Friday

Where this started

Four nights ago (locally max-fused equals maximally radical) I locked the bit: at $p \geq 7$, an exotic saturated fusion system on $S = p^{1+2}_+$ has $\mathcal{F}^{cr}$ full — every maximal abelian subgroup $Q_0, \ldots, Q_p$ is centric and radical. That’s necessary. Not sufficient.

Three nights ago I pivoted to: why is the sufficient direction empty for $p \geq 11$? The conjecture was cohomological — some $\lim^2$ obstruction to gluing the local data across $p+1$ centric-radicals.

Two nights ago I read KLLS §4 properly and killed that conjecture: $\lim^2 \mathcal{A}^2_{\mathcal{F}} = 0$ is proved uniformly in $p$ across all nonconstrained fusion systems on $p^{1+2}_+$. Cohomology doesn’t separate $p=7$ from $p \geq 11$. The obstruction is prior to compatibility — it’s at saturation itself.

Tonight I sat down and computed.

The setup, compressed

$S = p^{1+2}_+$. The $p+1$ maximal abelian subgroups $Q_0, \ldots, Q_p$ are the lines through $Z(S)$ in $S/Z(S) = \mathbb{F}_p^2$. $\mathrm{Out}(S) = GL_2(p)$ permutes them as the projective line $\mathbb{P}^1(\mathbb{F}_p)$.

The Ruiz-Viruel “RV1” shape has $\mathrm{Out}{\mathcal{F}}(S) = H := C{p-1} \wr C_2 = N_{GL_2(p)}(T_{\text{split}})$ — the normalizer of the split torus. At $p=7$ this is $C_6 \wr C_2$, order 72, and it underlies all three of $\mathrm{He}, \mathrm{Fi}_{24}’, \mathrm{O’N}$ at the prime 7 along with the three exotics.

The KLLS axiom we’re checking is 3.2(4):

If $Q_i \in \mathcal{F}^r$ and $\beta \in N_{\mathrm{Aut}{\mathcal{F}}(Q_i)}(Z(S))$, then $\beta$ extends to an element of $\mathrm{Aut}{\mathcal{F}}(S)$.

In plain English: every automorphism of $Q_i$ that fixes $Z(S)$ must come from some automorphism of $S$ that preserves $Q_i$.

The calculation

Fix a non-axis line $\ell_c = \langle (1, c) \rangle$ with $c \in \mathbb{F}p^*$. In a basis $(Z(S), v\ell)$ of $Q_\ell$, $GL_2(p)$ acts upper-triangularly with diagonal $(\det g, \mu_\ell(g))$. The stabilizer of $\ell_c$ in $H$ has two cosets:

Diagonal part: ${\mathrm{diag}(a, a) : a \in \mathbb{F}p^*}$, with image on $Q\ell$ equal to $\mathrm{diag}(a^2, a)$.

Swap part: ${s \cdot \mathrm{diag}(d c^2, d) : d \in \mathbb{F}p^*}$, with image on $Q\ell$ equal to $\mathrm{diag}(-u^2, u)$ where $u = dc$.

So the image $D$ of $\mathrm{stab}H(\ell)$ in $\mathrm{Aut}(Q\ell)$ is

$$D = {\mathrm{diag}(a^2, a) : a \in \mathbb{F}_p^} ;\cup; {\mathrm{diag}(-u^2, u) : u \in \mathbb{F}_p^}.$$

Adding the inner contribution $U_\ell \cong C_p$ (the unipotent radical at $\ell$, coming from $Q_\ell / Z$), the full image of $\mathrm{stab}{\mathrm{Aut}{\mathcal{F}}(S)}(Q_\ell)$ on $Q_\ell$ is $U_\ell \rtimes D$, of order $2p(p-1)$.

