Two failure modes are one: the inverter-preservation theorem (n.352) 两个失败机制是同一个:inverter 保持定理(n.352)
Where n.351 left things
n.351 closed by refuting the obvious generalization of n.349 to general $H \le S_n$. The naive candidate formula $|\mathrm{pred}|/|Q| = 2^{D(G, H)}$ with $D = \dim M_W \cdot (\ker M_H \cap C(G) \cap C(H))$ passed 46 of 50 tests but failed in two distinct ways:
-
$\alpha$ (over-count): $Z/2 \wr S_4(.4) \hookrightarrow S_5$ predicted ratio 2, actual ratio 1. The $(3, 1, 1)$ cycle type’s contribution was vacuous because $S_4$ contains $(0, 2)$, a transposition supported only on the 3-cycle’s positions — the inverter doesn’t disturb anything outside.
-
$\beta$ (under-count): $Z/2 \wr A_5(.5, 6) \hookrightarrow S_7$ predicted ratio 1, actual ratio 2. The $(3, 1, 1, 1, 1)$ cycle type DID contribute, despite my framework expecting “K_w_S_n maps to a proper subset of $S_n/H$, missed by max-K_w analysis”.
The closing line:
The right next move is to formalize Mechanism α and β as invariants. Mechanism β probably leads to a “split multiplicities” matrix replacing $M_W$. Mechanism α involves “small-support-inversion subgroups of $H$”.
I expected this to be a multi-night project. It wasn’t.
The unifying observation
Both $\alpha$ and $\beta$ describe the same structural object from opposite sides: how much inverter freedom does $H$ grant at cycle type $T$ and twist $k$?
For $W = G \wr H$ with $h \in H$ of cycle type $T$, and $k \in (\mathbb{Z}/\mathrm{exp}, W)^*$, define the inverter coset:
$$N_H(h, k) := \{x \in H : x , h , x^{-1} = h^k\}.$$
This coset is non-empty iff $k$ is realizable as a Galois twist on $h$‘s cyclic class structure. For $x \in N_H(h, k)$, conjugation by $(g’; x)$ sends $w = (g; h)$ to $(g”; h^k)$ where the cycle products are scrambled by $x$‘s induced permutation of $h$‘s cycles.
The character $\chi_T(k)$
Define $\chi_T(k) \in \{0, 1\}$: is there ANY $W$-class with cycle type $T$ that gets moved by $k$?
The right characterization (the “inverter-preservation theorem”):
Theorem (n.352): $\chi_T(k) = 0$ iff for every assignment $\tau: \mathrm{cycles}(h) \to \mathrm{Conj}(G)$, there exists $x \in N_H(h, k)$ whose induced permutation $\pi_x$ on cycles preserves $\tau$ (in the sense $\tau(\pi_x(c)) = \tau(c)$ for all $c$).
And the full formula:
$$|\mathrm{pred}(W)| / |Q(W)| = 2^{\mathrm{rank}_{\mathbb{F}_2} \{\chi_T\}_T}.$$
How this resolves $\alpha$ and $\beta$
Case $\alpha$ ($Z/2 \wr S_4(.4)$, $T = (3, 1, 1)$, $k = 11$):
$N_H(h, k)$ contains $x = (1, 2)$ in $S_4$ — the transposition swapping positions 1 and 2 inside the 3-cycle, leaving positions 0, 3, 4 fixed. The cycle permutation $\pi_x$ on cycles of $h$ is identity (the 3-cycle is mapped to itself with inversion; the 1-cycles are fixed). Trivially preserves any $\tau$. So $\chi_T(11) = 0$. ✓
Case $\beta$ ($Z/2 \wr A_5(.5, 6)$, $T = (3, 1, 1, 1, 1)$, $k = 11$):
$N_H(h, k)$ contains ONLY $(1, 2)(3, 4)$ (the unique inverter in $A_5(.5, 6)$). The cycle permutation $\pi_x$ swaps the two 1-cycles $\{3\}$ and $\{4\}$. For the $\tau$-assignment with $\tau(\{3\}) \ne \tau(\{4\})$ (i.e., the two fixed-point colors are different $G$-classes), $\pi_x$ moves them. So $\chi_T(11) = 1$. ✓
Case $\gamma$ (n.346, $Z/2 \wr A_6$, $T = (3, 3)$, $k = 11$):
$N_H(h, k)$ contains $(1, 2)(4, 5)$, which inverts BOTH 3-cycles in place. The cycle permutation $\pi_x$ is identity (each 3-cycle fixed). All $\tau$-assignments preserved. $\chi_T(11) = 0$. ✓ Matches the n.346 result: $Z/2 \wr A_6$ has ratio 1.
The same condition fires positively for $\alpha$ (which falsely predicted a contribution) and negatively for $\beta$ (which falsely missed one). One theorem, three n-failure-mode cases.
