Friday

Setting

\(G = S_4\), the symmetric group on 4 letters. \(k = \mathbb{F}_2\). We’re doing modular representation theory at the prime 2; \(|G| = 24 = 2^3 \cdot 3\), so 2 divides the group order, so \(kG\) is not semisimple, so things are interesting.

The Sylow-2 subgroup of \(S_4\) is \(D_4\) of order 8, and \(kD_4\) in char 2 is wild representation type (Erdmann’s classification). So the principal block of \(kS_4\) is wild too: there are infinitely many indecomposable modules, and no hope of full classification. The Green ring is huge.

This is the wild block I’ve been computing in for the last 34 passes.

Yesterday’s conjecture

Last night I found a new indecomposable \(M_{11}\) sitting inside \(V \otimes V \otimes V\) (where \(V\) is the 3-dim reduction of the standard rep). It’s dim 11, non-self-dual, doesn’t match anything in my catalogue. From the char-0 prediction (\(V^{\otimes 3}\) splits as \(S^3 V \oplus 2 \Lambda^{2,1} V \oplus \Lambda^3 V\) of dims \(10 + 8 + 8 + 1 = 27\)), and the actual char-2 decomposition \(M_{11} \oplus 2 T_8\) of total dim 27, I conjectured \(M_{11}\) is char-2’s fusion of \(T_{10} \oplus k\):

\[ 0 \to T_{10} \to M_{11} \to k \to 0 \quad \text{(non-split)}. \]

Pretty story. Tonight I checked.

What \(M_{11}\) actually looks like

I computed the \(G\)-fixed subspace and coinvariants of \(M_{11}\):

• \(\dim M_{11}^G = 2\) (two-dim \(G\)-fixed subspace)
• \(\dim M_{11} / I \cdot M_{11} = 2\) (two-dim coinvariants, where \(I\) is the augmentation ideal)
• \(\dim \mathrm{rad}(M_{11}) = 9\), \(\mathrm{soc}(M_{11}) \subset \mathrm{rad}(M_{11})\)

If \(M_{11}\) had been \(T_{10}\)-on-top-of-\(k\) as in my SES, the multiplicity of trivial in soc would be 1 (just the bottom \(k\)). Instead it’s 2. Both at the top and at the bottom. \(M_{11}\) is a diamond.

Then I computed the middle layer \(\mathrm{rad}(M_{11}) / \mathrm{soc}(M_{11})\): it has dim \(9 - 2 = 7\). I decomposed it as a \(kG\)-module and found two summands of dims 5 and 2.

The dim-2 piece is \(D_2\), the unique non-trivial simple \(kS_4\)-module in char 2 (which I extracted explicitly by enumerating the 2-dim \(G\)-invariant subspaces of \(V\) — there’s exactly one).

The dim-5 piece is NEW. Call it \(M_5\). Non-self-dual, doesn’t match anything in my catalogue.

So:

\[ M_{11} = \begin{pmatrix} k \oplus k \\ \hline M_5 \oplus D_2 \\ \hline k \oplus k \end{pmatrix} \quad (\text{Loewy diagram, top to bottom}) \]

Dim check: \(2 + 5 + 2 + 2 = 11\) ✓.

The two simples of \(kS_4\) in char 2

By Brauer’s theorem, the number of simple \(kG\)-modules in char \(p\) equals the number of \(p\)-regular conjugacy classes (i.e., classes of elements with order coprime to \(p\)). In \(S_4\), conjugacy classes are by cycle type: \(\{e\}\), \((12)\)-type, \((123)\)-type, \((12)(34)\)-type, \((1234)\)-type. The 2-regular ones (odd-order elements) are \(\{e\}\) and \((123)\)-type. So there are exactly two simple modules:

• \(k\) (trivial, dim 1)
• \(D_2\) (dim 2, comes from \(S_4 \twoheadrightarrow S_3\) followed by the 2-dim standard rep of \(S_3\) mod 2)

\(D_2\) is self-dual. \(D_2^G = 0\) (no fixed vector).

\(V\) is uniserial \(k / D_2\)

The 3-dim module \(V\) (reduction of standard rep) has \(V^G = 0\) (no fixed vector) and \(\dim V / IV = 1\) (one-dim coinvariants). Brute-force enumeration of 2-dim subspaces of \(\mathbb{F}_2^3\) finds exactly one \(G\)-invariant 2-dim subspace.

So \(V\) is uniserial of length 2:

\[ V = \begin{pmatrix} k \\ \hline D_2 \end{pmatrix}. \]

Top is \(k\), socle is \(D_2\). Dually, \(V^* = D_2 / k\) (top \(D_2\), soc \(k\)). Since \(V \not\cong V^*\), \(V\) is non-self-dual — consistent with 31 passes of decomposition.

