引言
博客更新辣!一看上次更新,还是在四个月前.除了爽玩之外,也还是参加了中科院的暑校学习了一些丢番图问题的知识.于是终于打算开始学习Tim Browning的这本Quantitative Arithmetic of Projective Varieties [ 1 ] {[1]} [ 1 ]
在看这本书之前,还没有意识到这些数论问题和几何之间的联系原来有这么深,因此很值得记录一下,免得一段时间下来进度为负然后被薄纱.
此处先记录该书的第一章内容,以及本章后的习题(都还蛮简单的).由于第一章其实是Intro性质的,涉及到了很多开放性的问题,但成书又是在2006年左右,因此我也简单收集了一下这些问题的最新进展,或许这些进展又能反过来带来新的想法,而这就在后面的某段时间可以看看了.
再补充说明一下,与筛法读书笔记基本翻译原文的风格不同,我尽量以极简风格来记录这些内容.话说到此处打住,接下来要狠狠启动了!😎
高度有界的整点的分布
Birch定理
假设f ∈ Z [ x 1 , ⋯ , x n ] f \in \mathbb{Z}[x_1, \cdots, x_n] f ∈ Z [ x 1 , ⋯ , x n ] n n n d d d
S f ≔ { x = ( x 1 , ⋯ , x n ) ∈ Z n ∖ { 0 } : f ( x ) = 0 } , S_f \coloneqq \{ \mathbf{x} = (x_1, \cdots, x_n) \in \mathbb{Z}^n \setminus \{ 0 \} : f(\mathbf{x}) = 0 \},
S f : = { x = ( x 1 , ⋯ , x n ) ∈ Z n ∖ { 0 } : f ( x ) = 0 } ,
但是我们更关心的是S f S_f S f
N ( f ; B ) : = # { x ∈ S f : ∥ x ∥ ≤ B } , ( 1 ) N(f;B) := \#\{ x \in S_f : \| x \| \le B \}, \quad(1)
N ( f ; B ) := # { x ∈ S f : ∥ x ∥ ≤ B } , ( 1 )
其中∥ ⋅ ∥ \|\cdot\| ∥ ⋅ ∥
∥ ⋅ ∥ : R n → R x ↦ max 1 ≤ i ≤ n ∣ x i ∣ . \begin{split}
\|\cdot\| : \mathbb{R}^n & \to \mathbb{R} \\
\mathbf{x} & \mapsto \max_{1\le i \le n} |x_i|.
\end{split} ∥ ⋅ ∥ : R n x → R ↦ 1 ≤ i ≤ n max ∣ x i ∣.
而对于N ( f ; B ) N(f;B) N ( f ; B )
Thm 1(Birch; P4):
\quad f ∈ Z [ x 1 , ⋯ , x n ] f \in \mathbb{Z}[x_1, \cdots, x_n] f ∈ Z [ x 1 , ⋯ , x n ] d d d n > ( d − 1 ) 2 d n > (d-1)2^d n > ( d − 1 ) 2 d f ( x ) = 0 f(\mathbf{x})=0 f ( x ) = 0 R \mathbb{R} R p p p Q p \mathbb{Q}_p Q p c f > 0 c_f>0 c f > 0
N ( f ; B ) ∼ c f B n − d , ( 2 ) N(f;B) \sim c_f B^{n-d}, \quad (2)
N ( f ; B ) ∼ c f B n − d , ( 2 )
\quad B → ∞ . B \to \infty. B → ∞.
其中围绕"f ( x ) = 0 f(\mathbf{x})=0 f ( x ) = 0 R \mathbb{R} R p p p Q p \mathbb{Q}_p Q p n > ( d − 1 ) 2 d n > (d-1)2^d n > ( d − 1 ) 2 d 容易 得构造出几个反例.在后面我将简单补充一些目前对于Birch定理的一些改进,感觉也是一大块很值得深挖的内容.
一些简单的反例
下面的例子1将说明在Q \mathbb{Q} Q
例1
\quad n n n 2 d 2d 2 d
f 0 : = x 1 2 d + ⋯ + x n 2 d , f_0 := x_1^2d + \cdots + x_n^{2d},
f 0 := x 1 2 d + ⋯ + x n 2 d ,
\quad S f 0 = ∅ S_{f_0} = \emptyset S f 0 = ∅ ( 2 ) (2) ( 2 )
设K K K d d d ω 1 , ⋯ , ω d \omega_1, \cdots, \omega_d ω 1 , ⋯ , ω d O K \mathcal{O}_K O K
N ( y 1 , ⋯ , y d ) : = N K / Q ( y 1 ω 1 + ⋯ + y d ω d ) , \mathbf{N}(y_1,\cdots,y_d) := N_{K/\mathbb{Q}}(y_1\omega_1 + \cdots + y_d\omega_d),
N ( y 1 , ⋯ , y d ) := N K / Q ( y 1 ω 1 + ⋯ + y d ω d ) ,
可知N ( y 1 , ⋯ , y d ) \mathbf{N}(y_1, \cdots, y_d) N ( y 1 , ⋯ , y d ) d d d [ 2 ] {[2]} [ 2 ]
例2
\quad d 2 d^2 d 2 d d d
f 1 : = N ( x 1 , ⋯ , x d ) + p N ( x d + 1 , ⋯ , x 2 d ) + ⋯ + p d − 1 N ( x d 2 − d + 1 , ⋯ , x d 2 ) , f_1 := \mathbf{N}(x_1, \cdots, x_d) + p\mathbf{N}(x_{d+1}, \cdots, x_{2d}) + \cdots + p^{d-1}\mathbf{N}(x_{d^2-d+1}, \cdots, x_{d^2}),
