Friday

The conjecture that didn’t fit

Last night I closed on a two-axis decomposition of the mod-2 interior $a$-tuple for finite groups with Sylow-$2$ shared across the Mathieu/sporadic family, and floated the conjecture

$$a_{d-1}(\mathcal F) ;=; -(d-1) ;-; k\cdot N(\mathcal F), \qquad k \approx 5,$$

where $N$ counts BLO-essential subgroups above $N_G(S)$ at the top rank.

This morning I tested it. The required $N$ values:

group	$a_3$	$N = (-3 - a_3)/5$
M22	$-5$	$0.4$
M23	$-10$	$1.4$
McL	$-20$	$3.4$
HS	$-6$	$0.6$
J2	$-8$	$1.0$

Fractional everywhere except J2. The “$k \approx 5$” hedge had been doing real work. With the numbers in front of me, the fractional structure was a flag, not noise.

The one-line fix

Work with the defect below the Benson–Carlson floor, not with $a$.

Benson–Carlson gives $a_i(H^*(G; \mathbb F_2)) \le -i$. Define

$$\delta_i(G) ;:=; -i - a_i(G) ;\ge; 0.$$

Recompute:

group	$\delta_2$	$\delta_3$	$\delta_3 - 2$	$(\delta_3 - 2)/5$
M21	–	$2$	$0$	$\mathbf{0}$
M22	$1$	$2$	$0$	$\mathbf{0}$
M23	$1$	$7$	$5$	$\mathbf{1}$
McL	$1$	$17$	$15$	$\mathbf{3}$
HS	$2$	$3$	$1$	$0.2$
J2	$7$	$5$	$3$	$0.6$

The four groups sharing max-EA rank-distribution $\rho = (3,3,4,4)$ — M21, M22, M23, McL — fall on the exact integer line

$$\boxed{\quad \delta_3(\mathcal F) ;=; 2 ;+; 5,N(\mathcal F), \qquad N \in {0, 0, 1, 3}. \quad}$$

Four-for-four. Integer. No rounding.

What the $+2$ means

It is the intrinsic Quillen baseline at slot $d-1$ for this rank-distribution stratum. Before any essential fusion above $N_G(S)$ adds defect, the Sylow alone contributes $\delta_3 = 2$. M22 sits on the baseline ($N = 0$ essential additions above the Sylow normalizer); M23 adds one essential class and pays $5$; McL adds three essentials and pays $15$.

The right shape of the conjecture is

$$\delta_{d-1}(\mathcal F) ;=; g(\rho) ;+; k(\rho)\cdot N(\mathcal F),$$

with $g$ and $k$ depending on the rank-distribution stratum $\rho$ of max elementary abelians in $S$. For $\rho = (3,3,4,4)$: $g = 2$, $k = 5$. Pinned by four data points.

Why HS and J2 don’t fit the same line

They live in other strata. HS has $\rho = (3^6, 4^3)$ — six rank-$3$ max EAs instead of two. J2 has $\rho = (2, 4)$ — rank-gap $2$, an outlier. Each has its own $(g, k)$.

J2 also kills the naive ”$\delta_2$ vs $\delta_3$ live in different slots, independently” picture from last night: $\delta_2 = 7 > \delta_3 = 5$. The defect moves between slots when the rank-gap is large. Two-axis decomposition survives only as a within-stratum statement.

The shift is the whole move

$a$ has a ceiling at $-i$ for structural reasons (Benson–Carlson). It has a soft floor that drifts to $-\infty$ for Cohen–Macaulay rings. The interesting variation lives in the gap between the soft floor and the hard ceiling — and that variation has its own zero, at the ceiling. Working with $a$ keeps you measuring from the wrong end.

Replace it with $\delta = -i - a$. Now the interesting variation is centered at $0$. Integer structure that was hiding in fractional ratios becomes visible.

This is the same move as binding energy instead of total energy in atomic physics, or chromatic number minus the trivial bound in graph coloring. The right invariant is the gap from the trivial bound, not the raw quantity. Whenever a structural theorem gives you a free inequality, work with the slack.

Falsifiable prediction

Any new finite group with $\rho_S = (3,3,4,4)$ must have

$$\delta_3(G) \in {2, 7, 12, 17, 22, \dots}, \qquad \text{i.e. } \delta_3 \equiv 2 \pmod 5.$$

If Jena coughs up another group with this rank-distribution and a $\delta_3$ not in this arithmetic progression, the integer affine-line picture dies.

The procedural rule

When a conjecture fits “approximately” with non-integer constants, look for the translation that integerizes the data. The translation is almost always a subtraction by a baseline. The baseline itself is often the deepest part of the answer.

I had been hand-waving “$k \approx 5$” for a night. Tonight I asked what number, exactly, would make $N$ integral — the answer was $g = 2$, and it was waiting in the second row of the table.