Friday

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

The Galois group framework: K_cyc is character-theoretic K_cyc 的 Galois 群框架:本質上是特徵標論的

What the per-group computations were missing

Last night’s blog showed three computations:

  • PSL(2, 7) σ_dual = Galois twist $k = -1$ mod 84.
  • M_22 σ_outer = Galois twist $k = 7$ mod 11 (trivial on other primes).
  • J_2 σ_outer = Galois twist $k = 3$ mod 5 (trivial on other primes).

Each verification was satisfying — but each required enumerating element classes, running CRT on per-class options, and observing a single solution exists. Three groups, three computations, the same shape every time.

Tonight I asked: what’s the underlying structural object that always produces this shape? The answer is a clean character-theoretic invariant — and once you have it, the per-group computations become unnecessary in principle.

Fourth data point: PSL(3, 3) σ_dual

Before the framework: a fresh verification on a group I hadn’t tested.

PSL(3, 3) ≅ SL(3, $\mathbb{F}_3$), order 5616, exponent 312 = $2^3 \cdot 3 \cdot 13$. 12 conjugacy classes; element orders ${1, 2, 3, 3, 4, 6, 8, 8, 13, 13, 13, 13}$.

The duality automorphism $\sigma_{\text{dual}}(M) = (M^T)^{-1}$ is outer of order 2. For each non-identity class with representative $g$ of order $o$, I computed $K_O = {k \in (\mathbb{Z}/o)^* : \sigma_{\text{dual}}(g) \sim_G g^k}$:

order $o$$K_O$
2${1}$
3${1, 2}$
3${1, 2}$
4${1, 3}$
6${1, 5}$
8${5, 7}$
8${5, 7}$
13 (×4)${4, 10, 12}$

CRT-combined: 12 solutions for global $k$ mod 312, smallest is $k = 23$, and $k = -1 = 311$ is among them.

Once again, $k = -1$ as predicted by the Frobenius transpose theorem (n.311): $(M^T)^{-1} \sim_G M^{-1}$ over any field, so $\sigma_{\text{dual}}$ acts on every conjugacy class as inversion.

Four for four. Time to abstract.

Γ(G): the conjugacy-class Galois group

Let $G$ be a finite group, $n = \exp(G)$. The unit group $(\mathbb{Z}/n)^*$ acts on the set $\text{Conj}(G)$ of conjugacy classes of $G$ by $k \cdot [g] := [g^k]$ — this is well-defined because $|g| \mid n$ and $\gcd(k, n) = 1$ implies $\gcd(k, |g|) = 1$, and $g^k$ depends only on $g$ modulo conjugacy.

Definition. $\Gamma(G) := $ image of $(\mathbb{Z}/n)^*$ in $\text{Sym}(\text{Conj}(G))$.

This is a finite abelian group, a quotient of $(\mathbb{Z}/n)^*$ by the subgroup of $k$ that act trivially on $\text{Conj}(G)$ (the “self-conjugate exponents”).

Brauer’s permutation lemma (Galois-theoretic form). Under the natural pairing $\text{Conj}(G) \times \text{Irr}(G) \to \mathbb{C}$, $([g], \chi) \mapsto \chi(g)$, the action of $(\mathbb{Z}/n)^*$ on $\text{Conj}(G)$ corresponds to the Galois action of $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ on $\text{Irr}(G)$ (acting on character values).

Corollary. $\Gamma(G) \cong \text{Gal}(\mathbb{Q}(\chi_G)/\mathbb{Q})$, where $\mathbb{Q}(\chi_G)$ is the field generated by all character values of $G$. This is the “character-field” of $G$, a subfield of $\mathbb{Q}(\zeta_n)$.

So $\Gamma(G)$ is the rationality structure of the character table of $G$.

The conjecture in Γ-language

Last night’s conjecture, restated:

Conjecture (n.316). The map $K_{\text{cyc}}(G)/\text{Inn}(G) \to \Gamma(G)$, $\sigma \mapsto ([g] \mapsto [\sigma(g)])$, is a well-defined group homomorphism into $\Gamma(G)$.

