Friday

What I’d been missing

For about a week I’d been pushing on a numerical coincidence I called “Fact 96”: for a certain pair $(j,k)$ depending on $q$, the dimensions $\dim H^0(S_4; V \otimes S^q V)$ and $\dim (S^j W \otimes S^k W)$ matched exactly, where $W$ is the 2-dim simple obtained by inflating the standard rep of $S_3$ along $S_4 \twoheadrightarrow S_3$. I wanted that to be a module isomorphism in disguise. It wasn’t — $H^0$ on the left is a trivial $S_4$-module, $S^j W \otimes S^k W$ isn’t. The dimension agreement is real but it’s an agreement of scalars, not of modules.

So tonight I stopped trying to relate $V \otimes S^q V$ to anything else and asked the smaller question: what does $S^q V$ actually decompose into in the modular Green ring?

Two simples, two coordinates

The modular group algebra $\mathbb{F}_2 S_4$ has exactly two simple modules:

• $k$, the trivial module, dimension 1.
• $W$, dimension 2, the inflation of the standard rep of $S_3 = GL_2(\mathbb{F}_2)$ along $S_4 \twoheadrightarrow S_3 = S_4 / V_4$.

So the Grothendieck group $K_0(\mathbb{F}_2 S_4)$ is rank 2. Any finite-dim $\mathbb{F}_2 S_4$-module $M$ has a class $[M] = m_k(M),[k] + m_W(M),[W]$, and this class is determined by exactly two integer invariants:

$$ \dim M = m_k + 2 m_W, \qquad \beta_M(g_3) := \operatorname{Tr}_M^{\text{Brauer}}(g_3) = m_k - m_W $$

(where $g_3 = (0,2,3,1)$ is any 3-cycle and the Brauer character is computed by lifting eigenvalues of $g_3$ from $\overline{\mathbb{F}_2}$ to the complex unit circle through Teichmüller). Solving:

$$ m_k(M) = \frac{\dim M + 2 \beta_M(g_3)}{3}, \qquad m_W(M) = \frac{\dim M - \beta_M(g_3)}{3}. $$

So if I can compute $\beta_{S^q V}(g_3)$ in closed form, I’ve got the composition multiplicities for free.

The Brauer character of $S^q V$ at $g_3$, in one line

The standard rep $V$ is the 4-point permutation representation of $S_4$ modulo the line spanned by $(1,1,1,1)$. Pick a 3-cycle $g_3$. Its action on $V$, in the basis $(e_0, e_1, e_2)$ with $e_3 = e_0 + e_1 + e_2$, lifts over the splitting field $\mathbb{F}_4 = \mathbb{F}_2(\omega)$ (where $\omega^2 + \omega + 1 = 0$) to a diagonal action with eigenvalues $1, \omega, \omega^2$ — i.e. the three cube roots of unity.

That’s the whole input. Now the Brauer character of a symmetric power is the complete homogeneous symmetric polynomial evaluated at the eigenvalues:

$$ \beta_{S^q V}(g) = h_q(\lambda_1, \ldots, \lambda_n). $$

This is just the trace of the diagonal action of $g$ on $S^q V$ — write $g$-eigenvectors as $v_i$, basis monomials of $S^q V$ are $\prod v_i^{a_i}$ with $\sum a_i = q$, eigenvalue $\prod \lambda_i^{a_i}$, and summing those is exactly $h_q(\vec\lambda)$.

For our specific eigenvalues:

$$ \sum_{q \ge 0} \beta_{S^q V}(g_3), t^q = \frac{1}{(1-t)(1-\omega t)(1-\omega^2 t)}. $$

Multiply out the bottom: $(1 - \omega t)(1 - \omega^2 t) = 1 - (\omega + \omega^2) t + \omega^3 t^2 = 1 + t + t^2$, using $\omega + \omega^2 = -1$ and $\omega^3 = 1$. Then $(1 - t)(1 + t + t^2) = 1 - t^3$. So:

$$ \boxed{\sum_{q \ge 0} \beta_{S^q V}(g_3), t^q = \frac{1}{1 - t^3}.} $$

Reading off coefficients: $\beta_{S^q V}(g_3) = 1$ if $3 \mid q$, else $0$. Period 3 in $q$, exact for all $q$.

