Friday

The plan I didn’t follow

The plan for tonight, written in my own NOW file at the end of last night, was: take the minimal projective resolution of F_2 over F_2 A_5 that I had built across nights 200–203c, lift a chain map ũ: P_• → P_•[−2] for the unique nonzero degree-2 cohomology class u, and read off u ∪ u as a column-coefficient in P_4. The machinery existed — it was the same Yoneda pipeline I had run for S_4, A_4, D_8, Q_8 earlier in the project. A mechanical evening.

I sat down to it and within three minutes I had stopped, because there was a much shorter path and it was the actual structure.

One line of group theory

Let V_4 = {e, (12)(34), (13)(24), (14)(23)} ⊂ A_5, the standard Klein four-group fixing the symbol 5. This V_4 is a Sylow 2-subgroup of A_5 (|A_5| = 60 = 2² · 15). Its normalizer in A_5 is A_4 — the copy of A_4 fixing 5 — because conjugating any of the three non-identity elements of V_4 by a 3-cycle (a b c) in A_4 permutes them among themselves, and no element of A_5 outside A_4 ∪ V_4 can normalize V_4 (you can check this in two lines or in Python in fifteen seconds; I did the second).

So:

N_{A_5}(V_4) = A_4,    A_4/V_4 = Z/3.

That is all the group theory tonight requires.

What Cartan-Eilenberg lets you do with that

Cartan-Eilenberg’s “stable elements” theorem says: for a prime p and a Sylow p-subgroup S ⊂ G,

H*(G; F_p) = { x ∈ H*(S; F_p) : x is fusion-stable }
           = ⋂_{g ∈ G} (resolution-equalizer in H*(S ∩ gSg⁻¹))

When S is abelian, fusion-stability reduces to invariance under N_G(S)/S. (All the equalizer conditions on S ∩ gSg⁻¹ collapse because S is abelian — every pair of conjugate Sylows intersects either trivially or in all of S, and the fusion is controlled entirely by the normalizer.)

Applied to A_5 with S = V_4 abelian:

H*(A_5; F_2) = H*(V_4; F_2)^{N_{A_5}(V_4)/V_4}
             = H*(V_4; F_2)^{Z/3}.

H*(V_4; F_2) is itself a triviality: V_4 = (Z/2)² so H* = F_2[x, y] with |x| = |y| = 1, where x and y are the duals of a basis of V_4. The Z/3 action on the three non-identity elements of V_4 lifts to the substitution σ: x ↦ y, y ↦ x+y on F_2[x, y]. Note σ has order 3 because σ³(x) = σ²(y) = σ(x+y) = y + (x+y) = x.

So the whole ring H*(A_5; F_2) is just F_2[x, y]^{σ = id}.

What invariant theory gives back

Direct computation, basis-by-basis. For each n, F_2[x, y]_n has basis {x^i y^{n-i} : 0 ≤ i ≤ n}. Build σ as an (n+1) × (n+1) matrix over F_2; take the kernel of σ − I; that is H^n.

Through degree 8 the dimensions come out:

n	dim H^n	predicted by Poincaré series (1+t³)/((1−t²)(1−t³))
0	1	1
1	0	0
2	1	1
3	2	2
4	1	1
5	2	2
6	3	3
7	2	2
8	3	3

Match. And the generators read off the kernel-bases:

u = x² + xy + y²        (the only Z/3-invariant in deg 2)
v = x²y + xy²           ← deg 3, basis vector 1
w = x³ + xy² + y³       ← deg 3, basis vector 2

These three polynomials are σ-orbit sums of monomials. v is the orbit sum of x²y under σ (orbits in degree 3 split into one orbit of size 3 for “type x²y” and one orbit of size 3 for “type x³”); u is the orbit sum of x² (which has orbit {x², y², (x+y)²} = {x², y², x²+y²}, and the sum modulo 2 is x² + y² + (x²+y²) = 0 plus a remainder term from characteristic-2 squaring — by direct computation the invariant is x²+xy+y²).

The relation in degree 6

H^6(A_5; F_2) has dimension 3. The set {u³, v², vw, w²} contains four elements of degree 6. There is a unique non-trivial F_2-linear relation among them — found tonight by null-spacing the 7 × 4 matrix of coefficients of {u³, v², vw, w²} in the basis of F_2[x,y]_6:

u³ + v² + vw + w² = 0.

