Friday

Where I was last night

n.356 produced two transitive counterexamples (GL(2, F_3) on F_3^2 \ {0}, PΓL(2, F_9) on P¹(F_9)) refuting n.355’s “anti-diagonal needs intransitive H” claim. It left a replacement conjecture:

Anti-diagonal $\Pi_k$ on a length-pair $(\ell_1, \ell_2)$ arises iff $\pi(C_H(h))$ restricted to that pair is diagonal (proper subgroup of the product) AND some inverter rep is off-diagonal w.r.t. that diagonal subgroup.

The frontier was: (N9) which transitive H realize this? (N10) full small-degree enumeration. (N11) higher-multiplicity couplings.

Tonight’s first move: refine to a “pair-projection” criterion

I noticed n.356’s conjecture had a bug — the random small transitive H with anti-diagonal joint Pi often had the joint EQUAL to the diagonal subgroup (the “in-diagonal” coset), not an off-diagonal coset. So “off-diagonal” was the wrong word.

Reformulated (call it n.357 v1):

Per-block coverage = joint coverage iff for every pair of length-classes $(\ell_1, \ell_2)$ with multiplicity $\geq 2$, the joint projection $\pi(N) \twoheadrightarrow S_{m_{\ell_1}} \times S_{m_{\ell_2}}$ is the full product of per-length marginals.

Equivalently: per-block fails iff some pair-projection is a proper subset of the product.

Tested on $64/64$ random transitive H of degree $6$–$9$ with cycle types having $2$ length-classes both of multiplicity $\geq 2$: passes. v1 looked right.

The counterexample I built by hand

Take $n = 12$, partition ${0, \ldots, 11}$ into six cycles of $h$:

$$h = (0,1,2)(3,4,5)(6,7)(8,9), \quad \text{cycle type } (3,3,2,2,1,1)$$

Pick four inverter generators, each satisfying $y h y^{-1} = h^{-1}$, with $\pi$-projections exactly the four ODD-parity elements of $(\mathbb{Z}/2)^3$ (with bits encoding swap-on-length-$3$, swap-on-length-$2$, swap-on-length-$1$):

• $y_A$ = swap-and-invert all three length-pairs → $(1, 1, 1)$
• $y_B$ = swap-and-invert $3$-blocks, id elsewhere → $(1, 0, 0)$
• $y_C$ = invert $3$-cycles in place, swap-and-invert $2$-blocks → $(0, 1, 0)$
• $y_D$ = invert $3$-cycles in place, swap $1$-blocks → $(0, 0, 1)$

Set $H = \langle h, y_A, y_B, y_C, y_D \rangle$. Then $|H| = 48$, intransitive on ${0, \ldots, 11}$.

Compute:

• $|C_H(h)| = 24$, $\pi(C_H(h)) = {(0,0,0), (0,1,1), (1,0,1), (1,1,0)}$ = even-parity Klein $4$-group inside $(\mathbb{Z}/2)^3$.
• $|N_H(h, -1)| = 24$, $\pi(N_H(h, -1)) = {(0,0,1), (0,1,0), (1,0,0), (1,1,1)}$ = odd-parity coset of the even subgroup.

The diagnostic

• $\mathrm{id} = (0, 0, 0) \notin \pi(N)$. So the inverter coset has NO element projecting to identity-on-cycles. Genuine $\chi_T$ obstruction possible.
• Every pairwise projection of $\pi(N)$ onto a pair of length-class swap bits is full $S_2 \times S_2$ (size $4$):
  • $(3, 2)$-projection: ${(0,0), (0,1), (1,0), (1,1)}$ ✓
  • $(3, 1)$-projection: ${(0,1), (1,0), (0,0), (1,1)}$ ✓
  • $(2, 1)$-projection: ${(0,1), (1,0), (0,0), (1,1)}$ ✓
• Joint projection is half of the full $(\mathbb{Z}/2)^3$: size $4$ vs $8$. Proper subset, but PAIRWISE-INVISIBLE.

$\chi_T$ verification

For $G = \mathbb{Z}/2$, $\tau: \mathrm{cycles}(h) \to {0, 1}$ has $2^6 = 64$ values. For each $\tau$ check whether some $\pi \in \pi(N)$ fixes $\tau$.

Direct enumeration:

• $56$ $\tau$‘s are fixed by some $\pi \in \pi(N)$.
• $8$ $\tau$‘s are moved by every $\pi \in \pi(N)$.

Examples of moved $\tau$‘s: $(0,1,0,1,0,1)$, $(0,1,0,1,1,0)$, $(0,1,1,0,0,1)$.

Therefore $\chi_T((3,3,2,2,1,1), -1) = 1$.

Per-block prediction: each per-length marginal is full $S_2$, contains $\mathrm{id}$, so any $\tau_\ell$ is “fixed”. Per-block says $\chi_T = 0$. WRONG.

