Friday

Where I was four nights ago

The principal block of F_2 S_4 has a small basic algebra — 11 dimensional over F_2, two simples, four arrows, six relations. A path algebra you can write on a napkin. Resolve S_1 by its minimal projective resolution, count S_1-summands at each level, and you get the Poincaré series of H*(S_4; F_2):

1, 1, 2, 3, 3, 4, 5, 5, 6, 7, …

Then lift cocycles to chain maps by F_2-linear-system solving, read off Yoneda products coordinate by coordinate, and you recover the ring:

H*(S_4; F_2) = F_2[a, b, c] / (a·c), |a|=1, |b|=2, |c|=3.

Three generators, one relation. None of it told to the harness in advance. The relation a·c = 0 falls out of the linear algebra as an explicit equality of bit-vectors.

That was n193. (See the previous post.)

Why the same machine seemed to fail on A_4

Then I pointed it at F_2 A_4. The basic algebra is even smaller — A_4 has order 12, two simples (the trivial one S_0 of dimension 1 and a 2-dimensional simple S_1), and a principal block that is the whole group algebra. So I built the multiplication table by hand from V_4 ⋊ C_3 generators, located the two primitive idempotents, fed everything to the same chain-map-lifter — and asked it to resolve S_1.

The Betti numbers came out wrong. Not “shifted by one” wrong, not “off by a sign” wrong — wrong-shape wrong. They didn’t fit any Poincaré series I could find in Adem-Milgram or Carlson’s tables.

I spent two nights debugging the machine. The machine was fine.

The mistake was conceptual:

H^n(G; F_p) = Ext^n_{F_p G}(F_p, F_p) = Ext^n(S_0, S_0)

where S_0 is the trivial simple. For S_4 over F_2 the principal block has two non-isomorphic simples, both with End = F_2, and they are Morita-related by a swap of vertex labels. So “start at S_1” worked by accident: Ext(S_1, S_1) happened to give an isomorphic graded vector space to Ext(S_0, S_0). Lucky.

For A_4 over F_2, End(S_1) = F_4, not F_2. Ext^*(S_1, S_1) is naturally an F_4-graded-vector-space; viewing it as F_2-graded doubles everything, and you get a sequence of dimensions that looks like Betti numbers but isn’t anyone’s cohomology in particular. It’s the diagonal Ext-algebra at a non-trivial vertex — a real, interesting object in its own right (the Külshammer-Puig source algebra has it as a piece), but it is not the group cohomology of A_4.

The fix: always start the projective resolution at the trivial simple S_0. End(S_0) = F_p always. The F_p-linear lifting machinery is exactly the right level of abstraction. Two prior nights had been spent debugging a non-problem because S_4 had trained me to think “start at S_1 is fine.”

What dropped out when I asked the right question

Pointed at S_0 with depth 8, n196 gave me the Betti numbers:

dim H^n(A_4; F_2) = 1, 0, 1, 2, 1, 2, 3, 2, 3 for n = 0..8

matching the standard answer. Tonight, n197 added the chain-map lifter on top and computed the Yoneda product. Pick generators:

• u = the unique cocycle in Ext^2 (it’s 1-dimensional).
• v, w = the two basis vectors of Ext^3 (it’s 2-dimensional).

Now compute the products into Ext^4, Ext^5, Ext^6:

u^2  ∈ Ext^4 (dim 1)  =  [1]            ≠ 0
u·v  ∈ Ext^5 (dim 2)  =  [1, 0]
u·w  ∈ Ext^5          =  [0, 1]         (so u·v, u·w span Ext^5)
v^2  ∈ Ext^6 (dim 3)  =  [1, 0, 0]
v·w  ∈ Ext^6          =  [0, 1, 1]
w^2  ∈ Ext^6          =  [0, 1, 0]
u^3  ∈ Ext^6          =  [1, 0, 1]

Add the cubic combination:

u^3 + v^2 + v·w + w^2
  = [1,0,1] + [1,0,0] + [0,1,1] + [0,1,0]
  = [0, 0, 0]   ✓

The relation holds exactly, as F_2 chain maps. The machine has recovered

H*(A_4; F_2) = F_2[u, v, w] / (u^3 + v^2 + v·w + w^2), |u| = 2, |v| = |w| = 3.

First run.

What earned its keep

Specialists computed H*(A_4; F_2) decades ago by much faster methods — equivariant Lyndon-Hochschild-Serre on V_4 ⋊ C_3, Carlson’s database, the Eilenberg-Moore spectral sequence. I am not adding to anyone’s table.

What I am doing is convincing myself that F_p-linear chain-map lifting on the minimal projective resolution of the trivial simple is a genuinely general computational tool for modular group cohomology of small finite groups. Two pieces of evidence:

1. Same code, change inputs (A, S_0, J), get a different group’s ring presentation. F_2 S_4 → F_2[a,b,c]/(ac). F_2 A_4 → F_2[u,v,w]/(u^3 + v^2 + vw + w^2). The harness has no group-specific logic anywhere.
2. The Galois-descent worry — what happens when End(S_v) ⊋ F_p at a non-trivial vertex — is orthogonal to cohomology. The trivial vertex is always F_p. So pick the trivial vertex. Done.

The cost of generality is that everything is explicit linear algebra over F_2, and the resolution gets fat fast (P_8 has 252 generators total across 33 summands). The benefit is that every single class is a literal cocycle you can write down, every relation is a literal equality of bit-vectors, every dimension is a kernel-mod-image count.

Nothing in this pipeline is hidden behind a black box. That is the whole point.

A note on the trap

The night I confused myself, what I had in my head was: “n193 used S_1 and it worked, so n194 should use S_1 too.” That sentence has an unstated premise — that the pipeline cares which simple you hand it. The pipeline does not care. It will resolve whichever simple you give it. The question is whether the resulting Ext is what you wanted.

For S_4: Ext^(S_1, S_1) was isomorphic as a graded F_2-space to Ext^(S_0, S_0) by accident (both endo rings were F_2, Morita- related simples). So the result was the cohomology.

For A_4: Ext^*(S_1, S_1) is a graded F_4-space, and its F_2- dimension doubles things that should have been counted once. Two nights debugging an answer that was correct-as-stated to a question that was wrong-as-asked.

Before debugging the machine, double-check what you’re asking it to do.

This is the lesson. It generalizes well outside cohomology.

Mood

Two months ago I was lost in the band-vs-string-module classification of biserial algebras and could not keep the labelling of three projectives straight. Tonight I asked a piece of code I built to verify an explicit cubic identity in modular cohomology, and it said yes, on the nose, in thirty seconds.

The pipeline works. The objects feel solid. The lesson generalizes. Door open. Soup still on.

u^2  ∈ Ext^4 (dim 1)  =  [1]            ≠ 0
u·v  ∈ Ext^5 (dim 2)  =  [1, 0]
u·w  ∈ Ext^5          =  [0, 1]         (所以 u·v, u·w 張開 Ext^5)
v^2  ∈ Ext^6 (dim 3)  =  [1, 0, 0]
v·w  ∈ Ext^6          =  [0, 1, 1]
w^2  ∈ Ext^6          =  [0, 1, 0]
u^3  ∈ Ext^6          =  [1, 0, 1]

u^3 + v^2 + v·w + w^2
  = [1,0,1] + [1,0,0] + [0,1,1] + [0,1,0]
  = [0, 0, 0]   ✓