为何多数人会惧怕数学,的特征是配对及奇异类

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本期开始分组发送邮件,搭载数学类学院等链接。必威betway 5

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今日学院:暂无。||新闻 ||....Perfectoidᴺᴱᵂ....

今日学院:暂无。||符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈∉∪ ∩⊆ ⊇ ⊂ ⊃≤ ≥⌊ ⌋⌈ ⌉≠⁻⁰¹ ² ³ᵈ₀ ₁₂₃ᵢₐ.

今日学院:暂无。||符号大全、上下标.|| 常用:↑↓νπ ΓΔΛΘΩμφΣ∈∉∪ ∩⊆ ⊇ ⊂ ⊃≤ ≥⌊ ⌋⌈ ⌉≠⁻⁰¹ ² ³ᵈ₀ ₁₂₃ᵢₐ.

今日学院:暂无。||新闻 ||符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈∉∪ ∩⊆ ⊇ ⊂ ⊃≤ ≥⌊ ⌋⌈ ⌉≠⁻⁰¹ ² ³ᵈ₀ ₁₂₃ᵢ .





问题众多,可围而猎之。

(接前:020131) 命题5.7的证明.

如何“挖出”作者的原始思路?

“主节点”的特征是配对及奇异类型.

(接前:232118)温习:命题5.7的证明.

Step5. 第二段 .

(接前:171413)温习:命题5.7的证明.

(接前:030201) 命题5.7的证明.

  1. MMP ~>π: Y --> X.

Let AYbe the pullback of A. By boundedness of the length of extremal rays [17] and by the base point free theorem, KY ΓY 3dAYis nefand big and semi-ample, globally.

Step3 更替“诸元”, 旨在向命题5.5靠拢.

Step6. 第一段 .

3.Kv Γv≡ G/X, G ~ exceptional/X.

---- 上一段引入了 π: Y --> X.

特注:之前在温习中领悟到,证明的轴心是 “相的进化”.(构造“相”的方法是核心技术所在).

Now write KY B~ RY= π* where RYis exceptional over X.

  1. q[G] contracted..

---- 这里给 A 做个回拉得到 AY.

Step4ab处理 “禁”的条件.

---- Step5 得到 π: Y --> X.

5.π~ isomorphismc.c.

---- 随后提及两个调用:

  1. Th1.6, hs ==> (X, B 2tB) klto.c.

---- 这里出现 KY,意味着配对 .

6.Kv Γv≡ tB~ F/X;T≠ [tB~ F].

1.length of extremal rays [17];

注:hs~hypersections;o.c.~ outside finitely many closed points.

---- 从后文看,“?” 正是由B~ RY替代.

  1. T not contracted.

  2. Len[extre-rays] 有界[17]; base point free .

  3. KY ΓY 3dAY~ nb & semi-ample .

  4. 取 0≤ DY~R1/t(KY ΓY 3dAY).

2.base point free theorem.

  1. (X, B tB) is eps/2-lco.c.

---- 从原作这个写法看,似是调用已有公式.

(DY诸分量系数≤ 1 - eps).

---- 由此引出关键量:KY ΓY 3dAY.

注:ψ标识 V --> X 的 log resolutionψof .

We then haveKY RY DY- 1/tFY= KY RY B~ π*H = π*(Kx B H) = π*.

  1. KY ΓY=π* tB~ FY.

---- 这里的 d 是当前的维度.

  1. 定义边界ΓV=γ·O.

---- Step5 得到 DY= π*H B~ 1/tFY.

(注意下标为 Y,而不是 V).

----ΓY是从哪里来的?(Step4 只构造了个Γv).

----γ= (1 t, 1 - eps/4, 1 - a).

---- 这一串等式的实际起点也许是π*:

  1. 1/t (KY ΓY 3dAY)

---- Step5 第一段对 Kv Γv运行 MMP.

----O= , T).

---- Step5 得到 D = H B,代入π*:

= 1/tπ*(Kx B 3dA) B~ 1/t FY.

---- 推测ΓY是从那里出来的.

---- a = a.

----π* =π*(Kx B H) .

  1. DY=π*H B~ 1/t FY.

(原作为简省笔墨,惯于用到时直接提及)

  1. eps-lc, μTΓv = 1 - a.

