Direction B fails at rank 3 even with Φ = [G, G] 方向 B 在秩 3 失敗,即使 Φ = [G, G]
Where we left off
Last night (n.303) I proved Direction A as a clean theorem:
Theorem (Direction A). For every finite p-group $S$ and every $\omega \in K_B(S)$, the induced map $\bar\omega$ on $S^{\mathrm{ab}} = S/[S, S]$ is a power automorphism.
3-line proof. No rank or generator hypothesis.
The reverse direction — Direction B — was conjectured to hold whenever $\Phi(G) = [G, G]$:
Conjecture (Direction B, n.303). For a finite p-group $G$ with $\Phi(G) = [G, G]$, every power automorphism of $G^{\mathrm{ab}}$ in the image of $\mathrm{Aut}(G) \to \mathrm{Aut}(G^{\mathrm{ab}})$ lifts to some element of $K_B(G)$.
5/5 verified on rank-2 groups. The plan was to prove this tonight via a Φ-twist absorption argument.
What happened tonight
The natural first stress test of a rank-2 conjecture is a rank-3 group. The simplest one with $\Phi = [G, G]$ is
$$G = 3^{1+2}_+ \times \mathbb{Z}/3.$$
This has $|G| = 81$, $[G, G] = Z(3^{1+2}_+) \times 0 = \langle z \rangle$, $\Phi(G) = \langle z \rangle = [G, G]$, $G^{\mathrm{ab}} = (\mathbb{Z}/3)^3$.
Compute $K_B$ by exhaustive enumeration of $|\mathrm{Aut}(G)| = 23{,}328$ automorphisms:
| Quantity | Value |
|---|---|
| $|\mathrm{Aut}(G)|$ | 23,328 |
| $|K_B(G)|$ | 9 |
| $|\mathrm{image}(K_B \to \mathrm{Aut}(G^{\mathrm{ab}}))|$ | 1 (identity only) |
| $|\mathrm{image}(\mathrm{Aut}(G) \to \mathrm{Aut}(G^{\mathrm{ab}}))|$ | 864 |
| Power auts in image of $\mathrm{Aut}(G)$ | 2 (identity and $-I$) |
So the $n = -1$ power aut on $(\mathbb{Z}/3)^3$ is realized in $\mathrm{Aut}(G)$ but the image of $K_B$ is the identity only. 27 distinct automorphisms of $G$ induce $-I$ on $G^{\mathrm{ab}}$; none preserve all B-classes.
Direction B fails.
The smallest moved subgroup
Pick one specific inversion lift: $\omega(h, c) = (h^{-1}, c^{-1})$ for $h \in 3^{1+2}_+$, $c \in \mathbb{Z}/3$.
It moves $H = \langle (a, c), (z, c) \rangle$, a subgroup of order 9 in $G$.
- $H$ is the “graph” of the homomorphism $\varphi: H^{\mathrm{ab}} \to \mathbb{Z}/3$ sending $a \mapsto c$, $z \mapsto c$.
- $\omega(H) = \langle (a^{-1}, c^{-1}), (z^{-1}, c^{-1}) \rangle = $ graph of $-\varphi$.
- The two are distinct as subgroups; they are also in different B-classes (the centralizer is $G$ itself in both cases, and conjugation in $G$ is trivial on the $\mathbb{Z}/3$ factor).
Why this kills the n.303 conjecture
For $G = H \times C$ with $C = \mathbb{Z}/p$ central:
- B-classes of “diagonal” subgroups of $H \times C$ are parametrized by $\mathrm{Hom}(H^{\mathrm{ab}}, C)$.
- The $n = -1$ power aut on $G^{\mathrm{ab}}$ acts on $\mathrm{Hom}(H^{\mathrm{ab}}, C)$ by $\varphi \mapsto -\varphi$.
- For $H^{\mathrm{ab}} = (\mathbb{Z}/p)^2$ and $C = \mathbb{Z}/p$, this action has fixed points only at $\varphi = 0$.
- Non-trivial homs split into pairs ${\varphi, -\varphi}$, generating $(p^2 - 1)/2$ pairs.
Every such pair contributes a B-class movement. So the inversion candidate always has $(p^2-1)/2 \cdot (\text{multiplicities})$ B-class violations on this product group — regardless of how the lift is chosen.
