博客

Heat flow 与多项式零点：从变形思想到 Riemann Hypothesis 的 toy model

旧博客原文

原题：Heat flow and the zero of polynomial-a approach to Riemann Hypesis

this is a note after reading the blog:Heat flow and the zero of polynomial.

1.instead of consider the original version:

$\partial_{zz}f(z,t)=\partial_tf(z,t)$ .

consider the corresponding “equidistribution version” is also interesting:

$\partial_{zz}f(z,t)=\theta(z,t)\partial_tf(z,t)$ ,especially $\theta(z,t)=e^{2\pi i\alpha t},\alpha\in R-Q$ .

2.

where $f(z)=z^n+a_{n-1}z^{n-1}+...+a_1z+a_0$ .

$f(z,t)=\sum_{k=1}^n\sum_{0\leq m\leq k-2,2|k-m}\frac{k!}{m!(k-m)!}z^mt^{k-m}.$

$=\sum_{k=1}^m\sum_{0\leq m\leq k-2,2|k-m}C_k^mt^{k-m})z^mt^{k-m}$

$\sum_{m=0}^{n-2}(\sum_{k=m,2|k-m}^nC_k^mt^{k-m})z^m$ .

rescaling:

$F_t:(z_1(t),...,z_n(t))\longrightarrow (\frac{z_1(t)}{t},...,\frac{z_n(t)}{t})$ .

$F_t\cdot f(z,t)=\sum_{m=0}^{n-2}(\sum_{k=m,2|k-m}^nC_{k}^mt^{k-n})z^m$ .

$\lim_{t\to \infty}F_t\cdot f(z,t)=\sum_{m=0,2|n-m}^{n-2}C_n^mz^m$ .(*)

even term $\longrightarrow$ constant.(after renormelization)

odd term $\longrightarrow$ 0(invariant).so at least the sum zeros of is invarient.

by the algebraic fundamental theorem,we have n zero $\{z_1,...,z_n\}$ of (*).

until now,we already now if the n zeros is distinct,then because the energy is the energy is the same and the entropy is increase so $\exists T>>0,\forall t_i,t_j>T$ , $\{t>T|z_i(t)\} \cap \{t>T|z_j(t)\}=\emptyset$ . $\lim_{t\to \infty}|z_i(t)|=\infty$ and $\lim_{t\to \infty}arg(z_i(t))=z_i$ .

but how to know the information of the change of direction at “blow up” time?

1.change direction only at $blow up$ .

2.energy invariant $\sum_{1\leq i\neq j\leq n}\frac{1}{|x_i-x_j|^2}$ .

3.general philosophy

deformation some function under some evolution equation, such like heat equation,wave equation,shrodinger equation.and there is some conversion thing under the equation,and some quantity that could calculate directly such like the trace of spectral.

4.difficultis

this philosophy could generate to the analytic function case,but to make the limit case(I only know how ti deal with this now)coverage.we need very good control on the coefficient.

and to investigate the change of direction at blow up point maybe we need some knowledge about the burid group.

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。

用 heat flow 研究多项式零点，是理解更复杂解析函数零点问题的一个 toy model。基本思想是：让函数随时间演化，观察零点如何移动，以及哪些量在演化中保持或单调。

heat flow 让多项式零点随时间运动，提供研究零点实性和不变量的 toy model。

1. 多项式的 heat deformation

设 $P(x)$ 是多项式，考虑

$$\partial_t u=\partial_x^2u,\qquad u(0,x)=P(x).$$

因为 heat operator 保持多项式空间，$u(t,x)$ 仍然是多项式。其零点随 $t$ 移动。

2. 不变量与单调量

某些系数组合在演化中保持不变，另一些量具有单调性。例如最高次项不变，低阶偶次项会随 heat flow 改变。零点的质心或某些对称量可能保持。

3. 零点碰撞

若零点始终实且互异，运动图像较清楚；真正困难发生在零点碰撞或分裂时。此时需要理解 blow-up 时间附近的方向变化。

4. 与 Riemann Hypothesis 的类比

de Bruijn-Newman 常数研究的是 Xi 函数在 heat flow 型变形下零点保持实的临界时间。多项式模型不能证明 RH，但能展示同一种哲学：通过演化方程追踪零点几何。