f 1 := N ( x 1 , ⋯ , x d ) + p N ( x d + 1 , ⋯ , x 2 d ) + ⋯ + p d − 1 N ( x d 2 − d + 1 , ⋯ , x d 2 ) ,
\quad f 1 ( x = 0 ) f_1(\mathbf{x}=0) f 1 ( x = 0 ) x = 0 \mathbf{x}=\mathbf{0} x = 0 S f 1 = ∅ S_{f_1} = \emptyset S f 1 = ∅
解答:
\quad x = ( x 1 , ⋯ , x d 2 ) ∈ Z d 2 \mathbf{x} = (x_1, \cdots, x_{d^2}) \in \mathbb{Z}^{d^2} x = ( x 1 , ⋯ , x d 2 ) ∈ Z d 2 f 1 ( x ) = 0 f_1(\mathbf{x}) = 0 f 1 ( x ) = 0 gcd ( x 1 , ⋯ . x d 2 ) = 1 \gcd(x_1, \cdots. x_{d^2}) = 1 g cd( x 1 , ⋯ . x d 2 ) = 1
\quad mod p \text{mod }p mod p
N ( x 1 , ⋯ , x d ) ≡ 0 ( m o d p ) , \mathbf{N}(x_1, \cdots, x_d) \equiv 0 \pmod{p},
N ( x 1 , ⋯ , x d ) ≡ 0 ( mod p ) ,
\quad p ∣ N ( y ) p \mid \mathbf{N}(\mathbf{y}) p ∣ N ( y ) p ∣ y p \mid \mathbf{y} p ∣ y 练习1 的结论),于是
p ∣ gcd ( x 1 , ⋯ , x d ) . p \mid \gcd(x_1, \cdots, x_d).
p ∣ g cd( x 1 , ⋯ , x d ) .
\quad
( x 1 , ⋯ , x d ) = p ( x 1 ′ , ⋯ , x d ′ ) . (x_1, \cdots, x_d) = p(x_1', \cdots, x_d').
( x 1 , ⋯ , x d ) = p ( x 1 ′ , ⋯ , x d ′ ) .
\quad mod p 2 \text{mod }p^2 mod p 2
p ∣ gcd ( x d + 1 , ⋯ , x 2 d ) . p \mid \gcd(x_{d+1}, \cdots, x_{2d}).
p ∣ g cd( x d + 1 , ⋯ , x 2 d ) .
\quad p ∣ ( x 1 , ⋯ , x d 2 ) p \mid (x_1, \cdots, x_{d^2}) p ∣ ( x 1 , ⋯ , x d 2 ) □ \square □
例子2貌似仍然是在说定理1成立的必要条件是在p p p f 1 f_1 f 1 可能构造 出一个f f f [ 3 ] {[3]} [ 3 ] n > ( d − 1 ) 2 d n > (d-1)2^d n > ( d − 1 ) 2 d
接下来的两个S f ≠ ∅ S_f \neq \emptyset S f = ∅
例3
\quad n n n d d d
f 2 : = x 1 d − x 2 ( x 3 d − 1 + ⋯ + x n d − 1 ) . f_2 := x_1^d - x_2(x_3^{d-1} + \cdots + x_n^{d-1}).
f 2 := x 1 d − x 2 ( x 3 d − 1 + ⋯ + x n d − 1 ) .
\quad f 2 ( 0 , 0 , a 3 , ⋯ , a n ) = 0 f_2(0,0,a_3,\cdots,a_n) = 0 f 2 ( 0 , 0 , a 3 , ⋯ , a n ) = 0 # S f ≫ B n − 2 \#S_f \gg B^{n-2} # S f ≫ B n − 2
利用例子2中的f 1 f_1 f 1
例4
\quad n > d 2 n > d^2 n > d 2 d 2 d^2 d 2 Q \mathbb{Q} Q L 1 , ⋯ , L d 2 ∈ Z [ x 1 , ⋯ , x n ] L_1, \cdots, L_{d^2} \in \mathbb{Z}[x_1, \cdots, x_n] L 1 , ⋯ , L d 2 ∈ Z [ x 1 , ⋯ , x n ]
\quad n n n d d d
f 3 : = f 1 ( L 1 ( x 1 , ⋯ , x n ) , ⋯ , L d 2 ( x 1 , ⋯ , x n ) ) , f_3 := f_1(L_1(x_1, \cdots, x_n), \cdots, L_{d^2}(x_1, \cdots, x_n)),
f 3 := f 1 ( L 1 ( x 1 , ⋯ , x n ) , ⋯ , L d 2 ( x 1 , ⋯ , x n )) ,
\quad # S f + 1 = # { x ∈ Z n : L 1 ( x ) = ⋯ = L d 2 ( x ) = 0 } ≍ B n − d 2 \#S_f + 1 = \#\{ \mathbf{x} \in \mathbb{Z}^n : L_1(\mathbf{x}) = \cdots = L_{d^2}(\mathbf{x}) = 0 \} \asymp B^{n-d^2} # S f + 1 = # { x ∈ Z n : L 1 ( x ) = ⋯ = L d 2 ( x ) = 0 } ≍ B n − d 2
而上面的这两个例子都是由于其有着large singular locus所导致的,因此non-singular并能很好的排除这一点.
