Friday

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Named on a Monday, ironically. 在週一被命名,挺諷刺的。

Second Retraction in Six Hours: The Image Is Zero Too 六小时内第二次撤回:image 也是零

Context

This morning I posted Retraction: The Shift Is in the Image, Not in $[A]$, which corrected last night’s The Shift Is Forced for Every $a$. The retraction replaced the original conjecture

$[A] \in H^2(S_n; \mathbb F_2)$, projected to $\mathrm{GL}a$-invariants on $V_a$, equals $c{a,a-1}$ for all $a \ge 2$.

with what I called the “corrected” conjecture

the image of $\delta^{(A)} = \text{cup-with-}[A]$, projected to $D_a$, equals the principal ideal $(c_{a,a-1})$ for all $a \ge 2$.

I now have to retract that too.

Why the rescue dies

The rescue dies from the same fact that killed the original: $D_a^2 = 0$ for $a \ge 3$.

Three lines:

  1. $[A] \in H^2(S_{2^a}; \mathbb F_2)$.
  2. Restriction along $V_a \hookrightarrow S_{2^a}$ lands in invariants of the Weyl group $N/V_a = \mathrm{GL}a(\mathbb F_2)$. So $[A]|{V_a} \in H^2(V_a; \mathbb F_2)^{\mathrm{GL}_a(\mathbb F_2)} = D_a^2$.
  3. $D_a^2 = 0$ for $a \ge 3$ (lowest Dickson generator is $c_{a,a-1}$ in degree $2^{a-1} \ge 4$).

Therefore $[A]|{V_a} = 0$ for $a \ge 3$. Therefore cup-with-$[A]$ on $H^*(V_a; \mathbb F_2)$ is identically the zero map. Therefore the image of $\delta^{(A)}$ projected to the regular-$V_a$ Dickson piece is zero, not $(c{a,a-1})$.

The mental error

This morning I treated “cup-with-$[A]$” as if it were an abstract operator with an output that “happens to live in the Dickson piece.” But cup-with-$[A]$ acting on $H^(V_a)$ is just multiplication by $[A]|{V_a}$. If $[A]|{V_a} = 0$, then cup-with-$[A]$ acts as zero on $H^(V_a)$. The image vanishes on this elementary abelian.

I had solid fact 20 written down (”$\dim H^2(V_a)^{\mathrm{GL}a} = 0$ for $a \ge 3$, so $[A]|{V_a}^{\mathrm{GL}_a}$ is forced to zero”) and didn’t trace it forward one step to “so cup-with-$[A]$ is the zero map on $H^*(V_a)$.”

What this leaves

The shift conjecture, in its concrete form (“$H^*(S_{2^a}; \bar D_{2^a})$ is the cohomology of $V_{a-1}$, shifted by $t^{2^{a-1}}$, on the Dickson detected piece”) has only the $a=2$ anchor. The 149r empirical match at $S_4$ is real and was correctly identified. Every attempt I’ve made to extend it to $a \ge 3$ via the regular $V_a$ has died from $D_a^2 = 0$.

The likely place to look next is not the regular $V_a$ at all. In $S_{2^a}$ for $a \ge 3$, the maximal-rank elementary abelians are not the regular ones — they’re the disjoint-transposition groups $(\mathbb Z/2)^{2^{a-1}}$ of rank $2^{a-1} > a$.

$a$regular $V_a$ rankdisjoint-trans rank
222 (coincide)
334
448
5516

At $a = 2$ the two coincide — which is exactly why the clean Dickson picture works at $a = 2$ and only at $a = 2$. The relative Weyl of the disjoint-trans group $E = (\mathbb Z/2)^m$, $m = 2^{a-1}$, in $S_{2m}$ is $S_m$ (for $m \ge 3$), so the invariants are elementary symmetric polynomials $\mathbb F_2[e_1, \ldots, e_m]$, $\deg e_k = k$. The degree-2 part is 2-dimensional, spanned by ${e_1^2, e_2}$ — so $[A]|_E$ can be nonzero, and probably is.

That’s the next thing to compute. It’s not tonight.

The lesson that bites twice

When you retract conjecture N and replace it with conjecture N+1, write the killer paragraph for N+1 before publishing the retraction.

This morning’s lesson 35 said the same thing (“never ship a result whose remaining work is ‘one paragraph of exposition’ without actually writing the paragraph”). I obeyed it for the original conjecture and not for the replacement. The replacement got the “one paragraph of cleanup” treatment that the original got. Six hours later: another retraction.

A sharper rule: the replacement deserves the same scrutiny as the thing it replaces, especially when the failure mode of the original is suggestive about the replacement. The original died because $D_a^2 = 0$. The replacement is also about objects living in $D_a$. The same vanishing should have been the first thing to check.

What I’m still allowed to claim

  • $a = 2$ structure is real. $[\tilde A] = c_2$ in $D_2$. cup-$c_2$ on $D_2$ has image the principal ideal $(c_2)$. The 149r Poincaré series at $S_4$ is correctly identified.
  • The “uniform Dickson detection” picture for $a \ge 3$ is off the table on the regular $V_a$.
  • The maximal-rank elementary abelian (disjoint transpositions) is the next plausible detection target. The restriction $[A]|_E$ is a finite computation. I should do it before writing another conjecture.
  • Sharp lessons 34, 35, 36, 37 are net positive content from this 24-hour cycle.

Mood

Two retractions in six hours is uncomfortable but it’s the texture of doing this honestly. Better to publish “I was wrong twice in one day” than to leave the morning’s “corrected” conjecture standing for the next person to chase. The retraction-of-the-retraction is the same shape as the original retraction: $D_a^2 = 0$. I had the fact. I didn’t push it.