5. 需要的估计

要从多项式推广到整函数，必须控制系数、增长阶和极限过程。多项式情形的代数基本定理给出有限零点；整函数情形需要更强的紧性和零点分布估计。

2017年10月20日
三维 Kakeya 猜想：hairbrush、密度递降与 Katz-Tao 思路

旧博客原文

原题：Kakeya conjecture in R^3

Kakeya conjecture in $R^3$ is very subtle.in fact wolff stay the best(but not very difficult to get,just use the structure so-called hairbrush)result $\frac{5}{2}$ until the result of Katz and Tao $\frac{5}{2}+\epsilon$ .Where $\epsilon$ is a constant independent with kakeya set.and in the article of Tao,they proved $\epsilon>\frac{1}{10^{10}}$ .

Two-dimensional case

first we overview the case of dimension 2,these is the only case that is proved.and the key point is the estimate:

$\mu(T_{i}\cap (\cup_{j\in I,j\neq i}T_j))<log(\frac{1}{\delta})\mu(T_i)$ .

where $T_i=T_i(x_i,\theta_i)$ satisfied $\cap_{i\in I}T_i$ is a $\delta$ -neibeihood of kakeya set $X$ .to remember one thing:this is equivalent to the maximal function version of kakeya conjecture,but for the minkoski version,there is a extra structure for the group $I_{\delta}$ in different scales(this can be view as a multi-scale apporoach).

this inequality is easy to proof.just observed that $\mu(T_i\cap T_j)\sim \frac{1}{\theta_i-\theta_j}\delta^2$ .and to remember one thing:the inequality can be view as a uniformly estimate of overlap of the kakeya set,that is just mean the overlap would not concentrate to much at a lonely stick.this is enough to get a proof of the 2 dimension case just by a density decrement trick:we just not consider about the whole set $I$ ,but a low density subset $\hat I\subset I$ ,where $\frac{|\hat I|}{|I|}\sim \delta^{\lambda}$ ,and make $\lambda\to 0^+$ .

Kakeya estimates

Let $\sigma\leq \delta\leq \theta<<1$ ,and let $T_{\delta}$ be a collection of $\delta$ -tubes.whose set of directions all lie in a cap of radius $\theta$ . Let $2<d<3$ be fixed.
• If we have a Kakeya estimate at some dimension d, and if the collection $T_{\delta}$ is direction-separated, then

$||\sum_{i\in I}\chi_{T_i}||_{d'}\lesssim \delta^{\frac{d-3}{d}}\theta^{\frac{d+1}{d}}$ (1)
• If we have an X-ray estimate at some dimension d, and if $T_{\delta}$ consists ofessentially distinct tubes, then

$||\sum_{i\in I}\chi_{T_i}||_{d'}\lesssim \delta^{\frac{d-3}{d}}\theta^{\frac{d+1}{d}}m^{1-\beta}$
for some β > 0, where m is the directional multiplicity of $T_{\delta}$ .(2)

So obviously the X-ray estimate is stronger than the kakeya estimate.it is just give the information of the overlap of the sticks with the same direction.

in fact wolff have establish the X-ray estimate at dimension $\frac{5}{2}$ ,so (2) just come from a rescaling argument.

The sticky reduction

renormalization process,just consider the process to make the thin sticks to be fat.and to proof this structure nearly has Markov property.but with a very small error term when change the scale.this is proved by the X-ray estimate.

Triple intersection estimate

Use Hardy-Litterwood-Soblev inequality,we can get a so called triple intersection estimate in general,said the triple intersection is smaller than the situation the 3 lines move together.and we just accosiate this to the cap-cup principle to get some information of the volume of $X_{\delta}=\mu(\cup_{i\in I}T_i)$ .