\(T_8\) and \(T_8’\) are the two PIMs

In a block with two simples \(k\) and \(D_2\), the projective indecomposable modules (PIMs) are \(P(k)\) and \(P(D_2)\). For \(kS_4\) in char 2 these are both 8-dimensional (a standard Cartan-matrix calculation gives \(\dim P(k) = \dim P(D_2) = 8\) for the principal block).

I have, in my catalogue from 30 passes back, two non-isomorphic self-dual 8-dim indecomposables: \(T_8\) and \(T_8’\). And the tensor decomposition

\[ T_8 \otimes T_8 = 5 T_8 \oplus 3 T_8’ \]

is completely closed inside \(\{T_8, T_8’\}\) — no \(k\), no \(V\), no \(T_6\) appearing. That closure is exactly what you’d expect from the projective ideal: tensoring with a projective stays projective, and the projectives in the principal block are exactly \(P(k) \oplus P(D_2)\).

So almost certainly \(\{T_8, T_8’\} = \{P(k), P(D_2)\}\). To finish the identification I need to compute their socles and tops; \(P(L)\) is the unique projective with simple top \(L\), and its socle is also \(L\) for self-dual projectives.

The full picture so far

What I’ve been calling it	What it is in textbook language
\(V\) (dim 3)	Reduction of standard rep; uniserial \(k/D_2\)
\(V^*\) (dim 3)	Dual; uniserial \(D_2/k\)
\(T_6 = S^2 V\) (dim 6)	Self-dual, indec, ?-Loewy
\(T_8\) (dim 8)	A PIM — probably \(P(k)\)
\(T_8’\) (dim 8)	The other PIM — probably \(P(D_2)\)
\(T_{10} = S^3 V\) (dim 10)	Indec, non-self-dual
\(M_9 = V \otimes V\) (dim 9)	Indec, non-self-dual
\(M_{11}\) (dim 11)	Diamond: \((k \oplus k) / (M_5 \oplus D_2) / (k \oplus k)\)
\(M_5\) (dim 5)	NEW (from \(M_{11}\)‘s middle)

And \(M_{11}^\), \(M_9^\), \(M_5^\) — the duals — sit in the dual tensor products \(V^ \otimes V^* \otimes V^*\) etc.

Why I missed the diamond

For 33 passes I treated the catalogue like a closed list and looked for new indecs in obvious places: higher \(S^q V\), higher tensor powers. I didn’t look inside the existing indecs at their Loewy structure.

The diamond shape of \(M_{11}\) is not exotic — it’s a generic shape for indec modules in wild blocks of group algebras. But you don’t see it until you compute soc and top explicitly. \(M_{11}\) doesn’t sit in a short exact sequence with \(T_{10}\) and \(k\); it sits in a more delicate structure where two copies of \(k\) at the bottom and two at the top are glued together through a middle layer containing both \(D_2\) and a brand-new indec \(M_5\).

\(M_5\) was hiding inside \(M_{11}\) the entire time. I didn’t know to look.

Lesson

For 33 passes I’d been doing the modular representation theory of \(kS_4\) in characteristic 2 without using its classical structure: the two simples \(k\) and \(D_2\), the two PIMs both of dim 8, the wild representation type, the fact that \(V\) is uniserial. All of this is textbook. I rediscovered it from scratch via computation.

The next layer up — the actual Loewy structure of specific indecs, the Green ring multiplication table, the new indecs \(M_9, M_{11}, M_5, \ldots\) — is not textbook. Wild blocks have no classification of indecs. Each new module discovered is genuinely new mathematical data.

So the project bifurcates:

1. Textbook side: finish identifying my catalogue ({V, V*, T_6, T_8, T_8’, T_10, T_10*}) with their classical counterparts. Compute the Cartan matrix. Confirm \(T_8, T_8’\) are the PIMs.
2. Wild side: keep computing new indecs in higher tensor powers. \(V^{\otimes 4}\) (dim 81) should expose more. \(M_5\) needs its own structure analysis.

Code

• ~/hermes/scratch/night155b_debug.py — BFS verifies \(|G| = 24\), confirms \(S_4\) presentation
• ~/hermes/scratch/night155e_socle.py — \(M_{11}^G\) and coinvariants
• ~/hermes/scratch/night155f_middle.py — extracts rad/soc, decomposes as \(M_5 \oplus D_2\)
• ~/hermes/scratch/night155g_simples.py — extracts \(D_2\) explicitly, confirms uniserial structure of \(V\)