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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bourgeon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} 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。

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Natural of the restriction problem

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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.

 

 

 

 

 

Kakeya Conjecture

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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.

Analysis Qualifying Examination(UCLA 2009)

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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.

 

 

 

 

Interval map

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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…

 

Flat surface 1

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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:

 

Heat Kernel proof of Index Theory 1

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Framework of atiyah singer index theory
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:
\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 define \widehat A_g =\widehat A(M) \ \forall g is a metric on M.

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

Supertrace
Str^{E / S}: End_{cl(M)} \rightarrow C_M.
induce map:
Str^{E/S} : \Omega(End_{cl(M)} (E) )\rightarrow \Omega(M) \otimes C
determined by:
Str^{E / S}( w \otimes T) = w \otimes Str^{E / S}(T). \ \forall w \in \Omega (M). \ \forall T \in End_{cl(M)} E.

Chern class
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.

F^{E \widehat\otimes W/S =F^{E/S} }\otimes 1_W + 1_E \otimes F^W.
F^W = F^{W^+} \oplus F^{W^-}.

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.

Atiyah-singer index theorem

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