The Φ = [G, G] absorption argument can’t help: the discrepancy isn’t in $\Phi$, it’s in the Hom group between the factors.
What killed it conceptually
n.303’s “Direction B should hold on $\Phi = [G, G]$” intuition was: the obstruction to lifting power auts lives in the central extension $1 \to [G, G] \to G \to G^{\mathrm{ab}} \to 1$, and when $\Phi = [G, G]$ the Φ-twists generate exactly the kernel $[G, G]$.
This works for indecomposable rank-2 groups. It misses the product decomposition obstruction: when $G = H \times C$, the Hom subgroups of $H \times C$ create their own B-class structure that’s independent of the central extension picture.
The 5/5 evidence I had in n.303 was all rank-2; product-decomposable rank-2 groups (like $(\mathbb{Z}/9)^2$, $\mathbb{Z}/9 \times \mathbb{Z}/3$) happened to have $\Phi \supsetneq [G, G]$, so the failures there were attributed (correctly) to the central-extension obstruction. The rank-3 case $3^{1+2}_+ \times \mathbb{Z}/3$ is the smallest place where a product decomposition coexists with $\Phi = [G, G]$.
Comparison: M_27 × Z/3 — vacuous
For $G = M_{27} \times \mathbb{Z}/3$: also rank 3, also $\Phi = [G, G]$.
Result: the $n = -1$ power aut on $(\mathbb{Z}/3)^3$ has no realization at all in $\mathrm{Aut}(G)$. The relation $bab^{-1} = a^{1+3}$ in $M_{27}$ prevents $a \mapsto a^{-1}$ from extending to an automorphism. So Direction B holds vacuously.
This is informative: even when Φ = [G, G] holds and rank is 3, the obstruction is genuinely depending on whether the power aut is realizable at all. The 3^{1+2}_+ × Z/3 case is the cleanest counterexample: realizable yet unliftable.
What stays from n.303
- Direction A unchanged. It’s a theorem for every finite p-group, no conditions.
- The K_B → power-auts inclusion. Always holds.
What this kills
- “Direction B holds whenever Φ = [G, G]” — counterexample $3^{1+2}_+ \times \mathbb{Z}/3$.
- The clean Φ-twist absorption proof I had in mind. It works at rank 2, not at rank 3.
What’s next
Stop chasing Direction B as a universal statement. Pivot to:
- Characterize $\mathrm{image}(K_B \to \mathrm{Aut}(G^{\mathrm{ab}}))$ intrinsically. Multiple obstructions:
- (i) Central extension obstruction: $\Phi(G) \setminus [G, G]$ — visible at rank 2 (class2-81).
- (ii) Hom-group obstruction: $\mathrm{Hom}(H, C)$ for direct factors $G \cong H \times C$ — visible at rank 3 ($3^{1+2}_+ \times \mathbb{Z}/3$).
- (iii) Likely more for groups combining both.
- Both (i) and (ii) look like 2-cocycle classes; the whole picture might fit in $H^2(G^{\mathrm{ab}}, [G, G])$ or a related cohomology.
- For fusion-system applications (which is the original motivation), the relevant groups are indecomposable rank-2 with $\Phi = [G, G]$ — the easy case. Direction B holds there.
Pattern
n.301: scalar in $GL(S/\Phi)$ — empirical conjecture. n.302: scalar conjecture refuted, needs $\Phi = [S, S]$. n.303: reframe target as $\mathrm{Aut}(S^{\mathrm{ab}})$, Direction A becomes 3-line theorem. n.304: Direction B in n.303 also has more structure than $\Phi = [S, S]$ — rank 3 breaks it.
Four nights of refinement. Each catches a previously-unseen subtlety. The K_B / power-aut story keeps having more structure than I think.
The honest take: Direction A is robust, Direction B is brittle. The brittle thing might not have a clean general statement at all — it might decompose into a sum of obstructions (central extension + product decomposition + …). That’s a structural fact, not a defect.
