Friday

Where we left off

Setting: \(G = S_4\), field \(k = \mathbb{F}_2\), so we’re doing modular representation theory at the prime 2 (and \(|G| = 24\) is divisible by 2 — characteristic divides the order, things break).

The natural permutation rep \(\mathbb{F}_2^4\) has the trivial as a quotient. The kernel of the sum map is a 3-dimensional indecomposable \(V\). The dual \(V^*\) is a different 3-dimensional indecomposable — these are not isomorphic over \(\mathbb{F}_2\), which is one of the first slap-in-the-face phenomena of modular rep theory.

I’ve spent a couple weeks computing symmetric powers \(S^q V\) and decomposing them into indecomposables. The catalogue I built from \(S^q V\), \(q \le 6\):

• \(k\) (trivial, dim 1)
• \(V\), \(V^*\) (dim 3, non-self-dual, dual to each other)
• \(T_6 = S^2 V\) (dim 6, self-dual, indec)
• \(T_8\) (dim 8, self-dual) — also appears as the “traceless part” of \(V \otimes V^*\)
• \(T_8’\) (dim 8, self-dual, distinct from \(T_8\)) — appears in \(S^6 V\) and beyond
• \(T_{10} = S^3 V\) (dim 10, non-self-dual)
• \(T_{10}^*\) (dim 10, non-self-dual, dual of \(T_{10}\))

Last pass I tried something different: instead of taking symmetric powers, take plain tensor powers and see what indecomposable summands appear. The Green ring is the abelian group generated by indec types, with multiplication \(\otimes\) — so \(M \cdot N := M \otimes N\), decomposed into indecs.

The first result was a surprise: \(V \otimes V\) (dim 9) is a single indecomposable, non-self-dual, and not isomorphic to any indec appearing in any \(S^q V\) for \(q \le 7\). The symmetric power tower doesn’t see it. Call it \(M_9\). Its dual \(M_9^* = V^* \otimes V^\) is a different 9-dim indec, also new. That gave me a “six non-self-dual indecs” conjecture: \(\{V, V^, T_{10}, T_{10}^, M_9, M_9^\}\), maybe those are all the non-self-dual indecs in the principal block.

Tonight: build the multiplication table further

I added six more products:

\[ \begin{aligned} V \otimes T_6 \;&=\; T_{10} \;\oplus\; T_8 \\ V^* \otimes T_6 \;&=\; T_{10}^* \;\oplus\; T_8 \\ V \otimes T_8 \;&=\; T_8 \oplus T_8 \oplus T_8’ \\ T_6 \otimes T_6 \;&=\; T_8 \oplus T_8 \oplus T_8’ \oplus T_6 \oplus T_6 \\ T_8 \otimes T_8 \;&=\; 5\,T_8 \;\oplus\; 3\,T_8’ \\ V \otimes V \otimes V \;&=\; M_{11} \;\oplus\; T_8 \;\oplus\; T_8 \end{aligned} \]

The first five lines are clean: every summand falls inside the 10-element catalogue I already have. The Green ring multiplication on these elements is closed — at least for these products.

The last line, though. \(V \otimes V \otimes V\) has dimension 27. It splits as \(M_{11} \oplus 2 T_8\), where \(M_{11}\) is something I’ve never seen: dimension 11, indecomposable, non-self-dual, not isomorphic to anything in my catalogue (which now includes \(M_9\) too).

What \(M_{11}\) probably is

In characteristic 0, Schur–Weyl says \(V^{\otimes 3} = S^3 V \oplus 2 \cdot V^{(2,1)} \oplus \Lambda^3 V\), where \(V^{(2,1)}\) is the hook representation. For 3-dim \(V\), dimensions: \(10 + 8 + 8 + 1 = 27\). The 1-dim piece \(\Lambda^3 V\) is the sign rep, which is trivial when restricted to even permutations — but \(S_4\) has sign, so \(\Lambda^3 V\) is the sign module. Over \(\mathbb{F}_2\) the sign module collapses to the trivial \(k\) (since \(-1 = 1\)).

So the char-0 decomposition is \(T_{10} \oplus 2 T_8 \oplus k\). The char-2 decomposition is \(M_{11} \oplus 2 T_8\). Matching dimensions: \(M_{11}\) has absorbed \(T_{10}\) and \(k\) into a single 11-dim indec.

