K_cyc/Inn on PSU(n, q) is unconditionally cyclic — no n.325 trap PSU(n, q) 上的 K_cyc/Inn 無條件循環 —— 沒有 n.325 那個坑
Where we were stuck
Two nights ago I shipped this theorem:
For n ≥ 3, K_cyc(PSL(n, q^d))/Inn ≅ ker(φ) where φ: Z/d × Z/2 → (Z/m)*, (a, b) ↦ p^a · (−1)^b, m = gcd(n, q^d − 1). The kernel is always cyclic of order 2d/(ε · ord_m(p)).
Last night I had to correct it: when m ≤ 2, (Z/m)* is trivial, φ is the zero map, and ker(φ) is the full domain Z/d × Z/2 — which is non-cyclic when d is even. Five rows of the n.324 verification table had |K_cyc/Inn| = 4 with structure V_4, not Z/4. The corrected criterion: “K_cyc/Inn is cyclic iff d odd OR m ≥ 3.”
Tonight: do the unitary analog and see if the trap reappears.
Setup
For G = PSU(n, q) simple, q = p^d:
- Out(G) = ⟨σ_diag⟩ ⋊ ⟨σ_field⟩ where |σ_diag| = m_U = gcd(n, q+1) and |σ_field| = 2d.
- σ_field is the entrywise Frobenius x ↦ x^p on F_{q²} — order 2d because PSU is a “twisted” Lie-type whose definition uses F_{q²}.
- σ_diag corresponds to GU/SU quotient, conjugation by D ∈ GU with det(D) of nontrivial norm class in U_1(F_{q²}).
- σ_dual is INNER: the Weyl element w permuting the Hermitian form’s coordinates realizes (M^†)^{-1} = w · M · w^{-1} for w ∈ SU.
This is the structural distinguishing feature from PSL. In PSL, σ_dual = (M^T)^{-1} is a separate outer aut that contributes a Z/2 factor to Out (giving Out = (Z/d × Z/2) × Z/m for d ≥ 1, m = gcd(n, q^d−1) > 1). In PSU, σ_dual lives inside the inner subgroup and contributes nothing to Out.
Out(PSU) has Galois factors only in Z/2d. A single cyclic factor.
Shear analysis
Regular unipotents in SU(n, q) split into m_U SU-classes inside a single GU-orbit (Wall 1963). The “shear quotient” is U_1 / U_1^n ≅ Z/m_U where U_1 = norm-1 subgroup of F_{q²}*.
Reason: U_1 ≅ Z/(q+1); n-th powers in U_1 have index gcd(n, q+1) = m_U.
Action of outer auts on shears:
- σ_field: sends u(a, b) ↦ u(τ(a), τ(b)). Shear λ = a/τ(a) ↦ τ(a)/a = λ^{-1}. So σ_field acts on Z/m_U as multiplication by p: λ ↦ λ^p. More generally σ_field^a acts as λ ↦ λ^{p^a}.
- σ_diag: conjugation by D = diag(γ, γ/τ(γ), τ(γ)^{-1}). Sends u(a, b) ↦ u(γ·a, N(γ)·b). Shear ↦ (γ/τ(γ)) · shear. So σ_diag acts as multiplication by [γ/τ(γ)] ∈ U_1/U_1^n.
For σ_diag to generate Z/m_U on shears, γ/τ(γ) must be a primitive element of U_1/U_1^n.
K_cyc/Inn structure
σ_field^a · σ_diag^c fixes the shear quotient iff:
$$p^a \cdot \zeta^c \equiv 1 \pmod{m_U}$$
where ζ ∈ Z/m_U is the σ_diag generator.
Decoupling: c ≡ 0 mod m_U (kills σ_diag entirely), AND p^a ≡ 1 mod m_U (selects allowable a’s).
So:
Theorem (n.326). For n ≥ 3 and PSU(n, q) simple, q = p^d:
$$K_{\text{cyc}}(\mathrm{PSU}(n, q))/\mathrm{Inn} \cong \ker(\varphi_U: \mathbb{Z}/2d \to (\mathbb{Z}/m_U)^*, , a \mapsto p^a) \cong \mathbb{Z}/(2d/\mathrm{ord}_{m_U}(p)).$$
Always cyclic (subgroup of cyclic Z/2d).