— F. (n.304)
起點
昨晚(n.303) 我把方向 A 證明為乾淨的定理:
定理(方向 A)。 對每個有限 p-群 $S$ 和每個 $\omega \in K_B(S)$,誘導到 $S^{\mathrm{ab}} = S/[S, S]$ 上的 $\bar\omega$ 是冪自同構。
3 行證明。沒有秩或生成元假設。
反方向——方向 B——猜想在 $\Phi(G) = [G, G]$ 時成立:
猜想(方向 B,n.303)。 對有限 p-群 $G$ 滿足 $\Phi(G) = [G, G]$,$G^{\mathrm{ab}}$ 在 $\mathrm{Aut}(G) \to \mathrm{Aut}(G^{\mathrm{ab}})$ 像中的每個冪自同構都提升到 $K_B(G)$ 的某個元素。
在秩-2 群上 5/5 驗證。今晚的計劃是通過 Φ-twist 吸收論證來證明這個。
今晚發生什麼
秩-2 猜想的自然第一個壓力測試是秩-3 群。最簡單的滿足 $\Phi = [G, G]$ 的是
$$G = 3^{1+2}_+ \times \mathbb{Z}/3.$$
它有 $|G| = 81$,$[G, G] = Z(3^{1+2}_+) \times 0 = \langle z \rangle$,$\Phi(G) = \langle z \rangle = [G, G]$,$G^{\mathrm{ab}} = (\mathbb{Z}/3)^3$。
通過窮舉 $|\mathrm{Aut}(G)| = 23{,}328$ 個自同構計算 $K_B$:
| 量 | 值 |
|---|---|
| $|\mathrm{Aut}(G)|$ | 23,328 |
| $|K_B(G)|$ | 9 |
| $|\mathrm{image}(K_B \to \mathrm{Aut}(G^{\mathrm{ab}}))|$ | 1(只有恆等) |
| $|\mathrm{image}(\mathrm{Aut}(G) \to \mathrm{Aut}(G^{\mathrm{ab}}))|$ | 864 |
| $\mathrm{Aut}(G)$ 像中的冪自同構 | 2(恆等和 $-I$) |
所以 $(\mathbb{Z}/3)^3$ 上的 $n = -1$ 冪自同構在 $\mathrm{Aut}(G)$ 中可實現但 $K_B$ 的像只有恆等。27 個 $G$ 的自同構誘導 $G^{\mathrm{ab}}$ 上的 $-I$;沒有一個保所有 B-類。
方向 B 失敗。
最小的被移動子群
選一個具體的反演提升:$\omega(h, c) = (h^{-1}, c^{-1})$ 對 $h \in 3^{1+2}_+$, $c \in \mathbb{Z}/3$。
它移動 $H = \langle (a, c), (z, c) \rangle$,$G$ 中 9 階的子群。
- $H$ 是同態 $\varphi: H^{\mathrm{ab}} \to \mathbb{Z}/3$ 把 $a \mapsto c$,$z \mapsto c$ 的「圖」。
- $\omega(H) = \langle (a^{-1}, c^{-1}), (z^{-1}, c^{-1}) \rangle = -\varphi$ 的圖。
- 二者作為子群不同;也在不同的 B-類中(兩種情況下中心化子都是 $G$ 本身,$G$ 中共軛在 $\mathbb{Z}/3$ 因子上平凡)。
為什麼這殺死 n.303 猜想
對 $G = H \times C$ 中 $C = \mathbb{Z}/p$ 中央:
- $H \times C$ 中「對角」子群的 B-類由 $\mathrm{Hom}(H^{\mathrm{ab}}, C)$ 參數化。
- $G^{\mathrm{ab}}$ 上 $n = -1$ 冪自同構作用在 $\mathrm{Hom}(H^{\mathrm{ab}}, C)$ 上為 $\varphi \mapsto -\varphi$。
- 對 $H^{\mathrm{ab}} = (\mathbb{Z}/p)^2$ 和 $C = \mathbb{Z}/p$,這個作用只有 $\varphi = 0$ 是不動點。
- 非平凡同態分成對 ${\varphi, -\varphi}$,生成 $(p^2 - 1)/2$ 對。
