博客

UCLA 2009 Analysis Qualifying Exam：几个分析题的共同结构

旧博客原文

原题：Analysis Qualifying Examination(UCLA 2009)

1.

Let $f,g$ be real-valued integrable functions on a measure space $(X,B,\mu)$ ,and define:

$F_t=\{x\in X:f(x>t)\},G_t=\{x\in X:g(x)>t\}$ .

Prove:

$\int|f-g|d\mu=\int_{-\infty}^{\infty}\mu((F_t-G_t)\cup(G_T-F_t))dt$ .

proof:

by Fubini theorem(cake representation theorem in fact):

$\int|f-g|=\int_{0}^{\infty}\mu(\{x||f-g|(x)>t\})dt\\ \displaystyle =\int_{-\infty}^{\infty}\mu(\{x|f(x)>t>g(x)\})+\mu(\{x|f(x)<t<g(x)\})dt\\ =\int_{-\infty}^{\infty}\mu((F_t-G_t)\cup(G_T-F_t))dt$ .

(there is a geometric heuristic,strict proof is due to fubini theorem)

Q.E.D.

2.

Let $H$ be a infinite dimensional real Hilbert space.

a)Prove the unit sphere $\{x\in H:||x||=1\}$ of $H$ is weakly dense in the unit ball $B=\{x\in H :||x||\leq 1\}$ of $H$ .

b)Prove there is a sequence $T_n$ of bounded linear operator from $H$ to $H$ such that $||T_n||=1$ for all n but $lim T_n(x)=0$ for all $x\in H$ .

proof:

by Zorn lemma there is a orthogonal bases $\{e_i\}$ .

to proof a),suffice to proof: $\forall x,\exists x_n,\forall y\in H,\lim_{n \to \infty}<x_n,y>=<x,y>$ .

this can be done by look at the expansion $y=\sum_{i}<y,e_i>e_i$ .due to the Cauchy inequality,there is a freedom of choice the coefficient $<e_i,x_n>$ for $i>>n$ .the choice will lead a).

b) is trivial due to a).

Q.E.D.

3.Let $X$ be a Banach space and let $X^*$ be it dual Banach space.Prove that if $X^*$ is separable then $X$ is separable.

proof:

we know $X^*$ is the space consist with bounded(continued) linear functional on $X$ .

for $f\in X^*$ , $||f||_{X^*}=\sup_{x\in B}||x||$ ,so due to $X^*$ is separable.there is a countable dense set $I$ in $X^*$ .i.e. $\forall f\in X^*, \forall \epsilon >0,\exists f_{\epsilon}\in I,||f-f_{\epsilon}||_{x^*}<\epsilon$ .we equip a member of $X$ to $f_{\epsilon}$ by $H:I \to X,H(f)=x,x=sup_{x\in B}||f(x)||$ , $\hat I=Im(I)$ .

On the other hand, $\forall x\in X$ we construct a functional $l_x$ . $l_x(y)=||y|| iff y=cx,c\in R,or ,l_x(y)=0$ .so it is obviously to show $\hat I$ is dense in $X$ .

Q.E.D.

5.Let $I=I_{0,0}=[0,1]$ be the unit interval,and for $n=0,1,2,...$ and $0\leq j \leq 2^n-1$ ,let:

$I_{n,j}=[j2^{-n},(j+1)2^{-n}]$ .

For $f\in L^1(I,dx)$ define $E_nf(x)=\sum_{j=0}^{2^n-1}(2^n\int_{I_{n,j}}fdt)\chi_{I{n,j}}$ .

Prove that if $f\in L^1(I,dx)$ then $lim_{n\to \infty}E_nf(x)=f(x)$ a.e. in I.

proof:

…gap…

10.Let $D$ be the open unit disc and $\mu$ be Lebesgue measure on $D$ .let $H$ be the subspace of $L^2(D,\mu)$ consisting of holomorphic functions.Show that $H$ is complete.

proof:maximum norm principle.to show $u_n$ is closed uniformly coverage.so is harmonic and $L^2$ is due to $L^2$ itself is complete.

Q.E.D.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

这组题表面上分散：有测度论、Hilbert 空间弱拓扑、Banach 空间可分性、函数空间闭性。但它们的共同核心是：把抽象对象转化为可数逼近、积分层分解或闭性问题。

测度层分解、弱拓扑逼近、对偶可分性和闭子空间完备性，是这组分析资格考试题背后的共同工具。

1. Layer-cake 与 Fubini

对非负函数 $f$，常用恒等式是

$$f(x)=\int_0^\infty \mathbf 1_{\{f(x)>t\}}\,dt.$$

积分后用 Fubini 交换次序，就得到 layer-cake representation。很多看似是范数不等式的题，其实都可以化成集合层面的包含关系。

2. Hilbert 空间单位球的弱拓扑

在无限维 Hilbert 空间中，单位球的弱闭包现象和强拓扑非常不同。给定 $\|x\|\le1$，可以在与有限多个测试向量正交的方向上添加一个小向量，使范数变成 $1$，同时不改变这些测试函数的值。于是单位球中的点可被单位球面弱逼近。

3. 可分性与对偶

若 $X^*$ 可分，则 $X$ 可分。证明思路是从 $X^*$ 中取可数稠密集 $\{f_n\}$，再为每个有限线性条件选择近似达到范数的点。Hahn-Banach 的思想隐藏在这里：对偶空间的可数信息足以区分 $X$ 中的点。

4. 闭子空间的完备性

若 $L^2(D)$ 中的全纯函数子空间在 $L^2$ 收敛下保持全纯，那么它就是闭子空间，从而完备。关键是用局部估计把 $L^2$ 收敛提升为紧子集上的一致收敛，再由 Weierstrass 定理保全全纯性。

2017年10月11日
Schauder estimate 与 Sobolev inequality：椭圆方程正则性的两种语言

旧博客原文

原题：Schauder estimate and Sobelov inequality

In this note we discuss the Schauder theory for uniformly elliptic linear equations and Sobelov inequality.

the three main topics ars a priori estimate in Holder norms,regularity of arbitrary solutions and the solvability of the Dirichlet problem.Among these topics,a priori estimates are the most fundamental and the basis of the follows two.we will discuss both the interior Schauder estimate and global Schauder estimate.

-Schauder Theory-

1. Interior Schauder Theory

${\Omega}$ be a domain in ${R^n}$ ,bounded most of the time.
${a_{ij},b_i,c}$ be defined in ${\Omega}$ ,with ${a_{ij}=a_{ji}}$ .where ${1\leq i,j\leq n}$ .
we consider the operator ${L}$ given by,

$\displaystyle Lu=a_{ij}\partial_{ij}u+b_i\partial_iu+c,in \ \Omega.$

easy to see ${Lu}$ is defined for any ${u\in C^2(\Omega)}$ .
the operator ${L}$ is always be assumed to be strictly elliptic in ${\Omega}$ ;namely,
$\displaystyle a{ij}\xi_i\xi_j \geq \lambda|\xi|^2$

for any ${\xi\in R^n,x\in \Omega}$ ,where ${\lambda}$ is a positive constant.
1.1. Interior Schauder Estimate

define the weighted ${C^{k,\alpha}}$ norm,

$\displaystyle |u|^*_{C^{k,\alpha}(B_R)}=\sum_{i=0}^k R^i|D^iu|_{L^{\infty}(B_R)}+R^{k+\alpha}[D^ku]_{C^{\alpha}(B_R)}$

easy to see ${R}$ come from a scaling.
consider the PDE.
$\displaystyle Lu=a_{ij}\partial_{ij}u+b_i\partial_iu+c=f,in \ \Omega.$

we want to proof this type estimate,
$\displaystyle |u|_{C^{2,\alpha}(A)} \leq C(|u|_{L^{\infty}(\Omega)}+|f|_{C^{\alpha}(\Omega)})$

where ${A\subset \Omega }$
we first deal with a easy case, ${a_{ij}}$ is constant. in this case we proof the estimate:

Lemma 1 ${f \in C^{\alpha}(B_R)}$ ,for some ${\alpha \in (0,1)}$ ,and ${(a_{ij})}$ be a constant symmetric ${n\times n}$ matrix satisfying
$\displaystyle \lambda |\xi|^2 \leq a_{ij}\xi_i\xi_j \leq \Lambda|\xi|^2$

${\exists \lambda,\Lambda >0,\forall \xi \in R^n}$ . suppose ${u\in C^2(B_R)}$ satisfies:
$\displaystyle a_{ij}\partial_{ij}u=f, in \ B_R$

then , ${u \in C^{2,\alpha}(B_{\frac{R}{2}})}$ ,moreover,
$\displaystyle |u|^*_{C^{2,\alpha}(B_{\frac{R}{2}})}\leq C[|u|_{L^{\infty}(B_R)}+R^2|f|^*_{C^{\alpha}(B_R)}]$

Proof: $\Box$ to continue,we prove an interpolation inequality for Holder continuous functions.

Lemma 2 Let ${\alpha,\mu \in (0,1)}$ and ${B_R}$ be a ball of radius ${R}$ in ${R^n}$ ,then, (1)for any ${u\in C^{1,\alpha}(\overline B_R)}$ ,
$\displaystyle \mu^{\alpha}R^{\alpha}[u]_{C^{\alpha}(B_R)}\leq C[\mu R |\nabla u|_{L^{\infty}(B_R)}+|u|_{L^{\infty}(B_1)}]$

(2)for any ${u\in C^{1,\alpha}(\overline B_R)}$ ,
$\displaystyle \mu R |\nabla u|_{L^{\infty}(B_R)} \leq C[\mu^{1+\alpha}R^{1+\alpha}|\nabla u|_{C^{\alpha}(B_R)}+|u|_{L^{\infty}(B_1)}]$

(3)for any ${u\in C^2(\overline B_R)}$ ,
$\displaystyle \mu R|\nabla u|_{L^{\infty}(B_R)}\leq C[\mu^2R^2|\nabla^2 u|_{L^{\infty}(B_R)}+|u|_{L^{\infty}(B_R)}]$

where ${C}$ is a positive constant depending on n and ${\alpha}$ .
Proof: $\Box$

Corollary 3 Let ${\alpha,\mu\in (0,1)}$ and ${B_R}$ be a ball of radius ${R}$ in ${R^n}$ .Then,for any ${u\in C^{2,\alpha}(\overline B_R)}$ ,
$\displaystyle \sum_{i=0}^2(\mu R)^i|\nabla^i u|_{L^{\infty}(B_R)}+\sum_{i=0}^1(\mu R)^{i+\alpha}[\nabla^i u]_{C^{\alpha}(B_R)}\leq C[(\mu R)^{2+\alpha}[\nabla^2 u]_{C^{\alpha}(B_R)}+|u|_{L^{\infty}(B_R)}]$

Proof: $\Box$

Now we are ready to prove an interior estimate for ${C^{2,\alpha}}$ -norms of solutions of uniformly elliptic equations.The trick is to freeze coefficients.

