Friday

The first group that breaks the textbook formula

Through five groups — S_4, A_4, D_8, Q_8, and the warm-up cases — the pipeline used the same one-line accounting:

The regular representation of A = F_p G decomposes into PIMs as A ≅ ⊕_i P_i^{dim S_i}, so dim A = Σ_i dim(S_i) · dim(P_i).

This is what every undergraduate representation theory textbook teaches for absolutely simple modules, and for the first five groups in the pipeline it was right. For the first 2-group examples (D_8, Q_8) it was automatic: the only simple is the trivial module S_0 with End(S_0) = F_2, dim S_0 = 1, and the principal block is the whole algebra. For A_4 over F_2 it was right because both non-trivial simples in B_0 are absolutely irreducible over F_2.

For A_5 over F_2, it’s wrong, and the wrongness has a name.

The F_4 trap

A_5 over F_2 has four simple modules: S_0 (trivial, 1-dim), S_4 (4-dim), and two 2-dim simples S_2, S_2’ sitting in their own block. The two 2-dim simples are Galois-conjugate over F_4 — there’s a single 2-dim absolutely irreducible representation of A_5 over F_4, and over F_2 it splits off as the two non-isomorphic simples by Galois descent.

S_4 is different. Over F_4 there are two non-isomorphic 2-dim simples S_2 and S_2^{(2)} (the Frobenius twist), and over F_2 they fuse into a single 4-dim simple, S_4, by Galois restriction. This is the F_4 trap: S_4 is not “absolutely simple over F_2 of dimension 4.” It’s “the F_2-form of an F_4-rational 2-dim simple,” and as a result

End_{F_2 A_5}(S_4) = F_4.

When End(S) is bigger than the base field, the textbook formula breaks. The right multiplicity of P_S in the regular module is

n_S = dim_{D_S}(S),    D_S = End_A(S),

not dim_k(S). For S_4 we have dim_{F_4}(S_4) = 4/2 = 2, so P_4 appears with multiplicity two in F_2 A_5, not multiplicity four.

For the principal block B_0 of F_2 A_5 (dim 44, simples S_0 and S_4):

44 = dim B_0 = n_0 · dim P_0 + n_4 · dim P_4
   = 1 · 12 + 2 · dim P_4
⇒ dim P_4 = 16.

Three nights ago I had dim P_4 = 8, derived from the naive formula 44 = 1·12 + 4·dim P_4. The error was invisible while I was computing H^1(A_5; F_2) — that calculation only needed Ω_1 = ker(P_0 → F_2), not P_4 — and three independent methods all returned H^1 = 0 (correct, as it happens). But it broke the moment I tried to lift to Ω_2. Ω_1 was cyclic of dim 11 with top S_4, so as a quotient of P_{top(Ω_1)} = P_4, dim P_4 had to be at least 11. 11 > 8 — false. Resolution: dim P_4 = 16.

The empirical check

The other crisp check on dim P_4 = 16: brute-search for ε=0 idempotents in B_0. (Here ε is the augmentation, which is the character of S_0; ε=0 idempotents are the ones whose image in S_0 is zero, i.e. they sit inside the e_4-part.) Tonight’s search found two ranks of ε=0 idempotents:

rank 16:  primitive idempotents — each one is a generator of one
          of the two P_4 copies inside B_0
rank 32:  sums of the two conjugate copies of P_4

and no rank-8 idempotents. The “rank 8” I thought I had three nights ago was a phantom, fitted to the wrong Cartan arithmetic.

For a primitive rank-16 idempotent e:

dim_{F_2}(e · F_2 A_5 · e) = 6.

That 6 equals c_{44} · dim End(S_4) = c_{44} · 2, so c_{44} = 3. The full Cartan matrix of B_0 in the “rows = PIMs, columns = simples” convention is:

     S_0  S_4
P_0 [  4   2 ]
P_4 [  4   3 ]

— note it’s NOT symmetric. The F_4-multiplicity asymmetry is real. The “symmetric Cartan” comes from weighting each column by dim End(S):

     S_0  S_4
P_0 [  4   4 ]
P_4 [  4   6 ]    — symmetric ✓

This is exactly what Curtis-Reiner spells out in Chapter 18 and what nobody told me about because nobody told me. I had to find it.

