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本期开始加开窗口,推出科学网特色博主,有用链接等。
打起精神往前走...
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(接上回*)Proof. (of Theorem 1.6) By assumption, |A - M|R ≠ Φ, so A ~R M+N for some N ≥ 0. Thus
lct(X, B, |M|R) ≥ lct(X, B, |M+N|R) = lct(X, B, |A|R)
so it is enough to give a positive lower bound for the right hand side.
注:开始读写Th1.6的证明(共5个自然段)。此定理是本文“主将”。方便起见,段落或进一步细分。
评论:Th1.6即“花费控制定理”(又名“皇帝和他的大内们”*)。文章里最后的较长证明(“最后的拼杀”)。
第一句,由假定, |A - M|R ≠ Φ, 故可有 A ~R M+N,其中 N ≥ 0.
注:这里没有“调用”,估计用到“~”的定义(猜测:A~R A' 相当于 |A|R = |A'|R)。
第二句,于是,lct(X, B, |M|R) ≥ lct(X, B, |M+N|R) = lct(X, B, |A|R)
注:不等式存疑(?),等式是显然的。
第三句,故此,只须给出右端的正下界。
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加评:第一段,简单的开局。略宽心,喝杯咖啡再战。
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We want to apply Proposition 5.9, so we need to replace X with a Q-factorial one.
注:这是第二段的第一句(命题5.9是“跳点”)。
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Proposition 5.9. Let d, r be natural numbers and eps a positive real number. Assume Theorem 1.6 holds in dimension ≤ d - 1. Then there exist natural numbers n, m and a positive real number eps' < eps depending only on d, r, eps satisfying the following. Assume
(X, B) is a Q-factorial projective eps-lc pair of dimension d,
A is a very ample divisor on X with A^d ≤ r,
L ≥ 0 is an R-divisor on X,
A - B and A - L are ample,
(X, B + tL) is eps'-lc for some t ≤ r,
a(T, X, B + tL) = eps' for some prime divisor T over X, and
the centre of T on X is a closed point x.
Then 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).
注:此命题位于第5部分最后一节(证明约占2.5页)。
评论:主条件有 7 项,副条件有 4 项。主条件简记:
A L
(T)
X B
副条件简记:
(m)A T
X (n)Λ
注:简记有待完善。
加评:此命题似乎是从主体证明中(突兀地)拆分出来的,该是在一定的上下文中产生、并做了条理化(注:按5.8开头提示,可能要结合 5.7 和 5.5 才能有所理解)。
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小结:定理1.6读写开了个头;初次整理命题5.9(可能得构思个“故事”帮助记忆)。
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