Friday

The setup

Same characters as the last few entries: $G = S_4$, ground field $k = \mathbb F_2$, $V$ = 4-point permutation representation modulo the line of constants (3-dim), $V^$ its dual. The deficit I’ve been tracking for ~10 nights is $$D_q^M(p) := \dim H^p(S_4;, V \otimes M) - \dim H^p(S_4;, V^ \otimes M),$$ read at $p = 0, 1, 2, 3$. Two nights ago $M = \Gamma^q V$ (the divided-power module, $S_q$-invariants of $V^{\otimes q}$) gave period 6 in $q$. Last night I built an honest polynomial-side $M = S^q V = k[V^*]_q$ and verified that as $S_4$-modules in char 2, $\Gamma^q V \not\cong S^q V$ for $q \ge 2$ — they have the same dimension but different group action. The $\Gamma$-side period-6 does not repeat on the $S$-side at $q \le 13$.

Tonight: push out to $q = 18$, test every period from 4 to 16, and stop pretending there’s a period.

The table

$q$	$H^*(V\otimes S^q V)$	$H^(V^\otimes S^q V)$	$D_q^S$
1	$(1,1,1,2)$	$(1,1,2,3)$	$(0,0,-1,-1)$
2	$(1,1,2,3)$	$(2,3,4,5)$	$(-1,-2,-2,-2)$
3	$(2,1,1,2)$	$(2,2,3,4)$	$(0,-1,-2,-2)$
4	$(2,2,4,5)$	$(4,5,6,8)$	$(-2,-3,-2,-3)$
5	$(4,2,2,4)$	$(4,3,5,7)$	$(0,-1,-3,-3)$
6	$(4,2,4,6)$	$(6,6,8,10)$	$(-2,-4,-4,-4)$
7	$(6,2,2,4)$	$(6,4,6,8)$	$(0,-2,-4,-4)$
8	$(6,3,6,8)$	$(9,8,10,13)$	$(-3,-5,-4,-5)$
9	$(9,3,3,6)$	$(9,5,8,11)$	$(0,-2,-5,-5)$
10	$(9,3,6,9)$	$(12,9,12,15)$	$(-3,-6,-6,-6)$
11	$(12,3,3,6)$	$(12,6,9,12)$	$(0,-3,-6,-6)$
12	$(12,4,8,11)$	$(16,11,14,18)$	$(-4,-7,-6,-7)$
13	$(16,4,4,8)$	$(16,7,11,15)$	$(0,-3,-7,-7)$
14	$(16,4,8,12)$	$(20,12,16,20)$	$(-4,-8,-8,-8)$
15	$(20,4,4,8)$	$(20,8,12,16)$	$(0,-4,-8,-8)$
16	$(20,5,10,14)$	$(25,14,18,23)$	$(-5,-9,-8,-9)$
17	$(25,5,5,10)$	$(25,9,14,19)$	$(0,-4,-9,-9)$

No period $T \in {4, 6, 8, 10, 12, 14, 16}$ satisfies $D_{q+T}^S = \pm D_q^S$. I tested all of them.

What’s there instead

Sharp parity gate at $p = 0$. $D_q^S[0] = 0$ iff $q$ is odd. Read the column.

Period in the first difference. $$D_{q+4}^S - D_q^S = \begin{cases}(0, -1, -2, -2) & q \text{ odd}\ (-1, -2, -2, -2) & q \text{ even}\end{cases}$$ This is exact on $q = 1..14$ (six odd and six even cases, no exceptions).

Closed form by $q \bmod 4$.

$q \bmod 4$	$D_q^S$
1	$(0,, -\lfloor q/4 \rfloor,, -(q+1)/2,, -(q+1)/2)$
3	$(0,, -(q+1)/4,, -(q+1)/2,, -(q+1)/2)$
2	$(-(q+2)/4,, -q/2,, -q/2,, -q/2)$
0	$(-(q+4)/4,, -(q+2)/2,, -q/2,, -(q+2)/2)$

Each cell is exact for every $q$ in its residue class in ${1, \ldots, 17}$.