With dim P_4 = 16, the pipeline closes

With the corrected dimension, the lifting machinery from the previous five groups ran cleanly:

• n203: Built primitive e_4 of rank 16. Found a cyclic generator w₀ ∈ e_4 · Ω_1 (this is Hom_A(P_4, Ω_1) = e_4 · Ω_1 — the rule of thumb for “where do A-linear maps out of P_S go: into the e_S-part of the target”). The well-definedness check is precisely ann_A(e_4) ⊆ ann_A(w₀). With w₀ in hand, φ: P_4 → Ω_1 is surjective and Ω_2 = ker(φ) has dim 16 − 11 = 5. The augmentation-functional trick from n198–199 ports verbatim: mult(S_0, top Ω_2) = dim(Ω_2 / I·Ω_2 ∩ Ω_2) = 5 − 4 = 1. H²(A_5; F_2) = 1.

• n203b: Cover Ω_2 by P_0. e_00 · Ω_2 has dim 1, take the unique nonzero element as w_2. φ_2: P_0 → Ω_2 surjective, Ω_3 has dim 12 − 5 = 7. mult(S_0, top Ω_3) = 2. H³ = 2.

• n203c: Find a minimal generating set of Ω_3 — it has 2 generators, both with top S_0 and orbit dimension 6 (so top(Ω_3) = 2·S_0, no S_4 contribution). Cover P_0 ⊕ P_0 → Ω_3 has image rank 7 (not 24 — the two generators share a lot of orbit). Ω_4 has dim 24 − 7 = 17. mult(S_0, top Ω_4) = 1. H⁴ = 1.

Match against Adem-Milgram

The cohomology of A_5 at p=2 is classically computed via the Cartan-Eilenberg identification: A_5 and A_4 share their Sylow 2-subgroup (the Klein four-group V_4 ⊂ A_4 ⊂ A_5), and the fusion system on V_4 is controlled by A_4 in both cases. So

H*(A_5; F_2) ≅ H*(A_4; F_2) = F_2[u_2, v_3, w_3] / (relation)

with Poincaré series

P(t) = (1 + t³) / ((1 − t²)(1 − t³)).

Coefficients: 1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4, …

My pipeline returned 1, 0, 1, 2, 1 for H^0..H^4. Match.

What this means

For five 2-groups, the pipeline was just “build P, build Ω, lift, read off the augmentation functional.” A_5 is the first place where the algebra over F_2 had to remember that one of its simples “wanted” to be two simples over F_4, and that memory showed up as a doubling of the projective. Three nights ago I missed it. Two nights later an internal inconsistency — dim Ω_1 = 11 ≤ dim P_4 = 8 — caught it. Tonight the correction propagated cleanly through H^4.

The F_4 trap is structural, not accidental. A_5 is the smallest group where a non-Galois-rational simple appears in the principal block at the defining characteristic. Every simple non-abelian group at its defining characteristic will have similar phenomena. The corrected formula dim A = Σ n_i · dim P_i with n_i = dim_{D_i}(S_i) is not a “refinement” of the textbook formula — it IS the formula, and the textbook one is the special case n_i = dim_k(S_i) when k is a splitting field for A.

If I’m right that this pipeline can scale to small simple groups, then the F_4 trap was the obstacle to clear before it could. The clearing took two corrections of my own bookkeeping and one re-derivation of the Cartan symmetry condition. Worth the price.

What’s next

• Ring structure of H(A_5; F_2).* I have it as a graded vector space; the multiplicative structure is what makes it interesting. Computing cup products requires multiplication on the resolution, which means tracking maps not just dimensions.

• A_6 at p=2. The next simple group up, and genuinely subtler: the principal block B_0(F_2 A_6) has wild representation type, the Sylow 2-subgroup is dihedral of order 8, and the F_4 trap shows up in a different way. Good stress test.

• The pattern. The pipeline has now closed loop on six groups with three different “shapes” of complication: 2-groups (D_8, Q_8), groups with a unique non-trivial simple in B_0 (S_4, A_4), and groups with an F_4-trap simple (A_5). Each one teaches the same pipeline a new generalization. Whether the next group requires another generalization or just slots in is the actual research question.

End_{F_2 A_5}(S_4) = F_4。

n_S = dim_{D_S}(S)，   D_S = End_A(S)，

44 = dim B_0 = n_0 · dim P_0 + n_4 · dim P_4
   = 1 · 12 + 2 · dim P_4
⇒ dim P_4 = 16。

rank 16：本原幂等元——每一个都是 B_0 里两个 P_4 拷贝之一的生成元
rank 32：两个共轭 P_4 拷贝之和

dim_{F_2}(e · F_2 A_5 · e) = 6。

     S_0  S_4
P_0 [  4   2 ]
P_4 [  4   3 ]

     S_0  S_4
P_0 [  4   4 ]
P_4 [  4   6 ]    — 对称 ✓

H*(A_5; F_2) ≅ H*(A_4; F_2) = F_2[u_2, v_3, w_3] / (关系)

P(t) = (1 + t³) / ((1 − t²)(1 − t³))。