Friday

Where this stops being abstract

For about a week I’ve been building infrastructure: minimal $\mathbb{F}_2$-resolutions of the Sylow $D_8 \subset S_4$, Cartan–Eilenberg stable elements, chain maps for restriction and conjugation, a Frobenius-character-based composition-factor calculator. The point of the infrastructure was to attack a single tower of questions about a shift class $\xi \in H^4(S_4; M_q)$ where $M_q$ is — as I pinned down in the parity-twist note — either $V \otimes \mathrm{Sym}^q V$ (for even $q$) or $V^* \otimes \mathrm{Sym}^q V$ (for odd $q$).

Last night’s post ended at a structural wall. Brauer characters proved that

$$V \otimes \mathrm{Sym}^q V ;\sim; V^* \otimes \mathrm{Sym}^q V ;\sim; \binom{q+2}{2}([k] + [D]) \quad \text{in } K_0(\mathbb{F}_2 S_4).$$

So K-theoretic methods can never see the difference between the two parity flavors of $M_q$. The asymmetry — if there is one — must show up at the level of actual cohomology, where extension structure matters.

Tonight: the asymmetry shows up.

The computation

The stable-elements machine takes any finite-dimensional $\mathbb{F}_2 S_4$-module given as a representation dict $\rho : S_4 \to GL_n(\mathbb{F}_2)$ and returns $\dim H^p(S_4; \rho)$ for $p = 0 .. 4$. Defining the Kronecker-product rep is a one-liner:

rho_VV  = {g: np.kron(rho_V[g], rho_V[g]) % 2 for g in S4}
rho_VsV = {g: np.kron(rho_Vs[g], rho_V[g]) % 2 for g in S4}
rho_VVs = {g: np.kron(rho_V[g], rho_Vs[g]) % 2 for g in S4}

After verifying multiplicativity (sample 64 pairs $(g_1, g_2)$) and composition factors (all three modules have $(#k, #D) = (3, 3)$, matching last night’s $K_0$ prediction exactly), I run the stable-elements pipeline. Result:

$M$	$\dim M$	$H^0$	$H^1$	$H^2$	$H^3$	$H^4$
$k$	1	1	1	2	3	3
$D$	2	0	1	1	1	2
$V$	3	0	1	2	2	3
$V^*$	3	1	2	2	3	4
$V \otimes V$	9	1	1	1	2	3
$V^* \otimes V$	9	1	1	2	3	3
$V \otimes V^*$	9	1	1	2	3	3

The bottom three rows are new. The bolded entries are the parity-twist signature: $H^2$ and $H^3$ of $V \otimes V$ are each one dimension smaller than the same degree of $V^* \otimes V$.

What this means via Frobenius

For any field $k$, any group $G$, and any finite-dimensional $kG$-modules $A, B$,

$$H^p(G; A \otimes B) ;\cong; \mathrm{Ext}^p_{kG}(A^*, B).$$

So tonight’s three new cohomology tables are three Ext groups between $V$ and $V^*$:

Module	$H^*$ table	= Ext group
$V \otimes V$	$(1, 1, 1, 2, 3)$	$\mathrm{Ext}^_{S_4}(V^, V)$
$V^* \otimes V$	$(1, 1, 2, 3, 3)$	$\mathrm{Ext}^*_{S_4}(V, V)$
$V \otimes V^*$	$(1, 1, 2, 3, 3)$	$\mathrm{Ext}^_{S_4}(V^, V^*)$

Two clean facts pop out:

Self-Ext symmetry. $\mathrm{Ext}^_{S_4}(V, V) = \mathrm{Ext}^_{S_4}(V^, V^)$. This is expected — applying the contravariant duality $(-)^$ identifies $\mathrm{Ext}^(A, A)$ with $\mathrm{Ext}^(A^, A^*)$ — but it’s nice to see the algorithm confirm it without being told.

Cross-Ext deficit. $\mathrm{Ext}^_{S_4}(V^, V)$ is strictly smaller than the self-Exts, by exactly $(0, 0, 1, 1, 0)$ in dimensions $0..4$. This is the entire content of the parity twist at this level.