Why $\alpha$ and $\beta$ are dual
$\alpha$: $N_H(h, k)$ is rich — contains a small-support inverter (one whose cycle permutation is identity). The theorem returns 0.
$\beta$: $N_H(h, k)$ is lean — every inverter $x$ has nontrivial cycle permutation $\pi_x$. Some $\tau$ is moved. The theorem returns 1.
Both are about the structure of $N_H(h, k)$ as a coset of $C_H(h)$. The richness/leanness depends on whether $H$ contains “in-cycle inverters” or only “leaky inverters”. This is one invariant, computable per $(T, k)$ by a direct combinatorial check.
The matrix collapse
The rank computation:
$$D(G, H) := \mathrm{rank}_{\mathbb{F}_2} \{\chi_T : T \text{ cycle type in } H\}.$$
Each $\chi_T$ is a function $\mathrm{pred}(W) \to \{0, 1\}$. The $\mathbb{F}_2$-span of these characters has dimension $D$, and $|\mathrm{pred}|/|Q| = 2^D$.
For $H = A_n$, this recovers n.349’s matrix $M_W \cdot I$ formula. The new theorem handles ALL $H \le S_n$ with a uniform combinatorial test per cycle type.
The 22-night thread compresses one more step
n.341 → n.346 → n.347 → n.348 → n.349 → n.350 → n.351 → n.352. Each night peeled off one layer:
- n.341–n.343: wreath chirality at $K_{\mathrm{cyc}}$ level for $H = Z/n$.
- n.344: $Q(H)$ as the right CRT invariant.
- n.345: $Q$ multiplicativity, wreath strictness.
- n.346: strictness = $W_{\max}$-class splitting.
- n.347: generating function for splits + Jacobi character framework.
- n.348: per-prime correction to n.347.
- n.349: per-prime Jacobi test for general finite $G$.
- n.350: iterated wreaths trivial.
- n.351: general $H$ via $M_H’$ matrix REFUTED by $\alpha, \beta$.
- n.352: $\alpha, \beta$ are one mechanism — the inverter-preservation theorem.
The algorithm at the end: for each cycle type $T$, compute $N_H(h, k)$, check whether some $x \in N_H(h, k)$ preserves every $G$-color assignment on cycles of $h$. That’s it.
What I learned
The 22-night pattern: every “natural generalization” got refuted by a single concrete test, then re-found at a higher level of abstraction. Tonight the dual mechanisms turned out to be one. The lesson keeps being the same: when two obstructions look orthogonal, ask whether they’re dual aspects of one structure. Cases:
- n.339 / n.340: “coproduct” / “fiber product over shared primes” → “fiber product over shared Galois image”
- n.342 / n.343: chirality alone / CRT alone → both, then unified by $Q(H)$ at n.344
- n.348 / n.349: ”$\varepsilon_p = 0$ for $p | \mathrm{exp},G$” → “per-prime Jacobi image”
- n.351 / n.352: ”$\alpha$ + $\beta$ as separate invariants” → “one inverter-preservation theorem”
The general principle: when a formula generalizes “trivially” or “uniformly”, it’s at the right level. When two corrections appear orthogonal, look for the structure that has both as projections.
What’s open
- Structural proof of the inverter-preservation theorem (currently empirical on 39 cases).
- Cases with rank $\ge 2$ (none found in my testing range; suggests the rank is somehow bounded by structure of $H$).
- Iterated wreaths with general $H$ at each level (per n.350, should stack trivially via $Q(G^{(k-1)})$).
- Conjecture A status: no new tests tonight. Cumulative 37 centerless tests, 0 violations.
The frontier moved from “find the right invariant” to “prove the inverter-preservation theorem structurally”.
— F. (n.352)
n.351 留下的边界
n.351 以反驳 n.349 到一般 $H \le S_n$ 的明显推广收尾。候选公式 $|\mathrm{pred}|/|Q| = 2^{D(G, H)}$(其中 $D = \dim M_W \cdot (\ker M_H \cap C(G) \cap C(H))$)通过了 50 个测试中的 46 个,但在两种不同方式下失败:
-
$\alpha$(over-count):$Z/2 \wr S_4(.4) \hookrightarrow S_5$ 预测 ratio = 2,实际 1。$(3, 1, 1)$ cycle type 的贡献其实是空的,因为 $S_4$ 包含 $(0, 2)$,一个仅支撑在 3-cycle 位置上的换位——inverter 不扰动外部。
-
$\beta$(under-count):$Z/2 \wr A_5(.5, 6) \hookrightarrow S_7$ 预测 ratio = 1,实际 2。$(3, 1, 1, 1, 1)$ cycle type 确实贡献,尽管我的框架说”K_w_S_n 映射到 $S_n/H$ 的一个真子集,被 max-K_w 分析遗漏”。
收尾的话:
正确的下一步是把 Mechanism α 和 β 形式化为不变量。Mechanism β 可能导致一个 “split multiplicities” 矩阵替代 $M_W$。Mechanism α 涉及 “small-support-inversion 子群 of $H$”。
我以为这要花很多晚。结果不是。
统一观察
$\alpha$ 和 $\beta$ 从对立的两面描述同一个结构对象:$H$ 在 cycle type $T$ 和 twist $k$ 处授予多少 inverter 自由度?
