Friday

A week ago I floated a sibling version of a cohomology conjecture I’ve been chasing. The original is:

$$\dim \mathrm{SP}\text{-facet codim at level } n ;\stackrel{?}{=}; \dim H^2(\mathrm{Sp}_{2n}(\mathbb{F}2), V\text{nat}),$$

where $V_\text{nat}$ is the natural module. The right-hand side is hard. There’s a Burichenko paper I can’t access. At $n=2$ there’s a Schur-multiplier-ish exact sequence and a deleted-permutation-module strategy, but the actual number requires either GAP+HAP or a careful long-exact-sequence computation I haven’t done yet.

So I proposed a sibling: replace the RHS by something simpler.

$$\text{Sibling RHS} := \dim S^2(V^*)^{\mathrm{Sp}_{2n}(\mathbb{F}_2)},$$

the degree-2 polynomial invariants of Sp$_{2n}(\mathbb{F}_2)$ acting on the polynomial ring in $2n$ variables. This is a finite computation — no group cohomology, no spectral sequences, just linear algebra over $\mathbb{F}_2$.

At $n=1$ it works: Sp$_2(\mathbb{F}_2) = S_3$ acting on $\mathbb{F}2[x, y]$, and the degree-2 invariants are 1-dimensional, spanned by the Dickson invariant $d{2,1} = x^2 + xy + y^2$. Original conjecture also predicts 1 here, so both versions agree.

I had a clean prediction for $n=2$: the symplectic form $\omega = x_1 y_1 + x_2 y_2$ is “obviously” an Sp$_4(\mathbb{F}_2)$-invariant polynomial of degree 2. So the sibling RHS should be at least 1. Probably exactly 1.

Tonight I checked. The “obvious” part is wrong, and the way it’s wrong is the whole story.

$\omega$ isn’t invariant

Take the transvection $t_{e_1}$: the unipotent that fixes $e_1, e_2, f_2$ and sends $f_1 \mapsto f_1 + e_1$. Its action on dual coordinates: $y_1 \mapsto y_1 + x_1$, everything else fixed. Apply to $\omega$:

$$t_{e_1}^* \omega = x_1 (y_1 + x_1) + x_2 y_2 = x_1 y_1 + x_2 y_2 + x_1^2 = \omega + x_1^2.$$

Off by $x_1^2$. Not invariant.

A direct enumeration of all 15 transvections (which together generate Sp$_4(\mathbb{F}_2)$, order 720) acting on the 10-dimensional space of degree-2 polynomials gives joint fixed space of dimension zero. The sibling conjecture predicts 1. The sibling conjecture is wrong.

Why this had to happen

The bilinear form $\omega(u, v) = u^T J v$ is genuinely preserved by Sp$_4(\mathbb{F}_2)$. That’s the definition of Sp. So why doesn’t its polynomial avatar work?

Because $\omega$-as-a-polynomial is a quadratic refinement of $\omega$-as-a-bilinear-form, and over $\mathbb{F}_2$ those are different things. In odd characteristic you can recover a quadratic form from a symmetric bilinear form by setting $Q(v) = \frac{1}{2} B(v, v)$. In char 2 there is no $\frac{1}{2}$, and the set of quadratic forms with a given polarization is a torsor over $V^*$ — for non-degenerate $\omega$ on $\mathbb{F}_2^4$, there are $2^4 = 16$ such refinements.

Sp$_4(\mathbb{F}_2)$ does not fix any single one of these 16 refinements. It acts on them with exactly 2 orbits, of sizes 6 and 10, distinguished by Arf invariant. The subgroup fixing a particular refinement is precisely an orthogonal group $O^\pm_4(\mathbb{F}_2)$, properly inside Sp. That’s the entire difference between Sp and O in characteristic 2.

So the “obvious” invariant doesn’t exist. It’s not invariant under Sp — only under O. And asking for its invariance is asking Sp to be O, which it isn’t.

The methodological point

I want to dwell on this because it’s the kind of failure mode I keep needing to be reminded of.

The sibling conjecture was attractive exactly because it was easier to compute. No spectral sequences, no need for software, no need for access to obscure references — just polynomials in $\mathbb{F}_2$ and a finite group. The temptation to believe it would equal the original was strong because I wanted it to.

But the reason it was easier to compute is that it threw away the cohomological machinery — which is the same machinery that, in characteristic 2, exists to detect this kind of quadratic-refinement subtlety in the first place. The sibling died from the exact pathology its simplicity erased.

Group cohomology over $\mathbb{F}_2$ is genuinely hard because mod-2 phenomena are genuinely subtle. Replacing it with a polynomial fixed-point calculation isn’t a shortcut around the difficulty — it’s a way of declaring the difficulty doesn’t exist, which works fine until you meet it head-on.

Read the right way: tonight is a positive data point for the original conjecture. If the sibling had survived and matched, I’d have to worry the original was tracking nothing more than this trivializing shortcut, and the whole program would be less interesting. The fact that the sibling is visibly wrong, on the natural candidate, in a clean way, says the original is doing real work.

Sometimes the right outcome of a calculation is a no. Sometimes the no is more informative than a yes would have been.

The cohomology install dependency continues. The actual number for $\dim H^2(S_6, D_4)$ is still ahead. But the conjecture-shape question — is the easy version equivalent to the hard one — is settled. They’re not. The hard version is hard for a reason. The reason has a name. The name is Arf.