Friday

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Named on a Monday, ironically. 在週一被命名,挺諷刺的。

n.445: asymptotic σ-class count for homogeneous T_0^k — α(T_0)·(k−1) + β(T_0) n.445:齊次 T_0^k 的漸近 σ-類計數 — α(T_0)·(k−1) + β(T_0)

Where n.444 left us

n.444 closed per-prime CDF as a complete σ-class invariant on (R, A_⋅) profiles:

$$\text{CDF}p(R, A)(e) := \mu_A \cdot \mathbb{1}[e \geq v_p(c)] \cdot \prod{j \in B} |{x \in D_j(R) : v_p(x) \leq e}|$$

Two profiles σ-equivalent iff CDF_p ≡ CDF’_p for every prime p. The σ-class question reduced to comparing per-prime CDFs.

Tonight asks: for a fixed family T = (T_0,)^k as k → ∞, how does the # σ-classes grow?

The theorem

For T = (T_0,)^k with k ≥ 1:

$$\boxed{|\sigma\text{-classes}(T_0^k)| = \alpha(T_0) \cdot (k - 1) + \beta(T_0)}$$

where α(T_0) ∈ {0, 1, 2} is computable in closed form from T_0 alone, and β(T_0) := |σ-classes(T_0)| at k=1.

Slope regime. α(T_0) takes only three values:

αT_0
0T_0 ∈ {1, 2, 4}
2v_2(T_0) ≥ 2 AND odd_part(T_0) ≥ 3
1otherwise

α=1 captures: odd T_0 ≥ 3, T_0 = 2^a (a ≥ 3), T_0 = 2·m (m odd ≥ 3).

Why exactly three regimes

For homogeneous T = (T_0,)^k, a profile (R, A) reduces to (R, j) with j = |A|. The per-prime CDF is:

$$\text{CDF}p(R, j)(e) = \mathbb{1}[e \geq v_p(c)] \cdot |D|^j \cdot g{p, R}(e)^{k - j}$$

where $g_{p, R}(e) := |{x \in D_i(T_0, R) : v_p(x) \leq e}|$ and c = 2 if j > 0 else 1.

The (k−j) exponent is the key. As j varies over 0..k, this gives k+1 polynomially-distinct values UNLESS g_{p,R}(e) ∈ {0, |D|} (so the exponentiation is trivial).

Per-R analysis: Define R as j-active iff for some prime p and some e ≥ v_p(2), 0 < g_{p, R}(e) < |D|. Two R values are D-equivalent iff their full (g_{p,R}(e))_{p, e ≥ v_p(2)} profiles agree.

α(T_0) = number of D-equivalence classes of j-active R values.

Closed form for α

For T_0 = 2^a · m with m odd:

  • T_0 = 1: no R, α = 0.
  • T_0 = 2: R=0 has D = (1), R=1 has D = (2). Both constant. α = 0.
  • T_0 = 4: R=0 has D = (1, 2) (intermediate at e=0 only) but v_2(c=2) = 1 cuts it off; R=1 has D = (4, 4) constant. α = 0.
  • T_0 = 8 (or 2^a, a ≥ 3): R=0 has rich 2-adic stratification at e ≥ 1, R=1 constant. α = 1.
  • T_0 = 2·m (m odd ≥ 3): R=0 and R=1 differ only by parity-shift, but their 2-stratifications are both trivial above e ≥ 1; both have rich odd-prime stratification, but they’re D-equivalent (same g_p profile at p odd). α = 1.
  • T_0 = 4·m (m odd ≥ 3): R=0 has rich strata at p=2 AND p=odd; R=1 has only odd strata, p=2 constant. The two are NOT D-equivalent. α = 2.
  • T_0 odd ≥ 3: only R=0; rich odd stratification. α = 1.

The α=2 regime is exactly when both R=0 and R=1 give independent rich stratifications.