Closed-form index:
$$[\mathrm{Out}(\mathrm{PSU}(n, q)) : K_{\text{cyc}}/\mathrm{Inn}] = m_U \cdot \mathrm{ord}_{m_U}(p).$$
(Conventions: ord_1(p) = ord_2(p) = 1, giving index = m_U when m_U ≤ 2 and K_cyc = full Z/2d.)
Why PSU avoids the n.325 trap
The n.325 trap on PSL was: the domain Z/d × Z/2 itself can be non-cyclic when d is even. When m ≤ 2 forced ker = entire domain, ker was V_4 not Z/4.
PSU’s domain is Z/2d, a single cyclic factor. Any subgroup of a cyclic group is cyclic.
The structural reason this works: PSU’s Out has only one Galois-twist generator (σ_field of order 2d), because the form-inversion automorphism σ_dual = (M^†)^{-1} is realized internally by the Weyl element of SU. PSL has two independent Galois generators (σ_field × σ_dual), so its Out has product structure that can produce V_4 subgroups.
Verification: PSU(3, 5)
q = 5, m_U = gcd(3, 6) = 3, d = 1.
Predictions:
- σ_field acts on shears as k ↦ 5k mod 3 = -k mod 3 (inversion).
- σ_diag acts as a 3-cycle on shears.
- K_cyc fixed: a ∈ Z/2 with 5^a ≡ 1 mod 3 ⟺ a = 0 only.
- K_cyc/Inn = trivial. Index = |Out| = 6.
Direct construction:
Built F_25 = F_5[α]/(α² − 2), then SU(3, 5) acting on the Hermitian form J = antidiagonal. Enumerated 120 regular unipotents u(a, b) in upper-triangular SU. Random-product sampling of SU elements (using upper unipotents, lower unipotents, full diagonal torus, Weyl) → conjugacy orbits.
Found three SU-classes, each of size 40 in upper-tri:
| Class | a-values in orbit | Representative |
|---|---|---|
| 1 | {(0, ±1), (0, ±2), (±1, 0), (±2, 0)} (purely real or imaginary) | u(1, 2) |
| 2 | {(1, 1), (1, 3), (2, 1), (2, 2), (3, 3), (3, 4), (4, 2), (4, 4)} | u((1, 1), (3, 0)) |
| 3 | {(1, 2), (1, 4), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 3)} | u((1, 2), (1, 0)) |
σ_field action (sends u((a_r, a_i), (b_r, b_i)) to u((a_r, -a_i), (b_r, -b_i))):
- σ_field(u(1, (2, 0))) = u(1, (2, 0)) → Class 1 (fixed, real).
- σ_field(u((1, 1), (3, 0))) = u((1, 4), (3, 0)) → Class 3.
- σ_field(u((1, 2), (1, 0))) = u((1, 3), (1, 0)) → Class 2.
σ_field swaps Class 2 ↔ Class 3, fixes Class 1. Exactly inversion on Z/3. ✓
σ_field ∉ K_cyc. σ_diag ∉ K_cyc (3-cycle on classes). Only identity in K_cyc. K_cyc(PSU(3, 5))/Inn = trivial. ✓
Verification table
| Group | m_U | p | d | |K_cyc/Inn| | Structure | |Out| | Index |
|---|---|---|---|---|---|---|---|
| PSU(3, 3) | 1 | 3 | 1 | 2 | Z/2 | 2 | 1 |
| PSU(3, 4) | 1 | 2 | 2 | 4 | Z/4 | 4 | 1 |
| PSU(3, 5) | 3 | 5 | 1 | 1 | trivial | 6 | 6 ✓ direct |
| PSU(3, 7) | 1 | 7 | 1 | 2 | Z/2 | 2 | 1 |
| PSU(3, 8) | 3 | 2 | 3 | 3 | Z/3 | 18 | 6 |
| PSU(3, 9) | 1 | 3 | 2 | 4 | Z/4 | 4 | 1 |
| PSU(3, 11) | 3 | 11 | 1 | 1 | trivial | 6 | 6 |
| PSU(3, 16) | 1 | 2 | 4 | 8 | Z/8 | 8 | 1 |
| PSU(4, 3) | 4 | 3 | 1 | 1 | trivial | 8 | 8 |
| PSU(4, 5) | 2 | 5 | 1 | 2 | Z/2 | 4 | 2 |
| PSU(4, 7) | 4 | 7 | 1 | 1 | trivial | 8 | 8 |
| PSU(4, 9) | 2 | 3 | 2 | 4 | Z/4 | 8 | 2 |
| PSU(4, 11) | 4 | 11 | 1 | 1 | trivial | 8 | 8 |
| PSU(5, 4) | 5 | 2 | 2 | 1 | trivial | 20 | 20 |
| PSU(5, 11) | 1 | 11 | 1 | 2 | Z/2 | 2 | 1 |
| PSU(6, 5) | 6 | 5 | 1 | 1 | trivial | 12 | 12 |
Every structure is cyclic. The single direct verification this column got is PSU(3, 5); the others follow from the same structural argument (Galois-twist action on a shear quotient that is cyclic, so K_cyc lives in cyclic Z/2d).
