导言：下文是群邮件的内容，邮件栏目"[March for reflection |Birkar]" 。基本上每天写一篇评论式学习笔记，通常花费4~6个小时读一段。这篇是今天凌晨 4:40许发出的，写了4个小时左右 (不含头尾的招牌阵)。原文是蓝色字体的三句话，有一句稍长分为几小段。这种学习方法称作"手读"，即 hand reading。这是第三轮学习，计划用一年时间做出完整的英文笔记。希望能对本系的学生有个带动的作用。

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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. 

If one enters a new world never known before, one has to enter that world through no where but its "hub"^.

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.

M          Λ           ♖          ♘

       S                          ☂

XF        pB            ♔          ♗

Note: the upper right/left corner is of output.

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

Para two ——

We claim that P' ≥ 0. 

---- Recall that P' takes a shape of μD'P' = - μD'⌊Γ' - Δ' + <(n + 1)Δ'>⌋,for D' ≠ S'.

---- There is a subtraction sign, somewhat contrast to the intuition.

---- One will see Γ' - Δ' be something.

---- "something" is generally derived from the re-arrangement, by certain principle.

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Note: recall that, d'n = nΔ'- ⌊(n + 1)Δ'⌋ = - Δ' + <(n + 1)Δ'>.

---- I use Δ'(d'n) to refer to the later form.

---- So, μD'P' = - μD'⌊Γ' + d'n⌋ = - μD'⌊Γ' + Δ'(d'n)⌋.

---- Well, yes, d'n = Δ'(d'n), nominally.

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Pick a component D' of P'.

---- This is a nice progress compared to v1.

---- "any" statement is abandoned.

---- Keep tight.

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Comment: to analyze the signs of P', one takes an component of P'.

---- components are everything for a divisor, like atoms.

---- In a sense, component is the hub of divisor.

---- So, D' is the hub of P'.

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Memo: component ~ hub.

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If D' is a component of E', then D' is not a component of Δ' ...

---- To apply a binary analysis to D' with the binary division of B' into E' and Δ'.

---- Recall that, the division B' = E' + Δ' is the key technical point to shape L' and P'.

---- That is, B' = E' + Δ' is the hub for L' and P'.

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Comment: the spirit is to apply a hub-to-hub analysis (peer to peer).

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... and μD'Γ' ∈ (0, 1) as B' - Γ' has small coefficients and μD'B' = 1, hence μD'P' = 0 ...

---- Recall μD'P' = - μD'⌊Γ' - Δ' + <(n + 1)Δ'>⌋.

---- As D' is in E', one has μD'B' = 1.

---- That is, B' = ... + 1·D' + ... .

---- As D' is not in Δ', it is left to check the coefficient of D' in Γ'.

---- Γ' is a "boundary" (see items of Pro4.1), so it's coefficients are not negative.

---- B' - Γ' has small coefficients (positive or negative), so Γ' can be viewed as a version of B', just with a slightly perturbed coefficients.

---- So, Γ' = ... + δ1·D' + ... where δ1 is a number close to 1 (slightly bigger or smaller).

---- Therefore, when D' is in E', one has μD'P' = - μD'⌊Γ' - Δ' + <(n + 1)Δ'>⌋ = - μD'⌊Γ'⌋ = -1 or 0.

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Special comment: it appears to expect an updated proof for this point.

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... on the other hand, if D' is not a component of E', then the absolute value of μD'(Γ' - Δ') = μD'(Γ' - B') is sufficiently small...

---- Clear.

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... and μS'<(n + 1)Δ'> ∈ [0, 1), hence μD'P' = 0 or μD'P' = 1, so in any case μD'P' ≥ 0.

---- Well, I view μS' as a typo. (S' is not in Δ'; see the end of Step 3).

---- That is, μD'<(n + 1)Δ'> ∈ [0, 1).

---- <(n + 1)Δ'> is fractional for any component (i.e. all the coefficients are in [0, 1)).

---- Recall μD'P' = - μD'⌊Γ' - Δ' + <(n + 1)Δ'>⌋.

----| If μD'<(n + 1)Δ'> = 0, i.e (n + 1)Δ' integral, one has μD'P' = - (-1) = 1 (for small negative coefficient of D' in B' - Γ') or μD'P' = 0 (for small positive coefficient of D' in B' - Γ').

----| if μD'<(n + 1)Δ'> is in (0, 1), the contribution from μD'(Γ' - Δ') = μD'(Γ' - B') can be always made negligible, hence μD'P' = 0.

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  Summary comment: I would consider E' be the sum of the components of B', which have coefficient 1/n (or m/n, with the positive integer m < n).

---- It appears ok to make sure nE' integral.

---- If the coefficients of B' are not in [0, 1], as I assume here, the coefficient 1 has nothing special.

---- I have the chance to check.

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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]

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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.