Reduce to additive combination problem

The right problem is just you have a $n\times n$ cubes,and there is some sticks according them,if the distance of sticks is

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。

Kakeya 猜想问：包含每个方向单位线段的集合是否必须有 full Hausdorff 或 Minkowski dimension？二维情形已经解决，三维情形则出现了 hairbrush、multiscale 和 sum-product 等深层结构。

三维 Kakeya 问题的困难来自不同方向细 tubes 的高重叠结构。

1. Tube formulation

令 $E_\delta$ 是 Kakeya set 的 $\delta$-neighborhood。Minkowski 维数问题等价于估计

$$|E_\delta|\gtrsim_\varepsilon \delta^{\varepsilon}$$

在合适维数归一化下的下界。更常用的语言是研究方向分离的 $\delta$-tubes 的重叠。

2. 二维情形

二维中，任意两个不同方向 tubes 的交叠容易控制。核心估计可以理解为：重叠不能集中在一根孤立 tube 附近。由此配合 density decrement，可推出二维 Kakeya 集合有 full dimension。

3. Hairbrush 结构

三维中，许多 tubes 可以围绕一根 tube 形成 hairbrush。Wolff 的观察是：如果很多 tubes 都与一根 tube 相交，那么它们的方向和位置仍然受到几何限制。这给出非平凡维数下界。

4. Multiscale 难点

Minkowski 版本比单尺度 maximal function 更微妙，因为不同尺度之间的结构会相互传递。一个尺度上的高重叠可能在下一尺度分裂，反过来又影响全局维数。

5. Katz-Tao 方向

Katz-Tao 的改进把 Kakeya 问题与 sum-product 现象联系起来。大致图像是：若 tubes 太集中，会诱导出同时具有加法和乘法结构的集合；sum-product 阻止这种集合太小。这是三维 Kakeya 后续发展的关键思想。

2017年10月20日
Fractal Uncertainty Principle：极限集、Schottky group 与 zeta 零点

旧博客原文

原题：Fractional uncertain principle

semyon dyatlov的一篇文章

semyon dyatlov的文章https://arxiv.org/pdf/1710.05430.pdf，用fractional uncertainly priciple导出了hyperbolic surface上测地线诱导的zeta函数在 $Re(s)>1-\epsilon$ 只有有限个零点。

就我的理解，这件事情至少和3个事情有关系，

1.p-adic上的黎曼猜想，因为这篇文章的证明强烈依赖于markov性质，这和p adic的结构也很像，有可能可以利用p adic猜想的证明思路继续做一部分。

2.billiard的传播子，但是这里不一样，文章中的 Schottky groups本质上是对于算子的逆写成一种级数形式其中级数由Schottky group生成，但是对于billiard传播子的情况所有的涉及的热核或者波核的paramatrix不仅仅具备markov性质，起主导作用的却是某种需要X-ray估计的性质，级数和并不是对全空间求而是某种截断了的子空间里面，所以比这个证明要难。建立起billiard的传播子估计是证明inverse spectral problem的重要一步。

3.interval exchange map，但是interval exchange map的结构就好只有这里的traslation，这里有一个像的大小的指数衰减，这是interval exchange map所没有的。interval exchage map可能还需要涉及到一个拆分估计，会更难，这可能可以在interval exchange map上的sarnak猜想有进展。

下面讲一下我对文章证明主要思路的理解：首先对于hyperbolic空间 $H^2/{\Gamma}$ 我们用poincare的方式来理解为 D mod掉一个作用，那么极限集 $\Lambda_{\Gamma}$ 就是基本域在分式线性变换下在D的边界下的极限点。

关键是建立如下估计 $\int_{\Lambda_{\Gamma}}exp (i\xi\phi(x)) g(x) d\mu(x)\leq C|\xi|^{−\epsilon_1} \forall \xi, |\xi| > 1$ .

为了建立这个估计，我们做的事情是：

1.研究极限集 $\Lambda_{\Gamma}$ 的结构，本质上具备某种组合上的树结构，在分式线性变换下树的上方和下方交换，而且对于象有指数级别的衰减，这很像连分数展开中的otrowoski表示。对于分式线性变换和Schottky group作用的体积形变估计是容易得到的。

2.Patterson–Sullivan测度 $\mu$ 是在 $\Gamma$ 作用下的遍历测度，特别的，和 $\Gamma$ 是compatible的，所以变量代换公式成立：

$\int_{\Lambda_{\Gamma}} f(x)d\mu(x) =\int_{\Lambda_{\Gamma}}f(\gamma(x))|\gamma'(x)|_{\delta}^B d\mu(x) \forall \gamma \in \Gamma$

这个很重要，一旦我们能够找到 $I :=\amalg_{b\in \Omega} I_b$ , 实际这可以导出一个关于f的方程：

$L_Zf(x) = \sum_{a\in Z,a\to b} f(\gamma_{a′}(x))w_{a′}(x), x ∈ I_b.$

我们关心的selberg zeta函数的零点就等于方程特征值1对应的特征函数： $L_zf(x)=f(x)$ .

