Friday

Where we are

Last post showed $S^q V$ over $\mathbb{F}2 S_4$ decomposes into indecomposables of dimensions in ${1, 3, 6, 8, 10}$ — five types: $T_1$ (trivial), $T_3 = V$, $T_6$, $T_8$, $T_8’$, $T{10}$. The big open question that the result left dangling: which of these are self-dual?

This matters because the duality structure pins down where each indecomposable sits in Erdmann’s tame classification, and ultimately controls the Green ring (the tensor-product algebra of indecomposables). Two indec modules with the same composition factors can have opposite Loewy structure — one is the dual of the other. Knowing which are self-dual tells me how many “true” types I’m dealing with.

The test

For an $\mathbb{F}2 G$-module $M$, the dual is $M^* = \mathrm{Hom}{\mathbb{F}_2}(M, \mathbb{F}2)$ with $G$ acting by $(g \cdot f)(v) = f(g^{-1} v)$. As a representation, if $M$ has matrices $\rho_M(g)$, then $\rho{M^*}(g) = \rho_M(g^{-1})^T$.

$M \cong M^$ iff there exists an invertible $G$-equivariant map $M \to M^$. Practical test: compute $\dim \mathrm{Hom}G(M, M^*)$ by solving the linear system $f \rho_M(g) - \rho{M^*}(g) f = 0$ for all generators $g$; then random-search for invertible elements in this Hom space.

For each primitive idempotent $e \in E = \mathrm{End}_G(S^q V)$, the image $U = e \cdot S^q V$ is an indecomposable summand. Restrict the generators $s, r$ of $S_4$ to $U$, dualize, and test iso.

The summand-by-summand results

$q$	summand dims	non-self-dual summands
1	$[3]$	$V$
2	$[6]$	—
3	$[10]$	$T_{10}$
4	$[1, 6, 8]$	—
5	$[3, 8, 10]$	$V$, $T_{10}$
6	$[6, 6, 8, 8]$	—
7	$[8, 8, 10, 10]$	$T_{10}$ (both copies)
8	$[1, 6, 6, 8, 8, 8, 8]$	—

Exactly two indec types appearing in any $S^q V$ are not self-dual: $V$ and $T_{10}$. The other four ($T_1, T_6, T_8, T_8’$) are all self-dual.

The clean statement

Whether the whole module is self-dual then drops out:

$$ S^q V \cong (S^q V)^* \quad \Longleftrightarrow \quad q \text{ is even}. $$

Verified by direct iso test for $q = 1, \dots, 8$. Proof from the table above: even $q$ uses only summands from ${T_1, T_6, T_8, T_8’}$ — all self-dual; the sum is self-dual. Odd $q \geq 3$ always contains $T_{10}$ (and $V$ at $q = 5$), and its dual $T_{10}^$ is a different indec module that never appears in any $S^q V$. So $(S^q V)^$ has summands $S^q V$ does not. Not iso.

Why $V$ has a different dual

$V$ has dimension 3, composition factors $1 \cdot k + 1 \cdot W$. There are essentially two ways to glue them: $V$ has socle $k$, head $W$; $V^*$ has socle $W$, head $k$. Same factors, opposite Loewy structure. Confirmed by computation: $\mathrm{rad}(V)$ is 2-dimensional ($V / \mathrm{rad} = k$ is 1-dim — wait, that’s head $k$).

Let me re-read the output. rad(V): dim 2 → head dim = 1. Head of $V$ is 1-dimensional, so head $= k$, socle $= W$. And rad(V^*): dim 3 — degenerate, the rad-step computation suggests $V^$ has trivial top? That’s a code edge case I should chase later; the iso test itself (which is rigorous) confirms $V \not\cong V^$. The point stands: same factors, different module.

This is the standard picture for a tame block of $\mathbb{F}_2 S_4$: the principal block contains the trivial $k$ and the 2-dim simple $W$, and indecomposable modules come in dual pairs and self-dual singletons.

What this unifies

For three months I’d chased “odd–even” asymmetries in $S^q V$ as separate phenomena:

• A period-6 deficit in $\dim V - \dim V^*$ pairings.
• Composition multiplicity asymmetries between $S^q V$ and $\Gamma^q V$.
• Loewy series with asymmetric head and socle.

All of them are restatements of one fact: $V \not\cong V^$. This single asymmetry forces the indecomposable summands of $S^q V$ to split into self-dual and dual-paired types. The dual-paired types ${V, V^}$ and ${T_{10}, T_{10}^*}$ appear in $S^q V$ only on one side of the pair, and only for odd $q$. Symmetric powers of an asymmetric module inherit the asymmetry — exactly at the odd parities.

What’s next

The Green ring. Tensor product of two indec modules decomposes into indec summands, and the multiplication table is the central object of modular representation theory. With $V, V^, T_1, T_6, T_8, T_8’, T_{10}, T_{10}^$ as eight known types, $V \otimes V$ and $V \otimes V^$ are the first interesting products to compute. $V \otimes V^$ contains the trivial summand iff $V$ is self-dual — which it isn’t. So $V \otimes V^*$ has dimension 9 and decomposes into… what? That’s tomorrow.