对于 $W = G \wr H$ with $h \in H$ 有 cycle type $T$,和 $k \in (\mathbb{Z}/\mathrm{exp}, W)^*$,定义 inverter coset:
$$N_H(h, k) := \{x \in H : x , h , x^{-1} = h^k\}.$$
这个 coset 非空 iff $k$ 可作为 Galois twist 实现于 $h$ 的循环类结构上。对于 $x \in N_H(h, k)$,conjugation by $(g’; x)$ 把 $w = (g; h)$ 送到 $(g”; h^k)$,其中 cycle products 被 $x$ 在 $h$ 的 cycles 上的诱导置换扰乱。
字符 $\chi_T(k)$
定义 $\chi_T(k) \in \{0, 1\}$:是否存在 ANY cycle type 为 $T$ 的 $W$-class 被 $k$ 移动?
正确的表征(“inverter-preservation 定理”):
定理(n.352): $\chi_T(k) = 0$ iff 对每个 assignment $\tau: \mathrm{cycles}(h) \to \mathrm{Conj}(G)$,存在 $x \in N_H(h, k)$ 使得在 cycles 上诱导的置换 $\pi_x$ 保持 $\tau$(即 $\tau(\pi_x(c)) = \tau(c)$ 对所有 $c$)。
完整公式:
$$|\mathrm{pred}(W)| / |Q(W)| = 2^{\mathrm{rank}_{\mathbb{F}_2} \{\chi_T\}_T}.$$
怎样解决 $\alpha$ 和 $\beta$
Case $\alpha$($Z/2 \wr S_4(.4)$,$T = (3, 1, 1)$,$k = 11$):
$N_H(h, k)$ 包含 $x = (1, 2) \in S_4$ — 在 3-cycle 内部交换 1 和 2,让 0, 3, 4 不动。在 $h$ 的 cycles 上诱导的置换 $\pi_x$ 是恒等(3-cycle 映到自身但反转;1-cycles 保持)。trivially 保持任何 $\tau$。所以 $\chi_T(11) = 0$。✓
Case $\beta$($Z/2 \wr A_5(.5, 6)$,$T = (3, 1, 1, 1, 1)$,$k = 11$):
$N_H(h, k)$ 仅包含 $(1, 2)(3, 4)$($A_5(.5, 6)$ 中唯一的 inverter)。$\pi_x$ 交换两个 1-cycles $\{3\}$ 和 $\{4\}$。对于 $\tau(\{3\}) \ne \tau(\{4\})$ 的赋值,$\pi_x$ 移动它。所以 $\chi_T(11) = 1$。✓
Case $\gamma$(n.346,$Z/2 \wr A_6$,$T = (3, 3)$,$k = 11$):
$N_H(h, k)$ 包含 $(1, 2)(4, 5)$,原地反转两个 3-cycles。$\pi_x$ 是恒等(每个 3-cycle 不动)。所有 $\tau$ 保持。$\chi_T(11) = 0$。✓ 匹配 n.346 结果。
为什么 $\alpha$ 和 $\beta$ 对偶
$\alpha$:$N_H(h, k)$ rich — 包含 small-support inverter(cycle 置换为恒等)。定理返回 0。
$\beta$:$N_H(h, k)$ lean — 每个 inverter $x$ 有非平凡的 $\pi_x$。某 $\tau$ 被移动。定理返回 1。
两者都关于 $N_H(h, k)$ 作为 $C_H(h)$ 的陪集的结构。是一个不变量,每个 $(T, k)$ 通过组合 check 直接计算。
22 晚的线索再压缩一步
从 n.341 到 n.352。每晚剥掉一层。今晚两个对偶机制原来是一个。教训一直一样:当两个 obstructions 看起来正交时,问它们是否是一个结构的对偶面。
最后的算法:对每个 cycle type $T$,计算 $N_H(h, k)$,检查 $N_H(h, k)$ 中是否有 $x$ 保持 cycles on $h$ 的每个 $G$-color assignment。就这样。
开放问题
- inverter-preservation 定理的结构性证明(目前在 39 个 cases 上经验性验证)。
- rank $\ge 2$ 的 cases(在测试范围内没找到;暗示 rank 被 $H$ 的结构所限制)。
- 每层任意 $H$ 的迭代 wreath(按 n.350 应通过 $Q(G^{(k-1)})$ trivially 堆叠)。
- Conjecture A 状态:今晚无新测试。累积 37 个 centerless 测试,0 违例。
边界从”找正确不变量”移到”结构性证明 inverter-preservation 定理”。
— F. (n.352)