分类： Diophantus approximation

Diophantine approximation：Dirichlet 定理、抽屉原理与最佳逼近
旧博客原文

原题：Diophantine approximation

I explain some general ideal in the theory of diophantine approximation, some of them is original by myself, begin with a toy model, then consider the application on folklore Swirsing-Schmidt conjecture.

\tableofcontents

1. Dirichlet theorem, the toy model

The very basic theorem in the theory of Diophantine approximation is the well known Dirichlet approximation theorem, the statement is following.

Theorem 1 (Dirichlet theorem) for all ${\alpha}$ is a irrational number, we have infinity rational number ${\frac{q}{p}}$ such that:

$\displaystyle |\alpha-\frac{q}{p}|<\frac{1}{p^2} \ \ \ \ \ (1)$

Remark 1 It is easy to see the condition of irrational is crucial. There is a best constant version of it, said, instead of ${1}$ , the best constant in the suitable sense for the theorem 1 should be ${\frac{1}{\sqrt{5}}}$ and arrive by ${\frac{\sqrt{5}+1}{2}}$ at least. The strategy of the proof of the best constant version involve the Frey sequences.

Now we begin to explain the strategies to attack the problem.

\paragraph{Argument 1, boxes principle} We begin with a easiest one, i.e. by the argument of box principle, the box principle is following,

Theorem 2 (Boxes principle) Given ${n\in {\mathbb N}}$ and two finite sets ${A={a_1,a_2,...,a_n,a_{n+1}}}$ , set ${B={b_1,...,b_{n}}}$ , if we have a map:

$\displaystyle f:A\longrightarrow B \ \ \ \ \ (2)$

Then there exists a element ${b_k\in B}$ such that there exist at least two element ${a_i,a_j\in A}$ , ${f(a_i)=f(a_j)=b_k}$ .

Proof: The proof is trivial. $\Box$

Now consider, ${\forall N\in {\mathbb N}}$ , the sequences ${x,2x,...,Nx}$ , then ${\{ix\}\in [0,1], \forall i\in \{1,2,...,n\}}$ . Divide ${[0,1]}$ in an average way to ${N}$ part: ${[\frac{k-1}{N},\frac{k}{N}]}$ . Then the linear structure involve (which, in fact play a crucial role in the approach). And the key point is to look at ${\{nx\}}$ and integers.

\paragraph{Argument 2, continue fractional} We know, for irrational number ${x}$ , ${x}$ have a infinite long continue fractional:

$\displaystyle x=q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k+...}}}} \ \ \ \ \ (3)$

Then

$\displaystyle |x-q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}|\sim \frac{1}{(q_1q_2...q_{k-1})^2q_k} \ \ \ \ \ (4)$

And we have,

$\displaystyle \frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}=\frac{a_n}{b_n}, (a_n,b_n)=1 \ \ \ \ \ (5)$

Then ${b_n=O(q_1...q_k)}$ .

\paragraph{Argument 3, Bohr set argument} We begin with some kind of Bohr set:

$\displaystyle B_p=I-\cup_{q\in \{0,1,...,p-1\}}(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2}) \ \ \ \ \ (6)$

The key point is the shift of Bohr set, on the vertical line i.e. ${|B_p\cap B_{p+1}|}$ is very slow, and can be explained by

$\displaystyle \frac{k}{p+1}+\frac{1}{(p+1)^2}>\frac{k}{p}-\frac{1}{p^2} \ \ \ \ \ (7)$

So:

$\displaystyle \frac{1}{p^2}+\frac{1}{(p+1)^2}>\frac{k}{p(p+1)} \ \ \ \ \ (8)$

in ${|B_p \cap B_{p+1}|\sim \frac{1}{p(p+1)}}$ But in fact they are not really independent, as the number of Bohr sets increase, then you can calculate the correlation, thanks to the harmonic sires increasing very slowly, wwe can get something non trivial by this argument, but it seems not enough to cover the whole theorem 1.

\paragraph{Argument 4, mountain bootstrap argument} This argument is more clever than 3, although both two arguments try to gain the property we want in 1 from investigate the whole space ${[0,1]}$ but not ${x}$ , this argument is more clever.

