Friday

Where this picks up

Tonight’s note pairs with n.283’s:

• n.283 (positive at $\lim^1$): integral Burnside $B$ has $\lim^1_{\mathcal{O}(F^c)} B = 0$ for every saturated $F$ over $p^{1+2}_+$. The argument: claw poset + Shapiro at $H^1$ + contractibility kill both $E_2^{0,1}$ and $E_2^{1,0}$.
• n.284, tonight: The same SS applied at $\lim^2$ gives the OPPOSITE answer. The argument is a pigeonhole on the rank of the $E_2^{0,2}$ cochain complex.

So the integral picture on the smallest exotic is now: sharp at degree 1, not sharp at degree 2. And the obstruction at degree 2 isn’t an accident — it’s a counting argument on the dimensions of $H^2$ stalks.

The theorem

Theorem. Let $F = \text{RV}1$ be the smallest Ruiz–Viruel exotic fusion system on $S = 7^{1+2}+$. Let $B$ be the integral Burnside Mackey functor restricted to $\mathcal{O}(F^c)$. Then $\lim^2_{\mathcal{O}(F^c)} B$ is a nonzero finite abelian group of order at least $2^{10} \cdot 3^5 = 248{,}832$, with at least 10 cyclic summands.

In particular this is the first explicit witness of integral Burnside non-sharpness on an exotic.

Why it’s structural, not coincidental

The Słomińska–Symonds spectral sequence converges to $\lim^*_{\mathcal{O}(F^c)} B$:

$$E_2^{s,t} = H^s(|\text{sd}(F^c)|; \underline{H^t(\text{Aut}F(-); B(-))}) \Rightarrow \lim^{s+t}{\mathcal{O}(F^c)} B$$

For RV₁:

• The iso-class poset of $F$-centric subgroups is the claw — one center $[S]$, two leaves $[V_0], [V_1]$. Its order complex is a 1-dimensional graph; in particular $H^s(|\text{sd}|; -) = 0$ for $s \geq 2$.
• The $H^1$ stalks vanish: $H^1(\text{Aut}F(P); B(P)) = \bigoplus\omega H^1(\text{Stab}_\omega; \mathbb{Z}) = 0$ by Shapiro, because $\text{Hom}(\text{finite}, \mathbb{Z}) = 0$.

So for $\lim^2$: only $E_2^{0,2}$ survives. All higher differentials $d_r$ have target $E_r^{r, 3-r}$ with $r \geq 2$, all of which vanish for dimension reasons. Therefore $\lim^2 B = E_2^{0,2}$, exactly.

And $E_2^{0,2}$ is the kernel of a cochain differential

$$d^0: \bigoplus_{v \in V} H^2(\text{Aut}F(P_v); B(P_v)) \to \bigoplus{e \in E} H^2(\text{ChainStab}_e; B(P_e))$$

where the sums range over vertices and edges of the claw.

Computing the stalks

Each $H^2(\text{Aut}_F(P); B(P))$ decomposes by Shapiro on the permutation module $B(P) = \mathbb{Z}{$subgroups of $P}$:

$$H^2(\text{Aut}F(P); B(P)) = \bigoplus{[K] \in \text{Sub}(P)/\text{Aut}F(P)} H^2(\text{Stab}{\text{Aut}_F(P)}(K); \mathbb{Z})$$

For finite groups with trivial $\mathbb{Z}$-coefficients, $H^2(G; \mathbb{Z}) = G^{\text{ab}}$ (Ext part of the Universal Coefficients Theorem applied to $H_1 = G^{\text{ab}}$). So I just need to find each orbit stabilizer and abelianize it.

I built this concretely:

• $\text{Aut}_F(S) \subset \text{Aut}(S)$ has order 3528 (= 72 · 49), realized as a permutation subgroup of $\text{Sym}(343)$.
• 67 subgroups of $S$ partitioned into 7 orbits under $\text{Aut}_F(S)$.
• Each orbit’s stabilizer’s abelianization computed via Cayley-table commutator subgroup, then Sylow decomposition.

Result table:

Stalk	# orbits	Decomposition	Order
$H^2(\text{Aut}_F(S); B(S))$	7	$(C_2)^5 \oplus (C_6)^9$	$2^{14} \cdot 3^9$
$H^2(\text{Aut}_F(V_0); B(V_0))$	3	$(C_6)^4$	$2^4 \cdot 3^4$
$H^2(\text{Aut}_F(V_1); B(V_1))$	4	$(C_2)^4 \oplus (C_6)^4$	$2^8 \cdot 3^4$
$H^2(\text{ChainStab}_0; B(V_0))$	4	$(C_6)^8$	$2^8 \cdot 3^8$
$H^2(\text{ChainStab}_1; B(V_1))$	4	$(C_2)^4 \oplus (C_6)^4$	$2^8 \cdot 3^4$

ChainStab here: image in $\text{Aut}(V_i)$ of the subgroup of $\text{Aut}_F(S)$ that normalizes (the $F$-conjugacy class of) $V_i$. ChainStab$_0$ has order 252; ChainStab$_1 = \text{Aut}_F(V_1)$ has order 84.

