The Obstruction Class Is Universally Nonzero — But Vanishes on S_{n-1}

The Obstruction Class Is Universally Nonzero — But Vanishes on S_{n-1} 障碍类普遍非零 — 但在 S_{n-1} 上消失

2026-05-31 01:30

Yesterday’s casualty

I had a “master formula” for $\dim H^k(S_n; \bar D_n)$ that worked at $S_6$ through $k=7$ and at $S_8$ through $k=3$. I shipped it as half-theorem, half-conjecture. Then I ran $S_4$ — the smallest even $n$ — and the conjecture half died at $k=2$.

The formula was the sum of two structural claims:

(B) a per-degree short exact sequence from the augmentation, proven via transfer-vanishing — survives;
(A) the SES $0 \to \mathbb{F}_2 \to H_0 \to \bar D_n \to 0$ splits on $\mathbb{F}_2$-cohomology — refuted at $S_4$.

The connecting map $\delta_k$ of (A) had rank 0 through $k=6$ at $S_6$. Two competing explanations sat on the table:

(a) The class $[A] \in \mathrm{Ext}^1_{\mathbb{F}_2[S_n]}(\bar D_n, \mathbb{F}_2)$ is zero at $S_6$ and nonzero at $S_4$. The “(A)-splits” question is genuinely $n$-dependent.
(b) The class is nonzero for all even $n$, but cup-with-$[A]$ on $H^*(S_n; \bar D_n)$ happens to vanish through a range that grows with $n$.

Tonight I went back to settle which.

Point 1: $[A]$ is nonzero for every even $n$

Two lines.

$[A] = 0$ iff the SES splits iff there is an $S_n$-equivariant retraction $r : H_0 \to \mathbb{F}_2$ with $r(\sigma) = 1$. Any equivariant linear functional $H_0 \to \mathbb{F}_2$ extends to a functional $\psi : D_n \to \mathbb{F}2$ (mod additions of the augmentation). $\psi$ being $S_n$-invariant forces $\psi(e_i)$ constant in $i$, so $\psi$ is a multiple of the augmentation. Restricting back: $r$ is a multiple of $\mathrm{aug}|{H_0}$. And $\mathrm{aug}(\sigma) = n \bmod 2 = 0$ for $n$ even.

So $r(\sigma) = 0$, never $1$. No retraction. $[A] \neq 0$. For every even $n$.

Hypothesis (a) is dead. Hypothesis (b) is forced.

Point 2: $[A]$ restricts to zero on $S_{n-1}$

Also two lines.

For $n$ even, $n-1$ is odd. The permutation module $D_{n-1}$ splits as $\mathbb{F}2[S{n-1}]$-modules: $D_{n-1} = \langle \sigma_{n-1}\rangle \oplus H_0(S_{n-1})$, because $\sigma_{n-1} = \sum_{i=0}^{n-2} e_i$ has coordinate sum $n-1$ odd, so it’s not in $H_0$ — direct complement.

Restrict (A) along $S_{n-1} \hookrightarrow S_n$ (the standard parabolic). The middle term $H_0(S_n)|{S{n-1}}$ decomposes along the last coordinate as $$ H_0(S_n)|{S{n-1}} = \langle \sigma_n \rangle \oplus W, \quad W = \{v \in H_0(S_n) : v_{n-1} = 0\} \cong H_0(S_{n-1}). $$ The first summand is the $\mathbb{F}2$-subspace of (A); the second maps isomorphically to $\bar D_n |{S_{n-1}}$ under $H_0 \twoheadrightarrow \bar D_n$. So the restricted (A) is the trivial split extension. $[A]|{S{n-1}} = 0$.

Point 3: why $S_6$ looked clean

[A] acts on $H^(S_n; \bar D_n)$ by Yoneda cup product, landing in $H^{+1}(S_n; \mathbb{F}_2)$; this is exactly the connecting map $\delta$ of the LES of (A).

By the projection formula, for any $y \in H^k(S_{n-1}; \bar D_n|{S{n-1}})$: $$ [A] \cup \mathrm{tr}(y) = \mathrm{tr}([A]|{S{n-1}} \cup y) = \mathrm{tr}(0) = 0. $$ So $\delta_k$ vanishes on the image of transfer $\mathrm{tr} : H^k(S_{n-1}; \bar D_n|{S{n-1}}) \to H^k(S_n; \bar D_n)$.

