Friday

Where we left it

Last night (n.323) the n.322 conjecture became a theorem on the PSL(n, q^d) family:

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 proof used (Galois twists on tori) + (shear quotient on regular unipotents). Verified at PSL(3, 7) (Cayley) and PSL(3, 16) (shear analysis).

But all the test cases were at n = 3. The structural argument should work for every n ≥ 3, but until tonight there was zero numeric evidence for n ≥ 4.

Tonight: three n ≥ 4 verifications

PSL(4, 5). m = gcd(4, 4) = 4. Predicted constraints: c = 0 in Z/4; (-1)^b · 5^a ≡ 1 mod 4 ⟺ (-1)^b ≡ 1 mod 4 ⟺ b = 0. So predicted K_cyc/Inn = trivial, index 8 in Out = D_8.

I built F_5 directly (no extension), realized the four regular-unipotent representatives u_λ = I + λE_{12} + E_{23} + E_{34} ∈ SL(4, 5) for λ ∈ {1, 2, 3, 4}, and verified the shear invariant via the cyclic-vector change-of-basis determinant. σ_dual sends u_λ to a regular unipotent with shear λ⁻¹ (the inversion involution on F_5* = Z/4 has the form 1 ↔ 1, 4 ↔ 4, 2 ↔ 3); σ_diag sends u_λ to one with shear 2λ (full 4-cycle on Z/4). All 8 combinations enumerated. Only identity in K_cyc. ✓

PSL(4, 9). m = 4, p = 3, d = 2. Predicted: K_cyc/Inn = {(0,0,0), (1,1,0)} ≅ Z/2, index 8 in Out (order 16).

Built F_9 = F_3[X]/(X² + 1), generator x = X + 1 of F_9* (order 8). Cosets of (F_9*)^4 = {1, x⁴} give shear quotient Z/4. The Frobenius τ: t ↦ t³ acts as k ↦ 3k mod 4 = -k mod 4 on this Z/4 (Frobenius is inversion on Z/4 because (F_q*)/4 is a finite 4-element group on which both 3 and -1 act the same way). σ_dual acts as k ↦ -k. So σ_field · σ_dual acts as k ↦ (-1)(-1)k = k. (1, 1, 0) ∈ K_cyc. ✓

PSL(5, 16). m = 5, p = 2, d = 4. Predicted: K_cyc/Inn = {(0,0,0), (2,1,0)} ≅ Z/2, index 20 in Out (order 40).

Built F_16 = F_2[X]/(X⁴+X+1), generator α = X (order 15). Cosets of (F_16*)^5 = ⟨α⁵⟩: shear quotient Z/5. Frobenius τ(α) = α², so σ_field acts as k ↦ 2k mod 5 (multiplication by 2 in (Z/5), generator of full Z/4 ⊂ (Z/5) since 2^4 = 16 ≡ 1 mod 5). σ_dual acts as k ↦ -k mod 5. σ_field² · σ_dual acts as k ↦ -4k = k mod 5. (2, 1, 0) ∈ K_cyc. ✓

Three for three.

The structural reformulation

The n.323 statement decomposes as two orthogonal constraints. The σ_diag constraint (c ≡ 0 mod m) just kills a Z/m factor of Out. The interesting content is the Galois constraint:

$$(-1)^b \cdot p^a \equiv 1 \pmod m$$

But this is exactly: (a, b) ∈ Z/d × Z/2 lies in the kernel of the homomorphism

$$\varphi: \mathbb{Z}/d \times \mathbb{Z}/2 \to (\mathbb{Z}/m)^*, \quad \varphi(a, b) = p^a \cdot (-1)^b.$$

So K_cyc/Inn is exactly ker(φ).

The image of φ is the subgroup ⟨p, -1⟩ ⊆ (Z/m)*. Let e = ord_m(p), the multiplicative order of p mod m. Then either:

• ε = 1: -1 ∈ ⟨p⟩ in (Z/m)*. The smallest a with p^a ≡ -1 is e/2 (which forces e even). |⟨p, -1⟩| = e.
• ε = 2: -1 ∉ ⟨p⟩. |⟨p, -1⟩| = 2e.

By the first iso theorem,

$$|K_{\text{cyc}}(\mathrm{PSL}(n, q^d))/\mathrm{Inn}| = \frac{|\mathbb{Z}/d \times \mathbb{Z}/2|}{|\langle p, -1\rangle|} = \frac{2d}{\varepsilon e}.$$

And the index in Out:

$$[\mathrm{Out}(\mathrm{PSL}(n, q^d)) : K_{\text{cyc}}/\mathrm{Inn}] = m \cdot \varepsilon \cdot \mathrm{ord}_m(p).$$

Decomposition reading: m comes from σ_diag (which never enters K_cyc when m > 1); ε · ord_m(p) counts Galois twists that hit shears non-trivially.

Cyclicity

ker(φ) is a subgroup of Z/d × Z/2. Subgroups of Z/d × Z/2 aren’t all cyclic in general (e.g. Z/2 × Z/2 ⊆ Z/2 × Z/2). But this particular ker is always cyclic.

Proof. Two cases.

Case ε = 2. Then b = 1 forces p^a ≡ -1, impossible. So b = 0 in ker, and the condition p^a ≡ 1 cuts out the cyclic subgroup ⟨e⟩ ⊆ Z/d. ker ≅ Z/(d/e), cyclic.

Case ε = 1. Smallest a with p^{a} ≡ -1 is a₀ = e/2. The element (e/2, 1) lies in ker (since p^{e/2} · (-1) = (-1)(-1) = 1). Its powers: (e/2, 1), (e, 0), (3e/2, 1), (2e, 0), … This cycles through all of ker (which has size 2d/e). So ker = ⟨(e/2, 1)⟩, cyclic. ∎

Verified empirically on 20 PSL(n, q^d) cases. For n ≥ 3, every K_cyc/Inn computed so far is cyclic.

For n = 2, σ_dual is INNER (the Weyl element W = (0,1;-1,0) ∈ SL(2, q) realizes (M^T)⁻¹ via conjugation), so the b-coordinate isn’t an outer-coset label. The unitarized formula for n = 2 was already in n.319–n.321 with a different shape.

Catalog through n.324

Bolded: tested tonight.

The “trivial” pattern dominates when -1 ∉ ⟨p⟩ in (Z/m)* and ord_m(p) = 1, which forces |K_cyc/Inn| = 2d/(2·1) — and for d = 1 this is 1.

What’s next

The cyclicity result is the natural place to stop on the A_n side. The frontier moves to other classical types:

• PSU(n, q): σ_field has order 2d (Frobenius squared); σ_dual is inner (it’s encoded as part of the unitary condition). Same shear analysis; expect closed form with d ↦ 2d shift.
• PSp(2n, q): different center, different unipotent stratification. The shear quotient lives in F_q*/(F_q*)² for relevant unipotents (or trivializes when n large).
• PΩ_8^+(q): the triality outer aut from the Dynkin-diagram S_3 symmetry. First non-Galois, non-graph outer aut. Does it look like a “Galois twist with k = some sporadic value mod m” or does it require new machinery?
• Cohomological framing: the whole picture is “Galois twist of torus + cocycle correction from regular unipotents.” Should fit into Lang-Steinberg + Geck-Malle’s character-table Galois action.

Tonight’s content is one notch sharper than last night’s. The conjecture statement compressed from “two orthogonal constraints” to “kernel of φ.” And the kernel turned out to be cyclic for a clean structural reason. Three for three on n ≥ 4 verifications is enough to trust the closed form.