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先看看故事的结尾~
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本期开始分组发送邮件，搭载数学类学院等链接。
先看看故事的结尾~
---- 之前读到Step2, 今番跳到Step7.
(此命题要构造 Λ, 想先看看它的样子).
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Step7. The pair (Y, T) is lc, by our choice of eps', because (Y, (1 - eps')T) is klt.
---- (Y, (1 - eps')T) klt ==> (Y, T) lc.(?)
---- eps' 最早出现在Step1 第一句, 在那里:
---- (Y, (1 - eps')S) Qfk ==> (Y, S) lc.
---- 但那里没说 Y 和 S 是什么.(?)
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Moreover, AY|T ~ 0 since T is mapped to the closed point x.
---- map(T) = x ==> AY|T = 0.
---- 但不知道map是哪个.(?)
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Now by Theorem 1.7, there is a natural number n depending only on d and there is 0 ≤ PY ~ (n + 1)lAY - n(KY + T) such that (Y, ΛY := T + 1/nPY) is lc near T.
---- 用Th1.7给出了 Λ 的构造 ΛY:=T + 1/nPY.
(在像空间给出了 Λ 的构造).
---- PY 用等价关系给出 (n + 1)lAY - n(KY + T).
---- 这里也反映出 T 起到微妙作用.
---- 按Th1.7, 该写成 T⁺ := T + 1/nPY.
---- 此处 T 既扮演边界, 也扮演 lc 的有效域.
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Let Λ be the pushdown of ΛY.
---- 用pushdown回到原像空间.
评论：有了Λ, 接下来要验证那4个“约束”.
(为何非要有那样的约束?)
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Since n(KY + ΛY) ~ (n + 1)lAY ~ 0/X, KY + ΛY is the pullback of Kx + Λ, and since T = φ⁻¹{x}, the pair (X, Λ) is lc near x.
---- 两个 since 有待验证.(?)
---- 推导也有待验证.(?)
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Moreover, nΛ is integral, and from Kx + Λ ~Q (n+1)l/n A we deduce 2lA - (Kx + Λ) is ample which in turn implies that there is a natural number m depending only on d, r, l such that mA - Λ is ample.
---- nΛ 是整的(直接带过去了).(该是显然的)
---- nΛ 整, Kx + Λ ~Q (n+1)l/n A ==> 2lA - (Kx + Λ) ample ==> mA - Λ ample. (?)
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Finally, a(T, X, Λ) = a(T, Y, ΛY) = 0, so T is a lc place of (X, Λ).
---- 待考.
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评论: 从方法上看，是用“补法”造相.
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小结：从 (Y, T) lc 出发, 在像空间给出了Λ的构造, 然后回到原像空间验证了4个约束.
https://blog.sciencenet.cn/blog-315774-1177652.html
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