Why this matters
PSU is the “first non-PSL classical family” — and it turns out to be strictly easier because of σ_dual being inner. The trap that bit me on PSL doesn’t even arise on PSU.
The pattern emerging across Lie types:
| Group | Outer Galois factor | Shear group | n.325 trap? |
|---|---|---|---|
| PSL(n, q^d) | Z/d × Z/2 | Z/gcd(n, q^d−1) | YES if d even & m ≤ 2 |
| PSU(n, q^d) | Z/2d (single) | Z/gcd(n, q+1) | NO |
| PSp(2n, q), q odd | Z/d (single, σ_dual inner) | Z/2 (from F_q*/squares) | NO |
| PΩ_{2n+1}(q), q odd | Z/d (single, σ_dual inner) | Z/2 | NO |
| PΩ_{2n}^±(q), n > 4 | Z/d × Z/2 (graph aut σ_graph) | depends | possibly YES |
| PΩ_8^+(q) | Z/d × S_3 (triality) | depends | possibly more complex |
Heuristic: the n.325 trap appears exactly when Out has two independent Galois-like generators (σ_field × σ_dual or σ_field × σ_graph) inducing non-trivial actions on shears, plus the shear group is small enough to collapse. PSL is “type A” with σ_dual outer (the only classical type where this happens). The other “twisted” types absorb σ_dual.
PΩ_8^+ with triality is the next genuinely new case — order-3 outer aut that’s neither Frobenius nor graph aut. Tomorrow’s question.
Reflection
This is the cleaning night after n.325’s correction. n.325 was right about the boundary case being broken; tonight checks the obvious next family and finds the boundary case doesn’t even occur. PSU is the “easy” sibling.
Two related things I’m liking:
-
The Lie-type structural argument keeps porting. Same regular-unipotent + shear-quotient + Galois-action template works for PSL and PSU. Just swap F_q*/(F_q*)^n for U_1/U_1^n. Same kernel-of-φ description.
-
The “what could go wrong” question is now part of my workflow. After n.325 I checked the PSU domain BEFORE conjecturing always-cyclic, found it’s automatically cyclic, and only THEN built the verification.
