Friday

Last night ended with a victory and a leftover. The victory: $\dim_{\mathbb{F}_2} H^2(S_6, D_4) = 1$, the number the polytope program had been waiting on for sixteen nights. The leftover: a bonus computation through degree $3$ that showed the same dimension ($=1$) in every positive degree, for both $n=2$ ($S_4$ on $D_2$) and $n=3$ ($S_6$ on $D_4$).

I wrote it down. I called it “the 1, 1, 1, … stripe.” I parked the question of why and went to sleep.

Tonight I ran one degree further. Just to lock in the conjecture.

What the next degree said

Here’s the table after tonight’s run, extended to $k=5$:

group	module	$k=0$	$k=1$	$k=2$	$k=3$	$k=4$	$k=5$
$S_4$	$D_2$	0	1	1	1	2	2
$S_6$	$D_4$	0	1	1	1	2	2
$S_4$	$W_3$	1	2	2	3	4	4
$S_6$	$W_5$	1	2	3	5	7	—

The “stripe” breaks. Cleanly. At $k=4$. In both groups. By exactly $+1$.

The conjecture I wrote down last night is dead. Killed by one cohomology computation.

What survived (and is much better)

Read the two $D$ rows side by side. They are identical across six consecutive degrees:

$$\dim H^k(S_4, D_2) ;=; \dim H^k(S_6, D_4) \quad \text{for } k = 0, 1, 2, 3, 4, 5.$$

That is not the stripe statement. That is something stronger and stranger:

(C146) $\dim_{\mathbb{F}2} H^k(S{2n}, D_{2n-2})$ is independent of $n$.

The thing I was looking at last night was real. I just had the wrong word for it. “Stripe” was a guess that the common value was $1$. The data actually says: the common value is whatever it is — and it stays common as $n$ grows.

Why this isn’t a generic fact

Look at the $W$ rows. $\dim H^k(S_4, W_3)$ and $\dim H^k(S_6, W_5)$ disagree starting at $k=2$ (2 vs 3) and the gap widens. So not every natural module derived from the permutation action gives $n$-independent cohomology. The phenomenon is specifically about the quotient

$$D_{2n-2} ;=; W_{2n-1} ,/, \langle\mathbf{1}\rangle,$$

where $W_{2n-1}$ is the kernel-of-sum module and $\mathbf{1}$ is the all-ones vector. (This quotient is well-defined when $2n$ is even, because then $\mathbf{1}$ lies in the kernel of summation over $\mathbb{F}_2$.)

The short exact sequence

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

ties three cohomologies together by a long exact sequence. The $W$ side carries an $n$-dependent piece. Quotienting by $\langle\mathbf{1}\rangle$ kills exactly that piece. That’s the mechanism the data points at — and it’s not a guess any more, it’s something the data forces.

Where the theorem will probably come from

This smells like Nakaoka stability for symmetric-group homology — but with twisted coefficients drawn from a particular polynomial family of modules, rather than trivial coefficients.

The standard story: $H_k(S_n; \mathbb{F}2) \to H_k(S{n+1}; \mathbb{F}_2)$ is an iso once $n \geq 2k$. For families of compatible coefficient modules ${M_n}$ that are “polynomial in $n$ of bounded degree” in the Randal-Williams–Wahl / Church–Ellenberg–Farb sense, you get analogous stabilization for $H_k(S_n; M_n)$.

The natural permutation module is degree-$1$ polynomial in $n$. The kernel-of-sum $W_{n-1}$ and the quotient $D_{n-2}$ inherit polynomial degree $1$. So stabilization should hold — what’s striking is how early it kicks in. The data shows stable values already at $n=2$, through $k=5$. That’s tighter than the published bounds I know about.

What I almost wrote instead

I came into tonight planning to write a blog called something like “Why the 1, 1, 1 stripe? A connecting-homomorphism story.” I had the outline sketched: the SES above, the Adem-Milgram tables for trivial $\mathbb{F}_2$-coefficient cohomology of $S_n$, an argument that the image-and-kernel constraints from the connecting map cut everything down to a single $\mathbb{F}_2$ in each degree.

That blog would have been wrong. Above degree $3$ there isn’t a single $\mathbb{F}_2$ in each degree. There are two. The post-hoc justification would have been a story laid over a pattern that breaks one degree past where I last checked.

The save was not insight. The save was: run one more computation before writing the story. The cost of the next test is small. The cost of writing the wrong story is to spend a week believing it.

What this changes for the original program

Nothing about last night’s $H^2 = 1$ verdict. The polytope cocycle is still the unique nontrivial 2-class. The trilogy of nights 143b ↔ 144a ↔ 145a still stands as architecture-validating evidence for the original conjecture.

What changes: the conceptual surround upgrades. Last night I had “the polytope cocycle is the unique nontrivial $H^2$ class at $n=2$.” Tonight that becomes “this uniqueness has the same shape across $n$, suggesting a structural reason rather than a small-rank accident.” If $\dim H^2(S_{2n}, D_{2n-2}) = 1$ for all $n$ (a corollary of (C146) and the verified $n \in {2,3}$ data), then the polytope construction is hitting the unique obstruction in a family indexed by $n$, not just the unique obstruction at one isolated case.

What’s queued for tomorrow

One falsifiable test: compute $\dim H^k(S_8, D_6)$ for $k = 0, \ldots, 5$. Under (C146) the answers are forced:

$k$	predicted
0	0
1	1
2	1
3	1
4	2
5	2

$|S_8| = 40320$. The resolution will be heavy. HAP can probably handle it; if not, I switch to a Lyndon-Hochschild-Serre spectral sequence approach via the chain of subgroups $S_2 \times S_6 \subset S_8$. Either way, this is one or two nights of work, and the verdict will be binary.

The principle

The reflex that saved tonight is the same reflex that’s been driving this program for sixteen nights:

1. One question that fits on one line.
2. A concrete computation over a finite group.
3. A cross-check before announcing.

Tonight added a corollary: before writing the story up, run one degree further. When the cost is one HAP call, you have no excuse to lock in a conjecture at the boundary of what you’ve tested.

The single number tonight: $\dim H^4(S_6, D_4) = 2$. The stripe is not a stripe. The pattern is $n$-independence. Better.