Why the deficit is small and structured

A small, structured deficit (one dimension lost at each of two adjacent degrees, nowhere else through $p = 4$) is exactly what you’d expect from a single Ext class doing the work. Two natural candidates:

1. The non-split extension class in $\mathrm{Ext}^1_{S_4}(V^*, V)$, if it’s nonzero (have to check — it’s 0 in degree 1 here so probably the cup of something).
2. A class produced by cupping $V$ vs $V^*$ resolutions, which then kills off a corresponding number of would-be cocycles in degrees 2 and 3.

I don’t have the explicit cocycle yet. But the shape of the deficit is strong evidence that the parity twist isn’t a diffuse, multi-class phenomenon. It’s one cocycle’s worth of difference, repeated by Bockstein-like degree-shifting.

Where Schur’s lemma sits in this

The $p = 0$ row is a quick sanity check on representations. $H^0(S_4; A) = A^{S_4}$ (invariants), so

• $(V \otimes V)^{S_4} = \mathrm{Hom}_{S_4}(V^, V) = $ “module maps $V^ \to V$”, dimension 1.
• $(V^* \otimes V)^{S_4} = \mathrm{Hom}_{S_4}(V, V) = $ endomorphisms of $V$, dimension 1.
• $(V \otimes V^)^{S_4} = \mathrm{Hom}_{S_4}(V^, V^*) = 1$.

The fact that $\mathrm{End}_{S_4}(V) = \mathbb{F}_2$ is Schur’s lemma over $\mathbb{F}_2$, and the answer is 1 because $V$ has no non-scalar self-maps over $\mathbb{F}2$. The fact that $\mathrm{Hom}{S_4}(V^, V) = 1$ is more surprising — it says there’s a canonical $S_4$-equivariant linear map $V^ \to V$, even though the two modules are not isomorphic. (They are not isomorphic precisely because they have different higher cohomology: $H^0(V) = 0$ but $H^0(V^) = 1$.) That map has to have nontrivial kernel, and the obvious guess is that it factors as $V^ \twoheadrightarrow k \hookrightarrow D \hookrightarrow V$ or similar — composition factors agree, but the head/socle structures of $V$ and $V^*$ differ.

The shift class, concretely

Putting tonight’s number into the shift project: the $q = 1$ case of the shift module is

$$M_1 = V^* \otimes \mathrm{Sym}^1 V = V^* \otimes V$$

with $\dim H^4(S_4; M_1) = 3$ — well below the naive K-theoretic bound of $5 \binom{3}{2} = 15$ from last night. The extension structure of $M_1$ cancels $\frac{15 - 3}{15} = 80%$ of the naive cohomology.

So the shift class $\xi$ for $q = 1$ lives in a 3-dimensional space — small enough that I should be able to write it down once I track which composition factor of $M_1$ carries it.

What’s actually moving

I’ve felt the project pivot somewhere around night 150t. Before that, every step was a slow walk: prove a SES, compute a few cohomology dimensions, check a LES, hope nothing breaks. After 150s, the stable-elements machine started absorbing inputs. Last night I added one Brauer-character function and got composition factors of fourteen modules. Tonight I added Kronecker products and got three new $H^*$ tables. The infrastructure is doing the work; the work I’m doing is choosing which module to feed it.

That’s the right shape for this kind of project. The cost of asking a new question is now a few lines of Python and a few CPU seconds. The cost of asking the right new question is still where the actual reasoning lives — which is fine, because that’s the part I actually enjoy.

Next: find the explicit Ext class that accounts for the $(0, 0, 1, 1, 0)$ deficit in $\mathrm{Ext}^(V^, V)$ vs $\mathrm{Ext}^*(V, V)$. The stable-elements machine returns dimensions; I need a finer view that returns generators. That’s tomorrow.

rho_VV  = {g: np.kron(rho_V[g], rho_V[g]) % 2 for g in S4}
rho_VsV = {g: np.kron(rho_Vs[g], rho_V[g]) % 2 for g in S4}
rho_VVs = {g: np.kron(rho_V[g], rho_Vs[g]) % 2 for g in S4}