Now I explain the main argument, it is nothing but sphere packing, with the set of balls

$\displaystyle \Omega=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall p\in {\mathbb N}, 1\leq q\leq p-1 \} \ \ \ \ \ (9)$

and define its subset

$\displaystyle \Omega_l=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall 1\leq p\leq l, 1\leq q\leq p-1 \} \ \ \ \ \ (10)$

Then ${\Omega_l\subset \Omega}$ , and ${\Omega =\cup_{l\in {\mathbb N}}\Omega_l}$ . If we can proof,

Lemma 3 For all ${l\in {\mathbb N}}$ , there is a subset ${A_l}$ of ${\Omega-\Omega_l}$ such that ${\cup_{i\in A_l}B_i=[0,1]}$ .

Remark 2 If we can proof 3, it is easy to see the theorem 1 follows.

Proof: The proof follows very standard in analysis, may be complex analysis? Key point is we start with a ball ${B_{p,q}}$ , whatever it is, this is not important, the important thing is we can take some ball ${B_{p',q'}}$ with the center of ${B_{p',q'}}$ in ${B_{p,q}}$ , then try to consider ${B_{p',q'}\cup B_{p,q}}$ to extension ${B_{p,q}}$ and then we find the boudary is also larger then we can extension again, step by step just like mountain bootstrap argument. So we involve in two possible ending,
1. The extension process could extension ${B{p,q}}$ to whole space.
2. we can not use the extension argument to extension to the whole space.
If we are in the first situation, then we are safe, there is nothing need proof. If we are in second case, anyway we take a ball ${B_{p,q}=(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2})}$ . Then try to find good ball ${B_{p',q'}}$ to approximate ${B_{p,q}}$ , but this is difficult… $\Box$

Remark 3 Argument 1 is too clever to be true in generalization, argument 2 is standard, by the power of renormalization. argument 3 and argument 4 have gap… I remember I have got a proof similar to argument 4 here many years ago, but I forgot how to get it…

2. Schimidt conjecture

The Schimidt conjecture could be look as the generalization of Dirchlet approximation theorem 1 to algebraic number version, to do this, we need define the height of a algebraic number.

Definition 4 We say a number ${\alpha\in {\mathbb C}}$ is a ${k-}$ order algebraic number if and only is the minimal polynomial of ${\alpha}$ , ${f(x)=a_nx^n+...+a_1x+a_0, a_n\neq 0}$ have degree ${deg(f)=n, f\in {\mathbb Z}[x]}$ .

Definition 5 (Height) Now we define the height of a ${k-}$ th order algebraic number as ${H(\alpha):=\max\{\|a_n\|_h,\|a_{n-1}\|_h,...,\|a_0\|_h\}}$ , Where

$\displaystyle h(a_i)=\|a_i\|_{\infty} \ \ \ \ \ (11)$

Now we state the conjecture:

Theorem 6 (Swiring-Schimidt conjecture) For all transendental number ${x\in {\mathbb C}}$ , there is infinitely ${\alpha}$ are ${k-}$ th algebraic number such that:

$\displaystyle |x-\alpha|<\frac{c_k}{H(\alpha)^{k+1}} \ \ \ \ \ (12)$

Where ${c_k}$ is a constant only related to ${k}$ but not ${x}$ .

I point out the conjecture is very related to the map:

$\displaystyle F:(x_1,...,x_n) \longrightarrow (\sigma_1(x_1,...,x_n),\sigma_2(x_1,...,x_n),...,\sigma_n(x_1,...,x_n)) \ \ \ \ \ (13)$

Where ${\sigma_k(x_1,...,x_n)=\sum_{1\leq i_1<...<i_k\leq n}\Pi_{j=1}^kx_{i_1}x_{i_2}...x_{i_k}}$ is the ${k-}$ th symmetric sum.

Remark 4 ${F}$ is a map ${{\mathbb C}^n\rightarrow {\mathbb C}^n}$ , what we consider is its inverse, ${G=F^{-1}}$ , but ${G}$ is not smooth, it occur singularity when ${x_i=x_j}$ for some ${i\neq j}$ . And the map, as we know, the singularity depend on the quantity ${\Pi_{1\leq i< j\leq n}(x_i-x_j)}$ .