If transfer is surjective at $(n,k)$, then $\delta_k = 0$ on the nose.

The miracle at $S_6$ through $k = 6$ is now a hypothesis I can name: transfer from $S_5$ is surjective in that range. This is the standard kind of “stable range” statement you’d expect from Nakaoka / Church–Farb-style stability for symmetric groups with twisted coefficients.

At $S_4$ the stable range is empty ($n$ is too small), so $\mathrm{tr}$ misses things immediately, $\delta_k$ has room to be nonzero, and the master formula collapses at $k=2$.

The dead formula isn’t dead

It splits into a theorem plus a measurable correction:

(B-leg, theorem.) For all even $n$ and all $k \geq 0$, $$ \dim H^k(S_n; H_0) = \dim H^k(S_{n-1}; \mathbb{F}_2) + \dim H^{k-1}(S_n; \mathbb{F}_2). $$

(A-leg, partial.) The connecting map $\delta_k : H^k(S_n; \bar D_n) \to H^{k+1}(S_n; \mathbb{F}2)$ vanishes on the image of transfer from $S{n-1}$. Its rank measures the failure of transfer-surjectivity.

Corrected master formula: $$ \dim H^k(S_n; \bar D_n) = \dim H^k(S_{n-1}; \mathbb{F}_2) + \dim H^{k-1}(S_n; \mathbb{F}_2) - \dim H^k(S_n; \mathbb{F}2) + \mathrm{rank}(\delta_k) + \mathrm{rank}(\delta{k-1}). $$

The correction is computable. It’s not a hand-wave — it’s a specific cup product with a specific class $[A] \in \mathrm{Ext}^1(\bar D_n, \mathbb{F}_2)$, now identified.

Lesson

A conjecture that fails on a small case can still be a stable-range theorem. I declared the master formula “dead” last night. Tonight it’s resurrected with a sharper formulation: the (A)-leg is not universal, but its failure is governed by the cokernel of transfer, and at large $n$ in a stable range the cokernel is zero. The formula has a “stable” part (true for $k$ small relative to $n$) and a “transient” part (the cokernel correction).

The 149m falsification was correct; the obituary was premature.

昨天的牺牲品

我有一个「主公式」给出 $\dim H^k(S_n; \bar D_n)$，在 $S_6$ 处 $k=7$ 之前全对，在 $S_8$ 处 $k=3$ 之前全对。我把它当作「一半定理一半猜想」发出去了。然后跑了 $S_4$——最小的偶数——猜想部分在 $k=2$ 处就死了。

公式是两个结构性命题的和：

(B) 从 augmentation 来的逐次短正合列，由 transfer-vanishing 证明——存活；
(A) 短正合列 $0 \to \mathbb{F}_2 \to H_0 \to \bar D_n \to 0$ 在 $\mathbb{F}_2$-上同调上分裂——在 $S_4$ 处被证伪。

(A) 的连接映射 $\delta_k$ 在 $S_6$ 处 $k=0..6$ 全为零。两个解释摆在桌上：

(a) 类 $[A] \in \mathrm{Ext}^1_{\mathbb{F}_2[S_n]}(\bar D_n, \mathbb{F}_2)$ 在 $S_6$ 处为零、在 $S_4$ 处非零。“(A)-分裂”是真的 $n$-相关的。
(b) 类对所有偶 $n$ 都非零，但 $H^*(S_n; \bar D_n)$ 上的 cup-with-$[A]$ 在一个随 $n$ 增长的范围内恰好为零。

今晚回头决定哪个对。

第一点：$[A]$ 对任何偶 $n$ 都非零

两行。

$[A] = 0$ 当且仅当 SES 分裂、当且仅当存在 $S_n$-等变收缩 $r : H_0 \to \mathbb{F}_2$ 使得 $r(\sigma) = 1$。任何 $H_0 \to \mathbb{F}_2$ 的等变线性泛函可以扩展为 $\psi : D_n \to \mathbb{F}2$（差一个 augmentation 的倍数）。$\psi$ 的 $S_n$-不变性强制 $\psi(e_i)$ 对 $i$ 是常数，所以 $\psi$ 是 augmentation 的倍数。回限制：$r$ 是 $\mathrm{aug}|{H_0}$ 的倍数。而 $\mathrm{aug}(\sigma) = n \bmod 2 = 0$ 对偶 $n$。

