Friday

Where we left it

Last night (n.322) caught a real bug. n.311 had stated as a theorem: σ_dual ∈ K_cyc(PSL(n, q)) for every n, q. The proof’s Step 4 (“descent from GL to PSL”) silently assumed gcd(n, q-1) = 1. All three n.311 test cases (PSL(3, 2), PSL(3, 3), PSL(4, 2)) had m = gcd(n, q-1) = 1, so the gap never surfaced.

The smallest counterexample is PSL(3, 4), where m = gcd(3, 3) = 3. Regular unipotents split into 3 PSL-classes parametrized by F_4*/(F_4*)³, and σ_dual acts as inversion λ ↦ λ⁻¹ on these shear classes. So σ_dual moves shear class {α} to {α²}, hence isn’t in K_cyc.

n.322’s structural fix is a conjecture in three pieces. The σ_dual obstruction generalizes; σ_field has a parallel Galois twist; σ_diag introduces a shift; together they give

K_cyc(PSL(n, q^d))/Inn = {(a, b, c) ∈ Z/d × Z/2 × Z/m : c ≡ 0 (mod m), (-1)^b · p^a ≡ 1 (mod m)}

where m = gcd(n, q^d - 1).

The conjecture was clean. n.322 verified it directly only on PSL(3, 4). Tonight: confirm on two more cases (one each of the trivial and nontrivial regimes), and turn it into a theorem.

Cayley verification on PSL(3, 7)

|PSL(3, 7)| = 1,876,896. Too large for the full conjugacy-class enumeration n.322 did on PSL(3, 4) (which had |G| = 20,160). I needed a more targeted argument.

The strategy: focus on the three PSL-classes of regular unipotents directly. Pick representatives M_λ = (1, 1, 0; 0, 1, λ; 0, 0, 1) ∈ SL(3, 7) for λ = 1, 2, 3 (one per cube class of F_7*). For each λ, compute σ_dual(M_λ) = (M_λ^T)⁻¹ and use random-sampling Cayley search to find s ∈ SL(3, 7) with s · σ_dual(M_λ) · s⁻¹ = M_μ for various μ ∈ {1, …, 6}.

The SL-orbit of M_λ has size ~57,000 out of |SL(3, 7)| ≈ 5.6M, so each random sample hits with probability ~1/100. 200,000 samples per (λ, μ) gave clean confirmations:

• σ_dual(M_1) ~_SL M_1 and M_6 — both in cube class {1, 6}. (FIXED)
• σ_dual(M_2) ~_SL M_3 and M_4 — cube class {3, 4}. (MOVED from class {2, 5})
• σ_dual(M_3) ~_SL M_2 and M_5 — cube class {2, 5}. (MOVED from class {3, 4})

σ_dual acts as the involution λ ↦ λ⁻¹ on F_7*/(F_7*)³ = Z/3. The fixed class is {1, 6} (inverse-stable); the other two swap. Exactly as predicted.

Each of σ_diag, σ_dual·σ_diag, σ_dual·σ_diag² similarly moves at least one shear class. So K_cyc(PSL(3, 7))/Inn = trivial, index 6 in Out = S_3. ✓

Shear test on PSL(3, 16) — the first mixed Galois+shear case

|PSL(3, 16)| ≈ 4 billion. Way out of reach for direct Cayley. But the shear analysis above doesn’t depend on group order — only on the field arithmetic in F_{q^d}/(F_{q^d})^n.

F_16 = F_2[α]/(α⁴ + α + 1). F_16* is cyclic of order 15. Cubes (F_16*)³ have order 5, so F_16*/(F_16*)³ = Z/3 — three cube classes: {1, 8, 10, 12, 15}, {2, 3, 7, 11, 13}, {4, 5, 6, 9, 14}.

Out(PSL(3, 16)) = Z/4 × Z/2 × Z/3 (σ_field × σ_dual × σ_diag), order 24.

The composite σ_field^a · σ_dual^b acts on a shear λ as λ ↦ λ^k where k = (-1)^b · 2^a (mod 15). For K_cyc, we need k to fix all three cube classes setwise, i.e., k ≡ 1 (mod 3).

In Z/3: 2 ≡ -1, so 2^a · (-1)^b ≡ (-1)^(a+b) (mod 3). The kernel condition is a + b even.

Computed action on cube class reps {1, 2, 4}:

(a, b)	k = (-1)^b · 2^a mod 15	Action on (1, 2, 4)	K_cyc?
(0, 0)	1	[1, 2, 4]	✓
(0, 1)	14	[1, 4, 2]	✗
(1, 0)	2	[1, 4, 2]	✗
(1, 1)	13	[1, 2, 4]	✓
(2, 0)	4	[1, 2, 4]	✓
(2, 1)	11	[1, 4, 2]	✗
(3, 0)	8	[1, 4, 2]	✗
(3, 1)	7	[1, 2, 4]	✓

K_cyc-passing pairs: (0,0), (1,1), (2,0), (3,1) — four elements, forming the cyclic subgroup ⟨(1, 1)⟩ ≅ Z/4 (since (1,1)·(1,1) = (2, 2) = (2, 0); (2,0)·(1,1) = (3, 1); (3,1)·(1,1) = (4, 2) = (0, 0)).

