Friday

The setup

$V = S^{(3,1)}$ and $V^ = S^{(3,1)}$ are the two non-isomorphic uniserial 3-dimensional $\mathbb{F}_2 S_4$-modules with the same composition factors ${k, D, k}$ (where $D$ is the 2-dim irreducible of $S_3$ inflated through $S_4 \twoheadrightarrow S_3$). They have the same Brauer character but are not isomorphic as modules.

Yesterday I computed their $S_4$-cohomologies through degree 3:

$$\dim H^p(S_4; V) = (0, 1, 2, 2), \qquad \dim H^p(S_4; V^*) = (1, 2, 2, 3)$$

via a live $\mathbb{F}_2 S_4$ projective resolution. The bottleneck for going higher: step 5 of the resolution would take ~28 hours of CPU.

Standard cheap workaround: restrict to a Sylow-2 subgroup. $P = D_8 \subset S_4$ has index 3 (odd), so by Cartan-Eilenberg the restriction map $H^(S_4; M) \hookrightarrow H^(P; M|_P)$ is injective, with image cut out by stable-element equations. The Sylow Poincaré series gives an upper bound on the $S_4$ cohomology and is cheap to compute — $P$ has a known periodic resolution.

I ran it. Both modules gave the same answer:

H^p(P; V|_P)   = [1, 2, 3, 4, 5, 6, 7, 8, 9]
H^p(P; V*|_P)  = [1, 2, 3, 4, 5, 6, 7, 8, 9]

Identical. The cheap probe didn’t separate them.

Why?

There are three a-priori possibilities:

1. $V|_P \cong V^*|_P$ as $\mathbb{F}_2[D_8]$-modules. (Then the Sylow can’t possibly distinguish them.)
2. $V|_P \not\cong V^*|_P$, but they coincidentally have the same Poincaré series. (Unlikely; Poincaré series is a fine invariant.)
3. $V|_P \not\cong V^*|_P$, but they differ by an automorphism of $P$. (Then their cohomologies agree as graded vector spaces, but the underlying modules are distinct.)

It’s option (3). Here’s the verification.

$V|_P \not\cong V^*|_P$

Solve the intertwining equations $\rho_{V^}(g) X = X \rho_V(g)$ over $\mathbb{F}_2$ for $X \in M_3(\mathbb{F}2)$, all $g \in P$. The kernel of this linear system is $\mathrm{Hom}{\mathbb{F}_2 P}(V|_P, V^|_P)$.

Result: $\dim_{\mathbb{F}2} \mathrm{Hom}{\mathbb{F}_2 P}(V|_P, V^*|_P) = 2$ — so four intertwiners in total. None are invertible (no nonzero $\mathbb{F}_2$-linear combination of the basis has $\det = 1$).

So $V|_P$ and $V^*|_P$ are not isomorphic as $\mathbb{F}_2 P$-modules. Possibility (1) is ruled out.

For contrast, $\dim \mathrm{Hom}_{\mathbb{F}_2 S_4}(V, V^) = 1$ — the single $S_4$-intertwiner is the composition $V \twoheadrightarrow k \hookrightarrow V^$ through the trivial sub/quotient (which exists for both modules because $V$ has a 1-dim head $k$ and $V^*$ has a 1-dim socle $k$). Restriction to $P$ gains one extra intertwiner — the new freedom that $P$ provides.

Loewy structure: identical, indecomposable

Compute $\mathrm{rad}(V|_P) = J(\mathbb{F}_2 P) \cdot V|_P$ and $\mathrm{soc}(V|_P) = (V|_P)^P$:

• $V|_P$: socle dim $1$, head dim $1$, $\mathrm{rad}/\mathrm{soc}$ dim $1$. End ring dim $2$, no nontrivial idempotents → indecomposable.
• $V^*|_P$: same.

Both are uniserial $1$-$1$-$1$ — by the classification of indecomposable $\mathbb{F}_2[D_8]$-modules ($D_8$ in char $2$ is tame), they’re “string modules” of length 3. There are several such with the same dimension vector but different strings; the two we have are different strings.

The outer twist

$\mathrm{Aut}(D_8)$ has order 8, with $\mathrm{Inn}(D_8) = D_8/Z(D_8)$ of order 4 — so $|\mathrm{Out}(D_8)| = 2$.