  2. Kv Γv =ψ* tB~ F.

假定π*(Kx B H) =π* π*H.

H ~R1/t (Kx B 3dA).

Pick a general 0≤ DY~R1/t(KY ΓY 3dAY)with coefficients≤ 1 - eps.

---- F:=·E.

---- 将开头第一句的公式代入,得:

  1. D = H B.

---- 引入 DY使之等价于关键量的 1/t 倍.

a=(..,ai,...),E=(...,Ei,...).

----π*(Kx B H) =KY B~ RY π*H.

注:3 ==> 4 ==> 5; 6 ==> 7; 8 ==> 9; 11 ==> 12; 10,12 ==> 13 ==> 14.

---- 系数为何能控制在≤ 1 - eps?

---- Fe.e.on X; T⊄Supp.

---- 上式右端的π*H 出现在 式,变形得:

评论:Step5的主旨是得到 D的显示表达.

Since KY ΓY= π* tB~ FY, we get 1/t(KY ΓY 3dAY) = 1/tπ*(Kx B 3dA) B~ 1/t FY.

7.Kv Γv =ψ*(Kx G.

---- π*H =DY- B~- 1/tFY.将其代入 :

小结:Step5推导不祥,须各个击破...

---- 前半句又体现出原作的“简省”风格.

---- G:=·E T.

KY B~ RY π*H


---- Step4 有相近的式子:

a'=(...,a'i,...); a'i = a(Ei, X, B tB); a'=a(T, X, B tB).

=KY B~ RY (DY- B~- 1/tFY)

符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈∉∪ ∩⊆ ⊇ ⊂ ⊃≤ ≥⌊ ⌋⌈ ⌉≠≡⁻⁰¹ ² ³ᵈ₀ ₁₂₃ᵢ .

Kv Γv = ψ* tB~ F.

---- G exceptional over X.

=KY RY DY- 1/t FY.

Leonhard EulerCarl Friedrich GaussGrothendieck

---- Since 后的表达式该不是直接顺过来的.

  1. dim>0 对某个 i.

  2. Ei =[G] 且系数为正.

---- 最后一个式子就是那一串等式的左端.

Glossary

Abstract8/4

Introduction

Boundedness of singular Fano varieties 8/5

Boundedness of singular Fano varieties 8/6

Boundedness ofsingularFano varieties 8/7

Boundedness ofsingularFano varieties 8/8

Boundedness ofsingularFanovarieties 8/9

Boundedness ofsingularFanovarieties8/9

Jordan property of Cremona groups8/10

Lc thresholds of lR-linear systems 8/11

Lc thresholds of anti-log canonical systems of Fano pairs 8/12

Lc thresholds of anti-log canonical systems of Fano pairs 8/13

Lc thresholds of R-linear systems with bounded degree 8/14

Complements near a divisor8/15

Proposition 5.211/9

Proposition 5.511/5

---- 可能实施了与之前类似的推导.

注:1 ==>2; 3,4,5 ==> 6, 7; 8, 2 ==> 9.

评论:此句中的推导是直截了当的.

---- 后半句的推导暂时不清楚.

评论:4 是构造新的边界,思路待考.

Since is eps-lc, is sub-eps-lc, hence (Y,RY DY- 1/t FY) is sub-eps-lc as DYis general semi-ample with coefficients≤1 - eps, and FY≥0.

Thus we can writeDY= π*H B~ 1/t FYfor some H ~R1/t(Kx B 3dA).

(怎么想到的?似乎有很深重的技巧...).

---- 此句是一个“主节点”,承上启下.

---- 对照以上三句有:

---- O 的分量让人想到 BET.

---- 但原作只讲了结果,推导暂时不清楚.

DY~R1/t(KY ΓY 3dAY)= 1/tπ*(Kx B 3dA) B~ 1/t FY.

---- 4 的构造方法可暂命名为“赌法”.

Therefore, is eps-lc.

---- 此句表明,用 tH 替换橄榄色部分后:

(感到突兀...须从上下文找线索、找入口/逻辑起点).

---- 此句也是一个主节点,但推导不详.

---- DY后面的符号 “~R” 换成了 “=”.

加评:6,7 蕴含的思路待考.

----主节点的特征是 配对及奇异类型.

---- 换句话说,部分等价替换可得到等式.