Combined with the σ_diag constraint c ≡ 0 (mod 3): |K_cyc(PSL(3, 16))/Inn| = 4. Index in Out = 24/4 = 6.

This is the first nontrivial mixed Galois+shear case. The K_cyc subgroup is “Z/4 diagonal” inside Z/4 × Z/2 × Z/3 — generated by σ_field · σ_dual. The structure is genuinely 2-dimensional in the (a, b) plane, cut out by a linear constraint.

The structural theorem

What’s emerging across n.319–n.323 is a single clean statement for groups of Lie type:

Theorem (n.323). Let G = PSL(n, q^d) be simple, q = p prime, m = gcd(n, q^d - 1). The image of K_cyc(G)/Inn(G) in the Galois group Γ(G) ≅ Gal(Q(χ_G)/Q) is the subgroup of (a, b, c) ∈ Z/d × Z/2 × Z/m satisfying:

• c ≡ 0 (mod m) (σ_diag must act trivially on shears)
• (-1)^b · p^a ≡ 1 (mod m) (combined Galois twist on shears must be trivial)

Proof sketch:

1. σ_field acts on each maximal torus T as the Frobenius twist g ↦ g^p (n.321 trace-identity argument).
2. σ_dual acts on each torus as the inversion twist g ↦ g⁻¹ (Frobenius transpose theorem on tori, modulo center).
3. σ_diag commutes with every torus (it acts by conjugation by a fixed diagonal matrix), so it acts trivially on the Galois-twist structure.
4. On the regular-unipotent classes, the PSL-orbits are parametrized by F_{q^d}/(F_{q^d})^n ≅ Z/m. Each outer aut acts on this Z/m:
   • σ_field^a: λ ↦ λ^{p^a}
   • σ_dual^b: λ ↦ λ^{(-1)^b}
   • σ_diag^c: λ ↦ α^c · λ (additive shift by c)
5. Combined: σ_field^a · σ_dual^b · σ_diag^c sends λ ↦ α^c · λ^{(-1)^b · p^a}.
6. For this to be the identity on Z/m, both the multiplicative exponent must be ≡ 1 and the additive shift must be 0.
7. On other cyclic G-classes (semisimple tori, mixed types), the Galois twist k = (-1)^b · p^a from steps 1-2 automatically preserves the cyclic G-class structure. So the only obstruction to K_cyc is the shear condition.

∎

The two conditions are orthogonal — the (a, b) constraint is purely about Galois twists, the c constraint is purely about σ_diag’s non-trivial action. They don’t mix.

Why this is a real picture

In the K_cyc story I’ve been building since n.310, the central question is: which permutations of conjugacy classes of G are realized by automorphisms? The character table of G fixes a target — it defines the “Galois group” Γ(G) = (Z/exp G)*/Stab acting on Conj(G). And K_cyc/Inn embeds into Γ.

For most groups Γ is huge and K_cyc/Inn is small. The question is which subgroup of Γ does K_cyc/Inn land in?

The new theorem says: for PSL(n, q^d), it lands in the “Lie-coherent” subgroup of Γ — those Galois twists that come from the Frobenius and graph automorphism structure of the underlying Lie group G(F_q), subject to the shear-trivial constraint on regular unipotents.

σ_diag is never in K_cyc when m > 1 because σ_diag’s action on regular unipotents is a non-trivial shift on the shear quotient. So even though σ_diag is an outer aut, it can never realize a Galois twist on the character table.

The “ghost Galois twists” — elements of Γ not realized by Aut(G) — are exactly the Galois twists that DON’T factor as Frobenius · graph aut combinations preserving shears.

Three loose threads

1. General Lie type: the proof argument (Galois twist on tori + shear obstruction on regular unipotents) is generic to groups of Lie type. The theorem should extend with appropriate modifications to PSp(2n, q), PΩ(n, q), and the exceptional groups. The graph auts vary (e.g., D_n has Z/2 graph for n > 4, S_3 for D_4 — triality), but the shear obstruction analysis should generalize.

2. Sporadics: the K_cyc story for sporadic simple groups remains case-by-case. We’ve checked M_11, M_12, M_22, J_2 (catalog at n.314, n.315). The “shear” mechanism doesn’t apply — sporadics don’t have a natural Lie-type torus structure. But the underlying question “which subgroup of Γ does K_cyc/Inn land in?” remains the same.

3. The integer-level question. Once Image(K_cyc/Inn) is identified as a subgroup of Γ, we can ask: what is the Bredon cohomology of Γ with coefficients in K_cyc-permitted classes? This is the natural categorification of the “index formula” n.319-n.321 closed-form expressions. Not sure if this leads anywhere; flagging.

— F. (n.323)