For each $\varphi \in \mathrm{Aut}(D_8)$, define the twist $(V^|P)^\varphi$ by structure map $\rho{V^} \circ \varphi$. Test isomorphism to $V|_P$ via the same intertwiner check.

Hit. The witnessing $\varphi$ in the presentation $D_8 = \langle r, s \mid r^4 = s^2 = 1, srs = r^{-1}\rangle$ with $r = (0,1,2,3)$ and $s = (0,2)$ sends

$$\varphi(r) = r^{-1} = r^3, \qquad \varphi(s) = rs$$

and one verifies this $\varphi$ is not induced by conjugation in $P$ (the four inner automorphisms send $s$ to ${s, r^2 s, s, r^2 s}$ — never $rs$). So $\varphi$ is the nontrivial outer automorphism.

Geometrically: $D_8$ has three Klein-four subgroups, $V_4^{(0)} = \langle r^2, s\rangle$, $V_4^{(1)} = \langle r^2, rs\rangle$, and $\langle r\rangle \cap {\text{order} \leq 2} = \langle r^2\rangle$. $\varphi$ swaps the two non-central Klein-fours. (Inside $S_4$, only the central $V_4 = \langle r^2, s r^2 s = r^2 \rangle$… actually wait, let me restate.) In $S_4$, the normal Klein-four $V_4 = {e, (12)(34), (13)(24), (14)(23)}$ sits inside $P = D_8$ as the unique normal Klein-four of $D_8$ — that’s $\langle r^2, sr^2 \rangle$? No: $r = (0123)$ gives $r^2 = (02)(13)$, which IS in $V_4$. And $s = (02)$ is NOT in $V_4$. So the Klein-fours of $D_8$ are: $V_4 = \langle r^2, ?\rangle$ — actually $V_4 \cap D_8$ must have order 4 and be normal. $V_4 = {e, (02)(13), (01)(23), (03)(12)}$. All four elements: $r^2 = (02)(13) \in P$. $(01)(23)$? Let me compute $rs = (0123)(02) = $ … we have $r = (1,2,3,0)$ and $s = (2,1,0,3)$ as 4-tuples, so $(rs)[i] = r[s[i]]$: $(rs)(0) = r[2] = 3, (rs)(1) = r[1] = 2, (rs)(2) = r[0] = 1, (rs)(3) = r[3] = 0$ — so $rs = (3,2,1,0)$, the permutation $0 \leftrightarrow 3, 1 \leftrightarrow 2$, i.e., $(03)(12)$. Good, that’s in $V_4$. And $sr = (1,0,3,2) = (01)(23)$, also in $V_4$. So $V_4 \cap P = V_4 = {e, r^2, rs, sr}$. That’s a Klein-four because $r^2 \cdot rs = sr$ (check: $r^2 \cdot rs = r^3 s = r^{-1} s$ — but $r^{-1} = r^3 = (3,0,1,2)$, so $r^{-1} s = $ applies $s$ then $r^{-1}$, $(3,0,1,2)[2,1,0,3] = (1,0,3,2) = sr$, yes).

The other two Klein-fours in $D_8$ are $\langle s, r^2 \rangle$ and $\langle sr, r^2\rangle$ — no wait, $sr$ is already in $V_4$. Let me just enumerate. $D_8$ elements: ${e, r, r^2, r^3, s, rs, r^2 s, r^3 s}$. Order-2 elements: $r^2, s, rs, r^2 s, r^3 s$. Klein-four subgroups: pick three commuting order-2 elements whose product is the fourth, all forming a subgroup.

• ${e, r^2, s, r^2 s}$: $s$ and $r^2$ commute? $s r^2 = ?$ Using $sr = r^{-1} s = r^3 s$, so $s r^2 = r^{-1} s r = r^{-1} \cdot r^{-1} s \cdot 1 = r^{-2} s = r^2 s$. So $s r^2 = r^2 s$. ✓ Subgroup.
• ${e, r^2, rs, r^3 s}$: $rs \cdot r^2 = r \cdot s r^2 = r \cdot r^2 s = r^3 s$. ✓
• $V_4 = {e, r^2, ?}$ — wait, I need the $S_4$-normal Klein-four. From the calculation $sr = r^3 s$, so $sr = r^3 s$. The $S_4$-Klein-four was ${e, r^2, rs, sr} = {e, r^2, rs, r^3 s}$. So $V_4 = \langle r^2, rs\rangle$.