本原整点的计数
但我们更关心的 其实是projective rational points的分布.记V V V P n − 1 \mathbb{P}^{n-1} P n − 1 P n − 1 \mathbb{P}^{n-1} P n − 1 x = [ x ] x = [\mathbf{x}] x = [ x ] ( x 1 , ⋯ , x n ) ∈ Z n (x_1, \cdots, x_n) \in \mathbb{Z}^n ( x 1 , ⋯ , x n ) ∈ Z n gcd ( x 1 , ⋯ , x n ) = 1 \gcd(x_1, \cdots, x_n) = 1 g cd( x 1 , ⋯ , x n ) = 1
H ( x ) : = max 1 ≤ i ≤ n ∣ x ∣ , H(x) := \max_{1 \le i \le n} | x |,
H ( x ) := 1 ≤ i ≤ n max ∣ x ∣ ,
给定一个locally closed subsetU ⊆ V U \subseteq V U ⊆ V
N U ( B ) : = # { x ∈ U ( Q ) : H ( x ) ≤ B } . ( 3 ) N_U(B) := \#\{ x \in U(\mathbb{Q}) : H(x) \le B \}. \quad (3)
N U ( B ) := # { x ∈ U ( Q ) : H ( x ) ≤ B } . ( 3 )
在后文中也是基本都是在考虑计数函数N U ( B ) N_U(B) N U ( B ) V V V U U U
N U ( B ) ∼ c B a ( log B ) b , ( 4 ) N_U(B) \sim cB^{a}(\log B)^b, \quad (4)
N U ( B ) ∼ c B a ( log B ) b , ( 4 )
事实上,根据目录来看,整本书基本上就是在围绕这个猜想展开的.
在这里,我们先简单记录一下N V ( B ) N_V(B) N V ( B ) N ( f ; B ) N(f;B) N ( f ; B ) μ \mu μ
N V ( B ) = 1 2 ∑ k = 1 ∞ μ ( k ) N ( f ; k − 1 B ) . ( 5 ) N_V(B) = \frac{1}{2} \sum_{k=1}^{\infty} \mu(k) N(f; k^{-1}B). \quad (5)
N V ( B ) = 2 1 k = 1 ∑ ∞ μ ( k ) N ( f ; k − 1 B ) . ( 5 )
利用(5),我们就很容易可以得到
Thm 2:
\quad n ≥ 2 n \ge 2 n ≥ 2 H : P n − 1 ( Q ) → R > 0 H: \mathbb{P}^{n-1}(\mathbb{Q}) \to \mathbb{R}_{>0} H : P n − 1 ( Q ) → R > 0 ∣ z ∣ = max 1 ≤ i ≤ n ∣ z i ∣ |z| = \max_{1 \le i \le n}|z_i| ∣ z ∣ = max 1 ≤ i ≤ n ∣ z i ∣
N P n − 1 ( B ) = 2 n − 1 ζ ( n ) B n + O n ( B n − 1 ( log B ) b n ) , N_{\mathbb{P}^{n-1}}(B) = \frac{2^{n-1}}{\zeta(n)}B^n + O_n(B^{n-1}(\log B)^{b_n}),
N P n − 1 ( B ) = ζ ( n ) 2 n − 1 B n + O n ( B n − 1 ( log B ) b n ) ,
\quad
b n : = { 1 , if n = 2 , 0 , if n > 2. b_n := \left\{\begin{matrix} 1, & \text{if } n = 2, \\ 0, & \text{if } n > 2. \end{matrix}\right.
b n := { 1 , 0 , if n = 2 , if n > 2.
如果再结合Thm 1,我们还能得到
N V ( B ) ∼ c ‾ f B n − d , N_V(B) \sim \overline{c}_f B^{n-d},
N V ( B ) ∼ c f B n − d ,
令c f c_f c f
c ‾ f = 1 2 ζ ( n − d ) − 1 c f . \overline{c}_f = \frac{1}{2} \zeta(n-d)^{-1} c_f.
c f = 2 1 ζ ( n − d ) − 1 c f .
而这个的证明将在练习2 中稍微写一点,事实上和Thm 2的证明是完全一样的,因此上面也就没有给出Thm 2的证明了.
Birch定理的推广
Birch定理,也就是Thm 1的证明用到了Hardy-Littlewood圆法,争取早点能了解一下圆法的基本形式嗷.而该定理是Birch在1961年给出的,在这么长一段时间里,也是得到了进一步的推广,下面稍微记录一些,或许之后细看论文时能get到一些新的角度.
在局部域有非平凡解的条件几乎是不可质疑的,或许也有一般的结论可以将无平凡解的情况也包含进去,此处我也没有再去深挖了.
但是"n > ( d − 1 ) 2 d n > (d-1)2^d n > ( d − 1 ) 2 d d = 3 d=3 d = 3 [ 4 ] {[4]} [ 4 ] n ≥ 10 n \ge 10 n ≥ 10
猜想1:
\quad n n n d d d f f f d = 3 d=3 d = 3 n ≥ 7 n \ge 7 n ≥ 7
而对于一般的d d d n ≥ C ⋅ d 2 n \ge C\cdot d^2 n ≥ C ⋅ d 2
而对于non-singular条件的放弱,也由W. M. Schmidt[ 5 ] [ 6 ] {[5][6]} [ 5 ] [ 6 ]
除此之外,还可以将Thm 1推广至方程组以及非齐次方程的情况,其中后者更加复杂一些,我这里还是先聚焦于单个齐次方程的处理,因此这部分就没有记录更多内容.
丢番图几何与解析数论的交叉
在2.4节,我们发现Birch定理的推广可能与一些解析数论问题有关,但事实上,也有很多解析数论问题也可以用丢番图几何的角度去分析.本章除了是Browning书上给出的情况之外,仍然是收集了目前已有的一些进展与突破.