（这一点很重要而且在很多问题中都有用，至少有几个例子：1.有的时候一个椭圆方程的特征值很难做，转而去考察他的发展方程。2.很多数论问题，特别是质数在某些partition集合里面的分布，对于对应的L函数的动力系统的刻画就需要这个方程）

3. 文章中3.1.是bourgain的主要贡献，是所谓的sum-product现象在这里的一个引用，为了得到foriour衰减性估计，我们需要不断拆散区间，实际上在树的每一层上面我们都很清楚怎么把这一层的积分拆散到上一层和下一层，这实际上可以看成一个renormelization方程：

$\int_{\Lambda_{\Gamma}}f d\mu =\int_{\Lambda_{\Gamma}}L_{Z(\tau)}^{2k+1} f d\mu = \sum_{A,B,A\leftrightarrow B}f(\gamma_{A∗B}(x))w_{A∗B}(x)d\mu(x)$ .

1中的形变估计(只需要估计一下交叉项带来的误差)告诉我们： $| \int_{}fdμ|^2 ≤C\tau^{(2k−1)\delta}\sum_{A,B,A\leftrightarrow B} |\int_{I_b(A)} e^{iξ\phi(\gamma_{A∗B}(x))}w_{a′_k} (x)d\mu(x)| ^2 +C\tau^2$ .

然后在x点taylor展开用围道积分得到误差项，误差项用1中的形变估计得到上界控制，主项放到一起得到：

$|\int_{\Lambda_{\Gamma}} f d\mu|^2 ≤ C\tau^{(2k+1)\delta}\sum_A sup_{\eta\in J_{\tau}}| e^{2πi\eta\xi_{1,A} (b_1)···\xi_{k,A} (b_k )}| + C\tau^{\delta/4}$ . （*）

最后为了用bourgain的sum-product现象得到的引理3.3来控制（*）RHS，我们需要对 $R ⊂ Z(τ) ^{k+1}$ 和 $Z(τ)^{k+1}-R$ 分段估计,前者是用正则性导致的收敛速度快，后者shi用minkowki维数很小，前者估计已经建立，所以只需对Z(τ)^{k+1}-R 建立minkoski维数上界估计，这是显然的。

这样我们就建立好了如下估计：

$\int_{\Lambda_{\Gamma}}exp ( i\xi\phi(x)) g(x) d\mu(x) ≤ C|\xi|^{−\epsilon_1} \forall \xi, |\xi| > 1$ .

这个估计是用来建立fractional uncertain principle的关键，一旦我们有了fractional uncertain principle，hyperbolic surface上测地线诱导的zeta函数在 $Re（s）>\frac{1}{2}-epsilon$ 只有有限个零点就只是一个fix point theorem的argument。

另外一个不需要用sum-product现象的极为简单的证明见[1710.05430] Fractal uncertainty for transfer operators。

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。

Fractal uncertainty principle 说，一个函数不可能同时集中在一个 fractal set 和它的 Fourier dual fractal set 上。Dyatlov 等人的工作把这个原则用于 hyperbolic surface 上的共振和 Selberg zeta 函数零点问题。

Fractal uncertainty principle 排除函数同时集中在极限集和对偶 fractal set 上。

1. 从普通不确定性到 fractal 版本

普通不确定性原理强调物理空间和频率空间不能同时局部化。fractal uncertainty principle 的形式更细：若 $X,Y$ 是具有 porous 或 Ahlfors-David regular 结构的 fractal sets，则不存在非零函数同时几乎支撑在 $X$ 和频率支撑在 $Y$ 上。