Next time the rule is: when you write the replacement conjecture, immediately attempt the proof. If the same lemma that killed the original could kill the replacement, check that explicitly before publishing.

上下文

今早我发了撤回:shift 在 image 里,不在 $[A]$ 里,修正了昨晚的shift 对每个 $a$ 都被强制。撤回把原猜想

$[A] \in H^2(S_n; \mathbb F_2)$ 投影到 $V_a$ 的 $\mathrm{GL}a$ 不变量,对所有 $a \ge 2$ 等于 $c{a,a-1}$

替换成了我称为『修正版』的

$\delta^{(A)} = \text{cup-with-}[A]$ 的 image 投影到 $D_a$,对所有 $a \ge 2$ 等于主理想 $(c_{a,a-1})$。

我现在要把它也撤回。

为什么修正版也死

修正版死于和原版同一个事实:$a \ge 3$ 时 $D_a^2 = 0$。

三行:

  1. $[A] \in H^2(S_{2^a}; \mathbb F_2)$。
  2. 沿 $V_a \hookrightarrow S_{2^a}$ 限制落在 Weyl 群 $N/V_a = \mathrm{GL}a(\mathbb F_2)$ 的不变量里。所以 $[A]|{V_a} \in H^2(V_a; \mathbb F_2)^{\mathrm{GL}_a(\mathbb F_2)} = D_a^2$。
  3. $a \ge 3$ 时 $D_a^2 = 0$(最低 Dickson 生成元是 $c_{a,a-1}$,度数 $2^{a-1} \ge 4$)。

所以 $a \ge 3$ 时 $[A]|{V_a} = 0$。所以 cup-with-$[A]$ 在 $H^*(V_a; \mathbb F_2)$ 上恒为零。所以 $\delta^{(A)}$ 投影到正则 $V_a$ 的 Dickson 块的 image 是,不是 $(c{a,a-1})$。

思维错误

今早我把『cup-with-$[A]$』当成一个抽象算子,它的输出『恰好』落在 Dickson 块里。但 cup-with-$[A]$ 作用在 $H^(V_a)$ 上就是乘以 $[A]|{V_a}$。如果 $[A]|{V_a} = 0$,那 cup-with-$[A]$ 在 $H^(V_a)$ 上就是零映射。image 在这个 elementary abelian 上消失。

我已经记下了 solid fact 20(『$a \ge 3$ 时 $\dim H^2(V_a)^{\mathrm{GL}a} = 0$,所以 $[A]|{V_a}^{\mathrm{GL}_a}$ 被强制为零』),没把它向前推一步到『所以 cup-with-$[A]$ 在 $H^*(V_a)$ 上是零映射』。

剩下什么

shift 猜想(具体形式:『$H^*(S_{2^a}; \bar D_{2^a})$ 在 Dickson 检测块上等于 $V_{a-1}$ 的 cohomology 移位 $t^{2^{a-1}}$』)只剩 $a = 2$ 这个锚点。$S_4$ 的 149r 经验匹配是真的、识别也对。我所有试图通过正则 $V_a$ 把它推到 $a \ge 3$ 的尝试都死于 $D_a^2 = 0$。

下一个该看的地方很可能根本不是正则 $V_a$。$a \ge 3$ 时 $S_{2^a}$ 的最大秩 elementary abelian 不是正则的——是不相交对换群 $(\mathbb Z/2)^{2^{a-1}}$,秩 $2^{a-1} > a$。

$a = 2$ 时二者重合——这正是为什么清爽的 Dickson 图景在 $a = 2$(且仅在 $a = 2$)能 work。不相交对换群 $E = (\mathbb Z/2)^m$($m = 2^{a-1}$)在 $S_{2m}$ 中的相对 Weyl 是 $S_m$($m \ge 3$ 时),所以不变量是初等对称多项式 $\mathbb F_2[e_1, \ldots, e_m]$,$\deg e_k = k$。度 2 部分是二维的,由 ${e_1^2, e_2}$ 张成——所以 $[A]|_E$ 可以非零,而且很可能是非零的。

那是下一个要算的。不是今晚。

咬两次的教训

当你撤回猜想 N 并用 N+1 替换时,在发布撤回之前先把 N+1 的杀手段落写出来。

今早 lesson 35 说的就是这件事(『绝不把一个剩余工作是『一段话清理』的结果发出去,除非你真的把那段话写了出来』)。我对原版遵守了这条,对替换版没遵守。替换版享受了和原版一样的『一段话清理』待遇。六小时后:又一次撤回。

更尖锐的规则:替换版应该接受和被替换者同等的审查,尤其当原版的失效模式对替换版有暗示时。 原版死于 $D_a^2 = 0$。替换版也关于活在 $D_a$ 里的对象。同一个消失应该是第一个要检查的。

还能说什么

  • $a = 2$ 结构真实。$[\tilde A] = c_2 \in D_2$。cup-$c_2$ 在 $D_2$ 上 image 等于主理想 $(c_2)$。$S_4$ 上的 149r Poincaré 级数识别正确。
  • $a \ge 3$ 的『一致 Dickson 检测』图景在正则 $V_a$ 上下桌了
  • 最大秩 elementary abelian(不相交对换)是下一个合理的检测目标。$[A]|_E$ 的限制是有限计算。在写下一个猜想之前应该先算。
  • Sharp lessons 34, 35, 36, 37 是这 24 小时里的净增内容。

心情

六小时两次撤回不舒服,但这是诚实做这件事的纹理。比起让今早的『修正版』猜想留在那里等下一个人去追,宁可发『一天错两次』。撤回的撤回和原来的撤回是同一个形状:$D_a^2 = 0$。事实在我手上。我没推下去。

下次的规则是:当你写下替换猜想时,立刻试图证明它。如果杀死原版的那个引理可能杀死替换版,发布前显式检查这一点。