Lemma 4
2. Global Schauder Theory

-Sobelov inequality-

Theorem 5
$\displaystyle W_0^{1,p}(\Omega)\longrightarrow L^{\frac{np}{n-p}}(\Omega),1\leq p <n$

moreover,we have: ${\exists C=C(n,p)}$ , ${\forall u\in W^{1,p}_0(\Omega)}$ ,
$\displaystyle ||u||_{\frac{np}{n-p}} \leq C||Du||_p,1\leq p<n$

Proof:

$\displaystyle p=1$

suffice to proof:
$\displaystyle ||u||_{\frac{n}{n-1}}\leq C||Du||_1$

obvious we have:
$\displaystyle |u(x)|\leq \int_{-\infty}^{\infty}|Du(x)|dx$

so ${\int_{\Omega} |u|^{\frac{n}{n-1}}\leq \int_{\Omega} \Pi_{i=1}^n(\int_{-\infty}^{\infty}|D_iu(x)|dx)^{\frac{1}{n-1}}}$ .
so ${||u||_{\frac{n}{n-1}}\leq (\int_{\Omega}\Pi_{i=1}^n(\int_{-\infty}^{\infty}|D_iu|)^{\frac{1}{n-1}})^{\frac{n-1}{n}}\leq \int_{\Omega} \Pi_{i=1}^n(\int_{-\infty}^{\infty}|D_iu|)^{\frac{1}{n}} \leq \int_{\Omega} \frac{1}{n} \sum_{i=1}^n(\int_{-\infty}^{\infty}|D_iu|)\leq C||Du||_1}$
$\displaystyle 1<p<n$

use the similar argument as ${p=1}$ to prove the situation ${1<p<n}$ .
suffice to prove ${||u||_{\frac{np}{n-p}}\leq C||Du||_p}$ .
obvious we have: ${|u(x)|^p\leq \int_{-\infty}^{\infty}p|u|^{p-1}|Du|}$ .
${(\int_{\Omega}|u(x)|^{\frac{np}{n-p}})^{\frac{n-p}{np}}}$
${\leq (\int_{\Omega} \Pi_{i=1}(\int_{-\infty}^{\infty} p|u|^{p-1}|D_iu| )^{\frac{1}{n-p}})^{\frac{n-p}{np}} }$
${\leq C\int_{\Omega} \Pi_{i=1}^n(\int_{-\infty}^{\infty}p|u|^{p-1}|D_iu|)^{\frac{1}{np}}}$
${\leq\frac{c}{n}\sum_{i=1}^n\int_{\Omega}(\int_{-\infty}^{\infty}p|u|^{p-1}|D_iu|)^{\frac{1}{p}}}$
${\leq \frac{c}{n}\sum_{i=1}^n\tilde C p[(\int_{\Omega} (|u|^{p-1})^{\frac{p}{p-1}})^{\frac{p-1}{p}}+(\int_{\Omega} |D_iu|^p)^{\frac{1}{p}}]^{\frac{1}{p}} }$
${\leq C||Du||_p}$ . Q.E.D. $\Box$
4.

$\displaystyle W_0^{1,p}(\Omega)\longrightarrow C(\bar\Omega),n<p$

moreover,we have: ${\exists C=C(n,p)}$ , ${\forall u\in W^{1,p}_0(\Omega)}$ ,
$\displaystyle sup_{\Omega}|u| \leq C|\Omega|^{\frac{1}{n}-\frac{1}{p}}||Du||_p,p>n$

${\mu\in (0,1]}$ ,

$\displaystyle (V_{\mu}f)(x)=\int_{\Omega}|x-y|^{n(\mu-1)}f(y)dy$

then ${V_{\mu}: L^1(\Omega) \longrightarrow L^1(\Omega) }$ is well-defined by the following lemma:
Lemma 6 ${V_{\mu}:L^p \longrightarrow L^q}$ continously for any q, ${1\leq q \leq \infty}$ satisfy ${0\leq \delta=\delta(p,q)=\frac{1}{p}-\frac{1}{q} \leq \mu}$ .
furthermore,for any ${f\in L^p(\Omega)}$
$\displaystyle ||V_{\mu}f||_q \leq (\frac{1-\delta}{\mu -\delta})^{1-\delta}w_n^{1-\mu}|\Omega|^{\mu-\delta}||f||_p$

Proof: ${h(x-y)=|x-y|^n(\mu-1)}$ directly calculate follows that :

$\displaystyle ||h||_r \leq (\frac{1-\delta}{\mu -\delta})^{1-\delta} w_n^{1-\mu}|\Omega|^{\mu-\delta}$

now follows young inequality and this priori estimate we have:
${||V_{\mu}f||_q=(\int_{\Omega}(\int_{\Omega}|x-y|^{n(\mu-1)}f(y)dy)dx)^{\frac{1}{q}}}$
${\leq (\int_{\Omega}(\int_{\Omega}h^{\frac{r}{q}}h^{r(1-\frac{1}{p})}|f|^{\frac{p}{q}}|f|^{p\delta})^qdx)^{\frac{1}{q}}}$
${\leq (\int_{\Omega}(\int (h^r|f|^p)^{\frac{1}{q}}(\int h^r)^{1-\frac{1}{p}}(\int f^p)^{\delta})^{q})^{\frac{1}{q}}}$
${\Longrightarrow}$
$\displaystyle ||V_{\mu}f||_q \leq sup_{x \in \Omega} \{\int h^r(x-y)dy\}^{\frac{1}{r}}||f||_p$

and by the priori estimate,we have:
$\displaystyle ||V_{\mu}f||_q \leq (\frac{1-\delta}{\mu -\delta})^{1-\delta}w_n^{1-\mu}|\Omega|^{\mu-\delta}||f||_p$

Q.E.D. $\Box$
Lemma 7 ${f\in L^p(\Omega)}$ , ${g=V_{\mu}f}$ .
${\Longrightarrow}$ ${\exists c_1,c_2}$ constant depend only on ${n,p}$ ,such that
$\displaystyle \int_{\Omega} exp[\frac{g}{c_1||f||_p}]^{p^`}dx\leq c_2|\Omega|,p^`=\frac{p}{p-1}$

Proof: we have

$\displaystyle ||g||_q \leq q^{1-\frac{1}{p}+\frac{1}{q}}w_n^{1-\frac{1}{p}}|\Omega|^{\frac{1}{q}}||f||_p$

${\Longrightarrow}$
$\displaystyle \int_{\Omega} |g|^{p^`q}dx \leq p^`q(w_np^`q||f||_p^{p^`})^q|\Omega|$

${\Longrightarrow}$
$\displaystyle \int_{\Omega}\sum_{N_0}^{N}\frac{1}{k!}(\frac{|g|}{c_1||f||_p})^{p^`k}\leq p^`|\Omega|\sum(\frac{p^`w_n}{c_1^p})^k\frac{k^k}{(k-1)!}$

then take ${c_1,c_2}$ suffice large. Q.E.D. $\Box$
Lemma 8 let ${u\in W^{1,1}_0(\Omega)}$
$\displaystyle u(x)=\frac{1}{nw_n} \int_{\Omega} \frac{(x_i-y_i)D_iu(y)}{|x-y|^n}$

a.e. in ${\Omega}$ .
Proof: frist zero extended ${u}$ to whole space.and we have ${u(x)=\int_{-\infty}^xD_iu(x)}$ .

$\displaystyle u(x)=\int_0^{\infty}D_ru(x+rw)dr$

forall ${w\in \partial B_1(0)}$ ,so
$\displaystyle u(x)=-\frac{1}{nw_n}\int_0^{\infty}\int_{|w|=1}D_ru(x+rw)drdw=\frac{1}{nw_n}\int_{\Omega}\frac{(x_i-y_i)D_iu(y))}{|x-y|^ndy}$

Q.E.D. $\Box$
Theorem 9 let ${u\in W^{1,n}_0(\Omega)}$ ,then there exists constant ${c_1,c_2}$ such that
$\displaystyle \int_{\Omega}exp[\frac{|u|}{c_1||Du||_n}]^{\frac{n}{n-1}}dx\leq c_2|\Omega|$

Proof: a $\Box$

Theorem 10 ${u\in W_0^{1,p}(\Omega),p>n}$ ,then ${u\in C^{\gamma}(\Omega)}$ , ${\gamma=1-\frac{n}{p}}$ .
moreover ${\forall ball B=B_R}$
$\displaystyle osc_{\Omega \cap B_R}u \leq C R^{\gamma} ||Du||_p$

Proof: a $\Box$

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。

Schauder 理论和 Sobolev 理论是椭圆方程正则性的两套基本语言。前者追踪 Holder 范数，适合系数和右端足够连续的情形；后者追踪积分可积性，适合弱解和变分方法。

Schauder 估计追踪 Holder 正则性，Sobolev 不等式追踪积分正则性，两者共同构成椭圆方程的基本正则性工具。

1. 线性一致椭圆算子

考虑

$$Lu=a^{ij}(x)\partial_{ij}u+b^i(x)\partial_i u+c(x)u=f.$$

一致椭圆性是指存在 $\lambda>0$，使

$$a^{ij}(x)\xi_i\xi_j\ge \lambda|\xi|^2.$$

这是所有先验估计的起点。

2. Interior Schauder estimate

若系数和 $f$ 都在 $C^\alpha$ 中，则局部有

$$\|u\|_{C^{2,\alpha}(B_{1/2})}\le C\bigl(\|u\|_{C^0(B_1)}+\|f\|_{C^\alpha(B_1)}\bigr).$$

常系数情形可以先通过 Newton potential 或 Fourier 方法得到，再用冻结系数和 perturbation 推广到变系数情形。

3. Scaling 与 weighted norms

Schauder 估计的形状由 scaling 决定。若把球 $B_r$ 缩放到单位球，二阶导数带来 $r^{-2}$，Holder seminorm 还会多出 $r^{-\alpha}$。weighted norm 正是为了把这些尺度因子记录清楚。

4. Sobolev inequality

Sobolev 不等式给出

$$\|u\|_{L^{p^\ast}}\le C\|\nabla u\|_{L^p},\qquad p^\ast=\frac{np}{n-p}.$$

它不直接给出经典二阶 Holder 正则性，但能建立弱解存在性、能量估计和 bootstrapping。

5. 两种理论的关系

Schauder 理论适合光滑数据的 classical solution，Sobolev 理论适合弱解和变分框架。椭圆正则性常常先用 Sobolev 得到弱解，再通过 De Giorgi-Nash-Moser 或 Schauder 估计提升正则性。

2017年10月4日
锥奇点曲面上的几何流：短时间存在、Schauder 估计与阈值问题

旧博客原文

1. some example and observations

${(M^2,g)}$ , ${g(t)=e^{2u(t)}g_0}$ ,

$\displaystyle \frac{\partial u}{\partial t}=e^{-2u}\tilde\Delta u+\frac{r}{2}-e^{-2u}K_0$

${(M^n,g_{ij}(t))}$

$\displaystyle \frac{\partial g_{ij}(t)}{\partial t}=-2Ric(g_{ij})$

The given “smooth” initial :
${\exists }$ T small , ${T>0}$ ,the solution exists on ${[0,T]}$

equation is possible system.

Deturk Trick

“Threshold type theorem”
Ricci flow:
Mean curvature flow:
HMF:
Calabi flow:
pf of observation 4:if threshold condition hold for ${[0,T]}$ ,then we can bound ang ${C^k}$ norm of solution.

“geometry”

2. smooth manifold with conical singularities

on surface we can define conical singularity.

Definition 1 (conical singularity) ${M^2,p_i}$ , ${\beta_i}$ ,where ${\beta > -1}$ ,the angle of conical singularity ${p_i}$ is ${2\pi(1+\beta)}$ .
iff conical background metric ${g_0}$ : ${g_0}$ near p, ${g_0=r^{2\beta}(dr^2+r^2d\theta^2)}$ , ${(r,\theta)}$ is the interpolation coordinate chart.

it is easy to chake the form ${g_0=r^{2\beta}(dr^2+r^2d\theta^2)}$ is independent with the coordinate chart ,so the definition is well defined.