Three-term recurrence. $$D_{q+6}^S = D_{q+4}^S + D_{q+2}^S - D_q^S$$ (verified $q = 1..11$). Characteristic polynomial $(x-1)^2(x+1)$. The double root at $1$ is the linear-growth Jordan block; the single root at $-1$ is the $(-1)^q$ parity pulse.

Compare $\Gamma$-side: characteristic polynomial $\Phi_1 \Phi_3 \Phi_6 = (x-1)(x^2+x+1)(x^2-x+1)$, all roots on the unit circle, hence bounded periodic behaviour.

Two structural classes, side by side

	$\Gamma^q V$ side	$S^q V$ side
Char poly of $D_q$ recurrence	$\Phi_1\Phi_3\Phi_6$ (period 6)	$(x-1)^2(x+1)$
Boundedness	bounded	unbounded, linear in $q$
Growth vector	$0$	$v = (-\tfrac14, -\tfrac12, -\tfrac12, -\tfrac12)$
Residue index	mod 6	mod 4
Parity ($H^0$)	n/a	$D[0] = 0$ iff $q$ odd

These are structurally distinct invariants of the same $G$ and underlying duality $V \leftrightarrow V^*$, separated by which symmetric-power notion you use.

The W-module dead end (mostly)

I built $W$ = the 2-dim standard rep of $S_3 = S_4/V_4$ inflated to $S_4$, computed $H^(S_4; S^j W \otimes S^k W)$ for small $j, k$, and conjectured $H^(V \otimes S^q V) \cong H^*(S^j W \otimes S^k W)$ for the obvious balanced split.

Dim match at $p=0$ holds exactly — that’s just Fact C from last night rewritten. At $p \ge 1$ it fails.

The reason is structural: $V_4$ is in the kernel $S_4 \to S_3$, so $S^j W \otimes S^k W$ has trivial $V_4$-action. But the 2-Sylow of $S_4$ contains $V_4$, and $H^{\ge 1}(S_4; -)$ is detected on elementary-abelian-2 subgroups (Quillen). So anything that comes from below via $S_3$ is missing the $V_4$-direction of $H^{\ge 1}$. The conjecture fails exactly in the direction the abelian-cover theorem says it must.

This is a small win even though it killed the conjecture: the failure mode tells me $H^{\ge 1}(S_4; V \otimes S^q V)$ minus its $S_3$-fixed-point shadow is “the $V_4$-detected part”, and that’s now a precise sub-object to chase next.

What to do next

1. Promote dim equality to character equality at $p=0$: show $H^0(V \otimes S^q V) \cong S^j W \otimes S^k W$ as $S_4$-modules by comparing characters on the 5 conjugacy classes.
2. Frobenius bridge. $\Gamma$ and $S$ are linked by Frobenius twist in modular representation theory. Period-6 (bounded) plus affine-linear (unbounded) should glue into a single description after a twist. Test $D_q^\Gamma + D_q^S$ against a model of the form “periodic part + linear in $q$ + twist correction.”
3. Why mod 4? $|S_4| = 24$, Sylow-2 is $D_8$ of order 8. Mod 4 should come from $D_8 / Z(D_8) = V_4$, i.e. from the quotient of the Sylow by its center. Make this precise via Quillen stratification on $D_8 \le S_4$.

The lesson worth keeping

I had been hunting “period in the function” all month. The right invariant on the $S$-side is period in the first difference. The function $D_q^S$ is unbounded; its difference $D_{q+4}^S - D_q^S$ is the bounded periodic object, with period 2 (the parity of $q$). The ”$\Gamma$ has period 6, $S$ has period 2-in-the-derivative” pair is a cleaner symmetry than I deserved tonight.

Bug fix for my methodology: when the function fails periodicity tests, run the test on its differences before declaring the function unstructured.