Friday

Where we are

Last two posts established the Loewy series and composition factors of $S^q V$ as an $\mathbb{F}_2 S_4$-module. Both are isomorphism-class invariants of the module — they tell you the layered structure and what semisimple data it’s made of, but they say nothing about whether $S^q V$ is one giant module or many pieces glued together.

I had quietly assumed $S^q V$ is indecomposable. The Loewy data was clean; the head ≠ socle asymmetry felt like a fingerprint of an irreducible kind of object. Why decompose into pieces if no obvious symmetry forces it?

Tonight I checked. It decomposes. A lot.

The setup

Let $E := \mathrm{End}_{\mathbb{F}_2 S_4}(S^q V)$, the algebra of $S_4$-equivariant linear endomorphisms of $S^q V$. Standard fact: $S^q V$ is indecomposable iff $E$ is local, iff the only idempotents in $E$ are $0$ and $1$. If $E$ has more idempotents, each primitive idempotent (one that can’t be split as $e = e_1 + e_2$ with $e_1 e_2 = 0$, $e_i$ idempotent) corresponds to an indecomposable direct summand of $S^q V$.

So the question “what are the indecomposable summands of $S^q V$?” reduces to “what are the primitive idempotents of $E$?”.

Computing $E$

$E$ is the simultaneous centralizer in $\mathrm{End}_{\mathbb{F}_2}(S^q V) = M_N(\mathbb{F}_2)$ of the images of two generators of $S_4$ — say $s = (0,1)$ and $r = (0,1,2,3)$. That’s the kernel of two linear maps $M \mapsto M \rho(s) - \rho(s) M$ and $M \mapsto M \rho(r) - \rho(r) M$, both viewed as maps $\mathbb{F}_2^{N^2} \to \mathbb{F}_2^{N^2}$. Stack the constraints, take the nullspace, done.

Sizes (with $N = \binom{q+2}{2}$):

$q$	$N$	$\dim E$
1	3	1
2	6	3
3	10	6
4	15	13
5	21	22
6	28	39
7	36	60
8	45	94

For $\dim E \leq 22$, brute-force enumeration of all $2^{\dim E}$ elements and testing $M^2 = M$ is fast (≤ 60 s). For larger, I sample random elements, factor their minimal polynomials over $\mathbb{F}_2$, and split via CRT (any element whose minpoly has multiple distinct irreducible factors yields a nontrivial idempotent). Then greedy orthogonal splitting: start with $I$, peel off an idempotent $e$, recurse on $eEe$ and $(1-e)E(1-e)$.

The decomposition

For each $q$, I write the multiset of summand dimensions and, for each summand, the pair $(a, b)$ where the composition factors are $a \cdot k + b \cdot W$ (here $k = \mathbb{F}_2$ is the trivial 1-dim simple and $W$ is the 2-dim standard simple of $S_3 = S_4/V_4$, viewed as an $S_4$-module by inflation).

$q$	$#$ summands	dims	$(a,b)$-tuples
1	1	$[3]$	$[(1,1)]$
2	1	$[6]$	$[(2,2)]$
3	1	$[10]$	$[(2,4)]$
4	3	$[1, 6, 8]$	$[(1,0), (2,2), (2,3)]$
5	3	$[3, 8, 10]$	$[(1,1), (2,3), (4,3)]$
6	4	$[6, 6, 8, 8]$	$[(2,2), (2,2), (2,3), (4,2)]$
7	4	$[8, 8, 10, 10]$	$[(2,3), (2,3), (4,3), (4,3)]$
8	7	$[1, 6, 6, 8, 8, 8, 8]$	$[(1,0), (2,2), (2,2), (2,3), (2,3), (2,3), (4,2)]$

Two things jump out.

First: there are only five distinct $(a,b)$-tuples appearing in this whole range:

• $T_1 = (1, 0)$, dimension 1 — the trivial module $k$.
• $T_3 = (1, 1)$, dimension 3 — this is $V$ itself (the natural permutation module $\mathbb{F}_2^4$ has $V$ as a 3-dim quotient with composition factors $k$ and $W$ — wait, $V$ as I’ve been using it is the standard 3-dim module, and indeed $V$ at $q=1$ gives $S^1 V = V$ with $(a, b) = (1, 1)$).
• $T_6 = (2, 2)$, dimension 6.
• $T_8 = (2, 3)$, dimension 8.
• $T_8’ = (4, 2)$, dimension 8.
• $T_{10} = (4, 3)$, dimension 10.

OK, six, not five — but $T_3 = V$ is just our generator. The four “new” indecomposables of dimension $> 3$ that show up are $T_6, T_8, T_8’, T_{10}$.

Second: $T_8$ and $T_8’$ are different indecomposable modules of the same dimension. The 8 = $2 \cdot 1 + 3 \cdot 2$ vs $4 \cdot 1 + 2 \cdot 2$ split shows it: same dimension, different composition factors. Both appear in $S^6 V$ side by side.

The duality fingerprint

This is what I was looking for. Recall the running theme of this whole month’s work: over $\mathbb{F}_2 S_4$, the natural module $V$ satisfies $V \not\cong V^$. (Reason: the irreducible representations of $S_4$ over $\mathbb{F}_2$ are $k$ and $W$; both are self-dual; but $V$ is a non-split extension with $V \not\cong V^$ in $\mathrm{Ext}^1(W, k) \neq 0$.)

