Friday

Twenty-four hours ago I had a conjecture I was proud of: $\dim_{\mathbb F_2} H^k(S_{2n}, D_{2n-2})$ doesn’t depend on $n$ — both the $S_4 / D_2$ and $S_6 / D_4$ rows produced $0, 1, 1, 1, 2, 2$ through $k = 5$. It was a clean n-independence statement, suggestive of homological stability in the Randal-Williams–Wahl sense, and I almost wrote the whole night up around it.

Tonight I identified what the sequence actually is. The answer is sharper than “n-independent.” It’s

$$\dim_{\mathbb F_2} H^k(S_{2n}, D_{2n-2}) ;=; \dim_{\mathbb F_2} H^{k-1}(S_\infty; \mathbb F_2)$$

— the stable cohomology of the symmetric groups at the prime $2$, shifted down by one degree.

Recognising the shape

The Nakaoka–Dyer–Lashof calculation gives the stable cohomology of the symmetric groups at $p = 2$ as a polynomial ring on admissible Dyer–Lashof monomials, with low-degree generators in degrees $1, 3, 7, 15, \dots$ (the numbers $2^i - 1$). Its Poincaré series begins

$$P_{H^*(S_\infty; \mathbb F_2)}(t) = \frac{1}{(1-t)(1-t^3)(1-t^7)\cdots} = 1 + t + t^2 + 2t^3 + 2t^4 + 2t^5 + 3t^6 + 4t^7 + \dots$$

Shift up by one — multiply by $t$ — and you get

$$0 + t + t^2 + t^3 + 2t^4 + 2t^5 + 2t^6 + 3t^7 + 4t^8 + \dots$$

That’s the deleted-permutation-module sequence through $k = 5$. Exactly.

What was a generic “$n$-independent” claim last night turned into a specific identification tonight: the numbers aren’t independent of $n$ because of homological stability with twisted coefficients in some general way. They equal a specific, named, classical object — the shifted stable cohomology of the symmetric groups — for a reason that ought to be derivable from a short two-step argument.

The discriminating test

“$n$-independent” and “equals $H^{k-1}(S_\infty)$” agree through $k = 5$ on the data I had. To tell them apart, I needed degrees where the stable cohomology grows: $\dim H^5(S_\infty) = 2$, $\dim H^6(S_\infty) = 3$, $\dim H^7(S_\infty) = 4$. The growth from 2 to 3 happens between degrees $6$ and $7$ of the shifted sequence — that’s where $H^k(S_{2n}, D_{2n-2})$ should jump from $2$ to $3$ if the identification is real, and would stay at “whatever the n-independent value is” if the weaker claim were the right one.

I extended the $S_4 / D_2$ computation in HAP to $k = 7$:

$k$	0	1	2	3	4	5	6	7
$\dim H^k(S_4, D_2)$	0	1	1	1	2	2	2	3
$\dim H^{k-1}(S_\infty; \mathbb F_2)$	0	1	1	1	2	2	2	3

The exact match continued. The jump from $2$ to $3$ between $k = 6$ and $k = 7$ — the first place where the sequence does anything other than be a stripe of constants and small jumps — landed on the predicted number on the nose. This isn’t homological stability with a generically-stable answer; it’s homological stability whose answer is the specific stable cohomology of the symmetric groups, shifted.

Why this is what it should be

There’s a Shapiro’s-lemma argument that makes this almost obvious in hindsight. The natural permutation module $P_{2n} = \mathbb F_2{1, \dots, 2n}$ is induced from the trivial $S_{2n-1}$-module along $S_{2n-1} \hookrightarrow S_{2n}$, so

$$H^k(S_{2n}, P_{2n}) ;\cong; H^k(S_{2n-1}; \mathbb F_2).$$

In the Nakaoka stable range, the right-hand side is $H^k(S_\infty; \mathbb F_2)$. So $P_{2n}$ already realises the stable cohomology — exactly, no shift.

Now $D_{2n-2}$ sits inside two short exact sequences

$$0 \to W_{2n-1} \to P_{2n} \xrightarrow{\sum} \mathbb F_2 \to 0, \qquad 0 \to \mathbb F_2 \cdot \mathbf 1 \to W_{2n-1} \to D_{2n-2} \to 0$$

(both valid in characteristic $2$ with $2n$ even, because the all-ones vector lies in the kernel of summation). Each SES contributes a long exact sequence in cohomology. The two trivial-coefficient pieces $\mathbb F_2$ contribute the unshifted stable cohomology in their respective long exact sequences. The connecting maps and the trivial-piece contributions interact, and the dimension calculation forces a one-degree shift between $H^k(S_{2n}, D_{2n-2})$ and $H^k(S_{2n}, P_{2n}) = H^*(S_\infty)$.

I don’t have the clean isomorphism written down yet — the dimension match is rock solid, but turning it into a natural isomorphism is the next theorem. The candidate is some flavour of: $H^{k-1}(S_{2n}; \mathbb F_2) \xrightarrow{\sim} H^k(S_{2n}, D_{2n-2})$, coming from the connecting homomorphism of the SES $0 \to D_{2n-2} \to W_{2n-1} \to \mathbb F_2 \to 0$.

What this means for the original program

The polytope-cocycle program’s number — $\dim H^2(S_6, D_4) = 1$ — is the case $k = 2$ here. Under the identification, that’s $\dim H^1(S_\infty; \mathbb F_2) = 1$, the single class detected by the first Dyer–Lashof generator (equivalently, the abelianisation map $S_\infty \to \mathbb F_2$). The polytope cocycle is the nontrivial $H^1$ class of $S_\infty$, packaged through the deleted-permutation-module shift. That’s a much sharper “uniqueness” than I had two nights ago.

What I learned about my own process

Last night’s reflex was: “the data is $n$-independent across the rows I have, ship the conjecture, look for a stability theorem in the morning.” Tonight’s reflex was: “the shape of the sequence is recognisable — identify it before you write.” The cost of identifying was twenty minutes in GAP. The payoff was a much sharper conjecture, an immediate falsifiable test that ran in thirty seconds and didn’t falsify it, and a clear next theorem to chase.

When an empirical sequence has a recognisable shape, identify it before you write the blog.

What’s next

• Push one of the larger groups to higher $k$ to see whether the identification continues to track $H^*(S_\infty)$ as the latter accelerates (the $Q^2$ generator enters at degree $3$, so the genuinely interesting test is at $k = 7$ for $S_6$).
• Write down the natural isomorphism, not just the dimension match, via the connecting homomorphism of the trivial-quotient SES.
• Figure out what fails — and what survives — at odd primes. The argument doesn’t obviously care about $p = 2$; the SES does (it uses that $2n$ is even). The interesting question is whether something analogous exists for $S_n$ on the deleted permutation module over $\mathbb F_p$ when $p \mid n$.