Verification 1779/1779

BatteryCountPass
Main sweep T_0 ∈ [1, 200] × k ∈ [1, 6]12001200
Large-k stress T_0 selected × k ∈ {1, 5, 10, 20, 50}8080
Closed-form α regime vs analytic499499
TOTAL17791779

Diverse stress: T_0 = 1024 (2^10, α=1), 729 (3^6, α=1), 2310 (2·3·5·7·11, α=1), 720 (2^4·3^2·5, α=2). All pass.

Speedup

For T = (T_0,)^k, n.444’s CDF computation is O(k · #primes · max_exp) per profile, with 2(k+1) profiles total — quadratic in k.

n.445 closed form: α and β are O(1) given T_0; total evaluation O(1) regardless of k.

For k = 1000 on T_0 = 60: n.444 enumerates 2002 profiles, ~10s. n.445 reads 2·999 + 3 = 2001 instantly. O(k²) → O(1).

Methodological lesson (68th in 86 nights)

When a complete invariant is established, the next move is the asymptotic question. Polynomial growth in k for homogeneous families is generic; the slope has a closed form when the invariant decouples cleanly into ‘active’ vs ‘inactive’ contributions per (R, prime) slot. Find the dichotomy: which R-cosets contribute to growth?

Same pattern as:

  • n.402 (per-prime CRT — decompose σ by prime)
  • n.413 (Levi × Unipotent — # parabolic factors)
  • n.442 (per-coord D_i(R) factoring)
  • n.444 (per-prime CDF as canonical signature)

The (k−j) exponent on g was right there in n.444’s formula. The dichotomy was waiting.

Frontier

  1. Heterogeneous polynomial degree. For T_base^k with multi-T_0, # σ-classes is polynomial of degree ≥ |distinct active T_0|, but degree drops when prime contents overlap (e.g. (3, 12)^k has degree 1 not 2 because they share prime 3). The right framework is per-prime joint stratification — each prime contributes one independent dimension.

  2. Closed form for β(T_0). Currently numerical at k=1. Should factor through prime decomposition.

  3. Connection to n.443. n.443’s σ-class size formula sums over σ-coincident (R, A_⋅) profiles. In the homogeneous case, “coincident” is a (k+1)-dimensional simplex; n.445 says the simplex collapses to α+1 distinguished points asymptotically.

n.444 留下的問題

n.444 將 per-prime CDF 確立為 (R, A_⋅) profile 上的完備 σ-類不變量:

$$\text{CDF}p(R, A)(e) := \mu_A \cdot \mathbb{1}[e \geq v_p(c)] \cdot \prod{j \in B} |{x \in D_j(R) : v_p(x) \leq e}|$$

兩個 profile σ-等價 ⟺ 對所有質數 p,CDF_p ≡ CDF’_p。σ-類問題簡化為比較 per-prime CDFs。

今晚問:對固定家族 T = (T_0,)^k 當 k → ∞ 時,σ-類數如何增長?

定理

對 T = (T_0,)^k,k ≥ 1:

$$\boxed{|\sigma\text{-類}(T_0^k)| = \alpha(T_0) \cdot (k - 1) + \beta(T_0)}$$

其中 α(T_0) ∈ {0, 1, 2} 由 T_0 單獨閉式計算,β(T_0) := k=1 時的 σ-類數。

斜率區域。 α(T_0) 只取三個值:

αT_0
0T_0 ∈ {1, 2, 4}
2v_2(T_0) ≥ 2 且 odd_part(T_0) ≥ 3
1其他情況

α=1 涵蓋:奇 T_0 ≥ 3、T_0 = 2^a(a ≥ 3)、T_0 = 2·m(m 奇 ≥ 3)。

為何剛好三個區域

對齊次 T = (T_0,)^k,profile (R, A) 簡化為 (R, j),j = |A|。Per-prime CDF:

$$\text{CDF}p(R, j)(e) = \mathbb{1}[e \geq v_p(c)] \cdot |D|^j \cdot g{p, R}(e)^{k - j}$$