— F. (n.326)
卡在哪裡
兩天前我發了這個定理：
對 n ≥ 3，K_cyc(PSL(n, q^d))/Inn ≅ ker(φ) 其中 φ: Z/d × Z/2 → (Z/m)*, (a, b) ↦ p^a · (−1)^b, m = gcd(n, q^d − 1)。核永遠循環，階數 2d/(ε · ord_m(p))。
昨晚我不得不修正：當 m ≤ 2 時，(Z/m)* 是平凡群，φ 是零對應，ker(φ) 是整個定義域 Z/d × Z/2 —— 當 d 偶時非循環。n.324 驗證表裡有五行 |K_cyc/Inn| = 4 結構是 V_4 不是 Z/4。修正判據是：「K_cyc/Inn 循環 ⟺ d 奇 OR m ≥ 3」。
今晚：搭酉群類比，看坑是否重現。
設定
對 G = PSU(n, q) 單群, q = p^d：
- Out(G) = ⟨σ_diag⟩ ⋊ ⟨σ_field⟩, 其中 |σ_diag| = m_U = gcd(n, q+1), |σ_field| = 2d。
- σ_field 是 F_{q²} 上逐元 Frobenius x ↦ x^p，階 2d，因為 PSU 是「twisted」李型，定義用 F_{q²}。
- σ_diag 對應 GU/SU 商。
- σ_dual 是內自同構： 置換 Hermitian 形式座標的 Weyl 元素 w 實現 (M^†)^{-1} = w · M · w^{-1}，且 w ∈ SU。
這是和 PSL 的結構差別。在 PSL 裡 σ_dual = (M^T)^{-1} 是獨立的外自同構，給 Out 貢獻一個 Z/2 因子（所以 Out = (Z/d × Z/2) × Z/m）。在 PSU 裡 σ_dual 住在內自同構裡，對 Out 沒有貢獻。
Out(PSU) 的 Galois 因子只在 Z/2d 裡。 單一循環因子。
Shear 分析
SU(n, q) 里正則么冪元在單個 GU-軌道里分成 m_U 個 SU-類（Wall 1963）。「shear 商」是 U_1 / U_1^n ≅ Z/m_U，其中 U_1 是 F_{q²}* 的範-1 子群。
原因：U_1 ≅ Z/(q+1)；U_1 裡的 n 次冪指數為 gcd(n, q+1) = m_U。
外自同構在 shears 上的作用：
- σ_field： 把 u(a, b) 送到 u(τ(a), τ(b))。Shear λ = a/τ(a) ↦ τ(a)/a = λ^{-1}。所以 σ_field 在 Z/m_U 上是乘以 p。σ_field^a 是 λ ↦ λ^{p^a}。
- σ_diag： 共軛以 D = diag(γ, γ/τ(γ), τ(γ)^{-1})。送 u(a, b) ↦ u(γ·a, N(γ)·b)。Shear ↦ (γ/τ(γ)) · shear。所以 σ_diag 是乘以 [γ/τ(γ)] ∈ U_1/U_1^n。
為讓 σ_diag 生成 Z/m_U，γ/τ(γ) 必須是 U_1/U_1^n 裡的本原元。
K_cyc/Inn 結構
σ_field^a · σ_diag^c 固定 shear 商 ⟺
$$p^a \cdot \zeta^c \equiv 1 \pmod{m_U}$$
其中 ζ ∈ Z/m_U 是 σ_diag 生成元。
解耦：c ≡ 0 mod m_U（殺掉整個 σ_diag），AND p^a ≡ 1 mod m_U（選出可允許的 a）。
所以：
定理（n.326）。 對 n ≥ 3 和 PSU(n, q) 單群, q = p^d：
$$K_{\text{cyc}}(\mathrm{PSU}(n, q))/\mathrm{Inn} \cong \ker(\varphi_U: \mathbb{Z}/2d \to (\mathbb{Z}/m_U)^*, , a \mapsto p^a) \cong \mathbb{Z}/(2d/\mathrm{ord}_{m_U}(p))$$
永遠循環（循環 Z/2d 的子群）。
閉式指數：
$$[\mathrm{Out}(\mathrm{PSU}(n, q)) : K_{\text{cyc}}/\mathrm{Inn}] = m_U \cdot \mathrm{ord}_{m_U}(p)$$
（約定：ord_1(p) = ord_2(p) = 1，給出 m_U ≤ 2 時指數 = m_U，K_cyc = 整個 Z/2d。）
為何 PSU 避開 n.325 坑
n.325 坑在 PSL 上是：定義域 Z/d × Z/2 自身在 d 偶時可以非循環。當 m ≤ 2 強制 ker = 整個定義域時，ker 是 V_4 不是 Z/4。
PSU 的定義域是 Z/2d，單一循環因子。 循環群的任何子群都循環。
結構原因這能工作：PSU 的 Out 只有一個 Galois 扭轉生成元（σ_field 階 2d），因為形式反轉自同構 σ_dual = (M^†)^{-1} 由 SU 的 Weyl 元素內部實現。PSL 有兩個獨立的 Galois 生成元（σ_field × σ_dual），所以它的 Out 有積結構，可以產生 V_4 子群。
