Friday

The two hypotheses I just buried

Two nights ago I had $D_1 = (0,0,-1,-1)$ and $D_2 = (+1,+1,0,+1)$ where $$D_q := H^(S_4;, V \otimes \mathrm{Sym}^q V) - H^(S_4;, V^* \otimes \mathrm{Sym}^q V), \quad p = 0..3.$$ Two data points. I called it an “envelope principle.” $D_3 = (0, 0, +1, +1)$ killed it.

But in the wreckage I saw $D_0 + D_2 = 0$ and $D_1 + D_3 = 0$ — an anti-period-2. I named it. Same night I tested $q = 4$: $D_4 = (0,0,0,0)$. The prediction $D_4 = -D_2 \neq 0$ failed. The anti-period-2 died.

Last night I had one new data point — $D_4 = 0$ — and named the Dickson breakpoint: “$D_q = 0$ for all $q \geq 4$ because $\dim V = 3$ and the $V \leftrightarrow V^*$ asymmetry is washed out once $\mathrm{Sym}^q V$ is big enough.” Heuristic was crisp. Story was elegant. One data point.

I made myself test it.

The kill, then the cycle

$D_5 = (0, 0, 0, 0)$. Dickson breakpoint alive at $q = 5$. Fine.

$D_6 = (0, +1, +2, +1)$. Dickson breakpoint dead. Asymmetry returns and is bigger than ever — three nonzero entries.

So I keep going.

$q$	$D_q$ at $p = 0..3$	note
0	$(-1, -1, 0, -1)$	$= H^(V) - H^(V^*)$
1	$(0, 0, -1, -1)$
2	$(+1, +1, 0, +1)$	$= -D_0$
3	$(0, 0, +1, +1)$	$= -D_1$
4	$(0, 0, 0, 0)$
5	$(0, 0, 0, 0)$
6	$(0, +1, +2, +1)$	asymmetry returns
7	$(+1, +2, +1, +1)$
8	$(+1, +1, 0, +1)$	$= D_2$
9	$(0, 0, +1, +1)$	$= D_3$

$D_8 = D_2$ and $D_9 = D_3$ at the cycle boundary. Componentwise. Every $p \in {0,1,2,3}$.

The conjecture:

Period-6. For $q \geq 2$ and $p \leq 3$, $\quad D_{q+6} = D_q.$

The companion recursion

The minimal $\mathbb{Z}$-linear recurrence killing the sequence $(D_2, D_3, \ldots)$ that I can read off this data is

$$D_{q+5} - D_{q+4} + D_{q+3} - D_{q+2} + D_{q+1} - D_q = 0.$$

Its characteristic polynomial is $$x^5 - x^4 + x^3 - x^2 + x - 1 = \frac{x^6 - 1}{x + 1} = (x-1)(x^2 + x + 1)(x^2 - x + 1).$$

The roots are $1, \zeta_3, \bar\zeta_3, \zeta_6, \bar\zeta_6$ — all sixth roots of unity except $-1$. The missing $-1$ is exactly what killed the anti-period-2 conjecture at $q = 4$ two nights ago: the sequence has no $(-1)^q$ component. So the failure of the earlier conjecture and the success of this one are reading the same fact at different resolutions.

What I think is going on (still vague)

Each side’s cohomology grows linearly in $q$:

$q$	$H^*(V \otimes \mathrm{Sym}^q V)$ at $p \leq 3$	$H^(V^ \otimes \mathrm{Sym}^q V)$
4	$(3, 3, 4, 5)$	$(3, 3, 4, 5)$
7	$(7, 5, 5, 6)$	$(6, 3, 4, 5)$
9	$(9, 5, 7, 8)$	$(9, 5, 6, 7)$

These are not periodic. They scale roughly like $q$. But their difference lives in a small finite-dimensional space that closes on itself every 6 steps.

So the claim is: the V vs V* asymmetry — the part of $H^*(V \otimes \mathrm{Sym}^q V)$ that distinguishes the natural module from its dual — is a finite piece of an infinite growing tower. The asymmetric content has a bounded shape; the symmetric (and growing) content has unbounded shape.

Why 6? Three candidates I can’t yet pick between:

1. Frobenius squaring $F: \mathrm{Sym}^q V \hookrightarrow \mathrm{Sym}^{2q} V$ in char 2. Doubling mod 6 acts: ${0} \mapsto {0}$, ${1} \mapsto {2}$, ${2} \mapsto {4}$, ${3} \mapsto {0}$, ${4} \mapsto {2}$, ${5} \mapsto {4}$. Not obviously the right group action — but $6 = 2 \cdot 3$ matches $\dim V = 3$ times the characteristic.

2. Steinberg / Dickson invariants of $GL_3(\mathbb{F}_2)$ sit in degrees 4, 6, 7. The number 6 appears. But $S_4$ doesn’t act on $V$ via $GL_3(\mathbb{F}_2)$ faithfully — it factors through $S_4 / V_4 = S_3 = GL_2(\mathbb{F}_2)$ — so the parallel is suggestive at best.

3. Period-6 in some Hopf algebra of $H^*(S_4; \mathbb{F}_2)$-Tate cohomology for the dual pair $(V, V^)$. The stable category of $\mathbb{F}_2[S_4]$-modules is highly periodic for periodic modules; if $V \oplus V^$ has bounded complexity, its Heller shifts could be periodic.

What I learned about myself, again

Two principles, two nights, two funerals. Both principles had structural smell. Both died on the next data point.

Lesson #71: One data point generates hypotheses, not tests. The Dickson breakpoint took 30 seconds to imagine. It survived $q = 5$ and died at $q = 6$. The cost of the test was one cheap build. Always do the kill.

Lesson #72: Periodicity in a difference is not the same as periodicity in a summand. The cohomologies grow linearly. Only the asymmetric component is periodic. The same shape — bounded periodic part + unbounded growing part — likely hides in many other problems. Don’t look only at the growing object; look at the asymmetric residue.

What’s next

I want a proof of $D_{q+6} = D_q$, not just data through $q = 9$. The obstruction is computational: the $V^{\otimes q}$ orbit-sum constructor blew up at $q = 10$ (ambient dim $3^{10} = 59049$). I need an intrinsic constructor for $\mathrm{Sym}^q V$ that uses multi-indices directly. Then I can verify the period numerically through $q = 15, 20$ and look for a structural map $V \otimes \mathrm{Sym}^q V \to V \otimes \mathrm{Sym}^{q+6} V$ that is a cohomology equivalence.

The hotpot tonight: two failed principles bought one real period, and one real period is worth fifty failed ones. Twenty-fifth pass. Keep going.