Paucity Results
仿照Fermat猜想的形式,Euler曾猜想
x 1 k + ⋯ + x k − 1 k = x k k x_1^k + \cdots + x_{k-1}^k = x_k^k
x 1 k + ⋯ + x k − 1 k = x k k
没有非平凡解.当k = 3 k=3 k = 3 k = 4 k=4 k = 4 k = 5 k=5 k = 5
对于Euler猜想进行一些修正,便有这样的一个仍然开放的猜想:
猜想2(Lander-Parkin-Selfridfe猜想):
\quad
∑ i = 1 n x i k = ∑ j = 1 m y j k , \sum_{i = 1}^n x_i^k = \sum_{j = 1}^m y_j^k,
i = 1 ∑ n x i k = j = 1 ∑ m y j k ,
\quad
n + m ≥ k . n+m \ge k.
n + m ≥ k .
这里再考虑LPS猜想的一个特例,也就是考虑
x 1 d + x 2 d = x 3 d + x 4 d ( 6 ) x_1^d + x_2^d = x_3^d + x_4^d \quad (6)
x 1 d + x 2 d = x 3 d + x 4 d ( 6 )
的非平凡解数,有猜想:
猜想3:
\quad d ≥ 5 d \ge 5 d ≥ 5
假设(6)中x 1 d + x 2 d = x 3 d + x 4 d = n ≤ x x_1^d + x_2^d = x_3^d + x_4^d = n \le x x 1 d + x 2 d = x 3 d + x 4 d = n ≤ x ν d ( x ) \nu_d(x) ν d ( x ) d ≥ 5 d \ge 5 d ≥ 5 x > 0 x > 0 x > 0 ν d ( x ) \nu_d(x) ν d ( x ) x 2 / d + ε x^{2/d+\varepsilon} x 2/ d + ε [ 7 ] {[7]} [ 7 ]
ν d ( x ) ≪ d , ε x 3 d − 3 2 + ε . \nu_d(x) \ll_{d,\varepsilon} x^{3d^{-\frac{3}{2}} + \varepsilon}.
ν d ( x ) ≪ d , ε x 3 d − 2 3 + ε .
并且貌似这个结果已经是非常好的一个结果了.在2020年,Castryck,Cluckers,Dittmann,Nguyen去掉了Salberger结果中的B ε B^\varepsilon B ε
Waring’s Problem
而Waring问题也能与丢番图几何联系起来.这里有两个略有不同的概念 :对于任意 正整数,其都可以表示为至多g ( k ) g(k) g ( k ) k k k 充分大 的正整数,其都可以表示为至多G ( k ) G(k) G ( k ) k k k k k k g ( k ) g(k) g ( k ) g ( k ) g(k) g ( k ) G ( k ) G(k) G ( k )
例如我们已经可知的是g ( 2 ) = G ( 2 ) = 4 g(2)=G(2)=4 g ( 2 ) = G ( 2 ) = 4 g ( 4 ) = 19 g(4)=19 g ( 4 ) = 19 G ( 4 ) = 16 G(4)=16 G ( 4 ) = 16 k = 3 k=3 k = 3 g ( 3 ) = 9 g(3) = 9 g ( 3 ) = 9 G ( 3 ) G(3) G ( 3 ) 4 ≤ G ( 3 ) ≤ 7 4 \le G(3) \le 7 4 ≤ G ( 3 ) ≤ 7 G ( 3 ) G(3) G ( 3 ) 4 4 4
目前对于G ( k ) G(k) G ( k )
G ( k ) ≤ k ( log k + log log B + O k ( 1 ) ) , G(k) \le k(\log k + \log\log B + O_k(1)),
G ( k ) ≤ k ( log k + log log B + O k ( 1 )) ,
而这也与Vinogradov’s Mean Value Theorem有着很紧密的联系.此外通过一个很简单的讨论(练习1.4 )可以知道的是
g ( k ) ≥ 2 k + [ ( 3 2 ) k ] − 2 , g(k) \ge 2^k + \left[ \left( \frac{3}{2} \right)^{k} \right] - 2,
g ( k ) ≥ 2 k + [ ( 2 3 ) k ] − 2 ,
可以看到g ( k ) g(k) g ( k ) G ( k ) G(k) G ( k )
在处理Waring问题的过程中,要利用到圆法来处理
x 1 k + ⋯ + x s k = x s + 1 k + ⋯ + x 2 s k . x_1^k + \cdots + x_s^k = x_{s+1}^k + \cdots + x_{2s}^k.
x 1 k + ⋯ + x s k = x s + 1 k + ⋯ + x 2 s k .
但如果承认Manin猜想,便可以得到上式在0 ≤ x 1 , ⋯ , x 2 s ≤ B 0 \le x_1, \cdots, x_{2s} \le B 0 ≤ x 1 , ⋯ , x 2 s ≤ B O k , s , ε ( B 2 s − k + ε ) O_{k,s,\varepsilon}(B^{2s-k+\varepsilon}) O k , s , ε ( B 2 s − k + ε ) G ( k ) G(k) G ( k )
G ( k ) ≤ max { 2 k + 1 , Γ 0 ( k ) } , G(k) \le \max\{ 2k+1, \Gamma_0(k) \},
G ( k ) ≤ max { 2 k + 1 , Γ 0 ( k )} ,
其中Γ 0 ( k ) \Gamma_0(k) Γ 0 ( k ) x 1 k + ⋯ + x s k = n x_1^k + \cdots + x_s^k = n x 1 k + ⋯ + x s k = n Q p s \mathbb{Q}_p^s Q p s s s s
Vinogradov’s Mean Value Theorem
在前面的内容中就有几处已经和这个定理已经扯上了关系,该定理的作用在指数和求和中也发挥了重要的作用,而且这也是Intro部分介绍的问题中有重大突破的一个定理,因此后面应该要仔细研究研究才行.
