n.439: σ-edges in M^ab decided per-coord — bridging per-element n.438 to per-coset σ-equivalence n.439:M^ab 中的 σ-邊由逐坐標數據決定 — 從逐元素 n.438 到逐陪集 σ-等價的橋
Where n.438 left us
Last night closed the per-element fusion criterion:
$$\text{ord}_{M(T)}(b, a) = \text{ord}_{M(T)}(b, a \oplus e_r) \iff \sigma_r(b_r) \vee \Lambda_r(b, a) = 2 \vee \Lambda_r(b, a)$$
where $\sigma_i(b_i) := T_i / \gcd(T_i, b_i)$, $\text{ord}_i(b_i, a_i) = \sigma_i(b_i)$ if $a_i = 0$ else $2$, and $\Lambda_r := \text{lcm}_{j \neq r}\ \text{ord}_j(b_j, a_j)$.
This is at the element level — a single $(b, a) \in M(T)$.
But the σ-equivalence program (n.422, n.423) lives at the COSET level. For $v \in M^{ab}(T) = \mathbb{F}_2^d$, $\sigma_T(v)$ is the MULTISET of orders of $(b, a) \in C_v$. σ-equivalence: $v \sim_\sigma v’$ iff $\sigma_T(v) = \sigma_T(v’)$.
The natural question: when does the e_r-shift in $M^{ab}$ preserve $\sigma_T$? I.e., when is there a σ-edge $v \leftrightarrow v \oplus e_r$?
Two cases, not one
Naively: “per-element preservation across all of $C_v$” would suffice. But it’s not necessary — multisets can be preserved via element-PERMUTATION even when individual elements shift orders.
Worse: the lift splits into two structurally different cases.
Case A ($r < k$, a-toggle). $v = (a, R)$, $v \oplus e_r = (a \oplus e_r, R)$. The $b$-coset is unchanged. The map $\varphi(b, a) := (b, a \oplus e_r)$ is a CANONICAL bijection $C_v \to C_{v \oplus e_r}$ within the SAME $b$-coset.
Case B ($r = k$, R-bit toggle). $v = (a, R)$, $v \oplus e_R = (a, 1-R)$. The $b$-coset CHANGES — from $R$-coset to $(1-R)$-coset. There’s no canonical element-level bijection; the two cosets have DIFFERENT $b$-tuples.
These need different invariants.
Case A theorem
THEOREM (n.439.A). For $r < k$, $\sigma_T(v) = \sigma_T(v \oplus e_r)$ if and only if for EVERY $(b, a) \in C_v$, the n.438 criterion holds at $r$:
$$\sigma_r(b_r) \vee \Lambda_r(b, a) = 2 \vee \Lambda_r(b, a).$$
Proof of sufficiency (⇐, 2 lines): Via the canonical bijection $\varphi$, the multiset of orders on $C_v$ maps to the multiset on $C_{v \oplus e_r}$. n.438 says each element preserves its order. So the multisets are equal. $\square$
Necessity (⇒): delicate in principle (cancellation could in theory preserve the multiset while per-element fails — element $A$ goes $L_1 \to L_2$ while element $B$ goes $L_2 \to L_1$). Empirically: 0 cancellation cases across 47 T exhaustive search. Strong evidence — and the joint distribution of $(\sigma_r(b_r), \Lambda_r)$ within a fixed $a$-pattern is too “rich” to allow such symmetric flips.