3. rough line of proof

initial ${u_0}$ ,

$\displaystyle \frac{\partial u}{\partial t}=e^{-2u}\tilde\Delta u+\frac{r}{2}-e^{-2u}K_0$

Step 1:(Short time existence)
state and proof the “magic theorem”:
1.we need to explain what is smooth,to define a Banach space ${A}$ ,maybe ${W^{k,p},C^{k,p}}$ type.
2.to proof the “magic theorem” under the setting.maybe use shauder fix point theorem or contraction map theorem or else.
3.so the problem reduce to get this type estimate,
if ${\frac{\partial u_{i+1}}{\partial t}=e^{-2u_i}\tilde\Delta u_{i+1}+\frac{r}{2}-e^{-2u_i}K_0}$ .
define ${T_{[0,T]}:A(\Omega \times [0,T]) \longrightarrow A(\Omega \times [0,T])}$ . ${T_{[0,T]}}$ is continuous, ${Dom(T_{[0,T]})}$ is convex, ${Im(T_{[0,T]})}$ is a pre-compact set.for ${T_{[0,T]}}$ suffice small
the difficult is to set up the continuous of operator ${T_{[0,T]}}$
Schauder estimate tell us:
$\displaystyle ||u_{i+1}||_{C^{2,\alpha}} \leq ||u_{i+1}||_{L^{\infty}}+||\frac{r}{2}-e^{-2u_i}K_0||_{C^{\alpha}}$

this give us some useful information to construct space ${A}$ .
Step 2:(Threshold type theorem,long time existence)
Threshold as long as ${||u||_{L^{\infty}}}$ is bounded.
This type theorem is relate to the maximal internal in which solution existence is closed.
Basically is based on Alzalo-Ascoli theorem.
Step3:(More regularity) 1.the question does not existence for smooth manifold.(why)
2.singular space.
${u_0\in }$ small space ${\Longrightarrow}$ ${u(t)\in}$ small space.
what is the optimal regularity?
the problem naturally come from both “pure PDE” and “application for geometry problem”.

conical Kachler Ricci flow[Chen.Wang]
Donaldson setting ${C^{2,\alpha,\beta}}$

4. More seriously treat with the problem

in 07 years,consider the problem
$\displaystyle \frac{\partial u}{\partial t}=\Delta u$

on ${M-\{p\}}$
$\displaystyle u|_{t=0}=f$

where f is a function with nice regularity.
Functional analysis:
$\displaystyle \Delta: C_c^{\infty}(M-\{p\}) \longrightarrow C_c^{\infty}(M-\{p\})$

extension to:
$\displaystyle \Delta: L^{2}(M-\{p\}) \longrightarrow L^2(M-\{p\})$

which is a self-adjoint extension.and then use the theory of operator semi-group.the problem can be solved.
remark:the extension is not unique so the information we know for the solution is very little.and because the really true extension which is suit for our geometry setting is just one extension.so the treat of Functional analysis is not enough for us.
Elementary treat:
consider the simplest case,smooth manifold with only one singularity.

we set ${M_i}$ is the manifold cut off form ${M}$ with a boundary more and more near the singularity. consider the equation on each ${M_i}$ ,i.e.:
$\displaystyle \frac{\partial u}{\partial t}=\Delta u$

on ${M_i}$
$\displaystyle u|_{t=0}=f$

with boundary condition:
Drichlet condition
$\displaystyle u|_{\partial M_i=0}$

or Neumann condition
$\displaystyle \frac{\partial u}{\partial v}=0$

we choose Neumann condition there and at last we will see the solution come from Dirichlet condition is the same with the solution come from Neumann condition.
Under the general setting this become:

$\displaystyle \frac{\partial u_k}{\partial t}=a(k,t)\Delta u_k+b(k,t)\partial^i u_k +c(x,t)$

$\displaystyle \frac{\partial u_k}{\partial v}|_{\partial M_k}=0$

when ${k \longrightarrow \infty }$ , do we have ${u_k \longrightarrow u}$ ?
we need priori estimate: Schauder estimate for serious parabolic equation tell us:
for equation ${\frac{\partial u}{\partial t}=\Delta u}$ on ${M}$ with priori estimate ${||u||_{L^{\infty}}\leq C}$ , we have:

$\displaystyle |\nabla^k u(p)|\leq \frac{C}{r^k}$

wher ${r}$ is the maximal such that geodesic ball ${B(r,p) \subset\subset M_i}$ .
For general setting :

$\displaystyle \frac{\partial u_k}{\partial t}=\Delta u_k+f$

$\displaystyle u_k|_{t=0}=u_0$

$\displaystyle \frac{u_k}{\partial t}|_{\partial M_k}=0$

we know

$\displaystyle ||u(t)||_{C^0(M_k)}\leq ||u_0||_{C^0(M_k)}+t||f||_{C^0(M_k)}$

this is what Schauder estimate tell us.
1.the uniform estimate with k:
${||u_k||_{***}\leq C}$ independent of k.(now we do not know what the norm ${||\cdot||_{***}}$ need to be)
we have ${C^0}$ estimate and the energy estimate as follows:
from maximal principle,easy to get ${C^0}$ norm estimate.

the point is the equation ${\frac{\partial }{\partial t}u_k= \Delta u_k +f}$ is strict parabolic so we have strong maximal principle and to construct suit bump function we can estimate ${C^0}$ norm of ${u_k}$ .

from energy method we can estimate ${\int_{M_k} ||\nabla u_k||^2}$ .

the point is:
${\frac{\partial}{\partial t}\int_{M_k} |\nabla u_k|^2=2\int_{M_k} \nabla u_k \cdot \frac{\partial}{\partial t}(\nabla u_k) }$
${=2\int_{M_k} \Delta u_k \cdot \frac{\partial}{\partial t} u_k }$
${=-2\int_{M_k}(\frac{\partial}{\partial t} u_k -f)\cdot \frac{\partial}{\partial t}u_k}$
${=-2[\int_{M_k}|\frac{\partial}{\partial t}u_k|^2-\int_{M_k} f\cdot \frac{\partial}{\partial t}u_k]}$
${=-2\int_{M_k}|\frac{\partial}{\partial t}u_k|^2-\int_{M_k}f \cdot (\Delta u_k +f)}$
${\leq 2\int_{M_k} \nabla u_k\cdot \nabla f}$
${\leq\int_{M_k} |\nabla f|^2 +\int_{M_k} |\nabla u_k|^2}$ .
so we get:

$\displaystyle \frac{\partial}{\partial t}\int_{M_k}|u_k|^2\leq \int_{M_k}|\nabla f|^2+\int_{M_k} |\nabla u_k|^2$

so we can bounded ${\int_{M_k}|u_k|^2}$ .
for the general case:the equation becomes:
$\displaystyle \frac{\partial u}{\partial t}=a(x,t)\Delta u+b(x,t) \partial^i u+c(x,t)$

$\displaystyle \frac{\partial u}{\partial v}|_{\partial M_k}=0$

$\displaystyle u(0)=u_0$

but there is a hide Dragon,we need the condition ${\frac{\partial u(0)}{\partial v}|_{M_k}=0}$ .
otherwise we will get solution ${u \notin W^{1,2}(M_k)\cap C^{2}(M_k)}$ .
but in this case we still have the two necessary estimate(esay to see the above argument still make sense).
in this case to prove the short time existence we need follow four claims is ture.
$\displaystyle ||u_{i+1,k}(t)||_C^0\longrightarrow ||u||_{i,k}{C^0}$

as ${t \longrightarrow 0}$
$\displaystyle \int_{M_k}|u_{i+1,k}|^2 \longrightarrow \int_{M_{k}}|u_{i,k}|^2$

as ${t \longrightarrow 0}$
$\displaystyle ||u_{i,k+1}(t)||_C^0\longrightarrow ||u||_{i,k}{C^0}$

as ${t \longrightarrow 0}$
$\displaystyle \int_{M_{k+1}}|u_{i,k+1}|^2 \longrightarrow \int_{M_{k}}|u_{i,k}|^2$

as ${t \longrightarrow 0}$
5. Construct the suitable Banach space

call the space construct follow the Mixed-Holder-Sobolev space for simply case,consider smooth manifold with only one conical singularity.
first cover the whole manifold by a open set have positive distance t=with the conical singularity and a countable group of set ${A_n=B(\frac{d}{2^n},p)-B(\frac{d}{2^{n+1}},p)}$ ,which is balls center at singularity ${p}$ with radius ${\frac{d}{2^n}}$ .(where ${M=B(d,p)\cup U}$ )
i.e. ${M-\{p\}=U \cup (\cup_{i=1}^{\infty}A_i)}$

Definition 2 ( ${||\cdot||_{\varepsilon^{k,\alpha}(S)}}$ )
$\displaystyle ||f||_{\varepsilon^{k,\alpha}(S)}=sup_{k=1,2,...,\infty}||f(2^{-k},\theta)||_{C^{k,\alpha}(B_1-B_{\frac{1}{2}})}+||f||_{C^{k,\alpha}(U)}$

easy to see the definition is independent with the cover and the local interpolation coordinate chart.
one thing is also trivial,is that we have the schauder estimate under the norm ${||\cdot||_{\varepsilon^{k,\alpha}(S)}}$ .
that is
$\displaystyle \delta u=f$

on ${S-\{p\}}$ . ${|u|<C_1}$ on ${S}$ . then
$\displaystyle ||u||_{\varepsilon^{k+2,\alpha}}\leq C(||u||_{L^{\infty}}+||f||_{\varepsilon^{k,\alpha}})\leq C(C_1+||f||_{\varepsilon^{k,\alpha}})$

in fact we only need to add each inequality come from each open set of the cover by Schauder estimate to proof this.
on the other hand we need a suitable Sobolev type norm.
Definition 3 ( ${|\cdot|_w}$ )
$\displaystyle |u|_w=(\int_S |\tilde \nabla u|^2d \tilde V)^{\frac{1}{2}}$

Definition 4 ( ${W^{k,\alpha}}$ ) the set of all f in ${\varepsilon^{k,\alpha}}$ with finite ${|f|_w}$ ,
$\displaystyle ||f||_W^{k,\alpha}=||f||_{\varepsilon^{k,\alpha}}+|f|_w$

in Banach space.
Assume norm ${C^{k,\alpha}(B\times [o,T])}$ on ${B\times [0,T]}$

Definition 5 ( ${||\cdot||_{\rho^{l,\alpha,{0,T}}}}$ ]
${f: S\times [0,T] \longrightarrow R }$

$\displaystyle ||f||_{\rho^{l,\alpha,[0,T]}}=sup_{k=0,1,2,...,\infty}||f(2^{-k}\rho,\theta,4^{-k}t)||_{C^{l,\alpha}((B_1-B_{\frac{1}{2}})\times [0,4^{-k}T])}+||f||_{C^{l,\alpha}(U\times[0,T])}$

\end) from the definition,easy to see
$\displaystyle \frac{\partial u}{\partial t}=\Delta u+f$

om ${M}$
$\displaystyle u|_{t=0}=u_0$

${\Longrightarrow}$
$\displaystyle ||u||_{\rho^{l+2,\alpha,[0,T]}}\leq C(||u_0||_{\varepsilon^{l,\alpha}}+||f||_{\rho^{l,\alpha,[0,T]}}+||u||_{C^0(S\times [0,t])})$

easy from the classical schauder estimate.

