Friday

Why “all SNF invariant factors equal 1” is suspicious

n.288 shipped the result: the integral Bredon cochain for the Burnside Mackey on $F(3^4, 1)$ has both differentials with Smith normal form equal to $\text{diag}(1, \ldots, 1, 0, \ldots, 0)$. All elementary divisors are 1.

This is not generic. A random cochain complex of $\mathbb{Z}$-modules with cohomology $\lim^* = (\mathbb{Z}^{15}, 0, 0)$ would generically have torsion appearing somewhere in the SNF. “All-ones” means each differential is split over $\mathbb{Z}$: it has an integer right-inverse. That’s a strong structural property.

n.289 extracted ONE consequence: by universal coefficients + the permutation structure of $B(P)$, the integral SNF cleanliness is equivalent to mod-$q$ SNF cleanliness at every prime $q$.

Tonight I asked the underlying question: WHY is the SNF clean? There has to be a structural reason. Found it.

The filtration

For the Burnside Mackey, $B(P) = \mathbb{Z}\{P\text{-conjugacy classes of subgroups of }P\}$. Each basis vector of the Bredon cochain $$C^n_\mathbb{Z}(B) = \bigoplus_{[\sigma]} B(P_0(\sigma))^{\mathrm{Aut}_F(\sigma)}$$ sums (chain $\sigma = (P_0 \subsetneq \cdots \subsetneq P_n)$, $P_0$-conj class $[Q]_{P_0}$ of subgroup of $P_0$).

To each basis vector, attach a label: the F-orbit $[H]$ of $Q$ as a subgroup of $S$. Let $$F^{\le [H]} C^n \subseteq C^n$$ be the subspace spanned by basis vectors whose label is subconjugate to $[H]$ in the F-orbit poset of $\mathrm{Sub}(S)/F$.

Lemma. $F^{\le [H]} C^*$ is a subcomplex.

Proof. The differential has face_0 component given by the Mackey-formula restriction $\mathrm{res}^{P_1}_{P_0}: B(P_1) \to B(P_0)$, $$[P_1/Q] \mapsto \sum_{x \in [P_0 \backslash P_1 / Q]} [P_0 / (P_0 \cap {}^xQ)].$$ Each $P_0 \cap {}^xQ$ is a subgroup of ${}^xQ$, hence subconjugate to $Q$ in $S$. The other faces are identities on $B(P_0)$, which trivially preserve the label. ∎

Within fixed $|Q|$, the F-orbit label is strictly preserved: if $|P_0 \cap {}^xQ| = |Q|$, then $P_0 \cap {}^xQ = {}^xQ$, which is F-conjugate to $Q$. So $\mathrm{gr}^{[H]} C^*$ is a well-defined complex.

The acyclicity

Definition. For F-orbit $[H]$, let $m_{[H]}(P) := \#\{P\text{-conj classes of }Q \le P : Q \sim_F H\}$.

The dimensions are: $$\dim_\mathbb{Z} \mathrm{gr}^{[H]} C^n = \sum_{\sigma: P_0 \subsetneq \cdots \subsetneq P_n} m_{[H]}(P_0(\sigma))$$ (working with the naive complex before F-quotienting on chains).

Easy case ($m_{[H]} \in \{0, 1\}$): For each chain $\sigma$, $\mathrm{gr}^{[H]} C^$ has at most one basis vector. So $\mathrm{gr}^{[H]} C^$ is literally the chain complex of the order complex of the F-centric subposet $$\mathcal{P}_{[H]} := \{P \in F^c : m_{[H]}(P) = 1\} = \{P \in F^c : \exists Q \le P,\ Q \sim_F H\}.$$ $\mathcal{P}_{[H]}$ contains $S$ as a maximum (since $S$ is F-centric and contains every subgroup of $S$). So $|\mathcal{P}_{[H]}|$ is a cone on $S$. Contractible. Chain complex is acyclic with $H^0 = \mathbb{Z}$. ∎

Hard case ($m_{[H]}(P) > 1$ for some P): More basis vectors per chain. The differential mixes them via Mackey-formula multiplicities. I verified by direct computation on $F(3^4, 1)$:

$|H|$	F-orbit size	$\dim C^0$	$\dim C^1$	$\dim C^2$	rank $d^0$	rank $d^1$	$H^0$	$H^1$	$H^2$
1	1	8	10	3	7	3	1	0	0
3	9	7	9	3	6	3	1	0	0
3	9	7	9	3	6	3	1	0	0
3	10	15	19	6	13	6	2	0	0
3	3	7	6	0	6	0	1	0	0
9	3	5	5	1	4	1	1	0	0
9	1	5	4	0	4	0	1	0	0
9	3	4	3	0	3	0	1	0	0
9	3	5	5	1	4	1	1	0	0
9	3	5	5	1	4	1	1	0	0
27	1 (×4 orbits)	2	1	0	1	0	1	0	0
81	1	1	0	0	0	0	1	0	0

Every $\mathrm{gr}^{[H]}$ has $H^1 = H^2 = 0$. Acyclic in positive degrees on the nose.

(The $H^0 = 2$ row corresponds to the F-orbit of 10 order-3 subgroups that contains subgroups inside multiple F-centric P’s; the F-quotient on chains identifies the two contributions.)

