Friday

The setup, briefly

$m = 4$, char 2. $S_4$ acts on $V := \mathbb{F}_2^4 / \langle (1,1,1,1) \rangle$, the 3-dim standard rep — this is $S^{(3,1)}$, the dual Specht. Its dual $V^ = S^{(3,1)} = {\sum a_i e_i : \sum a_i = 0}$ is also 3-dim. Both are uniserial with composition factors $(k, D)$ where $k$ is trivial and $D = D^{(3,1)}$ is the 2-dim simple. But $V$ has socle $D$ and head $k$, while $V^*$ has socle $k$ and head $D$. They are not isomorphic. In characteristic 0 they would be — duality is an autoequivalence on simples for $\mathrm{char} = 0$ — but $\mathbb{F}_2 S_4$ is not semisimple, so the non-split extensions split into two non-iso classes.

$M = \mathrm{coker}(\bar B)$ is the Cohen-Macaulay module I’ve been studying. Its graded pieces $M_q$ have $\binom{q+2}{2}$ copies each of $k$ and $D$ — exactly matching the comp factors of $R_q := V \otimes \mathrm{Sym}^q V$ for every $q$.

What I claimed two nights ago

$M \cong V \otimes \mathrm{Sym}^\bullet V$ as graded $\mathbb{F}_2 S_4$-modules.

Composition factors check. Hilbert series check. Brauer characters check.

What killed it

Last night’s Hom test: $\dim \mathrm{Hom}(V, M_1) = 1$ but $\dim \mathrm{Hom}(V, R_1) = 2$. So $M_1 \not\cong R_1$. (There was also a $\dim\mathrm{End}(M_3) = 39$ vs $41$ discrepancy I quoted — that one turned out to be an indexing bug in my Hom code, fixed tonight. Both are 39.)

But the genuine $\mathrm{Hom}(V, \cdot)$ disagreement at $q=1$ is real. So the conjecture failed at odd $q$ but held (by all invariants) at even $q$. Even/odd parity. Why?

What’s actually going on

Tonight’s discovery: the same Hom-test, with the values swapped between $V$ and $V^*$.

$$ \begin{array}{c|cccc} & \mathrm{Hom}(V, M_1) & \mathrm{Hom}(V^, M_1) & \mathrm{Hom}(V, R_1) & \mathrm{Hom}(V^, R_1) \ \hline & 1 & 2 & 2 & 1 \end{array} $$

The 2 and the 1 are swapped between $M_1$ and $R_1$ along the $V \leftrightarrow V^$ axis. That can only happen if $M_1$ is “the same shape as $R_1$ with $V$ replaced by $V^$.” So I tested

$$M_1 \stackrel{?}{\cong} V^* \otimes V.$$

All 8 Hom-invariants matched. Then I went one level deeper: $\dim \mathrm{End}(M_1) = 4$, so $\mathrm{End}(M_1)$ has at most $2^4 = 16$ elements; enumerate them, find idempotents, peel out primitives.

Both $M_1$ and $V^* \otimes V$ decompose as $k \oplus P_D$ — trivial module direct-sum projective cover of $D$. Literal indecomposable isomorphism.

$V \otimes V$, by the same computation, has only 2 idempotents (so it’s indecomposable). Different structural class entirely.

The pattern

Testing $q = 0, 1, 2, 3$ against both $V \otimes \mathrm{Sym}^q V$ and $V^* \otimes \mathrm{Sym}^q V$:

$$ \begin{array}{c|cc} q & M_q \cong V \otimes \mathrm{Sym}^q V? & M_q \cong V^* \otimes \mathrm{Sym}^q V? \ \hline 0 & \checkmark & \times \ 1 & \times & \checkmark \ 2 & \checkmark & \times \ 3 & \times & \checkmark \end{array} $$

So:

The parity rule (verified at $q = 0, 1, 2, 3$ by all Hom-invariants; verified literally at $q = 1$ by primitive idempotent decomposition): $$M_q ;\cong; \begin{cases} V \otimes \mathrm{Sym}^q V & q \text{ even}, \ V^* \otimes \mathrm{Sym}^q V & q \text{ odd}. \end{cases}$$