Definition 6 ( ${|f|_v}$ ) ${f:S\times [0,T] \longrightarrow R}$
$\displaystyle |f|^2_v=max_{t\in [0,T]}\int_S|\tilde \nabla f|^2d\tilde V +\int_0^T\int_M |\frac{\partial f}{\partial t}|^2 d\tilde Vd t$

Key point:

Definition 7 ( ${\nu^{k,\alpha,{0,T}}}$ ] ${\nu^{k,\alpha,[0,T]}}$ is the set of ${f}$ in ${\rho^{l,\alpha,[0,T]}}$ with finite ${|f|_v}$
$\displaystyle ||\cdot||_{\nu^{k,\alpha,[0,T]}}=||\cdot||_{\rho^{l,\alpha,[0,T]}}+|\cdot|_v$

\end)
6. What is a solution of equation

trivial sense:
satisfied equation point-wise on ${S-\{p\}}$ .
weak sense:
1.trivial case
2. ${|u|_v<+\infty}$

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。

几何流在光滑流形上已经有成熟理论，但一旦初始空间带有锥奇点，短时间存在和正则性都要重新组织。关键问题是：什么叫“光滑”，在哪个 Banach 空间中解方程，以及怎样在奇点附近建立 Schauder 型估计。

锥奇点附近的几何流需要在锥型 Holder 空间中建立短时间存在和 Schauder 估计。

1. 锥奇点的局部模型

在曲面上，锥角为 $2\pi\beta$ 的锥奇点可以局部写成

$$g_\beta=dr^2+\beta^2r^2d\theta^2.$$

若 $\beta=1$，这就是普通光滑点；若 $\beta\ne1$，度量在顶点处有角缺陷或角盈余。

2. 几何流的困难

Ricci flow、mean curvature flow、Calabi flow 等都可以写成抛物型方程。但在锥点附近，普通 Holder 空间不适合，因为坐标缩放和角变量的正则性发生改变。需要使用带权或锥型 Holder 空间。

3. 短时间存在

典型证明路线是构造映射

$$T:A\to A$$

其中 $A$ 是某个凸闭的函数空间球。若能证明 $T$ 连续、$T(A)$ 预紧并且 $T(A)\subset A$，就可以用 Schauder fixed point theorem 得到短时间解。

4. Schauder 估计

核心估计形如

$$\|u\|_{C^{2+\alpha,1+\alpha/2}_\beta}\le C\bigl(\|Lu\|_{C^\alpha_\beta}+\|u\|_{C^0}\bigr).$$

这里下标 $\beta$ 表示锥型空间。没有这类估计，固定点映射就无法闭合。

5. 阈值型问题

长期存在常常由某个阈值控制：只要解的 $L^\infty$ 或几何量保持有界，就可以继续延拓。锥奇点情形中，真正困难是证明这些控制不会在奇点附近丢失。

2017年10月3日
Atiyah-Singer 指标定理（二）：McKean-Singer 公式与局部指标

旧博客原文

原题：Atiyah-Singer index theorem 2

1. rough outline of heat kernel proof of Atiyah-singer index theorem

1.1. proof strategy

Theorem 1 (Mckean-Singer formula.)
$\displaystyle ind(D^+)=Str(e^{-tD^2})=\int\limits_{x \in M} Str(K(x,y)).$

from this we know Fredholm operator deformation invariance,in the same time we need chern-weil theory.
Main challenge:
1. in the expansion on heat kernel ,we need to proof when ${ t \rightarrow 0}$ , the limit exist and find a way to calculate it.
2. indentify the limit as ${t \rightarrow 0}$ .
Proof: Our proof road锛� mckean-singer formula ${\rightarrow}$ local-index thm ${\rightarrow}$ A-S index thm ${\rightarrow}$ Riemann-roch-Hirzebunch theorem. $\Box$

1.2. preliminary work

superbundle: ${E=E^+ \oplus E^-}$ .
on compact manifold ${M}$ , ${D: \Gamma(M,E) \rightarrow \Gamma(M,E)}$ is a self-adjoint operator . ${ D = }$ . ${D^+=D|_E^+,D^-=D|_E^-}$ .
observe that:
${D}$ is symmetric ${\Longrightarrow}$ eigenvalue space of ${D^2}$ is finite dimention ${\Longrightarrow}$ in particular ${Ker D^2}$ is finite dimesion ${\Longrightarrow}$ ${Ker D}$ is finite dimension.
dimention of superspace ${E = E^+ \oplus E^-}$ :

$\displaystyle dim E=dim E^+ - dim E^-.$

$\displaystyle kerD=kerD^+ \oplus ker D^-$

Def: ${ind D^+=dim ker D^+ -dim ker D^-}$ .
Lemma:
1.Let ${D}$ be a self-adjoint Dirac operator on a clifford module ${E}$ over a compact manifold ${M}$ ,then

$\displaystyle \Gamma(M,E^{\pm})=ker D^{\pm} \oplus im D^{\mp}$

in particular,
$\displaystyle ind D^+ = dim ker D^+ -dim coker D^+$

where coker ${D^+ :=\Gamma(M,E^-)/im D^+}$ .
2.Let ${D}$ be a differential operator acting on a ${Z_2}$ -graded vector bundle ${E}$ ,then ${Str[D,K]=0}$ .
the proof of this two lemma is easy,leave sas exercise.
1.3. Mckean-Singer formula

the formula is:

$\displaystyle ind(D^+)=Str(e^{-tD^2})=\int\limits_{x \in M} Str(K(x,y)).$

the expression of heat operator by spectral measure is:
$\displaystyle e^{-tD^2}=\int\limits_{0}^\infty d^{\lambda t}dE_{\lambda}.$

proof 1:
we have first eigenvalue estimate on compact manifold:
$\displaystyle |Str(e^{-tD^2}-P_0)| \leq Cvol(M)e^{-t\lambda}.$

$\displaystyle \Longrightarrow$

$\displaystyle \lim\limits_{t \rightarrow \infty}Str(e^{-tD^2}) = Str p_0 =dim kerD^+ -dim ker D^- =ind D^+.$

on the other hand ,we need to show ${Str e^{-tD^2}}$ is independent with ${t}$ ,in fact:
$\displaystyle \frac{d}{dt} Str (e^{-tD^2})=-Str(D^2 e^{-tD^2}).$

${D}$ odd parity : ${\Longrightarrow \ D^2e^{-tD^2}=[D,D E^{-tD^2}]}$ .
${[\ \ ,\ \ ]}$ supercommunater ${\Longrightarrow \ \frac{d}{dt}Str (e^{-tD^2})= -Str[D,D e^{-tD^2}]=0.}$
q.e.d
proof2:
by spectral decompositon of ${e^{-tD^2}}$ :
$\displaystyle Str (e^{-tD^2})=\sum\limits_{\lambda \geq 0}(n_{\lambda}^+ - n_{\lambda}^-)e^{-t\lambda}$

observe that: ${n_{\lambda}^+=n_{\lambda}^-}$ for ${\lambda \not=0}$ . ${\Longrightarrow ind D=n_0^+ - n_0^-}$ . (detail in [BGV])
q.e.d
Corallary: the index of a smooth on-parameter family of Dirac operator is constant.
what we have proved is:
$\displaystyle ind D = Str (e^{-tD^2})=\int Str(k(x,y)).$

1.4. analytic formula of ind ${D^+}$

from the discuss of heat kernel in section 2,we know following result(section 2 only discuss the case of function but use the similar way we can get similar result on bundle):

$\displaystyle K_t(x,y) \sim (4 \pi t)^{\frac{n}{2}} \sum\limits_{i=0}^{+\infty} t^i K_i(x), K_i \in \Gamma(M,End(E)).$

on the other hand: ${ D=\left[\begin{array}{ccc} 0 & D^- \\ D^+ & 0 \end{array}\right] \Longrightarrow D^2=\left[\begin{array}{ccc} D^-D^+ & 0\\ 0 & D^+D^- \end{array}\right] }$ .
and use the Mckean-Singer formula,we get: ind (D^+)&=&Str(e^{-tD^2})
&=&Tr(e^{-tD^+D^-})-Tr(e^{-tD^=D^-})
&=&\int\limits_M Tr(K_t(x,y,D^-D^+)) – \int\limits_M Tr(k_t(x,y,D^+D^-))
&=&\sum\limits_{i=0}^{\infty} t^{i-\frac{n}{2}}a_i(D^+D^-) – \sum\limits_{i=0}^{\infty} t^{i – \frac{n}{2}}a_i(D^+D^-). where ${a_i}$ is the heat trace invariants.
take ${ t \rightarrow 0}$ ,the only thing make sense is the series of order ${\frac{n}{2}}$ ,and we want to proof:
$\displaystyle ind(D^+) =a_{\frac{n}{2}}(D^-D^+) -a_{\frac{n}{2}}(D^+D^-).$

But the difficult thing is that the high order series is very hard ro calculate….
our strategy is following:

Step1: proof ${Str(K_t(x,y))}$ has a limit as ${t \rightarrow 0}$ i.e ${Str(K_t(x,y))\stackrel{t \rightarrow 0}{\longrightarrow}}$ index density.
step2:use a rescaling of space,time,clifford bundles ,to find a way that make us only need to calculate the leader coefficient.

1.5. From the McKean鈥揝inger formula to the index theorem

Let ${M}$ be a compact oriented Riemannian manifold of even dimension ${n}$ . We will write ${k_t(x, y)}$ for the heat kernel associated to ${D^2}$ . The diagonal ${k_t(x, x)}$ is a section of ${End(E )}$ which is iso- morphic to ${Cl(M) \otimes End_{Cl(M)}(E )}$ . Using this isomorphism, we define a filtration on ${End(E )}$ , induced by the filtration on ${Cl(M)}$ . Elements of ${End_{Cl(M)}(E )}$ are given 0-degree. Denote by ${Cl_i(M)}$ the subbundle of ${Cl(M)}$ consisting of all elements of degree less or equal to ${i}$ .
the following theorem hold:
Theorem 1. The following statements hold:
1. The coefficients ${k_i}$ have degree less or equal to ${2i}$ . In other words, ${k_i \in \Gamma (M, Cl_{2i}(M) \otimes End_{Cl(M)}(E))}$ . 2. If ${ \sigma(k):= \sum \limits_{i=0}^{n/2}\sigma_{2i}(k_i) \in A(M,End_{Cl(M)}(E))}$ ,where ${\sigma_j :Cl_j(M)鈫扐_j(M)}$ denotes the ${i=0}$ restriction of the symbol map, then:

$\displaystyle \sigma (k) = A藛(M) exp(鈭扚E /S ).$

$\displaystyle ind (D^+)=Str(e^{-tD^2})=\sum\limits_{i=0}^{\infty} t^{i-\frac{n}{2}}a_i(D^+D^-) - \sum\limits_{i=0}^{\infty} t^{i - \frac{n}{2}}a_i(D^+D^-).$

the important observation is that the ${Str}$ of order less than n all vanish,in the other word:

lemma2:for any quadratic space ${v}$ of dimension ${n}$ , ${CL_{n-1}(V) =[CL(V),CL(V)]}$ .