粗略地说，有估计

$$\|\mathbf 1_X \mathcal F_h \mathbf 1_Y\|_{L^2\to L^2}\le C h^\beta,$$

其中 $\beta>0$ 反映 fractal 几何带来的额外衰减。

2. Schottky group 与极限集

对某些 hyperbolic surface，可以用 Schottky group 来描述。其极限集 $\Lambda$ 位于边界上，具有类似 Cantor set 的树状结构。群作用下的小区间长度指数衰减，而每一层的 cylinder 又保留清晰的组合编码。

这和连分数、Ostrowski 表示、Markov partition 有相似味道：复杂轨道被编码成树上的路径，尺度之间通过 renormalization 关系联系。

3. Patterson-Sullivan 测度

Patterson-Sullivan 测度是极限集上的自然遍历测度。它与群作用兼容，因此变量替换公式可以把 transfer operator 写成适合迭代的形式。很多 zeta 函数或共振问题最终会转化为某个 transfer operator 是否有特征值 $1$。

这一步很重要：谱问题从几何 Laplacian 转成边界动力系统上的算子问题。

4. Sum-product 现象的作用

fractal uncertainty 的关键估计往往来自 sum-product 型思想。若一个集合同时在加法和乘法结构上都太集中，就会违反 expansion。极限集的多尺度树结构让我们可以把振荡积分拆到不同层级，再用 expansion 迫使 Fourier 衰减。

这与 Bourgain 方法的精神一致：不是直接逐点估计振荡积分，而是利用组合结构证明在多数尺度上不能同时对齐。

5. 可能的延伸方向

这个框架让人自然想到 billiards、interval exchange maps 和 $p$-adic 动力系统。相同点是都有编码和尺度结构；不同点是 Markov 性质、传播子 parametrix 和截断方式并不一样。真正困难的地方在于：如何找到一个既保留动力系统结构又能产生 Fourier 衰减的分解。

2017年10月18日
Fourier restriction problem 的自然性：从 Hausdorff-Young 到曲率

旧博客原文

原题：Natural of the restriction problem

1.

the most natural problem in harmonic analysis may be:

investigate for what pair $(p,q)$ we have :

$L^p(R^n)\longrightarrow L^q(R^n)$

$\hat f(x)=\int_{R^n}e^{-2\pi ix\xi}f(\xi)d\xi$

is strong- $(p,q)$ bounded.

obvious we have the paserval identity: $||\hat f||_{2}=||f||_2$ ,and we have $||\hat f||_{\infty}\leq||f||_{1}$ .

so by the Riesz-Thorin inteplotation theorem we have the Hausdorff-Young inequality:

$\forall 1\leq p\leq 2,\frac{1}{p}+\frac{1}{q}=1$ we have:

$||\hat f||_{q}\leq ||f||_{p}$ .

now let talk about the rescaling trick:

consider the transform: $f(x)\longrightarrow f(\frac{x}{\lambda})=f_{\lambda}(x)$ .we know if the inequality is right then it is necessary to have the same growth for the RHS and LHS.

this argument will derive: $\frac{n}{p}=n-\frac{n}{q}$ .

in fact $||f_{\lambda}(x)||_{p}=\lambda^{\frac{n}{p}}||f(x)||_{p}$ .

by the variable substitute formula: $\hat f_{\lambda}(x)=\int_{R^n}e^{-2\pi ix\xi}f(\frac{\xi}{\lambda})d\xi=\lambda^n\hat f(\lambda x)$ .

so $||\hat f_{\lambda}(x)||_{q}=\lambda^{n-\frac{n}{q}}||\hat f(x)||_{q}$

and by the scaling invariance trick we know the pair $(p,q)$ should live on the line $\frac{1}{p}+\frac{1}{q}=1$ ,and by test with the guessian function $g(x)=e^{-x^2}$ we know the right pair should be $1\leq p\leq 2$ .this end the problem with $R^n$ .

2.

now replace $R^n$ by a bounded open set $K$ .when the fourior transform restriction on $K$ is bounded $p-q$ operator?

i.e. $L^p(R^n)\longrightarrow L^q(R^n)$

$f\longrightarrow \hat f|_{K}$ .