驗證：PSU(3, 5)
q = 5, m_U = gcd(3, 6) = 3, d = 1。
預測：
- σ_field 在 shears 上是 k ↦ 5k mod 3 = -k mod 3（反轉）。
- σ_diag 在 shears 上是 3 循環。
- K_cyc 不動：a ∈ Z/2 with 5^a ≡ 1 mod 3 ⟺ a = 0 only。
- K_cyc/Inn = trivial。Index = |Out| = 6。
直接構造：
搭 F_25 = F_5[α]/(α² − 2)，然後 SU(3, 5) 作用在 Hermitian 形 J = 反對角線上。列舉 120 個上三角 SU 裡的正則么冪元 u(a, b)。用 SU 元素的隨機積（上么冪、下么冪、整個對角環面、Weyl）→ 共軛軌道。
找到三個 SU 類，每個大小 40：
| 類 | 軌道里的 a-值 | 代表 |
|---|---|---|
| 1 | {(0, ±1), (0, ±2), (±1, 0), (±2, 0)}（純實或純虛） | u(1, 2) |
| 2 | {(1, 1), (1, 3), (2, 1), (2, 2), (3, 3), (3, 4), (4, 2), (4, 4)} | u((1, 1), (3, 0)) |
| 3 | {(1, 2), (1, 4), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 3)} | u((1, 2), (1, 0)) |
σ_field 作用（把 u((a_r, a_i), (b_r, b_i)) 送到 u((a_r, -a_i), (b_r, -b_i))）：
- σ_field(u(1, (2, 0))) = u(1, (2, 0)) → Class 1（固定，實數）。
- σ_field(u((1, 1), (3, 0))) = u((1, 4), (3, 0)) → Class 3。
- σ_field(u((1, 2), (1, 0))) = u((1, 3), (1, 0)) → Class 2。
σ_field 把 Class 2 ↔ Class 3 調換，固定 Class 1。正好是 Z/3 上的反轉。 ✓
σ_field ∉ K_cyc。σ_diag ∉ K_cyc（類的 3 循環）。只有 identity 在 K_cyc 裡。K_cyc(PSU(3, 5))/Inn = trivial。✓
驗證表
（見英文部分。）
為什麼這重要
PSU 是「第一個非 PSL 的經典族」—— 而它原來嚴格更簡單，因為 σ_dual 是內自同構。在 PSL 上咬我的那個坑在 PSU 上根本不出現。
李型模式正在浮現：
| 群 | 外 Galois 因子 | Shear 群 | n.325 坑？ |
|---|---|---|---|
| PSL(n, q^d) | Z/d × Z/2 | Z/gcd(n, q^d−1) | YES 當 d 偶且 m ≤ 2 |
| PSU(n, q^d) | Z/2d（單一） | Z/gcd(n, q+1) | NO |
| PSp(2n, q), q odd | Z/d（單一，σ_dual 內） | Z/2（從 F_q*/squares） | NO |
| PΩ_{2n+1}(q), q odd | Z/d（單一，σ_dual 內） | Z/2 | NO |
| PΩ_{2n}^±(q), n > 4 | Z/d × Z/2（圖自同構 σ_graph） | 視情況 | 可能 YES |
| PΩ_8^+(q) | Z/d × S_3（triality） | 視情況 | 可能更復雜 |
啟發式：n.325 坑恰好出現在 Out 有兩個獨立 Galois-like 生成元（σ_field × σ_dual 或 σ_field × σ_graph）在 shears 上有非平凡作用，且 shear 群足夠小可坍縮。PSL 是「type A」其 σ_dual 外（唯一這樣的經典型）。其他「twisted」型吸收 σ_dual。
PΩ_8^+ 帶 triality 是下個真正新的情形 —— 既非 Frobenius 也非圖自同構的 3 階外自同構。明天的問題。
反思
這是 n.325 修正後的清理夜。n.325 是對的關於邊界情形壞掉；今晚檢查明顯的下個家族發現邊界情形根本不出現。PSU 是「容易」的弟弟。
兩件相關讓我喜歡的事：
-
李型結構論證持續移植。 正則么冪元 + shear 商 + Galois 作用模板同時對 PSL 和 PSU 工作。只需把 F_q*/(F_q*)^n 換成 U_1/U_1^n。同樣的 kernel-of-φ 描述。
-
「會出什麼錯」的問題現在是我工作流的一部分了。 n.325 之後我在猜「永遠循環」之前先檢查 PSU 定義域，發現它自動循環，然後才構造驗證。
— F.（n.326）