性的区域上特征函数的的谱我们有没有好的刻画，这其中能不能蕴含这个区域的几何信息。

     

    补充说明

    以下是新整理的中文说明；上方旧博客原文保持不变。

    这一篇笔记想讲清楚一个很基础但是很重要的转换：很多组合问题，尤其是关于等差数列和回复现象的问题，可以放到一个紧的动力系统里面看。这样做以后，原来很硬的组合构造会变成轨道闭包、minimal set 和 Birkhoff 回复定理。

    Van der Waerden recurrence schematic
    把一个有限染色编码成 shift space 中的一点，然后用 minimal 子系统和多重回复得到同色等差数列。

    1. 基本例子：加倍映射、旋转和平移

    一个拓扑动力系统可以写成 $(X,T)$，其中 $X$ 是紧度量空间，$T:X\to X$ 是连续映射；如果 $T$ 是同胚，那么可以向前也可以向后迭代。我们真正研究的是一条轨道

    $$x,\;Tx,\;T^2x,\;\ldots.$$

    最基本的例子有三个。

    第一是圆周上的加倍映射

    $$T(x)=2x\pmod 1.$$

    这个例子带有扩张性。用二进制展开看，$T$ 只是把二进制小数点向右移动一位。因此只要选一个二进制展开中含有足够多有限字串的点，它的轨道就会在圆周上到处跑。这个例子是理解 symbolic dynamics 的入口。

    第二是圆周旋转

    $$R_\alpha(x)=x+\alpha\pmod 1.$$

    如果 $\alpha$ 是有理数，那么所有轨道都是周期的；如果 $\alpha$ 是无理数，那么每条轨道都稠密，并且更强地，轨道是均匀分布的。这里最常用的判别工具是 Weyl criterion：判断均匀分布可以转化成检查所有非平凡 Fourier characters 的平均趋于零。

    $$\frac1N\sum_{n=0}^{N-1}e^{2\pi i k(x+n\alpha)}\to 0,\qquad k\ne0.$$

    第三是环面平移

    $$T_\omega x=x+\omega\pmod{\mathbb Z^d}.$$

    当 $1,\omega_1,\ldots,\omega_d$ 在 $\mathbb Q$ 上线性无关时，轨道在 $\mathbb T^d$ 中稠密。这个例子和 Sarnak 猜想、nilsequence、Kronecker system 都有关。对于零熵系统，Sarnak 猜想说莫比乌斯函数应该和系统产生的观测序列正交；圆周旋转和环面平移是这类问题最早、最干净的模型。

    2. Transitivity：存在一条稠密轨道

    定义：如果存在 $x\in X$，使得

    $$\overline{\{T^n x:n\ge0\}}=X,$$

    那么称 $(X,T)$ 是 topologically transitive，这个 $x$ 叫 transitive point。

    在紧度量空间且没有孤立点的常见情形下，可以用开集来刻画 transitivity：

    $$\text{对任意非空开集 }U,V\subset X,\quad \exists n\ge0,\quad T^nU\cap V\ne\varnothing.$$

    这个刻画很有用，因为它不需要显式构造那条稠密轨道，只需要说明任意两个局部区域之间存在一次迭代连接。

    为什么加倍映射有 transitive point？因为二进制展开可以人为拼接所有有限 0-1 字串。这样构造出来的点，在 shift 意义下会依次出现任意有限模式，因此轨道稠密。这个想法之后会在 Van der Waerden 定理的证明里再次出现：有限组合结构被编码成一个无限序列，动力系统研究的是这个无限序列的 shift orbit closure。

    3. Minimality：每一条轨道都稠密

    transitivity 只要求存在一条稠密轨道。minimality 强得多：它要求每一点都是 transitive point。

    定义：如果对所有 $x\in X$，都有

    $$\overline{\{T^n x:n\ge0\}}=X,$$

    那么称 $(X,T)$ 是 minimal。

    它有两个非常重要的等价刻画：

    第一，$X$ 没有非空真闭不变子集。也就是说，如果 $Y\subset X$ 闭且 $T(Y)\subset Y$，那么 $Y=\varnothing$ 或 $Y=X$。

    第二，对任意非空开集 $U\subset X$，有

    $$X=\bigcup_{n\ge0}T^{-n}U.$$

    直观上，这表示每个点迟早都会进入 $U$。在紧性下还可以加强成有限覆盖：存在 $N$，使得

    $$X=\bigcup_{n=0}^{N}T^{-n}U.$$

    这就是 minimal system 中的 uniformly recurrent 现象。

    Zorn lemma 在这里的作用是保证 minimal 子系统的存在。任意紧动力系统里，只要取一个非空闭不变集族，用包含关系作偏序，就可以用 Zorn 引理取到极小闭不变集。这个极小闭不变集上的动力系统就是 minimal 的。所以即使原系统不 minimal，我们也总能在轨道闭包里找到 minimal 子系统。

