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每段证明都有新鲜事~
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This is an in-mail from TYUST.
本期开始分组发送邮件，搭载数学类学院等链接。
Proof. Step 1. By ACC for lc tresholds [11], there is eps'∈(0, eps) depending only on d so that if (Y, (1 - eps')S) is a Q-factorial klt pair of dimension d where S is reduced, then (Y, S) is lc.
---- (Y, (1-eps')S) Qfk ==>(Y, S) lc.(S reduced)
---- 用到了 ACC 关于 lct 的现成结果.
---- ACC 该是[11] 提出的某种概念.(?).
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Cutting by general elements of |A| and applying Theorem 1.6 in lower dimension, we find a positive number v depending only on d, r, eps' such that (X, B + tL + 2v(B + tL)) is klt outside finitely many closed points.
---- 对 |A| 的一般元 切削加工, 并运用降维的定理1.6, 则有 v=v(d,r,eps') 使得 (X, B + tL + 2v(B + tL)) klt o.c.
---- Th1.6(≤ d) ==> (X, U + 2vU) klt o.c. (U:=B + tL).
---- 此关系之前已经出现过(Pro.5.7.Step4*).
---- lct 涉及B+tL形式. 若取L=B, 即有B+tB形式.
---- 本句只不过多套了一层(但仍是巧妙的).
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Thus (X, B + tL + v(B + tL)) is eps'/2-lc except outside finitely many closed points as (X, B + tL) is eps'-lc.
---- 这句中出现个“except”似是多余.
---- (X, U + vU) = 1/2(X, U) + 1/2(X, U + 2vU).
1/2eps'-lc eps'-lc klt or 0-lc
---- lc 的“前缀”也是凸组合关系.
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评论：第一句与后两句几乎“不搭”(只用来给出eps'?).
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小结：Step1 读写完毕.
https://blog.sciencenet.cn/blog-315774-1177243.html
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