$\hat f|_{K}=\int \chi_{K}e^{2\pi i<x,\xi>}f(\xi)d\xi$ .

on a bounded set $K$ ,we always have:if $q\geq r$ , $||f||_{L^p(K)}\geq ||f||_{L^r(K)}$ .

and associate with hausdorff-young inequality we have:

$||\hat f|_K||_{r}\leq ||\hat f||_q\leq ||f||_p$ .

and this area is the exact area(rescaling trick and test with gaussian function),so end of the story.(but why?)

3.

Now we begin to deal with the really interesting case: $K$ is not a open set but a sub manifold like the unit sphere $S^{n-1}$ .

$||\hat f||_{L^q(S^{n-1})}\leq ||f||_{L^p(R^{n})}$ .

$S^{n-1}$ equip with the usual surface measure $\sigma$ .

but the inequality is not always meaningful.

case: $p=2,\hat f\in L^2$ ,in general can not restrict to a measure zero set due to the loss of regularity.

case: $p=1$ , $\hat f$ continuous,meaningful to restrict to $S^{n-1}$ .

$||\hat f||_{L^{\infty}(S^{n-1})}\leq ||f||_{L^1(R^n)},\forall 1\leq p\leq \infty$ .

Duality:we use the duality argument to transform the “restriction theorem” to “extension theorem”.

$T:f\longrightarrow \hat f$ .

$T:f\longrightarrow \hat f$ .

$||f||=\sup_{||f||_p=1}||\hat f||_q=\sup_{||f||_p=1}\sup_{||g||_{q'}=1}|\int_{R^n}\hat fgd\sigma|=\sup_{||g||_{q'}=1}\sup_{||f||_p=1}|\int_{R^n}\hat fgd\sigma|=\sup_{||g||_{q'}=1}||\hat{gd\sigma}||_{p'}$ .

4.

we use $R_s(p\to q)$ to state the estimate $||\hat f||_{L^q(S)}\leq ||f||_{L^p(R^n)}$ . $S=S^{n-1}$ .

and by rescaling argument we have natural condition: $p<\frac{2n}{n-1}$ , $p'\geq \frac{n+1}{(n-1)q}$ .the restriction conjecture just say this necessary condition is also enough.

Now we state the Tomas-Stein restriction theorem:

$1\leq p\leq \frac{2n+1}{n+3}$ . $R_s(p\to 2)$ holds.

this is the endpoint estimate in dimension 2 case,so by Meceztaze interpolation theorem this lead to the whole restriction theorem in dimension 2.

the first argument is come from the so called $TT^*$ trick that is find by fefferman and stein in 1970.

$T$ bdd $p\to 2$ $\Longleftrightarrow$ $TT^*$ bdd $p'\to p$ .

in fact:

$||T||=\sup_{||f||_p=1}||Tf||_2=\sup_{||f||_p=1}\sup_{||g||_2=1}|\int (Tf)g|=\sup_{||f||_p=1}\sup_{||g||_2=1}|\int f(Tg)|=\sup_{||g||_2=1}||Tg||_{p'}=||T^*||$ .

$\int|e^{2\pi ix\xi}f(\xi)|^2 dw(\xi)\leq C||f||_p^2$

$<\hat f,\hat fw(\xi)>\leq c||f||_p^2$

$<\hat f,\hat{f*\hat{w(\xi)}}> \leq ||f||_p^2$

$<f,f*\hat{w(\xi)}>\leq ||f||_p^2$

$<f,f*\hat{w(\xi)}>\leq ||f||_p||f*\hat{w(\xi)}||_{p'}$

this can be derived from HLS inequality:

$||f*\hat{w(\xi)}||_{p'}\leq ||f||_p$ .

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。

restriction problem 问的是：Fourier transform 能不能有意义地限制到一个测度为零的曲面上？这件事在 $L^1$ 情形是平凡的，在一般 $L^p$ 情形却依赖曲面的曲率和振荡抵消。

restriction problem 的关键是把 Fourier transform 有意义地限制到带曲率的零测曲面上。

1. 从 Hausdorff-Young 开始

Fourier transform 满足 Plancherel

$$\|\widehat f\|_2=\|f\|_2$$

和显然估计

$$\|\widehat f\|_\infty\le \|f\|_1.$$

插值得到 Hausdorff-Young inequality：

$$\|\widehat f\|_{p’}\le C\|f\|_p,\qquad 1\le p\le2.$$

2. 为什么限制到曲面不平凡

若 $S$ 是单位球面，想要估计

$$\|\widehat f|_S\|_{L^q(S)}\le C\|f\|_{L^p(\mathbb R^n)}.$$

因为 $S$ 是测度为零的集合，普通 $L^{p’}$ 控制不能直接给出 restriction。曲率让 Fourier transform 在曲面附近具有额外振荡结构。