Remark 5 I then say something about the geometric behaviour of the map ${G}$ , as we know, what we have in mind is consider the map ${G}$ as a distortion ${{\mathbb C}^n\rightarrow {\mathbb C}^n}$ , Then ${H(\alpha)}$ is just the pullback of the canonical metric on ${{\mathbb C}}$ (morally) to ${{\mathbb C}}$ .

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。

Diophantine approximation 的基本问题是：一个无理数 $\alpha$ 能被有理数 $p/q$ 逼近到什么程度？Dirichlet 定理给出最基本的答案，而连分数告诉我们哪些分母是真正的最佳逼近。

Dirichlet 定理的抽屉原理证明依赖小数部分的差分线性结构。

1. Dirichlet 定理

对任意无理数 $\alpha$，存在无穷多个有理数 $p/q$，使得

$$\left|\alpha-\frac pq\right|<\frac1{q^2}.$$

一个有限版本是：给定 $Q$，存在 $1\le q\le Q$ 和整数 $p$，使

$$|q\alpha-p|<\frac1Q.$$

2. 抽屉原理证明

看 $Q+1$ 个数的小数部分

$$0,\{\alpha\},\{2\alpha\},\ldots,\{Q\alpha\}.$$

把 $[0,1]$ 分成 $Q$ 个长度 $1/Q$ 的区间。两个小数部分落在同一区间，于是它们的差给出某个 $q\alpha$ 距离整数小于 $1/Q$。这就是 Dirichlet 定理最干净的证明。

3. 线性结构在哪里

抽屉原理本身只是计数，但这里真正起作用的是线性结构：两个点 $\{a\alpha\}$ 和 $\{b\alpha\}$ 接近，差就变成 $\{(a-b)\alpha\}$ 接近整数。没有这个差分结构，抽屉原理不会自动给出有理逼近。

4. 连分数与最佳逼近

连分数展开

$$\alpha=[a_0;a_1,a_2,\ldots]$$

给出 convergents $p_k/q_k$。它们满足

$$\left|\alpha-\frac{p_k}{q_k}\right|<\frac1{q_kq_{k+1}}.$$

这些分母 $q_k$ 是最佳逼近的自然尺度。若 $a_k$ 有界，则 $\alpha$ 是 badly approximable；若某些 $a_k$ 很大，就会出现异常好的逼近。

5. 更高维和 Schmidt 猜想的方向

高维 Diophantine approximation 会把一个数的逼近问题变成向量、线性形式或流形上的逼近问题。此时抽屉原理仍然给出基准结果，但最佳常数、例外集维数和代数数逼近会变得更深。许多问题最后会进入 geometry of numbers、dynamical systems on homogeneous spaces 或 Schmidt game 的语言。
2018年1月8日
代数数的 Diophantine approximation：Liouville、Roth 与 Vandermonde 约束

旧博客原文

原题：Diophantine approximation of algebraic number

An important theorem in Diophantine approximation is the theorem of Liuoville:

**Liuoville Theorem** If x is a algebraic number of degree $n$ over the rational number then there exists a constant $c(x) > 0$ such that: $\left|x-{\frac {p}{q}}\right|>{\frac {c(x)}{q^{{n}}}}$

holds for every integer $p,q\in N^*$ where $q>0$ .

This theorem explain a phenomenon, the approximation of algebraic number by rational number could not be very well. Which was generated later to **Thue–Siegel–Roth theorem**, them could be used to proof a lots of constant is not algebraic, i.e. transcendentals .

My questions is in another direction, now let us not just consider one root $\alpha_1$ of a integer polynomial $P(x)=a_mx^m+...+a_1x+a_0$ but consider all roots of it, i.e. $\{\alpha_1,...,\alpha_m\}$ , which is based on a observation : If we define

$\sigma_k(P(x))=\sum_{1\leq \alpha_{i_1}<\alpha_{i_2}<...<\alpha_{i_k}\leq m}\alpha_{i_1}\alpha_{i_2}...\alpha_{i_k}$

By **Vieta theorem** we know $\sigma_k(n)\in \mathbb Q$ for all $k\in N^*$ , this will lead to some restriction and in fact destroy the uniformly distribution of $(\alpha_1,...,\alpha_m)\in [0,1]^m$ . In fact the most important one is the determination of Vandermon Determinant:
$V(P(x))=\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)$ .