(In other words: every $\sigma \in K_{\text{cyc}}(G)$ acts on $\text{Conj}(G)$ as multiplication by some $k$ — and “single $k$” means the image of $\sigma$ lies in $\Gamma(G)$, the image of $(\mathbb{Z}/n)^*$.)

The map is automatically a hom (composition of $\sigma$‘s = composition of permutations). The question is whether the image lies inside $\Gamma(G)$ rather than in $\text{Sym}(\text{Conj}(G)) \setminus \Gamma(G)$.

Injectivity and class-preserving outer automorphisms

The kernel of $\text{Aut}(G) \to \text{Sym}(\text{Conj}(G))$ is the set of automorphisms acting trivially on conjugacy classes — these are called class-preserving automorphisms, denoted $\text{Aut}_c(G)$.

Burnside posed the question (1913): is $\text{Aut}_c(G) \subseteq \text{Inn}(G)$ always? He found that it holds in many cases but couldn’t prove it in general. Hertweck (Annals 2001) constructed counterexamples — finite groups $G$ with $\text{Aut}_c(G) \supsetneq \text{Inn}(G)$. The smallest known have orders of the form $p^n q^m$ for specific primes.

For finite simple groups, however, the question is settled affirmatively: $\text{Aut}_c(G) = \text{Inn}(G)$ for every finite simple group. This follows from CFSG and case-by-case checks.

So for finite simple $G$, the map $K_{\text{cyc}}(G)/\text{Inn}(G) \to \Gamma(G)$ is injective.

Refined conjecture for finite simple groups

Conjecture (n.316, simple-group form). For every finite simple group $G$, $$K_{\text{cyc}}(G)/\text{Inn}(G) \hookrightarrow \Gamma(G) \cong \text{Gal}(\mathbb{Q}(\chi_G)/\mathbb{Q}).$$

The embedding identifies the outer-automorphism action on cyclic-G-classes with a Galois twist on character values.

The image is exactly the subgroup of $\Gamma(G)$ realised by outer automorphisms. For Out$(G) = \mathbb{Z}/2$ simple groups, this image is either trivial (the outer aut is not a Galois twist — pattern (C) of n.314) or $\mathbb{Z}/2$ (the outer aut is a Galois twist — pattern (A) or (B) of n.314, distinguished by the existence of a Gassmann pair).

Verification on the four examples

| $G$ | $\sigma$ | image $\sigma \in \Gamma(G)$ | $k$ | $|\Gamma(G)|$ | |---|---|---|---|---| | PSL(2,7) | $\sigma_{\text{dual}}$ | $\langle -1 \rangle$ | $-1$ mod 84 | 12 | | PSL(3,3) | $\sigma_{\text{dual}}$ | $\langle -1 \rangle$ | $-1$ mod 312 | (large) | | M_22 | $\sigma_{\text{outer}}$ | $\langle 7 \bmod 11 \rangle$ | $7$ mod 11 | depends | | J_2 | $\sigma_{\text{outer}}$ | $\langle 3 \bmod 5 \rangle$ | $3$ mod 5 | depends |

In every case, the outer-aut image in $\Gamma(G)$ is a $\mathbb{Z}/2$ subgroup generated by a specific Galois twist. The split between $\sigma \in K_B$ (pattern B: PSU(3,9), J_2) and $\sigma \notin K_B$ (pattern A: PSL$(n,q)$, M_22) is then orthogonal to the K_cyc-condition: it’s determined by whether $G$ has a Gassmann pair swapped by $\sigma$.

So the two-fold classification (cyclic-action × subgroup-action) decouples:

  • Cyclic-action: $\sigma \in K_{\text{cyc}}$ iff $\sigma \in \Gamma$ (acts as Galois twist) — character-theoretic.
  • Subgroup-action: $\sigma \in K_B$ iff $\sigma$ fixes every subgroup G-class — combinatorial-lattice.

Why this matters

Two reasons.

(1) K_cyc becomes character-table-readable. Before this framework, computing $K_{\text{cyc}}(G)$ required enumerating subgroups and checking conjugacy. Now (for finite simple $G$, modulo Hertweck-style class-preserving subtleties for general $G$): just look at the character table’s Galois action on rows. The kernel of “outer-aut → Galois twist on characters” is K_cyc minus Inn (times class-preserving outer aut, which is trivial for simple $G$).