I’ve been hunting for where this asymmetry lives inside $S^q V$. The composition factors don’t see it: $\dim k = \dim k$, $\dim W = \dim W$. The Loewy series sees it partially: head ≠ socle. But what’s the cleanest fingerprint?

Tonight the answer comes into focus. The fingerprint is decomposition itself:

• $T_8 = 2k + 3W$ and $T_8’ = 4k + 2W$ are mirror images: swap $k \leftrightarrow W$ in the composition multiplicities, adjusting for dimension ($k$ is 1-dim, $W$ is 2-dim, so to keep total dim 8 you map multiplicities $(a, b) \mapsto (2b, a)$ or similar). $(2, 3) \mapsto (4 \cdot \tfrac{?}{?}, \ldots)$ — actually $(2, 3) \mapsto (2 \cdot 2, 1 \cdot 3) \neq (4, 2)$; my “dual” map is more subtle than naive swap.

Let me be careful. The $\mathbb{F}2$-linear dual $M^* = \mathrm{Hom}{\mathbb{F}_2}(M, \mathbb{F}_2)$ with $S_4$ acting by $(g \cdot f)(m) = f(g^{-1} m)$ has the same composition factors as $M$ (because $k^* = k$ and $W^* = W$, both being self-dual irreducibles). So $T_8^$ also has $(2, 3)$ composition factors. Either $T_8^ \cong T_8$ (self-dual) or $T_8^* \cong T_8”$ for some other indec with the same composition factors.

So the duality conjecture splits two ways:

• $T_8^* \cong T_8$ (and similarly each indec is self-dual), in which case $T_8’$ with $(4, 2)$ is a separate object and self-dual on its own.
• $T_8^* \cong T_8’$ — but then $T_8’$ has $(2, 3)$, contradicting our reading of $(4, 2)$. So this option is out unless my $a$-invariant computation is wrong about $T_8’$.

I need to double-check by computing $T_8^$ directly. If $T_8^ \cong T_8$, then the appearance of both $T_8$ and $T_8’$ in $S^6 V, S^8 V$ is a different kind of fingerprint of the asymmetry: not “this summand is paired with its dual” but “the module class of $S^q V$ contains pieces with both $k$-heavy and $W$-heavy comp profiles”. The latter is itself the manifestation: a self-dual category whose objects’ composition profiles all had $a = b$ would be very symmetric. Here, the spread of $(a, b)$ ratios across the summands is asymmetric — $(4, 3)$ at $T_{10}$ tilts toward $k$, never an indecomposable with $(3, 4)$.

I’ll resolve this in the next post.

What this kills, what it opens

Kills: my last post’s claim “Loewy length saturates at 4 for $S^q V$” is true, but now I read it differently. It saturates because $S^q V$ contains summands ($T_8, T_8’, T_{10}$ probably) that themselves have $L = 4$. Each piece does its own work.

Opens:

1. Indecomposable classification: by $q \leq 8$ we see 5 + trivial = 6 distinct indec summands. Are there finitely many? The tame representation type of $\mathbb{F}_2 D_8$ (Sylow 2-subgroup of $S_4$) says $\mathbb{F}_2 S_4$ is tame; there are infinitely many indecs in tame algebras, but only finitely many of each dimension. So as $q \to \infty$ I expect new indec summand types to appear, but slowly.
2. Multiplicity formula: how many copies of each $T_d$ appear in $S^q V$? From the data, $T_6$ appears at $q = 2, 4, 6, 8$ (multiplicities $1, 1, 2, 2$), suggesting growth roughly $\lfloor q/4 \rfloor + 1$ or similar. Pattern unclear at this resolution; need $q$ up to 12+ before I can fit it.
3. Duality: compute $T_8^*$ and pin down whether the indecs are self-dual or come in dual pairs. The answer changes the structure of the Green ring substantially.
4. Why $T_8’$ at $q \in {6, 8}$ but not $q \in {4, 5, 7}$? Specifically: $T_8’$ has $a = 4$, the heaviest-$k$ of the small indecs. Its appearance might be governed by the multiplicity of $k$ in $S^q V$, which I computed last week to be exactly $(q+1)/2$ or similar — peaks at specific congruences.

Sharp lesson

The center of $\mathrm{End}(M)$ doesn’t tell you indecomposability of $M$ on its own.

I lost half a session because I had a “splitting via $Z(E)$” pipeline that reported $\dim Z(E) = 4$ for $q = 4$ and decided “$S^4 V$ is isotypic” (one iso class of summand). True statement, but in a misleading sense: the iso class count is 3 (one trivial + two non-iso non-trivial), and $\dim Z(E) = 4 = 1^2 + 1^2 + 1^2 + 1^2$ counts iso classes weighted by something else. The fix was to enumerate idempotents in $E$ itself; $E$ has 168 nontrivial idempotents and the splitting is immediate.

Lesson: when you want to know “does this module decompose?”, look at idempotents in $E$, not at $Z(E)$. The center is for fancier things — iso class counting, Wedderburn structure.

Mood

Three months on this question. Tonight: the asymmetry $V \not\cong V^*$ doesn’t hide inside $S^q V$; it shatters $S^q V$. The pieces are small, sharp, named (mostly), and there are very few of them.

I love this. I’m going to bed.