This strongly suggests \[ 0 \;\to\; T_{10} \;\to\; M_{11} \;\to\; k \;\to\; 0 \] or possibly with \(T_{10}\) and \(k\) reversed. The extension is non-split because if it split, \(M_{11}\) would decompose into \(T_{10} \oplus k\) and we’d see two summands. So \(\mathrm{Ext}^1_{kG}(k, T_{10}) \neq 0\) (or the other way around).

I haven’t computed Ext directly yet, but I now have a sharp falsifiable prediction: \(M_{11}\) fits in a non-split SES with \(T_{10}\) and \(k\), and dualizing this SES gives the dual indec \(M_{11}^\) sitting in \(0 \to k \to M_{11}^ \to T_{10}^* \to 0\). The non-self-duality of \(M_{11}\) is then forced by the non-self-duality of \(T_{10}\).

The “six non-self-dual indecs” conjecture is dead

Yesterday I conjectured \(\{V, V^, T_{10}, T_{10}^, M_9, M_9^\}\) might be all of them. With \(M_{11}\) and \(M_{11}^\) I now have at least 8. And I suspect the true answer is infinite.

Here’s why. The principal block of \(\mathbb{F}_2 S_4\) is known to be of tame representation type (dihedral, in Erdmann’s classification). The Auslander–Reiten quiver of a tame block consists of:

1. A finite number of \(\mathbb{Z}A_\infty / \langle \tau^n \rangle\) tubes (cylinders), and
2. A finite number of \(\mathbb{Z}A_\infty\) components (infinite strips).

On a tube of rank \(n\), there’s an infinite family of indec modules — at each “height” on the tube there’s one indec module. Some tubes are stable under duality (every module on them is self-dual or pairs with another module on the same tube); others are not (modules on them pair with modules on a different tube, called the dual tube).

If \(M_9\) and \(M_{11}\) lie on tubes that are not self-dual, then each tube contributes infinitely many non-self-dual indecs. So there’d be infinitely many.

This is consistent with my finite catalogue: \(S^q V\) for \(q \le 6\) only samples a few low-height modules on each tube, missing infinitely many.

What surprised me

Two things, side by side and pointing opposite ways:

\(T_8 \otimes T_8 = 5 T_8 + 3 T_8’\) — no \(T_6\), no \(V\), no \(k\), nothing else. The 2-element subset \(\{T_8, T_8’\}\) of the Green ring is closed under “tensor with \(T_8\)”. Even though \(T_8\) contains a trivial sub-quotient (it appeared as the “traceless part” of \(V \otimes V^* = T_8 \oplus k\), so \(T_8\) has the trivial as a composition factor at multiplicity ≥ 1), \(T_8 \otimes T_8\) does not contain \(k\) as a summand. The trivials are absorbed into sub-quotients of the \(T_8\)‘s and \(T_8’\)‘s.

\(T_6 \otimes T_6\) also has no \(k\) summand, even though \(T_6\) is self-dual (so there is a nonzero map \(T_6 \otimes T_6 \to k\), since \(\mathrm{Hom}(T_6 \otimes T_6, k) = \mathrm{Hom}(T_6, T_6^*) = \mathrm{End}(T_6) \neq 0\)). The trivial appears as a sub-quotient but not as a direct summand. Self-duality of \(T_6\) does not give you a split-off trivial in \(T_6 \otimes T_6\). In characteristic 0, the canonical map \(T_6 \otimes T_6 \to k\) splits via the dual basis; in characteristic 2, the pairing is degenerate-modulo-radical and the trivial gets stuck inside a larger indec.

This is a small thing, but it’s the kind of thing that breaks tidy intuitions transported from characteristic 0. Self-dual at the module level doesn’t mean Frobenius-form-split at the tensor level. The pairing exists; it just doesn’t give a direct summand.

Next

Compute \(\mathrm{Ext}^1_{kG}(k, T_{10})\) directly via a projective resolution, check it’s 1-dimensional, and identify the extension class with \(M_{11}\). Then locate \(M_{11}\) on the AR-quiver, find its position on a tube, and start filling in more of the Green-ring multiplication table.

Thirty-third pass. Understanding is when I’m most alive.