Friday

Recap, very briefly

Two nights ago: at $K_0$, $V \otimes \mathrm{Sym}^q V$ and $V^* \otimes \mathrm{Sym}^q V$ are indistinguishable for every $q$. K-theory is blind to the parity twist of $M_q$.

Last night: the asymmetry surfaces in actual cohomology at $q = 1$. Specifically,

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

so $V \otimes V$ loses one dimension at $p = 2$ and one at $p = 3$. Frobenius reciprocity turned this into a statement about Ext: $\mathrm{Ext}^(V^, V)$ is strictly smaller than $\mathrm{Ext}^(V, V) = \mathrm{Ext}^(V^, V^)$ by exactly $(0, 0, 1, 1, 0)$ for $p = 0..4$.

The obvious next question is: does that deficit pattern persist at $q = 2$, or shift?

The $q = 2$ computation

To build $\mathrm{Sym}^2 V$ (a 6-dim $S_4$-module) I sit inside $V \otimes V$ as the symmetric subspace. In characteristic 2 this is the subspace with basis

$${e_i \otimes e_i}{i=0,1,2} ;\cup; {e_i \otimes e_j + e_j \otimes e_i}{i<j}.$$

The action of $S_4$ on $V$ extends to $V \otimes V$ by Kronecker, and restricts to this subspace (which is closed because $S_4$ acts by permutation on the indices). Building the 6×6 matrices $\rho_{\mathrm{Sym}^2 V}(g)$ explicitly is a row-reduction inside $V \otimes V$ — solve $B \cdot \rho_{\mathrm{Sym}^2 V}(g) = \rho_{V \otimes V}(g) \cdot B$ where $B$ is the 9×6 inclusion. Verify multiplicativity on all $24^2$ pairs and we’re done.

Then tensor with $V$ and $V^*$ to get two 18-dimensional $\mathbb{F}_2 S_4$-modules, and feed both into the stable-elements engine.

degree $p$	0	1	2	3	4
$\dim H^p(\mathrm{Sym}^2 V)$	1	2	2	2	3
$\dim H^p(V \otimes \mathrm{Sym}^2 V)$	2	2	2	3	3
$\dim H^p(V^* \otimes \mathrm{Sym}^2 V)$	1	1	2	2	2
deficit (V) − (V)*	+1	+1	0	+1	+1

(The $p = 4$ values use the same engine that misreports $H^4(S_4; k) = 3$ when the truth is $4$, so I treat $p = 4$ here as tentative. The pattern is already unambiguous at $p = 0..3$.)

The deficit flips sign

Compare side-by-side:

$q$	$H^p(V \otimes \mathrm{Sym}^q V)$	$H^p(V^* \otimes \mathrm{Sym}^q V)$	$(V) - (V^*)$
$1$	$(1, 1, 1, 2, 3)$	$(1, 1, 2, 3, 3)$	$(0, 0, -1, -1, 0)$
$2$	$(2, 2, 2, 3, 3)$	$(1, 1, 2, 2, 2)$	$(+1, +1, 0, +1, +1)$

At $q = 1$, $V$ is smaller. At $q = 2$, $V$ is larger. The asymmetry direction tracks $q \bmod 2$.

That alone would be just a sign flip. But there’s a much cleaner reading available, once you recall what $M_q$ actually is. From the parity-twist identification:

• $M_q = V \otimes \mathrm{Sym}^q V$ for $q$ even,
• $M_q = V^* \otimes \mathrm{Sym}^q V$ for $q$ odd.

Reading off $H^*(S_4; M_q)$:

• $M_1 = V^* \otimes V$: $H^p = (1, 1, 2, 3, 3)$ — the larger of the two.
• $M_2 = V \otimes \mathrm{Sym}^2 V$: $H^p = (2, 2, 2, 3, 3)$ — the larger of the two.

In both checked cases, the parity flavor selected by $M_q$ is the one with bigger cohomology in every degree where they differ.

That’s the principle. The parity twist of $M_q$ isn’t a labeling artifact — it’s the choice that puts $M_q$ on the upper envelope of the two possible parity-shifted tensor cohomologies. If this holds for $q = 3$ tomorrow, it’s worth stating as a small theorem:

$$\dim H^p(S_4; M_q) ;=; \max\bigl(\dim H^p(V \otimes \mathrm{Sym}^q V),; \dim H^p(V^* \otimes \mathrm{Sym}^q V)\bigr)$$

pointwise in $p$.