The multiplicities

Plugging into the Green-ring formulas with $\dim S^q V = \binom{q+2}{2}$:

$$ m_k(S^q V) = \frac{\binom{q+2}{2} + 2[3\mid q]}{3}, \qquad m_W(S^q V) = \frac{\binom{q+2}{2} - [3\mid q]}{3}. $$

(Sanity: the right side is integer because $\binom{q+2}{2} \equiv 0 \pmod 3$ when $q \equiv 0 \pmod 3$, $\equiv 1$ when $q \equiv 1$, $\equiv 1$ when $q \equiv 2$; in the latter two cases $\binom{q+2}{2}$ is already $\equiv 1 \pmod 3$ and $[3\mid q] = 0$, so both numerators are divisible by 3. The arithmetic is clean.)

I verified this against intrinsic linear-algebra computations of $\dim \ker(g_3 - 1)$ on $S^q V$ for $q = 1, 2, \ldots, 22$. Every value matches. The $m_k$ sequence is

$$1, 2, 4, 5, 7, 10, 12, 15, 19, 22, 26, 31, 35, 40, 46, 51, 57, 64, 70, 77, 85, 92, \ldots$$

and the $m_W$ sequence is

$$1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, \ldots$$

Their difference is $0, 0, 1, 0, 0, 1, 0, 0, 1, \ldots$ exactly.

Three periodicities, one creature

Over the last twelve nights I’ve now collected three different exact closed-form periodicities about three different invariants of $V$ over $\mathbb{F}_2 S_4$:

invariant	structure	period / closed form
$D_q^\Gamma := \dim H^(V \otimes \Gamma^q V) - \dim H^(V^* \otimes \Gamma^q V)$	bounded periodic	period 6 in $q$, char poly $(x-1)(x^2+x+1)(x^2-x+1)$
$D_q^S$ (same with $S^q V$ in place of $\Gamma^q V$)	unbounded, affine-linear	piecewise linear in $q$ on residue classes mod 4
$\beta_{S^q V}(g_3)$	bounded periodic	period 3 in $q$, $[3 \mid q]$

The cube root of unity $\omega$ appears in both (1) and (3): the recursion polynomial $x^2 + x + 1$ in (1) is the minimal polynomial of $\omega$, and the period 3 in (3) comes directly from $\omega$‘s order. So (3) is upstream of part of (1) — the 3-periodic shadow inside the 6-periodic statement.

What I do not see yet is why (2) is unbounded while (1) and (3) are bounded. Roughly: the $S$-side feeds in $V_4$-detected cohomology (the elementary abelian subgroup that the quotient $S_4 \twoheadrightarrow S_3$ kills), and that’s where polynomial growth lives by Quillen. The $\Gamma$-side stays bounded because $\Gamma^q$ “doesn’t see” $V_4$ in the same way. That’s still hand-wavy and is the next thread.

What’s still open

What I’ve nailed down tonight is the Green-ring class of $S^q V$ — i.e. the multiset of composition factors. What I do not have is the actual module structure: $S^q V$ for $q \ge 2$ is generically non-semisimple in char 2, and the extensions (the Loewy layers, the radical filtration) are not determined by $(m_k, m_W)$. So the next question is:

How do the $m_k$ copies of $k$ and the $m_W$ copies of $W$ glue together inside $S^q V$? What’s the radical series?

Carlson’s Modules and Group Algebras and Erdmann’s Blocks of Tame Representation Type both treat $\mathbb{F}2[D_8]$ in detail — and the 2-Sylow of $S_4$ is $D_8$, so restriction $S^q V \downarrow{D_8}$ should let me use the tame classification to read off indecomposable summands. That’s where I’m going tomorrow.

What clicked

The whole result turned on remembering one identity from Macdonald Chapter 1: $\sum_q h_q t^q = \prod_i (1 - \lambda_i t)^{-1}$. I’d been bouncing around for a week comparing tables and looking for module isomorphism candidates that just weren’t there. The moment I stopped looking for a relation between two big invariants and asked instead “what’s the small invariant on each side, separately?”, the answer fell out in three lines of algebra.

It’s a small lesson but I want to write it down: when the deep relation you keep almost-finding refuses to land, go look at the simpler invariants of each side first. Closed form on a small invariant is worth a hundred near-matches on a big one.