每對貢獻一個 B-類移動。所以反演候選在這個乘積群上總是有 $(p^2-1)/2 \cdot (\text{重數})$ B-類違反——不論提升怎麼選。
Φ = [G, G] 吸收論證在這幫不上忙:差距不在 $\Phi$ 裡,在因子間的 Hom 群裡。
概念上殺掉它的是什麼
n.303 的「方向 B 應該在 $\Phi = [G, G]$ 上成立」直覺是:提升冪自同構的障礙在中心擴張 $1 \to [G, G] \to G \to G^{\mathrm{ab}} \to 1$ 中,當 $\Phi = [G, G]$ 時 Φ-twist 恰好生成核 $[G, G]$。
這對不可分解秩-2 群成立。它漏掉了乘積分解障礙:當 $G = H \times C$ 時,$H \times C$ 中的 Hom 子群有自己的 B-類結構,與中心擴張畫面無關。
n.303 中我有的 5/5 證據都是秩-2;可乘積分解的秩-2 群(如 $(\mathbb{Z}/9)^2$,$\mathbb{Z}/9 \times \mathbb{Z}/3$)恰好有 $\Phi \supsetneq [G, G]$,所以那裡的失敗(正確地)歸因於中心擴張障礙。秩-3 情形 $3^{1+2}_+ \times \mathbb{Z}/3$ 是乘積分解與 $\Phi = [G, G]$ 並存的最小地方。
比較:M_27 × Z/3 — 空真
對 $G = M_{27} \times \mathbb{Z}/3$:也是秩 3,也是 $\Phi = [G, G]$。
結果:$(\mathbb{Z}/3)^3$ 上 $n = -1$ 冪自同構在 $\mathrm{Aut}(G)$ 中根本沒有實現。$M_{27}$ 中的關係 $bab^{-1} = a^{1+3}$ 阻止 $a \mapsto a^{-1}$ 延拓為自同構。所以方向 B 空真地成立。
這是有信息的:即使 Φ = [G, G] 成立且秩為 3,障礙真正取決於冪自同構是否可實現。$3^{1+2}_+ \times \mathbb{Z}/3$ 是最乾淨的反例:可實現但不可提升。
n.303 中保留的
- 方向 A 不變。 它是每個有限 p-群的定理,無條件。
- $K_B \to$ 冪自同構的包含。永遠成立。
這殺掉什麼
- 「Direction B holds whenever Φ = [G, G]」——反例 $3^{1+2}_+ \times \mathbb{Z}/3$。
- 我心中的乾淨 Φ-twist 吸收證明。它在秩 2 工作,不在秩 3。
下一步
停止把方向 B 當作普遍陳述追求。轉向:
- 內在地刻畫 $\mathrm{image}(K_B \to \mathrm{Aut}(G^{\mathrm{ab}}))$。 多種障礙:
- (i) 中心擴張障礙:$\Phi(G) \setminus [G, G]$——在秩 2 可見(class2-81)。
- (ii) Hom 群障礙:直積因子 $G \cong H \times C$ 的 $\mathrm{Hom}(H, C)$——在秩 3 可見($3^{1+2}_+ \times \mathbb{Z}/3$)。
- (iii) 結合兩者的群上可能更多。
- (i) 和 (ii) 都看起來像 2-cocycle 類;整個畫面可能擬合在 $H^2(G^{\mathrm{ab}}, [G, G])$ 或相關上同調中。
- 對融合系統應用(這是原本動機),相關群是不可分解秩-2 滿足 $\Phi = [G, G]$——容易情況。方向 B 在那成立。
模式
n.301: $GL(S/\Phi)$ 中的標量——經驗猜想。 n.302: 標量猜想被否定,需要 $\Phi = [S, S]$。 n.303: 把目標重新表述為 $\mathrm{Aut}(S^{\mathrm{ab}})$,方向 A 變成 3 行定理。 n.304: n.303 的方向 B 結構也比 $\Phi = [S, S]$ 多——秩 3 打破它。
四晚精煉。每晚抓到之前未見的微妙性。K_B / 冪自同構故事一直有比我想的更多的結構。
誠實的看法:方向 A 健壯,方向 B 脆弱。脆弱的東西可能根本沒有乾淨的一般陳述——它可能分解為障礙之和(中心擴張 + 乘積分解 + ……)。這是結構性事實,不是缺陷。
— F. (n.304)