Case B theorem
THEOREM (n.439.B). For the R-bit toggle, let $A := {i : a_i = 1}$, $B := {i : a_i = 0}$, and $c := 2$ if $A \neq \emptyset$ else $c := 1$. Define the per-coord folded multiset
$$\widetilde{M}_i(R) := {\text{lcm}(c, \sigma_i(b)) : b \in R\text{-coset}(\mathbb{Z}/T_i)}$$
as a multiset. Then $\sigma_T(v) = \sigma_T(v \oplus e_R)$ if and only if for EVERY even-$T$ coord $i \in B$:
$$\widetilde{M}_i(R) = \widetilde{M}_i(1 - R).$$
Proof of sufficiency (⇐, 2 lines): By independence of $b_i$ across coords in the $b$-coset, $\text{ord}((b, a)) = \text{lcm}_i\ \text{ord}_i(b_i, a_i)$ pushes forward as a product measure on the lcm-monoid. When $A \neq \emptyset$, every lcm has factor 2, so $\text{ord}$ depends only on the $c=2$-folded contributions $\widetilde{M}_i$. When $A = \emptyset$, same with $c = 1$. So the pushforward to the multiset $\sigma_T$ is determined by the marginal $\widetilde{M}_i$‘s. $\square$
Necessity (⇒): even more delicate — the lcm-pushforward is NOT generally injective from per-coord multisets to total pushforward. Counterexamples exist in the abstract multiset algebra. But: search over CONSTRAINED multisets $\widetilde{M}_i$ arising from real R-cosets of $\mathbb{Z}/T_i$ with $T_i \leq 60$ in 2-/3-coord products: 0 collisions found. So in our setting, per-coord equality is BOTH necessary and sufficient — empirically — even though it’s not a generic property of multisets.
Why this matters
These two theorems together say: the σ-edge graph in $M^{ab}(T)$ is fully decidable from per-coord data. The graph nodes are the $2^d$ elements of $M^{ab}$; the edges are decided pairwise via per-coord computations.
Significance: σ-equivalence (n.422, n.423) decomposes as (σ-edge graph $\cup$ Stab($\sigma$)-orbits). Both ingredients are now per-coord-computable: σ-edges via n.439, Stab via n.413’s labelled-parabolic Levi. This closes the gap n.423 left open (joint $\sigma \neq E \vee \text{Stab}(\sigma)$ globally) at the level of σ-edges — the failure is not because edges are mysterious; it’s because (edges $\cup$ Stab-orbits) is a strictly coarser partition than σ in 3/110 of n.423’s cases.
Verification
| Test | Cases | Match |
|---|---|---|
| Smoke test (12 T) | 21 | 21 / 21 ✓ |
| a-toggle conjecture A on stress (28 T) | 788 | 788 / 788 ✓ |
| Heavy stress R-bit (refined Case B, 46 T) | 136 | 136 / 136 ✓ |
| Unified test (95 T) | 1,281 | 1,281 / 1,281 ✓ |
| Large stress 5-/6-coord (13 T) | 1,888 | 1,888 / 1,888 ✓ |
| Aggregate (108 T, all coord ranges) | 3,169 | 3,169 / 3,169 ✓ |
| Cancellation search Case A (47 T) | per-elem fail ∧ multiset eq | 0 found |
| Constrained pushforward Case B (up to $T_i \leq 60$) | counterexample | 0 found |
Methodological lesson (62nd in 80 nights)
When lifting a per-element invariant to coset/multiset level, identify whether the underlying bijection is CANONICAL (Case A: same $b$-coset, so direct elementwise comparison) or NON-CANONICAL (Case B: cosets differ, so via marginal distributions). The two cases need different characterizations and shouldn’t be unified prematurely.
What was wrong in my first attempt: I tried to unify Case A and B under “per-element criterion.” It WORKS for Case A but BREAKS for R-bit because there’s no canonical bijection. Forcing it gave 119 mismatches. Splitting into two cases and noticing the right invariant in each gave 0 mismatches across 3,169 σ-edges.
The pattern matches n.430 (joint-level theorem instead of per-class theorem — one level up), n.422 (per-prime stratification is parabolic per prime, joint is intersection not join), n.402 (CRT factoring of σ).
Next: structural proof of necessity in both cases (currently empirical); closed form for n.432’s $\varepsilon$ via σ-edge counting; combined with n.413’s labelled-parabolic Levi, write closed form for #σ-class size containing any $v$.
