今日学院:暂无。|| 新闻+ || 符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈ ∉ ∪ ∩ ⊆ ⊇ ⊂ ⊃ ≤ ≥ ⌊ ⌋ ⌈ ⌉ ≠ ⁻⁰ ¹ ² ³ ᵈ ₀ ₁ ₂ ₃ ᵢ ₐ. Step2 第一段 (逐句评论).
Replace A with a general member of |A|.
---- 此举的用意暂不清楚.
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Since nΛ is integral and degAΛ < Aᵈ ≤ r, the pair (X, Supp(Λ + A)) belongs to a bounded family of pairs P depending only on d, r, n.
---- 构造出有界族配对 (X, Supp(Λ + A)). 图解:
A ?
X Λ
---- (X, Λ) 局部 lc (见“附加4”).
---- nΛ 为整系数 (象征文明).
---- 对角线上有两个运算:
1. (度式) 乘法: degAΛ = Aᵈ⁻¹Λ ≤ Aᵈ ≤ r.
2. (支撑) 加法: Supp(Λ + A).
评论:方成于法,法现于方。法在对角,“出新”。
---- degAΛ ≤ r 预示相合乎法度(以侯的权力衡量).
---- Supp(Λ + A) 预示侯相联合,“摄领相事”.
---- (X, Supp(Λ + A)) 属于有界族,即“入格”.
---- 两物并立曰“方”,有“方”必有“法”. 法即映射.
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Thus there exist a log resolution φ: W --> X of (X, Λ) and a very ample divisor Aw ≥ 0 so that if Θw is the sum of the exceptional divisors of φ and the support of the birational transform of Λ, then (W, Θw + Aw) belongs to a bounded family Q of pairs depending only on d, r, n, P.
---- (X, Supp(Λ + A)) 入格 ==>
1. 存在映射φ.
2. 存在Aw(丰大).
---- φ 是回拉式,着意于 W 空间.
---- 对其exceptional divisors 求和,记作Θw.
---- 若Θw = Supp(bir(Λ)), 则(W, Θw + Aw) 入格.
图解:
Aw ?
W Θw
---- φ 的exceptional divisors 求和,我宁愿记作 E(φ).
---- 即 E(φ) = Θw = Supp(bir(Λ)).
---- 原作引入的记号Θw 会让人忍不住问:Θ 是什么角色? (带上下标w,也许只是为了提示与 W的联系?).
评论:总之,给定条件 E(φ) = Supp(bir(Λ)),则 (X, Supp(Λ + A)) 入格 ==> (W, Θw + Aw) 入格.