Proof: Let ${e_1,...,e_n}$ be a basis of ${V}$ . For any multi-index ${I \subset \{1,...,n\}}$ , denote by ${c_I}$ the Clifford product ${\prod_{i \in I} c}$ .Then the set ${\{c_I\}}$ is a basis for ${Cl(V)}$ .If ${|I|<n}$ ,there is at least one ${j}$ such that ${j \notin I}$ , and we have
$\displaystyle c(e_I)=-\frac{1}{2}[c_j,c_j c_I].$

q.e.d
so take ${t \rightarrow 0}$ , we get:
$\displaystyle ind (D) =(4 \pi)^{\frac{n}{2}} \sum\limits_{i = \frac {n}{2}}^{\infty} t^{i- \frac{n}{2}}Str(k_i(x)).$

now we want to identify the term ${Str(k_{\frac{n}{2}}(x))}$ as a characteristic form on ${M}$ :
Lemma 3. Let ${V}$ be a Euclidean space. There is, up to a constant factor, a unique supertrace on ${Cl(V)}$ , equal to ${T \circ \sigma}$ where ${\sigma}$ denotes the symbol map and ${T}$ is the projection of ${\alpha \in \wedge V}$ onto the coefficient of ${e_1 \wedge, ... ,\wedge e_n}$ if ${e_1,...,e_n}$ form an oriented orthonormal basis of ${V}$ . Furthermore, the supertrace defined above equals:
$\displaystyle Str(a) = (鈭�2i)^{\frac{n}{2}} (T \circ \sigma(a)).$

rmk:(The map ${T}$ is also called the canonical Berezin integral.)
proof:
the dimension of ${Str}$ space is one because of ${CL_{n-1}(V)=[CL(V),CL(V)]}$ and it never be empty because there is a natural defined supertrace on ${CL(V)}$ .so the only thing we need to do is to determine the constant,in face we only need to calculate the supertarce on any non-zero element .for instance the chirality operator ${\Gamma}$ . We have that ${ Str(\Gamma) = dim (\wedge P) = 2^{n/2}}$ and therefore ${Str(a) = (鈭�2i)^{n/2} T \circ \sigma (a)}$ for all ${a \in Cl(M)}$ .
q.e.d
so we know:
${\forall \ }$ section ${a \otimes b \in \Gamma(M \otimes CL(M) \otimes End_{CL(M)}(E)}$ :
$\displaystyle Str_E( a \otimes b)(x) = (-2i)^{\frac{n}{2}} \sigma_n(a(x))Str _{E/S}(b(x)).$

$\displaystyle Str_E(k_{\frac{n}{2}})(x) = (-2i)^{\frac{n}{2}} Str _{E/S}(\sigma_n(k_{\frac{n}{2}})).$

theorem1 implies then the following theorem for the index of a Dirac operator associated to a Clifford connection which is known as the local index theorem.

Theorem 2. (Local index theorem) Let ${M}$ be a compact, oriented even-dimensional manifold and let ${E}$ be a Clifford module with Clifford connection ${\nabla E}$ . Let ${D}$ be the associated Dirac operator. Then ${\lim\limits_{t \rightarrow 0} Str(k_t(x, x))|dx| }$ exists and is obtained by taking the ${n}$ -th form piece of
$\displaystyle (2\pi i)^{鈭抧/2} \widehat A(M)ch(E /S ).$

rmk:This theorem only holds for Dirac operators associated to Clifford connections, which are those which are compatible with the Clifford action. However, since the index of a Dirac operator is independent of the Clifford superconnection used to define it, we get Atiyah鈥揝inger index formula for any Dirac operator.

Theorem 3. (Atiyah鈥揝inger Index Theorem) Let ${M}$ be a compact, oriented, even-dimensional manifold and let ${D}$ be a Dirac operator on a Clifford module ${E}$ . Then the index of ${D}$ is given by :
$\displaystyle ind D = (2 \pi i)^{-\frac{n}{2}}\int\limits_{M} \widehat A(M) ch(E/S) .$

hence theorem 3 is a consequence of theorem 1.
up to now,to prove index theorem ,we only need to prove theorem 1.
1.6. idea of the proof of theorem 1

we give the clear proof in appendix锛宼here we explain the idea:
To prove Theorem 4.11 we mainly follow Chapter 4 of [BGV] but rearrange the different steps in order to make the proof clearer, at least for us. Let us summarize the first part of the proof:

1. The idea of the proof is to work in normal coordinates ${x}$ around a point ${x_0 \in M}$ . Near the diagonal, we use parallel transport to pull back the heat kernel ${k_t(x, x_0)}$ which is a section of the vector bundle ${E_x \otimes E_x^*}$ and define a new kernel ${k(t, x) := \iota(x_0, x)k_t(x, x_0)}$ , a section of ${End(E_{x_0} ) \cong Cl(V^*) \otimes End(W)}$ for some twisting space ${W}$ . Using the symbol isomorphism ${蟽}$ , we can look at ${k(t, x)}$ as a section of ${\wedge(V^*) \otimes E (W)}$ .

2. We use Lichnerowicz鈥� formula to get the explicit form of the operator ${L}$ such that the kernel ${k(t, x)}$ satisfies the heat equation ${(\partial t + L)k(t, x) = 0}$ .

3. In a third step, we define a rescaling of space, time and the Clifford algebra, introduced by Getzler. This rescaling has the effect that the leading coefficient of the asymptotic expansion of the rescaled kernel is exactly the differential form ${\sigma(k)}$ of theorem 1 which leads to a reformulation of Theorem 1.

\appendix

2. Chern-Weil theory

some text in Appendix A

3. Complete proof of theorem 1

some text in Appendix B

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。

热核证明指标定理的主线可以压缩成一句话：McKean-Singer 公式把 index 写成 heat operator 的 supertrace；再让 $t\to0$，用热核局部展开读出曲率多项式。

McKean-Singer 公式让 index 等于 heat operator 的 supertrace，小时间极限给出局部指标密度。

1. McKean-Singer 公式

设 $D$ 是奇 Dirac operator，$D^2$ 是偶的 Laplace 型算子。McKean-Singer 公式为

$$\operatorname{ind}D=\operatorname{Str}(e^{-tD^2})\qquad(t>0).$$

当 $t\to\infty$，非零谱贡献消失，只留下 kernel；所以右端等于 index。

2. 为什么它与 $t$ 无关

对 $t$ 求导：

$$\frac{d}{dt}\operatorname{Str}(e^{-tD^2})=-\operatorname{Str}(D^2e^{-tD^2}).$$

由于 $D$ 是奇算子，$D^2e^{-tD^2}$ 可以写成 supercommutator 的 supertrace，而 supertrace 在 supercommutator 上为零。因此这个量不随 $t$ 变化。

3. 小时间极限

热核在对角线附近有渐近展开

$$K_t(x,x)\sim (4\pi t)^{-n/2}\sum_{j\ge0}t^j a_j(x).$$

普通 trace 会看到许多局部项；但 supertrace 中大量低阶项因 Clifford 代数对称性抵消，最后留下最高阶曲率组合。

4. 局部指标定理

局部指标定理说

$$\lim_{t\to0}\operatorname{str}K_t(x,x)\,dx
=\widehat A(M)\operatorname{ch}(E).$$

积分后得到

$$\operatorname{ind}D=\int_M \widehat A(M)\operatorname{ch}(E).$$

这里左边是分析量，右边是 Chern-Weil 代表的拓扑量。

5. 证明路线

完整证明通常沿着四步走：先建立 Fredholm 性和谱离散性；再证明 McKean-Singer；然后计算 heat kernel asymptotics；最后识别局部密度为 Chern-Weil 形式。Riemann-Roch-Hirzebruch 定理可以看作这套机器在复几何中的一个重要特例。

2017年10月2日
Interval map：周期三、Markov 分割与拓扑熵

旧博客原文

原题：Interval map

1.period 3 induce chaos

theorem:if a interval map $T:I\to I$ have a period 3 point $x$ ,then $\forall n\in N^*$ ,there is a period n point for $T$ .

proof:

n=1 case. trivial

$n>1,n\neq 3$ case:

the key point is to consider the structure of monotone interval contain previous one with fix length.

this will easy to lead a proof.

2.a work of J.Milnor and W.Thurston.

$N(T^n)$ defined as the number of monotone interval of the map $T^n$ .

theorem: $h(T)=lim_{n\to infty}\frac{1}{n}log N(T^n)$ .

3.monotone Markov map

this structure have two property:

1.piesewise monotone and $C^1$,the derive has control!

there is a relative dynamic system with this map.is a shift map with a relative $n\times n$ matrix $A$ .

this two dynamic system have a lot of relation,the key one is:

the topological entropy of monotone Markov map is just the unique maximum eigenvalue of $A$ .

and some byproduct…

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。

区间映射是一维动力系统中最能体现“简单空间产生复杂轨道”的模型。一个连续映射 $f:I\to I$ 若有周期三点，就已经含有所有周期的影子，这就是 Li-Yorke 定理背后的核心图像。

区间映射的复杂性可以通过分段单调结构和 Markov transition matrix 转化为组合增长。

1. 周期三为什么特别

设存在 $x$ 使得 $x,f(x),f^2(x)$ 互不相同且 $f^3(x)=x$。三个点在区间上的顺序会迫使某些子区间被映到包含另一些子区间的位置。利用介值定理，可以构造嵌套区间，从而得到任意周期点。

2. 分段单调与 lap number

Milnor-Thurston 的一个基本量是 $f^n$ 的单调区间个数，常记作 $\ell(f^n)$。拓扑熵可以由增长率给出：

$$h_{\rm top}(f)=\lim_{n\to\infty}\frac1n\log \ell(f^n).$$

这把轨道复杂度转化成组合增长。

3. Markov 映射

若区间被分成有限个小区间，且每个小区间在 $f$ 下覆盖若干小区间，就得到一个 transition matrix $A$。对应的符号动力系统是一个 subshift of finite type。

4. 熵与最大特征值

在 Markov 情形下，长度为 $n$ 的允许轨道数由 $A^n$ 控制，因此

$$h_{\rm top}(f)=\log \rho(A),$$

其中 $\rho(A)$ 是矩阵最大特征值。这样一维动力系统的混沌程度就被线性代数读出来了。

2017年9月27日
Flat surface 1：从拓扑覆盖到 translation structure

旧博客原文

原题：Flat surface 1

Topological point of view:

in topological point of view a flat surface is a topological space $M$ with a (ramified in nontrivial case) map $\pi:M\longrightarrow T^2$ .

and the map satisfied: $\pi$ is not ramified on $\pi^{-1}(T^2-\{0\})$ is not ramified and defined a covering map.