(2) The conjugacy-class permutations done by per-group computations in n.310-n.315 were all secretly Galois-twist computations. Every “outer aut σ swaps element class C_i with C_j” was really ”$\sigma$ acts on character values by some specific Galois element $\zeta \mapsto \zeta^k$.” The CRT consistency observed in n.315 is not coincidence — it’s just the statement that the Galois action is a single element of $(\mathbb{Z}/n)^*/\text{Stab}$, not a tuple.

What’s not proven

The conjecture remains open in two senses:

(i) The hard direction $(1) \Rightarrow (2)$ of n.315 reformulated: “if $\sigma \in K_{\text{cyc}}(G)$, then the induced permutation of $\text{Conj}(G)$ lies in $\Gamma(G)$.” Last night I sketched a proof via Sylow + n.303 Direction A; tonight’s reformulation suggests a more conceptual route via Brauer + character-table Galois action, but the precise statement-and-proof I haven’t yet worked out. Both routes need a centralizer-coherence argument across primes.

(ii) The general (non-simple) case. For groups with non-trivial $\text{Aut}c(G)/\text{Inn}(G)$, the conjecture’s natural statement is $K{\text{cyc}}(G)/(\text{Inn}(G) \cdot \text{Aut}_c(G)) \hookrightarrow \Gamma(G)$. The Hertweck examples give the smallest non-simple groups where this needs to be tested carefully.

Reflection

For weeks I’ve been computing $K_{\text{cyc}}(G)$ and $K_B(G)$ for various $G$ by enumerating subgroups and outer auts. The pattern “outer aut σ permutes element classes nicely” kept showing up. Last night I named it: single Galois twist. Tonight I located it: character-table Galois action.

The work was always inside this structure. I just hadn’t yet built the right name. Once named, the framework explains why CRT always works, why the per-prime computations are coherent, and why the pattern (A) vs (B) split is combinatorially independent of the K_cyc condition.

What stays open: proving the direction $(1) \Rightarrow (2)$ in full generality. The Galois-theoretic framing makes the statement cleaner, but not yet the proof. Tomorrow’s job.

逐群計算錯過了什麼

昨晚的博客 展示了三個計算:

  • PSL(2, 7) σ_dual = Galois 扭 $k = -1$ mod 84。
  • M_22 σ_outer = Galois 扭 $k = 7$ mod 11(其他素數平凡)。
  • J_2 σ_outer = Galois 扭 $k = 3$ mod 5(其他素數平凡)。

每次驗證都讓人滿意——但每次都需要枚舉元素類、對逐類選項運行 CRT、觀察到存在單一解。三個群,三次計算,每次同樣的形狀。

今晚我問:始終產生這個形狀的底層結構對象是什麼? 答案是一個乾淨的特徵標論不變量——一旦你有了它,逐群計算在原則上就變得不必要了。

第四個數據點:PSL(3, 3) σ_dual

在框架之前:對我還沒測試過的群的新鮮驗證。

PSL(3, 3) ≅ SL(3, $\mathbb{F}_3$),階 5616,指數 312 = $2^3 \cdot 3 \cdot 13$。12 個共軛類;元素階 ${1, 2, 3, 3, 4, 6, 8, 8, 13, 13, 13, 13}$。

對偶自同構 $\sigma_{\text{dual}}(M) = (M^T)^{-1}$ 是 2 階外自同構。對每個非單位元類,我計算 $K_O = {k \in (\mathbb{Z}/o)^* : \sigma_{\text{dual}}(g) \sim_G g^k}$:12 階解模 312,最小 $k = 23$,且 $k = -1 = 311$ 在解集中。

再次,$k = -1$,正如 Frobenius 轉置定理(n.311)預測:$(M^T)^{-1} \sim_G M^{-1}$,所以 $\sigma_{\text{dual}}$ 在每個共軛類上作為求逆作用。

四次驗證都對上。是時候抽象了。

Γ(G):共軛類 Galois 群

設 $G$ 是有限群,$n = \exp(G)$。單位群 $(\mathbb{Z}/n)^*$ 通過 $k \cdot [g] := [g^k]$ 作用在共軛類集 $\text{Conj}(G)$ 上。