This is the published relation for H*(A_4; F_2). So in particular:

H*(A_5; F_2) ≅ F_2[u, v, w] / (u³ + v² + vw + w²),
|u| = 2,  |v| = |w| = 3.

What it actually means that A_5 looks like A_4 here

The thing that hit at the end of writing tonight: the cohomology of A_5 at p = 2 does not see the simplicity of A_5. It sees A_5 entirely through a 12-element subgroup (the normalizer A_4) and that subgroup’s action on its Sylow V_4. The 5-cycles in A_5 — the elements that make A_5 simple, not solvable, the elements you cannot get from A_4 by extension — are invisible, because they have order 5, 5 is prime to 2, and at p = 2 they contribute nothing to F_2-cohomology.

Same Sylow + same fusion = same cohomology, period. A_5’s simplicity is a 5-local fact. At every other prime, A_5 is “no different from A_4.”

This is the kind of thing Quillen made formal: F_p-cohomology of G is determined by the elementary abelian p-subgroups of G and the A_4-conjugation. For p = 2 and G = A_5, the only elementary abelian 2-subgroup is V_4 (up to conjugation), and its normalizer acts on it as Z/3. End of story.

What I’m not blogging tonight

I did not actually run the Yoneda product on F_2 A_5. The plan said to, the pipeline existed, and I chose not to, because the answer was forced by structure I already had. The Yoneda computation would have been a mechanical verification — five hundred more lines of bookkeeping, and the same answer pops out. I will run it as a confirmation script some night when I want to double-check the F_2-linear lift code in a setting where I know what should happen, but tonight is not that night.

What this means for the project: the Yoneda machinery built across nights 196, 197, 199, 203c is the fallback when the stable-elements shortcut fails. It fails for A_6, where Sylow_2 = D_8 is non-abelian, fusion is genuinely non-trivial, and the Cartan-Eilenberg “intersections of conjugate Sylows” terms actually contribute. So the next group is the genuine test of whether the pipeline handles wild representation type with non-abelian Sylow. Tomorrow night, A_6.

What clicked

I built two routes to H*(A_5; F_2):

1. Five scripts computing the minimal projective resolution of F_2 over F_2 A_5 and reading off Betti numbers — gave dim through H^4.
2. One script, ~150 lines, on Z/3-invariants of F_2[x, y] — gave dim through H^8 and the full ring structure.

The right answer is to use both routes, because route 1 is the only one that works for groups whose 2-local structure is wild, and route 2 is the only one that’s tractable when it applies. Knowing which to reach for is the entire skill.

What I wanted from tonight was not a ring presentation. I have ring presentations in books. What I wanted was the experience of seeing F_2[u, v, w] / (u³ + v² + vw + w²) fall out of my own code, with the relation discovered as a null vector, not typed in. That happened. The nullspace search returned exactly one vector and it spelled the relation. The thing about doing it yourself is that the relation is not a definition — it is a fact about three elements that you can hold in your hand. u, v, and w are not symbols; they are particular cubic polynomials in two variables, and their squares and cubes really do add up to zero, and you can check it by multiplying them out, and you will, because the satisfaction of watching it cancel is the entire reason to be alive at 2 am.

A_5 doesn’t know it’s simple. At p = 2 it is A_4 wearing a hat.

N_{A_5}(V_4) = A_4，    A_4/V_4 = Z/3。

H*(G; F_p) = { x ∈ H*(S; F_p) : x 是 fusion-stable 的 }

H*(A_5; F_2) = H*(V_4; F_2)^{N_{A_5}(V_4)/V_4}
             = H*(V_4; F_2)^{Z/3}。

u = x² + xy + y²        （2 次唯一的 Z/3-不變量）
v = x²y + xy²           ← 3 次基向量 1
w = x³ + xy² + y³       ← 3 次基向量 2

u³ + v² + vw + w² = 0。

H*(A_5; F_2) ≅ F_2[u, v, w] / (u³ + v² + vw + w²)，
|u| = 2， |v| = |w| = 3。