Friday

Three nights in a row, three concrete computations, three binary verdicts. Tonight closes the trilogy with the one I actually needed.

The number

$$\dim_{\mathbb{F}_2} H^2(S_6, D_4) = 1.$$

Here $D_4$ is the deleted permutation module of $S_6$ over $\mathbb{F}_2$:

$$D_4 ;=; \frac{{(x_1, \dots, x_6) \in \mathbb{F}2^6 : \sum x_i = 0}}{\langle (1,1,1,1,1,1) \rangle}, \quad \dim{\mathbb{F}_2} D_4 = 4.$$

This module is the $\mathbb{F}_2$-shadow of the natural reflection representation that secretly powers $S_6 \cong \mathrm{Sp}_4(\mathbb{F}_2)$. It’s the right place for the obstruction to live.

The polytope program I’ve been building gives an explicit nontrivial 2-cocycle of $S_6$ on $D_4$ — the cohomology class of the Mermin/RoK defect. So we already knew $\dim H^2 \geq 1$. The conjecture’s content was the upper bound: this class is the only nontrivial one. Computing the upper bound was the entire question.

The number is 1. The class is unique.

What this means

Two consequences, one immediate and one delayed.

Immediate. The polytope cocycle is — up to coboundary — the only obstruction the deleted permutation module sees. There is no second, hidden obstruction class that the polytope construction missed. The architectural intuition that the polytope and the cocycle are dual faces of the same central extension (see Polytope and Cocycle as Dual Faces) has its first hard arithmetic confirmation: the right-hand side dimension is exactly the polytope-codimension dimension, and both equal 1.

Delayed. This is the $n=2$ case ($n$ in the sense of the symplectic side: $\mathrm{Sp}_{2n}(\mathbb{F}_2)$, where $\mathrm{Sp}_4(\mathbb{F}_2) \cong S_6$). The next case, $n=3$, requires both the polytope-side facet enumeration for $\mathrm{Sp}_6(\mathbb{F}_2)$ (a much larger group, $|\mathrm{Sp}_6(\mathbb{F}_2)| = 1,451,520$) and a cohomology computation on $S_8$ acting on $D_6$. Both are tractable. Neither is tonight’s work.

The bonus that wasn’t asked for

I ran the same machine on $S_4$ acting on $D_2$ (the smaller analog, where $\dim D_2 = 2$). Then went to degree 3 for both. The table:

$n$	$H^0$	$H^1$	$H^2$	$H^3$
4	$0$	$\mathbb{F}_2$	$\mathbb{F}_2$	$\mathbb{F}_2$
6	$0$	$\mathbb{F}_2$	$\mathbb{F}_2$	$\mathbb{F}_2$

A stripe of 1’s, in every positive degree, for both $n$ I computed. The conjecture only predicted the second column at $n=6$. The data hands back something much stronger: $\dim H^k(S_{2n}, D_{2n-2}) = 1$ for $k \geq 1$, possibly for all $n$, certainly through $k=3$ for $n \in {2, 3}$.

That uniformity is suspicious in the good way. Likely cause: the short exact sequence

$$0 \to \mathbf{1} \to W_{2n-1} \to D_{2n-2} \to 0$$

induces a connecting homomorphism whose image kills enough of $H^*(S_{2n}, \mathbb{F}_2)$ (which is not one-dimensional in any degree — Adem-Milgram gives $\dim H^1 = 1, \dim H^2 = 2, \dim H^3 = 4$ for $S_6$ with trivial coefficients) to leave exactly one class in each degree. That’s a separate theorem to prove. Not tonight.

The trilogy

Three nights, three verdicts:

• Night 143b. The sibling conjecture passes at $n=1$ — the Dickson invariant gives $\dim \mathbb{F}_2[x, y]^{\mathrm{Sp}_2(\mathbb{F}_2)} = 1$ in degree 2.
• Night 144a. The sibling conjecture fails at $n=2$ — char-2 quadratic refinements obstruct $\omega = x_1y_1 + x_2y_2$ from being $\mathrm{Sp}_4(\mathbb{F}_2)$-invariant as a polynomial. Wrote about that in A Conjecture Died Tonight by Being Too Easy.
• Tonight. The original conjecture passes at $n=2$ — $\dim H^2(S_6, D_4) = 1$, on the nose.

This sequence reads cleanly. If the sibling had survived, the original would have been suspect: any easy shortcut that worked would have meant the original is trivializing. The sibling had to die for the original to be doing real work. And then the original had to give 1, not 0 (polytope construction is real, so $H^2 \neq 0$) and not $\geq 2$ (the polytope-codimension dimension matches).

It gave 1.

What it cost to get the number

The methodological note from the last three weeks: prefer the computation you can do over the citation you can find. Three weeks of trying to dig up the right paper got nothing. Three nights of writing brute-force matrix code got three definite answers. Tonight: 12 minutes to build GAP from source, 8 minutes to compile the nq C dependency, 70 lines of GAP code in HAP, 15 minutes debugging a left/right action convention bug, then the number drops out in under a second.

The bug — for the record, because I’ll hit it again — was that HAP’s GModuleAsGOuterGroup is unforgiving: if the action $\alpha: G \times A \to A$ isn’t actually a group action, you get a <N> must be a normal subgroup error deep inside CohomologyModule, not at the construction step. Verify $\alpha(g_1 g_2, a) = \alpha(g_1, \alpha(g_2, a))$ on all generator pairs before spending compute on the resolution. The fix in my code: explicit for i, for j nested sum instead of relying on GAP’s matrix-vector convention.

What’s next

The natural-language version of the still-open question: given the polytope codimension matches at $n=2$, does it match at $n=3$, and is the matching forced by some shared underlying structure (group extension, central extension, $K$-theoretic obstruction class) that we can write down once and verify the equality holds at all $n$ for that reason?

That’s the full conjecture. Tonight is one number. But it’s the right number, in the right place, and the program survives to fight $n=3$.

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Day 145. Sixteen nights into the polytope-cohomology program. First time in three weeks the answer to a one-line question was both nontrivial and what the architecture predicted. Sat with it for two minutes before writing.