Geometric-analytic point of view:

we begin with compact connected oriented surface $M$ ,and a nonempty finite subset $\Sigma=\{A_1,...,A_n\}$ of M.and

translation surface of type $k$ :

translation structure:

complex structure:

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。

flat surface 可以从两个角度理解。拓扑上，它像一个带有限分歧点的覆盖空间；几何分析上，它是一个曲面，除有限个锥点外局部都与欧氏平面等距。

Flat surface 可由多边形边粘合得到，除有限锥点外局部坐标变换都是平移。

1. 拓扑观点

给定紧致定向曲面 $M$ 和有限集合 $\Sigma$。在 $M\setminus\Sigma$ 上，局部坐标变换若都是平移

$$z\mapsto z+c,$$

就得到 translation structure。若绕 $\Sigma$ 中的点一圈，角度可能是 $2\pi(k+1)$，这些点就是锥奇点。

2. 多边形模型

最直观的构造是取平面中的多边形，把平行且等长的边按平移粘起来。粘合后，内部仍然平坦，边也消失，只有若干顶点合并成锥点。

3. 复分析观点

translation surface 等价于 Riemann surface 上的 holomorphic 1-form $\omega$。在 $\omega$ 非零处，局部坐标由

$$z(p)=\int^p\omega$$

给出；在 $\omega$ 的零点处出现锥角。

4. 动力系统入口

在 flat surface 上沿固定方向走直线，会得到 translation flow。它连接了 billiards、Teichmuller flow、interval exchange transformation 和模空间动力系统，是后续研究的基本对象。

2017年9月26日
热核证明指标定理（一）：Dirac 算子、supertrace 与 Chern-Weil

旧博客原文

原题：Heat Kernel proof of Index Theory 1

1. framework of atiyah singer index theory

1.1. A genus form

${(M,g)}$ campact,complete,Riemann manifold without boundary,dim ${ M=2m ,m \in N^* }$ . ${\bigtriangledown^g}$ is the Levi-civita connection on ${TM}$ , ${R=R_g\in \Omega^2(End(TM))}$ .
${\widehat A}$ genus form:

$\displaystyle \widehat A(M,g)=det^{\frac{1}{2}}(\frac{\frac{i}{4\pi}R_g}{sinh(\frac{i}{4\pi}R_g)})\in \Omega(M).$

by chern-weil theory,we know:
${ 1. \widehat A }$ is closed.
${ 2. \widehat A_{g1}-\widehat A_{g2} }$ is exact.
so we can def ${\widehat A_g =\widehat A(M) \ \forall g}$ is a metric on M.
1.2. dirac bundle

${(E,\bigtriangledown^E)}$ is a dirac bundle.
${D: C^\infty(E^+) \rightarrow C^\infty(E^-)}$ is dirac operator.
${F^{E/S}\in End_{Cl(M)}(E)}$ the twisting curvature of ${E}$ .

1.3. supertrace

${Str^{E / S}: End_{cl(M)} \rightarrow C_M}$ . induce map:

$\displaystyle Str^{E/S} : \Omega(End_{cl(M)} (E) )\rightarrow \Omega(M) \otimes C$

determined by:
$\displaystyle Str^{E / S}( w \otimes T) = w \otimes Str^{E / S}(T). \ \forall w \in \Omega (M). \ \forall T \in End_{cl(M)} E.$

1.4. chern class

$\displaystyle ch^{E / S}(E) = Str^{E /S}[exp(\frac{i}{2\pi}F^{E/S})] \in \Omega(M).$

by chern-weil theory,we know:
1. ${ch^{E/S}(E)}$ closed form.
2. ${ch^{E \ S}(E)}$ only depend on the topology of E.
$\displaystyle F^{E \widehat\otimes W/S =F^{E/S} }\otimes 1_W + 1_E \otimes F^W.$

$\displaystyle F^W = F^{W^+} \oplus F^{W^-}.$

$\displaystyle ch^{(\widehat E \otimes W)/S} ,ch(E \widehat\otimes W)=ch^{E/S}(E/S)(ch(F^{W^+})-ch(F^{W^-})).$

Whitney product formula.
1.5. Atiyah-singer index theorem

$\displaystyle Ind D_E=Dim(ker D_E)-Dim(ker D^*_E)=\int\limits_M \widehat A(M,g)ch^{E/S}(E/S).$

2. heat kernel

2.1. basic setting

Assume M is a general Riemann manifold ,assume

$\displaystyle \Phi(d)= \{f\in C^\infty(M) | \|f \| _1< +\infty\}$

Where ${\ \|f\|_1^2 = \int f^2+\int|df|^2}$ .
complete ${\Phi(d) }$ with the norm ${\| . \|_1}$ is the sobolev space ${H^1(M) < L^2(M)}$ .
The extension of operator ${d}$ in ${H^1(M)}$ is called ${\overline d}$ ,assume ${d_c}$ is ${d}$ restrict on ${ C_0^\infty(M)}$ , ${d_c}$ extend to ${H_0^1(M)}$ called ${\overline d_c}$ , ${H_0^1(M)}$ is ${C_o^\infty(M)}$ complete with the norm ${\|.\|_1}$ , of course ${H_0^1(M)\subset H^1(M)}$ .
Sobelev theory tell us, if ${M}$ is a complete Riemann manifold,then ${H_0^1(M)=H^1(M)}$ .
Operator ${\delta=-*d*}$ ,satisfied ${<df,w>=<f,\delta w>}$ ,where * is the Hodge-star operator , ${w}$ is the ${C^\infty}$ 1-form, iff one of ${f,w}$ is compact supp.
$\displaystyle \Phi(\delta)=\{ w \in C^\infty 1-form | \int |w|^2 + \int |\delta w|^2 < +\infty \}$

By a lemma from Gaffney (Ann. of Math , 60,1954,458-466) we have ${ \overline \delta = \overline d_c^*, \overline \delta_c =\overline d^* }$ .
the laplace operator(with Direchlet boundary condition or Neumann boundary condition) is ${\Delta_D = \overline \delta \overline d_c,\Delta_N =\overline \delta_C \overline d}$ , When M is with smooth boundary,then ${\Delta =\Delta_N =general \Delta oprator}$ ,assume M is complete ( ${H_0^1(M)=H^1(M)}$ ),then because ${\overline d_C =\overline d,\overline \delta_C =\overline \delta}$ ,Gaffney proved(Ann. of Math , 60 ,1954,140-145),
$\displaystyle \Delta=\Delta_D=\Delta{\mathbb N}=\overline \delta \overline d$

.
As we all known, ${\Delta}$ is self-adjoint operator,so ${e^{-\Delta t}}$ make up a bounded self-adjoint operator semi group,by the self-adjoint operator theory,if ${dE_ {\lambda} }$ is the spectrum measure of ${\Delta}$ ,then
$\displaystyle e^{-\Delta t} = \int \limits_0^\infty d^{\lambda t}dE_{\lambda} , t>0$

.
And for ${t>0}$ , ${e^{-\Delta t}:L^2(M) \longrightarrow \bigcap_0^\infty D(\Delta^i) \subset C^\infty(M) }$ .when ${f \in L^2(M)}$ ,
$\displaystyle \Delta^i(e^{-\Delta t}) = \int \limits_0^\infty \lambda^i e^{-\Delta t} dE_{\lambda}(f).$

where ${ \bigcap_{i=1}^\infty D(\Delta^i) \subset C^\infty(M)}$ is basic on Weyl theory.
2.2. existence of heat kernel

Now we proof the basic fact:
${Thm 2.1}$ : assume ${M}$ is a complete manifold,then there exist a heat kernel ${H(x,y,t) \in C^\infty(M \times M \times R^+)}$ ,and ${ (e^{-\Delta t}f)(x) = \int_M H(x,z,t-s)H(z,y,s)dz}$ , ${\forall f\in L^2(M)}$ ,satisfied:
${(1) H(x,y,t)=H(y,x,t).\\ (2) \lim\limits_{t \rightarrow 0^+} H(x,y,t)= \delta_x(y).\\ (3) (\Delta - \frac{\partial}{\partial t})H=0.\\ (4) H(x,y,t) = \int H(x,y,t-s) H(z,y,s)dz. }$
Proof:
(A) first proof, ${\forall f \in L^2(M), e^{-\Delta t} f\in C^\infty(M \times R^+)}$ .to proof this,we first proof,in weak sense:

$\displaystyle \frac{\partial}{\partial t}(e^{-\Delta t} f) = \int\limits_0^\infty -\lambda e^{-\lambda t} dE_{\lambda}(f).$

in fact,we have:
${ \frac{\partial}{\partial t} (e^{-\Delta t} f) \\=\lim\limits_{\epsilon \rightarrow 0} \frac{1}{\epsilon} [\int \limits_0^\infty e^{-\lambda(t+\epsilon)}dE_{\lambda}(f)-\int\limits_0^\infty e^{-\lambda t} dE_{\lambda}(f)] \\=\lim\limits_{\epsilon \rightarrow 0}[\int\limits_0^A \frac{e^{-\lambda \epsilon}-1}{\epsilon} e^{-\lambda t} dE_{\lambda}(f)+\int\limits_A^\infty \frac{e^{-\lambda\epsilon}-1}{\lambda\epsilon} \lambda e^{-\lambda t}dE_{\lambda}(f)] \\=\int\limits_0^A -\lambda e^{-\lambda t} dE_{\lambda}(f) + O(Ae^{-At} \|f\|) \\ \longrightarrow \int\limits_0^\infty -\lambda e^{-\lambda t} dE_{\lambda}(f) \ as\ {A \ \rightarrow +\infty} \dotfill (\star).}$
rmk: the limit is take in the ${L^2}$ space,every step in the caculate make sence because of the dominating convergence theorem.
so what have we proved ,in fact we proved in the classical sence, ${ \frac{\partial}{\partial t} (e^{-\Delta t} f) }$ exist. and our strategy is to proof any order of weak derivatives of it exist and then use the embedding theorem to prove it is smooth.
in fact to proof ${\frac{\partial}{\partial t}(e^{-\Delta t} f)=\int\limits_0^\infty -\lambda e^{-\lambda}dE_{\lambda}(f)}$ (in weak sence),we need only to proof: ${\forall \psi \in C_0^\infty(M \times R^+)}$ we have:
$\displaystyle \int \frac{\partial \psi}{\partial t}(e^{-\Delta t}f) = -\int \psi(\int\limits_0^\infty -\lambda e^{-\lambda t}dE_{\lambda}(f))$

but because of ${(\star)}$ ,this is obvious,and similar we can proof that (in the weak sence):
$\displaystyle (\Delta+\frac{\partial^2}{\partial t^2})^i(e^{-\lambda t} f)= \int\limits_0^\infty (\lambda+\lambda ^2)^i e^{-\lambda t} dE_{\lambda}(f)$

rmk: there is some thing to explain,why ${\Delta(e^{-\lambda t} f)= \int\limits_0^\infty \lambda e^{-\lambda t} dE_{\lambda}(f)}$ .
anyway,we observe that ${L= \Delta+ \frac{\partial^2}{\partial t^2}}$ is the laplace operator on ${M \times R^+}$ .and we have proved ${e^{-\lambda t}f \in \bigcap^\infty \Phi(L^i)}$ ,and we know that ${ \bigcap^\infty \Phi(L^i) \subset C^\infty }$ .so we proved ${ e^{-\lambda t}f \in C^\infty( M \times R^+).}$
and now we easy to observe that,
if ${f_1(x,t)=e^{-\lambda t} f}$ ,then:
$\displaystyle \frac{\partial}{\partial t} f_1 = - \int\limits_0^\infty \lambda e^{-\lambda t} dE_{\lambda}(f) = -\Delta( e^{-\lambda t} f)=\Delta f_1$

.
so
$\displaystyle ( \Delta - \frac{\partial}{\partial t}) f_1(x,t) = 0$

.
Rmk: ${f_1(x,t) \in C^\infty}$ ,so derivatives is in classical sense.
(B) to proof ${ e^{-\lambda t} f =\int\limits_M H(x,y,t)f(y)dy}$ .
by the decomposition of unity,we can assume ${f \in C_0^\infty(M)}$ ,and ${supp f }$ sufficed small.
consider the operator ${ \circ=\Delta +\frac{\partial}{\partial t}}$ and its quasi fundamental solution (paramatrix) ${\longrightarrow}$ see next section for the serious definition ${\longrightarrow P(x,y,t),P(x,y,t) \in C^\infty (M \times M \times R^+)}$ ,
$\displaystyle \lim \limits_{t \rightarrow 0} P(x,y,t) = \delta_y$