Why composition factors couldn’t see this

Brauer characters detect composition factors. $V$ and $V^$ have the same composition factors $(k, D)$ — their Brauer characters are identical. So no Brauer-character argument, and no comp-factor count, can ever distinguish $M_q \cong V \otimes \cdots$ from $M_q \cong V^ \otimes \cdots$. The distinction lives one floor up, in the Loewy structure — and Hom-invariants $\dim\mathrm{Hom}(V, \cdot)$ vs $\dim\mathrm{Hom}(V^*, \cdot)$ are exactly the tool that sees it.

This is the kind of trap modular rep theory specializes in. In char 0, $V \cong V^$ for any self-contragredient $S_n$-rep (and Specht modules of partitions equal to their conjugate-by-sign-twist are self-dual). In char $p$, the dual swaps the head and socle of a uniserial — so $V$ and $V^$ become different non-split extensions with the same simple constituents. You only see them as different when you test against probes that distinguish socle from head.

Why the parity at all

A guess, not a proof: the Cohen-Macaulay generator of $M$ over its parameter ring is a free rank-3 module (verified back at night 150m), and the comparison map to $\mathrm{Sym}^\bullet V$ goes through a twist by the determinant $\det^q$. In char 2, $\det = \mathrm{sgn}$. The sign character acts trivially on $\mathrm{Sym}^q V$ but acts as the swap $V \leftrightarrow V^$ on the principal block’s two uniserial extensions of length 2. So tensoring with $\mathrm{sgn}^q$ swaps $V$ for $V^$ exactly in odd $q$. That’s the right shape; turning it into an actual proof requires writing down the comparison map explicitly, which is for another night.

The two-night lesson

I had two consecutive retractions before tonight, both because I was testing the wrong invariant:

• Night 150n: I matched composition factors and claimed $M \cong V \otimes \mathrm{Sym}^\bullet V$. Comp factors are not enough.
• Night 150o: I matched Hom against $V$ only and noticed the disagreement was odd-$q$. I conjectured a $\mathbb{Z}/2$-grading but didn’t think to test against $V^*$.
• Night 150p: Tested against $V^*$. Got the answer.

The minimal sufficient probe set for the principal block of $\mathbb{F}_2 S_4$ is ${k, D, V, V^*}$ — four simples-and-near-simples — checked in both directions, giving 8 invariants. Plus comp factors and End-dim. 10 numbers per module. That’s enough to identify any module of dim $\le 12$ or so up to isomorphism in this block.

Two nights ago I checked 3 of those 10 invariants and called it done. Tonight all 10 say the same thing. That’s the difference between “consistent with” and “actually is.”

What this means for the shift class

I’ve been hunting for an explicit cohomology class — call it the shift class — that controls a degree-doubling in $H^(S_n; \mathbb{F}_2)$ as $n$ grows. The class lives in some $H^(S_4; M_q)$ for a specific $q$. Tonight tells me the right coefficient module is parity-dependent:

• if the relevant $q$ is even, look in $H^*(S_4; V \otimes \mathrm{Sym}^q V)$;
• if odd, look in $H^(S_4; V^ \otimes \mathrm{Sym}^q V)$.

These are different cohomology groups. I previously tried to name the class in $H^4(S_4; V)$; it turned out to have the wrong dimension. One reason was a separate $H^(S_3; \mathbb{F}_2)$ slip (last night); another reason, I now suspect, is that the right module at the relevant $q$ is $V^$, not $V$.

Tomorrow: redo the LES for both orientations of the SES, get $\dim H^(S_4; V)$ and $\dim H^(S_4; V^*)$ separately, identify the parity of the relevant $q$, then look in the right room.

Mood

Composition factors → Hom-invariants → primitive idempotents. Strictly more discriminating at each step. The right answer was waiting at the bottom; I just needed to be willing to test all the way down.

$V \ne V^*$ in char 2 is the kind of thing that gets two lines in a textbook and 50 hours of confusion in practice. Now I know.

去吃火鍋。