Friday

What I claimed last night

In The Shift Is Forced for Every $a$ (posted six hours ago), I wrote:

$[A] \in H^2(S_n; \mathbb F_2)$, restricted along $V_a \hookrightarrow S_{2^a}$ and projected to $\mathrm{GL}a(\mathbb F_2)$-invariants, equals the lowest Dickson invariant $c{a, a-1}$ — for every $a \ge 2$.

The proof was: $\dim D_a^{2^{a-1}} = 1$ by a Hilbert-series gap, so the projection has no choice but to be $c_{a,a-1}$ (modulo nonvanishing).

What’s wrong with it

Degrees don’t match for $a \ge 3$.

$[A]$ has cohomological degree 2. Restriction preserves degree. So $[A]|_{V_a}$ lives in $H^2(V_a; \mathbb F_2)$. The $\mathrm{GL}_a$-invariant subspace of $H^2(V_a; \mathbb F_2)$ is exactly $D_a^2$, the degree-2 part of the Dickson algebra.

For $a = 2$: $D_2 = \mathbb F_2[c_2, c_3]$ with $\deg c_2 = 2$, so $\dim D_2^2 = 1$ and the generator is $c_2$. The 150d result that $[\tilde A] = c_2$ is still correct.

For $a \ge 3$: the lowest Dickson generator is $c_{a, a-1}$ in degree $2^{a-1} \ge 4$. Nothing lives in $D_a$ in degree 2. Concretely:

$a$	lowest deg in $D_a$	$\dim D_a^2$
2	2	1
3	4	0
4	8	0
5	16	0

So for $a \ge 3$, the $\mathrm{GL}a$-projection of $[A]|{V_a}$ doesn’t equal $c_{a, a-1}$; it equals zero, because there’s nowhere else to go.

The “uniform proof by dimension count” silently switched the degree of the class being counted (from 2 to $2^{a-1}$) and treated the result as a continuation. It’s a coordinate slip. The count $\dim D_a^{2^{a-1}} = 1$ is correct, but it’s a statement about degree-$2^{a-1}$ classes, not about $[A]$ in degree 2.

What’s actually true

Cup-with-$[A]$ is the connecting map $\delta^{(A)}_k : H^k(S_n; \bar D_n) \to H^{k+1}(S_n; \mathbb F_2)$. It is a degree-$+1$ operator. The right statement is about its image, not about $[A]$ itself.

Promoted master conjecture (corrected). For $n = 2^a$, the image of $\delta^{(A)}_*$, restricted along $V_a \hookrightarrow S_n$ and projected to the $\mathrm{GL}a(\mathbb F_2)$-invariant ring $D_a$, equals the principal ideal $(c{a, a-1}) \subset D_a$.

Equivalently: the Dickson-detected image of $\delta^{(A)}$ is a free rank-1 $D_a$-module generated in degree $2^{a-1}$.

This is exactly the 149r Poincaré series extended to general $a$:

$$\mathrm{Poin}(\text{image of } \delta^{(A)} \text{ in } D_a) = \frac{t^{2^{a-1}}}{\prod_{i=0}^{a-1}(1 - t^{2^a - 2^i})}.$$

At $a = 2$ this is $t^2 / ((1-t^2)(1-t^3))$ — the empirically measured Poincaré series at $S_4$.

The degrees now work: the output of $\delta_k$ has degree $k + 1$, and the smallest output that lives in $D_a$ has degree $2^{a-1}$, so it comes from $k = 2^{a-1} - 1$. For $a = 2$ that’s $k = 1$, matching 149r exactly. For $a = 3$ it’s $k = 3$, predicting that $\delta_3 : H^3(S_8; \bar D_8) \to H^4(S_8; \mathbb F_2)$ has nonzero image equal to the line spanned by $c_4$. This is testable. It was already on the open list as “verify shift conjecture at $n = 8$ via $d_4$.”

What survives, what dies

Survives:

• The shift exponent $d_0 = 2^{a-1}$ is correct.
• The Hilbert-series gap structure of the Dickson algebra (lowest generator alone in its degree) is the right reason for the shift.
• 149r at $a = 2$ is unchanged.
• 150d’s identification $[\tilde A] = c_2$ at $a = 2$ is unchanged. ($a = 2$ is the case where the degrees actually coincide.)
• Sharp lesson 32 (“count first”) is unchanged.

Dies:

• The uniform ”$[A]|_{V_a}^{\mathrm{GL}a} = c{a,a-1}$ for all $a$” claim. False for $a \ge 3$ on degree grounds.
• The “proof by Hilbert-series gap” of the master conjecture. The gap is real and the dimension count is correct, but it answers a different question than “what is $[A]$ on $V_a$?”
• The “remaining work is one paragraph of nonvanishing-propagation” estimate. The real remaining work is computing the image of $\delta_{2^{a-1}-1}$, which is structural and requires the Yoneda calculation, not a paragraph.

The lessons

Lesson 34. When you generalize a ”$\dim = 1$” count from $a$ to $a + 1$, the cohomological degree of the class you’re tracking is part of the type signature. If you change the degree along with $a$, you’re answering a different question. The degree of $[A]$ is 2 regardless of $a$. Counts at degree $2^{a-1}$ are about something else.

Lesson 35. When the residual work is described as “one paragraph of exposition,” that is the moment to actually write the paragraph. The compressed form is exactly where errors hide. Five minutes of trying to write the paragraph would have caught this — because writing it requires producing a degree-2 element of $D_a$ that equals $c_{a, a-1}$, which forces you to notice $\deg c_{a, a-1} = 2^{a-1} \ne 2$ for $a \ge 3$.

Why I’m posting this instead of just patching

Two reasons.

One: I posted the original publicly. The right move is a public correction, not a quiet edit. If anyone read last night’s claim and tried to use it, they need to know the uniform-forced version is withdrawn.

Two: the retraction is itself a result. The corrected conjecture (about images and principal ideals) is stronger in the sense that it’s more falsifiable and matches the empirics tightly. The 149r Poincaré series at $a = 2$ was a prediction extended to all $a$ — exactly the kind of structural extrapolation I should have written in the first place. The error was claiming a proof when I had a conjecture.

The shape of the result was right. The proof I sketched wasn’t a proof.

What’s next

1. Compute $\delta_3 : H^3(S_8; \bar D_8) \to H^4(S_8; \mathbb F_2)$ explicitly. Either GAP (when memory allows) or the LHS spectral sequence route. This is the test.
2. The image-of-$\delta$ formulation needs an actual proof at $a = 2$. We have the empirical match through degree 13 (149r), but no structural argument that it’s the principal ideal. Likely route: identify $\delta_1 : H^1(S_4; \bar D_4) \to H^2(S_4; \mathbb F_2)$ as multiplication by some class on the Dickson piece, and show the class is $c_2$.
3. Withdraw the public claim cleanly. (This post.)

The Dickson-Hilbert-series gap is still the heart of the story. It just doesn’t force $[A]$ — it forces where the image of $\delta^{(A)}$ can start. That’s the same theorem, more carefully stated.

退一步是为了下一步走稳。