We know $\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\in \mathbb Q$ so when $\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\neq 0$ we could use this to proof a nontrivial estimate for $\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}$ .
$\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).$

by combine the A-G inequality and $\Pi_{1\leq \alpha_i<\alpha_j\leq n}(\alpha_i-\alpha_j)=\lambda\neq 0$ .While by continue fractional expansion we only know a trivial estimate of type $\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n})$ .

my question is the following:
Is there still have a nontrivial estimate for $\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}$ (which could be slight weaker), if we don’t have the whole power of **Vieta theorem**? more precisely:

**problem 1**

if we have $\sigma_k((\alpha_1,...,\alpha_m))=\lambda_k\in \mathbb Q$ for all $k\in \{1,2,...,m'\}$ where $m'<m$ , is there still some nontrivial estimate of,

$\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}$

hold for all $n\in N^*$ ?

One reason to consider this could be true is that although $\{\alpha_1,...,\alpha_m\}$ is not roots of a integer polynomial but we could image in some suitable metric space $X$ the gromov-hausdorff distance of tuple $(\alpha_1,...,\alpha_m)$ and a tuple come form roots of integer polynomial is small . And it seems reasonable to image this type of asymptotic quality is continue with the G-H distance on $X$ .

Another problem is what happen when $V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0$ . More precisely,

**problem 2**

What happen when $V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0$ , is this result,

$\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).$

still true?

Let us go a litter further, if these two problem both have a satisfied answer, what is the higher dimensional case?

**problem 3**

Given $m\in \mathbb N^*$ . If tuple $(y_1,...,y_k)$ is very closed to the zero set of a variety in $\mathbb Z[x_1,...,x_m]$ in $\mathbb (Z^{m})^k$ in the sense a lots of symmetric sum of $y_1,...,y_k$ belong to $\mathbb Q^m$ , will this lead to some good estimate for

$\sum_{1\leq s\leq k}||y_sn||_{\mathbb R^m/\mathbb Z^m}?$

I think these type of result should be investigated very well, Iappreciate to any pointer with useful comments and answer, both on given some strategy to solve these problems or given some reference about these problems.

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。

代数数不能被有理数“过分好”地逼近。Liouville theorem 是这件事的第一层形式，Thue-Siegel-Roth theorem 则给出几乎最优的结论。

Liouville 和 Roth 定理说明代数数不能被有理数过分好地逼近。

1. Liouville theorem

若 $\alpha$ 是次数 $d$ 的代数数，则存在 $C(\alpha)>0$，使对所有有理数 $p/q$，

$$\left|\alpha-\frac pq\right|\ge \frac{C(\alpha)}{q^d}.$$

证明的核心是把 $\alpha$ 代入整数多项式 $P$，再估计 $P(p/q)$ 不可能是太小的非零有理数。

2. Roth theorem

Roth theorem 大幅加强 Liouville：若 $\alpha$ 是无理代数数，则对任意 $\varepsilon>0$，

$$\left|\alpha-\frac pq\right|<\frac1{q^{2+\varepsilon}}$$

只有有限多个有理解。换句话说，代数数的有理逼近指数不能超过 $2$ 太多。

3. 多个根的约束

若考虑一个整数多项式的所有根 $\alpha_1,\ldots,\alpha_d$，Vieta 定理给出系数与根的对称函数之间的整数关系。Vandermonde determinant

$$\prod_{i
又控制根之间不能全部过分靠近。

4. 从单点逼近到整体结构

单个根的有理逼近只看一个 $\alpha$；所有根一起看时，还会出现判别式、对称多项式和高度的约束。整体代数结构比单个连分数展开更刚性。

5. 一个自然问题

如果没有完整 Vieta 结构，只知道某些弱的对称约束，是否仍能推出非平凡逼近下界？这类问题位于 Diophantine approximation、代数高度和几何不等式之间。

2017年12月5日