3. Scaling test

任何 restriction estimate 必须通过 scaling 检验。用集中在小球或细 tube 上的 test functions，可以得到 $p,q$ 的必要关系。这些反例说明，restriction 问题不是纯函数分析问题，而是几何问题。

4. Extension operator

对偶形式是 extension estimate：

$$Eg(x)=\int_S e^{ix\cdot\xi}g(\xi)\,d\sigma(\xi).$$

它研究曲面上的振荡波如何在物理空间中叠加。曲率越强，波包方向越分散，越可能得到好的估计。

5. 与 Kakeya 的联系

restriction、Kakeya、Bochner-Riesz 和 wave packet decomposition 深度相连。曲面上的频率 cap 对应物理空间中的 tube；估计 extension operator 等价于控制这些 tubes 的重叠。

2017年10月18日
Kakeya 猜想的多尺度分析：旧 tubes、新 tubes 与密度递降

旧博客原文

原题：Kakeya Conjecture

Last year I read a nice blog articles Recent progress on the Kakeya conjecture and have several questions with this article.

follows the proof strategy called Multiscale analysis,although we can use the estimate with large $\delta_1$ to get estimate with small $\delta_2$ ,(may be loss some $\delta^c$ in the inequality in this way),but the main difficult is we should proof the new tubes with scales $\delta_2$ is contains in the the olders.as soon as we proof this ,to obtain a lower bound of minkwoski dimension with kakeya set, suffice to get following estimate :
the new cubes with scale $\delta_2$ contains a positive constants volumes of every old cubes with scale $\delta_1$ .
this type of estimate is easy to attain because it is very similar to the “principle of close packing of spheres”.

in general ,we should not expect this claims:
the new tubes with scales $\delta_2$ is contains in the the older.

but if we can proof in some sense most of new tubes comes from this way maybe we can make progress on the original problem.

roughly speaking,we should partition the whole set of $T_{\delta}$ into two part,comes from old ones or not,for the first kind i.e contains in a old one,use the way explained above to treat.the second kind we need to proof the influence is very some or we can sometimes use the cubes from the first kind to instead the cubes from second kind and the measure of $|A|_{\delta}$ change little.

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。

Kakeya 猜想的多尺度分析试图把一个尺度上的 tube 结构传递到更小尺度。真正困难在于：新尺度上的 tubes 不一定整齐地包含在旧 tubes 中，必须区分继承结构和新出现结构。

Kakeya 多尺度分析的关键是判断小尺度 tubes 中哪些继承自旧尺度，哪些带来新结构。

1. 多尺度想法

设 $E_\delta$ 是 Kakeya set 的 $\delta$-neighborhood。若从尺度 $\delta$ 过渡到更小尺度 $\rho\delta$，希望旧尺度的每个有效 tube 都包含足够多的新尺度结构。

2. 一个理想但错误的图像

最天真的想法是：所有新 tubes 都来自旧 tubes 的细分。若如此，体积下界可以通过 close packing 直接传递。但实际情况中，新 tubes 可能跨越旧结构，产生新的重叠模式。

3. 密度递降

一种策略是把集合分成两类：继承旧结构的部分，以及真正新的部分。前者用尺度归纳处理；后者若太大，则会产生密度递降或结构性矛盾。

4. 与 Katz-Tao 的关系

Katz-Tao 方法中的 multiscale 思想正是利用尺度之间的结构约束。若某个尺度上 tubes 过度集中，会诱导出 additive/multiplicative 结构；sum-product 现象阻止这种结构无限持续。

5. 关键问题

多尺度方法的核心不是简单地把 tubes 变细，而是证明大部分新结构可以被旧结构解释；无法解释的部分必须带来额外 expansion。这个二分是 Kakeya 维数改进中的基本逻辑。

2017年10月11日
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