随机温习...
(接前: 02 01 31) “执行定理” (Th1.6)的证明(e). .
温习要用到的命题5.7(叙述). 先看图解:
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A L
Tx
X B|Λ
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注: 此图蕴含8个条件: 5 个常规, 3个扩展.
* (X, B) ~ proj., eps-lc. (高配)
* A ~ very ample, Aᵈ ≤ r. (有界重)
* L ≥ 0, R-divisor.
* Λ ≥ 0, Q-divisor, nΛ integral.
* A - B, Λ, L ~ ample.
. (X, Λ) ~ lc near x.
. T ~ lc place of (X, Λ), with center x on X.
. a(T, X, B) ≤ 1.
--- 结果: 对任何ν: U --> X 带 T (on U), 有 μTνL* ≤ q.
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前情回顾.
---- 第2~4段凑齐了5.9的7个条件, 这就“激活”了命题5.9.
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By Proposition 5.9, there is a Q-divisor Λ ≥ 0 such that nΛ is integral, mA - Λ is ample, (X, Λ) is lc near x, and T is a lc place of (X, Λ).
---- 命题5.9 就是为了构造如上 Λ.
注:此 Λ 是 5.7 (甲方) “发包”.
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Replacing A, C, L, s with 2mA, 2mC, 2mL, s/2m, respectively, and replacing r accordingly, we can assume A - B - sL and A - Λ are ample.
---- 替换即代入: A <~ 2mA, C <~2mC, L <~ 2mL, s <~ s/2m.
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讨论: ample 的推导.
---- ample 有点像比大小(大 - 小 = ample).
1) A - Λ <~ 2mA - Λ.
由 mA - Λ ample (见前句) ==> 2mA - Λ ample.
2) A - B -sL <~ 2mA - B - s/2m · 2mL
= 2mA - B - sL = (mA - B) + (mA -sL)
上面末尾两项加起来 ample, 因为:
A - B ample ==> mA - B ample
A - L ample ==> mA - L ample ==> mA -sL ample (因为 s ≤ 1).
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评论: 原作不提及推导, 而采用 “we can assume...” 或出于严谨或简明.(?)
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加评: 5.9 是给 5.7 做准备.
---- 上面这句是凑 ample 条件, 替换后:
---- A - L 和 A - Λ 都是 ample.
---- 原作特意提出 A - (B + sL) ample, 意味着 B + sL 相当于 5.7 中 B.
---- 恰巧, a(X, T, B + sL) = eps' ≤1.
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特评: 以上凑齐了5.7的条件, 这就激活了命题5.7.
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Applying Proposition 5.7 to (X, B + sL), there is a natural number q depending only on d, r, n, eps' such that if ν: U --> X is a resolution so that T is a divisor on U, then μTνL* ≤ q.
---- 命题5.7 作用于 (X, B + sL).
(通常来说, 命题的应用不会照搬, 而是套用).
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Pick such a resolution.
---- 补的这句很必要.
(澄清了上句中 resolution 的来源)
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评论: 这个第5段, 接连调用了命题 5.9 和 5.7, 得到一个不等式.
(5.9 是为 5.7 准备条件).
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小结: 此段没有实质困难.
符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈ ∉ ∪ ∩ ⊆ ⊇ ⊂ ⊃ ≤ ≥ ⌊ ⌋ ⌈ ⌉ ≠ ≡ ⁻⁰ ¹ ² ³ ᵈ ₀ ₁ ₂ ₃ ᵢ .