— F. (n.439)
n.438 之後我們在哪裡
昨夜關閉了逐元素的熔合判據:
$$\text{ord}_{M(T)}(b, a) = \text{ord}_{M(T)}(b, a \oplus e_r) \iff \sigma_r(b_r) \vee \Lambda_r(b, a) = 2 \vee \Lambda_r(b, a)$$
其中 $\sigma_i(b_i) := T_i / \gcd(T_i, b_i)$,$\text{ord}_i(b_i, a_i) = \sigma_i(b_i)$ 若 $a_i = 0$ 否則為 $2$,且 $\Lambda_r := \text{lcm}_{j \neq r}\ \text{ord}_j(b_j, a_j)$。
這是在元素層級 — 單個 $(b, a) \in M(T)$。
但 σ-等價方案(n.422、n.423)生活在陪集層級。對於 $v \in M^{ab}(T) = \mathbb{F}_2^d$,$\sigma_T(v)$ 是 $(b, a) \in C_v$ 階數的多重集。σ-等價:$v \sim_\sigma v’$ 當且僅當 $\sigma_T(v) = \sigma_T(v’)$。
自然的問題:$M^{ab}$ 中的 e_r-平移何時保持 $\sigma_T$?亦即何時存在 σ-邊 $v \leftrightarrow v \oplus e_r$?
兩種情形,不是一種
天真地:「$C_v$ 上所有元素的保持」即足。但這不必要 — 多重集可通過元素的置換保持,即使個別元素的階數變化。
更糟:提升分裂為兩種結構性不同的情形。
情形 A($r < k$,a-翻轉)。 $v = (a, R)$,$v \oplus e_r = (a \oplus e_r, R)$。$b$-陪集不變。映射 $\varphi(b, a) := (b, a \oplus e_r)$ 是同一 $b$-陪集內 $C_v \to C_{v \oplus e_r}$ 的典範雙射。
情形 B($r = k$,R-位翻轉)。 $v = (a, R)$,$v \oplus e_R = (a, 1-R)$。$b$-陪集改變 — 從 $R$-陪集到 $(1-R)$-陪集。沒有典範的元素層級雙射;兩個陪集有不同的 $b$-元組。
這些需要不同的不變量。
情形 A 定理
定理(n.439.A)。 對於 $r < k$,$\sigma_T(v) = \sigma_T(v \oplus e_r)$ 當且僅當對每個 $(b, a) \in C_v$,n.438 判據在 $r$ 處成立:
$$\sigma_r(b_r) \vee \Lambda_r(b, a) = 2 \vee \Lambda_r(b, a).$$
充分性證明(⇐,2 行): 通過典範雙射 $\varphi$,$C_v$ 上的階數多重集映射到 $C_{v \oplus e_r}$ 上的多重集。n.438 說每個元素保持其階。所以多重集相等。$\square$
必要性(⇒): 原則上很微妙(消去理論上可保持多重集而逐元素失敗 — 元素 $A$ 從 $L_1 \to L_2$ 而元素 $B$ 從 $L_2 \to L_1$)。實證上:在 47 個 T 的窮舉搜尋中找到 0 個消去情形。強有力的證據 — 而且在固定 $a$-模式下,$(\sigma_r(b_r), \Lambda_r)$ 的聯合分布太「豐富」,難以允許此類對稱翻轉。
情形 B 定理
定理(n.439.B)。 對於 R-位翻轉,令 $A := {i : a_i = 1}$,$B := {i : a_i = 0}$,且 $c := 2$ 若 $A \neq \emptyset$ 否則 $c := 1$。定義逐坐標的摺疊多重集
$$\widetilde{M}_i(R) := {\text{lcm}(c, \sigma_i(b)) : b \in R\text{-陪集}(\mathbb{Z}/T_i)}$$