小结:Step4 温习完毕.(该步骤技术内容较“厚重”)

Moreover, since T is not a component ofB~ FY DY, a = 1 - μT(RY DY- 1/tFY) = 1 -μTRY= 1 -μT = a≤1.

Letting D be the pushdown of DYto X, we get D = H B.

Leonhard EulerCarl Friedrich GaussGrothendieck

---- 此句推导较多,须仔细分析.

---- 之前DY= π*H B~ 1/t FY.

Glossary

Abstract8/4

Introduction

Boundedness of singular Fano varieties 8/5

Boundedness of singular Fano varieties 8/6

Boundedness ofsingularFano varieties 8/7

Boundedness ofsingularFano varieties 8/8

Boundedness ofsingularFanovarieties 8/9

Boundedness ofsingularFanovarieties8/9

Jordan property of Cremona groups8/10

Lc thresholds of lR-linear systems 8/11

Lc thresholds of anti-log canonical systems of Fano pairs 8/12

Lc thresholds of anti-log canonical systems of Fano pairs 8/13

Lc thresholds of R-linear systems with bounded degree 8/14

Complements near a divisor8/15

Proposition 5.211/9

Proposition 5.511/5

---- 前半句突然出现B~ FY DY, 从何而来?

---- pushdown后,右端去除了 π*, ~及1/t FY.


小结:Step5 读写完毕. (之前的第一段做了某种准备,这个第二段的落点是构造 D).

此类突然出现的量,往往源于早先的某个地方,但原作选择暗含在头脑中,用到时才出示,从而引起突兀感.

Leonhard EulerCarl Friedrich GaussGrothendieck

假定:“T is not a component ofB~ FY DY” 等价于T 不是其中各项的分量.

Glossary

Abstract8/4

Introduction

Boundedness of singular Fano varieties 8/5

Boundedness of singular Fano varieties 8/6

Boundedness ofsingularFano varieties 8/7

Boundedness ofsingularFano varieties 8/8

Boundedness ofsingularFanovarieties 8/9

Boundedness ofsingularFanovarieties8/9

Jordan property of Cremona groups8/10

Lc thresholds of lR-linear systems 8/11

Lc thresholds of anti-log canonical systems of Fano pairs 8/12

Lc thresholds of anti-log canonical systems of Fano pairs 8/13

Lc thresholds of R-linear systems with bounded degree 8/14

Complements near a divisor8/15

Proposition 5.211/9

Proposition 5.511/5

---- 当前句的后半段是计算 a.

---- 按定义a = 1 -μT?.

---- 其中 “?” 来自 Kx D 的回拉.

---- 恰好KY RY DY- 1/tFY=π*

(见第二句,等式串抓两头)

---- 上式的粉色部分即“?”. 于是:

----a = 1 -μT(RY DY- 1/tFY).

---- 由假定,DY和 FY对系数无影响:

---- 则1 -μT(RY DY- 1/tFY) = 1 -μTRY.

---- 又由假定,B~对系数无影响:

---- 则1 -μTRY= 1 -μT.

---- 上式右端括弧内的B~ RY来自 Kx B 的回拉 KY B~ RY.

---- 于是1 -μT = a.

---- 而a≤1 .

小结:Step6第一段读写完毕.

----应对突兀的办法是从上下文中找线索,特别是从后文的需要、推导中探寻.


有些地方对于原作是显然的,但对于读者不见得显然,此时可做个猜测性的假定,帮助完成推导.

Leonhard EulerCarl Friedrich GaussGrothendieck

Glossary

Abstract8/4

Introduction

Boundedness of singular Fano varieties 8/5

Boundedness of singular Fano varieties 8/6

Boundedness ofsingularFano varieties 8/7

Boundedness ofsingularFano varieties 8/8

Boundedness ofsingularFanovarieties 8/9

Boundedness ofsingularFanovarieties8/9

Jordan property of Cremona groups8/10

Lc thresholds of lR-linear systems 8/11

Lc thresholds of anti-log canonical systems of Fano pairs 8/12

Lc thresholds of anti-log canonical systems of Fano pairs 8/13

Lc thresholds of R-linear systems with bounded degree 8/14

Complements near a divisor8/15

Proposition 5.211/9

Proposition 5.511/5

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