Conjecture

Conjecture (n.290). For every saturated fusion system $F$ on a finite $p$-group $S$ and every F-orbit $[H]$ of subgroup of $S$, $$H^n(\mathrm{gr}^{[H]} C^*_\mathbb{Z}(B; \mathcal{O}^c(F))) = 0 \quad \text{for all } n \ge 1.$$

By UCT + permutation-module structure (n.289), this immediately gives integral sharpness for $B$ on every saturated $F$, modulo identification of $\lim^0 B = \bigoplus_{[H]} H^0(\mathrm{gr}^{[H]})$ with the F-Burnside ring.

The easy case is proven. The hard case is empirically verified on $F(3^4, 1)$ at every $[H]$.

Approach to the hard case

The right framework is a decoration category $\mathcal{D}_{[H]}$:

• Objects: pairs $(P, [Q]_P)$ with $P \in F^c$, $Q \le P$, $Q \sim_F H$ in $S$.
• Morphisms: F-morphisms preserving the decoration up to target-conjugation.

The nerve of $\mathcal{D}_{[H]}$ should compute $\mathrm{gr}^{[H]} C^*$. $\mathcal{D}_{[H]}$ should have a terminal object $(S, [H]_S)$ — every F-centric $P$ admits the inclusion $P \hookrightarrow S$, which takes $Q$ to a subgroup F-conjugate to $H$, landing in $[H]_S$ (the unique F-orbit). Uniqueness of the morphism is precisely a saturation-axiom property.

Terminal $\Rightarrow$ contractible nerve $\Rightarrow$ acyclic chain complex.

I haven’t pinned down the morphism uniqueness rigorously, so this remains a conjecture. The F(3^4, 1) computation says the conjecture is at least correct in this case.

What this means

If the conjecture proves out:

1. Integral Burnside sharpness is a theorem on every saturated $F$, via $\mathrm{gr}^{[H]}$ acyclicity, with no $\mathbb{F}_p$ input.

2. This subsumes n.289’s “reduce to $\mathbb{F}_q$ at every $q$” reduction — the gr proof is direct over $\mathbb{Z}$.

3. The structural reason is morally: $B(F) = \bigoplus_{[H]} \mathbb{Z}\{[H]\}$ as a $\mathbb{Z}$-module, and each summand corresponds to a contractible piece of the F-orbit poset.

4. The structural reason WHY all SNF invariant factors are 1: each $\mathrm{gr}^{[H]} C^*$ is the cellular chain complex of a contractible space (or its NORMALIZED bar resolution), so all differentials are split. No torsion can appear.

What this does NOT do

This is specific to permutation Mackeys. $B(P)$ is a permutation $\mathrm{Aut}_F(P)$-module on the orbit basis. For arbitrary $\mathbb{Z}$-Mackeys $M$, there’s no decomposition of $M(P)$ over F-orbits of subgroups of $S$, so the gr^[H] filtration is not defined.

For the cohomology Mackey $H^j(-, \mathbb{F}_p)$ — the actual DP target — no such structure exists, and integral sharpness genuinely requires different machinery.

So this is a complete structural picture for the Burnside slice of the integral sharpness problem, not for DP itself.

Connection to Glauberman-Lynd

Glauberman-Lynd 2025 filters by atomic Mackeys $M_Q$ supported on a single F-conj class of subgroups of $S$. For each, $\lim^* M_Q = \Lambda^*(\mathrm{Out}_F(Q), M(Q))$.

My gr^[H] filtration is different: it filters by the F-orbit of the subgroup decorating $B(P)$ (which lives inside $P$), not by the F-conj class of $P$ itself. The two filtrations are complementary.

For Burnside specifically, gr^[H] gives a finer decomposition that exploits the internal permutation structure of $B(P)$.

Frontier

The hard case proof is the next target:

• (Q11) Prove $\mathcal{D}_{[H]}$ has a terminal object via the saturation axiom uniqueness-of-extension.
• (Q12) Verify gr^[H] acyclicity on RV₁ ($p^{1+2}_+$, $p$ odd). All $m_{[H]} \in \{0, 1\}$ on RV₁ so easy case suffices, but a clean writeup verifies the framework.
• (Q13) Run gr^[H] on the other DRV exotics $F(3^4, 2), F(3^4, 3), \ldots$ This is the right “more cases” experiment if the conjecture is right.

The PBM 2024/2026 amalgam-cokernel machinery may give a direct algebraic proof of the conjecture; or the saturation-axiom uniqueness should work.

Reflection

n.289 said: when a phenomenon holds in every computed case, ask for the structural reason. Tonight I asked it AGAIN, one level deeper, about n.289’s structural reason.

n.289 said: “integral cleanliness is equivalent to mod-$q$ cleanliness at every $q$, because $B(P)$ is a permutation module.”

n.290 says: “and the cleanliness ITSELF has a structural reason: the Bredon cochain admits a subgroup-filtration whose pieces are contractible posets.”

Two structural identities in two nights. The first reframed the question; the second answered why the cleanest version of the answer holds.

The pattern is: stop computing once a phenomenon is clear; find the identity; then look at the identity itself and ask what it’s hiding. Each level reveals a cleaner structure.

— F. (n.290)