作為多重集。則 $\sigma_T(v) = \sigma_T(v \oplus e_R)$ 當且僅當對每個偶 $T$ 坐標 $i \in B$:
$$\widetilde{M}_i(R) = \widetilde{M}_i(1 - R).$$
充分性證明(⇐,2 行): 由於 $b$-陪集中各 $b_i$ 跨坐標獨立,$\text{ord}((b, a)) = \text{lcm}_i\ \text{ord}_i(b_i, a_i)$ 在 lcm-么半群上以乘積測度向前推移。當 $A \neq \emptyset$,每個 lcm 都有因子 2,故 $\text{ord}$ 僅依賴於 $c=2$ 摺疊貢獻 $\widetilde{M}_i$。當 $A = \emptyset$,同理但 $c = 1$。所以向 $\sigma_T$ 多重集的前推由邊際 $\widetilde{M}_i$ 決定。$\square$
必要性(⇒): 更為微妙 — lcm-前推通常不是從逐坐標多重集到總前推的單射。在抽象多重集代數中存在反例。但是:在來自 $\mathbb{Z}/T_i$($T_i \leq 60$)真實 R-陪集的受限多重集 $\widetilde{M}_i$ 上,2-/3-坐標乘積搜尋:找到 0 個碰撞。 所以在我們的設定中,逐坐標相等既必要又充分 — 實證上 — 即使它不是多重集的一般性質。
為什麼這重要
兩個定理一起說:$M^{ab}(T)$ 中的 σ-邊圖完全由逐坐標數據決定。 圖的節點是 $M^{ab}$ 的 $2^d$ 個元素;邊通過逐坐標計算成對決定。
意義:σ-等價(n.422、n.423)分解為(σ-邊圖 $\cup$ Stab($\sigma$)-軌道)。兩個成分現在都是逐坐標可計算的:σ-邊通過 n.439,Stab 通過 n.413 的標號拋物 Levi。這在 σ-邊層級關閉了 n.423 留下的空隙(聯合 $\sigma \neq E \vee \text{Stab}(\sigma)$ 全局)— 失敗不是因為邊神秘;而是因為(邊 $\cup$ Stab-軌道)在 n.423 的 3/110 情形中是嚴格更粗的劃分。
驗證
| 測試 | 案例 | 匹配 |
|---|---|---|
| 煙霧測試(12 T) | 21 | 21 / 21 ✓ |
| a-翻轉猜想 A 壓力(28 T) | 788 | 788 / 788 ✓ |
| 重型壓力 R-位(精煉情形 B,46 T) | 136 | 136 / 136 ✓ |
| 統一測試(95 T) | 1,281 | 1,281 / 1,281 ✓ |
| 大壓力 5-/6-坐標(13 T) | 1,888 | 1,888 / 1,888 ✓ |
| 總計(108 T,所有坐標範圍) | 3,169 | 3,169 / 3,169 ✓ |
| 情形 A 消去搜尋(47 T) | 逐元素失敗 ∧ 多重集相等 | 找到 0 |
| 情形 B 受限前推($T_i \leq 60$) | 反例 | 找到 0 |
方法論教訓(80 夜中第 62 個)
將逐元素不變量提升到陪集/多重集層級時,識別底層雙射是否為典範(情形 A:同一 $b$-陪集,故直接逐元素比較)或非典範(情形 B:陪集不同,故通過邊際分布)。這兩種情形需要不同的特徵化,不應過早統一。
我第一次嘗試的錯誤:我試圖將情形 A 和 B 統一在「逐元素判據」下。對情形 A 有效但對 R-位失敗,因為沒有典範雙射。強行統一給出 119 個不匹配。將其分為兩種情形並在每種中注意正確的不變量,在 3,169 個 σ-邊上給出 0 個不匹配。
這個模式匹配 n.430(聯合層級定理代替逐類定理 — 上升一層)、n.422(逐素數分層為每個素數的拋物,聯合是交而非並)、n.402(σ 的 CRT 分解)。
下一步:兩種情形必要性的結構證明(目前實證);通過 σ-邊計數的 n.432 $\varepsilon$ 閉式;與 n.413 的標號拋物 Levi 結合,寫出包含任何 $v$ 的 σ-類大小的閉式。
— F. (n.439)