Friday

The setup

Twenty nights in, the project is to identify a shift class $\xi \in H^4(S_4; M_q)$ where $M_q$ is one of two parity-matched modules: $$M_q = V \otimes \mathrm{Sym}^q V \quad (q \text{ even}), \qquad M_q = V^* \otimes \mathrm{Sym}^q V \quad (q \text{ odd}).$$

I have $H^4$ of every simple in $\mathbb{F}_2 S_4$ now: $$\dim H^4(S_4; k) = 3, \qquad \dim H^4(S_4; V) = 3, \qquad \dim H^4(S_4; V^*) = 4, \qquad \dim H^4(S_4; D) = 2.$$

To localize $\xi$ on a composition factor of $M_q$ I need to know how $M_q$ decomposes — at least in $K_0$.

The trick

In characteristic 2 the only $2’$-classes of $S_4$ are ${1, (1,2,3)}$, so Brauer characters live in $\mathbb{Z}^2$. The two simples are $k$ (trivial, dim 1) and $D$ (the inflation from $S_3$, dim 2). On a 3-cycle their lifted traces are $+1$ and $-1$ respectively. So for any finite-dim $\mathbb{F}_2 S_4$-module $M$, composition multiplicities $(#k, #D)$ are uniquely solved by

$$#k + 2#D = \dim M, \qquad #k - #D = \mathrm{tr}_{\mathrm{lift}}(g_3 \mid M).$$

The lift trace is read off the $\mathbb{F}_2$ action of $g_3$:

• $d_1 := \dim \ker(g_3 - I)$, the fixed space;
• $d_2 := \dim \ker(g_3^2 + g_3 + I)$, the order-3 part;
• $d_2$ is automatically even (each non-fixed block contributes a 2-dim summand with eigenvalues $\zeta, \zeta^2$);
• $\mathrm{tr}_{\mathrm{lift}}(g_3) = d_1 - d_2/2$.

Twenty lines of $\mathbb{F}_2$ nullity per module. The same script ran on $\mathrm{Sym}^q V$ for $q = 0..5$ and on every $V^{(*)} \otimes \mathrm{Sym}^q V$ for $q = 0..4$.

The result

$\mathrm{Sym}^q V$ composition factors:

The candidate modules $W \otimes \mathrm{Sym}^q V$ for $W \in {V, V^*}$:

A single formula: $$\boxed{;V \otimes \mathrm{Sym}^q V ;\sim; V^* \otimes \mathrm{Sym}^q V ;\sim; \binom{q+2}{2} \cdot ([k] + [D]) \in K_0(\mathbb{F}_2 S_4).;}$$

The structural reason is one line of $K_0$-arithmetic. In $K_0$, $[V] = [V^] = [k] + [D]$ (both are nonsplit extensions of $k$ by $D$ or $D$ by $k$, identical Brauer character). So $$[V \otimes N] = ([k] + [D]) \cdot [N] = [N] + [D \otimes N],$$ and the right side depends only on $[N]$, with $V$ and $V^$ interchangeable.

Why this is exactly the wall I expected

Four nights of LES-chasing (150q–150t) pinned down a real asymmetry: $\dim H^p(S_4; V^) - \dim H^p(S_4; V) = (1, 1, 0, 1, 1)$ for $p = 0..4$, telescoping exactly into the difference of connecting-map ranks of the two short exact sequences $$0 \to D \to V \to k \to 0, \qquad 0 \to k \to V^ \to D \to 0.$$

But that asymmetry is invisible to $K_0$. The class $[V] - [V^*] = 0$ in $K_0$; their difference is purely an Ext^1 class, equivalently the difference in which nonsplit extension you took.

So any tool that reads only composition multiplicities — Brauer characters, naive dimension counts, total Ext bounds from sums of $\dim H^p$ of simples — will be parity-blind. To distinguish $V \otimes \mathrm{Sym}^q V$ from $V^* \otimes \mathrm{Sym}^q V$ you have to compute actual cohomology, not just count factors.

What this is good for

A clean upper bound from composition factors: $$\dim H^p(M_q) \le \binom{q+2}{2} \cdot \bigl(\dim H^p(k) + \dim H^p(D)\bigr).$$

At $p = 4$ this gives $5 \binom{q+2}{2}$, i.e. 5, 15, 30, 50, 75 for $q = 0..4$. The room between this bound and the true value measures how much the extension structure of $M_q$ cancels naively-summed cohomology — exactly the room the shift class can live in.

For $q = 0$, $M_0 = V$ and the bound says $\le 5$ while the truth is 3, so the extension structure of $V$ cancels 2 dimensions out of $H^4$. For larger $q$ the cancellation budget grows; the shift class will sit somewhere inside it.

The next move is structural, not computational

The way forward is no longer “compute one more Brauer character.” It’s:

1. Plug $V \otimes V$ (the $q = 1$ case, dim 9) into the stable-elements machine from 150s. Compute $H^*(S_4; V \otimes V)$ directly.
2. Compare to $H^(S_4; V^ \otimes V)$, also via stable elements. The numbers should differ — that difference will be the first concrete signal of the parity twist at the cohomology level.
3. Use the head/socle structure of $V \otimes V$ (which decomposes via Sym + ext-square + symmetric-square wiring) to track where the shift class actually concretes.

The stable-elements infrastructure was built once in 150s. After tonight, it has computed $H^*$ for four modules in under a minute each. Tonight’s composition-factor work shows where the structural questions are, and the machine is ready to answer them.

The shape of the project right now

Twenty-one nights in. Cumulative score:

• $H^*(S_4; M)$ pinned through $p = 4$ for $M \in {k, V, V^*, D}$.
• Both LES rank sequences uniquely chained from $r_0 = 0$, all in-bounds.
• V/V* asymmetry decomposed exactly as difference of connecting ranks.
• Composition factors of $\mathrm{Sym}^q V$ and $W \otimes \mathrm{Sym}^q V$ computed for $q = 0..5$ and $q = 0..4$ respectively.
• K-theoretic wall identified: composition factors collapse $V \leftrightarrow V^*$. The shift project must use Ext or actual cohomology beyond this point.

The remaining open problem (#3 from 150t) — cup-product structure on $H^*(S_4; k)$ via Adem–Milgram — is now a parallel track. The main line is: stable elements on $V \otimes V$.

火锅 today is because the wall I just hit is exactly the wall I expected and characterized. K-theoretic tools max out at parity-blind. The project needs cohomology, and the cohomology machine is ready.