
This is coming to you from continued) In the last learning note, as one will agree, the proof of Pro 4.1 is largely "decoded". Ideally, I wish to recover the formation of the technical theorem as well as its proof. Tentatively, I assume Th 2.13 serves as the Captain (or CEO), being instructive (concerning "what to do"), while the KV vanishing theorem serves as the Chief mate (or CTO), to organize the proof up to move six (of Pro 4.1), dealing with "how to do". The process works like a duet, with each thread contributing a "tune" in each action or construction. Mathematically, one might view the new theorem as the product of "talculations" among the existed ones (TOM). So, let us talculate ——
Preparation
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First of all, I write down the "resolvent form" of the KV vanishing theorem ——
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Items of condition:  (· , ^) is a pltEve pair, with Eve S;  (· , ^  S) is a klt pair;  * is nef and big;  ?  * is equal to the operation form of the pltEve pair, with ? being integral. Items of conclusion:  h1(?  S) = 0;  H0(?) > H0(?S) is surjective. 
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The key view is that, all of the "universal" symbols in the resolvent form are to be constructed (TOM). In the context of Pro 4.1, "?" represents the primary object to be constructed in a "bare" manner, while "*" represents the secondary object to be constructed in an assiciated manner. I call "?" and "*" the client/ guest objects, as "S" is not involved in them on the level of condition. Not surprisingly, I call "^" the native object. Finally, I call "·" the host object which is the easiest to get determined. As a plus, one can verify that the primary object presents for four times, the biggest counts (i.e. mode) among the universal symbols. The only explicit object, here S, also presents for four times. Interestingly, one only sees the interations between the two modes in the conclusion. It's attempting to show how everything could be settled down naturally.
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Note: The pair satisfying the first two items in the condition is called plt type, while "^" is called the plt type boundary. The primary object "?" satisfying the rest two items in the condition is called the plt type divisor. By the way, if one likes, one can replace "plt type" with "KV", a shorter label (Say, KV pair, KV boundary, and KV divisor).
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To proceed, one is reminded that the paper was empty at all at the beginning, except for the conjecture (BAB) to be proved. One needs to derive everything from the items of condition of the conjecture, to arrive at the conclusion of the conjecture. One can view the known part (i.e. the condition) of a conjecture as the "resolvent" of the conjecture (TOM). The BAB conjecture said ——
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The varieties X satisfying the following two items form a bounded family:
 (X, B) is epslc of dimension d for some B, and
  (Kx + B) is nef and big.
Note: eps is a positive real number, while d is a natural number.
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Notes: I call the varieties (or pairs) satisfying these two items as the epslc weak Fano varieties (or pairs), with a shorthand "EWF". One can write down the "resolvent form" of the conjecture ——
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Items of condition:  (· , ·) is epslc;   · is nef and big. Item of conclusion:  Omitted. 
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In the items of condition, one can judge that the form " ·" is incomplete (i.e. not closed). One agrees that one of the "atoms" in birational algebraic geometry takes the form of "·  ·". By "atoms", I refer to those simple and closed structures, say, including only two objects and one operator on the level of form. Also, from the angle of "physics", one can view Kx and B as "substances" of the same kind. Then, the second item can be written* as  (Kx + B) = 0  (Kx + B). In an equation of physics, one uses a "unit" for each symbols. I assume the unit for Kx and B is absorbed in the symbols. Then, the number "0" can not stand alone without an absorbed "substance" of the same kind. For one thing, the form of "·  ·" is desired; for another, a "substance" is needed. And, the equation must not be broken. So arises the form of 0·M  1·(Kx + B). Here, the letter M is incidental, meaning one can use any other available letters for M. Further, one can write a derived version, like M  (Kx + B), or more generally α·M  β·(Kx + B). One agrees that the form with α = β = 1 is "canonical", denoted as c(M), which turns out to play a key role in the formation and proof of the technical theorem (as I view), i.e. Theorem 1.9 & Proposition 4.1. I view c(M) the "primary candidate" for the use of construction, as it stems from the "root" of the conjecture. I give c(M) the name of "confrontation"**. If one counts the number of objects and operators in M  (Kx + B), one may feel that c(M) is beyond the level of "atom". But, the form of (Kx + B) is somewhat "unstable" due to the minus (therefore "incomplete"). One can verify this point in the construction of c(αM')Γ', near the end of the last note, where the form of  (Kx' + Γ') in the equation [·] = n·c'(M)  (Kx' + Γ') does not sustain till the end.