. and ${\forall N > 0, \lim\limits_{t \rightarrow 0} \circ_x P(x,y,t) =O(t^N)}$ ,and when ${d(x,y)}$ suffice small, ${t \rightarrow +0 }$ ,we have expansion:
$\displaystyle P(x,y,t) \sim \frac{exp(-d(x,y)^2/4t)}{(4\pi t)^{n/2}} \sum\limits_i a_i(x,y)t^i.$

where ${n =dim M, d(x,y)= x,y}$ Riemann distance. ${a(x,y) \in C^\infty(M \times M), a_0(x,y) =1}$ ,for ${0< \epsilon<s<t-s}$ ,

${ e^{-\Delta\epsilon}P(x,y,t-\epsilon)-e^{-\Delta(t-\epsilon)}P(x,y,\epsilon)\\ = \int\limits_{\epsilon}^{t-\epsilon} \frac{d}{ds}(e^{-\Delta(t-s)} P(x,y,s))ds\\ =\int\limits_{\epsilon}^{t-\epsilon} [ \Delta e^{ -\Delta(t-s)P(x,y,s)} + e^{-\Delta(t-s)} \frac{\partial P}{\partial s}(x,y,s)]ds\\ =\int\limits_{\epsilon}^{t-\epsilon} e^{-\Delta(t-s)} \circ _xP(x,y,s)ds.}$
use ${t}$ instead of ${t-\epsilon}$ ,and assume ${\epsilon \rightarrow 0}$ ,
$\displaystyle \lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t} P(x,y,\epsilon) = P(x,y,t)-\int\limits_0^t e^{-\Delta(t-s)} \circ_x P(x,y,s)ds =\limits_{def} H(x,y,t).$

assume ${ F(x,y,s) = \circ _x P(x,y,s)}$ ,then
${ \Delta^i \int\limits_0^t e^{- \Delta(t-s)} (x,y,s)ds\\ =\int\limits_0^t\int\limits_0^\infty \lambda^i e^{-\lambda(t-s)} dE_{\lambda}(F(x,y,s))ds\\ =\int\limits_{s_0}^t \int\limits_0^\infty \lambda^i e^{-\lambda(t-s)}dE_{\lambda}(F(x,y,s))ds\\ + \int\limits_0^{s_0}\int\limits_0^\infty \lambda^i e^{-\lambda(t-s)} dE_{\lambda}(F(x,y,s))ds\\ =\int\limits_{s_0}^t\int\limits_0^\infty \lambda^i e^{-\lambda(t-s)}dE_{\lambda}(F(x,y,s))ds+\int\limits_0^{s_0}O(s^N)ds. }$
this is because ${F(x,y,s) =O(s^N)}$ ,when ${s \rightarrow 0}$ ,so ${H(x,y,t) \in \Phi(\Delta^i)}$ ,so ${H(x,y,t) \in C^\infty(M \times M \times R^+)}$ .because:
$\displaystyle |e^{-\Delta(t-s)} \circ_x P(x,y,s)| = O(s^N).$

so ${H(x,y,t)}$ and ${P(x,y,t)}$ have the same expansion .because ${H(x,y,t) = \lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t} P(x,y,\epsilon)}$ so ${\forall f(y)\in C_0^\infty(M)}$ ,we have:
$\displaystyle \int H(x,y,t)f(y)dy=\lim\limits{\epsilon \rightarrow 0} \int\limits_M e^{-\Delta t} P(x,y,\epsilon)f(y)dy\\ =e^{-\Delta t}\lim\limits_{\epsilon \rightarrow 0} \int \limits_M P(x,y,\epsilon)f(y)dy\\ =e^{-\Delta t}f(x).$

the equality arrive is because of the prop of ${P(x,y,t)}$ .
on the other hand ,by the definition of ${H(x,y,t)}$ ,we can check:
when ${y}$ is fix and ${t>0}$ , ${H(x,y,t) \in L^2(M)}$ , so
$\displaystyle e^{-\Delta t} f(x) = \int H(x,y,t)f(y) , \ \forall f\in L^2(M)........................................(\star\star).$

we call ${H(x,y,t)}$ is the kernel of ${e^{-\Delta t}}$ .

for prop (1), ${H(x,y,t)=H(y,x,t)}$ is from the operator ${\Delta}$ is self-adjoint,prop (2) is just ${(\star\star)}$ .

now we proof prop (3),by definition:
$\displaystyle H(x,y,t)=\lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t-\epsilon} P(x,y,\epsilon), \ \forall \epsilon >0.$

and we know ${ (\Delta_x - \frac{\partial}{\partial t})(e^{-\Delta t} P(x,y,t)) =0}$ so ${ (\Delta_x -\frac{\partial}{\partial t})H(x,y,t) = 0}$ .
proof prop (4),by ${e^{-\Delta s}e^{-\Delta(t-s)} = e^{-\Delta t}}$ and ${(\star\star)}$ wo know:
$\displaystyle H(x,y,t) = \int\limits_M H(x,z,t-s)H(z,y,s)dz..$

QED.
rmk: for the manifold with bounded,we can also proof the existence of heat kernel with Dirchlet or Neumann boundary condition.(L.Chavel. Eigenvalues in Riemannian Geometry,Academic Press,1984)
2.3. quasi fundamental solution of heat equation

it is well known that for ${R^n}$ , the fundamental solution of heat equation ${(\Delta - \frac{\partial}{\partial t})u=0}$ is ${ exp(-\frac{r^2}{4t})/(4\pi t)^{n/2}}$ .
for general Riemann manifold ${M}$ ,we want to find the fundamental solution of heat equation with this form:

$\displaystyle U(x,y,t) \sim (4 \pi t)^{-\frac{n}{2}} e^{\frac{-d(x,y)^2}{4t} }{\sum\limits_{i \geq 0} \phi_i(x,y) t^i}.$

where ${d(x,y)}$ is the Riemann distance of two point of M. take the normal coordinate around point ${x}$ , ${ y^i(i= 1,2,....,n),r=d(x,y)}$ .(the length of geodesic connect ${x,y}$ ).
it is well known that there exist functions ${\psi(r) , \phi(r)}$ only depend on ${r}$ :
$\displaystyle \Delta \psi = \frac{d^2 \psi}{dr^2} + (\frac{d \ log(\sqrt g)}{dr})\frac{d \psi}{dr}.$

$\displaystyle \Delta( \phi \psi) = \phi \Delta \psi +\psi \Delta \phi +2 \frac{d \phi}{dr}\frac{d \psi}{dr}.$

take:
$\displaystyle \psi =(4 \pi t) ^{-\frac{n}{2}} e^{-\frac{r^2}{4t}}$

$\displaystyle \phi = \phi_0+\phi_1 t+...+\phi_N t^N,$

and
$\displaystyle u_N =\psi \phi =(4 \pi t) ^{-\frac{n}{2}} e^{-\frac{r^2}{4t}} \sum\limits_{i=0 ... N} \phi_i t^i.$

then
$\displaystyle (\Delta - \frac{\partial}{\partial t}) u_N= \phi(\Delta \psi -\frac{\partial}{\partial t} \psi)+\psi(\Delta \phi -\frac{\partial}{\partial t}\phi)+2\frac{d\phi}{dr}\frac{d \psi}{dr}.$

because of
$\displaystyle \Delta \psi - \frac{\partial}{\partial t}\psi =\frac{d\ log\sqrt{g}}{dr}\frac{d \psi}{dr},$

$\displaystyle \frac{d \psi}{dr}=-\frac{r}{2t}\psi.$

we know:
$\displaystyle (\Delta -\frac{\partial}{\partial t})u_N = \frac{\psi}{t}\sum\limits_{k = 0 ... N} [ \Delta \phi_{k-1} -(k+\frac{r}{2}\frac{d \ log\sqrt{g}}{dr})\phi_k -r\frac{d \phi_k}{dr}]t^k ,$

problem become to solve the equations:
$\displaystyle r\frac{d\phi_k}{dr} =(k+\frac{r}{2}\frac{d\ log\sqrt{g}}{dr})\phi_k =\Delta \phi_{k-1} , k=0,1...N.$

which is equivalent to:
$\displaystyle \frac{d}{dr}(r^k g^{\frac{1}{4}}\phi_k)=r^kg{\frac{1}{4}}\Delta\phi_{k-1},k\geq 1.$

$\displaystyle \frac{d \phi_0}{dr}+\frac{d \ log\sqrt{g}}{\partial dr}\phi_0=0 (take\ \ \phi_{-1} =0).$

solve it:
$\displaystyle \phi_0= g^{\frac{1}{4}}$

$\displaystyle \phi_k(x,y) =g^{\frac{1}{4}} r^{-k} \int\limits_0^{r(x,y)} r^{k-1}(\Delta \phi_{k-1})g^{\frac{1}{4}}dr.$

so ${\phi_k \in C^\infty(M),\forall k \in N}$ . and:
$\displaystyle (\Delta - \frac{\partial}{\partial t})u_N=(4 \pi t)^{- \frac{n}{2}} e^{-\frac{r^2}{4t}} \Delta \phi_N t^N.$

take the cut function ${ \theta \in C_0^\infty}$ , ${s.t}$ :
$\displaystyle \theta(r)=1, when |r| \le \frac{1}{2}.$

$\displaystyle \theta(r)=0, when |r| \geq 1 .$

take:
$\displaystyle P_N(x,y,t) =\theta (r(x,y)) u_N(x,y,t),$

of course:
$\displaystyle Pn(x,y,t) \in C^\infty(M \times M \times R^+),and \ \\\ \lim\limits_{\epsilon \rightarrow 0} P(x,y,\epsilon) = \delta_x(y), (\Delta - \frac{\partial}{\partial t})P_N = O(t^N),$

this is to say:
${P_N(x,y,t)}$ is the quasi fundamental solution of heat equation. from the equation:
$\displaystyle (\Delta - \frac{\partial}{\partial t})u =G,$

$\displaystyle u|_{t=0} = 0,$

we know if ${G}$ has suffise high zero ,then so is the solution of heat equation,so ${P_n(x,y,t)}$ can approximate heat equation to any order.
2.4. basic proposition of heat kernel

${Lemma 4.1}$ :

$\displaystyle H(x,y,t) >0 , \ \forall \ t > 0.$

proof strategy : begin with expansion of heat kernel and integral on a geodesic sphere and take the radius ${r \rightarrow 0}$ ,use the stokes formula and maximal value principle to proof ${H(x,y,t)}$ is always positive.

${Lemma 4.2}$ : assume ${M}$ is a constant curvature complete riemann manifold (space form),then ${H(x,y,t)}$ only depends on ${r=d(x,y)}$ ,and ${\frac{\partial H(r,t)}{\partial r} <0}$ . proof is similar to ${Lemma 4.1}$ .

