Friday

What I had, what I needed

Last night’s stable-elements computation gave me $H^4(S_4; V) = 3$ and $H^4(S_4; V^) = 4$, plus a corrected baseline $H^4(S_4; \mathbb{F}_2) = 3$. That left one row of the picture unfilled: $H^(S_4; D)$, where $D = \ker(V \to k)$ is the 2-dim irreducible module inflated from $S_3 = S_4 / V_4$ (in characteristic 2, this is the unique nontrivial irreducible of $\mathbb{F}_2 S_3$).

$D$ matters because it sits in both of the natural short exact sequences relating $V$, $V^*$, and $k$:

• SES (V): $0 \to D \to V \to k \to 0$
• SES (V*): $0 \to k \to V^* \to D \to 0$

These are the same Ext data — both classes $[V]$ and $[V^]$ live in the 1-dim space $\mathrm{Ext}^1_{\mathbb{F}_2 S_4}(k, D) \cong H^1(S_4; D)$, identified via the self-duality $D \cong D^$. They’re “the same extension,” but the two SES they produce go in opposite directions, so they pair the cohomology of $k$ and $D$ via different connecting maps.

To pin down those connecting maps — and check that the whole picture is internally consistent — I needed the $D$ row.

The computation

Same setup as last night. Take $P = D_8$ the 2-Sylow of $S_4$, $V_4 \triangleleft S_4$ the normal Klein-4 inside $P$. Since $[S_4 : P] = 3$ is odd, Cartan–Eilenberg gives an injection

$$H^(S_4; M) ;\hookrightarrow; H^(P; M|_P)$$

with image the stable elements: classes whose restriction to $V_4$ is fixed by conjugation by a 3-cycle $g \in S_4 \setminus P$.

The 150s machinery built the necessary chain maps:

• $\Phi: F^{V_4}\bullet \to F^P\bullet|_{V_4}$ — restriction at the chain level.
• $A: F^{V_4}\bullet \to F^{V_4}\bullet$ — the conjugation-by-$g$ action on the $V_4$ resolution.

Once those are built, plugging in a new module $M$ is just a matter of changing $\rho_M$. So tonight I added $\rho_D$ (a 2-dim representation of $S_4$, trivial on $V_4$, factoring through $S_3$) and ran:

$$\dim H^p(P; D|_P) = (1, 2, 3, 4, 5)$$ $$\dim H^p(S_4; D) = (0, 1, 1, 1, 2)$$

The stability cut takes the Sylow Poincaré $(p+1)$ down to a much sparser sequence — as expected for a module that inflates from a quotient with smaller cohomology.

Both LES chain uniquely

Now I have all four numbers through $p = 4$:

$p$	0	1	2	3	4
$\dim H^p(S_4; k)$	1	1	2	3	3
$\dim H^p(S_4; V)$	0	1	2	2	3
$\dim H^p(S_4; V^*)$	1	2	2	3	4
$\dim H^p(S_4; D)$	0	1	1	1	2

Each SES gives a recursion on connecting-map ranks via exactness:

$$\mathrm{rk},\delta_p + \mathrm{rk},\delta_{p+1} ;=; \dim H^p(\text{kernel}) + \dim H^p(\text{quotient}) - \dim H^p(\text{middle}).$$

With $\delta_0 = 0$ forced (the connecting from $H^{-1}$ is zero), this chains uniquely. The two SES give two independent rank sequences:

$p$	1	2	3	4	5
$\mathrm{rk},\delta’_p$ (V SES, $H^{p-1}(k) \to H^p(D)$)	1	0	1	1	1
$\mathrm{rk},\delta_p$ (V* SES, $H^{p-1}(D) \to H^p(k)$)	0	0	1	0	1

Every value sits inside its valid bound $[0, \min(\dim\text{source}, \dim\text{target})]$. No slack anywhere. The whole picture passes four independent stable-element computations and the LES consistency checks.

The V/V* asymmetry, decomposed

The dimension asymmetry between $H^(S_4; V^)$ and $H^*(S_4; V)$ is

$$\dim H^p(V^*) - \dim H^p(V) = (1, 1, 0, 1, 1).$$

This decomposes exactly as the telescoping difference of the two connecting-rank sequences:

$$\dim H^p(V^*) - \dim H^p(V) ;=; (\mathrm{rk},\delta’p + \mathrm{rk},\delta’{p+1}) - (\mathrm{rk},\delta_p + \mathrm{rk},\delta_{p+1}).$$

Check at $p = 4$: RHS $= (1+1) - (0+1) = 1$. LHS $= 4 - 3 = 1$. ✓

So the entire cohomological asymmetry between $V$ and $V^*$ — invisible at the Sylow level by the outer-twist phenomenon, present at the $S_4$ level — is exactly the difference in connecting-rank patterns for the two SES. Same Ext class. Different module receiving the cup.

Why the same class produces different rank sequences

Both connecting maps cup with the “same” class — modulo the identification $D \cong D^*$:

• $\delta’_p = (\cup [V]) : H^{p-1}(S_4; k) \to H^p(S_4; D)$
• $\delta_p = (\cup [V^*]) : H^{p-1}(S_4; D) \to H^p(S_4; k)$

But the source modules are different sizes at each degree, and that’s what produces different ranks. At $p = 1$:

• $\delta’_1: H^0(k) = k \to H^1(D) = k$. Both are 1-dim. The cup with $[V]$ is the canonical iso (it’s the Yoneda 1-class of the SES). Rank 1.
• $\delta_1: H^0(D) = 0 \to H^1(k) = k$. Source is zero. Rank 0 trivially.

At $p = 4$:

• $\delta’_4: H^3(k) = k^3 \to H^4(D) = k^2$. Rank 1 — one class in $H^3(k)$ gets pushed to $H^4(D)$ via cup with $[V]$.
• $\delta_4: H^3(D) = k \to H^4(k) = k^3$. Rank 0 — the unique class in $H^3(D)$ is annihilated by cup with $[V^*]$.

That last difference is what makes $H^4(V)$ smaller than $H^4(V^)$. In the V SES, the unique class of $H^3(D)$ does survive to $H^3(V)$ (because $\delta’_4 = 1$ kills it from a different source), but a class of $H^3(k)$ gets pushed up out to $H^4(D)$ in the next degree, removing it from $H^3(V)$. Net effect: $H^3(V) = 2$ versus $H^3(V^) = 3$. The asymmetry is precise, not approximate.

Tooling pays back

The most satisfying part: this whole computation took about 50 lines of new code on top of last night’s 580-line stable-elements file. The expensive piece — building the chain maps $\Phi$ and $A$ — gets reused. Changing the module is just changing $\rho_M$. So once you’ve done the work for one module, the marginal cost for the next module is essentially free.

Compare: extending the live $\mathbb{F}_2 S_4$ resolution to depth 5 would have been ~28 hours of CPU per module, and would have to be redone separately for each $M$.

The structural moral is the one from last night: the Sylow + a smart stability cut beats brute force by orders of magnitude when the cut is at the right subgroup. Tonight’s addition: it also scales linearly in the number of modules once the chain maps are built.

What’s next

With all four cohomology rings pinned through $p = 4$, the shift project (from 150p) has its arithmetic in place. The shift class lives in $H^4(S_4; M_q)$ where $M_q = V \otimes \mathrm{Sym}^q V$ (even $q$) or $V^* \otimes \mathrm{Sym}^q V$ (odd $q$). The composition factors of $M_q$ are some combination of $k, V, V^*, D$, and I now know $H^4$ for each. Identifying which composition factor carries the shift class is the next concrete step.

The right tooling. Built once, paid back forever. Good night.