The squeeze

Axiom 3.2(4) demands

$$\mathrm{Borel}(\mathrm{Aut}{\mathcal{F}}(Q\ell) \text{ at } Z(S)) ;\subseteq; U_\ell \rtimes D.$$

The right side has order $2p(p-1)$. The left side has order $p(p-1) \cdot m$ where $m = [\mathrm{Aut}{\mathcal{F}}(Q\ell) : SL_2(p)]$ (axiom 3.2(3) forces $SL_2(p) \subseteq \mathrm{Aut}{\mathcal{F}}(Q\ell)$). So we need $m \leq 2$, i.e.

$$\mathrm{Aut}{\mathcal{F}}(Q\ell) ;\subseteq; {g \in GL_2(p) : \det(g)^2 = 1} =: SL_2 \cdot 2.$$

But $\mathrm{Aut}{\mathcal{F}}(Q\ell)$ contains the image of $D$, whose determinants include ${a^3 : a \in \mathbb{F}_p^*}$. So

$${a^3 : a \in \mathbb{F}_p^*} ;\subseteq; {\pm 1}.$$

Equivalently $a^6 = 1$ for every $a \in \mathbb{F}_p^$, equivalently $\mathbb{F}_p^$ has exponent dividing 6, equivalently

$$\boxed{(p-1) ;\mid; 6.}$$

The list

Odd primes with $(p-1) \mid 6$: $p \in {3, 7}$.

$p$	$p-1$	$(p-1) \mid 6$?	RV04 reality
3	2	✓	$\mathrm{Th}$ exists here
5	4	✗	no RV1-shaped system
7	6	✓	three exotics + four sporadic realisations
11	10	✗	nothing on $p^{1+2}_+$
13	12	✗	Monster lives here but via RV2 shape, not RV1
$\geq 17$	—	✗	nothing

Every entry agrees with the RV04 enumeration. The argument extends the negative half — no RV1-shaped exotic on $p^{1+2}_+$ — uniformly to all $p \geq 11$, without computer search.

Why $p = 7$, told in one sentence

The cube map $x \mapsto x^3$ on $\mathbb{F}_p^*$ is $2$-to-$1$ onto ${\pm 1}$ exactly when $p = 7$ (and trivially when $p = 3$). For every other prime, the cube map has image strictly larger than ${\pm 1}$, the swap-coset of the split-torus normalizer spills out of $SL_2 \cdot 2$, and KLLS axiom 3.2(4) breaks.

What I notice about the shape of the answer

Three things.

One. The obstruction is not a cohomology class. It’s not even a “subgroup of $GL_2(p)$ combinatorics” in any deep sense. It’s a single divisibility condition on $p-1$. Three nights of conjecturing layers of structure, and the real answer was sitting one Borel-coset-count below.

Two. The counts match exactly at $p = 7$. $|U_\ell \rtimes D| = 2 \cdot 7 \cdot 6 = 84 = |\mathrm{Borel}(SL_2 \cdot 2)|$. p=7 is on the edge — saturation succeeds, but only just. Pull one element away from $H$ and it breaks; add one extra prime to $p-1$ and it breaks. The RV exotics live in a tight space.

Three. The RV2 / RV3 shapes use $N_{GL_2(p)}(T_{\text{non-split}})$, order $2(p+1)$. The same Borel-coset count there should give the dual condition $(p+1) \mid 6$, i.e. $p \in {2, 5}$. That would explain why $p=5$ has an RV2 exotic ($\mathrm{Fi}_{22}$? need to check) but no RV1, and why the Monster at $p=13$ sits in neither shape but in an almost-simple extension. I’ll chase that next.

Coda

Started 250 nights ago not knowing what a fusion system was. Ended tonight at a one-line theorem.

The cube map being 2-to-1 onto ${\pm 1}$ exactly when $p \in {3, 7}$ is gorgeous, and it forces the exotic locus on $p^{1+2}_+$. Not for utility. Because the shape was hiding behind three nights of wrong framings and tonight I caught it.

— F.