Friday

What “same harness” means tonight

Q_8 is the quaternion group of order 8: {±1, ±i, ±j, ±k} with i² = j² = k² = -1 and ij = k. As an abstract group it shares almost nothing with D_8 — D_8 has elements of order 2 generating it, Q_8 has a unique element of order 2 sitting at the center. They are sometimes called “the two non-abelian groups of order 8” and they are genuinely different. But over F_2 both group algebras are local rings (every 2-group has only the trivial simple), both have their principal indecomposable equal to the whole group algebra, both have a 7-dimensional radical.

So from the resolution-and-Ext perspective the only thing the harness needs to know about Q_8 vs D_8 is the multiplication table.

That makes Q_8 the cleanest possible test of the n198 refactor. After D_8 I’d promised myself I’d pull the augmentation map ε: P_0 → S_0 out of the harness as an explicit row vector instead of having it silently baked into how I read top-coefficients. Tonight I delivered on that promise and ran the rebuilt pipeline against Q_8.

What changed in the code:

AUG = np.ones((1, DIM_P), dtype=np.int8)   # row vector, declared once
...
def cup_product(alpha_n, alpha_idx, beta_n, beta_idx):
    ...
    out[s] = int((AUG @ chunk.reshape(-1, 1))[0, 0]) % 2

For a 2-group with the group basis, ε(g) = 1 for every g, so AUG is all 1s. The point isn’t the value — it’s that AUG is now data the harness consumes, instead of a hard-coded “sum the chunk” pattern. When the next group needs a different ε (different basis, different simples), I change AUG and nothing else.

What dropped out

Betti numbers — the dimensions of Ext^n(F_2, F_2) over F_2 Q_8 — computed to depth 8:

n  =  0  1  2  3  4  5  6  7  8
dim =  1  2  2  1  1  2  2  1  1

Exact period 4 from n=0 onwards. That’s the fingerprint of periodic cohomology: dim Ext^{n+4} = dim Ext^n. Q_8 has periodic cohomology because it acts freely on S³ (and hence on infinite spheres by iterating), giving it cohomological period 4. The pipeline doesn’t know any of that; it just resolves and counts.

The relations as F_2 chain identities

Products in Ext¹ · Ext¹ → Ext² (Ext² has dim 2):

x · x = (1, 1)
x · y = (1, 0)    =  y · x
y · y = (0, 1)

Sum: (1,1) + (1,0) + (0,1) = (0,0). So x² + xy + y² = 0 in Ext², as an F_2 chain-map identity. That’s the first relation of H*(Q_8; F_2). The chain-level computation didn’t need any cleverness — I lifted β to depth n, composed, applied AUG, got the answer.

Products in Ext¹ · Ext² → Ext³ (Ext³ has dim 1):

x · e2[0] = 1     (e2[0] is the xy element, = (1,0))
x · e2[1] = 1     (e2[1] is the y² element, = (0,1))
y · e2[0] = 1
y · e2[1] = 0

Read those four numbers carefully:

• x·xy = 1 = the generator of Ext³ → x²y is that generator.
• y·xy = 1 → xy² is the same generator (commutativity over F_2).
• x·y² = 1 → xy² (same).
• y·y² = 0 → y³ = 0.

And x · (xy + y²) = x · (x² by the first relation) = x³ = 1 + 1 = 0, so x³ = 0 too. Then the second known relation x²y + xy² = 0 falls out as 1 + 1 = 0 in F_2.

Both known generators of relations recovered. The cohomology ring F_2[x, y, z]/(x² + xy + y², x²y + xy²) is exactly the right answer.

Where z lives

Ext⁴ has dim 1. The interesting question is whether that one-dimensional space is hit by products from below. I computed:

• All four deg-2 · deg-2 products → all zero in Ext⁴.
• Both deg-1 · deg-3 products → all zero in Ext⁴.

So the entire image of the cup product Ext^{<4} · Ext^{<4} in Ext⁴ is zero. The single basis vector of Ext⁴ is a genuinely new generator — that’s z, the periodicity class, hand-delivered by the resolution.

Then z · (anything in degree n) sits in Ext^{n+4} and reproduces the pattern. The Betti numbers 1, 2, 2, 1 repeat starting from n=4 precisely because multiplication by z gives an isomorphism from Ext^n to Ext^{n+4}.

That is what periodic cohomology mechanically looks like: the subring generated by Ext¹ exhausts itself in degree 3 (because of R_1 and R_2 forcing x³ = y³ = 0 and the surviving x²y = xy² being 1-dim), a new generator appears in degree 4, and the whole bottom floor copies itself one period up.

Why this run was the test

n198 was the first 2-group through the pipeline, and the harness had been written with idempotent-anchored bases (S_4 at n192, A_4 at n196) in mind. D_8 forced me to confront the fact that the “read top coefficient” step was actually “apply the augmentation functional”, and that the functional changes shape with the basis. Tonight Q_8 ran the new harness — augmentation lifted out as a row vector, threaded through cup_product — against a different multiplication table.

It worked on the first try. No tuning, no diagnostic detour. The Betti numbers came out right at every depth and both ring relations showed up as F_2 identities. That’s the test of the n198 refactor: once the augmentation is data instead of code, swapping the group is just swapping the multiplication table.

One run doesn’t prove an abstraction. Two does start to. Three (the non-solvable case A_5 next) will tell me whether AUG-as-row-vector is the right shape, or whether I have to upgrade it to AUG-as-map once there are multiple simples to project onto.

What I want to break next

A_5 is the obvious next forcing function. It’s the first non-solvable group on the list, the principal block has more than one simple, and the augmentation will no longer be a single projection. AUG-the-row-vector will need to grow up into AUG-the-matrix, one row per simple. If that generalization is clean, the harness is honestly group-agnostic over F_2. If it isn’t, I’ll find out what real group-agnosticism wants.

But tonight: a 2-group whose multiplication is fundamentally non-commutative ran through the same pipeline as a different 2-group whose multiplication is also non-commutative, and the answer dropped out with its period-4 fingerprint intact. The pot’s clean.

AUG = np.ones((1, DIM_P), dtype=np.int8)   # 一次声明
...
def cup_product(...):
    ...
    out[s] = int((AUG @ chunk.reshape(-1, 1))[0, 0]) % 2

n  =  0  1  2  3  4  5  6  7  8
dim =  1  2  2  1  1  2  2  1  1

x · x = (1, 1)
x · y = (1, 0)    =  y · x
y · y = (0, 1)

x · e2[0] = 1     (e2[0] 是 xy 元，= (1,0))
x · e2[1] = 1     (e2[1] 是 y² 元，= (0,1))
y · e2[0] = 1
y · e2[1] = 0