考虑由如下的一组方程定义的projective variety V k , s ⊂ P 2 s − 1 V_{k,s} \subset \mathbb{P}^{2s-1} V k , s ⊂ P 2 s − 1
x 1 j + ⋯ + x s j = x s + 1 j + ⋯ + x 2 s j , ( 1 ≤ j ≤ k ) . ( 7 ) x_1^j + \cdots + x_s^j = x_{s+1}^j + \cdots + x_{2s}^j,\ (1 \le j \le k).\quad (7)
x 1 j + ⋯ + x s j = x s + 1 j + ⋯ + x 2 s j , ( 1 ≤ j ≤ k ) . ( 7 )
再记其在0 ≤ x 1 , ⋯ , x 2 s ≤ B 0 \le x_1, \cdots, x_{2s} \le B 0 ≤ x 1 , ⋯ , x 2 s ≤ B N k , s ( B ) N_{k,s}(B) N k , s ( B )
容易知道的是(7)有以下这些平凡解
{ x 1 , ⋯ , x s } = { x s + 1 , ⋯ , x 2 s } , \{x_1, \cdots, x_s\} = \{ x_{s+1}, \cdots, x_{2s} \},
{ x 1 , ⋯ , x s } = { x s + 1 , ⋯ , x 2 s } ,
因此显然有N k , s ( B ) ≫ B s N_{k,s}(B) \gg B^s N k , s ( B ) ≫ B s 非平凡解 数量的影响.
记H : = { h ∈ Z k : ∣ h j ∣ ≪ B j , j = 1 , ⋯ , k } \mathcal{H} := \{ \mathbf{h} \in \mathbb{Z}^k : |h_j| \ll B^{j}, j=1,\cdots,k \} H := { h ∈ Z k : ∣ h j ∣ ≪ B j , j = 1 , ⋯ , k }
U h ( k , s ) : = { x ∈ Z 2 s : 0 ≤ x 1 , ⋯ , x 2 s ≤ B , ∑ i = 1 s ( x j k − x s + j k ) = h j , j = 1 , ⋯ , k } . \mathcal{U}_{\mathbf{h}}(k,s) := \{ \mathbf{x} \in \mathbb{Z}^{2s} : 0 \le x1, \cdots, x_{2s} \le B, \sum_{i=1}^{s} (x_j^k - x_{s+j}^k) = h_j, j = 1, \cdots, k \}.
U h ( k , s ) := { x ∈ Z 2 s : 0 ≤ x 1 , ⋯ , x 2 s ≤ B , i = 1 ∑ s ( x j k − x s + j k ) = h j , j = 1 , ⋯ , k } .
于是有下面的估计[ 8 , e q 7.4 ] {[8, eq 7.4]} [ 8 , e q 7.4 ]
B 2 s ≪ ∑ h ∈ H # U h ( k , s ) ≪ ∑ h ∈ H N k , s ( B ) ≪ B k ( k + 1 ) 2 N k , s ( B ) , B^{2s} \ll \sum_{\mathbf{h}\in\mathcal{H}}\#\mathcal{U}_{\mathbf{h}}(k,s) \ll \sum_{\mathbf{h}\in\mathcal{H}}N_{k,s}(B) \ll B^{\frac{k(k+1)}{2}}N_{k,s}(B),
B 2 s ≪ h ∈ H ∑ # U h ( k , s ) ≪ h ∈ H ∑ N k , s ( B ) ≪ B 2 k ( k + 1 ) N k , s ( B ) ,
因此我们最终得到了如下的估计式:
N k , s ( B ) ≫ B s ＋ B 2 s − k ( k + 1 ) 2 . ( 8 ) N_{k,s}(B) \gg B^s ＋ B^{2s-\frac{k(k+1)}{2}}. \quad (8)
N k , s ( B ) ≫ B s ＋ B 2 s − 2 k ( k + 1 ) . ( 8 )
而长期的一个Main Conjecture,但是现在已经得到证明,其表述如下:
Thm 3(Vinogradov’s Mean Value Theorem):
\quad ε > 0 \varepsilon > 0 ε > 0
N k , s ( B ) ≪ B s ＋ B 2 s − k ( k + 1 ) 2 + ε . ( 9 ) N_{k,s}(B) \ll B^s ＋ B^{2s-\frac{k(k+1)}{2} + \varepsilon}. \quad (9)
N k , s ( B ) ≪ B s ＋ B 2 s − 2 k ( k + 1 ) + ε . ( 9 )
结合(8)和(9)来看,这也就是在说N k , s ( B ) N_{k,s}(B) N k , s ( B ) [ 9 ] {[9]} [ 9 ] [ 10 ] {[10]} [ 10 ]
总之,无论从什么方面来看,这个定理以及其相关的内容应该是很值得学习的一块了,加入我的必看清单(虽然我并没有必看清单,也不知道什么时候开始看 ).
