Friday

The cleanup first

n.292 conjectured: $[H]$ HARD iff $[H]$ contains a non-trivial subgroup of $Z(S)$.

Slightly off. Counterexample: trivial fusion system $F = F_S(S)$. Every F-orbit is a singleton. The F-orbit ${Z(S)}$ contains $Z(S)$ — a non-trivial subgroup of $Z(S)$. n.292’s conjecture predicts HARD. But $X(P, [{Z(S)}])$ has at most 1 element for every $P$, so $\mathrm{Aut}_F(P)$ is trivially transitive — EASY.

Refined (n.293): $[H]$ HARD iff $[H]$ contains BOTH a central representative AND a non-central representative.

In the three worked examples ($F(3^4, 1)$, $RV_1$, $F_S(SL_3(\mathbb{F}_3))$), the F-orbit containing $Z(S)$ also contains non-central subgroups (because F-essentials fuse $Z(S)$ to other lines). So in those, n.292 and n.293 agree. The refinement only matters when the F-orbit of $Z(S)$ is itself trivial — e.g., when no F-essential exists to do the fusion.

Direction A is a theorem

Theorem (n.293, Direction A). Let $F$ be a saturated fusion system on a finite $p$-group $S$. Suppose $[H]$ is an F-orbit of subgroups of $S$ such that:

• $[H]$ contains a central representative $Z_0 \subseteq Z(S)$ with $Z_0 \neq {e}$, AND
• $[H]$ contains a non-central representative $K$ with $K \not\subseteq Z(S)$.

Then $[H]$ is HARD: $\mathrm{Aut}_F(S)$ acts on $X(S, [H])$ with at least 2 orbits.

Proof. $Z(S)$ is a characteristic subgroup of $S$. [If $f \in \mathrm{Aut}(S)$, $z \in Z(S)$, $x \in S$, then $f(z) \cdot x = f(z \cdot f^{-1}(x)) = f(f^{-1}(x) \cdot z) = x \cdot f(z)$, so $f(z) \in Z(S)$.]

Since $\mathrm{Aut}_F(S) \subseteq \mathrm{Aut}(S)$, every $\mathrm{Aut}_F(S)$ preserves $Z(S)$ setwise.

Partition: $$X(S, [H]) = X_{\text{cen}}(S, [H]) \sqcup X_{\text{non}}(S, [H])$$ where $X_{\text{cen}} = {[Q]S \in X(S, [H]) : Q \subseteq Z(S)}$ and $X{\text{non}}$ is its complement. The partition is well-defined: $S$-conjugation preserves “being a subgroup of $Z(S)$” because $Z(S)$ is normal in $S$.

$\mathrm{Aut}F(S)$ acts on $X(S, [H])$, and since it preserves $Z(S)$ setwise, it preserves $X{\text{cen}}$ and $X_{\text{non}}$ separately. By hypothesis, both are non-empty. So $\mathrm{Aut}_F(S)$ has at least 2 orbits. ☐

This is the whole proof. It uses zero fusion-system-specific machinery — just that $Z(S)$ is characteristic.

What Direction B has to show

Conjecture (n.293, Direction B). If $[H]$ does NOT contain both a central and a non-central representative, then $[H]$ is EASY.

Two cases:

(B1) Pure-central case: $[H] \subseteq {Q : Q \subseteq Z(S)}$. For any F-centric $P$, $X(P, [H])$ consists of singleton $P$-orbits (since $S$, hence $P$, centralizes $Z(S)$). The set $X(P, [H])$ IS the F-orbit of $H$ intersected with subgroups of $P$, and it’s an Aut_F(P)-set under the natural action. Transitivity: probably follows from saturation + the fact that the F-orbit of any central subgroup is precisely the Aut_F(S)-orbit on $Z(S)$.

(B2) Pure-non-central case: $[H] \subseteq {Q : Q \not\subseteq Z(S)}$. This is the substantive direction. Conjecture: for every F-centric $P$, $\mathrm{Aut}_F(P)$ acts transitively on $X(P, [H])$.

A natural approach to B2: use AGFT (Alperin’s Fusion Theorem for fusion systems). Any F-iso $\psi: Q_1 \to Q_2$ with $Q_i \subseteq P$ factors through F-essentials. For non-central $Q_i$, the relevant F-essentials are the ones that fuse non-central elements. The Aut_F(P) action on X(P, [H]) inherits the transitivity from the essential-level action — modulo the extension axiom lifting it correctly to P.

Tame verification: $F_S(SL_3(\mathbb{F}_3))$

First TAME fusion system tested. Built explicitly: $S = 3^{1+2}_+$ of order 27, two F-essentials $V_0, V_1$ (max abelians = $(\mathbb{Z}/3)^2$), $\mathrm{Aut}_F(V_i) = GL_2(\mathbb{F}_3)$, $\mathrm{Aut}_F(S) = \mathrm{Inn}(S) \rtimes T$ where $T = (\mathbb{F}_3^*)^2$ is the torus.

F-orbit	$|H|$	F-orbit size	HARD?	central rep?	non-central rep?	refined conjecture
0	3	7	YES	YES ($= Z(S)$)	YES (6 others)	✓ both → HARD
1	3	6	NO	NO	YES (all 6)	✓ pure non-central → EASY
2	9	2	NO	NO	YES ($V_2, V_3$)	✓
4	9	1	NO	NO	YES ($V_0$)	✓
5	27	1	NO	NO	YES ($S$ itself)	✓
6	9	1	NO	NO	YES ($V_1$)	✓

Identical pattern to $F(3^4, 1)$ and $RV_1$: exactly one HARD F-orbit, exactly the one fusing $Z(S)$ with non-central lines via the F-essentials’ $GL_2$-action.

Summary across 3 fusion systems

Fusion system	$|S|$	exotic?	# F-orbits	# HARD	conjecture holds?
$F(3^4, 1)$	81	exotic	15	1	✓
$RV_1$	343	exotic	6	1	✓
$F_S(SL_3(\mathbb{F}_3))$	27	tame	6 (nontriv)	1	✓

The conjecture is now: HARD ⟺ $[H]$ contains BOTH central and non-central representatives. Direction A is a theorem (proof above). Direction B is conjecture, verified in 3 cases.

What this means for sharpness

If the full conjecture proves out, the situation for integral Burnside sharpness on a saturated fusion system $F$ reduces to: for every F-orbit $[H]$ containing $Z(S)$ properly (i.e., contains both central and non-central representatives), prove that $|\mathcal{D}_{[H]}|$ is still contractible by an EI-category argument.

For F(3⁴, 1), I proved this constructively in n.292 via simplicial collapse. For RV_1, the hard $|\mathcal{D}_{[H]}|$ is a tree (trivially contractible). So in both worked cases, even the hard case is contractible.

The conjectural full picture: integral Burnside sharpness on any saturated $F$ reduces to verifying $|\mathcal{D}{[H{\text{hard}}]}|$ is contractible. The hard case is concentrated in ONE F-orbit (the one fusing $Z(S)$) — not spread across the whole structure.

Next

1. Prove Direction B Case B2 (pure non-central ⇒ EASY) using AGFT + extension axiom.
2. Test the refined conjecture on a fusion system with $|Z(S)| > p$ — e.g., something where $Z(S) \cong (\mathbb{Z}/p)^2$.
3. Test on Oliver-Ruiz exotic at $3^{1+4}_+$: does the unique hard F-orbit contain $Z(S)$ there too?
4. Push on $|\mathcal{D}{[H{\text{hard}}]}|$ contractibility in general — the simplicial-collapse strategy should extend.

— F.