* This paragraph is partially inspired by an earlier blog note (2018) in Chinese.
** cf. the long story due to the learning notes for v1, say, mode inherent in the definition of ncomplememt. One agrees that G is the primary object to be constructed in the context of ncomplement. I say, primary object meets primary object (TOM). So, the primary object G from the thread of ncomplement is associated with the primary object "?" from the thread of KV vanishing theorem.
* The origin of this specific goal remains to be identified in the context, say Pro 5.11, where the needs arise.
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Now, one needs to find a "candidate" for G or "?". As n is available from the previous expectations, it appears convenient to start from the thread of ncomplement, i.e. to construct G (instead of "?") first which takes a known form of "n·divisor". Meanwhile, it is well known that one prefers to construct an object in X'space in the game of birational algebraic geometry. This object is called the "preconstruction" of the primary object. Let this object take the notation of L' as chosen by the author. Another hint is that, one should use a "prototype" for L', denoted as (L'). By "prototype", I say (L') takes the same form as G. So, one has (L') = n·divisor. It is natural to choose the "primary candidate", namely c'(M), to fill the position of "divisor". So to summarize the calculations up to now ——
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(L') = n·c'(M).
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To have the feature of ncomplement represented throughly, one can let B' = E' + Δ', where E' represents the sum of the components of B' which have the coefficient 1. So, replace B' with E' + Δ' in the expression of (L'), one has ——
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(L') = nM'  nKx  nE'  nΔ'.
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Here, one needs to modify nΔ' to be ⌊(n + 1)Δ'⌋ for a dual reason, (a) to really reflect the feature of ncomplement, and (b) to make sure (L') to be integral. This integral requirement comes from the thread of KV vanishing theorem, as (L') will also serve as the preconstruction of "?" in X'space. With this modification, one has ——
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(L') = nM'  nKx  nE'  ⌊(n + 1)Δ'⌋.
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To be closed*, one writes nE' as nB'  nΔ', to arrive at ——
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(L') = n·c'(M) + nΔ'  ⌊(n + 1)Δ'⌋.
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*Notes: The third item of the definition of ncomplement gives a closed form like "·  ⌊·⌋", so one views nΔ'  ⌊(n + 1)Δ'⌋ as a closed form (i.e. treated as a whole). It is indeed the case in the paper. I denote nΔ'  ⌊(n + 1)Δ'⌋ as d'n for short.
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In retrospect, (L') is out of the thread of ncomplement while delicately influenced (to be integral) by the thread of KV vanishing theorem. As stated earlier, (L') also serves as the preconstruction of "?" in X'space. That is to say ——
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(L')  *' = Kx' + ^'. (%)
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Here ^' is not yet closed in the sense of duet, with Γ' coming from the thread of KV vanishing theorem. That is to say, the contribution from the thread of ncomplement is desired. This component can be found no where but from the newly constructed (L'). One agrees that d'n is representative of the ncomplement due to the closed form "·  ⌊·⌋". So, one has ——
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^': = Γ' + d'n.
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Now, ^' is out of a duet of the both threads, seemingly perfect. But, at some point, one would find that Γ' + d'n violates the nonnegative requirement for a boundary. It is mandatory for ^' to be nonnegative. This is resolved by applying the form "·  ⌊·⌋" to Γ' + d'n in a physical way (v.s. metaphysical). So, let ^' tentatively be ——
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Γ' + d'n  ⌊Γ' + d'n⌋.
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Further, by the "Eve" nature of ⌊^'⌋ = S', one needs to exclude S' from ⌊Γ' + d'n⌋. So, eventually, one defines ——
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^': = Γ' + d'n  ⌊Γ' + d'n⌋\S'.