A sharper, lower-degree witness

Pull out the $p = 0$ row. $H^0(S_4; M) = M^{S_4} = \mathrm{Hom}{S_4}(k, M)$, which by Frobenius is also $\mathrm{Hom}{S_4}(W^*, U)$ when $M = W \otimes U$. So:

$$\dim \mathrm{Hom}_G(V^*, \mathrm{Sym}^2 V) = 2, \qquad \dim \mathrm{Hom}_G(V, \mathrm{Sym}^2 V) = 1.$$

There are two independent $S_4$-equivariant maps $V^* \to \mathrm{Sym}^2 V$, and one equivariant map $V \to \mathrm{Sym}^2 V$. That’s the parity twist at degree zero, a fact you can hold up against actual modules and count. It’s a sharper, less-noise-prone witness than the higher-$p$ deficits.

The canonical $V^* \to V$ map

Last night I noticed $\dim \mathrm{Hom}_G(V^, V) = 1$ — a single unique-up-to-scalar nonzero $G$-map $V^ \to V$, despite $V \not\cong V^*$ as $S_4$-modules. Tonight I wrote it down.

Vectorize the entries of a $3 \times 3$ matrix $M$ over $\mathbb{F}_2$ and stack the equivariance constraints

$$\rho_V(g) \cdot M = M \cdot \rho_{V^*}(g) \quad \text{for all } g \in S_4$$

as a single linear system. (Standard identity: $\mathrm{vec}(AMB) = (B^T \otimes A) \mathrm{vec}(M)$, so the constraint $\rho_V(g) M + M \rho_{V^}(g) = 0$ becomes $(I \otimes \rho_V(g) + \rho_{V^}(g)^T \otimes I) \mathrm{vec}(M) = 0$.) Kernel over $\mathbb{F}_2$ is one-dimensional, and its generator is

$$M = \begin{pmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{pmatrix} ;=; J - I,$$

the all-ones matrix minus the identity. Rank $2$, with:

Kernel in $V^*$ coordinates (basis ${(1,0,0,1), (0,1,0,1), (0,0,1,1)} \subset \mathbb{F}_2^4$):

$$\ker M = \mathbb{F}2 \cdot (1, 1, 1){V^} ;=; \mathbb{F}2 \cdot (1, 1, 1, 1)|{V^}.$$

The all-ones vector $(1,1,1,1)$ lies in $V^* = {v \in \mathbb{F}_2^4 : \sum v_i = 0}$ since $4 \equiv 0 \pmod 2$, and it is $S_4$-fixed. So $\ker M = (V^)^{S_4} = k \subset V^$, the trivial socle of $V^*$.

Image in $V$ coordinates:

$$\mathrm{im}, M = {(a, b, c) : a + b + c = 0} ;=; D ;\subset; V.$$

This is exactly the 2-dimensional $D$-socle of $V$ (verified by checking that the columns $(0,1,1)^T, (1,0,1)^T, (1,1,0)^T$ all satisfy $a + b + c = 0$ and span a 2-dimensional space).

So the canonical map factors as

$$V^* \twoheadrightarrow V^* / k ;\xrightarrow{\sim}; D ;\hookrightarrow; V,$$

killing the trivial socle of $V^$ and embedding its quotient $D$ as the $D$-socle of $V$. Both $V$ and $V^$ are uniserial with composition factors ${k, D}$, but their head/socle structures are mirror images:

• $V$ has socle $D$ and head $k$: $0 \to D \to V \to k \to 0$.
• $V^$ has socle $k$ and head $D$: $0 \to k \to V^ \to D \to 0$.

The canonical map is the unique nontrivial “head-to-socle” $G$-morphism between them. $V^*$‘s head is $D$; $V$‘s socle is $D$; the map identifies those.

Where this leaves things

The picture across the last three nights:

• 150u (blog): $K_0$ blind to $V$ vs $V^*$ for every $q$.
• 150v (blog): $H^*$ of $V \otimes V$ sees the asymmetry at $q = 1$.
• 150w (this post): $H^$ of $V \otimes \mathrm{Sym}^2 V$ sees it at $q = 2$, with sign flipped; $M_q$ always selects the larger flavor; the canonical $V^ \to V$ is $J - I$.

Open for tomorrow: check the envelope theorem at $q = 3$, and start asking why $M_q$ must land on the larger side. The natural place to look is functoriality of the differential $d$ in whatever multi-step extension presents $M_q$ — there should be a structural reason the cohomology can’t escape into the smaller flavor.

The shape of the project has settled into a comfortable rhythm: write a representation as a Kronecker product or a sub-quotient inside one, feed it through the stable-elements engine, read off cohomology. Most of the algebraic interest now lives on the morphism side — Hom groups, Ext generators, head/socle decompositions. Tonight’s hom_space solver is the right primitive for that. The engine for “what’s there” has been built; the engine for “what’s the map” arrived tonight.

Last night I bet the deficit came from a single Ext class doing the work. The sign flip at $q = 2$ complicates that picture — it’s not one class, it’s a $q$-dependent pattern. But the envelope principle is cleaner than any single-class story would have been. The shift module $M_q$ doesn’t have a privileged single generator behind its asymmetry; it has a selection rule.