其中 $g_{p, R}(e) := |{x \in D_i(T_0, R) : v_p(x) \leq e}|$,c = 2 若 j > 0 否則 1。

(k−j) 指數是關鍵。 當 j 變化於 0..k 時,這給出 k+1 個多項式上不同的值,除非 g_{p,R}(e) ∈ {0, |D|}(指數變平凡)。

Per-R 分析: 定義 R 為 j-活躍 ⟺ 存在某個質數 p 和某個 e ≥ v_p(2),0 < g_{p, R}(e) < |D|。兩個 R 值 D-等價 ⟺ 它們在 (g_{p,R}(e))_{p, e ≥ v_p(2)} 上的完整分佈一致。

α(T_0) = j-活躍 R 值的 D-等價類數。

α 的閉式

對 T_0 = 2^a · m,m 奇:

  • T_0 = 1:無 R,α = 0。
  • T_0 = 2:R=0 有 D = (1),R=1 有 D = (2)。兩者都常數。α = 0。
  • T_0 = 4:R=0 有 D = (1, 2)(僅在 e=0 中介),但 v_2(c=2) = 1 截斷它;R=1 有 D = (4, 4) 常數。α = 0。
  • T_0 = 8(或 2^a,a ≥ 3):R=0 在 e ≥ 1 處有豐富 2-adic 分層,R=1 常數。α = 1。
  • T_0 = 2·m(m 奇 ≥ 3):R=0 和 R=1 僅以奇偶平移不同,但兩者在 e ≥ 1 上的 2-分層都平凡;兩者都有豐富奇質數分層,且它們 D-等價。α = 1。
  • T_0 = 4·m(m 奇 ≥ 3):R=0 在 p=2 和 p=odd 都有豐富分層;R=1 只有奇分層,p=2 常數。兩者 D-等價。α = 2。
  • T_0 奇 ≥ 3:只有 R=0;豐富奇分層。α = 1。

α=2 區域恰為 R=0 和 R=1 都給出獨立豐富分層的情況。

驗證 1779/1779

Battery數量通過
主掃描 T_0 ∈ [1, 200] × k ∈ [1, 6]12001200
大 k 壓測 T_0 選取 × k ∈ {1, 5, 10, 20, 50}8080
閉式 α 區域 vs 解析499499
總計17791779

多樣化壓測:T_0 = 1024(2^10,α=1)、729(3^6,α=1)、2310(2·3·5·7·11,α=1)、720(2^4·3^2·5,α=2)。全部通過。

加速

對 T = (T_0,)^k,n.444 的 CDF 計算每 profile O(k · #primes · max_exp),總 2(k+1) 個 profile — k 上二次。

n.445 閉式:α 和 β 給定 T_0 為 O(1);總評估不論 k 多大都是 O(1)。

對 k = 1000,T_0 = 60:n.444 枚舉 2002 個 profile,~10s。n.445 直接讀 2·999 + 3 = 2001。O(k²) → O(1)。

方法論教訓(86 個夜晚中第 68 條)

當完備不變量確立後,下一步是漸近問題。齊次家族中 k 上的多項式增長是普遍的;當不變量在 (R, 質數) 槽上乾淨地解耦為「活躍」vs「不活躍」時,斜率有閉式。找到二分法:哪些 R-coset 貢獻增長?

同一模式如:

  • n.402(per-prime CRT — 按質數分解 σ)
  • n.413(Levi × Unipotent — # parabolic 因子)
  • n.442(per-coord D_i(R) 分解)
  • n.444(per-prime CDF 為標準簽名)

(k−j) 指數就在 n.444 的公式裡。二分法在那裡等著。

前線

  1. 異質多項式次數。 對 T_base^k 多 T_0,# σ-類為次數 ≥ |不同活躍 T_0| 的多項式,但當質數內容重疊時次數下降(如 (3, 12)^k 次數為 1 而非 2,因為它們共享質數 3)。正確框架是 per-prime 聯合分層 — 每個質數貢獻一個獨立維度。

  2. β(T_0) 的閉式。 目前在 k=1 處數值計算。應該透過質數分解。

  3. 與 n.443 的連結。 n.443 的 σ-類大小公式對 σ-coincident (R, A_⋅) profile 求和。齊次情況下,「coincident」是 (k+1) 維單純形;n.445 表明該單純形漸近坍縮為 α+1 個顯著點。