# 菲文笔记 | Technical theorem (v2) ---- omission and puzzle

This is coming to you from continued) camera level one = {R, S, '} ==> camera of level two = {Rs, Rs', R', ...}.

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Step 7, Para three ——

Assume now that (X, B) is not lc over z = f(S).

---- This is to setup the assumption for a proof by contradiction.

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By the previous paragraph, (X, B) is lc near S and S is a non-klt centre of this pair.

---- This is a nice summary, so that one can judge what is "lc near S".

---- Basically, for certain pair, Say (X, B), one can define Ks + Bs by (Kx + B)|s.

---- Or, given Ks + Bs, one can recover the original pair by "inversion of adjunction", to show Ks + Bs = (Kx + B)|s.

---- The divisor S is required to be a "non-klt centre" of the original pair.

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New comment: These old comments appear strange to me.

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On the other hand, (X, Γ) is plt with ⌊ Γ ⌋  = S, so if u > 0 is sufficiently small, then (X, (1 - u)B + uΓ) is plt near S and S is a non-klt centre of this pair and no other non-klt centre intersects S.

---- In combination of last sentence, if (X, B + ) is lc near S and (X, Γ) is plt near S, then the pair of (small) convex combination is still plt near S.

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Comment (wrong intuition): At the first glance, one might feel ⌊(1 - u)B⌋ = 0 and ⌊uΓ⌋ = 0, for their coefficients are smaller than 1 ——

---- This is really the case. (No kidding).

---- But, there is a re-organization matter, however, for the sum of the two terms, before taking the floor operation.

---- That is, (1 - u)S + uS = S, while S has the coefficient 1.

---- For this aspect, u can be any number in [0, 1].

---- The small positive u is required by some other aspects.(to be hunted).

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New comment ⌊ (1 - u)B + uΓ ⌋ = S ?

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Then since (X, B) is not lc over z, the non-klt locus of (X, (1 - u)B + uΓ) has at least two connected components (one of which is S) near the fibre f⁻1 {z}.

---- Translation: if the original pair is not lc over z, then the (plt convex) combined pair has its non-klt locus possessed at least two connected components near f⁻1 {z}.

---- To illustrate this in a simple way, take the boundary as the pair ——

B      ~    u(B )Γ

not | lc          not | simple

z        ~      f⁻1 {z}

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Note:  u(B )Γ, understood in context, is the homemade notation for (X, (1 - u)B + uΓ).

Note: I call a (plt) pair is not simple, if its non-klt locus has least two connected components near f⁻1 {z}.

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New comment: The arise of (X, (1 - u)B + uΓ) is not apparent.

This contradicts the connectedness principle [25, Theorem 17.4] as - (Kx + (1 - u)B + uΓ) = - (1 - u) (Kx + B ) - u(Kx + Γ) ~ R -u(Kx + Γ) ~ R uαM - u(Kx + Γ) /Z is ample over Z.

---- That is, the defence form of u(B)Γ is ample over Z, which expects u(B)Γ simple near f⁻1 {z} by the connectedness principle.

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New comment: It is not apparent why - (1 - u) (Kx + B ) disappears and why uαM appears.

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Therefore, (X, B + ) is lc over z.

---- This is the close statement.

---- It's needed.

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New summary: This part appears over omitted.

↑↓ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ⌈ ⌉ ∨∧∞Φ⁻⁰ 1

Calling graph for the technical theorem (Th1.9) ——

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Th1.9

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[5, 2.13(7)]   Lem 2.26   Pro4.1   Lem2.7

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.....................................................Lem2.3

Note: Th1.9 is only called by Pro.5.11, one of the two devices for Th1.8, the executing theorem.

Pro4.1

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[5, ?]   [37, Pro3.8]   [5, Lem3.3]   Th2.13[5, Th1.7]   [16, Pro2.1.2]    [25, Th17.4]

Completed notes of the first round learning for v2 Pro.4.1 are * (v1).

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It is my hope that this action would not be viewed from the usual perspective that many adults tend to hold.

http://www.blog.sciencetimes.com.cn/blog-315774-1300526.html

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