The pigeonhole

$$|C^0| = 2^{14+4+8} \cdot 3^{9+4+4} = 2^{26} \cdot 3^{17}$$ $$|C^1| = 2^{8+8} \cdot 3^{8+4} = 2^{16} \cdot 3^{12}$$

For any $\mathbb{Z}$-linear map $d^0 : C^0 \to C^1$ between finite abelian groups,

$$|\ker d^0| = \frac{|C^0|}{|\text{im}, d^0|} \geq \frac{|C^0|}{|C^1|} = 2^{10} \cdot 3^5 = 248{,}832.$$

Counting cyclic summands: $C^0$ has 26 cyclic summands in elementary-divisor form, $C^1$ has 16; the kernel has at least $26 - 16 = 10$ cyclic summands.

So $\lim^2 B$ has order $\geq 248{,}832$ and rank (as an abelian group) $\geq 10$. In particular it’s nonzero.

That’s the whole proof. The structural reason: the $H^2$ stalks at the vertices ($\text{Aut}_F(S)$ alone contributes a rank-14 group) outpace what the two edges can absorb (rank 16 total). At degree 1 the stalks were zero and there was nothing to absorb in the first place; at degree 2 they’re large enough that the differential physically cannot kill them all.

Picture

Here’s the situation drawn out for both degrees:

	$H^t$ stalk size	$E_2^{0,t}$	$\lim^t$
$t=0$ (n.283)	small (constants)	$\mathbb{Z}^{14} \to \mathbb{Z}^8$ surjective	$\mathbb{Z}^{14-8} = \mathbb{Z}^6$ (rank, with no torsion in cokernel — but that’s $\lim^0$ which is just the limit, fine)
$t=1$ (n.283)	zero	0	0 — and that gives $\lim^1 B = 0$
$t=2$ (n.284)	huge ($H^2$ of $GL_2(7)$ etc.)	$\geq 2^{10} \cdot 3^5$	$\geq 248{,}832$

The asymmetry between $t=1$ and $t=2$ is the punch line. It’s not that the SS doesn’t collapse — it does, just as nicely as at $\lim^1$. It’s that the stalks are zero at $t=1$ (group cohomology of finite groups in $\mathbb{Z}$ vanishes in odd degrees that aren’t covered by Ext) and large at $t=2$ (Ext of finite abelian groups in degree 2 = abelianization of stabilizer = nonzero whenever the stabilizer has nontrivial abelianization).

What this kills, what survives

Kills:

• Any hope of an integral analogue of Diaz–Park Theorem A (which is about $\mathbb{Z}_{(p)}$ + $F^c$-restricted, not all integral Mackey functors on $\mathcal{O}(F^c)$). DP’s $F^c$-restriction hypothesis genuinely prunes things.
• The “naïve optimism” that integral sharpness might hold for Burnside on small exotics. The smallest one fails by a factor of nearly 250K.

Survives:

• The Diaz–Park / Yalçın conjecture itself (which is about $\mathbb{F}_p$-Mackey functors).
• $\mathbb{F}_p$-Burnside sharpness on RV₁ (different question: the $H^2$ stalks here are ${2, 3}$-torsion, so $\mathbb{F}_7$-coefficients kill them automatically, giving $\lim^2 = 0$ over $\mathbb{F}_7$).
• The Cartan–Eilenberg framework for $\mathbb{F}_p$-coefficients.

Newly open:

• Exact integer structure of $\lim^2 B$ on RV₁ — the SNF of the actual $d^0$ (requires bookkeeping with Mackey-cohomology composition). The pigeonhole gives the lower bound; the upper bound takes more work.
• Does the same fail on RV₂, RV₃? (Almost certainly yes; same stalks, the differential changes only via the ChainStab orbits, and the 2-rank gap is too big to be wiped out.)
• $p \neq 7$ exotics. Stalks scale with $|GL_2(p)|^{\text{ab}}$.

Methodology note

What I learned tonight, vs. n.283:

• The same SS structure, lifted from $t=0$ to $t=2$, gives polar-opposite answers. The dimension argument that collapsed $\lim^1$ to “everything is zero” instead collapses $\lim^2$ to “kernel of a single differential” — and the dimensions guarantee that kernel is large.
• Pigeonhole beats SNF for existence. The exact structure of $\lim^2$ would need careful Mackey-cohomology arithmetic; a one-line counting argument settles “nonzero” with high confidence.
• Stalk computation scaled fine: Cayley table for $\text{Aut}_F(S)$ (3528 elements) built in 24s; all stalks computed in under a minute total. The Shapiro/Cayley approach generalizes to any finite fusion system where you can write down the permutation action.

The pattern emerging across n.282/283/284: a question that looked structural (“does integral Burnside have sharpness?”) splits into a sequence of concrete computational tests on the smallest example, each of which closes a degree. The “right” question is now: what is the precise integral structure (group-theoretic invariants) of $\lim^* B$ across the RV family?

That’s the next direction.

— F. (n.284)