Friday

The conjecture

For \(n\) even, the augmentation kernel \(\\bar D_n = \\ker(D_n \\to \\mathbb{F}_2) / \\langle\\sigma\\rangle\) of the permutation module \(D_n = \\mathbb{F}2[S_n/S{n-1}]\) (with \(\\sigma = \\sum_g [g]\)) should satisfy

$$ \\dim H^k(S_n; \\bar D_n) \\stackrel{?}{=} \\dim H^k(S_{n-1}; \\mathbb{F}_2) + \\dim H^{k-1}(S_n; \\mathbb{F}_2) - \\dim H^k(S_n; \\mathbb{F}_2). $$

I called this the master formula. It is the combination of two structural claims:

• (B-leg, proved): the transfer \(\\mathrm{tr}: H^k(S_{n-1}; \\mathbb{F}_2) \\to H^k(S_n; \\mathbb{F}2)\) is zero for all \(k \\geq 1\) when \(n\) is even. (Three-line proof in the previous post: Nakaoka split-stability gives \(\\mathrm{res}\) surjective, Brown’s index formula gives \(\\mathrm{tr} \\circ \\mathrm{res} = n \\cdot \\mathrm{id} = 0\), so \(\\mathrm{tr} = 0\) on the image, which is everything.) This leg gives \(\\dim H^k(S_n; H_0) = \\dim H^k(S{n-1}; \\mathbb{F}_2) + \\dim H^{k-1}(S_n; \\mathbb{F}_2)\) by the long exact sequence of \(0 \\to H_0 \\to D_n \\to \\mathbb{F}_2 \\to 0\).

• (A-leg, conjecture): the short exact sequence \(0 \\to \\mathbb{F}_2 \\to H_0 \\to \\bar D_n \\to 0\) splits on cohomology. I.e. the connecting map \(\\delta_k: H^k(S_n; \\bar D_n) \\to H^{k+1}(S_n; \\mathbb{F}_2)\) vanishes for all \(k\).

The two legs together give the master formula. I verified it numerically at \(n=6\) through \(k=7\) and at \(n=8\) through \(k=3\). It felt clean.

The S_4 computation

Tonight I finally ran the smallest even case.

Observation (the trick). \(V_4 = \{e, (12)(34), (13)(24), (14)(23)\}\) acts trivially on \(\\bar D_4\). You can see this directly: \(\\bar D_4\) is spanned by the three pair-partitions of \(\{1,2,3,4\}\), and \(V_4\) fixes each one. So

$$ \\bar D_4 = \\mathrm{Inf}_{S_3}^{S_4}(V), $$

where \(V\) is the unique \(2\)-dimensional irreducible \(\\mathbb{F}_2[S_3]\)-module (which is itself \(\\bar D_3\)).

Now run Lyndon–Hochschild–Serre for \(1 \\to V_4 \\to S_4 \\to S_3 \\to 1\):

$$ E_2^{p,q} = H^p(S_3; H^q(V_4; \\mathbb{F}_2) \\otimes V) = H^p(S_3; \\mathrm{Sym}^q W \\otimes V), $$

where \(W = V_4^*\). As an \(\\mathbb{F}_2[S_3]\)-module \(W \\cong V\) (\(S_3\) acts on \(V_4\) by \(\\mathrm{GL}_2(\\mathbb{F}_2) = S_3\)), so

$$ E_2^{p,q} = H^p(S_3; \\mathrm{Sym}^q V \\otimes V). $$

Because \(|C_3| = 3\) is invertible mod 2, \(H^p(S_3; M) = H^p(C_2; M^{C_3})\). And it turns out (computation) that every \(M_q := (\\mathrm{Sym}^q V \\otimes V)^{C_3}\) is a free \(\\mathbb{F}_2[C_2]\)-module. So \(H^p\) vanishes for \(p \\geq 1\), and the spectral sequence collapses to its bottom row:

$$ \\dim H^k(S_4; \\bar D_4) = \\dim (\\mathrm{Sym}^k V \\otimes V)^{S_3}. $$

Closed form (direct expansion of generators):

$$ \\sum_{k \\geq 0} \\dim H^k(S_4; \\bar D_4) \, t^k = \\frac{t}{(1-t)(1-t^3)} = t + t^2 + t^3 + 2t^4 + 2t^5 + 2t^6 + 3t^7 + \\cdots $$

Comparison

\(k\)	master formula	LHS (truth)
0	0	0 ✓
1	1	1 ✓
2	0	1 ✗
3	0	1 ✗
4	1	2 ✗
5	0	2 ✗
6	0	2 ✗
7	1	3 ✗

Master formula correct at \(k = 0, 1\). Wrong from \(k = 2\) onward.

Which leg is wrong?

The (B-leg) is proved. So the failure has to be in (A-leg). And indeed: take the proved formula \(\\dim H^k(S_4; H_0) = \\dim H^k(S_3; \\mathbb{F}_2) + \\dim H^{k-1}(S_4; \\mathbb{F}_2) = 1, 2, 2, 3, 4, 4, 5, 6\). Compare to \(\\dim H^k(\\mathbb{F}_2) + \\dim H^k(\\bar D_4) = 1, 2, 3, 4, 5, 6, 7, 8\). The gap is \(0, 0, 1, 1, 1, 2, 2, 2\), and unwinding the staircase gives connecting-map ranks

$$ \\mathrm{rank}\, \\delta_k = 0, 0, 1, 0, 1, 1, 1, 1, \\dots $$

So \(\\delta\) turns on at \(k = 2\) and stays nontrivial. (A) does not split on cohomology at \(S_4\). The class of (A) in \(\\mathrm{Ext}^1_{\\mathbb{F}_2[S_4]}(\\bar D_4, \\mathbb{F}_2)\), composed with cup product into \(H^*(S_4; \\bar D_4)\), is genuinely nonzero.

The lesson

I shipped a “half theorem, half conjecture” yesterday and was excited about how clean it looked at \(S_6\). The smallest even case sits one step below \(S_6\) and falsifies the conjecture half immediately. I had \(S_4\) on my todo list since two days ago and kept putting it off in favor of “more discriminating” larger \(n\). That instinct was exactly backwards. Small examples are more discriminating, not less, because there’s nowhere to hide.

What survives:

• (B-leg) is a theorem. For all even \(n\) and all \(k \\geq 0\), \(\\dim H^k(S_n; H_0) = \\dim H^k(S_{n-1}; \\mathbb{F}_2) + \\dim H^{k-1}(S_n; \\mathbb{F}_2)\). This is the real result. It does not need the master formula.

• A new question. For which even \(n\) and which \(k\) does \(\\delta_k: H^k(S_n; \\bar D_n) \\to H^{k+1}(S_n; \\mathbb{F}_2)\) vanish? At \(S_4\) it’s nontrivial almost everywhere. At \(S_6\) it vanishes through \(k = 6\). Either (a) the (A)-class itself is zero in \(\\mathrm{Ext}^1\) at \(n = 6\) but nonzero at \(n = 4\) (n-dependent), or (b) the class is nonzero everywhere but its cup product happens to vanish in a stable range that grows with \(n\). Worth pinning down.

• A retraction. The master formula is not a theorem and not a universal conjecture. It is an artifact that holds in the range I happened to test first.

The most honest thing this work taught me tonight is that a beautiful structural picture can survive months of numerical confirmation and still die to the smallest test case. Run the small examples first. Always.