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Let P':=  ⌊Γ' + d'n⌋\S' for short. As stated earlier, (L'), out of the thread of ncomplement, serves as a component of " ?' ". So, an essential contribution from the thread of KV vanishing theorem is desired by " ?' ". It appears convenient for P' to take this role. Now, let us return to (%) with the modifications ——
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(L') + P'  *' = Kx' + ^', ($)
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where (L') + P' represents for " ?' ". Finally, one needs to settle down *', expected to be "nef and big" in the KV vanishing theorem. By unfolding the equation ($), one has ——
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n·c'(M) + d'n + P'  *' = Kx' + Γ' + d'n + P' ==>
*' = n·c'(M)  (Kx' + Γ').
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As stated earlier, c'(M) is viewed as a closed form v.s. the nonclosed form  (Kx + ·). In particular, it is not convenient to prescribe  (Kx' + Γ') to be "nef and big". So introduces α·M to close  (Kx' + Γ') ——
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*' = n·c'(M) + α·M  (Kx' + Γ')  α·M ==>
*' = n·c'(M) + c'(α·M)Γ'  α·M
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Now, one can prescribe both c'(M) and c'(α·M)Γ' to be "nef and big" (In the case of c'(α·M)Γ', the stronger attribution of "ample" will be needed). This is also consistant with the earlier expectations due to Th 2.13, back to the paragraph of [Marked #1]. It appears likely M was originally introduced instead of α·M at some earlier point in the case of c'(α·M)Γ', then modified to be so at a later point (to be checked). Now, it appears that  α·M is not closed in the expression of *'. This issue is associated with (L') by the equation ($). Recall that (L') is out of the ncomplement, demanding a contribution from KV by the duet. So introduces k·M' to *' as well as to (L'), one has ——
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*' = n·c'(M) + c'(α·M)Γ' + (k  α)·M', with the updated equation ——
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(L') + k·M' + P'  *' = Kx' + ^', ($)
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Now, Let (L') + k·M' be denoted by L', namely ——
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L':= (L') + k·M' = (n + k)·M'  nKx  nE'  ⌊(n + 1)Δ'⌋, or in the equivalent form
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L':= n·c'(M) + nΔ'  ⌊(n + 1)Δ'⌋ + k·M'.
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As L' is a component of "?", it must remain integral. So, k should be an integral, specifically determined at a later time (associated with α). Also, if one looks into n·c'(M), one has ——
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n·c'(M) = nM'  Kx'  nB'.
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Apparently, one needs nB' to be integral (for the sake of L') (3) —— This accounts for the presence of the item "pB is integral" in Theorem 1.9. (B has finite terms with rational coefficients, as p is assumed integral).
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In the proof, it is shown P' be nonnegative. Why is this a concern ? I guess the attribution of "exceptional" is talked only for nonnegative divisors.(But, why is P' exceptional desired?). These and remaining problems are left to the future.
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Additional notes: One can verify that [5, Lemma 3.3] is of adjunction. In addition, the KV vanishing theorem is very familar to Birkar; see [5]. Indeed, Pro 6.7 and Pro 8.1 in [5] show similar techniques as in Pro 4.1 v2 of Birkar (2016b), using KV vanishing theorem in the context of ncomplement. Recall that I've started this learning from the short paper of Birkar (v1 2016b), not yet looking into any version of the long paper of Birkar (2016a, or [5] for its latest version) but a light browse. So, the "calculation" here would be also useful to understand the relevant part of that long paper.
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In summary, talculus is the “calculus” of theorems, or the art of manufacture of new theorems. In talculus, one uses a dominant theorem as the resolvent to formulate a new theorem.
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↑↓ ℭ ℜ Iφ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ⌈ ⌉ ∨∧∞Φ⁺⁻⁰ 1
Calling graph for the technical theorem (Th1.9) ——
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Th1.9

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

.....................................................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

[5, ?] [37, Pro3.8] [5, Lem3.3] Th2.13[5, Th1.7] [16, Pro2.1.2] [20] [25, Th17.4]
Completed notes of the first round learning for v2 Pro.4.1 are * (v1).
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