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在方法上下功夫绝不会白费。

已有 1591 次阅读 2019-5-6 11:33 |个人分类:心路里程|系统分类:科研笔记

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在方法上下功夫绝不会白费。

(接前：05 04 02) 转到命题 5.9 的证明.

Step2. Let ψ: W --> X be a log resolution of (X, B + tL) so that T is a divisor on W.

---- 以(X, B + tL) 为参照做回拉.

---- 这是要“造相”的征兆/节奏.

Define a boundary Γw = (1 + v)(B~ + tL~) + (1 - eps'/4)ΣEi + (1 - eps')T where Ei are the exceptional divisors of ψ other than T, and ~ denotes birational transform.

---- 定义边界Γw, 但这仅是表象.

---- 实际上Γw是构造出来的(参Pro.5.7,Step4).

(待温习时再仔细推敲).

---- 此处有个变化 (1 - a)T -> (1 - eps')T.

Since (X, B + tL) is eps'-lc and since μTΓw = 1 - eps', we have Kw + Γw = Kw + B~ + tL~ + Σ(1 - ci)Ei + (1 - eps')T + v(B~ + tL~) + Σ(ci - eps'/4)Ei =ψ*(Kx + B + tL) + v(B~ + tL~) + F where F:= Σ(ci - eps'/4)Ei is effective and exceptional over X and its support does not contain T.

---- 此处的拆项并项也类似于Pro.5.7,Step4.

On the other hand, letting c'i = a(Ei, X, (1 + v)(B + tL)) and a' = a(T, X, (1 + v)(B + tL)), and recalling that (X, (1 + v)(B + tL)) is eps'/2-lc outside finitely many closed points, we can write Kw + Γw = Kw + (1 + v)(B~ + tL~) + Σ(1 - c'i)Ei + (1 - a')T + Σ(c'i - eps'/4)Ei + (a' - eps')T = ψ*(Kx + (1 + v)(B + tL)) + G where G:= Σ(c'i - eps'/4)Ei + (a' - eps')T is exceptional over X and if the image of Ei on X is positive-dimensional for some i, then Ei is a component of G with positive coefficient.

---- 上一句的参照配对是(X, U).

(U:= B + tL).

---- 此句的参照配对是 (X, U + vU).

小结：该步骤的计算完全类似于Pro.5.7,Step4.(将那里的 B 替换成了 U:= B + tL).