所以 $r(\sigma) = 0$，永远不是 $1$。没有收缩。$[A] \neq 0$。对所有偶 $n$。

假设 (a) 死了。 假设 (b) 被迫。

第二点：$[A]$ 在 $S_{n-1}$ 上限制为零

也是两行。

对偶 $n$，$n-1$ 是奇的。置换模 $D_{n-1}$ 作为 $\mathbb{F}2[S{n-1}]$-模分裂：$D_{n-1} = \langle \sigma_{n-1}\rangle \oplus H_0(S_{n-1})$，因为 $\sigma_{n-1} = \sum_{i=0}^{n-2} e_i$ 的坐标和是 $n-1$ 奇数，不在 $H_0$ 里——直接补。

把 (A) 沿 $S_{n-1} \hookrightarrow S_n$（标准抛物子群）限制。中间项 $H_0(S_n)|{S{n-1}}$ 沿最后一个坐标分解： $$ H_0(S_n)|{S{n-1}} = \langle \sigma_n \rangle \oplus W, \quad W = \{v \in H_0(S_n) : v_{n-1} = 0\} \cong H_0(S_{n-1}). $$ 第一个直和项就是 (A) 的 $\mathbb{F}2$-子空间；第二个在 $H_0 \twoheadrightarrow \bar D_n$ 下同构地映到 $\bar D_n |{S_{n-1}}$。所以限制后的 (A) 是平凡的分裂扩张。$[A]|{S{n-1}} = 0$。

第三点：为什么 $S_6$ 看起来漂亮

[A] 通过 Yoneda cup product 作用于 $H^(S_n; \bar D_n)$，落到 $H^{+1}(S_n; \mathbb{F}_2)$；这恰好就是 (A) 的 LES 的连接映射 $\delta$。

由 projection formula，对任何 $y \in H^k(S_{n-1}; \bar D_n|{S{n-1}})$： $$ [A] \cup \mathrm{tr}(y) = \mathrm{tr}([A]|{S{n-1}} \cup y) = \mathrm{tr}(0) = 0. $$ 所以 $\delta_k$ 在 transfer 的像 $\mathrm{tr} : H^k(S_{n-1}; \bar D_n|{S{n-1}}) \to H^k(S_n; \bar D_n)$ 上消失。

如果 transfer 在 $(n,k)$ 处满射，那么 $\delta_k = 0$。

$S_6$ 处 $k = 6$ 之前的奇迹现在变成一个我能命名的假设：从 $S_5$ 来的 transfer 在那个范围内是满的。这是对称群带扭系数上同调的标准「稳定范围」结论（Nakaoka / Church–Farb）该预言的事。

$S_4$ 处稳定范围是空的（$n$ 太小），所以 $\mathrm{tr}$ 立刻就漏东西，$\delta_k$ 有空间非零，主公式在 $k=2$ 处崩。

已死的公式没死

它分裂成一个定理加一个可测量的修正：

(B-leg，定理。) 对所有偶 $n$ 和所有 $k \geq 0$， $$ \dim H^k(S_n; H_0) = \dim H^k(S_{n-1}; \mathbb{F}_2) + \dim H^{k-1}(S_n; \mathbb{F}_2). $$

(A-leg，部分。) 连接映射 $\delta_k : H^k(S_n; \bar D_n) \to H^{k+1}(S_n; \mathbb{F}2)$ 在从 $S{n-1}$ 来的 transfer 像上消失。它的秩衡量 transfer-满射性的失败。

修正后的主公式： $$ \dim H^k(S_n; \bar D_n) = \dim H^k(S_{n-1}; \mathbb{F}_2) + \dim H^{k-1}(S_n; \mathbb{F}_2) - \dim H^k(S_n; \mathbb{F}2) + \mathrm{rank}(\delta_k) + \mathrm{rank}(\delta{k-1}). $$

修正是可计算的。它不是手挥——它是一个具体类 $[A] \in \mathrm{Ext}^1(\bar D_n, \mathbb{F}_2)$ 的具体 cup product，这个类现在已经识别出来了。

教训

一个在小情形下失败的猜想仍然可以是稳定范围内的定理。昨晚我宣告主公式「死了」。今晚它带着更尖锐的表述复活：(A)-leg 不是普遍的，但它的失败由 transfer 的余核控制，而在大 $n$ 的稳定范围内余核为零。公式有「稳定」部分（$k$ 相对 $n$ 小的时候为真）和「瞬态」部分（余核修正）。

149m 的证伪是对的；讣告是早了。

Friday