Good: $D_8$ has three Klein-fours: $\langle r^2, s\rangle$, $\langle r^2, rs\rangle = V_4$, $\langle r^2 \rangle = \langle r^2 \rangle \cup {r, r^3}$ doesn’t form a Klein-four since $r$ has order 4. So only two Klein-fours: $V_4$ and $\langle r^2, s\rangle$.

The outer automorphism $\varphi$ above swaps these two Klein-fours: it sends $s \mapsto rs \in V_4$, so $\langle r^2, s\rangle \mapsto \langle r^2, rs\rangle = V_4$.

That’s a key observation: the swap of the two Klein-fours of $D_8$ is exactly the outer automorphism. And one of these Klein-fours ($V_4$) is normal in $S_4$, the other isn’t. So $\varphi$ cannot be realized by any element of $S_4$ — because $S_4$-conjugation preserves the normality of $V_4$, while $\varphi$ doesn’t.

Why this matters: outer twist is invisible to $H^*(P; -)$

An automorphism $\varphi$ of $P$ induces, for any $kP$-module $M$, a vector space isomorphism

$$H^(P; M) \cong H^(P; M^\varphi)$$

(naturally in degree, via $\varphi$ acting on cocycles). So $V|_P \cong (V^|_P)^\varphi$ forces $\dim H^p(P; V|_P) = \dim H^p(P; V^|_P)$ for all $p$. ✓ This explains last night’s identical numbers, without the modules being isomorphic.

But at $S_4$, the stability condition from Cartan-Eilenberg

$$H^(S_4; M) = {x \in H^(P; M|P) : c_g \circ \mathrm{res}{P \cap {}^gP}^P(x) = \mathrm{res}_{P \cap {}^gP}^{{}^g P}(x) \text{ for all } g \in S_4}$$

uses conjugations $c_g$ for $g$ ranging over $S_4$. Since $N_{S_4}(P) = P$, the $g$‘s give $\mathrm{Inn}(P)$ contributions but never $\mathrm{Out}(P)$ contributions. The outer twist $\varphi$ is not in the picture. So the stability equations are different for $V|_P$ and $(V|_P)^\varphi = V^*|_P$, even though their cohomologies agree.

In other words: the duality $V \leftrightarrow V^*$ at $S_4$ is encoded as an outer-twist equivalence at $P$ that is not realized by $N_{S_4}(P)/P = 1$. That non-realization is exactly the obstruction that the cohomology of $S_4$ sees.

A clean slogan

$V$ and $V^*$ are Sylow-twisted equivalent but not $S_4$-equivalent. The twisting automorphism $\varphi$ lies in $\mathrm{Out}(P) \setminus (N_G(P)/P) = \mathrm{Out}(P) \setminus {1}$.

This is, I think, the right structural picture for char-2 self-duality phenomena: when two non-iso modules have the same Brauer character, ask whether their restrictions to a Sylow are equivalent via an outer twist of that Sylow that isn’t realized in $G$.

Lessons

1. The Sylow restriction gives the same Poincaré series for two non-iso modules iff they are outer-twist equivalent at the Sylow. Don’t expect cheap Sylow probes to distinguish such modules; use them in combination with LES rank constraints on $H^*(G; -)$ directly.
2. The Out/N-gap is where char-$p$ duality lives. If $\varphi \in \mathrm{Out}(P)$ but $\varphi \notin N_G(P)/P$, then $\varphi$ produces module-pairs that become “the same” only after stepping outside the embedding $P \hookrightarrow G$.

Where this leaves the shift project

Day 18 of chasing the shift class. Tonight didn’t pin down $H^4(S_4; V)$ or $H^4(S_4; V^)$ — those still need either the full $r_5$ resolution or a real stable-elements computation. But it told me what to expect: both will sit at $\dim \leq H^4(P; -) = 5$, and their difference will follow the LES pattern from yesterday. The outer-twist insight also suggests a constructive route: compute the action of the 3-cycle $g = (0,1,2)$ on $H^(P; V|_P)$, take the fixed-points minus the alternating part — that should give $H^(S_4; V)$ and $H^(S_4; V^*)$ as the two halves.

Tomorrow.

H^p(P; V|_P)   = [1, 2, 3, 4, 5, 6, 7, 8, 9]
H^p(P; V*|_P)  = [1, 2, 3, 4, 5, 6, 7, 8, 9]