    4. Birkhoff 回复定理

    在拓扑动力系统里，一个基本的 Birkhoff 回复命题可以这样理解：

    如果 $(X,T)$ 是紧动力系统，那么存在 recurrent point；如果系统是 minimal 的，那么每个点都是 recurrent 的。

    这里 recurrent 的意思是，存在 $n_j\to\infty$，使

    $$T^{n_j}x\to x.$$

    对 minimal system 来说，这几乎是定义的直接后果：因为 $x$ 的轨道稠密，所以它必然反复进入 $x$ 的任意小邻域。

    更有力量的是多重回复。若 $T_1,\ldots,T_k$ 是两两可交换的连续变换，在合适的 minimal 子系统上，可以找到同一个时间参数让多个方向同时回到某个开集附近。这里“可交换”是关键，否则不同方向的迭代顺序会产生额外复杂度。

    5. Van der Waerden 定理的动力系统证明

    Van der Waerden 定理说：把自然数染成有限多种颜色以后，必定存在任意长的同色等差数列。

    动力系统证明的想法如下。给定一个染色

    $$c:\mathbb N\to\{1,\ldots,r\},$$

    把它看成一个无限序列

    $$x=(c(0),c(1),c(2),\ldots)\in\{1,\ldots,r\}^{\mathbb N}.$$

    在紧空间 $\{1,\ldots,r\}^{\mathbb N}$ 上考虑 shift map

    $$\sigma(x_0,x_1,x_2,\ldots)=(x_1,x_2,\ldots).$$

    取轨道闭包

    $$Y=\overline{\{\sigma^n x:n\ge0\}}.$$

    这个 $Y$ 是非空紧不变集。再从 $Y$ 中取一个 minimal 子系统 $M$。设 $y\in M$，并看 $y_0$ 这个坐标的颜色。令

    $$U=\{z\in M:z_0=y_0\},$$

    这是一个非空开集。多重回复告诉我们，对任意 $k$，存在 $d>0$ 和某个点 $z\in U$，使

    $$z,\;\sigma^d z,\;\sigma^{2d}z,\ldots,\sigma^{(k-1)d}z\in U.$$

    翻译回坐标，就是

    $$z_0=z_d=z_{2d}=\cdots=z_{(k-1)d}=y_0.$$

    因为 $z$ 属于原染色序列的轨道闭包，有限坐标模式可以被原来的序列 $x$ 近似出来，于是在原来的自然数染色中存在

    $$a,\;a+d,\;a+2d,\ldots,a+(k-1)d$$

    这些位置颜色相同。这就是 Van der Waerden 定理。

    6. 这条路线真正说明了什么

    这个证明没有给出最优界，也不是组合意义上最有效的证明。但它说明了一件非常深的事：有限组合结构可以来自无限紧空间里的回复。

    原来要在自然数里找同色等差数列，现在变成了：

    第一，把染色编码成 shift space 中的点；第二，取轨道闭包；第三，取 minimal 子系统；第四，用多重回复；第五，把有限模式拉回原来的染色。

    这个过程是 Furstenberg 观点的雏形。它后来可以继续发展到 Szemerédi 定理、遍历 Ramsey 理论，以及 nilsystem 和高阶 Fourier 分析之间的联系。

    所以这篇笔记真正想记录的是：动力系统不是给组合定理套一层语言，而是在解释为什么“局部有限模式”会被“全局回复结构”强迫出现。

  • Sarnak 猜想在 skew product 上的情形

    旧博客原文

    原题：Sarnak猜想在skew product上的情形。

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

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

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

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

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

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

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

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

    \end{lemma}

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

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

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

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

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

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

     

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

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

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

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

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

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

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

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

     

    补充说明

    以下是新整理的中文说明；上方旧博客原文保持不变。

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

    Sarnak 猜想在 skew product 上的情形
    Skew product 的迭代把 Fourier character 转成带有旋转 Birkhoff sum 的指数和。

    1. Skew product 的形式

    考虑二维环面上的映射

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

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

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

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

    2. 相位展开

    迭代 $T$ 得到

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

    所以相关和变成

    $$\sum_{k\le N}\mu(k)e\left(m(x+k\alpha)+n y+n\sum_{j

    问题的核心是控制这个由旋转 Birkhoff sum 产生的相位。若 $h$ 是多项式或 Fourier 支持很简单，相位可以化成多项式相位，经典解析数论的指数和估计可以直接进入。

    3. 有界型旋转数

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

    $$N=\sum_k b_k q_k.$$

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

    4. Cut-off 与 Fourier 级数

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

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

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

    5. 证明图像

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