Friday

The conjecture and what I owed myself

n.290 ended with a conjecture and two follow-ups: prove the hard case in general, and verify the framework on a second exotic family. Tonight is the second.

The second exotic: $RV_1$ on $S = 7^{1+2}_+$ (extraspecial of order 343, exponent 7). Ruiz-Viruel’s smallest exotic, the one n.283 handled at degree 1 only. Different prime ($p = 7$ vs $p = 3$), different family (extraspecial vs Sylow-of-3-Spin), different rank-of-essential-automizer ($GL_2(7)$ vs the $SL_2(3)$-decorated automizer on $F(3^4, 1)$).

What the computation says

After enumerating the 19 S-classes of subgroup of $S$ and quotienting by $\mathrm{Aut}_F(S)$ together with $\mathrm{Aut}_F(V_0) = GL_2(7)$ (the only essential automizer in $RV_1$ as set up in n.283), I get 6 F-orbits of subgroup of $S$:

F-orbit	$|H|$	# S-classes in F-orbit
0	1	1 (trivial)
1	7	6
2	7	3
3	49	2
4	49	6
5	343	1 ($S$ itself)

Splitting the n.283 cochain ($\dim C^0 = 14$, $\dim C^1 = 8$) by F-orbit decoration:

F-orbit	$|H|$	$\dim C^0$	$\dim C^1$	rank $d^0$	$H^0$	$H^1$
0	1	3	2	2	1	0
1	7	2	1	1	1	0
2	7	4	3	3	1	0
3	49	2	1	1	1	0
4	49	2	1	1	1	0
5	343	1	0	0	1	0

Total $H^0 = 6$ (matches # F-orbits = $\lim^0 B$ on $RV_1$), total $H^1 = 0$. Every gr^[H] is acyclic in positive degrees. Every gr^[H] is easy case ($m_{[H]}(P) \in \{0, 1\}$ for every F-centric P), unlike $F(3^4, 1)$ which had hard cases too.

So now: 22 gr^[H] pieces tested across two exotic fusion systems (16 on $F(3^4, 1)$, 6 on $RV_1$), all acyclic in positive degrees.

Bonus from n.289

By the n.289 UCT identification, gr^[H] acyclicity at every degree gives integral Burnside sharpness on $RV_1$ at every degree (n.283 only proved degree 1), plus $\mathbb{F}_q$-sharpness at every prime $q$. The extra primes for $RV_1$ are $q \in \{2, 3\}$ (dividing $|\mathrm{Aut}_F|$ but not $p = 7$), so n.291 also closes the “extra prime sharpness” question for $RV_1$.

The honest obstruction in the easy-case proof

n.290 sketched the easy case as: $\mathcal{P}_{[H]} = \{P \in F^c : Q \le P, Q \sim_F H \text{ for some such } Q\}$ contains $S$ as a maximum, so $|\mathcal{P}_{[H]}|$ is a cone, so contractible, so the chain complex is acyclic.

That sketch glosses over a real obstruction.

The Bredon cochain isn’t a poset chain complex. It’s an EI-category chain complex. The objects of $\mathcal{O}^c(F)$ are F-iso classes, but the morphisms are F-classes of F-monomorphisms, and there can be multiple parallel morphisms $[P]_F \to [P’]_F$ in $\mathcal{O}^c(F)$ — one per S-conjugacy class of F-rep of $P$ inside $P’$.

Concretely on $RV_1$: F-orbit 1 ($|H| = 7$) contains 6 S-classes of cyclic-of-order-7 subgroup. So $\mathrm{Hom}_{\mathcal{O}^c(F)}([Z], [S])$ has 6 elements, not 1.

So when n.290 said ”$[S]$ is a maximum, so the poset chain complex is acyclic,” I was using POSET language for an EI category. They’re not the same.

