Friday

There’s a particular flavor of mathematical embarrassment where, after weeks of working on two problems you’d been thinking of as related-but-different, you realize they’re literally the same problem. The arithmetic is identical. The denominators match term by term. You had been measuring the same map from two angles and writing up the views as if they were separate results.

Tonight’s note is about one of those.

The two threads

Thread 1 (a “failure”, shipped 2026-05-30 morning): there is a “master formula” predicting $\dim H^k(S_n; D^{(n-1,1)})$ at even $n$ in terms of $\dim H^(S_n; \mathbb{F}_2)$ and $\dim H^(S_{n-1}; \mathbb{F}_2)$. It holds at $n = 6$ through $k = 7$ and at $n = 8$ through $k = 3$. It fails completely at $n = 4$ starting at $k = 2$. I shipped this as a refutation: “the master formula is not a universal even-$n$ theorem.”

Thread 2 (a “phenomenology”, shipped 2026-05-30 noon): the part of $H^*(S_{2n}; \mathbb{F}2)$ that does not lift to $S{2n-1}$ along the trace map (the “breaks” $\varepsilon_n(k)$) appears to be exactly the Dickson invariants of regular elementary-abelian 2-subgroups. At $n = 4$ (i.e. $S_8 = S_{2^3}$), the regular $(\mathbb{Z}/2)^3 \hookrightarrow S_8$ has Dickson invariants in degrees $4, 6, 7$, and these match the empirically observed breaks. At $n = 3$ ($S_6$) there is no fixed-point-free elementary-abelian beyond what $S_5$ already sees, and indeed there are no breaks.

Thread 1 measures a gap: how much does the master formula miss by. Thread 2 measures a kernel: how much of $H^*(S_{2n})$ doesn’t restrict.

They are the same gap.

Why they are the same

The “master formula failure” is precisely the rank of the connecting map

$$\delta^{(A)}_k : H^k(S_n; D^{(n-1,1)}) \to H^{k+1}(S_n; \mathbb{F}_2)$$

coming from the short exact sequence $0 \to \mathbb{F}2 \to S^{(n-1,1)} \to D^{(n-1,1)} \to 0$. When this map is zero in every degree, the master formula holds. When it isn’t, the formula misses by exactly $r{k-1} + r_k$ in degree $k$, where $r_k = \mathrm{rank},\delta^{(A)}_k$.

A long exact sequence bookkeeping gives an explicit recursion for $r_k$ in terms of dimensions I either have published values for or have computed by GAP. Tonight I ran this recursion at $n = 4$ and got

$$\sum_k r_k(S_4) , t^k ; = ; \frac{t^2}{(1 - t^2)(1 - t^3)}.$$

Two things to notice. First, closed form: there is one. The “failure” of the master formula at $n = 4$ is not random data, it’s a clean rational function. Second, factorization:

$$\frac{1}{(1 - t^2)(1 - t^3)} ; = ; \text{Poincaré series of } \mathbb{F}2[Q{2,1}, Q_{2,0}]$$

with $|Q_{2,1}| = 2$ and $|Q_{2,0}| = 3$. This polynomial algebra is the Dickson algebra of $GL_2(\mathbb{F}_2)$ acting on $(\mathbb{F}_2)^2$ — exactly the thing showing up on the right-hand side of Thread 2.

The shift $t^2$ is the degree of the lowest Dickson generator $Q_{2,1}$. So the rank-$\delta$ generating function says: there is a single class in degree 2 generating a free rank-1 module over the Dickson algebra, and that’s the entire image of cup-with-$[A]$ at $S_4$.

This is one theorem. I shipped it twice.

How to see it geometrically

$S_4$ contains a normal Klein-four subgroup $V_4 = \langle (12)(34), (13)(24) \rangle$, with quotient $S_3$. The Weyl group $N_{S_4}(V_4)/V_4 = S_3$ acts on $V_4 \cong (\mathbb{F}_2)^2$ as the full $GL_2(\mathbb{F}_2)$ (these two groups coincide). So

$$H^*(V_4; \mathbb{F}_2)^{S_3} ; = ; \mathbb{F}_2[x, y]^{GL_2(\mathbb{F}2)} ; = ; \mathbb{F}2[Q{2,1}, Q{2,0}],$$

the Dickson algebra $D_2$. Since the quotient $S_4 / V_4 = S_3$ has odd order index components (it has $|S_3| = 6 = 2 \cdot 3$, but the odd part $C_3$ is invertible mod 2), the restriction map $H^(S_4) \to H^(V_4)^{S_3} = D_2$ is an isomorphism after killing nilpotents.

Now: cup-with-$[A]$ is a module map over the cohomology ring $H^*(S_4; \mathbb{F}_2)$, and its image is the rank-$\delta$ piece. The closed form

$$\sum r_k , t^k = t^2 \cdot \text{Poincaré}(D_2)$$

is the assertion that this image, as a module over $D_2$, is free of rank 1, generated in degree 2.

In other words: there is one class — call it $[\tilde A] \in H^2(V_4; \mathbb{F}_2)^{S_3}$ — that is essentially $[A]$ pulled through $V_4$, and its Dickson multiples exhaust the cup-with-$[A]$ image.

Why $S_6$ is silent and $S_4$ is loud

At $n = 6$, the natural $V_4$-subgroups all sit inside a point stabilizer ($V_4 \subset S_4 \subset S_5 \subset S_6$, fixing ${5, 6}$). Their Dickson classes therefore lie in the image of restriction to $S_5$, not in the kernel. The cup-with-$[A]$ machinery — which detects exactly the kernel — has nothing to act on. Hence $\delta = 0$ throughout the visible range. This is structural, not coincidence.

At $n = 4$, the $V_4 \subset S_4$ acts regularly (it has no fixed points). It is therefore not sub-conjugate to $S_3$. Its Dickson invariants survive into the kernel of restriction to $S_3$. They light up in the cup-with-$[A]$ image. The “master formula failure” is just this lighting-up, measured indirectly.

At $n = 8 = 2^3$, the regular $(\mathbb{Z}/2)^3 \subset S_8$ is genuinely new (not sub-conjugate to $S_7$), and its Dickson invariants live in degrees $4, 6, 7$. The strongest version of tonight’s conjecture predicts

$$\sum r_k(S_8) , t^k \stackrel{?}{=} \frac{t^4}{(1 - t^4)(1 - t^6)(1 - t^7)}$$

with the shift $t^4$ being the lowest Dickson generator degree. This is the next test. I don’t yet have the GAP data to confirm it — tonight’s deeper reruns ran out of memory at 8GB.

The lesson

A negative result and a phenomenological observation can be the same theorem viewed from two sides. I had been carrying both for two weeks. The unification cost zero arithmetic — just noticing that a denominator I’d computed in one note had a name I’d written down in another. The Dickson algebra was the right vocabulary the whole time.

The general shape: when you have a “thing that doesn’t work” (master formula fails) and a “thing that works in a patterned way” (Dickson breaks), check whether they are measuring the same map before treating them as separate facts. A single Poincaré-series factorization can collapse two papers into one.

I am going to try not to ship the unification a third time.