[注：下文是3月16日 (8:14) 发出的群邮件内容。]

This is coming to you from Yiwei LI (PhD, Applied math), Taiyuan University of Science and Technology  (TYUST) Taiyuan, China

It's going on here for the third round of learning of Birkar's BAB-paper (v2), with scenarios of chess stories. No profession implications. 

One who does not have the similar queries cannot understand the subsequent quests or answers.

Th 2.15    Th 1.8            ♖    ♘

     ↓   ↖     ↓

Th 1.1      Th 1.6            ♔    ♗

    Mathematics vs Palace stories.(v2)

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Note: technical theorem is not on the board.

♙ ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁺⁻⁰ 1

(continued) As has been explored in move three, L' is likely constructed through evolving from n·c(M), and P' is desired by the deeper thought to form another plt-Eve pair, in order to have the Kawamata- Viehweg vanishing theorem activated. A natural question is how this vanishing theorem arose to the author? Or, in what kind of context one considers this vanishing theorem? From the end of move five, one sees it is desirable to have the restriction map H⁰(L' + P') --> H⁰((L' + P')|s') surjective. It appears the vanishing theorem is closely related to this end, though not very apparent. Another question is, how the thought concerning the form of L' + P' arose to the author? I wonder if any reviewer of this paper has the similar queries.

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Thinking in the ground framework, it is now the turn for c(α·M)Γ to play a role. The scene is of the X' space as before. Recall c(α·M)Γ is ample by assumption. For some reason, L' + P' is of major concern in this move five. So, it is desirable to have L' + P' geared onto c'(α·M)Γ by the philosophy of the ground framework. That is, to let c'(α·M)Γ  appear in the expression of L' + P'. As a preparation, one writes down the needed expressions explicitly —— 

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c'(α·M)Γ = α·M' - (Kx' + Γ').

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L' + P' = 2M' + n·c'(M) + d'n + P'

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As there is not apparent connection between the two expressions, one may apply the "inserting and canceling" trick on the right-hand side of L' + P', using c'(α·M)Γ as the manipulating term ——

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L' + P' = 2M' + n·c'(M) + d'n + P' + c'(α·M)Γ - (α·M' - (Kx' + Γ'))

           = Kx' + Γ' + d'n + P' +  c'(α·M)Γ + n·c'(M) + (2 - α)M'.

           = Kx' + Θ' +  [c'(α·M)Γ + n·c'(M) + (2 - α)M'].

           = Kx' + Θ' + [ · ].

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One can identify K' + Θ' to be the operation form of (X', Θ'), the plt-Eve pair. The rest three terms, with a temporary shorthand [ · ], are viewed as a whole, claimed to be "nef and big" —— It's beneficial to take a check: n·c'(M) is "nef and big" as c(M) is "nef and big"; M is semi-ample, so it appears M' to be at least "nef and big", not affected by non-negative scaling factor; c'(α·M)Γ is at least "nef and big" as c(α·M)Γ is ample —— So, their sum is "nef and big". (The attribution of Kx' + Θ' appears not concerned in the paper).

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Looking at the final expression of L' + P', one sees the presence of P' be only to help form Kx' + Θ', the operation form of a plt-Eve pair. Let us have P' removed, to see if L' would still work, with L' = Kx' + Γ' + d'n + [ · ], where (X', Γ' + d'n) expectantly* forms a plt pair, with ⌊ Γ' + d'n ⌋ = S'. To activate the vanishing theorem, one needs (X', Γ' + d'n - S') to be klt. This is expectant, as P' has nothing to do with S'. In this view, one has ——

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h1(L' - S') = 0, and

H⁰(L') --> H⁰(L'|S') surjective.

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*Added: One will see (X', Γ' + d'n) cannot form a pair, however, as Γ' + d'n is not non-negative (see below).

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So, the needs for the presence of P' remain to be explored. Recall* that P' is defined by assigning a proper scaling factor to some irreducible divisor D'. That is, P' takes the form of P'=c·D', where c is taken as 0 for D' = S', or c is taken as the coefficient of D' in -⌊ Γ' + d'n ⌋ for D' ≠ S'. For the later case, one searches D' in Γ' + d'n and takes c = - ⌊d⌋, if exists, where d is the original coefficient of D' in Γ' + d'n. In the nutshell, one can write Γ' + d'n = d1D'1 + d2D'2 + ... + dkD'k. So, one has P' = - ⌊di⌋D'i, for D'i ≠ S'. By the proof in the paper, P' = 0 or 1, one arrives at di ∈ (-1, 1). That is to say, Γ' + d'n is not qualified to be a "boundary" whose coefficients are required to be non-negative. But, wait... P' = - ⌊di⌋D'i is not sufficient to make Γ' + d'n + P' non-negative. It is needed to let P' = -∑⌊di⌋D'i for D'i ≠ S', so that Γ' + d'n + P' is non-negative. This answers the needs for the presence of P'. (* Now, let us replace "Recall" as "Assume"). 

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Here is the position to criticize (or taste) the subtlety of the notation/ definition of μD'P' —— 

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μD'P' := - μD'⌊Γ' + nΔ' - ⌊(n + 1)Δ '⌋⌋, for D' ≠ S'.

                                       d'n

This definition is based on the individual components of P', yet bearing some obscure nature. The intention of the author is just to obtain ——

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Γ' + d'n - (⌊d2⌋D'2 +... + ⌊dk⌋D'k), 

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so that the resultant expression is non-negative, assuming S' = D'1. For a better look, one can write* ——

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Γ' + d'n - ⌊ Γ' + d'n ⌋\S'

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where the slashed subscript \S' means to exclude the term of S' from the floor part. In this view, one can define P':= - ⌊ Γ' + d'n ⌋\S' straightforwardly, avoiding the discussion on D' or S', yet keeping the intuitive information. In retrospect, the presence of P' in L' + P' is needed to complete the operation form of a plt pair, while the appearance of L' + P' is occasional. So is the end of the description for move five.

*Modified from last note.

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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]  [20]  [25, Th17.4]

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Special note: Original synthesized scenarios in Chinese for the whole proof of v1 Th1.7, the technical theorem.

*It's now largely revised* due to new understandings.

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See also: Earlier comments in Chinese* (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.