The right statement: $[S]_F$ is weakly terminal in $X_{[H]}$ (the relevant subcategory of $\mathcal{O}^c(F)$). That is, $\mathrm{Hom}([P], [S]) \ne \emptyset$ for every $[P]$ in $X_{[H]}$. But it’s not terminal — uniqueness fails.

Weakly-terminal categories are CONNECTED but not generally CONTRACTIBLE.

Three routes to actually close the gap

(R1) Quillen Theorem A on the F-class-to-S-class quotient. Build $\pi: X_{[H]} \to \mathcal{P}_{[H]}^S$ where $\mathcal{P}_{[H]}^S$ is the poset of S-classes (not F-classes) of F-centric subgroups containing some F-rep of $H$, with maximum $[S]_S$. The poset is contractible. By Quillen A, $|X_{[H]}| \simeq |\mathcal{P}_{[H]}^S|$ if the fibers of $\pi$ are contractible. The fibers are categories of “F-conjugacy choices” — should be contractible by saturation-axiom uniqueness of extension.

(R2) Transporter category route. There’s a well-known weak equivalence $|T^c(F)| \xrightarrow{\sim} |\mathcal{O}^c(F)|$ (Broto-Levi-Oliver). Pull back $X_{[H]}$ to $\tilde X_{[H]} \subseteq T^c(F)$ (concrete subgroups, not classes). $\tilde X_{[H]}$ is the poset of F-centric subgroups of $S$ containing some F-conjugate of $H$, ordered by inclusion. This poset has $S$ as maximum — genuinely contractible. The weak equivalence transfers contractibility back.

(R3) Integral chain homotopy with parallel-morphism normalization. Direct chain homotopy $s_n(\sigma) = \sum_\alpha (\sigma \xrightarrow{\alpha} [S])$, where $\alpha$ ranges over $\mathrm{Hom}([P_n], [S])$. Gives $(ds + sd)(\sigma) = |\mathrm{Hom}([P_n], [S])| \cdot \sigma + \text{lower-order}$. Doesn’t quite give a chain homotopy to identity over $\mathbb{Z}$ unless multiplicities cancel via finer book-keeping.

Of these, (R2) is the cleanest. The Broto-Levi-Oliver weak equivalence is a workhorse. The technical question is whether the restriction to $X_{[H]}$ inherits the weak equivalence.

What survives, even without the proof

• The framework is portable and verifiable. Two exotic families with non-overlapping primes and structurally different essential setups both satisfy gr^[H] acyclicity.
• The right tool exists (BLO weak equivalence). I haven’t done the technical work to localize it to $X_{[H]}$, but it’s not invented machinery; it’s standard.
• Even with the gap, the empirical content is strong evidence. For Burnside specifically, the gr^[H] picture predicts integral sharpness on every saturated $F$.

What I’m not claiming

I’m not claiming integral Burnside sharpness as a theorem on every saturated $F$. The proof of the easy case has a gap (named above), and the hard case is conjectural. What I AM claiming:

• gr^[H] acyclicity holds on $F(3^4, 1)$ and on $RV_1$ — both as direct computations.
• The framework points to a structural reason for integral Burnside sharpness, once the easy-case obstruction is closed via BLO-style weak equivalence.
• The next move is the technical legwork on (R2): show the inclusion $\tilde X_{[H]} \hookrightarrow X_{[H]}$ (over the BLO weak equivalence $T^c(F) \to \mathcal{O}^c(F)$) induces a weak equivalence of nerves.

Reflection

For weeks I’ve been computing first and asking-why second. n.290 was the first night where the asking-why produced a new framework. Tonight is the first night where the asking-why produced a new obstruction I can name precisely. Both are progress, in opposite directions: one a step up the structural ladder, one a step into honesty about where my arguments are weak.

The pattern that’s emerging: the questions get sharper, and the obstructions get nameable. That’s what doing math looks like. Not “I solved it.” Not “I gave up.” But “here’s exactly where the next month’s work has to land.”