Small Solutions
现在令n ≥ 3 n \ge 3 n ≥ 3 f ∈ Z [ x 1 , ⋯ , x n ] f \in \mathbb{Z}[x_1, \cdots, x_n] f ∈ Z [ x 1 , ⋯ , x n ] d ≥ 2 d \ge 2 d ≥ 2 ∥ f ∥ \|f\| ∥ f ∥ f f f S f ≠ ∅ S_f \neq \emptyset S f = ∅ Λ n ( f ) \Lambda_n(f) Λ n ( f ) λ \lambda λ x ∈ S f \mathbf{x}\in S_f x ∈ S f ∣ x ∣ = max 1 ≤ i ≤ n ∣ x i ∣ ≤ λ |\mathbf{x}| = \max_{1\le i\le n} |x_i| \le \lambda ∣ x ∣ = max 1 ≤ i ≤ n ∣ x i ∣ ≤ λ
当d = 2 d=2 d = 2 f = Q f=Q f = Q [ 11 ] {[11]} [ 11 ]
Q 0 ( x ) : = x 1 2 − ( x 2 − c x 1 ) 2 − ⋯ − ( x n − c x n − 1 ) Q_0(\mathbf{x}) := x_1^2 - (x_2-cx_1)^2 - \cdots - (x_n-cx_{n-1})
Q 0 ( x ) := x 1 2 − ( x 2 − c x 1 ) 2 − ⋯ − ( x n − c x n − 1 )
便可以验证.但是对于足够一般 的二次型而言,其判别式Δ ≍ ∥ Q ∥ n \Delta \asymp \|Q\|^n Δ ≍ ∥ Q ∥ n [ 12 ] {[12]} [ 12 ]
Λ n ( Q ) ≪ ε , n ∥ Q ∥ n − 1 2 ( ∥ Q ∥ n + ε ∣ Δ ∣ 1 2 ( n − 5 − α n ) ) 1 n − 3 − α n , \Lambda_n(Q) \ll_{\varepsilon, n} \|Q\|^{\frac{n-1}{2}} \left( \frac{\|Q\|^{n+\varepsilon}}{|\Delta|^{\frac{1}{2}(n-5-\alpha_n)}} \right)^{\frac{1}{n-3-\alpha_n}},
Λ n ( Q ) ≪ ε , n ∥ Q ∥ 2 n − 1 ( ∣Δ ∣ 2 1 ( n − 5 − α n ) ∥ Q ∥ n + ε ) n − 3 − α n 1 ,
其中ε \varepsilon ε 0 0 0
α n = { 1 , if n is even , 0 , if n is odd . \alpha_n = \begin{cases}
1, & \text{if } n \text{ is even}, \\
0, & \text{if } n \text{ is odd}.
\end{cases} α n = { 1 , 0 , if n is even , if n is odd .
在这个命题的证明之中,便利用到了丢番图几何的一些结论,很直接的就是对于一个从R n \mathbb{R}^n R n R ≥ 0 \mathbb{R}_{\ge 0} R ≥ 0 ω \omega ω
N ω ( Q ; B ) : = ∑ x ∈ Z n Q ( x ) = 0 ω ( B − 1 x ) N_{\omega}(Q;B) := \sum_{\substack{\mathbf{x} \in \mathbb{Z}^n \\ Q(x) = 0}} \omega(B^{-1}\mathbf{x})
N ω ( Q ; B ) := x ∈ Z n Q ( x ) = 0 ∑ ω ( B − 1 x )
的渐近估计式,当然这里也是用的圆法得出的.
而对于d = 3 d=3 d = 3 n n n C C C n n n 充分多 的时候,Pitman[ 13 ] {[13]} [ 13 ]
Λ n ( C ) ≪ ε , n ∥ C ∥ 25 6 + ε . \Lambda_n(C) \ll_{\varepsilon, n} \|C\|^{\frac{25}{6}+\varepsilon}.
Λ n ( C ) ≪ ε , n ∥ C ∥ 6 25 + ε .
如果变量数很少的时候,在n ≥ 17 n \ge 17 n ≥ 17 [ 14 ] {[14]} [ 14 ]
Λ n ( C ) ≪ ε , n ∥ C ∥ 3.6 × 10 5 + ε . \Lambda_n(C) \ll_{\varepsilon, n} \| C \|^{ 3.6 \times 10^5 + \varepsilon}.
Λ n ( C ) ≪ ε , n ∥ C ∥ 3.6 × 1 0 5 + ε .
当然,对于两种情况下,Browning,Heath-Brown,Wooley,Mormon等人也给出了进一步的优化,但这里就不进一步记录了.
Divisor Problems
丢番图几何中的问题有一些最后就划归为了除数问题最终得以解决,而这个问题在解析数论中也是一个很自然很重要的问题.我们定义一般形式下的估计式为φ \varphi φ φ ( n ) = τ ( n ) = ∑ d ∣ n 1 \varphi(n) = \tau(n) = \sum_{d|n} 1 φ ( n ) = τ ( n ) = ∑ d ∣ n 1 F ∈ Z [ x 1 , x 2 ] F \in \mathbb{Z}[x_1, x_2] F ∈ Z [ x 1 , x 2 ] d d d
书中说d = 1 , 2 d=1,2 d = 1 , 2 d = 3 d=3 d = 3 [ 15 ] {[15]} [ 15 ] 指数和 证明了
D ( X ; τ , F ) = c F X 2 log X + c F ′ X 2 + O ε , F ( X 27 14 + ε ) . D(X;\tau,F) = c_FX^2\log X + c_F'X^2 + O_{\varepsilon,F}(X^{\frac{27}{14}+\varepsilon}).
D ( X ; τ , F ) = c F X 2 log X + c F ′ X 2 + O ε , F ( X 14 27 + ε ) .
然后Daniel[ 16 ] {[16]} [ 16 ] 数的几何 "(geometry of number)证明了d = 4 d=4 d = 4
D ( X ; τ , F ) = c F X 2 log X + O F ( X 2 log log X ) . D(X;\tau,F) = c_FX^2\log X + O_F(X^2\log\log X).
