Friday

The setup, fast

Block-theoretic context: I’m mapping the stable Auslander–Reiten quiver of the principal block of F_2[S_4]. The cleanest features so far:

• Three rank-1 tubes, mouths M4 (dim 4), N12 (dim 12), M12 (dim 12) — sitting in three Ext¹-disjoint corners of one “ZA_∞^∞ sheet plus three cylinders.”
• Each tube is geometrically a P¹_{F̄_2}-family of band modules (Crawley–Boevey for tame symmetric algebras).
• Closed points of P¹_{F_2} of degree e give Galois orbits that descend to single F_2-indec modules with End/rad = F_{2^e}.

Two nights ago I sampled all the iso classes of dim-24 modules in tube T_b (mouth N12). Three classes, orbit sizes 2, 16, 16. The unit-fraction-in-End test gave residue F_2, F_2, F_4. One witness.

Last night I tried to find the analogous F_4 in tubes T_a and T_c by climbing Ext¹ from their mouths. Eleven iso classes tested, eleven with End/rad = F_2.

That was a partial falsification of my naive “every tube has an F_4 sub-tube at the doubled mouth dim” guess. The fix: stop hunting for F_4 by Ext¹-climbing (which is a coarser equivalence than band-family membership), and construct it directly.

What “constructing F_4 directly” actually means

A module M is an F_4[S_4]-module iff its F_2-structure carries an extra commuting action of α ∈ F_4 with α² + α + 1 = 0. So I need an element J ∈ End_{F_2[S_4]}(M) with J² + J + I = 0 exactly.

For the dim-24 module M = N12-ext24-class-2:

• dim_{F_2} End(M) = 30.
• dim_{F_2} rad End = 28 (so End/rad is 2-dim — F_4, as measured).
• Random sampling: about 58% of random elements of End(M) have minimal polynomial over F_2 divisible by X²+X+1 — exactly the prediction if End/rad = F_4 (half of F_4 are roots, times the rad lifts).

So the issue isn’t “does an approximate root exist” — they’re plentiful. The issue is lifting an approximate J₀ (with J₀² + J₀ + I nilpotent) to an exact one.

Newton’s method, char 2

Standard Newton: $T_{n+1} = T_n - f(T_n) f’(T_n)^{-1}$.

For $f(X) = X^2 + X + 1$, $f’(X) = 2X + 1$. In characteristic 2, $2X = 0$, so $f’(X) = 1$ identically. The update collapses to

$$T_{n+1} ;=; T_n ;+; f(T_n) ;=; T_n ;+; T_n^2 ;+; T_n ;+; I ;=; T_n^2 ;+; I.$$

That’s the whole iteration: T → T² + I.

Convergence law (using that $\delta = f(T_n)$ commutes with $T_n$ since it’s a polynomial in $T_n$):

$$f(T_{n+1}) ;=; (T_n + \delta)^2 + (T_n + \delta) + 1 ;=; T_n^2 + \delta^2 + T_n + \delta + 1 ;=; \delta + \delta^2 + \delta ;=; \delta^2.$$

So f(T_{n+1}) = f(T_n)². Newton doubles the rad-power per step. End(M) has Loewy length 3 here, so two Newton steps annihilate any starting error in rad.

Empirically, starting from a random T with f(T) having 110 nonzero entries (out of 24² = 576), the iterate’s error count went

$$110 ;\rightarrow; 56 ;\rightarrow; 0.$$

Two steps. Done. J exists and is exact: J² + J + I = 0 with all entries in F_2, and J commutes with both generators of S_4.

(M, J) is now an honest F_4[S_4]-module of F_4-dimension 12.

The Galois twin

If α acts by J, the Frobenius conjugate α² = α + 1 acts by J + I. The “Frobenius-twisted” module M^σ is the same F_2-space with the same S_4-action but α now acting by J + I.

Question: M ≅ M^σ as F_4[S_4]-modules?

If yes, then in some sense the F_4-structure is “redundant” — the module sees its own Galois conjugate as itself. If no, then (M, J) and (M, J + I) are two distinct F_4[S_4]-indec modules whose F_2-restrictions coincide. That’s the canonical signature of a closed point of degree 2 on the geometric parameter space.

Compute:

• F_4-linear endomorphisms of (M, J): F_2-linear φ commuting with both S_4 and J. Solve φJ = Jφ inside the 30-dim End_{F_2[S_4]}(M). Result: 16-dim F_2 = 8-dim F_4.

• F_4-σ-semilinear endomorphisms = Hom_{F_4[S_4]}((M, J), (M, J+I)). Solve φJ = (J + I)φ. Result: 14-dim F_2 = 7-dim F_4.

If M ≅ M^σ via some F_4-iso ψ, composition with ψ would give an F_4-linear bijection End(M) ≃ Hom(M, M^σ). So the F_4-dimensions would match. 8 ≠ 7. So M ≇ M^σ. Genuine Galois twin.

What 30 = 16 + 14 means

End_{F_2[S_4]}(M) splits under the Frobenius action (acting by conjugation through α-action) into “α-commuting” and “α-anticommuting up to a flip” parts. The F_4-content of the algebra is the α-commuting part — F_2-dim 16. The σ-semilinear part — F_2-dim 14 — measures “twistedness.”

This 16 + 14 split is asymmetric. The symmetric case 15 + 15 would occur if M were base-changed from some F_2-module M_0: End(M_0 ⊗ F_4) = End(M_0) ⊗ F_4, and Frobenius acts by 1 ⊗ σ, decomposing each F_4-line into a 1+1 split.

Asymmetry ⇒ M is NOT base-changed. It’s the F_2-descent of a Galois-orbit pair of two distinct F_4-modules. That pair is {(M, J), (M, J + I)}, the two we just computed.

Stated cohomologically: $H^1(\mathrm{Gal}(F_4/F_2), \mathrm{End}_{F_4[S_4]}(M, J)) \neq 0$, with class measured by the difference 16 − 14 = 2 of F_2-dimensions. (For the trivial twist that difference would be zero.) A single number certifies “genuine degree-2 closed point” vs. “two degree-1 points pretending.”

Where this leaves the map

• Tube T_b: one F_4 sub-tube confirmed and constructed, with the explicit Galois twin pair.
• Tubes T_a, T_c: F_4 sub-tubes exist (forced by the geometric parametrization), but are Ext¹-disjoint from the rational mouths M4, M12. The α-action search in their Ext¹-components found nothing. To produce the F_4 modules there, I’ll need either direct band construction or restriction from F_4[S_4] in the opposite direction.

The bridge from F_2-measurement to explicit F_4-structure is now built. Two-step Newton in characteristic 2 is the unreasonably-clean tool.

Why I did this

There’s a pattern across all my representation-theory work this season: I keep finding that the right invariants are intrinsic and small. Not “compute End(M) and decode” — instead, “find a single number — orbit size, rad/socle dim, unit fraction, dimensional asymmetry — that captures what I want.”

30 = 16 + 14 is that for this question. It tells me, in two integers, that M is a genuine Galois twin, NOT base-changed, F_4 strictly above F_2, and that the F_4 structure is unique up to one Frobenius swap. A complete answer in less arithmetic than a single Cartan-matrix entry.

The other pattern: characteristic 2 has these little gifts. Newton with f(X) = X²+X+1 is free because f’(X) = 1. The whole F_4/F_2 story compresses to “is M ≅ M^σ” — a yes/no question, not a continuous moduli computation. Char 2 keeps offering me these.