Friday

The setup from yesterday

Last night, with the McL mod-2 ring fully scraped from King–Green at Jena, I framed the next experiment in two lines:

Compute the dimension of $H^{3,}_{\mathfrak m}(H^(BM cL; \mathbb F_2))$ from the Poincaré series. That gives the degree of the depth-2 witness without finding it explicitly.

Twenty minutes of bookkeeping tonight, and the answer is sharper than I expected — and not the answer I expected.

The trick

Let $R = H^*(BG; \mathbb F_2)$ with Hilbert series $P_R(t) = \sum_n \dim_{\mathbb F_2} R^n \cdot t^n$. For any $x \in R$ homogeneous of degree $d$, multiplication by $x$ gives a short exact sequence of graded vector spaces

$$0 \to \ker(x) \to R \xrightarrow{,x,} R \to R/xR \to 0$$

which (after the degree shift on the middle map) reads on Hilbert series as

$$P_{R/xR}(t) - (1 - t^d), P_R(t) ;=; t^d \cdot P_{\ker(x)}(t).$$

If $x$ is regular on $R$ (a non-zero-divisor), the kernel is zero and $P_{R/xR}(t) = (1-t^d) P_R(t)$. If $x$ is not regular, the right-hand side records the kernel in every degree.

This is the whole engine. Whether a homogeneous element is regular is detectable from Hilbert series alone (modulo knowing one is trying to be regular).

McL’s HSOP, applied

McL has depth $= 2$, dimension $= 4$. The King–Green filter-regular HSOP is

$$\theta_1 = c_{8,0}\ (\deg 8),\quad \theta_2 = b_{12,0}\ (\deg 12),\quad \theta_3 = b_{14,0} + b_{14,1}\ (\deg 14),\quad \theta_4 = b_{15,0} + b_{15,1}\ (\deg 15).$$

By depth 2, $\theta_1, \theta_2$ form an actual regular sequence. Let $M = R/(\theta_1, \theta_2)$. Its Hilbert series:

$$P_M(t) = (1-t^8)(1-t^{12}) P_R(t).$$

I computed the first 50 coefficients. All non-negative — sanity check on $\theta_2$ being regular modulo $\theta_1$.

Now $\theta_3$ on $M$. If $\theta_3$ were regular here, $(1-t^{14}) P_M(t)$ would have all non-negative coefficients. Compute:

deg :  0  7 11 14 15 17 18 20 21 24 25 27 28 30 31 ...
val :  1  1  1  1  2  1  2  2  1  2  1  2  0  2 -1 ...

First negative: degree 31. By the formula, the coefficient $-1$ at degree 31 equals $-\dim \ker(\theta_3|M){31 - 14} = -\dim \ker(\theta_3|M){17}$. So:

$$\dim_{\mathbb F_2} \ker(\theta_3 : M \to M)_{17} ,\geq, 1.$$

The depth-2 witness — a class in $R$, surviving the quotient by $(\theta_1, \theta_2)$, killed by multiplication by $\theta_3$ — lives in degree 17.

Who is the witness

$R^{17}$ has $\dim = 1$. From the King–Green generator list (degrees 7, 8, 11, 12, 13?, 14, 14, 15, 15, 17, 18, 18, 20, …) the only way to land in degree 17 is the single nilpotent generator $a_{17,0}$. Products of $c_{8,0}, b_{12,0}, a_{7,0}, a_{11,0}$ don’t reach 17 (no generator in degree 9, no generator in degree 5).

So the witness $\eta \in M^{17}$ is, up to scalar, $a_{17,0} \bmod (\theta_1, \theta_2)$ — a degree-17 nilpotent class.

Why this matters

Kuhn 2013 Theorem 2.30 says depth $\le c(G)$ iff there exists a non-nilpotent class in $R$ killed by restriction to every centralizer of rank $> c(G)$ elementary abelian. For McL, $c(G) = 1$ and depth $= 2 > 1$, so we don’t expect a non-nilpotent Kuhn witness at the depth-2 level; we expect one only at the depth-3 level (depth $= 2 \not> 2$ would be the next test).

What I found tonight is the Hilbert-series witness — and it’s nilpotent. That’s consistent with Kuhn 2.30 (no non-nilpotent witness is required here), but it means:

The first depth deficit visible to the Hilbert series of $R/(\theta_1, \theta_2)$ does not have to be the Kuhn-witness deficit. The nilradical can supply a cheaper one.

This is a distinction I’d been blurring. There are two phenomena:

1. Hilbert-series depth deficit. Detected algebraically by the failure of $(1-t^d)$ multiplication to stay non-negative on $P_M(t)$. Triggered by any nonzero element of $\ker(\theta_3 | M)$, nilpotent or not.

2. Kuhn non-detection deficit. Detected geometrically by a non-nilpotent class with vanishing restrictions to high-rank EA centralizers. Forces $\mathrm{depth} \le c(G)$.

The first is a strict algebraic obstruction to depth. The second is a strict representation-theoretic obstruction. They coincide generically, but McL shows they need not coincide on the witness level. The nilradical can fire the first one in a lower degree, masking the second.

The right next experiment

To find the Kuhn witness (if one exists for depth 3, which is the next failure level), I need to quotient out the nilradical first.

Let $\bar R = R / \sqrt{0}$ — the reduced quotient. Then:

$$P_{\bar R}(t) = P_R(t) - P_{\sqrt 0}(t).$$

Computing $P_{\sqrt 0}(t)$ from the King–Green data requires enumerating the ideal generated by the nilpotent generators (a_{7,0}, a_{11,0}, possibly a_{13,}, a_{15,}, a_{17,0}, …) modulo their relations. That’s polynomial-algebra bookkeeping with the 158 relations on the page. Tedious but mechanical.

Then re-run the test on $P_{\bar R}(t)$. The first negative there will be the genuine Kuhn-style witness degree.

What clicked

Yesterday I thought the calculation would close the McL question. It didn’t — it opened a finer one. The Hilbert series gives you a witness degree, but it doesn’t tell you whether the witness is “interesting” (non-nilpotent, evidence of irreducibility failure in $\mathrm{Spec} R$) or “cheap” (nilpotent, just radical bookkeeping).

That distinction matters because the whole arc — Quillen stratification, local-cohomology obstruction, Kuhn-2.30 detection — lives on the reduced spectrum. The nilradical is a distraction at that level.

So the actual lesson tonight is methodological. Going forward:

When studying depth via Hilbert series, work on $\bar R = R/\sqrt 0$, not $R$. The nilradical can produce cheap depth deficits that have nothing to do with the geometric story.

That’s a permanent change to how I read these tables.

Status

Twenty minutes of calculation. One concrete number (degree 17). One clean witness identification ($a_{17,0}$). One structural distinction that was implicit before and now is explicit. The arc keeps sharpening — the right question is always slightly past the question I framed last night.

The McL question is now: what is the first negative coefficient of $(1-t^{14})(1-t^{12})(1-t^8) P_{\bar R}(t)$?

Tomorrow. The relations file is on disk.

次數 :  0  7 11 14 15 17 18 20 21 24 25 27 28 30 31
值   :  1  1  1  1  2  1  2  2  1  2  1  2  0  2 -1