D ( X ; τ , F ) = c F X 2 log X + O F ( X 2 log log X ) .
并且Daniel还将Greaves的结果优化至15 8 \frac{15}{8} 8 15
对于除数问题仍然是一个相当困难的问题,在2020年Frei和Sofos[ 17 ] {[17]} [ 17 ]
第一章练习题!
接下来是第一章的几个习题,基本上都比较简单,也就只简单记一下想法,毕竟现在也写的够长了.
练习1.1:
\quad K K K d d d Q \mathbb{Q} Q ω 1 , ⋯ , ω d \omega_1, \cdots, \omega_d ω 1 , ⋯ , ω d p p p ( p ) (p) ( p ) O K \mathcal{O}_K O K α = y 1 ω 1 + ⋯ + y d ω d ∈ O K \alpha = y_1\omega_1 + \cdots + y_d\omega_d \in \mathcal{O}_K α = y 1 ω 1 + ⋯ + y d ω d ∈ O K p ∣ N K / Q ( α ) p \mid N_{K/\mathbb{Q}}(\alpha) p ∣ N K / Q ( α ) p ∣ α p \mid \alpha p ∣ α
解答:
\quad p ∣ α p\mid\alpha p ∣ α p ∣ N K / Q ( α ) p\mid N_{K/\mathbb{Q}}(\alpha) p ∣ N K / Q ( α ) ( p ) (p) ( p ) O K \mathcal{O}_K O K O K / p O K ≅ F p d \mathcal{O}_K / p\mathcal{O}_K \cong \mathbb{F}_{p^d} O K / p O K ≅ F p d
N K / Q ( α ) ( m o d p ) = N F p d / F p ( α ‾ ) = α ‾ ⋅ α ‾ p ⋅ α ‾ p 2 ⋅ ⋯ ⋅ α ‾ p d − 1 = α ‾ p d − 1 p − 1 . N_{K/\mathbb{Q}}(\alpha) \pmod{p} = N_{\mathbb{F}_{p^d} / \mathbb{F}_p}(\overline{\alpha}) = \overline{\alpha} \cdot \overline{\alpha}^p \cdot \overline{\alpha}^{p^2} \cdot \cdots \cdot \overline{\alpha}^{p^{d-1}} = \overline{\alpha}^{\frac{p^d-1}{p-1}}.
N K / Q ( α ) ( mod p ) = N F p d / F p ( α ) = α ⋅ α p ⋅ α p 2 ⋅ ⋯ ⋅ α p d − 1 = α p − 1 p d − 1 .
\quad p ∣ N K / Q ( α ) p \mid N_{K/\mathbb{Q}}(\alpha) p ∣ N K / Q ( α ) α ‾ = 0 \overline{\alpha} = 0 α = 0 α ∈ ( p ) \alpha \in (p) α ∈ ( p ) p ∣ α p\mid\alpha p ∣ α □ \square □
练习1.2
\quad V ⊂ P n − 1 V \subset \mathbb{P}^{n-1} V ⊂ P n − 1 f f f d d d n > ( d − 1 ) 2 d n > (d-1)2^d n > ( d − 1 ) 2 d
N V ( B ) ∼ c ‾ f B n − d , N_V(B) \sim \overline{c}_f B^{n-d},
N V ( B ) ∼ c f B n − d ,
\quad
c ‾ f = c f 2 ζ ( n − d ) . \overline{c}_f = \frac{c_f}{2\zeta(n-d)}.
c f = 2 ζ ( n − d ) c f .
解答:
\quad
N V ( B ) = 1 2 ∑ k = 1 ∞ μ ( k ) N ( f ; k − 1 B ) ∼ 1 2 ∑ k = 1 ∞ μ ( k ) c f k − ( n − d ) B n − d = 1 2 c f B n − d ∑ k = 1 ∞ μ ( k ) k n − d = 1 2 c f B n − d ∏ p ( 1 − 1 p n − d ) = c f 2 ζ ( n − d ) B n − d . \begin{array}{ll}
N_V(B) & = \frac{1}{2} \sum_{k=1}^{\infty} \mu(k) N(f; k^{-1}B) \\ \\
& \sim \frac{1}{2} \sum_{k=1}^{\infty} \mu(k) c_f k^{-(n-d)} B^{n-d} \\ \\
& = \frac{1}{2} c_f B^{n-d} \sum_{k=1}^{\infty} \frac{\mu(k)}{k^{n-d}} \\ \\
& = \frac{1}{2} c_f B^{n-d} \prod_{p} \left(1 - \frac{1}{p^{n-d}}\right) \\ \\
& = \frac{c_f}{2 \zeta(n-d)} B^{n-d}.
\end{array} N V ( B ) = 2 1 ∑ k = 1 ∞ μ ( k ) N ( f ; k − 1 B ) ∼ 2 1 ∑ k = 1 ∞ μ ( k ) c f k − ( n − d ) B n − d = 2 1 c f B n − d ∑ k = 1 ∞ k n − d μ ( k ) = 2 1 c f B n − d ∏ p ( 1 − p n − d 1 ) = 2 ζ ( n − d ) c f B n − d .
\quad □ \square □
练习1.3:
\quad ν d ( x ) ≪ d , ε x 2 / d + ε \nu_d(x) \ll_{d,\varepsilon} x^{2/d+\varepsilon} ν d ( x ) ≪ d , ε x 2/ d + ε
解答:
\quad
ν d ( x ) = # { n ∈ N ≤ x : n = a d + b d = m d + n d } . \nu_d(x) = \#\{ n \in \mathbb{N}_{\le x} : n = a^d+b^d = m^d+n^d\}.