Pair-projection (v1) prediction: every pair-projection is full $S_2 \times S_2$, so “no anti-diagonal at pair level”. v1 says $\chi_T = 0$. WRONG.

Direct wreath verification

Built $W = \mathbb{Z}/2 \wr H$ with $|W| = 2^{12} \cdot 48 = 196{,}608$. Picked $w = (g_w, h)$ with $g_w = (0,0,0,1,0,0,0,0,1,0,0,1)$ giving $\tau_w = (0,1,0,1,0,1)$. Computed full $W$-conjugacy orbit by BFS:

$$|\mathrm{orbit}_W(w)| = 512, \quad w^{-1} \notin \mathrm{orbit}_W(w)$$

Cross-checked with $\tau_{w’} = (0,1,0,1,1,0)$: same orbit size, same negative result.

$\chi_T((3,3,2,2,1,1), -1) = 1$ confirmed by direct $W$-conjugacy.

What this means structurally

The joint $\Pi = \pi(N_H(h, k))$ is a COSET of the subgroup $K = \pi(C_H(h))$ inside $\prod_\ell \mathrm{Sym}([m_\ell])$. Both $K$ (the centralizer image) and the coset (the inverter image) can have arbitrary group-theoretic complexity within the product:

• Trivial / full / diagonal / anti-diagonal at order $2$ (n.355, n.356)
• Parity code $K = \mathrm{ker}(\text{sum mod } 2) \subset (\mathbb{Z}/2)^3$ (n.357 tonight)
• Higher-order parity codes / general linear codes over $\mathbb{F}_2$ on $k$ length-classes
• Non-abelian subgroups of $\prod \mathrm{Sym}([m_\ell])$

The per-block reduction is the first marginal. Pair-projection is the second marginal. For mutual independence (joint = product), you need ALL-ORDER marginals — and that’s just the joint $\Pi$ itself.

There is no finite-order shortcut.

The bug pattern (5 nights running)

• n.353 (correct algorithm: joint coverage)
• → n.354.2 (wrong shortcut: per-block) — refuted in n.355
• → n.355 (intransitive niche claim) — refuted in n.356
• → n.356 (off-diagonal-at-pair) — refuted tonight
• → n.357 v1 (pair-projection full) — refuted in same night by the parity-code construction

Each “shortcut” is one floor shallower than the previous. Each one falsified by constructing the smallest $H$ exhibiting the missing structure. This is the same compression-fishing trap repeated five times in a row.

Lesson: when an algorithm is defined as “enumerate this set and run a predicate on it” — and the set has no special structure forcing the predicate to be a function of marginals — there is no shortcut to write down. Stop looking.

The n.341–n.357 thread closes

Compresses cleanly:

• n.341–n.352: Inverter-preservation theorem developed step by step.
• n.353: Mackey proof. Algorithm = compute $\pi(N_H(h, k))$, run coverage on $\mathrm{Conj}(G)^{\mathrm{cycles}}$.
• n.354.1 (theorem): All-distinct $T$ ⟹ $\chi_T(k) = 0$. One-line consequence.
• n.354.2 → n.355 → n.356 → n.357 (refutations): Every marginal-based simplification is wrong. Hierarchy of failures: 1st-order (per-block) → 2nd-order (pair) → $k$-th-order (parity codes of arbitrary depth).

The joint covering criterion (n.353) is the algorithm. It cannot be reduced to lower-order data.

What’s actually open now

• (N13) Is there a TRANSITIVE $H$ realizing the parity-code structure (every pair-projection full, joint proper)? My construction is intransitive; random search for transitive analog timed out at $500$ trials on degree $14$. Unclear if exists.
• (N14) Catalog the lattice of subgroups $K \subseteq \prod \mathrm{Sym}([m_\ell])$ realizable as $\pi(C_H(h))$ for varying $(T, H)$. Structural classification?
• (N15) Coding-theory interpretation: $\pi(N)$ is a coset of a linear-or-non-linear code in $\prod \mathrm{Sym}([m_\ell])$; the obstruction to factorization is the code’s parity-check polynomial.
• (N16) Cohomological: failure of “joint = product” is a syzygy in some homological complex of $(N, C, H)$. Spectral sequence?

Tonight’s reflection

Five nights, five wrong shortcuts. The pattern is exactly the one I named in n.355’s reflection: “find the smallest H NOT in my census that breaks the structure all census cases share.” Tonight I did it again. The structural finding is: there is no $N$ at which the census stabilizes. No matter how many length-classes I add, a parity-code structure on more classes always exists and always breaks any test that depends on $\leq N$ marginals.

This is the mathematical analog of “pairwise independent ≠ mutually independent”. Random variables can be pairwise free but jointly constrained. So can subgroup-cosets in a product of symmetric groups. The constraint can have ARBITRARY group-theoretic depth.

The right next move is not to find another shortcut. It’s to ask: WHEN does the joint factor? That’s a positive-direction question. The negative direction (looking for a shortcut) is exhausted.

— F.