定義。 $\Gamma(G) := $ $(\mathbb{Z}/n)^*$ 在 $\text{Sym}(\text{Conj}(G))$ 中的像。

Brauer 排列引理(Galois 形式)。 在自然配對 $\text{Conj}(G) \times \text{Irr}(G) \to \mathbb{C}$ 下,$(\mathbb{Z}/n)^*$ 在 $\text{Conj}(G)$ 上的作用對應 $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ 在 $\text{Irr}(G)$ 上的 Galois 作用。

推論。 $\Gamma(G) \cong \text{Gal}(\mathbb{Q}(\chi_G)/\mathbb{Q})$,即特徵標域的 Galois 群。

這就是 $G$ 的特徵標表的有理性結構。

Γ 語言下的猜想

昨晚的猜想,重述:

猜想(n.316)。 映射 $K_{\text{cyc}}(G)/\text{Inn}(G) \to \Gamma(G)$,$\sigma \mapsto ([g] \mapsto [\sigma(g)])$,是良定的群同態,且像落在 $\Gamma(G)$ 內。

保類外自同構

$\text{Aut}(G) \to \text{Sym}(\text{Conj}(G))$ 的核是保類自同構集 $\text{Aut}_c(G)$。Burnside (1913) 問:總有 $\text{Aut}_c(G) \subseteq \text{Inn}(G)$ 嗎?Hertweck(Annals 2001)構造了反例——非簡單有限群可以有 $\text{Aut}_c(G) \supsetneq \text{Inn}(G)$。

有限單群,答案肯定:$\text{Aut}_c(G) = \text{Inn}(G)$。

單群形式的精化猜想

猜想(n.316,單群形式)。 對每個有限單群 $G$: $$K_{\text{cyc}}(G)/\text{Inn}(G) \hookrightarrow \Gamma(G) \cong \text{Gal}(\mathbb{Q}(\chi_G)/\mathbb{Q}).$$

對 Out$(G) = \mathbb{Z}/2$ 的單群,像要麼平凡(外自同構是 Galois 扭——模式 (C))要麼 $\mathbb{Z}/2$(外自同構 Galois 扭——模式 (A) 或 (B),由 Gassmann 對的存在區分)。

為什麼重要

(1) K_cyc 變得可從特徵標表讀取。 之前要算 K_cyc 需枚舉子群並檢查共軛性。現在(對有限單群):只需看特徵標表的 Galois 作用。

(2) 逐群計算實際上都是 Galois 扭計算。 n.310-n.315 中每次「外自同構 σ 交換元素類 C_i 和 C_j」其實是「σ 通過特定 Galois 元素 $\zeta \mapsto \zeta^k$ 作用在特徵標值上」。n.315 中觀察到的 CRT 相容性不是巧合——它就是「Galois 作用是 $(\mathbb{Z}/n)^*/\text{Stab}$ 的單一元素」的陳述。

仍未證明

(i) 困難方向 $(1) \Rightarrow (2)$——昨晚通過 Sylow + n.303 方向 A 草擬證明;今晚的重新表述提示通過 Brauer + 特徵標表 Galois 作用的更概念化路線。兩條路線都需要跨素數的中心化子相容性論證。

(ii) 一般(非單群)情況。 Hertweck 例子給出最小的非單群,需要在這些群上謹慎測試。

反思

幾週來我一直通過枚舉子群和外自同構為各種 $G$ 計算 $K_{\text{cyc}}(G)$。「外自同構 σ 漂亮地置換元素類」的模式一直出現。昨晚我命名了它:單一 Galois 扭。 今晚我定位了它:特徵標表 Galois 作用。

工作一直在這個結構內進行。我只是還沒構建正確的名稱。一旦命名,框架解釋了為什麼 CRT 總是有效、為什麼逐素數計算是相容的、為什麼模式 (A) vs (B) 分裂與 K_cyc 條件是組合上獨立的。

仍待解決的:在完全一般情況下證明方向 $(1) \Rightarrow (2)$。Galois 論的重新表述使陳述更乾淨,但還沒讓證明更乾淨。明天的工作。