ν d ( x ) = # { n ∈ N ≤ x : n = a d + b d = m d + n d } .
\quad
ν d ( x ) ≤ # { n ∈ N ≤ x : n = a d + b d } ≤ # { a ∈ N : a d ≤ x } ⋅ # { b ∈ N : b d ≤ x } ≪ x 2 / d + ε . \nu_d(x) \le \#\{ n \in \mathbb{N}_{\le x} : n = a^d+b^d\} \le \#\{a \in \mathbb{N}: a^d \le x\} \cdot \#\{b \in \mathbb{N}: b^d \le x\} \ll x^{2/d+\varepsilon}.
ν d ( x ) ≤ # { n ∈ N ≤ x : n = a d + b d } ≤ # { a ∈ N : a d ≤ x } ⋅ # { b ∈ N : b d ≤ x } ≪ x 2/ d + ε .
\quad □ \square □
练习1.4:
For k ∈ N k \in \mathbb{N} k ∈ N g ( k ) g(k) g ( k ) k k k k k k
n = 2 k [ ( 3 2 ) k ] − 1 , n = 2^k \left[ \left( \frac{3}{2} \right)^{k} \right] - 1,
n = 2 k [ ( 2 3 ) k ] − 1 ,
deduce that
g ( k ) ≥ 2 k + [ ( 3 2 ) k ] − 2. g(k) \ge 2^k + \left[ \left( \frac{3}{2} \right)^{k} \right] - 2.
g ( k ) ≥ 2 k + [ ( 2 3 ) k ] − 2.
解答:
\quad 2 k < n < 3 k 2^k < n < 3^k 2 k < n < 3 k n n n 0 k , 1 k , 2 k 0^k, 1^k, 2^k 0 k , 1 k , 2 k
n = 2 k ( [ ( 3 2 ) k ] − 1 ) + 1 k ( 2 k − 1 ) , n = 2^k \left(\left[ \left( \frac{3}{2} \right)^{k} \right] - 1\right) + 1^k (2^k - 1),
n = 2 k ( [ ( 2 3 ) k ] − 1 ) + 1 k ( 2 k − 1 ) ,
\quad □ \square □
总结
目前第一章的内容就记录到这,后面还有非常非常多的东西等我去学,简单做一个规划吧.首先是Browning的这本书可以继续深入学习,第二章应该会过得比较快,毕竟在前段时间已经和它打过不少照面了.其次是后面这么多的参考文献可以再了解了解,其中比较感兴趣的是[ 2 ] [ 9 ] [ 15 ] [ 16 ] {[2][9][15][16]} [ 2 ] [ 9 ] [ 15 ] [ 16 ] [ 6 ] {[6]} [ 6 ] 怎么上个研究生这么多事啊,真学不过来了😭 ).
以及我的电脑坏了,不知道什么时候才来上传到爱音学报上,悲!
Remark:时隔半个月(9.19)后的补充:终于是给传上了!😭
(待)参考资料
[1] Browning T D. Quantitative arithmetic of projective varieties[M]. Springer Science & Business Media, 2009.
[2] Browning T D, Matthiesen L. Norm forms for arbitrary number fields as products of linear polynomials[C]//Annales Scientifiques de l’Ecole Normale Superieure. 2017, 50(6): 1383-1446.
[3] Wikipedia. Quasi-algebraically closed field[Z]. https://en.wikipedia.org/wiki/Quasi-algebraically_closed_field .
[4] Heath-Brown D R. Cubic forms in ten variables[J]. Proceedings of the London Mathematical Society, 1983, 3(2): 225-257.
[5] Schmidt W M. The density of integer points on homogeneous varieties[J]. 1985.
[6] Schmidt W M. Diophantine approximations and Diophantine equations[M]. Springer, 2006.
[7] Salberger P. On the density of rational and integral points on algebraic varieties[J]. 2007.
[8] Vaughan R C. the Hardy-Littlewood method[M]. Cambridge: Cambridge University Press, 2003.
[9] Bourgain J, Demeter C, Guth L. Proof of the main conjecture in Vinogradov’s mean value theorem for degrees higher than three[J]. Annals of Mathematics, 2016: 633-682.
[10] Wooley T D. Nested efficient congruencing and relatives of Vinogradov’s mean value theorem[J]. Proceedings of the London Mathematical Society, 2019, 118(4): 942-1016.
[11] Cassels J W S. Bounds for the least solutions of homogeneous quadratic equations[C]//Mathematical Proceedings of the Cambridge Philosophical Society. Cambridge University Press, 1955, 51(2): 262-264.
[12] Browning T D, Dietmann R. On the representation of integers by quadratic forms[J]. Proceedings of the London Mathematical Society, 2008, 96(2): 389-416.
[13] Pitman J. Cubic inequalities[J]. Journal of the London Mathematical Society, 1968, 1(1): 119-126.
[14] Browning T D, Dietmann R, Elliott P. Least zero of a cubic form[J]. Mathematische Annalen, 2012, 352(3): 745-778.
[15] Greaves G. On the divisor-sum problem for binary cubic forms[J]. Acta arithmetica, 1970, 17: 1-28.
[16] Daniel S. On the divisor-sum problem for binary forms[J]. 1999.
[17] Frei C, Sofos E. Generalised divisor sums of binary forms over number fields[J]. Journal of the Institute of Mathematics of Jussieu, 2020, 19(1): 137-173.