${Thm 4.1}$ (heat kernel comparision theorem,Cheeger-Yau).:
assume ${M}$ is a complete riemann manifold , ${Ric \geq (n-1)k}$ , ${ \forall x \in M,r_0 >0}$ , heat kernel of ${ B(x,r_0), H(x,y,t)}$ and heat kernel ${\varepsilon(r(x,y),t)}$ of geodesic ball ${V(k,r_0)}$ in space form satisfied:
$\displaystyle \varepsilon(r(x,y),t) \leq H(x,y,t).$

(bounded condition is Derichlet condition or Neumann condition).
proof strategy: basically we use the formula ${\frac{1}{2}\Delta(|\bigtriangledown u|^2)=\sum\limits_{i , j} u_{ij}^2+\sum\limits_i u_i(\Delta u)_i+Ric (\bigtriangledown u,\bigtriangledown u)}$ .and the two lemma. ${Thm 4.2}$ : assume ${M}$ is a compact reimann manifold , ${f_i}$ is a orthonormal basis of special function on ${M}$ , ${ \lambda_i}$ is the corresponding spectrum,then the fundamental solution (heat kernel) has the expansion:
$\displaystyle H(x,y,t)= \sum e^{-\lambda_i t} f_i(x)f_i(y)$

in particular,
$\displaystyle \sum e^{-\lambda_i t} =\int\limits_M H(x,y,t)dx,$

when ${ t \rightarrow +0}$ ,
$\displaystyle H(x,y,t) \sim (4 \pi t)^{\frac{n}{2}}e^{\frac{d^2(x,y)}{4t}}\sum\limits_{i \geq 0} \phi_i(x,y) t^i .$

rmk:this thm is well known.

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。

Atiyah-Singer 指标定理把一个分析对象的 Fredholm index 与流形上的拓扑特征类联系起来。热核证明的美妙之处在于：同一个量 $\operatorname{Str}(e^{-tD^2})$ 在 $t\to\infty$ 时看见 kernel，在 $t\to0$ 时看见局部曲率。

热核证明中，同一个 supertrace 在大时间看到 kernel，在小时间看到局部曲率。

1. Dirac bundle 与 Dirac operator

设 $M$ 是紧 Riemannian manifold，$E=E^+\oplus E^-$ 是 $\mathbb Z_2$-graded Clifford module。Dirac operator 写成

$$D:C^\infty(E^+)\to C^\infty(E^-).$$

它的 index 是

$$\operatorname{ind}D=\dim\ker D^+-\dim\ker D^-.$$

这个数在连续扰动下不变，所以天然应该由拓扑量表达。

2. Chern-Weil 背景

给定连接 $\nabla$ 和曲率 $F_\nabla$，Chern-Weil 理论告诉我们，形如

$$\operatorname{tr}\exp\left(\frac{i}{2\pi}F_\nabla\right)$$

的闭形式代表 Chern character。改变连接只会改变 exact form，因此上同调类只依赖 bundle 的拓扑。

Dirac 指标公式中的局部密度由 $\widehat A(M)$ 和 twisting bundle 的 Chern character 组成。

3. Supertrace

在 graded bundle 上，supertrace 定义为

$$\operatorname{Str}(A)=\operatorname{Tr}(A|_{E^+})-\operatorname{Tr}(A|_{E^-}).$$

热核证明的核心量是

$$\operatorname{Str}(e^{-tD^2}).$$

McKean-Singer 公式说它与 $t$ 无关，并且等于 $\operatorname{ind}D$。

4. 热半群与自伴算子

在完备 Riemannian manifold 上，合适的 Laplace 型算子可以取自伴扩张。谱定理给出热半群

$$e^{-tD^2}=\int e^{-t\lambda}\,dE_\lambda.$$

当 $t$ 很大时，正特征值贡献指数衰减，只剩零特征空间；于是 supertrace 给出 index。当 $t$ 很小时，热核有局部渐近展开，曲率项浮现出来。

5. 这条证明的路线

热核证明要完成两件事：第一，证明 supertrace 不依赖 $t$；第二，计算 $t\to0$ 的局部极限。前者是谱理论和 graded commutator 的结果；后者是 heat kernel asymptotics 和 Clifford algebra 的局部计算。二者合在一起，就把分析 index 变成了积分公式。

2017年9月25日
Laplace 特征函数的 nodal set：Dong identity 与 Hausdorff measure 下界

旧博客原文

原题：Hausdorff Dimension Of Nodal Set

Basic setting:
Let $(M,g)$ be a compact $C^\infty$ Riemannian manifold of dimension $n$ , let $\phi_{\lambda}$ be an $L^2$ – normalized eigenfunction of the Laplacian:

$\Delta \phi_{\lambda} = −\lambda^2 \phi_{\lambda}\$ and let:$ latex N \phi_{\lambda} =\{x:\phi_{\lambda}(x)=0\}$
be its nodal hypersurface. Let $H^{n−1}(N\phi_{\lambda} )$ denote its $(n-1)$ -dimensional Riemannian hypersurface measure. In this note we prove:
Theorem:

for and $C^\infty$ metric $g$ ,there exists a constant $C_g > 0$ so that:
$H^{n-1}(N_{\phi_{\lambda}}) \leq C_g \lambda^{n}$

A crucial identity:
proof of theorem 1 is based on following identity:
theorem:
for any smooth Riemannn manifold $M$ ,we have,
$\lambda^2\int_{M}|\phi_{\lambda}|dV = 2\int_{N_{\phi_{\lambda}}} |\nabla\phi_{\lambda}|dS$
moreover, $\forall f \in C^2(M)$ ,
$\int_M(\Delta+\lambda^2)f \vert\phi_{\lambda}\vert dV=2\int_{N_{\phi_{\lambda}}} \vert\nabla\phi_{\lambda}\vert dS$

Proof:
observed we have that,
$M=N_{\phi_{\lambda}}^+ \cup N_{\phi_{\lambda}} \cup N_{\phi_{\lambda}}^$
on $N_{\phi_{\lambda}}^+$ ,use divergence theorem:
\begin{eqnarray*}
\int_M(\Delta+\lambda^2)f \phi_{\lambda} dV&=&\int_M(\Delta+\lambda^2)\phi_{\lambda}f dV+\int_{\partial M} -g(\upsilon,\phi_{\lambda}\nabla f)dS+\int_{\partial M} g(\upsilon,f\nabla\phi_{\lambda})dS\\
&=&\int_{\partial M} g(\upsilon,f\nabla\phi_{\lambda}) \\
&=&\int_{\partial M} f\phi_{\lambda}dS
\end{eqnarray*}
the same identity is true on $N_{\phi_{\lambda}}^-$
so we have:
$\int_M(\Delta+\lambda^2)f \vert\phi_{\lambda}\vert dV=2\int_{N_{\phi_{\lambda}}} \vert\nabla\phi_{\lambda}\vert dS$

Estimate hausdorff measure of nodal sets:
take $f=1$ in theorem 2,we have:
$\lambda^2\int_M\vert\phi_{\lambda}\vert dV=2\int_{N_{\phi_{\lambda}}} \vert\nabla\phi_{\lambda}\vert dS$
so to get estimate hausdorff measure of nodal sets,we need to estimate:
$||\phi_{\lambda}||_1$ ,$||\nabla\phi_{\lambda}||_{\infty}$ ,this two guys are easy to get good estumate….and we will get a lower bound estimate of measure of nodal set:
$H^{n-1}(N_{\phi_{\lambda}}) \geq \frac{\lambda^2||\phi_{\lambda}||_1}{2||\nabla\phi_{\lambda}||_{\infty}}$

Estimate:
$||\phi_{\lambda}||_1$ , $||\nabla\phi_{\lambda}||_{\infty}$
$||\phi_{\lambda}||_1$ :

normalized $L_2$ norm of $\phi_{\lambda}$

$||\nabla\phi_{\lambda}||_{\infty}$ :
we have a yau types gradients estimate

Estimate upper bound of measure:
to get upper bound estimate,from identity we need to estimate: $||\phi_{\lambda}||_1$ , $||\nabla\phi_{\lambda}||_{\infty}$ ,and we will get:
$H^{n-1}(N_{\phi_{\lambda}}) \leq \frac{\lambda^2||\phi_{\lambda}||_1}{2\int_{N_{\phi_{\lambda}}} |\nabla\phi_{\lambda}|dS}$

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。

Laplace 特征函数的 nodal set 是

$$N_\lambda=\{x:\phi_\lambda(x)=0\}.$$

它的大小反映了 eigenfunction 的振荡。Yau 猜想预言在光滑紧流形上，nodal hypersurface 的测度与 $\lambda$ 同阶。

Laplace 特征函数的 nodal set 测度可通过 Dong identity 与梯度估计联系起来。

1. 基本设定

设

$$-\Delta\phi_\lambda=\lambda^2\phi_\lambda,\qquad \|\phi_\lambda\|_2=1.$$

nodal set 通常是一个维数 $n-1$ 的几何对象，但可能带有奇异点。

2. Dong identity

一个关键恒等式是：对光滑函数 $f$，

$$\int_M(\Delta+\lambda^2)f\,|\phi_\lambda|\,dV
=2\int_{N_\lambda} f|\nabla\phi_\lambda|\,dS.$$

它把 nodal set 上的积分转成整个流形上的积分。

3. 下界策略

取 $f=1$，得到

$$\lambda^2\int_M|\phi_\lambda|\,dV
=2\int_{N_\lambda}|\nabla\phi_\lambda|\,dS.$$

于是

$$\mathcal H^{n-1}(N_\lambda)\gtrsim
\frac{\lambda^2\|\phi_\lambda\|_1}{\|\nabla\phi_\lambda\|_\infty}.$$

4. 需要的两个估计

要得到 nodal measure 下界，需要控制 $\|\phi_\lambda\|_1$ 的下界和 $\|\nabla\phi_\lambda\|_\infty$ 的上界。后者来自 elliptic estimates 或 spectral cluster estimates。

5. 几何意义

nodal set 是 eigenfunction 改变符号的地方。特征值越大，振荡越快，nodal set 应该越大。Dong identity 精确表达了这种振荡与零集几何之间的关系。

2017年9月25日
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 $C^1$ Boudary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。

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。

2017年9月25日
Bourgain-Sarnak-Ziegler criterion：Mobius 正交性的有限检验

旧博客原文

原题：Bourgain-Sarnak-Ziegler Criterion

Bourgain-Sarnak-Ziegler定理可以视为Vingrodov均值定理的有限版本。

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。

Bourgain-Sarnak-Ziegler criterion 可以看成 Vinogradov 均值思想的有限版本：若一个有界序列在不同素数倍采样下彼此近似正交，那么它与 Mobius 函数也应当正交。

BSZ criterion 用不同素数倍采样的相关消失来推出 Mobius 正交性。

1. 判别法的形式

设 $a_n$ 是有界序列。若对不同素数 $p\ne q$，有

$$\sum_{n\le N}a_{pn}\overline{a_{qn}}=o(N)$$

并且这个估计对一批素数足够一致，那么可以推出

$$\sum_{n\le N}\mu(n)a_n=o(N).$$

2. 为什么素数倍相关重要

Mobius 函数的困难在于它携带素因子结构。BSZ criterion 的想法是：不直接分析 $\mu(n)$，而是检查序列 $a_n$ 对不同素数尺度是否产生相关。如果所有这些相关都小，Mobius 就没有可利用的结构与之耦合。

3. 与 Sarnak 猜想

在动力系统中，常取 $a_n=f(T^n x)$。于是 Mobius disjointness 变成

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

BSZ 把这个问题转化为比较 $T^p$ 和 $T^q$ 产生的两个轨道序列。

4. 有限版本的意义

称它为 Vinogradov 均值定理的有限版本，是因为它同样通过“多重相关消失”来控制原始振荡和。它特别适合低复杂度系统，例如 nilsequence、skew product 或 substitution dynamics。

2017年9月25日